Continuous Functions and Integrability
$begingroup$
Let $a<b$ $a in R$ and $f: [a,b] to R$ be a continuous function such that for all $k = 0,1,...,n$ we have $int_a^bt^kf(t)dt=0$. Find the minimum number of elements of the set ${xin (a,b)|f(x)=0}$
I came across this question in one of my homework. Intuitively I think that the solution of this question must be n+1.
Assume the interval is [-1,1], then it is obvious that for $int_a^bt^0f(t)dt=0$ we need at least 1 $x$ such that $f(x)=0$ because the integral of the function that is above f(x)=0 must be equal to the integral of the function that is below it. Now, multiplying this by t, the interval [-1,0) will shift signs and therefore there must be another point with $f(x)=0$, so to get $int_a^bt^1f(t)dt=0$ and $int_a^bt^0f(t)dt=0$ we need at least 2 points in the interval [-1,1] such that $f(x)=0$. Continuing inductively, I assume that the answer to this question must be $n+1$, however, I am having trouble showing my solution in a mathematically correct way.
Thank you!
real-analysis functions definite-integrals continuity
asked Jan 8 at 17:58
math1945math1945
22
$endgroup$
add a comment |
$begingroup$
Let $a<b$ $a in R$ and $f: [a,b] to R$ be a continuous function such that for all $k = 0,1,...,n$ we have $int_a^bt^kf(t)dt=0$. Find the minimum number of elements of the set ${xin (a,b)|f(x)=0}$
I came across this question in one of my homework. Intuitively I think that the solution of this question must be n+1.
Assume the interval is [-1,1], then it is obvious that for $int_a^bt^0f(t)dt=0$ we need at least 1 $x$ such that $f(x)=0$ because the integral of the function that is above f(x)=0 must be equal to the integral of the function that is below it. Now, multiplying this by t, the interval [-1,0) will shift signs and therefore there must be another point with $f(x)=0$, so to get $int_a^bt^1f(t)dt=0$ and $int_a^bt^0f(t)dt=0$ we need at least 2 points in the interval [-1,1] such that $f(x)=0$. Continuing inductively, I assume that the answer to this question must be $n+1$, however, I am having trouble showing my solution in a mathematically correct way.
Thank you!
real-analysis functions definite-integrals continuity
asked Jan 8 at 17:58
math1945math1945
22
$endgroup$
$begingroup$
What do we have to prove? What have you tried?
$endgroup$
– Shubham Johri
Jan 8 at 18:30
$begingroup$
Find is a better word here. My thought is that there must be at least n+1 elements with f(x)=0, but I do not know how to prove it.
$endgroup$
– math1945
Jan 8 at 18:36
add a comment |
$begingroup$
Let $a<b$ $a in R$ and $f: [a,b] to R$ be a continuous function such that for all $k = 0,1,...,n$ we have $int_a^bt^kf(t)dt=0$. Find the minimum number of elements of the set ${xin (a,b)|f(x)=0}$
I came across this question in one of my homework. Intuitively I think that the solution of this question must be n+1.
Assume the interval is [-1,1], then it is obvious that for $int_a^bt^0f(t)dt=0$ we need at least 1 $x$ such that $f(x)=0$ because the integral of the function that is above f(x)=0 must be equal to the integral of the function that is below it. Now, multiplying this by t, the interval [-1,0) will shift signs and therefore there must be another point with $f(x)=0$, so to get $int_a^bt^1f(t)dt=0$ and $int_a^bt^0f(t)dt=0$ we need at least 2 points in the interval [-1,1] such that $f(x)=0$. Continuing inductively, I assume that the answer to this question must be $n+1$, however, I am having trouble showing my solution in a mathematically correct way.
Thank you!
real-analysis functions definite-integrals continuity
asked Jan 8 at 17:58
math1945math1945
22
$endgroup$
Let $a<b$ $a in R$ and $f: [a,b] to R$ be a continuous function such that for all $k = 0,1,...,n$ we have $int_a^bt^kf(t)dt=0$. Find the minimum number of elements of the set ${xin (a,b)|f(x)=0}$
I came across this question in one of my homework. Intuitively I think that the solution of this question must be n+1.
Assume the interval is [-1,1], then it is obvious that for $int_a^bt^0f(t)dt=0$ we need at least 1 $x$ such that $f(x)=0$ because the integral of the function that is above f(x)=0 must be equal to the integral of the function that is below it. Now, multiplying this by t, the interval [-1,0) will shift signs and therefore there must be another point with $f(x)=0$, so to get $int_a^bt^1f(t)dt=0$ and $int_a^bt^0f(t)dt=0$ we need at least 2 points in the interval [-1,1] such that $f(x)=0$. Continuing inductively, I assume that the answer to this question must be $n+1$, however, I am having trouble showing my solution in a mathematically correct way.
