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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
2 years, 1 month 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.

Kartik Sharma - 2 years, 1 month ago

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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 !

Kartik Sharma - 2 years, 1 month ago

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Do we assume \(a\) and \(b\) to be positive integers?

Samuel Jones - 2 years, 1 month ago

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

Kartik Sharma - 2 years, 1 month ago

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In \(f_a^b(x)\), is \(a\le b\)? If no, how can we interpret \(3\) in base \(2\)?

Micah Wood - 2 years, 1 month ago

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@Micah Wood Okay, you may assume that.

Kartik Sharma - 2 years, 1 month ago

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

Aditya Kumar - 2 years, 1 month 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 - 2 years, 1 month ago

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Ok. I'll give it a try again.

Aditya Kumar - 2 years, 1 month ago

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