As in the last posting, we begin by asserting the existence of a function from the positive reals onto a set `L`,
`\qquad\qquad\qquad "lg":RR^+ \rightarrow L`.
Also, as stated in the last posting, "the log of a product is the sum of the logs of the factors" is not the only way to characterise the generic logarithm.
We can instead simply assert the end result of the last posting, and let that be our entire starting condition.
That is, for all positive real `x` and all real `p`, we assert that
`\qquad\qquad\qquad "lg"(x^p) = p\cdot"lg"(x).`
This defines multiplication of quantities in `L` by arbitrary real numbers. Let us call this, with some degree of anticipation, the "scalar multiplication" condition. As before, the existence of models ensures the consistency of the definition.
Consider now arbitrary positive real numbers `u` and `v`. Then for arbitrary positive real `x \ne 1`, we can find `p` and `q` such that `u = x^p` and `v = x^q`. Then
`\qquad\qquad\qquad "lg"(u v) = "lg"(x^p x^q)`,
`\qquad\qquad\qquad\qquad = "lg"(x^{p+q})`,
`\qquad\qquad\qquad\qquad = (p+q)\cdot"lg"(x)`,
`\qquad\qquad\qquad\qquad = p\cdot"lg"(x) + q\cdot"lg"(x)`,
`\qquad\qquad\qquad\qquad = "lg"(x^p) + "lg"(x^q)`,
`\qquad\qquad\qquad\qquad = "lg"(u) + "lg"(v)`.
and we have established the "the log of a product is the sum of the logs of the factors" condition with which we began the last posting.
From this we could derive all subsequent results just as we did there---but that would be wasteful, as most of those results are simply instances of the scalar multiplication condition.
We can also use the scalar multiplication condition to define division in `L`, at least for `x \ne 1` (with the consequence that `"lg"(x) \ne 0)`. In that case, we have
`\qquad\qquad\qquad \frac{"lg"(x^p)}{"lg"(x)} = p.`
Since, given any positive real `x` and `y`, we can find a unique real number `p` such that `y = x^p`, we can divide any quantity in `L` by any nonzero quantity in `L`.
Thursday, March 4, 2010
Additive characterisation of the generic logarithm
At this point, let's define the generic logarithm a little more carefully. The generic logarithm is a function from the positive real numbers onto a set `L`,
`\qquad\qquad\qquad "lg": RR^+ \rightarrow L,`
satisfying certain conditions. We have some choice as to the conditions we use.
One possibility is to require that, for all positive real numbers `x` and `y`,
`\qquad\qquad\qquad "lg"(x y) = "lg"(x) + "lg"(y)`,
where this defines the addition on the right hand side. This is the "the log of a product is the sum of the logs of the factors" condition. (We shall need another condition eventually, but let's not introduce it until the need is clear.)
Then it is equally true that
`\qquad\qquad\qquad "lg"(w z) = "lg"(w) + "lg"(z)`.
Letting `w = x y` and substituting,
`\qquad\qquad\qquad "lg"(x y z) = "lg"(x y) + "lg"(z)`,
`\qquad\qquad\qquad\qquad = "lg"(x) + "lg"(y) + "lg"(z)`,
and we have the rather obvious-looking result that
`\qquad\qquad\qquad "lg"(x y z) = "lg"(x) + "lg"(y) + "lg"(z)`,
By repeating this argument, we can extend it to finite products of any length.
For `n` a counting number, we can break the logarithm of `n` factors of positive real `x` into `n` copies of the logarithm of `x` :
`\qquad\qquad\qquad "lg"(x^n) = "lg"(x\cdotx\cdotxcdot\cdots\cdotx)`,
`\qquad\qquad\qquad\qquad = "lg"(x) + "lg"(x) + \cdots + "lg"(x)`,
`\qquad\qquad\qquad\qquad = n\cdot"lg"(x)`.
The last two lines define multiplication of quantities in `L` by counting numbers.
Substituting `y = 1` in `"lg"(x y) = "lg"(x) + "lg"(y)`, we get
`\qquad\qquad\qquad"lg"(x 1) = "lg"(x) + "lg"(1)`,
i.e.
