Odd, even, plus, minus

Relevant wiki: Multiplying and Dividing Monomials

$\Large\dfrac{\color{#D61F06}{\cancel{x^2}}(1+x^2+x^4+x^6+x^{8})}{\color{#D61F06}{\cancel{x^2}}(1-x^2+x^4-x^6+x^{8})}=\dfrac{\color{#EC7300}{\cancel{x^3}}(1+x^2+x^4+x^6+x^{8})}{\color{#EC7300}{\cancel{x^3}}(1-x^2+x^4-x^6+x^{8})}$

We see that both the sides are identical $\forall x\in\mathbb R-\{0\}$ and hence result hold for all non zero real x.

$\large \text {RHS =}\dfrac{x^3+x^5+x^7+x^9+x^{11}}{x^3-x^5+x^7-x^9+x^{11}} = \dfrac {x(x^2 + x^4+x^6+x^8 +x^{10})}{x(x^2 - x^4 + x^6 -x^8 + x^{10})}$

$\large \dfrac { \cancel{x}(x^2 + x^4+x^6+x^8 +x^{10})}{\cancel{x}(x^2 - x^4 + x^6 -x^8 + x^{10})} = \dfrac {x^2 + x^4 + x^6 + x^8 + x^{10}}{x^2 - x^4 + x^6 - x^8 +x^{10}}$

$\large \Rightarrow \dfrac {x^2 + x^4 + x^6 + x^8 + x^{10}}{x^3 - x^5 + x^7 - x^9 +x^{11}} = \dfrac {x^2 + x^4 + x^6 + x^8 + x^{10}}{x^2 - x^4 + x^6 - x^8 +x^{10}}$ , which is always true for non zero numbers.

$\dfrac{x^2+x^4+x^6+x^8+x^{10}}{x^2-x^4+x^6-x^8+x^{10}} = \dfrac{x^3+x^5+x^7+x^9+x^{11}}{x^3-x^5+x^7-x^9+x^{11}}\\ \dfrac{x^2(1+x^2+x^4+x^6+x^8)}{x^2(1-x^2+x^4-x^6+x^8)} = \dfrac{x^3(1+x^2+x^4+x^6+x^8)}{x^3(1-x^2+x^4-x^6+x^8)}$

Since $x \neq 0, x^2 \neq 0, x^3 \neq 0$ . This means that we can cancel out $x^2$ and $x^3$ , which gives us

$\dfrac{1+x^2+x^4+x^6+x^8}{1-x^2+x^4-x^6+x^8} = \dfrac{1+x^2+x^4+x^6+x^8}{1-x^2+x^4-x^6+x^8}$

Obviously, LHS and RHS are the same, therefore the equation should hold for all real $x\neq 0$ . However, for a more complete proof, we should show that the denominator $1-x^2+x^4-x^6+x^8\neq 0$ for all $x \neq 0$ .

There are three methods to do this.

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Method 1 (as suggested by @Calvin Lin )

Note that for all real values of $x \neq 0$ , $x^2 > 0,\;x^4> 0,\;x^6 > 0,\; x^8 > 0$

Now, if $x \geq 1$ or $x \leq -1,\;x^2 \geq 1,\;x^8 \geq x^6 \geq x^4 \geq x^2$

This implies that $x^8-x^6 \geq 0,\;x^4-x^2 \geq 0$ . Therefore,

$x^8-x^6+x^4-x^2+1 \geq 1,\;x^2 \geq 1$

If $-1 < x < 1,\;0 < x^2 < 1,\;x^8 < x^6 < x^4 < x^2$

This implies that $1-x^2>0, x^4-x^6 > 0$ . Therefore,

$1-x^2+x^4-x^6+x^8 > 0,\;0 < x^2 < 1$

From here, we can see that the denominator $>0$ for all real $x \neq 0$

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Method 2 (my preferred method)

We let $y=x^2$ . This yields a quartic function

$1-y+y^2-y^3+y^4$

We can then determine the roots of this function with the quartic discriminant :

$\Delta = 256 - 192 -128+144-27+144-6-80+18+16-4-27+18-4-4+1=125 > 0\\ P = 8-3 = 5 > 0\\ D=64-16+16-16-3=45 > 0$

$\Delta > 0,\;P >0,\; D> 0$ , therefore the quartic function has no real roots.

There are no real roots of $y$ , therefore no real $x^2$ satisfies $1-x^2+x^4-x^6+x^8=0$ , which also implies that the denominator $\neq 0$ for all real $x \neq 0$

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Method 3 (graphical/calculus)

For an eighth degree polynomial with a positive leading coefficient, both ends of the graph head towards $+\infty$

From here, we can deduce that if all the minimum points of the graph have a positive $y$ -coordinate, we can say that the graph of the polynomial never intersects the $x$ -axis.

We can let $f(x) = 1-x^2+x^4-x^6+x^8$ , and find the values of $x$ that satisfy $f'(x) = 0$

$f'(x) = 8x^7-6x^5+4x^3-2x\\ 8x^7-6x^5+4x^3-2x = 0\\ 2x(4x^6-3x^4+2x^2-1)=0\\ x=0, 4x^6-3x^4+2x^2-1=0$

For the second factor, you can let $z=x^2$ and solve the cubic polynomial $4z^3-3z^2+2z-1=0$ . This method is very tedious without a calculator, but it's doable.

