in this video I used Bernoulli's inequality to show that a sequence in monotonic increasing. I took this approach because not all classes allow the use of derivatives to show that a sequence in monotone
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.
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.
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
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 . . .
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.
@@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))
(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.
