Notes for this video are available for my patrons: 👉 www.patreon.com/posts/notes-proof-of-82385985

@annaclarafenyo8185

Жыл бұрын

This video is nonsense. The proof that the multiplication function is continuous is trivial, and the proof that the composition of continuous functions is continuous is also trivial. (x+epsilon)(y+delta) = x y + epsilon*x + delta*y +epsilon*delta, and all quantities are infinitesimal aside from x*y. This observation can be turned into an epsilon delta proof automatically by rote. It's ridiculous to call this difficult, it's obvious.

@blackpenredpen

Жыл бұрын

@@annaclarafenyo8185 f and g might be discontinuous

@annaclarafenyo8185

Жыл бұрын

@@blackpenredpen Continuous AT THE POINT means the limit of f equals M and the limit of g equals N, you can set the values to be M and N, and then they are continuous, and the continuous composition of continuous functions is continuous. These methods are formalizations of infinitesimal arguments that are obvious: (f+df)(g+dg)=fg + infinitesimal. There is no work in formalizing this. Making this obvious nonsense sound difficult just serves to mystify epsilon-delta, and make it a hazing ritual rather than a method of proof you internalize.

@blackpenredpen

Жыл бұрын

@@annaclarafenyo8185 f and g don’t need to be continuous at a

@annaclarafenyo8185

Жыл бұрын

@@blackpenredpen IF YOU DEFINE THEM SO THAT THE LIMIT IS EQUAL, THEY ARE BY DEFINITION CONTINUOUS AT A. You should not be teaching matheamatics.

You know something is wrong when bprp doesn't use whiteboard for proof

@andreasxfjd4141

Жыл бұрын

Simply if he doesn’t use a whiteboard (than something has changed)

@akasunanosasori7547

Жыл бұрын

Actually my teacher showed us this proof in grade 11, I don’t think it’s too hard (now we weren’t supposed to learn the proof but showed us nonetheless)

@blackpenredpen

Жыл бұрын

I needed to keep everything on the board for this video thus I chose the iPad route 😆

@Paul-ob2hy

Жыл бұрын

@@blackpenredpenwhat app did you use on your ipad? would it be good for notes at university?

@blackpenredpen

Жыл бұрын

@@Paul-ob2hy I never took notes on an iPad when I was a student. However, I have been using the app "Good Notes" and I have been loving it!

27:05 to avoid the 0/0 case, we can simply have epsilon*|L|/[2(|L| + 1)] < epsilon*(|L| + 1)/[2(|L| + 1)] = epsilon/2, and the solution is unaffected in any way ^^

@blackpenredpen

Жыл бұрын

Ahhhh. I didn’t think of that 😆

@christopheriman4921

Жыл бұрын

Or you can observe that 0

@user-en5vj6vr2u

Жыл бұрын

Yeah i was thinking he would do that

Not to kiss your behind or whatever, but this is the most helpful channel I’ve ever found when it comes to helping me with my EXTREMELY hard math class

@blackpenredpen

Жыл бұрын

I am happy to hear! Thank you!

This is one of the more complicated proofs of basic analysis facts and I've never really had a good intuition for the quantities that come up in the proof, so thank you for doing a good job of talking through it.

26:07 I saw some people trying to explain how we don't have to bother about the |L|=0 case in a certain way, but in the end : We know that |L| 0) guarantees us the result ! Edit : Nevermind it has been said by multiple people already, my bad

1:28 This is the best explanation of the epsilon-delta definition of limits I have ever seen. Everything for me just clicked once I saw this. You're an awesome teacher!

@blackpenredpen

Жыл бұрын

Thank you!

26:27 You don't have to cancel two zeros if you add 1 to |L|, proving that step rigorously! |L| * (epsilon)/(|L|+1) < (|L| + 1) * (epsilon)/(|L|+1) = (epsilon)/2

@keedt

Жыл бұрын

was going to write this 👍.

@blackpenredpen

Жыл бұрын

Ah yes. I didn’t think of that. Thanks for letting me know.

