Orthogonal vectors

**alexmahone** · May 17th 2011, 07:04 AM

Find a vector $(w_1,w_2,w_3,...)$ that is orthogonal to $v=(1,\frac{1}{2},\frac{1}{4},...)$ . Compute its length $||w||$ .

**tonio** · May 17th 2011, 07:26 AM

Originally Posted by alexmahone

Find a vector $(w_1,w_2,w_3,...)$ that is orthogonal to $v=(1,\frac{1}{2},\frac{1}{4},...)$ . Compute its length $||w||$ .

Orthogonal...in what vector space and with respect to what inner product??

Tonio

**alexmahone** · May 17th 2011, 07:29 AM

Originally Posted by tonio

Orthogonal...in what vector space and with respect to what inner product??

Tonio

Hilbert space with respect to dot product.

**tonio** · May 17th 2011, 07:46 AM

Originally Posted by alexmahone

Hilbert space with respect to dot product.

There are infinite Hilbert spaces ( to add "with inner product" is futile: ANY Hilbert space is a vector space with an inner product). I can think of at least

two different Hilbert spaces with their inner products to which your vector belongs, so again: what vector space with what iiner product?

Tonio

**alexmahone** · May 17th 2011, 07:51 AM

Originally Posted by tonio

There are infinite Hilbert spaces ( to add "with inner product" is futile: ANY Hilbert space is a vector space with an inner product). I can think of at least

two different Hilbert spaces with their inner products to which your vector belongs, so again: what vector space with what iiner product?

Tonio

I think the question merely requires us to find $w_1,w_2,w_3,...$ such that $w_1+\frac{w_2}{2}+\frac{w_3}{4}+...=0$ .

**tonio** · May 17th 2011, 08:13 AM

Originally Posted by alexmahone

I think the question merely requires us to find $w_1,w_2,w_3,...$ such that $w_1+\frac{w_2}{2}+\frac{w_3}{4}+...=0$ .

That's what I also thought, but then you already know what to look for...

...

Idea: $\sum\limits^\infty_{n=0}\frac{1}{4^n}=\frac{4}{3} \, , \, \, \,\sum\limits^\infty_{n=1}\frac{1}{2^{2n-1}}=2\sum\limits^\infty_{n=1}\frac{1}{4^n}=2\cdot \frac{1}{3}=\frac{2}{3}$

Tonio

**alexmahone** · May 17th 2011, 08:26 AM

Originally Posted by tonio

That's what I also thought, but then you already know what to look for...

...

Idea: $\sum\limits^\infty_{n=0}\frac{1}{4^n}=\frac{4}{3} \, , \, \, \,\sum\limits^\infty_{n=1}\frac{1}{2^{2n-1}}=2\sum\limits^\infty_{n=1}\frac{1}{4^n}=2\cdot \frac{1}{3}=\frac{2}{3}$

Tonio

I'm afraid I don't follow your hint.

**Opalg** · May 17th 2011, 11:54 AM

Originally Posted by alexmahone

I think the question merely requires us to find $w_1,w_2,w_3,...$ such that $w_1+\frac{w_2}{2}+\frac{w_3}{4}+...=0$ .

You also need to ensure that $\sum|w_n|^2<\infty$ , and in fact you need to be able to compute that sum in order to do the second part of the question.

Big hint: Can you find a solution in which only the first two coordinates of w are nonzero?

**alexmahone** · May 17th 2011, 12:27 PM

Originally Posted by Opalg

You also need to ensure that $\sum|w_n|^2<\infty$ , and in fact you need to be able to compute that sum in order to do the second part of the question.

Big hint: Can you find a solution in which only the first two coordinates of w are nonzero?

Thanks. $w=(1,-2,0,...)$ and its length is $\sqrt{5}$ .

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