Wikipedia:Reference desk/Archives/Mathematics/2009 September 12

= September 12 =

Simplify integral
Can this be simplified? $$\frac{d}{dv}\int_{-\infty}^{\infty}\phi_1(v_1) \left(\int_{-v_1}^{v-v_1}\phi_2(v_2)dv_2\right) dv_1$$

Probably some bad choice of variables here, I'll rewrite it first.

$$\frac{d}{dv}\int_{-\infty}^{\infty}f(y) \left(\int_{-y}^{v-y}g(x)dx\right) dy$$

Thanks in advance--Yanwen (talk) 03:54, 12 September 2009 (UTC)
 * I get f*g where * means convolution. I'm kinda rusty with this type of problem so take for what it's worth.--RDBury (talk) 05:05, 12 September 2009 (UTC)


 * This appears to be an exercise (I'm guessing homework) in using the Leibniz integral rule and the first fundamental theorem of calculus. 67.122.211.205 (talk) 19:09, 12 September 2009 (UTC)


 * Ahha. Leibniz integral rule was what I was missing.
 * $$\frac{d}{dv}\int_{-\infty}^{\infty}f(y)

\left(\int_{-y}^{v-y}g(x)dx\right) dy$$
 * $$\int_{-\infty}^{\infty}\frac{\partial}{\partial v}f(y)

\left(G(v-y)-G(-y)\right) dy$$
 * $$\int_{-\infty}^{\infty}f(y)g(v-y)dy$$--Yanwen (talk) 01:39, 13 September 2009 (UTC)

what is the value of k that must be added to 7,16,43,79 so that they are in proportion?
what is the value of k that must be added to 7,16,43,79 so that they are in proportion? —Preceding unsigned comment added by 122.168.68.145 (talk) 09:32, 12 September 2009 (UTC)
 * 5.--RDBury (talk) 14:16, 12 September 2009 (UTC)
 * only if 16=19... Tinfoilcat (talk) 16:19, 12 September 2009 (UTC)
 * In proportion to what? --Tango (talk) 16:20, 12 September 2009 (UTC)
 * My interpretation was that you're supposed to add a constant k to each term and end up with a bit of a geometric series (i.e. al, al^2, al^3, al^4). This problem does not have a solution Tinfoilcat (talk) 16:28, 12 September 2009 (UTC)
 * Perhaps they don't have to be consecutive terms from a geometric series? Or could it just be that one must divide the next, which divides the next, etc.? --Tango (talk) 16:41, 12 September 2009 (UTC)
 * The OP probably means that the first and second number should be in the same ratio as the third and fourth. In this case, RDBury's answer is correct. -- Meni Rosenfeld (talk) 19:49, 12 September 2009 (UTC)


 * As Meni stated, RDBury's answer is correct BUT, doesn't this strike anyone as a HW problem? hydnjo (talk) 21:08, 12 September 2009 (UTC)

Fourier Transforms
Could anyone let me know whether I'm doing this wrong?

For $$f(x)=\left\{ \begin{array}{lr} cos(x):|x|<\frac{\pi}{2}\\ 0\,\,\,\,\,\,\,\,\,\,\,\,\, :|x|>\frac{\pi}{2}\end{array} \right.$$, do we have $$\tilde{f}(k)=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}f(x)e^{-ikx} \, dx$$ (that's the definition I'm using, so I know that bit's correct) $$=Re(\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}e^{ix}e^{-ikx} \, dx) = Re[\frac{e^{(1-k)ix}}{(1-k)i}]_{\frac{-\pi}{2}}^{\frac{\pi}{2}}=(*)\,Re[\frac{-i \cdot 2i}{1-k}\,sin(1-k)(\frac{\pi}{2})]=2\frac{sin((1-k)(\frac{\pi}{2}))}{1-k}$$. Is that correct?

Because I thought $$\tilde{f'}(k)=ik\tilde{f}(x)$$, but then $$f'(x)=-sin(x)$$ so $$\tilde{f'}(k)=\tilde{-sin(x)}=\frac{-1}{i}Im[(\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}e^{ix}e^{-ikx} \, dx]$$ which unless I'm being stupid becomes 0 at $$(*)$$, doesn't it? But then that doesn't equal $$ik \cdot 2\frac{sin((1-k)(\frac{\pi}{2}))}{1-k}$$, does it? Perhaps I'm being dense.

Following that, I'm trying to show 'by considering the F.T. of the above function and its derivative', that $$\int_{0}^{\infty}\frac{\frac{\pi^2}{4}-cos^2(t)}{(\frac{\pi^2}{4}-t^2)^2}=\int_{0}^{\infty}\frac{t^2-cos^2(t)}{(\frac{\pi^2}{4}-t^2)^2}=\frac{\pi}{4}$$, but I haven't a clue how to do it! I know it's related to Parseval's relation, but other than possibly moving the 2 integrals together and canceling a $$(\frac{\pi^2}{4}-t^2)$$, then showing it's equal to 0 (which still doesn't solve the problem fully), i'm not sure how to move along, or whether that's even a good idea. Could anyone suggest anything perhaps?

Thanks a lot,

Delaypoems101 (talk) 23:50, 12 September 2009 (UTC)


 * Let me point out one error you made in computing the Fourier Transform of $$ f(x)$$. It is true that $$ \cos (x) = Re \{e^{ix}\}$$; however
 * $$ \int \cos (x) e^{-ikx} dx = \int Re\{e^{ix}\} \; e^{-ikx} dx \ne Re \{\int e^{ix} \; e^{-ikx} dx \}$$.
 * Instead use the relation $$ 2\cos (x) = e^{ix}+e^{-ix}$$ to compute the Fourier transform integral (Hint: $$\tilde{f}(k)$$ is sum of two shifted sinc functions). Abecedare (talk) 00:11, 13 September 2009 (UTC)


 * Yeah, we can caluculate the Fourier transform directly using integration by parts twice.
 * $$ \hat{f}(\xi) = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \!\! \cos(x)e^{-ix\xi} \ dx = \frac{2\cos\left(\frac{\pi}{2}\xi\right)}{1-\xi^2} $$
 * As for the first integral, well, you have some problems. For example, the integrand has an even-ordered pole when $$t = \pm\frac{\pi}{2}$$. This singularity is non-removable since the numerator of the integrand is always positive for real t. In fact
 * $$ \int_0^y \frac{\frac{\pi^2}{4}- \cos^2(t)}{(\frac{\pi^2}{4}-t^2)^2} \ dt \to \infty \ \mbox{as} \ y \to \infty \ . $$
 * As for the second integral, well, again:
 * $$ \int_0^y \frac{t^2 - \cos^2(t)}{(\frac{\pi^2}{4}-t^2)^2} \ dt \to \infty \ \mbox{as} \ y \to \infty \ . $$  Dr Dec  ( Talk )    01:19, 13 September 2009 (UTC)

Oh god, how stupid of me, there's no minus in the numerator - sorry about wasting your time! Do things work better with $$\int_0^y \frac{\frac{\pi^2}{4} \cos^2(t)}{(\frac{\pi^2}{4}-t^2)^2} \ dt$$ and $$\int_0^y \frac{t^2 \cos^2(t)}{(\frac{\pi^2}{4}-t^2)^2} \ dt$$?Delaypoems101 (talk) 02:08, 13 September 2009 (UTC)


 * No matter - got it sorted: thanks all! Delaypoems101 (talk) 03:41, 13 September 2009 (UTC)