Existence of polynomial p such that $|f(x) − p(x^2)| < epsilon$
$begingroup$
Let $f$ be a real valued continuous function on $left[−1, 1right]$ such that $f(x) = f(−x)$ for all $x in left[−1, 1right]$.
Show that for every $epsilon > 0$ there is a polynomial $pleft(xright)$ with rational coefficients such that for every $x in left[−1, 1right]$,
$$|fleft(xright) − pleft(x^2right)| < epsilon$$.
I was trying something like this
$$f(x)=f(0)+frac{f^{'}(0)}{1!}x+frac{f^{''}(0)}{2!}x^2+frac{f^{'''}(0)}{3!}x^3+frac{f^{''''}(0)}{4!}x^4+dots dots$$
$$f(-x)=f(0)-frac{f^{'}(0)}{1!}x+frac{f^{''}(0)}{2!}x^2-frac{f^{'''}(0)}{3!}x^3+frac{f^{''''}(0)}{4!}x^4-dots dots$$
adding these two we get
$$f(x)+f(-x)=2f(0)+2frac{f^{''}(0)}{2!}x^2+2frac{f^{''''}(0)}{4!}x^4+2frac{f^{''''''}(0)}{6!}x^6+dots dots$$
Because $f(x)=f(-x)$
we have $$2f(x)=2f(0)+2frac{f^{''}(0)}{2!}x^2+2frac{f^{''''}(0)}{4!}x^4+2frac{f^{''''''}(0)}{6!}x^6+dots dots$$
Hence $$f(x)=f(0)+2frac{f^{''}(0)}{2!}x^2+frac{f^{''''}(0)}{4!}x^4+frac{f^{''''''}(0)}{6!}x^6+dots dots$$
Thus taking $$p(x)=f(x^{frac12})=f(0)+frac{f^{''}(0)}{2!}x+frac{f^{''''}(0)}{4!}x^2+2frac{f^{''''''}(0)}{6!}x^3+dots dots$$
This gives me $$|fleft(xright) − pleft(x^2right)|=0$$
But I want a polynomial $p(x)$ which should of finite degree. So I have trouble with this part
Can Someone write the answer of this question formally. I am not sure about presenting my answer.
real-analysis calculus polynomials power-series taylor-expansion
$endgroup$
add a comment |
$begingroup$
Let $f$ be a real valued continuous function on $left[−1, 1right]$ such that $f(x) = f(−x)$ for all $x in left[−1, 1right]$.
Show that for every $epsilon > 0$ there is a polynomial $pleft(xright)$ with rational coefficients such that for every $x in left[−1, 1right]$,
$$|fleft(xright) − pleft(x^2right)| < epsilon$$.
I was trying something like this
$$f(x)=f(0)+frac{f^{'}(0)}{1!}x+frac{f^{''}(0)}{2!}x^2+frac{f^{'''}(0)}{3!}x^3+frac{f^{''''}(0)}{4!}x^4+dots dots$$
$$f(-x)=f(0)-frac{f^{'}(0)}{1!}x+frac{f^{''}(0)}{2!}x^2-frac{f^{'''}(0)}{3!}x^3+frac{f^{''''}(0)}{4!}x^4-dots dots$$
adding these two we get
$$f(x)+f(-x)=2f(0)+2frac{f^{''}(0)}{2!}x^2+2frac{f^{''''}(0)}{4!}x^4+2frac{f^{''''''}(0)}{6!}x^6+dots dots$$
Because $f(x)=f(-x)$
we have $$2f(x)=2f(0)+2frac{f^{''}(0)}{2!}x^2+2frac{f^{''''}(0)}{4!}x^4+2frac{f^{''''''}(0)}{6!}x^6+dots dots$$
Hence $$f(x)=f(0)+2frac{f^{''}(0)}{2!}x^2+frac{f^{''''}(0)}{4!}x^4+frac{f^{''''''}(0)}{6!}x^6+dots dots$$
Thus taking $$p(x)=f(x^{frac12})=f(0)+frac{f^{''}(0)}{2!}x+frac{f^{''''}(0)}{4!}x^2+2frac{f^{''''''}(0)}{6!}x^3+dots dots$$
This gives me $$|fleft(xright) − pleft(x^2right)|=0$$
But I want a polynomial $p(x)$ which should of finite degree. So I have trouble with this part
Can Someone write the answer of this question formally. I am not sure about presenting my answer.
real-analysis calculus polynomials power-series taylor-expansion
$endgroup$
2
$begingroup$
Do you know Weierstrass approximation theorem?
