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## How do I sum the series sum_(n=1)^oon(3/4)^n?

First of all: "How much is the sum of a geometric series?".

${\sum}_{n = 0}^{+ \infty} {r}^{n} = {a}_{0} / \left(1 - r\right)$, where r (such as $| r | < 1$) is called "ratio" ($r = {a}_{k} / {a}_{k - 1}$) and ${a}_{0}$ is the first term of the series (in this case ${a}_{0} = {r}^{0} = 1$).

Let's list all the terms of the series:

$1 \cdot \frac{3}{4} + 2 \cdot {\left(\frac{3}{4}\right)}^{2} + 3 \cdot {\left(\frac{3}{4}\right)}^{3} + 4 \cdot {\left(\frac{3}{4}\right)}^{4} + \ldots$, but we can also say:

$\frac{3}{4} + \frac{9}{16} + \frac{9}{16} + \frac{27}{64} + \frac{27}{64} + \frac{27}{64} + \frac{81}{256} + \frac{81}{256} + \frac{81}{256} + \frac{81}{256} + \ldots$,

or, better:

$\frac{3}{4} + \frac{9}{16} + \frac{27}{64} + \frac{81}{256} + \ldots +$

$\frac{9}{16} + \frac{27}{64} + \frac{81}{256} + \ldots +$

$\frac{27}{64} + \frac{81}{256} + \ldots +$

$\frac{81}{256} + \ldots +$

$\ldots$

The first sum is:

${\sum}_{n = 1}^{+ \infty} {\left(\frac{3}{4}\right)}^{n}$,

The second sum is:

${\sum}_{n = 2}^{+ \infty} {\left(\frac{3}{4}\right)}^{n}$,

The third sum is:

${\sum}_{n = 3}^{+ \infty} {\left(\frac{3}{4}\right)}^{n}$,

The fourth sum is:

${\sum}_{n = 4}^{+ \infty} {\left(\frac{3}{4}\right)}^{n}$,

$\ldots$

So:

${\sum}_{n = 1}^{+ \infty} n {\left(\frac{3}{4}\right)}^{n} = {\sum}_{n = 1}^{+ \infty} {\left(\frac{3}{4}\right)}^{n} + {\sum}_{n = 2}^{+ \infty} {\left(\frac{3}{4}\right)}^{n} + {\sum}_{n = 3}^{+ \infty} {\left(\frac{3}{4}\right)}^{n} + {\sum}_{n = 4}^{+ \infty} {\left(\frac{3}{4}\right)}^{n} + \ldots =$.

$= \frac{3}{4} \cdot \frac{1}{1 - \frac{3}{4}} + \frac{9}{16} \cdot \frac{1}{1 - \frac{3}{4}} + \frac{27}{64} \cdot \frac{1}{1 - \frac{3}{4}} + \frac{81}{256} \cdot \frac{1}{1 - \frac{3}{4}} + \ldots =$

$= \frac{3}{4} \cdot 4 + \frac{9}{16} \cdot 4 + \frac{27}{64} \cdot 4 + \frac{81}{256} \cdot 4 + \ldots =$

$= 4 \left(\frac{3}{4} + \frac{9}{16} + \frac{27}{64} + \frac{81}{256} + \ldots\right) =$

$= 4 \cdot {\sum}_{n = 1}^{+ \infty} {\left(\frac{3}{4}\right)}^{n} = 4 \cdot \frac{3}{4} \cdot \frac{1}{1 - \frac{3}{4}} = 4 \cdot \frac{3}{4} \cdot 4 = 12$.

## Water is leaking out of an inverted conical tank at the rate of 10,000cm^3/min cm/min at the same time that water is being pumped into the tank at a constant rate. The tank has a height of 6m and the diameter at the top is 4m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2m, how can I find the rate at which water is being pumped into the tank.?

(see below for solution method)

#### Explanation:

Start by ignoring the leakage and determine the rate of inflow required to achieve the specified rate of height (depth) of water increase.

