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Finding the convergence of a binomial expansion

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  • #1
chwala
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Homework Statement


Expand ##(1+3x-4x^2)^{0.5}/(1-2x)^2## find its convergence value


Homework Equations




The Attempt at a Solution


on expansion
##(1+3/2x-3.125x^2+4.6875x^3+...)(1+4x+12x^2+32x^3+...)##
##1+5.5x+14.875x^2+42.1875x^3+... ##
how do i prove for convergence here?
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement


Expand ##(1+3x-4x^2)^{0.5}/(1-2x)^2## find its convergence value


Homework Equations




The Attempt at a Solution


on expansion
##(1+3/2x-3.125x^2+4.6875x^3+...)(1+4x+12x^2+32x^3+...)##
##1+5.5x+14.875x^2+42.1875x^3+... ##
how do i prove for convergence here?
The series will converge in the common convergence regions of ##\sqrt{1+3x-4x^2}## and ##(1-2x)^{-2}##. So, for the first factor, convergence occurs if ##-1 < 3x-4x^2 < 1## and for the second factor if ##-1 < 2x < 1##. After that, finding the actual series does not look very easy, although getting the first few terms is not too hard.
 
  • #3
chwala
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The series will converge in the common convergence regions of ##\sqrt{1+3x-4x^2}## and ##(1-2x)^{-2}##. So, for the first factor, convergence occurs if ##-1 < 3x-4x^2 < 1## and for the second factor if ##-1 < 2x < 1##. After that, finding the actual series does not look very easy, although getting the first few terms is not too hard.
I am not getting sir, still seeing blanks. are you using ratio test or what? whats the difference in using this tests and using... say, Mean value theorem or liapunov function in checking for convergence?
 
  • #4
Ray Vickson
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I am not getting sir, still seeing blanks. are you using ratio test or what? whats the difference in using this tests and using... say, Mean value theorem or liapunov function in checking for convergence?
We know when the binomial series for ##(1+u)^{1/2}## converges---for example, by using the ratio test on ##|u|##. We know when the binomial series for ##(1+v)^{-2}## converges---for example, by the ratio test on ##|v|##. Now just put ##u = 2x - 4 x^2## and ##v = -2x.##
 
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  • #5
chwala
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ok by ratio theorem,
on ## |v|= |2x|##
using the ratio test , arent we going to have ##L= Lim n→∞ |2(n+1)/(2n)|=1?##
test fails?
 
  • #6
Ray Vickson
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ok by ratio theorem,
on ## |v|= |2x|##
using the ratio test , arent we going to have ##L= Lim n→∞ |2(n+1)/(2n)|=1?##
test fails?
I asked what the convergence region was for ##(1+v)^{-2}##; there is no ##x## anywhere in that expression!

Anyway, the ratio test must involve the value of ##v## somehow. Surely there is a difference between ##\sum_n c_n 0^n## and ##\sum_n c_n 100^n##, is there not? Whether or not the series ##\sum_n c_n v^n## converges will almost always depend on what is ##v##.
 
  • #7
chwala
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now from my research,
## {(1+3x-4x^2)^{0.5}}/{(1-2x)^2}= (1-x)^{0.5} (1+4x)^{0.5}(1-2x)^-2##
now the terms on the rhs are similar to the maclaurin series , now on using ratio test.....## (1+x)^k## will have the interval of convergence being
##(1+x)^k = -1<x<1##
similarly,
##(1-x)^{0.5}= -1<(-x)<1##
##(1+4x)^{0.5}= -1<(4x)<1##
##(1-2x)^-2= -1<(-2x)<1##
the radius of convergence of all the three expansions coming to
##-0.25<x<0.25## all centred at ##0##
confirm whether this is true, bingo from Africa.
 
  • #8
chwala
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I asked what the convergence region was for ##(1+v)^{-2}##; there is no ##x## anywhere in that expression!

Anyway, the ratio test must involve the value of ##v## somehow. Surely there is a difference between ##\sum_n c_n 0^n## and ##\sum_n c_n 100^n##, is there not? Whether or not the series ##\sum_n c_n v^n## converges will almost always depend on what is ##v##.
maybe you should have hinted the maclaurin series, its straight forward and to the point..
 
  • #9
Ray Vickson
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maybe you should have hinted the maclaurin series, its straight forward and to the point..
It is not up to me to contradict you: YOU used the term "binomial expansion" in the title of your thread, and I just went along with that.

Anyway, your region ##-1/4 < x < 1/4## is partly correct: it is correct on one side, but too narrow on the other side. The actual convergence region is not symmetric about ##x = 0##.
 
