## 230 Reputation

10 years, 143 days
Beijing, China

## Factorization and ascending order of an ...

Maple 2015

Dear Users!
I hope are fine here. I got the following expression after a lot of computations

((1/2)*r*(r-1)+(1/6)*r*(r-1)*(r-2))*`&Delta;y`[-1]^3+(1/2)*r*(r-1)*`&Delta;y`[-1]^2+(1/120)*r*(r-1)*(r-2)*(r-3)*(r-4)*`&Delta;y`[-2]^7+((1/6)*r*(r-1)*(r-2)+(1/12)*r*(r-1)*(r-2)*(r-3)+(1/120)*r*(r-1)*(r-2)*(r-3)*(r-4))*`&Delta;y`[-2]^5+((1/6)*r*(r-1)*(r-2)+(1/24)*r*(r-1)*(r-2)*(r-3))*`&Delta;y`[-2]^4+((1/24)*r*(r-1)*(r-2)*(r-3)+(1/60)*r*(r-1)*(r-2)*(r-3)*(r-4))*`&Delta;y`[-3]^7+((1/24)*r*(r-1)*(r-2)*(r-3)+(1/60)*r*(r-1)*(r-2)*(r-3)*(r-4))*`&Delta;y`[-3]^6+r*`&Delta;y`[0]+y[0]

Actually, for the above, I want the factorization of each coefficient of `&Delta;y`[0], `&Delta;y`[-1], `&Delta;y`[-2] etc and the above expression shoud be in descending order given as:

y[0]+r*`&Delta;y`[0]+(1/2)*r*(r-1)*`&Delta;y`[-1]^2+(1/6)*r*(r-1)*(1+r)*`&Delta;y`[-1]^3+(1/24)*r*(r-1)*(r-2)*(1+r)*`&Delta;y`[-2]^4+(1/120)*r*(r-1)*(r-2)*(r+2)*(1+r)*`&Delta;y`[-2]^5+(1/120)*r*(r-1)*(r-2)*(r-3)*(r-4)*`&Delta;y`[-2]^7+(1/120)*r*(r-1)*(r-2)*(r-3)*(-3+2*r)*`&Delta;y`[-3]^6+(1/120)*r*(r-1)*(r-2)*(r-3)*(-3+2*r)*`&Delta;y`[-3]^7

I am waiting for your positive response. Thanks

## How can I construct the difference table...

Maple 2015

Dear Users! Hope everyone is fine here. I have some points x_0,x_1,..,x_N and corresponding to these number have values y_0, _1, ..., y_N as,

restart
a := 1; b := 5; h := 1; f := 1/x; N := (b-a)/h;
for i from 0 while i <= N do x[i] := h*i+a; y[i] := eval(f, x = x[i]) end do;

Now I want to develope the following difference table using the values of y_0, y_1, ..., y_N as,

where difference column generated using the following concept

## Uses of chain rule to compute the deriva...

Maple 2015

Dear Users!

Hope everyone is fine here. Let me explain my problem first for this consider
diff(Y(xi), xi) = mu*(1-Y(xi)^2)
Then the derivative of a function U=u(Y(xi)) using chain rule (and expression menstiones as red) is given as,
diff(U, xi) = (diff(diff(Y, xi), Y))*U and (diff(diff(Y, xi), Y))*U = mu*(1-Y(xi)^2)*(diff(U, Y))
Similarly the second-order derivaitve of U=u(Y(xi)) using chain rule (and expression menstiones as red) is given as,
((&DifferentialD;)^(2))/(&DifferentialD; xi^(2))U=(&DifferentialD;)/(&DifferentialD; xi)(mu (1-Y^(2)(xi))*(&DifferentialD;)/(&DifferentialD; Y)U)=((&DifferentialD;)/(&DifferentialD; Y)*(&DifferentialD;)/(&DifferentialD; xi)Y)(mu (1-Y^(2)(xi))*(&DifferentialD;)/(&DifferentialD; Y)U)=(&DifferentialD;)/(&DifferentialD; Y)(mu^(2) (1-Y^(2)(xi))^(2)*(&DifferentialD;)/(&DifferentialD; Y)U)=-2 Y(xi) mu^(2) (1-Y^(2)(xi))*(&DifferentialD;)/(&DifferentialD; Y)U+ mu^(2) (1-Y^(2)(xi))^(2)*((&DifferentialD;)^(2))/(&DifferentialD; Y^(2))U;
In the similar way I want to compute the higher-order (like 5th order) derivaitve of U w.r.t. xi using the chain rule  (and expression menstiones as red) explained in above. Kindly help me soolve my problem

I am waiting for positive response.

## Separate Real and Imaginary Parts of Com...

Maple 2015

Dear Users!

Hope you are doing well. I have a funtion give bellow:
beta[1]*exp(x*alpha[1]+y*beta[1]-z*sqrt(-alpha[1]^2-beta[1]^2))/(1+exp(x*alpha[1]+y*beta[1]-z*sqrt(-alpha[1]^2-beta[1]^2)));
For any value of alpha[1] and beta[1] the term highlighted red becomes the imaginary form. I want to separate the real and imaginary parts of this function. Kindly help me in this matter, thanks

## Problem in comparing the coefficient of ...

Maple 2015

Dear Useres!

Hope everyone is fine here! I want to compare the coeficient of exp(k*eta[3]+m*eta[1]+n*eta[2]) for k=0,1,2,3,...,N,n=0,1,2,3,...,N and m=0,1,2,3,...,N for N=10 in the following attached file. But I got some error, please have a look and try to fix it as early as possible. Please take care and thanks

Compare_coeff.mw

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