Engineering Mathematics: Apr 2, 2011

Taylor and Maclaurin Series

TS.ID.1: Taylor Series Definition

Sigma representation

$\sum_{n=0} ^ {\infty } \frac {f^{(n)}(a)}{n!} \, (x-a)^{n}$

TS.ID.2: Maclaurin Series

TS.ID.3: Trigonometric Functions

$\sin x = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} x^{2n+1} = x - \frac{x^3}{3!} + \frac{x^5}{5!}- \cdots\text{ for all } x$

proof

$\cos x = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n)!} x^{2n} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots\text{ for all } x$
proof

$\tan x = \sum^{\infty}_{n=1} \frac{B_{2n} (-4)^n (1-4^n)}{(2n)!} x^{2n-1} = x + \frac{x^3}{3} + \frac{2 x^5}{15} + \cdots\text{ for }|x| < \frac{\pi}{2}$

proof

TS.ID.4: Hyberbolic Functions

$\sinh x = \sum^{\infty}_{n=0} \frac{x^{2n+1}}{(2n+1)!} = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots\text{ for all } x\$

proof

$\cosh x = \sum^{\infty}_{n=0} \frac{x^{2n}}{(2n)!} = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots\text{ for all } x$

proof

$\tanh x = \sum^{\infty}_{n=1} \frac{B_{2n} 4^n (4^n-1)}{(2n)!} x^{2n-1} = x-\frac{1}{3}x^3+\frac{2}{15}x^5-\frac{17}{315}x^7+\cdots \text{ for }|x| < \frac{\pi}{2}$

proof

TS.ID.5: Geometric Series

$\frac{1}{1-x}=\sum _{n=0}^{\infty }x^{n}=1+x+x^{2}+\cdot \cdot \cdot +x^{n}$

TS.ID.6: Logarithemtic Function

$\ln (1-x)=-\sum _{n=0}^{\infty }\frac{x^{n}}{n}=-(1+x+\frac{x^{2}}{2}+\cdot \cdot \cdot +\frac{x^{n}}{n})$

Exponential Function

E.ID.1: Taylor Series

$e^x\ =\ 1\ +\ x\ +\ \frac{x^2}{2} +\ \frac{x^3}{6} +\ \frac{x^4}{24} +\ \frac{x^5}{120}\ \dots\ = \sum_{n\,=\,0}^{\infty}\frac{x^n}{n!}$

E.ID.2: Euler's Formula

$e^{ix}\ =\ cosx\ +\ i sinx$

$cosx\ =\ \frac{1}{2}\left(e^{ix}\ +\ e^{-ix}\right)$

$i sinx\ =\ \frac{1}{2}\left(e^{ix}\ -\ e^{-ix}\right)$

E.ID.3: Bernouli's Formula

$e^x\ =\ \lim_{n\rightarrow\infty}\left(1\ +\ \frac{x}{n}\right)^n$

Proof

power expansion

$\left ( 1+\frac{x}{n} \right )^{n}=1+\binom{n}{1}\frac{x}{n}+\binom{n}{2}\frac{x^{2}}{n^{2}}+\binom{n}{3}\frac{x^{3}}{n^{3}}+\cdot \cdot \cdot +\binom{n}{n}\frac{x^{n}}{n^{n}}$

$\left ( 1+\frac{x}{n} \right )^{n}=1+\frac{n!}{1!(n-1)!}\frac{x}{n}+\frac{n!}{2!(n-2)!}\frac{x^{2}}{n^{2}}+\frac{n!}{3!(n-3)!}\frac{x^{3}}{n^{3}}+\cdot \cdot \cdot +\frac{n!}{n!0!}\frac{x^{n}}{n^{n}}$
after dividing

$\left ( 1+\frac{x}{n} \right )^{n}=1+\frac{n}{1!}\frac{x}{n}+\frac{n(n-1)}{2!}\frac{x^{2}}{n^{2}}+\frac{n(n-1)(n-2)}{3!}\frac{x^{3}}{n^{3}}+\cdot \cdot \cdot +\frac{n(n-1)(n-2)\cdot \cdot \cdot \cdot 1}{n!}\frac{x^{n}}{n^{n}}$

again , after dividing

$\left ( 1+\frac{x}{n} \right )^{n}=1+x+\frac{1(1-\frac{1}{n})}{2!}x^{2}+\frac{1(1-\frac{1}{n})(1-\frac{2}{n})}{3!}x^{3}+\cdot \cdot \cdot +\frac{1(1-\frac{1}{n})(1-\frac{2}{n})\cdot \cdot \cdot \cdot \frac{1}{n}}{n!}x^{n}$

