The problem consists of several calculus questions. We are asked to: a. Find the tangent plane to $z = \ln(2x+y)$ at $(1,2)$. b. Show the conditions for the convergence and divergence of a p-series $\sum_{x=1}^{\infty} \frac{1}{x^p}$. c. Find $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2}$ given $u = (x^2+y^2+z^2)^{1/2}$. d. If $a_n > 0$ for all $n$ and $\frac{a_{n+1}}{a_n} < k < 1$, show $\sum a_n$ converges, and then show $\sum \frac{2^n}{n^{3+1}}$ diverges. (Typo: $n^{3+1}$ likely means $n^3 + 1$). e. Use 5-segment Mid-ordinate and Simpson's one-third rule to integrate $f(x) = 2+5x+3x^2+2x^3$ from $a=0$ to $b=2.5$ and determine the relative error. f. Find the slopes of the traces to $z = 10-4x^2-y^2$ at the point $(1,2)$. g. Find the radius of convergence of the power series $\sum_{n=0}^{\infty} \frac{n(-1)^n}{4^n}(x+3)^n$.

1. Problem Description

The problem consists of several calculus questions. We are asked to:

a. Find the tangent plane to $z = \ln(2x+y)$ at $(1,2)$.

b. Show the conditions for the convergence and divergence of a p-series $\sum_{x=1}^{\infty} \frac{1}{x^p}$.

c. Find $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2}$ given $u = (x^2+y^2+z^2)^{1/2}$.

d. If $a_n > 0$ for all $n$ and $\frac{a_{n+1}}{a_n} < k < 1$, show $\sum a_n$ converges, and then show $\sum \frac{2^n}{n^{3+1}}$ diverges. (Typo: $n^{3+1}$ likely means $n^3 + 1$).

e. Use 5-segment Mid-ordinate and Simpson's one-third rule to integrate $f(x) = 2+5x+3x^2+2x^3$ from $a=0$ to $b=2.5$ and determine the relative error.

f. Find the slopes of the traces to $z = 10-4x^2-y^2$ at the point $(1,2)$.

g. Find the radius of convergence of the power series $\sum_{n=0}^{\infty} \frac{n(-1)^n}{4^n}(x+3)^n$.

2. Solution Steps

a. Tangent plane to $z = \ln(2x+y)$ at $(1,2)$.

First, find $z_0 = \ln(2(1)+2) = \ln(4)$.

Next, calculate the partial derivatives:

$\frac{\partial z}{\partial x} = \frac{2}{2x+y}$, so $\frac{\partial z}{\partial x}(1,2) = \frac{2}{2(1)+2} = \frac{2}{4} = \frac{1}{2}$.

$\frac{\partial z}{\partial y} = \frac{1}{2x+y}$, so $\frac{\partial z}{\partial y}(1,2) = \frac{1}{2(1)+2} = \frac{1}{4}$.

The equation of the tangent plane is:

$z - z_0 = \frac{\partial z}{\partial x}(x-x_0) + \frac{\partial z}{\partial y}(y-y_0)$

$z - \ln(4) = \frac{1}{2}(x-1) + \frac{1}{4}(y-2)$

$z = \frac{1}{2}x + \frac{1}{4}y - \frac{1}{2} - \frac{1}{2} + \ln(4)$

$z = \frac{1}{2}x + \frac{1}{4}y - 1 + \ln(4)$

b. p-series $\sum_{x=1}^{\infty} \frac{1}{x^p}$.

This is a standard result. The p-series converges if $p > 1$ and diverges if $p \le 1$. This can be proven using the integral test.

c. $u = (x^2+y^2+z^2)^{1/2}$. Find $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2}$.

