| Joel
Kammet |
In my calculus II class last semester, we sped through the section on series too fast to think much about the actual meaning of what we were doing, so now I'm trying to make some sense of it all. OK, so I'm given this function f(x)=(7x-1)/ (3x2+2x-1) and I have to find a power series representation of it. So, simplify the function using partial fractions to get f(x)=1/(3x-1)+2/(x+1). Then monkey around with it some more and I get:
. And it appears that the interval of convergence is (-1/3, 1/3). So, after all that, although there is a whole universe of possible values of x > 1/3 and x < -1/3 (x =/=-1) for the original function, the series is a valid representation of the function only within the interval (-1/3,1/3). Yes?? Meaning, I guess, that if you want to use a series to simplify working with some function in a "real" application, the first thing you must do (after finding the series) is determine that the series' interval of convergence coincides with the domain of values you want/need/expect/observed, etc.? Is that the only conclusion to be drawn from all of this, or is there some profound truth that I'm completely missing? ?? |
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| Arun
Iyer |
Joel, In your example above you have explicitly used the power series.. 1/(1-x) = 1+x+x2 +x3 +.... Now here are some questions for you to think about: 1> Can i find a power series for a function?? 2> Is there more than one series possible for a particular function??? 3> does each of the power series found share the same interval of convergence?? 4> Are the limits posed upon these series' mathematical or physical?? love arun P.S-> look at the next post on Uniqueness of Power series to give yourself some ideas. |
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| Joel
Kammet |
Arun, I was already thinking along those lines, but your questions provide a helpful frame of reference. So: 1> I have a very limited repertoire of ways to find a power series OTHER THAN A TAYLOR SERIES for a given function. It has to be a function that I can manipulate (by factoring, differentiating, integrating, etc) into something that looks like 1/1-x or ex or sin x or cos x. I assume there are others; but these are the only ones I know. 2> Clearly yes. At some point last night I finally realized that the whole point of finding a Taylor series about some "a" other than 0 is to be able to modify the domain of x-values for which the series can be used. Duh! Last night I tried unsuccessfully to find another (NON-TAYLOR) series for that function (7x-1)/(3x2 +2x-1). Eventually I gave up on that, and came up with a Taylor series that does the trick:
3> Now, by changing the value of a, the interval of convergence apparently can be changed to any desired range (although for each a, the radius of convergence remains the same, 2/3). It's also clear, as Andre pointed out in Marcos' thread, that the coefficients remain constant for all values of a (very convenient for programming). 4> I'm not exactly sure what you're getting at here. I would say that the limitations of the series are mathematical, while the limitations of the applications (whatever they may be) are physical. And the Taylor series lets us tailor (I couldn't resist that) the series to the constraints of the application. Would it be accurate to say that the Taylor series technique is THE way to find multiple series for a given function, or is it just one of several ways? Where do I go from here? Thanks for your help. Joel PS: in case anyone's interested, here's a cute little program that evaluates that series. I left the summation expression in 2 steps to make it easier to alter it for different series. // Taylor.cpp // evaluates Taylor series about a for F(x) = f(x)+g(x) = 1/(3x-1) + 2/(x+1) // for other functions, modify a, x, u and v accordingly #include < iostream.h> #include < stdlib.h> #include < math.h> #define DEGREE 10 int main() { int n; double x,a,u,v,sum=0.0; a = 8; x = 8; u = 1/(3*a-1); // u represents f(a) v = 1/(a+1); // v represents (1/2)*g(a) (don't want to apply exponent to the numerator) for(n=0;n < DEGREE;n++) { sum += pow(u,(n+1)) * pow(-1,n) * pow(3.0,n) * pow((x-a),n); sum += 2 * pow(v,(n+1)) * pow(-1,n) * pow((x-a),n); cout < < "partial sum " < < sum < < endl; } cout < < sum < < endl; system("PAUSE"); return 0; } |
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| Arun
Iyer |
1> OK 2> Right. 3> It is not necessary that the series converges for all a's.So one has to be a careful here. 4> Right. Now as for Other ways of getting a series.One way could be of finding a fourier series. love arun |
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| Andre
Rzym |
