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Math Help - TAYLOR-function null

  1. #1
    J.R
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    TAYLOR-function null

    hello,

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

    We suppose :

    \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 ]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 u \in [a,b]
    |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 f^{n+1} from [a,b].

    so, here, we have:

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

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

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

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

    so we have:

    |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 C^{\infty} from [a,b] to \mathbb{R}.

    We suppose :

    \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 ]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 u \in [a,b]
    |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 f^{n+1} from [a,b].

    so, here, we have:

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

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

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

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

    so we have:

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

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