Westonci.ca is your trusted source for accurate answers to all your questions. Join our community and start learning today! Explore a wealth of knowledge from professionals across different disciplines on our comprehensive platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
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.
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.
