The Bernoulli inequality actually is (1 + x)ⁿ > 1+ nx for x≥-1, x≠0, n≥2. So the sequence is strictly increasing. Proof by induction. For the base n=2, this is 1 + 2x + x² > 1 + 2x. For the step of induction, since (1+x)≥0, (1+x)ⁿ⁺¹ ≥ (1+nx)(1+x) = 1 + (n+1)x + nx² > 1 + (n+1)x.
In analysis every injective continuous function over a given interval [a,b] is monotonic. The proof is simple. For simplicity we can suppose that f(b) ≥ f(a) and prove that f is increasing. Let's suppose that f is Not increasing. Then there are 2 points X and y where a
Bernoulli's approximation helps figure out sequences like an and an+1 to see if it is increasing for an arbitrary n. You explained using the >= 1 + nx so clear and clean in your proof that I'll find whenever I have to check a sequence by Bernoulli's Principle I can also see if an n+1 term is still greater than a n term result. 👍
@@robertveith6383 ... There is no technical English classes agreeing to your statements in math. Keep going to school in the higher math courses. I doubt you will be using parentheses like what you propose on your homework, tests and understanding what professors write on chalkboards and class viewing screens.
0:52 it is not obvious that the sequence is increasing, the only thing that is obvious is that the first several terms are increasing, anything can happen after that . . .
@@tulsaken2754 It's not about that function. This is the rule that a differentiable function is increasing when its derivative is non-negative, meaning greater than or equal to zero. Back to this function The derivative of the function f(x) = g(x)^h(x) is calculated using the identity g(x)^h(x) = e^(ln(g(x)^h(x))) = e^(h(x)*ln(g(x))), and by applying the chain rule to e^z function. (e^z)'=z'*e^z, z=h(x)*ln(g(x))
Oh, shoot, now I get it. The whole rational expression is the "x" and you're making it look like the x term in the right hand side of Bernoulli's inequality. I feel foolish. Luckily it's a familiar feeling for me.
(1+1/n)^n=exp(n*log(1+1/n) and you can prove that limit n*log(1+1/n)=1 Easy way; n*log(1+1/n)=(log(1+1/n)-log(1+0))/1/n so the limit is the derivative of x->log(1+x) evaluated in x=0.
The Bernoulli inequality actually is
(1 + x)ⁿ > 1+ nx for x≥-1, x≠0, n≥2. So the sequence is strictly increasing.
Proof by induction.
For the base n=2, this is
1 + 2x + x² > 1 + 2x.
For the step of induction, since (1+x)≥0,
(1+x)ⁿ⁺¹ ≥ (1+nx)(1+x)
= 1 + (n+1)x + nx²
> 1 + (n+1)x.
Newton's binomial gives a direct proof. (1+x)^n=1+nx+something positive if x>=0
By the way, there is a fourth way to prove it: by contradiction. If the sequence is NOT monotone increasing while we definitely have a₁
For all yx then y
As a bonus show that this limit is bounded
and we will be able to conclude that this sequence is convergent
In analysis every injective continuous function over a given interval [a,b] is monotonic.
The proof is simple.
For simplicity we can suppose that f(b) ≥ f(a) and prove that f is increasing.
Let's suppose that f is Not increasing. Then there are 2 points X and y where a
My favourite quasi-monotonic function is f(x) = x^3. It’s increasing almost all the time, except when turning the corner at zero. 😂
Bernoulli's approximation helps figure out sequences like an and an+1 to see if it is increasing for an arbitrary n. You explained using the >= 1 + nx so clear and clean in your proof that I'll find whenever I have to check a sequence by Bernoulli's Principle I can also see if an n+1 term is still greater than a n term result. 👍
Write a(n) and a(n + 1) or a_n and a_(n + 1), instead of what you wrote.
@@robertveith6383 ... There is no technical English classes agreeing to your statements in math. Keep going to school in the higher math courses. I doubt you will be using parentheses like what you propose on your homework, tests and understanding what professors write on chalkboards and class viewing screens.
In fact the inequality you use is (1-x)^n>1-nx ,x
How can I show it if I have calculus
Cool and experienced
0:52 it is not obvious that the sequence is increasing, the only thing that is obvious is that the first several terms are increasing, anything can happen after that . . .
He said this :)
Whats an example of non-monotone sequence.
Sequence of cosines
@@PrimeNewtons Do you mean taylor series expansion or just a normal cosine sequence?
@@Orillians cos 1, cos 2, cos 3, .........
The derivative of f(x) must be greater than or equal to 0.
Can you do this derivative?
@@tulsaken2754
It's not about that function. This is the rule that a differentiable function is increasing when its derivative is non-negative, meaning greater than or equal to zero.
Back to this function
The derivative of the function f(x) = g(x)^h(x) is calculated using the identity g(x)^h(x) = e^(ln(g(x)^h(x))) = e^(h(x)*ln(g(x))), and by applying the chain rule to e^z function.
(e^z)'=z'*e^z, z=h(x)*ln(g(x))
In the sixth step you casually replaced the 1 in the numerator with n. It's not clear to me why you can do that. Will someone explain? ☺️
Multiply by n using Bernoulli's inequality
Oh, shoot, now I get it. The whole rational expression is the "x" and you're making it look like the x term in the right hand side of Bernoulli's inequality. I feel foolish. Luckily it's a familiar feeling for me.
How to prove that equal e when n tends to ifinity
This equation defines e
(1+1/n)^n=exp(n*log(1+1/n) and you can prove that limit n*log(1+1/n)=1 Easy way; n*log(1+1/n)=(log(1+1/n)-log(1+0))/1/n so the limit is the derivative of x->log(1+x) evaluated in x=0.
Longest intro ever?
Almost. This is not the longest.
asnwer=1x