Binomial expansion infinite series: why need form $a^n(1+frac{b}{a}x)^n$?
$begingroup$
I suppose my question is about limits and infinities. I have been considering the expansion of $(a+bx)^n$ for negative and fractional $n$, and my text book says to always factorise out the $a^n$ like so:
$(a+bx)^n =a^n(1+frac{b}{a}x)^n= a^n[1+nfrac{b}{a}x+frac{n(n-1)}{2!}(frac{b}{a}x)^2+...]$
And then the expansion is valid for $|frac{bx}{a}|<1$
My question is, why is the $a$ term involved in thevalidity requirement? Why is it incorrect to simply expand the binomial as
$(a+bx)^n = a^n+a^{n-1}n(bx)+frac{a^{n-2}n(n-1)}{2!}(bx)^2+...]$
and say it is valid for $|bx|<1$?
sequences-and-series infinity binomial-theorem negative-binomial
asked Sep 14 '16 at 13:46
21joanna1221joanna12
1,0671616
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add a comment |
$begingroup$
I suppose my question is about limits and infinities. I have been considering the expansion of $(a+bx)^n$ for negative and fractional $n$, and my text book says to always factorise out the $a^n$ like so:
$(a+bx)^n =a^n(1+frac{b}{a}x)^n= a^n[1+nfrac{b}{a}x+frac{n(n-1)}{2!}(frac{b}{a}x)^2+...]$
And then the expansion is valid for $|frac{bx}{a}|<1$
My question is, why is the $a$ term involved in thevalidity requirement? Why is it incorrect to simply expand the binomial as
$(a+bx)^n = a^n+a^{n-1}n(bx)+frac{a^{n-2}n(n-1)}{2!}(bx)^2+...]$
and say it is valid for $|bx|<1$?
sequences-and-series infinity binomial-theorem negative-binomial
asked Sep 14 '16 at 13:46
21joanna1221joanna12
1,0671616
$endgroup$
add a comment |
$begingroup$
I suppose my question is about limits and infinities. I have been considering the expansion of $(a+bx)^n$ for negative and fractional $n$, and my text book says to always factorise out the $a^n$ like so:
$(a+bx)^n =a^n(1+frac{b}{a}x)^n= a^n[1+nfrac{b}{a}x+frac{n(n-1)}{2!}(frac{b}{a}x)^2+...]$
And then the expansion is valid for $|frac{bx}{a}|<1$
My question is, why is the $a$ term involved in thevalidity requirement? Why is it incorrect to simply expand the binomial as
$(a+bx)^n = a^n+a^{n-1}n(bx)+frac{a^{n-2}n(n-1)}{2!}(bx)^2+...]$
and say it is valid for $|bx|<1$?
sequences-and-series infinity binomial-theorem negative-binomial
asked Sep 14 '16 at 13:46
21joanna1221joanna12
1,0671616
$endgroup$
I suppose my question is about limits and infinities. I have been considering the expansion of $(a+bx)^n$ for negative and fractional $n$, and my text book says to always factorise out the $a^n$ like so:
$(a+bx)^n =a^n(1+frac{b}{a}x)^n= a^n[1+nfrac{b}{a}x+frac{n(n-1)}{2!}(frac{b}{a}x)^2+...]$
And then the expansion is valid for $|frac{bx}{a}|<1$
My question is, why is the $a$ term involved in thevalidity requirement? Why is it incorrect to simply expand the binomial as
$(a+bx)^n = a^n+a^{n-1}n(bx)+frac{a^{n-2}n(n-1)}{2!}(bx)^2+...]$
and say it is valid for $|bx|<1$?
sequences-and-series infinity binomial-theorem negative-binomial
sequences-and-series infinity binomial-theorem negative-binomial
asked Sep 14 '16 at 13:46
21joanna1221joanna12
1,0671616
asked Sep 14 '16 at 13:46
21joanna1221joanna12
1,0671616
asked Sep 14 '16 at 13:46
21joanna1221joanna12
1,0671616
asked Sep 14 '16 at 13:46
21joanna1221joanna12
1,0671616
asked Sep 14 '16 at 13:46
21joanna1221joanna12
1,0671616
1,0671616
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This is first of all a matter of book-keeping. The $(1+bx/a)^n$ is
considered a geometric series
$$
(1+y)^n=sum_{ige 0} binom{n}{i}y^i
$$
where $y=bx/a$ is the free variable, and where
the radius of convergence is known to be $|y|<1$. So instead of
remembering a lot of different subcases of how to approach this,
only a single criterion (parameter) comes into play.
