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In the sequence of positive number x1, x2, x3,. . . what is

by Guest Fri Jun 22, 2007 3:46 pm

In the sequence of positive number x1, x2, x3,. . . what is the value of x1??


Image


I didn't know where to begin with this one! Any advice would be appreciated.
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by StaceyKoprince Fri Jun 22, 2007 9:04 pm

Hi, looks like you tried to load an image but can't see anything. Can you either upload again or try to describe via text?
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Original Problem

by Guest Sat Jun 23, 2007 9:47 pm

(1) X (sub j) = X(sub j - 1) / 2 for all integers j > 1

(2) X (sub 5) = X(sub 4) / ((X sub 4) + 1)


I hope this makes sense - it's a series problem with X sub 5 representing the 5th member of the series, etc,, etc. Thanks!
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by StaceyKoprince Mon Jun 25, 2007 7:21 pm

Thanks, I think I get the statements now. Can you also just double-check the wording of the question itself for me? You say in your second post that it is a "series" problem, but the text you've shown for the problem does not indicate a series, it indicates a sequence (these are two different, though related, things). I'm not sure if you're just using the terms interchangeably because you weren't aware of the difference between the two terms or if there's some wording missing from the question. So just make sure that the question is worded exactly the way it is on the test.
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by Guest Tue Jun 26, 2007 8:30 am

Yes, the wording on the question is correct. I used series interchangeably with sequence by accident.
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by Guest Wed Feb 13, 2008 6:15 pm

can some one solve this one.
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by StaceyKoprince Fri Feb 15, 2008 2:06 am

We're asked to find the first term in the sequence and given a couple of recursive equations. To start, know that you can't solve recursive equations with just one recursive equation and nothing else (no starting point, no other equation, etc.). One piece of info isn't enough.

(1) can't do anything with this by itself. Eliminate A and D.

(2) Again, with just this equation, we can't calculate anything. Eliminate B.

(1) + (2) Now maybe we can do something. We can combine our equations here. If we use j = 5 for the first equation, we get x-sub-5 = (x-sub-4)/2. The second equation tells us that x-sub-5 = (x-sub-4)/(x-sub-4 + 1). Combine parts to get 2 = x-sub-4 + 1 or 1 = x-sub-4.

Now that I have a value for x-sub-4, I have a starting point! If you completely understand how recursive sequences work (ie, a starting point is sufficient), you can stop here. The actual math is below.
From eqn 1: x-sub-4 = (x-sub-3)/2 = 1. Therefore, x-sub-3 = 2
repeat: x-sub-3 = (x-sub-2)/2 = 2. Therefore, x-sub-2 = 4
repeat: x-sub-2 = (x-sub-1)/2 = 4. Therefore, x-sub-1 = 8
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Re:

by sudaif Thu Jul 22, 2010 6:08 am

StaceyKoprince Wrote:We're asked to find the first term in the sequence and given a couple of recursive equations. To start, know that you can't solve recursive equations with just one recursive equation and nothing else (no starting point, no other equation, etc.). One piece of info isn't enough.

(1) can't do anything with this by itself. Eliminate A and D.

(2) Again, with just this equation, we can't calculate anything. Eliminate B.

(1) + (2) Now maybe we can do something. We can combine our equations here. If we use j = 5 for the first equation, we get x-sub-5 = (x-sub-4)/2. The second equation tells us that x-sub-5 = (x-sub-4)/(x-sub-4 + 1). Combine parts to get 2 = x-sub-4 + 1 or 1 = x-sub-4.

Now that I have a value for x-sub-4, I have a starting point! If you completely understand how recursive sequences work (ie, a starting point is sufficient), you can stop here. The actual math is below.
From eqn 1: x-sub-4 = (x-sub-3)/2 = 1. Therefore, x-sub-3 = 2
repeat: x-sub-3 = (x-sub-2)/2 = 2. Therefore, x-sub-2 = 4
repeat: x-sub-2 = (x-sub-1)/2 = 4. Therefore, x-sub-1 = 8


