If f(x-y). f(y)=f(x), f(5)=32, then what is f(7)?

[littlemathquark]

Homework Statement

If f(x-y). f(y) =f(x), f(5) =32, then what is f(7)?

Relevant Equations

If f(x-y). f(y) =f(x), f(5) =32, then what is f(7)?

I can find f(0)=1 and f(1)=2 and f(7)=128.
I Wonder how can I find f function?

 

[fresh_42]

If you write $z=x-y$ then the condition becomes $f(z)\cdot f(x)=f(z+x).$ What do you know about this functional equation?

 

[littlemathquark]

Yes, Cauchy exponansiyel equation, I mean $f(x) =a^x$. But how can I find this?

 

[fresh_42]

You can use $f(x)=a^x$ with $f(5)=a^5=32$ and find $a=2.$ This is mainly by inspection.

We do not have enough information to uniquely determine $f(x)$. All we know is that $f(x)=a^{c\cdot x}$ since these are all functions that satisfy the functional equation $(*).$ Hence, $f(5)=32=2^5=\left(a^c\right)^5 \Longrightarrow 2=a^c$ and all these functions are solutions. $(a,c)=(2,1)$ is only the easiest one.

Now, how do we know $(*)$? Honestly, I am not sure. I would set $f(x)=\sum_{k=0}^\infty a_k x^k$ and try to figure out the coefficients $a_k.$ However, this requires that $f(x)$ can be written as a power series and I'm not sure whether this can be deduced only by the functional equation. I guess so, but I don't know the answer without trying.

 

[WWGD]

Maybe just remembering the Exponential satisfies $a ^{x+y}=a^x *a^y$.

 

[Demystifier]

@fresh_42 determined that $f(x)=a^{cx}$ and $a^c=2$. Using the formula $a^{cx}=(a^c)^x$, we have $f(x)=2^x$. Or am I missing something?

 

[littlemathquark]

I can find answer without finding f function:
$f(5)=f(4)f(1)=f(3)f(1)^2=\cdots=f(1)^5\implies f(1)=2$
$f(7)=f(6)f(1)=f(5)f(1)^2=32\cdot 2^2=128$

 

[Gavran]

$f(1-1)f(1)=f(1)$
$f(0)f(1)=f(1)$
$f(0)=1$

By using
$\begin{align} f(n x)&=f((n-1)x)f(x)=f((n-2))f^2(x)=f((n-3))f^3(x)=...\nonumber\\ &=f(1)f^{n-1}(x)=f(0)f^n(x)=f^n(x)\nonumber \end{align}$
and $f(5)=32$
for $x=1$ it can be shown that $f(n)=2^n$ with only whole numbers included into a domain of the function.

By using the next proof
$f(2x-x)f(x)=f(2x)=f(x)f(x)$
$f(x-(-x))f(-x)=f(x)=f(2x)f(-x)\implies f(2x)=f(x)/f(-x)$
$f(x)f(x)=f(x)/f(-x)$
$f(x)f(-x)=1$
all integers can be included into the domain of the function.

 

[D H]

@fresh_42 determined that $f(x)=a^{cx}$ and $a^c=2$. Using the formula $a^{cx}=(a^c)^x$, we have $f(x)=2^x$. Or am I missing something?

The expression $a^{cx}=(a^c)^x$ is valid if $a$ is a positive real number and $c$ is $c$ any real number. The expression is not necessarily valid for negative values or complex values of $a$ or complex values of $c$. However, this is a "Baby Rudin" level class. So it's probably a real analysis class, where the multiple vagaries regarding the complex exponential function simply don't come into play.

 

[D H]

$f(1-1)f(1)=f(1)$
$f(0)f(1)=f(1)$
$f(0)=1$

Note that one could also plug in $x=y=0$, yielding $f(0)^2 = f(0)$. There are two solutions, $f(0)=0$ and $f(0)=1$. From $f(0)f(x) = f(x)$ and with $f(0)=0$, one can deduce that $f(x) \equiv 0\, \forall\, x$. You can however rule that other solution out due to the fact that $f(5) = 32$.

You may need to add the requirement that $f(x)$ is continuous for all real $x$ to prove the already-found function is unique. Obviously, $f(x) = 2^x$ satisfies $f(x-y)f(y) = f(x)$ for all real $x$ and $y$.

 

[pasmith]

You can use $f(x)=a^x$ with $f(5)=a^5=32$ and find $a=2.$ This is mainly by inspection.

