# Alice and Bob are flipping coins…

Alice and Bob are playing a game. They randomly determine who starts, then they take turns flipping a number of coins (N) and adding them to a growing pile. The first one to collect their target number of tails (T) wins.

When Alice's variables are equal to Bob's ($N_{A} = N_{B}$, $T_{A} = T_{B}$), the odds of her winning are obviously 50%. However, for

$N_{A} = 2, T_{A} = 20, N_{B} = 1, T_{B} = 10$

Alice's chances of victory appear to be slightly lower than 50%. This is based on running a few hundred thousand simulations of the game in Python. This outcome is, unfortunately, unintuitive to me. What is the mathematical reason for it?

Note: This is a specific example chosen to highlight an issue I'm having in a more general problem. In the general problem, the players each have: Odds of an attempt getting them a point (O), number of attempts they get to make on their turn (N), and total number of collected points needed to win (T). If someone could also provide an equation that predicts the probability of Alice or Bob winning, given $O_{A}, N_{A}, T_{A}, O_{B}, N_{B}$, and $T_{B}$, I would be grateful.

• Just so I'm clear on what's going on here: In the posited problem, Alice repeatedly (on her turn) flips two coins, and adds the number of tails to her tally, while Bob repeatedly (on his turn) flips a single coin, and adds the number of tails to his tally. If Alice reaches $20$ before Bob reaches $10$, she wins; otherwise, Bob wins. Is that right? – Brian Tung Nov 30 '16 at 0:28
• @Brian Tung: Yes, that's right. – Teller Nov 30 '16 at 0:31
• This isn't a complete answer, but note that if we let $X_A$ and $X_B$ be random variables representing the number of turns required for Alice and Bob to reach their totals, respectively, then $X_A$ is the half the sum (rounded up) of $20$ geometrically distributed random variables with parameter $1/2$, and $X_B$ is the sum of $10$ such random variables. Two observations: One, such small counts are not enough for the distributions of $X_A$ and $X_B$ to become indistinguishable from normal, and $X_B$ will retain more of a heavy (right-side) tail than $X_A$. (cont'd) – Brian Tung Nov 30 '16 at 0:47
• And two, $X_A$ has to be rounded up. I think that half a turn is probably not negligible either. – Brian Tung Nov 30 '16 at 0:47
• Part of the intuition is that when Alice wins she may have 20 or 21 tails, therefore one tail may be wasted. Bob, on the contrary, will never waste a flip. – A.G. Nov 30 '16 at 0:53

You're right! This is rather non-intuitive.

Since the player who starts is chosen at random, we can ignore this in our calculations since it will average out (you can show this more formally).

On each turn let $A$ be the number of tails that Alice flips, and $B$ the number of tails for Bob.

$A \sim Binomial(2,1/2)$

$B \sim Binomial(1,1/2)$

Now let $A_j$ be the number of tails that Alice has after turn j. Define $B_j$ similarly for Bob. We can see that $A_j = A_{j-1} + A$ and $B_j = B_{j-1} + B$. So at each turn, $A_j$ is the sum of iid binomials, and $B_j$ is the sum of different iid binomials. Hence $A_j$ and $B_j$ are both still binomial. I'll omit this proof but if you're interested I can add it. Now we have:

$A_j \sim Binomial(2j, 1/2)$

$B_j \sim Binomial(j, 1/2)$

So for each turn j, we should compare the probabilities $P(A_j \geq 20)$ and $P(B_j \geq 10)$. These can be found in terms of the Binomial CDF.

For your problem I have plotted these probabilities for turns 1 through 30.

Interestingly, as the game goes on longer, it appears the Alice has a higher chance of winning! But on average, the game will end after 20 turns and Bob still has a slightly higher probability at this point (.58 vs .56). Bob has a higher chance of winning early in the game, which ends up working in his favor!

UPDATE: We can actually derive an equation for P(Alice Wins). Since it is in terms of the Binomial CDF it has to be computed numerically, as the Binomial CDF depends on the Incomplete Beta Function.

THE EQUATION:

$P(\text{Alice Wins}) = \sum_{j=1}^\infty P(\text{Alice Wins On Turn j})$

$P(\text{Alice Wins On Turn j}) = \\ P(A_j \geq T_A, \ A_{j-1} < T_A)\bigl[P(B_j < T_B) + \frac{1}{2}P(B_j \geq T_B, \ B_{j-1} < T_B)\bigr]$

Let's break it down.

$P(\text{Alice Wins on Turn j}) = P_1(P_2 + \frac{1}{2}P_3)$

$P_1$ is just the probability that Alice reaches her Target on turn j (not before). $P_2$ is the probability that Bob hasn't yet reached his target, and $P_3$ is the probability that Bob also JUST reached his target. If this happens, then Alice wins only if she went first, which was a 50-50 chance, hence the $\frac{1}{2}$.

Now we must simply calculate the $P_i$.The easiest of which is $P_2$, just the cdf $F_{B_j}(T_B-1)$. $P_1$ and $P_2$ are somewhat more involved, although there calculations are the same exact idea.

$P(A_j \geq T_A, \ A_{j-1} < T_A) = \sum_{m=1}^{N_A}\sum_{n=m}^{N_A}P(A_{j-1} = T_A - m)P(A = n)$

I skipped the details, but the above is due to the fact that we can write what we want in terms of $A_{j-1}$ and $A$ (instead of $_j$), and these variable are independent.

