Mathematics Challenge
Oxford University Calculus Challenge
2023 Oxford Mathematics Admission Test
![](https://cdn-images-1.readmedium.com/v2/resize:fit:800/1*x458kZUc4W9ix97qyBWDTg@2x.jpeg)
We looked at a geometry problem on the Oxford Mathematics Admission Test yesterday. If you think you have a chance to get into Oxford University, take a look at today’s problem.
![](https://cdn-images-1.readmedium.com/v2/resize:fit:800/1*i7QJ67XKOErJvENRIzcvng.png)
Here’s a hint: while solving the challenge, you should be able to find a geometric sum …
I recommend you to pause the article, grab your pen and paper, and give this a go. When you are ready, keep reading for the solution! ✒️
Solution
The tricky part about this question is the nature of the floor function applied to a logarithm.
![](https://cdn-images-1.readmedium.com/v2/resize:fit:800/1*1D6gSGDamS1XhoOqKTY8pQ.png)
The floor function, as stated, takes the largest whole number that is less than or equal to x. For example ⌊3.6⌋ = 3 and ⌊-1.5⌋ = -2.
What happens when we apply the floor function to a logarithm of base 2? In fact, we are interested in its behavior for 0 < x < 2.
![](https://cdn-images-1.readmedium.com/v2/resize:fit:800/1*uHfCOjM9tI8AjEzEp_xjyA.png)
I have picked out some values of x to plug into the function. When x = 2, the resulting output is 1 and when x = 1, the result output is 0. What about values in between 2 and 1? Say 3/2?
In fact this is equivalent to asking
![](https://cdn-images-1.readmedium.com/v2/resize:fit:800/1*ZkRwHJf2grWlyalqeQk3gw.png)
Knowledge of logarithms tells us x must be less than 1 and greater than 0.
This means ⌊log2(3/2)⌋ must be 0. And this holds true for all values 1 ≤ x < 2.
What about when x = 1, x = 1/2, x = 1/4 and so on?
![](https://cdn-images-1.readmedium.com/v2/resize:fit:800/1*9_h5zEQHjOWfBXxtfvKN2w@2x.jpeg)
We partition the whole interval [0, 2] into infinitely many sections where the length of each section is from 2^n to 2^(n+1). The biggest section is when n = 0 such that we have 2⁰ = 1 and 2¹ = 2.
If we pay close attention now, we see that for the interval starting with 2^n, the output from the floor function of logarithm with base 2 results in n.
![](https://cdn-images-1.readmedium.com/v2/resize:fit:800/1*px0Evg3N8hZA0ZCASAFPRg@2x.jpeg)
I have given an example above. The integral considers the function from 1/4 to 1/2, or 2^(-2) to 2^(-1).
Here’s a task for you. Try and find the integral when x ranges from 1/8 to 1/4. What about from 1/16 to 1/8? And so on?
![](https://cdn-images-1.readmedium.com/v2/resize:fit:800/1*Wj0IezSbIz36mNLBf2Gg1g.png)
Recall the aim of the question, which is to evaluate the integral of the function from 0 to 2.
Now the integral is equal to the sum of infinitely many integrals with 2^n to 2^(n+1) each for each segment, or geometrically, the total area of infinitely many rectangles.
Let’s write out the whole sum.
![](https://cdn-images-1.readmedium.com/v2/resize:fit:800/1*xb5L56I7IUdIOLMVNf98hw@2x.jpeg)
It goes all the way to infinity, where the bounds become arbitrary close to 0 and 1/2^n.
![](https://cdn-images-1.readmedium.com/v2/resize:fit:800/1*AG7xT7-Chkn6o0pq8LM8lQ@2x.jpeg)
Now from the set up above, we should see this pattern for every integral, and in fact, upon computing the first few, we see the following geometric series!
![](https://cdn-images-1.readmedium.com/v2/resize:fit:800/1*QErL3-eYjY0wFKTo0gtV2g@2x.jpeg)
Convince yourself that the highlighted terms are equivalent.
![](https://cdn-images-1.readmedium.com/v2/resize:fit:800/1*ACHVSKeDnmSSxOVKIU1CMg@2x.jpeg)
This pattern continues ad infinitum and so the common ratio is in fact 2/3.
Finally, the geometric sum is convergent.
![](https://cdn-images-1.readmedium.com/v2/resize:fit:800/1*9iiChO0Z3Jojo1HxUn6Y1g.png)
The answer is c.
Thank you for reading. Don’t forget to clap the article if you find it insightful.
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