Question

approximation of a binomial series to 8 decimal places. details inside

Answer 1

\[\int\limits_{0}^{0.1}\frac{ 1 }{ \sqrt{1+x^3} }\]

Answer 2

\[compare \to (1+x)^k=\sum_{n=0}^{\infty}\left(\begin{matrix}k \\ n\end{matrix}\right)x^n= 1+kx+ \frac{ k(k-1))x^2 }{ 2! }+\frac{ k(k-1)(k-2)x^3 }{ 3! }+...\]
where k= -1/2 and x= -x^3 in this case

Answer 3

i am really stuck on this one!

Answer 4

Due to the binomial theorem,
\[(1+x^3)^{-1/2}=1-\frac{x^3}{2}+\frac{3x^6}{8}-\frac{5x^9}{16}+\cdots\]

Answer 5

ok, yea i did get that! i had to check what i got... from here, do i do the integral?

Answer 6

im not sure how to get to the "evaluation" part with the .1

Answer 7

Yes, integrate term by term. You only need a few terms to get that 8 decimal approximation. Notice that \(0.1^9=0.000000001\), so you can truncate the integration to the first four terms.

Answer 8

they gave me a crazy example where they plugged x=1 into the summation formation after the integration and it really looked confusing!

Answer 9

x=.1 i menat

Answer 10

\[\int_0^{0.1}(1+x^3)^{-1/2}\,dx\approx \left[x-\frac{x^4}{8}+\frac{3x^7}{56}-\frac{5x^{10}}{160}\right]_0^{0.1}\]

Answer 11

ok, that is what i was hoping i could do~! they plugged it into the \[\sum_{n=1}^{\infty}(-1)^n1*3*5...\] part and i was like whoa

Answer 12

i was looking at what you got there... sorry for the pause!

Answer 13

is it .099987505?

Answer 14

Yes I'm getting the same

Answer 15

ok, i will make that zero a 1 and call it 8 places!

Answer 16

thank you sooo much! as usual