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Infinite Series Expansion
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Quickster94
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#1
Infinite Series Expansion
Wondering what people think of this problem:

The fraction \dfrac{2}{x^2-3x+2} can be written as an infinite series. Find the sum of the first four terms of the series expansion for x=-1 and x=-2
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PostPosted: Mon Nov 02, 2009 6:11 pm  Back to top 
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#2
Probably a nicer way...

Solution

\frac{a}{1-r} is the formula for infinite geometric series...a=2 and r=-x^2+3x-1, which is the common ratio.

If x=-1, then this is -1-3-1=-5, so 2+2(-5)+2(-5)^2+2(-5)^3=-208.

If x=-2, then this is -4-6-1=-11, so 2+2(-11)+2(-11)^2+2(-11)^3=-2440.

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PostPosted: Mon Nov 02, 2009 6:25 pm  Back to top 
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Quickster94
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#3
Right, but that formula only works for -1<r<1 right? At both of those cases r isn't within those values...

Does anyone think the wording is sort of ambiguous?
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PostPosted: Wed Nov 04, 2009 7:08 pm  Back to top 
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randomguy64
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#4
Erm...if r isn't within those values, the sum would be either positive or negative infinity (or zero, if a is zero).

PostPosted: Wed Nov 04, 2009 8:05 pm  Back to top 
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AndrewTom
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#5
Hi. Using partial fractions, or by inspection,

\frac{2}{x^{2}-3x+2} = \frac{2}{(x-1)(x-2)} = \frac{2}{x-2} -\frac{2}{x-1}

= 2(1-x)^{-1} - (1-\frac{x}{2})^{-1}, valid for -1 < x < 1,

= 2+2x+2x^{2} +2x^{3} + ... - (1+ \frac{x}{2} + \frac{x^{2}}{4} + \frac{x^{3}}{8} + ...)

= 1+\frac{3x}{2} + \frac{7x^{2}}{4} + \frac{15x^{3}}{8} + ....

But this is not valid for x=-1 or x=-2

???

PostPosted: Thu Nov 05, 2009 9:37 am  Back to top 
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fishythefish
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#6
How about this?

When x=-1, the fraction simplifies to \frac{1}{3}, which can be written:

\displaystyle\sum_{x=1}^{\infty}\left(\frac{1}{4}\right)^x

a_1=\frac{1}{4}

a_2=\frac{1}{16}

a_3=\frac{1}{64}

a_4=\frac{1}{256}

a_1+a_2+a_3+a_4=\frac{85}{256}
-----------------------------------------------------------------------------------------------------------------

When x=-2, the fraction simplifies to \frac{1}{6}, which can be written:

\displaystyle\sum_{x=1}^{\infty}\left(\frac{1}{7}\right)^x

a_1=\frac{1}{7}

a_2=\frac{1}{49}

a_3=\frac{1}{343}

a_4=\frac{1}{2401}

a_1+a_2+a_3+a_4=\frac{400}{2401}
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PostPosted: Thu Nov 05, 2009 4:10 pm  Back to top 
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#7
@ confused audience: read the question carefully. It's not the whole series, hence convergence has nothing to do with this problem. Wink
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PostPosted: Thu Nov 05, 2009 4:28 pm  Back to top 
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randomguy64
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#8
Convergence has EVERYTHING to do with this problem.

The only way the question makes sense is if 2/(x^2-3x+2)=a+ar+ar^r+... and 0 < |r|< 1. If not, 2/(x^2-3x+2) must equal zero, and obviously x=-1 and x=-2 aren't roots, so that can't be it. Also, the fraction obviously doesn't form an infinite series, since there is absolutely no way to get the next term while the "x=-1 and x=-2" part makes sense.

@fishy: How does that work? More accurately, how did you get that formula out of the fraction given?

PostPosted: Thu Nov 05, 2009 6:22 pm  Back to top 
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#9
randomguy64 wrote:
Convergence has EVERYTHING to do with this problem.

The only way the question makes sense is if 2/(x^2-3x+2)=a+ar+ar^r+... and 0 < |r|< 1. If not, 2/(x^2-3x+2) must equal zero, and obviously x=-1 and x=-2 aren't roots, so that can't be it. Also, the fraction obviously doesn't form an infinite series, since there is absolutely no way to get the next term while the "x=-1 and x=-2" part makes sense.

@fishy: How does that work? More accurately, how did you get that formula out of the fraction given?

Quickster94 wrote:
Find the sum of the first four terms of the series expansion for x = - 1 and x = - 2

You absolutely do not need the concept of convergence to answer this question. Read the question, it's asking for the sum of the FIRST FOUR TERMS. Did I make myself clear?
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PostPosted: Thu Nov 05, 2009 11:08 pm  Back to top 
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azjps
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#10
That point is moot because the question is vague as written, as an "infinite series" implies that the desired fraction is the convergence of the partial sums of a certain sequence, of which there are certainly many sequences that could apply (eg, one can easily make the first four terms sum to whatever one likes - what's wrong with, say, the infinite sequence \pi,e,\pi^e,5,f(x) - (\pi + e + \pi^e + 5),0,0,0,\ldots?). I would presume that something like "power [Maclaurin] series" was intended, in which case see AndrewTom's post (though that series indeed does not converge).
Last edited by azjps on Fri Nov 06, 2009 3:21 pm; edited 1 time in total 
PostPosted: Fri Nov 06, 2009 8:51 am  Back to top 
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randomguy64
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#11
10000th User wrote:
randomguy64 wrote:
quote

You absolutely do not need the concept of convergence to answer this question. Read the question, it's asking for the sum of the FIRST FOUR TERMS. Did I make myself clear?


