Mixing
Is $f$ that satisfies $int_a^bf(x)dx=int_a^bxf(x)dx=int_a^bx^2f(x)dx=0$ a constant?
My question is:
Define $fin C^{0}[a,b]$ satisfying $int_a^bf(x)dx=int_a^bxf(x)dx=int_a^bx^2f(x)dx=0$. Must this function be a constant function?
I made a easy analysis, which can prove that $f$ has at least three points, as follows
we have $int_{a}^{b}g(x)fleft(xright)dx=0$ , in which $g(x)=mx^2+nx+p$. $forall m,n,p in mathbb{R}$
If $f$ has less than three zero points, we can select appropriate $m$,$n$,$p$ satisfying $g(x)f(x)>0$ for all $x$ which is not the zero point of $f$. And it's contradictory to $int_{a}^{b}g(x)fleft(xright)dx=0$. Thus $f$ must have at least three zero points.
So can we make a further analysis that indicates $f$ is a constant function?
Any ideas would be highly appreciated!
real-analysis definite-integrals
asked Nov 21 '18 at 15:11
Zero
3419
|
show 1 more comment
My question is:
Define $fin C^{0}[a,b]$ satisfying $int_a^bf(x)dx=int_a^bxf(x)dx=int_a^bx^2f(x)dx=0$. Must this function be a constant function?
I made a easy analysis, which can prove that $f$ has at least three points, as follows
we have $int_{a}^{b}g(x)fleft(xright)dx=0$ , in which $g(x)=mx^2+nx+p$. $forall m,n,p in mathbb{R}$
If $f$ has less than three zero points, we can select appropriate $m$,$n$,$p$ satisfying $g(x)f(x)>0$ for all $x$ which is not the zero point of $f$. And it's contradictory to $int_{a}^{b}g(x)fleft(xright)dx=0$. Thus $f$ must have at least three zero points.
So can we make a further analysis that indicates $f$ is a constant function?
Any ideas would be highly appreciated!
real-analysis definite-integrals
asked Nov 21 '18 at 15:11
Zero
3419
1
I think you can find non-zero cubic polynomials.
– Arthur
Nov 21 '18 at 15:12
1
Look up orthogonal polynomials. As Arthur promised, there will be a non-trivial cubic $f(x)$ satisfying all these constraints. And if you add the requirement $int_a^bx^3f(x),dx=0$ then you are guaranteed to find a non-trivial quartic. Et cetera.
– Jyrki Lahtonen
Nov 21 '18 at 15:19
2
Only if you add $int_a^b x^n f(x){rm }dx = 0$ for all $ninmathbb{N}$ can you guarantee that $f(x) equiv 0$ (this follows from Weierstrass approximation theorem).
– Winther
Nov 21 '18 at 15:26
1
All explicit examples below are of polynomials, but note that there are also non-polynomial examples like $f(x) = sin left(frac{2 pi (x-a)}{b-a}right)-2 sin left(frac{4 pi (x-a)}{b-a}right)$.
– Winther
Nov 21 '18 at 15:36
2
@HenningMakholm Yes, it's a very standard argument. See e.g. If $int_0^1 f(x)x^n dx=0$ for every $n$, then $f=0$.
– Winther
Nov 21 '18 at 15:47
|
show 1 more comment
My question is:
Define $fin C^{0}[a,b]$ satisfying $int_a^bf(x)dx=int_a^bxf(x)dx=int_a^bx^2f(x)dx=0$. Must this function be a constant function?
I made a easy analysis, which can prove that $f$ has at least three points, as follows
we have $int_{a}^{b}g(x)fleft(xright)dx=0$ , in which $g(x)=mx^2+nx+p$. $forall m,n,p in mathbb{R}$
If $f$ has less than three zero points, we can select appropriate $m$,$n$,$p$ satisfying $g(x)f(x)>0$ for all $x$ which is not the zero point of $f$. And it's contradictory to $int_{a}^{b}g(x)fleft(xright)dx=0$. Thus $f$ must have at least three zero points.
So can we make a further analysis that indicates $f$ is a constant function?
Any ideas would be highly appreciated!
real-analysis definite-integrals
asked Nov 21 '18 at 15:11
Zero
3419
My question is:
Define $fin C^{0}[a,b]$ satisfying $int_a^bf(x)dx=int_a^bxf(x)dx=int_a^bx^2f(x)dx=0$. Must this function be a constant function?
