Editing Dimensional analysis (section)

=== Polynomials and transcendental functions ===
Bridgman's theorem restricts the type of function that can be used to define a physical quantity from general (dimensionally compounded) quantities to only products of powers of the quantities, unless some of the independent quantities are algebraically combined to yield dimensionless groups, whose functions are grouped together in the dimensionless numeric multiplying factor.<ref>{{harvnb|Bridgman|1922|loc=2. Dimensional Formulas pp. 17–27}}</ref><ref>{{cite journal |first1=Mário N. |last1=Berberan-Santos |first2=Lionello |last2=Pogliani |title=Two alternative derivations of Bridgman's theorem |journal=Journal of Mathematical Chemistry |volume=26 |pages=255–261, See §5 General Results p. 259 |date=1999 |doi=10.1023/A:1019102415633 |s2cid=14833238 |url=https://core.ac.uk/download/pdf/22873054.pdf}}</ref>  This excludes polynomials of more than one term or transcendental functions not of that form.

[[Scalar (physics)|Scalar]] arguments to [[transcendental function]]s such as [[Exponential function|exponential]], [[Trigonometric function|trigonometric]] and [[logarithm]]ic functions, or to [[inhomogeneous polynomial]]s, must be [[dimensionless quantities]].  (Note: this requirement is somewhat relaxed in Siano's orientational analysis described below, in which the square of certain dimensioned quantities are dimensionless.)

<!--see discussion page/transcendental functions This requirement is clear when one observes the [[Taylor expansion]]s for these functions (a sum of various powers of the function argument). For example, the logarithm of 3&nbsp;kg is undefined even though the logarithm of&nbsp;3 is nearly&nbsp;0.477. An attempt to compute ln&nbsp;3&nbsp;kg would produce, if one naively took ln&nbsp;3&nbsp;kg to mean the dimensionally meaningless "ln(1&nbsp;+&nbsp;2&nbsp;kg)",
: <math>\mathrm{2\,kg} - \frac{\mathrm{4\,kg}^2}{2} + \cdots ,</math>
which is dimensionally incompatible – the sum has no meaningful dimension – requiring the argument of transcendental functions to be dimensionless.

Another way to understand this problem is that the different coefficients ''scale'' differently under change of unit – were one to reconsider this in grams as "ln&thinsp;3000&nbsp;g" instead of "ln&thinsp;3&nbsp;kg", one could compute ln&thinsp;3000, but in terms of the [[Taylor series]], the degree 1 term would scale by 1000, the degree-2 term would scale by 1000<sup>2</sup>, and so forth – the overall output would not scale as a particular dimension.
-->
While most mathematical identities about dimensionless numbers translate in a straightforward manner to dimensional quantities, care must be taken with logarithms of ratios: the identity {{math|1=log(''a''/''b'') = log&thinsp;''a'' − log&thinsp;''b''}}, where the logarithm is taken in any base, holds for dimensionless numbers {{math|''a''}} and {{math|''b''}}, but it does ''not'' hold if {{math|''a''}} and {{math|''b''}} are dimensional, because in this case the left-hand side is well-defined but the right-hand side is not.<ref>{{harnvb|Berberan-Santos|Pogliani|1999|page=256}}</ref>

Similarly, while one can evaluate [[monomials]] ({{math|''x''<sup>''n''</sup>}}) of dimensional quantities, one cannot evaluate polynomials of mixed degree with dimensionless coefficients on dimensional quantities: for {{math|''x''<sup>2</sup>}}, the expression {{nowrap|1=(3 m)<sup>2</sup> = 9 m<sup>2</sup>}} makes sense (as an area), while for {{math|''x''<sup>2</sup> + ''x''}}, the expression {{nowrap|1=(3 m)<sup>2</sup> + 3 m = 9 m<sup>2</sup> + 3 m}} does not make sense.

However, polynomials of mixed degree can make sense if the coefficients are suitably chosen physical quantities that are not dimensionless.  For example,
: <math> \tfrac{1}{2} \cdot (\mathrm{-9.8~m/s^2}) \cdot t^2 + (\mathrm{500~m/s}) \cdot t. </math>

This is the height to which an object rises in time&nbsp;{{math|''t''}} if the acceleration of [[gravity]] is 9.8 {{nowrap|metres per second per second}} and the initial upward speed is 500 {{nowrap|metres per second}}.  It is not necessary for {{math|''t''}} to be in ''seconds''.  For example, suppose {{math|''t''}}&nbsp;=&nbsp;0.01&nbsp;minutes.  Then the first term would be
: <math>\begin{align}
      &\tfrac{1}{2} \cdot (\mathrm{-9.8~m/s^2}) \cdot (\mathrm{0.01~min})^2 \\[10pt]
  ={} &\tfrac{1}{2} \cdot -9.8 \cdot \left(0.01^2\right) (\mathrm{min/s})^2 \cdot \mathrm{m} \\[10pt]
  ={} &\tfrac{1}{2} \cdot -9.8 \cdot \left(0.01^2\right) \cdot 60^2 \cdot \mathrm{m}.
\end{align}</math>