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Thread: power series derivative

  1. #1
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    power series derivative

    Hi, can anyone please help me with this question?

    Consider the powerseries:
    $\displaystyle f(z)=\displaystyle \sum_{n=0}^{\infty} a_n z^n$
    with the property that
    $\displaystyle f(\frac{1}{a})=\frac{1}{a^3} \forall n\in\mathbb N$
    Prove that
    $\displaystyle f'(0)=0$

    I had hint to use Uniqueness Theorem of Power Series, but I still have no idea how to do it. Please help me, thanks a lot.
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  2. #2
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    You mean $\displaystyle f(\frac{1}{n})= \frac{1}{n^3}\forall n\in \mathbb N$.

    For the power series $\displaystyle \displaystyle f(z)= \sum_{n=0}^\infty a_nz^n$, $\displaystyle f'(z)= \displaystyle\sum_{n=0}^\infty} na_nz^{n-1}$ and, in particular, $\displaystyle f'(0)= a_1$. You are really asked to prove that $\displaystyle a_1= 0$.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by tsang View Post
    Hi, can anyone please help me with this question?

    Consider the powerseries:
    $\displaystyle f(z)=\displaystyle \sum_{n=0}^{\infty} a_n z^n$
    with the property that

    $\displaystyle f(\frac{1}{a})=\frac{1}{a^3} \forall n\in\mathbb N$

    Prove that

    $\displaystyle f'(0)=0$

    I had hint to use Uniqueness Theorem of Power Series, but I still have no idea how to do it. Please help me, thanks a lot.
    If $\displaystyle f(\frac{1}{a})= \frac{1}{a^{3}}\ , \ \forall a \in \mathbb{N}$ that means that is...

    $\displaystyle \displaystyle \sum_{n=0}^{\infty} \frac{a_{n}}{a^{n}} = \frac{1}{a^{3}}\ , \ \forall a \in \mathbb{N}$ (1)

    ... so that is...

    $\displaystyle a_{n}=\left\{\begin{array}{ll}1 ,\,\, n=3\\{}\\0 ,\,\, n \ne 3\end{array}\right.$ (2)

    ... and from (2) You conclude that is $\displaystyle f^{'}(0)=0$...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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