I made a easy analysis, which can prove that $f$ has at least three points, as follows
we have $int_{a}^{b}g(x)fleft(xright)dx=0$ , in which $g(x)=mx^2+nx+p$. $forall m,n,p in mathbb{R}$
If $f$ has less than three zero points, we can select appropriate $m$,$n$,$p$ satisfying $g(x)f(x)>0$ for all $x$ which is not the zero point of $f$. And it's contradictory to $int_{a}^{b}g(x)fleft(xright)dx=0$. Thus $f$ must have at least three zero points.
So can we make a further analysis that indicates $f$ is a constant function?
Any ideas would be highly appreciated!
real-analysis definite-integrals
real-analysis definite-integrals
asked Nov 21 '18 at 15:11
Zero
3419
asked Nov 21 '18 at 15:11
Zero
3419
edited Nov 21 '18 at 15:12
asked Nov 21 '18 at 15:11
Zero
3419
asked Nov 21 '18 at 15:11
Zero
3419
asked Nov 21 '18 at 15:11
Zero
3419
3419
1
I think you can find non-zero cubic polynomials.
– Arthur
Nov 21 '18 at 15:12
1
Look up orthogonal polynomials. As Arthur promised, there will be a non-trivial cubic $f(x)$ satisfying all these constraints. And if you add the requirement $int_a^bx^3f(x),dx=0$ then you are guaranteed to find a non-trivial quartic. Et cetera.
– Jyrki Lahtonen
Nov 21 '18 at 15:19
2
Only if you add $int_a^b x^n f(x){rm }dx = 0$ for all $ninmathbb{N}$ can you guarantee that $f(x) equiv 0$ (this follows from Weierstrass approximation theorem).
– Winther
Nov 21 '18 at 15:26
1
All explicit examples below are of polynomials, but note that there are also non-polynomial examples like $f(x) = sin left(frac{2 pi (x-a)}{b-a}right)-2 sin left(frac{4 pi (x-a)}{b-a}right)$.
– Winther
Nov 21 '18 at 15:36
2
@HenningMakholm Yes, it's a very standard argument. See e.g. If $int_0^1 f(x)x^n dx=0$ for every $n$, then $f=0$.
– Winther
Nov 21 '18 at 15:47
|
show 1 more comment
1
I think you can find non-zero cubic polynomials.
– Arthur
Nov 21 '18 at 15:12
1
Look up orthogonal polynomials. As Arthur promised, there will be a non-trivial cubic $f(x)$ satisfying all these constraints. And if you add the requirement $int_a^bx^3f(x),dx=0$ then you are guaranteed to find a non-trivial quartic. Et cetera.
– Jyrki Lahtonen
Nov 21 '18 at 15:19
2
Only if you add $int_a^b x^n f(x){rm }dx = 0$ for all $ninmathbb{N}$ can you guarantee that $f(x) equiv 0$ (this follows from Weierstrass approximation theorem).
– Winther
Nov 21 '18 at 15:26
1
All explicit examples below are of polynomials, but note that there are also non-polynomial examples like $f(x) = sin left(frac{2 pi (x-a)}{b-a}right)-2 sin left(frac{4 pi (x-a)}{b-a}right)$.
– Winther
Nov 21 '18 at 15:36
2
@HenningMakholm Yes, it's a very standard argument. See e.g. If $int_0^1 f(x)x^n dx=0$ for every $n$, then $f=0$.
– Winther
Nov 21 '18 at 15:47
1
1
I think you can find non-zero cubic polynomials.
– Arthur
Nov 21 '18 at 15:12
I think you can find non-zero cubic polynomials.
– Arthur
Nov 21 '18 at 15:12
1
1
Look up orthogonal polynomials. As Arthur promised, there will be a non-trivial cubic $f(x)$ satisfying all these constraints. And if you add the requirement $int_a^bx^3f(x),dx=0$ then you are guaranteed to find a non-trivial quartic. Et cetera.
– Jyrki Lahtonen
Nov 21 '18 at 15:19
Look up orthogonal polynomials. As Arthur promised, there will be a non-trivial cubic $f(x)$ satisfying all these constraints. And if you add the requirement $int_a^bx^3f(x),dx=0$ then you are guaranteed to find a non-trivial quartic. Et cetera.
– Jyrki Lahtonen
Nov 21 '18 at 15:19
2
2
Only if you add $int_a^b x^n f(x){rm }dx = 0$ for all $ninmathbb{N}$ can you guarantee that $f(x) equiv 0$ (this follows from Weierstrass approximation theorem).