Later we'll use the fact that
Actual inflow rate
= Inflow Rate for Increased Depth + Leakage Rate For the given cone the ratio of r adius to h eight is $\frac{1}{3}$
so
$r = \frac{1}{3} h$

The formula for the volume of a cone:
$V = \frac{\pi {r}^{2} h}{3}$ becomes $V = \frac{\pi {h}^{3}}{27}$

$\frac{d V}{\mathrm{dh}} = \frac{\pi {h}^{2}}{9}$

We are interested in the change in Volume with respect to time and note that
$\frac{d V}{\mathrm{dt}} = \frac{d V}{\mathrm{dh}} \cdot \frac{d h}{\mathrm{dt}}$

Using the value we've already calculated for $\frac{d V}{\mathrm{dh}}$ and the supplied value of $20$ cm/min (at a height of $h = 200$ cm)
we get:

$\frac{d V}{\mathrm{dt}} = \frac{\pi {\left(200 c m\right)}^{2} \cdot \left(20 c m\right)}{9 \min}$
$= \frac{800000 \pi}{9}$ $c {m}^{3}$/min
or roughly
$279 , 252.7$ $c {m}^{3}$/min

This is the Inflow Rate Required to Cause Height Increase and
ignores the Rate of Leakage

The Actual Inflow Rate needs to be the sum of these two:
$279 , 252.7$ $c {m}^{3}$/min $+ 10 , 000$ $c {m}^{3}$/min
$= 289 , 252.7$ $c {m}^{3}$/min

An interesting example related to this one is to consider the piecewise-defined function $f$ given by the equations $f \left(x\right) = {x}^{2} \sin \left(\frac{1}{x}\right)$ for $x \ne 0$ and $f \left(0\right) = 0$. Even though this function has infinitely many oscillations near the origin, the derivative $f ' \left(0\right)$ exists. On the other hand, the derivative $f '$ is not continuous at $x = 0$.

#### Explanation:

Again, let $f$ be defined in a piecewise way by $f \left(x\right) = {x}^{2} \sin \left(\frac{1}{x}\right)$ when $x \ne 0$ and $f \left(0\right) = 0$.

Certainly, when $x \ne 0$ the function is differentiable and continuous and the Product Rule and Chain Rule can be used to say that:

$f ' \left(x\right) = 2 x \cdot \sin \left(\frac{1}{x}\right) + {x}^{2} \cdot \cos \left(\frac{1}{x}\right) \cdot \left(- 1\right) {x}^{- 2}$

$= 2 x \sin \left(\frac{1}{x}\right) - \cos \left(\frac{1}{x}\right)$ for $x \ne 0$.

But $f$ is continuous everywhere (${\lim}_{x \to 0} f \left(x\right) = 0 = f \left(0\right)$ by the Squeeze Theorem) and $f ' \left(0\right)$ happens to exist as well! Do a limit calculation:

$f ' \left(0\right) = {\lim}_{h \to 0} \frac{f \left(0 + h\right) - f \left(0\right)}{h} = {\lim}_{h \to 0} \frac{{h}^{2} \cdot \sin \left(\frac{1}{h}\right) - 0}{h}$

$= {\lim}_{h \to 0} h \sin \left(\frac{1}{h}\right) = 0$, where the last equality follows by the Squeeze Theorem.

In other words, $f ' \left(0\right) = 0$.

Here's a graph of the function $f$ (blue) and its derivative $f '$ (red). Even though the derivative exists everywhere, it is not well-behaved near the origin. Not only does it have infinitely many oscillations as $x \to 0$, but the oscillations never decrease below 1 in amplitude (and ${\lim}_{x \to 0} f ' \left(x\right)$ fails to exist so that $f '$ is not continuous at $x = 0$). On the other hand, the graph of the derivative contains the point $\left(0 , 0\right)$ as well.

Pretty amazing, isn't it?!?! Here a closer view near the origin and also making sure we can still see the blue curve. It's oscillating infinitely often as $x \to 0$ as well, though the oscillations are rapidly decreasing in amplitude as $x \to 0$. ## How do you use the Squeeze Theorem to find lim Sin(x)/x as x approaches zero?

For a non-rigorous proof, please see below.

