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System of n equations of Real Analysis , I.S.I Entrance 2008, Solution to Subjective Problem No. 9.

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Understand the problem

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 For \(n\ge3 \), determine all real solutions of the system of \(n\) equations :

                                               \(x_1+x_2+\cdots+x_{n-1}=\frac{1}{x_n}\)

                                                                 ...............................

                      \(x_1+x_2+\cdots+x_{i-1}+x_{i+1}+\cdots+x_{n-1}+x_n=\frac{1}{x_i}\)

                                                                 ................................

                                               \(x_2+x_3+\cdots+x_{n-1}+x_n=\frac{1}{x_1}\).

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I.S.I. (Indian Statistical Institute) B.Math.(Hons.) Entrance Examination 2008. Subjective Problem no. 9.

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System of \(n\) Equations

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9 out of 10

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Mathematical Olympiad Challenges by Titu Andreescu & Razvan Gelca.

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Start with hints

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Do you really need a hint? Try it first!

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 Let \( s = x_1+x_2+\cdots+x_i+\cdots +x_{n-1}+x_n\).

Then the system of \(n\) equations are equivalent to \(x_i^2-sx_i+1=0\) for \(i=1,2,....,n\).

It follows that \(x_1,x_2,.....,x_{n-1},x_n\) are solutions to the quadratic equation: \(u^2-su+1=0\).                                                   \(\cdots\cdots\cdots\cdots (i)\)

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Now we have two possible cases.

Case - I :  Two roots of the equation (i) are equal, then , all \(x_i\) are equal. 

i.e. , \(x_1=x_2=\cdots=x_{n-1}=x_n=u\).

\(\Rightarrow s=nu\).

Putting this value of \(s\) in equation (i) we get, \((n-1)u^2=1\).

\(\Rightarrow u=\frac{1}{\pm \sqrt{n-1}}\).

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Case - II : Two roots of equation (i) are not equal. Then let these two roots are \(u_1 \) and \(u_2\). Also let among {\(x_1,x_2,......,x_{n-1},x_n\)}   \( k\)  of them equal to \(u_1\) and \((n-k) \) of them equal to \(u_2\), where \(0<k<n\).

      In this case, we have , (a) \(u_1+u_2=s\).    [Sum of roots of equation (i)]

                                              (b) \(u_1\cdot u_2=1\).    [ Product of roots of equation (i)].

Now,.   \(s=x_1+x_2+\cdots+x_{n-1}+x_n=k\cdot u_1+(n-k)\cdot u_2\)

        \(\Rightarrow s=(u_1+u_2)+(k-1)\cdot u_1+(n-k-1)\cdot u_2\) 

        \(\Rightarrow (k-1)\cdot u_1+(n-k-1)\cdot u_2=0\)   [using (a)]

        \(\Rightarrow  (k-1)\cdot u_1^2=(k+1-n)\cdot u_1\cdot u_2=k+1-n\le 0\) [using (b) and since ,\( k<n\Rightarrow k+1\le n\)].

      \(\Rightarrow u_1^2\le 0\)

But \(u_1\neq 0\) as the coefficient of \(u^0\) is 1(\(\neq 0\)) in equation (i).

        \(\Rightarrow u_1^2<0\)

\(\Rightarrow u_1\) is not real. So we have no real solution to the system of \(n\) equations in this case.

 

 

 

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Thus the total no. of real solutions to the system of \(n\) equations is two and these two solutions are:

                        \(x_1=x_2=\cdots=x_{n-1}=x_n=\frac{1}{\sqrt{n-1}}\)      and

                        \(x_1=x_2=\cdots=x_{n-1}=x_n=-\frac{1}{\sqrt{n-1}}\).(Ans.)

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Connected Program at Cheenta

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Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

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