Taylor Series Approximation for e^-x

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MICHAEL RICHARDS
MICHAEL RICHARDS 2021-2-19
评论： Walter Roberson 2022-9-4
I'm trying to write a taylor series code for e^-x without using the taylor function in matlab. Each time I run the code I end up with an empty variable for my answer and I dont know whats wrong. Please help!
Here is my code:
syms ff(x)
normTrueError = TaylorSeries(0.25, 1, 1)
function [ans] = ff(x)
ans = exp(-x);
end
function [normTrueError] = TaylorSeries(xi, xiplus1, n)
h = (xiplus1 - xi);
fXiplus1 = ff(xi);
for i = 1:n
fXiplus1 = fXiplus1 + (diff(ff(xi), i)/factorial(i))*h^i;
end
trueValue = ff(xiplus1);
normTrueError = fXiplus1 - trueValue;
end
  2 个评论
James Tursa
James Tursa 2021-2-19
What is the point of the syms ff(x)? Aren't you just trying to calculate a numeric Taylor series approximation and compare it to the MATLAB exp( ) function? What is the actual wording of your assignment?
MICHAEL RICHARDS
MICHAEL RICHARDS 2021-2-19
I added the syms after trying to troubleshoot my code. Yes I am trying to calculate the numeric taylor series approximation with inputs xi, xi+1 and n. The code should output the normalized true error at xi+1.

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回答（2 个）

David Hill
David Hill 2021-2-19
编辑：David Hill 2021-2-19
function x=TaylExp(x)
x=sum((-x).^(0:18)./factorial(0:18));
end
  1 个评论
MICHAEL RICHARDS
MICHAEL RICHARDS 2021-2-19
That is not the taylor series approximation. Taylor series is:
f(xi+1)=f(xi)+f'(xi)*h+f''(xi)/2!*h^2+f'''(xi)/3!*h^3....etc

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dasari
dasari 2022-9-4
Ran in:
syms ff(x)
normTrueError = TaylorSeries(0.25, 1, 1)
normTrueError = []
function [ans] = ff(x)
ans = exp(-x);
end
function [normTrueError] = TaylorSeries(xi, xiplus1, n)
h = (xiplus1 - xi);
fXiplus1 = ff(xi);
for i = 1:n
fXiplus1 = fXiplus1 + (diff(ff(xi), i)/factorial(i))*h^i;
end
trueValue = ff(xiplus1);
normTrueError = fXiplus1 - trueValue;
end
  1 个评论
Walter Roberson
Walter Roberson 2022-9-4
syms ff(x)
That does not tell matlab to make the local function ff take symbolic inputs and return a symbolic result! When the local function ff is invoked, the exp() in it will return a result that is the same datatype as the input passed to it. Your function is passing xi to it but xi is numeric 0.25. exp(-0.25) is going to be a numeric result and you would then be taking numeric diff() of the scalar results, which is going to return [] because numeric diff() has to do with the difference between adjacent elements.
Your code is using diff() to take derivatives. You need to pass symbolic x to ff(), take the derivative of the result, and subs() xi for x in the result.

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