20 Taylor Approximation, Higher Derivative Test for Local Extrema, Riemann Integral, Darboux Sums and Integrals

Lecture from: 08.05.2024 | Video: Video ETHZ

Review: Power Series and Their Derivatives

Flashback: Power Series and Their Coefficients A key fact we’ve learned: if a function $f(x)$ can be represented by a power series $f(x)={\sum }_{k=0}^{\infty }{c}_k(x-x_0{)}^k$ (that converges in some region), then:

1. $f$ is infinitely differentiable (smooth) in that region.
2. We can find all its derivatives by differentiating the series term by term.
3. Crucially, the coefficients $c_k$ of the series are directly determined by the function’s derivatives at the center point $x_0$: $c_j=\frac{f^{(j)}(x_0)}{j!}$. This is a HUGE link!

If $f(x)={\sum }_{k=0}^{\infty }{c}_k(x-x_0{)}^k$ is a power series with radius of convergence $\rho >0$, then $f$ is smooth (infinitely differentiable) in the open interval $(x_0-\rho ,x_0+\rho )$. Furthermore, for any non-negative integer $j\in \mathbb{N}$ (where $f^{(0)}=f$), its $j$-th derivative is given by differentiating term-by-term: $f^{(j)}(x)={\sum }_{k=j}^{\infty }{c}_k\cdot k(k-1)\dots (k-j+1)(x-x_0{)}^{k-j}$ A crucial consequence is obtained by evaluating this at $x=x_0$. All terms in the sum become zero except for the first term (where $k=j$): $f^{(j)}(x_0)=c_j\cdot j(j-1)\dots (1)\cdot (x_0-x_0{)}^0=c_j\cdot j!$.

This gives us the formula for the coefficients: $c_j=\frac{f^{(j)}(x_0)}{j!}$

Without Summation Notation (to see the pattern more clearly): If $f(x)=c_0+c_1(x-x_0)+c_2(x-x_0{)}^2+c_3(x-x_0{)}^3+\dots$

• $f(x_0)=c_0⟹c_0=\frac{f^{(0)}(x_0)}{0!}$
• $f^{\prime }(x)=c_1+2c_2(x-x_0)+3c_3(x-x_0{)}^2+\cdots ⟹f^{\prime }(x_0)=c_1⟹c_1=\frac{f^{(1)}(x_0)}{1!}$
• $f^{\prime \prime }(x)=2c_2+3\cdot 2c_3(x-x_0)+\cdots ⟹f^{\prime \prime }(x_0)=2c_2⟹c_2=\frac{f^{\prime \prime }(x_0)}{2!}$
• $f^{\prime \prime \prime }(x)=3\cdot 2\cdot 1c_3+\cdots ⟹f^{\prime \prime \prime }(x_0)=6c_3⟹c_3=\frac{f^{\prime \prime \prime }(x_0)}{3!}$
• And in general, $f^{(n)}(x_0)=n!c_n$, so $c_n=\frac{f^{(n)}(x_0)}{n!}$.

Taylor Approximation

Using Derivatives to Build Polynomial Approximations The beautiful relationship $c_k=\frac{f^{(k)}(x_0)}{k!}$ from power series gives us a brilliant idea. If a function $f$ is smooth (or at least has enough derivatives) near a point $x_0$, we can use its derivatives at that single point $x_0$ to construct a polynomial that should approximate $f(x)$ well for $x$ values near $x_0$. This is the essence of Taylor approximation.

The relationship $c_k=\frac{f^{(k)}(x_0)}{k!}$ for power series makes it plausible that if a function $f$ is smooth (or at least sufficiently differentiable) near a point $x_0$, we can approximate it using a polynomial constructed from its derivatives at $x_0$.

Indeed, if $f(x)$ can be represented as a power series ${\sum }_{k=0}^{\infty }{c}_k(x-x_0{)}^k$, then substituting $c_k=\frac{f^{(k)}(x_0)}{k!}$ gives $f(x)={\sum }_{k=0}^{\infty }\frac{f^{(k)}(x_0)}{k!}(x-x_0{)}^k$ This means $f(x)$ is the limit of the partial sums: $f(x)={\lim }_{n\to \infty }{\sum }_{k=0}^n\frac{f^{(k)}(x_0)}{k!}(x-x_0{)}^k$ This convergence is uniform on any closed interval around $x_0$ that is strictly within the radius of convergence.

