Maclaurin Series Help (1st Post)

[jasonleroy]

1. Find the Maclaurin series for f(x) = cos(x³) and use it to determine f⁶(0)
2.I know what the series expansion is. My question is, are they asking what the 6th term is with x set equal to 0? If so, all terms would be equal to zero. According to the book, the solution is -360

 

[DryRun]

It's a good method to list the derivatives and its corresponding value at x=0. Work your way up to the 6th power of f(x). Add them all up. Voila!

f(x)=\cos (x^3) then, f(0)=1
f'(x)=-3x^2\sin (x^3) then, f'(0)=0
Etc.

 

[jasonleroy]

Thanks! What is the point of finding the series expansion if you have to do it manually anyway?

 

[DryRun]

Thanks! What is the point of finding the series expansion if you have to do it manually anyway?

Your problem in post #1 consists of 2 parts actually. The first part asks that you find the Maclaurin series, but it didn't tell you up to which term (it goes on to infinity), and as it turns out that missing bit of information is given as part 2, which asks you to stop at the 6th derivative.

 

[jasonleroy]

That makes sense. This section has been expressing functions as a series in summation notation. If I were to just find the summation, it doesn't get me to the correct answer. What is the summation even used for if it doesn't get you to the same solution?

 

[jasonleroy]

\Sigma((-1)ⁿx⁶ⁿ)/2n! If I were to use this, substituting 0 for x, I'd get nowhere.

 

[DryRun]

That makes sense. This section has been expressing functions as a series in summation notation. If I were to just find the summation, it doesn't get me to the correct answer. What is the summation even used for if it doesn't get you to the same solution?

The Maclaurin theorem (it's actually based on Taylor's theorem with x=0) provides an estimate for the error involved when the function f(x) is approximated near the point x=0 by the polynomial of degree n in (x-0) which best describes the behavior of the function f(x) near that point.

So, the more terms (higher derivatives) you add, the closer you get to the actual answer.

 

[jasonleroy]

So you can't actually use the summation to find the correct answer?

 

[DryRun]

\Sigma((-1)ⁿx⁶ⁿ)/2n! If I were to use this, substituting 0 for x, I'd get nowhere.

This is a power series with center 0.
\sum_{n=0}^{\infty} \frac{(-1)^nx^{6n}}{2n!}
Put n=0 to get your first term, then put n=1 for the second term, etc. Add them up to get your power series.

So you can't actually use the summation to find the correct answer?

The answer to your problem in post #1, is only to find the term f^6(0). There is no summation required. Consider it as an indirect way of getting the answer.

By the way, you haven't show any calculations. That's not how things work on this forum. I've been very generous with the replies, but most people won't bother if you don't show your efforts.

 

[jasonleroy]

I'm taking this class online, so I think I may be missing the big picture. In my thinking, if I use the summation to find the 6th term I would have:

(-1)⁶x³⁶/12!

But this isn't the same as the 6th derivatine of cos(x³).

I think my question may actually be, how are these 2 methods related? Or are they?

I appreciate your help.

 

[DryRun]

I'm taking this class online, so I think I may be missing the big picture. In my thinking, if I use the summation to find the 6th term I would have:

(-1)⁶x³⁶/12!

But this isn't the same as the 6th derivatine of cos(x³).

I think my question may actually be, how are these 2 methods related? Or are they?

I appreciate your help.

To put simply, the method for working out a power series is different from the Maclaurin theorem, as i have explained in the replies above.