$endgroup$
– Thomas Shelby
Dec 10 '18 at 4:54
$begingroup$
is my approach wrong?
$endgroup$
– user408906
Dec 10 '18 at 4:59
2
$begingroup$
@Suraj The function is assumed to be continuous this does not guarantee a Taylor expansion.
$endgroup$
– clark
Dec 10 '18 at 5:01
$begingroup$
@clark I see. Thanks. How do I proceed ahead then?
$endgroup$
– user408906
Dec 10 '18 at 5:02
1
$begingroup$
You can use Weierstrass approximation theorem and the denseness of polynomials with rational coefficients in $C[a,b] $.
$endgroup$
– Thomas Shelby
Dec 10 '18 at 5:04
add a comment |
$begingroup$
Let $f$ be a real valued continuous function on $left[−1, 1right]$ such that $f(x) = f(−x)$ for all $x in left[−1, 1right]$.
Show that for every $epsilon > 0$ there is a polynomial $pleft(xright)$ with rational coefficients such that for every $x in left[−1, 1right]$,
$$|fleft(xright) − pleft(x^2right)| < epsilon$$.
I was trying something like this
$$f(x)=f(0)+frac{f^{'}(0)}{1!}x+frac{f^{''}(0)}{2!}x^2+frac{f^{'''}(0)}{3!}x^3+frac{f^{''''}(0)}{4!}x^4+dots dots$$
$$f(-x)=f(0)-frac{f^{'}(0)}{1!}x+frac{f^{''}(0)}{2!}x^2-frac{f^{'''}(0)}{3!}x^3+frac{f^{''''}(0)}{4!}x^4-dots dots$$
adding these two we get
$$f(x)+f(-x)=2f(0)+2frac{f^{''}(0)}{2!}x^2+2frac{f^{''''}(0)}{4!}x^4+2frac{f^{''''''}(0)}{6!}x^6+dots dots$$
Because $f(x)=f(-x)$
we have $$2f(x)=2f(0)+2frac{f^{''}(0)}{2!}x^2+2frac{f^{''''}(0)}{4!}x^4+2frac{f^{''''''}(0)}{6!}x^6+dots dots$$
Hence $$f(x)=f(0)+2frac{f^{''}(0)}{2!}x^2+frac{f^{''''}(0)}{4!}x^4+frac{f^{''''''}(0)}{6!}x^6+dots dots$$
Thus taking $$p(x)=f(x^{frac12})=f(0)+frac{f^{''}(0)}{2!}x+frac{f^{''''}(0)}{4!}x^2+2frac{f^{''''''}(0)}{6!}x^3+dots dots$$
This gives me $$|fleft(xright) − pleft(x^2right)|=0$$
But I want a polynomial $p(x)$ which should of finite degree. So I have trouble with this part
Can Someone write the answer of this question formally. I am not sure about presenting my answer.
real-analysis calculus polynomials power-series taylor-expansion
$endgroup$
Let $f$ be a real valued continuous function on $left[−1, 1right]$ such that $f(x) = f(−x)$ for all $x in left[−1, 1right]$.
Show that for every $epsilon > 0$ there is a polynomial $pleft(xright)$ with rational coefficients such that for every $x in left[−1, 1right]$,
$$|fleft(xright) − pleft(x^2right)| < epsilon$$.
