Splitting function into odd and even parts

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Homework Help Overview

The discussion revolves around splitting the function f(x) = e^x + πe^(-x) into its odd and even components, with a focus on expressing the results in terms of hyperbolic functions, specifically cosh(x) and sinh(x).

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the method of identifying the even and odd parts of the function using the definitions of hyperbolic functions. There are attempts to verify the correctness of the derived expressions for a(x) and b(x). Questions about the properties of even and odd functions are raised, particularly regarding how to check these properties.

Discussion Status

The discussion appears to be ongoing, with some participants expressing confidence in their calculations while others seek confirmation and clarification on the properties of even and odd functions. There is a focus on self-verification and understanding the underlying concepts.

Contextual Notes

Participants mention the importance of checking their work independently, especially in the context of exams, indicating a learning environment where self-assessment is encouraged.

j3dwards
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Homework Statement


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.

Homework Equations


f(x) = 0.5[f(x) + f(-x)] +0.5[f(x) - f(-x)]

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?
 
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j3dwards said:

Homework Statement


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.

Homework Equations


f(x) = 0.5[f(x) + f(-x)] +0.5[f(x) - f(-x)]

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?

What do YOU think? Have you made any errors?
 
Ray Vickson said:
What do YOU think? Have you made any errors?
From my workings, I believe I am correct? I can't see any errors myself, so was wondering if you could help me.
 
j3dwards said:
From my workings, I believe I am correct? I can't see any errors myself, so was wondering if you could help me.

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.
 
LCKurtz said:
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.
Yes it does add up. So I am correct?
 
j3dwards said:
Yes it does add up. So I am correct?

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.
 
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.
 
Ray Vickson said:
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.
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.
 
Thewindyfan said:
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.
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
j3dwards said:
I'm not really sure how to use a(-x) = a(x) or -a(x) to check.

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
Thewindyfan said:
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
Okay perfect, thank you so much.
 

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