Mixing
Finding orthonormal basis. Is there error on textbook?
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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.
linear-algebra orthogonality
asked Jan 12 at 4:24
4charwon4charwon
274
$endgroup$
|
show 6 more comments
$begingroup$
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.
linear-algebra orthogonality
asked Jan 12 at 4:24
4charwon4charwon
274
$endgroup$
$begingroup$
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 $I$, $$left<f,gright>=int_I f(x)g(x)w(x),dx$$ 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
|
show 6 more comments
$begingroup$
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.
linear-algebra orthogonality
asked Jan 12 at 4:24
4charwon4charwon
274
$endgroup$
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.
linear-algebra orthogonality
linear-algebra orthogonality
asked Jan 12 at 4:24
4charwon4charwon
274
asked Jan 12 at 4:24
4charwon4charwon
274
edited Jan 12 at 4:37
4charwon
asked Jan 12 at 4:24
4charwon4charwon
274
asked Jan 12 at 4:24
4charwon4charwon
274
asked Jan 12 at 4:24
4charwon4charwon
274
274
$begingroup$
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 $I$, $$left<f,gright>=int_I f(x)g(x)w(x),dx$$ 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
|
show 6 more comments
$begingroup$
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 $I$, $$left<f,gright>=int_I f(x)g(x)w(x),dx$$ 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
$begingroup$
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
$begingroup$
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
3
$begingroup$
For any interval $I$ and any positive function $w$ on $I$, $$left<f,gright>=int_I f(x)g(x)w(x),dx$$ 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
$begingroup$
For any interval $I$ and any positive function $w$ on $I$, $$left<f,gright>=int_I f(x)g(x)w(x),dx$$ 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
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
$begingroup$
Off topic: on Amazon, the number of negative reviews that this book has received is staggering.
$endgroup$
– user1551
Jan 12 at 4:43
1
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
$begingroup$
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
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
$begingroup$
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
|
show 6 more comments
1 Answer
1
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oldest
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$begingroup$
The book's answer isn't orthogonal: $(x,x^2-frac12x)=int_0^1x(x^2-frac12x)operatorname dx=[x^4/4-x^3/6]_0^1neq0$.
Yours, on the other hand, appears to be correct.
answered Jan 12 at 5:02
Chris CusterChris Custer
13.1k3827
$endgroup$
add a comment |
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$begingroup$
The book's answer isn't orthogonal: $(x,x^2-frac12x)=int_0^1x(x^2-frac12x)operatorname dx=[x^4/4-x^3/6]_0^1neq0$.
Yours, on the other hand, appears to be correct.
answered Jan 12 at 5:02
Chris CusterChris Custer
13.1k3827
$endgroup$
add a comment |
$begingroup$
The book's answer isn't orthogonal: $(x,x^2-frac12x)=int_0^1x(x^2-frac12x)operatorname dx=[x^4/4-x^3/6]_0^1neq0$.
Yours, on the other hand, appears to be correct.
answered Jan 12 at 5:02
Chris CusterChris Custer
13.1k3827
$endgroup$
add a comment |
$begingroup$
The book's answer isn't orthogonal: $(x,x^2-frac12x)=int_0^1x(x^2-frac12x)operatorname dx=[x^4/4-x^3/6]_0^1neq0$.
Yours, on the other hand, appears to be correct.
answered Jan 12 at 5:02
Chris CusterChris Custer
13.1k3827
$endgroup$
The book's answer isn't orthogonal: $(x,x^2-frac12x)=int_0^1x(x^2-frac12x)operatorname dx=[x^4/4-x^3/6]_0^1neq0$.
Yours, on the other hand, appears to be correct.
answered Jan 12 at 5:02
Chris CusterChris Custer
13.1k3827
edited Jan 12 at 5:10
answered Jan 12 at 5:02
Chris CusterChris Custer
13.1k3827
answered Jan 12 at 5:02
Chris CusterChris Custer
13.1k3827
answered Jan 12 at 5:02
Chris CusterChris Custer
13.1k3827
13.1k3827
add a comment |
add a comment |
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$begingroup$
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 $I$, $$left<f,gright>=int_I f(x)g(x)w(x),dx$$ 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