Not understanding textbook solution: Mary Boas mathematical methods

kq6up
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Homework Statement



4. Write the Maclaurin series for 1/√(1 + x) in ∑ form using the binomial coefficient
notation. Then find a formula for the binomial coefficients in terms of n as we did
in Example 2 above.

Homework Equations



{ \left( 1+x \right) }^{ P }=\sum _{ n=0 }^{ \infty }{ \left( \underset { n }{ P } \right) } { x }^{ n }

The Attempt at a Solution



This is what I got, \frac { 1 }{ \sqrt { (1+x) } } =\sum _{ n=0 }^{ \infty }{ \left( \underset { n }{ -1/2 } \right) } { x }^{ n }

This is the book's solution \left( \overset { -1/2 }{ n } \right) =\frac { { (-1) }^{ n }(2n-1)! }{ (2n)! } I am not understanding the whole double factorial.
 
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They want you to write it out using the binomial coefficients. n!=1*3*5*...*n where n is odd.
 
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kq6up said:


This is the book's solution \left( \overset { -1/2 }{ n } \right)


Use the \binom{}{} tex command for binomial coefficients:$$
\binom{-\frac 1 2}{n}$$
 
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kq6up said:

Homework Statement



4. Write the Maclaurin series for 1/√(1 + x) in ∑ form using the binomial coefficient
notation. Then find a formula for the binomial coefficients in terms of n as we did
in Example 2 above.


Homework Equations



{ \left( 1+x \right) }^{ P }=\sum _{ n=0 }^{ \infty }{ \left( \underset { n }{ P } \right) } { x }^{ n }

The Attempt at a Solution



This is what I got, \frac { 1 }{ \sqrt { (1+x) } } =\sum _{ n=0 }^{ \infty }{ \left( \underset { n }{ -1/2 } \right) } { x }^{ n }

This is the book's solution \left( \overset { -1/2 }{ n } \right) =\frac { { (-1) }^{ n }(2n-1)! }{ (2n)! } I am not understanding the whole double factorial.

Write out the formula for ##{-1/2 \choose n}##. For example,
{-1/2 \choose 0} = 1 \\<br /> {-1/2 \choose 1} = -1/2 \\<br /> {-1/2 \choose 2} = (-1/2)(-1/2 \:-1)/2! = \frac{1 \cdot 3}{2^2 \, 2!}
etc.

The notation ##x!## means ##x(x-2)(x-4) \cdots ##, ending at a final factor of 2 or 1 according as x is even or odd.
 
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Ah, now I see it (2k-1)! = \prod_{i=1}^k (2i-1)

I have never seen that notation before. Thanks for the tip. Also, thanks for the LaTeX tip.

Regards,
Chris Maness
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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