Euler Totient Function
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- čas přidán 30. 06. 2024
- This video is dedicated to finding an expression for the number of positive integers between 1 and a positive integer n that have no common factors with n. This function of n is known as the Euler Totient Function and is used throughout number theory.
#EulerTotientFunction #NumberTheory #PrimeNumbers
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Very good content. Kindly keep making such Quality Content..
Appreciations from India!!
Thanks!
This video is really helpful. Thank you!
I love Euler's Totient Function! So happy you made a video about this!
Crypto!!
No entiendo nada de ingles y aun asi entendi perfectamente la explicacion que no encontre este tema en mi idioma, muchas gracias por compartir, su explicacion es genial.
Congrats on 8K!!!
Thank you!!
We can simplify this formula to phi(n)=(p1-1)(p2-1)••••••(pk-1) ! It's super super simple!
Where n=p1^e1•p2^e2•••pk^ek where pk is a prime.
Very informative and well explained
Thanks!
Excellent.thanks ProfOmar
Thanks Yaseen!
VERY NICE!
Wow. Nice. I love challenging solving calculus,geometry,algebra and trigonometry question here on my.....
Phi, is a really neat function. One neat thing about it is that there are still some very basic things we don't know about it. For example, for any prime p, obviously phi(p) = p-1, and thus, in particular phi(p) divides p-1. Lehmer asked if there was any composite n where phi(n)|n-1. This problem is still open. We do know that any counterexample has to be very large. (Greater than 10^20 and must have at least 14 distinct prime factors.)
I didn’t know about this problem wow
@@ProfOmarMath If you want to look it up the problem is generally called "Lehmer's totient question" or "Lehmer's totient conjecture" (to distinguish it form Lehmer's conjecture which is often used to mean the Lehmer conjecture for polynomials).
Another fun one is Carmichael's conjecture which is that for any positive n, there is an m not equal to n, such that phi(m) = phi(n). (Essentially saying that phi fails the horizontal line test for its entire range.)
And ultimately i have done it what actually happened there .Thanks sir,
If we're talking about the sets Ai. Shouldn't the union of all these not contain overlaps because a set has no duplicate elements
Just to be sure: up to 10:31, we are adding up fractions coresponding to all the different combinations of the primes with each other, right? So for the pairs we have k choose 2 fractions, for the triples k choose 3, and so on, correct?
That’s totally it!
Why do we switch between plus and minus (9:40)?
What does p|n mean (in the thumbnail with Π notation)? Very curious.
Ah yes the vertical bar means divides
If m and n are not coprime, phi is not multiplicative because we double-count the (1-1/p) factors corresponding to the prime factors they share, right?
The (1-1/p) factors in the product as which we can express phi all only appear once but if we split a suitable number n into two numbers x and y who are not coprime and multiply phi(x) with phi(y), the factors corresponding to the shared prime factors will appear more than once.
It’s a little different as to why. The difficulty is more readily seen with actual numbers. Say we want to compute phi(8*12). We want to eliminate things from 1 to 96 that have common factors with 96. From the 8 we can eliminate multiples of 2. From the 12 we eliminate multiples of 2 and 3, but we have already eliminated multiples of 2 so there is some overlapping happening. Hard to explain on here haha
@@ProfOmarMath I think that's what I mean, actually.
If you multiply phi(8) with phi(12) you get 8*12*(1-1/2)*(1-1/2)*(1-1/3) instead of phi(96) which is only 96*(1-1/2)*(1-1/3).
In the product of phi(8) and phi(12) the (1-1/2) appears more than once which doesn't match phi(96).
Can you do one on tensors
That’s a great idea.....” What is a tensor?”
Still working on the highly requested “What is a manifold?”
So so so good!!!!!!!!!!!!!!!!!!!!!!
Thanks Aash 😀
@@ProfOmarMath thank you
@@ProfOmarMath u are welcome