5.1 Even and odd functions  (Page 3/3)

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Square of an even or odd function

The square of even or odd function is always an even function.

Properties of derivatives

1: If f(x) is an even differentiable function on R, then f’(x) is an odd function. In other words, if f(x) is an even function, then its first derivative with respect to "x" is an odd function.

2: If f(x) is an odd differentiable function on R, then f’(x) is an even function. In other words, if f(x) is an odd function, then its first derivative with respect to "x" is an even function.

Composition of a function

Every real function can be considered to be composed from addition of an even and an odd function. This composition is unique for every real function. We follow an algorithm to prove this as :

Let f(x) be a real function for x $\in$ R. Then,

$f\left(x\right)=f\left(x\right)+f\left(-x\right)\right\}-f\left(-x\right)\right\}$

Rearranging,

$f\left(x\right)=\frac{1}{2}\left\{f\left(x\right)+f\left(-x\right)\right\}+\frac{1}{2}\left\{f\left(x\right)-f\left(-x\right)\right\}=g\left(x\right)+h\left(x\right)$

Now, we seek to determine the nature of functions “g(x)” and “h(x). For “g(x)”, we have :

$⇒g\left(-x\right)=\frac{1}{2}\left[f\left(-x\right)+f\left\{-\left(-x\right)\right\}\right]=\frac{1}{2}\left\{f\left(-x\right)+f\left(x\right)\right\}=g\left(x\right)$

Thus, “g(x)” is an even function.

Similarly,

$⇒h\left(-x\right)=\frac{1}{2}\left[f\left(-x\right)-f\left\{-\left(-x\right)\right\}\right]=\frac{1}{2}\left\{f\left(-x\right)-f\left(x\right)\right\}=-h\left(x\right)$

Clearly, “h(x)” is an odd function. We, therefore, conclude that all real functions can be expressed as addition of even and odd functions.

Even and odd extensions of function

A function has three components – definition(rule), domain and range. What could be the meaning of extension of function? As a matter of fact, we can not extend these components. The concept of extending of function is actually not a general concept, but limited with respect to certain property of a function. Here, we shall consider few even and odd extensions. Idea is to complete a function defined in one half of its representation (x>=0) with other half such that resulting function is either even or odd function.

Even function

Let f(x) is defined in [0,a]. Then, even extension is defined as :

|f(x); 0≤x≤a g(x) = || f(-x); -a≤x<0

The graphical interpretation of such extension is that graph of function f(x) is extended in other half which is mirror image of f(x) in y-axis i.e. image across y-axis.

Odd extension

Let f(x) is defined in [0,a]. Then, odd extension is defined as :

| f(x); 0≤x≤a g(x) = || -f(x); -a≤x<0

The graphical interpretation of such extension is that graph of function f(x) is extended in other half which is mirror image of f(x) in x-axis i.e. image across x-axis.

Exercises

Determine whether f(x) is odd or even, when :

$f\left(x\right)={e}^{x}+{e}^{-x}$

The function “f(x)” consists of exponential terms. Here,

$⇒f\left(-x\right)={e}^{-x}+{e}^{-\left(-x\right)}={e}^{-x}+{e}^{x}={e}^{x}+{e}^{-x}=f\left(x\right)$

Hence, given function is even function.

Determine whether f(x) is odd or even, when :

$f\left(x\right)=\frac{x}{{e}^{x}-1}+\frac{x}{2}$

The function “f(x)” consists of exponential terms. In order to check polarity, we determine f(-x) :

$f\left(-x\right)=-\frac{x}{{e}^{-x}-1}+\frac{-x}{2}=-\frac{x}{1/{e}^{x}-1}-\frac{x}{2}$

$⇒f\left(-x\right)=-\frac{x{e}^{x}}{1-{e}^{x}}-\frac{x}{2}$

We observe here that it might be tedious to reduce the expression to either “f(x)” or “-f(x)”. However, if we evaluate f(x) – f(-x), then the resulting expression can be easily reduced to simpler form.

$f\left(x\right)-f\left(-x\right)=\frac{x}{{e}^{x}-1}+\frac{x}{2}+\frac{x{e}^{x}}{1-{e}^{x}}+\frac{x}{2}$

$⇒f\left(x\right)-f\left(-x\right)=\frac{x}{{e}^{x}-1}-\frac{x{e}^{x}}{{e}^{x}-1}+x=\frac{x\left(1-{e}^{x}\right)}{{e}^{x}-1}+x=0$

Hence,

$f\left(x\right)=f\left(-x\right)$

It means that given function is an even function.

) How to check whether a pulse equation of the form

$y=\frac{a}{\left\{{\left(3x+4t\right)}^{2}+b\right\}}$

is symmetric or asymmetric, here "a" and "b" are constants.

Posted by Dr. R.K.Singhal through e-mail

The pulse function has two independent variables “x” and “t”. The function needs to be even for being symmetric about y-axis at a given instant, say t =0.

We check the nature of function at t = 0.

$⇒y=\frac{a}{\left(9{x}^{2}+b\right)}$

$⇒f\left(-x\right)=\frac{a}{\left\{9{\left(-x\right)}^{2}+b\right\}}=\frac{a}{\left(9{x}^{2}+b\right)}=f\left(x\right)$

Thus, we conclude that given pulse function is symmetric.

Determine whether f(x) is odd or even, when :

$f\left(x\right)={x}^{2}\mathrm{cos}x-|\mathrm{sin}x|$

The “f(x)” function consists of trigonometric and modulus functions. Here,

$⇒f\left(-x\right)={\left(-x\right)}^{2}\mathrm{cos}\left(-x\right)-|\mathrm{sin}\left(-x\right)|$

We know that :

${\left(-x\right)}^{2}={x}^{2};\phantom{\rule{1em}{0ex}}\mathrm{cos}\left(-x\right)=\mathrm{cos}x;\phantom{\rule{1em}{0ex}}|\mathrm{sin}\left(-x\right)|=|-\mathrm{sin}x|=|\mathrm{sin}x|$

Putting these values in the expression of f(-x), we have :

$⇒f\left(-x\right)={\left(-x\right)}^{2}\mathrm{cos}\left(-x\right)-|\mathrm{sin}\left(-x\right)|={x}^{2}\mathrm{cos}x-|\mathrm{sin}x|=f\left(x\right)$

Hence, given function is an even function.

Determine whether f(x) is odd or even, when :

$f\left(x\right)=x{e}^{-{x}^{2}{\mathrm{tan}}^{2}x}$

The “f(x)” function consists of exponential terms having trigonometric function in the exponent. Here,

$⇒f\left(-x\right)=\left(-x\right){e}^{-\left\{{\left(-x\right)}^{2}{\mathrm{tan}}^{2}\left(-x\right)\right\}}$

We know that :

${\left(-x\right)}^{2}={x}^{2};\phantom{\rule{1em}{0ex}}{\mathrm{tan}}^{2}\left(-x\right)={\left(-\mathrm{tan}x\right)}^{2}={\mathrm{tan}}^{2}x$

$⇒f\left(-x\right)=\left(-x\right){e}^{-\left\{{\left(-x\right)}^{2}\mathrm{tan}{}^{2}\left(-x\right)\right\}}=-x{e}^{-{x}^{2}\mathrm{tan}{}^{2}x}=-f\left(x\right)$

Hence, given function is an odd function.

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