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$x$ln$x$ quadratic approximation at $x=1$

2 ways :

1) quadratic approx whole thing : $(x-1)^2 + \dfrac {(x-1)^2} 2$

This isn't same as
2) split $x$ and take quad approx of ln$x$ at $x=1$

$x((x-1) - \dfrac 1 2 (x-1)^2) = x(2x - \dfrac 3 2)$

Why are they not same?

Bernard
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zcahfg2
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  • Recheck your second simplification. There is an error – Shailesh Sep 27 '15 at 09:52
  • if 1) is the first 2 terms of Taylor series, the first term should be $(x-1)$ – Alex Sep 27 '15 at 09:54
  • @Shailesh $x$(ln $x$) = $x((x-1)-\dfrac {1} {2} (x-1)^2)$
    $= x(x-1 - \dfrac 1 2 x^2 + x - \dfrac 1 2)$
    dropping $x^3$ terms... $x(2x-\dfrac 3 2)$     – zcahfg2 Sep 27 '15 at 10:04
  • You can't drop (x -1)^3. There is an x term and a constant term in there. – Shailesh Sep 27 '15 at 10:14
  • @Shailesh So I cant drop third term of ln$x$ and onwards? i.e. $(1-x)^4$ and onwards..

    But this is what the solution paper did.. similar but he did this by substituting $x = h + 1$. so approx. at $h = 0$ instead.

    $(1+h)$ ln$(1+h) = (1+h)(h-\dfrac {h^2} 2) = h + \dfrac {h^2} 2$ so substituing back gives 1).

    I don't see how this should be different to my second method.. He only took quad approx to ln$x$ too

     – zcahfg2 Sep 27 '15 at 10:24
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    No. You can only do it when you expand around zero. For all others, the lower terms will always be there and have to be accounted – Shailesh Sep 27 '15 at 10:27
  • Finally made sense and why he used substitution. Thanks a lot! – zcahfg2 Sep 27 '15 at 10:28

1 Answers1

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Change variable: if $x=1+h$, then \begin{align*}x\ln x&=(1+h)\ln(1+h)=(1+h)\Bigl(h-\frac{h^2}2+o(h^2)\Bigr)=h+\frac{h^2}2+o(h^2)\\ &=x-1+\frac{(x-1)^2}2+o\bigl((x-1)^2\bigr)=\frac{x^2-1}2+o\bigl((x-1)^2\bigr). \end{align*}

Bernard
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