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### Topic: summation of exponential function  (Read 13831 times)

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#### xiankai

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« on: August 09, 2006, 07:26:59 PM »
n
?   r2 = ?
r=1

i can't figure this out, at least for now...
one learns best by teaching

#### pantone159

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« Reply #1 on: August 10, 2006, 01:51:13 AM »
it equals (1/6)(n)(n+1)(2n+1), or (1/3)(n)(n+1/2)(n+1).

#### xiankai

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« Reply #2 on: August 10, 2006, 02:02:46 AM »
how do u get that? one learns best by teaching

#### pantone159

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« Reply #3 on: August 10, 2006, 02:45:50 PM »
how do u get that? I got it from 'Concrete Mathematics, A foundation for computer science', by Graham,Knuth,Patashnik.
This is a really cool book that goes in great depths into this stuff.  I wish I had taken a class in this to learn it better, but oh well.

Section 2.5 covers several different methods for evaluating this exact sum.
Method 0 : Look it up.  This is the method I used in answering your question at first. Method 1 : Guess the answer, then prove it.  Not much more interesting.
Method 2 is clever and nifty, however:

Consider the sum of the *cubes* from 1 to N, which I'll call Cn since I'm too lazy to put in the subscript codes.

Cn = sum(k=1..n, k3)               // note more laziness with notation

I'll also define
Sn = sum(k=1..n, k2) which is what you want to find.

Also note that sum(k=1..n, k) = (1/2)(n)(n+1), I assume you've already figured that one out.

then...

Cn+1 = Cn + (n+1)3 = sum(k=1..n+1, k3)

Next change variables on the sum, to l where l = k-1, and so

Cn + (n+1)3 = sum(l=0..n, (l+1)3)
= sum(l=0..n, l3 + 3*l2 + 3*l + 1)
= 1 + sum(l=1..n, l3 + 3*l2 + 3*l + 1)
= 1 + sum(l=1..n, l3) + 3 * sum(l=1..n, l2) + 3 * sum(l=1..n, l) + sum(l=1..n, 1)
= 1 + Cn + 3 * Sn + 3 * (1/2)(n)(n+1) + n

The Cn can be subtracted off both sides of this equation, thus removing it.  This is convenient since we didn't actually care about it in the first place. Rearrange a bit, to solve for Sn which is what you want,

Sn = (1/3)*((n+1)3 - 1 - (3/2)(n)(n+1) - n)

a little more algebra then beats this into the final form

Sn = (1/6)(n)(n+1)(2n+1)

#### xiankai

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« Reply #4 on: August 11, 2006, 12:17:12 PM »
mm thats very innovative, of late more and more obscure problems that i encounter have their solutions in cubic equations last thing; i want to clarify on how to replace k with l

method 1: by algebraic replacement

Replace k = k-1

n+1
?   k3
k=1

n+1
?   (k-1)3
k-1=1

n+1
?   l3
l=1

n
?   (l+1)3
l=0

n+1
?   k3
k=1

n+1-1
?   (l+1)3
k-1=0

n
?   (l+1)3
l=0
one learns best by teaching

#### pantone159

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« Reply #5 on: August 11, 2006, 02:47:12 PM »
The way I think about it, is to replace the thing you are summing *and* the limits at the same time, the same way you do for definite integrals.  So, for

sum (k=1..n+1, k^3)

I want to replace k by l+1, and the limit k=1 by l+1=1, or l=0, and k=n+1 by l+1=n+1, or l=n, giving:

sum (l=0..n, (l+1)^3)