7
$\begingroup$

Let us consider Takagi-function defined by

$T(x) \colon\!= \sum_{n=0}^{\infty}s(2^nx)/2^n$,

where $s(x) \colon\!\!= \underset{n \in {\Bbb Z}}{\mathrm{min}} \,|x-n|$.

$T(x)$ has its period $1$, so I am trying to develop its Fourier expansion

$S(T)(x) \colon\!=\sum_{n= -\infty}^{\infty}\left(a_n\,sin(2 \pi n x) + b_n\,cos(2 \pi n x)\right)$.

Q: Where do we have $T(x) = S(T)(x)$?

For the case of Weierstrass function $W(x) \colon\!= \sum_{n=0}^{\infty}a^n cos(b^n \pi x)$ with $ab > 1 + 3\pi/2$, $ 0 < a < 1$ and $b$ an odd integer, it is already expanded in Fourier series with period $2$ by shape.

By Carlson, we know $T(x) = S(T)(x)$ except for sets of Lebesgue measure $0$. I would like to know whether $T(x) = S(T)(x)$ holds everywhere for Takagi-function as well.

$\endgroup$

1 Answer 1

6
$\begingroup$

Yes. The triangle wave $s(x):=\min_{k\in\mathbb{Z}}\big|x-k\big|$ has an absolutely convergent Fourier series

$$s(x)=\frac{1}{4}-\frac{2}{\pi^2}\sum_{k=0}^\infty\frac{1}{(2k+1)^2}\cos\big(2\pi (2k+1)x\big)\, ,\qquad x\in\mathbb{R}.$$

Therefore $$T(x):=\sum_{n=0}^\infty2^{-n}s(2^nx)= \frac{1}{4}\sum_{n=0}^\infty2^{-n}-\frac{2}{\pi^2}\sum_{n=0}^\infty\sum_{k=0}^\infty \frac{1}{2^n(2k+1)^2}\cos\big(2\pi 2^n(2k+1)x\big)\, .$$ By absolute convergence we can reorder the double sum. Since every positive integer $m\ge1$ writes uniquely as $m=2^n(2k+1)$ for nonnegative integers $n$ and $k$, we then obtain an absolutely convergent Fourier series for $T(x)$:

$$T(x)=\sum_{m=0}^\infty a_m\cos(2\pi m x)$$ with $a_0=1/2$ and for $m\ge 1$ $$a_m:=-\frac{2^{\nu(m)+1}}{\pi^2m^2}\, , $$ where $2^{\nu(m)}$ is the maximum power of $2$ that divides $m$.

$\endgroup$
1
  • $\begingroup$ (I added it to the wiki article on $T(x)$ ) $\endgroup$ Mar 11, 2014 at 17:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.