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
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.
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 :
$\displaystyle f'(x)=\lim_{h \to 0} ~ f(x) \cdot \frac{g(0+h)-g(0)}{h}+\lim_{h \to 0} ~ g(x) \cdot \frac{f(0+h)-f(0)}{h}$
$\displaystyle f'(x)=f(x) \cdot \underbrace{g'(0)}_{=0}+g(x) \cdot \underbrace{f'(0)}_{=1}$
$\displaystyle \implies \boxed{f'(x)=g(x)}$
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)?
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)
$