Very cool theorem. Interesting to note that the opposite is not true. Let x in [a,b] and f be continious. Then f(x) is not necessarlly in [f(a),f(b)]. The proof pretty much uses the bissection method to find the root of f tilde.
You define \tilde{f} to be -g if g(a) > 0. Then it is very clear that always \tilde{f}(a) ≤ 0, but to me, it is not directly obvious why also \tilde{f}(b)≥0. Let's say instead of the drawn function f, f is a parabola that is flipped at the minimum to get g. Wouldn't then both g(a) and g(b) be less or equal than zero? Thanks in advance! I'm probably missing something obvious.
Thanks for the question! Don't forget that we first translate the function by y. And because y is intermediate point, it is not possible to have both end points on the same side of the x-axis. Just look again how we have to choose y and how we define g. Best wishes! :)
when you define f tilde in 4:29, shouldn't it be f tilde = -g when g(a) > g(b) and f tilde = g otherwise ? Because at 3:58 you said that you wanted the value on the right to be larger than the value on the left.
Are {an} and {bn} Cauchy sequences because these are monotonic and bounded (and thus convergent)? Just like in Bolzano-Weierstrass theorem? Also, what books are you using for the videos if any?
Yeah, you could argue like this or simply look at the Cauchy property | a_n - a_m |. This distance can be immediately calculated and one sees that we have Cauchy sequences.
With this construction, what if \tilde{f} has multiple 0's? Couldn't you then miss the \tilde{y} you're looking for, and end up converging to a different 0? Well actually now after writing this I suppose it's enough that there exist sequences which converge to \tilde{y}...
A discontinuous function can have intermediate value property It was a long debate whether IVT defined continuity and unfortunately we discovered that does doesn’t and in fact we get various levels of continuity (Lipschitz and absolutely continuity)
@@chainetravail2439 because just plain old vanilla continuity isn't enough for a lot of things e.g. continuity is not enough to ensure that the set of all functions is complete (see uniform limit theorem)
I'm happy to say I came up with such a function myself: f(x) = sin(1/x) for x>0, f(0)=0 Discontinuous at 0, but will attain every value on [0,b] for b>0
![Real Analysis 32 | Intermediate Value Theorem [dark version]](/img/1.gif)
One of my favorite theorems from Analysis, together with MVT :)
no way bruh
Very cool theorem. Interesting to note that the opposite is not true. Let x in [a,b] and f be continious. Then f(x) is not necessarlly in [f(a),f(b)]. The proof pretty much uses the bissection method to find the root of f tilde.
Helps me a lot. Great proof process!
Glad it helped! And thanks for your support :)
Thanks for sharing such an insightful proof.
Glad it was helpful!
You define \tilde{f} to be -g if g(a) > 0. Then it is very clear that always \tilde{f}(a) ≤ 0, but to me, it is not directly obvious why also \tilde{f}(b)≥0. Let's say instead of the drawn function f, f is a parabola that is flipped at the minimum to get g. Wouldn't then both g(a) and g(b) be less or equal than zero?
Thanks in advance! I'm probably missing something obvious.
Thanks for the question! Don't forget that we first translate the function by y. And because y is intermediate point, it is not possible to have both end points on the same side of the x-axis. Just look again how we have to choose y and how we define g.
Best wishes! :)
Funny that the "Intermediate value theorem" is the halfway point of the playlist
Haha :D
Nice illustration and explanation.
Thank you! Cheers!
when you define f tilde in 4:29, shouldn't it be f tilde = -g when g(a) > g(b) and f tilde = g otherwise ?
Because at 3:58 you said that you wanted the value on the right to be larger than the value on the left.
Yeah, but remember we shifted the function to the bottom. Therefore g(a) > 0 has the same meaning as g(a) > g(b).
Interesting Theorem. Keep them coming! :D
Proof, suggest a textbook to read along with.
Are {an} and {bn} Cauchy sequences because these are monotonic and bounded (and thus convergent)? Just like in Bolzano-Weierstrass theorem? Also, what books are you using for the videos if any?
Yeah, you could argue like this or simply look at the Cauchy property | a_n - a_m |. This distance can be immediately calculated and one sees that we have Cauchy sequences.
With this construction, what if \tilde{f} has multiple 0's? Couldn't you then miss the \tilde{y} you're looking for, and end up converging to a different 0?
Well actually now after writing this I suppose it's enough that there exist sequences which converge to \tilde{y}...
Yes, we only have existence here :)
A discontinuous function can have intermediate value property
It was a long debate whether IVT defined continuity and unfortunately we discovered that does doesn’t and in fact we get various levels of continuity (Lipschitz and absolutely continuity)
The function f(x) = x (x= 0.5) on [0,1] is discontinuous but IVT holds, no?
Why unfortunately?
@@chainetravail2439 because just plain old vanilla continuity isn't enough for a lot of things
e.g. continuity is not enough to ensure that the set of all functions is complete (see uniform limit theorem)
@@chair547 what??
a=0.25; b=1
By IVT we should have f(c)=0.1, but it’s not in the interval.
You seem to be wrong
I'm happy to say I came up with such a function myself:
f(x) = sin(1/x) for x>0, f(0)=0
Discontinuous at 0, but will attain every value on [0,b] for b>0
Hi, is there problem with the website? It keeps telling me that it can't be found :c
Thanks! It should work again. I changes some links without changing them in the description.
We have shown [f(a), f(b)] is an interval but is this enough to show f[[a,b]] is an interval?
No, that is not enough :)
Ah, so we have to apply IVT to [xmin,xmax]
@antoniocleibo