I like this style of video!! Its nice that you speed up the 'uninteresting' parts, and you are good at explaining your arguments in a short and concise way
@@salem1599 Example: f(x) = -1/x + 3, x>1. This has high bound of 3, even increases monotonically, yet it still converges to 3, not e, when x - > infinity. By your comment, you have no clue about limit and convergence criterias...
Since limit of (1+1/x)^n, as n→∞, is e. And, 2 < e < 3. Since n is finite (n < ∞), intuitively it could be concluded that even for larger n’s, it is less than 3, so for smaller n's it ofcourse follows that.
As I suspect you're aware, a similar argument shows (1+1/n)^n is an increasing sequence. Together with the result (1+1/n)^(n+1) is a decreasing sequence, the first sequence is bounded above by any element of the second sequence and vice versa, so both sequences have a limit, and since the ratio of the two sequences goes to 1, they have a common limit.
With a little calculus you can get a simpler argument: It suffices to show that 3^(1/n) > 1/n + 1 for any n in Z+. But this follows from 1) 3^(1/x) is monotonically decreasing on (0,inf) since its derivative (-ln 3)(3^(1/x))/x^2 is negative on positive R. And 2) for n=1, 3^(1/1) > 1/1+1 = 2, which is obviously true. Thus 3^(1/n) > 1/n+ 1 for n>=1, => 3= (3(1/n))^n) > (1+1/n)^n since 1+1/n and 3 are both positive real numbers greater than one. QED
Unless I am misunderstanding something your proof it doesn’t work. I will use the same reasoning as you are and show something obviously false with it. With that reasoning we can say 3^(1/x) is decreasing and for n = 1, 3^(1/n) > n therefore 3^(1/n) > n for all n in Z+ which is obviously false. I think showing that 3^(1/n) > 1 + 1/n is a lot harder than that.
Simply do the binomial expansion of (1+x/n)^n = 1 + nx/n + n(n-1)(x/n)^2/2! + n(n-1)(n-2)(n-3)(x/n)^3/3! + ... + n!(x/n)^n/n!. That is term-by-term less than 1 + x + x^2/2! + x^3/3! + ... x^n/n! which is less than e^x. So setting x = 1, we find (1+1/n)^n < e^1 = e < 3.
You don't actually need the Bernoulli inequality for proving the lemma. You can also do it this way: ((n+1)/n)^(n+1) / ((n+2)/(n+1))^(n+2) = ((n²+2n+1)/(n²+2n))^(n+1) (n+1)/(n+2) = (n²+2n+1)/(n²+2n) (n²+2n+1)/(n²+2n) ..... (n²+2n+1)/(n²+2n) (n+1)/(n+2), where the first factor appears (n+2) times in the product. Now we use that this factor is obviously > 1 and that for such a fraction (with positive numerator and denominator), the value of the fraction becomes greater if we decrease both the numerator and the denominator by 1. So we have ((n+1)/n)^(n+1) / ((n+2)/(n+1))^(n+2) > (n²+2n+1)/(n²+2n) (n²+2n)/(n²+2n-1) ..... (n²+n+1)/(n²+n) (n+1)/(n+2), and now in this product of all the (n+1) fractions, lots of the denominators and numerators will cancel, leaving us with ((n+1)/n)^(n+1) / ((n+2)/(n+1))^(n+2) > (n²+2n+1)/(n²+n) (n+1)/(n+2) = (n²+2n+1)(n²+2n) > 1.
I suggest a different proof: consider f[x]= ln[1+1/x)^x=x*ln(1+1/x] .It is not difficult to show f[x] is increasing in x for x≥ 1. If we denote y = 1/x then we can use De L'Hopital's rule to see that lim f[y] for y -> 0 = 1 =lim f[x], x-> inf. Hence ln((1+1/x)^x )1 that means that for any n>1 (1+1/n)^n < e
Mathematics often puts up a lot of scaffolding which they then remove and just magically pull out og the hat in the proof. Short and sweet but it hides the chain of though for us. In a previous video he used the term "scratch work" and this is exactly the kind of scaffolding I'm talking about.
why did you create a seperate case for numbers less then 5? is there some rule that didn't work for them? edit: ah right they are more then three when you increase the power by 1 so your proof doesn't work for them
Why is n=5? The statements seem to be chosen completely random: e.g. it's somehow ^(n+1) and not ^n in the lemma. I always feel discouraged to learn math after seeing such things...
