Mastering Mathematics Smartly
by Wee Wen Shih

It seems that the forum features of Freewebs are a little problematic. I've decided to move our discussions over to Delphi Forums.

Please continue with your questions on the new forum platform. Thank you!

Let me answer a question from the old forum, on which I could not post a response.

Query:

Im quite puzzled about something. Was sharing with my friend this question and she asked me why can't we subsitute x as 13/14.

13/14 is derived from below:

(1+x)/(1-x) = 27

The final answer is incorrect though but the steps seem logical. Furthermore 13/14 is less within |x|<1..

Response:

Hi,

  The choice of x is not just a value that satisfies the condition of the validity range only.

  The value of x should also be chosen such that x^n becomes very small as n increases. If we let x = 13/14, (13/14)^2 and (13/14)^3 are still quite large (0.8 plus). However, if we take x as 1/26, then (1/26)^2 and (1/26)^3 are very small values.

  Hope it clarifies your doubt. Thanks!

Cheers,
Wen Shih

Query:

This is a question from Prelim.

The numbers X_n satisfy the relation X_n+1 = -sqrt(((2X_n)+1)/5) for n=1,2,3...
As n --> infinity, Xn --> k
(i) Find the exact value of k
(ii) Show that if X_n < k, then X_n < X_n+1

For part (ii),the solution given by the teacher is:

X_n< (1-sqrt(6))/5 ---> 5((X_n)^2)-(2X_n)-1>0

--> ((5X_n)^2) > 2(X_n) + 1
---> X_n < - Sqrt((2X_n)+1)/5)
--> X_n < X_n+1

What is the logic between this 2 steps below?
--> ((5X_n)^2) > 2(X_n) + 1
---> X_n < - Sqrt((2X_n)+1)/5)


How did the inequality change? And how is it possible to Sqrt both sides in an inequality?

Response:

First, X_n = k satisfies the equation 5(X_n)^2 - 2X_n - 1 = 0.

So if X_n < k, 5(X_n)^2 - 2X_n - 1 > 0. This can be seen by means of a U-shaped graph cutting the x-axis at x = k.

Rearranging, 5(X_n)^ 2 > 2X_n + 1.

=> (X_n)^2 > 1/5 (2X_n + 1)

We reject X_n > sqrt[ 1/5 (2X_n + 1) ] and take the other, i.e.

X_n < - sqrt[ 1/5 (2X_n + 1) ], since X_n is negative as a result of X_n < k and k is negative.

Thus, X_n < X_(n + 1), since X_(n + 1) = - sqrt[ 1/5 (2X_n + 1) ] by definition.