Apostol volume 1 calc? what do I need

[tripsynth]

Hi, I'm curious in learning calculus again but through a more rigorous fashion than plug and chug that I did last year.

I'm trying to go slowly through Apostol's calculus but the way he's deriving the integral for a parabolic function x^2 is kinda troubling me.

I'm fine until he reaches a point where he wants to prove that

1^2 + 2^2 + ... + n^2 = n^3/3 +n^2/2 + n/6

he ends up proving it by examining

(k+1)^3 = k^3 + 3k^2 + 3k + 1

my problem is how did he decide to use (k+1)^3?

one more issue is he later goes on to state

1^2 + 2^2 + ... + (n-1)^2 < n^3/3 < 1^2 + 2^2 + ... + n^2
and has a proof in the later sections for it.
where did the n^3/3 come from ?? why is he using it?


I feel like I'm missing something in my math education. Should I go through a number theory book before continuing ?

 

[sachinism]

well that's something practice would teach u

i suggest u better go through some elementary induction problems , that should do

 

[deluks917]

(k+1)^3 = k^3 + 3k^2 + 3k + 1

my problem is how did he decide to use (k+1)^3?

one more issue is he later goes on to state

1^2 + 2^2 + ... + (n-1)^2 < n^3/3 < 1^2 + 2^2 + ... + n^2
and has a proof in the later sections for it.
where did the n^3/3 come from ?? why is he using it?

The reason he uses (k+1) ^3 is to set up a telescoping series:(k+1)^3 - k^3= 3k^2 + 3k + 1. the idea is to get an expression for the sum of squares.

(2)^3 - (1)^3 + (3)^3 - (2)^3 + ... (k+1)^3 - (k)^3 = (k+1)^3 -1 (as all the other terms cancel). Do you see how this gives (K+ 1) ^3 - 1 = 3(1^2 + 2^2 + ... + k^2) + 3(1 + ... + k) + k? Since we know what 3(1 + ... +k) is we can now solve for (1^2 + ... + k^2).

You've taken calculus so you know the integral of x^2 is (x^3)/3. Apostal is trying to derive this formula with the fundamental theorem of calculus. That's where the n^3/3 comes from. The point of the inequality is to use the squeeze theorem to prove the upper and lower sum (Riemann sums or inf/sup definition I don't know what Apostal uses) converge to (x^3)/3 as n-> infinity. Do you understand the later parts of his derivation?

 

[tripsynth]

The reason he uses (k+1) ^3 is to set up a telescoping series:(k+1)^3 - k^3= 3k^2 + 3k + 1. the idea is to get an expression for the sum of squares.

(2)^3 - (1)^3 + (3)^3 - (2)^3 + ... (k+1)^3 - (k)^3 = (k+1)^3 -1 (as all the other terms cancel). Do you see how this gives (K+ 1) ^3 - 1 = 3(1^2 + 2^2 + ... + k^2) + 3(1 + ... + k) + k? Since we know what 3(1 + ... +k) is we can now solve for (1^2 + ... + k^2).

You've taken calculus so you know the integral of x^2 is (x^3)/3. Apostal is trying to derive this formula with the fundamental theorem of calculus. That's where the n^3/3 comes from. The point of the inequality is to use the squeeze theorem to prove the upper and lower sum (Riemann sums or inf/sup definition I don't know what Apostal uses) converge to (x^3)/3 as n-> infinity. Do you understand the later parts of his derivation?

I'll look over the telescoping series and read over that section again later.

yeah I knew that integral but on that page it just appears with that inequality. he doesn't develop n^3/3 it just appears at that moment. I think I understand the rest of the proof though. I just was confused about where n^3/3 came from.

Thank you for your help.