Limit and factorization

Calculus Level 3

lim x 1 ( 1 x m + 1 m x + ( m 1 ) 1 x n + 1 n x + ( n 1 ) ) = 521 \lim_{x\to 1} \left( \frac{1}{x^{m+1}-mx+(m-1)} - \frac{1}{x^{n+1}-nx+(n-1)}\right) = 521

Given that m m and n n are positive integers satisfying the above limit. Find the minimum value of n n .


Inspiration


The answer is 261.

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2 solutions

Chan Lye Lee
Apr 26, 2018

Note that x m + 1 m x + ( m 1 ) = ( x 1 ) ( x m + x m 1 + + x 2 + x ( m 1 ) ) x^{m+1}-mx+(m-1) = \left( x-1 \right) \left( x^m + x^{m-1} + \ldots + x^2 + x -\left(m-1\right) \right) .

Then

lim x 1 ( 1 x m + 1 m x + ( m 1 ) 1 x n + 1 n x + ( n 1 ) ) = lim x 1 ( x n + x n 1 + + x 2 + x ( n 1 ) ) ( x m + x m 1 + + x 2 + x ( m 1 ) ) ( x 1 ) ( x m + x m 1 + + x 2 + x ( m 1 ) ) ( x n + x n 1 + + x 2 + x ( n 1 ) ) . \begin{aligned} & \lim_{x\to 1} \left( \frac{1}{x^{m+1}-mx+(m-1)} - \frac{1}{x^{n+1}-nx+(n-1)}\right) \\ &=\lim_{x\to 1} \frac{\left( x^n + x^{n-1} + \ldots + x^2 + x -\left(n-1\right) \right) -\left( x^m + x^{m-1} + \ldots + x^2 + x -\left(m-1\right) \right) }{\left( x-1 \right) \left( x^m + x^{m-1} + \ldots + x^2 + x -\left(m-1\right) \right) \left( x^n + x^{n-1} + \ldots + x^2 + x -\left(n-1\right) \right) }. \end{aligned}

Let f ( x ) = ( x n + x n 1 + + x 2 + x ( n 1 ) ) ( x m + x m 1 + + x 2 + x ( m 1 ) ) f(x)=\left( x^n + x^{n-1} + \ldots + x^2 + x -\left(n-1\right) \right) -\left( x^m + x^{m-1} + \ldots + x^2 + x -\left(m-1\right) \right) and g ( x ) = ( x 1 ) ( x m + x m 1 + + x 2 + x ( m 1 ) ) ( x n + x n 1 + + x 2 + x ( n 1 ) ) g(x)=\left( x-1 \right) \left( x^m + x^{m-1} + \ldots + x^2 + x -\left(m-1\right) \right) \left( x^n + x^{n-1} + \ldots + x^2 + x -\left(n-1\right) \right) .

As f ( 1 ) = g ( 1 ) = 0 f(1)=g(1)=0 , we may use L'Hôpital's Rule.

It is not difficult to see that f ( 1 ) = n ( n + 1 ) 2 m ( m + 1 ) 2 f^\prime(1)= \frac{n(n+1)}{2}-\frac{m(m+1)}{2} .

For g ( x ) g(x) , let g ( x ) = ( x 1 ) h ( x ) g(x)=(x-1) h(x) where h ( x ) = ( x m + x m 1 + + x 2 + x ( m 1 ) ) ( x n + x n 1 + + x 2 + x ( n 1 ) ) h(x)=\left( x^m + x^{m-1} + \ldots + x^2 + x -\left(m-1\right) \right) \left( x^n + x^{n-1} + \ldots + x^2 + x -\left(n-1\right) \right) . Now g ( x ) = ( x 1 ) h ( x ) + h ( x ) g ^\prime (x)=(x-1)h^\prime(x) + h(x) and so g ( 1 ) = h ( 1 ) = 1 g ^\prime (1)=h(1)=1 as we know that h ( 1 ) h^\prime(1) exists.

Now,

lim x 1 ( 1 x m + 1 m x + ( m 1 ) 1 x n + 1 n x + ( n 1 ) ) = lim x 1 ( x n + x n 1 + + x 2 + x ( n 1 ) ) ( x m + x m 1 + + x 2 + x ( m 1 ) ) ( x 1 ) ( x m + x m 1 + + x 2 + x ( m 1 ) ) ( x n + x n 1 + + x 2 + x ( n 1 ) ) = lim x 1 f ( x ) g ( x ) = lim x 1 f ( x ) g ( x ) = f ( 1 ) g ( 1 ) = n ( n + 1 ) 2 m ( m + 1 ) 2 \begin{aligned} & \lim_{x\to 1} \left( \frac{1}{x^{m+1}-mx+(m-1)} - \frac{1}{x^{n+1}-nx+(n-1)}\right) \\ &=\lim_{x\to 1} \frac{\left( x^n + x^{n-1} + \ldots + x^2 + x -\left(n-1\right) \right) -\left( x^m + x^{m-1} + \ldots + x^2 + x -\left(m-1\right) \right) }{\left( x-1 \right) \left( x^m + x^{m-1} + \ldots + x^2 + x -\left(m-1\right) \right) \left( x^n + x^{n-1} + \ldots + x^2 + x -\left(n-1\right) \right) } \\ &=\lim_{x\to 1} \frac{f(x)}{g(x)}\\ &=\lim_{x\to 1} \frac{f^\prime(x)}{g^\prime(x)}\\ &= \frac{f^\prime(1)}{g^\prime(1)}\\ &=\frac{n(n+1)}{2}-\frac{m(m+1)}{2} \end{aligned}

