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Math Help - Prove f(x)=||x||^2 is differentiable. Help plz?

  1. #1
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    Prove f(x)=||x||^2 is differentiable. Help plz?

    Can anyone help me with this question???

    prove f(x)=||x||^2 is differentiable using the definition of differentiability
    where x is a vector in R^n. and ||x|| is the length of the vecotr(Not allowed to use use components of x)
    x=(x1,x2,...xn)

    The definitoin is lim as h->0, [f(a+h)-f(a) -c*h]/h =0
    where c is the gradient of f,the function is differentiable at point a


    I have absolutely no idea where to start, it's not like you have two distinct vairables..x,y....and prove differentiable at a particular point, and I am not allowed to use the components of x.How do i prove it's differentiable everywhere...

    Anyone give me some hints?


    Thanks so much
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  2. #2
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    It's easy to see that c*h = 2<x,h> ie. 2<x,h>=D_f(x)h use this and the fact that \Vert x+h \Vert ^2 = ^\Vert x \Vert ^2 + 2<x,h> + \Vert h \Vert ^2 and substitute in the definition: \vert \frac{ f(x+h) - f(x) - D_f(x)h}{ \Vert h \Vert } \vert and see that this goes to zero as \Vert h \Vert \rightarrow 0. ( <,> denotes the interior product, I'm assuming you're using the usual norm in \mathbb{R} ^n)
    Last edited by Jose27; October 12th 2009 at 08:23 PM. Reason: Missed a 2
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  3. #3
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    Quote Originally Posted by Jose27 View Post
    It's easy to see that c*h = <x,h> ie. <x,h>=D_f(x)h use this and the fact that \Vert x+h \Vert ^2 = ^\Vert x \Vert ^2 + 2<x,h> + \Vert h \Vert ^2 and substitute in the definition: \vert \frac{ f(x+h) - f(x) - D_f(x)h}{ \Vert h \Vert } \vert and see that this goes to zero as \Vert h \Vert \rightarrow 0. ( <,> denotes the interior product, I'm assuming you're using the usual norm in \mathbb{R} ^n)
    Thanks so much
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