Calculus: Difference between revisions
Jump to navigation
Jump to search
(→Series) |
|||
(One intermediate revision by the same user not shown) | |||
Line 25: | Line 25: | ||
|$$\sum_{i=1}^n i^2$$ |
|$$\sum_{i=1}^n i^2$$ |
||
|$$\frac{1}{6}n(2n-1)(n-1)$$ |
|$$\frac{1}{6}n(2n-1)(n-1)$$ |
||
|See [http://polysum.tripod.com/Alternative.pdf]. Basically, develop \( |
|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 |
||
|} |
|} |
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 |