– Winther
Nov 21 '18 at 15:26
Only if you add $int_a^b x^n f(x){rm }dx = 0$ for all $ninmathbb{N}$ can you guarantee that $f(x) equiv 0$ (this follows from Weierstrass approximation theorem).
– Winther
Nov 21 '18 at 15:26
1
1
All explicit examples below are of polynomials, but note that there are also non-polynomial examples like $f(x) = sin left(frac{2 pi (x-a)}{b-a}right)-2 sin left(frac{4 pi (x-a)}{b-a}right)$.
– Winther
Nov 21 '18 at 15:36
All explicit examples below are of polynomials, but note that there are also non-polynomial examples like $f(x) = sin left(frac{2 pi (x-a)}{b-a}right)-2 sin left(frac{4 pi (x-a)}{b-a}right)$.
– Winther
Nov 21 '18 at 15:36
2
2
@HenningMakholm Yes, it's a very standard argument. See e.g. If $int_0^1 f(x)x^n dx=0$ for every $n$, then $f=0$.
– Winther
Nov 21 '18 at 15:47
@HenningMakholm Yes, it's a very standard argument. See e.g. If $int_0^1 f(x)x^n dx=0$ for every $n$, then $f=0$.
– Winther
Nov 21 '18 at 15:47
|
show 1 more comment
4 Answers
4
active
oldest
votes
The constraints on $f$ only mean that $f$ is orthogonal to the vector space of polynomials of degree less than 2. But there do exist loads of functions who are not constants and satisfy this conditions.
Take a look at the Gram Schmidt orthonormalisation procedure and try to find a polynomial (for example of degree 3) orthogonal to the polynomials $1$, $x$ and $x^2$.
answered Nov 21 '18 at 15:20
dallonsi
1187
Yup, agreed. If you use Legendres then you can look up the answer because the orthogonalisation has been done for you, which is nice
– Richard Martin
Nov 21 '18 at 15:21
add a comment |
$f$ is not necessarily constant. To see this, note that the vector space $C^0[a,b]$ has infinite dimension, and
$$ f mapsto left (int_a^b f(x)dx, int_a^b x f(x)dx, int_a^b x^2 f(x)dxright) $$
is a linear transformation $C^0[a,b] to mathbb R^3$. Since the domain has larger dimension than the codomain, the kernel of the transformation (that is, the set of all functions for which all three integrals are zero) must be nontrivial.
(So there's a qualifying $f$ that is not the zero function, which means that it is not constant, because the first integral for a nonzero constant function would not be $0$).
answered Nov 21 '18 at 15:28
Henning Makholm
238k16303539
add a comment |
No. WLOG $a=-1$ and $b=1$ (otherwise rescale). Then let $f$ be a Legendre polynomial of degree $ge 3$.
answered Nov 21 '18 at 15:19
Richard Martin
1,61118
add a comment |
To construct a counter-example, take $[a,b]=[0,1]$ and consider a cubic polynomial
$$f(x)=x^3+cx^2+dx+e$$
Then:
$$int_0^1 f(x)dx=frac14+frac{c}{3}+frac{d}{2}+e=0$$
$$int_0^1 xf(x)dx=frac15+frac{c}{4}+frac{d}{3}+frac{e}{2}=0$$
$$int_0^1 x^2f(x)dx=frac16+frac{c}{5}+frac{d}{4}+frac{e}{3}=0$$
We have three equations in three unknowns, so we can hope for a solution. And indeed if we do the algebra (or have Wolfram Alpha do it for us), we get $(c,d,e)=(-frac32,frac35,-frac{1}{20})$. Multiplying by $20$ to get rid of the fractions gives
$$f(x)=20x^3-30x^2+12x-1$$
answered Nov 21 '18 at 15:35
TonyK
41.7k353132
Thank you, by the way I think the last term is $-1$
– Zero
Nov 21 '18 at 15:53
@Zero: yes, thank you. I have corrected it now.
– TonyK
Nov 21 '18 at 15:59
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
The constraints on $f$ only mean that $f$ is orthogonal to the vector space of polynomials of degree less than 2. But there do exist loads of functions who are not constants and satisfy this conditions.
