System of n equations of Real Analysis , I.S.I Entrance 2008, Solution to Subjective Problem No. 9.
Join Trial or Access Free ResourcesJoin Trial or Access Free ResourcesFor \(n\ge3 \), determine all real solutions of the system of \(n\) equations :
\(x_1+x_2+\cdots+x_{n-1}=\frac{1}{x_n}\)
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\(x_1+x_2+\cdots+x_{i-1}+x_{i+1}+\cdots+x_{n-1}+x_n=\frac{1}{x_i}\)
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\(x_2+x_3+\cdots+x_{n-1}+x_n=\frac{1}{x_1}\).
[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.22.4"][et_pb_column type="4_4" _builder_version="3.22.4"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.22.4" body_font="Raleway||||||||" toggle_font="||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.22.4" title_text_shadow_horizontal_length="0em" title_text_shadow_vertical_length="0em" title_text_shadow_blur_strength="0em" body_text_shadow_horizontal_length="0em" body_text_shadow_vertical_length="0em" body_text_shadow_blur_strength="0em"]I.S.I. (Indian Statistical Institute) B.Math.(Hons.) Entrance Examination 2008. Subjective Problem no. 9.
System of \(n\) Equations
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[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.22.4"]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: \(u^2-su+1=0\). \(\cdots\cdots\cdots\cdots (i)\)
[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.22.4"]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, \((n-1)u^2=1\).
\(\Rightarrow u=\frac{1}{\pm \sqrt{n-1}}\).
[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.22.4"]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.
[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.22.4"]
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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