`\qquad\qquad\qquad"lg"(x) = "lg"(x) + "lg"(1)`.
Subtracting the quantity `"lg"(x)` from each side and rearranging, we have that
`\qquad\qquad\qquad "lg"(1) = 0`.
This is useful in its own right.
Now `0` is the number (or quantity) that does nothing when you add it. `1` is the number that does nothing when you multiply it. For any positive real number `x`, multiplying by no factors of `x` is the same as multiplying by `1`, i.e.
`\qquad\qquad\qquad x^0 = 1`.
So if `"lg"(1) = 0`, then
`\qquad\qquad\qquad "lg"(x^0) = 0 = 0\cdot"lg"(x)`,
since `0` times any quantity is still `0`.
We have therefore established that
`\qquad\qquad\qquad "lg"(x^n) = n\cdot"lg"(x)`
is true not only for `n` a counting number, but also when `n = 0`.
Now let `y = 1//x` in `"lg"(x y) = "lg"(x) + "lg"(y)`. We can do this, because if `x` is positive real, so is `1//x`. Then
`\qquad\qquad\qquad"lg"(x\cdot1/x) = "lg"(x) + "lg"(1/x)`.
Since `x\cdot1//x = 1` and `"lg"(1) = 0`
`\qquad\qquad\qquad 0 = "lg"(x) + "lg"(1/x)`.
The astute reader will notice that this defines a suitably behaved `0` in `L`.
Subtracting`"lg"(x)` from each side and rearranging, we get that
`\qquad\qquad\qquad "lg"(1/x)= -"lg"(x)`,
which can also be written
`\qquad\qquad\qquad "lg"(x^{-1})= -1\cdot"lg"(x)`.
This, too, is a definition---it tells us what negation means in `L`.
So now we have established that
`\qquad\qquad\qquad "lg"(x^n) = n\cdot"lg"(x)`
also holds when `n = -1`.
We can find the logarithm of a product of a counting number `m` copies of `x^{-1}` as we did for multiple copies of `x` :
`\qquad\qquad\qquad "lg"((x^{-1})^m) = "lg"((x^{-1})\cdot(x^{-1})\cdot(x^{-1})cdot\cdots\cdot(x^{-1}))`,
`\qquad\qquad\qquad\qquad = "lg"((x^{-1})) + "lg"((x^{-1})) + \cdots + "lg"((x^{-1}))`,
`\qquad\qquad\qquad\qquad = m\cdot"lg"((x^{-1}))`.
Then, since "lg"(x^{-1})= -1\cdot"lg"(x)`, we have that
`\qquad\qquad\qquad "lg"((x^{-1})^m) = m\cdot(-1\cdot"lg"(x))`,
`\qquad\qquad\qquad "lg"((x^{-1})^m) = -m\cdot"lg"(x)`.
So now, setting `n = -m`, we have established that
`\qquad\qquad\qquad "lg"(x^n) = n\cdot"lg"(x)`
for any integer `n`. i.e., we can always move an integer power outside the logarithm as a multiplier.
Now let `p` be any rational number, which we write as `p = r/s`, where `s` is a counting number (and therefore not zero), and `r` is an integer. Let `y = x^p = x^{r/s}`. Then, raising this to the power of `s`, we find that
`\qquad\qquad\qquad y^s = x^r`,
and so
`\qquad\qquad\qquad "lg"(y^s) = "lg"(x^r)`,
Since `r` and `s` are both integers, we know we can move the powers outside the logarithms to get
`\qquad\qquad\qquad s\cdot"lg"(y) = r\cdot"lg"(x)`.
We now divide both sides by `s`---we know `s` is nonzero---and write
`\qquad\qquad\qquad "lg"(y) = {r/s}\cdot"lg"(x)`,
or
`\qquad\qquad\qquad "lg"(x^p) = p\cdot"lg"(x)`.
This, again, is a definition. It tells us what it means to multiply a quantity in `L` by a rational number. But it also allows us to assert that
`\qquad\qquad\qquad "lg"(x^n) = n\cdot"lg"(x)`
is true for any rational number `n`, not just integers.
It is reasonable to hope that this would be true for any real `n`, not just rationals. Practical uses of mathematics should be insensitive to whether a number is rational or not.