It's a complicated problem, but we have all the peices we need to compute the exact probability that Alice wins, given parameters $N_A, T_A, O_A, N_B, T_B, O_B$ and even $P_A$, the probability that Alice goes first.

$P(\text{Alice Wins}) = \sum_{j=1}^\infty\\ \biggl[\sum_{m=1}^{N_A}\sum_{n=m}^{N_A}P(A_{j-1} = T_A - m)P(A = n)\biggr]\biggl[1-F_{B_j}(T_B-1) + \\ P_A\sum_{m=1}^{N_B}\sum_{n=m}^{N_B}P(B_{j-1} = T_B - m)P(B = n)\biggr]$

Indeed for the problem you described above, we see that Alice has only a 46.32907 % chance of winning. If you want to see more details, or play around with different values, the R-code which calculates the equation I've just described can be found here.


# Probability : At least one boy vs Exactly one boy [duplicate]

Part A. a company is holding a dinner for working mothers with at least on son. Ms. Smith, a moher of two children is invited. What is the probability that both children are boys?

ANS：P(both boys | at least 1 boy ) = 1/3

Part B. Your new colleague Ms. Parker also have two children. You see her walking with one of her children and that child is a boy. What is the probability that both of them are boys?

ANS: P(both boys | one boy ) = 0.5

I don't understand Part B. If you know one of her children is a boy, does this implies that Ms. Parker has $\geq1$ boy? In mathematical terms, this means that Ms.Parker has at least one boy, true? How come it is different from Part A. Confuse!

## marked as duplicate by JMoravitz, user91500, GNUSupporter 8964民主女神 地下教會, A Blumenthal, VladhagenFeb 27 '17 at 22:57

• Because in part (A) we don't have any information as to which child was the boy, but in part (B) we have information that the child in our sight specifically is a boy. This is a stronger condition than her having at least one boy since there exist times where her having at least one boy is true but we see a girl instead. – JMoravitz Feb 27 '17 at 4:50
• Part A outcome are - boy-boy, boy-girl, girl-boy. So having boy-boy is 1/3. Now, part B outcomes - boy-boy and boy-girl, since first one is definitely a boy. So 1/2. – Kaster Feb 27 '17 at 4:58
• Thanks for your reply. I still couldn't see the difference between seeing child A is a boy and just know at least one of them is boy... – wrek Feb 27 '17 at 4:58
• Knowing child $A$ is a boy implies that at least one child is a boy. Knowing at least one child is a boy does not imply child $A$ is a boy. Knowing at least one child is a boy is a weaker piece of information – JMoravitz Feb 27 '17 at 5:00
• At the risk of confusing you further, part (B) really should have an extra condition in order to ensure the probability is in fact $0.5$ and that is the knowledge that the specific child she is seen walking with was picked uniformly at random from her two children. (the problem changes further if we know something about her selection process for which child to take. For extreme example, if we know that if she has a daughter that she keeps her daughter with her at all times this would imply that since we don't see a daughter with her that she has no daughters) – JMoravitz Feb 27 '17 at 5:17

Let boy be denoted by $B$ and girl by $G$. If a family has two children, then there are the following possibilities for their kids $$BB,BG,GB,GG,$$ each with equal probability, e.g., $P(BB) = P(BG) = 1/4$.
Now, it should be clear for part A, that all we know is that the family is picked randomly from families that are $$BB,BG,GB.$$ Now, the probability that they have two boys is $$P(BB|BB\cup BG \cup GB) = \frac{P(BB\cap (BB\cup BG \cup GB))}{P(BB\cup BG \cup GB)} = \frac{P(BB)}{P(BB\cup BG \cup GB)} = \frac{1/4}{3/4} = \frac{1}{3}.$$
Now, in part B, we have something simlilar, but slightly different. What we have is that we saw a child and it's a boy. Now, the underlying assumption is that we randomly selected a child from the parent's two children. Thus, we have two equally likely scenarios, either we saw child 1 and it was a boy, or we saw child 2 and it was a boy. Let us focus on the former, and note that the probabilities are the same. We note that the event that "child one was a boy" is $BB \cup BG$.
$$P(BB|\text{child one was a boy}) = \frac{P(BB \cap (BB,BG))}{P(BB \cup BG)} = \frac{P(BB)}{P(BB \cup BG)} = \frac{1/4}{2/4} = 1/2.$$ Now, this happened with probability $1/2$ so the total probability that the family in part B has two boys is $$P(BB|\text{(child one was selected AND a boy) OR (child two was selected AND a boy)}).$$ Because it can't happen that child one was selected and child two was selected, these two events are disjoint, hence the probability is equal to $$P(BB|\text{child one was selected AND a boy)} +P(BB|\text{child two was selected AND a boy}) \\ =P(BB|\text{child one was a boy})P(\text{child one was selected}) \\ +P(BB|\text{child two was a boy})P(\text{child two was selected})\\ = \frac{1}{2}\cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{2}.$$ Where the first equality comes from the fact that whether or not child one was selected was independent of the family composition. Now, this was a bit wordy, but I hope it is helpful in its own right.
Note, throughout it is assumed that you know that a conditional probability of event $A$ given event $B$, denoted $P(A|B)$, is given by $$P(A|B) = \frac{P(A,B)}{P(B)}.$$


# Covering a set with geometric progressions

Consider the set $S_n=\{1,2,\cdots ,n\}$. What is the minimum number of distinct geometric progressions that cover $S_n$? Let us call this number $a_n$. I was wondering about this number after doing a problem from the Allrussian MO, 1995.

