Rational Zeros of a Function

Normally zeros of a function mean when we plug the values for the variables the functions values tends to be zero.  Let us consider if we are having a function with variable x and we have a set of solution p(x) if we plug the solution for the given variables present in the function we will get the function f(x) = 0. To find the rational zeros we have to use the rational zeros theorem.

I like to share this Define Function Notation with you all through my article.

Rational Zeros of a Function – Examples:

The rational theorem states that if we are having the polynomial p(x) wit the integer coefficients and we are having the zeros of the polynomial `p / q` then we can say `p(p / q) = 0` . Here p is nothing but the constant term of the polynomial and the q I nothing but the leading coefficient of the polynomial p(x). We will see some examples for finding the rational zeros of a function.

Example 1 for rational zeros of a function:

Find the rational zeros of the following function.

2x2 + 12x + 10

Solution:

The given function is

2x2 + 12x + 10

First we have to find the rots of the constant term. ±1, ±2, ±5, ±10

Now the leading co – efficient of the constant term is 2. So we have to divide by 2.

±`p / q` = ±`1/ 2` , ±`2/ 2` , ±`5/ 2` , ±`10/ 2`

= ±`1/2` , ±1, ±`5/2` , ±5

Now we have to use the synthetic division method to find the rational zeros.

1 / 2 | 2           12        10

|               1         13/2

|_____________________

2           13        33/2      = not a zero

-1 / 2 | 2           12        10

|               -1         -11/2

|_____________________

2           11        9/2      = not a zero



1 | 2           12        10

|               2         24

|_____________________

2           24        34      = not a zero

-1 | 2           12        10

|               -2        -10

|_____________________

2           10        0      = is a zero

5/2 | 2           12        10

|               5        85/2

|_____________________

2           17        105/2      = not a zero

- 5/2 | 2           12        10

|               -5        -35/2

|_____________________

2           7        15/2      = not a zero

5 | 2           12        10

|              10        110

|_____________________

2           22        120      = not a zero

-5 | 2           12        10

|              -10      -10

|_____________________

2           2        0      = is a zero

So from the above the rational zeros of the functions are p(x) is -1 and -5

I am planning to write more post on finding equivalent fractions and how to subtract decimals. Keep checking my blog.

Rational Zeros of a Function – more Examples:

Example 2 for rational zeros of a function:

Find the rational zeros of the following function.

x2 + 4x + 3

Solution:

The given function is

x2 + 4x + 3

First we have to find the rots of the constant term. ±1, ±3

Now the leading co – efficient of the constant term is 1. So we have to divide by 1.

±`p / q` = ±`1/ 1` , ±`3 / 1`

= ±1, ±3

Now we have to use the synthetic division method to find the rational zeros.

1 | 1           4        3

|               1       5

|_____________________

1           5       8      = not a zero

-1| 1           4        3

|              -1      -3

|_____________________

1            3        0      = is a zero

3 | 1           4        3

|              3        21

|_____________________

1          7       24      = not a zero

-3| 1           4        3

|             -3      -3

|_____________________

1            1        0      = is a zero

So from the above the rational zeros of the functions are p(x) is -1 and -3

Is this topic algebra questions and answers hard for you? Watch out for my coming posts.

Interval of Convergence for Taylor Series

The interval of convergence for the given series is the set of all values such that the series converges if the values are within the interval and diverges if the value exceeds the interval.
The interval of convergence series must have the interval a - R < x < a + R since at this interval power series will converge.

In this article, we are going to see few example and practice problems of Taylor series to find interval of convergence which help you to learn interval of convergence.
Example Problems to Find the Interval of Convergence for Taylor Series:

Example problem 1:

Solve and determine the interval of convergence for Taylor seriessum_(n = 0)^oo(x^n)/(n!) .

Solution:

Step 1: Given series

sum_(n = 0)^oo (x^n)/(n!)  .

Step 2: Find L using the ratio test

L =  lim_(n->oo) | ((x^(n+1))/((n+1)!))/((x^n)/(n!)) |

= lim_(n->oo) |   (xn!)/((n + 1)(n!))  |

= lim_(n->oo) |   x/(n + 1)  |

= 0

So, this series converge for all value of x. Therefore, the interval of convergence is (-∞, ∞).

Step 3: Solution

Hence, the interval of convergence for the given Taylor series is (-∞, ∞).

Example problem 2:

Solve and determine the interval of convergence for Taylor series sum_(n = 0)^oo(-1)n (x^(2n + 1))/((2n + 1)!) .

Solution:

Step 1: Given series

sum_(n = 0)^oo(-1)n (x^(2n + 1))/((2n + 1)!)  .

Step 2: Find L using the ratio test

L =  lim_(n->oo) | (-1)^(n + 1)(x^(2(n + 1)+1))/((2n + 1)!) |

= lim_(n->oo) |  (-1)^(n + 1)(x^(2n + 3))/((2n + 1)!)   |

= 0

So, this series converge for all value of x. Therefore, the interval of convergence is (-∞, ∞).


Step 3: Solution

Hence, the interval of convergence for the given Taylor series is (-∞, ∞).

Practice Problems to Find the Interval of Convergence for Taylor Series:

1) Determine the interval of convergence for Taylor series sum_(n = 0)^oo(-1)n (x^(2n))/(2n!)  .

2) Determine the interval of convergence for Taylor series sum_(n = 0)^ooxn.

Solutions:

1) The interval of convergence for the given Taylor series is (-∞, ∞).

2) The interval of convergence for the given Taylor series is |x| < 1.