Question

Maclauren series

Answer 1

Find your derivatives:
\[\begin{align*}
f(x)&=(1-x)^{-1/2}\\
f'(x)&=\frac{1}{2}(1-x)^{-3/2}\\
f''(x)&=\frac{3}{2}\times\frac{1}{2}(1-x)^{-5/2}\\
&\vdots\\
f^{(n)}(x)&=\frac{(2n+1)\times\cdots\times3\times1}{2^n}(1-x)^{-(2n+1)/2}
\end{align*}\]
The Mclaurin series for a function \(f(x)\) has the form
\[\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n\]

Answer 2

Typo: the \(n\)th derivative should have \((2n-1)\) in place of the \((2n+1)\) in the coefficient.

Answer 3

where did you get the 2n-1

Answer 4

because i calculated the first 4 derivatives and f^n(0) is always like a decimal

Answer 5

and i thought about maybe doing (n-.5)! but i guess that doesnt work

Answer 6

because its a maclauren and x is zero, the value of the derivatives would just be the exponent of all the previous derivatives multiplied. But the exponents are decimals

Answer 7

It's a matter of pattern recognition. For each successive derivative, you are subtracting 1 from the exponent:
\[\begin{array}{c|c}
\text{order of derivative}\,(n)&\text{coefficient}&\text{exponent}\\
\hline
1&\frac{1}{2}&-\frac{1}{2}-1=-\frac{3}{2}\\
2&\frac{3}{2}\times\frac{1}{2}&-\frac{3}{2}-1=-\frac{5}{2}\\
3&\frac{5}{2}\times\frac{3}{2}\times\frac{1}{2}&-\frac{5}{2}-1=-\frac{7}{2}\\
\vdots&\vdots&\vdots
\end{array}\]
For each \(n\), the leading fraction of the coefficient is \(\dfrac{2n-1}{2}\), and the exponent is \(-\dfrac{2n+1}{2}\).

Answer 8

but the coefficient for the third term would be 15/2 correct?

Answer 9

based off of that (2n-1)/2 thing, it would be (2(3)-1)/2=5/2

Answer 10

which isnt correct

Answer 11

and it would still all be under n! because of the maclauren definition

Answer 12

I think you're forgetting about the denominator. For the \(n\)th derivative, you are also dividing by \(2^n\).

Answer 13

I mentioned "leading fraction" without clarifying that it is not the entire coefficient. Sorry.

Answer 14

Im just very bad with recognizing patterns between series

Answer 15

It comes with practice.

Answer 16

You'll find that the Maclaurin series for \((1-x)^{-1/2}\) is thus
\[\sum_{n=0}^\infty \frac{(2n+1)\times\cdots\times3\times1}{2^nn!}x^n\]
To check the interval of convergence, the ratio test should do the trick.
\[\lim_{n\to\infty}\left|\frac{\dfrac{(2(n+1)+1)\times(2n+1)\times\cdots\times3\times1}{2^{n+1}(n+1)!}x^{n+1}}{\dfrac{(2n+1)\times\cdots\times3\times1}{2^nn!}x^n}\right|=|x|\lim_{n\to\infty}\frac{2n+3}{2(n+1)}\]
The series converges if this limit is less than 1.