How Do You Find the Maclaurin Polynomials for cos(πx)?

[Mosaness]

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.

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

As for the sigma notation:

Here is what I obtained:
Ʃⁿ_{k = 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 .

 

[LCKurtz]

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.

f^k(x) = (∏)^kcos(∏x)'
and f^k(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.