For example, suppose that we wish to obtain the projections on the xy-plane of the asymptotic lines or the lines of curvature of a helicoid, the axis Oz being the axis of helicoidal movement in the sliding of the surface upon itself. It is clear that if a curve C of the xy-plane is a solution of the problem, then all the curves which we obtain by making C turn through any angle about the origin are also solutions.
So far we have supposed the group C? We are now led to examine the following problem : A differential equation being given, to recognize whether or not it admits one or more one-parameter continuous groups of transformations, and to determine these groups. This is a very important question, which cannot be developed here in detail. We shall limit ourselves to a few particulars. Infinitesimal transformations.
It may be shown, as above, that such a group is obtained by integrating a system of differential equations dx dx dx' -! In this new form we see immediately that these transformations form a group. Let us consider the problem of developing this function according to increasing, powers of t. F is regular in the neighborhood of the values Xj, x 2 , , Xn, the series on the right is convergent as long as is sufficiently small.
To every one- parameter group corresponds an infinitesimal transformation, and conversely. If we choose at pleasure n functions, 1? The introduction of infinitesimal transformations has made it possible to apply the methods of the differential calculus to the theory of groups. Besides, in many questions concerning groups it is the infinitesimal transformation which ig concerned, as we shall see from a few examples. Let us consider x v x 2 ,, x n as the coordinates of a point in space of n dimensions, and t as an independent variable which denotes the time.
If t varies, the point with the coordinates x[, X 2 , -, x' n describes in a space of n dimensions a curve, or trajectory, starting from the point x 1? The space of n dimensions, or at least a region of that space, is thus decomposed into an infinite number of one-dimensional manifolds, and each point of the given region belongs to a single one-dimensional manifold.
It is easy to obtain all the invariants of a group. We say, in this case, that the function F admits the infinitesimal transformation of the group. Let us notice that if two groups have for infinitesimal transformations X f and IlX f respectively, where II x x , x 2 , -, x n is any function whatever, these two groups have the same invariants, even though they are not identical. If we apply to the same point the transformations of the two groups, this point will indeed describe the same path, but with different velocities.
We shall now introduce another important concept. These calculations lead to formulae of the form ; a. We shall assume this fact, the proof of which presents no other difficulties than the writing of rather long expressions. We shall merely show how the infinitesimal transformation of the extended group can be obtained. Let be the infinitesimal transformation of the given group. We can write the equations of this group in the form and from them we derive The coefficient of t on the right, after expansion in a single power series, is the only thing we need to know.
It is obtained by a division and is equal to. Let be the infinitesimal transformation of the group G. In particular, T f must be an integral of the equation This condition is sufficient. This new method requires only the knowledge of the infinitesimal transfor- mation of the group. As there exist an infinite number of integrating fac- tors, we see that every equation of the first order admits an infinite number of infinitesimal transformations.
Let X f be the corresponding linear equation and T f the symbol of an infinitesimal transformation. To show this, it suffices to prove that the coefficients of a derivative of the second order are the same in X[T f ] and T[X f ]. The equa- tion of the second order being given, if we wish to find the infinitesimal trans- formations which it admits, we have at our disposal the unknown functions x, y , ij x, y , which do not contain y'. Writing the condition that the preced- ing relation is independent of?
In general these equations will be incom- patible, and we see that an equation of the second order, taken arbitrarily, does not admit any infinitesimal transformation. The same thing is true of equations of higher order, and it is seen from this how Sophus Lie was able to classify differential equations according to the number of independent infini- tesimal transformations which they admit. II, Exs. Find a group of transformations for the differential equation where a is constant, and deduce from it an integrating factor.
We then obtain x and y by the integration of a differ- ential equation of the second order. They possess a group of characteristic proper- ties which distinguish them sharply and at the same time simplify their study. Moreover, they appear in a great number of important applications of Analysis, and a preliminary study of them is very useful before undertaking the study of differential equations of the most general form.
Except when otherwise expressly stated, we shall study here only those equations whose coefficients are analytic functions of the independent variable. Singular points of a linear differential equation. Let us suppose that the coefficients a f are analytic in a circle C with the radius R and with its center at the point x , and let?
Applying to the equations 2 a general result established above 23 , we see that the equation 1 has an integral analytic in the circle C Q , taking on the value? Let L be a path joining two non-singular points X Q and A", and not passing through any singular point ; the integral which is defined by the initial conditions X Q , y ,? We can follow, by means of this series, the variation of the integral along the path L as long as the path does not go out of the circle C. If the path L leaves the circle C at a point a, let us take a point x l on the path within the citcle C Q and near enough to a so that the circle C l with the center x l passing through the nearest singular point does not lie entirely within the circle C.
From the series P x and from those which we obtain by suc- cessive differentiations, we can derive the values of the integral and of its first Ti 1 derivatives at the point a? At the end of a. In fact, let S be the length of the path L and 8 the lower limit of the distance of any point of L to the singular points. We know, therefore, a priori, what are the only points which can be singular points for the integrals of a linear equation.
It may, how- ever, happen that a point a is a singular point for some of the coeffi- cients a t without being a singular point for all the integrals. In the particular case where the coefficients are all polynomials or integral functions, all the integrals are analytic functions in the whole plane ; that is, they are integral functions and they may reduce to polynomials. The reasoning may be extended also to the case where the coefficients a t - have any singularities whatever, it being possible for these functions to be multiple- valued. If we start from a point x , where these coefficients are analytic, and if we cause the variable x to describe a path i, along the whole length of which we can continue the analytic extension of the coefficients a t , we can like- wise continue the analytic extension of the integrals along this path.
The power series which represent the integrals are convergent in the same circles as the series which represent the coefficients. These results are entirely in accord with those which we have deduced from the method of successive approximations Fundamental systems. Let us consider a linear equation which is also homogeneous, that is, not containing a term independent of y, 3 where F y denotes no longer a function of the variable y but the result of an operation carried out on a function y of the variable x.
If we know n particular integrals? We cannot conclude from this that the expression 4 really rep- resents the general integral of the equation 3 ; we must first assure ourselves that we can dispose of the constants C v 7 2 , , C n in such a way that, for a particular value X Q of x, different from a singular point, y and its first n 1 derivatives take on any valo. Setting the values of the integral y y and of its first n 1 derivatives at the point X Q , equal to these arbitrary quantities, we obtain a system of n linear equations to determine the constants C v C 2 , -, C n.
The determinant of the coefficients of these unknowns must be different from zero. We can then determine the constants C t so that y and its first? Every inte- gral of the equation 3 is therefore included in the formula 4. We say, for brevity, that this formula represents the general inte- gral of the equation 3. C 3 , , C n are constants not all of which are zero. The condition is first necessary. Let A be a region of the plane of the variable x where the functions? In order that this may be the case, it is necessary and sufficient that all the determinants analogous to A which can be formed with p -f- 1 of these integrals shall be zero, one at least of the determinants formed with p integrals being different from zero.
The same lemma enables us also to prove that the general integral of the equation 3 is represented by an expression of the form 4. For, let y x ,? Finally, C, the coefficient of y, is certainly different from zero, since the integrals y lt? Every linear equation of the nth order has an infinite number of funda- mental systems of integrals. We have, in fact, by the rule for the multiplication of determinants, A F 1?
We shall verify this by cal- culating this quotient. If we give to x an increment A, and if we replace each elemeni. There remains, after developing with respect to the elements of the last row and taking account of the determinants which have two rows identical, 11 dA The quotient which we wish to calculate is therefore equal to a 1? This expression for A shows that this determinant is different from zero at every non-singular point, if it is not identically zero a result which we could also have obtained from the preceding properties. This shows that any n linearly independent functions y lt y 2 , , y n can always be regarded as forming a fundamental system of integrals of a linear equation.
