Sum Slopes at infinity

Calculus Level pending

The function x 3 + y 3 = 1 { x }^{ 3 }+{ y }^{ 3 }=1 has a slope of m as x approaches negative infinity, and a slope of n as y approaches negative infinity.

If:

∑ i = 0 ∞ m i + ∑ j = 2 ∞ n j = q \displaystyle \sum _{ i=0 }^{ \infty }{ { m }^{ i } } +\sum _{ j=2 }^{ \infty }{ { n }^{ j } } = q ;

Then what is the largest value that q can have?


The answer is 2.

Seth Lovelace by Seth Lovelace , 23, USA

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1 solution

Seth Lovelace
Nov 28, 2014

The function x 3 + y 3 = 1 { x }^{ 3 }+{ y }^{ 3 }=1 has derivative d y d x = − x 2 1 − 2 x 3 + x 6 3 \frac { dy }{ dx } =\frac { { -x }^{ 2 } }{ \sqrt [ 3 ]{ 1-{ 2x }^{ 3 }+{ x }^{ 6 }}} . By taking the limit as x goes to negative infinity, it can be seen that the slope will be -1. By symmetry, the slope as y approaches negative infinity must also be -1. Thus m=-1, and n=-1.

Now, if we look at the series, it can be seen that they are in fact the same series:

1-1+1-1+1-1...

Now this series can have three potential values, depending on how you place your parentheses. If you evaluate likewise the series will equal:

(1-1)+(1-1)+(1-1)... = 0

1+(-1+1)+(-1+1)... = 1

1-1+1-1+1-1+1... = S ; By rearranging terms by subtracting 1, we get:

S-1=-1+1-1+1-1+1-1... ; -1+1-1+1-1... = -S, thus:

S-1 = -S ; 2S = 1 ; S = 1 2 \frac { 1 }{ 2 }

We are looking for the largest value of q, which was determined to be equal to 2S. The largest value of S is 1. Thus, q = 2(1) = 2 \boxed { 2 }

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