Hermite Polynomials

The Hermite polynomials which are a classical orthogonal polynomial sequence arises in mathematics in probability, such as the Edgeworth series; in combinatorials, as an example of an Appell sequence, obeying the umbral calculus; and in physics, where they give rise to the eigenstates of the quantum harmonic oscillator. They are named in honor of Charles Hermite.

Definition

Hn(x) = (-1)nex2/2 [`d/dx`(e-x2/2)]n

(the "probabilists' Hermite polynomials"), or sometimes by

Hn(x) =(-1)nex2 [`d/dx`(e-x2)]n

(the "physicists' Hermite polynomials"). The above two definitions are not exactly equivalent; either is a rescaling of the other, to wit

Hnphys(x) = 2n/2Hnprob(`sqrt(2)`x )

Usually follow the first convention is followed. That convention is often preferred by probabilists because

`1/sqrt(2Pi)` e-x2/2

is the probability density function for the normal distribution in which expected value 0 and standard deviation 1.

Properties of Hermite polynomials

Hn is a polynomial of degree n. According to probabilists' definition it has leading coefficient 1, while according to physicists' definition it has leading coefficient 2n.

Orthogonality

Hn(x) is an nth-degree polynomial for n = 0, 1, 2, 3, .... These polynomials show orthogonality with respect to the weight function     (measure)

W(x) = e-x2/2   (probabilist)

or

W(x) = e-x2   (physicist)

Completeness

The Hermite polynomials (probabilist or physicist) form the basis of orthogonality of the Hilbert space of functions satisfying
`int_-oo^oo|f(x)|^2w(x)dx` < `oo`


Hermite's differential equation

The probabilists' version of Hermite polynomials are solutions of the differential equation

(e-x2/2 u')' + `lambda`e-x2/2 u = 0

Recursion relation

Hermite polynomials' sequences also satisfies the recursion

Hn+1(x) = xHn(x) - H'n(x) (probabilist)

Hn+1(x) = 2xHn(x) - H'n(x) (physicist)


Applications of Hermite polynomials

Hermite functions

One can define the Hermite functions from the physicists' version of polynomials:

`psi`n(x)  =   `1/sqrt(n!2^nsqrt(Pi))`  e-x2/2 Hn(x)


Recursion relation

Following recursion relations of Hermite polynomials, the Hermite functions satisfy

`psi`'n(x) = `sqrt(n/2)``psi`n-1(x) - `sqrt((n+1)/2)``psi`n+1(x)

Cramér's inequality

The Hermite functions satisfy the following inequality due to Harald Cramér

|`psi`n(x)|   `<=` K

for x real, where the constant K is less than 1.086435.

Hermite functions act as eigenfunctions of the Fourier transform

Area using Perimeter

Area:
Area is a measure of  the outermost surface of an object. Area is a bounded space with its surrounding. Area is the  measure of  plane region or  amount at surface which any object occupies is called the area for that object. Area for any object is a measure ofbounded surface to its bounded space

Perimeter:
The length of the boundary of any closed figure is called its perimeter. The distance around a two-dimensional shape,for a circle perimeter is a circumference.closed surface for inner space occupied it said to be the perimeter for an object.

Finding Area using Perimeter For Square:

Area of Square =a2
Perimeter of Square =4a
Example 1:
For square board the perimeter is 84cm find the area for square board.

Solution:
we know perimeter is 84cm
for square perimeter,4a=84
a=84/4
a=21
area of square =a2
=212
area of square =441cm

Example 2:
The perimeter of a square playground is 1200m. Find its area using perimeter.

Solution :
Perimeter of the square ground, p = 4a
4a = p
a =p/4

Hence, a = 1200
4 m [since p = 1200 m ]
∴ a = 300 m
Area of the square ground A = a2
= 300 m × 300 m
Area = 90000 sq.m.

Finding Area using Perimeter For Rectangle:

Area of Rectangle =Length `xx` Breadth
Perimeter of rectangle =2(Length+Breadth)

Example 1:
If perimeter of rectangle is 18 m and its breadth 4m. find its area using perimeter.

Solution:
2 length + 2 breadth = perimeter
2l + 2`xx` 4 = 18
2l + 8 = 18
2l = 18– 8
2 l = 10
l =5
l=5,b=4
Now the area of the rectangle A = l `xx` b
= 5 m × 4 m
= 20 sq.m.

Example 2:
The perimeter of the floor of a rectangular room is 24 m and the  length is 7 m. Find its area.