Thank you!
real-analysis functions definite-integrals continuity
real-analysis functions definite-integrals continuity
asked Jan 8 at 17:58
math1945math1945
22
asked Jan 8 at 17:58
math1945math1945
22
edited Jan 9 at 14:10
math1945
asked Jan 8 at 17:58
math1945math1945
22
asked Jan 8 at 17:58
math1945math1945
22
asked Jan 8 at 17:58
math1945math1945
22
22
$begingroup$
What do we have to prove? What have you tried?
$endgroup$
– Shubham Johri
Jan 8 at 18:30
$begingroup$
Find is a better word here. My thought is that there must be at least n+1 elements with f(x)=0, but I do not know how to prove it.
$endgroup$
– math1945
Jan 8 at 18:36
add a comment |
$begingroup$
What do we have to prove? What have you tried?
$endgroup$
– Shubham Johri
Jan 8 at 18:30
$begingroup$
Find is a better word here. My thought is that there must be at least n+1 elements with f(x)=0, but I do not know how to prove it.
$endgroup$
– math1945
Jan 8 at 18:36
$begingroup$
What do we have to prove? What have you tried?
$endgroup$
– Shubham Johri
Jan 8 at 18:30
$begingroup$
What do we have to prove? What have you tried?
$endgroup$
– Shubham Johri
Jan 8 at 18:30
$begingroup$
Find is a better word here. My thought is that there must be at least n+1 elements with f(x)=0, but I do not know how to prove it.
$endgroup$
– math1945
Jan 8 at 18:36
$begingroup$
Find is a better word here. My thought is that there must be at least n+1 elements with f(x)=0, but I do not know how to prove it.
$endgroup$
– math1945
Jan 8 at 18:36
add a comment |
1 Answer
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$begingroup$
The polynomials of degree $le n+1$ form a vector space $V_{n+1}$ of dimension $n+2$. The linear map $T_n: f mapsto [int_a^b x^0 f(x); dx, ldots, int_a^b x^{n} f(x); dx]$ takes $V_{n+1}$ into $mathbb R^{n+1}$.
Since $n+2 > n+1$ it must have nontrivial kernel, i.e. there is some nonzero $f in V_{n+1}$ such that $int_a^b x^j f(x); dx = 0$ for all $j = 0 ldots n$. Since it has degree $le n+1$ and is not identically $0$,
such $f$ has at most $n+1$ zeros in the interval.
On the other hand, suppose $f$ has fewer than $n+1$ zeros in the interval. Let $p_i$ be the points in the interval where $f$ changes sign. There are at most $n$ of them. Thus $P(x) = prod_i (x - p_i)$ is a polynomial of degree $le n$ such that $P(x) f(x)$ is either always $ge 0$ or always $le 0$ for $x$ in the interval. Thus $int_a^b P(x) f(x); dx ne 0$. Conclude that $int_a^b x^j f(x); dx ne 0$ for some $j in {0,1,ldots,n}$.
answered Jan 8 at 18:30
Robert IsraelRobert Israel
330k23219473
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$begingroup$
The polynomials of degree $le n+1$ form a vector space $V_{n+1}$ of dimension $n+2$. The linear map $T_n: f mapsto [int_a^b x^0 f(x); dx, ldots, int_a^b x^{n} f(x); dx]$ takes $V_{n+1}$ into $mathbb R^{n+1}$.
Since $n+2 > n+1$ it must have nontrivial kernel, i.e. there is some nonzero $f in V_{n+1}$ such that $int_a^b x^j f(x); dx = 0$ for all $j = 0 ldots n$. Since it has degree $le n+1$ and is not identically $0$,
such $f$ has at most $n+1$ zeros in the interval.
On the other hand, suppose $f$ has fewer than $n+1$ zeros in the interval. Let $p_i$ be the points in the interval where $f$ changes sign. There are at most $n$ of them. Thus $P(x) = prod_i (x - p_i)$ is a polynomial of degree $le n$ such that $P(x) f(x)$ is either always $ge 0$ or always $le 0$ for $x$ in the interval. Thus $int_a^b P(x) f(x); dx ne 0$. Conclude that $int_a^b x^j f(x); dx ne 0$ for some $j in {0,1,ldots,n}$.
answered Jan 8 at 18:30
Robert IsraelRobert Israel
330k23219473
$endgroup$
add a comment |
$begingroup$
The polynomials of degree $le n+1$ form a vector space $V_{n+1}$ of dimension $n+2$. The linear map $T_n: f mapsto [int_a^b x^0 f(x); dx, ldots, int_a^b x^{n} f(x); dx]$ takes $V_{n+1}$ into $mathbb R^{n+1}$.
