# Coefficient for a Term in a Taylor Expansion for Cosine

1. Jan 15, 2017

### Drakkith

Staff Emeritus
1. The problem statement, all variables and given/known data
The coefficient of the term (x−π)2 in the Taylor expansion for f(x)=cos(x) about x=π is:

2. Relevant equations
$cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!}...$

3. The attempt at a solution

Unless my taylor series for cosine is incorrect, I'm not seeing any (x-π)2 term. What am I missing here? I see that the 2nd term is a squared term and that the coefficient is -1. Is that the answer?

2. Jan 15, 2017

### Ray Vickson

You need to apply the correct definition of a Taylor series. You are using the Maclaurin expansion, not the Taylor expansion.

3. Jan 15, 2017

### Drakkith

Staff Emeritus
I see. Do you have a good reference that has the correct series? Everything I've found so far has the taylor series for cosine exactly as I have it. Very frustrating.

4. Jan 15, 2017

### Staff: Mentor

5. Jan 15, 2017

### Drakkith

Staff Emeritus
Thanks Fresh. I usually avoid wikipedia when doing homework as I rarely find it helpful, but it appears this time I was incorrect.

6. Jan 15, 2017

### Staff: Mentor

They are usually quite correct in the definition section and fast to look up. I mostly look it up in two languages, as it's not a 1:1 translation.

7. Jan 15, 2017

### Drakkith

Staff Emeritus
Okay, wiki's definition with a = π:
$F(π) + \frac{F'(π)}{1!}(x-π) + \frac{F''(π)}{2!}(x-π)^2 + \frac{F'''(π)}{3!}(x-π)^3 + ...$

The first few derivatives of cos(x) are:
$F(x) = cos(x)$
$F'(x) = -sin(x)$
$F''(x) = -cos(x)$
$F'''(x) = sin(x)$

So that would be:
$cos(π) - sin(π)(x-π) - \frac{cos(π)}{2}(x-π)^2 + \frac{sin(π)}{6}(x-π)^3 + ...$

or

$-1 - 0 - \frac{1}{2}(x-π)^2 + 0 + ...$

So the coefficient would be -1/2.

8. Jan 15, 2017

### Staff: Mentor

I haven't done it, but something like this, yes. Maybe you'll have to take a few terms more, as they all include lower powers of $x$.

9. Jan 15, 2017

### Staff: Mentor

As already mentioned what you have is the Maclaurin series for the cosine function. A Maclaurin series is a series in powers of x (or powers of (x - 0)). A Taylor's series is one that is in powers of x - a. A Maclaurin series is a special case of a Taylor's series.

Your textbook should have definitions of both types of series.

10. Jan 15, 2017

### Drakkith

Staff Emeritus
Unfortunately I don't have my old textbook with me, so I can't look it up, and I didn't look at what the homework would be over before I left my house. I've had a really bad start to the semester thanks to some sleep issues and I'm still trying to get everything together and planned out.

Thanks guys.

11. Jan 16, 2017

### Ray Vickson

Since you have already solved the problem I guess it is OK to now point out that we can write $\cos(x) = -\cos(y)$, where $y = \pi - x$; that is a standard trigonometric identity (and is obvious geometrically). When $x$ is near $\pi$, $y$ is near 0, expanding in powers of $y$ makes sense.