# How to do a Taylor expansion with a matrix

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kuroshiba le 26 Mai 2023
Commenté : Paul le 1 Juin 2023
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!
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kuroshiba le 1 Juin 2023
I see, I had mistakenly thought that Taylor expansion could be done with constants!

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### Réponse acceptée

Torsten le 26 Mai 2023
Modifié(e) : Torsten le 26 Mai 2023
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 =
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kuroshiba le 1 Juin 2023
That is what I wanted to know! Thank you!
I'm sorry for the lack of words and understanding that bothered you.

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### Plus de réponses (2)

KSSV le 26 Mai 2023
syms x
f = exp(x)
f =
T = taylor(f)
T =
In place of x substitue each value of G.
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John D'Errico le 26 Mai 2023
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?
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Paul le 1 Juin 2023
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

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