$\frac{\partial u}{\partial x} = \frac{1}{2}(x^2+y^2+z^2)^{-1/2}(2x) = \frac{x}{\sqrt{x^2+y^2+z^2}}$

$\frac{\partial^2 u}{\partial x^2} = \frac{\sqrt{x^2+y^2+z^2} - x \frac{x}{\sqrt{x^2+y^2+z^2}}}{x^2+y^2+z^2} = \frac{x^2+y^2+z^2 - x^2}{(x^2+y^2+z^2)^{3/2}} = \frac{y^2+z^2}{(x^2+y^2+z^2)^{3/2}}$

Similarly, $\frac{\partial^2 u}{\partial y^2} = \frac{x^2+z^2}{(x^2+y^2+z^2)^{3/2}}$ and $\frac{\partial^2 u}{\partial z^2} = \frac{x^2+y^2}{(x^2+y^2+z^2)^{3/2}}$

$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2} = \frac{y^2+z^2 + x^2+z^2 + x^2+y^2}{(x^2+y^2+z^2)^{3/2}} = \frac{2(x^2+y^2+z^2)}{(x^2+y^2+z^2)^{3/2}} = \frac{2}{\sqrt{x^2+y^2+z^2}}$

d. If $a_n > 0$ and $\frac{a_{n+1}}{a_n} < k < 1$, show $\sum a_n$ converges.

By the ratio test, if $\lim_{n\to\infty} \frac{a_{n+1}}{a_n} < 1$, the series converges. Since $\frac{a_{n+1}}{a_n} < k < 1$, $\lim_{n\to\infty} \frac{a_{n+1}}{a_n} \le k < 1$. Therefore, the series $\sum a_n$ converges.

Show $\sum \frac{2^n}{n^3+1}$ diverges.

Let $a_n = \frac{2^n}{n^3+1}$. Then $\frac{a_{n+1}}{a_n} = \frac{2^{n+1}}{(n+1)^3+1} \cdot \frac{n^3+1}{2^n} = 2 \frac{n^3+1}{(n+1)^3+1} = 2 \frac{n^3+1}{n^3+3n^2+3n+2}$.

$\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = \lim_{n\to\infty} 2 \frac{n^3+1}{n^3+3n^2+3n+2} = 2$. Since the limit is greater than 1, by the ratio test, the series diverges.

e. Integrate $f(x) = 2+5x+3x^2+2x^3$ from $a=0$ to $b=2.5$ using 5 segments.

The width of each segment is $h = \frac{b-a}{n} = \frac{2.5-0}{5} = 0.5$.

Exact value of the integral: $\int_0^{2.5} (2+5x+3x^2+2x^3)dx = [2x + \frac{5}{2}x^2 + x^3 + \frac{1}{2}x^4]_0^{2.5} = 2(2.5) + \frac{5}{2}(2.5)^2 + (2.5)^3 + \frac{1}{2}(2.5)^4 = 5 + 15.625 + 15.625 + 4.8828125 = 41.1328125$.

[i] Mid-ordinate rule:

Midpoints: $x_i = 0.25, 0.75, 1.25, 1.75, 2.25$.

$f(0.25) = 2 + 5(0.25) + 3(0.25)^2 + 2(0.25)^3 = 2 + 1.25 + 0.1875 + 0.03125 = 3.46875$

$f(0.75) = 2 + 5(0.75) + 3(0.75)^2 + 2(0.75)^3 = 2 + 3.75 + 1.6875 + 0.84375 = 8.28125$

$f(1.25) = 2 + 5(1.25) + 3(1.25)^2 + 2(1.25)^3 = 2 + 6.25 + 4.6875 + 3.90625 = 16.84375$

$f(1.75) = 2 + 5(1.75) + 3(1.75)^2 + 2(1.75)^3 = 2 + 8.75 + 9.1875 + 10.71875 = 30.65625$

$f(2.25) = 2 + 5(2.25) + 3(2.25)^2 + 2(2.25)^3 = 2 + 11.25 + 15.1875 + 22.78125 = 51.21875$

Mid-ordinate approximation: $h \sum f(x_i) = 0.5(3.46875 + 8.28125 + 16.84375 + 30.65625 + 51.21875) = 0.5(110.46875) = 55.234375$.

