The Christmas Equation

-
I posted the following in December 2021 on my first-year pure module's Piazza forum.

Hi everyone

I saw this a few years ago on QI, and found another version on the web at http://mathandmultimedia.com/2014/12/02/merry-christmas-equation/

If we are told that

\(\qquad\displaystyle y=\frac{\ln(\frac{x}m-sa)}{r^2}\)

multiplying by \(r^2\) gives

\(\qquad yr^2=\ln(\frac{x}m-sa)\,.\)

Then taking \(e\) to the power of both sides give

\(\qquad\displaystyle e^{yr^2}=\frac{x}m-sa\,,\)

and multiplying both sides by \(m\) gives

\(\qquad me^{yr^2}=x-msa\,,\)

which can be rewritten as

\(\qquad me^{rry}=x-mas\,.\)

(And a Happy New Year!)

Best wishes,

Dr Feinstein

Comments

Post a Comment

Popular posts from this blog

A rule of thumb for when to use the Ratio Test

-
The Ratio Test is a powerful test for both sequences and series of non-zero real numbers. It comes in various forms, but here is one commonly used version. Theorem (Ratio Test) Let $(x_n)_{n\in\N}$ be a sequence of non-zero real numbers. Suppose that \[\frac{|x_{n+1}|}{|x_n|}\to L \text{ as } n\to\infty\,,\] where $L$ is either a non-negative real number or $+\infty$. (a) If $L>1$, then $|x_n| \to +\infty$ as $n \to + \infty$, the sequence $(x_n)$ is divergent, and the series $\displaystyle \sum_{n=1}^\infty x_n$ diverges. (b) If $L \in [0,1)$, then the series $\displaystyle \sum_{n=1}^\infty x_n$ is absolutely convergent (and hence convergent), and the sequence $(x_n)$ converges to $0$. (c) If $L=1$, then the Ratio Test is inconclusive , and you need to use a different test . (The Ratio Test tells you nothing in this case about the convergence or otherwise of either the sequence or the series.) Notes Here (b) does not tell you anything about the value of the sum of the series ...
Read more

The Extreme Value Theorem: a bootstrap approach

-
At the very end of last week I was teaching the first-year mathematics students about the Extreme Value Theorem. Here is a reminder of what the theorem says. Theorem (Extreme Value Theorem) Let $a,b$ be real numbers with $a<b$, and let $f:[a,b]\to\R$ be continuous. Then there exist points $c$ and $d$ in $[a,b]$ such that, for all $x \in [a,b]$, we have $f(c)\leq f(x) \leq f(d)$. Here we have \[f(c)=\min\{f(x):x\in[a,b]\}=\inf\{f(x):x\in[a,b]\}\] and  \[f(d)=\max\{f(x):x\in[a,b]\}=\sup\{f(x):x\in[a,b]\}\,.\] Unfortunately I only had a few minutes left for this, ran out of time, and didn't finish the proof. As I am on a very tight schedule this year, I had to refer the students to the the typed notes for the rest of the proof. On the other hand, I did finish the first part of the proof, showing that a continuous real-valued function $f$ on a closed and bounded interval $[a,b]$ is always bounded above, i.e., the set $f([a,b])$ is bounded above. In this post I want to see how you ca...
Read more

An introduction to the Hahn-Banach extension theorem: Part VI

-
In this post we sketch a second proof of the Hahn-Banach Extension Theorem for continuous linear functionals on real normed spaces, this time using transfinite induction. We omit some of the details, which the reader can fill in. In particular, some of the details are similar to those involved in proving the claim made in Part V (when we obtained an upper bound for a chain). For a non-zero ordinal $\gamma$, recall that, with ordinal interval notation, we have $\gamma=[0,\gamma)$. That is, $\gamma$ is equal to the set of all ordinals which are strictly less than $\gamma$. However, we use ordinal interval notation throughout. If you prefer to use $\gamma$ instead of $[0,\gamma)$ that is fine, of course! Looking at the Wikipedia page on the Hahn-Banach Theorem , apparently both Hahn and Banach (independently) used transfinite induction in their proofs, rather than Zorn's Lemma. Theorem (Hahn-Banach Extension Theorem for linear functionals on normed spaces over $\R$)  Let $(E,\jnorm)$ ...
Read more