Mathematical tools for natural sciences

Suppose we want to estimate a quantity Q which is a function of variables x,y,z,...

We write,

\(~~~~~~~~~~~~~~~~~~~~~\small{Q = f(x,y,z,...)}\)

In the previous section, we derived an expression for the uncertainity df of a function \(\small{f(x,y,z,...)}\) in terms
of errors (dx, dy, dz,...) in the variables as,

\(\small{ dQ~=~ \dfrac{\partial f}{\partial x} dx~+~\dfrac{\partial f}{\partial y} dy~+~\dfrac{\partial f}{\partial z} dz ~+~....}\)

where \(\small{dx, dy, dz, ... }\) are the actual erros on the quantities.

In general, we may not get the values of actual errors on measured quantities, since their correct values are not known.
For each variable, we generally get a mean value and its standard deviation computed from certain number of samples.
**We can use the standard deviations of dependent variables as a measure of their uncertainity **.

**The question is, given the standard deviations of the variables x, y, z,..., can we estimate the standard deviation in Q as a measure of the uncertainity on its computed value?.** We will derive a methodology that uses the standard deviations \( \sigma_x, \sigma_y, \sigma_z, ....\) of the underlying distributions.

If we want to skip the derivation, we can jump to the summary formula box that follows to see the result and proceed from that point.

** Derivation of a generalized error propagation formula:**

We start with the function,

\(\small{Q = f(x,y,x,...)}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1) \)

We already know that

\(\small{ dQ~=~ \dfrac{\partial f}{\partial x} dx~+~\dfrac{\partial f}{\partial y} dy~+~\dfrac{\partial f}{\partial z} dz ~+~....}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2) \)

Let \(\small{x_i, y_i, z_i, ... }\) denote the individual samples from their respective population

Let \(\small{\overline{x}, \overline{y}, \overline{z},... }\) denote the mean values over the

We have, for a single set of data points \(\small{x_i, y_i, z_i, ... }\),

\(\small{Q_i = f(x_i, y_i, z_i) }\)

We also assume that the best estimate of Q is when the variables have their average values (this may not be true always). We can write,

\(\small{\overline{Q} = f(\overline{x}, \overline{y}, \overline{z}) }\)

We can write,

\(\small{dQ = Q_i - \overline{Q},~~~~dx = x_i -\overline{x},~~~~dy=y_i-\overline{y},~~~~dz = z_i-\overline{z} }\)

Substituting these expressions in equation (2) we get,

\(\small{dQ~=~Q_i-\overline{Q}~=~(x-x_i)\dfrac{\partial f}{\partial x}~+~(y-y_i)\dfrac{\partial f}{\partial y}~+~(z-z_i)\dfrac{\partial f}{\partial z}~+~....}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(3) \)

Using these, we will write down the expression for the variance \(\small{\sigma^2_Q }\) in Q:

\(\sigma_Q^2~=~\small{\dfrac{1}{N}\sum_\limits {i=1}^N (Q_i - \overline{Q})^2 }\)

\(~~~~~~=~\small{\dfrac{1}{N} \sum_\limits{i=1}^N \left( (x_i - \overline{x})\dfrac{\partial f }{\partial x } + (y_i - \overline{y})\dfrac{\partial f }{\partial y } + (z_i - \overline{z})\dfrac{\partial f }{\partial z } + \right )^2 }\)

\(~~~~~~~ \begin{split} ~=~~\small{ \dfrac{1}{N} \sum_\limits{i=1}^N \left( (x_i - \overline{x})^2\left(\dfrac{\partial f }{\partial x }\right)^2 + (y_i - \overline{y})^2\left(\dfrac{\partial f }{\partial y }\right)^2 + (z_i - \overline{z})^2\left(\dfrac{\partial f }{\partial z }\right)^2 \\ \quad + 2(x_i - \overline{x})(y_i - \overline{y})\left(\dfrac{\partial f}{\partial x}\right) \left(\dfrac{\partial f}{\partial y}\right) + 2(y_i - \overline{y})(z_i - \overline{z})\left(\dfrac{\partial f}{\partial y}\right) \left(\dfrac{\partial f}{\partial z}\right) \\ \quad + 2(z_i - \overline{z})(x_i - \overline{z})\left(\dfrac{\partial f}{\partial z}\right) \left(\dfrac{\partial f}{\partial x}\right)~+~....... \right )~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(4) \\ } \end{split} \)

We know that,

\( \sigma_x^2 = \small{ \dfrac{1}{N} \sum\limits_{i=1}^{N}(x_i - \overline{x})^2 ~~~~~~~~ }\) Variance in x

\( \sigma_y^2 = \small{ \dfrac{1}{N} \sum\limits_{i=1}^{N}(y_i - \overline{y})^2~~~~~~~~ }\) Variance in y

\( \sigma_z^2 = \small{ \dfrac{1}{N} \sum\limits_{i=1}^{N}(z_i - \overline{z})^2 ~~~~~~~~ }\) Variance in z

\( \sigma_{xy}^2 = \small{ \dfrac{1}{N} \sum\limits_{i=1}^N (x_i - \overline{x})(y_i - \overline{y})~~~~ }\) Covariance between x and y.

