# Identity for cosh

1. Feb 14, 2013

### aaaa202

Show that cosh(x) = 1 => x = 0

I am only allowed to use the definition of cosh, the algebraic rules for the exponential function, that exp(x2)>exp(x1) for x2 > x1, and the fact that we have defined it with the requirement:

exp(x) ≥ 1 + x

The exp(x) term of course is not trouble.

What I lack prooving is that:
exp(x) > 1 - x

Any ideas how I might approach it? I don't need answers, just hints

2. Feb 14, 2013

### CompuChip

If x > 0, then 2x > 0 so 1 + x > 1 + x - 2x.

If x < 0, then it's not true (exp goes to 0 while 1 - x goes to infinity for $x \to -\infty$).

3. Feb 14, 2013

### aaaa202

oh sorry I can see I made a fatal error.
I meant i need to show:
exp(-x) > 1 - x

4. Feb 14, 2013

### Staff: Mentor

A quick sketch of the graphs of y = e-x and y = 1 - x shows that e-x ≥ 1 - x for all real x, and that we get equality at only one value of x.

Assuming you can use techniques of calculus, let f(x) = e-x - (1 - x), and find the minimum value and where it occurs.

5. Feb 14, 2013

### aaaa202

I know calculus but this is an analysis course, where it has not yet been introduced. Thus I am to find another way of proving it.

6. Feb 14, 2013

### Staff: Mentor

Really? Are you saying that you haven't learned differentiation, and you're in an analysis course?

7. Feb 14, 2013

### aaaa202

I'm talking a rigorous analysis course where everything is built up slowly. So far we have covered continuity and the properties of limits. I know differential calculus, but I am not allowed to use it for this handout.

8. Feb 14, 2013

### vela

Staff Emeritus
If you know that $e^x \ge 1+x$ for all $x$, you can plug anything you want in for $x$. In particular, you can plug in $-x$.

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