$n \to \infty$

$\left ( 1+\frac{x}{n} \right )^{n}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\cdot \cdot \cdot +\frac{x^{n}}{n!}=e^{x}$
E.ID.4: Logarithms

$e^{ln(x)}\ =\ x$

$a^x\ =\ \left(e^{ln(a)}\right)^x\ =\ e^{x\,ln(a)}$

$y\ =\ a^x \Leftrightarrow\ x\ =\ log_a(y)\ \equiv\ \frac{ln(y)}{ln(a)}$

$\frac{da^x}{dx}\ =\ ln(a)\,e^{x\,ln(a)}\ =\ ln(a)\,a^x$

E.ID.5: Hyberbolic Functions

$e^{x}\ =\ coshx\ +\ sinhx$

$coshx\ =\ \frac{1}{2}\left(e^{x}\ +\ e^{-x}\right)$

$sinhx\ =\ \frac{1}{2}\left(e^{x}\ -\ e^{-x}\right)$

$tanhx\ =\ \frac{sinhx}{coshx}\ =\ \frac{e^x\ -\ e^{-x}}{e^x\ +\ e^{-x}}$

$tanh\frac{1}{2}x\ =\ \frac{e^x\ -\ 1}{e^x\ +\ 1}$

$e^x\ =\ \frac{1\ +\ tanh\frac{1}{2}x}{1\ -\ tanh\frac{1}{2}x}$

Geometric Series

GS.ID.1: Geometric Series Formula

$a + ar + a r^2 + a r^3 + \cdots + a r^{n-1} = \sum_{k=0}^{n-1} ar^k= a \, \frac{1-r^{n}}{1-r}$

Proof

$s=a + ar + a r^2 + a r^3 + \cdots + a r^{n-1}$

Multiply s with r , therefore

$rs=ar + ar^2 + a r^3 + a r^4 + \cdots + a r^{n}$

Subtract rs from s , therefore

$s-rs = s(1-r) = a-ar^n$

therefore,

$s= a\frac{1-r^{n}}{1-r}$

As n goes to infinity, the absolute value of r must be less than one for the series to converge, therefore ,

$s= a\frac{1}{1-r}$

GS.ID.2: Geometric Series of Exponential

$S= e^{-p}+e^{-2p}+e^{-3p}+e^{-4p}+\cdots+e^{-np} = \sum_{k=1}^ne^{-kp}$

As n goes to infinity, the absolute value of e converges

$S=\frac{e^{-p}}{1-e^{-p}}=\frac{1}{e^p-1}$

GS.ID.3: Geometric Series Derivatives

$\sum _{k=0}^{\infty }k^{2}q^{k-1}=\frac{2q}{(1-q)^{3}}+\frac{1}{(1-q)^{2}}$

Given

$\sum _{k=0}^{\infty }q^{k}=\frac{1}{1-q}$

(1)

First derivative

$\sum _{k=0}^{\infty }kq^{k-1}=\frac{1}{(1-q)^{2}}$

(2)

Second derivative

$\sum _{k=0}^{\infty }k(k-1)q^{k-2}=\frac{2}{(1-q)^{3}}$

(3)

$\sum _{k=0}^{\infty }k(k-1)q^{k-2}=\sum _{k=0}^{\infty }k^{2}q^{k-2}-\sum _{k=0}^{\infty }kq^{k-2}$

(4)

$\sum _{k=0}^{\infty }k(k-1)q^{k-2}=\frac{1}{q}\sum _{k=0}^{\infty }k^{2}q^{k-1}-\frac{1}{q}\sum _{k=0}^{\infty }kq^{k-1}$

(5)

from (3) and (5)

$\frac{2}{(1-q)^{3}}=\frac{1}{q}\sum _{k=0}^{\infty }k^{2}q^{k-1}-\frac{1}{q}\sum _{k=0}^{\infty }kq^{k-1}$

(6)

from (2) and (6)

