HaHa , we can apply magic differentiation in this problem Again !

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- Aug 22nd 2009, 10:04 PMsimplependulum
- Aug 22nd 2009, 10:16 PMBruno J.
- Aug 23rd 2009, 12:45 AMluobo
- Aug 23rd 2009, 04:10 AMsimplependulum
luobo : i am looking forward to seeing your solution , actually , I have been shocked to see how you (or someone ) applied magic differentiation in the Integral Bee !! It is really a powerful tool (Clapping)

i think the solution is two straight lines passing throguh the origin (at,bt) and (-at,bt) (Happy) - Aug 23rd 2009, 11:13 AMPaulRS
- Aug 23rd 2009, 01:20 PMKrizalid
Here's another solution to my proposed problem using a powered tool:

Now since and is an integrable function on besides on we get and on we have hence the Dominated Convergence Theorem applies and - Aug 23rd 2009, 09:39 PMynj
- Aug 23rd 2009, 10:20 PMNonCommAlg
finding all for which the above equality holds requires considering several cases. i'm not going to do that! haha so i will only evaluate the integral.

let then:

1) therefore applying integration by parts twice gives us:

2) we know that where is the sign function.

3) using some routine trigonometry identities we get: .

4) applying 1) to 3) and then using 2) will finally give us:

- Aug 23rd 2009, 10:56 PMsimplependulum
- Aug 24th 2009, 02:33 AMluobo

The problem would be easier to solve if it changes this way:

Solve for real values of

or even

Step (1): Show for a fixed , the following integral is proportional to , i.e.

Step (2): Solve for real values of such that

For Step (1), can we use the "differentiation"? hahaha... - Aug 24th 2009, 02:38 AMNonCommAlg
- Aug 24th 2009, 05:12 AMluobo
- Aug 24th 2009, 05:20 AMluobo
- Aug 24th 2009, 05:28 AMmr fantastic