Question

write down the expansion in ascending powers of x up to the term x^3 of
i)(1+x)^1/2
ii)(1-x)^-1/2
Hence or otherwise, expand ((1+x)/(1-x))^1/2 in ascending powers of x to the term x^2
by using x=1/10 obtain an estimate to three decimal place for square root of 11

Answer 1

i)1+x/2-x^2/8+x^3/16
ii)1+x/2+3x^2/8 +5x^3/16

Answer 2

but how can i estimate the value of square 11 using this expansion

Answer 3

You're talking about Taylor expansions, right?

Answer 4

i dont know whats taylors expansion.

Answer 5

It's a method of expressing a function with continuous derivatives as a series of other functions.

For example, you could write \(\sin x\) as an infinite-term polynomial,
\[\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots=\sum_{n=0}^\infty \frac{(-1)^{n}x^{2n+1}}{(2n+1)!}\]
Chances are you haven't seen this unless you've taken a course in calculus.

Perhaps you know about the binomial theorem?

Answer 6

yes i do know about binomial theorm

Answer 7

Okay, in that case, you must be using the expansion formula,
\[(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+\cdots\]
for \(-1<x<1\).

Answer 8

i did solve part i and part ii and i wrote both answers at the top using this method but iam stuck on the estimation part

Answer 9

Right, and the expansion of \(\left(\dfrac{1+x}{1-x}\right)^{1/2}\) shouldn't be hard to find. Do you have that too?

Answer 10

i know that would expand (1+x) and (1-x) seperately but (x-1)^1/2 they are both similar to the expansion of i and ii

Answer 11

1+x/2-x^2/8/1-x/2-x^2/8

Answer 12

Substitute x=1/10 and i got 0.9975

Answer 13

You'll need the expansion for the estimate part.

Notice that for \(x=\dfrac{1}{10}\), you have
\[\left(\dfrac{1+\dfrac{1}{10}}{1-\dfrac{1}{10}}\right)^{1/2}=\left(\dfrac{\dfrac{11}{10}}{\dfrac{9}{10}}\right)^{1/2}=\left(\frac{11}{9}\right)^{1/2}=\frac{1}{3}\sqrt{11}\]
This means that when you plug in \(x=\dfrac{1}{10}\) into the expansion, you have
\[\frac{1}{3}\sqrt{11}\approx 1+(\cdots)\frac{1}{10}+(\cdots)\frac{1}{10^2}\]
which means
\[\sqrt{11}\approx 3\left(1+(\cdots)\frac{1}{10}+(\cdots)\frac{1}{10^2}\right)\]

Answer 14

why dont we plug x=1/10 directly in the expansion and then do the square root and dividing.

Answer 15

I don't think your third expansion is right. The first two,
\[(1+x)^{1/2}\approx 1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}\\
(1-x)^{-1/2}\approx 1+\frac{x}{2}+\frac{3x^2}{8}+\frac{5x^3}{16}\]
are correct.

The third expansion is the product of the two above:
\[\left(\frac{1+x}{1-x}\right)^{1/2}=(1+x)^{1/2}(1-x)^{-1/2}\]
so
\[\left(\frac{1+x}{1-x}\right)^{1/2}\approx \left(1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}\right)\left(1+\frac{x}{2}+\frac{3x^2}{8}+\frac{5x^3}{16}\right)\]
Up to the \(x^2\) term, you can ignore the \(x^3\) terms of the component expansions:
\[\begin{align*}\left(\frac{1+x}{1-x}\right)^{1/2}&\approx \left(1+\frac{x}{2}-\frac{x^2}{8}\right)\left(1+\frac{x}{2}+\frac{3x^2}{8}\right)\\
&\approx 1+\frac{x}{2}+\frac{3x^2}{8}+\frac{x}{2}+\frac{x^2}{4}+\frac{3x^3}{16}-\frac{x^2}{8}-\frac{x^3}{16}-\frac{3x^4}{64}\\
&\approx 1+\frac{x}{2}+\frac{3x^2}{8}+\frac{x}{2}+\frac{x^2}{4}-\frac{x^2}{8}\\
&\approx 1+x+\frac{x^2}{2}
\end{align*}\]
So,
\[\begin{align*}\frac{1}{3}\sqrt{11}&\approx 1+\frac{1}{10}+\frac{1}{200}\\
\sqrt{11}&\approx 3.315
\end{align*}\]

Answer 16

Thank you so much :)

Answer 17

You're welcome, I hope it helped!

Answer 18

yes it did help but one more question iam sorry how did you get 1/3(11)^1/2

Answer 19

That's the value you get for plugging \(x=\dfrac{1}{10}\) in the product expression, \((1+x)^{1/2}(1-x)^{-1/2}\).

Answer 20

oh, thank you again :)

Answer 21

You're welcome