Coefficient for a Term in a Taylor Expansion for Cosine

[Drakkith]

Homework Statement

The coefficient of the term (x−π)² in the Taylor expansion for f(x)=cos(x) about x=π is:

Homework Equations

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

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?

 

[Ray Vickson]

Homework Statement

The coefficient of the term (x−π)² in the Taylor expansion for f(x)=cos(x) about x=π is:

Homework Equations

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

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?

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

 

[Drakkith]

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

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.

 

[fresh_42]

You could use Wiki's first definition here: https://en.wikipedia.org/wiki/Taylor_series with $a= \pi$

 

[Drakkith]

You could use Wiki's first definition here: https://en.wikipedia.org/wiki/Taylor_series with $a= \pi$

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

 

[fresh_42]

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

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.

 

[Drakkith]

You could use Wiki's first definition here: https://en.wikipedia.org/wiki/Taylor_series with $a= \pi$

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.

 

[fresh_42]

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 $x$.

 

[Mark44]

Everything I've found so far has the taylor series for cosine exactly as I have it.

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.

 

[Drakkith]

Your textbook should have definitions of both types of series.

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.

 

[Ray Vickson]

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.

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.