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.
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.
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.
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.
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.
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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.
One of my favorite parts of 1st semester college calculus, 1983.
0:49 Special guest
1:02 Special guest leaving
1:06 Special guest climbing stairs
12:02
I love the special guest
That is actually kids ( his/sibling's ) being forced to attend calulus lectures.
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.
Where things which seem obvious need to be proove, too. I like this perfectly clear demonstration.
Thank you for all this content. Greetings from Germany!
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.
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.
What's name of that theorem?
@@filipbaciak4514 Probably too later to respond now, but it's called Cauchy's Mean Value Theorem or the Generalised Mean Value Theorem.
Perfect. My calculus class just went over this topic this week
Please sovle hard problems on LMVT ( Rolls theorem )
These videos are great
Is this real analysis or calc1
what was that on 0:48?
Kids watching the proof 😂
how would we go about proving for the case d(a) not equal to d(b)
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.
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
0:48 lol
First comment about the typo in the thumbnail (the=then)!
Hey Michael, your videos are superb.
Even I do have a channel regarding education, do watch it.
🔥🔥🔥
Nobody cares about nice value theorem 😭
Hi,
The camera gets more and more down, now we can't read the first lines you wrote at the top of the board.