Solution :
2 length + 2 breadth = perimeter
2 × 7 + 2b = 24
14 + 2b = 24
2b = 24 – 14
2 b = 10
b =`10/2`
= 5
Now the area using perimeter of  rectangle A = l` xx` b
= 7 m × 5 m
= 35 sq.m.

Dividing Fractions Solver

As division of whole numbers shows how often one integer is contained in another integer, to division of fraction shows how often one fraction is contained in another. If the numerator of any fraction be made a denominator, and the denominator a numerator, the fraction of made is called the reciprocal of the former. Thus 4 / 3 is the reciprocal of 3 / 4.

In division * as the divisor is to the dividend, of is a unit of the quotient (both in whole numbers and fraction).

Dividing fractions solver – Steps and methods:

To dividing algebraic fraction, follow these steps:

Write the given fractions.
Change the division sign to a multiplication sign and invert the second fraction.
Write the given fraction and cancel any common factors.
Multiply the numerators.
Multiply the denominators.

Dividing fractions solver – Methods:

Dividing common fraction
Dividing a fraction and a whole number
Dividing mixed number


Dividing fractions solver – Three methods and example problem:

Dividing common fractions:

To divide common fraction, follow these steps:

Invert the divisor.
Then multiply the numerator by the numerator and the denominator by the denominator.
Reduce to lowest terms when possible.

Example:

Dividing the fractions using fraction solver

(5 / 6) / (3 / 4)

Solution:

(5 / 6) / (3 / 4) (divisor)

= (5 /6) * (4 / 3)

= (5 * 4) / (6 * 3)

= 20 / 18

= 10 / 9

Dividing a fractions and a whole number:

To divide a fraction and a whole number, follow these steps:

Change the whole number to a fraction by the placing the whole number over one.
Invert the divisor.
Then multiply the numerator by the numerator and the denominator by the denominator.
Reduce to lowest terms when possible.

Example:

Dividing the fractions using fraction solver

16 / (5 / 8)

Solution:

16 / (5 / 8)

= (16 / 1) / (5 / 8)

= (16 / 1) * (8 / 5)

= (16 * 8) / (1 * 5)

= 128 / 5

= 25 3/5

Dividing mixed number fractions:

To divide mixed numbers, follow these steps:

Change the mixed number to an improper fraction
Invert the divisor.
Then multiply the numerator by the numerator and the denominator by the denominator.
Reduce to lowest terms when possible.

Example:

Dividing the fractions using fraction solver

(2 7/8) / (4 / 5)

Solution:

(2 7/8) / (4 / 5)

= (2 7/8) = ((2 * 8 + 7) / 8) = 23 / 5

= (23 / 8) / (4 / 5)

= (23 / 8) * (5 / 4)

= (23 * 5) / (8 * 4)

= (115 / 32)

= 3 19/ 32.

Hard Subtraction Problems

Subtraction is one of the arithmetic operations; it is the inverse of addition, meaning that if we start with any number and add any number and then subtract the same number we added, we return to the number we started with. Subtraction is denoted by a minus sign in infix notation. The traditional names for the parts of the formula

c − b = a

(Source: Wikipedia)

Hard subtraction example problems:

Hard subtraction problem 1:

Dylan made five hundred twenty-three cupcakes. Two hundred ten of them have already been put into packages. How many cupcakes are left to be packaged?

Solution:

Dylan made five hundred twenty-three cupcakes.

Two hundred ten of them have already been put into packages.

Cupcakes are left to be packaged = 523 – 210

= 313 cupcakes.

Answer: 313 cupcakes are left to be packaged.

Hard subtraction problem 2:

There are four hundred fifty employees working in an office building. Hundred an sixteen of them are about to leave to go out to lunch. How many employees will be left in the building?

Solution:

There are four hundred fifty employees working in an office building.

Hundred an Sixteen of them are about to leave to go out to lunch.

= 450 – 116

= 334

Answer: 334 employees will be left in the building.

Hard subtraction problem 3:

The pet store had three hundred twenty-eight bags of bird feed. Two hundred six will be eaten today. How many bags will be left?

Solution:

The pet store had three hundred twenty-eight bags of bird feed.

Two hundred six will be eaten today.

= 328 – 206

= 122 bags.

Answer: 122 bags will be left.

Hard subtraction practice problem:

Practice problem 1:

There were twenty-three magazines on a bookshelf this morning. Students came and borrowed ten of them. How many magazines are left?