The example you have given shows why math is not just about learning formula and applying them but also involves a great deal of creativity and "outside the box thinking" which only comes from extended study and solving many problems. If you look at the olympiad problems, most of them require some form of creative thinking, be it some extra geometric construction or considering a seemingly unrelated case which ends up being relevant to the problem at hand. The point is not to feel discouraged but embrace the fact that not every solution is straightforward and may involve some seemingly unrelated (but mathematically sound and valid) steps which tie in together beautifully at the end. That is one of the biggest joys of doing math. Happy learning 😊😊
Can we do log base n on both sides then expand the log that remains using a Taylor series, and say something about the left side looking like a truncated Taylor series?
I think your lemma counts as the induction step and then you just need to show the base case of n = 1 giving us less than 3 and you can claim by induction hypothesis the theorem is true.
btw (1+1/n)^n is always less than 3 is also because of how it converges to one constant the higher the value of n. Here, as n approaches infinity, the expression is equal to e(about 2.71) and is less than 3.
I don’t think it’s a valid proof. Basically you’re saying that this expression converges to e which is less than 3, which sounds like circular reasoning
@@ziem24 Moreover, even if you use the fact that the sequence converges to e < 3, that doesn't guarantee that every element of the sequence is less than 3. You need something more to show that there isn't some small, finite n such that (1 + 1/n)^n > 3. The proof in the video achieves this, but there are many other approaches.
Oh, I see why you had to split the proof for n>5. 5 is the smallest n such that (1+1/n)^(n + 1) is less than 3. All other n < 5 make (1+1/n)^(n + 1) be larger than 3.
This is a bad video because it does NOT show the proving process. How the equations were figured out, how Bernoulli was selected, why the magic n=5, etc. It is just a video presentation of a completed proof, and unfortunately a big missed opportunity to actually teach math solving approach to a problem. Always teach the method, the approach, not the solution.
This is a bad comment for an excellent video. A math proof works exactly as shown in the video. The solving process is impossible to be shown in a UA-cam video. You have to study Math for years to get the knowledge needed for even only think to try and prove a math theorem like this. Mathematicians spent years or even centuries to solve some Math problems and now you pretend a video that shows the thinking process of a mathematician. LOL
I like the shorter format though. The inequality itself can be another video eventually we'll have the whole of Analysis in bite size videos here. ❤ Sure I would have liked a bit of expansion on why 5. But that can also be another video. Keep me watching. 😂
My proof: It's the limit definition of e without the limit. Plug in bigger and bigger numbers and it will approach e from the left so it will never get bigger than it, and e ≈ 2.718 < 3.
This is circular. One usually proves "it is the limit definition of e" by first establishing that the sequence is bounded and increasing, and thus converges to some number that we call e.
@coreyleander7911 No it is not circular because that is not the actual reason why this limit was discovered to approach e, or at least not the only way. You can prove it kind of like finding compound interest to compute e and recognizing a pattern that turns into this formula. And finding that's it's bounded and increasing doesn't make it circular either, we know those things right away in proving it.
I like this style of video!! Its nice that you speed up the 'uninteresting' parts, and you are good at explaining your arguments in a short and concise way
The limit is e as n approaches infinity. Great proof
For that to count as proof, You also need to demonstrate that the function is increasing over that entire interval.
Having a high bound is not proof of convergence. See sine. You need also monotonic increasing. Also having a limit does not prove that that is e.
@@tiborfutotabletare you sure? prove it
@@salem1599 Example: f(x) = -1/x + 3, x>1. This has high bound of 3, even increases monotonically, yet it still converges to 3, not e, when x - > infinity. By your comment, you have no clue about limit and convergence criterias...
@@tiborfutotablet bro he never said the contrary, but he is right it does converge to e
Since limit of (1+1/x)^n, as n→∞, is e. And, 2 < e < 3. Since n is finite (n < ∞), intuitively it could be concluded that even for larger n’s, it is less than 3, so for smaller n's it ofcourse follows that.
As I suspect you're aware, a similar argument shows (1+1/n)^n is an increasing sequence. Together with the result (1+1/n)^(n+1) is a decreasing sequence, the first sequence is bounded above by any element of the second sequence and vice versa, so both sequences have a limit, and since the ratio of the two sequences goes to 1, they have a common limit.
Nice proof, indeed! I expected you to apply induction.
