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Thread: Power series expansion of (1-x)^-1/2. Approximating the value of square root 2.

  1. #1
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    Power series expansion of (1-x)^-1/2. Approximating the value of square root 2.

    Hello, first post,
    This is the question everyone is stuck on for our Year 11 Maths C assignment on Matrices and Applications. Imgur: The most awesome images on the Internet
    Question 4a and b.
    Looking at 4a, its really simple with basic surds and I have already verified that it equals Square Root 2.
    For question 4b, I used (1-x)^-1/2=SquareRoot(2) and solved x for 1/2. I then subbed in x=1/2 into the first 4 terms of the Power Series Expansion and got 1.383. I seriously thought I got it cos 1.383 is a pretty decent approximation for SquareRoot(2). I worked absolute error for this to be 0.087311.

    Our teacher said this is wrong. She said you aren't meant to solve for X but apply the equation you have verified in 4a into the power series for question 4b. I'm so confused and I don't really get what she means by that.

    First post so I don't know what to expect but I was pretty desperate to write this and come here haha. Any help would be very appreciated.
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    Re: Power series expansion of (1-x)^-1/2. Approximating the value of square root 2.

    As $\displaystyle \begin{align*} \left( 1 - x \right) ^{-\frac{1}{2}} = \frac{1}{\sqrt{1 - x}} \end{align*}$, if you sub in $\displaystyle \begin{align*} x = -1 \end{align*}$ you will get $\displaystyle \begin{align*} \frac{1}{\sqrt{2}} \end{align*}$, so you can use the first four terms of your series to get an approximation for $\displaystyle \begin{align*} \frac{1}{\sqrt{2}} \end{align*}$ and then solve for $\displaystyle \begin{align*} \sqrt{2} \end{align*}$.
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    Re: Power series expansion of (1-x)^-1/2. Approximating the value of square root 2.

    Quote Originally Posted by Bumdwarf View Post
    Hello, first post,
    This is the question everyone is stuck on for our Year 11 Maths C assignment on Matrices and Applications. Imgur: The most awesome images on the Internet
    Question 4a and b.
    Looking at 4a, its really simple with basic surds and I have already verified that it equals Square Root 2.
    For question 4b, I used (1-x)^-1/2=SquareRoot(2) and solved x for 1/2. I then subbed in x=1/2 into the first 4 terms of the Power Series Expansion and got 1.383. I seriously thought I got it cos 1.383 is a pretty decent approximation for SquareRoot(2). I worked absolute error for this to be 0.087311.

    Our teacher said this is wrong. She said you aren't meant to solve for X but apply the equation you have verified in 4a into the power series for question 4b. I'm so confused and I don't really get what she means by that.

    First post so I don't know what to expect but I was pretty desperate to write this and come here haha. Any help would be very appreciated.
    A badly worded question in my opinion. If you are required to use your result from 4(a) it should say so.

    I'd say you are supposed to use the expansion with x=1/50. Find the sum of the first 4 terms. Then multiply the result by 7/5.
    That is, work out the RHS of the expression in 4(a).

    The question should say:
    "Using your result from 4(a) together with the first four terms of the power series......blah, blah, blah". (If that's the way they want you to do it!)
    Last edited by Debsta; Feb 22nd 2017 at 06:56 PM.
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    Re: Power series expansion of (1-x)^-1/2. Approximating the value of square root 2.

    Thanks, this makes sense to me. I'll ask my teacher if this is right.
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    Re: Power series expansion of (1-x)^-1/2. Approximating the value of square root 2.

    Quote Originally Posted by Bumdwarf View Post
    Thanks, this makes sense to me. I'll ask my teacher if this is right.
    I'd be interested to know what they say.
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    Re: Power series expansion of (1-x)^-1/2. Approximating the value of square root 2.

    I thought of a better way.

    Notice that $\displaystyle \begin{align*} \left( 1 - x \right) ^{-\frac{1}{2}} = \frac{1}{\sqrt{1 - x}} \end{align*}$ and if you let $\displaystyle \begin{align*} x = \frac{1}{2} \end{align*}$ you will have $\displaystyle \begin{align*} \frac{1}{\sqrt{1 - \frac{1}{2}}} = \frac{1}{ \sqrt{ \frac{1}{2} } } = \frac{1}{ \frac{1}{ \sqrt{2} } } = \sqrt{2} \end{align*}$, this should get you the first four terms quite quickly.
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    Re: Power series expansion of (1-x)^-1/2. Approximating the value of square root 2.

    Quote Originally Posted by Debsta View Post
    I'd be interested to know what they say.
    She said we are on the right track so I'll take that as a yes. Thanks heaps.
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