# Approximation of $e^{-x^2}$

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.

question 2A-12 c

Thanks

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

We have: $$e^x\approx 1+x\Rightarrow e^{-x^2}\approx 1-x^2$$

• 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? – Martin May 15 at 14:58
• Because $-x^2$ is the exponent of e. – richard1941 May 15 at 15:00
• would you not use 1+x+1/2(x^2) as that is the quatratic approximation then sub in -x^2 into that? – Martin May 15 at 15:08
• 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. – richard1941 May 15 at 15:09
• I thought quadratic approximation meant specifically a taylor series up to the second derivative? ocw.mit.edu/courses/mathematics/… – Martin May 15 at 15:14

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.

• 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. – richard1941 May 15 at 15:13

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,

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