Uni Math

[nubs]

Hey, I've been sick for the last two weeks and have missed the first 8 lectures for Accelerated Mathematics :/
So yeah I thought I'd post some questions I'm having trouble with every now and again, at least until I catch up

Where x is an element of R, and n is an element of N, let E_n(x) = (1 + x/n)ⁿ

Prove 1-x²/n E_n(x)E_n(-x) 1 if n |x|

Thanks guys

[mark_alec]

What have you tried doing?

[nubs]

The lecture notes don't cover it very well so I am quite clueless on where to even begin

But I let n approach infinity and I got 0 < 1 <= 1
Where does that leave me?

[John President]

First half of the proof: First note than for all natural numbers, n^2 is greater than or equal to n.
Also, by expanding En(x)*En(-x), we get En(x)*En(-x) = 1 - {[x/n]^2)^n}.
Therefore, (x^2)/(n^2) (let's call this term a) is less than or equal to (x^2)/n (= term b)
Therefore, 1-(term b) is less than or equal to 1-(term a). As n is greater than or equal to 1, 1-(term a) is less than or equal to [(1-term a)^n].

Note that we're trying to prove that (1-term b) is less than or equal to [(1-term a)^n]. By combining these two expressions, we achieve this.



Second half of the proof: This part is simpler. En(x)*En(-x) = 1 - {[x/n]^2)^n}. The term {[x/n]^2)^n}is clearly greater than 0, so 1-{[x/n]^2)^n} < or equal to 1.

Hope this helped

[VivaTequila]

oh nibs