\( \sigma_{yz}^2 = \small{ \dfrac{1}{N} \sum\limits_{i=1}^N (y_i - \overline{y})(z_i - \overline{z})~~~~~~ }\) Covariance between y and z

\( \sigma_{zx}^2 = \small{ \dfrac{1}{N} \sum\limits_{i=1}^N (z_i - \overline{z})(x_i - \overline{x})~~~~~~ }\) Covariance between z and x

................... similarly for other variables ............................

Substituting the above expressions into equation(4), we get,

\(\small{ \sigma_Q^2~=~\sigma_x^2 \left( \dfrac{\partial f}{\partial x} \right)^2 + \sigma_y^2 \left( \dfrac{\partial f}{\partial y} \right)^2 + \sigma_z^2 \left( \dfrac{\partial f}{\partial z} \right)^2 + \sigma_{xy}^2 \left(\dfrac{\partial f}{\partial x} \right) \left(\dfrac{\partial f}{\partial y} \right) + \sigma_{yz}^2 \left(\dfrac{\partial f}{\partial y} \right) \left(\dfrac{\partial f}{\partial z} \right) + \sigma_{zx}^2 \left(\dfrac{\partial f}{\partial z} \right) \left(\dfrac{\partial f}{\partial x} \right) +... } \)

\(~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-------- (5) \)

In this case, the above expression for the variance in Q reduces to,

\(\small{ \sigma_Q^2~\approx~\sigma_x^2 \left( \dfrac{\partial f}{\partial x} \right)^2 + \sigma_y^2 \left( \dfrac{\partial f}{\partial y} \right)^2 + \sigma_z^2 \left( \dfrac{\partial f}{\partial z} \right)^2 + .... }~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(6) \)

We summarize the important reuslt from the above derivation here:

For a function \(\small{Q = f(x,y,z,...)}\) of

\(\small{ \sigma_Q^2~\approx~\sigma_x^2 \left( \dfrac{\partial f}{\partial x} \right)^2 + \sigma_y^2 \left( \dfrac{\partial f}{\partial y} \right)^2 + \sigma_z^2 \left( \dfrac{\partial f}{\partial z} \right)^2 + .... } \)

\(\small{\sigma_Q \approx \sqrt{ \sigma_x^2 \left( \dfrac{\partial f}{\partial x} \right)^2 + \sigma_y^2 \left( \dfrac{\partial f}{\partial y} \right)^2 + \sigma_z^2 \left( \dfrac{\partial f}{\partial z} \right)^2 + .... } }\)

Note that the variances of the independent varibales add

Let us now derive expressions for the uncertainity in a quantity Q which is a function of two independent (uncorrelated) variables x and y. These formulas are handy in computing the uncertainity in many simple estimations. In every case, we will use the propagation formula derive before to arrive at a handy expression.

Let \(~~\small{Q~=~ax \pm by }~~\), where a and b are constants

Then, \(~~~\small{\left(\dfrac{\partial Q}{\partial x}\right)~=~a,~~~~~~ \left(\dfrac{\partial Q}{\partial y}\right)~=~\pm b }\)

The propagation formula \(~~\small{\sigma_Q^2 = \sigma_x^2 \left(\dfrac{\partial Q}{\partial x}\right)^2 + \sigma_y^2 \left(\dfrac{\partial Q}{\partial y}\right)^2 }~~\) becomes,

Note that whether two quantities are added or subtracted, their uncertainities always add quardratically.

The circumference of the rectangular field is given by\(~~~\small{S = 2(L+W) = 2L+2W}\)

Given that \(~~\small{\sigma_L = 0.2~m,~~~a = 2,~~~\sigma_W = 0.15~m,~~~b = 2 }\),

\(\small{\sigma_S^2 = \sigma_L^2 a^2 + \sigma_W^2 b^2~~=~~0.2^2\times 2^2 + 0.15^2 \times2^2~=~4.182 }\)

Therefore, \(~~\small{\sigma_S~=~\sqrt{4.182} = 2.045~m~~~ }\) is the uncertainity in the circumference.