The general linear equation. A non-homogeneous linear equation can be written in the form where the term independent of y has been isolated on the right-hand side. It often happens in practice that we can easily obtain a particular integral of a linear non-homogeneous equa- tion, and in this case we are led to the integration of the homogeneous equation.
In general, if we know the general integral of the homogeneous equation, we can always obtain by quadratures the general integral of the non-homogeneous equation supposing, of course, that the left-hand side is the same for the two equations. The following process, due to Lagrange, is called the method of the variation of constants. Let y l9? We can evidently establish between these n functions n 1 relations chosen at pleas- ure, provided that they are not inconsistent with the equation Substituting the preceding values of? C a , , C n by quadratures. We can also make use of the following method, due to Cauchy.
From this it follows that the function Y represented by the definite integral 18 is a particular integral of the non-homogeneous linear equation. It will be noticed that this integral, as well as its first n 1 derivatives, is zero for the lower limit x , which is supposed different from a singular point. Depression of the order of a linear equation. If we know a certain number of particular integrals of a linear equation, we can make use of them to diminish the order of the equation. Th6 integration of a linear homogeneous equation of which p inde- pendent particular integrals are known reduces, therefore, to the inte- gration of a linear homogeneous equation of order np, followed by quadratures.
There exists, in fact, a very close relation between these two kinds of differential equations. Legendre's polynomial X n I, 90, 2d ed. In reality this is not the cape, or, rather, the quadrature reduces to the calcula- tion of fpdx. We can therefore calculate this integral by rational operations I, , 2d ed. Analogies with algebraic equations. The preceding properties establish an evident analogy between the theory of linear differential equations and the theory of algebraic equations.
This analogy persists in a large number of questions. As an example of this we shall show how we can extend to linear equations the theory of the greatest common divisor. In general, let be a symbolic polynomial where a , a 1? If a is not zero, we shall say for brevity that F y is of the nth order.
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We can find a third polynomial G y of order n m such that F y G [F l y ] is at most of order m 1 a poly- nomial of order zero is of the form ay, where a is a function of x. Continuing in this way, we see that we can determine, step by step, the coefficients X , X 1? This operation is entirely analogous to the division of one algebraic polynomial by another. If F k y is of the degree zero, the two equations have no other common integral than the trivial solution y 0. We can deduce from this observation another solution of a problem already treated. We can form a linear equation of the pth order.
The reduction is the same as by the first method, but the new process is more symmetric. Appel, Laguerre, Halphen, E. Pi card, and many others after them have ex- tended to linear equations the theory of symmetric functions of the roots, the theory of invariants, and the very fundamental work of Galois relative to the group of an algebraic equation. We can sometimes make use of this transformation to simplify a linear equa- tion. For example, if we wish to make the coefficient of the derivative of order n 1 disappear, we find that it suffices to put retaining the variable x.
The adjoint equation. Lagrange extended the theory of integrating factors to linear equations in the following way. Let us try to find a function z of x such that the product zF y shall be the derivative with respect to x of another function linear in y and in its derivatives up to those of order n 1. The general for- mula for integration by parts I, 87, 2d ed. We see, then, that if we know an integral of the adjoint equation, the inte- gration of the given equation is reduced to the integration of a linear equation of order n 1 whose right-hand side is an arbitrary constant.
If we know p independent integrals, z 1? Eliminating the derivativ. II, Bk. IV, chap. See also Exercise 17, p. Linear differential equa- tions with constant coefficients were integrated by Euler. By the general theory 37 none of the integrals of this equation have a singular point in the finite plane ; that is, they are integral functions of x. The series which repre- sent the successive derivatives have an analogous form.
We shall show that this integral can be expressed in terms of expo- nential functions when it does not reduce to a polynomial. The equation 37 is a recurrent formula with constant coefficients which connects the n -f- 1 consecutive coefficients. Now it is easy to find particular solutions of that equation. If r is a root of this equation, it is clear that the relation 37 is satisfied, what- ever may be the value of the integer 7? Before studying the case in which the auxiliary equation has multiple roots, we shall prove a lemma. Let the k distinct roots of f r be r 1?
From these roots we can form n particular integrals of the linear equation. Such a relation is impossible if the k numbers r v r a , -, r k are distinct. It is understood that any term in the identity is simply omitted if the corresponding polynomial is zero. But this is evidently absurd. If this equation is not solved, the recurrent relations 37 enable us always to calculate, step by step, as many as we wish of the coefficients of the power series which represents the integral corresponding to the given initial conditions.
We can determine in advance the number of coefficients which it suffices to calculate in order to obtain the value of the integral with a certain degree of approximation. More generally, suppose that the deriva- tive of the lowest order which appears in the left-hand side is the derivative of the pth order.
According to the case which has just been treated, this equation in z has a poly- nomial of the mth degree for a particular integral. The coefficients of this polynomial can again be determined by a direct substitution. Suppose in particular that P x reduces to a constant factor C. By virtue of a general remark 38 we can there- fore find a particular integral directly whenever the right-hand side is the sum of products of exponentials and polynomials. We must next find a particular integral of each of the four equations obtained by taking successively for right-hand sides ae x , 6e 2x , c sin x, g cos "2 x.
The sum of these two integrals is of the form x m cosx -f n sinx , and we can determine the coefficients m and n by substituting in F y and equating the result identi- cally to c sin x. This method avoids the use of the symbol i.
Adding all these par- ticular integrals to the right-hand side of the equation 45 , we obtain the general integral of the given equation A large number of methods have been devised for the integration of linear equations with constant coeffi- cients, particularly in the case where the auxiliary equation has mul- tiple roots. Let r lt r 2 , , r q be these q values of r such that the functions..
If for certain par- ticular values of these parameters the fj values r v r 2 , , r q are not distinct, the number of the known integrals is diminished. Suppose, for example, that? This leads again to the results which we obtained before directly. Euler's linear equation. The given linear equation is therefore transformed by this change of variable into an equation with constant coefficients.
To obtain the general integral of the equation 47 , it is not necessary to carry out the calculations of this change of variable, for we know that the transformed equation has integrals of the form e rt. The given equation has therefore a certain number of integrals of the form e i y x r. If, in the equation 47 , we replace the right-hand side by an expression of the form a- m Q Logx , where Q denotes a polynomial, it can be shown, as in the case of the equations with constant coeffi- cients, that the new equation thus obtained has as a particular inte- gral an expression of the same form, whose unknown coefficients can be calculated by a substitution.
Laplace's equation. We can sometimes represent the integrals of a linear equation by definite integrals in which the independent variable appears as a parameter under the integral sign. L where Z is a function of the variable z and where L is a definite path of inte- gration independent of x. The function under the integral sign in the expression 61 is the derivative with respect to z of Ze zx Q, provided that we have 52 We derive from this condition where the lower limit z does not cause Q z to vanish.
The function Z having thus been determined, the definite integral 51 is equal to the variation of the auxiliary function along the path L. It will suffice, therefore, in order to obtain an integral of the given equation 49 , to choose the path of integration L in such a way that the function V takes on the same value at the end as at the beginning, and so that the integral 50 has a finite value different from zero. The function V is multiplied by e' 27ria when z describes the loop.
The definite integral 50 , taken over this path. It gives a particular integral of the given equation. We do not find n particular integrals in this way. In order to obtain others, we may look for the paths L having their extremities at certain of the singular points a, 6, c, - , I and such that the function V vanishes at the two extremities. If a is an m-f old root of Q z 0, the rational function R z contains a term of the form A m.. Proceeding in the same way with the other points 6, c, , I, we see that we can determine new paths X, closed or not, giving other particular integrals.
