How can one show that$f(x^n)=nf(x)$

Question

How can one show that if $f(xy)=f(x)+f(y)$ holds for all for real $x$ and $y$ that

$$f(x^n)=nf(x).$$ How can i prove that $f(\frac{1}{x})=-f(x)$

To calculate f(1) do i need to pute x=1 ?

Do i need to use induction?

Answer 1

To find $f(1)$, take $x=y=1$. You get that $$f(1)=f(1)+f(1)$$ Subtract $f(1)$ from each side, and you're left with $$f(1) = 0$$

To show that $$f\left(\frac{1}{x}\right)=-f(x)$$ (for $x\neq 0$) take $y=\frac{1}{x}$ in the functional equation. You get $$0=f(1)=f\left(x\cdot\frac{1}{x}\right)=f(x)+f\left(\frac{1}{x}\right)$$ Subtract $f(x)$ from each side and you get $$f\left(\frac{1}{x}\right)=-f(x)$$

To show that $$f(x^n)=nf(x)$$ for all $n$, you would use induction. I'll leave you to fill in the details, but if $f(x^n) = nf(x)$, then what does the functional equation tell you the value of $f(x^{n+1}) = f(x\cdot x^n)$ is?

Once you have shown that $f(x^n)=nf(x)$ for natural values of $n$, you can extend it to negative values of $n$ as well, also using induction.

If $f(x^n)=nf(x)$, then $$f(x^{n-1})=f\left(\frac{1}{x}\cdot x^n\right) = f\left(\frac{1}{x}\right) + f(x^n)$$ You can then use the results from above to conclude that $f(x^{n-1})=(n-1)f(x)$, which gives you that $f(x^n)=nf(x)$ for all integers $n$ by induction.

You can even extend the result to rational values of $n$. Once you have that $f(x^n)=nf(x)$ for all integers $n$, you can consider the value of $$f\left(x^{\frac{m}{n}}\right)$$ for integers $m$ and $n$. We know that $$mf(x) = f(x^m) = f\left(\left(x^{\frac{m}{n}}\right)^n\right) = nf\left(x^{\frac{m}{n}}\right)$$ Now divide each side by $n$.

It is not possible to go any further than this without knowing something about whether $f$ is continuous or not.

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If $0$ is in the domain of $f$ then things become even simpler.

Take $y=0$. Then for any $x$ we have that $$ f(0) = f(x\cdot 0) = f(x) + f(0) $$ and so $f(x)=0$ for all $x$.

The statement that $f(x^n)=nf(x)$ then just becomes $0=0$.

Answer 2

We proceed by induction.

The base case is clearly true, since $f(x)=f(x^1)=1\cdot f(x)=f(x).$

Assuming the claim to be true for all integers $j \leq k$, we have: $$f\left(x^{k+1}\right)=f\left(x^{k}x\right)=f(x^k)+f(x)=k \cdot f(x)+f(x)=(k+1)\cdot f(x).$$

Hence, by the principle of mathematical induction, the claim is true for all integers $n \geq 1$.