I was trying something like this
$$f(x)=f(0)+frac{f^{'}(0)}{1!}x+frac{f^{''}(0)}{2!}x^2+frac{f^{'''}(0)}{3!}x^3+frac{f^{''''}(0)}{4!}x^4+dots dots$$
$$f(-x)=f(0)-frac{f^{'}(0)}{1!}x+frac{f^{''}(0)}{2!}x^2-frac{f^{'''}(0)}{3!}x^3+frac{f^{''''}(0)}{4!}x^4-dots dots$$
adding these two we get
$$f(x)+f(-x)=2f(0)+2frac{f^{''}(0)}{2!}x^2+2frac{f^{''''}(0)}{4!}x^4+2frac{f^{''''''}(0)}{6!}x^6+dots dots$$
Because $f(x)=f(-x)$
we have $$2f(x)=2f(0)+2frac{f^{''}(0)}{2!}x^2+2frac{f^{''''}(0)}{4!}x^4+2frac{f^{''''''}(0)}{6!}x^6+dots dots$$
Hence $$f(x)=f(0)+2frac{f^{''}(0)}{2!}x^2+frac{f^{''''}(0)}{4!}x^4+frac{f^{''''''}(0)}{6!}x^6+dots dots$$
Thus taking $$p(x)=f(x^{frac12})=f(0)+frac{f^{''}(0)}{2!}x+frac{f^{''''}(0)}{4!}x^2+2frac{f^{''''''}(0)}{6!}x^3+dots dots$$
This gives me $$|fleft(xright) − pleft(x^2right)|=0$$
But I want a polynomial $p(x)$ which should of finite degree. So I have trouble with this part
Can Someone write the answer of this question formally. I am not sure about presenting my answer.
real-analysis calculus polynomials power-series taylor-expansion
real-analysis calculus polynomials power-series taylor-expansion
asked Dec 10 '18 at 4:52
user408906
2
$begingroup$
Do you know Weierstrass approximation theorem?
$endgroup$
– Thomas Shelby
Dec 10 '18 at 4:54
$begingroup$
is my approach wrong?
$endgroup$
– user408906
Dec 10 '18 at 4:59
2
$begingroup$
@Suraj The function is assumed to be continuous this does not guarantee a Taylor expansion.
$endgroup$
– clark
Dec 10 '18 at 5:01
$begingroup$
@clark I see. Thanks. How do I proceed ahead then?
$endgroup$
– user408906
Dec 10 '18 at 5:02
1
$begingroup$
You can use Weierstrass approximation theorem and the denseness of polynomials with rational coefficients in $C[a,b] $.
$endgroup$
– Thomas Shelby
Dec 10 '18 at 5:04
add a comment |
2
$begingroup$
Do you know Weierstrass approximation theorem?
$endgroup$
– Thomas Shelby
Dec 10 '18 at 4:54
$begingroup$
is my approach wrong?
$endgroup$
– user408906
Dec 10 '18 at 4:59
2
$begingroup$
@Suraj The function is assumed to be continuous this does not guarantee a Taylor expansion.
$endgroup$
– clark
Dec 10 '18 at 5:01
$begingroup$
@clark I see. Thanks. How do I proceed ahead then?
$endgroup$
– user408906
Dec 10 '18 at 5:02
1
$begingroup$
You can use Weierstrass approximation theorem and the denseness of polynomials with rational coefficients in $C[a,b] $.
$endgroup$
– Thomas Shelby
Dec 10 '18 at 5:04
2
2
$begingroup$
Do you know Weierstrass approximation theorem?
$endgroup$
– Thomas Shelby
Dec 10 '18 at 4:54
$begingroup$
Do you know Weierstrass approximation theorem?
$endgroup$
– Thomas Shelby
Dec 10 '18 at 4:54
$begingroup$
is my approach wrong?
$endgroup$
– user408906
Dec 10 '18 at 4:59
$begingroup$
is my approach wrong?
$endgroup$
– user408906
Dec 10 '18 at 4:59
2
2
$begingroup$
@Suraj The function is assumed to be continuous this does not guarantee a Taylor expansion.
$endgroup$
– clark
Dec 10 '18 at 5:01
$begingroup$
@Suraj The function is assumed to be continuous this does not guarantee a Taylor expansion.
$endgroup$
– clark
Dec 10 '18 at 5:01
$begingroup$
@clark I see. Thanks. How do I proceed ahead then?
$endgroup$
– user408906
Dec 10 '18 at 5:02
$begingroup$
@clark I see. Thanks. How do I proceed ahead then?
