I am working on a homework problem that asks the following: "Find an expression for the general term of each of the series below. Use $n$ as your index, and pick your general term so that the sum giving the series starts with $n=0$." $$x^5sin(x^2)=x^7-\frac{x^{11}}{3!}+\frac{x^{15}}{5!}-\frac{x^{19}}{7!}+...$$ general term=?
I am at a complete loss as far as where to start for this problem. Any assistance would be greatly appreciated.
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$\begingroup$You have probably covered that $$\sin(x)=\sum\limits_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}$$ Substituting $x^2$ instead, we get $$\sum\limits_{n=0}^\infty\frac{(-1)^nx^{4n+2}}{(2n+1)!}$$ Multiplying that by $x^5$ yields $$\sum\limits_{n=0}^\infty\frac{(-1)^nx^{4n+7}}{(2n+1)!}$$
$\endgroup$ $\begingroup$Notice that the powers of $x$ go up by $4$, and begin with $7$. The formula for an arithmetic progression with common difference $4$ and initial term $7$ is given by $7 + 4n$. These will be your powers of $x$.
Notice also that we have alternation between negatives and positives. That tells me that we want either a factor of $(-1)^n$ or $(-1)^{n+1}$, depending on whether we start with a negative or positive. The series above starts with a positive at $n = 0$, and so does $(-1)^n$, so $(-1)^n$ will also be a factor.
Finally, the factorials, once you fill in the $1!$ under the first term, also follow an arithmetic progression, beginning at $1$, but increasing by $2$ each time. So, on the denominator, we'll see $(2n + 1)!$.
Putting everything together, we get the series, $$\sum_{n=0}^\infty \frac{(-1)^nx^{7+4n}}{(2n+1)!}.$$ Hope that helps!
$\endgroup$ $\begingroup$The idea is to write $$x^7 - \frac{x^{11}}{3!} + \frac{x^{15}}{5!} - \cdots = \sum_{n \geq 0}a_nx^{b(n)}.$$Since we start with $x^7$ and we go on in powers of $4$, we must have $$\sum_{n \geq 0}a_n x^{7+4n}.$$The signs keep changing, so we'll have some $(-1)^n$ there. Also, note that the factorials in the denominators go like $1!$, $3!$, $5!$, etc, ranging over odd numbers. We end up with: $$x^5\sin(x^2) = \sum_{n \geq 0}\frac{(-1)^n}{(2n+1)!}x^{7+4n}.$$ This allows us to find out $$\frac{{\rm d}^k}{{\rm d}x^k}\bigg|_{x = 0}(x^5\sin(x^2))$$by just looking at the series: find $n$ such that $7+4n = k$ and compute the coefficient for this $n$.
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