0.1 Summation, Sampled Trajectories, and Aggregate Metrics

CCD objectives and constraints often aggregate quantities across time samples, operating conditions, load cases, or uncertain scenarios. Summation notation lets us express those aggregates compactly and unambiguously.

Here we review the properties of finite sums, then connect the mean, variance, and standard deviation to trajectory metrics and uncertainty studies.

Introduction¶

For example, if we take $x_i = i^2$ , then $\displaystyle \sum_{i = 1}^6 i^2$ represents the sum of the squares of all integers from 1 to 6:

\sum_{i = 1}^6 i^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2

Notice that both the starting and ending indices (1 and 6, respectively) are included in the sum. Summation notation allows us to express sums conveniently – the left-hand side above is more compact than the right-hand side.

Often, we’ll take the sum of the first $n$ terms of a sequence. For example, the sum of the squares of the first $n$ positive integers is:

\sum_{i = 1}^n i^2

Note that the index of summation can be any variable name ( $i$ is just a typical choice). That is, $\displaystyle \sum_{j = 1}^n j^2$ , $\displaystyle \sum_{i = 1}^n i^2$ , and $\displaystyle \sum_{\text{zebra} = 1}^n \text{zebra}^2$ all represent the same sum.

Summation notation can be thought of in terms of a for-loop. In Python, to compute the sum $\displaystyle \sum_{i = a}^b i^2$ , we could write:

total = 0
for i in range(a, b + 1):
	total = total + i ** 2

As we mentioned above, the ending index is inclusive in summation notation. This is in contrast to Python, where the ending index is exclusive, which is why we provided b + 1 as the second argument to the range function instead of b.

Activity 0.1.1¶

Activity 0.1.1

Identify what makes each of the following expressions invalid or ambiguous, and why.

$\displaystyle \sum_{k = 1}^{13} k^2 + k$
$\displaystyle i \sum_{i = 1}^5 i$
$\displaystyle \sum_{i = 1}^n x_i + x_i$
$\displaystyle \sum_{i = 5}^3 i^2$
$\displaystyle \sum_{i = 2}^\pi x_i$

Properties and Examples¶

To illustrate various properties of summation notation, we’ll use the following fact:

\sum_{i = 0}^n 2^i = 2^0 + 2^1 + 2^2 + ... + 2^n = 2^{n+1} - 1

The expression $2^{n+1} - 1$ is called a closed form for the summation. When closed forms exist, they make it easy to compute the value of a summation. You’ll also notice that in addition to writing the sum using summation notation, I also showed the first few and last terms of the sum, with a ... to indicate that the pattern continues in between. As convenient as summation notation is, explicitly writing the first few and last terms of a sum can sometimes make it easier to understand what exactly is being summed.

For example, $\displaystyle \sum_{i = 0}^3 2^i = 2^0 + 2^1 + 2^2 + 2^3 = 1 + 2 + 4 + 8 = 15$ , which is indeed $2^{3+1} - 1 = 15$ . The sequence $2^0, 2^1, 2^2, \ldots, 2^n$ is called a geometric sequence, and the resulting sum is called a geometric series.

For convenience, we’ll define $\displaystyle S(n) = \sum_{i = 0}^n 2^i$ ; again, $S(n) = 2^{n+1} - 1$ .

The best way to learn through these examples is to try to solve them yourself before looking at the solution.

Example: Partial Sums¶

Determine the value of $\displaystyle \sum_{i = 8}^{19} 2^i$ . (Find an answer that doesn’t involve summation notation or a sum over many terms.)

Solution

We can’t use the formula for $S(n)$ directly, because the starting index is 8, not 0. But, if we look at the expansion of $S(19)$ , we’ll see that it contains the sum we’re looking for:

\begin{align*} S(19) &= \sum_{i = 0}^{19} 2^i \\ &= 2^0 + 2^1 + 2^2 + \cdots + 2^{19} \\ &= (2^0 + 2^1 + 2^2 + \cdots + 2^7) + (2^8 + 2^9 + 2^{10} + \cdots + 2^{19}) \\ &= \underbrace{\sum_{i = 0}^{7} 2^i}_{S(7)} \:\:\:\:\:\: + \underbrace{\sum_{i = 8}^{19} 2^i}_{\text{what we're looking for}} \end{align*}

So, the piece we’re looking for is $S(19) - S(7)$ , or:

\sum_{i = 8}^{19} 2^i = S(19) - S(7) = (2^{20} - 1) - (2^8 - 1) = 2^{20} - 2^8

The specific value of the sum, $2^{20} - 2^8$ , is $1048576 - 256 = 1048320$ , if you’re curious.

