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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.

Note by Kartik Sharma
1 year, 4 months ago

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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. · 1 year, 4 months ago

There can be more generalizations to this as well. For example, considering $$a,b$$ to have non-integer values or changing the limits of the problem and so on. Take these up as challenges now!

Check out @Aditya Kumar ! · 1 year, 4 months ago

Do we assume $$a$$ and $$b$$ to be positive integers? · 1 year, 4 months ago

Oh yeah I forgot to mention that. I apologize for that. · 1 year, 4 months ago

In $$f_a^b(x)$$, is $$a\le b$$? If no, how can we interpret $$3$$ in base $$2$$? · 1 year, 4 months ago

Okay, you may assume that. · 1 year, 4 months ago

Can u provide the solution?? I couldn't get any. · 1 year, 4 months ago

It's one of my favorites. I will surely post solutions for all the unanswered challenges as soon as the contest ends. · 1 year, 4 months ago