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

AI Thread Summary
The discussion revolves around finding the function f(x) given the functional equation f(x-y)f(y) = f(x) and the condition f(5) = 32. It is established that f(x) can be expressed in the form f(x) = a^(c*x), leading to the conclusion that f(1) = 2 and f(7) = 128. The participants explore the implications of the functional equation, noting that it resembles Cauchy's exponential equation, which suggests that f(x) could be an exponential function. Ultimately, they conclude that f(x) = 2^x is a valid solution, satisfying the given conditions and the functional equation.
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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?
 
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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?
 
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Yes, Cauchy exponansiyel equation, I mean ##f(x) =a^x##. But how can I find this?
 
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.
 
Maybe just remembering the Exponential satisfies ##a ^{x+y}=a^x *a^y##.
 
@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?
 
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##
 
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## 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.
 
Demystifier said:
@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.
 
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  • #10
Gavran said:
## 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##.
 
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  • #11
fresh_42 said:
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, <br /> 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 <br /> \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&#039;(x) = f(x)f&#039;(0).
 
  • #12
Well, from the functional equation, it's a homeomorphism between an additive ring and a multiplicative one.
 
  • #13
pasmith said:
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, <br /> 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 <br /> \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&#039;(x) = f(x)f&#039;(0).
So if I integrate ##f'(x) =f(x) f'(0)## I can find f function. Thank you.
 
  • #14
pasmith said:
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, <br /> 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 <br /> \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&#039;(x) = f(x)f&#039;(0).
I know that if ##f## continuous one point, it's continuous everywhere. But ##f## is differentiable at zero why it is differentiable everywhere?
 
  • #15
littlemathquark said:
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*}
 
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