Taylor Polynomial Series

In mathematics, the Taylor series is a representation of a function as an infinite sum of terms calculated from the values of its derivatives at a single point. It is named after the English mathematician Brook Taylor. If the series is centered at zero, the series is also called a Maclaurin series, named after the Scottish mathematician Colin Maclaurin. It is common practice to use a finite number of terms of the series to approximate a function. The Taylor series may be regarded as the limit of the Taylor polynomials.  In the article we shall discuss about Taylor polynomial series.(Source: Wikipedia)

Taylor polynomial series:

Find the Taylor series for sin x.

    Sin x = `sum_(0)^oo` `((-1)^k)/((2k + 1)!)` x^2k+1

The remainder term is not expressible in any simple way but can be estimated by using the Lagrange's form of the remainder. The coefficients

   `((-1)^k)/((2k + 1)!)`

are easily verified by calculating successive derivatives of f(x) = sin x and using the formulas

 a_k=`(f^(k)(0))/(k!)`

To check convergence of the series, apply Lagrange's form for R_a(x); For each x`in` R. there exists Z such that

R_n(x) = `(f^(n + 1)(z))/((n + 1)!)` xⁿ⁺¹

Now |fⁿ⁺¹(z)| equals either |cos z| or |sin z| So, in either case,|fⁿ⁺¹ (z)|`<=`1 ,and

       |R_n(x) |`<=` | x |ⁿ⁺¹ /(n+1)!

Since | x |ⁿ⁺¹ /(n+1)!`->`0 as n `|->` `oo`  for all x`in` R, we can see that the remainder term |R_n(x)|`|->` 0 as n `|->` `oo`

 for all x`in` R. Thus the series representation is completely justified for all real x.

  Observe that our estimate for |R_n (x)|,

                                                    |R_n (x)|`<=`|x|^_n+1/(n+1)!

gives also a sense of the rate of convergence of the series for fixed x, for example, for | x|`<=` 1, we find

                                                    |R_n (x)|`<=`1/(n+1)!

Thus, if we want to calculate sin x on (-1, 1) to within .01, we need take only the first five terms of the series (n = 4) to achieve that degree of accuracy.

 Had we used the integral form for R_n (x) we would have obtained a similar estimate.

Sample problem for Taylor polynomial series:

Pro:  Evaluate the definite integral `int_0^1` Sin (x) dx

Sol:  The integrand has no anti derivative expressible in the terms of familiar functions. Howebver, we know how to find its Taylor series. we know that

      Sin t = t -`(t^3)/(3!)` + `(t^5)/(5!)` - `(t^7)/(7!)` + ----

Now if we substitute t = x, we have

     Sin (x) = X - `(x^3)/(3!)` + `(x^5)/(5!)` - `(x^7)/(7!)` + ----

In spite of the fact that we cannot antidifferentiate the function, we can antidifferentiate the Taylor series:

    `int_0^1` Sin (x) dx = `int_0^1` (X - `(x^3)/(3!)` + `(x^5)/(5!)` - `(x^7)/(7!)` + ----) dx

                            =(`(x^2)/(2)` -`(x^)/(4*3!)` + `(x^6)/(6*5!)`-`(x ^ 7)/(8*7!)`+ -----) |₀¹

                            = (`(x^2)/(2)` -`(x^)/(4*3!)` + `(x^6)/(6*5!)`+`(x^7)/(8*7)` + ---)

Notice that this is an alternating series so we know that it converges. if we add up the first four terms, the pattern becomes ckear: the series converges to 0.2871

Negative Integer Exponents

The exponents are which integer is placed in the power, of base numbers. It can be easily represent as, “a small number to the right side and above of base number”. It is called as exponents. These exponents have some of important rules and laws. Power with negative integer exponents is also one of the rules. Here we are going to explain about this negative integer exponent rule.

If we are having variables, which is containing the exponents and it have equal bases means, we can do some mathematical operations with the exponents. Those operations are called as the “laws of exponents” or “rules of exponents”. In this rule based negative integer rule of exponent is defined as following ways,

Definition for negative integer exponents:

It is otherwise called as power with negative exponent rules. This negative exponent rule is defined as, if m is a positive integer and x is a non-zero rational number, then it can be denoted as,

X-m = (1/x)^m (or)

= (1/x)^m

Which is (x)^-m is the reciprocal of (x)^m

And we adopt the same rule for rational exponents also. If p/q is a positive rational number means, and x>0 is a rational number, then

 X^ - (p/q) = (1/x)^ (p/q) = (1/x)^ (p/q) .

