I recently read a proof on the transcendental nature of pi. Niven's function, differentiation, Taylor's series, and Integration by parts, and a bit of combinatorics [trivial] and finally proof by contradiction. Very nice because it brings together 5 core concepts, to make a proof. Niven's function was interesting because it was a designed with certain properties, and that in itself is interesting as a challenge. Can you construct a function that has certain characteristics?
Yes! one of my favourite identities binomial series!! The case alpha not an postive integer is kinda not used alot which is what makes solution 1 awesome
There is another way of calculating in this problem, which starts by differentiating the original integral J(x) by x as a parameter. At each next kth step of differentiation (k=1,2,...n), we will get that Jk(0)=0, while the n-th order derivative of this integral will be equal to Jn(x)=n! *[1/(1-x)^(n+1) -1] . At the same time Jk(x)=k! *[1/(1-x)^(k+1)], if k= n+1,n+2, ...... and Jk(0)=k!, if k= n+1,n+2,...... Note that J(0)=0 , and the Taylor-Maclaurin series has the form J(x)= J(0)+ Σ (from k=1 to k=∞) Jk(0)*x^ k/k!=Σ (from k=n+1 to k=∞)x^k =x^(n+1 )/(1- x).
Using the substitution "u=(x-t)/(1-t)" on the starting integral turns this into single-line calculation.
Thank you random netizen
I recently read a proof on the transcendental nature of pi.
Niven's function, differentiation, Taylor's series, and Integration by parts, and a bit of combinatorics [trivial] and finally proof by contradiction. Very nice because it brings together 5 core concepts, to make a proof.
Niven's function was interesting because it was a designed with certain properties, and that in itself is interesting as a challenge. Can you construct a function that has certain characteristics?
Yes! one of my favourite identities binomial series!! The case alpha not an postive integer is kinda not used alot which is what makes solution 1 awesome
At 6:36 why is it the integral from 0 to 1?
Yes it should be from 0 to x and after the sub u=t/x that he was talking about you will get integral from 0 to 1.he just skiped this step.
from Morocco thank you very much
We can use substitution v = (x-1)*1/(1-t)+1
but integration by parts was my first idea
There is another way of calculating in this problem, which starts by differentiating the original integral J(x) by x as a parameter.
At each next kth step of differentiation (k=1,2,...n), we will get that Jk(0)=0, while
the n-th order derivative of this integral will be equal to Jn(x)=n! *[1/(1-x)^(n+1) -1] .
At the same time Jk(x)=k! *[1/(1-x)^(k+1)], if k= n+1,n+2, ......
and Jk(0)=k!, if k= n+1,n+2,......
Note that J(0)=0 , and the Taylor-Maclaurin series has the form
J(x)= J(0)+ Σ (from k=1 to k=∞) Jk(0)*x^ k/k!=Σ (from k=n+1 to k=∞)x^k =x^(n+1 )/(1- x).
Isn't that more or less just the integral formula of the remainder of the series for 1/(1-x)??
THANKS PROFESOR !!!!!, VERY INTERESTING. !!!!!!
16:07 and that's a good place to stop
Very nice video.
I've been honing my skills. The first thing I thought was "aha! Beta-function".