Lesson 26: Taylor series and asymptotic series restart;

I don't think there's an easy way in general to tell whether a series solution to a functional equation will have a positive radius of convergence. Here's one where the radius does turn out to be positive. eq := A(x) = A(2*x)^2/4 + x; Again, 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 when LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2Ji1GLDYlUSJ4RicvJSdpdGFsaWNHUSV0cnVlRicvJSxtYXRodmFyaWFudEdRJ2l0YWxpY0YnLUkjbW9HRiQ2LVEiPUYnL0Y4USdub3JtYWxGJy8lJmZlbmNlR1EmZmFsc2VGJy8lKnNlcGFyYXRvckdGQi8lKXN0cmV0Y2h5R0ZCLyUqc3ltbWV0cmljR0ZCLyUobGFyZ2VvcEdGQi8lLm1vdmFibGVsaW1pdHNHRkIvJSdhY2NlbnRHRkIvJSdsc3BhY2VHUSwwLjI3Nzc3NzhlbUYnLyUncnNwYWNlR0ZRLUkjbW5HRiQ2JFEiMEYnRj5GPkYrRj4=, so a series about LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2Ji1GLDYlUSJ4RicvJSdpdGFsaWNHUSV0cnVlRicvJSxtYXRodmFyaWFudEdRJ2l0YWxpY0YnLUkjbW9HRiQ2LVEiPUYnL0Y4USdub3JtYWxGJy8lJmZlbmNlR1EmZmFsc2VGJy8lKnNlcGFyYXRvckdGQi8lKXN0cmV0Y2h5R0ZCLyUqc3ltbWV0cmljR0ZCLyUobGFyZ2VvcEdGQi8lLm1vdmFibGVsaW1pdHNHRkIvJSdhY2NlbnRHRkIvJSdsc3BhY2VHUSwwLjI3Nzc3NzhlbUYnLyUncnNwYWNlR0ZRLUkjbW5HRiQ2JFEiMEYnRj5GPkYrRj4= might work. For LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2Ji1GLDYlUSJ4RicvJSdpdGFsaWNHUSV0cnVlRicvJSxtYXRodmFyaWFudEdRJ2l0YWxpY0YnLUkjbW9HRiQ2LVEiPUYnL0Y4USdub3JtYWxGJy8lJmZlbmNlR1EmZmFsc2VGJy8lKnNlcGFyYXRvckdGQi8lKXN0cmV0Y2h5R0ZCLyUqc3ltbWV0cmljR0ZCLyUobGFyZ2VvcEdGQi8lLm1vdmFibGVsaW1pdHNHRkIvJSdhY2NlbnRHRkIvJSdsc3BhY2VHUSwwLjI3Nzc3NzhlbUYnLyUncnNwYWNlR0ZRLUkjbW5HRiQ2JFEiMEYnRj5GPkYrRj4= we have eval(eq,x=0); solve(%,A(0)); I'll try for LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYpLUkjbWlHRiQ2JVEiQUYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JKG1mZW5jZWRHRiQ2JC1GIzYkLUkjbW5HRiQ2JFEiMEYnL0YzUSdub3JtYWxGJ0Y+Rj4tSSNtb0dGJDYtUSJ+RidGPi8lJmZlbmNlR1EmZmFsc2VGJy8lKnNlcGFyYXRvckdGRi8lKXN0cmV0Y2h5R0ZGLyUqc3ltbWV0cmljR0ZGLyUobGFyZ2VvcEdGRi8lLm1vdmFibGVsaW1pdHNHRkYvJSdhY2NlbnRHRkYvJSdsc3BhY2VHUSYwLjBlbUYnLyUncnNwYWNlR0ZVLUZBNi1RIj1GJ0Y+RkRGR0ZJRktGTUZPRlEvRlRRLDAuMjc3Nzc3OGVtRicvRldGZm5GQC1GOzYkUSI0RidGPkY+. Aseries:= unapply(4 + add(a[j]*x^j,j=1..20),x); eval(eq,A=Aseries); eqs:= {seq(coeff(lhs(%)-rhs(%),x,j),j=1..20)}: solve(%); As:= unapply(eval(Aseries(x),%),x); Aapprox:= evalf(As(x)); What's the radius of convergence? with(plots): pointplot([seq([j,abs(coeff(Aapprox,x,j))^(1/j)], j=1..20)]); pointplot([seq([j, ln(abs(coeff(Aapprox,x,j)))], j=1..20)]); It looks very much like these are approaching a straight line. The slope should be approximately slope:= ln(abs(coeff(Aapprox,x,20)))-ln(abs(coeff(Aapprox,x,19))); So the radius of convergence should be approximately exp(-slope); This is probably not very accurate, though.

