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$T(a)=\begin{cases} 0 & a\leq 0\\ e^{-1/a} & a>0 \end{cases}$

$Z(a)=\begin{cases}0 & a\geq 1\\ (1+a) e^{-1/(1-a)} & a<1 \end{cases}$

$p=\int_{-1}^{1}Z(x)dx$

$Y(a)=\frac{1}{p}\int_{-1}^{a}Z(x)dx$

$R(a)=Y(a-1)Y(2-a)$

(Why all four functions are in $C^{\infty}(R)$)

My real question is that -->

how to prove the Taylor series of $R(a)$ centered at zero converges to $R$ however its sum equals to $R(a)$ for $a$ in a much smaller interval. What is the interval?

Please Help me for how to solve this problem? It is so difficult. I dont know how to solve. Thank you:)

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To clarify: you want the interval of convergence of the Taylor series of $R$ at $0$, and the subset where its sum is equal to $R$, right? –  1015 Mar 3 '13 at 21:41
    
@julien Yes this is right. –  B11b Mar 3 '13 at 21:44
    
Something tells me you have a typo and $p=\int_{-1}^1Z(x)dx$. –  1015 Mar 3 '13 at 22:35
    
@julien I have calculated this. And I found approximately 0.420158 –  B11b Mar 3 '13 at 22:37
    
Can you solve or produce an idea? @julien ? –  B11b Mar 3 '13 at 23:10

2 Answers 2

up vote 4 down vote accepted

First part: show by induction that $$ T^{(n)}(x)=\frac{P_n(x)}{x^{2n}}e^{-1/x} \qquad\forall x>0 $$ whith $P_n$ a polynomial.

It follows that $$ \lim_{0^-}T^{(n)}(x)=\lim_{0^+}T^{(n)}(x)=0 $$ for all $n$.

So $T$, which is obviously $C^\infty$ on $\mathbb{R}^*$, is also $C^\infty$ at $0$.

Now $Z,Y,R$ are built from $T$ in such a way that they are also $C^\infty$.

Second part: Note that $$ 1=Y(1)=Y(x)\qquad \forall x\geq 1. $$ It follows that $$ R(x)=Y(x-1)\qquad \forall x\leq 1. $$

Now observe that the complex function $$ f(z)=(1+z)e^{-\frac{1}{1-z}} $$ is holomorphic on $\mathbb{C}\setminus\{1\}$.

In particular, $f$ is analytic on $(-3,1)$, so $Z$ is analytic on $(-3,1)$. Both are equal to their Taylor series on $(-3,1)$, by restriction of the holomorphic function $f$ to the reals.

Therefore the antiderivative $$ g:x\longmapsto \int_{-1}^xZ(t)dt $$ is analytic and equal to its Taylor series at $0$ on $(-3,1)$, so $$ g(-1+x)=\sum_{n\geq 0}\frac{g^{(n)}(-1)}{n!}x^n \qquad \forall x\in (-2,2). $$

Now a fortiori $$ R(x)=Y(x-1)=\frac{1}{p}g(-1+x) \qquad \forall x\in (-1,1) $$ is analytic and equal to its Taylor series at $0$ on $(-1,1)$.

Recall the radius of convergence of the Taylor series of a holomorphic function at $z_0$ is the distance between $z_0$ and the complement of the domain of analyticity. It follows that the radius of convergence of the Taylor series above is $2$.

Also, $Y(x-1)$ is equal to its Taylor series at $0$ on $(-2,2)$. So it is equal to $$ Y(2-x)Y(x-1)=R(x) \qquad \forall x\in (-2,1] $$ since $Y(2-x)=1$ there.

And on $(1,2)$, we have $Y(2-x)\neq 1$ and $Y(x-1)\neq 0$ so the Taylor series does not equal $R(x)$ there.

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Ok!! You are really excallent mathematician!!!! I will think deeplly on your question. Thank you for your help. And hopefully you Will gain best successes in your real life in tour math career. God bless you. Thank you:) –  B11b Mar 3 '13 at 23:18
    
@B11 I know some really good mathematicians so I can assure you I am not that good... Thank you though. Note I've completed my answer. –  1015 Mar 4 '13 at 0:04
    
but you will think fine while solving a question. Your math career Will be excellent, i believe.. Thank you for your solution. Now I'm trying to understand the solution deeply. Thanksfully –  B11b Mar 4 '13 at 0:28
    
Don't have to prove $f$ is holomorphic since its the main part of the entire proof? –  Pratyush Sarkar Mar 4 '13 at 0:30
    
@PratyushSarkar The function $z\to 1/(1-z)$ is holomorphic on $\mathbb{C}\setminus\{1\}$. So $f$ is holomorphic on the same domain, by composition/product with entire functions. Trivially. –  1015 Mar 4 '13 at 0:33

For the first one we obviously have any order derivative of $T(a)$ is differentiable for $a < 0$ and $a > 0$ so you are left with the case $a = 0$. For that you can just get a formula for the $n^\text{th}$ derivative of the exponential part of the function and show that the limit as $a \to 0$ is $0$. So $T \in C^\infty$.

$Z(a)$ can be written as $Z(a) = (1 + a)T(1 - a)$ which shows $Z \in C^\infty$.

$Y$ is defined as an integral of a continuous function, so it must be differentiable and its derivative is $Y'(a) = \frac{1}{p}Z(a)$ whose higher order derivatives are obviously differentiable and so $Y \in C^\infty$.

For $R$, any order derivatives are just a sum of products of differentiable functions, so $R \in C^\infty$.

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@partyushSarkar thank you:) I understand. But you did not say anything about the second part of the question; that's, Taylor question. Help me please. Because my real question is this. Thank you again:) –  B11b Mar 3 '13 at 21:43
    
@B11 Hehe. Sorry, that was because I didn't know how to do that yet as I am in first year and series is our next and last topic. I know Taylor series and a little bit about analytic functions but not enough to answer the question. –  Pratyush Sarkar Mar 4 '13 at 0:27

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