The question is:
Show that f(x + π) = (-e^(-1/5)* π)*f(x)
Background info:
f(x) = (e^(-1/5)* x)*sin x
I would really appreciate a walkthrough answer because I've been trying for hours to get their solution with no such luck.
Thanx
The question is:
Show that f(x + π) = (-e^(-1/5)* π)*f(x)
Background info:
f(x) = (e^(-1/5)* x)*sin x
I would really appreciate a walkthrough answer because I've been trying for hours to get their solution with no such luck.
Thanx
The only thing I can think to suggest is just to do it!
Since f(x) is defined as $\displaystyle e^{-x/5} \sin (x)$, $\displaystyle f(x+ \pi) = e^{-(x+\pi)/5} \sin (x+\pi)$. Now you should know that $\displaystyle \sin (x+\pi)= -\sin (x)$ and, by the "laws of exponents", $\displaystyle e^{-(x+ \pi)/5}= e^{-x/5- \pi/5}= e^{-x/5}e^{-\pi/5}$. Putting those together,
$\displaystyle f(x+\pi)= e^{-x/5}e^{-\pi/5}(-\sin (x))= -e^{-\pi/5}e^{-x/5} \sin (x)= -e^{-\pi/5}f(x)$.