Is this equality true?
(Figure might not show well if using darkmode)
Previous iteration: #704390 Guest iteration: #704571 (@cryotosensei)
Visual representation of the surprising validity of this equality...
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It took me a few minutes to figure out how to understand this image. I love feeling the neural pathways reconnect!
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Happy to hear it
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200 sats \ 1 reply \ @Murch 30 Sep
Yes, it's true. I suspect that m is supposed to be a natural number greater than or equal to 1.
Assuming it's true for any one value in the sequence, we can prove that it's true for all succeeding values per induction.
To go from m to m+1, we show that both sides grow the same:
\sum{n=1}^{m+1} n^3 = \left( \sum_{n=1}^{m+1} n \right)^2
(m+1)^3 + \sum{n=1}^m n^3 = \left( m+1+ \sum_{n=1}^{m} n \right)^2
(m+1)^3 + \sum{n=1}^m n^3 = (m+1)^2 + 2*(m+1)*\left(\sum_{n=1}^{m} n\right) + \left( \sum_{n=1}^{m} n \right)^2
Use Gauss sum: (m+1)^3 + \sum{n=1}^m n^3 = (m+1)^2 + 2*(m+1)m(m+1)/2 + \left( \sum_{n=1}^{m} n \right)^2
Simplify: (m+1)^3 + \sum{n=1}^m n^3 = (m+1)^2 + m*(m+1)^2 + \left( \sum_{n=1}^{m} n \right)^2
1x+nx=(1+n)x: (m+1)^3 + \sum{n=1}^m n^3 = 1(m+1)^2 + m*(m+1)^2 + \left( \sum_{n=1}^{m} n \right)^2
(m+1)^3 + \sum{n=1}^m n^3 = (1+m)*(m+1)^2 + \left( \sum_{n=1}^{m} n \right)^2 (m+1)^3 + \sum{n=1}^m n^3 = (m+1)^3 + \left( \sum_{n=1}^{m} n \right)^2
Deduct (m+1)^3 from both sides: \sum{n=1}^m n^3 = \left( \sum_{n=1}^m n \right)^2
For m = 1: sum(1^3) = (sum(1))^2
It's true for m=1, and we have shown that if it's true for any m, it's also true for m+1. QED.
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Another one by induction...
We do need MathJax support~~
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Yes, assuming it's true, it's true for both m and m-1. Lets substract both-sides of equations for m and m-1 and we get:
m^3 = ( 1+2+3+...m)^2 - (1+2+3...(m-1))^2
Rth side is two arithmetic series. We can simplify:
m^3 = ( (1+m) * (m) / 2 )^2 - ( (1+m-1) * (m-1) / 2) ) ^ 2 m^3 = ( (m+1) * m / 2 )^2 - ( (m-1) * m / 2 )^2
Divide both sides by (m / 2)^2
4m = ( (m+1) )^2 - ( (m-1) )^2 4m = ( (m+1 + m-1) ) * ( (m+1-m+1) ) 4m = ( 2m ) * ( 2 ) 4m = 4m
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A proof by induction, I like it...
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For reference, I just posted the video that I promised in the context of our infinity problem from 2 days ago: #705385
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My intuition says no, but I don't recall how to demonstrate that.
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Intuition can be misleading...
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I forgot how to read some of the notation. I haven't done much math in such a long time. I like reading these and the solutions people create.
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