use taylor series result to find value?

4 Ansichten (letzte 30 Tage)
Raphael Hatami
Raphael Hatami am 12 Sep. 2019
Beantwortet: GURU PRASAD am 16 Sep. 2021
Hello, I think this is relatively simple but I can't figure it out. I am computing the taylor series of a function and simply want to evaluate that taylor series result at a specific value x. How can I do this automatically (without copy/pasting the result from the command window)? Thank you
%y = taylor(f,xx,0,1)
syms x
f=@(x) 25*x.^3-6*x.^2+7*x-88;
truevalue = f(3)
T0 = f
T1 = taylor(f,x,1,'Order',1)
T2 = taylor(f,x,1,'Order',2)
T3 = taylor(f,x,1,'Order',3)
result = T3(3)
%T2result = @(x) 70*x - 132
%T2estimate = T2result(3)
%T3result = @(x) 70*x + 69*(x - 1)^2 - 132
%T3estimate = T3result(3)
The result part is what I'm trying to do...

Antworten (2)

David Hill
David Hill am 12 Sep. 2019
result=subs(T3,x,3);
  2 Kommentare
John D'Errico
John D'Errico am 12 Sep. 2019
Or, you could use matlabFunction to create a function handle that will evaluate the function, converting the symbolic expression into one that will work as a simple function.
MINATI
MINATI am 4 Jan. 2020
how to apply the values f(0)=0;f'(0)=1;f"(0)=a;(syms a) in
taylor(f(x),x,'order',3) to find x+a*x^2/2;
Help please @
minatipatra456@gmail.com

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GURU PRASAD
GURU PRASAD am 16 Sep. 2021
f = @(x) 25*x^3-6*x^2+7*x-88;
sample_at = 3;
trueval = f(sample_at);
syms xx;
fs = f(xx);
for n = 0:4
y = double( subs( taylor(fs, n, 1), xx, sample_at ) );
disp([n, y]);
%now calculate the relative error
end

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