Can the set $\{1,2,\cdots,100\}$ be covered with $12$ geometric progressions?

It becomes straightforward after observing the fact no three primes can be in a geometric progression. Hence the problem is restated to the obvious contradiction $$\pi(100)\le 24$$ Now I had searched a bit and here it is proven that $a_{100}\ge 24$. Now I would like some asymptotics, or references, or better bounds on $a_n$ .

I have also found the fact that $$a_n\ge \left\lfloor{\frac{3n}{\pi^2}}\right\rfloor$$

Which is obvious since any geometric progression contains at most $2$ squarefree numbers and there are about $\dfrac{6n}{\pi^2}$ squarefree numbers less than $n$. Note it surpasses the bound given. But is something better possible? Thanks in advance.

• You mean minimum, not maximum. – Richard Stanley Jul 1 '14 at 15:17
• The $n/2$ upper bound holds even for progressions with integer coefficients: consider progressions with ratio 2 and odd initial values. – Emil Jeřábek Jul 1 '14 at 15:41
• You can get the upper bound of $3n/8$ by pairing off odd numbers in $(n/2,n)$. Then for each odd number below $n/2$ use the geometric progression with ratio $2$ starting from it. – Lucia Jul 1 '14 at 15:53
• One can improve the lower bound slightly. Let $S$ be the set of integers which are either square-free or have the form $2^3 3^2 m$, where $m$ is square-free and not divisible by $2$ or $3$. Then $S$ has density $\delta>6/\pi^2$ and still contains no more than two elements in any one geometric progression, so you get $a_n/n\geq \delta/2$. I do not know the maximal density of a subset of $\mathbf{N}$ with no more than two elements in any geometric progression. – Sean Eberhard Jul 1 '14 at 17:47
• arxiv.org/pdf/1311.4331v1.pdf may also be of some interest. – Gerry Myerson Jul 2 '14 at 0:47

## 1 Answer

We can reduce Lucia's upper bound of $3/8$ a little further as follows. Begin by taking the $n/4$ geometric progressions of common ratio $2$ beginning at each odd number at most $n/2$. Then for each odd number $x$ in the range $[n/2,25n/49]$ divisible by $25$ include the progression $\{x,(7/5)x,(7/5)^2x\}$. Pair off the remaining odd numbers. If I've added this up correctly I get $3/8 - 1/(4900)$ as an upper bound.

As I explained in my comment one can improve the lower bound $3/\pi^2$ slightly by considering a set of integers $S$ larger than just the square-frees but still not containing any three terms of a geometric progression. Here is a somewhat general construction of such a set. Let $Q\subset\mathbf{Z}_{\geq0}^2$ be any set without three points on a line. Now let $p_1,p_2,\dots$ be the primes and let $$S = \left\{\text{integers }n= \prod p_i^{e_i} \text{ such that } (e_1,e_2)\in Q, (e_3,e_4)\in Q,\dots\right\}.$$ Then $S$ has no three terms of any geometric progression, and one could write down an Euler-product-like expression for the density of $S$ in terms of $Q$. Note if $Q$ contains $$\{(0,0),(1,0),(0,1),(1,1)\}$$ then $S$ contains all square-free integers, but $Q$ could be larger as well. Taking $$Q = \{(0,0),(1,0),(0,1),(1,1),(2,3)\},$$ the density of $S$ is $$\prod_{i=1}^\infty \left(1-\frac{1}{p_{2i-1}}\right) \left(1-\frac{1}{p_{2i}}\right)(1 + p_{2i-1}^{-1} + p_{2i}^{-1} + p_{2i-1}^{-1} p_{2i}^{-1} + p_{2i-1}^{-2} p_{2i}^{-3}),$$ which in any case is bounded below by $$\frac{1 + 2^{-1} + 3^{-1} + 2^{-1} 3^{-1} + 2^{-2} 3^{-3}}{1 + 2^{-1} + 3^{-1} + 2^{-1} 3^{-1}} \prod_{p} \left(1-\frac{1}{p^2}\right) = \frac{217}{36\pi^2}.$$ Thus $\frac{217}{72\pi^2}$ is a lower bound.

This problem is mentioned as problem 2014.2.1 in http://arxiv.org/pdf/1406.3558v2.pdf, though presumably others have pondered it as well.

• You might as well use odd x which are multiples of 25 and are greater than n/4 instead of waiting till n/2. You can treat other multiples of squares this way as well, especially the odd multiples of 9 greater than n/4 and less than 9n/25. – The Masked Avenger Jul 2 '14 at 5:24


Compare growth rate of functions (exponential vs. polynomial)

I have to compare the growth rate of the following sequences

$a_n=a_{n-1}=10$

$b_n=\sum_{k=1}^n k^2$

$c_n =\frac{n^2}{10}$

$d_n=\left( \frac{3}{2} \right)^n$

I've rewritten $b_n$ and $d_n$ to

$$b_n = \frac{1}{6} (2n^3+3n^2+n) \quad \text{and} \quad d_n=1.5^n$$

and would say that $a_n = \mathcal{O}(1)$, $b_n = \mathcal{O}(n^3)$, $c_n = \mathcal{O}(n^2)$ and $d_n = \mathcal{O}(1.5^n$).