#### Explanation:

For a positive central angle of $x$ radians ($0 < x < \frac{\pi}{2}$) (not degrees) Source: The geometric idea is that

$\text{Area of "Delta KOA < "Area of " "Sector KOA" < "Area of } \Delta L O A$

$\text{Area of } \Delta K O A = \frac{1}{2} \left(1\right) \left(\sin x\right) \setminus \setminus \setminus$ ($\frac{1}{2} \text{base"*"height}$)

$\text{Area of " "Sector KOA} = \frac{1}{2} {\left(1\right)}^{2} x \setminus \setminus \setminus$ ($x$ is in radians)

$\text{Area of } \Delta L O A = \frac{1}{2} \tan x \setminus \setminus \setminus$ ($A L = \tan x$)

So we have:

$\sin \frac{x}{2} < \frac{x}{2} < \tan \frac{x}{2}$

For small positive $x$, we have $\in x > 0$ so we can multiply through by $\frac{2}{\sin} x$, to get

$1 < \frac{x}{\sin} x < \frac{1}{\cos} x$

So

$\cos x < \sin \frac{x}{x} < 1$ for $0 < x < \frac{\pi}{2}$.

${\lim}_{x \rightarrow {0}^{+}} \cos x = 1$ and ${\lim}_{x \rightarrow {0}^{+}} 1 = 1$

so ${\lim}_{x \rightarrow {0}^{+}} \sin \frac{x}{x} = 1$

We also have, for these small $x$, $\sin \left(- x\right) = - \sin x$, so $\frac{- x}{\sin} \left(- x\right) = \frac{x}{\sin} x$ and $\cos \left(- x\right) = \cos x$, so

$\cos x < \sin \frac{x}{x} < 1$ for $- \frac{\pi}{2} < x < 0$.

${\lim}_{x \rightarrow {0}^{-}} \cos x = 1$ and ${\lim}_{x \rightarrow {0}^{-}} 1 = 1$

so ${\lim}_{x \rightarrow {0}^{-}} \sin \frac{x}{x} = 1$

Since both one sided limits are $1$, the limit is $1$.

Note

This proof uses the fact that ${\lim}_{x \rightarrow 0} \cos x = 1$. That can also be stated "the cosine function is continuous at $0$".

That fact can be proved from the fact that ${\lim}_{x \rightarrow 0} \sin x = 0$. (The sine function is continuous at $0$.)
Which can be proved using the squeeze theorem in a argument rather like the one used above.

Furthermore: Using both of those facts we can show that the sine and cosine functions are continuous at every real number.

## Can a point of inflection be undefined?

See the explanation section below.

#### Explanation:

A point of inflection is a point on the graph at which the concavity of the graph changes.

If a function is undefined at some value of $x$, there can be no inflection point.

However, concavity can change as we pass, left to right across an $x$ values for which the function is undefined.

Example

$f \left(x\right) = \frac{1}{x}$ is concave down for $x < 0$ and concave up for $x > 0$.

The concavity changes "at" $x = 0$.

But, since $f \left(0\right)$ is undefined, there is no inflection point for the graph of this function.

graph{1/x [-10.6, 11.9, -5.985, 5.265]}

Example 2

$f \left(x\right) = \sqrt{x}$ is concave up for $x < 0$ and concave down for $x > 0$.

$f ' \left(x\right) = \frac{1}{3 {\sqrt{x}}^{2}}$ and $f ' \left(x\right) = \frac{- 2}{9 {\sqrt{x}}^{5}}$

The second derivative is undefined at $x = 0$.

But, since $f \left(0\right)$ is defined, there is an inflection point for the graph of this function. Namely, $\left(0 , 0\right)$

graph{x^(1/3) [-3.735, 5.034, -2.55, 1.835]}

## For f(x)=(2x+1)/(x+2)  what is the equation of the tangent line at x=1?

I found: $y = \frac{1}{3} x + \frac{2}{3}$

#### Explanation:

First you need to find the slope $m$ of the tangent line. This is found deriving your function and calculating it at $x = 1$:

