1. ## Orthogonalisation

How do I need to solve this question? Do I need to use the Gram-Schmit orthogonalization process to transform the basis $\{ 1,x,x^2 \}$ into an orthogonal basis, and then normalize the orthogonal basis vectors to obtain an orthonormal basis?

I mean letting $w_1=1, w_2=x, w_3 =x^2$ then

$v_1 = 1$

$v_2 = x- \frac{w_2 . v_1}{\| v_1 \|^2}v_1$

$= x - \int^1_0 x = x-\frac{x^2}{2}1 = \frac{1}{2}$

$v_3 = x^2-\frac{1}{2}- \frac{x^2. \frac{1}{2}}{\| \frac{1}{2} \|^2} \frac{1}{2}$

$= x^2-\frac{1}{2}-\frac{1}{16} \int_0^1 x^2=x^2- \frac{1}{2}-\frac{1}{54}$

Then we normalize each element like this

$q_1 = \frac{v_1}{\|v_1\|}= \frac{1}{\sqrt{1}}$

and so on...

But I doubt this is the correct method of doing this...

2. Originally Posted by demode

How do I need to solve this question? Do I need to use the Gram-Schmit orthogonalization process to transform the basis $\{ 1,x,x^2 \}$ into an orthogonal basis, and then normalize the orthogonal basis vectors to obtain an orthonormal basis?

I mean letting $w_1=1, w_2=x, w_3 =x^2$ then

$v_1 = 1$

$v_2 = x- \frac{w_2 . v_1}{\| v_1 \|^2}v_1$

$= x - \int^1_0 x = x-\frac{x^2}{2}1 = \frac{1}{2}$

Uh? You have $x-\frac{1}{2}$ ...why $\frac{1}{2}$ alone? Perhaps a typo?

$v_3 = x^2-\frac{1}{2}- \frac{x^2. \frac{1}{2}}{\| \frac{1}{2} \|^2} \frac{1}{2}$

$= x^2-\frac{1}{2}-\frac{1}{16} \int_0^1 x^2=x^2- \frac{1}{2}-\frac{1}{54}$

Then we normalize each element like this

$q_1 = \frac{v_1}{\|v_1\|}= \frac{1}{\sqrt{1}}$

and so on...

But I doubt this is the correct method of doing this...

Well, if you doubt it why don't you CHECK it?

Tonio

3. Hi Tonio, yes that was a typo. But my point is that I'm not sure if my approach to this problem is even correct. I can't check anything because I have no worked examples of Fourier approximations with piece-wise functions anywhere in my textbook.

4. Originally Posted by demode
Hi Tonio, yes that was a typo. But my point is that I'm not sure if my approach to this problem is even correct. I can't check anything because I have no worked examples of Fourier approximations with piece-wise functions anywhere in my textbook.

What does this have to do with Fourier series? You only have to check whether the vectors you got, after you normalized to vectors of length 1, are pairwise orthogonal...and they must be, since normalizing vectors don't change orthogonality. Gram-Schmidt process usually gives orthoNORMAL basis, which is orthogonal and of length 1.

When I wrote "check it" I meant you check the above, which you can.

Tonio

5. Sorry...

Okay, here I've corrected my working:

Let $w_1=1, w_2=x, w_3 =x^2$ then

$v_1 = w_1 = 1$

$v_2 = x- \frac{w_2 . v_1}{\| v_1 \|^2}v_1$

$= x - \int^1_0 x dx= x-\frac{x^2}{2} = x- \frac{1}{2}$

$v_3 = x^2- (x - \frac{1}{2})- \frac{w_3.v_2}{\| v_2 \|^2} v_2$

$\frac{w_3.v_2}{\| v_2 \|^2} v_2 = \frac{x^2.(x-\frac{1}{2})}{\sqrt{(x^2 + \frac{1}{4}} )^2}=\frac{\int_0^1 x^3-\frac{x^2}{2}}{x^2+\frac{1}{4}}(x-\frac{1}{2})$ $= \frac{\frac{1}{12}}{x^2+ \frac{1}{4}}(x-\frac{1}{2})=\frac{2x-1}{24x^2-6}$

