Odd and even functions | Identify Integrate. Hello friends, today I’ll talk about the odd and even functions. Have a look!!

The odd and even functions and some of their properties are very important in the Fourier series of functions.

**Odd and even functions – identify them**

Let’s suppose I have a function . Now will be an odd function if . And will be an even function if . Next, I’ll give some examples.

Suppose . Now I’ll replace with . So it will be . And that means . Also, I know that . Therefore I can say that . Thus is an even function.

Now I’ll give another example. Let’s say . If I replace with , I get . And that means . Also, I know that . Therefore I can say that . Thus is an odd function.

Now these odd and even functions have some interesting properties such as:

**odd function × odd function ⇒ even function****odd function × even function ⇒ odd function****even function × even function ⇒ even function**

Next, I’ll give some examples of how to identify the odd and even functions.

**Solved examples of how to identify the odd and even functions**

Disclaimer: None of these examples is mine. I have chosen these from some book or books. The references are at the end of the post.

**Example 1**

According to Croft et al. (2000), “By using the properties of odd and even functions state whether the following function is odd, even or neither .”

**Solution**

Now here the function is, say,

As I can see, is a product of two functions and . First of all, I’ll check whether is an odd or even function.

So let’s say, . Next, I’ll replace with . So that gives

Thus I can say that . So is an odd function.

Now I’ll check whether is an odd or even function.

So let’s say, . Next, I’ll replace with . So that gives

Since , I can say that

Thus I can say that . So is an odd function.

As I can see, both and are odd functions. Thus the product of these two odd functions will be an even function. So is an even function.

Hence I can conclude that this is the solution to the given example. Now I’ll give you another example.

##### Example 2

According to Croft et al. (2000), “By using the properties of odd and even functions state whether the following function is odd, even or neither .”

**Solution**

Now here the function is, say,

As I can see, is a product of two functions and . First of all, I’ll check whether is an odd or even function.

So let’s say, . Next, I’ll replace with . So that gives

Thus I can say that . So is an odd function.

Now I’ll check whether is an odd or even function.

So let’s say, . Next, I’ll replace with . So that gives

Since , I can say that

Thus I can say that . So is an even function.

As I can see, is an odd function but and is an even function. Thus the product of one odd function and an even function will be an odd function. So is an odd function.

Hence I can conclude that this is the solution to the given example. Now I’ll give you another example.

##### Example 3

According to Croft et al. (2000), “By using the properties of odd and even functions state whether the following function is odd, even or neither .”

**Solution**

Now here the function is, say,

As I can see, is a product of two functions and . First of all, I’ll check whether is an odd or even function.

So let’s say, . Next, I’ll replace with . So that gives

Thus I can say that or . So is neither an odd nor an even function.

Now I’ll check whether is an odd or even function.

So let’s say, . Next, I’ll replace with . So that gives

Since , I can say that

Thus I can say that . So is an odd function.

As I can see, is neither an odd nor an even function but is an odd function. Thus the product of these two functions is neither odd nor an even function. So is neither odd nor even function.

Hence I can conclude that this is the solution to the given example.

Dear friends, this is the end of today’s post. Thank you very much for reading this. Please let me know how you feel about it. Soon I will be back again with a new post. Till then, bye, bye!!

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