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I have just seen a lecture about linear approximation, in which it was established that when $x$ is near $0$:

  1. The linear approximation of $e^x$ is $1+x$

  2. The linear approximation of $(1+x)^r$ is $1+rx$

Afterwards, the lecturer explained how to find—using 1 and 2—the linear approximation of:

$e^{-3x} \times (1+x)^{-1/2}$.

His next line is:

$(1-3x)(1-0.5x)$

So it seems that he was acting under the assumption that the linear approximation of the product of two functions is the product of the linear approximations of the functions.

Is that true in general for any order of approximations? Maybe just for linear ones? Maybe just when x is near zero? Maybe only when the approximation is linear and x near 0?

Here is a link to the lecture. The comments were made at 26:30.

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  • $\begingroup$ $(1-3x)(1-0.5x) = 1.5x^2 -3.5x+1$ is a quadratic approximation. The correct linear approximation is given by the product rule $\endgroup$ Commented Dec 27, 2020 at 11:47
  • $\begingroup$ thanks for the correction, but I still don't understand why is it allowed to take the product of the linear approximations of the inner functions as any order approximations of the whole function. $\endgroup$ Commented Dec 27, 2020 at 11:49
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    $\begingroup$ Sorry -- I meant to be dismissive of the professor, not of your question. You're right to be skeptical of this, because it's incorrect (though, as an answer now points out, it is correct up to first order. That is, if you take a linear approximation to this approximation, you get a correct linear approximation of the original function). $\endgroup$ Commented Dec 27, 2020 at 11:52
  • $\begingroup$ $(1-3x)(1-0.5x) = 1.5x^2 -3.5x+1$ and the linear approximation is $ -3.5x+1$. So the question in the title has answer "no". But we can multiply and then drop extra terms, so this is a fast method to do it. As remarked by Halia, it is the same result as using the product rule. $\endgroup$
    – GEdgar
    Commented Dec 27, 2020 at 12:01
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    $\begingroup$ This is always correct. If we approximate $f$ and $g$ near $a$ by $f(a) + f'(a)x$ and $g(a) + g'(a)x$ respectively, then the product of these approximations is $f(a)g(a) + (f(a) g'(a) + f'(a) g(a)) x + f'(a)g'(a)x^2$. You can see the linear term agrees with the formula for the product rule, which is the "correct" way to compute the linear approximation. $\endgroup$ Commented Dec 27, 2020 at 12:36

4 Answers 4

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This is commonly used, and that is because the difference between the two is negligible.

In general, if we have a function $f(x)$ approximated by $f(x_0)+f'(x_0)(x-x_0)$ and another function $g(x)$ approximated the same then you can do some algebra to multiply the two approximations in a part and multiply the two functions then approximate in the other part and compare the results, it won't be the same but the difference would be pretty negligible as long as $x$ is close to $x_0$ so you can use the two methods interchangeably, to be exact the difference would be $f'(x_0)g'(x_0)(x-x_0)^2$.

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  • $\begingroup$ thanks for your input! I do not understand it, though. what do you mean by "in a part"? I find it all hard to follow. $\endgroup$ Commented Dec 27, 2020 at 12:24
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Let's consider the function in question, $f(x)=e^{-3x} \cdot (1+x)^{-1/2}$. To find the linear approximation we need to compute $f(0)$ and $f'(0)$:

  • $f(0)=1$
  • Using the product rule, we find that $$f'(x)=-\frac{e^{-3x}(6x+7)}{2(1+x)^{3/2}} \, .$$Hence, $f'(0)=-7/2$.

The linear approximation formula is $$f(a+h) \approx f(a)+f'(a)h \, .$$ At $a=0$, it is \begin{align} f(h)&\approx f(0)+f'(0)h \\ f(h)&\approx 1-\frac{7}{2}h \, . \end{align} Since you used the variable $x$ in your question, we'll rewrite this as $f(x) \approx 1-\frac{7}{2}x$. This is not the same as the approximation given in the lecture. As HallaSurvivor notes in the comments, the approximation given is a quadratic approximation. If we expand the brackets, it is $$ f(x) \approx 1.5x^2 - 3.5x + 1 \, . $$ But this approximation is only very rough and ready. If we use the formula $$ f(a+h) \approx f(a) + f'(a)h + \frac{f''(a)}{2}h^2 $$ we obtain a different result, which is better than the one obtained by the lecturer. Still, the quadratic approximation obtained by the lecturer is better than the linear approximation we computed earlier.

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  • $\begingroup$ thanks! it did make sense to me that the product of the linear approximation will give a good approximation, but now that I know that it is not the best quadratic approximation, it all makes more sense to me. $\endgroup$ Commented Dec 27, 2020 at 12:26
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You can prove that the approximation at each order, is exactly equal: Let $H(x)=f(x)g(x)$, by approximating $f$ and $g$ by their taylor series: $$H(x)\approx \left(f(x_0)+f^{\prime}(x_0)(x-x_0)+\frac{f^{\prime \prime}(x_0)}{2}(x-x_0)^2+\dots\right)\left(g(x_0)+g^{\prime}(x_0)(x-x_0)+\frac{g^{\prime \prime}(x_0)}{2}(x-x_0)^2+\dots\right)$$ Multiplying and grouping by $(x-x_0)^k$: $$H(x)\approx f(x_0)g(x_0)+(x-x_0)\left(f(x_0)g^{\prime}(x_0)+f^{\prime}(x_0)g(x_0)\right)+(x-x_0)^2\left(\frac{f(x_0)g^{\prime\prime}(x_0)}{2}+\frac{g(x_0)f^{\prime\prime}(x_0)}{2}+f^{\prime}(x_0)g^{\prime}(x_0)\right)$$ which, using the product rule, you can check that equals the taylor expansion of $H(x)$. So, yes, approximating each function separately up to order $k$ and then multiplying them will give you the same answer as approximating the function as a whole up to the same order $k$. In your example, you are approximating the functions up to linear terms, therefore, your approximation of $e^{-3x}(1+x)^{-1/2}$ will be correct up to linear terms, and the quadratic one will be incorrect and should be discarded.

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    $\begingroup$ nice, really what I was looking for. I think the professor should have talked about that - I hate it when those kinds of steps are skipped. $\endgroup$ Commented Dec 27, 2020 at 12:39
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The linear approximation of a product equals the product of the linear approximations modulo terms of degree $2$.

Define the operator $M_1$ that takes a differentiable function and giving the linear approximation of it: $$ M_1 f(x) := f(0) + f'(0) x $$

Then, $$\begin{align} M_1 f(x) \, M_1 g(x) &= (f(0)+f'(0)x)(g(0)+g'(0)x) \\ &= f(0)g(0) + f(0)g'(0)x + f'(0)g(0)x + f'(0)g'(0)x^2 \\ &= (fg)(0) + (fg)'(0)x + f'(0)g'(0)x^2 \\ &= M_1(fg)(x) + O(x^2) . \end{align}$$

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