$\frac{2}{(1-q)^{3}}=\frac{1}{q}\sum _{k=0}^{\infty }k^{2}q^{k-1}-\frac{1}{q}\frac{1}{(1-q)^{2}}$

(7)

$\frac{1}{q}\sum _{k=0}^{\infty }k^{2}q^{k-1}=\frac{2}{(1-q)^{3}}+\frac{1}{q}\frac{1}{(1-q)^{2}}$

(8)

$\sum _{k=0}^{\infty }k^{2}q^{k-1}=\frac{2q}{(1-q)^{3}}+\frac{1}{(1-q)^{2}}$

Bernoulli polynomial

1. Introduction

The generating function of Bernoulli polynomial is defined as:

When x =0, (known as Bernoulli number)

When x =1,

2. Identities

2.1. Theorem of complement

Proof

(1)

(2)

(3)

from (1) and (3),

(4)

therefore,

2.2. Theorem of arguments addition

Proof

$\sum _{n=0}^{\infty }B_{n}(x+y)\frac{t^{n}}{n!}= \frac{te^{t(x+y)}}{e^{t}-1}= \frac{te^{tx}e^{ty}}{e^{t}-1}$

(1)

from E.ID.1

$\frac{te^{tx}}{e^{t}-1}e^{ty}=(\sum _{s=0}^{\infty }B_{s}(x)\frac{t^{s}}{s!})(\sum _{r=0}^{\infty }\frac{t^{r}y^{r}}{r!})$

(2)

the convolution of the two series is (SC.ID.2)

$(\sum _{s=0}^{\infty }B_{s}(x)\frac{t^{s}}{s!})(\sum _{r=0}^{\infty }\frac{t^{r}y^{r}}{r!}) = \sum _{n=0}^{\infty}\sum _{s=0}^{n }\binom{n}{s}B_{s}(x)y^{n-s}\frac{t^{n}}{n!}$

(3)

from (1) and (3),

$\sum _{n=0}^{\infty}B_{n}(x+y)\frac{t^{n}}{n!} = \sum _{n=0}^{\infty}\sum _{s=0}^{n }\binom{n}{s}B_{s}(x)y^{n-s}\frac{t^{n}}{n!}$

(4)

therefore,

2.3. Theorem of multiplication

$\frac{1}{m}\sum _{k=0}^{m-1}B_{n}(z+\frac{k}{m})=m^{-n}B_{n}(mz)$

Proof

to prove the above identity we use the following geometric identity (GS.ID.1):

$\frac{1}{e^{t}-1}= \frac{1+e^{t}+\cdot \cdot \cdot +e^{(m-1)t}}{e^{mt}-1}$

therefore,

$\sum _{n=0}^{\infty}B_{n}(mz)\frac{t^{n}}{n!}=\frac{te^{mtz}}{e^{t}-1}=\frac{te^{mtz}(1+e^{t}+\cdot \cdot \cdot +e^{(m-1)t})}{e^{mt}-1}$

(1)

multiplying and dividing by m , therefore

$\frac{te^{mtz}}{e^{t}-1}=\frac{1}{m}\frac{mte^{mtz}(1+e^{t}+\cdot \cdot \cdot +e^{(m-1)t})}{e^{mt}-1}$

(2)

$\frac{1}{m}\frac{mte^{mtz}(1+e^{t}+\cdot \cdot \cdot +e^{(m-1)t})}{e^{mt}-1}= \frac{1}{m}\sum _{k=0}^{m-1}\frac{mte^{mt(z+\frac{k}{m})}}{e^{mt}-1}$

(3)

$\frac{1}{m}\sum _{k=0}^{m-1}\frac{mte^{mt(z+\frac{k}{m})}}{e^{mt}-1}= \frac{1}{m}\sum _{k=0}^{m-1}\sum _{n=0}^{\infty}B_{n}(z+\frac{k}{m})\frac{m^{n}t^{n}}{n!}$

(4)

from (1) and (4),

$\sum _{n=0}^{\infty}B_{n}(mz)\frac{t^{n}}{n!}= \frac{1}{m}\sum _{k=0}^{m-1}\sum _{n=0}^{\infty}B_{n}(z+\frac{k}{m})\frac{m^{n}t^{n}}{n!}$

