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June 15th, 2018, 01:05 AM | #1 |
Member Joined: Aug 2017 From: India Posts: 50 Thanks: 2 | 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, 08:37 AM | #2 |
Global Moderator Joined: Dec 2006 Posts: 20,274 Thanks: 1959 |
Consider the consequences of x$_1$ = 0.
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June 15th, 2018, 09:43 AM | #3 |
Senior Member Joined: Sep 2015 From: USA Posts: 2,299 Thanks: 1220 |
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, 10:04 AM | #4 |
Global Moderator Joined: Dec 2006 Posts: 20,274 Thanks: 1959 | |
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June 15th, 2018, 10:09 AM | #5 | |
Senior Member Joined: Sep 2015 From: USA Posts: 2,299 Thanks: 1220 | Quote:
Actually it's completely wrong. Let $x_1=0$ The resulting sequence is derived without complication and the product does equal 0. Last edited by romsek; June 15th, 2018 at 10:12 AM. | |
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June 15th, 2018, 10:58 AM | #6 |
Global Moderator Joined: Dec 2006 Posts: 20,274 Thanks: 1959 |
The product doesn't include x$_1$, so it isn't zero.
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June 15th, 2018, 11:47 PM | #7 |
Member Joined: Aug 2017 From: India Posts: 50 Thanks: 2 |
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 01:47 AM. |
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June 16th, 2018, 01:52 AM | #8 |
Global Moderator Joined: Dec 2006 Posts: 20,274 Thanks: 1959 |
Your answer for (A) is expressed unclearly. What about (E)? |
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June 16th, 2018, 02:03 AM | #9 |
Member Joined: Aug 2017 From: India Posts: 50 Thanks: 2 | |
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June 16th, 2018, 02:52 AM | #10 |
Global Moderator Joined: Dec 2006 Posts: 20,274 Thanks: 1959 |
(A) specifies x$_1$ is not equal to 1. However, (A) is incorrect.
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