Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This question already has an answer here:

I have been given an optional challenge problem at the end of one of my earlier tutorials that I am unsure how to solve. It is a question with three parts, but I would like to tackle them separately with a little help on part a.

a function $f$ is even if $f(-x) = f(x)$ for all $x$, and odd if $f(-x) = -f(x)$ for all $x$.

Now I am told to show that every polynomial is the sum of both an even and an odd function, but have no idea how to go about this.

Is the key to this question that $f(-x) = x^2$ will always $= f(x)$ $\therefore$ $x^2$ is even? I understand that, and if a secondary function was $g(-x) = x$, where x is only positive, then it would make sense that this is a odd function, but it is for all x, and therefore a negative x value in g(-x) would become positive and it wouldn't be an odd or an even function.

share|improve this question

marked as duplicate by Marc van Leeuwen, Daniel Fischer, azimut, Dominic Michaelis, mau Oct 14 '13 at 7:34

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

2  
Not just polynomials but rather every function can be written as a sum of an even and an odd function. –  Fixed Point Oct 14 '13 at 5:10
    
3  
@Display Name:- Not only for polynomials but every function can be written as "the sum of both an even and an odd function" indeed $f(x)=\frac{f(x)+f(-x)}2 + \frac {f(x)-f(-x)}2$ –  Souvik Dey Oct 14 '13 at 5:10
1  
DisplayName: You must be able to understand the formula in the comment by @SouvikDey, why it is true, and why its terms are even respectively odd. If this is still too advanced for you, then I am afraid that you cannot understand the question, so there is no point asking for an answer. –  Marc van Leeuwen Oct 14 '13 at 5:19
1  
@Marc van Leeuwen: Thank you , but I don't think I need to post it as an answer any more , there are lots of nice answers already posted and Secret Math's answer follows the same line as that of mine. –  Souvik Dey Oct 14 '13 at 5:31

4 Answers 4

up vote 11 down vote accepted

Hint: $x$, $x^3$, $x^5$, ... are all odd, while $1$, $x^2$, $x^4$, ... are all even (prove this!). So try decomposing a polynomial into a sum of terms only involving even powers of $x$, and terms only involving odd powers of $x$.

share|improve this answer

Think about the easiest even (or odd) functions. The first one I will think of is $f(x) = x$ and $f(x) = x^2$. Next you will realize that for $f(x) = x^n$ it is even or odd depending on if $n$ is even or odd. (That also helps you to understand why they are called even (or odd) functions.)

Now that we are only looking at polynomials, we can naturally divide a polynomials into odd and even degrees. The rest would be easy.

A more general construction works for any functions: $\frac{f(x) + f(-x)}{2}$ is always even, and $\frac{f(x) - f(-x)}{2}$ is always odd, and they add up to $f(x)$.

share|improve this answer
    
Well you got the idea from other people's comment. –  Secret Math Oct 14 '13 at 5:13

I think T. Bongers's answer is the most natural way to solve this problem, seeing as your given function is a polynomial. Here is an alternative method. You have a given function $P(x)$ which you want to express as the sum of an odd function, call it $D(x)$, and an even function, call it $E(x)$. So $P(x),D(x),E(x)$ satisfy the identity $P(x)=D(x)+E(x)$. That's one equation in two unknowns, $D(x)$ and $E(x)$. Now, substituting $-x$ for $x$ in the identity $P(x)=D(x)+E(x)$, you get another identity, $P(-x)=D(-x)+E(-x)=-D(x)+E(x)$. So now you have two equations in two unknowns: $D(x)+E(x)=P(x)$ and $-D(x)+E(x)=P(-x)$. Solving this system of equations, you get formulas for $D(x)$ and $E(x)$ in terms of $P(x)$ and $P(-x)$. You can then verify that $D(x)$ and $E(x)$ are indeed odd and even respectively, and that (assuming $P(x)$ is a polynomial, or even an infinite series in powers of $x$) they are the same functions that T. Bongers hinted at in his answer.

share|improve this answer

This is actually true of all functions $f$ defined on $\mathbb{R}$. Let $$f(x) = f_e(x) + f_o(x),$$ then $$f(-x) = f_e(-x) + f_o(-x) = f_e(x) - f_o(x).$$ From these two equations you can determine what $f_e(x)$ and $f_o(x)$ are rather explicitly. In the case where $f(x)$ is a polynomial, you get the decomposition that user61527 describes (you should check this).


Note, the first equation seems to assume $f$ can be written as the sum of an even and odd function. What we are really doing is supposing that such a decomposition exists, then determining what $f_e$ and $f_o$ must be if they do exist. The last thing to check is that $f_e$ and $f_o$ add up to $f$ and are even and odd respectively. This shows that the decomposition does exist.

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.