Distance between two polynomials (inner product)
$begingroup$
I don't know how I've gotten this question wrong. I have to compute the distance between:
$f(t) = 2t + 3$ and $g(t) = 3t^2 -1$
Their inner product is defined as $int_{0}^{1}f(t)g(t)dt$
So I figured the distance would be $sqrt{(f-g,f-g)}$
Where $(f-g,f-g)$ is the inner product of f-g with itself.
I got answer answer of $sqrt{frac{242}{15}}$ but my book says $sqrt{frac{123}{10}}$ and I don't understand why. I've checked that the integral evaluates to my answer so I don't think I've made a calculation error so maybe the error is in my setup?
linear-algebra inner-product-space
asked Dec 13 '18 at 22:34
dj1121dj1121
274
$endgroup$
add a comment |
$begingroup$
I don't know how I've gotten this question wrong. I have to compute the distance between:
$f(t) = 2t + 3$ and $g(t) = 3t^2 -1$
Their inner product is defined as $int_{0}^{1}f(t)g(t)dt$
So I figured the distance would be $sqrt{(f-g,f-g)}$
Where $(f-g,f-g)$ is the inner product of f-g with itself.
I got answer answer of $sqrt{frac{242}{15}}$ but my book says $sqrt{frac{123}{10}}$ and I don't understand why. I've checked that the integral evaluates to my answer so I don't think I've made a calculation error so maybe the error is in my setup?
linear-algebra inner-product-space
asked Dec 13 '18 at 22:34
dj1121dj1121
274
$endgroup$
1
$begingroup$
I agree with your answer, not with the one from your book. Maybe some more details about the problem description can clarify? Or maybe the book just has a mistake in it.
$endgroup$
– SmileyCraft
Dec 13 '18 at 22:43
$begingroup$
Texts are full of errors, especially in numerical results like this. It may well be that the text is wrong and you are right. Your setup looks right to me, and using that, your computation is definitely right. My recommendation to you is to relax.
$endgroup$
– Lubin
Dec 13 '18 at 22:46
add a comment |
$begingroup$
I don't know how I've gotten this question wrong. I have to compute the distance between:
$f(t) = 2t + 3$ and $g(t) = 3t^2 -1$
Their inner product is defined as $int_{0}^{1}f(t)g(t)dt$
So I figured the distance would be $sqrt{(f-g,f-g)}$
Where $(f-g,f-g)$ is the inner product of f-g with itself.
I got answer answer of $sqrt{frac{242}{15}}$ but my book says $sqrt{frac{123}{10}}$ and I don't understand why. I've checked that the integral evaluates to my answer so I don't think I've made a calculation error so maybe the error is in my setup?
linear-algebra inner-product-space
asked Dec 13 '18 at 22:34
dj1121dj1121
274
$endgroup$
I don't know how I've gotten this question wrong. I have to compute the distance between:
$f(t) = 2t + 3$ and $g(t) = 3t^2 -1$
Their inner product is defined as $int_{0}^{1}f(t)g(t)dt$
So I figured the distance would be $sqrt{(f-g,f-g)}$
Where $(f-g,f-g)$ is the inner product of f-g with itself.
I got answer answer of $sqrt{frac{242}{15}}$ but my book says $sqrt{frac{123}{10}}$ and I don't understand why. I've checked that the integral evaluates to my answer so I don't think I've made a calculation error so maybe the error is in my setup?
linear-algebra inner-product-space
linear-algebra inner-product-space
asked Dec 13 '18 at 22:34
dj1121dj1121
274
asked Dec 13 '18 at 22:34
dj1121dj1121
274
asked Dec 13 '18 at 22:34
dj1121dj1121
274
asked Dec 13 '18 at 22:34
dj1121dj1121
274
asked Dec 13 '18 at 22:34
dj1121dj1121
274
274
1
$begingroup$
I agree with your answer, not with the one from your book. Maybe some more details about the problem description can clarify? Or maybe the book just has a mistake in it.
$endgroup$
– SmileyCraft
Dec 13 '18 at 22:43
$begingroup$
Texts are full of errors, especially in numerical results like this. It may well be that the text is wrong and you are right. Your setup looks right to me, and using that, your computation is definitely right. My recommendation to you is to relax.
$endgroup$
– Lubin
Dec 13 '18 at 22:46
add a comment |
1
$begingroup$
I agree with your answer, not with the one from your book. Maybe some more details about the problem description can clarify? Or maybe the book just has a mistake in it.
$endgroup$
– SmileyCraft
Dec 13 '18 at 22:43
$begingroup$
Texts are full of errors, especially in numerical results like this. It may well be that the text is wrong and you are right. Your setup looks right to me, and using that, your computation is definitely right. My recommendation to you is to relax.
$endgroup$
– Lubin
Dec 13 '18 at 22:46
1
1
$begingroup$
I agree with your answer, not with the one from your book. Maybe some more details about the problem description can clarify? Or maybe the book just has a mistake in it.
$endgroup$
– SmileyCraft
Dec 13 '18 at 22:43
$begingroup$
I agree with your answer, not with the one from your book. Maybe some more details about the problem description can clarify? Or maybe the book just has a mistake in it.
$endgroup$
– SmileyCraft
Dec 13 '18 at 22:43
$begingroup$
Texts are full of errors, especially in numerical results like this. It may well be that the text is wrong and you are right. Your setup looks right to me, and using that, your computation is definitely right. My recommendation to you is to relax.
$endgroup$
– Lubin
Dec 13 '18 at 22:46
$begingroup$
Texts are full of errors, especially in numerical results like this. It may well be that the text is wrong and you are right. Your setup looks right to me, and using that, your computation is definitely right. My recommendation to you is to relax.
$endgroup$
– Lubin
Dec 13 '18 at 22:46
add a comment |
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$begingroup$
I agree with your answer, not with the one from your book. Maybe some more details about the problem description can clarify? Or maybe the book just has a mistake in it.
$endgroup$
– SmileyCraft
Dec 13 '18 at 22:43
$begingroup$
Texts are full of errors, especially in numerical results like this. It may well be that the text is wrong and you are right. Your setup looks right to me, and using that, your computation is definitely right. My recommendation to you is to relax.
$endgroup$
– Lubin
Dec 13 '18 at 22:46