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Part 1 Let $f(x) = ax^n$, where $a$ is any real number. Prove that $f$ is even if $n$ is an even integer. (Integers can be negative too)

Part 2 Prove that if you add any two even functions, you get an even function

I'm confused as to how you would prove adding two even functions would get you an even function.

user112533
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2 Answers2

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Part 1) $$f(x)=ax^n$$ So, let $n=2k$. Then, $$f(x)=ax^n=a(x^2)^k$$ Obviously, $x^2$ is an even function ($(-x)^2=x^2$). So, $$f(x)=ax^n=a(x^2)^k$$ is an even function.

Part 2) Let $f(x),g(x)$ be two even functions. Then, $$f(x)=f(-x),\qquad g(x)=g(-x)$$ Adding the two gives $$f(x)+g(x)=h(x)$$ Using the above relation then gives $$f(x)+g(x)=h(x)=f(-x)+g(-x)$$ The last expression is $h(-x)$. We have therefore shown that $h(x)$ is even. Q.E.D.

  • Thanks for answering! Just out of curiosity, if you composed any two odd functions, what would you get and how would you prove it? – user112533 Mar 06 '14 at 02:17
  • Say $f(x),g(x)$ are odd. Then, $f(-x)=-f(x)$, and the same holds for $g(x)$. Thus, $f(g(-x))=f(-g(x))=-f(g(x))$, or $(f\circ g)(-x)=-(f\circ g)(x)$. –  Mar 06 '14 at 02:19
  • Oh, I see. Thank you very much! Your answers were a great help to my understanding – user112533 Mar 06 '14 at 02:23
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    @user112533 Anything for a fellow mathematician! –  Mar 06 '14 at 02:24
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The first part is simple. Suppose $f(x) = ax^n$for an even integer $n$. $f(-x) = a(-x)^n = ax^n = f(x)$ since $n$ is even.

As for the second part, suppose $h(x) = f(x) + g(x)$ for each $x$ in the domain of $h$, where $f$ and $g$ are even functions. Then, $h(-x) = f(-x) + g(-x) = f(x) + g(x) = h(x) $

If you are unsure about the definition of an even function then check this out.

Ishfaaq
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