Calculus: Difference between revisions
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(Taylor and McLaurin series - sqrt(1+x), (1+x)^(-1)) |
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|$$= 1+x+x^{2}+x^{3}+\cdots \!$$ |
|$$= 1+x+x^{2}+x^{3}+\cdots \!$$ |
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=== Summation === |
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|$$\sum_{i=1}^n i^2$$ |
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|$$\frac{1}{6}n(2n-1)(n-1)$$ |
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|See [http://polysum.tripod.com/Alternative.pdf]. Basically, develop \(\sum_{i=1}^n (i^3+3i^2+3i+1)=\sum_{i=1}^n (i+1)^3=\sum_{i=2}^{n+1} i^3=\sum_{i=1}^n i^3+(n+1)^3-1\), then isolate and simplify |
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Latest revision as of 00:47, 30 November 2014
Series
Taylor and McLaurin Series
Frequently used series:
| $$\sqrt{1+x}$$ | 0 ≤ x ≤ 1 | $$= \sum _{ {n=0} }^{\infty }{\frac {(-1)^{n}(2n)!}{(1-2n)(n!)^{2}(4^{n})}}x^{n}$$ | $$= 1+\textstyle {\frac {1}{2}}x-{\frac {1}{8}}x^{2}+{\frac {1}{16}}x^{3}-{\frac {5}{128}}x^{4}+\dots ,\!$$ | [1] |
| $$\frac{1}{1+x}$$ | -1 < x < 1 | $$= \sum _{ {n=0} }^{\infty }x^{n}$$ | $$= 1+x+x^{2}+x^{3}+\cdots \!$$ |
Summation
| $$\sum_{i=1}^n i^2$$ | $$\frac{1}{6}n(2n-1)(n-1)$$ | See [2]. Basically, develop \(\sum_{i=1}^n (i^3+3i^2+3i+1)=\sum_{i=1}^n (i+1)^3=\sum_{i=2}^{n+1} i^3=\sum_{i=1}^n i^3+(n+1)^3-1\), then isolate and simplify |