The decomposition you use in the first argument is also useful for simplifying the infinite polynomial of coefficients i³, or in general any infinite polynomial whose coefficients are polynomials of the summation index. Basically with these decaying power you can interprete the terms of the new infinite polynomial (the one with decaying powers) as a derivative (not necessarily of first order) of a geometric infinite polynomial, which we know how to calculate, and now you can conclude with one or multiple integrations. However I didn't know that this decomposition was also useful for calculating the sum you showed. The second argument was also very pretty and unique, although proving it doesn't seem that easy since the arrangement of the cubes seems hard to generalize formally.
Indeed, discrete calculus is truly an incredible tool to have under your belt...the third method in this video is related to what you mentioned at first. Generating functions are also incredible useful. If you're interested in more, please check out my ongoing "series" on them! ruclips.net/p/PL1f-n4ubT347s3sE4RHP5yz72eB-vprQN
we want to prove that, if the hypothesis if true for some number, it will also be true for the next one. that way, if we prove that the hypothesis works if n=1, then it will also work when n=2. if it works with n=2, it will also work when n=3, etc... this way we can prove that this will work all the way to infinity without having to check all possible numbers.
Great question! I assume you are asking about the fundamental theorem of discrete calculus; since we take the sum over Δu at each step of the sequence, where Δu is defined as = u_(i+1) - u_i, when we sum from i=1 to n, the first term is u_(2) - u_1, while the last term (i=n) would be u_(n+1) - u_n. If you expand the sum out and expand each Δu_i term, you'll see that all of the terms except the u_(n+1) and u_1 cancel! This is called a "telescoping sum," and if you'd like to see it more in action please check out my video on the sum of the first N Fibonacci numbers: ruclips.net/video/0joLJ8697PU/видео.html
You forgot the dumb way. We observe sum i^0=n Sum i=n(n+1)/2 Sum i^2=1/6n(n+1)(2n+1) Conjecture that sum i^k =P(n) where rank of polynomial P is k+1. Set up a system of equations using P(0)0, P(1) ... P(k+1) to calculate coefficients In this case e=0 a+b+c+d+e=1 16a+8b+4c+2d+e=9 81a+27b+9c+3d+e=36 256a+64b+16c+4d+e=100 Prove it works by mathematical induction
The decomposition you use in the first argument is also useful for simplifying the infinite polynomial of coefficients i³, or in general any infinite polynomial whose coefficients are polynomials of the summation index. Basically with these decaying power you can interprete the terms of the new infinite polynomial (the one with decaying powers) as a derivative (not necessarily of first order) of a geometric infinite polynomial, which we know how to calculate, and now you can conclude with one or multiple integrations. However I didn't know that this decomposition was also useful for calculating the sum you showed. The second argument was also very pretty and unique, although proving it doesn't seem that easy since the arrangement of the cubes seems hard to generalize formally.
Indeed, discrete calculus is truly an incredible tool to have under your belt...the third method in this video is related to what you mentioned at first. Generating functions are also incredible useful. If you're interested in more, please check out my ongoing "series" on them! ruclips.net/p/PL1f-n4ubT347s3sE4RHP5yz72eB-vprQN
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Thanks for the sub! Exciting content coming up 😀
Could you explain please why we go to n+1 and not just n 🙏
we want to prove that, if the hypothesis if true for some number, it will also be true for the next one.
that way, if we prove that the hypothesis works if n=1, then it will also work when n=2. if it works with n=2, it will also work when n=3, etc... this way we can prove that this will work all the way to infinity without having to check all possible numbers.
this is called mathematical induction if you want to learn more
Great question! I assume you are asking about the fundamental theorem of discrete calculus; since we take the sum over Δu at each step of the sequence, where Δu is defined as = u_(i+1) - u_i, when we sum from i=1 to n, the first term is u_(2) - u_1, while the last term (i=n) would be u_(n+1) - u_n. If you expand the sum out and expand each Δu_i term, you'll see that all of the terms except the u_(n+1) and u_1 cancel! This is called a "telescoping sum," and if you'd like to see it more in action please check out my video on the sum of the first N Fibonacci numbers: ruclips.net/video/0joLJ8697PU/видео.html
You forgot the dumb way.
We observe sum i^0=n
Sum i=n(n+1)/2
Sum i^2=1/6n(n+1)(2n+1)
Conjecture that sum i^k =P(n) where rank of polynomial P is k+1.
Set up a system of equations using P(0)0, P(1) ... P(k+1) to calculate coefficients
In this case
e=0
a+b+c+d+e=1
16a+8b+4c+2d+e=9
81a+27b+9c+3d+e=36
256a+64b+16c+4d+e=100
Prove it works by mathematical induction
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