The various ways to ensure reasonable behaviour amount to various assertions of that `lg` is a continuous function, i.e. that numbers that are close in `RR+` have logarithms that are close in `L`.
The simplest way is to ensure that we get what we want is simply to assert that
`\qquad\qquad\qquad "lg"(x^n) = n\cdot"lg"(x)`
is true for all real numbers `n`, not just rational `n`. One needs to establish that no contradictions arise, but the existence of models (such as ordinary logarithms to a particular base) guarantees contradictionlessness in this less specific setting.
The real question here is how to justify the equation
`\qquad\qquad\qquad "lg"(x^n) = n\cdot"lg"(x)`
as a definition for multiplication of quantities in `L` by irrational numbers `n`.
One way to do this is to import into `L` the notion of order already found in `RR^+`. One can then transfer the definition of an irrational in `RR` using, say, Dedekind cuts over the rationals, into a definition of irrational powers.
To import that order, simply assert that whenever positive real numbers `x` and `y` are such that `x \le y`, then `"lg"(x) \le "lg"(y)`. This defines the meaning of `\le` in `L`. Since "`le`" is, in fact, a total order in `RR^+`, the version of "`\le`" defined for `L` is also a total order, and the function `lg` is monotonic, and therefore continuous.
Consider, now, the quantities `x^u` and `x^w`, with `u` and `w` rational, `u < w`, and `x` a positive real number. There are three cases :
In all three cases, `x^v` lies between `x^u` and `x^w`, i.e.
`\qquad\qquad\qquad \min(x^u, x^w) \le x^v \le \max(x^u, x^w).
Then, by the monotonicity of `lg`, the quantity `"lg"(x^v)` lies between "lg"(x^u)` and "lg"(x^w)`, and so `"lg"(x^v)` lies between "u\cdotlg"(x)` and w\cdot"lg"(x)`.
Let `v` now be irrational, and for definiteness, assume the first case, `x > 1`. Consider the Dedekind cut for `v`, i.e. the partition of the rational numbers into two subsets, those that are below `v` and those that are above `v`. Then here `u` belongs to the lowside set, and `w` to the highside set.
Since we know how find rational multiples of `lg(x)`, the logarithms of the lowside and highside sets are also well defined in `L`. Since `lg` is monotonic, the order also transfers, so just as every member of the lowside set is lower than every member of the highside set in `RR`, so also is every member of the image of the lowside set lower---in the sense that is used in `L`---than every member of the image of the highside set. So, for the given `x`, we can define a Dedekind cut in `L`, and can use this to define multiplication of quantities `lg(x)` in `L`, when `x > 1`.
The argument for the case `x < 1` switches the ends, but is otherwise the same.
For the case `x = 1`,
`\qquad\qquad"lg"(x^u) = "lg"(x^v) = "lg"(x^w) = "lg"(1) = 0.`
Here, the Dedekind cut argument fails, but it still seems reasonable to define the product of `v` and `0` by
`"lg"(1^v) = "lg"(1) = 0 = v\cdot0`,
i.e. that `n\cdot0 = 0`, whether `n` is rational or irrational.
`\qquad\qquad\qquad "lg": RR^+ \rightarrow L,`
satisfying certain conditions. We have some choice as to the conditions we use.
One possibility is to require that, for all positive real numbers `x` and `y`,
`\qquad\qquad\qquad "lg"(x y) = "lg"(x) + "lg"(y)`,
where this defines the addition on the right hand side. This is the "the log of a product is the sum of the logs of the factors" condition. (We shall need another condition eventually, but let's not introduce it until the need is clear.)
Then it is equally true that
`\qquad\qquad\qquad "lg"(w z) = "lg"(w) + "lg"(z)`.
Letting `w = x y` and substituting,
`\qquad\qquad\qquad "lg"(x y z) = "lg"(x y) + "lg"(z)`,
`\qquad\qquad\qquad\qquad = "lg"(x) + "lg"(y) + "lg"(z)`,
and we have the rather obvious-looking result that
`\qquad\qquad\qquad "lg"(x y z) = "lg"(x) + "lg"(y) + "lg"(z)`,
By repeating this argument, we can extend it to finite products of any length.
