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### Derivative of Logarithm

Figure 3: {$y = \log_b x$}

#### Observations

From inspection of the graphs in Figure 3 (which probably is cheating since the calculator that drew the graph knows some results we are trying to find), it appears that for {$b\gt1$}, logarithms have positive slopes, approaching vertical as b approaches 1. The logarithms are all 1 for {$x=0$}, which can be derived directly from the definition of logarithm. Likewise, the reciprocal of b is found where {$log_b x = -1$}.

Apparently, b can be any number greater than one, and {$log_b x$} can assume any value greater than zero. Again, as the algebra insists, every curve passes through {$(1,0)$}, but this time the inverses of b are found when {$log_b x = 1$}. This last difference is owing to inverting the base being equivalent to negating the exponent.

A slightly different but symmetrical set of conclusions can be drawn when {$0 \lt b \lt 1$}. The slopes are now everywhere negative, approaching vertical as b approaches 1.

Figure 3: logs with various bases.

Figure 4: logs and exponentials with fractional bases.

#### Argument

I will take the derivative of the general logarithm and show that it depends upon the existence of a certain limit. I will then (attempt) to show that the limit exists, whereupon the derivative will have been found.

#### Taking the derivative

Definition derivative:

{$$f^\prime (x) = \lim_{ h \rightarrow0} {{f(x + h) -f(x)} \over { h}}$$}

To find the derivative of {$\; \log_b x, b \ne 1, b \gt 0$}:

{\begin{align} f^\prime \left( \log_b x \right) &= \lim_{ h \rightarrow0} {{ \log_b (x + h) - \log_b x } \over { h}} \cr &= \lim_{ h \rightarrow0}{ \left( {1 \over h} \right) \left( \log_b (x + h) - \log_b x \right)} \cr &= \lim_{ h \rightarrow0}{ \left( {1 \over h} \right) \log_b \left( {{x + h} \over x }\right)} \cr &= \lim_{ h \rightarrow0}{ \left( {1 \over h} \right) \log_b \left( {{1 + {h \over x}} \over 1 }\right)} \cr &= \lim_{ h \rightarrow0}{ \left( {1 \over x} \right) \left( {x \over h} \right) \log_b \left( {{1 + {h \over x}} \over 1 }\right)} \cr &= \lim_{ h \rightarrow0}{ \left( {1 \over x} \right) \log_b \left( {{1 + {h \over x}} \over 1 }\right)^{x \over h}} \cr &= {1 \over x}\log_b \left[ \lim_{ h \rightarrow0}{ \left( {1 + {h \over x}} \right)^{x \over h}} \right] \cr &= {1 \over x} \log_b \left[ \lim_{ n \rightarrow \infty }{ \left( 1 + {1 \over n} \right) ^n } \right] \end{align}}

In this last step, {$n = {x \over h}$} to simplify the expression, but the limit variable has to be adjusted as the limit of x/h is approaching infinity as h approaches zero.

If this limit exists, the derivative has been found. For the moment, this is nicknamed e for the sake of the following headline, without prejudice to the issue of whether it exists or has a value.

##### Lemma: e exists

Figure 5: Does

{$$\lim_{ n \rightarrow \infty }{ \left( 1 + {1 \over n} \right) ^n}$$} exist?

###### Observations

That the limit exists is obvious from the graph, but proving it does is another thing. It appears that {$$y = \left( 1 + {1 \over n} \right) ^n$$} is increasing for n positive, but also {$$y = \left( 1 + {1 \over n} \right) ^n < y = \left( 1 + {1 \over n} \right) ^{n+1}$$} where they are both defined and n is positive. Moreover, {$$y = \left( 1 + {1 \over n} \right) ^{n+1}$$} is ever decreasing in the same range, and the difference between them is decreasing over the same range, but is never quite zero.