Finally, we can also take, for the paths of integration, curves going off to in- finity. We are again led to determine a path L having an infinite branch such that the function V approaches zero when the point z goes off indefinitely on this branch. For the second path of integration we can take next a curve surrounding one of the singular points i and having two infinite branches with an asymptotic direction such that the real part of zx approaches oo.
When this is the case, we can also take for the path of integration the straight line joining the two points -f i and i. More- over, the integral thus obtained is identical with the first except for a constant factor. In order to reduce this integral to the usual form, let us put z it. But if the path L is a closed curve, the definite integral 55 is always zero. It seems, then, that in this case the application of the general method gives only one particular integral. However, in this apparently unfavorable case we can express the general inte- gral in terms of elementary functions.
For, let us make the inverse transfor- mation to the preceding, so that n shall be half of a negative odd number. Taking for the path L a circle having one of the points i for center, we see that the residue of the function with respect to each of these poles is an integral of the linear equation. These two particular integrals are independent, for their quotient is equal to the product of e 2ix and a rational function. It is clear that their sum is a real integral, as is also the product of their difference and i.
The linear equation with constant coefficients is a particular case of Laplace's equation, which is obtained by supposing all the coefficients 6,- zero. The general method appears to fail, since the expression for Z becomes illusory. But it requires only a little care to recognize how the method must be modified.
If, therefore, 11 2 denotes any analytic function in a region R of the plane, the definite integral taken along any closed curve L lying in this region, is a particular integral of the linear equation with constant coefficients. We see how this result, due to Cauchy, is thus easily brought into close relation with Laplace's method. As a verification, it is easy to find the known particular integrals. This agrees with the result already known. Mathematicians have therefore been led to study the properties of these integrals directly from the equation itself, instead of trying to express them somewhat at random as combinations of a finite num- ber of known functions.
We have already seen Chap. Ill, Part I that the nature of the singular points of an analytic function is an essential element enabling us in certain cases to characterize these functions completely. We know a priori 37 the singular points of the integrals of a linear equation. We shall now show how we can make a complete study of the integrals in the neighborhood of a singular point in a special case, which is nevertheless rather general and very important.
Let C be a circle with the center a in the interior of which p v p 2 , , p n have no other singular point than a and are otherwise analytic. Let X Q be a point within C near a. All the integrals are analytic in the neighborhood of the point X. Let y lt y 2 , , y n be n particular integrals of a fundamental system.
If the variable x describes in the positive sense a circle passing through the point x about a as center, we can follow the analytic extension of the integrals y v y 2 , -. It is easy to obtain the value of the determinant D formed by these n 2 coefficients. For we have, by 38, If x describes the circle y with the center a in the positive sense, yt changes into Yi ; hence we have A r,, r a ,. This determi- nant is therefore never zero.
Since the coefficients in the equations 58 depend upon the fundamental system chosen, it is natural to seek a particular system of integrals such that these expressions are as simple as possible. Such a relation cannot exist between the n integrals unless the coefficients of y l9 y a , , y n all vanish separately.
This being the case, let us suppose first that the characteristic equation has n distinct roots ,s t , s 2 , , s n. These n integrals u v W 2 , u n form a fundamental system. After one, two, , n 1 circuits, we should have the relations of the same form, C n s n u n 0,.
It is easy to form an analytic function which is multiplied by a constant factor s different from zero after a circuit around the point a. In a circle C with the radius R about the point a is center and in which the coefficients p lf , p n are analytic except at the point a, the integral Uk cannot have any other singular point than a.
We can dismiss the possi- bility that a is a pole. Examination of the general case. It remains to examine the case where the iharacteristic equation has multiple roots.
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We shall show that we can always ind n integrals forming a fundamental system and breaking up into a certain minber of groups such that if y 1? If the n roots are distinct, vhich is the case we have just examined, each group is composed of a single ntegral. Since the theorem is assumed to hold for n 1 variables, we may uppose this auxiliary substitution reduced to the canonical form. This amounts o replacing y 2 , y 8 ,. If we carry out the same transformations on the equations 65 , it will be necessary to add to the right-hand side of the preceding relations terms con- taining u as a factor.
We shall first try to make as many as possible of these coefficients disappear. Suppose, for definiteness, that there are two such groups, containing respectively p and q variables. The theorem stated is therefore true in general. Sauvage Annales de, I'Ecole Normale superieure, , p. Formal expressions for the integrals. It remains for us to find a formal expression which will show clearly the law of permutation of the integrals of the same group after a circuit around the point a.
In order that it may be satisfied identically, it will suffice if it is satisfied by another value of i, for example, by t i 2, since the two sides are polynomials of degree i 2 in t. The determination of the numbers s lt s 2 ,. We can obtain these exponents r t - by algebraic calculations whenever all the integrals of the equation considered are regular in the neighborhood of the point a. It may even happen that the point a is an ordinary point for some of the coefficients p t -.
We have to show that every equation which has two independent integrals of the form I or of the form II in the neighborhood of the origin belongs to the Fuchs type. The first part of the proposition is therefore established. Having taken a root of this equation for r, we can choose c arbitrarily.
Let us take, for example, c 1. Discarding this case for the moment, we shall obtain a particular integral represented by a series of the form 73 , the convergence of which will be demonstrated later. If the equation D r has two distinct roots r, r', whose difference is not an integer, the preceding method enables us to obtain two independent integrals. This is no longer the case if the two roots of the equation 74 are equal or if their difference is an integer. Let r and r p be these two roots, where p is a positive integer or zero.
A second integral y 2 is given by the general formula 23 , which becomes here ,. If A is the coefficient of x? This result agrees precisely with the general theory. As a particular case, it may happen that we have. For this it is necessary that b Q be zero. Since their diff erence is not an integer, it follows that the general integral of the equation 72 X does not contain any logarithmic term in the neighborhood of the origin. The same thing is therefore true of the equation We can therefore determine a positive number fj.
Extension to the general case. The proof of Fuchs 1 theorem for the general case can be based on the same principles by showing that if it is true for an equation of order n 1 , it is also true for an equation of order n. The theorem being supposed true for a linear equation of order n 1, this equation in u' is of the Fuchs form ; the same thing is evidently true of the equation in u and therefore of the equation in y. Let r be the largest root of one of these groups. We have just seen that the equation 71 has a particular integral of the form x r 0 x , where 0 x is an analytic function in the neighborhood of the origin and such that 0 is not zero.
If or is not an integer, we easily see, by a succession of integrations by parts, that fv dx is an expres- sion of the same kind as v. If a is an integer, fvdx contains also a logarithmic term where C is a constant coefficient. Fuchs 1 theorem is therefore true for an equation of the nth order. IV, Gauss's equation. This series is convergent in the circle T with unit radius about the origin as center.
If 7 is an integer, the difference between two roots of the characteristic equation is zero or equal to an integer, and the integral contains in general a logarithmic term in the neighborhood of the origin. Bessel's equation. Likewise, if B and A is different from zero, we can suppose A 1. These formulse are valid in the whole plane. If 7 is an integer, the general integral of the equation 92 always contains a logarithmic term. If n is not an integer, the preceding development shows that the general integral of BessePs equation 96 is We have shown above 46 that if n is half an odd integer, the general integral of the equation 95 can be expressed in terms of elementary transcendental functions.
Hence the transcendental function J y, x is expressible in terms of exponential functions if 7 is half of an odd integer. Picard's equations. Given a linear differential equation with coefficients analytic except for poles, we can determine by Fuchs' method whether the general integral is itself an analytic function except for poles. For this it is necessary and sufficient : 1 that the integrals shall be regular in the neigh- borhood of each of the singular points ; 2 that all the roots of the charac- teristic equation, relative to each of these singular points, shall be integers ; finally, 3 that all the logarithmic terms shall disappear from the expression for the general integral in the neighborhood of a singular point.
