Bernoulli's Inequality

In this video, I used Bernoulli's inequality to solve a size comparison problem. i also showed the basic derivation of the inequality using Newton's generalized binomial expansion theorem

[taegul]

Your smile really makes us happy. God bless you, brother!

[MadaraUchihaSecondRikudo]

Intuitively: This is the difference between normal percentage math and compounding interest. If I add 2% to 100 twice I get 100 + 2 + 2, this is the 1+nx, if I do the compounding version it's 100 + 2 + (2 and a bit more) because it's now 2% of 102 and not 100, this is the (1 + x)^n. So each term would have a little bit more than x added to it (and increasing with each pass)

[okemefulachidera]

If this man had been my maths teacher in my life maths would have never been a burden to me

Thanks you sir
I love your teaching method
Keep it up 👍

[kornelviktor]

The way I did it:
1.02=1+1/50 if you raise 1+1/50 to the 50th power you will approximately get e (by definition) which is (2.7182818...) but we have to square it since we have the 100th power. So on the left hand side we have approximately e^2 and the right hand side is just 2.8 so 1.02^100 is bigger by far surprisingly.

[majora]

I fully recognize that your "challenge" to answer the question without using a calculator or brute forcing it and without Bernoulli's inequality was rhetorical, but I gave it a go anyway and this is what I came up with. Taking the natural log of both sides preserves the inequality, so we're really looking at comparing ln(1.02^100) = 100*ln(1.02) and ln(2.8)

By using Taylor's Theorem we can derive an inequality to set an upper and lower bound on ln(1 + x). More specifically, we can show that x - x^2/2 <= ln(1 + x) <= x for any real number x >= 0. Plugging in x = 0.02 and x = 1.8 respectively yields:

ln(1.02) >= 0.0198 and thus 100*ln(1.02) >= 1.98
ln(2.8) <= 1.8

Since our lower bound on the left-hand side is strictly greater than our upper bound on the right-hand side, we can conclude that 1.02^100 > 2.8

[electric_chris]

10:34 That's an evil laugh if I've ever heard one

[dragonking]

Great videos to watch before sleep! Thanks! Love your work.

[artursoliyev]

In my collage we are asked to prove Bernoulli's inequality by using induction . Can I write this down to paper

[kethanraman]

Rule of 72. Theoretically 1.02^72 would be around 4. So we can safely assume 1.02^100 is much bigger

[eriliken]

I would legit pay to have you as my professor, I feel like i love math here and hate it in class

[marcgriselhubert]

The Bernoulli's inequality can also be proofed using convexity.

[temporarytemporary-fhdf]

When x is negative you should take three not two from left. And it's worth putting it because positive reals are just one case among two all the toough work is hidden in the other case.

[chaosredefined]

When you are proving Bernoulli's inequality, you show that (1 + x)^n = 1 + nx + stuff. You then claim that if we remove the stuff, it's smaller, so therefore (1 + x)^n >= 1 + nx. How do you know that the stuff is non-negative? The next term is non-negative (it's n(n-1) x^2 / 2, which is at least 0), but the term after that is n(n-1)(n-2) x^3 / 6, which may be negative if x is negative, and you did allow x's from -1 > x > 0.

Better approach:
Obviously, it's true for n=1, as that gives us (1 + x)^1 >= 1 + 1x, which is 1 + x >= 1 + x, which is the equality case.

Next, assume it's true for some natural number n=k. Therefore (1+x)^k >= 1 + kx. Multiply through by 1+x, we get (1+x)^(k+1) >= (1 + kx)(1+x) = 1 + kx + x + kx^2 = 1 + (k+1)x + kx^2. As x^2 >= 0 and k is a natural number, and therefore k > 0, this means (1+x)^(k+1) >= 1 + (k+1)x. Thus, if it's true for n=k, then it's true for n=k+1.

Thus, since it's true for n=1, set k=1 and we get that it's true for n=2. Set k=2 and we get that it's true for n=3. Thus, it's true by induction.

[fangliren]

Great video! The binomial expansion gives us the result we want for n>=2 (and sort of n=0, 1 for two special cases of 1=1 and 1+x=1+x). But do we know anything about negative integer values of n? It certainly seems true that the result holds for (x+1)^-1 and 1+(-1)x, ie that 1/(x+1) >= 1-x as long as x>-1. We won’t be able to explore these negative exponents as easily using the trick with the binomial expansion, but is there another way?

[RyanLewis-Johnson-wqxs]

I did it in my head without a calculator.

[pauljackson]

Can you use a similar method, for N=>2, and get (1 + x)^N >= 1 + N*x + N*(N+1)/2*x^2?
And can it be generalized to fractional N's using the gamma function to get factorials?

[BartBuzz]

Great refresher on Bernoulli's contributions! But which is larger !.02^100 or 4? There must be another math "trick" to show this without using a calculator.

[joaomane]

Hi, professor. Could you please make a video on Darboux's theorem? Thanks :D

[kimsanov]

How to show that left side is >= and not just > ?

[sciphyskyguy]

7:25 You didn’t forget to write the summation. You were using the Einstein convention in which a repeated index simply implies summation over that index. 😉