answered Sep 14 '16 at 13:55
R. J. MatharR. J. Mathar
211
$endgroup$
add a comment |
$begingroup$
It doesn't matter, if $n$ is an integer, as then the expansion has finite termage; and when $n$ isn't an integer, the convergence of the series is contingent on the ratio of $bx$ to $a$; so factoring-out the $a$ & gathering all the variables & constants into a single term makes whether it satisfies the convergence criterion more explicit.
answered Dec 3 '18 at 19:24
AmbretteOrriseyAmbretteOrrisey
54210
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
This is first of all a matter of book-keeping. The $(1+bx/a)^n$ is
considered a geometric series
$$
(1+y)^n=sum_{ige 0} binom{n}{i}y^i
$$
where $y=bx/a$ is the free variable, and where
the radius of convergence is known to be $|y|<1$. So instead of
remembering a lot of different subcases of how to approach this,
only a single criterion (parameter) comes into play.
answered Sep 14 '16 at 13:55
R. J. MatharR. J. Mathar
211
$endgroup$
add a comment |
$begingroup$
This is first of all a matter of book-keeping. The $(1+bx/a)^n$ is
considered a geometric series
$$
(1+y)^n=sum_{ige 0} binom{n}{i}y^i
$$
where $y=bx/a$ is the free variable, and where
the radius of convergence is known to be $|y|<1$. So instead of
remembering a lot of different subcases of how to approach this,
only a single criterion (parameter) comes into play.
answered Sep 14 '16 at 13:55
R. J. MatharR. J. Mathar
211
$endgroup$
add a comment |
$begingroup$
This is first of all a matter of book-keeping. The $(1+bx/a)^n$ is
considered a geometric series
$$
(1+y)^n=sum_{ige 0} binom{n}{i}y^i
$$
where $y=bx/a$ is the free variable, and where
the radius of convergence is known to be $|y|<1$. So instead of
remembering a lot of different subcases of how to approach this,
only a single criterion (parameter) comes into play.
answered Sep 14 '16 at 13:55
R. J. MatharR. J. Mathar
211
$endgroup$
This is first of all a matter of book-keeping. The $(1+bx/a)^n$ is
considered a geometric series
$$
(1+y)^n=sum_{ige 0} binom{n}{i}y^i
$$
where $y=bx/a$ is the free variable, and where
the radius of convergence is known to be $|y|<1$. So instead of
remembering a lot of different subcases of how to approach this,
only a single criterion (parameter) comes into play.
answered Sep 14 '16 at 13:55
R. J. MatharR. J. Mathar
211
answered Sep 14 '16 at 13:55
R. J. MatharR. J. Mathar
211
answered Sep 14 '16 at 13:55
R. J. MatharR. J. Mathar
211
answered Sep 14 '16 at 13:55
R. J. MatharR. J. Mathar
211
211
add a comment |
add a comment |
$begingroup$
It doesn't matter, if $n$ is an integer, as then the expansion has finite termage; and when $n$ isn't an integer, the convergence of the series is contingent on the ratio of $bx$ to $a$; so factoring-out the $a$ & gathering all the variables & constants into a single term makes whether it satisfies the convergence criterion more explicit.
answered Dec 3 '18 at 19:24
AmbretteOrriseyAmbretteOrrisey
54210
$endgroup$
add a comment |
$begingroup$
It doesn't matter, if $n$ is an integer, as then the expansion has finite termage; and when $n$ isn't an integer, the convergence of the series is contingent on the ratio of $bx$ to $a$; so factoring-out the $a$ & gathering all the variables & constants into a single term makes whether it satisfies the convergence criterion more explicit.
answered Dec 3 '18 at 19:24
AmbretteOrriseyAmbretteOrrisey
54210
$endgroup$
add a comment |
$begingroup$
It doesn't matter, if $n$ is an integer, as then the expansion has finite termage; and when $n$ isn't an integer, the convergence of the series is contingent on the ratio of $bx$ to $a$; so factoring-out the $a$ & gathering all the variables & constants into a single term makes whether it satisfies the convergence criterion more explicit.
answered Dec 3 '18 at 19:24
AmbretteOrriseyAmbretteOrrisey
54210
$endgroup$
It doesn't matter, if $n$ is an integer, as then the expansion has finite termage; and when $n$ isn't an integer, the convergence of the series is contingent on the ratio of $bx$ to $a$; so factoring-out the $a$ & gathering all the variables & constants into a single term makes whether it satisfies the convergence criterion more explicit.
answered Dec 3 '18 at 19:24
AmbretteOrriseyAmbretteOrrisey
54210
answered Dec 3 '18 at 19:24
AmbretteOrriseyAmbretteOrrisey
54210
answered Dec 3 '18 at 19:24
AmbretteOrriseyAmbretteOrrisey
54210
answered Dec 3 '18 at 19:24
AmbretteOrriseyAmbretteOrrisey
54210
54210
add a comment |
add a comment |
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