The problem is finding orthonormal basis for W=span{u1=x,u2=x^2}

And as lots of people think, it is not very difficult problem

My answer is

${ \sqrt{3}x,\sqrt{80}(x^2-\frac{3}{4}x) }$

But, textbook says

$ \sqrt{3}x,\sqrt{30}(x^2-\frac{1}{2}x)$

I checked another edition of the textbook but it was same.

I use elementary linear algebra by koleman

Is there any error I have made? Because of this problem I can’t believe my answer for similar problems.enter image description here

  • $\begingroup$ Please, if you really need to include an image, could you ensure that it is the right way up. $\endgroup$ – Lord Shark the Unknown Jan 12 at 4:31
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    $\begingroup$ For any interval $I$ and any positive function $w$ on $I$, $$\left<f,g\right>=\int_I f(x)g(x)w(x)\,dx$$ defines an inner product. The choice of interval $I$ and weight function $w$ does affect what the inner product is. $\endgroup$ – Lord Shark the Unknown Jan 12 at 4:41
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    $\begingroup$ Off topic: on Amazon, the number of negative reviews that this book has received is staggering. $\endgroup$ – user1551 Jan 12 at 4:43
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    $\begingroup$ Oh... my professor taught me that interval is always 0 to 1 and weighted function is 1. $\endgroup$ – 4charwon Jan 12 at 4:45
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    $\begingroup$ In answer to your question: you are right. Whoever did the solutions probably forgot that the second term needed to be multiplied by $x$ too when making sure they are orthogonal. $\endgroup$ – spaceisdarkgreen Jan 12 at 4:51

The book's answer isn't orthogonal: $(x,x^2-\frac12x)=\int_0^1x(x^2-\frac12x)\operatorname dx=[x^4/4-x^3/6]_0^1\neq0$.

Yours, on the other hand, appears to be correct.


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