Answer: 13 magazines are left

Practice problem 2:

There are fifty-two colored pens in a desk. There are twenty students and each will need one. How many colored pens will be left?

Answer: 32 colored pens will be left.

Practice problem 3:

There are forty-two birds sitting in two trees. One tree has twenty-two birds in it. How many birds are in the other tree?

Answer: 20 birds are in the other tree.

Parallelograms

Parallelogram is defined as the " four sided closed planar figure whose opposite sides are parallel to each other  " .

           Here ABCD is a parallelogram

There are some conditions to be satisfied by a quadrilateral so that it could be identified as parallelograms and they are as follows :
1 . The two pair of sides it consist of should be parallel to each other.
2 . The two pairs of opposite sides should be of same length .
3 . The angles of the two pair of opposite sides must be same .
4 . The diagonals drawn inside the parallelogram should bisect each other .
5 . One pair of opposite sides should be parallel and equal in length .

Types of Parallelograms

There are three types of parallelograms and they are as follows :

•  Square :

                    
                  
It is considere to be a type of parallelogram because it satisfies the condition required for a figure to be considered as the parallelogram such as
1 . The opposite pair of sides are parallel.
2 . The opposite sides are of same length .
3 . The opposite pair of angles are same .
4 . The diagonals inside the square bisect each other .
5 . One pair of opposite sides are parallel and they are equal in length .

• Rectangle :

                         
                          Rectangle is considered to be a type of parallelogram as it satisfies all the condition required to be considered as the parallelogram .

Types of parallelograms : Rhombus

• Rhombus :

                       
                             Rhombus is also considered as a type of parallelogram as it satisfies all the conditions of the parallelogram .

Secant Tangent Sine

Trigonometric functions are commonly defined as ratios of two sides of a right triangle containing the angle, and can equivalently be defined as the lengths of various line segments from a unit circle. The trigonometric functions are functions of an angle. It is also called as circular function. Trigonometric functions are important in the study of triangles and modeling periodic phenomena, among many other applications. The familiar trigonometric functions are the sine, cosine, and tangent. The other trigonometric function like cosecant, secant and cot are related to the familiar function.
                                                                                                                                                                             Source Wikipedia.

Some Trigonometric related functions:

1. `sin^2 theta + cos^2 theta = 1`
2. `tan theta = sin theta / cos theta`
3.` cot theta = 1/tan theta = cos theta/ sin theta `
4. `1+ tan^2 theta = sec^2 theta`
5.`sec theta ` = `1/cos theta`
6.` cosec theta = 1/ sin theta`
7. `sin 2theta = 2 sin theta cos theta`
8.`sin^2 theta ` = `(1 - cos 2theta)/2`

More details about trigonometric functions:

           In Right-angle triangle
               
Sine (Sin):
       In right-angle triangle, Ratio of the opposite side length and the hypotenuse of an angle is called as sine. It is reciprocal of cosecant.
             ` sin theta` = (opposite side) / (hypotenuse side)
Tangent (Tan):
         In right-angle triangle, Ratio of the opposite side length and the adjacent side length of an angle is called as tangent. In trigonometry relation is the ratio of sin and cosine. The tangent is reciprocal of cot.
                ` tan theta` = (opposite side) / (adjacent side)
Secant (Sec) = `1/cos` :
             In right-angle triangle, Ratio of the hypotenuse and the adjacent side length of an angle is called as secant. The secant is reciprocal of cosine.
                 `Sec theta` = (hypotenuse)/(adjacent) <br>
Cosine (Cos):
      In right-angle triangle, Ratio of the adjacent side length and the hypotenuse of an angle is called as cosine. It is a reciprocal of secant.
             .`cos theta` = (adjacent side)/(hypotenuse) 
Cosecant (cosec) = `1/sin` :
          In right-angle triangle, Ratio of  the hypotenuse and the opposite side length of an angle is called as cosecant. The cosecant is reciprocal of sine.
                ` cosec theta` = (hypotenuse)/(opposite side)
Cot` (1/ tan)` :
       In right-angle triangle, Ratio of the adjacent side length and the opposite side of an angle is called as cot. In trigonometry relation is the ratio of cosine and sin. The cot is reciprocal of tangent.
              `cot theta ` = (adjacent side)/(opposite side)

Trigonometric problems:

Trigonometric problem 1:
      If x = a cos t + b sin t and y = a sin t − b cos t,  show that   x² + y² = a² + b²
   Solution:
              x² + y² = (a cos t+ b sin t)² + (a sin t − b cos t)²                       
                          = a² cos²t + b² sin²t + 2ab cos t sin t+ a² sin²t + b² cos²t − 2ab sint cost                        
                          = a² (cos²t + sin²t) + b²(sin²t + cos²t)+ 2ab cos t sin t − 2ab sint cost
                          = a² + b²
                     Hence it has been proved.