You should get $3$ points, two minimum with a positive $y$ -coordinate, and a maximum point

$\begin{aligned} \dfrac{x^2 + x^4 + x^6 + x^8 + x^{10}}{x^2 - x^4 + x^6 - x^8 + x^{10}} & = y (say) \rightarrow \boxed{1}\\ \\ \dfrac{(x^2 + x^6 + x^{10}) + (x^4 + x^8)}{(x^2 + x^6 + x^{10}) - (x^4 + x^8)} & = \dfrac{y}{1}\\ \\ \dfrac{(x^2 + x^6 + x^{10}) + (x^4 + x^8) + (x^2 + x^6 + x^{10}) - (x^4 + x^8)}{(x^2 + x^6 + x^{10}) + (x^4 + x^8) - (x^2 + x^6 + x^{10}) + (x^4 + x^8)} & = \dfrac{y + 1}{y - 1} (\text{By componendo and dividendo})\\ \\ \dfrac{2(x^2 + x^6 + x^{10})}{2(x^4 + x^8)} & = \dfrac{y + 1}{y - 1}\\ \\ \dfrac{x(x^2 + x^6 + x^{10})}{x(x^4 + x^8)} & = \dfrac{y + 1}{y - 1} (\text{Multiplying and dividing x in numerator and denominator})\\ \\ \dfrac{(x^3 + x^7 + x^{11})}{(x^5 + x^9)} & = \dfrac{y + 1}{y - 1}\\ \\ \dfrac{(x^3 + x^7 + x^{11}) + (x^5 + x^9)+ (x^3 + x^7 + x^{11}) - (x^5 + x^9)}{(x^3 + x^7 + x^{11}) + (x^5 + x^9) - (x^3 + x^7 + x^{11}) + (x^5 + x^9)} & = \dfrac{y + 1 + y - 1}{y + 1 - y + 1} (\text{By componendo and dividendo})\\ \\ \dfrac{2(x^3 + x^7 + x^{11})}{2(x^5 + x^9)} & = \dfrac{2y}{2}\\ \\ \dfrac{(x^3 + x^7 + x^{11})}{(x^5 + x^9)} & = y \rightarrow \boxed{2}\\ \\ \text{From 1 and 2}:-\\ \dfrac{x^2 + x^4 + x^6 + x^8 + x^{10}}{x^2 - x^4 + x^6 - x^8 + x^{10}} & = \dfrac{x^3 + x^5 + x^7 + x^9 + x^{11}}{x^3 - x^5 + x^7 - x^89+ x^{11}} \end{aligned}$

$\boxed{\text{Hence, proved}}$

Why does every one do complicated stuff. Is it not enough to multiply the left side with x/x to get the right side?

$\Large\frac{x^{2}+x^{4}+x^{6}+x^{8}+x^{10}}{x^{2}-x^{4}+x^{6}-x^{8}+x^{10}}$ = $\Large\frac{x^{3}+x^{5}+x^{7}+x^{9}+x^{11}}{x^{3}-x^{5}+x^{7}-x^{9}+x^{11}}$

Let's look at the two sides of the equality separately. Beginning with the left hand side:

$\Large\frac{x^{2}+x^{4}+x^{6}+x^{8}+x^{10}}{x^{2}-x^{4}+x^{6}-x^{8}+x^{10}}$

We can factor $x^{2}$ out of both the numerator and the denominator:

$\Large\frac{x^{2}(1+x^{2}+x^{4}+x^{6}+x^{8})}{x^{2}(1-x^{2}+x^{4}-x^{6}+x^{8})}$

$\Large\frac{\cancel{x^{2}}(1+x^{2}+x^{4}+x^{6}+x^{8})}{\cancel{x^{2}}(1-x^{2}+x^{4}-x^{6}+x^{8})}$

$\Large\frac{1+x^{2}+x^{4}+x^{6}+x^{8}}{1-x^{2}+x^{4}-x^{6}+x^{8}}$

Cool! Let's do the right hand side now:

$\Large\frac{x^{3}+x^{5}+x^{7}+x^{9}+x^{11}}{x^{3}-x^{5}+x^{7}-x^{9}+x^{11}}$

We can factor out $x^{3}$ :

$\Large\frac{x^{3}(1+x^{2}+x^{4}+x^{6}+x^{8})}{x^{3}(1-x^{2}+x^{4}-x^{6}+x^{8})}$

$\Large\frac{\cancel{x^{3}}(1+x^{2}+x^{4}+x^{6}+x^{8})}{\cancel{x^{3}}(1-x^{2}+x^{4}-x^{6}+x^{8})}$

$\Large\frac{1+x^{2}+x^{4}+x^{6}+x^{8}}{1-x^{2}+x^{4}-x^{6}+x^{8}}$

So both sides of the expression are equivalent. Therefore the equation holds true for all $x\ne 0$ .

we also can check it by putting x=1, then satisfied the equation. so equation is true..

(x²+x⁴+x^6+x^8+x^10)/(x²-x⁴+x^6-x^8+x^10)*x/x=(x³+x^5+x^7+x^9+x^11)/(x³-x^5+x^7-x^9+x^11)

(for any x≠0, x/x=1. So multipling any of the sides by x/x shouldn't change the equation. Multipling the left side of the equation by x/x gives you the right side of the equation, so it is true to any x≠0