Here's another use of Taylor series. What is the arc length of the ellipse 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? y1 := b*sqrt(1-x^2); g := sqrt(1+diff(y1,x)^2); L := 4*Int(g,x=0..1); EE:=value(L); This integral is actually the definition of the special function EllipticE, so in itself that's not saying much. FunctionAdvisor(definition,EllipticE(p)); plot(EE,b=0..5); We might try writing the integrand as a series in powers of b. I'll stick that 4 inside the integral. S:= taylor(4*g,b,10); That won't work: the integrals of each term over 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 will diverge. How about in powers of LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2Ji1GLDYlUSJiRicvJSdpdGFsaWNHUSV0cnVlRicvJSxtYXRodmFyaWFudEdRJ2l0YWxpY0YnLUkjbW9HRiQ2LVEoJm1pbnVzO0YnL0Y4USdub3JtYWxGJy8lJmZlbmNlR1EmZmFsc2VGJy8lKnNlcGFyYXRvckdGQi8lKXN0cmV0Y2h5R0ZCLyUqc3ltbWV0cmljR0ZCLyUobGFyZ2VvcEdGQi8lLm1vdmFibGVsaW1pdHNHRkIvJSdhY2NlbnRHRkIvJSdsc3BhY2VHUSwwLjIyMjIyMjJlbUYnLyUncnNwYWNlR0ZRLUkjbW5HRiQ2JFEiMUYnRj5GPkYrRj4=? Note that LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2Ji1GLDYlUSJiRicvJSdpdGFsaWNHUSV0cnVlRicvJSxtYXRodmFyaWFudEdRJ2l0YWxpY0YnLUkjbW9HRiQ2LVEiPUYnL0Y4USdub3JtYWxGJy8lJmZlbmNlR1EmZmFsc2VGJy8lKnNlcGFyYXRvckdGQi8lKXN0cmV0Y2h5R0ZCLyUqc3ltbWV0cmljR0ZCLyUobGFyZ2VvcEdGQi8lLm1vdmFibGVsaW1pdHNHRkIvJSdhY2NlbnRHRkIvJSdsc3BhY2VHUSwwLjI3Nzc3NzhlbUYnLyUncnNwYWNlR0ZRLUkjbW5HRiQ2JFEiMUYnRj5GPkYrRj4= makes our ellipse into a circle. S:= taylor(4*g,b=1,10); At first sight this looks just as bad, but it really isn't: the square root gives us an improper integral that converges. Now we'll want to integrate each term for 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. A convenient way of doing something to every term of a series is with the map command. map(int,%,x=0..1); What map does is take a function or command and a Maple object and produce a new object with the function applied to each operand of the object. Thus for a list: map(t -> t^2, [a,b,c]); Or for a sum of terms: map(sin,a+b+c); If there are extra arguments, they are included in each function call. For example: map(F, [a,b,c], d,e); In our example, the operands of the series structure were the coefficients of the series, and so for each term Maple integrated the coefficient on 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 and made that the corresponding coefficient of the new series. Of course it's quicker to just use the taylor command. LE:= taylor(EE,b=1,40); By the way, does it have a Maclaurin series? taylor(EE,b=0,20); series(EE,b=0,10); What's the radius of convergence of the series around b=1? pointplot([seq([n, ln(abs(coeff(LE,b-1,n)))],n=1..39)]); P:= %: This doesn't look very much like a straight line. Actually it's approximately a constant plus a linear term plus a logarithmic term. There is a command Fit in the Statistics package that can be used to fit a curve to data. Data:= evalf([seq(ln(abs(coeff(LE,b-1,n))),n=3..39)]); Statistics[Fit](a*ln(x)+b+c*x,[$3..39],Data,x); display([plot(%,x=1..39),P]); If this is right, the limiting slope would be mlimit:= limit(diff(%%,x),x=infinity); and the radius of convergence would be exp(mlimit); Actually I think R is exactly 1. I'm quite sure it is at most 1, because EllipticE is not twice differentiable at 0. plot(diff(EE, b),b=-2..2); plot(diff(EE,b$2),b=-2..2); limit(diff(EE,b$2),b=0); Cutoff for Midterm 2 is here!