$f ' \left(x\right) = \frac{2 \left(x + 2\right) - \left(2 x + 1\right)}{x + 2} ^ 2 = \frac{2 x + 4 - 2 x - 1}{x + 2} ^ 2 = \frac{3}{x + 2} ^ 2$

at $x = 1$
$f ' \left(3\right) = \frac{3}{1 + 2} ^ 2 = \frac{1}{3} = m$

Now we need the $y$ value of the tangence point as well; we find it setting $x = 1$ into the original function:
$f \left(1\right) = \frac{2 + 1}{1 + 2} = \frac{3}{3} = 1$

The equation of a line passing through ${x}_{0} = 1 \mathmr{and} {y}_{0} = 1$ with slope $m = \frac{1}{3}$ is given as:
$y - {y}_{0} = m \left(x - {x}_{0}\right)$
$y - 1 = \frac{1}{3} \left(x - 1\right)$
$y = \frac{1}{3} x - \frac{1}{3} + 1$
$y = \frac{1}{3} x + \frac{2}{3}$

Graphically: ## What is the derivative of arc csc(sqrt(x+1))?

$\frac{d}{\mathrm{dx}} \text{arccsc} \left(\sqrt{x + 1}\right) = - \frac{1}{2 \left(x + 1\right) \sqrt{x}}$

#### Explanation:

To evaluate inverse trigonometric functions, a valuable technique is implicit differentiation:
$\frac{d}{\mathrm{dx}} f \left(y\right) = \frac{d}{\mathrm{dy}} f \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}}$

In this case, we also will use the chain rule:
$\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = f ' \left(g \left(x\right)\right) g ' \left(x\right)$

as well as the following derivatives:
$\frac{d}{\mathrm{dx}} \csc \left(x\right) = - \csc \left(x\right) \cot \left(x\right)$
$\frac{d}{\mathrm{dx}} {x}^{n} = n {x}^{n - 1}$

Now, let $y = \text{arccsc} \left(\sqrt{x + 1}\right)$
$\implies \csc \left(y\right) = \csc \left(\text{arccsc} \left(\sqrt{x + 1}\right)\right) = \sqrt{x + 1}$
$\implies \frac{d}{\mathrm{dx}} \csc \left(y\right) = \frac{d}{\mathrm{dx}} \sqrt{x + 1}$

Through implicit differentiation:
$\frac{d}{\mathrm{dx}} \csc \left(y\right) = \frac{d}{\mathrm{dy}} \csc \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = - \csc \left(y\right) \cot \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}}$

And through the chain rule
$\frac{d}{\mathrm{dx}} {\left(x + 1\right)}^{\frac{1}{2}} = \frac{1}{2} {\left(x + 1\right)}^{- \frac{1}{2}} \left(\frac{d}{\mathrm{dx}} \left(x + 1\right)\right) = \frac{1}{2 \sqrt{x + 1}}$

So
$- \csc \left(y\right) \cot \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{x + 1}}$
$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{2 \sqrt{x + 1} \cdot \csc \left(y\right) \cot \left(y\right)}$

But we want our final answer entirely in terms of $x$. To fix this, we remember that $\csc \left(y\right) = \sqrt{x + 1}$ and draw a corresponding right triangle: From this we can see that $\cot \left(y\right) = \sqrt{x}$
Thus we have
$- \frac{1}{2 \sqrt{x + 1} \cdot \csc \left(y\right) \cot \left(y\right)} = - \frac{1}{2 \sqrt{x + 1} \cdot \sqrt{x + 1} \cdot \sqrt{x}}$

From this, we can get our final answer

$\frac{d}{\mathrm{dx}} \text{arccsc} \left(\sqrt{x + 1}\right) = - \frac{1}{2 \left(x + 1\right) \sqrt{x}}$

## How do you find int (x^2) / (x^2 -3x +2) dx using partial fractions?

1) make sure that the degree of your denominator is greater than the degree of your numerator
2) factorize the denominator
3) perform the partial fraction decomposition
4) solve the integral!

Solution: $x + 8 \ln | x - 2 | - 5 \ln | x - 1 | + c$

#### Explanation:

1) Check degrees of denominator and numerator

First of all, you can't immediately start building partial fractions since the degree of your numerator is equal to the degree of your denominator (the strongest power in both expressions is ${x}^{2}$).