For `n` a counting number, we can break the logarithm of `n` factors of positive real `x` into `n` copies of the logarithm of `x` :
`\qquad\qquad\qquad "lg"(x^n) = "lg"(x\cdotx\cdotxcdot\cdots\cdotx)`,
`\qquad\qquad\qquad\qquad = "lg"(x) + "lg"(x) + \cdots + "lg"(x)`,
`\qquad\qquad\qquad\qquad = n\cdot"lg"(x)`.
The last two lines define multiplication of quantities in `L` by counting numbers.
Substituting `y = 1` in `"lg"(x y) = "lg"(x) + "lg"(y)`, we get
`\qquad\qquad\qquad"lg"(x 1) = "lg"(x) + "lg"(1)`,
i.e.
`\qquad\qquad\qquad"lg"(x) = "lg"(x) + "lg"(1)`.
Subtracting the quantity `"lg"(x)` from each side and rearranging, we have that
`\qquad\qquad\qquad "lg"(1) = 0`.
This is useful in its own right.
Now `0` is the number (or quantity) that does nothing when you add it. `1` is the number that does nothing when you multiply it. For any positive real number `x`, multiplying by no factors of `x` is the same as multiplying by `1`, i.e.
`\qquad\qquad\qquad x^0 = 1`.
So if `"lg"(1) = 0`, then
`\qquad\qquad\qquad "lg"(x^0) = 0 = 0\cdot"lg"(x)`,
since `0` times any quantity is still `0`.
We have therefore established that
`\qquad\qquad\qquad "lg"(x^n) = n\cdot"lg"(x)`
is true not only for `n` a counting number, but also when `n = 0`.
Now let `y = 1//x` in `"lg"(x y) = "lg"(x) + "lg"(y)`. We can do this, because if `x` is positive real, so is `1//x`. Then
`\qquad\qquad\qquad"lg"(x\cdot1/x) = "lg"(x) + "lg"(1/x)`.
Since `x\cdot1//x = 1` and `"lg"(1) = 0`
`\qquad\qquad\qquad 0 = "lg"(x) + "lg"(1/x)`.
The astute reader will notice that this defines a suitably behaved `0` in `L`.
Subtracting`"lg"(x)` from each side and rearranging, we get that
`\qquad\qquad\qquad "lg"(1/x)= -"lg"(x)`,
which can also be written
`\qquad\qquad\qquad "lg"(x^{-1})= -1\cdot"lg"(x)`.
This, too, is a definition---it tells us what negation means in `L`.
So now we have established that
`\qquad\qquad\qquad "lg"(x^n) = n\cdot"lg"(x)`
also holds when `n = -1`.
We can find the logarithm of a product of a counting number `m` copies of `x^{-1}` as we did for multiple copies of `x` :
`\qquad\qquad\qquad "lg"((x^{-1})^m) = "lg"((x^{-1})\cdot(x^{-1})\cdot(x^{-1})cdot\cdots\cdot(x^{-1}))`,
`\qquad\qquad\qquad\qquad = "lg"((x^{-1})) + "lg"((x^{-1})) + \cdots + "lg"((x^{-1}))`,
`\qquad\qquad\qquad\qquad = m\cdot"lg"((x^{-1}))`.
Then, since "lg"(x^{-1})= -1\cdot"lg"(x)`, we have that
`\qquad\qquad\qquad "lg"((x^{-1})^m) = m\cdot(-1\cdot"lg"(x))`,
`\qquad\qquad\qquad "lg"((x^{-1})^m) = -m\cdot"lg"(x)`.
So now, setting `n = -m`, we have established that
`\qquad\qquad\qquad "lg"(x^n) = n\cdot"lg"(x)`
for any integer `n`. i.e., we can always move an integer power outside the logarithm as a multiplier.
Now let `p` be any rational number, which we write as `p = r/s`, where `s` is a counting number (and therefore not zero), and `r` is an integer. Let `y = x^p = x^{r/s}`. Then, raising this to the power of `s`, we find that
`\qquad\qquad\qquad y^s = x^r`,
and so
`\qquad\qquad\qquad "lg"(y^s) = "lg"(x^r)`,
Since `r` and `s` are both integers, we know we can move the powers outside the logarithms to get
`\qquad\qquad\qquad s\cdot"lg"(y) = r\cdot"lg"(x)`.
