Answer :
Sure! Let's break down the process step by step to find the Taylor series centered at [tex]\( C = 1 \)[/tex] for the function [tex]\( f(x) = \frac{2}{x} \)[/tex], and then determine its radius and interval of convergence.
### Step 1: Define the Function and its Derivatives
The function we are dealing with is [tex]\( f(x) = \frac{2}{x} \)[/tex]. We'll need to find the derivatives of [tex]\( f(x) \)[/tex] evaluated at [tex]\( x = 1 \)[/tex].
First, let's calculate the function value and its derivatives:
[tex]\[ f(x) = \frac{2}{x} \][/tex]
#### First Derivative:
[tex]\[ f'(x) = -\frac{2}{x^2} \][/tex]
#### Second Derivative:
[tex]\[ f''(x) = \frac{4}{x^3} \][/tex]
#### Third Derivative:
[tex]\[ f'''(x) = -\frac{12}{x^4} \][/tex]
#### Fourth Derivative:
[tex]\[ f^{(4)}(x) = \frac{48}{x^5} \][/tex]
#### Fifth Derivative:
[tex]\[ f^{(5)}(x) = -\frac{240}{x^6} \][/tex]
### Step 2: Evaluate the Derivatives at [tex]\( x = 1 \)[/tex]
Now we substitute [tex]\( x = 1 \)[/tex] in each of these derivatives.
#### Function value at [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = \frac{2}{1} = 2 \][/tex]
#### First Derivative at [tex]\( x = 1 \)[/tex]:
[tex]\[ f'(1) = -\frac{2}{1^2} = -2 \][/tex]
#### Second Derivative at [tex]\( x = 1 \)[/tex]:
[tex]\[ f''(1) = \frac{4}{1^3} = 4 \][/tex]
#### Third Derivative at [tex]\( x = 1 \)[/tex]:
[tex]\[ f'''(1) = -\frac{12}{1^4} = -12 \][/tex]
#### Fourth Derivative at [tex]\( x = 1 \)[/tex]:
[tex]\[ f^{(4)}(1) = \frac{48}{1^5} = 48 \][/tex]
#### Fifth Derivative at [tex]\( x = 1 \)[/tex]:
[tex]\[ f^{(5)}(1) = -\frac{240}{1^6} = -240 \][/tex]
### Step 3: Construct the Taylor Series
The Taylor series for a function [tex]\( f(x) \)[/tex] centered at [tex]\( C = 1 \)[/tex] is given by:
[tex]\[ f(x) = f(1) + f'(1)(x - 1) + \frac{f''(1)}{2!}(x - 1)^2 + \frac{f'''(1)}{3!}(x - 1)^3 + \frac{f^{(4)}(1)}{4!}(x - 1)^4 + \frac{f^{(5)}(1)}{5!}(x - 1)^5 + \cdots \][/tex]
Substituting the values we have calculated:
[tex]\[ f(x) \approx 2 - 2(x-1) + \frac{4}{2!}(x-1)^2 - \frac{12}{3!}(x-1)^3 + \frac{48}{4!}(x-1)^4 - \frac{240}{5!}(x-1)^5 \][/tex]
Simplifying the factorials:
[tex]\[ f(x) \approx 2 - 2(x-1) + 2(x-1)^2 - 2(x-1)^3 + 2(x-1)^4 - 2(x-1)^5 \][/tex]
Therefore, the Taylor series up to the 5th degree term is:
[tex]\[ f(x) \approx 2 - 2(x-1) + 2(x-1)^2 - 2(x-1)^3 + 2(x-1)^4 - 2(x-1)^5 \][/tex]
### Step 4: Find the Radius of Convergence
The radius of convergence, [tex]\( R \)[/tex], of the Taylor series can be found by identifying the distance from the center of the series to the nearest singularity. For [tex]\( f(x) = \frac{2}{x} \)[/tex], there is a singularity at [tex]\( x = 0 \)[/tex].
Since the series is centered at [tex]\( x = 1 \)[/tex], the distance to the nearest singularity at [tex]\( x = 0 \)[/tex] is:
[tex]\[ R = |1 - 0| = 1 \][/tex]
### Step 5: Find the Interval of Convergence
The interval of convergence is determined by the radius of convergence. Since [tex]\( R = 1 \)[/tex], we check the interval around the center [tex]\( x = 1 \)[/tex]. Given the nature of the function and the singularity at [tex]\( x = 0 \)[/tex]:
[tex]\[ \text{Interval of Convergence}: (0, \infty) \][/tex]
### Summary
- Taylor Series: [tex]\( f(x) \approx 2 - 2(x-1) + 2(x-1)^2 - 2(x-1)^3 + 2(x-1)^4 - 2(x-1)^5 \)[/tex]
- Radius of Convergence: [tex]\( R = 1 \)[/tex]
- Interval of Convergence: [tex]\( (0, \infty) \)[/tex]
These calculations and conclusions provide a comprehensive solution to the problem.