$endgroup$
– user408906
Dec 10 '18 at 5:02
1
1
$begingroup$
You can use Weierstrass approximation theorem and the denseness of polynomials with rational coefficients in $C[a,b] $.
$endgroup$
– Thomas Shelby
Dec 10 '18 at 5:04
$begingroup$
You can use Weierstrass approximation theorem and the denseness of polynomials with rational coefficients in $C[a,b] $.
$endgroup$
– Thomas Shelby
Dec 10 '18 at 5:04
add a comment |
1 Answer
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$begingroup$
Your approach is not valid because $f$ need not admit a Taylor series expansion. $f(sqrt {|x|})$ is a continuous function. By Wierstrass Theorem there is a polynomial $p$ such that $|f(sqrt {|x|})-p(x)| < epsilon$ for all $x$. Replacing $x$ by $x^{2}$ we get $|f(x)-p(x^{2})| < epsilon$ for all $x in [0,1]$. Now use the fact that $f(-x)=f(x)$ to conclude that the inequality holds for $x in [-1,0] $ also.
$endgroup$
add a comment |
Your Answer
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1 Answer
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$begingroup$
Your approach is not valid because $f$ need not admit a Taylor series expansion. $f(sqrt {|x|})$ is a continuous function. By Wierstrass Theorem there is a polynomial $p$ such that $|f(sqrt {|x|})-p(x)| < epsilon$ for all $x$. Replacing $x$ by $x^{2}$ we get $|f(x)-p(x^{2})| < epsilon$ for all $x in [0,1]$. Now use the fact that $f(-x)=f(x)$ to conclude that the inequality holds for $x in [-1,0] $ also.
$endgroup$
add a comment |
$begingroup$
Your approach is not valid because $f$ need not admit a Taylor series expansion. $f(sqrt {|x|})$ is a continuous function. By Wierstrass Theorem there is a polynomial $p$ such that $|f(sqrt {|x|})-p(x)| < epsilon$ for all $x$. Replacing $x$ by $x^{2}$ we get $|f(x)-p(x^{2})| < epsilon$ for all $x in [0,1]$. Now use the fact that $f(-x)=f(x)$ to conclude that the inequality holds for $x in [-1,0] $ also.
$endgroup$
add a comment |
$begingroup$
Your approach is not valid because $f$ need not admit a Taylor series expansion. $f(sqrt {|x|})$ is a continuous function. By Wierstrass Theorem there is a polynomial $p$ such that $|f(sqrt {|x|})-p(x)| < epsilon$ for all $x$. Replacing $x$ by $x^{2}$ we get $|f(x)-p(x^{2})| < epsilon$ for all $x in [0,1]$. Now use the fact that $f(-x)=f(x)$ to conclude that the inequality holds for $x in [-1,0] $ also.
$endgroup$
Your approach is not valid because $f$ need not admit a Taylor series expansion. $f(sqrt {|x|})$ is a continuous function. By Wierstrass Theorem there is a polynomial $p$ such that $|f(sqrt {|x|})-p(x)| < epsilon$ for all $x$. Replacing $x$ by $x^{2}$ we get $|f(x)-p(x^{2})| < epsilon$ for all $x in [0,1]$. Now use the fact that $f(-x)=f(x)$ to conclude that the inequality holds for $x in [-1,0] $ also.
answered Dec 10 '18 at 5:24
Kavi Rama MurthyKavi Rama Murthy
60.9k42261
60.9k42261
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2
$begingroup$
Do you know Weierstrass approximation theorem?
$endgroup$
– Thomas Shelby
Dec 10 '18 at 4:54
$begingroup$
is my approach wrong?
$endgroup$
– user408906
Dec 10 '18 at 4:59
2
$begingroup$
@Suraj The function is assumed to be continuous this does not guarantee a Taylor expansion.
$endgroup$
– clark
Dec 10 '18 at 5:01
$begingroup$
@clark I see. Thanks. How do I proceed ahead then?
$endgroup$
– user408906
Dec 10 '18 at 5:02
1
$begingroup$
You can use Weierstrass approximation theorem and the denseness of polynomials with rational coefficients in $C[a,b] $.
$endgroup$
– Thomas Shelby
Dec 10 '18 at 5:04