Generate two equivalent fractions for each fraction. Use fraction tikes or number lines.<br> 1/3

(a) Suppose anxn has finite radius of convergence R and an ≥ 0 for all n. Show that if the series converges at R, then it also c

valina [46]

Answer:

a) See the proof below.

b) $\sum \frac{(-x)^n}{n}$

Step-by-step explanation:

Part a

For this case we assume that we have the following series $\sum a)n x^n$ and this series has a finite radius of convergence $R$ and we assume that $a_n \geq 0$ for all n, this information is given by the problem.

We assume that the series converges at the point $x= R$ since w eknwo that converges, and since converges we can conclude that:

$\sum a)n R^n < \infty$

For this case we need to show that converges also for $x=-R$

So we need to proof that $\sum a_n (-R)^n < \infty$

We can do some algebra and we can rewrite the following expression like this:

$\sum a_n (-R)^n = \sum (-1)^n a)n R^n$ and we see that the last series is alternating.

Since we know that $\sum a_n x^n$ converges then the sequence { $a_n R^n$ } must be positive and we need to have $lim_{n\to \infty} a^n R^n = 0$

And then by the alternating series test we can conclude that $\sum a_n (-R)^n$ also converges. And then we conclude that the power series $a_n x^n$ converges for $x=-R$ ,and that complete the proof.

Part b

For this case we need to provide a series whose interval of convergence is exactly (-1,1]

And the best function for this $\frac{(-x)^n}{n}$

Because the series $\sum \frac{(-x)^n}{n}$ converges to $-ln(1+x)$ when $|x|$ using the root test.

But by the properties of the natural log the series diverges at $x=-1$ because $\sum \frac{1}{n} =\infty$ and for $x=1$ we know that converges since $\sum \frac{-1}{n}$ is an alternating series that converges because the expression tends to 0.

6 0

3 years ago

Identify the polygon that has vertices A(−10,−1), P(−7,3), E(−3,0), and X(−6,−4), and then find the perimeter and area of the po

Bogdan [553]

Answer:

square; perimeter 20 units; area 25 square units.

Step-by-step explanation:

As the attachment shows, each side of the polygon is the hypotenuse of a 3-4-5 right triangle, so has length 5 units. The perimeter is the sum of those lengths, 4×5 = 20; the area is the product of the lengths of adjacent sides, 5×5 = 25.

The figure is a square of side length 5 units.

The perimeter is 20 units; the area is 25 square units.

4 0

3 years ago

For any number the polynomial x ^ 2 - c ^ 2 can be factored as

ivann1987 [24]

Answer:

(x + c) (x - c)

Step-by-step explanation:

For any number the polynomial $x ^ 2 - c ^ 2$ can be factored as (x + c) (x - c)

4 0

3 years ago

Read 2 more answers

Please help? Graph <img src="https://tex.z-dn.net/?f=-3x%5E2%2B12y%5E2%3D84" id="TexFormula1" title="-3x^2+12y^2=84" alt="-3x^2+

mariarad [96]

$-3x^2+12y^2=84$
$12y^2=3x^2+84$
$y^2=\dfrac{x^2}4+7$
$y=\pm\sqrt{\dfrac{x^2}4+7}$

For either square root to exist, you require that $\dfrac{x^2}4+7\ge0$ . This is true for all $x$ , since $\dfrac{x^2}4$ is always non-negative. This means the domain of $y$ as a function of $x$ is all real numbers, or $x\in\mathbb R$ or $(-\infty,infty)$ .

Now, because $\dfrac{x^2}4+7$ is non-negative, and the smallest value it can take on is 7, it follows that the minimum value for the positive square root must be $\sqrt7$ , while the maximum value of the negative root must be $-\sqrt7$ . This means the range is $y\in\mathbb R\setminus(-\sqrt7,\sqrt7)$ , or $|y|\ge\sqrt7$ , or $(-\infty,-\sqrt7]\cup[\sqrt7,\infty)$ .

6 0

4 years ago

If each month in reno had the same average rainfall as in august, what would the total number of millimeters be after 12 months

Inessa [10]

Answer:

12 times the August rainfall

Step-by-step explanation:

The total of 12 of the same value is 12 times that value. Multiply the August average by 12 to get your answer.

3 0

3 years ago

Generate two equivalent fractions for each fraction. Use fraction tikes or number lines. 1/3