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).
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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 |
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Practice
Unless otherwise instructed, evaluate these sums. Give your answers in exact form.
Evaluate \(\displaystyle{ \sum_{n=1}^{5}{ (2n+3) } }\)
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Evaluate \(\displaystyle{ \sum_{n=1}^{5}{ (2n+3) } }\)
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2590 video
video by The Organic Chemistry Tutor |
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Evaluate \(\displaystyle{ \sum_{n=1}^{6}{ n^2 } }\)
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Evaluate \(\displaystyle{ \sum_{n=1}^{6}{ n^2 } }\)
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2591 video
video by The Organic Chemistry Tutor |
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Evaluate \(\displaystyle{ \sum_{k=0}^{4}{ \frac{1}{k^2+1} } }\)
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Evaluate \(\displaystyle{ \sum_{k=0}^{4}{ \frac{1}{k^2+1} } }\)
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158/85
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Evaluate \(\displaystyle{ \sum_{k=0}^{4}{ \frac{1}{k^2+1} } }\)
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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
video by The Organic Chemistry Tutor |
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Final Answer |
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158/85
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Evaluate \(\displaystyle{ \sum_{n=1}^{3}{ 2^n } }\)
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Evaluate \(\displaystyle{ \sum_{n=1}^{3}{ 2^n } }\)
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2593 video
video by The Organic Chemistry Tutor |
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Evaluate \(\displaystyle{\sum_{k=1}^{6}{\frac{1}{k^2}}}\)
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Evaluate \(\displaystyle{\sum_{k=1}^{6}{\frac{1}{k^2}}}\)
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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.
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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.
Solution |
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880 video
video by Krista King Math |
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Evaluate \(\displaystyle{\sum_{r=1}^{8}{(r-1)(r+2)}}\)
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Evaluate \(\displaystyle{\sum_{r=1}^{8}{(r-1)(r+2)}}\)
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881 video
video by Krista King Math |
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Evaluate \(\displaystyle{\sum_{i=1}^{4}{(-1)^i i^2}}\)
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Evaluate \(\displaystyle{\sum_{i=1}^{4}{(-1)^i i^2}}\)
Solution |
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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
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
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Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
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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Practice Instructions
Unless otherwise instructed, evaluate these sums. Give your answers in exact form.
