×

# Mirror image equation

Just observed this , I founded it nice so I am sharing it

Now suppose this is the question -

$$2x^{8} - 9x^{7} + 20x^{6} - 33x^{5} + 46x^{4} - 66x^{3} + 80x^{2} - 72x + 32 =0$$

Let's go in a generalized manner, if we had this equation

$$a,b,c...,y>0 , z \in R - \{0\}$$

$$ax^{2m} + bx^{2m - 1} + cx^{2m -2} + .... + zx^{m} + ...... + cx^2 + bx + a =0$$

Keeping $$x=0 , ~gives~a=0~so~we~can't~take~x=0$$

Now see the beauty,

Dividing the whole equation by $$x^m$$

$$ax^m + bx^{m-1} + cx^{m-2} + ..... + z + ....... + \dfrac{c}{x^{m-2}} + \dfrac{b}{x^{m-1}} + \dfrac{a}{x^m}$$

$$= a(x^m + \dfrac{1}{x^m}) + b(x^{m-1} + \dfrac{1}{x^{m-1}}) + c(x^{m-2} + \dfrac{1}{x^{m-2}}) + .....+ z =0$$

$$x + \dfrac{1}{x} = y$$

Example(Application) -

$$2x^4 + x^3 + x^2 + x + 2 = 0$$

See here the coefficients of $$x^n$$ are a kind of mirror image after $$x^2$$

Dividing by $$x^2$$ ( $$x \neq 0 ~as~ 2\neq 0$$)

$$2(x^2 + \dfrac{1}{x^2}) + (x + \dfrac{1}{x}) + 1 =0$$

Thus

$$2(y^2 - 2) + y + 1 =0$$

$$2y^2 + y - 3 =0$$

$$(2y + 3)(y - 1) = 0$$

$$y = \dfrac{-3}{2} , y =1$$

Further we have to substitute y and solve the quadratic

First few

$$x^2 + \dfrac{1}{x^2} = y^2 - 2$$ (See how lovely this ratio is! $$~ (x + \dfrac{1}{x})^2 = x^2 + \dfrac{1}{x^2} + \boxed{2x\times\dfrac{1}{x}}$$)

$$x^3 + \dfrac{1}{x^3} = y(\boxed{y^2-2}) - \boxed{y} = y^3 - 3y$$

$$x^4 + \dfrac{1}{x^4} = y(\boxed{ y^3 - 3y}) - \boxed{y^2-2}$$

Thus, we can generalize it as -

$$x^{n + 1} + \dfrac{1}{x^{n+1}} = \Big(x^n + \dfrac{1}{x^n} \Big)\Big(x + \dfrac{1}{x}\Big) - \Big( x^{n - 1} + \dfrac{1}{x^{n-1}}\Big)$$

Writing in terms of t(as most of us write)

$$t_{n} = x^n + \dfrac{1}{x^n}$$

$$\huge{\boxed{t_{n+1} = t_{1}(t_{n}) - t_{n-1}}}$$

The question first I asked , solve it yourself - it can be solved by the same method

Note by Megh Choksi
2 years, 2 months ago

Sort by:

Nice Note , Megh , Thanks For Sharing This ! $$\ddot\smile$$ · 2 years, 2 months ago

FYI, This is called as Reciprocal Equation. Good find! :) · 2 years, 2 months ago

Thanks! I used this trick in a problem in Hall&Knight. · 2 years, 2 months ago

Yay! Me too :) · 2 years, 2 months ago

nice... · 2 years, 2 months ago

Nice note @megh choksi . I never used this method in general. For quartics, I love this! It would help me a lot! · 2 years, 2 months ago

See how this method is useful in solving this question · 2 years, 2 months ago

Nice note! Did you mean $$t_{n+1} = t_{1}t_{n} - t_{n-1}$$? · 2 years, 2 months ago

Yes , thanks for pointing out and reading it · 2 years, 2 months ago

I just realized that it can be proven easily by $$a^{n+1} + b^{n+1} = (a+b)(a^{n}+b^{n}) - ab(a^{n-1}+b^{n-1})$$. · 2 years, 2 months ago

Great , recently I was solving problems of these kinds on brilliant.org , so just I generalized it , nothing more is here · 2 years, 2 months ago

You're missing an x in the third equation, it should be $$cx^{2m-2}$$, not just $$c^{2m-2}$$ unless that was intentional. · 2 years, 2 months ago