How are you planning to find the first four terms without realizing that the infinite series is a converging one, and therefore use the formula for it?

@fishythefish: ignore my previous question; I figured it out.

@azjps: you're right...

PostPosted: Fri Nov 06, 2009 3:14 pm  Back to top 
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#12
azjps wrote:
That point is moot because the question is vague as written, as an "infinite series" implies that the desired fraction is the convergence of the partial sums of a certain sequence, of which there are certainly many sequences that could apply (eg, one can easily make the first four terms sum to whatever one likes - what's wrong with, say, the infinite sequence \pi,e,\pi^e,5,f(x) - (\pi + e + \pi^e + 5),0,0,0,\ldots?). I would presume that something like "power [Maclaurin] series" was intended, in which case see AndrewTom's post (though that series indeed does not converge).
You do have a valuable point and this is perhaps what randomguy64 may be trying to say in a not-so-clear way.

Using power series, we could define a new function f(x), where f is the first four terms of the power series p(x). Finding f(-1) and f(-2) is now a feasible thing.

Quickster94: if the question is from an old book, it is possibly implied that the infinite series 'converges wherever its base point might be at'. Power series is usually the choice for representing a real function and thus, you can follow AndrewTom's series expansions.
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PostPosted: Fri Nov 06, 2009 8:11 pm  Back to top 
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atomicwedgie
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#13
Re: Infinite Series Expansion
Quickster94 wrote:
Wondering what people think of this problem:

The fraction \dfrac{2}{x^2 - 3x + 2} can be written as an infinite series. Find the sum of the first four terms of the series expansion for x = - 1 and x = - 2


There is only one ambiguity in this question, and that is the nature of the "infinite series." A function could have a variety of infinite series representations.
That said, however, it may be assumed that the problem is asking for the appropriate power series representation; i.e.,
f(x) = \sum_{k=0}^\infty a_k x^k
for suitable real constants a_k.
If we make this assumption, then the problem is completely unambiguous. Since
x^2 - 3x + 2 = (x-1)(x-2)
it follows that
\frac{2}{x^2 - 3x + 2} = 2\left(\frac{1}{x-2} - \frac{1}{x-1}\right) = \frac{2}{1-x} - \frac{1}{1-(x/2)}.
Since
\frac{1}{1-x} = 1+x+x^2+\cdots = \sum_{k=0}^\infty x^k,
we find
\frac{2}{x^2 - 3x + 2} = 2 \sum_{k=0}^\infty x^k - \sum_{k=0}^\infty (x/2)^k = \sum_{k=0}^\infty (2-2^{-k}) x^k.
Thus as a formal power series the given function has coefficients of the form a_k = 2-2^{-k}. We are not concerned with convergence because we are not interested in evaluating the power series in its entirety. All we are being asked to do is compute a partial sum of the formal power series:
g(x) = \sum_{k=0}^3 a_k x^k = \sum_{k=0}^3 (2-2^{-k}) x^k.
It is a straightforward exercise to compute g(-1) and g(-2). Again, it doesn't matter that the series is divergent for these values, because we are not trying to equate it to the value of 2/(x^2 - 3x + 2) at x = -1 or x = -2.

PostPosted: Sat Nov 07, 2009 1:30 am  Back to top 
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fishythefish
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#14
Debate over convergence aside, I'm still wondering what you all think of my solution. The question did not imply that we had to find the general expansion for the fraction, so I plugged in the values of x first, and then used infinite geometric series that I know to equal the results. After that, finding the first four terms is trivial.

BTW: Using this method, you can make the series converge. Razz
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PostPosted: Sat Nov 07, 2009 2:05 pm  Back to top 
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Quickster94
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#15
This is interesting

So it seems there's a lot of ambiguity about the question, which is what I thought. Anyway, "official solution" is problem one of below link:

http://www.vtmathcoalition.org/talent-search/TS0910Solutions1.pdf

Can anyone prove that this is the only solution or that more than one solution exists?
I feel like there must be more than one infinite series that sums to that in closed form...
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PostPosted: Sat Nov 07, 2009 8:47 pm  Back to top 
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atomicwedgie
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#16
Quickster94 wrote:
This is interesting

So it seems there's a lot of ambiguity about the question, which is what I thought. Anyway, "official solution" is problem one of below link:

http://www.vtmathcoalition.org/talent-search/TS0910Solutions1.pdf

Can anyone prove that this is the only solution or that more than one solution exists?
I feel like there must be more than one infinite series that sums to that in closed form...


Well, as I already mentioned, there are any number of "infinite series" representations of the given function--Fourier series, Laurent series, power series...the particular series expansion that was employed in the solution is a type of power series. One could see it as an expansion in powers of the reciprocal of x.

Their expansion is no more "correct" than the one I provided in my previous post. If the question had originally asked for x=-2 only, then some justification could be made in that the series provided in their solution is convergent for such a value. But at x = -1, neither power series is convergent.

It's a badly worded question, no doubt.

PostPosted: Sat Nov 07, 2009 9:47 pm  Back to top 
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#17
Can somebody explain how the infinite series in the "real solution" is equal to the fraction?

PostPosted: Sun Nov 08, 2009 3:46 pm  Back to top 
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Quickster94
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#18
They arrive at it using polynomial long division.
I don't really have the LaTeX skills to show out the work for that, so if more explanation is needed, some1 else will have to step in Mr. Green
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PostPosted: Wed Nov 11, 2009 8:02 pm  Back to top 
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