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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|$$\sum_{i=1}^n i^2$$
|$$\frac{1}{6}n(2n-1)(n-1)$$
|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