Another type of series tells us about the behaviour of a function as a variable goes to infinity. The asympt command will do this. In our example, what happens to the length of the ellipse as LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYnLUkjbWlHRiQ2JVEiYkYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JI21vR0YkNi1RIn5GJy9GM1Enbm9ybWFsRicvJSZmZW5jZUdRJmZhbHNlRicvJSpzZXBhcmF0b3JHRj0vJSlzdHJldGNoeUdGPS8lKnN5bW1ldHJpY0dGPS8lKGxhcmdlb3BHRj0vJS5tb3ZhYmxlbGltaXRzR0Y9LyUnYWNjZW50R0Y9LyUnbHNwYWNlR1EmMC4wZW1GJy8lJ3JzcGFjZUdGTC1GNjYtUSgmc3JhcnI7RidGOUY7Rj5GQEZCRkRGRkZIRkpGTUY1Rjk=LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2JC1GLDYlUSgmIzg3MzQ7RicvJSdpdGFsaWNHUSV0cnVlRicvJSxtYXRodmFyaWFudEdRJ2l0YWxpY0YnL0Y4USdub3JtYWxGJ0YrRjo= ? asympt(EE,b); This 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 isn't to be taken quite literally: there's probably a term in 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 there. If we ask for more terms: asympt(EE,b,7); Here's another example: what does the following expression look like as LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYnLUkjbWlHRiQ2JVEibkYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JI21vR0YkNi1RIn5GJy9GM1Enbm9ybWFsRicvJSZmZW5jZUdRJmZhbHNlRicvJSpzZXBhcmF0b3JHRj0vJSlzdHJldGNoeUdGPS8lKnN5bW1ldHJpY0dGPS8lKGxhcmdlb3BHRj0vJS5tb3ZhYmxlbGltaXRzR0Y9LyUnYWNjZW50R0Y9LyUnbHNwYWNlR1EmMC4wZW1GJy8lJ3JzcGFjZUdGTC1GNjYtUSgmc3JhcnI7RidGOUY7Rj5GQEZCRkRGRkZIRkpGTUY1Rjk=LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2JC1GLDYlUSgmIzg3MzQ7RicvJSdpdGFsaWNHUSV0cnVlRicvJSxtYXRodmFyaWFudEdRJ2l0YWxpY0YnL0Y4USdub3JtYWxGJ0YrRjo=? q := 1/(n^2+1)-n/(n^3+3); asympt(q,n,10); In this case asympt is essentially computing a Maclaurin series. Change variable to 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 , which goes to 0 as n goes to infinity . qt:= eval(q,n=1/t); A bit of simplification might be useful. normal(qt); This has a Maclaurin series. taylor(%, t, 10); And now change variables back to n. eval(%, t=1/n); This is exactly what asympt gave us. 1/q also has an asymptotic expression. asympt(1/q, n, 10); Note that this involves positive as well as negative powers of n. How did that come about? normal(1/qt); taylor(%, t, 10); It doesn't have a Maclaurin series because it has a singularity at t = 0 (look at the LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2JC1JJW1zdXBHRiQ2JS1GLDYlUSJ0RicvJSdpdGFsaWNHUSV0cnVlRicvJSxtYXRodmFyaWFudEdRJ2l0YWxpY0YnLUkjbW5HRiQ2JFEiNEYnL0Y7USdub3JtYWxGJy8lMXN1cGVyc2NyaXB0c2hpZnRHUSIwRidGQUYrRkE= in the denominator). What it does have is called a Laurent series (in Math 300). In this case you can just take the Maclaurin series of LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkmbWZyYWNHRiQ2KC1GIzYkLUklbXN1cEdGJDYlLUkjbWlHRiQ2JVEidEYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JI21uR0YkNiRRIjRGJy9GO1Enbm9ybWFsRicvJTFzdXBlcnNjcmlwdHNoaWZ0R1EiMEYnRkEtRiM2JC1GNDYlUSNxdEYnRjdGOkZBLyUubGluZXRoaWNrbmVzc0dRIjFGJy8lK2Rlbm9tYWxpZ25HUSdjZW50ZXJGJy8lKW51bWFsaWduR0ZQLyUpYmV2ZWxsZWRHUSZmYWxzZUYnRkE= , and then divide by LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2JC1JJW1zdXBHRiQ2JS1GLDYlUSJ0RicvJSdpdGFsaWNHUSV0cnVlRicvJSxtYXRodmFyaWFudEdRJ2l0YWxpY0YnLUkjbW5HRiQ2JFEiNEYnL0Y7USdub3JtYWxGJy8lMXN1cGVyc2NyaXB0c2hpZnRHUSIwRidGQUYrRkE= . taylor(t^4/qt, t, 10); convert(%,polynom)/t^4; expand(%); eval(%, t=1/n); Or we could use series instead of taylor, since series can do Laurent series. series(1/qt,t,10); But not all asymptotic expressions arise in this way. The name asympt is actually a bit unfortunate, because it confuses two quite different ideas: a series at infinity, and an asymptotic series.

