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?
From my workings, I believe I am correct? I can't see any errors myself, so was wondering if you could help me.Ray Vickson said:What do YOU think? Have you made any errors?
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.
Yes it does add up. So I am correct?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.
j3dwards said:Yes it does add up. So I am correct?
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.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'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?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.
j3dwards said:I'm not really sure how to use a(-x) = a(x) or -a(x) to check.
Okay perfect, thank you so much.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
Splitting a function into odd and even parts allows us to separate the function into two parts that each have different properties. This can make it easier to analyze the function and understand its behavior.
In order to split a function into odd and even parts, we first need to identify if the function is odd, even, or neither. If the function is odd, we can use the property f(-x) = -f(x) to split the function into odd and even parts. If the function is even, we can use the property f(-x) = f(x) to split the function. If the function is neither odd nor even, we cannot split it into odd and even parts.
The graph of an odd function is symmetric about the origin, meaning that if we fold the graph in half along the y-axis, the two halves will be a mirror image of each other. The graph of an even function is symmetric about the y-axis, meaning that if we rotate the graph 180 degrees around the origin, it will look the same as the original graph.
No, not all functions can be split into odd and even parts. Only functions that satisfy the properties of odd and even functions can be split into odd and even parts. If a function does not satisfy these properties, it cannot be split into odd and even parts.
Splitting a function into odd and even parts can be useful in signal processing, where odd and even parts can represent different types of signals. It can also be useful in solving differential equations, as splitting the function can simplify the equations and make them easier to solve.