The general integral is then a single-valued analytic function except for poles in the whole plane. If the coefficients of the equation are rational functions, there are only a finite num- ber of singular points a lf a 2 , , a n. If this last condition is satisfied, we can obtain the general integral by equating coefficients according to the method of undetermined coefficients. Since we know an upper bound for the degree of this polynomial, the coefficients can be determined by replacing y by an expression of the form P x IT x a; - m , where P x is the most general polynomial of this degree, in the left-hand side of the given equation, and then equating the result identically to zero.
Picard has given another very important case where the general integral can be expressed in terms of the classic transcendental functions. Given a linear homogeneous differential equation, whose coefficients are elliptic functions of the independent variable with identical periods, if its general integral is an analytic function except for poles, that integral can be expressed in terms of the standard transcendental functions of the theory of elliptic functions. For simplicity in writing, let us develop the proof for an equation of the second order only.
The integrals 0j x , 2 x are therefore analytic functions except r poles,. In the first case the integrals 0j x , 2 x are again doubly periodic functions of the second kind. In the second case the integral t x alone is a doubly periodic function of the second kind.
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A x Bx is therefore an elliptic function. Then the difference is again an elliptic function. The integration of this equation by Hermite was the starting point for the preceding theory. The gen- eral integral of this equation is a function analytic except for poles. Their difference is an odd integer, and the coefficient of y is an even function; therefore the expression for the general integral does not contain any logarithmic term see ftn.
Equations with periodic coefficients. In many important questions of mechanics, linear equations with periodic coefficients occur. We shall indicate rapidly their more important properties. The equations define a linear substitution with constant coefficients, whose determinant is not zero. We are therefore led to a study entirely similar to the one which has already been made in detail in 48, It follows from that study that we can always choose a fundamental system of integrals such that the relations reduce to a simple canonical form.
All the roots' of this equation are different from zero, since their product is equal to the determinant JEf, whose value we have just written down. The real parts of these exponents, which are determined without ambiguity, are called the characteristic numbers. Taking r for a new variable, the problem is reduced to one solved above We see again here, as in 49, that all these integrals can be deduced from the last one of the group.
Let us observe that z p t is a polynomial in t of degree p 1, whose coefficients are periodic functions of t. From the theory of finite differences, we know that the successive differences A t z p , A 2 Zj,, , The integrals of the group , corresponding to the characteristic onent a, approach zero when t becomes infinite passing through positive ues, if and only if the real part of a is negative. In order that all the inte- 1s of the equation shall approach zero as t becomes infinite, it is there- e necessary and sufficient that all the characteristic numbers shall be negative, what amounts to the same thing, that the absolute value of each of the roots the equation is less than unity.
If the coefficients of the equa- n are real, the same thing will evidently be true in this case of the egrals y v? It is ar that by combining these 2p integrals linearly in pairs we can derive from iin a system of 2p real integrals. Finally, suppose that s is a real negative root. If the coefficients of s real, it is clear that the real part and the coefficient of V 1 must each isfy separately the linear equation. We would proceed similarly with the ler integrals of the group if p is greater than unity. Moreover, the case where s is real and negative reduces to the case where s -eal and positive by considering the period 2 w instead of the period w.
Note 3. The reasoning used under the supposition that the iable t moves along the real axis applies without modification to the case which that variable moves in the strip R. It follows that the functions 0,- t , ich appear in the expressions of , are periodic analytic functions in the ip R. II, Part I, The ribbons are solid enough, but somehow seem a bit lacking in how they are implemented, as if I was expecting something else to pop out throughout the set.
Battle Mode and Futeztsu are both used pretty well, especially Battle Mode, which is probably the most interesting thing in the set in my opinion, due to the timing it requires and descision making it entails. Aside from that, she kind of feels just like a spacer with some nice moves thrown in, like most of the throws and the pummel. It's not exactly bad, just sorta It fits somewhat sleepily in after Shana and Khamsin, I think.
And it somehow seemed less enjoyable than either. I am sure some people will find better things in this set, however. For me I'm afraid to say I don't think it measured up to the other two, Kibble. Smash Hero Jun 1, Incomplete Set 8 Gintoki Sakata. Gintoki is the main character of Gintama, the same series this guy comes from. Gintoki is a former samurai who makes a living with his little business "Odd Jobs" where he'll take any job alongside his two teenage sidekicks, though things tend to go badly for them most of the time when they're constantly struggling to pay the rent to their understandably impatient landlord.
As you might have expected from the first impression image, Gintoki tends to act uncaring and can be a bit of a jerk, but he's actually a lot wiser and compassionate than he lets off and possesses frightening skills as a samurai that generally remain locked away along with his dark past For now however, let's just say that say that an odd job has been taken by an ODD man And I'm the client. Ho boy. You know that guy in Brawl who's name is a sexual innuendo? Gintoki's kinda like him.
He stands close to johnny's height, but that's only because our wavy-haired friend stands upright instead of hunched down - Gintoki's the shorter one at the still manly cm. Gintoki might have a lot of bad habits, but at least he doesn't smoke and thus has pretty average stats, being a former samurai and all. That's kinda all you need to know, unless you want me to include some extra mechanic to make you want to play as him You were expecting a sword-based move first up? Screw that, Gintoki would rather encourage the consumption of strawberry milk!
Feeling the need to quench his high blood sugar level, Gintoki takes out the good stuff and holds it as an item because that's a lot more convenient than just drinking it right away. Holding B makes Gintoki down all that sugary goodness over a period of 1. The milk carton disappears upon being emptied, but if there was still some left it can be tossed as a weak throwing item that randomly makes anyone stepping over it trip Gintoki can use standard attacks while holding the carton, and he cannot take out another for 15 seconds because sadly he isn't rich There's something else you should know about this attack, however.
You see, Gintoki was dumb enough to pick this milk up past its expiry date because he thought it would taste better that way The victim suffers the reverse effects of the milk's healing and take up to 1. And that's more than likely to happen too, as not only is there no visual indication as to whether the milk has gone off but its effects will only be apparent right after being drank!
The longer milk stays bad, the more severe the effects become as time is continually added onto the negative effects and their severity increases by 1. Gintoki takes to his trademark motorized scooter which he uses to drive all around Edo This is similar to Wario's motorbike, only while Gintoki doesn't hop as far when jumping off and can't turn around or even stop for that matter he can in fact use his ground game while cruising along the ground and his aerial game in the air, providing him with the awesome potential he needs to impress you he can't shield or dash however, and automatically gets off the bike upon air dodging.
The scooter can be knocked off the stage and hit foes along the way, but will rarely hit Gintoki due to being sent flying at a lower trajectory than him - without his scooter Gintoki's Side Special is replaced with a desperate leap similar to Diddy Kong's Monkey Flip that gets him back on his scooter on contact with it, which you really should try to do instead of letting it fall given he won't get it back for another 5 seconds. When Gintoki gets off his scooter it will continue driving forward and can hit him when being knocked back, something you might want to save a few brain cells to be aware of lest you die a stupid death.
Also, since the Side Special input serves no function on the scooter it is replaced with Gintoki placing whatever item he has on him next to his feet, which will be sent flying as if it was thrown when the scooter is knocked away. You won't stop the scooter from moving just by attacking it, but grabbing it works like a charm The scooter has a total of 50HP and breaks apart upon destruction, exploding dramatically and sending out debris. Also, if any character is every hit by a scooter from behind, it'll ram up their butt for massive pain!
