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Here 'n'is a constant number between 1 and 2 and we know that $V<<c$. Then how do you show that:$$\frac{1}{1+\frac{V}{nc}}$$ can be approximated by this expression:$$(1-\frac{V}{nc})$$

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You can simplify the problem by asking how to estimate

$$ \frac{1}{1+x} $$

You can apply all of the various methods you learned in single variable calculus to produce an estimate; additional methods apply too, such as recognizing it as a geometric series:

$$ \frac{1}{1+x} = 1 - x + x^2 - \ldots $$


However, you're already been given the estimate directly; it might be easier to verify it rather than try to figure out how you would derive it had you not known about it. The absolute difference happens to have a nice form which makes it easy to see that the absolute error is small:

$$ \frac{1}{1+x} - (1 - x) = \frac{x^2}{1+x} $$

In fact, you could even use this approach to derive the estimate, which I will demonstrate since it's an uncommon, but neat technique.

If you recognized that the value is approximately $1$, you can write:

$$ \frac{1}{1+x} - 1 = -\frac{x}{1+x} $$

and then recognize the right hand side is approximately $-x$:

$$ \left( -\frac{x}{1+x} \right) - (-x) = \frac{x^2}{1+x}$$

and combine the estimates to get

$$ \frac{1}{1+x} = 1 - x + \frac{x^2}{1+x} $$

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$$\frac{1}{1+k} = \frac{1-k}{(1-k)(1+k)}= \frac{1-k}{1-k^2}\approx 1-k$$ if $k$ is very small. $k=\frac{V}{nc}$ with $V\ll c$, so this is exactly what you want.

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Another way of presenting this is to telescope the expression to arbitrary precision generating the geometric series in the process

Thus we start with

$$\frac{1}{1+\frac{V}{nc}}=\frac{1}{1+\frac{V}{nc}}\frac{1-\frac{V}{nc}}{1-\frac{V}{nc}}=\frac{1-\frac{V}{nc}}{1-\left(\frac{V}{nc}\right)^2}$$

and the next iteration $$\frac{1}{1+\frac{V}{nc}}=\frac{1-\frac{V}{nc}}{1-\left(\frac{V}{nc}\right)^2}\frac {1+\left(\frac{V}{nc}\right)^2} {1+\left(\frac{V}{nc}\right)^2}=\frac{1-\frac{V}{nc}+ \left(\frac{V}{nc}\right)^2-\left(\frac{V}{nc}\right)^3} {1-\left(\frac{V}{nc}\right)^4}$$

and so on.

We can stop at any stage once we are happy to approximate the denominator as $1$.

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The inverse of the first, multiplied by the second, is $$1-\frac{V^2}{n^2c^2}$$ very near from $1$.

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