This motivates the following definition:

Definition: Taylor Polynomial

The Taylor Polynomial: A Function’s “Best Fit” Polynomial For a function $f$ that’s $n$-times differentiable at a point $x_0$, its Taylor polynomial of order $n$ centered at $x_0$ is a specific polynomial of degree at most $n$. Its magic property? This polynomial $T_n(x)$ and its first $n$ derivatives perfectly match the values of $f$ and its first $n$ derivatives at the point $x_0$. It’s designed to be the best possible polynomial approximation of degree $n$ to $f$ right around $x_0$.

Let $I\subseteq \mathbb{R}$ be an interval with more than one point. Let $n\in \mathbb{N}$ (a non-negative integer), and let $f:I\to \mathbb{R}$ be $n$-times differentiable at $x_0\in I$.

The Taylor polynomial of order $n$ (or degree $n$) of $f$ at (or around, or centered at) $x_0$ is: $T_n(f;x;x_0)={\sum }_{k=0}^n\frac{f^{(k)}(x_0)}{k!}(x-x_0{)}^k$ This polynomial is often denoted $T_{n,f,x_0}(x)$ or simply $T_n(x)$ if $f$ and $x_0$ are clear from context. It is the unique polynomial $P(x)$ of degree at most $n$ such that $P(x)$ and its first $n$ derivatives match $f(x)$ and its first $n$ derivatives at the point $x_0$: $P(x_0)=f(x_0)$, $P^{\prime }(x_0)=f^{\prime }(x_0)$, $\dots$, $P^{(n)}(x_0)=f^{(n)}(x_0)$.

Taylor Polynomials for $f(x)=\cos (x)$ at $x_0=0$

Example: Approximating Cosine with Polynomials Let’s build some Taylor polynomials for $f(x)=\cos (x)$ centered at $x_0=0$ (these are also called Maclaurin polynomials). We need the derivatives of $\cos (x)$ at $0$: $\cos (0)=1$, $-\sin (0)=0$, $-\cos (0)=-1$, $\sin (0)=0$, $\cos (0)=1$, and so on. The polynomials will use these values to try and “hug” the cosine curve near $x=0$.

The derivatives of $f(x)=\cos (x)$ evaluated at $x_0=0$ are: $f(x)=\cos (x)⟹f(0)=1$ $f^{\prime }(x)=-\sin (x)⟹f^{\prime }(0)=0$ $f^{\prime \prime }(x)=-\cos (x)⟹f^{\prime \prime }(0)=-1$ $f^{\prime \prime \prime }(x)=\sin (x)⟹f^{\prime \prime \prime }(0)=0$ $f^{(4)}(x)=\cos (x)⟹f^{(4)}(0)=1$ (The pattern of derivatives at $0$ is $1,0,-1,0,1,0,-1,0,\dots$)

The first few Taylor polynomials for $\cos (x)$ at $x_0=0$ are: $T_0(f;x;0)=\frac{f(0)}{0!}{x}^0=\frac{1}{1}\cdot 1=1$ $T_1(f;x;0)=f(0)+\frac{f^{\prime }(0)}{1!}x=1+\frac{0}{1}x=1$ (Since $f^{\prime }(0)=0$, $T_1$ is the same as $T_0$).

$T_2(f;x;0)=f(0)+f^{\prime }(0)x+\frac{f^{\prime \prime }(0)}{2!}{x}^2=1+0x+\frac{-1}{2}{x}^2=1-\frac{x^2}{2}$ $T_3(f;x;0)=T_2(f;x;0)+\frac{f^{\prime \prime \prime }(0)}{3!}{x}^3=1-\frac{x^2}{2}+\frac{0}{6}{x}^3=1-\frac{x^2}{2}$ (Since $f^{\prime \prime \prime }(0)=0$, $T_3$ is the same as $T_2$).

