Why summation of n^2 up to infinity is similar to n^3

[scenic frigate]

Hello, can someone explained why s is assumed to be n^3 please.

[scenic frigate]

I'm completely loss about this, anyone knows what we are trying to show/prove pls

[livid tendon]

Well summation of 1 is n
And summation of n is (n^2+n)/2

So a guess would be summation of n^2 is has the form an^3+ bn^2 +cn+d

[scenic frigate]

Well summation of 1 is n
wdym pls

how

[livid tendon]

1+1+1+1..+1 with n numbers of 1 is n

[scenic frigate]

ah yeah

And summation of n is (n^2+n)/2
what about this one pls

[livid tendon]

Another way is to think about (k+1)^3-k^3

[livid tendon]

Say we have S=1+2+3+4+...+n
Rewrite. S=n+(n-1)+(n-2)+...+1

[scenic frigate]

yep

[livid tendon]

So we have 2S=(n+1)+(n+1)+...+(n+1)

[scenic frigate]

yeah

s = n(n+1) / 2

(I think)

[scenic frigate]

in the video the professor make used of this technique but I didn't understand, what is the point of using this technique, I mean how do we know this apply or we should do that?

[scenic frigate]

it's a cubic polynomial? But the values a, b, c and d, Im confused of what the user is doing, like he assume at one time when n = 0, this equation to = 0, at another time when n = 1, equation = 1

[livid tendon]

This is the sum of squares, so of course at n=0 the sum of squares is 0, same for 1

[scenic frigate]

but at one point we have d, which is a constant, d isn't left instead of 0?

tbh, Im confused about what we say is an "educated guess", is there anything I can read to better understand this?

[livid tendon]

when n=0, an^3=0 same for bn^2 and cn, we only have d left, and since at n=0 the sum is 0 we have d=0

[livid tendon]

Not entirely sure

[livid tendon]

There is a way without guessing

[scenic frigate]

hmm is that because we have 0 term?

like when n = 0, this mean our equation don't have any terms?

[livid tendon]

https://brilliant.org/wiki/sum-of-n-n2-or-n3/

This might help you

[scenic frigate]

will have a look, thanks !

By the way, it's something we assume to be true?

like summation of n^2 lead to n^3 ?

like some kind of tautology?

[livid tendon]

Not really, it's just a guess

It could have been false

It's just that in this case it always is true

[scenic frigate]

yeah I see, and here, the number 0,1,5 and 14

for do we come out with the numbers 5 and 14?

[polar bramble]

hm no we can know/assume that in advance, it doesn’t need to be just a guess

[livid tendon]

1^2+2^2=1+4=5

Add 3^2 and you get 14

[scenic frigate]

hmm but where the numbers come from pls :c

I guess by this summation formula?

[livid tendon]

Ye you just manually calculate it for n=0,1,2,3

[long anvil]

Is this not just the partial sum formula $\sum\limits_{k=1}^n = \frac16 n(n+1)(2n+1)$?

[jade cosmos]

nicolytic

[livid tendon]

Yes?

[long anvil]

Oh ok

Why is there an ansatz involved?

[scenic frigate]

yeah I see, when we have the bunch of equations, how do we solve for the values? I only know how to solve 2 simultaneous equations, is there any method involved there pls

[livid tendon]

What's an ansatz

[long anvil]

An educated guess that hepls you find the solution, the word is in the original image

[livid tendon]

Gaussian elimination

[polar bramble]

the function $$f: n \mapsto \sum_{k=1}^n 1$$ exhibits constant growth (it grows by $1$ when we increase $n$ by $1$), so $f$ itself should be a degree 1 polynomial (since these are the functions that have constant growth, or that have a constant derivative, as learned in e.g. calculus). similarly, function $$g: n \mapsto \sum_{k=1}^n k$$ exhibits linear growth (it grows by $n$ when we incrase $n$ by 1), so $g$ itself should be a degree 2 polynomial. and so on

[jade cosmos]

sIayIa

[scenic frigate]

I don't have the notion of how we would prove it using calculus yet 😭 , any resource where I can learn that? :c

[livid tendon]

There no calculus anywhere in this entire conversation

[polar bramble]

hm it's not really needed

[scenic frigate]

okk

I will refresh about this discussion, and come back if I have other questions, ty !

[scenic frigate]

just if someone can pinpoint this pls

[livid tendon]

.

You should Google it yourself I'm kinda too lazy to teach you it

[scenic frigate]

yep will do that, ty !

[jaunty grove]

https://en.wikipedia.org/wiki/Faulhaber's_formula# for the exact value of the sum

[jaunty grove]

if you only need an approximation, we know that $$ \int_{k-1}^{k} tdt \leq k \leq \int_k^{k+1} tdt$$ Summing from $k = 1$ to $n$ gives $$\frac{n^3}3 =\int_0^n t dt \leq 1^2 + 2^2 + \dots + n^2 \leq \int_1^{n+1} = \frac{(n+1)^3}3-1$$ So the sum is $\approx \frac{n^3}3$

[jade cosmos]

bloubbloub

[scenic frigate]

I only have knowledge about high school maths but the calculus used here is really interesting. I want to understand what is happening, is there anything I should know before trying to understand it pls, like what part of calculus

[jaunty grove]

Integrals i guess

[scenic frigate]

noted, ty !

.solved