Example: Constant Multiples¶

Determine the value of $\displaystyle \sum_{i = 0}^{7} 5 \cdot 2^i$ .

Solution

\begin{align*} \sum_{i = 0}^{7} 5 \cdot 2^i &= 5 \cdot 2^0 + 5 \cdot 2^1 + 5 \cdot 2^2 + 5 \cdot 2^3 + 5 \cdot 2^4 + 5 \cdot 2^5 + 5 \cdot 2^6 + 5 \cdot 2^7 \\ &= 5 (2^0 + 2^1 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7) \\ &= 5 \sum_{i = 0}^{7} 2^i \\ &= 5 (2^{8} - 1) \end{align*}

This last line is what’s most important; its specific value is $5 \cdot (2^8 - 1) = 5 \cdot 255 = 1275$ .

Example: Sum of a Constant¶

Determine the value of $\displaystyle \sum_{i = 7}^{20} 5$ .

Solution

\begin{align*} \sum_{i = 7}^{20} 5 &= \underbrace{5 + 5 + 5 + \cdots + 5}_{14 \text{ times}} \\ &= 5 \cdot 14 \\ &= 70 \end{align*}

Where did 14 come from? The starting index is 7, and the ending index is 20. So, the number of terms is $20 - 7 + 1 = 14$ .

Example: Shifting Indices¶

Determine the value of $\displaystyle \sum_{i = 10}^{20} 2^{i - 5}$ .

Solution

As we did in the very first example, let’s try and write the sum as a difference of two calls to $S(n)$ . $S(n)$ involves the sum of terms of the form $2^i$ , and the sum we’re looking for involves the terms $2^{i - 5}$ , so we’ll need to shift the indices.

When $i = 10$ , $i - 5 = 5$ , and when $i = 20$ , $i - 5 = 15$ . So, we can rewrite the sum in question as:

\sum_{i = 10}^{20} 2^{i - 5} = \sum_{i = 5}^{15} 2^i

This looks like $S(15) - S(4)$ , or $2^{16} - 1 - (2^5 - 1) = 2^{16} - 2^5 = 65536 - 32 = 65504$ .

Example: Separating Sums¶

Given that $\displaystyle \sum_{i = 1}^{n} i = \frac{n(n+1)}{2}$ , determine the value of $\displaystyle \sum_{i = 0}^{12} (2^i + 5i)$ .

Solution

\begin{align*} \sum_{i = 0}^{12} (2^i + 5i) &= ({\color{3d81f6}{2^0}} + {\color{orange}{5 \cdot 0}}) + ({\color{3d81f6}{2^1}} + {\color{orange}{5 \cdot 1}}) + ({\color{3d81f6}{2^2}} + {\color{orange}{5 \cdot 2}}) + \ldots + ({\color{3d81f6}{2^{12}}} + {\color{orange}{5 \cdot 12}}) \\ &= ({\color{3d81f6}{2^0}} + {\color{3d81f6}{2^1}} + {\color{3d81f6}{2^2}} + \ldots + {\color{3d81f6}{2^{12}}}) + ({\color{orange}{5 \cdot 0}} + {\color{orange}{5 \cdot 1}} + {\color{orange}{5 \cdot 2}} + \ldots + {\color{orange}{5 \cdot 12}}) \\ &= \sum_{i = 0}^{12} {\color{3d81f6}{2^i}} + \sum_{i = 0}^{12} {\color{orange}{5i}} \\ &= 2^{13} - 1 + 5 \cdot \frac{12 \cdot 13}{2} \\ &= 8581 \end{align*}

The key here is that we can split the sum into two sums.

