Mirror image equation

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

Now suppose this is the question -

2x89x7+20x633x5+46x466x3+80x272x+32=02x^{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,zR{0} a,b,c...,y>0 , z \in R - \{0\}

ax2m+bx2m1+cx2m2+....+zxm+......+cx2+bx+a=0 ax^{2m} + bx^{2m - 1} + cx^{2m -2} + .... + zx^{m} + ...... + cx^2 + bx + a =0

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

Now see the beauty,

Dividing the whole equation by xmx^m

axm+bxm1+cxm2+.....+z+.......+cxm2+bxm1+axm ax^m + bx^{m-1} + cx^{m-2} + ..... + z + ....... + \dfrac{c}{x^{m-2}} + \dfrac{b}{x^{m-1}} + \dfrac{a}{x^m}

=a(xm+1xm)+b(xm1+1xm1)+c(xm2+1xm2)+.....+z=0 = 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+1x=y x + \dfrac{1}{x} = y

Example(Application) -

2x4+x3+x2+x+2=0 2x^4 + x^3 + x^2 + x + 2 = 0

See here the coefficients of xnx^n are a kind of mirror image after x2x^2

Dividing by x2x^2 ( x0 as 20x \neq 0 ~as~ 2\neq 0)

2(x2+1x2)+(x+1x)+1=0 2(x^2 + \dfrac{1}{x^2}) + (x + \dfrac{1}{x}) + 1 =0

Thus

2(y22)+y+1=0 2(y^2 - 2) + y + 1 =0

2y2+y3=0 2y^2 + y - 3 =0

(2y+3)(y1)=0 (2y + 3)(y - 1) = 0

y=32,y=1 y = \dfrac{-3}{2} , y =1

Further we have to substitute y and solve the quadratic


First few

x2+1x2=y22x^2 + \dfrac{1}{x^2} = y^2 - 2 (See how lovely this ratio is!  (x+1x)2=x2+1x2+2x×1x~ (x + \dfrac{1}{x})^2 = x^2 + \dfrac{1}{x^2} + \boxed{2x\times\dfrac{1}{x}})

x3+1x3=y(y22)y=y33yx^3 + \dfrac{1}{x^3} = y(\boxed{y^2-2}) - \boxed{y} = y^3 - 3y

x4+1x4=y(y33y)y22x^4 + \dfrac{1}{x^4} = y(\boxed{ y^3 - 3y}) - \boxed{y^2-2}

Thus, we can generalize it as -

xn+1+1xn+1=(xn+1xn)(x+1x)(xn1+1xn1) 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)

tn=xn+1xn t_{n} = x^n + \dfrac{1}{x^n}

tn+1=t1(tn)tn1\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

Its the same question I posted earlier

Note by U Z
4 years, 9 months ago

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1 vote

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Comments

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Nice Note , Megh , Thanks For Sharing This ! ¨\ddot\smile

Deepanshu Gupta - 4 years, 9 months ago

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Thanks! I used this trick in a problem in Hall&Knight.

Satvik Golechha - 4 years, 9 months ago

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Yay! Me too :)

Krishna Ar - 4 years, 9 months ago

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FYI, This is called as Reciprocal Equation. Good find! :)

Krishna Ar - 4 years, 9 months ago

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

Pranjal Jain - 4 years, 9 months ago

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nice...

Vanessa Aisadora Vinan - 4 years, 9 months ago

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Nice note! Did you mean tn+1=t1tntn1t_{n+1} = t_{1}t_{n} - t_{n-1}?

Samuraiwarm Tsunayoshi - 4 years, 9 months ago

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Yes , thanks for pointing out and reading it

U Z - 4 years, 9 months ago

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I just realized that it can be proven easily by an+1+bn+1=(a+b)(an+bn)ab(an1+bn1)a^{n+1} + b^{n+1} = (a+b)(a^{n}+b^{n}) - ab(a^{n-1}+b^{n-1}).

Samuraiwarm Tsunayoshi - 4 years, 9 months ago

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@Samuraiwarm Tsunayoshi Great , recently I was solving problems of these kinds on brilliant.org , so just I generalized it , nothing more is here

U Z - 4 years, 9 months ago

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You're missing an x in the third equation, it should be cx2m2cx^{2m-2}, not just c2m2c^{2m-2} unless that was intentional.

tytan le nguyen - 4 years, 9 months ago

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@Tytan Le Nguyen Thanks for pointing out as well as for reading it

U Z - 4 years, 9 months ago

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See how this method is useful in solving this question

U Z - 4 years, 9 months ago

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