# Even function about a point over a restricted range

Why is $f(x)=(x-1)^2$sin$(n\pi x)$ even about $x=1$ for $0\leq x \leq2$?

I understand that $(x-1)^2$ is even about $x=1$ and I can plot the graph for various values of $n$ on wolfram alpha, but how do I go about proving the above is correct?

• How do you define being even about a point? – user84413 Jul 29 '14 at 16:52
• $f(a+x)=f(a-x)$ – user144895 Jul 29 '14 at 16:53
• do you mean with $a=1$? – Avitus Jul 29 '14 at 17:03
• @Avitus yes with $a=1$ – user144895 Jul 29 '14 at 17:49

$f(1+a)=a^2\sin(n\pi(1+a))=a^2\sin(n\pi+n\pi a)=a^2(\sin n\pi\cos n\pi a+\cos n\pi\sin n\pi a)$

$=a^2\cos n\pi\sin n\pi a$, but

$f(1-a)=a^2\sin(n\pi(1-a))=a^2\sin(n\pi-n\pi a)=a^2(\sin n\pi\cos n\pi a-\cos n\pi\sin n\pi a)$

$=-a^2\cos n\pi\sin n\pi a$; so the function does not appear to be even about 1.

For example, if $n=1$, $f(3/2)=-1/4$ while $f(1/2)=1/4$.

• But if you sketch it, it is even in the range $0\leq x \leq 2$, which is the region I am integrating over, hence the integral is zero over $0\leq x \leq 2$. – user144895 Jul 29 '14 at 17:48
• Even about x=1 in $0\leq x \leq 2$ – user144895 Jul 29 '14 at 17:58
• @lemony9201 I think the function is actually odd about x=1, if that is defined in the analogous way: $f(1-a)=-f(1+a)$. – user84413 Jul 29 '14 at 18:34
• Apologies you are correct, how can i prove this? – user144895 Jul 29 '14 at 18:43
• @lemony9201 I believe you can use the equalities above to show that $f(1-a)=-a^2\cos n \pi\sin n\pi a=-f(1+a)$. – user84413 Jul 29 '14 at 21:28