Calc 2 exam 2

Lecture 14: Work

Work= force x distance

Force= -k(x)

In math we use F=k(x)

x represents how much the spring is stretched from its natural length

HOW TO APPROACH SPRING PROBLEMS

identify measurements given in the equation
find k, the spring constant, by plugging in given force and the difference between the spring’s final length and its natural length
1. given F= $k\left(x_2-x_1\right)$
use found spring constant in the work equation.
1. W= $\int kx$
2. since we know k, the new variable is x
integrate the equation to find amount of work required
1. the unit of measurement will be Joules

HOW TO APPROACH WATER PUMP PROBLEMS

identify measurements given in the equation
1. most often given side length of tank and density
2. IT IS COMMON THAT GRAVITY IS THE FORCE OF ACCELERATION AND IS EXPECTED TO REMAIN AS THE VARIABLE “g”
Calculate the force exerted by the water, taking into account the flow rate and density of the fluid.
1. W= Force x distance
2. F= mass x acceleration
3. F= Density x volume x acceleration
  1. use mass = density x volume, where volume can be expressed as flow rate multiplied by time. This will allow us to determine the total weight of the water contributing to the force.
Work= density x volume x acceleration x distance
1. be sure that you know the shape of the tank! this is important and necessary for calculation the volume
2. PROFESSOR LOVES TO USE CONES
  1. the volume of a cone is $\frac{\pi r^3}{4}$
Determine the height the water needs to be pumped and any friction losses in the system.
1. use variable depictions like $\Delta x$ and $x_{i}$
Integrate the final work equation to find total work in Joules
1. final answer should still have $\pi$ and g

Lecture 15: sequences

Sequences are lists of numbers like

a1, a2, a3, …, an

HOW TO TELL IF THE SEQUENCE CONVERGES OR DIVERGES

To determine if a sequence converges or diverges, we can use several tests, including the Limit Test, where we calculate the limit of the sequence as n approaches infinity. If the limit exists and equals a finite number, the sequence converges. If the limit is infinite or does not exist, the sequence diverges.

EXAMPLES)

(-1, 1, -1, 1, -1, 1, …)

this series oscillates and therefore diverges

(1, 1/2, 1/3, 1/4, …)

This series has an increasing denominator as it grows closer to infinity, therefore this series converges to 0

ABSOLUTE CONVERGENCE THEOREM

$\lim_{n\rightarrow\infty}\left\vert a_{n}\right\vert=0$

then

$\lim_{n\rightarrow\infty}a_{n}=0$

This theorem establishes that if the limit of the absolute values of the series terms approaches zero, then the series itself converges to zero, thereby confirming the conditions required for absolute convergence.

THIS IS NECESSARY FOR SERIES THAT SEEM TO OSCILLATE, meaning they bounce back and forth from negative to positive

Lecture 16: Series

SN= a1+a2+ … +an

This is the same as

$\Sigma\left(a_{n}\right)=\Sigma\left(n\right)$

Partial sums

the addition of the an and all of the prior an sums
S5= a1+a2+a3+a4+a5

Partial sums tend to cancel to show final SN equation

manipulate the SN to allow for cancellations, but make sure to end with a SN on the left and a new equation on the right
plug the new equation in the summation, then test its limit approaching infinity
- this will show a new form allowing for convergence to be found in a simpler manner

REMEMBER: a series is the sum of an infinite list of numbers

you can not have a series without a sequence
series uses summation notation, sequences use brackets and commas

TEST FOR DIVERGENCE

If the limit of the sequence does not approach zero, the series diverges.

important to note that infinity is not 0, so even though it feels wrong it works

BUT

if the series approaches 0 the test does not apply, THIS DOES NOT MEAN IT CONVERGES, it simply can not be shown through the TFD

this requires further tests

Lecture 17: Summing series

Geometric series!!!!

$\Sigma a\left(r\right)^{n}$

r is the common ratio

Series converges if…

\left\vert r\right\vert<1

Series diverges if…

$\left\vert r\right\vert\ge1$

If the series converges, it converges to

$\frac{a\cdot r^{c}}{1-2}$

the numerator represents the first term!

TO TURN A SERIES INTO A GEOMETRIC FORM

undistributed exponents
pull the a value out of the fraction if unaffected by n exponent
multiply by fraction equal to one
- this removes the possibility of n-1 in the exponent

WHEN USING THE RULES OF A GEOMETRIC SERIES, MUST SAY THE SERIES CONVERGES BY THE GEOMETRIC SERIES TEST

Telescoping series!!!

$\Sigma\left(b_{n}-b_{n+1}\right)$ converges if $\lim_{n\to\infty}S_{N}$

*this equation shows that the values are cancelling in pairs

add up all of the partial sums, ending with n=N

This sum is the values of $S_{N}$

*the limit might diverge. this could be due to oscillation from a trig function

IMPORTANT: might need to use partial fractions in order to put a series into telescoping form

Lecture 18: integral Test

To determine convergence, remember to apply the integral test appropriately, ensuring that the function meets the criteria for monotonicity and continuity.

