Recent content by iceman713

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    Find the 100th Derivative of e^(-x^2)

    Thanks very much. It's a little awkward to think that the 100th term in the series is not the 100th term in the original Taylor/MacLaurin series but I suppose that's what I didn't realize.
  2. I

    Find the 100th Derivative of e^(-x^2)

    Homework Statement Find the exact value for the 100th Derivative of f(x) = e^{-x^{2}} evaluated at x = 0 Homework Equations e^{x} = \sum^{\infty}_{0} \frac{x^{n}}{n!} e^{-x^{2}} = \sum^{\infty}_{0} \frac{(-1)^{n}x^{2n}}{n!} The Attempt at a Solution The only thing I can think of to do is to...
  3. I

    Finding Maclaurin Series Expansions of Functions

    I also have another question but don't feel like making another thread for it, I've already typed it up on mathbin: http://mathbin.net/2926
  4. I

    Finding Maclaurin Series Expansions of Functions

    I tried doing that, the issue I ran into was that the question asks for the first four non-zero terms of the series. I attempted to take a few derivatives, but by the 3rd one I was already dealing with 4 separate sums of functions(that in turn have product rule applied and each become 2 more...
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    Finding Maclaurin Series Expansions of Functions

    Hey, here's is my problem as the exam states it: A) Write out the first four non-zero terms of the Maclaurin Series for F(x) = (1+X^{7})^{-4} Give all of the coefficients in exact form, simplified as much as possible. B) Find the exact value of the 21st order derivative of F(x) =...
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    Limits, shrinking circles, points of intersection and love

    Look at the picture, it is indeed problem 60. :-p And I did completely forget that I wasn't allowed to just graph it, as I type this I'm trying to force my eyes to not look at your post so I won't cheat. Nice tip too, thanks. Edit: whoooo, I did it
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    Limits, shrinking circles, points of intersection and love

    Homework Statement http://www.nellilevental.com/calc1.jpg Homework Equations C1 = \left(x - 1\right) ^{2} + y^{2} = 1 C2 = x^{2} + y^{2} = r^{2} The Attempt at a Solution Initially I thought that the "value" of R was just going to increase without bound since the slope of the line was...
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