They don't want to get hemorrhoids, now do they? Gintoki takes out his scooter yet again The scooter has similar controls to ROB's recovery whilst letting Gintoki attack, only you cannot stop moving upwards If you stay on the scooter when it reaches the top blast zone like an idiot, you'll just get yourself killed in an incredibly stupid manner, so try to avoid that. The scooter will continue flying upwards even when it leaves the screen, and will continue doing so for 5 seconds Just don't get too carried away however, as Gintoki only has one scooter he has to share between both his Side and Up Specials and he can't simply alternate between the two formations when riding.
Without his scooter Gintoki performs a leap identical to the Side Special's that makes the scooter fly upwards upon getting on it, regardless of whether it was brought out with the Side Special and vice versa. You might think Gintoki already sucks given how sparsely he'll get to use his scooter, but he already has that delicious strawberry milk for healing he can even use it to get a speed boost on his scooter, remember?
For once, Gintoki decides to brandish his beloved wooden katana that has Lake Toya engraved on the side This is basically Ike's Counter, with Gintoki parrying the attack back at 1. Aside from the obvious uses you all know and love from Brawl's swordsmen, Gintoki's counter serves a more practical use in defending himself from his own scooter should it be knocked towards or fall on him, and he can even use the impending debris as a formidable weapon.
Gintoki can even use the debris from his scooter to counter enemies if they destroying it while at low HP, making them think twice about messing with this former samurai. Instead of beating on the enemy, Gintoki delves into one of his unhygienic habits and starts picking his nose! He'll alternate between each finger every 0. Boogers on Gintoki's hand do a number of things, namely increasing grab time by 0.
Hey, Gintoki once destroyed a car with his boogers Gintoki leaps forward impulsively, putting on a comically enraged face whilst making the foe taste the sole of his foot! This actually serves a purpose other than fractional stalling, as Gintoki can propel himself off the foe before kicking them away if the control stick is tapped in a certain direction. Up is essentially a footstool jump that puts the foe into prone directly beneath Gintoki, forward makes him bound forward for an aerial approach akin to a horizontal jump and kick the foe behind him a little while backwards has the same effect.
You can also press down to knock the foe directly beneath Gintoki and follow up like a jerk, but there's no use more satisfying for this move over jumping on a minion to reach their master - Gintoki will even bound off projectiles if they're big enough as in you can jump off a fully charged Aura Sphere but no way in hell something like Link's arrows. This move also serves as an alternate way for Gintoki to get back on his scooter without leaving himself so open, though if it's being sent flying back he'll bound off it instead.
Having had a little too much sugar, Gintoki projectile vomits! Gintoki unzips his john and lets out his restless pee. No I am not kidding. The only downside to this attack is that it has punishable lag as Gintoki unzips the little fellow back in, which all the while is exposed to the outside world - if Gintoki receives a blow to the crotch, he'll grovel over dramatically and take a huge amount of hitstun along with 1.
Kind of bad for the most part, but against melee attacks this actually stalls the opponent for several purposes at the cost of Gintoki's jewels. Especially bad if Gintoki is interrupted while peeing, as he'll leave that vulnerable spot hanging out until he uses this attack again to withdraw it. Gintoki gets serious with his Smashes, drawing his sword as he squats in an utmost prepared stance. Then, in a typical samurai maneuver, he'll rush forward in a sudden burst of speed that positions him SBBs ahead of where he was he'll stop at ledges thankfully.
The Smash is similar to Snake's in that the pre-charge is laggy but the attack can be released anytime with seemingly no forewarning - for this reason foes might want to jump when avoiding this attack unless they're confident in their dodging skills. You can combine this attack with Gintoki's scooter for some interesting attacking opportunities - rushing forward with a scooter in-between yourself and a foe can result in it being destroyed with low enough HP and scattering debris to hit foes if they tried to jump, charging this attack atop the scooter will have Gintoki get off and rush ahead as usual to the point where you can corner a foe from both sides or even run them over with the scooter and slash away at them it's likely foes will be hit by the scooter first, especially when taking strawberry milk's speed boost into account, but even so the slash will still take effect afterwards Gintoki holds his sword close to him with one hand before thrusting it upwards.
Surprisingly quick to execute for an attack of such power, but suffers some ending lag which leaves Gintoki open if he misses. It's very useful for keeping foes on-guard however, and does a top-notch job at repelling or even using your scooter as a weapon. Only a vulgar main character such as Gintoki would dare do such a thing aside from Wario. Gintoki can prolong the fart for as long as he likes, even being able to spread it as he rides his scooter in the air!
It's best if foes try to put up with the whiff when they're caught in it, as being flinched every time they enter isn't a very nice thing when there's a lot around them they have to worry about. Gintoki is also damaged by his own fart however, but he can use the flinching to save his own life by setting it up earlier! If you're daring enough you could deliberately drink bad milk and spread the fart across the stage on your scooter, showing foes the insane lengths you're willing to go to stay on-air - the fart cloud will deal that strong knockback before staying out as a damaging cloud for 3 seconds, and if foes re-enter it during that time they'll flinch.
Gintoki does what no other Shonen hero would have the balls to do Eh, he ordered that sword online and broke it many times, so it's not like he can't just get another. This is identical to throwing a Beam Sword upwards, and Gintoki will automatically catch it if he isn't in the middle of an attack or holding another item yes, he can be hit by it due to carelessness.
The sword will plant itself into the ground upon landing there and act as a Beam Sword for other characters. Gintoki can still use his sword-based attacks without his sword, but he'll be locked in place at the end of their starting lag until you release the input or wait for the sword to appear - the only exception to this is the F-Smash, in which Gintoki will rush forward anyway but will not execute the attack unless he managed to pick up his sword before moving past the foe.
Gintoki will automatically draw another sword when attacking with it if the one he drew out was thrown offstage, so don't worry about ever outright losing them. This has a few subtle interactions, namely bouncing the sword in-between a scooter above Gintoki to damage rack it and delaying the throw of the sword with boogers to catch foes off-guard.
Predicting what Gintoki will do just got more difficult. Gintoki holds his sword to the ground and plummets with it. At least your scooter won't try to escape, and can be used as a shield while approaching or to simply get it back onto ground. If you hit a grounded foe with an impaled scooter they'll take the knockback of the attack as usual AND be damaged by the debris beforehand if the scooter would be destroyed.
The scooter will deal its falling and spiking damage to those hit in midair, and if it's destroyed up there the debris will fly out downwards as numerous projectiles of pain. It's only natural that Gintoki would return to his comical roots for a grab of all things, reaching out for the foe's collar with both hands in ANGER.
Sticky, yucky anger if Gintoki picked his nose and made his hands more adhesive with boogers You can thank me for making you remember that later. If Gintoki tries grabbing while on his scooter he'll jump off ahead of it in a grabbing attempt that sees him crash to the floor comically if he misses and take a scooter to the butt. If Gintoki grabs the foe however, the scooter will eventually ram up his butt anyway but he'll be knocked into the foe and take them flying with him!
This throw is very fast, and with your strawberry milk it is in fact possible to slam a foe into your runaway scooter before you're both hit. Gintoki is a hard mixture of comical AKA vulgar and serious elements, effectively representing the nature of his series to the core. He's deceptively unpredictable, yet fairly straightforward - good at surviving, yet low on resources. Kat: a bit of confirmation, the Dsmash hits with energy spines, so it extends corruption as the mechanic states Edit edit: The pummel is more or less a way to have that unique "throw" while still having the other 4 available Glad you liked it :D.
ForwardArrow Smash Journeyman Jun 1, Joined Aug 17, Messages Katapultar Smash Ace Jun 2, Hercule Barton, nicknamed Elly by her friends, is one of the four main characters of Tantei Opera Milky Holmes, a loli-detective Japanese franchise that revolves around detectives and gentlemen-thieves with superpowers called "Toys". Elly's Toy gives her super strength, which happens to contrast with her withdrawn personality; she generally gives little input on a situation and gets embarrassed very easily.