We now divide both sides by `s`---we know `s` is nonzero---and write
`\qquad\qquad\qquad "lg"(y) = {r/s}\cdot"lg"(x)`,
or
`\qquad\qquad\qquad "lg"(x^p) = p\cdot"lg"(x)`.
This, again, is a definition. It tells us what it means to multiply a quantity in `L` by a rational number. But it also allows us to assert that
`\qquad\qquad\qquad "lg"(x^n) = n\cdot"lg"(x)`
is true for any rational number `n`, not just integers.
It is reasonable to hope that this would be true for any real `n`, not just rationals. Practical uses of mathematics should be insensitive to whether a number is rational or not.
The various ways to ensure reasonable behaviour amount to various assertions of that `lg` is a continuous function, i.e. that numbers that are close in `RR+` have logarithms that are close in `L`.
The simplest way is to ensure that we get what we want is simply to assert that
`\qquad\qquad\qquad "lg"(x^n) = n\cdot"lg"(x)`
is true for all real numbers `n`, not just rational `n`. One needs to establish that no contradictions arise, but the existence of models (such as ordinary logarithms to a particular base) guarantees contradictionlessness in this less specific setting.
The real question here is how to justify the equation
`\qquad\qquad\qquad "lg"(x^n) = n\cdot"lg"(x)`
as a definition for multiplication of quantities in `L` by irrational numbers `n`.
One way to do this is to import into `L` the notion of order already found in `RR^+`. One can then transfer the definition of an irrational in `RR` using, say, Dedekind cuts over the rationals, into a definition of irrational powers.
To import that order, simply assert that whenever positive real numbers `x` and `y` are such that `x \le y`, then `"lg"(x) \le "lg"(y)`. This defines the meaning of `\le` in `L`. Since "`le`" is, in fact, a total order in `RR^+`, the version of "`\le`" defined for `L` is also a total order, and the function `lg` is monotonic, and therefore continuous.
Consider, now, the quantities `x^u` and `x^w`, with `u` and `w` rational, `u < w`, and `x` a positive real number. There are three cases :
- if `x > 1`, then `x^u < x^w`,
- if `x = 1`, then `x^u = x^w`, and
- if `x < 1`, then `x^u > x^w`.
In all three cases, `x^v` lies between `x^u` and `x^w`, i.e.
`\qquad\qquad\qquad \min(x^u, x^w) \le x^v \le \max(x^u, x^w).
Then, by the monotonicity of `lg`, the quantity `"lg"(x^v)` lies between "lg"(x^u)` and "lg"(x^w)`, and so `"lg"(x^v)` lies between "u\cdotlg"(x)` and w\cdot"lg"(x)`.
Let `v` now be irrational, and for definiteness, assume the first case, `x > 1`. Consider the Dedekind cut for `v`, i.e. the partition of the rational numbers into two subsets, those that are below `v` and those that are above `v`. Then here `u` belongs to the lowside set, and `w` to the highside set.
Since we know how find rational multiples of `lg(x)`, the logarithms of the lowside and highside sets are also well defined in `L`. Since `lg` is monotonic, the order also transfers, so just as every member of the lowside set is lower than every member of the highside set in `RR`, so also is every member of the image of the lowside set lower---in the sense that is used in `L`---than every member of the image of the highside set. So, for the given `x`, we can define a Dedekind cut in `L`, and can use this to define multiplication of quantities `lg(x)` in `L`, when `x > 1`.
The argument for the case `x < 1` switches the ends, but is otherwise the same.
For the case `x = 1`,
`\qquad\qquad"lg"(x^u) = "lg"(x^v) = "lg"(x^w) = "lg"(1) = 0.`
Here, the Dedekind cut argument fails, but it still seems reasonable to define the product of `v` and `0` by
`"lg"(1^v) = "lg"(1) = 0 = v\cdot0`,
i.e. that `n\cdot0 = 0`, whether `n` is rational or irrational.
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