# Thread: please help with difficult functions

1. ## please help with difficult functions

Pure Mathematics 1 by Backhouse
Ex:2f q13
The real function f, defined for all $\displaystyle x \epsilon \mathbb{R}$, is said to be multiplicative if, for all $\displaystyle y \epsilon \mathbb{R}$, $\displaystyle x \epsilon \mathbb{R}$,

f(xy) = f(x)f(y)

Q: Prove that if f is multiplicative function then
a) either f(0) =0 or f(x)=1
b) either f(1) = 1 or f(x) = 0
c) $\displaystyle f(x)^{n}$ = $\displaystyle \left \{ f(x) \right \}^{n}$ for all positive integers n.

Give example of a non- constant multiplicative function

My attempt:
I have tried to apply what I have learn from the chapter of functions from this book but there is no bit on multiplicative functions, I have only learnt about odd & even functions. I tried but failed please help

let f(x) = even functions thus f(a) = f(-a)
f(y) = odd functions thus - f(b) = f(-b)

f(ab) = f(a)f(b)
= f(-a) (-f(b)) = -f(-a)f(-b) = -f(a)f(-b)

hence i am stuck please help. I have tried to look at youtube and search for multiplicative functions but it seems to be advanced number theory. I dont have knowledge in that area thus please can an expert give an answer that does not have advance mathematical symbols as my maths level is at high school

Thank you very much for your help in advance

2. ## Re: please help with difficult functions

Use the information you are given.
\begin{align}&\text{Given} & f(xy) &= f(x)f(y) \\
&\text{set $y=0$ so that} & f(0) &= f(x)f(0) \\
&& f(0) - f(0)f(x) &= 0 \\
&& f(0)\big(1-f(x)\big) &= 0
\end{align}
Follow a similar idea for the second part.

For the third part, I think the question should say $$f(x^n) = f^n(x)$$ or $$f(x^n) = \{f(x)\}^n$$ which mean the same thing. Your proof will probably be via induction.

3. ## Re: please help with difficult functions Originally Posted by Archie Use the information you are given.
\begin{align}&\text{Given} & f(xy) &= f(x)f(y) \\
&\text{set $y=0$ so that} & f(0) &= f(x)f(0) \\
&& f(0) - f(0)f(x) &= 0 \\
&& f(0)\big(1-f(x)\big) &= 0
\end{align}
Follow a similar idea for the second part.

For the third part, I think the question should say $$f(x^n) = f^n(x)$$ or $$f(x^n) = \{f(x)\}^n$$ which mean the same thing. Your proof will probably be via induction.

thank you

so i try to use proof via induction

prove

$\displaystyle f(x^n)=\left \{ f(x) \right \}^n$ for all positive integers n

first show that the step hold for n= 1

$\displaystyle f(x^1)=\left \{ f(x) \right \}^1$
$\displaystyle \therefore$
$\displaystyle f(x)=\left \{ f(x) \right \}$

$\displaystyle f(x) = f(x)$

Suppose it holds for n = k

$\displaystyle f(x^n)=\left \{ f(x) \right \}^n$

$\displaystyle f(x^k)=\left \{ f(x) \right \}^k$
$\displaystyle f(x^1 . x^{k-1})=\left \{ f(x) \right \}^{1} . \left \{ f(x) \right \}^{k-1}$
$\displaystyle f(x^1 . x^{k-1})= f(x)^1 . f(x)^{k-1}$
$\displaystyle f(x^1) . f(x^{k-1})= f(x)^1 . f(x)^{k-1}$
$\displaystyle f(x) . f(x^{k-1})= f(x) . f(x)^{k-1}$

$\displaystyle ( f(x) . f(x^{k-1})) - (f(x) . f(x)^{k-1})=0$

$\displaystyle f(x)[( f(x^{k-1})) - ( f(x)^{k-1})]=0$

$\displaystyle f(x)=0$ or $\displaystyle [( f(x^{k-1})) - ( f(x)^{k-1})]=0$

$\displaystyle f(x) = 0$ is invalid since n represents all positive integers

$\displaystyle f(x^{k-1}) = f(x)^{k-1}$
since $\displaystyle f(x^{k-1}) = f(x)^{k-1}$

thus $\displaystyle f(x^{k}) = f(x)^{k}$

then let n = k+1

$\displaystyle f(x^{k+1})=\left \{ f(x) \right \}^{k+1}$

$\displaystyle f(x^{k+1})=\left \{ f(x) \right \}^{k+1}$
$\displaystyle f(x^{k+1})= f(x)^{k+1}$
$\displaystyle f(x^{k}).f(x^{1}) = f(x)^{k}.f(x)^{1}$

$\displaystyle [f(x^{k}).f(x^{1})]- [f(x)^{k}.f(x)^{1}] = 0$
$\displaystyle [f(x^{k}).f(x)]- [f(x)^{k}.f(x)] = 0$
$\displaystyle f(x)[f(x^{k})-f(x)^{k}] = 0$
$\displaystyle f(x) =0$ or $\displaystyle [f(x^{k})-f(x)^{k}]=0$

$\displaystyle f(x) = 0$ is invalid since n represents all positive integers

$\displaystyle [f(x^{k})-f(x)^{k}]=0$
$\displaystyle f(x^{k})=f(x)^{k}$

since $\displaystyle f(x^{k}) = f(x)^{k}$

thus $\displaystyle f(x^{k+1}) = f(x)^{k+1}$

So we have shown that if it is true for some n=k it is also true for n=k+1. We have shown that it is true for n=1, therefore by the principle of mathematical induction it is true for all the positive integers n.

Please can you tell me if I am right

and the last point of the question is to give an example of a non-constant multiplicative function but i tried to read wikipedia but it is a bit difficult to understand please can you share some light?

thank you

4. ## Re: please help with difficult functions Originally Posted by bigmansouf so i try to use proof via induction prove
$\displaystyle f(x^n)=\left \{ f(x) \right \}^n$ for all positive integers n
first show that the step hold for n= 1
$\displaystyle f(x^1)=\left \{ f(x) \right \}^1$
$\displaystyle \therefore$
$\displaystyle f(x)=\left \{ f(x) \right \}$
$\displaystyle f(x) = f(x)$

Suppose it holds for n = k
$\displaystyle f(x^n)=\left \{ f(x) \right \}^n$
Why do you muck thing up by not just using simple equivalences.
You have suppose that $\displaystyle f(x^N)=\left \{ f(x) \right \}^N$ so use it.
\displaystyle \begin{align*}f(x^{N+1}&=f(x^N\cdot x) \\&=f(x^N)\cdot f(x)\text{ why?}\\&=[f(x)]^N\cdot[f(x)]\text{ HOW?}\\&=[f(x)]^{N+1}\text{ HOW & WHY?} \end{align*}
How are we done?