Real Analysis | The Mean Value Theorem

[@jonaskoelker]

At 8:25 a function is pulled out of a hat, explained by a secant line. Here's a piecemeal way to build it: Let L(x) be such that L(a) = 0, in particular let L(x) = f(x) - f(a). Let R(x) be such that R(b) = 0, in particular let R(x) = f(x) - f(b). Just like a DJ who turns down the volume of one song while turning up the volume of the next, we want to mix L and R. Let mix(a, x, b) be how far x is between a and b, such that mix(a, a, b) = 0 and mix(a, b, b) = 1. In particular, let mix(a, x, b) = (x - a)/(b - a). Then let g(x) = (1 - mix(a, x, b))*L(x) + mix(a, x, b)*R(x). Then g(x) = d(x) for all x, i.e. this is how to build the 'd' function. This is easiest to show if you put both expression on a /(b-a) fraction with three terms in the numerator: one with f(a), one with f(x) and one with f(b). [That is, simplify towards a common target form.] In particular: g(a) = (1 - mix(a, a, b))*L(a) + mix(a, a, b)*R(a) = (1-0)*0 + 0*R(a) = 0. g(b) = (1 - mix(a, b, b))*L(b) + mix(a, b, b)*R(b) = (1 - 1)*L(b) + 1*R(b) = 0*L(b) + 1*0 = 0. We chose a linear mix(*, *, *) function because we wanted to prove something about a linear-looking formula. In video games, non-linear "mix"-like functions are used for transitions of visual effects.

[@goodplacetostop2973]

0:49 Special guest 1:02 Special guest leaving 1:06 Special guest climbing stairs 12:02

[@hydraslair4723]

I love the special guest

[@ujwalsmanhas1093]

That is actually kids ( his/sibling's ) being forced to attend calulus lectures.

[@argonwheatbelly637]

One of my favorite parts of 1st semester college calculus, 1983.

[@get2113]

It is important to note that the MVT only works if f is a scalar valued function. The is no result of the form f(y)-f(x)=D(y-x) where D is a Jacobian matrix evaluated at point on a line segment between y and x. Its easy to forget this. There is a simple workaround with slightly messy notation.

[@jonaskoelker]

Theorem: let f and g be continuous on [a, b] and differentiable on (a, b). If f(a) = g(a) and f(b) = g(b) there exists some c in (a, b) such that f'(c) = g'(c). Proof: apply Rolle's theorem to f - g. Corollary: Let g be the secant going through f at a and b, then the mean value theorem follows. Observation: if g = 0 then my theorem is exactly Rolle's theorem. Observation: if g is any other constant function then my theorem is a slight generalization of Rolle's theorem.

[@egillandersson1780]

Where things which seem obvious need to be proove, too. I like this perfectly clear demonstration.

[@jabunapg1387]

Thank you for all this content. Greetings from Germany!

[@timurpryadilin8830]

Tgere is a beautiful generalisation of this result. If f(x) abd g(x) are differentiable and continuous on [a, b] functions with g'(x) being non-zero everywhere, then there is a point c, such that f'(c) /g'(c) =(f(b) - f(a)) /(g(b) - g(a)). If we take g(x) =x, we get the mean value theorem.

[@filipbaciak4514]

What's name of that theorem?

[@adamc973]

@@filipbaciak4514 Probably too later to respond now, but it's called Cauchy's Mean Value Theorem or the Generalised Mean Value Theorem.

[@amberdeshbhratar9342]

Please sovle hard problems on LMVT ( Rolls theorem )

[@kylenason]

Perfect. My calculus class just went over this topic this week

[@25thturtle48]

Nobody cares about nice value theorem 😭

[@andresarroyo1809]

These videos are great

[@chrisk9902]

0:48 lol

[@filipchris245]

what was that on 0:48?

[@Thewardra]

Kids watching the proof 😂

[@thayanithirk1784]

Is this real analysis or calc1

[@MrRyanroberson1]

for which functions can we guarantee that (in the interval [a,b]) there is some f'(n) = (f(b) - f(a))/(b-a)? if there is some f(a),f(b) -> c,d, from this we know that there is some f'(n) = 0. is there some h(x) such that h'(n) = (c-d)/(b-a), in other words the secant's slope for h with x values a and b. Choose a and b such that f(a) = f(b) so we look: h'(n) = f'(n) + j'(n) = (f(b) - f(a) + j(b) - j(a))/(b-a) = (j(b) - j(a))/(b-a) = j'(n). this shows that kx is a sufficient choice; let g(x) = f(x)+kx. g(b) - g(a) = f(b) - f(a) + kb - ka = k(b-a), and so (g(b) - g(a))/(b-a) = g'(n) = k.

[@MrRyanroberson1]

ok i paused the video halfway through because i thought of this and pressed play... jeez. you really think of everything, don't you? XD

[@dielaughing73]

First comment about the typo in the thumbnail (the=then)!

[@arvindsrinivasan424]

🔥🔥🔥

[@datsmydab-minecraft-and-mo5666]

how would we go about proving for the case d(a) not equal to d(b)

[@TheBoardTutor]

Hey Michael, your videos are superb. Even I do have a channel regarding education, do watch it.

[@CM63_France]

Hi, The camera gets more and more down, now we can't read the first lines you wrote at the top of the board.