Take a look at the Gram Schmidt orthonormalisation procedure and try to find a polynomial (for example of degree 3) orthogonal to the polynomials $1$, $x$ and $x^2$.
answered Nov 21 '18 at 15:20
dallonsi
1187
Yup, agreed. If you use Legendres then you can look up the answer because the orthogonalisation has been done for you, which is nice
– Richard Martin
Nov 21 '18 at 15:21
add a comment |
The constraints on $f$ only mean that $f$ is orthogonal to the vector space of polynomials of degree less than 2. But there do exist loads of functions who are not constants and satisfy this conditions.
Take a look at the Gram Schmidt orthonormalisation procedure and try to find a polynomial (for example of degree 3) orthogonal to the polynomials $1$, $x$ and $x^2$.
answered Nov 21 '18 at 15:20
dallonsi
1187
Yup, agreed. If you use Legendres then you can look up the answer because the orthogonalisation has been done for you, which is nice
– Richard Martin
Nov 21 '18 at 15:21
add a comment |
The constraints on $f$ only mean that $f$ is orthogonal to the vector space of polynomials of degree less than 2. But there do exist loads of functions who are not constants and satisfy this conditions.
Take a look at the Gram Schmidt orthonormalisation procedure and try to find a polynomial (for example of degree 3) orthogonal to the polynomials $1$, $x$ and $x^2$.
answered Nov 21 '18 at 15:20
dallonsi
1187
The constraints on $f$ only mean that $f$ is orthogonal to the vector space of polynomials of degree less than 2. But there do exist loads of functions who are not constants and satisfy this conditions.
Take a look at the Gram Schmidt orthonormalisation procedure and try to find a polynomial (for example of degree 3) orthogonal to the polynomials $1$, $x$ and $x^2$.
answered Nov 21 '18 at 15:20
dallonsi
1187
answered Nov 21 '18 at 15:20
dallonsi
1187
answered Nov 21 '18 at 15:20
dallonsi
1187
answered Nov 21 '18 at 15:20
dallonsi
1187
1187
Yup, agreed. If you use Legendres then you can look up the answer because the orthogonalisation has been done for you, which is nice
– Richard Martin
Nov 21 '18 at 15:21
add a comment |
Yup, agreed. If you use Legendres then you can look up the answer because the orthogonalisation has been done for you, which is nice
– Richard Martin
Nov 21 '18 at 15:21
Yup, agreed. If you use Legendres then you can look up the answer because the orthogonalisation has been done for you, which is nice
– Richard Martin
Nov 21 '18 at 15:21
Yup, agreed. If you use Legendres then you can look up the answer because the orthogonalisation has been done for you, which is nice
– Richard Martin
Nov 21 '18 at 15:21
add a comment |
$f$ is not necessarily constant. To see this, note that the vector space $C^0[a,b]$ has infinite dimension, and
$$ f mapsto left (int_a^b f(x)dx, int_a^b x f(x)dx, int_a^b x^2 f(x)dxright) $$
is a linear transformation $C^0[a,b] to mathbb R^3$. Since the domain has larger dimension than the codomain, the kernel of the transformation (that is, the set of all functions for which all three integrals are zero) must be nontrivial.
(So there's a qualifying $f$ that is not the zero function, which means that it is not constant, because the first integral for a nonzero constant function would not be $0$).
answered Nov 21 '18 at 15:28
Henning Makholm
238k16303539
add a comment |
$f$ is not necessarily constant. To see this, note that the vector space $C^0[a,b]$ has infinite dimension, and
$$ f mapsto left (int_a^b f(x)dx, int_a^b x f(x)dx, int_a^b x^2 f(x)dxright) $$
is a linear transformation $C^0[a,b] to mathbb R^3$. Since the domain has larger dimension than the codomain, the kernel of the transformation (that is, the set of all functions for which all three integrals are zero) must be nontrivial.
(So there's a qualifying $f$ that is not the zero function, which means that it is not constant, because the first integral for a nonzero constant function would not be $0$).
answered Nov 21 '18 at 15:28
Henning Makholm
238k16303539
add a comment |
$f$ is not necessarily constant. To see this, note that the vector space $C^0[a,b]$ has infinite dimension, and
$$ f mapsto left (int_a^b f(x)dx, int_a^b x f(x)dx, int_a^b x^2 f(x)dxright) $$
is a linear transformation $C^0[a,b] to mathbb R^3$. Since the domain has larger dimension than the codomain, the kernel of the transformation (that is, the set of all functions for which all three integrals are zero) must be nontrivial.
