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Thread: TAYLOR-function null

  1. #1
    J.R
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    TAYLOR-function null

    hello,

    f is a function $\displaystyle C^{\infty}$ from [a,b] to $\displaystyle \mathbb{R}$.

    We suppose :

    $\displaystyle \exists x_0 \in ]a,b[ \ and \ k \ real \ >0 | \forall n\in \mathbb{N},f^{(n)}(x_0)=0 \ \ and \ \ sup_{[a,b]}\{|f^{(n)}(x)|\}\le k^nn!$

    1) Show that f is the function null from $\displaystyle ]x_0-\frac{1}{k},x_0+\frac{1}{k}[$.

    I think that we must use Taylor's inegalities which say that (I don't know if it's the same name ...) :


    for $\displaystyle u \in [a,b]$
    $\displaystyle |f(b)-\sum_{k=0}^n\frac{(x-u)^k}{k!}f^{(k)}(u)|\le M\frac{(b-u)^{n+1}}{(n+1)!}$

    where M majors $\displaystyle f^{n+1}$ from [a,b].

    so, here, we have:

    for $\displaystyle x\in ]x_0-\frac{1}{k},x_0+\frac{1}{k}[$

    $\displaystyle |f(x)|\le \frac{|x-x_0|^{n+1}}{(n+1)!}M$

    howerver we can take $\displaystyle M=k^{n+1}(n+1)!$

    and $\displaystyle |x-x_0|\le \frac{1}{k}$

    so we have:

    $\displaystyle |f(x)|\le 1$ ...

    and I have no idea to keep on the demonstration.

    thanks

    do you understand what I said ?
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  2. #2
    J.R
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    if I am mistaken tell me (In first, I want improve my english and do maths ) ...

    thanks
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  3. #3
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    Quote Originally Posted by J.R View Post
    hello,

    f is a function $\displaystyle C^{\infty}$ from [a,b] to $\displaystyle \mathbb{R}$.

    We suppose :

    $\displaystyle \exists x_0 \in ]a,b[ \ and \ k \ real \ >0 | \forall n\in \mathbb{N},f^{(n)}(x_0)=0 \ \ and \ \ sup_{[a,b]}\{|f^{(n)}(x)|\}\le k^nn!$

    1) Show that f is the function null from $\displaystyle ]x_0-\frac{1}{k},x_0+\frac{1}{k}[$.

    I think that we must use Taylor's inegalities which say that (I don't know if it's the same name ...) :


    for $\displaystyle u \in [a,b]$
    $\displaystyle |f(b)-\sum_{k=0}^n\frac{(x-u)^k}{k!}f^{(k)}(u)|\le M\frac{(b-u)^{n+1}}{(n+1)!}$

    where M majors $\displaystyle f^{n+1}$ from [a,b].

    so, here, we have:

    for $\displaystyle x\in ]x_0-\frac{1}{k},x_0+\frac{1}{k}[$

    $\displaystyle |f(x)|\le \frac{|x-x_0|^{n+1}}{(n+1)!}M$

    howerver we can take $\displaystyle M=k^{n+1}(n+1)!$

    and $\displaystyle |x-x_0|\le \frac{1}{k}$

    so we have:

    $\displaystyle |f(x)|\le 1$ ...

    and I have no idea to keep on the demonstration.
    Your argument is correct, and it shows that $\displaystyle |f(x)|\leqslant k^{n+1}|x-x_0|^{n+1}$. But you are told that x belongs to the open interval $\displaystyle (x_0-\tfrac1k,x_0+\tfrac1k)$ (or in the horrible French notation $\displaystyle ]x_0-\tfrac1k,x_0+\tfrac1k[$ ), so you know that $\displaystyle k|x-x_0|$ is strictly less than 1, so its (n+1)th power can be made arbitrarily small and hence $\displaystyle f(x)=0$.
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  4. #4
    J.R
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    Thumbs up Problem solve

    message removes
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  5. #5
    J.R
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    Thumbs up Problem solve



    Sincerely,
    thank you very much.

    @+
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