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If $x,y,z$ increase by $\%1$ , $\%2$ , $\%3$ respectively , then $w=\dfrac{x^2y^3}{z^4}$ ............. approximately.

$1)\ \%\ 3\text{ decrease}$

$2)\ \%\ 4\text{ decrease}$

$3)\ \%\ 3\text{ increase}$

$4)\ \%\ 4\text{ increase}$

I denote the new $w$ by $w'$,

$$w'-w=\left( \frac{1.01^2.1.02^3}{1.03^4}-1\right)\times w=\left(\frac{101^2\times102^3}{103^4\times100}-1\right)\times w$$

From here should I evaluate $\dfrac{101^2\times102^3}{103^4\times100}-1$ by hand (calculator is not allowed) or there is a quicker method to get to the correct answer?

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    $\begingroup$ $1.01^2\cdot 1.02^3/1.03^4$. Is the fraction of $w$ after the increase in $(x,y,z)$. $\endgroup$
    – David P
    Sep 3, 2021 at 14:57

4 Answers 4

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We can use approximations here

$$w=\dfrac{x^2y^3}{z^4}$$

Taking $\ln()$ on both sides

$$\ln w=2\ln x+3\ln y-4\ln z$$

Differentiating both side gives

$$\displaystyle\frac{dw}{w}=2\frac{dx}{x}+3\frac{dy}{y}-4\frac{dz}{z}$$

Substituting values

$$\displaystyle\frac{dw}{w}=2\frac{0.01x}{x}+3\frac{0.02y}{y}-4\frac{0.03z}{z}$$

$$\displaystyle\frac{dw}{w}=0.02+0.06-0.12=-0.04$$

Therefore $$\displaystyle\frac{dw}{w}*100=-4\text{%}$$

Hence there's a decrease of $4\text{%}$ in $w$

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    $\begingroup$ Thanks for the answer! What if $w,y<0$ we can't take $\ln()$ from $w$ and $y^3$? $\endgroup$
    – Etemon
    Sep 3, 2021 at 15:13
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    $\begingroup$ @MasterShifu Suppose value of $y$ is $-9$ and then it increases by $1\text{%}$ i.e. becomes $-8.9$ , then we can see from expression that value of $w$ will also increase and this increase will be equal to the increase in $w$ when value of $y$ is $8.9$ and increase by $1\text{%}$ i.e. becomes $9$ because of symmetry. $\endgroup$ Sep 3, 2021 at 15:30
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A rule of thumb: $1.001^n$ is about $1.00n$. More precisely, $(1+\alpha)^n\approx 1+n\alpha$ for small $\alpha$. This helps with a lot of estimation problems.

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  • $\begingroup$ You lost a $1$ in $1.001n$ and it is $n\alpha$ that needs to be small. $\endgroup$ Sep 3, 2021 at 15:17
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Dividing the numerator and denominator both by $100^5$ gives $$\frac{1.01^2 . 1.02^3}{1.03^4} -1$$ Now, we use $(1+x)^n=1+nx$ for $x<<1$.

Using this, we get $(1+0.02)(1+0.06)(1-0.12)-1$ which is approximately equal to $-0.04$.

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    $\begingroup$ It is $nx$ that needs to be $\ll 1$ for the approximation to be valid. MathJax hint: you can use \ll and \gg to get $\ll$ and $\gg$ $\endgroup$ Sep 3, 2021 at 15:18
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You can use partial elasticities, also know as condition numbers:

\begin{align*} \varepsilon_w \approx & \frac{x w'_x}{w} \varepsilon_x+\frac{y w'_y}{w} \varepsilon_y + \frac{z w'_z}{w} \varepsilon_z\\ = & 2\varepsilon_x + 3 \varepsilon_y - 4 \varepsilon_z= 2 + 3\times 2-4 \times 3 = -4 \end{align*}

So, the answer is an approximate decreasing by $\text{4%}$.

This comes from Taylor's formula: \begin{align*} f(\tilde x) \approx f(x) + f'(x)(\tilde x -x) \Rightarrow \\ f(\tilde x) - f(x) \approx f'(x)(\tilde x -x) \Rightarrow \\ \frac{f(\tilde x) - f(x)}{f(x)} \approx \frac{f'(x)(\tilde x -x)}{f(x)} \Rightarrow\\ \frac{f(\tilde x) - f(x)}{f(x)} \approx \frac{xf'(x)}{f(x)} \frac{\tilde x -x}{x} \end{align*}

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  • $\begingroup$ @GEdgar No, they are not. The elasticity, or condition number of a function $f$ at a point $x$ is given by $\frac{x f'(x)}{f(x)}$. $\endgroup$ Sep 3, 2021 at 17:40

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