We do not have enough information to uniquely determine $f(x)$. All we know is that $f(x)=a^{c\cdot x}$ since these are all functions that satisfy the functional equation $(*).$ Hence, $f(5)=32=2^5=\left(a^c\right)^5 \Longrightarrow 2=a^c$ and all these functions are solutions. $(a,c)=(2,1)$ is only the easiest one.

Now, how do we know $(*)$? Honestly, I am not sure. I would set $f(x)=\sum_{k=0}^\infty a_k x^k$ and try to figure out the coefficients $a_k.$ However, this requires that $f(x)$ can be written as a power series and I'm not sure whether this can be deduced only by the functional equation. I guess so, but I don't know the answer without trying.

From $f(x + y) = f(x)f(y)$ we know that $f(nx) = f(x)^n$ for every positive integer $n$. That is sufficient to solve the question, but setting $x = 1$ we find $f(n) = f(1)^n$. Also for $n \neq 0$, $$f(1) = f(nn^{-1}) = f(n^{-1})^n \Rightarrow f(n^{-1}) = f(1)^{1/n}$$ and for positive integer $m$ we have $$f(n/m) = f(1)^{n/m}.$$ so that $f(x) = f(1)^x$ for positive rational $x$. Values for negative $x$ then follow from $$f(|x|)f(-|x|) = f(|x| - |x|) = f(0) = 1.$$

It follows from the functional equation that $$\frac{f(x+h) - f(x)}{h} = f(x) \frac{f(h) - 1}{h} = f(x)\frac{f(h) - f(0)}{h}$$ so that if $f$ is differentiable at zero then it is differentiable everywhere, with $f'(x) = f(x)f'(0).$

 

[WWGD]

Well, from the functional equation, it's a homeomorphism between an additive ring and a multiplicative one.

 

[littlemathquark]

From $f(x + y) = f(x)f(y)$ we know that $f(nx) = f(x)^n$ for every positive integer $n$. That is sufficient to solve the question, but setting $x = 1$ we find $f(n) = f(1)^n$. Also for $n \neq 0$, $$f(1) = f(nn^{-1}) = f(n^{-1})^n \Rightarrow f(n^{-1}) = f(1)^{1/n}$$ and for positive integer $m$ we have $$f(n/m) = f(1)^{n/m}.$$ so that $f(x) = f(1)^x$ for positive rational $x$. Values for negative $x$ then follow from $$f(|x|)f(-|x|) = f(|x| - |x|) = f(0) = 1.$$

It follows from the functional equation that $$\frac{f(x+h) - f(x)}{h} = f(x) \frac{f(h) - 1}{h} = f(x)\frac{f(h) - f(0)}{h}$$ so that if $f$ is differentiable at zero then it is differentiable everywhere, with $f'(x) = f(x)f'(0).$

So if I integrate $f'(x) =f(x) f'(0)$ I can find f function. Thank you.

 

[littlemathquark]

From $f(x + y) = f(x)f(y)$ we know that $f(nx) = f(x)^n$ for every positive integer $n$. That is sufficient to solve the question, but setting $x = 1$ we find $f(n) = f(1)^n$. Also for $n \neq 0$, $$f(1) = f(nn^{-1}) = f(n^{-1})^n \Rightarrow f(n^{-1}) = f(1)^{1/n}$$ and for positive integer $m$ we have $$f(n/m) = f(1)^{n/m}.$$ so that $f(x) = f(1)^x$ for positive rational $x$. Values for negative $x$ then follow from $$f(|x|)f(-|x|) = f(|x| - |x|) = f(0) = 1.$$

It follows from the functional equation that $$\frac{f(x+h) - f(x)}{h} = f(x) \frac{f(h) - 1}{h} = f(x)\frac{f(h) - f(0)}{h}$$ so that if $f$ is differentiable at zero then it is differentiable everywhere, with $f'(x) = f(x)f'(0).$

I know that if $f$ continuous one point, it's continuous everywhere. But $f$ is differentiable at zero why it is differentiable everywhere?

 

[fresh_42]

I know that if $f$ continuous one point, it's continuous everywhere. But $f$ is differentiable at zero why it is differentiable everywhere?

\begin{align*}
\lim_{h \to 0}\dfrac{f(a+h)-f(a)}{h}&=f(a)\lim_{h \to 0} \dfrac{f(h)-1}{h}\\
&=f(a)\lim_{h \to 0}\dfrac{f(0+h)-f(0)}{h}\\
&=f(a)\cdot f'(0)
\end{align*}