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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,316

the answer seems too odd for me -n(n+1)(2n+1)/6, and what mathematical induction can do is just to prove other than to derive. Hoping some genius could handle this out, with ease and simplisity, like the derivation of 1+a+a²+a³+...+an

it's a huge challenge!! Cuz Google doesn't provide a derivation!!:P

**X'(y-Xβ)=0**

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

I don't see why you don't want it to be using induction, but whatever. I remember that question being asked here before, so there'll be a topic here somewhere with a proof. I remember that it has an induction proof and a different proof as well, so you get two answers!

I'll try to find the link now. Hang on.

Edit: That was easier than I thought! Here it is.

*Last edited by mathsyperson (2006-03-13 01:43:42)*

Why did the vector cross the road?

It wanted to be normal.

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,316

Many thanks!

**X'(y-Xβ)=0**

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,316

Thank you for the link. The link finally reminded me of the proof i once read in Thomas' Calculus book, it uses a neighbor elimation trick:

key thought -- (k+1)³-k³ =3k²+3k+1

3(1²+2²+3²+...+n²)

3(1 +2 +3 +... +n)

+) 1 +1 +1 +... +1

___________________

(-1³+2³)+(-2³+3³)+(-3³+4³)+...+(-n³+(n+1)³)

=-1+(n+1)³=n³+3n²+3n

3x + 3n(n+1)/2+n=n³+3n²+3n

3x=n³+3/2 n²+1/2 n= 1/2 n(2n²+3n+1)

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

**X'(y-Xβ)=0**

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

Hi.

The function:

is very useful for calculating sums.

That's because:

If you want to calculate this:

, you may express the f(x) = x^i as:

,

where r(x) is the "remainder", which has less power than f(x).

Example for f(x)=x^1:

S_2 (x) = (x+1)^2-x^2=x^2+2x+1-x^2=2x+1

(S_2 (x))/2=x+1/2

f(x)=x^1=x=0.5 S_2(x) - 0.5 = 0.5(S_2(x)-1).

So

.

There's also and binomial proof, which is more usable and universal, but it's harder too.

IPBLE: Increasing Performance By Lowering Expectations.

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,316

Good proof! Thanks a lot

a small error, though

f(x) = x^i should be x^k instead, same to the following

in order to know sum of i^k, we need to do know k-1 sums of i^l ,for l=1,2,...k-1

but to me, sum of squares is enough to derive Spearman's Coefficient

*Last edited by George,Y (2006-03-19 14:51:28)*

**X'(y-Xβ)=0**

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

Yes, thank you.

IPBLE: Increasing Performance By Lowering Expectations.

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