# Prove derivative of an odd function is even and vise versa.

• yungman
In summary, the student is trying to prove that if a function is even, then the derivative is cosine and if the function is odd, then the derivative is sine. However, the proof is using series expansions and limits which may not be the most efficient way to go about proving the equation.

## Homework Statement

Prove if f(x) is defined on -L< x < L, and if f(x) is odd function on (-L,L), the f'(x) is even function. and vise versa.

## Homework Equations

Using Fourier series expansion, f(x) is odd function on (-L,L) can be represented by Fourier sine series expansion. If f(x) is even function on (-L,L), the it can be represented by cosine series.

## The Attempt at a Solution

By series expansion of odd and even function, if f(x) is odd and represented by sine series, then the derivative of the sine series is cosine series and it become an even function. And the reverse is true from even to odd by derivative.

Anyone have a better way of proving this than what I have instead of referring to series expansion?

You could use the limit definition of f'(-x) to show that it is equal to f'(x)

Let $n(x) = -x$ so basically just a "negation function". To say that f is even is to say f(x)=f(-x) or in other words $f = f \circ n$, and to say it's odd is to say -f(x)=f(-x) or in other words $-f = f\circ n$.

Now use:
$$(f \circ n)(x)' = f'(n(x)) n'(x) = -f'(-x)$$
in both cases.

rasmhop said:
Let $n(x) = -x$ so basically just a "negation function". To say that f is even is to say f(x)=f(-x) or in other words $f = f \circ n$, and to say it's odd is to say -f(x)=f(-x) or in other words $-f = f\circ n$.

Now use:
$$(f \circ n)(x)' = f'(n(x)) n'(x) = -f'(-x)$$
in both cases.

Is $$f\circ n = f(n(x))=f(-x)?$$

Last edited:
Yes it is

rasmhop said:
Let $n(x) = -x$ so basically just a "negation function". To say that f is even is to say f(x)=f(-x) or in other words $f = f \circ n$, and to say it's odd is to say -f(x)=f(-x) or in other words $-f = f\circ n$.

Now use:
$$(f \circ n)(x)' = f'(n(x)) n'(x) = -f'(-x)$$
in both cases.

Let me clarify:

A) Let f(x) be even:

$$\Rightarrow f(-x)=f(x)$$

$$f(-x)= f(n(x)) =f(x)\Rightarrow [f(n(x))]'=f'(x)$$ (1)

$$[df(n(x))]' =f'(nx)n'(x)=-f'(n(x))=-f'(-x)$$ (2)

(1) and (2)$$\Rightarrow f'(x)=-f'(-x)$$

Therefore if f(x) is even, f'(x) is odd.

B) Let f(x) be odd:

$$\Rightarrow f(-x)=-f(x)$$

$$f(-x)= f(n(x)) =-f(x)\Rightarrow [f(n(x))]'=-f'(x)$$ (3)

$$[df(n(x))]' =f'(nx)n'(x)=-f'(n(x))=-f'(-x)$$ (4)

(3) and (4)$$\Rightarrow -f'(x)=-f'(-x)$$

Therefore if f(x) is odd, f'(x) is even

Tell me whether I got this right? Does this apply to integration that integrate of an odd function is an even function plus a constant and vise versa?

Thanks

Alan

yungman said:
Tell me whether I got this right?
Yes you did.

Does this apply to integration that integrate of an odd function is an even function plus a constant and vise versa?

Given a function f(x) that is integrable on the appropriate intervals define:
$$F(x) = \int_0^x f(t) dt$$
Then you wish to show that f(x) is odd imply F(x) even, and f(x) even imply F(x) odd. In general we have the identity:
$$\int_a^b f(t) dt = -\int_{-a}^{-b}f(-t)dt$$
You can use that to write:
$$F(-x) = \int_0^{-x} f(t)dt = -\int_0^x f(-t) dt$$
Now you can use f(-t)=f(t) or f(-t)=-f(t) depending on whether you're in the odd or even case to investigate how F(-x) relates to F(x).

rasmhop said:
Yes you did.

Given a function f(x) that is integrable on the appropriate intervals define:
$$F(x) = \int_0^x f(t) dt$$
Then you wish to show that f(x) is odd imply F(x) even, and f(x) even imply F(x) odd. In general we have the identity:

$$\int_a^b f(t) dt = -\int_{-a}^{-b}f(-t)dt$$
You can use that to write:
$$F(-x) = \int_0^{-x} f(t)dt = -\int_0^x f(-t) dt$$
Now you can use f(-t)=f(t) or f(-t)=-f(t) depending on whether you're in the odd or even case to investigate how F(-x) relates to F(x).

Thanks for your help, I need to verify again.

Let u=-t, therefore du=-dt. Also the integration of t from -a to -b imply integration of u from a to b.

$$\Rightarrow -\int_{-a}^{-b}f(-t)dt = \int_{a}^{b}f(u)du = F(b)-F(a)$$

$$\int_{a}^{b}f(t)dt = F(b)-F(a)\Rightarrow \int_a^b f(t) dt = -\int_{-a}^{-b}f(-t)dt = F(b)-F(a)$$

$$\Rightarrow F(x)=\int_0^x f(t)dt = -\int _0^{-x}f(-t)dt$$

If f(x) is even, f(-t)=f(t):

$$\Rightarrow F(x)=\int_0^x f(t)dt = -\int _0^{-x}f(-t)dt = -\int _0^{-x}f(t)dt = -[F(-x)-F(0)] = -F(-x)$$

This give F(x)=-F(-x) which imply if f(x) is even, F(x) is odd. Therefore integration of an even function gives an odd function.

There other way would be the same that integrate of an odd function give an even function.

Am I correct?

## 1. What is an odd function?

An odd function is a mathematical function that satisfies the property f(-x) = -f(x) for all values of x. This means that the function is symmetric about the origin and has a graph that is rotated 180 degrees about the origin.

## 2. What is an even function?

An even function is a mathematical function that satisfies the property f(-x) = f(x) for all values of x. This means that the function is symmetric about the y-axis and has a graph that is unchanged when reflected over the y-axis.

## 3. How do you prove that the derivative of an odd function is even?

To prove that the derivative of an odd function is even, we can use the definition of an odd function and the properties of derivatives. We start by using the definition of an odd function, f(-x) = -f(x), and then take the derivative of both sides. This will result in -f'(x) = f'(-x). We can then use the property of even functions, f(-x) = f(x), to simplify the right side to f'(x). Therefore, we have shown that -f'(x) = f'(x), which means that the derivative of an odd function is even.

## 4. Can you provide an example of an odd function?

Yes, one example of an odd function is f(x) = x^3. We can see that it satisfies the property f(-x) = -f(x) by plugging in -x for x and simplifying to -(-x)^3 = -x^3. The graph of this function is also symmetric about the origin, confirming that it is an odd function.

## 5. Can you prove that the derivative of an even function is odd?

Yes, we can prove that the derivative of an even function is odd by following a similar process as in question 3. We start with the definition of an even function, f(-x) = f(x), and take the derivative of both sides. This will result in f'(-x) = f'(x). We can then use the property of odd functions, f(-x) = -f(x), to simplify the left side to -f'(x). Therefore, we have shown that f'(x) = -f'(x), which means that the derivative of an even function is odd.