Gram-Schmidt for 1, x, x^2 Must find orthonormal basis

• shreddinglicks
In summary, the problem was to find an orthonormal basis for the functions 1, x, and x^2 from -1 to 1 using the Gram-Schmidt equations. The attempt at a solution involved using the Gram-Schmidt process, but the final answer needed to be normalized again to obtain the correct result.

Homework Statement

Find orthonormal basis for 1, x, x^2 from -1 to 1.

Homework Equations

Gram-Schmidt equations

The Attempt at a Solution

I did the problem. My attempt is attached. Can someone review and explain where I went wrong? It would be much appreciated.

Attachments

• CCI08312018.jpg
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It looks like it might be mostly correct, but you got a little sloppy at the end, and your answer for ## v_3 ## should be of the form ## v_3= Ax^2+Bx+C ## where ## A , B ##, and ## C ## are also closed form expressions, and not a long decimal number. ## \\ ## Edit: And the denominator is unnecessary on the last two terms of ## v_3 ## because the functions ## v_1 ## and ## v_2 ## are normalized. ## \\ ## Additional edit: I think you got it right, but the last term you need to leave it as ## C=-\frac{\sqrt{10}}{6} ##, instead of -.52704.. which I'm sure started out as ##-\frac{\sqrt{10}}{6} ##. ## \\ ## Additional edit: I think it is necessary to normalize this final ## v_3 ## one more time. Also, I think it is not necessary to normalize the vector you are working with before you subtract out the other projections of the previous unit vectors. It makes for an extra step.

Last edited:
shreddinglicks said:

Homework Statement

Find orthonormal basis for 1, x, x^2 from -1 to 1.

Homework Equations

Gram-Schmidt equations

The Attempt at a Solution

I did the problem. My attempt is attached. Can someone review and explain where I went wrong? It would be much appreciated.

vela
It looks like it might be mostly correct, but you got a little sloppy at the end, and your answer for ## v_3 ## should be of the form ## v_3= Ax^2+Bx+C ## where ## A , B ##, and ## C ## are also closed form expressions, and not a long decimal number. ## \\ ## Edit: And the denominator is unnecessary on the last two terms of ## v_3 ## because the functions ## v_1 ## and ## v_2 ## are normalized. ## \\ ## Additional edit: I think you got it right, but the last term you need to leave it as ## C=-\frac{\sqrt{10}}{6} ##, instead of -.52704.. which I'm sure started out as ##-\frac{\sqrt{10}}{6} ##. ## \\ ## Additional edit: I think it is necessary to normalize this final ## v_3 ## one more time. Also, I think it is not necessary to normalize the vector you are working with before you subtract out the other projections of the previous unit vectors. It makes for an extra step.

I went through the problem again. It seems if I normalize it after I use the Gram-Schmidt process I get the correct answer. I'm not sure why though.