Are there any functions other than $a^{kx}$ that satisfy the relation $f(x+y)=f(x)\cdot f(y)$? Actually I have a question that states that there exists a function such that f(x+y)=f(x)f(y) and f(5) =2 ,f'(0)=3,then f'(5)=?......I thought of assuming the function as a^(kx) and proceeded but I don't get my solution right
Question on Functional equation
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exponential-function
functional-equations
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1Yes, for example let $f(x)=\begin{cases} e^x & x\in\mathbb Q\\0& x\in \mathbb R-\mathbb Q\end{cases}$. – 2017-01-06
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0http://math.stackexchange.com/questions/423492/overview-of-basic-facts-about-cauchy-functional-equation – 2017-01-06
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0If you add the condition that f be *continuous* then the answer is "no". – 2017-01-06
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0Actually I have a question that states that there exists a function such that f(x+y)=f(x)f(y) and f(5) =2 ,f'(0)=3,then f'(5)=?......I thought of assuming the function as a^(kx) and proceeded but I don't get my solution right – 2017-01-06
1 Answers
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There is no need to solve this equation. Just compute: $$ \frac{\partial}{\partial x} f(x+y) = f'(x+y) $$ and $$ \frac{\partial}{\partial x} \left[f(x)f(y)\right] = f'(x)\cdot f(y) $$ Equate these two and set $x=0$, $y = 5$ to find: $$ f'(5) = f'(0)\cdot f(5) = 3\cdot 5 = 15 $$
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0Why did you put f'(y)=0? And why is my method incorrect when I have enough conditions to generate the function itself? – 2017-01-06
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0I didn't do that. Reread my response. Your method would work if you did all the algebra correctly, but it is overkill. – 2017-01-06
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0Thanks Damian...I get your point – 2017-01-06
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0I thought about this some more and the fact that continuity is not mentioned is crucial. If $f$ was continuous then the exponential is the unique solution. But this f is NOT continuous. Let's see why. Assume the function is continuous. Then $f$ is an exponential, $a^{kx}$. Plugging $y=x$ we get $f(2x)=f(x)^2$. Differentiating we find that $f'(2x) = f(x)\cdot f'(x)$. Plugging in $x=0$ we find that $f(0)=0$. But $a^0=1$, a contradiction. Hence $f$ is NOT continuous. – 2017-01-06
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0Yeah...that makes sense now...thanks a lot sir! – 2017-01-07