Which is, (x)^-(p/q) is the reciprocal of (x) ^(p/q) or the number obtained by raising the reciprocal of x to the exponent p/q.

For example: 1). 3^-2

= (1/3)^2

= (1/3)^-2

= -6 .

2). (4)^-(2/3)

= (1/4)^-(2/3)

= (1/4)^ (2/3)

This kind of exponentiation used for discovers the negative integer exponents and simplify the problems.


Example problems for negative integer exponents:

1) Solve: (8)^-(2/3)

Solution:

Given: (8)^-(2/3)

= (1/8)^ (2/3)

= [(1/8)^ (1/3)]^ 2

= (1/2)^2, since (1/2)^3 = 1/8

= 1/4 .

2) Solve: (32/243)^-(4/5)

Solution:

Given: (32/243)^-(4/5)

= (243/32)^(4/5)

= [(243/32)^(1/5)]^4

= [(3^5/2^5)^(1/5)]^4

= [((3/2)^5)^(1/5)]^4

= (3/2)^4

= 81/16.

3) Evaluate, and find the following negative integer exponent value:

Evaluate: (27/125)^-(2/3)  xx    (27/125)^-(4/3)

Solution:

 (27/125)^- (2/3) xx (27/125)^-(4/3)

= (125/25)^(2/3) xx (125/27)^(4/3)

= [(5^3/3^3)^(1/3)]^ 2 xx [(5^3/3^3)^(1/3)]^4

= [((5/3)^3)^(1/3)]^2 xx [((5/3)^3)^(1/3)]^4

= (5/3)^2 xx (5/3)^4

= (5/3)^6

= 15625/729.

These all are the explanations and example problems may clear about the negative integer exponents.

Math Absolute Value Inequalities

In math, the absolute value |x| of a real number x is x's arithmetical value lacking view to its symbol. So, for example, 86 is the absolute value of both 86 and −86. Generalization of the absolute value for real numbers occurs in a extensive diversity of math settings. Consider an absolute value is also definite for the complex numbers, the quaternions, prepared rings, fields and vector spaces. The absolute value is strictly associated to the ideas of magnitude, distance, and norm in different math and physical contexts.

Properties of the Math absolute value inequalities

The absolute value fundamental properties are:

 |x| = sqrt(x^2)                                      (1) Basic

 |x| \ge 0                                            (2)     Non-negativity

 |x| = 0 \iff x = 0                            (3)     Positive-definiteness

 |xy| = |x||y|\,                                  (4)     Multiplicativeness

 |x+y| \le |x| + |y|                            (5)     Subadditivity

Another important property of the absolute value includes these are the:

 |-x| = |x|\,                                      (6)     Symmetry

|x - y| = 0 \iff x = y                     (7)     Identity of indiscernible (equivalent to positive-definiteness)

|x - y| \le |x - z| +|z - y|                 (8)     Triangle inequality (equivalent to sub additivity)

|x/y| = |x| / |y| \mbox{ (if } y \ne 0) \,                   (9)      Preservation of division (equivalent to multiplicativeness)

|x-y| \ge ||x| - |y||                           (10)     (Equivalent to sub additivity)

Math absolute value inequalities – Examples:

Math absolute value inequalities – Example 1:

|x - 5| < 3

Set up the two times inequality -3 < x -5 < 3 and then solve.

-3 < x – 5 < 3

2 < x < 8

In interval notation, the answer is (2, 8).
Math absolute value inequalities – Example 2:

|2x + 3| <=-8

This solution will involve setting up two separate inequalities and solving each.

2x + 3 <= -8

2x <=-11

x<=-11/2

Else

2x+3 >=8

2x >= 5

x>=5/2

In interval notation, the answer is (-oo, -11/2) U (5/2, oo) .
Math absolute value inequalities – Example 3:

|x - 9| < 4

Set up the two times inequality -4 < x -9 < 4 and then solve.

-4 < x – 9 < 4

5 < x < 13

In interval notation, the answer is (5, 13).

Math absolute value inequalities – Example 4:

|3x + 3| <= - 8

This solution will involve setting up two separate inequalities and solving each.

3x + 3 <= -8

 3x <=-11

x<=-11/3

Else

3x+3 >=8

 3x >= 5

x>=5/3

In interval notation, the answer is (-oo, -11/3) U (5/3, oo).

Negative Number Calculator

In this article we are discussing about subtracting negative number by using calculator. Negative number also real number. The negative number is represented as minus symbol “- “. Negative number or elements are less than zero such as -4, -7, -/3. A negative number may be parenthesized with its symbol, For example a subtracting is clearer if written (-7) + (−5) = -13 Using calculator we can the negative number.
Negative number calculator:

Let us see how to negative number using calculator.

