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I'm doing the applications of differentiation problem sheet from MIT single variable calculus and I don't understand the solution given in the question. I can solve the question using the Taylor approximation, however, I don't think that is what you're meant to do judging by the solutions.

https://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/unit-2-applications-of-differentiation/part-a-approximation-and-curve-sketching/problem-set-3/MIT18_01SC_pset2sol.pdf

question 2A-12 c

Thanks

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    $\begingroup$ I don't think you mean $e^{(-x)^2}$, do you? That would be the same as $e^{x^2}$. $\endgroup$ – TonyK May 15 at 14:40
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    $\begingroup$ And surely it's $2A$-$12c$, not $2A$-$12b$? Make an effort! $\endgroup$ – TonyK May 15 at 14:42
  • $\begingroup$ sorry wrote this question in a rush before I left the house $\endgroup$ – Martin May 15 at 14:51
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We have: $$e^x\approx 1+x\Rightarrow e^{-x^2}\approx 1-x^2$$

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  • $\begingroup$ I don't know if this is a stupid question or not, but, how come you're allowed to sub in -x^2 into the linear approximation to get the quadratic approximation? $\endgroup$ – Martin May 15 at 14:58
  • $\begingroup$ Because $-x^2$ is the exponent of e. $\endgroup$ – richard1941 May 15 at 15:00
  • $\begingroup$ would you not use 1+x+1/2(x^2) as that is the quatratic approximation then sub in -x^2 into that? $\endgroup$ – Martin May 15 at 15:08
  • $\begingroup$ And it all depends on what you mean by a quadratic approximation. Some might say that the quadratic approximation is what you get by substituting $-x^2$ into $e^x = 1+x+\frac{x^2}{2}$, which would result in a fourth degree polynomial. $\endgroup$ – richard1941 May 15 at 15:09
  • $\begingroup$ I thought quadratic approximation meant specifically a taylor series up to the second derivative? ocw.mit.edu/courses/mathematics/… $\endgroup$ – Martin May 15 at 15:14
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The solution actually tells you to use the Taylor Series approximation for $e^x$ which is $\sum_{k=0}^{\infty}x^n/n!$ and plug in $-x^2$ for $x$ to get the approximation for $e^{-x^2}$.

$$e^x\approx 1+x \land x \mapsto -x^2 \implies e^{-x^2}\approx 1-x^2$$

Turns out that this approximation looks good for $x\in \left[-0.5, 0.5 \right]$ which obviously depends on what use this approximation is being put to and what restrictions on permissible error are imposed.

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    $\begingroup$ And for better approximations far from x=0, go to the chapter on probability functions in AMS 55, the Handbook of Mathematical Functions. Alas, there is no single approximation that is good everywhere. $\endgroup$ – richard1941 May 15 at 15:13
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If $x$ is small (in absolute value), then

$$e^x\approx 1+x.$$

For instance, with $x=-0.01$,

$$e^{-0.01}=0.99004983\approx 1-0.01.$$

But it makes no difference if we write

$$e^{-x^2}\approx 1-x^2$$ and try $x=0.1$.

Just two ways to write the same thing.


By the way, maths don't go wrong.

The derivatives of $e^x$ are $$e^x,e^x,e^x,e^x,e^x,e^x,\cdots$$ which evaluate as $1,1,1,1,1,1,\cdots$ at $x=0$, giving the Taylor coefficients $1,1,\dfrac12,\dfrac16,\dfrac1{24},\dfrac1{120},\cdots$.

On the other hand, the derivatives of $e^{-x^2}$ are

$$e^{-x^2},-2xe^{-x^2},(4x^2-2)e^{-x^2},(12x-8x^3)e^{-x^2},(16x^4-48x+12)e^{-x^2},(-35x^5+160x^3-120)e^{-x^2},\cdots$$

giving

$$1,0,-2,0,12,0,\cdots$$

and as should,

$$1,0,-1,0,\frac12,0,-\frac16,0,\frac1{24},\cdots$$

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