Infinite tangents

Calculus Level 5

Consider a point P 1 P_1 on the curve y = x 3 y=x^3 such that the tangent on P 1 = ( 1 , 1 ) P_1 = (1,1) meets the curve again at P 2 P_2 . And the tangent at P 2 P_2 meets the curve at P 3 P_3 and so on.

Let ( x n , y n ) (x_n,y_n) be the coordinates of P n P_n then evaluate:

lim ⁡ n → ∞ ∑ r = 1 n 1 x r lim ⁡ n → ∞ ∑ r = 1 n 1 y r \displaystyle \dfrac{\displaystyle \lim_{n \to \infty} \sum_{r=1}^{n} \dfrac{1}{x_r}}{\displaystyle \lim_{n \to \infty} \sum_{r=1}^{n}\dfrac{1}{y_r}}

If the answer is of the form A B \dfrac AB , where A A and B B are coprime positive integers , find A + B A+B ..

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The answer is 7.

Neelesh Vij by neelesh vij , 21, India

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1 solution

Let any point on the curve be ( h , h 3 ) (h, h^{3})
The equation of tangent at this point is,
y − h 3 = 3 h 2 ( x − h ) y - h^{3} = 3h^{2}(x-h)
y = 3 h 2 x − 2 h 3 y = 3h^{2}x - 2h^{3}
Equating it with the curve again,
x 3 = 3 h 2 x − 2 h 3 x^{3} = 3h^{2}x - 2h^{3}
x 3 − 3 h 2 x + 2 h 3 = 0 x^{3} -3h^{2}x + 2h^{3} = 0
( x − h ) 2 ( x + 2 h ) = 0 (x-h)^{2}(x+2h) = 0
This forms a relation,
x r + 1 = − 2 x r → y r + 1 = − 8 y r x_{r+1} = -2x_{r} \rightarrow y_{r+1} =-8y_{r}
∑ r = 1 ∞ 1 x r ∑ r = 1 ∞ 1 y r = 1 1 − ( − 1 2 ) 1 1 − ( − 1 8 ) = 3 4 \dfrac{\displaystyle \sum_{r=1}^{\infty}\frac{1}{x_{r}}}{\displaystyle \sum_{r=1}^{\infty} \frac{1}{y_{r}}} = \dfrac{\frac{1}{1-(-\frac{1}{2})}}{\frac{1}{1-(-\frac{1}{8})}} = \dfrac{3}{4}
A + B = 7 A + B = 7


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Yess.......Just the same!!!!

Aaghaz Mahajan - 3 years, 3 months ago

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