I've stumbled over an interesting question. In $\cos(x)$, $x$ is measured in, say, radians. When I expand cosine in Taylor's series, I have the terms with $x^3$, $x^5$ etc. so I am summing up $\text{rad}^3$, $\text{rad}^5$ etc. to get the dimensionless value of cosine. Obviously something is not right here.

I couldn't find any clarification on this issue, so I would appreciate some explanation. I've got one idea myself, but I am quite unsure about it:

If $f = \cos(x)$, $x$ can be expressed as $|x|\cdot u$, where $|x|$ is the value and $u$ the unit of the argument, in our case $u = \text{rad}$. In the Taylor's expansion we have $f^{(n)}(x_0)\cdot(x-x_0)^n$.

$f'(x) = \dfrac{\partial f}{\partial x}$, unit of $\partial\cos$ is 1, since cosine is dimensionless, while the unit of $\partial x$ is radian. Thus the unit of the $n$-th derivative of cosine is $1/\text{rad}^n$.

We then come to each term in the Taylor's series for cosine being dimensionless, thus resulting in the dimensionless sum.

Does it make sense?

EDIT: Since the discussion in the comments has concentrated around radians being dimensionless units, please consider any function with any unit of the argument. The whole consideration remains the same. For example, energy can be expressed as $E=mc^2 = f(m)$. The expansion will then have the terms including $\text{kg}$, $\text{kg}^2$, $\text{kg}^3$ etc. Something should “compensate” these powerful kilograms, and I think it's $f^{(n)}$ that does. But I'd like to hear a confirmation from someone on better terms with mathematics than I am.

  • 1
    $\begingroup$ Radians are dimensionless. Function arguments are just (real or complex, or other) numbers (for $\cos,\,\exp$ and its ilk). But irrespective of that, yes, your argument makes sense. $\endgroup$ – Daniel Fischer Jun 27 '14 at 12:08
  • $\begingroup$ To add to what Daniel said, remember that $\pi \, \text{rad}$ is the ratio of the circumference (length) to the diameter (length), so it is dimensionless. $\endgroup$ – M. Vinay Jun 27 '14 at 12:11
  • $\begingroup$ @DanielFischer, I really have troubles with the expression "some unit is dimensionless". Anyway, you can replace cosine with any other function where $f$ and $x$ have different units. $\endgroup$ – texnic Jun 27 '14 at 12:13
  • $\begingroup$ A dimensionless unit is just a tag to indicate where a number comes from. In the case of radians, as M. Vinay said, $x$ radians is the quotient of two lengths, and $\frac{x\text{ parsec}}{1\text{ parsec}} = x$ is just a number. $\endgroup$ – Daniel Fischer Jun 27 '14 at 12:18
  • $\begingroup$ I understand the argumentation but I still do not like it :) I am fully on the side of the author of this paper. However, as said, forget the radians, take anything else, such as $E=mc^2=f(m)$. Mass is definitely not dimensionless, and neither is the energy :) $\endgroup$ – texnic Jun 27 '14 at 12:33

Note, that the Taylor series also involves derivatives. And what is a derivative? Let us denote the units of $x$ as $[x] = \text{s}$ and $[f(x)] = \text{m}$ (You can think of time, and space, respectively).

Then a derivative is $f'(x_0) = \lim_{x\rightarrow x_0} \frac{f(x_0)-f(x)}{x_0-x}$. So, plugging in the units, we get $[f'(x)] = \text{m}/\text{s}$. Similarly, $[f''(x)]=\text{m}/\text{s}^2$, and you can continue this onwards. Again, intuitively the derivative is the velocity, and the second derivative is the acceleration, so you can see that if you think of $f$ and $x$ as space/time, the units again check out.

Now, the Taylor expansion looks like this:

$$f(x) = f(x_0) + f'(x_0)(x-x_0) + \frac12f''(x_0)(x-x_0)^2 + ... \approx \text{m} + \frac{\text{m}}{\text{s}} \text{s} + \frac {\text{m}}{\text{s}^2}\text{s}^2 + ... = \text{m}+\text{m}+\text{m} ... = \text{m}$$

(I have used the $\approx$ sign to transfer from precise expression to the measurement units. Hope that this is clear).

  • $\begingroup$ Lovely, exactly what I was looking for! Thanks. $\endgroup$ – texnic Jun 27 '14 at 12:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.