Key Takeaways¶

In the order they were introduced in the examples, here are some useful properties of summations:

Partial Sums:
$\sum_{i = 1}^{n} x_i = \sum_{i = 1}^{k} x_i + \sum_{i = k+1}^{n} x_i$
Constant Multiples:
$\sum_{i = 1}^{n} c x_i = c \sum_{i = 1}^{n} x_i$
Sum of a Constant:
$\sum_{i = 1}^{n} c = c \cdot n$
Shifting Indices:
$\sum_{i = k}^{n} x_i = \sum_{i = 0}^{n-k} x_{i+k}$
Separating Sums:
$\sum_{i = 1}^{n} (x_i + y_i) = \sum_{i = 1}^{n} x_i + \sum_{i = 1}^{n} y_i$

For more practice, work through Activity 0.1.2 (which has five parts).

Activity 0.1.2¶

Activity 0.1.2

Activity 2.1

Given that $\displaystyle \sum_{i = 1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$ , determine the value of $\displaystyle \sum_{j = 3}^{10} (j^2 - 7j)$ .

Activity 2.2

Show that $\displaystyle \sum_{n = 1}^{100} \frac{1}{n(n+1)} = 1 - \frac{1}{101}$ .

Hint: Try writing $\frac{1}{n(n+1)}$ as a difference of two fractions.

Activity 2.3

Recall, $n! = n \cdot (n-1) \cdot (n-2) \cdot \ldots \cdot 1$ .

The Taylor Series expansion for the function $f(x) = e^x$ , at the point $x = 0$ , is given by:

e^x = \sum_{n = 0}^{\infty} \frac{x^n}{n!}

Find a closed form expression for the infinite series:

\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \frac{1}{6!} - \frac{1}{7!} + \cdots

Activity 2.4

${n \choose k}$ , pronounced “n choose k”, is the number of ways to choose $k$ items from a set of $n$ items.

{n \choose k} = \frac{n!}{k!(n-k)!}

For example, ${5 \choose 2} = 10$ , because there are 10 ways to choose 2 items from a set of 5 items, and $\frac{5!}{2!(5-2)!} = \frac{120}{2 \cdot 6} = 10$ .

Using the fact introduced in Activity 2.1, find a closed form expression for the following sum:

\sum_{k = 2}^n {k \choose 2}

Activity 2.5

Argue why the following equality holds (it’ll be hard to prove this algebraically):

\sum_{k = 0}^n {n \choose k} = 2^n

Mean and Standard Deviation¶

We will frequently work with $n$ scalar^[1] samples $x_1, x_2, \ldots, x_n$ . The index might identify time nodes along a trajectory, wind-speed bins, road profiles, Monte Carlo realizations, or experimental trials.

Mean¶

The mean, or average, of all $n$ values is given the symbol $\bar{x}$ (pronounced “x-bar”) and is defined as follows:

\bar{x} = \frac{x_1 + x_2 + \ldots + x_n}{n} = \frac{1}{n} \sum_{i = 1}^n x_i

You’ve likely seen this definition before. But, an often-forgotten property of the mean is that the sum of the deviations from the mean is zero. By that, I mean (no pun intended) that if you:

compute the mean of a sequence of numbers,
compute the signed difference between each number and the mean, and then
sum all of those differences, the result will be zero.

Let’s first see this in action, then show why it is true in general. Suppose actuator force is sampled at four time nodes, giving 72, 63, 68, and 65 N. The mean force is:

\bar{x} = \frac{72 + 63 + 68 + 65}{4} = 67

The deviations from the mean are:

\begin{align*} 72 - 67 &= 5 \\ 63 - 67 &= -4 \\ 68 - 67 &= 1 \\ 65 - 67 &= -2 \end{align*}

The sum of the four deviations, then, is:

5 + (-4) + 1 + (-2) = 0

So, the mean deviation from the mean is zero in this example.

This is also true in general. Precisely, I’m claiming that if $x_1, x_2, ..., x_{n-1}, x_n$ are any $n$ numbers, and $\bar{x}$ is their mean, then $\displaystyle \sum_{i = 1}^n (x_i - \bar{x}) = 0$ .