Joel, You have definitely got the idea. There is more, though, that you will meet in university maths. Note that the radius of convergence of your first series is 1/3 not 2/3, and if you change 'a', the radius will usually change. 1) The definition of f(x) is perfectly valid for complex x as well. So is the series expansion. It turns out that the series converges for |x| < 1/3. So it is actually a circle of convergence (of radius 1/3) 2) You will have noticed that f(x) is singular at x=-1, x=1/3. The power series is a (Taylor) expansion about x=0. The nearest singularity to x=0 is x=1/3. It is this singularity that causes the radius of convergence to be 1/3. This is a consequence of the Cauchy-Taylor Theorem . Therefore if you compute a Taylor expansion about, say, x=10, the radius of convergence will be different. 3) The series diverges beyond x=1/3. E.g., if you substutute x=10, the series diverges. Nevertheless, by a process called Analytic Continuation (have a look on Mathworld) we can take the original series, and keep generating new series (about different centres) until one is valid for x=10. It will agree with your original definition of f(x) for x=10! I'll leave you to judge whether any of this is (as you put it) profound or not! Andre |
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| Joel
Kammet |
3> Are you referring here only to a's for which the original expression is undefined if x=a, i.e. 1/3 and -1 in my example, or might there be others? I haven't had any contact with Fourier series yet, although I believe I will this semester in Differential Equations. Anything else I should be looking into along the lines of the series I listed in "1> "? |
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| Joel
Kammet |
I just noticed Andre's post which appeared while I was responding to Arun's. I see that there is much there to think about as well. I can say right away that yes, of course the radius is 1/3 & I have no idea why I said 2/3 (inventing the "diameter of convergence", I guess). The rest will take much more time. Thanks again to both of you. |
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| Joel
Kammet |
I see now that each time I change a, not only does the interval of convergence change, but the radius changes as well. But the relationship that Andre points out between the singularities and the radius of convergence is even more interesting. It seems that for every value of a, R = |a-1/3|. A bit surprising that this is true even for a < -1. So the other singularity seems to have no impact on R. I looked at series for several other functions, and it seems that in each case where singularities exist in the function, the same relationship exists in the series. And where there were no singularities in the function, R = infinity . Is this universally true, or was I just lucky in my choice of functions? I couldn't find much on the Cauchy-Taylor theorem. Does it go by another name? |
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| Andre
Rzym |
Joel, If I read your second paragraph correctly, it?s not quite correct. Suppose we compute a Taylor expansion about x=a. Then the radius of convergence is governed by the distance to the nearest singularity. Thus suppose we expand about x = -2. The nearest singularity is at x = -1. Therefore the radius of convergence is |(-2)-(-1)| = 1. The singularity at x = 1/3 has no bearing on R. I probably ought to issue a health warning here: If you are not expected to know the C-T theorem, you probably ought not to use it to answer exam questions on convergence of power series! It?s a good check on your results, though. Andre |
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| Joel
Kammet |
Regarding R for the expansion about -2: that's what I expected based on your post, but somewhere in the course of all those ratio tests I got too "efficient" & started testing only the .../(3a-1)n+1 part of the expression, ignoring the .../(a+1)n+1 part. By the time I got to testing negative values I had forgotten that I was abbreviating the test. Thanks for making me look again. Regarding the C-T theorem, why is it barely mentioned on the web? Google comes up with only 3 references, and even those are very brief. Regarding using it: never occurred to me. O |
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| Ben
Tormey |
Joel, Another useful theorem: Let
be analytic in some region containing the origin, let a singularity of f(z) of smallest modulus be at a point z0 ¹ 0, and let e > 0 be given. Then there exists N such that for all n > N we have |an| < (1/|z0| + e)n. Further for infinitely many n we have |an| > (1/|z0| - e)n. ... So if
, then |an| = |3n - 2(-1)n| < (3+e)n. |
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| Joel
Kammet |
I'm sure it is Ben, & someday I plan to understand what you posted. But when I got to the 3rd level of words to look up in order to understand the definitions of the words I was looking up, I decided that today is not that day. Thanks anyway. Joel |