(5)

therefore,

$\frac{1}{m}\sum _{k=0}^{m-1}B_{n}(z+\frac{k}{m})=m^{-n}B_{n}(mz)$

2.4. Theorem of hyperbolic-Bernoulli function

$\frac{t}{e^{t}-1}=\frac{t}{2}\coth (\frac{t}{2})-\frac{t}{2}$

proof

$\frac{t}{e^{t}-1}= \sum _{n=0}^{\infty}B_{n}\frac{t^{n}}{n!}$

$\frac{t}{e^{t}-1}= \frac{t}{e^{t/2}(e^{t/2}-e^{-t/2})}$

$\frac{t}{e^{t/2}(e^{t/2}-e^{-t/2})} = \frac{te^{-t/2}}{2\sinh (\frac{t}{2})}$

$\frac{te^{-t/2}}{2\sinh (\frac{t}{2})} = \frac{t}{2}\frac{\cosh (\frac{t}{2})-\sinh (\frac{t}{2})}{\sinh (\frac{t}{2})}$

$\frac{t}{2}\frac{\cosh (\frac{t}{2})-\sinh (\frac{t}{2})}{\sinh (\frac{t}{2})}=\frac{t}{2}\coth (\frac{t}{2})-\frac{t}{2}$

therefore;

$\frac{t}{e^{t}-1}=\frac{t}{2}\coth (\frac{t}{2})-\frac{t}{2}$

2.5. Theorem of exponential-Bernoulli function

based on (UI.ID.1)

$\frac{t}{e^{t}+1}=\frac{t}{e^{t}-1} -\frac{2t}{e^{2t}-1}$

and based on

therefore,

$\frac{t}{e^{t}+1}=-\sum _{n=0}^{\infty }(2^{n}-1)B_{n}(0)\frac{t^{n}}{n!}$

2.5. Theorem of Bernoulli polynomial in term of Bernoulli number

$B_{n}(x)=\sum _{k=0}^{n}\binom{n}{k}B_{k}x^{n-k}$

Proof

$\sum _{n=0}^{\infty }B_{n}(x)\frac{t^{n}}{n!}=\frac{te^{xt}}{e^{t}-1}=e^{xt}\frac{t}{e^{t}-1}$

(1)

$e^{xt}\frac{t}{e^{t}-1}=\left ( \sum _{n=0}^{\infty }x^{n}\frac{t^{n}}{n!} \right )\left (\sum _{k=0}^{\infty }B_{n}(0)\frac{t^{k}}{k!} \right )$

(2)

based on (SC.ID.2)

$e^{xt}\frac{t}{e^{t}-1}=\sum _{n=0}^{\infty }\left ( \sum _{k=0}^{n}\binom{n}{k}B_{k}x^{n-k} \right )\frac{t^{n}}{n!}$

(3)

from (1) and (3) , therefore

$\sum _{n=0}^{\infty }B_{n}(x)\frac{t^{n}}{n!}=\sum _{n=0}^{\infty }\left ( \sum _{k=0}^{n}\binom{n}{k}B_{k}x^{n-k} \right )\frac{t^{n}}{n!}$

equating the similar coefficients

$B_{n}(x)=\sum _{k=0}^{n}\binom{n}{k}B_{k}x^{n-k}$

2.6. Theorem of sum of nth powers

$\sum _{n=0}^{\infty }\left (B_{n}(k)-B_{n}(0) \right )\frac{t^{n}}{n!}=\frac{te^{xt}}{e^{t}-1}-\frac{t}{e^{t}-1}=\frac{t(e^{xt}-1)}{e^{t}-1}$

(1)

based on geometric series of exponential and taylor series of exponential function (GS.ID.2) (E.ID.1)

$\frac{t(e^{xt}-1)}{e^{t}-1}=t\sum _{j=0}^{k-1}e^{tj}=t\sum_{n=0}^{\infty } \left ( \sum _{j=0}^{k-1}j^{n} \right )\frac{t^{n}}{n!}$

(2)

from (1) and (2)

$B_{n}(k)-B_{n}(0)=t\sum _{j=0}^{k-1}j^{n}$

(3)

from the definitions of Bernoulli polynomials and number

$t\sum _{n=0}^{\infty }\left (\frac{ B_{n+1}(k)-B_{n+1}(0)}{n+1} \right )\frac{t^{n}}{n!}=\sum _{n=0}^{\infty }\left ( B_{n}(k)-B_{n}(0) \right )\frac{t^{n}}{n!}$