When sorting them, it should be

$$a_n < c_n < b_n < d_n$$

Here is the part, that I do not get. When graphing the functions or simply calculating, I get that $\mathcal{O}(1.5^n)<\mathcal{O}(n^2)$. Since $(k^n=)1.5^n$ is an exponential function with $k>1$, it should have a growth greater than the polynomial function $n^2$. I can't seem to figure out which of the 2 options is the right one.

Any help with explanation would be appreciated.

• You got it wrong for $b_n$ Sep 17, 2016 at 12:19
• @polfosol is it the explicit expression for the sum or the growth rate? Sep 17, 2016 at 12:21
• Just to clarify, in the description of $b_n$ is it $k=1$ as the lower bound of the summation rather than $k-1$? Sep 17, 2016 at 12:21
• @ParclyTaxel Yes, it should be $k=1$. Thanks. Corrected. Sep 17, 2016 at 12:22
• @Labbiqa: The explicit expression and consequently, the growth rate are both wrong Sep 17, 2016 at 12:29

That $\mathcal O(1.5^n)$ seems to grow slower than $\mathcal O(n^2)$ (or $\mathcal O(n^3)$) is an illusion. Exponential functions with exponents greater than 1 grow faster than all polynomial functions, but only eventually.
This is the key word: growth rates describe limiting behaviour, not behaviour at a particular point. In particular, the point where $1.5^n$ overtakes $n^2$ in growth was off your graph, and you missed it.
• Thanks for the clarification. I thought it was odd since no matter how much I zoomed out, it still didn't grow faster than $\mathcal{O}(n^2)$. Sep 17, 2016 at 14:06
• @Labbiqa, the graphs for $n^2$ and $1.5^n$ cross at about $n\approx 12.5$. Take a look at their graphs over an interval like $1 < n < 20$. Sep 18, 2016 at 13:39


# Linear algebra system of equations

Given the following linear system:

$x - y + 2z + 3w = 12$

$x - 2y + z + w = 8$

$4x + y + 2z - w = 3$

1) find all solutions of the system. Write the solution in Vector Form
2) find all solutions of the corresponding homogeneous system

I have turned this into a matrix and worked it out two ways resulting in different outcomes, neither of which is solvable...

I'll try to show my work, but I am new to this site and the formatting for matrices is fairly hard to figure out...

1 -1 2 3 12
1 -2 1 1 8
4 1 2 -1 3 (-4 row1)

1 -1 2 3 12
1 -2 1 1 8 (-row1)
0 5 -6 -13 -45

1 -1 2 3 12
0 -1 -1 -2 -4
0 5 -6 -13 -45(+5row2)

1 -1 2 3 12
0 -1 -1 -2 -4
0 0 -11 -23 -65

from here I don't think there's anything else I can do, I'm fairly sure there are infinitely many solutions.

• Let's see the work done so we can help. – Vincent Sep 8 '14 at 20:18
• Wow thanks, but of course. – Daniel Charry Sep 8 '14 at 20:19
• By this computation, it appears that there should be a non-empty solution set. If you show us what you did, we could try to figure out where you went wrong. – Omnomnomnom Sep 8 '14 at 20:24
• I edited the question with my work, any thoughts? I'm thinking that It just needs to be put in vector form? – donsavage Sep 8 '14 at 20:43

Oh of course there is more that can be done!

So starting from where you left, I would first multiple $R_2$ with the scalar $(-1)$ and then add the new $R_2$ to $R_1$ and replace it. Then I get,

$$\left({\begin{matrix} 1 & 0 & 3 & 5 & 16 \\ 0 & 1 & 1 & 2 & 4 \\ 0 & 0 & -11 & -23 & -65 \\ \end{matrix}}\right)$$

$$\underrightarrow{\left({\frac{-1}{11}}\right) R_3 \ \text{and} -1 R_3 + R_2 \ \text{and} -3R_3 + R_1 }$$

$$\left({\begin{matrix} 1 & 0 & 0 & \frac{-14}{11} & \frac{-19}{11}\\ 0 & 1 & 0 & \frac{-1}{11} & \frac{-21}{11} \\ 0 & 0 & 1 & \frac{23}{11} & \frac{65}{11} \\ \end{matrix}}\right)$$

Now translating back to equations you have,

$$x = \frac{14}{11}w + \frac{19}{11}$$ $$y = \frac{1}{11}w + \frac{21}{11}$$ $$z = \frac{-23}{11}w + \frac{-65}{11}$$

Now any value you take for $w$ will solve the system. The method underlined here is made formal in Linear Algebra by Hoffman, Kunze which you should definitely read.

Now the solution set in vector form would be,

$$\left({\begin{matrix} x \\ y \\ z \\ w \\ \end{matrix}}\right) = w\left({\begin{matrix} \frac{14}{11} \\ \frac{1}{11} \\ \frac{-23}{11} \\ 1 \\ \end{matrix}}\right) + \left({\begin{matrix} \frac{19}{11} \\ \frac{21}{11} \\ \frac{-65}{11} \\ 0 \\ \end{matrix}}\right)$$

where $w$ is any scalar in $\Bbb R$.

• Could you explain why when converting it back into the equations the sign of the last two rows change? Shouldn't the last column keep it's sign? for example: z = -23/11w + 65/11 – donsavage Sep 10 '14 at 0:20
• Actually your partly right. The equation should be $z = \frac{-23}{11}w - \frac{65}{11}$ since the last row in the matrix $0 \; 0 \; 1 \; \frac{23}{11} \; \frac{-65}{11}$ represents the equation $z + \frac{23}{11}w = \frac{-65}{11}$. So lslight alteration needed. – Ishfaaq Sep 10 '14 at 1:35


# Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the same?

