Calculus: Difference between revisions

Revision as of 00:46, 30 November 2014

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	$$= \sum _{ {n=0} }^{\infty }x^{n}$$	$$= 1+x+x^{2}+x^{3}+\cdots \!$$

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

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 |$$= 1+x+x^{2}+x^{3}+\cdots \!$$
 |
+|}
+=== Summation ===
+{| class=wikitable width=100%
+|-
+|$$\sum_{i=1}^n i^2$$
+|$$\frac{1}{6}n(2n-1)(n-1)$$
+|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
 |}