This is the magic of ramanujan’s master theorem. As long as the expansion of the function is valid in some open disc of radius>0, ramanujan’s master theorem will hold, as long as the integral itself converges
For the last problem, it would have been easier to replace ln(1+x) with 1/(1+x). Then, Phi(n)=Gamma(n+1), making the value of the integral Gamma(s)*Gamma(1-s), and thus the final answer pi*csc(pi*s).
@@krisbrandenberger544 but if I simply replace ln(1+x) with 1/(1+x) then I am solving an entirely different problem, not the one I had set out to solve.
@@danielrosado3213 But if you look at the series expansion for ln(1+x), the sum would start at n=1 instead of n=0, because division by 0 is not allowed. And Ramanujan's Master Theorem only applies to sums that start at n=0.
@@krisbrandenberger544 true, I should have shifted the index of the power series of ln(1+x) in order to account for that. Thank you for pointing this out.
By RMT, the integration of (x^n) dx from 0 to infinity is (1/n)gamma(1/n)sin(pi/(2n)), let n=1 results 1. However, by using Wolfram Apha or other similar software, the answer is divergent. Why?
This integral only converges for certain values of n. It is easy for verify graphically that sinx oscillates infinitely, hence its integral does not converge to a value. For this reason, it’s important to first analyze whether or not an integral converges before applying RMT, because RMT will give you an answer even if the integral diverges
Thanks for your speedy reply. I do not have the ability to read the criteria proved by someone for RMT to be valid. Could you make a video how to check if RMT is valid under the given conditions?
@@HuiSengKheong RMT will always be valid if the integral is convergent. I may make a video on this in the future, but I can’t promise anything soon, sorry.
In the last example, may I know where to get the power expansion of ln(1+x)=-Sum((-1)^n x^n)/n, where n is from 0 to infinity?! Since the denominator is n, if n=0, the sum will blow out!!
@@danielrosado3213 Thank you for the examples you presented and resolved using Ramanujan's master theorem. You should do a series of videos with more examples on this very important theorem in Number Theory.
I think you present very well but an improvement would be to write more clearly. Please regard this as an improvement rather than a criticism
Your posts r unique man ❤❤❤❤love
this is even more satisfying than the feynman trick 😮💨
I know! The way it destroys crazy integrals in seconds is unmatched
Why can we use the expansion of 1/(1+t) even though we are outside of its domain of convergence ?
This is the magic of ramanujan’s master theorem. As long as the expansion of the function is valid in some open disc of radius>0, ramanujan’s master theorem will hold, as long as the integral itself converges
For the last problem, it would have been easier to replace ln(1+x) with 1/(1+x). Then, Phi(n)=Gamma(n+1), making the value of the integral Gamma(s)*Gamma(1-s), and thus the final answer
pi*csc(pi*s).
Do you mean using integration by parts? Because otherwise it would be a different problem.
No, I meant by using Ramanujan's Master Theorem.
@@krisbrandenberger544 but if I simply replace ln(1+x) with 1/(1+x) then I am solving an entirely different problem, not the one I had set out to solve.
@@danielrosado3213 But if you look at the series expansion for
ln(1+x), the sum would start at n=1 instead of n=0, because division by 0 is not allowed. And Ramanujan's Master Theorem only applies to sums that start at n=0.
@@krisbrandenberger544 true, I should have shifted the index of the power series of ln(1+x) in order to account for that. Thank you for pointing this out.
i learned what an exponent was today!
F**k....overkill but awesome nonetheless
as long as it gets me a beautiful solution 😉
very good！
By RMT, the integration of (x^n) dx from 0 to infinity is (1/n)gamma(1/n)sin(pi/(2n)), let n=1 results 1. However, by using Wolfram Apha or other similar software, the answer is divergent. Why?
Typo mistake, i mean integration of sin(x^n) not x^n!
This integral only converges for certain values of n. It is easy for verify graphically that sinx oscillates infinitely, hence its integral does not converge to a value. For this reason, it’s important to first analyze whether or not an integral converges before applying RMT, because RMT will give you an answer even if the integral diverges
Thanks for your speedy reply. I do not have the ability to read the criteria proved by someone for RMT to be valid. Could you make a video how to check if RMT is valid under the given conditions?
@@HuiSengKheong RMT will always be valid if the integral is convergent. I may make a video on this in the future, but I can’t promise anything soon, sorry.
Thank you once again 😊
In the last example, may I know where to get the power expansion of ln(1+x)=-Sum((-1)^n x^n)/n, where n is from 0 to infinity?! Since the denominator is n, if n=0, the sum will blow out!!
Sorry, the summation should be n=1 to infinity
But RMT requests the n starts from 0 to infinity!
@@HuiSengKheong yes, I did the last problem wrong here, it should’ve been done using integration by parts followed by application of the theorem
I still not able to solve the problem using integration by part followed by RMT! Can make a video to show? Thanks.
At 11:29 I think that you forgot the π in the csc().
Yes, I did but pointed it out in editing. Thank you for pointing out this mistake.
@@danielrosado3213 Thank you for the examples you presented and resolved using Ramanujan's master theorem.
You should do a series of videos with more examples on this very important theorem in Number Theory.
@@divergentmaths is there a use for this theorem in number theory? That doesn’t really make much sense to me.
Daniel Rosado I love you ❤❤❤
This video is so amazing my honey
His name is pronounced "Rama-New-Gin"
this is wayyy off, the way he pronounced was fine/better
very inseresting video, but write more clearly, please.
You need to improve your writting. Is not clear enough
Kindly stop faking american accents, it reduces the quality of your videos
I’m literally American living in the United States 💀
💀
Love to mumble. Best to drift each sentence down to inauditability.