Since $n+2 > n+1$ it must have nontrivial kernel, i.e. there is some nonzero $f in V_{n+1}$ such that $int_a^b x^j f(x); dx = 0$ for all $j = 0 ldots n$. Since it has degree $le n+1$ and is not identically $0$,
such $f$ has at most $n+1$ zeros in the interval.
On the other hand, suppose $f$ has fewer than $n+1$ zeros in the interval. Let $p_i$ be the points in the interval where $f$ changes sign. There are at most $n$ of them. Thus $P(x) = prod_i (x - p_i)$ is a polynomial of degree $le n$ such that $P(x) f(x)$ is either always $ge 0$ or always $le 0$ for $x$ in the interval. Thus $int_a^b P(x) f(x); dx ne 0$. Conclude that $int_a^b x^j f(x); dx ne 0$ for some $j in {0,1,ldots,n}$.
answered Jan 8 at 18:30
Robert IsraelRobert Israel
330k23219473
$endgroup$
add a comment |
$begingroup$
The polynomials of degree $le n+1$ form a vector space $V_{n+1}$ of dimension $n+2$. The linear map $T_n: f mapsto [int_a^b x^0 f(x); dx, ldots, int_a^b x^{n} f(x); dx]$ takes $V_{n+1}$ into $mathbb R^{n+1}$.
Since $n+2 > n+1$ it must have nontrivial kernel, i.e. there is some nonzero $f in V_{n+1}$ such that $int_a^b x^j f(x); dx = 0$ for all $j = 0 ldots n$. Since it has degree $le n+1$ and is not identically $0$,
such $f$ has at most $n+1$ zeros in the interval.
On the other hand, suppose $f$ has fewer than $n+1$ zeros in the interval. Let $p_i$ be the points in the interval where $f$ changes sign. There are at most $n$ of them. Thus $P(x) = prod_i (x - p_i)$ is a polynomial of degree $le n$ such that $P(x) f(x)$ is either always $ge 0$ or always $le 0$ for $x$ in the interval. Thus $int_a^b P(x) f(x); dx ne 0$. Conclude that $int_a^b x^j f(x); dx ne 0$ for some $j in {0,1,ldots,n}$.
answered Jan 8 at 18:30
Robert IsraelRobert Israel
330k23219473
$endgroup$
The polynomials of degree $le n+1$ form a vector space $V_{n+1}$ of dimension $n+2$. The linear map $T_n: f mapsto [int_a^b x^0 f(x); dx, ldots, int_a^b x^{n} f(x); dx]$ takes $V_{n+1}$ into $mathbb R^{n+1}$.
Since $n+2 > n+1$ it must have nontrivial kernel, i.e. there is some nonzero $f in V_{n+1}$ such that $int_a^b x^j f(x); dx = 0$ for all $j = 0 ldots n$. Since it has degree $le n+1$ and is not identically $0$,
such $f$ has at most $n+1$ zeros in the interval.
On the other hand, suppose $f$ has fewer than $n+1$ zeros in the interval. Let $p_i$ be the points in the interval where $f$ changes sign. There are at most $n$ of them. Thus $P(x) = prod_i (x - p_i)$ is a polynomial of degree $le n$ such that $P(x) f(x)$ is either always $ge 0$ or always $le 0$ for $x$ in the interval. Thus $int_a^b P(x) f(x); dx ne 0$. Conclude that $int_a^b x^j f(x); dx ne 0$ for some $j in {0,1,ldots,n}$.
answered Jan 8 at 18:30
Robert IsraelRobert Israel
330k23219473
answered Jan 8 at 18:30
Robert IsraelRobert Israel
330k23219473
answered Jan 8 at 18:30
Robert IsraelRobert Israel
330k23219473
answered Jan 8 at 18:30
Robert IsraelRobert Israel
330k23219473
330k23219473
add a comment |
add a comment |
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$begingroup$
What do we have to prove? What have you tried?
$endgroup$
– Shubham Johri
Jan 8 at 18:30
$begingroup$
Find is a better word here. My thought is that there must be at least n+1 elements with f(x)=0, but I do not know how to prove it.
$endgroup$
– math1945
Jan 8 at 18:36