Not too much of a detective eh? As you'd expect from a shy little girl, Elly has weak below-average stats in terms of movement and is a floaty lightweight Jigglypuff. She's not very tall either. On the other hand, her super strength allows her to carry hefty items like crates as if they were item capsules and throw them with the strength of a giant giant; in other words, she throws them at insanely high speeds and distances which can make quick work out of enemies if they're hit by them, and can even be used as "traps" of sorts if you throw them towards the top of the screen in which they won't come down for a good number of seconds.
Elly timidly holds her hands above her head for as long you hold B, which halves all damage she takes from above and allows her to instantly grab hold of any items that'd falls on her and wield them. If you use this move while Elly is holding an item however she'll instead "wind-up" and attempt to throw it in the same way a character would throw a Trophy Stand in SSE! This triples the already ridiculous speed, distance and power at which Elly can throw an item to the point where even a simple item capsule could probably KO most characters at very low damage percentages, though nobody's gonna actually be stupid enough to leave themselves open for that all that prep-time.
Elly bends down behind her whilst timidly yet comically looking around for any pebbles that might be lying around, with she picking one off the ground for every 0. Elly can throw out as many pebbles as she likes as to litter the stage with them, though she suffers some rather punishing ending lag in doing so due to her timidity; that, and most of the time the pebbles will simply travel all the way to the other side of the stage if they don't hit a wall or enemy - even if they do however, there's still the chance that one could end up bouncing back and hitting Elly Given her super strength, Elly might be able to recover by stomping on the air to gain momentum!
Once unearthed, the sign becomes a rather lengthy weapon in Elly's hands that leans into the background as to prevent it from acting as a mobile wall. Once the sign's reached its peak, it'll end up staying embedded in any part of the stage it would've made contact with as an obstruction that'll act like a wall or a solid platform with grabbable ledges depending on which way it was embedded, albeit one that players can roll past or even drop through with a dodge given how the sign is not a "true wall".
The sign cannot be destroyed but rather can be bent out of shape with non-projectile attacks, with smaller portions of the sign being easier to bend than larger ones; for specifics, it'd take a move as powerful as Dedede's F-Smash to bend the sign into an L-shape with one blow. Elly can remove the sign by using the Down Special next to it in order to re-wield it and straighten it out in the process, though otherwise it can be made to stay embedded for the rest of the match - only one sign can be out at a time.
From there it'll either fall off the stage or sit on the ground as a pseudo-platform that can't be destroyed or picked up by other characters, but can be pushed off the ledge by them at a rate of half their dashing speed or more if they have good physical strength. Needless to say, the sign can end up spiking or knocking Elly down if it would be knocked out of her but the knockback in question would not be enough to send her flying out of the sign's range quickly enough.
Because of the attack's surprising power for a standard it can quite easily out-prioritize a good deal of quick attacks and projectiles in order to cover Elly from the front, with enemies generally needing to jump or roll behind Elly if they don't have anything they can overpower or out-range her with since this can become quite annoying and even dangerous at higher percentages or in tight-spaces where enemies can be chained for at least a fair bit of damage on walls. This is a good albeit laggy way to keep distant enemies on their toes or spike offstage enemies, but if the edge of the sign would end up striking a wall or another part of the stage on a higher plane, Elly will be able to wedge it in there by keep A held down for the entire move, which in itself causes the sign to act as a solid obstacle that can potentially cage enemies between Elly and the aforementioned wall with only the area Elly was occupying to use as an exit.
Gerund or Infinitive? Quiz
Poor Elly goes into a fit of embarrassment as she wildly swings both her arms like windmills whilst dashing for as long as you continue to hold A, which causes her to fling any items she'd make contact with her upwards as if she'd thrown them there herself; a great way to knock your pebbles and street sign into the air or approach your enemy whilst countering their attempts to use your stones against you.
Speaking of which, foes that make contact with Elly's flailing arms are flung right up to the top of the screen, though they don't take any hazardous damage or knockback of sorts unless an item would end up falling beneath where they are that could inflict some rather fatal damage on them. Elly will stop running about if she'd meet a wall or a ledge, though in any case she'll suffer some punishable ending lag anyway as she tries to catch her breath.
Elly normally suffers some punishable ending lag after using the move, but if you continued to hold down A afterwards she will in fact "push" her area of the sign upwards so it ends up leaning perfectly upright 3 SBBs high with Elly on the very top and the sides acting as an obstructive wall to enemies albeit one they can roll around for a while before slowly leaning and falling as a spiker after no more than 2 seconds; Elly can and will keep hanging on to the top of the sign even when it's falling, though she can cancel out of this and transition into her aerial game with a jump, make the sign lean backwards by holding backwards on the control stick, let go and fall back down immediately with a downwards input, or even make the sign lean over and fall prematurely with a forward input that makes Elly fly forward 2 Platform's worth of distance where she's allowed to use her aerials at that given time.
While the entire process is rather extreme on a whole in that it allows Elly to attack and defend at the same time, quite a lot of space is required to pull it off in the first place and enemies can make the sign lean away from them prematurely to mess up Elly with a strong attack like a F-tilt, though Elly can still transition into her aerial game from this none-the-less. This has some lag and is rather difficult to hit with but downsizes your hitbox and you can keep the attack out for as long as you like for juggling purposes in terms of both enemies and items in order to stay safe.
Elly removes a decent-sized rock from the ground which deals similar damage to Samus' parts when thrown. You know the drill, this is a simple throwing item that Elly can use to destroy the enemy, especially if you wind it up with the Neutral Special. If Elly was already holding an item she'll place the rock in front of her. Elly suddenly tears out a chunk of the stage in the background and holds it in the form of a boulder that's as large as Kirby or 1. Uncharged, the item can be picked up, thrown and rolled around like a barrel to deal damage based on the attack think F-tilt , with Elly being able to freely toss it around with twice the speed and distance but 1.
Scattered pebbles make this all the easier for enemies to make this happen, so you'll really have to pick your time and place when trying to pull off the big move. Note that Elly can in-fact use her Neutral Special whilst holding the boulder to chuck it forward and drastically increase its power for a OHKO move, or simply throw it up into the air as a rather deadly trap that'll spell doom for whoever happens to be beneath it once it comes down.
After all, you could keep using this move over and over to make quick work of your opponents given how powerful even the un-charged boulders are, and the fact that you can juggle all the large boulders you make Elly attempts to unearth something in a similar vain to her Down Special, except this time she tears off a chunk of the stage!
The torn area is then thrown all the way up to the top of the screen at Sonic's dashing speed and will obviously Star KO any players standing on the torn area, which can range from being a Platform wide and a Kirby deep or thrice that area if fully charged to the point where a piece of the stage can end up going missing; the move naturally suffers from obscene lag however, twice that of Warlock Punch without taking the charging time into account.
In a sudden panic, Elly stamps the ground repeatedly to cause a tremor that affects the entire area she's on, occurring while the attack is being charged. Grounded objects with HP will be demolished very quickly given they take damage every 0. To dodge this attack at all foes will have to stay in the air or try to attack the defenseless Elly - remember that since this attack occurs during the charge she'll take more knockback from attacks.
The attack comes out unpredictably fast but lasts for a lengthy 2 seconds afterwards, reflecting projectiles that come Elly's way. Using this move with a sign produces the same effect as the Standard. Elly reaches above her with both hands in the usual grab manner, but if she gets someone she'll squeal in fright and chuck her foe 4 SBBs into the air with non-KO'ing pushback; on the other hand the move is deadly via throwing the enemy into falling items, though the grab suffers from a bit of lag and bad reach so its harder to pull off than you'd think.