1. Dec 29, 2012

### tahayassen

This is a discussion we had in another part of the forum and I'm wondering who is correct. The discussion is becoming increasingly confusing and annoyingly (regardless of the posts in between the discussion), no one with a "Science Advisor", "Homework Helper", or "PF Mentor" title is stepping in to end the argument, so I would appreciate it if you would end the argument.

2. Dec 29, 2012

### Number Nine

Re: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s

The cardinalities of the intervals [0 1] and [0 2] are the same. In fact, there are as many numbers in the interval [0 1] as there are real numbers.

This is a pretty poor explanation of what's happening. Taking limits in calculus is a process; we examine what happens as a quantity increases without bound. The limit of that function as x goes to infinity is zero because x! increases faster as x increases, not because it is a "larger quantity". "Infinity" in this context has a completely different meaning than the "infinity" used to describe the size of sets.

3. Dec 29, 2012

### rbj

Re: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s

there is an infinite, but countably infinite amount of rational numbers between 0 and 1. this is also the case for the interval from 0 to 2.

but there is an uncountably infinite amount of real numbers between 0 and 1. this is also the case for the interval from 0 to 2.

4. Dec 29, 2012

### micromass

Staff Emeritus
Re: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s

Jack212222 is wrong. Cantor's theory is nothing like what he describes. He seems to be confusing things with limits.

And yes, the cardinality of [0,1] is exactly the same as the cardinality of [0,2]. So both sets have the same size.

EricVT is wrong too, since he assumes that $\frac{+\infty}{+\infty}$ is defined when it is not. So the ratio doesn't even make sense.

5. Dec 29, 2012

### micromass

Staff Emeritus
Re: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s

Yes, this is true. But the example of [0,1] and [0,2] is not a good example since both infinities are the same here.

However, we can look at the sets $\mathbb{N}$ and $\mathbb{R}$. Those are infinite sets, but the latter set is much larger than the former.

See the following FAQ post: https://www.physicsforums.com/showthread.php?t=507003 (also check out the sequels whose link is at the bottom of the thread).

6. Dec 29, 2012

### HallsofIvy

Staff Emeritus
Re: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s

You also have to be careful how you define "larger". Micromass is completely correct about cardinality and that will be appropriate if 'larger' means "has more numbers in the set" as was originally asked. But one can also argue that the interval [0, 2] is twice as long, and so twice as large, as the interval [0, 1]. It depends upon what you are comparing.

7. Dec 29, 2012

### tahayassen

Re: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s

Thanks everyone. The FAQ was very helpful. I also found this helpful:

Last edited by a moderator: Sep 25, 2014
8. Dec 31, 2012

### alan6459

Re: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s

Two sets have the same cardinaility if there exists a 1-1 correspondence between them.
The function f(x) = 2x establishes such a 1-1 correspondence between the rationals in
[0,1] and [0,2]. It also establishes a 1-1 correspondence between the reals in [0,1] and [0,2].

9. Dec 31, 2012

### 1MileCrash

Re: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s

Some infinities are larger than others. Namely, uncountable infinities are larger than countable infinities and that is all there is to it.

The number of real numbers between 0 and 1 is uncountably infinite. The number of real numbers between 0 and 2 is uncountably infinite. The number of real numbers period is uncountably infinite. These all describe the same cardinality.

The natural numbers are countably infinite, the real numbers are uncountably infinite. Therefore the cardinality of natural numbers is smaller than that of the real numbers.

There is no differentiation between any two uncountably infinite, or two countably infinite sets, that I am aware of, and such an idea doesn't really make sense to me.

10. Dec 31, 2012

### micromass

Staff Emeritus
Re: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s

Sure there is. There are many types of uncountably infinite sets. For example, the set $\mathcal{P}(\mathbb{R})$ (power set of the reals) has a strictly larger cardinality than $\mathbb{R}$. and $\mathcal{P}(\mathcal{P}(\mathbb{R}))$ is even larger!! This process continues indefinitely.

11. Dec 31, 2012

### tahayassen

Re: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s

Ah, thanks for pointing that out.

Edit: I take my thank you back.

Last edited: Dec 31, 2012
12. Jan 2, 2013

### Tac-Tics

Re: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s

13. Jan 2, 2013

### dextercioby

Re: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s

Just as a curiosity, what's the connection between the discrete/countable infinite sets (sequences) $\mathcal{P}^{n}(\mathbb{R})$ and $\aleph_{n}$ ?

Last edited: Jan 2, 2013
14. Jan 2, 2013

### 1MileCrash

Ah.. I failed to think of deriving a set from an uncountable set in such a way that the cardinality must be greater. It doesn't make much sense at all to say that a set and it's power set have the same cardinality. That is easy to see.

Thank you.

Is this limited to the idea of power sets? It feels like this could only occur when you "build" a "higher order" uncountable set from a previously uncountable set.

Also- the power set of the naturals would certainly be of higher cardinality than the naturals and thus not at a one-to-one correspondence with the naturals and therefore uncountable, right?

Does that make P(N) have the same cardinality as R, and P(P(N)) have the same cardinality as P(R)?

Or am I oversimplifying the idea?

