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September 12

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Simplify integral

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Can this be simplified?

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

Thanks in advance--Yanwen (talk) 03:54, 12 September 2009 (UTC)[reply]

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)[reply]
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)[reply]
Ahha. Leibniz integral rule was what I was missing.
--Yanwen (talk) 01:39, 13 September 2009 (UTC)[reply]

what is the value of k that must be added to 7,16,43,79 so that they are in proportion?

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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)[reply]

5.--RDBury (talk) 14:16, 12 September 2009 (UTC)[reply]
only if 16=19...Tinfoilcat (talk) 16:19, 12 September 2009 (UTC)[reply]
In proportion to what? --Tango (talk) 16:20, 12 September 2009 (UTC)[reply]
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)[reply]
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)[reply]
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)[reply]
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)[reply]

Fourier Transforms

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Resolved

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

For , do we have (that's the definition I'm using, so I know that bit's correct) . Is that correct?

Because I thought , but then so which unless I'm being stupid becomes 0 at , doesn't it? But then that doesn't equal , 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 , 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 , 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)[reply]

Let me point out one error you made in computing the Fourier Transform of . It is true that ; however
.
Instead use the relation to compute the Fourier transform integral (Hint: is sum of two shifted sinc functions). Abecedare (talk) 00:11, 13 September 2009 (UTC)[reply]
Yeah, we can caluculate the Fourier transform directly using integration by parts twice.
As for the first integral, well, you have some problems. For example, the integrand has an even-ordered pole when . This singularity is non-removable since the numerator of the integrand is always positive for real t. In fact
As for the second integral, well, again:
~~ Dr Dec (Talk) ~~ 01:19, 13 September 2009 (UTC)[reply]

Oh god, how stupid of me, there's no minus in the numerator - sorry about wasting your time! Do things work better with and ?Delaypoems101 (talk) 02:08, 13 September 2009 (UTC)[reply]

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