How to do a Taylor expansion with a matrix

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kuroshiba 2023 年 5 月 26 日
コメント済み: Paul 2023 年 6 月 1 日
I have tried the official matlab website that describes the Taylor expansion, but it doesn't work!
G = [0,4;4,0];
T = taylor(exp(G));
error message "Function 'taylor' (input argument of type 'double') is undefined."
I would like to know the result of infinite convergence separately.
I would be glad if you could tell me!
  2 件のコメント
kuroshiba 2023 年 6 月 1 日
I see, I had mistakenly thought that Taylor expansion could be done with constants!
Thanks for your comments pointing that out!



Torsten 2023 年 5 月 26 日
編集済み: Torsten 2023 年 5 月 26 日
Maybe you mean:
G = [0,4;4,0];
Gexp = expm(G)
Gexp = 2×2
27.3082 27.2899 27.2899 27.3082
syms t
G = [0,4;4,0]*t;
Gexp = simplify(taylor(expm(G),t,'ExpansionPoint',1))
Gexp = 
Gexp = simplify(subs(expm(G),t,1))
Gexp = 
  1 件のコメント
kuroshiba 2023 年 6 月 1 日
That is what I wanted to know! Thank you!
I'm sorry for the lack of words and understanding that bothered you.


その他の回答 (2 件)

KSSV 2023 年 5 月 26 日
syms x
f = exp(x)
f = 
T = taylor(f)
T = 
In place of x substitue each value of G.

John D'Errico
John D'Errico 2023 年 5 月 26 日
You cannot compute the Taylor series of a constant. You CAN compute a Taylor series, and then evaluate it at that constant value, since the truncated series is then a polynomial.
Will only a few terms from that Taylor series be close to yielding a convergent result? This is something you need to consider, and that is a big part of your homework where you have shown no effort. The eigenvalues of G will be an important factor.
G = [0,4;4,0];
ans = 2×1
-4 4
So, given that, will a simple Taylor series for exp(x) converge well for x==4 (or x==-4, for that matter)? How many terms would you expect that to require? Why did I compute the eigenvalues of G here? How are they pertinent?
  1 件のコメント
Paul 2023 年 6 月 1 日
Isn't the Taylor series of function that's a constant just the constant?
syms f(x)
f(x) = sym(8);
ans = 



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