This move can also be used to grapple onto the edge of an embedded street sign and bend it until you let go of A in order to manipulate it, generally for defensive purposes or projectile bouncing. It can also be used to toss items back up the same height they were originally thrown without having to redo it from scratch, namely stuff like boulders. Elly puts on a brave face as she reaches out with one hand to apprehend the foe, who escapes from her with thrice the difficulty due to her strength The way the Pummel is executed also has relevance; on the first use Elly will slam her foe into the ground in front of her, where she will keep them positioned until she Pummels them again to slam them on the other side of her.
This allows Elly to directly control her foe's positioning or even use them as a kind of shield against items that'd fall on top of her by Pummeling the foe at the exact time the item would fall on Elly so that the foe is struck by it instead. This leaves them invulnerable to enemy attacks but forces them to escape with 3. For those unfamiliar with the franchise or anime at hand , the premise of Tantei Opera Milky Holmes actually revolves around the main girls losing their powers and being forced to cope with the problems it brings while trying to find a way to restore them.
The Milky Holmes were once renown detectives, but after losing their Toys on the first episode at that they become nothing more than bumbling failures. The group's only saving grace is that their archenemy actually liked fighting with them and thus secretly infiltrates the academy they attend to try and help get their powers back, pushing and motivating them to do so all the while under the guise of the Student Body President.
Unfortunately, Elly will get no such sympathy from her foes when this happens to her in Brawl At the start of the match, Elly begins as a rather fearsome character that makes most other fighters shake in their boots. Finally, if Elly tries to grab a foe she'll end up holding them for as long as a normal grab note that the foe is struggling far less , and will find that her Pummel and Throws are now little more than exercises of futility. Essentially speaking, Elly has become a helpless character who stands at the mercy of her foes.
How cruel. But Elly's an detective. She shouldn't give up so easily; what would her ancestors think!? Keep trying Elly, keep trying, because with sheer determination your strength might come back to you for one attack! This chance occurs on each hit of Elly's attack, so it's a good thing she has many multi-hitting attacks and once that chance appears her attack will be powered up for the rest of its duration - this is also good with her grab. The chances of Elly being able to use her Toy for one attack also increases significantly when she spams an attack, so if the foe sees her using the same attack over and over again they'd better put their guard up.
If the player is particularly paranoid about Elly losing her Toy, they might want to throw out a whole bunch of items for her to use so she has something to defend herself with when she becomes defenseless. Otherwise, they might want to go for a KO before it can happen. As a proud detective who wields a Toy, Elly -should- have no need for a Final Smash. On the other hand, she'll become absolutely desperate for it the moment she loses her Toy, as her Final Smash will instantly give it back to her good as new.
Do note that the Smash Ball is rather difficult for Elly to obtain without her Toy as she only has her "trap" projectiles and weak rocks to smash it open with, but it'll be worthwhile in the end due to it being twice as hard for her to lose her Toy afterwards this only happens once however. Additionally, being 3 places behind the other players will give Elly the Pity Final Smash and consequently her Toys back for further use.
ForwardArrow Smash Journeyman Jun 2, I will admit I'm a bit surprised to see an entirely negative comment coming from you of all people, Kat. Sort of glad you are actually willing to make them now that you're commenting, though that my first set in a while had to be the victim of it makes me a little sad. Haven't felt too confident in my set making abilities lately Admittedly, the soul steal is used for "stalling" but the fact is Falz is so terrifying with a proper set-up that the foes pretty much have to try and pressure him regardless of this.
And of course there's a bit risk-reward system to the souls in that you can plant them in other foes and such, since while that depletes your defense it makes you way more powerful offensively. Not to mention his method of lockdown is a bit different than other bosses. Not to mention he can't really stall at the top blast zone since he naturally falls down if he teleports that high okay yes I forgot to put a limit on that move, I'll concede to you there and edit one in , and stalling at the other blast zones the most he can really do is make a Megid cloud, since any other tactics he performs from there are highly predictable, and it won't be -THAT- long until he has to return to the stage and becomes unable to stall anyway.
I happen to agree with you on the Grab's visual I dunno why I did it like that, but I kind of forgot the visual for that attack anyway The odd thing was though, I was trying to be more character relevant here than Vol Opt. I'm pretty annoyed about how much of the stuff in Vol's set I just made up, so I kind of wanted to repair that by making as much of Falz's set as possible come from in game.
Maybe that wasn't the right route to go and some of the stuff I had to make up was kind of tacky, though. Either way, I'm not necessarily saying you're wrong about the set, I just wanted to try and defend it since my current frontrunner is Doc freaking Scratch and he's not exactly the follow up to Praetors I was hoping he'd be.
Joined Jun 23, Messages 6, Location Texas. Welp, I'm being bullied into this again Comment blocks, how do they work. And if you don't care about my opinion, feel free to not even read the block. But being a generic "class" character, I imagine it's probably fine. I don't know why, but 30HP sounds to me like too much HP for zombies that you spawn five at a time.
Even though, it IS balanced out by like 10 or 20 second buffer times. Even though, I don't know. I think it could be better if maybe the zombies were weaker and the buffer times were slower. The mind control thing seems interesting, and sort of reminds me of move descriptions out of Luxord I don't see any glaring issues at a read, not a lot to say on the matter. The neutral special was probably actually my favorite part of this, I find the concept of all head attacks being used by the disembodied head to be a funny concept.
Other than that, I don't really know how I feel about the up special. It felt a little "meh". A lot of attacks seem to help out the disembodied head thing nicely enough, and overall I find the set amusing. To begin with, I like the concept of using stage entrance as part of the play strategy. That's a lot of stuff to use for an advantage, and the fight hasn't even started. The rest of the set is a pretty huge chunk. I don't have anything in particular to complain about, but not much of the rest of the set really stuck out to me except for the length and amount of detail.
But I think it turned out well. Okay then. I didn't peg this guy as a character who could really At first glance, this seems pretty proppy with the coach thing. And the whole gradual transformation thing sort of reminds me of Tempura Wizard. I don't know how versatile I'd call this, but I'd call it something different.
And I can see that it works with what it has. Yes, the metal coat would be great for reflecting projectiles, but then there are things like Thief that only sound useful when items are involved Feint also seems pretty out of place, seeming kind of useless to the anti-camping playstyle considering most campers would be dodging to open gaps. I can't say I like this one a huge amount, but it's alright.
Not a lot of moves come across to me as "anti-camping" past the major specials, either. Also, all white text is not the best thing ever. This is weird. I find the concept awkward and a little confusing, but maybe I just don't get it. It seems sort of like a puppet set, with one amp focusing on movement while the other three attack; I don't personally see that as a bad thing, do with it what you will. I do like when taunts are beneficial to a set, though. So good job. I wouldn't be able to make a set for an enemy like this. A dual-weapon variation set.
I'm not sure I get the light saber though. Does it spawn as an item that benefits her when she picks it up? Or what? If so, that seems to nerf her down in matches where items are off, which is sort of a common setting. Looking at this, I like the concept of the Growing Man item It seems pretty impending and nasty. As amusing as it would be to have several of them on stage, it's for the best that there's only ever one. The up special nearly scared me; I thought it was going to be a repeat of Emidius' down special. These two are a Zombie match made in-- heaven?
That colloquialism is ill-fitting. Moving on. Sacrifice the crawler to untap Grimmy and guess what!!? Grimmy is a Zombie and so the crawler can be cast again from your 'yard. If you do use Rooftop Storm , it's fun to note that Grimgrin can be cast for zero from your command zone, and the crawler can be cast from your graveyard for zero too. Pretty win-win. These go together so much, that I've combined their exposition into one block. Essentially both Alesha and Kaalia are very, very similar though different.
