Use the Maclaurin series to solve the following: $$ \frac{d^6}{dx^6}(x^4e^{x^2}) $$ I got about halfway through the problem before getting stuck. I am not sure how to solve it... Any advice? Also, sorry for my formatting mistakes, computer nube here. Feel free to correct them.
Please help! I'm really stuck, and I need an answer soon!
On
Firstly, I like this question. So I will tell you how to solve it without actually solving it for you. Because that would spoil the fun.
Step 1: Compute the Maclaurin series for $e^{x^2}$. Remember that the formula for a Maclaurin series is given by the following. $$f(x)=f(0)+xf^{\prime}(0)+\frac{x^2}{2!}f^{\prime\prime}(0)+\frac{x^3}{3!}f^{\prime\prime\prime}(0)+\ldots$$ So let $f(x)=e^{x^2}$. Then $f^{\prime}(x)=2xe^{x^2}$ (why?), and $f^{\prime\prime}(x)=2xe^{x^2}+2e^{x^2}$. Continuing, you should realise that not all of the terms are zero when you substitute in $x=0$. You should also spot a pattern, which is that $f^n(x)=f^{n-1}(x)+2f^{n-2}(x)$. EDIT: As 7raiden7 has pointed out in the comments, it is much easier to recall the Maclaurin series for $e^z$ and then substitute in $z=x^2$. This will save you a good few minutes!
Step 2: Realise that $x^4e^{x^2}=x^4f(0)+x^5f^{\prime}(0)+\frac{x^6}{2!}f^{\prime\prime}(0)+\frac{x^7}{3!}f^{\prime\prime\prime}(0)+\ldots$
Step 3: Differentiate this new series six times, using the fact that its derivative is the sum of the derivative of the terms of the new series.
Exercise: Use your working to evaluate $\frac{d^{104}}{dy^{104}}(x^4e^{x^2})$.
Expand the function as a series, then differentiate term by term, and try to identify the resulting series. Looking at the function to differentiate, you'll have to decompose into a sum of series of the form $x^k \mathrm{e}^{x^2}$.