(i) \(~~\) Let \(\small{Q~=~a x y,~~ }\) where a is a constant

We write,\(~~~\small{\left(\dfrac{\partial Q}{\partial x}\right)=ay, }~~~\)\(~~~\small{\left(\dfrac{\partial Q}{\partial y}\right)=ax }\)

The propagation formula \(~~\small{\sigma_Q^2 = \sigma_x^2 \left(\dfrac{\partial Q}{\partial x}\right)^2 + \sigma_y^2 \left(\dfrac{\partial Q}{\partial y}\right)^2 }~~\) becomes,

\(\small{ \sigma_Q^2~=~\sigma_x^2 a^2 y^2 + \sigma_y^2 a^2 x^2 }\)

Dividing throught by Q, we get

\(\small{\sigma_Q~=~ Q \sqrt{ \dfrac{\sigma_x^2}{x^2} + \dfrac{\sigma_y^2}{y^2} } }\)

(ii) \(~~\) Let \(~~\small{Q = \dfrac{ax}{y}}\)

with this,\(~~~\small{\left(\dfrac{\partial Q}{\partial x}\right)=\dfrac{a}{y}, }~~~\)\(~~~\small{\left(\dfrac{\partial Q}{\partial y}\right)=-\dfrac{ax}{y^2} }\)

Substituting in the propagation formula we get,

\(\small{\sigma_Q^2~=~\sigma_x^2 \dfrac{a^2}{y^2}~+~\sigma_y^2 \dfrac{a^2x^2}{y^4} }\)

Dividing throught by Q, we get

\(\small{\sigma_Q~=~ Q \sqrt{ \dfrac{\sigma_x^2}{x^2} + \dfrac{\sigma_y^2}{y^2} } }\)

We notice that during both multiplication and division, the uncertainities add in the same way.

We have, area of triangle \(~~\small{A = \dfrac{1}{2}bh~=~\dfrac{1}{2} \times 14 \times 23.5~=~ 164.5~cm^2 }\)

\(\small{\sigma_A~=~A \sqrt{\dfrac{\sigma_b^2}{b^2} + \dfrac{\sigma_h^2}{h^2}}~~=~~164.5 \times \sqrt{ \dfrac{0.7^2}{14^2} + \dfrac{0.5^2}{23.5^2} }~~=~~8.94~cm^2 }~~~~\) is the uncertainity in the area.

We get\(~~\small{\dfrac{\partial Q}{\partial x} = \pm ab x^{\pm b-1} = \dfrac{bQ}{x} }\)

\( \small{\sigma_Q^2 = \sigma_x^2 \left(\dfrac{\partial Q}{\partial x}\right)^2~=~\sigma_x^2 \dfrac{Q^2 b^2}{x^2} }\)

\(\small{\sigma_Q~=~Q b \dfrac{\sigma_x}{x} }\)

Volume of the sphere = \(\small{ V = \dfrac{4}{3} \pi r^3 = \dfrac{4}{3} \times 3.14 \times 7.61^3 = 1846.0~cm^3 }\)

The uncertainity in volume = \(\small{\sigma_V = Vb \dfrac{\sigma_r}{r} = 1846.0 \times 3 \times \dfrac{0.2}{7.61} = 145.5~cm^3 }\)

Consider the exponential relation \(~~\small{Q = ae^{\pm bx}} \)

\(\small{\dfrac{\partial Q}{\partial x} = \pm ab e^{\pm bQ} = \pm bQ }\)

\(\small{\sigma_Q^2 = \sigma_x^2 \left(\dfrac{\partial Q}{\partial x}\right)^2 = \sigma_x^2 (\pm bQ)^2 = \sigma_x^2 b^2Q^2 }\)

\(\small{\sigma_Q = Qb\sigma_x }\)

In the case when the constant raised to the poer is not exponential e, we can estimate error propagation by the following manipulation:

Let \(~~\small{Q~=~a^{\pm bx} }\)

Writing \(\small{a}\) as \(\small{e^{log(a)} }\), we get

\(~~~\small{Q~=~(e^{log(a)})^{\pm bx}~=~e^{\pm(b~ log(a)) x} }\)

This is an exponential form. Using the previously derived formmula for propagation of errors in exponential function, we can write,

Let \(~~\small{Q~=~a~log(x)}\)

We get \(~~\small{ \dfrac{\partial Q }{\partial x} = \dfrac{a}{x} }\)

With this,\(~~\small{\dfrac{\sigma_Q^2}{Q^2} = \sigma_x^2 \left (\dfrac{\partial Q }{\partial x}\right )^2~=~ a^2 \dfrac{\sigma_x^2}{x^2} }\)