Todai Entrance Exam: Math 2018 – Problem 2

Problem link

(1) Base case: Show that f_{1}(x) = c_{1}x^{a_{1}} holds true. We have: f_{1}(x) = c_{1}x^{a_{1}} = cx^{0} = c.

Induction step: Show that f_{n}(x) = c_{n}x^{a_{n}} holds true leads to the fact that f_{n+1}(x) = c_{n+1}x^{a_{n+1}} holds true.

    \[ f_{n+1}(x) = p\int_{0}^{x}(f_{n}(t))^{1/q}dt = p\int_{0}^{x}(c_{n}t^{a_{n}})^{1/q}dt = p (c_{n})^{1/q}\frac{t^{q^{-1}a + 1}}{q^{-1}a + 1} \Big |^{x}_{0} \]

    \[ f_{n+1}(x) = \frac{p(c_{n})^{1/q}}{q^{-1}a+ 1}x^{q^{-1}a + 1} = c_{n+1}x^{a_{n+1}} \]

So the statement follows.

(2) This problem is interestingly tricky.

First we find the critical points of g_{n}(x):

    \[ \frac{\mathrm{d} g_{n}(x)}{\mathrm{d} x} = a_{n}x^{a_{n} - 1} - px^{p-1} = 0 \Leftrightarrow x_{c} = \left ( \frac{p}{a_{n}} \right )^{\frac{1}{a_{n}-p}} \]

If p > a_{n} then g_{n}(x) < 0, \forall x \in (0, 1) (due to the fact that x^{p} > x^{a_{n}}), so x_{n} \in \{ 0, 1 \} as g_{n}(x_{n}) = 0.

If p = a_{n} then g_{n}(x) = 0, \forall x \in [0, 1].

If p < a_{n} then the found critical point x_{c} = \left ( \frac{p}{a_{n}} \right )^{\frac{1}{a_{n}-p}} < 1. We have g_{n}(0) = 0 and g_{n}(1) = 0, so if g_{n}(x_{c}) > 0 then x_{c} is the global maximum. It is easy to see that \forall x \in (0, 1), p < a_{n}, g_{n}(x) > 0 (due to the fact that x^{p} < x^{a_{n}}). So in this case, x_{n} = x_{c} = \left ( \frac{p}{a_{n}} \right )^{\frac{1}{a_{n}-p}}.

To summarize:

    \[ \begin{Bmatrix} x_{n} \in { 0, 1 } & \text{if} & p > a_{n}\\  x_{n} \in [ 0, 1 ] & \text{if} & p = a_{n}\\  x_{n} = \left ( \frac{p}{a_{n}} \right )^{\frac{1}{a_{n}-p}} & \text{if} & p < a_{n}  \end{matrix} \]

(3) First we prove that a_{n} = \sum_{i = 0}^{n-2}\frac{1}{q^{i}} (the sum notation here is a little bit tricky for n = 1, reference):

Base case: a_{1} = \sum_{i = 0}^{-1}\frac{1}{q^{i}} = 0.

Induction step:

    \[ a_{n+1} = q^{-1}a_{n} + 1 = q^{-1}\sum_{i = 0}^{n-2}\frac{1}{q^{i}} + 1 = \sum_{i = 1}^{n-1}\frac{1}{q^{i}} + \frac{1}{q^{0}} = \sum_{i = 0}^{n-1}\frac{1}{q^{i}} \]

So the statement follows.

Second we prove that \lim_{n \rightarrow \infty} g_{n}(x) = 0.

    \[ \begin{matrix} \lim_{n \rightarrow \infty} g_{n}(x) = x^{\frac{1}{1 - \frac{1}{q}}} - x^{p}  & \text{as} & \lim_{n \rightarrow \infty}a_{n} = \frac{1}{1 - \frac{1}{q}}\\   \lim_{n \rightarrow \infty} g_{n}(x) = 0 & \text{as} & \frac{1}{p} + \frac{1}{q} = 1 \Leftrightarrow p = \frac{1}{1-\frac{1}{q}} \end{matrix} \]

So the statement follows.

Here I don’t understand why the problem only requires x \in [0, 1] but not for any x?

(4) We have:

    \[ \frac{d_{n+1}}{d_{n}} = (c_{n+1})^{q^{n+1}}\frac{1}{(c_{n})^{q^{n}}} = \left ( \frac{p(c_{n})^{1/q}}{a_{n+1}} \right )^{q^{n+1}}\frac{1}{(c_{n})^{q^{n}}} = \frac{(p)^{q^{n+1}}}{(a_{n+1})^{q^{n+1}}} \]

It is shown from (3) that \lim_{n \rightarrow \infty}a_{n} = \frac{1}{1 - \frac{1}{q}} = p, so:

    \[ \lim_{n \rightarrow \infty}\frac{d_{n+1}}{d_{n}} = 1 \]

I don’t know why I didn’t use the fact given by the examination though… (Something wrong?)