$T_4(f;x;0)=T_3(f;x;0)+\frac{f^{(4)}(0)}{4!}{x}^4=(1-\frac{x^2}{2})+\frac{1}{24}{x}^4=1-\frac{x^2}{2!}+\frac{x^4}{4!}$

Notice how these polynomials start to resemble the well-known Maclaurin series for $\cos (x)$: $\cos (x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-\dots$

Satz: Taylor’s Theorem with Lagrange Remainder (Taylor Approximation)

How Good is the Taylor Approximation? Enter the Remainder! Taylor’s Theorem is a cornerstone result. It doesn’t just say that $T_n(x)$ approximates $f(x)$; it gives us an explicit formula for the error or remainder term, $R_n(x)=f(x)-T_n(x)$. The Lagrange form of this remainder tells us the error is $\frac{f^{(n+1)}(\xi )}{(n+1)!}(x-x_0{)}^{n+1}$, where $\xi$ is some mysterious point between $x_0$ and $x$. This means the error depends on the $(n+1)$-th derivative of $f$ and how far $x$ is from $x_0$.

Let $a<b$ in $\mathbb{R}$. Let $f:[a,b]\to \mathbb{R}$ be a function that is $n$-times continuously differentiable on the closed interval $[a,b]$ and $(n+1)$-times differentiable on the open interval $(a,b)$. Let $x_0\in [a,b]$. Then for every $x\in [a,b]$ with $x\ne x_0$, there exists a point $\xi$ strictly between $x_0$ and $x$ such that: $f(x)=T_n(f;x;x_0)+\frac{f^{(n+1)}(\xi )}{(n+1)!}(x-x_0{)}^{n+1}$ Remark: The term $R_n(f;x;x_0)=f(x)-T_n(f;x;x_0)=\frac{f^{(n+1)}(\xi )}{(n+1)!}(x-x_0{)}^{n+1}$ is called the remainder term. This specific formula is the Lagrange form of the remainder. It quantifies the error when approximating $f(x)$ by its $n$-th Taylor polynomial. The existence of such a $\xi$ is guaranteed, though its exact value is usually unknown.

Korollar: Qualitative Version of Taylor Approximation (Peano Form of Remainder)

The Error Shrinks Super Fast! (Little-o Notation) This corollary gives us another way to think about the error. It says that the difference $f(x)-T_n(f;x;x_0)$ goes to zero faster than $(x-x_0{)}^n$ does as $x$ approaches $x_0$. We use “little-o” notation for this: $f(x)-T_n(f;x;x_0)=o((x-x_0{)}^n)$. This essentially means that $T_n$ is an exceptionally good polynomial approximation of degree $n$ to $f$ very close to $x_0$.

Let $a<b$ in $\mathbb{R}$, and $f:[a,b]\to \mathbb{R}$ be $n$-times continuously differentiable on $[a,b]$ (this implies $f^{(n)}$ is continuous). Let $x_0\in [a,b]$. Then: ${\lim }_{x\to x_0}\frac{f(x)-T_n(f;x;x_0)}{(x-x_0{)}^n}=0$ This means that the error $f(x)-T_n(f;x;x_0)$ diminishes more rapidly than $(x-x_0{)}^n$ as $x\to x_0$. We write this using little-o notation as $f(x)-T_n(f;x;x_0)=o((x-x_0{)}^n)$ as $x\to x_0$. The Taylor polynomial $T_n$ is the “best” polynomial approximation of degree at most $n$ to $f$ near $x_0$ in this specific sense.

Exercise: Find the corresponding statement for the tangent line approximation (which is $T_1(f;x;x_0)=f(x_0)+f^{\prime }(x_0)(x-x_0)$) that we encountered at the beginning of our discussion of differentiability. (Hint: The definition of differentiability is ${\lim }_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=f^{\prime }(x_0)$. Rearranging this gives ${\lim }_{x\to x_0}\frac{f(x)-(f(x_0)+f^{\prime }(x_0)(x-x_0))}{x-x_0}=0$. This is exactly the statement for $n=1$.)

