Why Sigma Notation Helps
A Riemann sum records contributions from several subintervals. When there are only a few, it is natural to write every contribution separately. When there are many, a compact notation makes the pattern easier to see and the expression easier to work with.
Sigma notation is that compact notation. Its main benefit is not merely saving space: it makes the correspondence between a term and its index explicit. For a Riemann sum, that correspondence must keep three items aligned: the interval, the sample point chosen in that interval, and the width of that interval.
This tutorial develops the notation for tagged sums, including partitions whose widths may differ and uniform partitions whose widths are equal. The focus is on translating and checking expressions, not on taking limits of sums.
Finite Sums in Sigma Notation
The symbol \(\sum\), read “the sum of,” gives a compact instruction to add a sequence of terms. In \(\sum_{i=1}^{n} a_i\), the letter \(i\) is the index, \(1\) is its starting value, and \(n\) is its ending value. The expression \(a_i\) tells us what term to use for each index.
For example, \(\sum_{i=1}^{4}(2i+1)\) instructs us to substitute \(i=1,2,3,4\) into \(2i+1\), then add the results:
The lower and upper limits tell us exactly how many terms to include. The index is not an extra factor and is not a variable that remains after the addition is performed.
Connecting Indices to a Tagged Partition
Let \(a=x_0<x_1<\cdots<x_n=b\) be a partition of the interval \([a,b]\). Its \(i\)th subinterval is \([x_{i-1},x_i]\), so the width of that subinterval is \(\Delta x_i=x_i-x_{i-1}\). A tag \(\xi_i\) for that subinterval is a chosen point in it: \(x_{i-1}\leq \xi_i\leq x_i\).
At index \(i\), the three pieces work together: \(\xi_i\) is in the \(i\)th subinterval, \(f(\xi_i)\) is the sampled function value there, and \(\Delta x_i\) is the width of that same subinterval. The index starts at \(1\) because the first subinterval runs from \(x_0\) to \(x_1\). It ends at \(n\) because there are \(n\) subintervals.
The width must be present. A sample value \(f(\xi_i)\) is a height; multiplying by \(\Delta x_i\) weights it by the width of its interval. For unequal subintervals, each term may therefore have a different width.
Deriving the Sigma Form from the Expanded Sum
To see why the notation represents the intended sum, write out the first few terms. At \(i=1\), the term is \(f(\xi_1)\Delta x_1\), and \(\Delta x_1=x_1-x_0\). At \(i=2\), it is \(f(\xi_2)\Delta x_2\), with \(\Delta x_2=x_2-x_1\). Continuing the same rule gives the last term at \(i=n\).
This is a direct consequence of the definition of a finite sum: each integer index from \(1\) through \(n\) contributes one term. The subscripts in the width are shifted deliberately. The interval numbered \(i\) has endpoints \(x_{i-1}\) and \(x_i\), not \(x_i\) and \(x_{i+1}\).
Worked Example: Write an Unequal-Width Sum in Sigma Notation
Worked Example: Write an Unequal-Width Sum in Sigma Notation
Consider the partition \(0=x_0<x_1=1<x_2=3<x_3=6\) and the function \(f(x)=x\). Choose tags \(\xi_1=\frac12\), \(\xi_2=2\), and \(\xi_3=5\). Write the expanded sum in sigma notation and evaluate it.
First find each width from its own pair of endpoints:
Now evaluate the function at each corresponding tag: \(f(\xi_1)=\frac12\), \(f(\xi_2)=2\), and \(f(\xi_3)=5\). The expanded sum is therefore
The index \(i\) runs from \(1\) to \(3\). For each \(i\), the tag is the midpoint of its interval, so \(\xi_i=\frac{x_{i-1}+x_i}{2}\), and the width is \(x_i-x_{i-1}\). Thus the sum in sigma notation is
Checking by expanding this expression gives \(\frac12(1-0)+2(3-1)+5(6-3)=\frac12+4+15=\frac{39}{2}\), exactly the value found above.
Expanding a Given Riemann Sum
To expand a sum, substitute each index value in order. Do not change the tag rule or width rule while substituting. If the summand is \(f(\xi_i)\Delta x_i\), replace both \(\xi_i\) and \(\Delta x_i\) according to their definitions for that index.
Worked Example: Expand and Evaluate a Tagged Sum
On the partition \(0=x_0<x_1=1<x_2=3<x_3=4\), let \(f(x)=x^2\), and choose the right endpoint as the tag on each subinterval: \(\xi_i=x_i\). Expand and evaluate the tagged sum.