Stacey: i had a follow-up question, although this is an old post.
Since these are recursive sequences, would it be okay if in the statement 1 + statement 2 scenario, I took statement 1 as x-sub-2=x-sub-1/2 and statement 2 as x-sub-2=x-sub-1/((x-sub-1)+1).
The reason why I ask that is because then we could directly get x-sub-1. I feel I may be missing something here. Thanks in advance.
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Re: In the sequence of positive number x1, x2, x3,. . . what is

by maaz_gmat Fri Jul 23, 2010 3:13 am

Hi,

Could you explain how u got this part -
Combine parts to get 2 = x-sub-4 + 1 or 1 = x-sub-4.
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Re: In the sequence of positive number x1, x2, x3,. . . what is

by sudaif Fri Jul 23, 2010 3:43 am

your question is for stacey?
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Re: In the sequence of positive number x1, x2, x3,. . . what is

by brent.carpenetti Fri Jul 23, 2010 8:41 pm

For ease of entering calculations into this question, please note that all X-sub-4 will be displayed as x.

After combining the statements and setting j=5 so that statement 1 can be made equal to statement two, I arrive at this equation:

(x)/(x+1)=(x)/(2)

Stacey, I think you set the denominators equal to each other because the numerators were equivalent. However, if we do so, it neglects the possibility of x = 0 to solve the equation.

If we cross multiply the equation, we arrive at the following:

x^2 + x = 2*x

Simplified:

x^2 = x

x = -1,0, or 1

However, x = -1 is impossible because (x)/(x+1) would become:
(-1)/(-1+1) == (-1)/(0)
Division by zero is impossible.

Thus we are left with x = 0 or 1.
The answer is E, both statements together are not sufficient to answer the question. We cannot be certain where to start the recursive formula.

Stacey I could be wrong, I often am. Is there something I missed in this problem?

-Brent Carpenetti
Carpenetti Calculus
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Re: In the sequence of positive number x1, x2, x3,. . . what is

by rohit801 Sat Jul 24, 2010 4:39 pm

Brent,
the question stems says -"In the sequence of positive number x1, x2, x3,. . . what is the value of x1??

Your comment -" However, if we do so, it neglects the possibility of x = 0 to solve the equation" will imply that all the numbers in the sequence will be 0.

That is, if x4=0, then given the recursive relation in the sequence, all x[i] will be 0, which can't be the case. Hence, we are ok with this assumption.

Hope it helps!
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Re: In the sequence of positive number x1, x2, x3,. . . what is

by brent.carpenetti Sun Jul 25, 2010 4:57 pm

Rohit,

Yes, of X-sub-4 = (0,1) we can eliminate the possibility of zero.

The positive number required by the problem stem does remove zero from our possible solutions.

I negelcted the positive number requirement. Leaving only X-sub-4 = 1 . (Even if -1 worked.)

Hopefully I can pay more attention on test day!

-Good looking out.

Brent Carpenetti
Carpenetti Calculus
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Re: In the sequence of positive number x1, x2, x3,. . . what is

by sudaif Mon Jul 26, 2010 3:09 am

maybe the instructors missed my query. would appreciate some feedback. thank you!
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Re: In the sequence of positive number x1, x2, x3,. . . what is

by RonPurewal Thu Aug 05, 2010 8:17 am

sudaif Wrote:maybe the instructors missed my query. would appreciate some feedback. thank you!


don't do this -- i.e., don't post a message that says "please answer my question".
this is called "bumping" the thread; it brings the thread up to the most recent position in the folder.

the problem, of course, is that we answer the posts strictly in order from oldest to newest. therefore, if you post a message, with no content, that says "please answer this post", then you are moving the thread to the LAST place in the queue.

please be patient -- we will get to all of the threads. if you make posts like this one, you're just making it take longer.
thanks.

--

now, for the actual question:

Since these are recursive sequences, would it be okay if in the statement 1 + statement 2 scenario, I took statement 1 as x-sub-2=x-sub-1/2 and statement 2 as x-sub-2=x-sub-1/((x-sub-1)+1).


you can do that with statement 1, but NOT with statement 2.

statement 2 is not a recursive statement -- notice that there are no index variables (such as "j" or "n") in that statement.
statement 2 is a statement that deals only with the specific 4th and 5th terms of the sequence, and is in fact false if you insert x1 (= 8) and x2 (= 4) into it.