Proof of the Corollary (using Taylor’s Theorem for order $n-1$): We need $f$ to be $n$-times differentiable for $T_n$ to be defined, and $f^{(n)}$ to be continuous at $x_0$ for the limit step. By Taylor’s Theorem applied with order $n-1$ (so we need $f$ to be $n$-times differentiable for the remainder term $f^{(n)}({\xi }_x)$ to make sense): $f(x)=T_{n-1}(f;x;x_0)+\frac{f^{(n)}({\xi }_x)}{n!}(x-x_0{)}^n$, for some ${\xi }_x$ between $x_0$ and $x$. Recall the definition of $T_n(f;x;x_0)$: $T_n(f;x;x_0)=T_{n-1}(f;x;x_0)+\frac{f^{(n)}(x_0)}{n!}(x-x_0{)}^n$. Subtracting these two equations: $f(x)-T_n(f;x;x_0)=\left(T_{n-1}(f;x;x_0)+\frac{f^{(n)}({\xi }_x)}{n!}(x-x_0{)}^n\right)-\left(T_{n-1}(f;x;x_0)+\frac{f^{(n)}(x_0)}{n!}(x-x_0{)}^n\right)$ $f(x)-T_n(f;x;x_0)=\frac{f^{(n)}({\xi }_x)}{n!}(x-x_0{)}^n-\frac{f^{(n)}(x_0)}{n!}(x-x_0{)}^n$ $f(x)-T_n(f;x;x_0)=\frac{1}{n!}(f^{(n)}({\xi }_x)-f^{(n)}(x_0))(x-x_0{)}^n$. So, for $x\ne x_0$: $\frac{f(x)-T_n(f;x;x_0)}{(x-x_0{)}^n}=\frac{1}{n!}(f^{(n)}({\xi }_x)-f^{(n)}(x_0))$ As $x\to x_0$, we also have ${\xi }_x\to x_0$ (since ${\xi }_x$ is squeezed between $x$ and $x_0$). Since $f$ is $n$-times continuously differentiable, its $n$-th derivative $f^{(n)}$ is continuous at $x_0$. Therefore, ${\lim }_{{\xi }_x\to x_0}{f}^{(n)}({\xi }_x)=f^{(n)}(x_0)$, which means ${\lim }_{x\to x_0}(f^{(n)}({\xi }_x)-f^{(n)}(x_0))=0$. This implies: ${\lim }_{x\to x_0}\frac{f(x)-T_n(f;x;x_0)}{(x-x_0{)}^n}=\frac{1}{n!}\cdot 0=0$ This completes the proof. $\square$

Consequences for Local Extrema: Higher Derivative Test

Taylor’s Idea Helps Find Maxima and Minima! Taylor approximations give us a powerful tool to analyze a function’s behavior near a critical point (where $f^{\prime }(x_0)=0$). If the first few derivatives are zero, the first non-zero derivative at $x_0$ can tell us if we have a local maximum, minimum, or neither.

Korollar (Higher Order Derivative Test - slightly reformulated)

The Test: Using the First Non-Zero Derivative Suppose at a point $x_0$, the function’s derivative is zero, and maybe even the next few are also zero: $f^{\prime }(x_0)=f^{\prime \prime }(x_0)=\cdots =f^{(n-1)}(x_0)=0$. But the $n$-th derivative, $f^{(n)}(x_0)$, is NOT zero.

• If $n$ is odd: $f$ has no local extremum at $x_0$. (It’s typically an inflection point with a horizontal tangent if $n\ge 3$). Intuition: The term $(x-x_0{)}^n$ changes sign as $x$ passes $x_0$, so $f(x)-f(x_0)$ also changes sign locally.
• If $n$ is even and $f^{(n)}(x_0)>0$: $f$ has a strict local minimum at $x_0$. Intuition: $(x-x_0{)}^n$ is positive, so $f(x)-f(x_0)$ is positive locally; $f(x)>f(x_0)$.
• If $n$ is even and $f^{(n)}(x_0)<0$: $f$ has a strict local maximum at $x_0$. Intuition: $(x-x_0{)}^n$ is positive, so $f(x)-f(x_0)$ is negative locally; $f(x)<f(x_0)$.

Let $n\ge 1$ be an integer, $a<b$ in $\mathbb{R}$. Let $f:[a,b]\to \mathbb{R}$ be $n$-times continuously differentiable, and let $x_0\in (a,b)$ be an interior point of the interval. Assume: $f^{\prime }(x_0)=f^{\prime \prime }(x_0)=\cdots =f^{(n-1)}(x_0)=0$, but $f^{(n)}(x_0)\ne 0$. (This means $f^{(n)}(x_0)$ is the first non-zero derivative at $x_0$).

