@@CM63_France the actual difference is an increasing positive sequence that goes to infinity plus an increasing negative term that approaches 0. So, since the first difference is 1, all differences are greater than 1.
What we can notice from this proof is that when we negate the definition we can construct sequence since delta derived from the negation is an arbitrary posive number.
I was going to ask if uniform continuity means the same as "continuous everywhere" but clearly it does not. That final proof (for f(x)=x^3 not being uniformly continuous on A does not work if A has finite limits, since you can no longer generate the two infinite sequences {x_n} and {y_n}. Essentially, the problem stems from the "Application: f(x)=x^3 is not uniformly continuous" which should have specified that the domain A is the entire set of real numbers. The property of being uniformly continuous requires specification of the domain to be meaningful. I suppose one could argue that by not specifying the domain, it is implicitly the real numbers, but given that up to that point the discussion involved a specific domain A, suddenly changing to an implied domain requires at least a comment to that effect. Or perhaps this is some carry-over context from the previous video where he was unable to show uniform continuity for f(x)=x^3. It looks sort of like uniform continuity over A requires that the first derivative be finite in A. Otherwise one can always generate a sequence pair as was done here, except instead of approaching infinity, the sequences would approach a point where the first derivative becomes infinite.
> So, is there a connection between the derivative and uniform continuity? Yes. Let f: A -> R be differentiable on the interior of A, with f' bounded and A connected. Then f is uniformly continuous. Intuition: If f is not uniformly continuous there exists a rate of change (epsilon_0) such that no matter how near we force points to be, there's a pair of points x, y such that the rate of change between them is at least epsilon_0. The derivative is the rate of change. If it is bounded then f can only grow so fast, i.e. the x and y must be far enough apart for f to grow by the epsilon_0 amount. Proof: We want to show that for all epsilon > 0 there exists a delta such that when |y - x| < delta then |f(y) - f(x)| < epsilon. So let epsilon be given. Since f' is bounded there exists an M such that -M < f'(x) < M for all x. For every x and y with x < y we have [f(y) - f(x)] / (y - x) = f'(c) for some c, but f' is bounded. Thus -M < [f(y) - f(x)] / (y-x) < M for all x < y. But then -M*(y-x) < f(y) - f(x) < M*(y-x). "Let" M * (y-x) < epsilon, i.e. let delta = epsilon/M. If y - x < delta then M*(y-x) < M*delta = M*epsilon/M = epsilon. But then f(y) - f(x) < epsilon. Likewise -M*(y - x) < f(y) - f(x); negating both sides gives f(x) - f(y) < M*(y-x) < epsilon. Note that |f(x) - f(y)| is either f(y) - f(x) or f(x) - f(y), both of which are less than epsilon, so |f(y) - f(x)| < epsilon. But that's exactly what we wanted to show. Note that the x < y assumption is WLOG: if x = y then |f(x) - f(y)| = |f(x) - f(x)| = 0 < epsilon, and if y < x we can swap the names of x and y-they are used symmetrically. We need A to be connected for [x, y] to be contained in A for all x and y. We need interior differentiability for the proper f'(c) to exist with c in the [x, y] interval. See also math.stackexchange.com/questions/291166/prove-bounded-derivative-if-and-only-if-uniform-continuity and math.stackexchange.com/questions/118665/why-if-f-is-unbounded-then-f-isnt-uniformly-continuous. I guess this is a good place to stop this comment.
Hi, Could you please rise up a little bit your camera? When a sub-title appears on the top of the screen, it may hide what you wrote. And the concrete band in the bottom is not usefull. Thanks.
Please help me in this question If each of the algebraic expressions lx^2 + mx + n mx^2 + nx + l nx^2 + lx + m Are perfect squares, then proove that (l+m)/n=-4 Plz help😢😢
@@sahilbaori9052 Thnks for the solution though, but i still didn't understood b^2=4ac. Shouldn't the discriminant equal to a perfect square other than 0? So b^2-4ac=k^2
@@utsav8981 discriminant will be equal to k^2 when the solution is real and distinct. But in this case it is real and identical (perfect square) so it has to be 0.
should have been |3n-3/n+1/n^3| at the end there.
same doubt buddy
I'm revisiting analysis and this is a great channel to follow in the process.
You should have used xn = n+1/n, yn = n. You have a sign error at the very end, and your conclusion was that 1^3-0^3 > 3
You are right, I saw that too, may be we can say the quantity becomes greater than 3 after a certain value of n.
@@CM63_France the actual difference is an increasing positive sequence that goes to infinity plus an increasing negative term that approaches 0. So, since the first difference is 1, all differences are greater than 1.
One of the best youtube channels, digging the new haircut ❤️
This channel is such a gem.
Hello all! I don't get the part why we've set delta to be 1/n?
