Cosine function problem

[Meital]

I am trying to solve the following problem:

Show that | cos x | <= 1, that is, I am trying to show that the absolute value of cos x is less than or equal to 1 for all values of x, assuming that the only thing I know is the series of the function cosine, that is, just using the following I need to be able to solve the problem:

 

cos x = 1 + Sum ( -1)^n (x^2n)/2n! the sum for n = 1 to infinity

I am also trying to prove the same results for

sin x = sum (-1)^n-1 x^(2n-1) / (2n - 1)!

 

For the cos x, I broke it into 3 parts, one for x < 1, x = 1, x >1.

For x < or = 1, we have cos x = 1 + sum (-1)^n* x^2n/2n!, and we know this is the alternating series and it gives 1 + ( the value of the sum = -1 + e^x), but for x < or = 1, the e^x goes to 0, so we end up having |cos x| < or = to 1 for x < or = to 1. Is this right? Also, how do I show that for x > 1. Please help me. Thanks in advance.

[5614]

Draw the cosine graph....

[Meital]

I am not allowed to use graphs or anything, it is for Advanced Cal class..all we know is the series of the cosine and sine functions, then using that and convergence of series I guess, we need to show | cos x| < or = to 1. I know it would be so obvious to use graphs! But unfortunately I cannot use them here.

 

assuming that the only thing I know is the series of the function cosine' date=' that is, just using the following I need to be able to solve the problem:

 

cos x = 1 + Sum ( -1)^n (x^2n)/2n! the sum for n = 1 to infinity

I am also trying to prove the same results for

sin x = sum (-1)^n-1 x^(2n-1) / (2n - 1)!.[/quote']

[Dave]

My thoughts would be to first use the triangle inequality to split up the sum and the 1. Beyond that, I haven't really had many ideas.

[Johnny5]

I am trying to solve the following problem:

Show that | cos x | <= 1' date=' that is, I am trying to show that the absolute value of cos x is less than or equal to 1 for all values of x, assuming that the only thing I know is the series of the function cosine.[/quote']

 

In the XY plane, think about the cosine squared graph using calculus. It's maxima have y coordinate 1, and its minima have y coordinate 0.

 

I think that if you can prove that the cosine squared series is necessarily greater than zero, and less than or equal to one, it will necessarily follow that the cosine graph is trapped between y= -1 and y=1, and there is a formula for multiplication of one series by another, so that is what you need.

 

Since you are given the cosine series, you can multiply the cosine series by itself using the formula, and get the series for the cosine squared. So try approaching the problem from this direction... series multiplication.

 

After you have the formula for the cosine squared series, prove the following statement using the pythagorean theorem:

 

Sine^2 + cosine^2 = 1

 

Therefore the cosine series squared will be equivalent to one minus the sine series squared, i.e.

 

(cos X)^2 = 1 - (sin X)^2

 

Somewhere in the logic of this suggestion, is an answer, I almost guarantee it.

 

Regards

[dryga]

I think Cauchy series multiplication would get messy very fast.

 

The easiest way to show it, I think, is to first show that [math] \cos x = \frac{e^{ix} + e^{-ix}}{2}[/math] and [math] \sin x = \frac{e^{ix} - e^{-ix}}{2i}[/math] (you just use the Taylor series for [math]e^x[/math]). Once you have the functions written that way, it's easy to show that [math]\sin^2 x + \cos^2 x = 1[/math], and it'll follow that the absolute values for both sine and cosine are smaller than or equal to one.

[dryga]

Okay, so maybe I should start checking how old the threads I reply to are.

[Dave]

Not a problem That one didn't seem to have a solid answer, and it's always good for these kind of threads to have one.