Not understanding textbook solution: Mary Boas mathematical methods

[kq6up]

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.

 

[crownedbishop]

They want you to write it out using the binomial coefficients. n!=1*3*5*...*n where n is odd.

 

[LCKurtz]

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

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

 

[Ray Vickson]

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.

 

[kq6up]

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