@Apollorion

Жыл бұрын

tzovgo, you made a little mistake in the formula, which I didn't notice at first, neither did, I presume, BPRP. We both understood what you meant, but you forgot to put the 2 in the denominator whenever |L| was present in the quotient. This is what it should be: epsilon*|L|/[2(|L| + 1)] < epsilon*(|L| + 1)/[2(|L| + 1)] = epsilon/2

This is blackpenredpenbluepengreenpen and yellow highlighter

@blackpenredpen

Жыл бұрын

😆

@Maths_3.1415

Жыл бұрын

​@@blackpenredpen 😂

extremely good job and explanations. As a new online teacher, I find your content really inspiring. I now recognize that I mimic a lot your teaching style. Keep up the good job !

@blackpenredpen

Жыл бұрын

Thank you for your kind words. Wishing you all the best! : )

Thank you very much!!! I Found this proof in Spivak's book but I was not getting how it was done and after watching your video now is all clear to me. Love you!

@blackpenredpen

9 күн бұрын

Glad it helped!

Bprb, WHAT AN AWESOME PROOF! Even though it turned out to be a very difficult one, I understood every step of it, and that's because you are en EXCELLENT TEACHER. Thank you! 🙏🏻

I'm graduating soon in Mechanical Engineering and I just have to say your channel is the best! You've helped me so much!

@blackpenredpen

Жыл бұрын

Thank you. : )

I have always loved working out these fundamental proofs. I actually did this particular one out just a few months ago working through my Schaum's Calculus workbook. I didn't think it was particularly difficult- the really key insight is that addition/subtraction trick that allows the breaking apart of the functions, but it is a common technique in algebra, so came to me almost instantly.

I prefer to use epsilon-delta arguments only in the simplest cases. For example, - A sum of two functions approaching 0 approaches 0 - A function approaching 0 times a bounded function approaches 0 Now f can be expressed as L plus a function approaching 0, and g can be expressed as M plus a function approaching 0. You can just multiply it out and see that the product is LM plus a function approaching 0 i.e. the limit of the product is LM.

Excellent explanation 👍

You are looking CLEAN on the thumbnail man. Nice

The thing is you see, for the initial limit isn't using the product rule for limits, it's just substituting that c value for every instance of x

I remember getting this question in my university practice and getting the proof myself is so satisfying!

@blackpenredpen

Жыл бұрын

the moment we draw that box at the very end!!

Wow - this takes me back to my modern analysis course in college. I'll never forget how brutal that was. But good job with the proof!

@blackpenredpen

Жыл бұрын

Thanks, Scott!

Amazing video, thank you so much

You could also just prove the rule for limits of sequences, and then use the sequential definition of the limit to get the same result for limits of functions. Another approach would be to prove the result in the special case L = 0 = M. This is quite simple as you just choose δ such that |x - a| Then for the more general case define new functions: F = f - L, G = g - M Now the limit of both F and G is just zero, so the limit of FG is zero. Do a bit of expanding and out comes that the limit of fg is LM.

Thank u so muchhhhhhhhhhh!!!!!!!!! U save me life by this wonderful video, thank uuuuuuuuuuuuuuuuuuuuuuuuuu!!!!!!!!!!!!!!

Main trick is writing g(x) as g(x) -M+M after this there is no need to worry about the term f(x) g(x). It is all about being careful. We have to adjust things.

Seeing math makes me remembers when i find the answer Like "I have 9 numbers, the first 5 numbers has a mean of 12, while the last 4 numbers has a mean of 3, what is the mean of all 9 numbers" And thats when i realise i can just do (5 × 12 + 4 × 3) ÷ 9 and got 8

Great explanation.

I know the one for sequences and it is quite hard. I had to make my own one because I couldn't remember the one from the lecture. Anyway, from that you can easily arrive at the one for functions thanks to the sequence definition of a limit

Sir if we have to prove cone curve surface area then its can be prove by like this ,we take a triangle in it and move about dy angle from central then on cover surface area there would be a thin triangle and bease would be r× dy(it approximately be taken as straight line in calculus)and height lateral height L then area of that triangle would be 1/2×r×dy×L and when we intergrate it with limits 0 to 2pie our answer is pie ×r×L, sir does this is correct , and dy×r is straight line

Alternative idea: Prove that the multiplication map (x,y) -> xy, is continuous R^2 -> R, which isn't that hard. Then, the result follows easily.

@justinyoung4381

Жыл бұрын

On second thought, it's more or less the same as your proof, but seems simpler because of the removal of f(x) and g(x).

