Calculus Challenge 5!

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

For example,

f210(0.5){f}_{2}^{10}(0.5) will mean first changing 0.50.5 to base 22 i.e. 0.10.1 and then interpreting 0.10.1 as a base 1010 number. That's it!

So, find a general formula for

01fba(x) dx\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
3 years, 11 months ago

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So, now I guess I must post the solution.

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

fba(x)=k=1bkx(mod2)ak\displaystyle f_b^a(x) = \sum_{k=1}^{\infty}{\frac{\left \lfloor b^k x \right \rfloor \pmod{2}}{a^k}}

01k=0bkx(modb)ak dx\displaystyle \int_{0}^{1}{\sum_{k=0}^{\infty}{\frac{\left \lfloor b^k x \right \rfloor \pmod{b}}{a^k}} \ dx}

k=11ak01bkx(modb) dx\displaystyle \sum_{k=1}^{\infty}{\frac{1}{a^k} \int_{0}^{1}{\left \lfloor b^k x \right \rfloor \pmod{b} \ dx}}

Lemma:01bkx(modb) dx=b12\displaystyle \text{Lemma} : \int_{0}^{1}{\left \lfloor b^k x \right \rfloor \pmod{b} \ dx} = \frac{b-1}{2}

Proof:\displaystyle \text{Proof} : First substitute u=bkxu = b^k x and then draw a graph for u(modb)\displaystyle u \pmod{b} from u=0u = 0 to u=bku = b^k. You will find that it is periodic with period bb and in each period, there are 1×k1 \times k rectangles where k=[1,2,,b1]\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 1×1+1×2+1×3++1×(b1)=b(b1)2\displaystyle 1 \times 1 + 1 \times 2 + 1 \times 3 + \cdots + 1 \times (b-1) = \frac{b(b-1)}{2} and there are bk1\displaystyle {b}^{k-1} "oscillations". As a result our total area becomes bk1b(b1)2=bk(b1)2\displaystyle {b}^{k-1} \frac{b(b-1)}{2} = \frac{b^k(b-1)}{2}. So, our integral which is just 1bk\displaystyle \frac{1}{b^k} of this total area becomes b12\displaystyle \frac{b-1}{2}.

Now, we get

k=11akb12\displaystyle \sum_{k=1}^{\infty}{\frac{1}{a^k} \frac{b-1}{2}}

=b12(a1)\large \displaystyle = \boxed{\frac{b-1}{2(a-1)}}

This is original.

Kartik Sharma - 3 years, 11 months ago

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There can be more generalizations to this as well. For example, considering a,ba,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 !

Kartik Sharma - 3 years, 11 months ago

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Can u provide the solution?? I couldn't get any.

Aditya Kumar - 3 years, 11 months ago

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It's one of my favorites. I will surely post solutions for all the unanswered challenges as soon as the contest ends.

Kartik Sharma - 3 years, 11 months ago

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Ok. I'll give it a try again.

Aditya Kumar - 3 years, 11 months ago

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Do we assume aa and bb to be positive integers?

Samuel Jones - 3 years, 11 months ago

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Oh yeah I forgot to mention that. I apologize for that.

Kartik Sharma - 3 years, 11 months ago

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In fab(x)f_a^b(x), is aba\le b? If no, how can we interpret 33 in base 22?

Micah Wood - 3 years, 11 months ago

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@Micah Wood Okay, you may assume that.

Kartik Sharma - 3 years, 11 months ago

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