Last edited: Jan 2, 2013
15. Jan 2, 2013

### dextercioby

Re: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s

P^n (N) (=P(P(P...P(N)))) for arbitrary n has the same cardinality with N.

16. Jan 2, 2013

### 1MileCrash

I don't understand how that could be the case.

17. Jan 2, 2013

### dextercioby

Re: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s

Who's P(N) ? Write an explicit formula using the 3 symbols: ..., { and }. Then calculate its cardinality.

18. Jan 2, 2013

### 1MileCrash

Re: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s

Forgive me if I'm not understanding but,

The number of elements of P(N) is the number of possible subsets of natural numbers that can be formed.

That is surely uncountable, as I can combine any number of any natural numbers I want to form some subset. If I call one subset, the non-proper subset, the entire set of natural numbers, one subset, what's another subset? The subset of 1 million of any natural numbers? 2 million any natural numbers? Considering all singleton sets of each natural number is already a countably infinite number of subsets, and there are uncountably many subsets besides those, it doesn't work in my brain to call P(N) a set of countable cardinality.

EDIT:

I found this

http://www.earlham.edu/~peters/writing/infapp.htm#thm3

Which I guess is a much better and more formal version of what I'm thinking.

Last edited: Jan 2, 2013
19. Jan 2, 2013

### Number Nine

Re: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s

You're right, P(N) is uncountable. An easy way to see this is to identify each element of P(A) with an infinite binary string (easy to show that this is a bijection) and note that the set of all such strings is uncountable.

20. Jan 2, 2013

### 1MileCrash

Re: Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the s

Okay, I'm not sure how to construct these infinite binary strings but I do understand why P(N) is of uncountable cardinality. I have a few more questions.

I have been reading about "beth numbers" and this seems to be the name for the "order of infinity."

So here are my questions:

Is the beth number the sole indicator of cardinality of uncountable sets? If so, this would imply that P(N) and R have the exact same cardinality, right?
Is the power set concept the only source of constructing a "higher order" infinity?

Are there any sets that we naturally consider that aren't derived from taking the power set of some other set with a beth number greater than two?

I do realize that this question is pretty much identical to my previous one, but I guess what I'm asking is if there is a set with a beth number greater than or equal to 2 that we could describe in some other way than a power set of some other set?

I'm sorry for kind of derailing the topic but I find this really cool, as a grad, what focuses on this type of thing? Is it just set theory?


# irrational numbers?!

what are rational and irrational numbers??
i know that irrational numbers are numbers that cannot be expressed as a fraction, but i'm still confused.

there is this question:

which of the following are irrational numbers: √2, √8, 22/7, pi, 2√3

i know that 22/7 is rational, pi is irrational, but what about those square roots?

1. Square roots of numbers that are not the squares of integers are all irrational. There is a way to prove that but I forgot the details. You have to assume that a fraction works and then prove that the assumption leads to a contradiction

Any number that does not meet the definition of being rational is irrational.

posted by drwls
2. - √2 is either rational or irrational.

- Assume that √ is rational so that
√2 = a/b, with a/b in lowest terms
- Square both sides to get
2 = a^2/b^2
then a^2 = 2b^2
- the right side of this equation is clearly an even number, since anything multiplied by 2 is even
- so a^2 must be even. We also know that if we square an odd number the result is odd, and if we square an even number the result is even
so a must be even
so a could be written as 2k
- rewriting our equation as
2b^2= (2k)^2
2b^2 = 4k^2
b^2 = 2k^2

by the same argument as above 2k^2 is even , so b has to be even

which means a and b are both even, therefore a/b is not in lowest terms

BUT that contradicts my assumption that a/b was a fraction in lowest terms

so √2 = a/b is a false statement
therefore √2 cannot be rational, and
must then be irrational

the same argument could be used for √3 and all other square roots

posted by Reiny
3. true or false does this number represent a rational number 0.20200200020000200000

posted by brandy
4. Is √4 Irrational?

posted by Amber
5. is this seriouly the answer for the LEAP test *feecepum*
you're not going to get ALL 50 answers -_-

posted by Cereal.... is life. -Life Cereal

First Name

## Similar Questions

1. ### THE HOMEWORK IS DUE TODAY- HELP! HELP! PLEASE

There exist irrational numbers a and b so that a^b is rational. look at the numbers √2^√2 and (√2^√2)^√2 a) use similar idea to prove that there exists a rational number a and an irrational number b
2. ### math

Which statements are true for irrational numbers written in decimal form? A. Irrational numbers are nonterminating. B. Irrational numbers are repeating. C. Irrational numbers are nonrepeating. D. Irrational numbers are

There exist irrational numbers a and b so that a^b is rational. look at the numbers √2^√2 and (√2^√2)^√2 a) use similar idea to prove that there exists a rational number a and an irrational number b
4. ### math

There exist irrational numbers a and b so that a^b is rational. look at the numbers √2^√2 and (√2^√2)^√2 a) use similar idea to prove that there exists a rational number a and an irrational number b
5. ### math

Which statement is true? A. All irrational numbers are also rational numbers. B. Half of the irrational numbers are also rational numbers. C. One-third of the irrational numbers are also rational numbers D. Irrational numbers
6. ### Algebra

Order the following numbers from least to greatest. √5, -0.1, (-5/3), 0.7, √2 a. 0.7, √2, (-5/3), √5, -0.1 b. √5, √2, 0.7, (-5/3), -0.1 c. (-5/3), -0.1, 0.7, √2, √5 ? d. -0.1, 0.7,
7. ### Math