I find it easier in the general case of a finite dimension vectorial space like R^n and considering a subordinate norm which is sub-multiplicative... Of course you need a bit of topology but the proof is just much nice

Once you have delta-1 and delta-2, do you even need delta-3? That is, can’t you take min(delta1, delta2) and leave it at that?

@blackpenredpen

Жыл бұрын

Yes, bc I used the given limits for three different ineq in the blue part.

What is the name of the whiteboard program??

Flashbacks to Real Analysis I where our professor wanted us to figure out how to do this proof on our own *shudders*

Next video: 100 matrix transformation and eigenvalue questions

Can you make Differentiation and integration basics concepts for beginners. I'm unable to understand it in my school.

I think the easiest way to think about the 0/0 part is: if |L|=0, then |L|*epsilon/(2*(|L|+1))=0 which is definitely smaller than epsilon/2.

@Apollorion

Жыл бұрын

Not greater, smaller: 0 is the smallest non-negative real number!

@rmlu9767

Жыл бұрын

@@Apollorion thank you. Corrected.

This video is filled with tasty _rigor_ I love it... Make more of these please

Using 1 is nice. I used to introduce another constant something epsilon2

Now proof that the proof is 1618 times harder than the calculation

@Apollorion

Жыл бұрын

..and not 1619 or 1617 times harder.

Omg I can’t believe I actually understood all of this!

That is why you derive the sequence criterion for limits, and then you can use the easier to prove limit theorems.

I was just trying to remember this proof! (with some difficulty after ~15 years 😅)

PF of "as x→a, f(x)/g(x)→L/M" directly is also "hard". "directly" i mean without proving the special case "as x→a, 1/g(x)→1/M" and using prada law for f(x)/g(x)=f(x)*[1/g(x)]

26:30 I think it would be a clearer explanation to simply state that for all values of L, |L| < |L| + 1. Diving both sides by |L| + 1 gives [ |L| / (|L| +1) ] < 1, so you have [ |L|*eps / 2*(|L| + 1) ] < eps/2. Great video! Haven’t seen this proof since my undergrad in 2018.

@Ninja20704

Жыл бұрын

That’s what i was thinking too. If we regroup the terms as [eps/2][|L|/|L|+1]

Fun fact we in Algeria we study real analysis in the first year of bachelor degree

Woahh......I feel smarter

awesome!!

Could you prove that integral of e^(-x^2) has no solution in elementary functions?

Do more proofs please. ❤

Hi Blackpenredpen. Make a video demonstrating why π is irrational. I am subscribed to your channel, very good videos. Greetings from Mexico

Please make a tutorial on how you switch your markers on this board

@epsilia3611

Жыл бұрын

I didn't know you were a genius in humour too mister

@Maths_3.1415

Жыл бұрын

​@@epsilia3611 thank you 😅

@blackpenredpen

Жыл бұрын

waiting for someone to buy me another apple pencil lol

@Maths_3.1415

Жыл бұрын

​@@blackpenredpen 😂

Zero pairs are evil period for me - this is why I liked when Dr Peyam did the proof for derivative of product rule he drew a bunch of quadrilaterals within quadrilaterals and subtracted them out. - only caveat is you need to suspend disbelief that this does not work just for positive but negative area (and functions as well).

You can simplify the proof by using the lemma : lim_(x->a)f(x)=L iff there exists K>0 s.t. for all epsilon >0 there exists delta s.t |x-a| |f(x)-L|

Hi dear Professor, I have a topic that might interest you and I would like to see a video about it: PROVE THAT FOR EVERY REAL NUMBER NOT A MULTIPLE OF 2 CONTAINS IN ITS DECOMPOSITION AT LEAST ONE ODD NUMBER

The example needs not the propiety, the function is continous therefore the limit converges to the value

Can I ask, how do we use the epsilon-delta defn to prove a limit than involves x-> +/- infinity? Because infinity is not a number so we cant say a=+/-inf. Or is there a different definition we have to use? Thank you in advance

@thatapollo7773

Жыл бұрын

I will give an example Say we need to prove lim x->+inf 1/x = 0 We need to show that for all epsilon > 0 There exists delta st. For all x>delta(THE KEY DIFFRENCE) f(x) - 0 < epsilon

@SatanicNerfd

Жыл бұрын

One way to see this is by substituting y=1/x and take the limit y->0

@Ninja20704

Жыл бұрын

@@thatapollo7773 Thank you. But it should be |f(x)-0|

@vonneumann6161

Жыл бұрын

Yes, there are different definitions for all of those. For example, if you want to prove lim[x->a]f(x) = +infinity, you have to show that for any M there exists some delta>0 such that if |x-a| M. It’s similar to epsilon-N definition. It’s a bit different when a = +/- infinity and all the other versions but they’re all basically the same.