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  • #10
chwala
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It is not up to me to contradict you: YOU used the term "binomial expansion" in the title of your thread, and I just went along with that.

Anyway, your region ##-1/x < x < 1/4## is partly correct: it is correct on one side, but too narrow on the other side. The actual convergence region is not symmetric about ##x = 0##.
just check ## (-1/x)?## probably you meant ##(-1/4)## why is it wrong just explain ...
 
  • #11
chwala
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It is not up to me to contradict you: YOU used the term "binomial expansion" in the title of your thread, and I just went along with that.

Anyway, your region ##-1/x < x < 1/4## is partly correct: it is correct on one side, but too narrow on the other side. The actual convergence region is not symmetric about ##x = 0##.
i appreciate bro, as an expert maybe you should concentrate more on the body of the problem rather than title, i am learning this things....thats why i come to forum for guidance...
 
  • #12
chwala
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upload_2018-6-24_15-53-55.jpeg
 

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  • #14
Ray Vickson
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i appreciate bro, as an expert maybe you should concentrate more on the body of the problem rather than title, i am learning this things....thats why i come to forum for guidance...
If you know about the Maclauren expansion you do not need me to tell you about it. Besides, I have already given you all the hints you need in post #2. The rest is up to you!
 
  • #15
Ray Vickson
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now from my research,
## {(1+3x-4x^2)^{0.5}}/{(1-2x)^2}= (1-x)^{0.5} (1+4x)^{0.5}(1-2x)^-2##
now the terms on the rhs are similar to the maclaurin series , now on using ratio test.....## (1+x)^k## will have the interval of convergence being
##(1+x)^k = -1<x<1##
similarly,
##(1-x)^{0.5}= -1<(-x)<1##
##(1+4x)^{0.5}= -1<(4x)<1##
##(1-2x)^-2= -1<(-2x)<1##
the radius of convergence of all the three expansions coming to
##-0.25<x<0.25## all centred at ##0##
confirm whether this is true, bingo from Africa.
Yes, it is true. The method I told you before is mistaken, so I gave you bad advice. Sorry.
 
  • #16
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Anyway, your region ##-1/4 < x < 1/4## is partly correct: it is correct on one side, but too narrow on the other side. The actual convergence region is not symmetric about ##x = 0##.
A conundrum. A power series in x should have a symetric radius of convergence around the point of expansion (x=0). It should diverge outside of the radius. I guess that the substitution and rearrangement of terms changes the nature of the convergence.
 
  • #17
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A conundrum. A power series in x should have a symetric radius of convergence around the point of expansion (x=0). It should diverge outside of the radius. I guess that the substitution and rearrangement of terms changes the nature of the convergence.
The situation may be more complicated than that.

Certainly, a product of power series is convergent in the intersection of the individual convergence regions, so a sufficient condition for convergence of a product of series is ##|x| < R,## where ##R## is the minimum of all the individual convergent radii. However, that is not a necessary condition, so the actual convergence region of the product may be larger. For example, the three functions
$$f_1(x) = \frac{e^x}{1-x}, \hspace{2ex} f_2(x) = (1-x)^{1/4} e^x, \hspace{2ex} f_3(x) = (1-x)^{3/4}$$
all have radii of convgence equal to 1. However, their product is the series for ##e^{2x}##, which has radius of convergence equal to ##\infty##.
 
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  • #18
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However, their product is the series for ##e^{2x}##, which has radius of convergence equal to ##\infty##.
Right. The radius might change. But the range of convergence of the final power series will still be a symetric circle centered at the point of expansion.
 
  • #19
chwala
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A conundrum. A power series in x should have a symetric radius of convergence around the point of expansion (x=0). It should diverge outside of the radius. I guess that the substitution and rearrangement of terms changes the nature of the convergence.
can you re arrange so that we can see the changes? regards
 
  • #20
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can you re arrange so that we can see the changes? regards
I am referring to the rearrangement that one does when one starts with a power series expansion of, for instance, ##(1+u)^{1/2}##, substitutes ##u=2x-4x^2##, and rearranges it to a power series of x. The two power series of ##u## and of ##x## have radii of convergence centered at ##u=0## and ##x=0##, respectively. Their convergence ranges do not necessarily correspond as expected directly from ##u=2x-4x^2##
 
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  • #21
chwala
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Yes, it is true. The method I told you before is mistaken, so I gave you bad advice. Sorry.
sorry i meant "interval of convergence" and not "radius of convergence" my apologies......
 

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