exam 2 memorize

growth rates

(ln(n))^p << n^q<< aⁿ<< n!<<nⁿ

test for divergence

diverges: if lim as n—>∞ of a_n≠0, then ∑a_n diverges

This test cannot be used to show convergence

Geometric series converges and diverges when?

Series: ∑arⁿ

^Converges: ∣r∣<1

Diverges: ∣r∣≥1

Formula for the sum of geometric series

S: a₁(first term)/1-r.

Σarⁿ

telescoping series

series: Σ(b_n-b_n+1)

converges= lim as n—>∞ of b_n=L

diverges: cant be determined

telescoping series sum formula

S=b₁-L

Generalized p-Test

series:∑1/n^p(ln(n))^q

Converges: p>1, and if p=1 and q>1 when there is ln(n)

Diverges: p≤1, p=1 and q ≤1

Just like with the Integral Test the p-Series Test does not tell us
what a convergent series sums to.

Alternating series

∑(-1)ⁿb_n

Converges: 0<b_n+1≤b_n, and lim as n—>∞ of b_n=0

Diverges: cant be determined

an alternating series is convergent when its terms (1) decrease in absolute value, and (2) approach zero in the limit.

Integral (f continuous, positive, and decreasing)

Series ∑a_n, a_n=f(n) ≥0

Converges: ∫ 1 to ∞ converges

Diverges: ∫ from 1 to ∞ diverges

Warning: The Integral Test only shows whether a series converges or diverges.
It does not tell us what a series converges to. In particular, the value of the
improper integral is not the sum of the series

Root

converges: lim n—>∞ of √∣a_n∣<1

diverges: lim n—>∞ of √∣a_n∣>1

Test is inconclusive if √∣a_n∣=1

Ratio

convergence: lim n—>∞ ∣a_n+1/a_n∣ < 1

divergence: lim n—>∞ ∣a_n+1/a_n∣ > 1

test is inconclusive if ∣a_n+1/a_n∣= 1

Notes:
-If the Ratio Test shows a series converges, then that series will converge
absolutely.
- If a series converges conditionally, then the Ratio Test will always be in-
conclusive.
-The Ratio Test is always inconclusive if an is a rational function. In this
case use the Limit Comparision Test.

Direct comparison test

convergence: b_n is > a_n, and b_n converges, so both converge

divergence: b_n is < a_n, and b_n diverges, so both diverge

Warning: Both series must be series of positive terms in order to use the
Direct Comparison Test.

limit comparison test

convergence: a_n/b_n= L, and L is positive and b_n converges, both converge

divergence: a_n/b_n= L and L is positive and b_n diverges, both diverge

In general, if X is a continuous random variable, then the probability that
a ≤ X ≤ b is given by

P(a ≤ X ≤ b) = ∫_a^b f(x)

to be a probability density function, what conditions must be satisfied:

1. f (x) ≥ 0 for all x, and
2. ∫∞_-∞ f(x)=1

The mean of a probability density function f (x) is given by

μ =∫∞_-∞ x*f(x) dx

The median is the number m such that

∫_m∞ f(x) dx = 1/2

work formula

W = F d
W_i = F (x_i)∆x

W = ∫_a^b F(x) dx

Alternating series test conditions:

There are three conditions that must be satisfied in order to use the Alternating Series Test:

1. bn ≥ 0 for all n (positive)
2. {bn} is decreasing
3. lim n→∞ bn = 0

Absolute vs conditional convergence

A series is absolutely convergent if the series |an| converges.

If an converges but |an| diverges, then a_n is conditionally convergent

Alternating Series Estimatation Theorem

: Let bn ≥ 0 for all n such
that {bn} is decreasing and l’m n→∞ bn = 0. Let ∑ (−1)ⁿbn = S and let S_N =
NX
n=0
(−1)nbn. Then |S − SN | ≤ bN +1

The Alternating Series Remainder Estimate says that the error obtained by us-
ing the partial sum SN to estimate the true sum of the convergent alternating
series ∑ (−1)ⁿbn is never more than bN +1, the next term in the series.

useful limits

1. lim
n→∞ (1 + a/n)^bn= e^ab
2. lim
n→∞ n√|p(n)| = 1

lim to infinity of bigger expo/smaller exp

lim= infinity

lim to infinity of smaller expo/bigger expo

lim=0

sequence is monotonic and bounded, then…

sequence is monotonic and bounded, then the sequence converges.

(ln(n))^p << n^q<< aⁿ<< n!<<nⁿ

^{something farther right/farther left}

^{bigger/smaller}

Infitinity

(n))^p << n^q<< aⁿ<< n!<<nⁿ

^{something farther left/farther right}

^{smaller/bigger}

0

spring problem formulas

F=kx

1 cm= 0.01 m, 25/0.1=250

F=k(difference between natural and stretched length).

W=f*d

integrate using f=kx with found k value and the given stretch lengths

work chain problem equations

w=f*d

work= mass*deltax*acceleration*xi

or….

top:

mass*deltax*acceleration*xi

bottom (chain left over):

work= mass*deltax*acceleration*distance

2*deltax*g*30(lower value of integration bounds)

cubic tank questions

w=f*d

mass*acceleration*distance

mass=volume*density

volume=area*delta x

so…

integrate using the bounds where water is:

volume*deltax*density*g*xi

Area in cubic tank=area of bottom (4×4)

cone problems

use simiar triangles to find radius to height

radi= 2, xi=4, so….. r=xi/2

w=f*d

volume*density*g*xi

Volume=Area*deltaX

area = pi(r)²*deltax= area = pi(xi/2)²*deltax

so………..

pi(xi/2)²*deltax*density*g*xi

probabiliy density function conditions

1: positive for all x values

2. integration from negative to pos. infinity =1

mean= neg in post inf= integration of x(fx)

median= itegation from m to infinirt= 1/2