So, I guess like all humans, both in game and out. They are both Mardu, and both of their abilities, while still different, are dependent on them attacking, which means they need to survive said attack phases in order to continue with their chicanery. Now, let's start with Master of Cruelties. Both Alesha and Kaalia can cheat him into play, bringing him in tapped and attacking, and overwriting his latent effect of "Attacks only once.
Alesha usually has a component of graveyard recursion and sacrifice outlets, so I am certain you can tap attack, attack for reals, sacrifice it somehow, tap attack and so forth. Kaalia is less combo-y about it, but you can definitely sneak him in and attack another time. Both added evasion and preventing combat damage when attacking are perfect for both Commanders here. On top of that, add a Dolmen Gate for frak's sake. Iroas also helps keep the Master around too. While we are at it, in case you are curious, you can return Kaalia with Alesha, but Kaalia's trigger doesn't go off unless you declare her as an attacker, which can't happen if she enters the battlefield tapped and attacking.
It reanimates her though. Incidentally, you can always try to make either of these Commanders unblockable, which means you might consider Dauthi Embrace or Shizo, Death's Storehouse or Trailblazer's Boots or Whispersilk Cloak Usually, I am a huge proponent of Trostani and the first Ajani's ultimate working amazingly together, and I admit, I still uber-ultra-super-mega-love that combo, but I happen to love this interaction even more.
Two mana for seven life is pretty boss. Having this out allows you to block a threat and be able to score seven life for two mana also. You can add in Pangosaur too, which self-bounces every time someone drops a land, netting you six life for four mana-- not the best deal, but still pretty fun. There's also always the option of just putting in something like Blood Clock and Indomitable Ancients. Ok, now we are really entering into the territory of my favorite combos ever. I'm pumped.
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Two life, five mana, kill anything. Selenia can also use some of the tricks Obzedat does, with mass destruction but has the added bonus of being able to evade sorcery-branded removals. So attack, return her, Wrath, then play her. A good board position. But my favorite, is the Faithmender. It's simple, elegant, beautiful, and makes me smile. You cannot tell me you haven't seen Horde as a Commander of a five-color control deck. Its vanilla keyword abilities for five mana make it perfect after Wrathing-- but the Ashes make it even more perfect, because now you can recast all your stuff for five mana from your graveyard.
Two other protips for you: 1. Ashes works great with Lord of the Undead , and Soulshift. Horde works great with Nameless Inversion. In a Wanderer deck, you want to maximize your potential for Cascades by stacking your deck. You also want to stack your deck, because you don't want to Cascade into a Preordain , but a Bribery or a Keranos, God of Storms. I love Soothsaying here because you can dig deep on the end of the turn preceding yours in order to Cascade into useful things.
If all else fails, you can shuffle your library. My point is Soothsaying is underrated. Covered this in my Land Week theme article , but it's still epic, and I share it now in hopes you one day employ it. Karametra's out, and the Lion can bounce itself. It is a repeatable Rampant Growth , that, when paired with some tricky Landfall stuff, can completely overpower your side of the board.
Seriously, read the article, and enjoy the fun deck I put together. I can't wait for that landfall-not-labeled-landfall new enchantment in Origins. When you are cheating things into play with Zirilan, you want to make sure you can keep that Dragon from being exiled. Another viable option is a sacrifice outlet, and graveyard recursion.
Red doesn't have much, but there is some, and there are artifacts and lands that can help. I'm including this in for speculation's sake, but I already know I'm going to love this combination. As I said in one of the FFTR podcasts, Pia and Kiran turn into an artifact token, thus, before they fizzle out due to the clause in Feldon's ability, can be sacrificed to themself for two damage, and they leave behind two thopters.
My Feldon article is located here , and it is loaded to the gills with juicy combos that work alongside Feldon. Heck, if we reversed roles, and made Pia and Kiran your Commander, Feldon would go great in their deck! Any artifact creature you made with the token could still be sacrificed to them before it expires without being really useful. This is just such an awesome frakkin' combo. The Champion has always held a special place in my heart, even before I knew it looked like the Mandalorian-hatin', Sith-aligned, bounty huntin' Durge.
For any deck employing black and the Champion, he works wonders with False Cure , and the upcoming Origins' Tainted Remedy. If you had out the Remedy and some form of flickering, that, would be sweet. Also, since it says player and not opponent, you can target yourself with a Torpor Orb out, and then blink him for your own shenanigans. I love Dredge and Delve with Sidisi.
I love Fact or Fiction and cards of its ilk, which put cards into your graveyard from your library, giving you zombies and card advantage In a Sidisi deck you should be a Buddhist, and embrace the concept of Impermanence. Cards in your yard will be exiled, cards in your library will be milled. Brago can't pull this off unless he also has out a Flickerwisp or something, since Brago is creatures you control. Roon holds no impediment to flickerblinking out your Drake to steal more creatures. Lots of great things pair with Oloro: Exquisite Blood , Drogskol Reaver , Phyrexian Arena , and a whole host of things that will someday be the subject of a booshy article.
Paradox Haze , besides being a card I love in card decks, is a simplistic and fun combo. Instead of 2 life every turn, you gain 4. Adding one simple card to that, Angelic Accord , and it becomes ultra-mega-awesome. Add in all three of the viable Hondens for even more fun. Another beautiful interaction: Paying the 3 life helps enable Dethrone, and makes Marchesa come back to life via her own ability. Just Stellar. I've resisted talking about Sakashima until now, but legit-- Sakashima is my favorite card to combo with most any Commander that does really cool stuff. My favorite though, is to copy Edric, when Edric is my Commander.
Double Card draw Elves! I played a game where the first creature I cast was Tandem Lookout , the second Edric, third Sakashima! Throw in a Lorescale Coatl and a Chasm Skulker in there'd be just fine. Thanks for joining me on this ride. Did anything inspire you to make a new deck? Did I miss anything you would've included? Lemme know in the comments below, or maybe shoot me a message in Twitter: CmdrJohnnyBoosh. The additional cost should still get added, if you follow the rules. That is, the commander is free the first time, cost 2 the second and 4 the third and so on.
First of all, Brago cannot target himself if he got Unquestioned Authority. One of the things protection stops is targeting. If we considered Pentarch Ward assuming blue or white was not choosen or better yet Traveler's Cloak instead we would run into the problem that if we flickered both Brago and the aura, the aura could not go back on Brago. The problem is that you put everything into play at once and you need to put auras on something which is in play just before the aura enters play that is, the aura is never not attached to anything and therefore you need to put it on something just before it enters play.
Hence, you cannot flicker Brago AND the aura. If you want a combo with Brago pick Strionic Resonator. That plus 2 artifact mana gives you infinite flickers. Artifact mana works great with Brago anyway. Regarding Skullbriar and Varolz it should say that you could put it back into command zone not hand. I should have mentioned that the Rooftop storm adds two each time, but pay two mana for your second casting of a 7 mana commander is still epic.
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In fact, it has to get hosed 4 times for it to cost only one mana more than its original CMC, which is epic. Brago has pro creatures yes, but you can still flicker the aura, just not brago. Maybe I misspoke about that, but the actual AURA doesnt have procreatures, so you can flicker it and reattach it to brago for the next time it attacks, as you say.
As for skullbriar, these articles are long, and get written and re-written, and I proof them tons of times, but as the author, am apt to miss something here or there. I've been enjoying Vish Kal and Illusionist's Bracers recently. Any commander with a powerful activated ability being copied for free is good times. Frankenstein and his Monster, this seemingly innocent inclusion in a horror-trope-theme-set that features zombies in blue and black is actually pretty epic in Commander, if your Commander happens to be a Zombie.