Then:

1. If $n$ is odd, $f$ has no local extremum at $x_0$. (If $n\ge 3$, $x_0$ is a saddle/inflection point with a horizontal tangent. If $n=1$, this condition means $f^{\prime }(x_0)\ne 0$, so $x_0$ wasn’t a critical point to begin with unless we are just classifying points).
2. If $n$ is even and $f^{(n)}(x_0)>0$, then $f$ has a strict local minimum at $x_0$. (This means there exists some $\delta >0$ such that for all $x\in (x_0-\delta ,x_0+\delta )$ with $x\ne x_0$, we have $f(x)>f(x_0)$).
3. If $n$ is even and $f^{(n)}(x_0)<0$, then $f$ has a strict local maximum at $x_0$.

(Proof idea: Near $x_0$, $f(x)-f(x_0)\approx \frac{f^{(n)}(x_0)}{n!}(x-x_0{)}^n$. The sign of this difference determines if $f(x)$ is above or below $f(x_0)$.)

Korollar (The Special Case $n=2$: Second Derivative Test)

The Familiar Second Derivative Test This is the higher order test when $n=2$, which we often use: If $f^{\prime }(x_0)=0$ (critical point):

• If $f^{\prime \prime }(x_0)>0⟹$ cup up $⟹$ strict local minimum.
• If $f^{\prime \prime }(x_0)<0⟹$ cup down $⟹$ strict local maximum. (If $f^{\prime \prime }(x_0)=0$, this test is inconclusive, and we’d need to check higher derivatives or other methods).

Let $a<b$ in $\mathbb{R}$, $f:[a,b]\to \mathbb{R}$ be twice continuously differentiable ($C^2$), and $x_0\in (a,b)$. Assume: $f^{\prime }(x_0)=0$ (i.e., $x_0$ is a critical point) and $f^{\prime \prime }(x_0)\ne 0$. Then:

1. If $f^{\prime \prime }(x_0)>0$, then $f$ has a (strict) local minimum at $x_0$.
2. If $f^{\prime \prime }(x_0)<0$, then $f$ has a (strict) local maximum at $x_0$.

Beispiel: Finding Local Extrema for $f(x)=x^4-x^2+1$

Putting the Test to Work Let’s find local maxima/minima for $f(x)=x^4-x^2+1$.

1. Find critical points (where $f^{\prime }(x)=0$).
2. Use the second derivative test $f^{\prime \prime }(x)$ at these points.

What are the local extrema of $f(x)=x^4-x^2+1$?

This function is smooth (it’s a polynomial). Candidates for local extrema occur where $f^{\prime }(x)=0$.

$f^{\prime }(x)=4x^3-2x$. Set $f^{\prime }(x)=0⟹4x^3-2x=0⟹2x(2x^2-1)=0$.

The critical points (candidates) are: $x_0=0$, or $2x^2-1=0⟹x^2=1/2⟹x_{1,2}=\pm \frac{1}{\sqrt{2}}$.

To determine the nature of these extrema, we use the second derivative: $f^{\prime \prime }(x)=\frac{d}{dx}(4x^3-2x)=12x^2-2$.

• At $x_0=0$: $f^{\prime }(0)=0$. $f^{\prime \prime }(0)=12(0{)}^2-2=-2$. Since $f^{\prime \prime }(0)=-2<0$, by the Second Derivative Test, $f$ has a local maximum at $x_0=0$. The value is $f(0)=0^4-0^2+1=1$.

• At $x_1=\frac{1}{\sqrt{2}}$ (and similarly for $x_2=-\frac{1}{\sqrt{2}}$ due to symmetry, since $x^2=1/2$ in both cases): $f^{\prime }(\pm \frac{1}{\sqrt{2}})=0$. $f^{\prime \prime }(\pm \frac{1}{\sqrt{2}})=12\left({\left(\pm \frac{1}{\sqrt{2}}\right)}^2\right)-2=12\left(\frac{1}{2}\right)-2=6-2=4$. Since $f^{\prime \prime }(\pm \frac{1}{\sqrt{2}})=4>0$, by the Second Derivative Test, $f$ has local minima at $x=\frac{1}{\sqrt{2}}$ and $x=-\frac{1}{\sqrt{2}}$. The value is $f(\pm \frac{1}{\sqrt{2}})=(\frac{1}{2}{)}^2-(\frac{1}{2})+1=\frac{1}{4}-\frac{1}{2}+1=\frac{3}{4}$.