1a) Prove that there exist irrational numbers a and b so that a^b is rational. look at the numbers √2^√2 and (√2^√2)^√2 1b) use similar idea to prove that there exists a rational number a and an
8. ### Math

What are irrational, rational, and natural numbers? A rational number is a number that can be written as a ratio. That means it can be written as a fraction, in which both the numerator (the number on top) and the denominator (the
9. ### Algebra

28. Use the Distributive Property to simplify x(4x^2 + x + 4) Is it 4x^3 + x^2 + 4x? 42. To which set of numbers does 0 not belong? 46. Pi belongs to which group of real numbers? a. Odd numbers b. rational numbers c. irrational
10. ### Algerba II

Hi Do decimals such as 2.718 represent rational numbers or irrational numbers. Explain. Do repeating decimals such as 2.3333 . . . represent rational numbers or irrational numbers? Explain. So I look back in my textbook and find

More Similar Questions


# matrix multiplication associative proof

Property 1: Associative Property of Multiplication A(BC) = (AB)C where A,B, and C are matrices of scalar values. Proof We will concentrate on 2 × 2 matrices. 2. A professor I had for a first-year graduate course gave us an example of why caution might be required. How do you multiply two matrices? Hence, associative law of sets for intersection has been proved. Matrix addition and scalar multiplication satisfy commutative, associative, and distributive laws. Matrix multiplication is indeed associative and thus the order irrelevant. Theorem 2 Matrix multiplication is associative. 14 minutes ago #3 TheMercury79. (This can be proved directly--which is a little tricky--or one can note that since matrices represent linear transformations, and linear transformations are functions, and multiplying two matrices is the same as composing the corresponding two functions, and function composition is always associative, then matrix multiplication must also be associative.) Please Write The Proof Step By … It turned out they are the same. ible n×n matrices with entries in F under matrix multiplication. As examples of multiplication modulo 6: 4 * 5 = 2 2 * 3 = 0 3 * 9 = 3 The answer … Answer Save. 3 Answers. The Associative Property of Multiplication of Matrices states: Let A , B and C be n × n matrices. However, this proof can be extended to matrices of any size. Theorem 7 If A and B are n×n matrices such that BA = I n (the identity matrix), then B and A are invertible, and B = A−1. Since Theorem MMA says matrix multipication is associative, it means we do not have to be careful about the order in which we perform matrix multiplication, nor how we parenthesize an expression with just several matrices multiplied togther. Then (AB)Ce j = (AB)c j … Likes TheMercury79. Then the following properties hold: a) A(BC) = (AB)C (associativity of matrix multipliction) b) (A+B)C= AC+BC (the right distributive property) c) C(A+B) = CA+CB (the left distributive property) Proof: We will prove part (a). Zero matrix on multiplication If AB = O, then A ≠ O, B ≠ O is possible 3. But for other arithmetic operations, subtraction and division, this law is not applied, because there could be a change in result.This is due to change in position of integers during addition and multiplication, do not change the sign of the integers. $$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \cdot \begin{pmatrix} e & f \\ g & h \end{pmatrix} = \begin{pmatrix} ae + bg & af + bh \\ ce + dg & cf + dh \end{pmatrix}$$ Lecture 2: Fun with matrix multiplication, System of linear equations. Lv 4. In standard truth-functional propositional logic, association, or associativity are two valid rules of replacement. This preview shows page 33 - 36 out of 79 pages. What are some of the laws of matrix multiplication? SAT Math Test Prep Online Crash Course Algebra & Geometry Study Guide Review, Functions,Youtube - Duration: 2:28:48. A+(B +C) = (A+B)+C (Matrix addition is associative.) Question: Prove The Associative Law For Matrix Multiplication: (AB)C = A(BC). That is if C,B and A are matrices with the correct dimensions, then (CB)A = C(BA). Clearly, any Kronecker product that involves a zero matrix (i.e., a matrix whose entries are all zeros) gives a zero matrix as a result: Associativity. but composition is associative for all maps, linear or not. Learning Objectives. Multiplicative identity: For a square matrix A AI = IA = A where I is the identity matrix of the same order as A. Let’s look at them in detail We used these matrices For any matrix A, ( AT)T = A. 2 The point is you only need to show associativity for multiplication by vectors, i.e. Matrix-Matrix Multiplication is Associative Let A, B, and C be matrices of conforming dimensions. ... the same computational complexity as matrix multiplication. The Organic Chemistry Tutor 1,739,892 views B. Except for the lack of commutativity, matrix multiplication is algebraically well-behaved. Prove the associative law of multiplication for 2x2 matrices.? Special types of matrices include square matrices, diagonal matrices, upper and lower triangular matrices, identity matrices, and zero matrices. for matrices M,N and vectors v, that (M.N).v = M.