Power Series Represenation Problem... Sort of

[nolematterwhat]

I'm stuck on a particular homework problem, and I'm not even sure how to go about finding a solution.

I'm given a power series represenation of some function:

f(x) = SUM[n=0-->inf] (-1^n * 3^n * x^5n)/n!

I'm supposed to find the "exact" value of f(1) and the "exact" value of the 100th order derivative of f(0).

Wouldn't the exact value of f(1) simply be the power series minus the x value since x^5n would become 1 for all n?

Also, how would I go about finding the 100th derviative of x w/out differentiating every order from 1-99? I know it must have something to do with the definition of the Maclaurin series, but for the life of me, I just can't see it.

Thanks.

 

[ChaoticLlama]

Does the Maclaurin Series (Taylor series about x=0) ring any bells?

\(\displaystyle \L\ f(x) \approx f(0) + f'(0)x + \frac{{f''(0)}}{{2!}}x^2 + \frac{{f'''(0)}}{{3!}}x^3 + ... + \frac{{f^n (0)}}{{n!}}x^n\)

 

[nolematterwhat]

Yes... but that's the problem. I'm (either because I'm dense or tired) missing the connection of how to relate general series to my particular problem.

It looks as if the 100th derivative of f should be (for my particular problem stated above) f[100](0) = -1^100 * 3^100 but I know that just doesn't seem right.

 

[soroban]

Hello, nolematterwhat!

Here's part (a) . . . it requires a bit of Recognition.

Recall that: \(\displaystyle \L\,e^u\;=\;1\,+\,u\,+\,\frac{u^2}{2!}\,+\,\frac{u^3}{3!}\,+\,\frac{u^4}{4!}\,+\,\cdots\;\) [1]

\(\displaystyle \L f(x)\;=\;\sum^{\infty}_{n=0}\frac{(-1)^n\cdot3^n\cdot x^{5n}}{n!}\)

I'm supposed to find (a) the exact value of \(\displaystyle f(1)\)
\(\displaystyle \;\;\)and (b) the exact value of the 100th order derivative of \(\displaystyle f(0).\)

\(\displaystyle \L f(1)\;=\;\sum^{\infty}_{n=0}\frac{(-1)^n\cdot3^n}{n!} \;= \; 1\,-\,3\,+\,\frac{3^2}{2!}\,-\,\frac{3^3}{3!}\,+\,\frac{3^4}{4!}\,-\,\cdots\)

If you're lucky, you recognize the right side as [1] with \(\displaystyle u\,=\,-3\)

Therefore: \(\displaystyle \L\,f(1)\;=\;e^{-3}\)

 

[nolematterwhat]

OF COURSE! Thank you.

Dang it... I should've seen that.

Not to diminish my gratitude, but any ideas on the 100th derivative of x? Still lost on that one.

 

[nolematterwhat]

Just in case anyone was curious... I figured out the second part of my question.

Just needed to find Cn (which is f[n](0)/n!) then solve for f[n](0) to get the exact value of Cn.