Math / Calculus

QuestionFind a power series representation for the function. (Give $y$ $\begin{array}{c} f(x)=\ln (5-x) \\ f(x)=\ln (5)-\sum_{n=1}^{\infty}(\square) \end{array}$
Determine the radius of convergence, $R$ . $R=$ $\square$ Need Help? Read It Watch It Master It Submit Answer

Studdy Solution

STEP 1

What is this asking? We need to rewrite the function $f(x) = \ln(5 - x)$ as a power series and then find the radius of convergence for that power series. Watch out! Remember that the standard power series for $\ln(1 + x)$ is only valid when $|x| < 1$ , so we'll need to manipulate our function to get it into that form and be careful with the interval of convergence.

STEP 2

1. Rewrite the function
2. Find the derivative
3. Express as a geometric series
4. Integrate to find the power series
5. Determine the radius of convergence

STEP 3

We want to rewrite our function $f(x) = \ln(5 - x)$ to look a bit more like the standard form $\ln(1 + u)$ .
Let's factor out a 5: $f(x) = \ln\left(5\left(1 - \frac{x}{5}\right)\right)$

STEP 4

Using the logarithm property $\ln(ab) = \ln(a) + \ln(b)$ , we can rewrite this as: $f(x) = \ln(5) + \ln\left(1 - \frac{x}{5}\right)$ Now we have a $\ln(1+u)$ form, where $u = -\frac{x}{5}$ .

STEP 5

Let's take the derivative of $f(x)$ with respect to $x$ : $f'(x) = \frac{d}{dx}\left[\ln(5) + \ln\left(1 - \frac{x}{5}\right)\right]$ Since the derivative of a constant is zero, and using the chain rule, we get: $f'(x) = 0 + \frac{1}{1 - \frac{x}{5}} \cdot \left(-\frac{1}{5}\right) = -\frac{1}{5 - x}$

STEP 6

We can rewrite $f'(x)$ as: $f'(x) = -\frac{1}{5}\cdot\frac{1}{1-\frac{x}{5}}$ This looks like the sum of a geometric series!

STEP 7

Recall that $\frac{1}{1 - r} = \sum_{n=0}^{\infty} r^n$ for $|r| < 1$ .
In our case, $r = \frac{x}{5}$ , so: $f'(x) = -\frac{1}{5} \sum_{n=0}^{\infty} \left(\frac{x}{5}\right)^n = -\sum_{n=0}^{\infty} \frac{x^n}{5^{n+1}}$ This is valid when $\left|\frac{x}{5}\right| < 1$ , or $|x| < 5$ .

STEP 8

Now, let's integrate both sides with respect to $x$ to get back to $f(x)$ : $\int f'(x)\,dx = \int \left(-\sum_{n=0}^{\infty} \frac{x^n}{5^{n+1}}\right) dx$ $f(x) = -\sum_{n=0}^{\infty} \frac{x^{n+1}}{(n+1)5^{n+1}} + C$

STEP 9

To find $C$ , we can plug in $x = 0$ : $f(0) = \ln(5 - 0) = \ln(5)$ $f(0) = -\sum_{n=0}^{\infty} \frac{0^{n+1}}{(n+1)5^{n+1}} + C = C$ So, $C = \ln(5)$ .

STEP 10

We can rewrite the series starting from $n=1$ by shifting the index.
Let $k = n+1$ , then $n = k-1$ : $f(x) = \ln(5) - \sum_{k=1}^{\infty} \frac{x^k}{k \cdot 5^k}$ Replacing $k$ with $n$ , we get: $f(x) = \ln(5) - \sum_{n=1}^{\infty} \frac{x^n}{n \cdot 5^n}$

STEP 11

Remember, our geometric series representation was valid for $|x| < 5$ .

STEP 12

Therefore, the radius of convergence is $R = 5$ .

STEP 13

The power series representation for $f(x) = \ln(5-x)$ is $\ln(5) - \sum_{n=1}^{\infty} \frac{x^n}{n \cdot 5^n}$ The radius of convergence is $R = 5$ .

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QuestionFind a power series representation for the function. (Give $y$ $\begin{array}{c} f(x)=\ln (5-x) \\ f(x)=\ln (5)-\sum_{n=1}^{\infty}(\square) \end{array}$
Determine the radius of convergence, $R$ . $R=$ $\square$ Need Help? Read It Watch It Master It Submit Answer

Studdy Solution

STEP 1

STEP 2

1. Rewrite the function
2. Find the derivative
3. Express as a geometric series
4. Integrate to find the power series
5. Determine the radius of convergence

STEP 3

We want to rewrite our function $f(x) = \ln(5 - x)$ to look a bit more like the standard form $\ln(1 + u)$ .
Let's factor out a 5: $f(x) = \ln\left(5\left(1 - \frac{x}{5}\right)\right)$

STEP 4

Using the logarithm property $\ln(ab) = \ln(a) + \ln(b)$ , we can rewrite this as: $f(x) = \ln(5) + \ln\left(1 - \frac{x}{5}\right)$ Now we have a $\ln(1+u)$ form, where $u = -\frac{x}{5}$ .

STEP 5

STEP 6

We can rewrite $f'(x)$ as: $f'(x) = -\frac{1}{5}\cdot\frac{1}{1-\frac{x}{5}}$ This looks like the sum of a geometric series!