Let’s prove it:

\begin{align*} \sum_{i = 1}^n (x_i - \bar{x}) &= \sum_{i = 1}^n x_i - \sum_{i = 1}^n \bar{x} \\ &= \sum_{i = 1}^n x_i - n \bar{x} \\ &= \sum_{i = 1}^n x_i - n \left( \frac{x_1 + x_2 + \ldots + x_n}{n} \right) \\ &= \sum_{i = 1}^n x_i - (x_1 + x_2 + \ldots + x_n) \\ &= 0 \end{align*}

So, the deviations sum to zero in general. The positive and negative deviations therefore cancel. In the actuator-force example, the positive deviations are 5 and 1 N and the negative deviations are -4 and -2 N. The mean can be viewed as the balance point of the sampled trajectory.

Standard Deviation¶

Because signed deviations always average to zero, they cannot measure how strongly a trajectory varies around its mean. A nearly constant response and a highly oscillatory response could both have zero mean deviation.

To measure the spread of sampled values, we remove the sign cancellation by squaring each deviation:

Compute the sample mean
Compute the deviation of each value from the mean
Square each deviation
Take the average of the squared deviations

The result of this process is called the variance, denoted $s^2$ or $\sigma^2$ ; its square root is called the standard deviation.

Following the above steps, the variance is given by:

\sigma^2 = \frac{1}{n} \sum_{i = 1}^n (x_i - \bar{x})^2

The same squared-error structure appears in ride-comfort objectives, tracking penalties, fatigue proxies, state-estimation covariances, and stochastic CCD formulations.

Activity 0.1.3¶

Activity 0.1.3

What is the variance of the actuator-force samples 72, 63, 68, and 65 N? What physical information is lost if only the mean is reported?

As a final note, the variance has a convenient, equivalent form:

\sigma^2 = \frac{1}{n} \sum_{i = 1}^n (x_i - \bar{x})^2 = \frac{1}{n} \sum_{i = 1}^n x_i^2 - \bar{x}^2

In English, we might say the variance is “the mean of the squares of $x$ minus the square of the mean of $x$ ”.

Your turn: prove this equivalent form of the variance.

Solution

\begin{align*} \sigma^2 &= \frac{1}{n} \sum_{i = 1}^n (x_i - \bar{x})^2 \\ &= \frac{1}{n} \sum_{i = 1}^n (x_i^2 - 2x_i\bar{x} + \bar{x}^2) \\ &= \frac{1}{n} \sum_{i = 1}^n x_i^2 - \frac{2}{n} \sum_{i = 1}^n x_i\bar{x} + \frac{1}{n} \sum_{i = 1}^n \bar{x}^2 \\ &= \frac{1}{n} \sum_{i = 1}^n x_i^2 - 2\bar{x} \frac{1}{n} \sum_{i = 1}^n x_i + \bar{x}^2 \\ &= \frac{1}{n} \sum_{i = 1}^n x_i^2 - 2\bar{x}^2 + \bar{x}^2 \\ &= \frac{1}{n} \sum_{i = 1}^n x_i^2 - \bar{x}^2 \end{align*}

RMS Metrics and Discretized Objectives¶

For a sampled signal, the root-mean-square value is

x_{\mathrm{RMS}} = \sqrt{\frac{1}{n}\sum_{i=1}^{n}x_i^2}.

Unlike standard deviation, RMS is measured relative to zero rather than relative to the sample mean. It therefore captures both a nonzero offset and oscillation. Ride acceleration, tracking error, actuator effort, voltage, and structural response are often summarized with RMS or related quadratic measures.

A time integral is also commonly replaced by a weighted sum on a mesh:

J = \int_{t_0}^{t_f}L(t)\,\mathrm{d}t \;\approx\; \sum_{k=0}^{N} w_k L(t_k).

The weights $w_k$ depend on the integration or transcription scheme. Consequently, changing the mesh without checking convergence can change the apparent CCD optimum.

Activity 0.1.4¶

Activity 0.1.4

Suppose a sampled sensor signal $x_1, x_2, \ldots, x_n$ has mean $\bar{x}$ and variance $\sigma^2$ .

A calibrated sensor reports $y_i=-4x_i+3$ . Find the mean and variance of $y_1, y_2, \ldots, y_n$ , and justify your answer using the definitions of mean and variance.

Footnotes¶

“Scalar” just means “individual number”, as opposed to a vector or matrix which can contain multiple numbers, as we’ll see later in the course.
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