So $v_3 = x^2-(x-\frac{1}{2})- \frac{2x-1}{24x^2-6}$

Then we normalize them

$q_1 = \frac{v_1}{\|v_1\|}= \frac{1}{\sqrt{1}}$

$q_2 = \frac{v_2}{\|v_2\|}= \frac{x-\frac{1}{2}}{\sqrt{x^2-\frac{1}{4}}}$

$q_3 = \frac{v_3}{\|v_3\|}= \frac{x^2-(x-\frac{1}{2})- \frac{2x-1}{24x^2-6}}{\sqrt{(x^2-(x-\frac{1}{2})- \frac{2x-1}{24x^2-6})^2}}$

So, is this correct then? I'm not sure if I know how to check the conditions you mentioned that they are pair-wise orthogonal & of lenght 1...

6. Originally Posted by demode
Sorry...

Okay, here I've corrected my working:

Let $w_1=1, w_2=x, w_3 =x^2$ then

$v_1 = w_1 = 1$

$v_2 = x- \frac{w_2 . v_1}{\| v_1 \|^2}v_1$

$= x - \int^1_0 x dx= x-\frac{x^2}{2} = x- \frac{1}{2}$

$v_3 = x^2- (x - \frac{1}{2})- \frac{w_3.v_2}{\| v_2 \|^2} v_2$

$\frac{w_3.v_2}{\| v_2 \|^2} v_2 = \frac{x^2.(x-\frac{1}{2})}{\sqrt{(x^2 + \frac{1}{4}} )^2}=\frac{\int_0^1 x^3-\frac{x^2}{2}}{x^2+\frac{1}{4}}(x-\frac{1}{2})$ $= \frac{\frac{1}{12}}{x^2+ \frac{1}{4}}(x-\frac{1}{2})=\frac{2x-1}{24x^2-6}$

So $v_3 = x^2-(x-\frac{1}{2})- \frac{2x-1}{24x^2-6}$

Then we normalize them

$q_1 = \frac{v_1}{\|v_1\|}= \frac{1}{\sqrt{1}}$

$q_2 = \frac{v_2}{\|v_2\|}= \frac{x-\frac{1}{2}}{\sqrt{x^2-\frac{1}{4}}}$

This can't possibly be correct since the norm of a vector, just as the inner product of two vectors, must be a number...So:

$\left|\left|x-\frac{1}{2}\right|\right|=\sqrt{\int^1_0\left(x-\frac{1}{2}\right)^2dx}$ $=\sqrt{\frac{1}{3}\left[\left(x-\frac{1}{2}\right)^3\right]^1_0}=\sqrt{\frac{1}{3}\left(\frac{1}{8}-\left(-\frac{1}{8}\right)\right)}=\sqrt{\frac{1}{12}}$ $=\frac{1}{2\sqrt{3}}$ .

Now take it from here correcting the following.

Tonio

$q_3 = \frac{v_3}{\|v_3\|}= \frac{x^2-(x-\frac{1}{2})- \frac{2x-1}{24x^2-6}}{\sqrt{(x^2-(x-\frac{1}{2})- \frac{2x-1}{24x^2-6})^2}}$

So, is this correct then? I'm not sure if I know how to check the conditions you mentioned that they are pair-wise orthogonal & of lenght 1...
.

7. Before I continue, I see that for finding the norm you have used the definition that $\| v \| = \sqrt{\left\langle v,v \right\rangle} = \sqrt{\int_0^1 v^2}$.

So does this mean that my $V_1$ and $v_2$ are incorrect?? ...Because I didn't use this definition to find the norms in $\frac{w_2 . v_1}{\| v_1 \|^2}v_1$ and $\frac{w_3.v_2}{\| v_2 \|^2} v_2$!