That is a nice one. Another way to evaluate the infinite product is by multiplying both the numerator and denominator with 1 - x^(2^k). Then by expanding the infinite product we can observe that most of the terms are canceled sequentially and the only term that is left, is the numerator of the first term: 1 - x^(2^0) = 1 - x. Though this intuition holds in a finite product of n terms, but when we deal with an infinite product things are always tricky. So in order to prove that the infinite product converges and is equal to 1 - x, we have to take the limit of the product from k = 0 until k = +oo. But for any positive integer n the product inside the integral from k = 0 until k = n can be expanded and after the cancellation the only term that is left is (1 - x) / (1 - x^(2^n)). Taking the limit of the product to n = +oo considering that the x values are all between 0 < x < 1, the term in the denominator x^(2^n) converges to 0. Hence the infinite product after taking the limit equals with 1 - x.
Yeah i think you need combinatorics. The reciprocal is GF for number of partitions into distinct parts that are powers of. Hence, 1/(1-x). The expression is then 1-x, and then FTC can be used. One problem is idk how to find when this infinite product converge.
This problem shows that Calculus is easy--integrating 1-x is trivial! The hard part of calculus is precalulus/algebra. The ONLY reason a student (who tries) will fail Calculus is because of their arithmetic and algebra skills
That is a nice one. Another way to evaluate the infinite product is by multiplying both the numerator and denominator with 1 - x^(2^k). Then by expanding the infinite product we can observe that most of the terms are canceled sequentially and the only term that is left, is the numerator of the first term: 1 - x^(2^0) = 1 - x. Though this intuition holds in a finite product of n terms, but when we deal with an infinite product things are always tricky. So in order to prove that the infinite product converges and is equal to 1 - x, we have to take the limit of the product from k = 0 until k = +oo. But for any positive integer n the product inside the integral from k = 0 until k = n can be expanded and after the cancellation the only term that is left is (1 - x) / (1 - x^(2^n)). Taking the limit of the product to n = +oo considering that the x values are all between 0 < x < 1, the term in the denominator x^(2^n) converges to 0. Hence the infinite product after taking the limit equals with 1 - x.
i love how you teach!!
love this guys quote
Great video! Love the shirt!
What a amazing question ❤
Thanks for an other video master...
You've made calculus pretty simple huh
❤
Yeah i think you need combinatorics.
The reciprocal is GF for number of partitions into distinct parts that are powers of. Hence, 1/(1-x). The expression is then 1-x, and then FTC can be used.
One problem is idk how to find when this infinite product converge.
Now I'm going to hide under a blanket all day.
This problem shows that Calculus is easy--integrating 1-x is trivial! The hard part of calculus is precalulus/algebra. The ONLY reason a student (who tries) will fail Calculus is because of their arithmetic and algebra skills
how do we know that |x |< 1 ? to write the sum like ( 1/1-x )
Gli estremi di integrazione sono (0-1), quindi necessariamente |x|
You forgot about algebra or you finished it ?
he finishes the algebra part at around 11:27
@@5gallonsofwater495 No, no it was about his video series
In my opinion in his video series on algebra still is missing few subjects
@@holyshit922 oh lol, i don't know anything about that
How do we know that |r| < 1 here.
Because the integral is from 0 to 1