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I am following MIT's calculus videos and I have noticed that when dealing with linear approximations, the professor calculates a set of approximation "formulas" for $x$ near $0$ like $1+x$ for $e^x$ , and when he has to calculate say, $e^{3x}$, he uses the same plugging the factor of $x$ instead of $x$ to get $1+3x$, but he doesn't talk about wether this is a valid thing to do with precalculated approximations other than the ones near 0, and at least with some examples I have tried it doesn't work. I do find logical that this works near $0$ because as $x$ goes to $0$ $x$ multiplied by some factor will also go to $0$, but I'm not sure if that is the actual reason why that works, or even if it works for all approximations near $0$.

So the question is wether this does not work for $x$ not near $0$, and if it works for all approximations near $0$

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He uses Maclaurin series for $x$ near $0$ or Taylor series for $x=a$ non necessary $0$.

Generally it's convenient a substitution such that you have to work near $0$.

For example let $f(x)=e^{3x}$, putting $3x=y$ and using $e^y\approx 1+y+\cdots$ and substituting back you'll find $f(x)\approx 1+3x+\cdots$

Another example: let $f(x)=e^{x}$; the approximation of $f$ near $x=-4$ can be found putting $x+4=y$ so that you can use the approximation near $y=0$; because $e^{x}=e^{y-4}=e^{-4}e^y$ and using $e^y\approx 1+y+\cdots$ and substituting back you'll find $f(x)\approx e^{-4}(1+(x+4)+\cdots)$.

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  • $\begingroup$ Alright, thanks! In the lectures just a linear approximation formula is given and explained as a tangent line equation and I had not realised that linear and quadratic approximations are actually the first terms of the polynomial expansions. $\endgroup$ – Aitor Ormazabal Dec 24 '13 at 12:06
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Let consider that you know the expansion of Exp[Y] when Y is small (1 + Y + Y^2 /2 + ...). Now, you need the expansion of Exp[3*X] when X is small; if X is small, 3*X is still small. So, replace in the first development Y by 3*X and you get (1 + 3*X + 9^*X^2 / 2 +...).

So, in practice, you only need to know a few simple expansions to be able to extend and apply them to similar cases.

Please tell me if this clarifies things to you.

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  • $\begingroup$ The "linearization" of a function, f, at x= a, is given by f'(a)(x- a)+ f(a). With $f(x)= e^{3x}$, $f'(x)= 3e^{3x}$ so "linearization near x= -4" means that a= -4. $f(-4)= e^{-12}$ and $f'(-4)= 3e^{-12}$ so the "linearization near x= -4" is $y= 3e^{-12}(x+ 4)+ e^{-12}$. $\endgroup$ – user247327 Oct 14 '16 at 14:37

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