# Homework Help: Splitting function into odd and even parts

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1. Jan 13, 2016

### j3dwards

1. The problem statement, all variables and given/known data
Split the function f(x) = ex + πe−x into odd and even parts, and express your result in terms of cosh x and sinh x.

2. Relevant equations
f(x) = 0.5[f(x) + f(-x)] +0.5[f(x) - f(-x)]

3. The attempt at a solution
So i know that:

ex = 0.5[ex - e-x] + 0.5[ex + e-x] = sinh(x) + cosh(x)

So let the even part be a(x) and the odd be b(x).

a(x) = cosh(x) + πcosh(x) = (1 + π) cosh(x)
b(x) = sinh(x) - πsinh(x) = (1 - π) sinh(x)

Is this correct?

2. Jan 13, 2016

### Ray Vickson

What do YOU think? Have you made any errors?

3. Jan 13, 2016

### j3dwards

From my workings, I believe I am correct? I can't see any errors myself, so was wondering if you could help me.

4. Jan 13, 2016

### LCKurtz

All you have to do to check it yourself is check whether a(x) is even, b(x) is odd, and they add to your original function.

5. Jan 13, 2016

### j3dwards

Yes it does add up. So I am correct?

6. Jan 13, 2016

### Ray Vickson

Why do you need to ask? Is a(x) even? Is b(x) odd? Is it true that f(x) = a(x) + b(x)?

You should get in the habit of checking these things for yourself because in many situations (such as in exams) you cannot ask anybody else! Have some confidence in your own work.

7. Jan 13, 2016

### Thewindyfan

You can double check by seeing if a(-x) = a(x) or -a(x).

Same for b(x).

Make sure to go through your steps again and see if they make sense to you.

8. Jan 13, 2016

### j3dwards

I did check originally, but I'm just quite unsure about odd and even functions so I was just asking to make sure so that for that exam, I knew the correct method.

9. Jan 13, 2016

### j3dwards

I'm not really sure how to use a(-x) = a(x) or -a(x) to check. Do I literally just make the rhs of the equation equal to negative of what it is a see if it comes out with the same answer?

10. Jan 13, 2016

### Thewindyfan

Yep, you plug in -x and simplify the function as best as you can until you get either the original function or the opposite of the function

11. Jan 13, 2016

### j3dwards

Okay perfect, thank you so much.