Post History
#2: Post edited
by
celtschk
·
2021-08-03T13:08:00Z (over 3 years ago)
Added explanation of dimensional analysis
- Assuming you copied the formula from the book correctly, your mistake is simply to assume that your book is right.
- As a physicist, my first instinct was to make a dimensional analysis. The formula for $I$ gives a dimensionally valid expression as long as $a$ and $b$ have the inverse unit of $x$ (so that $ax$ and $bx$ are unitless), giving $[I]=[x]^{-1}$. Therefore the second derivative must have the unit $[I]/([a][b]) = [x]$.
- Your result indeed has that property, while the book's integral has the unit $[x]^3$, which the correct result cannot have.
- By the way, in your result you've got an $x$ in the numerator and another one in the denominator; you certainly can cancel those out. Therefore when completely cancelled, your result doesn't have an $x$ in the denominator either.
- Assuming you copied the formula from the book correctly, your mistake is simply to assume that your book is right.
- As a physicist, my first instinct was to make a dimensional analysis. The formula for $I$ gives a dimensionally valid expression as long as $a$ and $b$ have the inverse unit of $x$ (so that $ax$ and $bx$ are unitless), giving $[I]=[x]^{-1}$. Therefore the second derivative must have the unit $[I]/([a][b]) = [x]$.
- Your result indeed has that property, while the book's integral has the unit $[x]^3$, which the correct result cannot have.
- By the way, in your result you've got an $x$ in the numerator and another one in the denominator; you certainly can cancel those out. Therefore when completely cancelled, your result doesn't have an $x$ in the denominator either.
- Explanation of the dimensional analysis:
- Dimensional analysis is a method used in physics to quickly check physics equations for errors (and also to guess physical relations, but that is clearly outside the scope here). Basically it means to check that all units of an equation are correct.
- Now your problem probably doesn't have an origin in physics (at least you didn't mention any), but still we can profit from dimensional analysis by assuming suitable units (this of course only works when the starting formula allows it).
- In dimensional analysis, the unit of a quantity $q$ is denoted as $[q]$. For example, if $l$ is a length, then $[l]$ is a length unit (since unit-less factors are ignored in dimensional analysis, it doesn't matter which one). Such “unconcrete units” are called dimensions (e.g. $l$ has the dimension of a length).
- The rules of a valid expression are then:
- * In a sum or difference, all terms must have the same dimension, and then the sum has the same dimension, too (e.g. you cannot add length and mass; if you add lengths, you get a length, if you add masses, you get a mass).
- * The dimension of a product/quotient is the product/quotient of the dimensions.
- * A dimension of a quantity to some power (whose exponent must be dimensionless) is that power of the dimension.
- * The arguments of any other mathematical function must be dimensionless (there are some complications with the logarithm which are irrelevant for this post). The function value then is dimensionless as well.
- * In an equality or inequality, both sides must have the same dimension.
- * In dimensional analysis, $\mathrm dx$ is treated the same as if it were $x$.
- Looking at the definition for $I$, we see three variables that might get dimensions (units) attached: $a$, $b$ and $x$. We further see that $ax$ and $bx$ are arguments for $\tan^{-1}$ and therefore must be dimensionless, that is, $[a]=[b]=[x]^{-1}$. But there is no restriction for the dimension of $x$, therefore we can apply dimensional analysis using an arbitrary dimension for $x$.
- Since the $\tan^{-1}$-terms are dimensionless factors, we can ignore them here; the only terms that contribute to the dimension are the $x^2$ in the denomuniator, and the $dx$ from the integral that is treated like a factor $x$.
- We therefore get
- $$[I]=\frac{[x]}{[x]^2}=[x]^{-1}.$$
- Now the nice thing is that if an equation does conform to dimensional analyis, this still stays true for derivatives. Therefore since
- $$\left[\frac{\mathrm d^2I}{\mathrm da\\,\mathrm db}\right] = \frac{[I]}{[a][b]} =\frac{[x]^{-1}}{[x]^{-1}[x]^{-1}} = [x]$$
- we know that whatever formula we get on the right hand side must also have that dimension.
- As you should by now be able to check, your formula has that property, but the formula you quote from the book doesn't. Therefore the formula from the book is wrong.
#1: Initial revision
by
celtschk
·
2021-08-03T11:49:33Z (over 3 years ago)
Assuming you copied the formula from the book correctly, your mistake is simply to assume that your book is right.
As a physicist, my first instinct was to make a dimensional analysis. The formula for $I$ gives a dimensionally valid expression as long as $a$ and $b$ have the inverse unit of $x$ (so that $ax$ and $bx$ are unitless), giving $[I]=[x]^{-1}$. Therefore the second derivative must have the unit $[I]/([a][b]) = [x]$.
Your result indeed has that property, while the book's integral has the unit $[x]^3$, which the correct result cannot have.
By the way, in your result you've got an $x$ in the numerator and another one in the denominator; you certainly can cancel those out. Therefore when completely cancelled, your result doesn't have an $x$ in the denominator either.