As a fun project, I want to create an interesting, insightful, yet educational video about some fascinating topic/concept/idea in mathematics that only requires a basic understanding of mathematics so that anyone who watches it can learn and appreciate something from it.
What are your ideas? What is something simple yet captivating about mathematics?
I was thinking perhaps showcasing a few of the interesting proofs of the Pythagorean theorem, but what are your ideas?
FYI: This is just for my own personal fun and I hope someone out there is able to learn something. I am not a big YouTuber and will not make any money off this video. I will credit whoever's idea I use if I use it.
Here are two idea, I hope it is good:
The mean-value theorem states that Given a function $f(x)$ that is continuous on a closed interval $[a,b]$ and differentiable on the open interval $(a,b)$ then there is a point $c \in (a,b)$ such that $f'(c)=\frac{f(b)-f(a)}{b-a}$.
One proof uses Roll's theorem, Roll's theorem says that if we have a function $g(x)$ that is continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$ and $g(b)=g(a)=0$, then there is a point $c \in(a,b)$ such that $g'(c)=0$.
A proof of the mean-value theorem uses the Roll's theorem where they transform the function $f(x)$ into a function $g(x)$ having the required properties specified in the Roll's theorem and apply Roll's theorem on $g(x)$. It would be cool to see that transformation visually, this way it would be easier to understand what the proof is doing rather than trying to memorize the transformation.
The transformation is as follow: $g(x)= f(x)-f(a)-\frac{f(b)-f(a)}{b-a}(x-a)$.
what this transformation is doing can be understood in the following steps: the step $f(x)-f(a)$ is to shift the original function $f(x)$ downwards so that its left end point $f(a)$ lie on the x-axis as required by the Roll's theorem that $f(a)-f(a)=0$.
Then the step for subtracting $\frac{f(b)-f(a)}{b-a}(x-a)$ by $f(x)-f(a)$ from $f(x)-f(a)$ is to shift the entire function $f(x)-f(a)$ down by different amount at each point in a way that the function evaluated at $b$ is $0$. Namely, the right end point is also on the x-axis while the left endpoint stays unchanged. But why does subtracting $\frac{f(b)-f(a)}{b-a}(x-a)$ work? Let's understand what is this function that requires so much trouble to write: $\frac{f(b)-f(a)}{b-a}(x-a)$ is the function of the line connecting $f(b)$ and $f(a)$ but shifted to the right along the x-axis a little bit so that its x-intersect is $a$. This will therefore bring the right end point of $(f(x)-f(a))$ on the x-axis!
The result $g(x) = f(x)-f(a)-\frac{f(b)-f(a)}{b-a}(x-a)$will then have the form that Roll's theorem specify, namely, $g(a)=g(b)=0$, $g(x)$ is continuous on $[a,b]$, differentiable on $(a,b)$.
Finally applying Roll's theorem, we get that there is a point $c \in (a,b)$ such that $g'(c)=0$. But $g'(x)=f'(x)-\frac{f(b)-f(a)}{b-a}$ and $0=g'(c)=f'(c)-\frac{f(b)-f(a)}{b-a} \iff f'(c)=\frac{f(b)-f(a)}{b-a}$ which is the claim of the mean-value theorem.
The fact that between any two real numbers, there is a rational number, you can try to make visualization that if $a<b$, then multiply them by large enough integer $m$ so that there is a integer n such that $am<n<bm$ which gives the desired result.