Existence of polynomial p such that $|f(x) − p(x^2)| < epsilon$












2












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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
















2












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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














2












2








2





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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














  • 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










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.






share|cite|improve this 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.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 10 '18 at 5:24









        Kavi Rama MurthyKavi Rama Murthy

        56.5k42159




        56.5k42159






























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