Trigonometric problem 2:
      Prove that trigonometric relation: `(sec theta + 1/ (cot theta))^2 ` = `(1 + sin theta)/(1 - sin theta)`    
      Solution:
         Take left side, `sec theta + 1/cot theta `
         Squared the term, `sec theta + 1/cot theta` = `(sec theta + 1/cot theta)^2`   
       Now,
          `(sec theta + 1/cot theta)^2 = ( (1/cos theta) + (sin theta / cos theta) )^2`    ; we know, `sec theta = (1/cos theta) ; 1/cot theta = (sin theta /cos theta)`
                                     = `(1 + sin theta )^2/ (cos^2theta)`
                                     = `(1 + sin theta)^2 / (1 - sin^2 theta)`                                ; sin²θ + cos²θ = 1;  so,cos²θ = 1- sin²θ
                                     = `((1 + sin theta)(1 + sin theta)) / ((1 + sin theta) (1 - sin theta))`
      `(sec theta + tan theta)^2`     = `(1 + sin theta)/(1 - sin theta)`  
                                                  = Right side.
                  Hence, the given trigonometric relation is proved.

Value Chain Functions

Value chain Rule:

The derivative of the composition of the functions  which is computed as a product of their derivatives. To be precise:
Let f(x) and g(x) be two functions and let (g. f) (x) = g (f(x)) be their composition.

Value Chain rule 1:

If h(x) = g(f(x)), then h'(x) = g'(f(x))f'(x)

Note: In computing the composition g(f(x)), we apply f to x first, and then we apply g to f(x). According to the chain rule, when doing the derivative, we proceed in the opposite order: g is done first, and then f. One more thing, the f(x) part in g'f(x)) states that, while doing the derivative of g, we do not change f(x)(the f(x) term which is usually called the “inside”).

Sometimes we think of the composition in the following way:

y= g(u) and u =f(x) (i.e., y depends io u, and u depends on x). In that case, y depends on x and its derivative is,

Value Chain Rule 2:

If y= g(u) and u=f(x), then dy/dx = dy/du*du/dx

Examples of value chain function:

Value Chain Rule - Example1:

Compute the derivatives of the following functions.

(a)  Y= (x2+1)14

(b) Y = `sqrt (sin x +1)`

(c)  Y = `1/(e^x +2)`

(d) Y = sin(x2 + 1)

(e)  Y = cos(sec x)

( f )   Y = (sin x)2 + sin (x2).

Solution:

(a)  We start by computing the derivative of the power of 14:

Y' = 14(x2+1)13(x2+1)`

= 14(x2+1)132x

= 28(x2+1)13

(b) Writing y = `(sin x+1)^(-1/2)` , we get

Y` =`1/2` `(sinx + 1)^(-1/2)` (sinx+1)`

=  `1/2`  (sinx + 1)-1/2 cos x

=  `1/2` cos x(sin x +1) -1/2.

(c)  Y = (ex+2)-1.thus,

Y` = (-1)(ex+2)-2(ex+2)`

= -ex(ex+2)-2

= -ex/(ex+2)2.

(d) We start by computing the derivative if sin:

Y` = cos (x2+1)(x2+1)`

= 2x cos(x2+1)

(e)  Y` =-sin(sec x) (sec x)`

= - sin (sec x ) sec x tan x.

( f )   We have to be careful about the order:

Y` = 2(sin x)1(sin x)` + cos(x2)(x2)`

= 2sin x cos x +2x cos(x2)

= sin 2x +2x cos(x2)

In simplifying, we used the formula 2 sin x cos c  = sin 2x.

Value chain Rule - Example 2:

Find `dy/dx` for the following functions,

(a)  Y = 4u2 – 3u +2, u = ex+2e2x

(b) Y  = ln u, u=sin x + cos x

Solution:

(a)  By the chain Rule ,

`dy/dx = (dy/(du)) . ((du)/dx)`

= (8u -3)(ex + 4e2x)

= 8((ex + 2e2x) – 3) (ex + 4e2x).

(b) As in (a),

`dy/dx = dy/(du) . (du)/dx`   = `1/u(cos x - sin x) =(cos x -sin x)/(sin x + cos x)`