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Thanks

I use Wolframalpha some time to verify my final answer.

Thanks

I am glad to hear that.

I never read that.

Thanks for sharing

Right. I missed it only at this instant. Thanks for catching that.

What it is?

ruclips.net/user/shortsCL0VKs-Ub2M?si=A9rvBN1OL4bPiYDb

Sir in this video this is question no 12 please solve this

@@Kamlesh-x7u I am sorry, I have no idea what you are asking.

ruclips.net/user/shortsCL0VKs-Ub2M?si=A9rvBN1OL4bPiYDb

Thanks

A simple problem

It does not matter . Prod_(k=1 to n) e^(s/k) =Prod_(b=1 to n) e^(s/b)

That is correct.

My English skill is low level,but I'm glad if you reply me.

@@太田陽介-o3d It is okay. I understand what you are asking.1/(1/4)=4 and 1/(1/3)=3. Therefore every number between the interval (1/(1/4), 1/(1/3)) will have an integral value of 3. Same can be said about the other interval.

@@cipherunity Is it okay that the group of [1/x], x∈(3≦x<4) contains 1/4？ The definition of [x] is x -1≦ x <x +1, isn't it? I appreciate it if you reply to me🙇‍

@@太田陽介-o3d That is correct.

I am working on all kind of questions.

@@cipherunity 研究的問題是什麼?

@@user-ub1cg8cd9k Sorry I do not get your question

@@cipherunity 你目前的功課（進度）到哪了

@@user-ub1cg8cd9k Your question is not clear.

You are welcome. I am glad to hear that.

very good

@@cipherunity have you read over the problems yet? Are they of good enough quality for this channel

Tenho 13 anos

@@qwertyman123 Still busy with other things

Thanks

Thanks

Thanks

welcome

The 2nd substitution is very powerful. I used it in many places. It helps you to divide the logarithmic term in the numerator to suitable factors.

That is correct. It can be easily done without using Mellin transform

I shall look into it. You might see another video with alternate solution

Thanks

It is already there

I shall look into it.

I can look into it

​@@cipherunity Let n be a positive integer and p a prime, both greater than or equal to 2. Suppose n divides p-1 and p divides n^3 -1. Prove 4p-3 is a perfect square. n divides p-1 -> n<=p-1. n^3-1 = (n-1)(n^2+n+1). Suppose p divides n-1. Then this implies that p<=n-1 -> n > p. But this contradicts that n <= p-1. Hence p divides n^2+n+1. n dividing p -1 implies p = nk+1 for some positive integer n. We want to find when (n^2+n+1)/(nk+1) is a positive integer. Observe that gcd(k,nk+1) = 1: if we set the gcd to D, we see D must divide nk+1 -n(k) = 1, forcing D to be one. Hence, (n^2+n+1)/(nk+1) is an integer if and only if (k^2*n^2 + k^2*n + k^2)/(nk+1). Using algebraic manipulations, the expression (k^2*n^2 + k^2*n + k^2) is equal to (nk+1)^2 - 2(nk+1) + k^2-k+1, which is k^2 -k + 1 modulo nk+1. Observe that gcd(n,nk+1) = 1 using the same argument we did earlier. Hence k^2 - k + 1 = 0 mod(nk+1) if and only if nk^2 - nk + n = 0 mod(nk+1). We have that this is equal to k(nk+1) - (nk+1) + n - k - 1 mod (nk+1), which reduces to n-k-1 = 0mod(nk+1). Note |n-k-1| < nk+1 for all n>=2, k>=1: we may conclude n-k-1 = 0, or k = n-1. Then 4p-3 = 4(n(n-1)+1) - 3 = 4n^2 - 4n + 4-3 = 4n^2 - 4n + 1 = (2n-1)^2. The proof is complete.

@@qwertyman123 saved

@@cipherunity is the algebra problem good

@@qwertyman123 I just saved it. But still I need to read it.

There are other ways to solve this integral. May be you see another video with an alternate solution

Definitely, however the use of the Mellin Transform is very informational to viewers.

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I learn it only two days before

​@@cipherunityIt would be useful to lay down and prove the properties of the Mellin Transform when you get a chance.

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If you see the table for Laplace transform. You shall find a formula ׆ (t^p) = Γ(p+1)/s^(p+1) implies (׆)^(-1)(1/s^(p+1)) = t^p/Γ(p+1) or (׆)^(-1)(1/s^(p)) = t^p -1/Γ(p) Similarly ׆ sinwt =w/(s^2+w^2). In our case w= 1

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Very good

I shall check it out

I just attach a card(video) to this video. It tells about the other method this integral was done.

@@cipherunity hmm

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Music name?

@@williammartin4416 "Mirage" by Chris Haugen

I have used this technique many time. You shall see it in my next video too.

@@cipherunitynice, video request try this problem (1-sinx)^1/2 from 0 to pi Hint : u sub x=2theta

@@user-lu6yg3vk9z It is done. You may watch the video

What you wrote is not clear.

Where can I send you a picture of it sir?

@@baijantisatyal5718 Please write it again using parentheses

I= dx/((√3x - √3/2)^2 -23/4) Is it clear sir?

@@baijantisatyal5718 sqrt(3x) or sqrt(3)x?