### Step 1: Define the Function and its Derivatives
The function we are dealing with is [tex]\( f(x) = \frac{2}{x} \)[/tex]. We'll need to find the derivatives of [tex]\( f(x) \)[/tex] evaluated at [tex]\( x = 1 \)[/tex].
First, let's calculate the function value and its derivatives:
[tex]\[ f(x) = \frac{2}{x} \][/tex]
#### First Derivative:
[tex]\[ f'(x) = -\frac{2}{x^2} \][/tex]
#### Second Derivative:
[tex]\[ f''(x) = \frac{4}{x^3} \][/tex]
#### Third Derivative:
[tex]\[ f'''(x) = -\frac{12}{x^4} \][/tex]
#### Fourth Derivative:
[tex]\[ f^{(4)}(x) = \frac{48}{x^5} \][/tex]
#### Fifth Derivative:
[tex]\[ f^{(5)}(x) = -\frac{240}{x^6} \][/tex]
### Step 2: Evaluate the Derivatives at [tex]\( x = 1 \)[/tex]
Now we substitute [tex]\( x = 1 \)[/tex] in each of these derivatives.
#### Function value at [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = \frac{2}{1} = 2 \][/tex]
#### First Derivative at [tex]\( x = 1 \)[/tex]:
[tex]\[ f'(1) = -\frac{2}{1^2} = -2 \][/tex]
#### Second Derivative at [tex]\( x = 1 \)[/tex]:
[tex]\[ f''(1) = \frac{4}{1^3} = 4 \][/tex]
#### Third Derivative at [tex]\( x = 1 \)[/tex]:
[tex]\[ f'''(1) = -\frac{12}{1^4} = -12 \][/tex]
#### Fourth Derivative at [tex]\( x = 1 \)[/tex]:
[tex]\[ f^{(4)}(1) = \frac{48}{1^5} = 48 \][/tex]
#### Fifth Derivative at [tex]\( x = 1 \)[/tex]:
[tex]\[ f^{(5)}(1) = -\frac{240}{1^6} = -240 \][/tex]
### Step 3: Construct the Taylor Series
The Taylor series for a function [tex]\( f(x) \)[/tex] centered at [tex]\( C = 1 \)[/tex] is given by:
[tex]\[ f(x) = f(1) + f'(1)(x - 1) + \frac{f''(1)}{2!}(x - 1)^2 + \frac{f'''(1)}{3!}(x - 1)^3 + \frac{f^{(4)}(1)}{4!}(x - 1)^4 + \frac{f^{(5)}(1)}{5!}(x - 1)^5 + \cdots \][/tex]
Substituting the values we have calculated:
[tex]\[ f(x) \approx 2 - 2(x-1) + \frac{4}{2!}(x-1)^2 - \frac{12}{3!}(x-1)^3 + \frac{48}{4!}(x-1)^4 - \frac{240}{5!}(x-1)^5 \][/tex]
Simplifying the factorials:
[tex]\[ f(x) \approx 2 - 2(x-1) + 2(x-1)^2 - 2(x-1)^3 + 2(x-1)^4 - 2(x-1)^5 \][/tex]
Therefore, the Taylor series up to the 5th degree term is:
[tex]\[ f(x) \approx 2 - 2(x-1) + 2(x-1)^2 - 2(x-1)^3 + 2(x-1)^4 - 2(x-1)^5 \][/tex]
### Step 4: Find the Radius of Convergence
The radius of convergence, [tex]\( R \)[/tex], of the Taylor series can be found by identifying the distance from the center of the series to the nearest singularity. For [tex]\( f(x) = \frac{2}{x} \)[/tex], there is a singularity at [tex]\( x = 0 \)[/tex].
Since the series is centered at [tex]\( x = 1 \)[/tex], the distance to the nearest singularity at [tex]\( x = 0 \)[/tex] is:
[tex]\[ R = |1 - 0| = 1 \][/tex]
### Step 5: Find the Interval of Convergence
The interval of convergence is determined by the radius of convergence. Since [tex]\( R = 1 \)[/tex], we check the interval around the center [tex]\( x = 1 \)[/tex]. Given the nature of the function and the singularity at [tex]\( x = 0 \)[/tex]:
[tex]\[ \text{Interval of Convergence}: (0, \infty) \][/tex]
### Summary
- Taylor Series: [tex]\( f(x) \approx 2 - 2(x-1) + 2(x-1)^2 - 2(x-1)^3 + 2(x-1)^4 - 2(x-1)^5 \)[/tex]
- Radius of Convergence: [tex]\( R = 1 \)[/tex]
- Interval of Convergence: [tex]\( (0, \infty) \)[/tex]
These calculations and conclusions provide a comprehensive solution to the problem.