Caveat: It’s possible for a smooth function to have a strict local extremum at a point even if all higher derivatives are zero at that point. The Higher Order Derivative Test (Korollar) only applies if some $f^{(n)}(x_0)$ is non-zero.

Example: The function $f(x)=\{\begin{array}{ll}{e}^{-1/x^2}& \text{if }x\ne 0\\ 0& \text{if }x=0\end{array}$ (which is a classic example of a $C^{\infty }$ function that is not analytic at $x=0$) has a strict local minimum at $x=0$ (since $f(x)>0$ for $x\ne 0$ and $f(0)=0$).

However, it can be shown that $f^{(k)}(0)=0$ for all $k\in \mathbb{N}$. The Taylor series for this function around $x=0$ is identically zero and does not represent the function anywhere but at $x=0$.

The Riemann Integral

New Adventure: From Slopes to Areas! We’re shifting gears from derivatives (which tell us about rates of change and slopes) to a new concept: the integral. The Riemann integral is a way to formally define and calculate the area under a curve.

Motivation: Calculating Areas

Motivation: Calculation of areas. For example:

• What is the area of a circle of radius $r$? (Answer: $\pi r^2$)
• What is the area of the region bounded by the graph of $y=1/x$, the x-axis, and the vertical lines $x=1$ and $x=2$? This area will turn out to be $\ln (2)$.

Definition and Integrability Criteria

Laying the Groundwork: Slicing and Summing To find the area under a potentially complicated curve, the basic idea is to:

1. Chop the interval on the x-axis into many small pieces (a “partition”).
2. Over each small piece, approximate the area under the curve with a simple rectangle.
3. Sum the areas of these rectangles.
4. See what happens to this sum as we make the pieces smaller and smaller. This section introduces the formal machinery: partitions, Darboux sums, and finally, the definition of Riemann integrability.

Basic Setup for Integration Interval

In the following, $a<b$ in $\mathbb{R}$ and $I=[a,b]$ is a closed, bounded interval.

Definition: Partition of an Interval

Chopping Up the Interval: Partitions A “partition” of an interval $[a,b]$ is just a fancy way of saying we pick a finite set of points in the interval, starting with $a$ and ending with $b$, that divide the interval into smaller subintervals.

Formal Definition of a Partition

A partition $P$ of an interval $I=[a,b]$ is a finite, ordered subset of $I$ that includes the endpoints $a$ and $b$.

We usually write $P=\{x_0,x_1,\dots ,x_n\}$ where $a=x_0<x_1<\cdots <x_n=b$.

Resulting Subintervals and Their Lengths

Remark: A partition $P=\{x_0,x_1,\dots ,x_n\}$ (consisting of $n+1$ points) divides the interval $I=[a,b]$ into $n$ subintervals: $[x_0,x_1],[x_1,x_2],\dots ,[x_{n-1},x_n]$. The $i$-th subinterval is $I_i=[x_{i-1},x_i]$, and its length is $\Delta x_i=x_i-x_{i-1}$. (The ${\delta }_i$ in the original text is also commonly used for $\Delta x_i$).

Refinements of Partitions

Definition of a Refinement

If $P$ and $P^{\prime }$ are partitions of $I$, then $P^{\prime }$ is called a refinement of $P$ if $P\subseteq P^{\prime }$ (meaning $P^{\prime }$ contains all the points of $P$, and possibly more). Intuition: A refinement means you’re chopping the interval into even smaller (or at least, not larger) pieces.

Common Refinement

Remark: Any two partitions $P,P^{\prime }$ of $I$ always possess a common refinement, for example, their union $P\cup P^{\prime }$. This common refinement contains all the points from both $P$ and $P^{\prime }$.

Prerequisite: Boundedness of the Function

Now, let $f:[a,b]\to \mathbb{R}$ be a bounded function. This is a crucial prerequisite: $f$ must not shoot off to $\pm \infty$ within the interval. Formally, this means there exists a real number $M_{glob}\ge 0$ such that $|f(x)|\le M_{glob}$ for all $x\in [a,b]$.

