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In the derivation of showing that a certain condition is sufficient and necessary, of which the context is given at the end, I couldn't rigorously resolve an intermediate step of technical nature.

Given three non-negative numbers $a$, $b$ and $c$ which satisfy $a\gg$ $\frac{1}{\frac{1}{b}+\frac{1}{c}}$, are we justified to put $a \gg b$ so that we have $$ \frac{1}{\frac{1}{b}+\frac{1}{a}} \approx b\quad ? $$

I'm tempted to just follow the line of \begin{align} \frac{1}{\frac{1}{b}+\frac{1}{a}} &= \frac{1}{\frac{1}{b}+ (\frac{1}{c} - \frac{1}{c}) + \frac{1}{a}} \\ &=\frac{1}{(\frac{1}{b}+ \frac{1}{c} + \frac{1}{a}) - \frac{1}{c}} \\ &\approx \frac{1}{(\frac{1}{b}+ \frac{1}{c}) - \frac{1}{c}} \\ &= b\ . \end{align}

I don't find this reasoning very convincing, since we don't seem to make a judgement on a possible differing order of magnitude between $b$ and $c$ and its resulting effect.


I've tried considering the following to shed some light.

Set $a = 10^{x}, b=10^y, c = 10^{z}$ and require $a \gg \frac{1}{\frac{1}{b} + \frac{1}{c}}$.

Taking the base 10 logarithm of both sides,

$$x \gg y+z-\log(b+c)\ .$$

Which doesn't clarify much.


Context. [All following quantities are, indeed, functions of frequency $\omega$. For readability, we omit explicitly denoting this dependence $f = f(\omega).]$

We consider the system of a coupled low-pass filter circuit.

Two low-pass filters are said to not load eachother if the input impedance $Z_{in,2}$ of the second one is much greater than the output impedance $Z_{out,1}$ of the first one.

If this is the case, we can approximate the system transfer function $H$ as the product of the respective transfer functions of the individual filters $H \approx H_1 \cdot H_2$.

It can be shown that the system transfer function is given by $$ H = \bigg( 1 - \frac{Z_{R_1}}{Z_{R_1} +\frac{1}{ \frac{1}{Z_{C_1}} + \frac{1}{Z_{R_2} + Z_{C_2}} } } \bigg) \cdot \frac{Z_{C_2}}{Z_{R_2} + Z_{C_2}}\ . $$ Similarly, we have $$ H_1 = \frac{Z_{C_1}}{Z_{C_1}+Z_{R_1}}\quad \text{ and }\quad H_2 = \frac{Z_{C_2}}{Z_{C_2} + Z_{R_2}} \ . $$

Now, $$Z_{in,2} \equiv Z_{C_2} + Z_{R_2} \gg \frac{1}{\frac{1}{Z_{R_1}} + \frac{1}{Z_{C_1}} } \equiv Z_{out,1}$$ seems to be a necessary and sufficient condition to justify $H \approx H_1 \cdot H_2$ .

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  • $\begingroup$ Your condition is equivalent to $\frac1a\ll\frac{1}{b}+\frac{1}{c}$. I think reasoning will be easier from there, generally. We can easily see that we have no idea whether $a$ and $b$ are close or far apart just from that. It could be $c$ that was very small. But clearly, either $c\ll a$ or $b\ll a$. $\endgroup$ – Arthur Jul 12 '17 at 11:53
  • $\begingroup$ Yes, I noted that. How does this observation help us justify $a\gg b$? Namely, we also have $\frac{1}{b} + \frac{1}{c} \geq \frac{1}{b} = b$. If $\geq$ would be a $\leq$ in this case I would be convinced of $a \gg b$, but this isn't the case. $\endgroup$ – Mussé Redi Jul 12 '17 at 11:58
  • $\begingroup$ @Arthur Never mind my last comment. Indeed, we either have $c \ll a$ or $b \ll a$. How can we make it clear that the distance between $b$ and $c$ is irrelevant? $\endgroup$ – Mussé Redi Jul 12 '17 at 12:04
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    $\begingroup$ Well, if the distance between $b$ and $c$ is great, then only one of $\frac1b$ and $\frac1c$ really contributes to $\frac1b+\frac1c$. And if they're close, then $\frac1b+\frac1c$ is at most twice the largest one of $\frac1b$ and $\frac1c$. So if a factor of $2$ is negligible to you, then there you have it, only the largest of $\frac1b$ and $\frac1c$ contributes, and whether the other one is far away or close, it doesn't matter. $\endgroup$ – Arthur Jul 12 '17 at 12:06
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In my opinion, the fact that $x \gg y+z-\log(b+c),$ where $a = 10^x,$ $b=10^y,$ and $c = 10^z,$ clarifies the problem almost completely.

We need only consider an example such as $y=100,$ $z=-100,$ where $\log(b+c) \approx \log(b) = y,$ to see that in this case $$y+z-\log(b+c) \approx z,$$ and therefore in this case the fact that $a\gg1/\left(\frac{1}{b}+\frac{1}{c}\right)$ has told us only that $a \gg c$ and has told us nothing about the relative sizes of $a$ and $b.$ We could have either $x = 0$ or $x = 200$ and all of the formulas above would still be true.

When I write that this "clarifies" the problem, by the way, I mean that this formula shows that the method of proof you are attempting does not work.

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As mentioned by @Arthur in the comments, we have $c\ll a$ or $b \ll a$.

The distance between $b$ and $c$ is irrelevant because if they're far apart, one of the terms dominates over the other. On the other hand, if they're close, we can approximate the two terms by twice the bigger one, where 2 is indeed considered negligible.

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