However, the partial fraction decomposition will only work if the degree of the denominator is greater than the degree of the numerator.

If the expression was more complicated, normally, at this point, a polynomial division is in order. Here we can achieve the goal in an easier way:

${x}^{2} / \left({x}^{2} - 3 x + 2\right) = \frac{{x}^{2} \textcolor{b l u e}{- 3 x + 2} \textcolor{red}{+ 3 x - 2}}{{x}^{2} - 3 x + 2}$

$\textcolor{w h i t e}{\times \times \times \times x} = \frac{{x}^{2} - 3 x + 2}{{x}^{2} - 3 x + 2} + \frac{3 x - 2}{{x}^{2} - 3 x + 2}$

$\textcolor{w h i t e}{\times \times \times \times x} = 1 + \frac{3 x + 2}{{x}^{2} - 3 x + 2}$

So, now we can build the partial fractions decomposition of the latter fraction.

2) Complete factorization of the denominator

To do so, let's factorize the denominator completely first:

${x}^{2} - 3 x + 2 = \left(x - 2\right) \left(x - 1\right)$

[You can do so by setting ${x}^{2} - 3 x + 2 = 0$, finding the solutions ${x}_{1}$ and ${x}_{2}$ and factorizing with $\left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$.]

3) Partial fraction decomposition

So, our goal is to find $A$ and $B$ so that

$\frac{3 x + 2}{\left(x - 2\right) \left(x - 1\right)} = \frac{A}{x - 2} + \frac{B}{x - 1}$

First, multiply both sides of the equation with $\left(x - 2\right) \left(x - 1\right)$.

$\iff 3 x + 2 = A \cdot \left(x - 1\right) + B \cdot \left(x - 2\right)$
$\iff \textcolor{b l u e}{3 x} + \textcolor{red}{2} = \textcolor{b l u e}{A \cdot x} \textcolor{red}{\text{ "- A) + color(blue)(B * x) color(red)(" } - 2 B}$

To find the solution of this equation, we need to "collect" the $\textcolor{b l u e}{x}$ terms and the $\textcolor{red}{\text{constant}}$ terms:

$\left\{\begin{matrix}3 = A + B \textcolor{w h i t e}{\times \times \times x} \textcolor{b l u e}{x} \text{ terms" \\ 2 = -A - 2B color(white)(xxxx) color(red)("constant")" terms}\end{matrix}\right.$

The solution of this linear equation is $B = - 5$ and $A = 8$.

So, our partial fraction decomposition is:

$\frac{3 x + 2}{\left(x - 2\right) \left(x - 1\right)} = \frac{8}{x - 2} - \frac{5}{x - 1}$

In total:

${x}^{2} / \left({x}^{2} - 3 x + 2\right) = 1 + \frac{3 x + 2}{{x}^{2} - 3 x + 2} = 1 + \frac{8}{x - 2} - \frac{5}{x - 1}$

4) Solving the integral

The last thing left to do is solve the integral!

$\int {x}^{2} / \left({x}^{2} - 3 x + 2\right) \text{d" x = int (1 + 8 / (x-2) - 5 / (x-1)) "d} x$

$\textcolor{w h i t e}{\times \times \times \times \times \times} = \int 1 \text{d"x + int 8/(x-2) "d" x - int 5/(x-1) "d} x$

$\textcolor{w h i t e}{\times \times \times \times \times \times} = x + 8 \ln | x - 2 | - 5 \ln | x - 1 | + c$

Hope that this helped!

## How do you differentiate f(x)=cose^(4x) using the chain rule.?

$f ' \left(x\right) = - 4 \sin \left({e}^{4 x}\right) \cdot {e}^{4 x}$

#### Explanation:

Your chain can be defined as follows:

$f \left(x\right) = \textcolor{g r e e n}{\cos} \textcolor{b l u e}{{e}^{\textcolor{red}{4 x}}}$

$\textcolor{w h i t e}{\times x} = \cos {e}^{w \left(x\right)} \textcolor{w h i t e}{\times x} \text{ where } w \left(x\right) = \textcolor{red}{4 x}$

