Problem with a matlab question

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Joe
Joe on 5 Aug 2013
Currently doing revision for a MATLAB module next year. Was attempting a question and was hoping for some help it.
I am asked to consider this Taylor series expansion:
S(x,n)=(x-1) - 1/2(x-1)^2 + 1/3(x-1)^3 - 1/4(x-1)^4 +...+ ((-1)^(n-1))*(1/n)*(x-1)^n
From this write a matlab function that, given the values of x and n, returns the value of S(x,n).
Obviously, I do not want someone to do the problem for me and just copy it. I would just like some advise and how to represent this as a script.
Thank you in advance for your time

Accepted Answer

Evan
Evan on 5 Aug 2013
Edited: Evan on 5 Aug 2013
There are multiple ways you can do this. The first involves a for loop and, while it might require more lines of code, is probably a more basic route to follow and might be beneficial for getting into the swing of MATLAB.
The second way, however, is much more compact and takes advantage of MATLAB's strengths: array operations. A hint: in order to do this, the element-by-element array operators (.* ./ .^) will be needed instead of the normal matrix arithmetical operators (* / ^).
For either method, what you are needing to do is the same: you want to apply some arithmetic operation to all integer values from 1 to n, then you want to add up the results.
If that's too vague or you think an example might help, just let me know.
  5 Comments
Jan
Jan on 6 Aug 2013
Edited: Jan on 6 Aug 2013
Btw, the power operation is very expensive. Although runtime will most likely not matter in your case, this is the general approach to save time:
c = 1;
sgn = -1;
for n = 1:m
c = c * (x - 1);
sgn = -sgn;
S = S + sgn .* (c ./ n);
end
In the next step you can omit sgn also by injecting it into c: c = c * (1 - x)

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More Answers (1)

Iain
Iain on 5 Aug 2013
Its hard to give you help without giving you the answer to this one.
Option 1. Recursively call the script.
S = func(x,n)
function s = func(x,n) if n >0 s = "this term" + func(x,n-1); else s = "this term"; end
Option 2. Vectorize the summation
n = 1:n;
S = sum((-1).^(n-1) .* 1./n .*(x-1).^2 ))
  1 Comment
Joe
Joe on 5 Aug 2013
Yeah I had started to do that. What I had done is S = ((-1).^(n-1) .* 1./n .*(x-1).^2 )).
As this would just provide you with a function where it would solve the nth order. I'm assuming that by using the 'sum' function it will add up all orders up to the desired one. From the help you and Evan have provided I feel I have a much better understanding of the problem. Thank you

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