Relative error: $|\frac{41.1328125 - 55.234375}{41.1328125}| = |\frac{-14.1015625}{41.1328125}| = 0.34281 \approx 34.28\%$

[ii] Simpson's one-third rule:

Simpson's rule: $\frac{h}{3} [f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + f(x_4)]$.

$x_0 = 0, x_1 = 0.5, x_2 = 1, x_3 = 1.5, x_4 = 2, x_5=2.5$

Note that since we are given 5 segments, that indicates 6 points to be used, not 5 as mentioned earlier. Thus the limits of integration must have equal number of segments.

$f(0) = 2$, $f(0.5) = 2 + 5(0.5) + 3(0.5)^2 + 2(0.5)^3 = 2 + 2.5 + 0.75 + 0.25 = 5.5$

$f(1) = 2 + 5(1) + 3(1)^2 + 2(1)^3 = 2 + 5 + 3 + 2 = 12$

$f(1.5) = 2 + 5(1.5) + 3(1.5)^2 + 2(1.5)^3 = 2 + 7.5 + 6.75 + 6.75 = 23.0$

$f(2) = 2 + 5(2) + 3(2)^2 + 2(2)^3 = 2 + 10 + 12 + 16 = 40$

$f(2.5) = 2 + 5(2.5) + 3(2.5)^2 + 2(2.5)^3 = 2 + 12.5 + 18.75 + 31.25 = 64.5$

Simpson's rule with 5 segments (6 points) where h = 0.5

$\frac{h}{3} [f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + 2f(x_4) + f(x_5)] = \frac{0.5}{3} [2 + 4(5.5) + 2(12) + 4(23) + 2(40) + 64.5] = \frac{0.5}{3} [2+22+24+92+80+64.5] = \frac{0.5}{3} [284.5] = 47.41666666666667$.

Relative error: $|\frac{41.1328125 - 47.41666666666667}{41.1328125}| = |\frac{-6.283854166666665}{41.1328125}| = 0.152773 \approx 15.28\%$

f. Slopes of the traces to $z = 10-4x^2-y^2$ at the point $(1,2)$.

Trace 1: $y=2$, $z = 10 - 4x^2 - 4$. $z = 6 - 4x^2$.

$\frac{dz}{dx} = -8x$. At $(1,2)$, $\frac{dz}{dx} = -8(1) = -8$.

Trace 2: $x=1$, $z = 10 - 4 - y^2$. $z = 6 - y^2$.

$\frac{dz}{dy} = -2y$. At $(1,2)$, $\frac{dz}{dy} = -2(2) = -4$.

g. Radius of convergence for power series $\sum_{n=0}^{\infty} \frac{n(-1)^n}{4^n}(x+3)^n$.

Let $a_n = \frac{n(-1)^n}{4^n}(x+3)^n$. Using the ratio test:

$\lim_{n\to\infty} |\frac{a_{n+1}}{a_n}| = \lim_{n\to\infty} |\frac{(n+1)(-1)^{n+1}(x+3)^{n+1}}{4^{n+1}} \cdot \frac{4^n}{n(-1)^n(x+3)^n}| = \lim_{n\to\infty} |\frac{n+1}{n} \cdot \frac{1}{4} \cdot (x+3)| = \frac{|x+3|}{4}$.

For convergence, $\frac{|x+3|}{4} < 1$. $|x+3| < 4$. Therefore, the radius of convergence is

4.

3. Final Answer

a. $z = \frac{1}{2}x + \frac{1}{4}y - 1 + \ln(4)$

b. Converges if $p > 1$, diverges if $p \le 1$

c. $\frac{2}{\sqrt{x^2+y^2+z^2}}$

d. Proof shown in solution steps, $\sum \frac{2^n}{n^3+1}$ diverges.

e. [i] 55.234375, relative error: 34.28%

[ii] 47.4167, relative error: 15.28%

f. -8, -4

g. 4