STEP 7

Recall that $\frac{1}{1 - r} = \sum_{n=0}^{\infty} r^n$ for $|r| < 1$ .
In our case, $r = \frac{x}{5}$ , so: $f'(x) = -\frac{1}{5} \sum_{n=0}^{\infty} \left(\frac{x}{5}\right)^n = -\sum_{n=0}^{\infty} \frac{x^n}{5^{n+1}}$ This is valid when $\left|\frac{x}{5}\right| < 1$ , or $|x| < 5$ .

STEP 8

Now, let's integrate both sides with respect to $x$ to get back to $f(x)$ : $\int f'(x)\,dx = \int \left(-\sum_{n=0}^{\infty} \frac{x^n}{5^{n+1}}\right) dx$ $f(x) = -\sum_{n=0}^{\infty} \frac{x^{n+1}}{(n+1)5^{n+1}} + C$

STEP 9

To find $C$ , we can plug in $x = 0$ : $f(0) = \ln(5 - 0) = \ln(5)$ $f(0) = -\sum_{n=0}^{\infty} \frac{0^{n+1}}{(n+1)5^{n+1}} + C = C$ So, $C = \ln(5)$ .

STEP 10

We can rewrite the series starting from $n=1$ by shifting the index.
Let $k = n+1$ , then $n = k-1$ : $f(x) = \ln(5) - \sum_{k=1}^{\infty} \frac{x^k}{k \cdot 5^k}$ Replacing $k$ with $n$ , we get: $f(x) = \ln(5) - \sum_{n=1}^{\infty} \frac{x^n}{n \cdot 5^n}$

STEP 11

Remember, our geometric series representation was valid for $|x| < 5$ .

STEP 12

Therefore, the radius of convergence is $R = 5$ .

STEP 13

The power series representation for $f(x) = \ln(5-x)$ is $\ln(5) - \sum_{n=1}^{\infty} \frac{x^n}{n \cdot 5^n}$ The radius of convergence is $R = 5$ .

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Studdy Solution

STEP 1

STEP 2

1. Rewrite the function2. Find the derivative3. Express as a geometric series4. Integrate to find the power series5. Determine the radius of convergence

STEP 3

We want to **rewrite** our function f(x)=ln⁡(5−x)f(x) = \ln(5 - x)f(x)=ln(5−x) to look a bit more like the standard form ln⁡(1+u)\ln(1 + u)ln(1+u).Let's **factor out** a 5: f(x)=ln⁡(5(1−x5))f(x) = \ln\left(5\left(1 - \frac{x}{5}\right)\right)f(x)=ln(5(1−5x​))

STEP 4

STEP 5

STEP 6

We can **rewrite** f′(x)f'(x)f′(x) as: f′(x)=−15⋅11−x5f'(x) = -\frac{1}{5}\cdot\frac{1}{1-\frac{x}{5}}f′(x)=−51​⋅1−5x​1​ This looks like the sum of a **geometric series**!

STEP 7

STEP 8

STEP 9

To **find** CCC, we can plug in x=0x = 0x=0: f(0)=ln⁡(5−0)=ln⁡(5)f(0) = \ln(5 - 0) = \ln(5)f(0)=ln(5−0)=ln(5) f(0)=−∑n=0∞0n+1(n+1)5n+1+C=Cf(0) = -\sum_{n=0}^{\infty} \frac{0^{n+1}}{(n+1)5^{n+1}} + C = Cf(0)=−n=0∑∞​(n+1)5n+10n+1​+C=CSo, C=ln⁡(5)C = \ln(5)C=ln(5).

STEP 10

STEP 11

Remember, our geometric series representation was **valid** for ∣x∣<5|x| < 5∣x∣<5.

STEP 12

Therefore, the **radius of convergence** is R=5R = 5R=5.

STEP 13

The power series representation for f(x)=ln⁡(5−x)f(x) = \ln(5-x)f(x)=ln(5−x) is ln⁡(5)−∑n=1∞xnn⋅5n\ln(5) - \sum_{n=1}^{\infty} \frac{x^n}{n \cdot 5^n}ln(5)−n=1∑∞​n⋅5nxn​ The radius of convergence is R=5R = 5R=5.

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1. Rewrite the function
2. Find the derivative
3. Express as a geometric series
4. Integrate to find the power series
5. Determine the radius of convergence

We want to rewrite our function $f(x) = \ln(5 - x)$ to look a bit more like the standard form $\ln(1 + u)$ .
Let's factor out a 5: $f(x) = \ln\left(5\left(1 - \frac{x}{5}\right)\right)$

We can rewrite $f'(x)$ as: $f'(x) = -\frac{1}{5}\cdot\frac{1}{1-\frac{x}{5}}$ This looks like the sum of a geometric series!

To find $C$ , we can plug in $x = 0$ : $f(0) = \ln(5 - 0) = \ln(5)$ $f(0) = -\sum_{n=0}^{\infty} \frac{0^{n+1}}{(n+1)5^{n+1}} + C = C$ So, $C = \ln(5)$ .

Remember, our geometric series representation was valid for $|x| < 5$ .

Therefore, the radius of convergence is $R = 5$ .

The power series representation for $f(x) = \ln(5-x)$ is $\ln(5) - \sum_{n=1}^{\infty} \frac{x^n}{n \cdot 5^n}$ The radius of convergence is $R = 5$ .