(So there's a qualifying $f$ that is not the zero function, which means that it is not constant, because the first integral for a nonzero constant function would not be $0$).
answered Nov 21 '18 at 15:28
Henning Makholm
238k16303539
$f$ is not necessarily constant. To see this, note that the vector space $C^0[a,b]$ has infinite dimension, and
$$ f mapsto left (int_a^b f(x)dx, int_a^b x f(x)dx, int_a^b x^2 f(x)dxright) $$
is a linear transformation $C^0[a,b] to mathbb R^3$. Since the domain has larger dimension than the codomain, the kernel of the transformation (that is, the set of all functions for which all three integrals are zero) must be nontrivial.
(So there's a qualifying $f$ that is not the zero function, which means that it is not constant, because the first integral for a nonzero constant function would not be $0$).
answered Nov 21 '18 at 15:28
Henning Makholm
238k16303539
edited Nov 21 '18 at 15:34
answered Nov 21 '18 at 15:28
Henning Makholm
238k16303539
answered Nov 21 '18 at 15:28
Henning Makholm
238k16303539
answered Nov 21 '18 at 15:28
Henning Makholm
238k16303539
238k16303539
add a comment |
add a comment |
No. WLOG $a=-1$ and $b=1$ (otherwise rescale). Then let $f$ be a Legendre polynomial of degree $ge 3$.
answered Nov 21 '18 at 15:19
Richard Martin
1,61118
add a comment |
No. WLOG $a=-1$ and $b=1$ (otherwise rescale). Then let $f$ be a Legendre polynomial of degree $ge 3$.
answered Nov 21 '18 at 15:19
Richard Martin
1,61118
add a comment |
No. WLOG $a=-1$ and $b=1$ (otherwise rescale). Then let $f$ be a Legendre polynomial of degree $ge 3$.
answered Nov 21 '18 at 15:19
Richard Martin
1,61118
No. WLOG $a=-1$ and $b=1$ (otherwise rescale). Then let $f$ be a Legendre polynomial of degree $ge 3$.
answered Nov 21 '18 at 15:19
Richard Martin
1,61118
answered Nov 21 '18 at 15:19
Richard Martin
1,61118
answered Nov 21 '18 at 15:19
Richard Martin
1,61118
answered Nov 21 '18 at 15:19
Richard Martin
1,61118
1,61118
add a comment |
add a comment |
To construct a counter-example, take $[a,b]=[0,1]$ and consider a cubic polynomial
$$f(x)=x^3+cx^2+dx+e$$
Then:
$$int_0^1 f(x)dx=frac14+frac{c}{3}+frac{d}{2}+e=0$$
$$int_0^1 xf(x)dx=frac15+frac{c}{4}+frac{d}{3}+frac{e}{2}=0$$
$$int_0^1 x^2f(x)dx=frac16+frac{c}{5}+frac{d}{4}+frac{e}{3}=0$$
We have three equations in three unknowns, so we can hope for a solution. And indeed if we do the algebra (or have Wolfram Alpha do it for us), we get $(c,d,e)=(-frac32,frac35,-frac{1}{20})$. Multiplying by $20$ to get rid of the fractions gives
$$f(x)=20x^3-30x^2+12x-1$$
answered Nov 21 '18 at 15:35
TonyK
41.7k353132
Thank you, by the way I think the last term is $-1$
– Zero
Nov 21 '18 at 15:53
@Zero: yes, thank you. I have corrected it now.
– TonyK
Nov 21 '18 at 15:59
add a comment |
To construct a counter-example, take $[a,b]=[0,1]$ and consider a cubic polynomial
$$f(x)=x^3+cx^2+dx+e$$
Then:
$$int_0^1 f(x)dx=frac14+frac{c}{3}+frac{d}{2}+e=0$$
$$int_0^1 xf(x)dx=frac15+frac{c}{4}+frac{d}{3}+frac{e}{2}=0$$
$$int_0^1 x^2f(x)dx=frac16+frac{c}{5}+frac{d}{4}+frac{e}{3}=0$$
We have three equations in three unknowns, so we can hope for a solution. And indeed if we do the algebra (or have Wolfram Alpha do it for us), we get $(c,d,e)=(-frac32,frac35,-frac{1}{20})$. Multiplying by $20$ to get rid of the fractions gives
$$f(x)=20x^3-30x^2+12x-1$$
answered Nov 21 '18 at 15:35
TonyK
41.7k353132
Thank you, by the way I think the last term is $-1$
– Zero
Nov 21 '18 at 15:53
@Zero: yes, thank you. I have corrected it now.
