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
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.
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)
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 = 0.0198 and thus 100*ln(1.02) >= 1.98 ln(2.8) 2.8
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?
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.
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?
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.
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.
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. 😉
I don't know how Bernoulli gets credit for Newton's Calculus that was doing the same thing including the creation of Pascal's Triangle in his difference tables on difference derivatives ∆^kf(x) written out in tables on Newton's differential Calculus. I am guessing because the "Probability" Parentheses to expand the Binomial Distribution is a different topic than the Newton Differential Table that would find the same results especially Pascal's Triangle results.
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.
That's very clever
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)
Exactly, that's what I did as well.
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 👍
Your smile really makes us happy. God bless you, brother!
10:34 That's an evil laugh if I've ever heard one
No, it's a laugh of joy!
Great videos to watch before sleep! Thanks! Love your work.
love your enthusiasm
The Bernoulli's inequality can also be proofed using convexity.
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 = 0.0198 and thus 100*ln(1.02) >= 1.98
ln(2.8) 2.8
wow!
good job!
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?
Hi, professor. Could you please make a video on Darboux's theorem? Thanks :D
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.
Yes. 1.02¹⁰⁰ = (1.02⁵⁰)². We know that 1.02⁵⁰ is very close to e (by its definition). Since e²>4 by a margin, 1.02¹⁰⁰ is gt 4 as well.
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?
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.
Rule of 72. Theoretically 1.02^72 would be around 4. So we can safely assume 1.02^100 is much bigger
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.
I usually prefer induction in such cases.
How to show that left side is >= and not just > ?
Trivially so. 3:06 set x=0.
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. 😉
I don't know how Bernoulli gets credit for Newton's Calculus that was doing the same thing including the creation of Pascal's Triangle in his difference tables on difference derivatives ∆^kf(x) written out in tables on Newton's differential Calculus. I am guessing because the "Probability" Parentheses to expand the Binomial Distribution is a different topic than the Newton Differential Table that would find the same results especially Pascal's Triangle results.
Newton stole many things like whole calculus from Leibniz
(1.02)^100 or 7.2 which is larger
Use calculator, bro. 1.02^100~7.22.
But we can calculate approximately ((1+1/50)^50)^2 =~e^2=7.4
@@ewavr I know but in my opinion 7.2 would be better than 2.8
1.02
Too many bernaullis