• Sep 27th 2008, 06:14 AM
Hweengee
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, 07:58 AM
Showcase_22
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.
• Sep 27th 2008, 08:20 AM
Hweengee
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, 08:27 AM
Moo
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 :

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

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

$\implies \boxed{f'(x)=g(x)}$ (Tongueout)
• Sep 27th 2008, 08:39 AM
pankaj
A simple trick is required to prove this.

$
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}
$

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

$
=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}
$

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

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

$
=g(x)
$

Do we need values of f(0) and g(0)?
• Sep 27th 2008, 08:42 AM
Hweengee
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, 08:57 AM
pankaj
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

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

$Put a=0$

$
f'(b)=g(b)
$
• Sep 27th 2008, 09:04 AM
Hweengee
yea i get it now. thanks.