In the Symbolic toolbox,how to get the results like in the old version

Question

Xizeng Feng 2021-11-22

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1592724-in-the-symbolic-toolbox-how-to-get-the-results-like-in-the-old-version

评论： Xizeng Feng 2021-11-23

采纳的回答： Paul

In the symbolic toolbox, the results in newer version are different from the old version. For example:

>> syms x

>> int(1/(x-x^2))

In new version：

ans =

2*atanh(2*x - 1)

In old version：

ans=

log(x)-log(x-1)

______________________________________________

and：

>> dsolve('Dx=x-x^2')

In new version：

ans =

1

0

-1/(exp(C2 - t) - 1)

In old version：

ans=

1/(1+exp((-t)*C1)

________________________________________________

And also:

>> int(x-x^2)

In new version：

ans =

-(x^2*(2*x - 3))/6

>> pretty(ans)

2

x (2 x - 3)

>> collect(ans)

ans =

- x^3/3 + x^2/2

- ------------

6

In old version：

ans=

1/2*x^2-1/3*x^3

______________________________________________________-

There even a bigger difference：

>>syms a b c d x;

>>solve('a*x^3+b*x^2+c*x+d=0')

In Matlab(V6.5), x can be expressed by a,b,c,d though it is a long expression.

But in Matlab R2018, the result is as follow：

>>syms a b c d x

>>eqn=a*x^3+b*x^2+c*x+d==0

>> solve(eqn,x)

ans =

root(a*z^3 + b*z^2 + c*z + d, z, 1)

root(a*z^3 + b*z^2 + c*z + d, z, 2)

root(a*z^3 + b*z^2 + c*z + d, z, 3)

It did nothing indeed!

I want to have the results like the ones in the old version, what can I do?

Thanks in advance for the help!

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Walter Roberson 2021-11-23

Note: in MATLAB 6.5, the symbolic toolbox was based around Maplesoft's Maple software; in current versions, it is based on the MuPAD symbolic engine.

Xizeng Feng 2021-11-23

thank you for the reminding!

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Answer 1

Paul 2021-11-22

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1592724-in-the-symbolic-toolbox-how-to-get-the-results-like-in-the-old-version#answer_837809

编辑：Paul 2021-11-22

在 MATLAB Online 中打开

% first example

syms x

g(x) = 1/(x - x^2);

f(x) = int(g(x))

f(x) =

simplify(rewrite(f(x),'log'),100)

ans =

% as it turns out, f(x) returned by 2021b is the same as the "old

% solution." But this didn't have to be the case; they could have differed

% by a constant because this usage of int returns an anti-derivative.

% second example

clear

syms x(t)

sol = dsolve(diff(x(t))==x-x^2)

sol =

% the second and third entries are valid solutions to the differential

% equation (for specific boundary condtions) as can be seen by direct subustitution.

% Don't know why the "old

% version" did not return these solutions. Also, we see that 2021b returns

% sol(1) in a different from than the "new version" in the Question. But

% it's the same answer as the old version

sol = sol(1);

[num,den] = numden(sol);

sol = 1/simplify(den/num)

sol =

% but C1 is arbitrary, so we are allowed to negate it

syms C1

sol = subs(sol,C1,-C1)

sol =

% we can also manipulate the expression in the Question

syms C2

g(t) = -1/(exp(C2 - t) - 1)

g(t) =

g(t) = expand(g(t))

g(t) =

% but C2 is an arbitrary constant, so we can sub it for a different

% arbitrary consant, which returns us to the form of sol given above.

g(t) = subs(g(t),exp(C2),C1)

g(t) =

% third example

clear

syms x

f(x) = int(x-x^2)

f(x) =

f(x) = expand(f(x))

f(x) =

% Fourth example. Use the MaxDegree option

clear

syms a b c d x

eqn=a*x^3+b*x^2+c*x+d==0;

sol = solve(eqn,'MaxDegree',3)

sol =

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Xizeng Feng 2021-11-23

thank you very much for your answer !

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Answer 2

Yongjian Feng 2021-11-22

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Seems like the same answer but presented in different form. The new version factorizes the result for example for

int(x-x^2)

It should not matter, right?

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