TargetIIT

Consider the polynomials [im]http://codecogs.izyba.com/gif.latex?f%28x%29%20%3Da_1%20+a_2x+a_3x%5E2+a_4x%5E4[/im] and

[im]http://codecogs.izyba.com/gif.latex?g%28x%29%20%3Db_1%20+b_2x+b_3x%5E2+b_4x%5E4[/im] where all coefficients are real

It is known that [im]http://codecogs.izyba.com/gif.latex?%5Cforall%20%5C%20x%20%5Cin%20%5Cmathbb%7BR%7D%20%5C%20%5C%20%5Bf%28x%29%5D%20%3D%20%5Bg%28x%29%5D[/im]

Is it necessary that f(x) = g(x)?

ne hint sir.

f[p]-1[/p](n)=g[p]-1[/p](n) for all n in integers

OOPS! [2] I did not see this!


[im]http://codecogs.izyba.com/gif.latex?f%28x%29%3Da_1%5Cleft%28x-%5Calpha%20%5Cright%29%28x-%5Cbeta%29%28x-%5Cgamma%29%28x+%5Calpha+%5Cbeta+%5Cgamma%29[/im].

[im]http://codecogs.izyba.com/gif.latex?g%28x%29%3Da_2%5Cleft%28x-p%20%5Cright%29%28x-q%29%28x-r%29%28x+p+q+r%29[/im]

From the fact that [im]http://codecogs.izyba.com/gif.latex?f%28%5Calpha%29%3Df%28%5Cbeta%29%3Df%28%5Cgamma%29%3Df%28%5Calpha+%5Cbeta+%5Cgamma%29%3D0[/im], we arrive at the conclusion that the roots of g(x) and f(x) are same.

Now we need to evaluate l and m where it is given that [im]http://codecogs.izyba.com/gif.latex?%5Bln%5D%3D%5Bmn%5D[/im]

Where [im]http://codecogs.izyba.com/gif.latex?n%3D%5Cprod%28%7Bx-%5Calpha%7D%29[/im]....Needless to mention that l and m are just the constants.

WE see that l=m is the only solution in this case.
Proving this is no big deal if l and m are rationals. However if they are irrationals then one can say,

[im]http://codecogs.izyba.com/gif.latex?%5Bl%5D%3Dx_1x_2x_3...x_n[/im] & [im]http://codecogs.izyba.com/gif.latex?%5Cleft%5C%7Bl%20%5Cright%5C%7D%3Dy_1y_2y_3...y_m...[/im] - i.e let it reccur.

Similarly let [im]http://codecogs.izyba.com/gif.latex?%5Bm%5D%3Dx%27_1x%27_2...x%27_k[/im] & [im]http://codecogs.izyba.com/gif.latex?%5Cleft%5C%7Bm%20%5Cright%5C%7D%3Dy%27_1y%27_2...y%27_s...[/im]

I chhose [im]http://codecogs.izyba.com/gif.latex?n%3D10%5Et[/im], where t is any arbitary large natural.

That itself completes the story proving [im]http://codecogs.izyba.com/gif.latex?%5Cboxed%7Bf%28x%29%3Dg%28x%29%7D[/im] [im]http://codecogs.izyba.com/gif.latex?%5Cforall[/im] x belonging to reals. [4]

The main thing that I forgot to mention was that x[ss]1[/ss]'x'[ss]2[/ss]....are the digits of [m] and not nos, multiplied together. Same for y'[ss]i[/ss],   x[ss]i[/ss]  & y[ss]i[/ss]'.