Find a valid derivative bound
A ceiling over the derivative
The chalk curve shows \(\left|f^{(4)}(t)\right|=\dfrac{6}{t^4}\), the size (absolute value) of the fourth derivative of \(f(x)=\ln x\).
The remainder of the degree 3 polynomial at the center 1 uses \(f^{(4)}(c)\). The point \(c\) is unknown, but it lies in the faint band from 1 to the input \(x=1.5\).
The orange line is a bound \(M\), which is valid if it is at or above the curve on the whole band. Move the slider, and the curve turns red where \(M\) is too small.
At \(M=1.2\), most of the curve in the band is red, so this \(M\) is not valid.
At \(M=6\), no part is red. The curve falls from left to right, so its largest value in the band is at \(t=1\), and 6 is the smallest valid \(M\).
Any derivative, any interval
The chalk curve now shows the size of the derivative of order \(n+1\) for any function. Drag the input to change the band.
The curve can be highest at an end of the band or in the middle. The line must be at or above it on the whole band.
Any number above a valid M is also valid, so you can round M up, but a smaller M gives a smaller error bound.
Notation
The bound \(M\) is at least as large as the size of the derivative at every \(t\) between \(a\) and \(x\), ends included.
\[\left|f^{(n+1)}(t)\right|\le\cy{M}\]To find M, sketch the size of the derivative on the interval. If it rises, the largest value is at the right end. If it falls, it is at the left end. If it rises and then falls, it can be inside the interval.
A sine or a cosine is between \(-1\) and 1, so for any constants \(k\) and \(b\), \(|k\sin bt|\le|k|\) and \(|k\cos bt|\le|k|\).
When it fails
If you read the curve only at the input, you find \(M=6/1.5^4\approx1.19\). Near \(t=1\) the curve is far above this line, so it turns red.
The point \(c\) can be anywhere in the band, so \(M\) must be at or above the curve at every point of the band.
Example 1: A sine of 2x
The Taylor polynomial of degree 4 at 0 approximates \(f(x)=\sin 2x\) at \(x=2\). Find a valid \(M\) for the interval from 0 to 2.
The degree is 4, so the remainder uses the derivative of order 5.
\[n+1=5\]Each derivative brings out a factor 2 from the chain rule, so the fifth derivative has the factor \(2^5=32\).
\[f^{(5)}(x)=32\cos 2x\]Because the cosine is between \(-1\) and 1, the size of the derivative is at most 32. The orange line at 32 touches the tops of the curve, so no part is red.
\[\left|f^{(5)}(t)\right|\le\cy{32}\]
Example 2: A falling exponential
The Taylor polynomial of degree 3 at 0 approximates \(f(x)=e^{-x}\) at \(x=-1\). Find a valid \(M\) for the interval from \(-1\) to 0.
The degree is 3, so the remainder uses the derivative of order 4. It is positive, so it is equal to its size.
\[f^{(4)}(t)=e^{-t}\]A tempting wrong move: The center 0 is the known point, so use the value there.
\[\cr{M=e^{0}=1}\]The curve rises to \(e\approx2.72\) at \(t=-1\), so the line \(M=1\) is below it, and the curve is red there.
The function \(e^{-t}\) falls as \(t\) increases, so its largest value in the band is at the left end \(t=-1\). That value is \(e\), and \(e<3\), so 3 is valid.
\[\cy{M=3}\]
Practice
The Taylor polynomial of degree 1 at 0 approximates \(f(x)=\dfrac{1}{3-x}\) at \(x=2\). Find the smallest valid \(M\) for the interval from 0 to 2.
Move M and read the red part. Is the curve highest at the left end or at the right end of the band?
Solution
The degree is 1, so the remainder uses the derivative of order 2. Differentiating \(f'(t)=(3-t)^{-2}\) gives a positive function on the band.
\[f''(t)=\dfrac{2}{(3-t)^3}\]As \(t\) increases toward 3, the denominator becomes smaller, so the derivative rises. Its largest value is at the right end \(t=2\).
\[f''(2)=\dfrac{2}{1^3}=2\]At M = 2 no part is red, and the line touches the curve at the right end, so 2 is the smallest valid M.
\[\cy{M=2}\]
Questions
In questions 1 and 2, a polynomial of degree 3 at the center 0 approximates \(f\), and the curve shows \(\left|f^{(4)}(t)\right|\). The faint band covers the interval from the center to the input, here 3.
Suppose the input is 1, so the band covers the interval from 0 to 1. What is the smallest valid \(M\)?
- 4
- 1
- 5
Answer
4
The input is 3, so the band covers the interval from 0 to 3. What is the smallest valid \(M\)?
- 5
- 4
- 1
Answer
5
The Taylor polynomial of degree 3 at 0 approximates \(g(x)=x^5\) at \(x=2\). What is the smallest valid \(M\) for \(\left|g^{(4)}\right|\) on the interval from 0 to 2?
- 240
- 0
- 120
Answer
240
Summary
You find the largest value of the size of the derivative of order \(n+1\) over the whole interval between the center and the input.
The remainder uses this derivative at an unknown point of the interval, so \(M\) must be at least its size at every point.
You use M in the error bound, and you can round it up to a convenient number.
This is one of four skills in topic 10.12 of Which & Why: AP Calculus.