J := Int(exp(-t)/t, t=x..infinity); value(J); asympt(%,x,10); Let's see if we can reproduce this. It's really telling us about a series for LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2Ji1JJW1zdXBHRiQ2JS1JI21vR0YkNi1RLyZFeHBvbmVudGlhbEU7RicvJSxtYXRodmFyaWFudEdRJ25vcm1hbEYnLyUmZmVuY2VHUSZmYWxzZUYnLyUqc2VwYXJhdG9yR0Y9LyUpc3RyZXRjaHlHRj0vJSpzeW1tZXRyaWNHRj0vJShsYXJnZW9wR0Y9LyUubW92YWJsZWxpbWl0c0dGPS8lJ2FjY2VudEdGPS8lJ2xzcGFjZUdRJjAuMGVtRicvJSdyc3BhY2VHUSwwLjExMTExMTFlbUYnLUYsNiVRInhGJy8lJ2l0YWxpY0dRJXRydWVGJy9GOVEnaXRhbGljRicvJTFzdXBlcnNjcmlwdHNoaWZ0R1EiMEYnLUY1Ni1RMSZJbnZpc2libGVUaW1lcztGJ0Y4RjtGPkZARkJGREZGRkhGSi9GTkZMLUYsNiVRIkpGJ0ZTRlZGOEYrRjg=. EJ:= exp(x)*J; We'll use the IntegrationTools package. with(IntegrationTools): Start with a change of variables, to make the integral go from 0 to infinity. Change(J,t=x+u,u); J1:= expand(%); Now 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 has a series in powers of u. taylor(1/(x+u),u,10); Using this in our integral should bother you, since the series only converges for 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, and we're integrating LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEidUYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy9GM1Enbm9ybWFsRic= from 0 to LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2JC1GLDYlUSgmIzg3MzQ7RicvJSdpdGFsaWNHUSV0cnVlRicvJSxtYXRodmFyaWFudEdRJ2l0YWxpY0YnL0Y4USdub3JtYWxGJ0YrRjo=. But it's possible to justify this (in Math 301, you might find this called Watson's lemma). eval(J1,1/(x+u)=convert(%,polynom)); value(%); expand(%); It would be nicer to see this sorted by powers of x. The sort command does that. JSFH sort(%,x); That's what asympt gave us, with one more term. Do you recognize the numbers in the numerators? What if we use the whole series for 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, rather than just a partial sum? eq := 1/(x+u) = convert(1/(x+u),FormalPowerSeries,u); I want to multiply each term by 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, and integrate from 0 to LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEoJmluZmluO0YnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy9GM1Enbm9ybWFsRic=. Here's a useful integral: Int(exp(-u)*u^k,u=0..infinity); value(%) assuming k >= 0; This is actually one definition of the Gamma