## 17Calculus - Sigma Notation

Before working with definite integrals and infinite series, you need to understand sigma notation. Sigma notation is just a compact way to write sums, i.e. adding a sequence of numbers or terms.
A term using sigma notation is called a series or summation (sum for short).

For example, if we have $$1 + 2 + 3 + 4 + 5$$, we can write this as $$\displaystyle{ \sum_{i=1}^{5}{i} }$$. Let's look at each part of this notation.
1. First, $$\sum$$ just means to add. This is the upper-case Greek letter sigma.
2. The letter i is called the index. It is usually an integer. You will also see other variables used here like n or k.
3. The assignment $$i=1$$ under the sigma gives you the starting value of the index and the index letter.
4. The number above the sigma tells you the ending value of the index. We don't need to repeat 'i=' here. The index is incremented by 1 until it reaches in the ending value. It includes the ending value. For infinite series, you will find $$\infty$$ above the sigma.
5. In this sum, we just have i, so we iterate over the index and add up the results.

In general, we add up more interesting things than just integers. So you will sometimes see the notation $$\displaystyle{ \sum_{i=1}^{n}{a_i} }$$ where $$a_i$$ is some term involving the index i.

Before we go on, let's watch a video. This is a good overview of sigma notation. In this video, he starts by explaining the general notation and then he works several examples.

### PatrickJMT - Summation Notation [10min-15secs]

video by PatrickJMT

Practice

Unless otherwise instructed, evaluate these sums. Give your answers in exact form.

Evaluate $$\displaystyle{ \sum_{n=1}^{5}{ (2n+3) } }$$

Problem Statement

Evaluate $$\displaystyle{ \sum_{n=1}^{5}{ (2n+3) } }$$

Solution

### 2590 video

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Evaluate $$\displaystyle{ \sum_{n=1}^{6}{ n^2 } }$$

Problem Statement

Evaluate $$\displaystyle{ \sum_{n=1}^{6}{ n^2 } }$$

Solution

### 2591 video

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Evaluate $$\displaystyle{ \sum_{k=0}^{4}{ \frac{1}{k^2+1} } }$$

Problem Statement

Evaluate $$\displaystyle{ \sum_{k=0}^{4}{ \frac{1}{k^2+1} } }$$

158/85

Problem Statement

Evaluate $$\displaystyle{ \sum_{k=0}^{4}{ \frac{1}{k^2+1} } }$$

Solution

A minute or two into this video he realizes that he didn't write down one of the terms and he posts a note about it. Even though the final answer in the video is incorrect, his procedure is good to watch.

### 2592 video

158/85

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Evaluate $$\displaystyle{ \sum_{n=1}^{3}{ 2^n } }$$

Problem Statement

Evaluate $$\displaystyle{ \sum_{n=1}^{3}{ 2^n } }$$

Solution

### 2593 video

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Evaluate $$\displaystyle{\sum_{k=1}^{6}{\frac{1}{k^2}}}$$

Problem Statement

Evaluate $$\displaystyle{\sum_{k=1}^{6}{\frac{1}{k^2}}}$$

Solution

### 879 video

video by Krista King Math

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Write$$\displaystyle{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+}$$ $$\displaystyle{\frac{1}{16}+}$$ $$\displaystyle{\frac{1}{32}+}$$ $$\displaystyle{\frac{1}{64}}$$ in sigma notation.

Problem Statement

Write$$\displaystyle{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+}$$ $$\displaystyle{\frac{1}{16}+}$$ $$\displaystyle{\frac{1}{32}+}$$ $$\displaystyle{\frac{1}{64}}$$ in sigma notation.

Solution

### 880 video

video by Krista King Math

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Evaluate $$\displaystyle{\sum_{r=1}^{8}{(r-1)(r+2)}}$$

Problem Statement

Evaluate $$\displaystyle{\sum_{r=1}^{8}{(r-1)(r+2)}}$$

Solution

### 881 video

video by Krista King Math

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Evaluate $$\displaystyle{\sum_{i=1}^{4}{(-1)^i i^2}}$$

Problem Statement

Evaluate $$\displaystyle{\sum_{i=1}^{4}{(-1)^i i^2}}$$

Solution

### 882 video

video by PatrickJMT

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You CAN Ace Calculus

### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia] Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

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