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Finding the MacLaurin Series of a function

  • Thread starter Badmouton
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  • #1
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I have to find the Maclaurin series of:
(1) f(x)=cos(x)+x,
(2) g(x)= cos(x^2)+x^2
(3) h(x)=x*sin(2x).


I'm stuck at the first one, I kind of understand the concept of how P(0)=f(0)+f'(0)x+(f''(0)x^2)/2+. . .
What it gave me when I started calculating the value of the fn was this:
f(0)=cos(0)+0=1
f'(0)=-sin(0)=0
f''(0)=-cos(0)=0

And the pattern kept repeating as follows: 1,0,-1,0,1,0,-1,0.

So when I want to write the mclaurin series, should it come out as?
P(x)=Ʃ(x2n(-1)n)/n!

As for the other problems, I really don't know how to start
 

Answers and Replies

  • #2
326
3
I have to find the Maclaurin series of:
(1) f(x)=cos(x)+x,
(2) g(x)= cos(x^2)+x^2
(3) h(x)=x*sin(2x).


I'm stuck at the first one, I kind of understand the concept of how P(0)=f(0)+f'(0)x+(f''(0)x^2)/2+. . .
What it gave me when I started calculating the value of the fn was this:
f(0)=cos(0)+0=1
f'(0)=-sin(0)=0
f''(0)=-cos(0)=-1

And the pattern kept repeating as follows: 1,0,-1,0,1,0,-1,0.

So when I want to write the mclaurin series, should it come out as?
P(x)=Ʃ(x2n(-1)n)/n!

As for the other problems, I really don't know how to start
You mean -cos(0) = -1.

Yes, you should get somewhat that pattern. Actually, you are close with the answers you have. It's not x^(2n)(-1)ⁿ/n! since x^(2n) doesn't occur in (cos(x) + x) altogether! You can only express cos(x) as the Maclaurin series.

See: http://www.wolframalpha.com/input/?i=cos(x)

You should get the answer.
 

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