Splitting function into odd and even parts

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SUMMARY

The discussion focuses on splitting the function f(x) = e^x + πe^(-x) into its odd and even components, utilizing the definitions of hyperbolic functions. The even part is correctly identified as a(x) = (1 + π) cosh(x), while the odd part is b(x) = (1 - π) sinh(x). Participants emphasize the importance of verifying that a(x) is even and b(x) is odd, ensuring that f(x) equals the sum of a(x) and b(x). The method for checking these properties involves substituting -x into the functions and simplifying.

PREREQUISITES
  • Understanding of hyperbolic functions, specifically cosh(x) and sinh(x)
  • Familiarity with the concepts of odd and even functions
  • Basic knowledge of function manipulation and algebraic simplification
  • Experience with mathematical proofs and verification techniques
NEXT STEPS
  • Study the properties of odd and even functions in depth
  • Learn how to derive and manipulate hyperbolic functions
  • Explore function decomposition techniques in calculus
  • Practice verifying function properties through substitution and simplification
USEFUL FOR

Students studying calculus, particularly those focusing on function analysis, as well as educators teaching mathematical concepts related to odd and even functions and hyperbolic functions.

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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