Notation for Subinterval Lengths in Darboux Sums

For a partition $P=\{x_0,x_1,\dots ,x_n\}$ of $I=[a,b]$, let ${\delta }_i=\Delta x_i=x_i-x_{i-1}$ be the length of the $i$-th subinterval $[x_{i-1},x_i]$ (for $i=1,\dots ,n$).

Definition: Lower and Upper Darboux Sums

Approximating Area: Underestimates and Overestimates For each small slice (subinterval) created by our partition, we can find the lowest value ($m_i$) and highest value ($M_i$) the function $f$ takes on that slice.

• Lower Darboux Sum ($s(f,P)$): Use $m_i$ as the height of a rectangle on the $i$-th slice. The sum of these “under-rectangles” gives an underestimate of the total area.
• Upper Darboux Sum ($S(f,P)$): Use $M_i$ as the height of a rectangle on the $i$-th slice. The sum of these “over-rectangles” gives an overestimate.

Local Infima ($m_i$) and Suprema ($M_i$) on Subintervals

For a bounded function $f$ and a partition $P$:

• Let $m_i=\inf \{f(x)\mid x\in [x_{i-1},x_i]\}$ (the infimum, or greatest lower bound, of $f$ on the $i$-th subinterval).
• Let $M_i=\sup \{f(x)\mid x\in [x_{i-1},x_i]\}$ (the supremum, or least upper bound, of $f$ on the $i$-th subinterval).

Since $f$ is bounded on $[a,b]$, $m_i$ and $M_i$ are guaranteed to be finite real numbers for each subinterval.

The Lower Darboux Sum ($s(f,P)$)

The lower Darboux sum of $f$ with respect to the partition $P$ is: $s(f,P):={\sum }_{i=1}^n{m}_i{\delta }_i={\sum }_{i=1}^n{m}_i(x_i-x_{i-1})$ This represents an approximation to the area under the graph of $f$ using rectangles that are inscribed under the graph.

The Upper Darboux Sum ($S(f,P)$)

The upper Darboux sum of $f$ with respect to the partition $P$ is: $S(f,P):={\sum }_{i=1}^n{M}_i{\delta }_i={\sum }_{i=1}^n{M}_i(x_i-x_{i-1})$ This represents an approximation using rectangles that are circumscribed over (or contain the part of) the graph of $f$.

Remark: Bounds on Darboux Sums

Relationship to Global Function Bounds

If $m_{glob}=\inf \{f(x)\mid x\in [a,b]\}$ and $M_{glob}=\sup \{f(x)\mid x\in [a,b]\}$ are the global infimum and supremum of $f$ on $I$, then for any subinterval $[x_{i-1},x_i]$: $m_{glob}\le m_i\le M_i\le M_{glob}$.

Overall Inequality Chain

Then, for any partition $P$: $m_{glob}(b-a)=\sum m_{glob}{\delta }_i\le \sum m_i{\delta }_i=s(f,P)$.

Similarly, $S(f,P)=\sum M_i{\delta }_i\le \sum M_{glob}{\delta }_i=M_{glob}(b-a)$. So, $m_{glob}(b-a)\le s(f,P)\le S(f,P)\le M_{glob}(b-a)$.

The set of all lower sums and the set of all upper sums are bounded.

Lemma: Properties of Darboux Sums under Refinement

Squeezing In on the True Area This lemma tells us two important things:

1. Refining helps: If you add more points to your partition (refine it), your lower sum can only get bigger (or stay the same), and your upper sum can only get smaller (or stay the same). They get closer to each other and, intuitively, closer to the true area.
2. Lower sums are always below upper sums: Any lower sum (for any partition) is always less than or equal to any upper sum (for any, possibly different, partition). This is crucial for defining the integral.

Property 1: Effect of Refinement on Individual Sums

1. If $P^{\prime }$ is a refinement of a partition $P$ (i.e., $P\subseteq P^{\prime }$), then: $s(f,P)\le s(f,P^{\prime })\text{and}S(f,P^{\prime })\le S(f,P)$ Combining these, we always have $s(f,P)\le s(f,P^{\prime })\le S(f,P^{\prime })\le S(f,P)$.

   Intuition: Imagine you have a single subinterval in $P$. If $P^{\prime }$ splits this subinterval into two smaller ones, the minimum values on the smaller intervals can only be greater than or equal to the minimum on the larger original interval (and similarly for maximums). This pushes the lower sum up (or keeps it level) and the upper sum down (or keeps it level).