Negative number is the similar as subtracting the corresponding negative number:

Example: (-7) + (-7) = -14

Negative number calculator – Example problems:

Example 1:

(-3) + (-4) = ?

Solution:

First enter the negative number in the calculator.

Then press the equal to button.

The solution is displayed on the calculator.

negative number calculator

The subtract solution is -7.

Example 2:

(-5) - (-4) = ?

Solution:

First enter the negative number in the calculator.

Then press the equal to button.

The solution is displayed on the calculator.

negative number calculator

The subtract solution is -1.

Example 3:

(-4) x (-4) = ?

Solution:

First enter the negative number in the calculator.

Then press the equal to button.

The solution is displayed on the calculator.

negative number calculator

The subtract solution is 16.

Example 4:

(-9) `-:` (-3) = ?

Solution:

First enter the negative number in the calculator.

Then press the equal to button.

The solution is displayed on the calculator.

negative number calculator

The subtract solution is 3.

Example 5:

(-10) x (-20) = ?

Solution:

First enter the negative number in the calculator.

Then press the equal to button.

The solution is displayed on the calculator.

negative number calculator

The subtract solution is 200.

Negative number calculator – practice problems:

Problem 1: (-5) + (-6)

Problem 2: (-4) – (-2)

Problem 3: (-9) x (-18)

Problem 4: (-5) `-:` (-20)

Negative number calculator – answer key:

Problem 1: -11

Problem 2: -2

Problem 3: 162

Problem 4: -0.25

Solving Monotonicity and Concavity

A function is said to be rising on an interval [a, b] = I if f (z1) < f (z2) whenever z1 < z2 for z1, z2 `in` I. A function is said to be decreasing on an interval [a, b] = I if f (z1) > f (z2) whenever z1 < z2 for z1, z2 `in` I. A function which is strictly increasing or strictly decreasing on an interval is said to be monotonic on that interval. A function is said to be curved in up on an interval [a, b] if f ’’ (z) > 0 on [a, b]. A function is said to be curved in down on an interval [a, b] if f ’’ (z) < 0 on [a, b]. A point (c, f(c)) on the graph of y = f (z) is called an inflection point if the concavity changes at the point. (I.E., it is curved in up on some interval (a, c) and curved in down on some interval (c, b) or vice versa.)

Solving monotonicity and concavity - Examples:

Solving monotonicity and concavity - Example 1:

Find inflection points & determine concavity for f (x)

f ‘(x) = `(x^5)/20 + (x^4)/12 - (3x^3)/3 - 10`

f ’(x) = `(x^4)/4 + (x^3)/3 - 3(x^2)`

f ‘’ (x) = x3 + x2 – 3x

= x(x2 + x - 3)

f ‘‘(x) = 0

x = 0, -2.3, 1.3

Inflection pts: x= -2.3, 0, 1.3

Curved in up: (-2.3,0), (1.3,infinity)

Curved in down: (-oo,-2.3), (0,1.3)

Solving monotonicity and concavity

Solving monotonicity and concavity - More Examples:

Solving monotonicity and concavity - Example 1:

Find inflection points & determine concavity for f (x)

f (x) = `(x^5)/20 + (x^4)/12 - (x^3)/3 +10`

f ‘ (x) = `(x^4)/4 + (x^3)/3 - x^2`

f ‘’ (x) = x3 + x2 – 2x = x(x2 + x - 2)

= x (x+2) (x-1)

f ‘‘ (x) = 0

x = 0, -2, 1

f ’’(-5) < 0, f’’(-1) > 0, f ’’(.5) < 0, f ’’(10) > 0

Inflection pts: x=-2,0,1

Curved in up: (-2,0), (1,`oo` )

Curved in down: (`-oo` .,-2), (0,1)

Solving monotonicity and concavity

Trinomial Multiplication

In algebraic expression consisting of only one term is called a monomial, two terms is called a binomial and when there are three terms separated by an addition or a subtraction operation is called a trinomial. ‘Tri’ in the word trinomial means three and hence the name. It is also referred to as a polynomial as ‘poly’ means more than two terms. For example, a+3b-c consists of three terms a, b and c and hence a trinomial. In 3b, 3 is called the coefficient of b. In a given polynomial we come across like terms and unlike terms.