function. FunctionAdvisor(definition,GAMMA(k)); You know this better as LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2JkYrLUYjNiUtRiw2JVEia0YnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JI21vR0YkNi1RIiFGJy9GOlEnbm9ybWFsRicvJSZmZW5jZUdRJmZhbHNlRicvJSpzZXBhcmF0b3JHRkQvJSlzdHJldGNoeUdGRC8lKnN5bW1ldHJpY0dGRC8lKGxhcmdlb3BHRkQvJS5tb3ZhYmxlbGltaXRzR0ZELyUnYWNjZW50R0ZELyUnbHNwYWNlR1EsMC4xMTExMTExZW1GJy8lJ3JzcGFjZUdGU0ZARitGQEYrRkA=. convert(%%,factorial); So our series should be LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2KEYrLUYjNiYtSSVtc3VwR0YkNiUtSSNtb0dGJDYtUS8mRXhwb25lbnRpYWxFO0YnLyUsbWF0aHZhcmlhbnRHUSdub3JtYWxGJy8lJmZlbmNlR1EmZmFsc2VGJy8lKnNlcGFyYXRvckdGPy8lKXN0cmV0Y2h5R0Y/LyUqc3ltbWV0cmljR0Y/LyUobGFyZ2VvcEdGPy8lLm1vdmFibGVsaW1pdHNHRj8vJSdhY2NlbnRHRj8vJSdsc3BhY2VHUSYwLjBlbUYnLyUncnNwYWNlR1EsMC4xMTExMTExZW1GJy1GLDYlUSJ4RicvJSdpdGFsaWNHUSV0cnVlRicvRjtRJ2l0YWxpY0YnLyUxc3VwZXJzY3JpcHRzaGlmdEdRIjBGJy1GNzYtUTEmSW52aXNpYmxlVGltZXM7RidGOkY9RkBGQkZERkZGSEZKRkwvRlBGTi1JKG1mZW5jZWRHRiQ2JC1GIzYpLUkobXN1YnN1cEdGJDYnLUY3Ni9RJiZpbnQ7RicvJStmb3JlZ3JvdW5kR1EuWzE0NCwxNDQsMTQ0XUYnRjovSSttc2VtYW50aWNzR0YkUSZpbmVydEYnRj1GQEZCRkRGRkZIRkpGTEZqbkZSLUYsNiVRKCZpbmZpbjtGJ0ZVRlgvRmVuUSIyRicvJS9zdWJzY3JpcHRzaGlmdEdGZm4tSSZtZnJhY0dGJDYoLUYjNiQtRjQ2JUY2LUYjNiUtRjc2LVEqJnVtaW51czA7RidGOkY9RkBGQkZERkZGSEZKL0ZNUSwwLjIyMjIyMjJlbUYnL0ZQRmBxLUYsNiVRInRGJ0ZVRlhGOkZaRjotRiM2JEZicUY6LyUubGluZXRoaWNrbmVzc0dRIjFGJy8lK2Rlbm9tYWxpZ25HUSdjZW50ZXJGJy8lKW51bWFsaWduR0Zcci8lKWJldmVsbGVkR0Y/RistSSdtc3BhY2VHRiQ2Ji8lJ2hlaWdodEdRJjAuMGV4RicvJSZ3aWR0aEdRJjAuM2VtRicvJSZkZXB0aEdGZnIvJSpsaW5lYnJlYWtHUSVhdXRvRictRjc2L1EwJkRpZmZlcmVudGlhbEQ7RidGZm9GOkZpb0Y9RkBGQkZERkZGSEZKRkxGam5GYnFGOkY6RjotRjc2LVEiPUYnRjpGPUZARkJGREZGRkhGSi9GTVEsMC4yNzc3Nzc4ZW1GJy9GUEZmcy1GIzYnLUkrbXVuZGVyb3ZlckdGJDYnLUY3Ni1RJiZTdW07RidGOkY9RkAvRkNGV0ZEL0ZHRlcvRklGV0ZKRkwvRlBRLDAuMTY2NjY2N2VtRictRiM2Ji1GLDYlUSJrRidGVUZYRmJzLUkjbW5HRiQ2JEZmbkY6RjpGXHBGSi8lLGFjY2VudHVuZGVyR0Y/RistRmJyNiZGZHIvRmhyUSQ1LjBGJ0ZqckZccy1GZHA2KC1GIzYnLUY0NiUtRlxvNiQtRiM2JUZccS1GW3U2JEZpcUY6RjpGOkZndEZaRmduLUYjNiVGZ3QtRjc2LVEiIUYnRjpGPUZARkJGREZGRkhGSi9GTUZRRk9GOkYrRjotRiM2JC1GNDYlRlItRiM2JkZndC1GNzYtUSIrRidGOkY9RkBGQkZERkZGSEZKRl9xRmFxRl12RjpGWkY6RmdxRmpxRl1yRl9yRjpGK0Y6RitGOg== This matches the result from the partial sum. Do you believe this series? For what x does it converge? Nevertheless, it is useful. Here's another