The widths are \(\Delta x_1=1-0=1\), \(\Delta x_2=3-1=2\), and \(\Delta x_3=4-3=1\). The tags are \(\xi_1=1\), \(\xi_2=3\), and \(\xi_3=4\). Hence
Substituting the values gives
To check, the three intervals contribute \(f(1)(1-0)=1\), \(f(3)(3-1)=18\), and \(f(4)(4-3)=16\). Their sum is \(35\), as required. In particular, the middle term uses the width \(2\), not \(1\): its interval is \([1,3]\).
Uniform Partitions and a Common Width
A uniform partition divides \([a,b]\) into \(n\) subintervals of equal width. Its partition points can be written as \(x_i=a+i\Delta x\), where \(\Delta x=(b-a)/n\). Substitution shows that the \(i\)th interval has width \(x_i-x_{i-1}=\Delta x\). This notation is useful here because it lets every term display the same width.
If \(\xi_i\) is the tag chosen in the \(i\)th interval, the tagged sum for this uniform partition is \(\sum_{i=1}^{n} f(\xi_i)\Delta x\). The width is outside the index-dependent part of the summand, but it is still multiplied into every term. The tag rule must still place \(\xi_i\) in \([x_{i-1},x_i]\).
Worked Example: Write and Expand a Uniform-Partition Sum
Divide \([0,3]\) into \(3\) equal subintervals, use each midpoint as its tag, and let \(f(x)=2x+1\). Write the sum in sigma notation, expand it, and evaluate it.
The common width is \(\Delta x=(3-0)/3=1\). The partition points are \(x_i=0+i\), so \(x_0=0\), \(x_1=1\), \(x_2=2\), and \(x_3=3\). The midpoint tags are \(\xi_i=(x_{i-1}+x_i)/2=i-\frac12\). Thus
Substitute the three index values. The tags are \(\frac12,\frac32,\frac52\), and the corresponding function values are \(2,4,6\):
Checking directly from \(f(x)=2x+1\) gives \(f(\frac12)=2\), \(f(\frac32)=4\), and \(f(\frac52)=6\); multiplying each by its interval width \(1\) again gives \(2+4+6=12\).
A Reliable Translation Method
When translating between expanded and sigma notation, it helps to separate the bookkeeping from the arithmetic. First identify how many subintervals or terms there are. Then write down what the index represents, and only then substitute or compress the terms.
If there are \(n\) subintervals, the index for the tagged sum runs from \(1\) to \(n\).
At index \(i\), use \([x_{i-1},x_i]\), so the width is \(x_i-x_{i-1}\).
Check \(x_{i-1}\leq \xi_i\leq x_i\) before evaluating \(f(\xi_i)\).
For expansion, substitute \(i=1,2,\ldots,n\) in order. For sigma notation, make one general term that follows the same rule at every index.
Common Mistake: Shifting the Width or Dropping It
A frequent error is to pair the tag from one subinterval with the width of another. Another is to write only the sampled function values and forget the widths. For example, in the unequal partition from the worked example with \(f(x)=x^2\), the middle interval is \([1,3]\), its tag is \(3\), and its width is \(2\). The correct middle contribution is \(f(3)(3-1)=3^2(2)=18\).
A wrong calculation might use the previous interval's width and write \(3^2(1)=9\), or omit the width and write \(3^2=9\). Neither represents the area contribution from \([1,3]\). The correction is to read the index as a matched set: for \(i=2\), the tag is \(x_2=3\) and the width is \(x_2-x_1=2\). The complete sum is \(1^2(1)+3^2(2)+4^2(1)=35\).
Why This Notation Matters
Writing a Riemann sum with sigma notation makes its structure visible even when the number of subintervals is large. The notation also provides a precise way to describe how every term is generated: choose a subinterval, evaluate the function at its tag, and multiply by that interval’s width.
For unequal widths, retaining \(\Delta x_i\) in the summand reminds us that interval sizes can vary. For a uniform partition, writing the common \(\Delta x\) makes the repeated width explicit. In either case, the key discipline is the same: the index, tag, and width must refer to the same subinterval.
Check Your Understanding
For each question, pay attention to the index range as well as the term being added.
- Expand \(\sum_{i=1}^{3}(i+2)\).
- A partition has \(x_0=0\), \(x_1=2\), and \(x_2=5\). Write the tagged Riemann sum using tags \(\xi_1\) and \(\xi_2\).
- For the interval \([x_{i-1},x_i]\), express its width using the partition points.
- A uniform partition of \([2,8]\) has \(3\) subintervals. What is its common width?
- In \(\sum_{i=1}^{n}f(\xi_i)\Delta x_i\), what must be true of \(\xi_i\), and which subinterval does \(\Delta x_i\) measure?