Like terms are the terms which have the same variable or literal but a different coefficient. For example, (3b, -7b);(5x2, 12x2);(-4xy, 7xy) are some of the like terms. Unlike terms as the name suggests are the terms which have different variables. For example, (7x, 8y); (4ab, -4ac); (2y2, -2x2) etc are some of the unlike terms.

It is not necessary that there would be only two like or unlike terms, it depends on the number of terms in the given polynomial. Now let us learn as to How do you multiply trinomials. There are two methods in which we can multiply trinomials, one is the horizontal method the other is the vertical method.

The steps to be followed in a Horizontal method of multiplication are as follows:
Example: Multiply (4x2-3x+5)(x2+5x-3)
Solution: First and second terms are chosen irrespective of the order
Let (4x2-3x+5) be the first term and (x2+5x-3) be the second term

Arrange the two polynomials horizontally
(4x2-3x+5) X (x2+5x-3)
Now distribute each of the terms of the first trinomial with each of the terms of the second trinomial
=4x2(x2+5x-3)- 3x(x2+5x-3) +5(x2+5x-3)
=4x4+20x3-12x2 - 3x3-15x2+9x +5x2+25x-15

Combine the like terms and simplify
=4x4+x3(20-3)+x2(-12-15+5)+ x(9+25) – 15
= 4x4+17x3-22x2+34x-15

The steps to be followed in a Vertical method of multiplication are as follows:
Example: Multiply (4x2-3x+5)(x2+5x-3)
The first and the second terms are chosen irrespective of the order
Let (4x2-3x+5) be the first term and (x2+5x-3) be the second term
Arrange the two polynomials in a vertical form
4x2-3x+5
X  x2+5x-3

Working from right towards left each term of the lower trinomial is multiplied with each of the terms of the upper trinomials. Then the products are written underneath the second trinomial in three rows in the order of the degree of each term and then the like terms simplified as shown below
4x2-3x+5
X  x2+5x-3
-12x2+ 9x – 15
+20x3-15x2 + 25x
4x4 - 3x3  + 5x2                      .
4x4+17x3-22x2+ 34x-15 is the final product!

Solve Explicit Differentiation

In calculus,Explicit is a function which the independent variable. The function f explicitly is to provide a preparation for determining the output of the given function y in terms of the input value x: y = f(x). Derivative of an explicit function is called as explicit differentiation. for example  y = x3 + 5. The process of finding the differentiation of the independent variable in an explicit function by differentiating each term separately, by expressing the derivative of the independent variable as a symbol, and by solving the resulting expression for the symbol.

Solve explicit differentiation problems:

Let us see some problems and its helps to solve an explicit differentiation.

Solve explicit differentiation problem 1:

Find the differentiation of given explicit function   y = x2 - 15x + 3.

Solution:

Given explicit function is  y = x2 - 15x + 3.

Differentiation of explicit function is   dy/dx  = d/dx ( x2 - 15x + 3)

Separate the each term, so, we get

= d/dx (x2) - d/dx (15x) + d/dx (3)

= d/dx (x2) - 15d/dx(x) + d/dx (3).

= 2 x(2-1) - 15 + 0

= 2x - 15

The differentiation of an explicit function is  2x - 15

Solve explicit differentiation problem 2:

Find the differentiation of an explicit function  x2 +  cot x = - y

Solution:

Given explicit function is  x2 +  cot x = - y

Multiply by (-1) on both sides,  y = - x2 -  cot x

Differentiation of an explicit function,

y =   - x2 -  cot x

dy/dx = d/dx -x2   - d/dx (cot x) .

=  - 2x - (-cosec2 x) .

= - 2x + cosec2x

The Differentiation of an explicit function is  - 2x + cosec2x

Solve explicit differentiation problem 3:

Find the differentiation of given explicit function  2x2 + y2 = 1

Solution:

Given explicit function is  2x2 + y2 = 1

Subtract 2x2 on both sides.we get,

2x2 + y2 - 2x2 = 1 - 2x2

y 2 = 1 - 2x2

Take square root on both sides, we get

 sqrt (y^2) =  +- sqrt (1 - 2x^2)

y = +- sqrt (1 - 2x^2) .

Differentiate the function,  Let u = 1 - 2x2            and           y = sqrt u

(du)/(dx) = - 4x                              dy/(du) = 1/(2sqrtu)

So,   dy/dx =  ((dy)/(du)) ((du)/(dx)) .

=  1/(2sqrtu) . (-4x )

=  (- 2x) / sqrtu .

Substitute u = y, So we get

= - (2x)/y   .

The differentiation of an explicit function is  - (2x)/y    .