How to do a Taylor expansion with a matrix

20 ビュー (過去 30 日間)
kuroshiba
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
Ashutosh
Ashutosh 2023 年 5 月 26 日
I am not sure I understand your query. A Taylor expansion can be constructed for a function of some variable x. What you are feeding into the Taylor function taylor(), seems to be a constant. You can't have a Taylor expansion, approximation or anything for a constant.
kuroshiba
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!

サインインしてコメントする。

採用された回答

Torsten
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
or
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
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.

サインインしてコメントする。

その他の回答 (2 件)

KSSV
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];
eig(G)
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
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);
taylor(f(x))
ans = 
8

サインインしてコメントする。

カテゴリ

Help Center および File ExchangeCalculus についてさらに検索

製品


リリース

R2022b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by