way to get it, which makes the "asymptotic" character of the series clearer, and gives us control over the remainders. I'll use integration by parts. JS[0] := J1; JS[1] := Parts(JS[0],1/(x+u)); Note that for LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2JVEieEYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JI21vR0YkNi1RIj5GJy9GM1Enbm9ybWFsRicvJSZmZW5jZUdRJmZhbHNlRicvJSpzZXBhcmF0b3JHRj0vJSlzdHJldGNoeUdGPS8lKnN5bW1ldHJpY0dGPS8lKGxhcmdlb3BHRj0vJS5tb3ZhYmxlbGltaXRzR0Y9LyUnYWNjZW50R0Y9LyUnbHNwYWNlR1EsMC4yNzc3Nzc4ZW1GJy8lJ3JzcGFjZUdGTC1JI21uR0YkNiRRIjBGJ0Y5Rjk=, 0 < 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 Thus we can say 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. JS[2] := Parts(JS[1],1/(x+u)^2); Similarly, 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 so LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2Jy1JJW1zdWJHRiQ2JS1GLDYlUSJKRicvJSdpdGFsaWNHUSV0cnVlRicvJSxtYXRodmFyaWFudEdRJ2l0YWxpY0YnLUYjNiQtSSNtbkdGJDYkUSIyRicvRjtRJ25vcm1hbEYnRkMvJS9zdWJzY3JpcHRzaGlmdEdRIjBGJy1JI21vR0YkNi1RIj1GJ0ZDLyUmZmVuY2VHUSZmYWxzZUYnLyUqc2VwYXJhdG9yR0ZOLyUpc3RyZXRjaHlHRk4vJSpzeW1tZXRyaWNHRk4vJShsYXJnZW9wR0ZOLyUubW92YWJsZWxpbWl0c0dGTi8lJ2FjY2VudEdGTi8lJ2xzcGFjZUdRLDAuMjc3Nzc3OGVtRicvJSdyc3BhY2VHRmduLUYjNiktSSZtZnJhY0dGJDYoLUYjNiQtRkA2JFEiMUYnRkNGQy1GIzYkLUYsNiVRInhGJ0Y3RjpGQy8lLmxpbmV0aGlja25lc3NHRmNvLyUrZGVub21hbGlnbkdRJ2NlbnRlckYnLyUpbnVtYWxpZ25HRl1wLyUpYmV2ZWxsZWRHRk4tRkk2LVEoJm1pbnVzO0YnRkNGTEZPRlFGU0ZVRldGWS9GZm5RLDAuMjIyMjIyMmVtRicvRmluRmZwLUZdbzYoRl9vLUYjNiZGKy1GIzYkLUklbXN1cEdGJDYlRmZvRj8vJTFzdXBlcnNjcmlwdHNoaWZ0R0ZHRkNGK0ZDRmlvRltwRl5wRmBwLUZJNi1RIitGJ0ZDRkxGT0ZRRlNGVUZXRllGZXBGZ3AtRiM2Ji1GLDYlUSJPRicvRjhGTkZDLUZJNi1RMCZBcHBseUZ1bmN0aW9uO0YnRkNGTEZPRlFGU0ZVRldGWS9GZm5RJjAuMGVtRicvRmluRmByLUkobWZlbmNlZEdGJDYkLUYjNiQtRl1vNihGX28tRiM2JkYrLUYjNiQtRl9xNiVGZm8tRkA2JFEiM0YnRkNGYXFGQ0YrRkNGaW9GW3BGXnBGYHBGQ0ZDRkNGK0ZDRitGQ0YrRkM= . And so on. for count from 3 to 8 do JS[count]:= Parts(JS[count-1],1/(x+u)^count) end do; Let's see how well the partial sums of our asymptotic series do at approximating the original integral J. for count from 1 to 20 do PS[count]:= exp(-x)*add((-1)^k*k!/x^(1+k),k=0..count-1) end do: plot([value(J),seq(PS[count],count=1..20)],x=0..5,-5..5);

map Fit (in the Statistics package) asympt

LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYjLUkjbWlHRiQ2I1EhRic=