I watched a video where a problem involved recognizing that $\sin x$ is an odd function and $\sin^3 x$ is also odd. But the presenter didn't explain why $\sin^3 x$ is also odd. Why does the fact that the function is odd not change when it is cubed? Is there a rule where for every even power the odd function is even and for every odd power the odd function remains odd? What about for even functions?
Even and odd functions and whether even/odd characteristics change with powers
even-and-odd-functions
Related Solutions
Well, by definition $f(x)+g(x)$ is even if and only if $f(x)+g(x)=f(-x)+g(-x)$ for all $x$. If $g(x)$ is even, it satisfies $g(x)=g(-x)$ for all $x$. Inserting this into the previous equation gives $f(x)=f(-x)$ for all $x$, hence $f$ must be even.
Why did it work for odd? Here the definition requires $f(x)+g(x)=-f(-x)-g(-x)$. Since $g(x)$ is even, this transforms to $f(x)+g(x)=-f(-x)-g(x)$ i.e. $g(x) = -\frac 12 (f(x)+f(-x))$, which your example satisfies. In other words: if $f$ is not even, subtracting its even part [*] makes it odd.
[*] This refers to the fact that you can decompose every function $f(x)$ into its even part $f_e(x)$ and its odd part $f_o(x)$ such that $f(x) = f_e(x)+f_o(x)$, where $f_e(x) =\frac 12 (f(x)+f(-x))$ and $f_o(x) =\frac 12 (f(x)-f(-x))$.
The author of the solution states that the RHS of both equations are odd (i.e. $2y$ is odd, which makes sense) therefore $g(y)$ is odd. This confuses me a bit because from what I can see $f′(x)$ is neither even or odd so is it a rule that an odd function times a neither equals an odd?
No, there's no such rule because that would be a false statement: "odd" times "neither" isn't going to be odd (e.g., consider $x\cdot(x+1)$). But note that you do NOT have "a function times another function" here; instead, the second factor here is just a number, either $f'(a)$ or $f'(-a)$ in the two equations, obtained by plugging in a fixed value $a$ or $-a$ into $f'(x)$. So this statement is simply an observation that if you multiply or divide an odd function by a nonzero number, you get an odd function: for a constant $c\ne0$, $g(y)$ is odd iff $cg(y)$ is odd.
Then the author states that $$g(y)(f′(a)−f′(−a))=0$$ hence $f′(a)$ is even. This also confuses me because how was the above equation made and how can it be stated that $f′(a)$ is even?
To be honest, I only partially understand this part. Here's what I do understand. First of all, this equation comes from subtracting the two equations above from each other: $$g(y)f′(a)−g(y)f′(−a)=g(y)(f′(a)−f′(−a))=0$$ and subtracting the right-hand sides gives $0$. I presume the context suggests that $g(y)$ is not identically zero (it depends on $y$, and even if its value is zero for some $y$, it's not zero for all $y$), therefore $f′(a)−f′(−a)=0$ has to be true, which is equivalent to $f′(a)=f′(−a)$. This looks precisely like the definition of being even …
… but to me it looks that this equality $f′(a)=f′(−a)$ has been established for one input value $x=a$ only, while an even function must satisfy it for all $x$ in the domain. Unless $a$ can vary, I don't see how this argument is valid.
Or maybe it's just a poor exposition for an actually true conclusion? In fact, we don't need $f'$ to be an even or odd function. The condition $f′(a)=f′(−a)$ does not imply that, but it does imply what we want: that $C_3=0$. Here's how: $$\begin{gather} f′(a)=f′(−a) \\ C_3\sinh(ka)+C_4\cosh(ka)=−C_3\sinh(ka)+C_4\cosh(ka) \\ C_3\sinh(ka)=−C_3\sinh(ka) \end{gather}$$ from which $C_3\sinh(ka)=0$. Note that $\sinh(x)=0$ only at $x=0$. I presume from the context we know that $a\ne0$, therefore $\sinh(a)\ne0$, therefore $C_3=0$.
Best Answer
Note that $$(-1)^{2k}=(+1)$$ and $$(-1)^{2k+1}=(-1)$$ $$(+1)^k = (+1)$$
Thus if a function is odd we have $$f^{2k}(-x) = (-1)^{2k}f^{2k}(x)=f^{2k}(x)$$ and $$f^{2k+1}(-x) = (-1)^{2k+1}f ^{2k+1}(x)=-f^{2k+1}(x)$$
Thus odd functions to the odd powers are odd and to the even powers are even.
Even functions to any power stay even.