0
$\begingroup$

I want to show that a sequence $\sin(n\pi x)$ where $x$ is in $[0,1]$ is decreasing? is there any test I can apply and if there is no test can somebody please tell me how to show it?

since I think that i am doing wrong. what I did:

$\sin((n+1)\pi x)=\sin(n\pi x+\pi x)=\sin(n\pi x)\cos(n\pi)+\sin(n\pi)\cos(n\pi x)<\sin(n\pi x)$ but why???

thanks.

$\endgroup$
1
$\begingroup$

It's not. Let $x = \frac{1}{2}$. Then your sequence is $0, 1, 0, -1, \dots$.

$\endgroup$
  • $\begingroup$ actually i was trying to show that a sequence of series uniform convergence by using Abel's Test and function in the sum includes term of sin(nπx) and i have to show that this is decreasing but how?any comment. $\endgroup$ – ruud Mar 19 '14 at 1:25
  • $\begingroup$ Well, if you want an answer to that question, then I think it would be good for you to ask it. $\endgroup$ – wckronholm Mar 19 '14 at 2:10
  • $\begingroup$ ok. the series is Un(x)=A(x+ (sum sign ((-1)^n/n) e^(-n^2 π^2 t) sin(nπx)) i need to use Abel test I checked everything except for monotonic decreasing. $\endgroup$ – ruud Mar 19 '14 at 2:37
  • $\begingroup$ I meant ask it as a new question. And when you do, take the time to format the math symbols properly. $\endgroup$ – wckronholm Mar 19 '14 at 2:51
0
$\begingroup$

As wckronholm says, you can see that the statement is wrong because $x=\frac12$ produces an oscillating sequence. In fact, any nonzero rational number will produce an oscillating sequence. Irrational numbers will exhibit similar up-down-up behavior but they will not actually repeat numerically.

As for the argument, it has a typographical error which turns out to be fatal: it states that $\sin(n\pi x + \pi x) = \sin(n\pi x)\cos (n\pi) + \sin(n\pi)\cos(n\pi x)$ but what is actually true is that $$\sin(n\pi x + \pi x) = \sin(n\pi x)\cos (\pi x) + \sin(\pi x)\cos(n\pi x)$$ from which the final inequality no longer follows.

$\endgroup$

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.