Calculus 2 MAT 217 Problem Set 7 Brian Reece DMACC Problem 1 Find the Taylor series for f x e5 x e3 x centered at x0 0 Problem 2 Find a power series that has 23 as an interval of convergence Problem 3...

### Problem 1: Taylor Series for $f(x)=e^{5x}-e^{3x}$ #### Solution By Steps ***Step 1: Identify the Taylor series formula*** The Taylor series of a function $f(x)$ centered at $x_0$ is given by $f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n$. ***Step 2: Calculate derivatives at $x_0=0$*** For $f(x)=e^{5x}-e^{3x}$, the derivatives at $x_0=0$ are: - $f(0) = e^0 - e^0 = 0$ - $f'(x) = 5e^{5x} - 3e^{3x}$, so $f'(0) = 5 - 3 = 2$ - $f''(x) = 25e^{5x} - 9e^{3x}$, so $f''(0) = 25 - 9 = 16$ - Generalizing, $f^{(n)}(0) = 5^n - 3^n$. ***Step 3: Construct the Taylor series*** Substitute the derivatives into the Taylor series formula: $f(x) = \sum_{n=0}^{\infty} \frac{5^n - 3^n}{n!}x^n$. #### Final Answer The Taylor series for $f(x)=e^{5x}-e^{3x}$ centered at $x_0=0$ is $\sum_{n=0}^{\infty} \frac{5^n - 3^n}{n!}x^n$. #### Key Concept Taylor Series #### Key Concept Explanation The Taylor series represents a function as an infinite sum of its derivatives at a single point. It's a powerful tool for approximating function values and studying function behavior analytically. --- ### Problem 2: Power Series with Interval of Convergence $(-2,3)$ #### Solution By Steps ***Step 1: Identify a suitable power series*** A simple power series with a variable interval of convergence is $\sum_{n=1}^{\infty} a_n (x-c)^n$. Adjusting $a_n$ and $c$ can tailor the interval of convergence. ***Step 2: Choose a series with known convergence*** The series $\sum_{n=1}^{\infty} \frac{x^n}{n^2}$ converges for $|x|<1$. To adjust the interval to $(-2,3)$, we transform $x$ to fit this range. ***Step 3: Transform the series*** Let $y = \frac{x-c}{r}$, where $c$ is the center and $r$ is the radius of the interval. For $(-2,3)$, $c=\frac{3-2}{2}=0.5$ and $r=2.5$. Thus, $x = 2.5y + 0.5$. ***Step 4: Substitute and adjust the series*** Substitute $x = 2.5y + 0.5$ into the series: $\sum_{n=1}^{\infty} \frac{(2.5y + 0.5)^n}{n^2}$. #### Final Answer A power series with interval of convergence $(-2,3)$ is $\sum_{n=1}^{\infty} \frac{(2.5y + 0.5)^n}{n^2}$, where $y$ is transformed to fit the interval. #### Key Concept Interval of Convergence #### Key Concept Explanation The interval of convergence is the set of all real numbers for which a power series converges. Adjusting the series terms can tailor this interval to specific ranges. --- ### Problem 3: Series Representation and Sum #### Solution By Steps ***Step 1: Identify the series terms*** Given $f(x)=\frac{1}{6}+\frac{2}{6^2}x+\frac{3}{6^3}x^2+\ldots$, the general term is $\frac{n+1}{6^{n+1}}x^n$. ***Step 2: Integrate term by term*** Integrating $f(x)$ from 0 to 1 gives $\int_{0}^{1} \frac{n+1}{6^{n+1}}x^n dx = \left[\frac{n+1}{6^{n+1}}\frac{x^{n+1}}{n+1}\right]_0^1 = \frac{1}{6^{n+1}}$. ***Step 3: Write the first three terms and the general term*** First three terms: $\frac{1}{6}, \frac{1}{6^2}, \frac{1}{6^3}$. General term: $\frac{1}{6^{n+1}}$. ***Step 4: Find the sum of the series*** The sum is a geometric series with $a=\frac{1}{6}$ and $r=\frac{1}{6}$: $S = \frac{a}{1-r} = \frac{1/6}{1-1/6} = \frac{1}{5}$. #### Final Answer The sum of the series is $\frac{1}{5}$. #### Key Concept Series Sum #### Key Concept Explanation The sum of an infinite series is found by summing its terms. For geometric series, the sum is given by $S = \frac{a}{1-r}$, where $a$ is the first term and $r$ is the common ratio.