graphing software

A graphing software is available at padowan.dk

Another software is winplot available at math.exeter.edu/rparris/

An interactive geometry graphing system where input can be in the form of equations or co ordinates which can be manipulated even after the graph has been drawn is available at geogebra.org

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proving a trigonometry identity

show that [( tanx -sinx ) / (2tanx) ] = sin² (x/2)
divide term by term and change
every thing in terms of sinx and cosx and
then use half angle formulae
use this link for trigonometric identities formulae
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trigonometric identities

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Integration formulae

note that ln(x) refers to natural log, that is log to the base e
Integration formulae of functions like
x^n , e^(ax ) exp(ax), sin(ax), cos(ax), tan(x), 1 / sqrt(a^2 - x^2), 1 / (a^2 - x^2) ...

integration guide on certain topics

PAGE 1 BASIC INTEGRATION
 
PAGE 2 INTEGRATION BY SUBSTITUTION

 PAGE 3 INTEGRATION BY COMPLETION OF SQUARES
 
PAGE 4 INTEGRATION BY PARTS
 
PAGE 5 INTEGRATION BY MANIPULATION OF 
NUMERATOR IN TERMS OF DENOMINATOR

PAGE 6 INTEGRATION USING PARTIAL FRACTIONS
 
PAGE 7 INTEGRATION OF 1 / [ACOSX +BSINX +C]


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equation of tangent at (e,e) on y = x lnx

y = x lnx

dy/dx = x(1/x) + lnx (1) = 1 + lnx

slope of tangent , m = (dy/dx) at (e,e) = 1 + ln(e) = 1 + 1 = 2

equation of tangent at (e,e) is
(y - y1) = m (x-x1)

y - e = 2(x-e)

or y = 2x - e
other examples
equation of tangent to y =x^4 at the point where x=2


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problem on automorphism in a group

If G is a group and f :G -->G is defined as f(x) = a ֿ¹ x a for x in G and a is a fixed element of G
then show that f is an isomorphism of G onto G
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sum of n terms of a geometric progression

find the nth term of the series 1 + 2 + 4 + 8 + 16 + ...
Also find the sum of the first n terms

The sequence 1, 2 , 4 , 8 , 16 , ... is a geometric progression (G.P.) with
first term a=1 and common ratio, r = 2
since 2/1 = 4/2 = 8/4 =16/8 = ...

nth term = a {r ^ ( n-1) } = 2^(n-1)

try the following page for formulae on sequences and series AP and GP
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idempotent law in a boolean algebra

show that a + a = a in a boolean algebra

a = a + 0
= a + (a. a' )
= (a + a ). (a + a ' )
= (a + a ). 1
= (a + a )

therefore a + a = a
note that the dual is a . a = a

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boundedness law in a boolean algebra

Show that a +1 = 1 in a boolean algebra

1 = a + a'
= a + (a' .1 )
= (a + a' ) . ( a+1) distributive law
= 1 . ( a+1)
=a +1

therefore a + 1 = 1

note that the dual is a . 0 = 0

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problem on integration using trigonometry formulae

evaluate integral of sin(5x)sin(8x)
first use the trigonometric formulae for sinAsinB to split the integral into two terms
it is easy to use the bigger angle as A
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telescoping series

simplify 1/12 + 1/20 +1/30 +1/42 +1/9900 using partial fraction
try to express in the form 1 / n(n+1)
use partial fraction to split into two terms
then sum
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trigonometric identities

trigonometric identities
addition formulae , double angle formulae, half angle formulae , formulae for removing square and cube from sin and cos



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integration by trigonometric substitution

evaluate integral of 1 /sqrt(1-9x^2) using trigonometric substitution


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real and imaginary parts of ln(x+iy)

find the real and imaginary parts of ln(x+iy) or logarithm of a complex number,
first express x+iy in modulus amplitude form
then take log

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basic result of complex number for plus two student

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to construct truth tables for [~ (p ^ q ) ]v (~r)

since there are three variables there will be 2^3 =8 rows
start the first column with 4 T's and 4F's
second column with 2 T's and 2F's ...
third column T and F alternatively
then use the definition of ^ , ~ to get the truth table of the components and finally that of the question
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solving a fourth degree equation given one of its complex roots

solve 2x^4-8x^3+19x^2-22x+5=0 given that 1+2i is one of its roots

complex roots occur in conjugate pairs if the coefficients are real
so 1-2i is also a root
{x - (1+2i)}{x - (1-2i)} is a factor
multiply into a quadratic
divide the given equation by this quadratic factor to extract the other quadratic factor and solve that


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problem on sequences

If Sn is the sum of the cubes of the first n natural numbers, find the sum of 1 / sqrt(Sn) where n varies from 1 to 2000
first use the formula for sum of the cubes of the first n natural numbers
take the square root and the reciprocal then adjust into a telescoping series


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