calc PROOFS?

Question

PROVE that 1-(x^2 / 2) < cosx for all x epsilon (0, infinity)

(i took the derivs of each and then set x>1 ..... is there another way to do this?)

Answer 1

The McLauren expansion for cos x about x = 0 is

cos x = 1 - (x^2/2!) + (x^4/4!) - .... + (-1)^(n-1)(x^2(n-1)/(2(n-1)!) + .....

Now Inth term| =(x^2(n-1)/(2(n-1)!) < 0 for all x (as x is raised always to an even power and k! is always positive.

Also |nth term| < |(n-1)th term| (can show this by a simple inductive proof)

Therefore (2n)th term - (2n+ 1)th > 0
Therefore cos x = 1 - (x^2/2!) + (x^4/4!) - .... + (-1)^(n-1)(x^2(n-1)/(2(n-1)!) + .....
= [1 - (x^2/2!)] +{[(x^4/4!) - x^6/6!] ... + [(-1)^(n-1)(x^2(n-1)/(2(n-1)! + (-1)^(n)(x^(2n)/(2n)!] + ... }
> 1 - (x^2/2!)
= 1 - x^2/2 QED

Answer 2

You know that the Taylor series expansion of cos(x) is:

cos(x) = 1 - x^2/2 + x^4/(4!) - x^6/(6!) +- . . .

This is an alternating series whose terms decrease to zero. This means that the difference between 1 - x^2/2 and cos(x) is bound by x^4/(4!). In other words, cos(x) is x^4/(4!) "higher" than 1-x^2/2. For x \in (0,\infty), x^4/(4!) > 0. Thus,

1 - x^2/2 < 1 - x^2/2 + x^4/(4!) - x^6/(6!) +- . . .

Therefore,

1 - x^2/2 < cos(x)

There you go.

Answer 3

prepare top here: ??(a² - u²) du = ½u ?(a² - u²) + ½a² arcsin(u/a) + C this could be a Trigonometric Substitution concern: ?(a² - b²x²) ? x = (a/b)sin? ??(a² - u²) du = ??(a² - (asin?)²)acos? d? = ??(a² - a²sin²?)acos? d? = ??(a²(a million - sin²?))acos? d? = ??(a²cos²?)acos? d? = ?(acos?)acos? d? = ?(a²cos²?) d? = a²?(a million - sin²?) d? = a²[? - ½(? - cos?sin?)] + C = a²[arcsin(u/a) - ½(arcsin(u/a) - (?(a² - u²)/a)u/a)] + C = a²[arcsin(u/a) - ½arcsin(u/a) + ½u?(a² - u²)/a²] + C = a²[½arcsin(u/a) + ½u?(a² - u²)/a²] + C = ½a² arcsin(u/a) + ½u?(a² - u²) + C [This has been shown] u = asin? ? sin? = u/a = O/H ? ? = arcsin(u/a) du = acos? d? cos? = A/H = ?(a² - u²)/a