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Math Help - Proof by use of Maclaurin's Expansion or Binomial Expansion

  1. #1
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    Proof by use of Maclaurin's Expansion or Binomial Expansion

    Prove that if x is so small that x^3 and higher powers of x may be neglected that

    {[1-x]/[1+x]}^1/2 = 1 - x + 0.5x^2

    I just can't get it out. Help please! I have A levels beginning next week!!!!!!!!
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by presando View Post
    Prove that if x is so small that x^3 and higher powers of x may be neglected that

    f(x)=\sqrt{\frac{1-x}{1+x}} = 1 - x + \frac{x^2}{2}

    I just can't get it out. Help please! I have A levels beginning next week!!!!!!!!
    Yes, use MacLaurin. The first term is \frac{f(0)}{0!}. The next term is \frac{f'(0)}{1!}x. The next term is \frac{f''(0)}{2!}x^2.

    So f(x)=f(0)+f'(0)x+\frac{f''(0)}{2}x^2 = 1-x+\frac{x^2}{2}

    I verified on my calculator that this is correct, so I'm not going to do out all the derivatives as it would take too long.
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  3. #3
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    Hi

    You can also separate like this

    \sqrt{\frac{1-x}{1+x}} = \sqrt{(1-x)(1-x+x^2+o(x^2))} = \sqrt{1+(-2x+2x^2+o(x^2))}

    -2x+2x^2+o(x^2))-\frac{1}{2!}\:\frac14\-2x+2x^2+o(x^2))^2" alt="\sqrt{\frac{1-x}{1+x}} = 1 + \frac12\-2x+2x^2+o(x^2))-\frac{1}{2!}\:\frac14\-2x+2x^2+o(x^2))^2" />

    \sqrt{\frac{1-x}{1+x}}= 1-x+x^2-\frac12\:x^2 + o(x^2)= 1-x+\frac12\:x^2+o(x^2)
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  4. #4
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    Quote Originally Posted by running-gag View Post
    Hi

    You can also separate like this

    \sqrt{\frac{1-x}{1+x}} = \sqrt{(1-x)(1-x+x^2+o(x^2))} = \sqrt{1+(-2x+2x^2+o(x^2))}

    -2x+2x^2+o(x^2))-\frac{1}{2!}\:\frac14\-2x+2x^2+o(x^2))^2" alt="\sqrt{\frac{1-x}{1+x}} = 1 + \frac12\-2x+2x^2+o(x^2))-\frac{1}{2!}\:\frac14\-2x+2x^2+o(x^2))^2" />

    \sqrt{\frac{1-x}{1+x}}= 1-x+x^2-\frac12\:x^2 + o(x^2)= 1-x+\frac12\:x^2+o(x^2)
    Thanks!

    However, I have never seen that method used before. I'm going to write it out and see if I understand it.
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  5. #5
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    Quote Originally Posted by redsoxfan325 View Post
    Yes, use MacLaurin. The first term is \frac{f(0)}{0!}. The next term is \frac{f'(0)}{1!}x. The next term is \frac{f''(0)}{2!}x^2.

    So f(x)=f(0)+f'(0)x+\frac{f''(0)}{2}x^2 = 1-x+\frac{x^2}{2}

    I verified on my calculator that this is correct, so I'm not going to do out all the derivatives as it would take too long.
    It's the simplification of f'(x) and f''(x) that's killing me.
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  6. #6
    Super Member redsoxfan325's Avatar
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    You don't need to simplify them, just evaluate them. Even if it's some crazy expression, plugging in x=0 shouldn't be too hard.

    f'(0)=-1
    f''(0)=1
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  7. #7
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    Quote Originally Posted by redsoxfan325 View Post
    You don't need to simplify them, just evaluate them. Even if it's some crazy expression, plugging in x=0 shouldn't be too hard.

    f'(0)=-1
    f''(0)=1
    I'll try.
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