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June 15th, 2018, 12:05 AM   #1
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Failing to understand the problem

I am failing to understand the problem to make any attempt. How to solve it ? and what is the significance of these kind of problems?
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June 15th, 2018, 07:37 AM   #2
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Consider the consequences of x$_1$ = 0.
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June 15th, 2018, 08:43 AM   #3
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for (a) consider $x_1 = 2$

for (b) consider values where $x = \dfrac{1}{x-1}$

for (c) count how many solutions there are to the equation mentioned for (b)

for (d), suppose there was a set of $x_k$ such that the product equalled $0$.

well clearly one of the $x_k=0$. It can only be the 20007th value or we're going to try and divide by $0$ to find $x_{k+1}$

But there is no value $x$ such that $\dfrac{1}{x-1}=0$ and thus there is no way for the recursion to produce $x_{2007}=0$
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June 15th, 2018, 09:04 AM   #4
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Quote:
Originally Posted by romsek View Post
(d) . . . it can only be the 20007th value or we're going to try and divide by $0$ to find $x_{k+1}$
That reasoning is incorrect, as x$_{2006}$ = 0 implies x$_{2007}$ = -1.
However, no zero occurs in the product, because 1/(x - 1) cannot be zero.
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June 15th, 2018, 09:09 AM   #5
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Quote:
Originally Posted by skipjack View Post
That reasoning is incorrect, as x$_{2006}$ = 0 implies x$_{2007}$ = -1.
However, no zero occurs in the product, because 1/(x - 1) cannot be zero.
Yes, I had just noticed that and was correcting it. Thank you.

Actually it's completely wrong. Let $x_1=0$

The resulting sequence is derived without complication and the product does equal 0.
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Last edited by romsek; June 15th, 2018 at 09:12 AM.
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June 15th, 2018, 09:58 AM   #6
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The product doesn't include x$_1$, so it isn't zero.
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June 15th, 2018, 10:47 PM   #7
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Thank you very much. From your support, I finally arrived like this

A.
$\displaystyle
x1 = 2, x2 = 1 (2,1),
x1=1, \text{ Invalid Condition}
$

B.
Solving for $\displaystyle x = 1/(x-1); $

$\displaystyle
x = (1 \pm \sqrt{5})/2
$

C. There are only two choices for x as above in B.
D. There does not exist a choice.

Last edited by skipjack; June 16th, 2018 at 12:47 AM.
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June 16th, 2018, 12:52 AM   #8
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Your answer for (A) is expressed unclearly.

What about (E)?
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June 16th, 2018, 01:03 AM   #9
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Quote:
Originally Posted by skipjack View Post
Your answer for (A) is expressed unclearly.

What about (E)?
A.
$\displaystyle
x1=2,x2=1 (2,1)
$
x1 = 1; The equation is not valid.

OK. I understand now the answer should be A.
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June 16th, 2018, 01:52 AM   #10
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(A) specifies x$_1$ is not equal to 1. However, (A) is incorrect.
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