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Math Help - Stuck bad on this question

  1. #1
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    Stuck bad on this question

    Let f(x) and g(x) be functions s.t.

    f(a+b)=f(a)g(b)+g(a)f(b) for all a,b in R

    Given that f(0)=0,f'(0)=1,g(0)=1 and g'(0)=0, show that

    f'(x)=g(x) for all x in R

    Would appreciate any hints that could be given.

    Thanks
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  2. #2
    Super Member Showcase_22's Avatar
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    hmm, I tried this ("tried" implies that it fails but you might be able to work some magic on it):

    f'(x)=\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}

    f'(x)=\lim_{h\rightarrow 0} \frac{f(x)g(h)+g(x)f(h)-f(x)}{h}

    f'(x)=\frac{f(x)-f(x)}{h}

    I'm clearly not getting g(x). I think this method is the way to go, except it doesn't use the information f'(0)=1 and g'(0)=0.

    I haven't tried induction yet but it works for x=0. This method might take you somewhere.
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  3. #3
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    i'll try it out, but i've been getting

    f'(a+b)=f'(a)g(b)+g'(b)f(a)+g'(a)f(b)+g(a)f'(b)
    if b=0,
    f'(a+0)=f'(a).1+g(a).1
    f'(a)=f'(a)+g(a)
    g(a)=0 ?

    so maybe the definition of the derivative would be the way to go.
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  4. #4
    Moo
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    Hello,

    I think I got it

    f'(x)=\lim_{h \to 0} ~ \frac{f(x)g(h)+f(h)g(x)-f(x)}{h}

    f'(x)=\lim_{h \to 0} ~ \frac{f(x)(g(h)-1)+f(h)g(x)}{h}

    Assuming the limits exist, blah blah, we have :

    f'(x)=\lim_{h \to 0} ~ \frac{f(x)(g(h)-1)}{h}+\lim_{h \to 0} ~ \frac{f(h)g(x)}{h}

    g(h)-1=g(0+h)-g(0) and f(h)=f(0+h)-0=f(0+h)-f(0)

    Hence :

    f'(x)=\lim_{h \to 0} ~ f(x) \cdot \frac{g(0+h)-g(0)}{h}+\lim_{h \to 0} ~ g(x) \cdot \frac{f(0+h)-f(0)}{h}

    f'(x)=f(x) \cdot \underbrace{g'(0)}_{=0}+g(x) \cdot \underbrace{f'(0)}_{=1}

    \implies \boxed{f'(x)=g(x)}
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  5. #5
    Senior Member pankaj's Avatar
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    A simple trick is required to prove this.

     <br />
f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x+0)}{h}=\lim_{h\to 0}\frac{f(x)g(h)+g(x)f(h)-(f(x)g(0)+f(0)g(x))}{h}<br />

     <br />
=\lim_{h\to 0}(f(x)\frac{g(h)-g(0)}{h}+g(x)\frac{f(h)-f(0)}{h})<br />

     <br />
=f(x)\lim_{h\to 0}\frac{g(0+h)-g(0)}{h}+g(x)\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}<br />

     <br />
=f(x)g'(0)+g(x)f'(0)<br />

     <br />
=f(x)(0)+g(x)(1)<br />

     <br />
=g(x)<br />

    Do we need values of f(0) and g(0)?
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  6. #6
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    thanks alot guys. i wonder why differentiating f(a+b) the way i did wouldn't work. could anyone offer some explanation?

    thanks.
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  7. #7
    Senior Member pankaj's Avatar
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    O.K. Differentiating implies that you have assumed that the function is differentiable for all x whereas you have been provided values of the derivatives at only 2 points viz. x=a and x=b .That is why you can use only the definition of the derivative.By doing this we are actually checking whether derivatives exist for all x or not if they exist at the given points.

    However, had it been given that the function is differentiable I would not have any qualms by proceeding in the way you have but with a slight difference.
    Differentiating with respect to a treating b as a constant

    f'(a+b)=f'(a)g(b)+f(b)g'(a)

    Put a=0

     <br />
f'(b)=g(b)<br />
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  8. #8
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    yea i get it now. thanks.
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