– TonyK
Nov 21 '18 at 15:59
add a comment |
To construct a counter-example, take $[a,b]=[0,1]$ and consider a cubic polynomial
$$f(x)=x^3+cx^2+dx+e$$
Then:
$$int_0^1 f(x)dx=frac14+frac{c}{3}+frac{d}{2}+e=0$$
$$int_0^1 xf(x)dx=frac15+frac{c}{4}+frac{d}{3}+frac{e}{2}=0$$
$$int_0^1 x^2f(x)dx=frac16+frac{c}{5}+frac{d}{4}+frac{e}{3}=0$$
We have three equations in three unknowns, so we can hope for a solution. And indeed if we do the algebra (or have Wolfram Alpha do it for us), we get $(c,d,e)=(-frac32,frac35,-frac{1}{20})$. Multiplying by $20$ to get rid of the fractions gives
$$f(x)=20x^3-30x^2+12x-1$$
answered Nov 21 '18 at 15:35
TonyK
41.7k353132
To construct a counter-example, take $[a,b]=[0,1]$ and consider a cubic polynomial
$$f(x)=x^3+cx^2+dx+e$$
Then:
$$int_0^1 f(x)dx=frac14+frac{c}{3}+frac{d}{2}+e=0$$
$$int_0^1 xf(x)dx=frac15+frac{c}{4}+frac{d}{3}+frac{e}{2}=0$$
$$int_0^1 x^2f(x)dx=frac16+frac{c}{5}+frac{d}{4}+frac{e}{3}=0$$
We have three equations in three unknowns, so we can hope for a solution. And indeed if we do the algebra (or have Wolfram Alpha do it for us), we get $(c,d,e)=(-frac32,frac35,-frac{1}{20})$. Multiplying by $20$ to get rid of the fractions gives
$$f(x)=20x^3-30x^2+12x-1$$
answered Nov 21 '18 at 15:35
TonyK
41.7k353132
edited Nov 21 '18 at 15:49
answered Nov 21 '18 at 15:35
TonyK
41.7k353132
answered Nov 21 '18 at 15:35
TonyK
41.7k353132
answered Nov 21 '18 at 15:35
TonyK
41.7k353132
41.7k353132
Thank you, by the way I think the last term is $-1$
– Zero
Nov 21 '18 at 15:53
@Zero: yes, thank you. I have corrected it now.
– TonyK
Nov 21 '18 at 15:59
add a comment |
Thank you, by the way I think the last term is $-1$
– Zero
Nov 21 '18 at 15:53
@Zero: yes, thank you. I have corrected it now.
– TonyK
Nov 21 '18 at 15:59
Thank you, by the way I think the last term is $-1$
– Zero
Nov 21 '18 at 15:53
Thank you, by the way I think the last term is $-1$
– Zero
Nov 21 '18 at 15:53
@Zero: yes, thank you. I have corrected it now.
– TonyK
Nov 21 '18 at 15:59
@Zero: yes, thank you. I have corrected it now.
– TonyK
Nov 21 '18 at 15:59
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1
I think you can find non-zero cubic polynomials.
– Arthur
Nov 21 '18 at 15:12
1
Look up orthogonal polynomials. As Arthur promised, there will be a non-trivial cubic $f(x)$ satisfying all these constraints. And if you add the requirement $int_a^bx^3f(x),dx=0$ then you are guaranteed to find a non-trivial quartic. Et cetera.
– Jyrki Lahtonen
Nov 21 '18 at 15:19
2
Only if you add $int_a^b x^n f(x){rm }dx = 0$ for all $ninmathbb{N}$ can you guarantee that $f(x) equiv 0$ (this follows from Weierstrass approximation theorem).
– Winther
Nov 21 '18 at 15:26
1
All explicit examples below are of polynomials, but note that there are also non-polynomial examples like $f(x) = sin left(frac{2 pi (x-a)}{b-a}right)-2 sin left(frac{4 pi (x-a)}{b-a}right)$.
– Winther
Nov 21 '18 at 15:36
2
@HenningMakholm Yes, it's a very standard argument. See e.g. If $int_0^1 f(x)x^n dx=0$ for every $n$, then $f=0$.
– Winther
Nov 21 '18 at 15:47