Engineering Mathematics

Sunday, April 17, 2011

examples of Geometric Series

Perquisites :

1. Geometric Series

Example (1)

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if x=1

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if x=-1

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Saturday, April 16, 2011

Cauchy Product (1 − 1 + 1 − 1 + ... )

Perquisites :

1. Series convolution (Cauchy product)

given

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Cauchy product

$\begin{array}{rcl} c_n & = &\displaystyle \sum_{k=0}^n a_k b_{n-k}=\sum_{k=0}^n (-1)^k (-1)^{n-k} \\[1em] & = &\displaystyle \sum_{k=0}^n (-1)^n = (-1)^n(n+1). \end{array}$

then, the product is

$\sum_{n=0}^\infty(-1)^n(n+1) = 1-2+3-4+\cdots.$

therefore,

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(1)

based on geometric series (GS.ID.1)

$\sum_{k=0}^{n} a r^k = \frac{a}{1-r}.$

(2)

a=1 , r=-1 in the following equation

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(3)

from (2) and (3)

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(4)

from (1) and (4)

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Monday, April 11, 2011

Riemann zeta function

firstly , I recommend you to read these Riemann zeta function , Euler product to have fundamental understanding of the Riemann zeta function.

RZ.ID.1:

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p is a prime number

Proof

given

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(1)

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(2)

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(3)

cancelling out the similarities, therefore

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(4)

RZ.ID.2:

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Proof

derived by multiplying (1) and (4)

RZ.ID.3:

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proof:

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by differentiating the above equation

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RZ.ID.4:

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Proof

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(1)

replacing

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therefore,

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(2)

from (1) and (2)

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multiplying and dividing by -1

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Sunday, April 10, 2011

Bernoulli-Zeta functions

BZ.ID.1: Bernoulli-Zeta relation

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Proof

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(1)

based on the following identity

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(2)

substituting (2) into (1)

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(3)

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(4)

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derivative of constant , which is 1, equals = 0

(5)

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(6)

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(7)

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(8)

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if=k=2k

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if n=k+1

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Wednesday, April 6, 2011

Cauchy Product (Series convolution )

This post will help students understand the derivation of harmonic series through Cauchy Product

Given identities:

Logarithmic Function

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(1)

Geometric Series

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(2)

Harmonic Series

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(3)

product of (1) and (2)

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based on (SC.ID.1)

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therefore,

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therefore,

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(4)

if x=1/2

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Integration of (4)

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