Calculus Challenge 5!

Define ${f}_{a}^{b}(x)$ as a function which converts $x$ into base $a$ and then interprets it as a number in base $b$ .

For example,

${f}_{2}^{10}(0.5)$ will mean first changing $0.5$ to base $2$ i.e. $0.1$ and then interpreting $0.1$ as a base $10$ number. That's it!

So, find a general formula for

$\displaystyle \int_{0}^{1}{{f}_{b}^{a}(x) \ dx}$

I will not be giving the form for this problem as then the difficulty will be halved so try yourself and then if required I will post the solution as the contest ends.

Easy Math Editor

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Comments

So, now I guess I must post the solution.

First of all what does $f_b^a (x)$ means? For $x<1$ , $\displaystyle {f}_{b}(x) = \sum_{k=1}^{\infty}{\left \lfloor b^k x \right \rfloor \pmod{b}}$ and

$\displaystyle f_b^a(x) = \sum_{k=1}^{\infty}{\frac{\left \lfloor b^k x \right \rfloor \pmod{2}}{a^k}}$

$\displaystyle \int_{0}^{1}{\sum_{k=0}^{\infty}{\frac{\left \lfloor b^k x \right \rfloor \pmod{b}}{a^k}} \ dx}$

$\displaystyle \sum_{k=1}^{\infty}{\frac{1}{a^k} \int_{0}^{1}{\left \lfloor b^k x \right \rfloor \pmod{b} \ dx}}$

$\displaystyle \text{Lemma} : \int_{0}^{1}{\left \lfloor b^k x \right \rfloor \pmod{b} \ dx} = \frac{b-1}{2}$

$\displaystyle \text{Proof} :$ First substitute $u = b^k x$ and then draw a graph for $\displaystyle u \pmod{b}$ from $u = 0$ to $u = b^k$ . You will find that it is periodic with period $b$ and in each period, there are $1 \times k$ rectangles where $\displaystyle k = [1,2,\cdots, b-1]$ .

Hence, the area under the graph will just be the area of these rectangles i.e. in one period the area is $\displaystyle 1 \times 1 + 1 \times 2 + 1 \times 3 + \cdots + 1 \times (b-1) = \frac{b(b-1)}{2}$ and there are $\displaystyle {b}^{k-1}$ "oscillations". As a result our total area becomes $\displaystyle {b}^{k-1} \frac{b(b-1)}{2} = \frac{b^k(b-1)}{2}$ . So, our integral which is just $\displaystyle \frac{1}{b^k}$ of this total area becomes $\displaystyle \frac{b-1}{2}$ .

Now, we get

$\displaystyle \sum_{k=1}^{\infty}{\frac{1}{a^k} \frac{b-1}{2}}$

$\large \displaystyle = \boxed{\frac{b-1}{2(a-1)}}$

This is original.