We all know from school that arguments of transcendental functions such as exponential, trigonometric and logarithmic functions, or to inhomogeneous polynomials, must be dimensionless quantities. But is there a simple way to prove it?
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I don't think this concerns with math: this is just a consequence of the way we measure things in real life. Anyway: this holds by definition, there is nothing to prove. – Crostul Mar 15 '16 at 15:08
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Moreover, there exist algebraic functions which must have dimensionless argument: for example take $f(x) = x^3+x$. – Crostul Mar 15 '16 at 15:10
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yes thats right! There is nothing very peculiar for transcendental functions from this point of view. – Orbifold Mar 15 '16 at 15:15
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answered in first answer here http://math.stackexchange.com/questions/238390/units-of-a-log-of-a-physical-quantity – snulty Mar 15 '16 at 15:25
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@Crostul If $f$ is a potential and its units are Joules ($J$), in physics, the coefficients can carry units. The $1$ in front of the $x^3$ term is in units of $J/m^3$ and the $1$ in front of the $x$ term is in $J/m$, assuming that $x$ is given in metres ($m$) – snulty Mar 15 '16 at 16:25
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Analytic expressions $\Psi(x)$, like $$\Psi(x)={\log(1+x^2)-\cos(3x+5)\over 2\pi(e^{2x}+e^{-5x})}\ ,$$ take "pure" real or complex numbers $x$ as input, and are simply not defined for $x:=$"the distance between earth and moon", or similar.
Dimensional quantities, e.g., $1.5$ volt, $\ 3.75\,\$$, $\ 1{1\over8}$ in, $384\,000$ km, are not elements of ${\mathbb R}$, but elements of some one-dimensional real vector space, and only obtain a numerical value by choosing a basis in that space. This value is then subject to the familiar laws of transformation under change of basis, i.e., of the respective "unit of measure". Of course you are allowed to consider $e^{3.75}$ or $e^{384\,000}$, but these numbers have no meaning whatsoever in connection with the respective problem at hand.
Christian Blatter
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It's worth adding that the set, $\mathbb{R}+$, of all strictly positive real numbers forms a one-dimensional real vector space, in which the operation of 'addition' is the usual operation of multiplication on $\mathbb{R}$ (restricted to $\mathbb{R}+$), the 'zero' element is $1$, the 'negative' of $x$ is $1/x$, the operation of scalar multiplication is $\mathbb{R} \times \mathbb{R}+ \to \mathbb{R}+$, $(\lambda, x) \mapsto x^\lambda$, and the coordinate of $x$ with respect to the basis ${b}$ is $\log_b x$. – Calum Gilhooley Mar 15 '16 at 21:04
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One simple way to see it is the following. Transcendental functions typically have a series representation. Now if the argument is a dimensionful quantity then from the series representation it would mean that we are adding quantities of different dimensions. But this we cant do. We can't add apples with oranges. For dimensionless quantities there is no problem.
But there could ofcourse be more rigorous arguments!
Orbifold
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Without that, it wouldn't be possible to handle a change of unit with mere factors.
For instance, $e^{2.54 x}$ can't be expressed in terms of $fe^x$ where $f$ would be a suitable conversion factor.
This is by contrast with a power function like $x^3$, such that $(2.54x)^3=(2.54)^3x^3$.
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Just to reiterate a previous answer by glebovg here which has a link to a very good answer to your question.
Just to summarise it roughly, the argument of a transcendental function needs to be dimensionless, but not for one the reasons you might think regarding power series expansions.
Here's an example from physics:
Suppose $x(t)=\alpha t^3-\beta t$ gives the position of a particle moving in $1d$, with time $t$ given in seconds $s$ and position $x$ in metres $m$, and $\alpha,\beta$ constants.
Now for our answer to come out in metres, $\alpha$ and $\beta$ although constants must also have units. The units of which we'll denote $[\alpha]=m\cdot s^{-3}$, and $[\beta]=m\cdot s^{-1}$.
Another way which lends more to the idea of taylor expansion is the following. We call velocity $v$ the rate of change of position with respect to time. Naturally in the above case this is $\dfrac{dx}{dt}$, and its units are $\dfrac{m}{s}$, the units of $x$ divided by the units of $t$.
So we can also notice in the above example that $\alpha=\dfrac{1}{3!}\dfrac{d^3x}{d t^3}$ (in principle at $t=0$) and $\beta=\dfrac{1}{1!}\dfrac{dx}{d t}$ at $t=0$, and so for this reason they also must have the units as above.
If $f(x)$ has units then $f(x)=f(a)+f'(a)(x-a)+\ldots$ all have the same units as $f$ for this same reason.
To address the idea of transcendental function having units, take for example $\log(10\text{ grams})$. If we could expand this as $$\log(10*1\text{gram})=\log(10)+\log(1 \text{ gram}),$$ we'd have to come up with an interpretation of what it means to take a log of 1 gram, in particular it should be something $x$ which if $e$ is the base of the log, we have $e^x=1$ gram. Now there's no real sensible answer to this. I mean suppose we make a new unit called 'log-gram' with this property, then when we took our above equation, the lhs is in 'log-grams' and the right hand side is a pure number, log(10) and something which is in 'log-grams' also, but this breaks our rule that we can't add two things with different units.
Well maybe we could say the $1$ in front of log(10) carries the units and so on $\ldots$ It becomes very unreasonable at this stage however, and a much nicer solution which doesn't change the numerical outcome of calculations is to introduce a scale.
So in our argument for the function log as above should be read $\log\left(\frac{10 \text{ grams}}{1 \text{ gram}}\right)$
