Odd function

Algebra Level 3

An odd function $f:\mathbb{R} \to \mathbb{R}$ satisfies $f(x)=f(x+2)$ .

Find the value of $f(1)+f(2)+f(3)+ ... + f(2017)$ .

The answer is 0.

1 solution

Lee Dongheng
Aug 10, 2017

Since $f$ is an odd function, $f(-x)=-f(x)$

Let $x=0$ , we have $f(0)=0$

$\because f(x)=f(x+2)$

$\therefore f(0)=f(2)=f(4)= ... =f(2016)=0$

Let $x=1$ , we have $f(-1)=-f(1)$

From $f(x)=f(x+2)$ , $f(-1)=f(-1+2)=f(1)$

$\therefore f(1)=0, f(1)=f(3)=f(5)= ... =f(2017)=0$

$\therefore f(1)+f(2)+f(3)+ ... f(2017)=\boxed{0}$

Is f(x)=0 an odd function though?

JD Money - 3 years, 10 months ago

$f(x)=0$ is both an even and odd function. Also in this problem $f(x)=0$ is not the only function that respects the constraints, take for example $f(x)=\sin(\pi x)$ . We can just deduce that the function outputs $0$ when the input is an integer number and that's all we need for this problem

Marco Brezzi - 3 years, 10 months ago

Odd function

The answer is 0.

1 solution

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