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

[shreddinglicks]

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.

 

[Charles Link]

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.

 

[Ray Vickson]

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.

Most helpers will skip your question and not even attempt to read your posted image. If you are serious about wanting help you ought to type out your work--not nearly as horrible a procedure as you might think. See the Guidelines for more information on this issue.

 

[shreddinglicks]

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.

 

[Charles Link]

This way is quicker. In your first paper, you still need to normalize your answer of $v_3=\frac{\sqrt{10}}{2}x^2-\frac{\sqrt{10}}{6}$. If you work with that, it gives you a denominator of 2/3 as what will make it a unit vector, so that the answer for $v_3$ in your first paper needs to be multiplied by 3/2 to make it a unit vector. It then gives the same correct answer as your second paper.

 

[shreddinglicks]

This way is quicker. In your first paper, you still need to normalize your answer of $v_3=\frac{\sqrt{10}}{2}x^2-\frac{\sqrt{10}}{6}$. If you work with that, it gives you a denominator of 2/3 as what will make it a unit vector, so that the answer in your first paper needs to be multiplied by 3/2. It then gives the same correct answer as your second paper.

I see what you mean. Thanks a bunch!