2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ Off topic: on Amazon, the number of negative reviews that this book has received is staggering. $\endgroup$ – user1551 Jan 12 at 4:43
  • 1
    $\begingroup$ Oh... my professor taught me that interval is always 0 to 1 and weighted function is 1. $\endgroup$ – 4charwon Jan 12 at 4:45
  • 2
    $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.