With a little calculus you can get a simpler argument: It suffices to show that 3^(1/n) > 1/n + 1 for any n in Z+. But this follows from 1) 3^(1/x) is monotonically decreasing on (0,inf) since its derivative (-ln 3)(3^(1/x))/x^2 is negative on positive R. And 2) for n=1, 3^(1/1) > 1/1+1 = 2, which is obviously true. Thus 3^(1/n) > 1/n+ 1 for n>=1, => 3= (3(1/n))^n) > (1+1/n)^n since 1+1/n and 3 are both positive real numbers greater than one. QED
Unless I am misunderstanding something your proof it doesn’t work. I will use the same reasoning as you are and show something obviously false with it.
With that reasoning we can say 3^(1/x) is decreasing and for n = 1, 3^(1/n) > n therefore 3^(1/n) > n for all n in Z+ which is obviously false. I think showing that 3^(1/n) > 1 + 1/n is a lot harder than that.
Last step is incorrect.
@@HoSza1 right both of you are...let me try again
@@ryznak4814 back to the drawing board
@@SigmaChuck by the way i also tried to find a more simple solution, but it remained elusive to my silly self this time.
Simply do the binomial expansion of (1+x/n)^n = 1 + nx/n + n(n-1)(x/n)^2/2! + n(n-1)(n-2)(n-3)(x/n)^3/3! + ... + n!(x/n)^n/n!.
That is term-by-term less than 1 + x + x^2/2! + x^3/3! + ... x^n/n! which is less than e^x.
So setting x = 1, we find (1+1/n)^n < e^1 = e < 3.
I like this solution!
thanks for the vids, in my final year of math undergrad and love just keeping my thunker fresh
You don't actually need the Bernoulli inequality for proving the lemma. You can also do it this way:
((n+1)/n)^(n+1) / ((n+2)/(n+1))^(n+2) = ((n²+2n+1)/(n²+2n))^(n+1) (n+1)/(n+2)
= (n²+2n+1)/(n²+2n) (n²+2n+1)/(n²+2n) ..... (n²+2n+1)/(n²+2n) (n+1)/(n+2),
where the first factor appears (n+2) times in the product. Now we use that this factor is obviously > 1 and that for such a fraction (with positive numerator and denominator), the value of the fraction becomes greater if we decrease both the numerator and the denominator by 1. So we have
((n+1)/n)^(n+1) / ((n+2)/(n+1))^(n+2)
> (n²+2n+1)/(n²+2n) (n²+2n)/(n²+2n-1) ..... (n²+n+1)/(n²+n) (n+1)/(n+2),
and now in this product of all the (n+1) fractions, lots of the denominators and numerators will cancel, leaving us with
((n+1)/n)^(n+1) / ((n+2)/(n+1))^(n+2) > (n²+2n+1)/(n²+n) (n+1)/(n+2) = (n²+2n+1)(n²+2n) > 1.
Nice proof.
Great proof my man, you are the goat ❤❤
I suggest a different proof: consider f[x]= ln[1+1/x)^x=x*ln(1+1/x] .It is not difficult to show f[x] is increasing in x for x≥ 1.
If we denote y = 1/x then we can use De L'Hopital's rule to see that lim f[y] for y -> 0 = 1 =lim f[x], x-> inf. Hence ln((1+1/x)^x )1 that means that for any n>1 (1+1/n)^n < e
Why do these videos kinda give me a 2000s vibe? And why do I like it?
Fantastic proof, thank you.
Mathematics often puts up a lot of scaffolding which they then remove and just magically pull out og the hat in the proof. Short and sweet but it hides the chain of though for us. In a previous video he used the term "scratch work" and this is exactly the kind of scaffolding I'm talking about.
why did you create a seperate case for numbers less then 5? is there some rule that didn't work for them?
edit: ah right they are more then three when you increase the power by 1 so your proof doesn't work for them
Great Video
Are you sure this is the easiest way to prove the theorem? 😁
Loved it, did you choose 5 because it's the first number where (1 + 1/n)^{n + 1} is less then 3 ?
Yes he did
Why is n=5? The statements seem to be chosen completely random: e.g. it's somehow ^(n+1) and not ^n in the lemma. I always feel discouraged to learn math after seeing such things...