*INTEGRAL TEST SHOWS CONVERGENCE, BUT NOT WHAT THE VALUE CONVERGES TO OR THE SUM OF THE SERIES

f(x) must be continuous, positive, and decreasing on [1, $\infty$ ] and an=f(n)

If $\int_0^{\infty}\!f\left(x\right)\,dx$ converges, then $\Sigma a_{n}$ converges

If $\int_0^{\infty}\!f\left(x\right)\,dx$ diverges, then $\Sigma a_{n}$ diverges

THESE CONDITIONS MUST BE SATISDIED TO USE THE TEST

continuous on [1, $\infty$ ]
positive on [1, $\infty$ ]
decreasing on [1, $\infty$ ]

*MUST ALWAYS TEST THE CONDITIONS

harmonic series: $\Sigma\frac{1}{n}$

TFD: THE HARMONIC SERIES HAS A LIMIT EQUAL TO 0 BUT STILL DIVERGES, THIS IS IMPORTANT TO KNOW

p-series test

$\Sigma\left(\frac{1}{n^{p}}\right)$

Series converges if p>1

series diverges if p $\le1$

generalized p-series test

$\Sigma\left(\frac{1}{n^{p}\left(\ln\left(n\right)\right)^{q}}\right)$

same rules as p series, except when p=1

when p=1, you must now look at q

Series converges if q>1

series diverges if q $\le1$ or if p < 1, as it shifts the behavior of the series towards divergence.

Lecture 19: Direct Comparison

$\Sigma\left(a_{n}\right),\Sigma\left(b_{n}\right)$

both has only positive terms

MUST FOLLOW CRITERIA

If $\Sigma b_{n}$ converges and $a_{n}\le b_{n}$ for all of n, then $\Sigma a_{n}$ also converges

If $\Sigma b_{n}$ diverges and $a_{n}\ge b_{n}$ for all of n, then $\Sigma a_{n}$ also diverges

IF THE GREATER THAN/LESS THAN SYMBOL IS INCORRECT, THEN THE DCT DOES NOT APPLY

* $\frac{1}{n!}$ can be compared to 1/n²

ANSWER MUST INCLUDE

if bn converges or diverge
comparison of size of bn and an
used DCT to prove convergence/divergence

Lecture 20: Limit Comparison

Analyzes the behavior of series using the Limit Comparison Test (LCT) with a known series to determine convergence or divergence.

$\Sigma\left(a_{n}\right),\Sigma\left(b_{n}\right)$

both have only positive terms

$\lim_{n\to\infty}\frac{\left(a_{n}\right)}{\left(b_{n}\right)}$ = c

If c is positive and finite, then both series either converge or diverge together.
- the series act the same as long as c $\ge0$
It is essential to choose a comparison series $b_{n}$ that is simpler and for which the convergence is already known.
- there is no requirement that $a_{n}\le b_{n}$
answer must state that because b diverges/converges, a diverges/converges by the LCT
FOR TRIG: $b_{n}$ can be simplified as the value inside of the sine or cosine functions, this is because each trig function converges to 1 as n approaches infinity.

Lecture 21: Alternating Series

An alternating series has the form of

$\Sigma\left(-1\right)^{n}b_{n}$

Alternating Series Test (AST)

The series converges if…

$b_{n}\ge0$ for all n

{ $b_{n}$ } is decreasing
$\lim_{n\to\infty}b_{n}=0$

The alternating series test takes no specific math, only proving that the series fits all the criteria.

only place for partial credit is in the limit

*ALL 3 CRITERIA MUST BE STATED AS SATISFIED OR NOT SATISFIED IN THE ANSWER

the harmonic series $\Sigma\frac{1}{n}$ diverges, but the alternating harmonic series $\Sigma\frac{\left(-1\right)^{n+1}}{n}$ converges

IMPORTANT RULE: $\Sigma a_{n}$ is absolutely convergent if $\Sigma\left\vert a_{n}\right\vert$ converges. Conversely, a series is conditionally convergent, if $\Sigma a_{n}$ converges but $\Sigma\left\vert a_{n}\right\vert$ diverges.

this rule is important to showing the convergence of a non-alternating series with positive and negative terms

Alternating Series Estimation Theorem

Let $\Sigma\left(-1\right)^{n}b_{n}$ = S (summation approaches infinity) and $S_{N=}\Sigma\left(-1\right)^{n}b_{n}$ (summation approaches N)

Then, $\left\vert S-S_{N}\right\vert\le b_{N+1}$

The following must still be true, just proving the series alternates in general:

$b_{n}\ge0$ for all n

{ $b_{n}$ } is decreasing
$\lim_{n\to\infty}b_{n}=0$
The remainder estimate says that the error of the partial sum estimate is of series $\Sigma\left(-1\right)^{n}b_{n}\le b_{N+1}$ , $b_{N+1}$ is the upper bound (next term in the series)

Lecture 22: Ratio and Root Tests

*PLEASE NOTE: THESE TESTS CAN NOT BE ABBREVIATED IN THE ANSWERS

BOTH TESTS ALLOW FOR CANCELLATIONS, MAKING COMPLICATED EXPONENTS OR FACTORIALS EASIER TO MANAGE AND SIMPLER WHEN FINDING CONVERGENCE

Ratio Test

$\Sigma a_{n}\ne0$ and $\lim_{n\to\infty}\left\vert\frac{\left(a_{n+1}\right)}{a_{n}}\right\vert=L$

Converges absolutely when

L<1

Diverges when

L>1

Test is inconclusive when

L=1

the ratio test is always inconclusive when a series converges conditionally
RATIO TEST IS BEST TO USE WHEN YOU SEE A FACTORIAL (!)

Root Test

$\Sigma a_{n}\ne0$ and $\lim_{n\to\infty}\left(\left\vert a_{n}\right\vert\right)^{\frac{1}{n}}=L$

Converges absolutely when

L<1

Diverges when

L>1

Test is inconclusive when

L=1

ROOT TEST IS BEST TO USE WHEN YOU SEE n IN THE EXPONENT