Question

(PDE)(Gram-Schmidt Function Orthogonalization) I was wondering if somebody else can figure out where I'm (I think) making a mistake, I'm trying to make a given set of functions orthogonal via Gram-Schmidt.

Answer 1

@Pompeii00

Answer 2

@freckles , are you at all good with PDE/Gram-Schmidt?

Answer 3

I can look it up and try to help you with it.

Answer 4

But I will have to do so after lunch.

Answer 5

Alright, cool; thanks nonetheless.

Answer 6

This was used in quantum mechanics to find a convenient basis. I'm not sure how much I remember about it though.

Answer 7

i'm back. like did you want to show the problem.

Answer 8

Yeah, just wanted to make sure somebody was sticking around before I committed. What I'm trying to do is, given three functions, f_0, f_1, and f_2, each being

Answer 9

\[f_0=1,\ \ \ f_1=x \ \ \ , f_2=x^2\]Respectively, where the relationship between them is \[f_n=x^n, \ \ \ n \in \mathrm{N}\]

Answer 10

Making those three functions orthogonal by the Gram-Schmidt process.

Answer 11

@dan815

Answer 12

The issue that I'm having is when I try to take the inner product of f_1 and g_2 as in the Gram-Schmidt procedure, I don't get a value of zero, which I need to get, AFAIK, in order for the functions to be demonstrably orthogonal.

Answer 13

what is g_2?

Answer 14

Wait a minute, one sec. Trying to find a bunch of new orthogonal vectors, \[g_0=f_0, \ \ \ \ \ \ \ \ g_1=f_1-\frac{(g_0,f_1)}{(g_0,g_0)}g_0\]

Answer 15

Where the things in parentheses are the inner product of the two functions.

Answer 16

\[(f,g)=\int\limits_0^1f \cdot g dx \text {\right }\]

Answer 17

or is it -1 to 1 instead of 0 to 1

Answer 18

Yep, exactly, so for the first one, it's just equal to itself (at least to be orthogonal), for the second one, it should be x minus the inner product of x and 1.

It's 0 to 1.

Answer 19

\[g_0=1\]\[g_1=x-\frac{\int\limits_{}^{}(1)(x)dx}{\int\limits_{}^{}(1)(1)dx}(1)\]

Answer 20

g1=x-1/2

Answer 21

I would have gotten x-1/2, not sure how you got 3/2-yeah, and then, since those are orthogonal, if we take the inner product of the two of them, the result should be zero, right?

Answer 22

i did the integral of g1*g1 dx on bottom

Answer 23

mistakenly

Answer 24

i mean f1*f1 lol

Answer 25

\[\int\limits_{}^{}(1)\bigg(x-\frac{1}{2}\bigg)dx=\int\limits_{}^{}\bigg(x-\frac{1}{2}\bigg)dx=\frac{x^2}{2}-\frac{x}{2}\]

Answer 26

you get 1/2-1/2

Answer 27

Wait, nevermind, that makes sense now, doesn't it?

1/2-1/2=0. Alright, not sure where I screwed up originally, lol. But now I need to make them orthonormal as well.

Answer 28

So I need to divide each by its inner product with itself, which in the first case, we don't need to do (it already has magnitude 1) and need to do the second one as well.

Answer 29

\[\frac{x-1/2}{\int\limits_{}^{}(x-1/2)(x-1/2)dx}\]

Answer 30

\[\frac{x-\frac{1}{2}}{\frac{1}{3}-\frac{1}{2}+\frac{1}{4}}=\frac{12x-6}{4-6+3}=\frac{12x-6}{-2+3}=\frac{12x-6}{1}=12x-6\]
if i didn't make a mistake

Answer 31

\[\int\limits_{0}^{1}\bigg(x^2-\frac{1}{4}x+\frac{1}{4}\bigg)dx=\frac{x^3}{3}-\frac{1}{8}x^2+\frac{1}{4}\bigg]_0^1\]

Answer 32

1/4x, that last part

Answer 33

\[(x-\frac{1}{2})(x-\frac{1}{2})=x(x-\frac{1}{2})-\frac{1}{2}(x-\frac{1}{2}) \\ =x^2-\frac{1}{2}x-\frac{1}{2}x+\frac{1}{4} \\ =x^2-\frac{1+1}{2}x+\frac{1}{4}\]

Answer 34

or recall (x-b)(x-b)=x^2-2xb+b^2

Answer 35

Ah, I gotcha, I see my mistake

Answer 36

Alright, but the problem is, that inner product needs to be zero.