$\textcolor{w h i t e}{\times x} = \cos v \left(w\right) \textcolor{w h i t e}{\times \xi i i} \text{ where } v \left(w\right) = \textcolor{b l u e}{{e}^{w}}$

$\textcolor{w h i t e}{\times x} = u \left(v\right) \textcolor{w h i t e}{\times \times \times i} \text{ where } u \left(v\right) = \textcolor{g r e e n}{\cos \left(v\right)}$

Thus, to compute the derivative, you need to build the derivatives of $w \left(x\right)$, $v \left(w\right)$ and $u \left(v\right)$:

$w \left(x\right) = 4 x \textcolor{w h i t e}{\times x} \implies \textcolor{w h i t e}{\times} w ' \left(x\right) = 4$

$v \left(w\right) = {e}^{w} \textcolor{w h i t e}{\times x} \implies \textcolor{w h i t e}{\times} v ' \left(w\right) = {e}^{w} = {e}^{4 x}$

$u \left(v\right) = \cos v \textcolor{w h i t e}{\times} \implies \textcolor{w h i t e}{\times} u ' \left(v\right) = - \sin v = - \sin \left({e}^{w}\right) = - \sin \left({e}^{4 x}\right)$

Now, the only thing left to do is to multiply the three derivatives!

$f ' \left(x\right) = u ' \left(v\right) \cdot v ' \left(w\right) \cdot w ' \left(x\right)$
$\textcolor{w h i t e}{\times \times} = - \sin \left({e}^{4 x}\right) \cdot {e}^{4 x} \cdot 4$
$\textcolor{w h i t e}{\times \times} = - 4 \sin \left({e}^{4 x}\right) \cdot {e}^{4 x}$

The "why" depends on how you've defined ${e}^{x}$.

#### Explanation:

Define $\ln x$ first

One approach is to define $\ln x = {\int}_{1}^{x} \frac{1}{t} \mathrm{dt}$ and
then to define $\exp \left(x\right)$ to be the inverse of $\ln x$ and,
finally, define ${e}^{x} = \exp \left(x\right)$.

In this case $y = {e}^{x}$ if and only if $\ln y = x$.

Differentiating implicitly gets us $\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = 1$.

So, $\frac{\mathrm{dy}}{\mathrm{dx}} = y = {e}^{x}$

Define ${e}^{x}$ independently of $\ln x$

Definition 1

For positive $a$, define

${a}^{x} = {\lim}_{r \rightarrow x} {a}^{x}$ with $r$ in the rational numbers
(We owe you a proof that this is well-defined.)

Then, using the definition of derivative:

$\frac{d}{\mathrm{dx}} \left({a}^{x}\right) = {\lim}_{h \rightarrow 0} \frac{{a}^{x + h} - {a}^{x}}{h} = \lim \left(h \rightarrow 0\right) \left({a}^{x} \frac{{a}^{h} - 1}{h}\right)$

We then define $e$ to be the number that satisfies ${\lim}_{h \rightarrow 0} \frac{{e}^{h} - 1}{h} = 1$

With this definition we get
$\frac{d}{\mathrm{dx}} \left({e}^{x}\right) = {\lim}_{h \rightarrow 0} \frac{{e}^{x + h} - {e}^{x}}{h} = {\lim}_{h \rightarrow 0} \left({e}^{x} \frac{{e}^{h} - 1}{h}\right)$

$= {e}^{x} {\lim}_{h \rightarrow 0} \left(\frac{{e}^{h} - 1}{h}\right) = {e}^{x}$

Definition 2

e^x = 1+x/1+x^2/(2*1)+x^3/(3*2*1) + x^4/(4!)+x^5/(5!)+ * * *

(n! = n(n-1)(n-2)* * * (1))
(We owe you a proof that this is well defined.)

Differentiating term by term (we owe you a proof that this is possible), we get

d/dx(e^x) = 0+1+(cancel(2)x)/(cancel(2)*1)+(cancel(3)x^2)/(cancel(3)*2*1) + x^3/(3!)+x^4/(4!)+* * *

Which simplifies to 1+x/1+x^2/(2*1)+x^3/(3*2*1) + x^4/(4!)+x^5/(5!)+ * * *  which is again ${e}^{x}$