Property 2: Comparison of Any Lower Sum with Any Upper Sum

1. For any two arbitrary partitions $P_1,P_2$ of the interval $I$: $s(f,P_1)\le S(f,P_2)$ (The lower sum for any partition is less than or equal to the upper sum for any other partition, or even the same partition).

   Proof Sketch for Property 2

   Proof sketch for (2): Let $P^{\prime }=P_1\cup P_2$ be the common refinement of $P_1$ and $P_2$.

   Using part (1): $s(f,P_1)\le s(f,P^{\prime })$ (since $P^{\prime }$ refines $P_1$) We also know that for any single partition $P^{\prime }$, $s(f,P^{\prime })\le S(f,P^{\prime })$ (since $m_i\le M_i$ for all $i$).

   And again using part (1): $S(f,P^{\prime })\le S(f,P_2)$ (since $P^{\prime }$ refines $P_2$) Combining these inequalities: $s(f,P_1)\le s(f,P^{\prime })\le S(f,P^{\prime })\le S(f,P_2)$. Thus, $s(f,P_1)\le S(f,P_2)$.

Definition: Lower and Upper Darboux Integrals

The Best Possible Underestimate and Overestimate Now we consider all possible partitions of our interval.

• The lower Darboux integral ($s(f)$) is the “highest” of all possible lower Darboux sums. It’s the supremum (least upper bound) of all the underestimates.
• The upper Darboux integral ($S(f)$) is the “lowest” of all possible upper Darboux sums. It’s the infimum (greatest lower bound) of all the overestimates.

The Lower Darboux Integral ($s(f)$)

Let $\mathcal{P}(I)$ be the set of all possible partitions of the interval $I=[a,b]$.

The lower Darboux integral of $f$ over $I$ is defined as: $s(f):=\sup \{s(f,P)\mid P\in \mathcal{P}(I)\}\text{(also denoted }\underline{{\int }_a^b}f(x)dx\text{ or "Unterintegral")}$

The Upper Darboux Integral ($S(f)$)

The upper Darboux integral of $f$ over $I$ is defined as: $S(f):=\inf \{S(f,P)\mid P\in \mathcal{P}(I)\}\text{(also denoted }\overline{{\int }_a^b}f(x)dx\text{ or "Oberintegral")}$

Fundamental Inequality: $s(f)\le S(f)$

Remark: From Lemma 5.1.2 (part 2), we know that any $s(f,P_1)$ is a lower bound for the set of all upper sums $\{S(f,P_2)\}$. Therefore, the supremum of lower sums, $s(f)$, must be less than or equal to any particular upper sum $S(f,P_2)$. This makes $s(f)$ a lower bound for the set of all upper sums. Thus, $s(f)$ must be less than or equal to the infimum of the upper sums, $S(f)$.

So, for any bounded function $f$, we always have: $s(f)\le S(f)$

Riemann Integrability

When Does the Area Make Sense? Riemann Integrability! A bounded function $f$ is said to be Riemann integrable on $[a,b]$ if its lower Darboux integral and its upper Darboux integral are equal. That is, if the best possible underestimate matches the best possible overestimate. If they are equal, this common value is defined to be the Riemann integral of $f$ from $a$ to $b$, denoted ${\int }_a^{b}f(x)dx$. This is what we intuitively think of as the “area under the curve.”

Condition for Riemann Integrability

A bounded function $f:[a,b]\to \mathbb{R}$ is Riemann integrable (or simply “integrable”) on $[a,b]$ if its lower Darboux integral equals its upper Darboux integral: $s(f)=S(f)$

The Riemann Integral ${\int }_a^{b}f(x)dx$

In this case, this common value is called the Riemann integral of $f$ over $[a,b]$ (or from $a$ to $b$) and is denoted by: ${\int }_a^{b}f(x)dx:=s(f)=S(f)$

Consequence of Non-Integrability

If $s(f)<S(f)$, the function is not Riemann integrable on $[a,b]$. Intuitively, this means there’s an unresolvable “gap” between the total area of inscribed and circumscribed rectangles, no matter how fine the partition.

Continue here: 21 Dirichlet’s Function, Integrability