The example you have given shows why math is not just about learning formula and applying them but also involves a great deal of creativity and "outside the box thinking" which only comes from extended study and solving many problems. If you look at the olympiad problems, most of them require some form of creative thinking, be it some extra geometric construction or considering a seemingly unrelated case which ends up being relevant to the problem at hand.
The point is not to feel discouraged but embrace the fact that not every solution is straightforward and may involve some seemingly unrelated (but mathematically sound and valid) steps which tie in together beautifully at the end. That is one of the biggest joys of doing math. Happy learning 😊😊
Can we do log base n on both sides then expand the log that remains using a Taylor series, and say something about the left side looking like a truncated Taylor series?
Good one. Did anybody work out a proof by induction? I tried for a while and gave up and theorems about convergent monotonic sequences to show it.
I think your lemma counts as the induction step and then you just need to show the base case of n = 1 giving us less than 3 and you can claim by induction hypothesis the theorem is true.
btw (1+1/n)^n is always less than 3 is also because of how it converges to one constant the higher the value of n. Here, as n approaches infinity, the expression is equal to e(about 2.71) and is less than 3.
I don’t think it’s a valid proof. Basically you’re saying that this expression converges to e which is less than 3, which sounds like circular reasoning
@@ziem24 Moreover, even if you use the fact that the sequence converges to e < 3, that doesn't guarantee that every element of the sequence is less than 3. You need something more to show that there isn't some small, finite n such that (1 + 1/n)^n > 3. The proof in the video achieves this, but there are many other approaches.
The real proof: this limit f of n as n approches infinity is famous and it tends to e, which is approximately 2,71... which is less than 3.
I LI EK? Shouldn't that be I LI KE?
I used concept of limits and expansion of e to prove that its value wpuld tend to e which is less than zero
e is less than zero? Seriously? :D You probably meant "less than 3"?
i’d rather show it’s lower than e…
Oh, I see why you had to split the proof for n>5.
5 is the smallest n such that (1+1/n)^(n + 1) is less than 3. All other n < 5 make (1+1/n)^(n + 1) be larger than 3.
If you hadn't split it and applied the lemma with a=1 and b=n you could have only proved that (1+1/n)^n is less than 4.
You like complex stuff.
This is a bad video because it does NOT show the proving process. How the equations were figured out, how Bernoulli was selected, why the magic n=5, etc. It is just a video presentation of a completed proof, and unfortunately a big missed opportunity to actually teach math solving approach to a problem. Always teach the method, the approach, not the solution.
This is a bad comment for an excellent video. A math proof works exactly as shown in the video. The solving process is impossible to be shown in a UA-cam video. You have to study Math for years to get the knowledge needed for even only think to try and prove a math theorem like this. Mathematicians spent years or even centuries to solve some Math problems and now you pretend a video that shows the thinking process of a mathematician. LOL
man really just said LOL because he was so unsatisfied
Goofy@@DanishHafiz-gt4sq
@@johan2346u i aint at that time bro
I like the shorter format though. The inequality itself can be another video eventually we'll have the whole of Analysis in bite size videos here. ❤
Sure I would have liked a bit of expansion on why 5. But that can also be another video. Keep me watching. 😂
Nice proof, but you can also prove it using lhopitals by rephrasing the functions into a form of 0/0
My proof:
It's the limit definition of e without the limit. Plug in bigger and bigger numbers and it will approach e from the left so it will never get bigger than it, and e ≈ 2.718 < 3.
Exactly what I thought 😂
@@TheGreggv yeah that was pretty obvious just looking at the thumbnail: "of course e
This is circular. One usually proves "it is the limit definition of e" by first establishing that the sequence is bounded and increasing, and thus converges to some number that we call e.
@coreyleander7911 No it is not circular because that is not the actual reason why this limit was discovered to approach e, or at least not the only way. You can prove it kind of like finding compound interest to compute e and recognizing a pattern that turns into this formula. And finding that's it's bounded and increasing doesn't make it circular either, we know those things right away in proving it.
@@maxhagenauer24 he's not wrong
You wanna say the definition of e figures into your proof then you still have to prove e is less than 3
n=-1, dividing by zero??? Wtf is this channel bro
n=-1 is not a positive integer. The theorem only holds for positive integers. No one is dividing by zero. This channel is great.
go to school kid
Bro who in the world writes Z+ , write N instead, holy crap. Read it is Z*, my bad but really poor choice of letters
@@adamfurlong4979 just use the fact that ln(1+1/n) < = 1/ n and the exercice is over (because e