We define a real and continuous function by $f(x+1)=f(x)$. I know that if $x\in \mathbb{Z}$, the function f is a constant function. I was wondering if $x \in \mathbb{R}$, the function is constant or not. It seems that it is obviously constant and the question may be naive. But I just want to be sure. Thanks.
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3$\begingroup$ Let $f(x) = \sin (2\pi x)$ then $f(x+1) = \sin(2\pi (x+1) ) =\sin (2\pi x + 2\pi) = \sin(2\pi x) = f(x)$. $\endgroup$– fleabloodCommented Sep 17, 2019 at 4:56
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1$\begingroup$ Are you familiar with the concept of a "periodic function"? That's a function where $f(x) = f(x+c)$ but that doesn't mean if $x< y < x+c$ that $f(y) = f(x)$. $\endgroup$– fleabloodCommented Sep 17, 2019 at 4:59
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$\begingroup$ okay. Thanks. I got it. $\endgroup$– tomriddle99Commented Sep 17, 2019 at 5:00
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3 Answers
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Hint: Take any periodic function with period 1.
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It's true if $f$ is a polynomial, which might be what you're thinking of. But there are many more interesting functions than polynomials out there!
answered Sep 17, 2019 at 5:00
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Induction implies that $$f(x) = f(x - \lfloor x \rfloor), $$ where $\lfloor x \rfloor$ denotes the integer part of $x$, so that $x - \lfloor x \rfloor \in [0, 1)$ is the fractional part of $x$. Thus, $f$ is determined by its restriction $f\vert_{[0, 1)}$ to $[0, 1)$.
Conversely, given any function $g : [0, 1) \to \Bbb R$, setting $$f(x) := g(x - \lfloor x \rfloor)$$ defines a function satisfying $f(x + 1) = f(x)$, and by construction this process is the inverse of the above restriction.
answered Sep 17, 2019 at 5:01