Apparently the difference between them can be found in the function P where:

{$$\color{purple}{P = \left[ {{n+1} \over n} \right] ^n \left( {{n+1} \over n} - 1 \right)}$$}

For convenience in reference:

{$$\color{red}{R = y = \left( 1 + {1 \over n} \right) ^n}$$} {$$\color{darkgreen}{G = y = \left( 1 + {1 \over n} \right) ^{n+1} }$$}

The claims we want to enter are:

1. R is always increasing, n > 0.
2. G is always decreasing, n > 0.
3. R never exceeds G, n > 0.
4. Therefore, R has a limit as n approaches ∞.

In light of P, claim c. is easy. For positive n, P is always positive, abeit eventually small. 0 < P, and P = G - R, so R < G.

The problems are claims a. and b. They are sufficiently similar that one expects if a method can be found for a., it will work for b.

Now there are plenty of demonstrations that for integer n > 0 b. is true. These go through induction on n and rely on the (integer) binomial theorem or known series. While it would be perverse to think that the general case would not provide similar results to those of the integer case, still these demonstrations are ultimately unsatisfying. On the other hand, most of the tools for dealing with the general case are beyond me at the moment, and for that matter, I am not sure which of them, if any, might introduce circularity. For example, one cannot use the derivative of the functions because that depends on e, the existence of which is in question.

Integer n case a.

To show that R is increating for all positive n greater than 0, Love et al. invoke the binomial theorem. {$$R_{n+1} = \left( 1+{1 \over {n+1}}\right) ^{n+1} = \sum_{k=0}^{n+1} {{n+1} \choose k} \left( {1 \over {n+1}} \right) ^k$$}

{$$\begin{gather} R_{n+2} = \left( 1+{1 \over {n+2}}\right) ^{n+2} = \sum_{k=0}^{n+2} {{n+2} \choose k} \left( {1 \over {n+2}} \right) ^k \cr = \sum_{k=0}^{n+1} {{n+2} \choose k} \left( {1 \over {n+2}} \right) ^k + {{n+2} \choose {n+2}} \left( {1 \over {n+2}} \right) ^{n+2} \end{gather}$$}

The last term the n+2 is broken out of the sum. Now the two sums each have k terms.

Now the kth term of each sum in turn:

{$$\begin{gather} _kR_{n+1} = \left( {{(n+1)^\underline{k}} \over {k!}} \right) \left( {1 \over {n+1}} \right) ^k \end{gather}$$}

{$$= \left( {(n+1)((n+1)-1)((n+1)-2)\cdots((n+1)-k+1) \over {k!}} \right) {1 \over {(n+1)^k}}$$}

{$$= \left( {{ { \overbrace {(n+1)((n+1)-1)((n+1)-2)\cdots((n+1)-k+1) }^{ k \text{ terms}} } \over {(n+1)^k}} \over {k!}} \right)$$}

{$$_kR_{n+1} = \left( {{ (1) (1 - {1 \over {n+1}} ) (1- {2 \over{n+1}} ) \cdots (1 - {{k-1}\over{n+1}}) } \over {k!}} \right)$$}

Likewise,

{$$_kR_{n+2} = \left( {{ (1) (1 - {1 \over {n+2}} ) (1- {2 \over{n+2}} ) \cdots (1 - {{k-1}\over{n+2}}) } \over {k!}} \right)$$}

Now, the denominator "k!" is the same for both. But every factor in the numerator of the term for {$_kR{n+1}$} is less than or equal to the corresponding factor in {$_kR_{n+2}$}. (It is less than or equal to include the k = 0.)

Thus,

{$$_kR{n+1} \le _kR_{n+2}$$}

Moreover, _kR_{n+2} has an extra positive value term that was broken out of the sum.

Therefore,

{$$R{n+1} \lt kR_{n+2}$$}

This concludes the defense of claim a. through the integer values of n from 1 to infinity.

Sources:

1. Love, Clyde E., Earl David Rainville Differential and Integral Calculus (Macmillian, 1916) Google Books §45. "The derivative of the logarithm" pp. 58-9.
2. Bernoulli's inequality - Wikipedia, the free encyclopedia
3. FooPlot: Online graphing calculator and function plotter

Recommended:

Category: Math Calculus Logarithms Exponentials

This is a student's notebook. I am not responsible if you copy it for homework, and it turns out to be wrong.

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### March 16, 2018

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