Maclaurin Polynomials

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1. Find the Maclaurin polynomials of order n = 0, 1, 2, 3, and 4, and then find the nth MacLaurin polynomials for the function in sigma notation.

cos(∏x)





2. Here is what I did:
p0x = cos (0∏) = 1
p1x = cos(0∏) - ∏sin(0∏)x = 1
p2x = cos(0∏) - ∏sin(0∏)x -[itex]\frac{∏2(cos∏x)(x2)}{2!}[/itex](

and so on.....

And the pattern that forms are that the odd values for k are 0.

So p0x = p1x
and p2x = p3x

etc etc.

fk(x) = (∏)kcos(∏x)'
and fk(0) = (∏)k(-1)k

As for the sigma notation:

Here is what I obtained:
Ʃnk = 0([itex]\frac{(∏2k(-1)k}{k!}[/itex](x)2k

According to the solution manual however, there is a n/2 on top of the sigma notation instead of a n.

I don't understand .
 

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  • #2
LCKurtz
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1. Find the Maclaurin polynomials of order n = 0, 1, 2, 3, and 4, and then find the nth MacLaurin polynomials for the function in sigma notation.

cos(∏x)





2. Here is what I did:
p0x = cos (0∏) = 1
p1x = cos(0∏) - ∏sin(0∏)x = 1
p2x = cos(0∏) - ∏sin(0∏)x -[itex]\frac{∏2(cos∏x)(x2)}{2!}[/itex](

and so on.....

And the pattern that forms are that the odd values for k are 0.

So p0x = p1x
and p2x = p3x

etc etc.

fk(x) = (∏)kcos(∏x)'
and fk(0) = (∏)k(-1)k

No. That isn't right if k is odd.

As for the sigma notation:

Here is what I obtained:
##\sum_{k=0}^n\frac{(-1)^k\pi^{2k}x^{2k}}{k!}##
[Fixed your latex. Don't mix other symbols with tex]
According to the solution manual however, there is a n/2 on top of the sigma notation instead of a n.

I don't understand .

I think the only thing wrong with your final answer is you should have a ##(2k)!## in the denominator. Then note that your general sum gives only the even powers, which are the only non-zero terms. Your book's answer probably has something to do with indexing it differently. Hard to say without seeing it.
 

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