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Math Help - Differentiability Proof

  1. #1
    OXF
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    Differentiability Proof

    Let g:R->R be a twice differentiable function satisfying g(0)=g'(0)=1 and g''(x)=g(x)=0 for all x in R.
    (i) Prove g has derivatives of all orders.
    (ii)Let x>0. Show that there exists a constant M>0 such that |g^n(Ax)|<=M for all n in N and A in (0,1).

    Any help would be appreciated; not really sure where to start with this one.
    OXF
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  2. #2
    Super Member girdav's Avatar
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    I guess you meant g''(x)+g(x)=0. You can try to show the result by induction on the order of the derivative.
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  3. #3
    OXF
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    Yeah, I meant g''(x)+g(x)=0.
    How would I go about an inductive proof?

    Would you assume the kth derivative exists (therefore we know the (k-1)th, (k-2)th, ... etc exist also), then apply the defintion of the derivative? I don't see how it would work?
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  4. #4
    Super Member girdav's Avatar
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    If we assume that the k-th derivative of g exists and is continuous, then so is the k-th derivative of g'', which is the (k+2)-th derivative of g.
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  5. #5
    OXF
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    is it valid to say that as the k-th derivative exists, so does the (k+2)th derivative? I thought the rule stated would only apply to g and g'', rather than g^(k) and g^(k+2).
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