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After conducting a series of experiments, a physicist concluded that the pressure around an object placed in a moving fluid is given by $$P(M) = P_0 \left( 1 + \frac{k - 1}{2}M \right)^{k/(k-1)},$$where $M$ is the square of the ratio of the speed of the fluid to the speed of sound, $P_0$ is a positive constant, and $k$ is a positive integer greater than 1. Prove that the pressure is approximately $P_0 \left(1 + \frac{k}{2} M \right)$ for small values of $M$.

My initial thought was to try and apply linear approximation, but I didn't know what to do with the given function as it seemed very messy.

questionasker
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2 Answers2

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hint

You might know that if $ |x|<1$, then

$$(1+x)^a=1+ax+\frac{a(a-1)}{2!}x^2+\frac{a(a-1)(a-2)}{3!}x^3+....$$

$$\approx 1+ax\; \text{ for very small } x$$

So, for small $ M$,

$$(1+\frac{k-1}{2}M)^{\frac{k}{k-1}}$$ $$\approx 1+\frac{k}{k-1}\frac{k-1}{2}M$$ $$=1+\frac k2 M$$

hamam_Abdallah
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$$P = P_0 \left( 1 + \frac{k - 1}{2}M \right)^{\frac k{k-1}}$$ $$\log(P)=\log(P_0)+\frac k{k-1}\log\left( 1 + \frac{k - 1}{2}M \right)$$ Now, by Taylor $$\frac k{k-1}\log\left( 1 + \frac{k - 1}{2}M \right)=\frac k{k-1}\Bigg[ \frac{1}{2} (k-1) M-\frac{1}{8} (k-1)^2 M^2+O\left(M^3\right)\Bigg]$$ $$\frac k{k-1}\log\left( 1 + \frac{k - 1}{2}M \right)=\frac{k }{2}M+\frac{(1-k) k}{8} M^2+O\left(M^3\right)$$ Now, using $y=e^{\log(y)}$ $$\left( 1 + \frac{k - 1}{2}M \right)^{\frac k{k-1}}=1+\frac{k M}{2}+\frac{k M^2}{8}+O\left(M^3\right)$$ Multiply by $P_0$ and truncate to $O\left(M^2\right)$

Claude Leibovici
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