Why bound these increments?
An exponential or logarithmic difference can be difficult to evaluate exactly. Derivative bounds provide a way to estimate such a difference using only the interval endpoints. The Mean Value Theorem is the key: it expresses an increment as the interval length multiplied by a derivative value at some intermediate point.
For the exponential function, the derivative is \(e^x\), which increases with \(x\). For the natural logarithm, the derivative is \(1/x\), which decreases on its positive domain. These derivative behaviours make it possible to identify explicit upper and lower bounds.
The Mean Value Theorem and derivative bounds
The Mean Value Theorem states that if a function \(f\) is continuous on the closed interval \([a,b]\) and differentiable on the open interval \((a,b)\), where \(a<b\), then there is a number \(c\in(a,b)\) such that
Suppose, in addition, that \(m\leq f'(t)\leq M\) for every \(t\in(a,b)\). Since \(b-a>0\), multiplying by the interval length preserves the inequalities. Therefore
The bounds \(m\) and \(M\) must bound the derivative throughout the interval. They are not merely values chosen because they look plausible at an endpoint.
Bounds for exponential increments
The exponential function is continuous and differentiable on all of \(\mathbb R\), and its derivative is \(e^t\). Since \(e^t\) is increasing, on an interval \([a,b]\) with \(a<b\), its derivative values lie between \(e^a\) and \(e^b\). Applying the theorem gives:
This bounds the increase in the exponential using the interval length and the exponential values at its endpoints. For an increment written from \(a\) to \(a+h\), where \(h>0\), the same result is
There is also a useful inequality that applies for every real \(x\), including negative \(x\). For \(x\neq 0\), apply the Mean Value Theorem to \(e^t\) on the interval with endpoints \(0\) and \(x\). It gives a \(c\) strictly between those endpoints such that
If \(x>0\), then \(1<e^c<e^x\). Multiplying by \(x>0\) gives \(x<e^x-1<xe^x\). If \(x<0\), then \(e^x<e^c<1\); multiplying by \(x<0\) reverses both inequalities and again gives \(x<e^x-1<xe^x\). At \(x=0\), all three expressions are zero. Thus, for every real \(x\),
Worked Example: Bound an exponential increment on \([1,3]\)
We want to bound \(e^3-e\). The interval length is \(3-1=2\), and the derivative of \(e^t\) on \((1,3)\) lies between \(e\) and \(e^3\). The Mean Value Theorem derivative bounds therefore give
so
To check the application, the Mean Value Theorem gives \(e^3-e=2e^c\) for some \(c\in(1,3)\). Because \(e^t\) is increasing, \(e<e^c<e^3\), and multiplying by \(2\) yields the stated bounds. The derivative range and the interval length have both been used in the correct order.
Worked Example: Use the increment inequality at a negative input
Take \(x=-1\). The general inequality \(x\leq e^x-1\leq xe^x\) gives
Here \(e^{-1}-1\) is negative, and \(-1<-e^{-1}\) because \(e>1\). For a direct check of the sign handling, the Mean Value Theorem gives \(e^{-1}-1=-e^c\) for some \(c\in(-1,0)\). Thus \(e^{-1}<e^c<1\), so \(-1<-e^c<-e^{-1}\). This agrees with the displayed bound.
Bounds for logarithmic increments
The natural logarithm is defined for positive inputs. Its derivative is \(1/t\), which is positive and decreasing for \(t>0\). If \(0<a<b\), then throughout \((a,b)\) the derivative lies between \(1/b\) and \(1/a\). The Mean Value Theorem therefore gives
Because \(\ln b-\ln a=\ln(b/a)\), this also bounds the logarithm of a ratio. The condition \(a>0\) is essential: the logarithm and its derivative are being used only on the positive real numbers.
A particularly useful form results by taking \(a=1\) and \(b=1+x\), where \(x>-1\) and \(x>0\). It yields
The same displayed inequality holds when \(-1<x<0\), although it is helpful to check the sign rather than apply the positive-increment formula mechanically. For \(x\neq0\), the Mean Value Theorem on the interval between \(1\) and \(1+x\) gives \(\ln(1+x)-\ln 1=x/c\) for some \(c\) between those endpoints. Since \(\ln 1=0\),
If \(x>0\), then \(1<c<1+x\), so \(1/(1+x)<1/c<1\); multiplying by \(x>0\) gives the stated bounds. If \(-1<x<0\), then \(1+x<c<1\), so \(1<1/c<1/(1+x)\); multiplying by \(x<0\) reverses the inequalities and again gives the stated bounds. When \(x=0\), both sides and the logarithm are zero.
Worked Example: Bound \(\ln(5/2)\)
Apply the logarithmic increment bound with \(a=2\) and \(b=5\). Both inputs are positive and \(2<5\), so its conditions hold. The interval length is \(5-2=3\), giving
Using \(\ln 5-\ln 2=\ln(5/2)\), simplify the fractions:
To check the endpoint derivative bounds, \(1/t\) on \((2,5)\) is greater than \(1/5\) and less than \(1/2\). Multiplication by the positive length \(3\) gives \(3/5\) and \(3/2\), as in the result.
Worked Example: Bound \(\ln(1/2)\) using a negative increment
Set \(x=-1/2\). This satisfies \(x>-1\), so \(1+x=1/2\) is positive. The inequality for \(\ln(1+x)\) gives
Since \(1-1/2=1/2\), the left-hand fraction is \(-1\). Therefore
For a check, the Mean Value Theorem gives \(\ln(1/2)=(-1/2)/c\) for some \(c\in(1/2,1)\). Consequently \(1<1/c<2\). Multiplication by \(-1/2\) reverses the inequalities, giving \(-1<(-1/2)/c<-1/2\), consistent with the claimed bounds.
Why the endpoint choices matter
The derivative bounds depend on how the derivative behaves across the interval. For \(e^t\), the smaller derivative value is at the left endpoint and the larger one at the right endpoint. For \(1/t\), the order is reversed: the smaller derivative value is at the right endpoint and the larger one at the left endpoint. In both cases, one must bound the derivative on the whole interval before multiplying by its positive length.
Summary
The Mean Value Theorem turns an increment into an interval length times a derivative value somewhere inside the interval. Bounding that derivative throughout the interval gives explicit bounds for the increment. For the exponential, increasing derivative values lead to bounds using the left and right endpoint exponentials. For the logarithm, the positive decreasing derivative \(1/t\) gives bounds using the reciprocal of the right and left endpoints, respectively.
For increments that start at \(1\), the resulting inequalities are \(x\leq e^x-1\leq xe^x\) for every real \(x\), and \(x/(1+x)\leq\ln(1+x)\leq x\) for \(x>-1\). In either case, attention to the sign of the interval length or increment is essential when multiplying inequalities.
Check Your Understanding
Use the derivative bounds and sign conditions, rather than decimal approximations, to answer these questions.
- Apply the exponential increment bound to \(e^4-e^2\). What lower and upper bounds result?
- Use the logarithmic increment bound to bound \(\ln(7/3)\).
- For what values of \(x\) is \(\ln(1+x)\) defined, and what bounds does the derivative argument give?
- Why does multiplying an inequality by a negative \(x\) reverse its order?
- On the interval \([2,4]\), what are a lower and an upper bound for the derivative of \(\ln t\)?