What we can notice from this proof is that when we negate the definition we can construct sequence since delta derived from the negation is an arbitrary posive number.
I was going to ask if uniform continuity means the same as "continuous everywhere" but clearly it does not.
That final proof (for f(x)=x^3 not being uniformly continuous on A does not work if A has finite limits, since you can no longer generate the two infinite sequences {x_n} and {y_n}.
Essentially, the problem stems from the "Application: f(x)=x^3 is not uniformly continuous" which should have specified that the domain A is the entire set of real numbers. The property of being uniformly continuous requires specification of the domain to be meaningful. I suppose one could argue that by not specifying the domain, it is implicitly the real numbers, but given that up to that point the discussion involved a specific domain A, suddenly changing to an implied domain requires at least a comment to that effect. Or perhaps this is some carry-over context from the previous video where he was unable to show uniform continuity for f(x)=x^3.
It looks sort of like uniform continuity over A requires that the first derivative be finite in A. Otherwise one can always generate a sequence pair as was done here, except instead of approaching infinity, the sequences would approach a point where the first derivative becomes infinite.
yess, finally
finishing off x^3 for real, so satisfying
Well done video!
So, is there a connection between the derivative and uniform continuity?
> So, is there a connection between the derivative and uniform continuity?
Yes.
Let f: A -> R be differentiable on the interior of A, with f' bounded and A connected. Then f is uniformly continuous.
Intuition:
If f is not uniformly continuous there exists a rate of change (epsilon_0) such that no matter how near we force points to be, there's a pair of points x, y such that the rate of change between them is at least epsilon_0. The derivative is the rate of change. If it is bounded then f can only grow so fast, i.e. the x and y must be far enough apart for f to grow by the epsilon_0 amount.
Proof:
We want to show that for all epsilon > 0 there exists a delta such that when |y - x| < delta then |f(y) - f(x)| < epsilon. So let epsilon be given.
Since f' is bounded there exists an M such that -M < f'(x) < M for all x.
For every x and y with x < y we have [f(y) - f(x)] / (y - x) = f'(c) for some c, but f' is bounded.
Thus -M < [f(y) - f(x)] / (y-x) < M for all x < y. But then -M*(y-x) < f(y) - f(x) < M*(y-x).
"Let" M * (y-x) < epsilon, i.e. let delta = epsilon/M. If y - x < delta then M*(y-x) < M*delta = M*epsilon/M = epsilon.
But then f(y) - f(x) < epsilon. Likewise -M*(y - x) < f(y) - f(x); negating both sides gives f(x) - f(y) < M*(y-x) < epsilon.
Note that |f(x) - f(y)| is either f(y) - f(x) or f(x) - f(y), both of which are less than epsilon, so |f(y) - f(x)| < epsilon.
But that's exactly what we wanted to show.
Note that the x < y assumption is WLOG: if x = y then |f(x) - f(y)| = |f(x) - f(x)| = 0 < epsilon, and if y < x we can swap the names of x and y-they are used symmetrically. We need A to be connected for [x, y] to be contained in A for all x and y. We need interior differentiability for the proper f'(c) to exist with c in the [x, y] interval.
See also math.stackexchange.com/questions/291166/prove-bounded-derivative-if-and-only-if-uniform-continuity and math.stackexchange.com/questions/118665/why-if-f-is-unbounded-then-f-isnt-uniformly-continuous.
I guess this is a good place to stop this comment.
x[n]=cbrt(n)
y[n]=cbrt(n+1)
Hi,
Could you please rise up a little bit your camera? When a sub-title appears on the top of the screen, it may hide what you wrote. And the concrete band in the bottom is not usefull. Thanks.
Nice haircut, as nice as Dedekind cut.
HAHAHHAHAH
How is it that youtube did not suggest me your channel before!(? ) Great video, thanks a lot!
RUclips recommend bullshit content first... such great content like of Michael Penn remains hidden!!!
18:34
Please help me in this question
If each of the algebraic expressions
lx^2 + mx + n
mx^2 + nx + l
nx^2 + lx + m
Are perfect squares, then proove that (l+m)/n=-4
Plz help😢😢
n^2 = 4lm ... (1)
m^2 = 4ln ... (2)
l^2 = 4mn ... (3)
Subtract 2 and 3
(m+l)(m-l) = 4n(l-m)
m+l = -4n
(m+l)/n = -4
@@sahilbaori9052 How did you arrive at the first step that is n^2=4lm?
@@utsav8981 they are perfect squares so b^2 = 4ac.
@@sahilbaori9052 Thnks for the solution though, but i still didn't understood b^2=4ac. Shouldn't the discriminant equal to a perfect square other than 0? So b^2-4ac=k^2
@@utsav8981 discriminant will be equal to k^2 when the solution is real and distinct. But in this case it is real and identical (perfect square) so it has to be 0.
nice
18:34