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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- Sep 27th 2008, 05:14 AMHweengeeStuck 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 - Sep 27th 2008, 06:58 AMShowcase_22
hmm, I tried this ("tried" implies that it fails but you might be able to work some magic on it):

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

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

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

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. - Sep 27th 2008, 07:20 AMHweengee
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. - Sep 27th 2008, 07:27 AMMoo
Hello,

I think I got it :)

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

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

Assuming the limits exist, blah blah, we have :

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

$\displaystyle g(h)-1=g(0+h)-g(0)$ and $\displaystyle f(h)=f(0+h)-0=f(0+h)-f(0)$

Hence :

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

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

$\displaystyle \implies \boxed{f'(x)=g(x)}$ (Tongueout) - Sep 27th 2008, 07:39 AMpankaj
A simple trick is required to prove this.

$\displaystyle

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}

$

$\displaystyle

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

$

$\displaystyle

=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}

$

$\displaystyle

=f(x)g'(0)+g(x)f'(0)

$

$\displaystyle

=f(x)(0)+g(x)(1)

$

$\displaystyle

=g(x)

$

Do we need values of f(0) and g(0)? - Sep 27th 2008, 07:42 AMHweengee
thanks alot guys. i wonder why differentiating f(a+b) the way i did wouldn't work. could anyone offer some explanation?

thanks. - Sep 27th 2008, 07:57 AMpankaj
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 $\displaystyle a$ treating $\displaystyle b$ as a constant

$\displaystyle f'(a+b)=f'(a)g(b)+f(b)g'(a)$

$\displaystyle Put a=0$

$\displaystyle

f'(b)=g(b)

$ - Sep 27th 2008, 08:04 AMHweengee
yea i get it now. thanks.