Another Functional Equation #2

Algebra Level 5

$\large{f(x+y) = 3^yf(x) + 2^xf(y)}$

Let $f(x)$ be a function where $f: \mathbb R \to \mathbb R$ for which $f(1) = 1$ such that for all real $x,y$ , the functional equation above satisfies.

Then find the value of $f(20) - f(15)$ .

Note : To avoid tediousness, first generalize $f(x)$ , then use a calculator to find the value asked.

The answer is 3471419686.

2 solutions

คลุง แจ็ค
Aug 15, 2015

Consider we substitute y to x and x to y.

$\displaystyle f(y+x)=f(x+y)=3^{x}f(y)+2^{y}f(x)=3^{y}f(x)+2^{x}f(y)$

$\displaystyle \left ( 3^{x}-2^{x} \right )f(y)=\left ( 3^{y}-2^{y} \right )f(x)$

$\displaystyle \left ( \dfrac{3^{x}-2^{x}}{f(x)} \right )=\left ( \dfrac{3^{y}-2^{y}}{f(y)} \right )$

This relation has to be true for all x,y and since x,y are independent variables, so, these ratio have to be constant.

$\displaystyle\left ( \dfrac{3^{x}-2^{x}}{f(x)} \right )=\dfrac{1}{C}$

$\displaystyle f(x)=C(3^{x}-2^{x})$ and from $f(1)=1$ then $C=1$

$\displaystyle f(x)=(3^{x}-2^{x})$

$\displaystyle f(20)-f(15)=3^{20}-2^{20}-3^{15}+2^{15}=\boxed{3471419686}$

Since we require values at only integral points, a much simpler approach would be to substitute $x=n, y=1$ . This gives the following recursion $f(n+1)=3f(n)+2^n$ Which is simpler to solve or just compute.

Abhishek Sinha - 5 years, 9 months ago

Chew-Seong Cheong
Aug 20, 2015

It is given that: $\begin{cases} f(x+y) = 3^y f(x) + 2^x f(y) \\ f(1) = 1 \end{cases}$

$\Rightarrow \begin{cases} f(n+1) = 3f(n) + 2^nf(1) = 3f(n) + 2^n \\ f(1+n) = 3^n f(1) + 2f(n) = 3^n + 2f(n) \end{cases}$

$\begin{aligned} \Rightarrow 3f(n) + 2^n & = 3^n + 2f(n) \\ f(n) & = 3^n - 2^n \\ \Rightarrow f(20) - f(15) & = 3^{20} - 2^{20} - 3^{15} + 2^{15} \\ & = \boxed{3471419686} \end{aligned}$

Cool Solution + Nice approach an upvote given

Department 8 - 5 years, 9 months ago

Another Functional Equation #2

The answer is 3471419686.

2 solutions

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