@thatapollo7773

Жыл бұрын

​@@Ninja20704 yes, I forgot to put the absolute value

What videos is he referring to, where he has explained the epsilon-delta.

@Theraot

Жыл бұрын

kzread.info/dash/bejne/dpiop7Okj6eugdY.html

@blackpenredpen

Жыл бұрын

See links in description 😃

Thanks PROF)

The last step can be easily justified by noticing ε|L|/[2(|L|+1)] = (ε/2)[|L|/(|L|+1)] and also |L|/(|L|+1) This is true even when L=0. Therefore ε|L|/[2(|L|+1)] QED lol Edit: I realized afterwards many others explained this already or similar arguments haha but yeah the proof is sound, was really nice following it till the end

Back in the day when I used to have to do such proofs, I was irritated by the need to "edit" the proof after completing it, so choosing epsilon/3 for example, as the last line added three such terms. It makes the proof more "magical" for those trying to read it ("Why epsilon/3!!!!") After all, anyone who understands the limit/continuity/whatever definitions understands that "< 3*epsilon" is as good as "< epsilon" given the arbitrary choice of epsilon in the first place.

@theproofessayist8441

Жыл бұрын

Lol I was lazy and sometimes said you see how all the lines of deduction are in order when thinking from scratch and stuff - told the professor in comments please read from bottom to top in reverse. They said no - you have to rewrite the entire proof again. I found that was annoying.

Using nonstandard analysis, this theorem becomes quite easy to prove: Start with lim(x->a) f(x)g(x) =: A. For any nonzero infinitesimal h, we have A = st(f(a+h)g(a+h)). But since L = lim(x->a) f(x) = st(f(a+h)) and M = lim(x->a) g(x) = st(g(a+h)) for any nonzero infinitesimal h, there are infinitesimals k,l for which f(a+h) = L + k and g(a+h) = M + l. This directly implies A = st((L+k)(M+l)) = st(LM +kM + lL + kl) = LM. QED

Blackpenredpen can you do a proof of descartes rule of signs. I'm starving for that proof. You would explain the proof very easily. 😢

@calculus988

Жыл бұрын

Im willing to pay you

Give suggestion to study PDE , Its hard to grasp its concepts

I'll ask you a question What is -ln(-1) =? A . iπ B . -iπ C. A ans B both D. Can't possible E. None of the above I am not testing you but by watching just your videos i got that question in my mind

THANK YOU FOR ALL YOUR SUPPORT AND I HAVE GOT 83 PERCENTAGE AND 99 IN CS VERY THANKFUL TO THE WONDERFUL TEACHERS

@robertveith6383

Жыл бұрын

Stop yelling your post in all caps. It is rude.

try finding the values for a, b and c when f(x)=ax^2+bx+c f(A)=B f(C)=D f(E)=F I did this equation myself and the answer is very long

I liked how U say ah ah fells so good😂😂😂

I feel like its kind of the engineer way how you explain the first example. A (for me nicer way of justifying this is by setting: h(x)=f(x)g(x), and assume they are continuous, which implies, that the limit for all a (exept infinity) exists. Now we can use sequence continuity by saying: lim ( f(x)g(x) )= lim h(x)=h( lim x ) = f( lim x )g( lim x). And we are done. I am of course assuming that you need to know about sequence continuity, but this should be more elementary than limit to some random point a, where you often have some 0/0 or infinity/infinity problems :)

this is very easily taught in asia. its just that NA students arent exposed to inequality too much.

Here is how I broke up the functions: |f(x)g(x)-LK| Add and subtract 2LK, and add and subtract [Kf(x) + Lg(x)] to get |f(x)g(x)-Kf(x)-Lg(x)+LK+Kf(x)-LK+Lg(x)-LK| |[f(x)-L][g(x)-K]+K[f(x)-L]+L[g(x)-K]| From here, you break it apart using the triangle inequality which gives eventually |f(x)g(x)-LK|< ε1ε2 + |K|ε1 + |L|ε2

Let ɛ < 0

0:45 Does the limit of x as x approaches 3 exist? Yes, it does. How do we know? Epsilon-delta time!