Thanks to my labmate, maybe I was wrong. It is not so trivial to prove that a_{n} = \frac{q^{2-n} - q}{1 - q} (this comes from the fact that a_{n+1} + \frac{q}{1-q} = \frac{1}{q}\left ( a_{n} + \frac{q}{1-q} \right )), p = \frac{q}{q - 1}.

    \[ c_{n + 1} = \frac{p(c_{n})^{1/q}}{a_{n+1}} = \frac{\frac{q}{q - 1}(c_{n})^{1/q}}{\frac{q^{2-n} - q}{1 - q}} = \frac{(c_{n})^{1/q}}{1-q^{1-n}} \]

Therefore:

    \[ \frac{d_{n+1}}{d_{n}} = (c_{n+1})^{q^{n+1}}\frac{1}{(c_{n})^{q^{n}}} =   \frac{1}{(1-q^{1-n})^{q^{n+1}}} =  \left( \frac{1}{(1-q^{1-n})^{q^{1-n}}} \right )^{q^{2}} \]

So:

    \[ \lim_{n \rightarrow \infty} \frac{d_{n+1}}{d_{n}} = \lim_{n \rightarrow \infty} \left( \frac{1}{(1-q^{1-n})^{q^{1-n}}} \right )^{q^{2}} = \frac{1}{e^{q^{2}}}\]

(5) We proved \lim_{n \rightarrow \infty}a_{n} = \frac{1}{1 - \frac{1}{q}} = p in (3). So:

    \[ \lim_{n \rightarrow \infty} c_{n} = \frac{p\lim_{n \rightarrow \infty} (c_{n-1})^{1/q}}{\lim_{n \rightarrow \infty} (a_{n})} = \lim_{n \rightarrow \infty} (c_{n-1})^{1/q} =  \left ( \frac{p\lim_{n \rightarrow \infty} (c_{n-2})^{1/q}}{\lim_{n \rightarrow \infty} (a_{n-1})}  \right )^{1/q} = \lim_{n \rightarrow \infty} (c_{n-2})^{(1/q)^{2}} \]

    \[  \lim_{n \rightarrow \infty} c_{n} = \text{...} = \left ( \frac{p\lim_{n \rightarrow \infty} (c_{n-i})^{1/q}}{\lim_{n \rightarrow \infty} (a_{n-i})}  \right )^{(1/q)^{i}} \]

When n approaches infinity, a_{n} approaches p, so does a_{n-i} given small enough i (you can factor a_{n-i} into a sum and argue that the subsum \frac{1}{q^{n}} + ... + \frac{1}{q^{n-i}} approaches 0 when n approaches infinity, so no much difference in value between a_{n} and a_{n-i}). So (p)^{\frac{1}{q^{i}} and (a_{n-i})^{\frac{1}{q^{i}} will cancel each other out when n approaches infinity and given small enough i.

In the case i becomes large enough that makes the difference between a_{n} and a_{n-i} noticeable, \frac{1}{q^{i}} also starts to approaches 0 and makes both (p)^{\frac{1}{q^{i}} and (a_{n-i})^{\frac{1}{q^{i}} approaches 1. So (p)^{\frac{1}{q^{i}} and (a_{n-i})^{\frac{1}{q^{i}} will cancel each other out given big enough i.

Therefore:

    \[ \forall i \in [1, n-1], \lim_{n \rightarrow \infty} c_{n} = \left ( \frac{p\lim_{n \rightarrow \infty} (c_{n-i})^{1/q}}{\lim_{n \rightarrow \infty} (a_{n-i})}  \right )^{(1/q)^{i}} = \lim_{n \rightarrow \infty} c^{(1/q)^{n-1}} = 1 \]

As hinted by (6), I’m pretty sure \lim_{n \rightarrow \infty} c_{n} = 1 though.

(6) We proved \lim_{n \rightarrow \infty}a_{n} = \frac{1}{1 - \frac{1}{q}} = p in (3) and we proved \lim_{n \rightarrow \infty} c_{n} = 1 in (5). So:

    \[ \lim_{n \rightarrow \infty} f_{n}(x) = \lim_{n \rightarrow \infty} c_{n}x^{a_{n}} = x^{p} \]

So the statement follows.


Facebooktwittergoogle_plusredditpinterestlinkedinmail

1 comment on “Todai Entrance Exam: Math 2018 – Problem 2”

Leave a Reply

Your email address will not be published. Required fields are marked *