This means that n ( n + 1 ) 2 m ( m + 1 ) 2 = 521 \frac{n(n+1)}{2}-\frac{m(m+1)}{2} = 521 . As 521 is a prime number, there are only two pairs of positive integer solutions ( m , n ) (m,n) to this Diophantine Equation: ( n m ) ( n + m + 1 ) = 2 × 521 = 1 × 1042 (n-m)(n+m+1)=2\times 521 = 1\times 1042

Namely: n m = 2 , n + m + 1 = 521 n-m=2, n+m+1=521 or n m = 1 , n + m + 1 = 1042 n-m=1, n+m+1=1042 . Solve the equations and we obtain ( m , n ) = ( 259 , 261 ) (m,n)=(259, 261) or ( 520 , 521 ) (520, 521) . Hence the minimum value of n n is 261 \boxed{261} .

Pi Han Goh
May 9, 2018

Let f L ( x ) = x L + 1 L x + ( L 1 ) f_L (x) = x^{L+1} - Lx + (L-1) . Then, f L ( x ) = ( L + 1 ) x L L f L ( x ) = L ( L + 1 ) x L 1 f_L ' (x) = (L+1)x^L - L \Rightarrow f_L ''(x) = L(L+1)x^{L-1} .

And f L ( 1 ) = 0 , f L ( 1 ) = 1 , f L ( 1 ) = L ( L + 1 ) f_L (1) = 0 , f_L '(1) = 1 , f_L ''(1) = L(L+1) .

Given, 521 = lim x 1 ( 1 x m + 1 m x + ( m 1 ) 1 x n + 1 n x + ( n 1 ) ) = lim x 1 ( 1 f m ( x ) 1 f n ( x ) ) = lim x 1 ( f n ( x ) f m ( x ) f m ( x ) f n ( x ) ) Apply L’hopital rule because 0 / 0 = lim x 1 ( f n ( x ) f m ( x ) f m ( x ) f n ( x ) + f m ( x ) f n ( x ) ) Apply L’hopital rule again because 0 / 0 = lim x 1 ( f n ( x ) f m ( x ) f m ( x ) f n ( x ) + f m ( x ) f n ( x ) + f m ( x ) f n ( x ) + f m ( x ) f n ( x ) ) = n ( n + 1 ) m ( m + 1 ) 0 + 2 × 1 2 + 0 521 × 8 = 4 n ( n + 1 ) 4 m ( m + 1 ) = ( 2 n + 1 ) 2 ( 2 m + 1 ) 2 = ( 2 n 2 m ) ( 2 n + 2 m + 2 ) 521 × 2 = ( n m ) ( n + m + 1 ) \begin{array} { r l l } \displaystyle 521 =& \displaystyle \lim_{x\to 1} \left( \frac{1}{x^{m+1}-mx+(m-1)} - \frac{1}{x^{n+1}-nx+(n-1)}\right) \\ \displaystyle =& \displaystyle \lim_{x\to 1} \left( \dfrac1{f_m(x)} - \dfrac1{f_n(x)} \right) \\ \displaystyle =& \displaystyle \lim_{x\to 1} \left( \dfrac{f_n(x) - f_m (x) }{f_m(x) \cdot f_n (x) } \right) & \qquad \text{Apply L'hopital rule because }0/0 \\ \displaystyle =& \displaystyle \lim_{x\to 1} \left( \dfrac{f_n'(x) - f_m '(x) }{f_m'(x) \cdot f_n (x) + f_m(x) \cdot f_n' (x) } \right) & \qquad \text{Apply L'hopital rule again because }0/0 \\ \displaystyle =& \displaystyle \lim_{x\to 1} \left( \dfrac{f_n''(x) - f_m ''(x) }{f_m''(x) \cdot f_n (x) + f_m'(x) \cdot f_n' (x) + f_m'(x) \cdot f_n' (x) + f_m(x) \cdot f_n'' (x) } \right) \\ =& \dfrac{ n(n+1) - m(m+1) }{0 + 2\times1^2 + 0} \\ 521 \times8 =& 4n(n+1) - 4m(m+1) = (2n + 1)^2 - (2m+1)^2 = (2n - 2m)(2n + 2m + 2) \\ 521 \times2 =& (n-m)(n + m + 1) \\ \end{array}

Then, ( n m , n + m + 1 ) = ( 2 , 521 ) , ( 1 , 521 × 1 ) ( n , m ) = ( 521 , 520 ) , ( 261 , 259 ) (n-m, n+m+ 1) = (2,521) , (1,521\times1) \Rightarrow (n,m) =(521,520), (261,259) . The answer is 261 \boxed{261} .

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