Answer 37

and it will be

Answer 38

\[\int\limits_{0}^{1}(12x-6)(1) dx=\frac{12x^2}{2}-6x|_0^1=6-6=0\]

Answer 39

Do you evaluate the entire expression at zero to one? What's the basis for that? That would equal zero, but I don't see where that operation (evaluating that final amount) suddenly comes from.

Answer 40

i thought you said the inner product still needs to be zero

Answer 41

so that is what i was showing that the inner product of 1 and (12x-6) is 0

Answer 42

this means that they are orthogonal

Answer 43

Wait, what? I'm just getting lost/disoriented in the math, give me a moment.

Answer 44

Alright, so if I read correctly, our result 12x-6 is the ortho*normal* version of g_1?

It's the unit vector version of (x-1/2)?

Answer 45

yeah

Answer 46

Alright, so since that is orthonormal and is also obviously orthogonal, the inner product with that and g_1 should also be zero.\[\int\limits_{}^{}(1)(12x-6)dx=\int\limits_{}^{}(12x-6)dx=(6x^2-6)\bigg]_0^1=0\]

Answer 47

Yay! Lol, alright, cool.

Answer 48

\[f_0=1,f_1=x,f_2=x^2 \\ g_1=f_1-\frac{(f_1 ,f_0)}{(f_0,f_0)} f_0=x-\frac{\frac{x^2}{2}|_0^1}{x|_0^1}(1)=x-\frac{\frac{1}{2}}{1}=x-\frac{1}{2} \\ \frac{g_1}{||g_1||}=12(x-\frac{1}{2})=12x-6\\ \text{ but I think we also need \to do } g_2=f_2-\frac{(f_2,f_0)}{(f_0,f_0)}f_0-\frac {(f_2,g_1)}{(g_1,g_1)}g_1 \text{ and then find } \\ \frac{g_2}{||g_2||}\]

Answer 49

like the orthogonal basis should be g0,g1,g2

Answer 50

and the orthornormal basis should be g0/||g0||,g1/||g_1||,g2/||g_2||

Answer 51

\[(f_2,f_0)=(x^2,1)=\frac{x^3}{3}|_0^1=\frac{1}{3} \\ (f_0,f_0)=(1,1)=x|_0^1=1 \\ (f_2,g_1)=(x^2,x-\frac{1}{2})=\frac{x^4}{4}-\frac{1}{2}\frac{x^3}{3}|_0^1=\frac{1}{4}-\frac{1}{6}=\frac{3}{12}-\frac{2}{12}=\frac{1}{12} \\ (g_1,g_1)=(x-\frac{1}{2},x-\frac{1}{2})=\frac{x^3}{3}-\frac{x^2}{2}+\frac{1}{4}x|_0^1=\frac{1}{3}-\frac{1}{2}+\frac{1}{4}=\frac{4}{12}-\frac{6}{12}+\frac{3}{12} \\\text{ so } (g_1,g_1)=\frac{1}{12}\]

Answer 52

\[g_2=f_2-\frac{(f_2,f_0)}{(f_0,f_0)}f_0-\frac {(f_2,g_1)}{(g_1,g_1)}g_1 \\ g_2=x^2-\frac{\frac{1}{3}}{1}(1)-\frac{\frac{1}{12}}{\frac{1}{12}}(x-\frac{1}{2}) \\ g_2=x^2-\frac{1}{3}-(x-\frac{1}{2}) \\ g_2=x^2-x-\frac{1}{3}+\frac{1}{2} \\ g_2=x^2-x-\frac{2}{6}+\frac{3}{6} \\ g_2=x^2-x+\frac{1}{6}\]

Answer 53

\[\frac{g_2}{||g_2||}=\frac{x^2-x+\frac{1}{6}}{\int\limits_{0}^{1}(x^2-x+\frac{1}{6})(x^2-x+\frac{1}{6}) dx}\]

Answer 54

which should be just 180x^2-180x+30

Answer 55

Alright, cool; yeah, I understand this now, which is as much as I wanted to go. I have one last question (going to open up a new thing for it) just relating to doing a fourier expansion given some function and an orthonormal basis.

Answer 56

\[\text{ answer is } \\ (\frac{g_0}{||g_0||},\frac{g_1}{||g_1||},\frac{g_2}{||g_2||}) \\ =(1,12x-6,180x^2-180x+30)\]

Answer 57

if there was no mistake made

Answer 58

i can try

Answer 59

Alright, yeah, no worries. In any case, thank you!

Answer 60

honestly how to whip out my linear algebra book for that last problem
a bit rusty i am

Answer 61

I want to make one more note here
in case of confusion on notation
I was taking ||g|| to mean (g,g) whch you know when int_0^1 g^2 dx