Ooh, one I know for sure!

Kay, so the Maclaurin series for f(x) is Sum_n=0^inf (f⁽ⁿ⁾(0)*xⁿ/n!)

So if you can get the series in that form, which might be a pain, f⁽⁴⁾(0)/4! would be the fifth coefficient, and you can then easily find f⁽⁴⁾(0).

I've done that, and that's where the problem really hits, it evaluates to this.

f⁽⁴⁾(0) = x⁴ + 4x³ + 6x² + 4x + 1

I know that the answer should be 25, but I can't figure out a way to work backwards from 25, or forwards from the fourth derivative.

By the way, next time I have a Calculus question, can I just pm you? Since you're the only one answering anyways.

Sounds good, but isn't it fun displaying all this crazy calculus to the world?

All right:

e^{x+x^2}~=1+(x+x^2)+(x+x^2)^2/2!+(x+x^2)^3/3!+(x+x^2)^4/4!

No way I'm expanding that by hand...TI-89 time!

Okay, expanding those five terms should come out like this:

e^{x+x^2}~=1+x+3x^2/2!+7x^3/3!+25x^4/4!+ blahdeyblah (the rest isn't important since we are looking for f⁽⁴⁾(0) )

So using the general definition of a Maclaurin series for a function that I mentioned in the first comment, the fifth term in the series should be f⁽⁴⁾(0)*x⁴/4!. The fifth term looks to be 25x⁴/4!, so that means f⁽⁴⁾(0) is 25.

But isn't the fifth coefficient just (x+x^2)^4/4! ?

Or do you have to put it in the form f⁽ⁿ⁾(0)*xⁿ/n! before it's even valid?

Yeah, you have to expand it and rewrite it into that form. :X

And coefficient =/= term.

Doh!

One more question for the night, what about the function e^sin(x)? With the same criteria as above.

How would you rewrite that into the above form?

O_O

Do you have to use it to find f⁽⁴⁾(0) again? Or can you build the series by finding the derivatives separately and plugging them into the f⁽ⁿ⁾(0)*xⁿ/n! form?

The problem want's me to use the Maclaurin series for e^x to find the series for e^sin(x), and then find the fourth derivative.

Actually, I bet it wants me to then use the Maclaurin series for sin(x) after plugging it into the series for e^x. That sound about right?

I think so. Going the 1+sin(x)+(sin(x))^2/2!+(sin(x))^3/3!+... route sounds the best. You would be in worlds of pain trying to convert that to Sum f⁽ⁿ⁾(0)*xⁿ/n! form.

Alrighty then, thanks for all your help!

You're welcome! I enjoy revisiting my Calc BC days.