How about: dy/d(dy/dx) = y

When I did an online real analysis class the professor only proved this for limits of sequences. Then after that he didn’t bother proving it explicitly for limits of functions, rather he used the sequential characterization of limits to be like “we did this already” lol. I don’t blame him for not wanting to go through this again honestly!

@blakedylanmusic

Жыл бұрын

Also for the |L|/(1+|L|) thing, you could also break it into cases, ie if |L| = 0 then of course ε|L|/2(1+|L|) = 0 < ε/2. Then in the case |L| is nonzero you do what you showed in the video.

@blackpenredpen

Жыл бұрын

In fact, I should have done it as (like many people who have already pointed out) |L|/(|L|+1) I couldn't believe I didn't think of that when I was working this out lol

@blakedylanmusic

Жыл бұрын

@@blackpenredpen honestly as a grad student in math, I feel like that a lot 😂😂😂 I’ll spend hours trying to come up with a certain proof and then when I look at my professors solution I’m like “why didn’t I think of that” hahaha

I remember when I first watched bprp I was in high school and didn't know many things. Now I can be more rigor than bprp.

if |L| = 0 the expression is 0 and is definitely less than epsilon/2 (because epsilon is positive and half of it is still positive), otherwise |L|/(|L|+1) is less than 1 so the second expression is also less than epsilon/2.

Sir,I try this as a undergraduate(12th) student but I can't. And also want to prove derivative of u/v and uv rule. Love from India.

Great explanation. Technical requirements explained so well that a 14 year old could understand it 👍.

@blackpenredpen

Жыл бұрын

Glad to hear 😃

I don't remember L'Hopital's Rule being so difficult needing f(x)/h(x) when there is a numerator and denominator limit of the form f(x>a)g(x>a) 😬🤣

This is why we have these theorems in the first place, right? You don't want to go through this ε-δ stuff every single time, so you go through it *once,* to prove the theorem, and afterwards just use the theorem.

@tobybartels8426

Жыл бұрын

@LeftRight : That's true, but even those are closer to the general proof for multiplication that we did, than to the specific proof for x√(x+1) that we avoided. (Besides, if we really prove all the theorems that calculus students implicitly rely on, for addition, multiplication, powers, trig functions, etc, then we'll have gotten plenty of practice.)

Please integrate this hard problem : İntegral ln(x)ln(1-x)/(1+x²) dx borders 0 to 1 😓😓

1+1=2 also hard to prove if you look in uni books

27:00 this part is simple. Notice |L|/(|L|+1) < 1

PROF thinked this better than blackboard.

We allwayse proof the limits using the defenition of limits

Why do you always indicate that |x-a| > 0? Is that only to specify that x=/= a?

@Zephei

Жыл бұрын

Well, x ≠ a is equivalent to |x-a| > 0, which is part of the definition of a limit.

for me the worst in maths is when you have to prove an evidence (1+1=2)... Great video by the way

@blackpenredpen

Жыл бұрын

Thanks!

👍

I always though this was obvious,and therefore non provable😅

Now do the Chain Rule 😅

Nice proof! We can also let |g(x)-M|

Next video: Polynomial division?

"The proof is left as an exercise for the professor."

10:18

way better than whiteboard

tried that during class. wasted 30 minutes, not sure if my proof works, there might be an easy proof that doesn't even rely on delta epsillon:(. guess imma post my proof on reddit and hope for the best.

1618 = 2×809

If |L| = 0 then obviously |L|*|g(x)-M| =0 since this is 0 multiplied with something. And 0 is always less than ε/2 since ε is positive. For |L| >0 the shown method works.

😂Is that a Monty Python joke? 1618, a movie about the Spanish Inquisition? ETA: |L|/(|L|+1) < 1 since |L| < |L|+1. No shenanigans necessary. Great video.

@blackpenredpen

Жыл бұрын

No, 1618 is 1000*φ, the golden ratio 😃

@Daniel-oy2he

Жыл бұрын

@@blackpenredpen Ah, nice. I didn't expect that or the Spanish Inquisition.

Bring a calculus intro video for a 8 grader