GR9367 #18, 19

Forum for the GRE subject test in mathematics.
ParsleyX
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Joined: Tue Oct 30, 2012 3:46 pm

GR9367 #18, 19

Postby ParsleyX » Mon Nov 05, 2012 5:04 pm

Link to test: http://www.math.ucla.edu/~cmarshak/GRE2.pdf

#18 - I figure this ultimately comes down to knowing power series expansions for some common mathematical expressions, but I am weak in this area. How would you approach this? What should I brush up on? Please explain it like I'm 5.

#19 - "Which of the following is a general solution of the differential equation:
y''' - 3y'' + 3y' - y = 0

And then each option is a bunch of e^t terms.

This is a type of problem where I'd probably waste a lot of time trying to find the derivatives and plugging things together. It's easy to make a mistake going that route. Are there any facts or approaches to this problem which make it less mechanical or more obvious what *isnt* the answer?

Thanks!

TheYoungin
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Joined: Mon May 30, 2011 10:18 pm

Re: GR9367 #18, 19

Postby TheYoungin » Mon Nov 05, 2012 6:43 pm

For #18, notice that the expression inside the sum can be written as (-x^2)^n so the infinite sum is a geometric series with rate r=-x^2. Then using the formula for the infinite series here http://en.wikipedia.org/wiki/Geometric_series#Formula should get you the solution.


For #19, I just solved the r^3 - 3r^2 + 3r - 1 = 0 and noticed r=1 was a solution. After using that, I figured out that r=1 was a solution with multiplicity three. So the solution is of the form c1*e^t + c2*t*e^t + c3*t^2*e^t. Not sure of any other way to do it.

math_applicant
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Re: GR9367 #18, 19

Postby math_applicant » Mon Nov 05, 2012 7:26 pm

TheYoungin wrote:For #19, I just solved the r^3 - 3r^2 + 3r - 1 = 0 and noticed r=1 was a solution. After using that, I figured out that r=1 was a solution with multiplicity three. So the solution is of the form c1*e^t + c2*t*e^t + c3*t^2*e^t. Not sure of any other way to do it.


for #19, use the coefficients of t in the power of e in every solution and check which one works

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redcar777
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Re: GR9367 #18, 19

Postby redcar777 » Mon Nov 05, 2012 8:04 pm

18) Its nothing fancy when it comes to series-- its a geometric series, as @TheYoungin points out.

So its probably easier to sum and then take the derivative. Sometimes its easier to do the term-by-term derivative and then to sum the series, but not in this case.

f(x) = 1/(1+x^2)
f'(x) = -2x/(1+x^2)^2

19) Let D be the derivative, considered as an operator, and then factor the equation they way you would a polynomial

D^3 y - 3D^2 y +3 Dy - y =0
or (D^3 - 3D^2 + 3D - 1)y = 0
or (D-1)^3 y = 0

So one solution is given by solving (D-1)y = 0, which is yields y=e^t (in general, a factor of (D-r) gives a solution of e^rt )

So given a solution of e^t and a repeated factor, the next is given by solving (D-1)y = e^t, which happens to a have a solution te^t.

Given a solution of the form te^t and a factor which appears three times, the next
is given by solving (D-1)y = te^t, which happens to have a solution 1/2 t^2 e^t

In general, if you have k+1 repeated factors like (D-r) in the equation, then you can iteratively solve starting with e^rt to get a general solution like 1/k! t^k e^rt (though the leading constant factor doesn't really matter). You can see this taking the solution e^rt and using the technique of variation of parameters to try a solution of the form y=u(t) e^rt to get an equation like u'(t) = t^(k-1)/(k-1)!, which you can directly integrate.




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