1
$\begingroup$

My Calc 2 teacher did not really teach us how to do these and I have a test on it tomorrow. Any help you can provide to help me understand this type of problem would be much appreciated.
The question:

Show that the given power series is a solution to the differential equation $$y''+y=0$$ $$y=\sum_{n=0}^{\infty}(-1)^n\frac { x^{2n+1}}{(2n+1)!}$$ Thank you in advance!!

$\endgroup$
4
  • $\begingroup$ That‘s just the $\sin$ function. $\endgroup$ – Frieder Jäckel Apr 8 '18 at 22:27
  • $\begingroup$ How do you know that it is the sin function? My teacher did a really poor job at teaching us how to even recognize what function it is. $\endgroup$ – Chad Bowman Apr 8 '18 at 22:31
  • $\begingroup$ You can differentiate power series termwise $\endgroup$ – Frieder Jäckel Apr 8 '18 at 22:34
  • $\begingroup$ I did that and got ((2n+1)(-1)^n(x)^2n)/(2n+1)! but I am not sure where to start with my index. I have tested 0,1, and 2 but none of them work. $\endgroup$ – Chad Bowman Apr 8 '18 at 22:37
1
$\begingroup$

Hint

$$y=\sum_{n=0}^{\infty}(-1)^n\frac { x^{2n+1}}{(2n+1)!}$$ $$y'=\sum_{n=0}^{\infty}(-1)^n\frac { x^{2n}}{(2n)!}$$ $$y''=\sum_{n=1}^{\infty}(-1)^n\frac { x^{2n-1}}{(2n-1)!}$$ Substitute $n=m+1$ $$y''=\sum_{m=0}^{\infty}(-1)^{m+1}\frac { x^{2m+1}}{(2m+1)!}$$ $$y''=-\sum_{m=0}^{\infty}(-1)^{m}\frac { x^{2m+1}}{(2m+1)!}$$ Therefore $$y''=-y$$ And $$y''+y=0$$

$\endgroup$
1
  • $\begingroup$ Thank you so much $\endgroup$ – Chad Bowman Apr 9 '18 at 1:23
1
$\begingroup$

Suppose you just write out the first few terms

\begin{eqnarray} y&=&x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots\\ y^\prime&=&1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots\\ y^{\prime\prime}&=&-x+\frac{x^3}{3!}-\frac{x^5}{5!}+\frac{x^7}{7!}-\\ \end{eqnarray}

Then notice that

$$ y^{\prime\prime}+y=0$$

$\endgroup$
2
  • $\begingroup$ Thank you for the help!! $\endgroup$ – Chad Bowman Apr 9 '18 at 1:24
  • $\begingroup$ You are welcome. $\endgroup$ – John Wayland Bales Apr 9 '18 at 5:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.