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I was hoping someone could show or explain why it is that a function of the form $f(x) = ax^d + bx^e + cx^g+\cdots $ going on for some arbitrary length will be an odd function assuming $d, e, g$ and so on are all odd numbers, and likewise why it will be even if $d, e, g$ and so on are all even numbers. Furthermore, why is it if say, $d$ and $e$ are even but $g$ is odd that $f(x)$ will then become neither even nor odd?

Thanks.

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3 Answers

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An even function is one for which $f(-x) = f(x)$ for all values of $x$ (e.g. evaluating at -6 is the same as evaluating at 6). If $n$ is an even exponent, then $(-x)^n = (-1)^nx^n = x^n$, since an even number of negative signs will cancel out. If all the exponents are even, then this happens for every term in the polynomial, so the result is the same as the original polynomial.

An odd function is one for which $f(-x) = -f(x)$ for all values of $x$ (i.e. the minus sign factors out). If $n$ is an odd exponent, then $(-x)^n = (-1)^nx^n = -x^n$, since an odd number of negative signs leaves just one negative sign remaining. If all the exponents are odd, then we get: $$ f(-x) = ax^d + bx^e + cx^g + \cdots = -ax^d - bx^e - cx^g - \cdots = -(ax^d + bx^e + cx^g + \cdots) = -f(x). $$

If there is a mixture of odd an even exponents, then neither of these nice properties will hold, so the function will be neither even nor odd.

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Thanks, that's exactly what I was looking for. –  user48226 Nov 17 '12 at 1:41
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If the exponents are all odd, then $f(x)$ is the sum of odd functions, and hence is odd.

If the exponents are all even, then $f(x)$ is the sum of even functions, and hence is even.

As far as your last question, the sum of an odd function and even function is neither even nor odd.


Proof: Sum of Odd Functions is Odd:

Given two odd functions $f$ and $g$. Since they are odd functions $f(-x) = -f(x)$, and $g(-x) = -g(x)$. Hence: \begin{align*} f(-x) + g(-x) &= -f(x) - g(x) \\ &= -(f+g)(x) \\ \implies (f+g) & \text{ is odd if $f$ and $g$ are odd.} \end{align*}

Proof: Sum of Even Functions is Even:

Given two even functions $f$ and $g$, then $f(-x) = f(x)$, and $g(-x) = g(x)$. Hence: \begin{align*} f(-x) + g(-x) &= f(x) + g(x) \\ \implies (f+g) &\text{ is even if $f$ and $g$ are even.} \end{align*}

Proof: Sum of an odd function and even function is neither odd or even.

If $f$ is odd, and $g$ is even \begin{align*} f(-x) + g(-x) &= -f(x) + g(x) \\ &= -(f-g)(x) \\ \implies (f+g)&\text{ is neither odd or even if $f$ is even, $g$ is odd.} \end{align*}

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Thanks, that's a cool way to prove it. –  user48226 Nov 17 '12 at 2:05
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It’s just a matter of checking the definitions of even function and odd function. Let’s look at a simple case that displays all of the possible behaviors of the more general case: consider $f(x)=ax^m+bx^n$.

  • Suppose that $m$ and $n$ are even, say $m=2k$ and $n=2\ell$; then $$(-x)^m=(-x)^{2k}=\left((-x)^2\right)^k=\left(x^2\right)^k=x^{2k}=x^m\;,$$ and similarly $(-x)^n=x^n$, so $$f(-x)=a(-x)^m+b(-x)^n=ax^m+bx^n=f(x)\;,$$ and by definition $f$ is an even function.

  • Now suppose that $m$ and $n$ are odd, say $m=2k+1$ and $n=2\ell+1$. Then $$(-x)^m=(-x)^{2k+1}=(-x)^{2k}(-x)=x^{2k}(-x)=-x^{2k+1}=-x^m\;,$$ and similarly $(-x)^n=-x^n$, so $$f(-x)=a(-x)^m+b(-x)^n=-ax^m-bx^n=-\left(ax^m+bx^n\right)=-f(x)\;,$$ and by definition $f$ is an odd function.

  • Finally, suppose that $m$ is odd and $n$ is even. (The argument is the same if $m$ is even and $n$ is odd, just reversing the rôles of $m$ and $n$.) Then by what we’ve already seen we know that $$f(-x)=a(-x)^m+b(-x)^n=-ax^m+bx^n\;,$$ which is neither $f(x)$ nor $f(-x)$. Thus, $f$ is neither even nor odd.

Nothing essentially different happens if there are more than two terms.

In fact the names even function and odd function come from the behavior of even and odd powers, respectively.

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Thanks, also a solid explanation. –  user48226 Nov 17 '12 at 1:49
    
@ascii: You’re welcome. –  Brian M. Scott Nov 17 '12 at 1:50
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