(N.v). Even if matrix A can be multiplied with matrix B and matrix B can be multiplied to matrix A, this doesn't necessarily give us that AB=BA. The first is that if $$r= (r_1,\ldots, r_n)$$ is a 1 n row vector and $$c = \begin{pmatrix} c_1 \\ \vdots \\ c_n \end{pmatrix}$$ is a n 1 column vector, we define $rc = r_1c_1 + \cdots + r_n c_n. The associative property holds: Proof. c i  j = ∑ 1 ≤ k ≤ m a i  k  b k  … 1 decade ago. In general, if A is an m n matrix (meaning it has m rows and n columns), the matrix product AB will exist if and only if the matrix B has n rows. Cool Dude. I just ended up with different expressions on the transposes. Example 1: Verify the associative property of matrix multiplication for the following matrices. On the RHS we have: and On the LHS we have: and Hence the associative … In other words, unlike the integers, matrices are noncommutative. Subsection DROEM Determinants, Row Operations, Elementary Matrices. It is easy to see that GL n(F) is, in fact, a group: matrix multiplication is associative; the identity element is I n, the n×n matrix with 1’s along the main diagonal and 0’s everywhere else; and the matrices are invertible by choice. That is, a double transpose of a matrix is equal to the original matrix. 3. r(A+B) = rA+rB (Scalar multiplication distributes over matrix addition.) Then, ( A B ) C = A ( B C ) . Distributive law: A (B + C) = AB + AC (A + B) C = AC + BC 5. In Maths, associative law is applicable to only two of the four major arithmetic operations, which are addition and multiplication. Proof Let be a matrix. So the ij entry of AB is: ai1 b1j + ai2 b2j. Theorem 2 matrix multiplication is associative proof. Properties of Matrix Multiplication: Theorem 1.2Let A, B, and C be matrices of appropriate sizes. Pages 79. I am working with Paul Halmos's Linear Algebra Problem Book and the seventh problem asks you to show that multiplication modulo 6 is commutative and associative. Second Law: Second law states that the union of a set to the union of two other sets is the same. Favorite Answer. A+B = B +A (Matrix addition is commutative.) Properties of Matrix Arithmetic Let A, B, and C be m×n matrices and r,s ∈ R. 1. School Georgia Institute Of Technology; Course Title MATH S121; Uploaded By at1029. The -th ... , by applying the definition of Kronecker product and that of multiplication of a matrix by a scalar, we obtain Zero matrices. Then (AB)C=A(BC). Associative law: (AB) C = A (BC) 4. Proof: Suppose that BA = I … If B is an n p matrix, AB will be an m p matrix. As a final preparation for our two most important theorems about determinants, we prove a handful of facts about the interplay of row operations and matrix multiplication with elementary matrices with regard to the determinant. A matrix is full-rank iff its determinant is non-0; Full-rank square matrix is invertible; AB = I implies BA = I; Full-rank square matrix in RREF is the identity matrix; Elementary row operation is matrix pre-multiplication; Matrix multiplication is associative; Determinant of upper triangular matrix e.g (3/2)*sqrt(1/2) … Parts (b) and (c) are left as homework exercises. Because matrices represent linear functions, and matrix multiplication represents function composition, one can immediately conclude that matrix multiplication is associative. Matrix multiplication is Associative Let A be a m\times n matrix, B a n\times p matrix, and C a p\times q matrix. So this is where we draw the line on explaining every last detail in a proof. 2.2 Matrix multiplication. What are some interesting matrices which lead to special products? Then (AB)C = A(BC): Proof Let e j equal the jth unit basis vector. Relevance. We are going to build up the definition of matrix multiplication in several steps. Proposition (associative property) Multiplication of a matrix by a scalar is associative, that is, for any matrix and any scalars and . Propositional logic Rule of replacement. Square matrices form a (semi)ring; Full-rank square matrix is invertible; Row equivalence matrix; Inverse of a matrix; Bounding matrix quadratic form using eigenvalues; Inverse of product; AB = I implies BA = I; Determinant of product is product of determinants; Equations with row equivalent matrices have the same solution set; Info: Depth: 3 16 5. fresh_42 said: Then you have made a mistake somewhere. Matrix arithmetic has some of the same properties as real number arithmetic. What is a symmetric matrix? Let the entries of the matrices be denoted by a11, a12, a21, a22 for A, etc. it then follows that (MN)P = M(NP) for all matrices M,N,P. Solution: Here we need to calculate both R.H.S (right-hand-side) and L.H.S (left-hand-side) of A (BC) = (AB) C using (associative) property. Matrix multiplication Matrix inverse Kernel and image Radboud University Nijmegen Matrix multiplication Solution: generalise from A v A vector is a matrix with one column: The number in the i-th rowand the rst columnof Av is the dot product of the i-th row of A with the rst column of v. So for matrices A;B: Proof: (1) Let D = AB, G = BC Informal Proof of the Associative Law of Matrix Multiplication 1. Thanks. Therefore, the associative property of matrices is simply a specific case of the associative property of function composition. Please Write The Proof Step By Step And Clearly. What is the inverse of a matrix? Since matrix multiplication obeys M(av+bw) = aMv + bMw, it is a linear map. Then, (i) The product A  B exists if and only if m = p. (ii) Assume m = p, and define coefficients. Corollary 6 Matrix multiplication is associative. The argument in the proof is shorter, clearer, and says why this property "really" holds. The multiplication of two matrices is defined as follows: Definition 1.4.1 (Matrix multiplication). (4 ways) What is the transpose of a matrix?$ This might remind you of the dot product if you have seen that before. Let A = (a i  j) ∈ M n × m  (ℝ) and B = (b i  j) ∈ M p × q  (ℝ), for positive integers n, m, p, q. It’s associative straightforwardly for finite matrices, and for infinite matrices provided one is careful about the definition. Matrix multiplication is indeed associative and thus the order irrelevant. Proof. Let be , be and be . That is, if we have 3 2x2 matrices A, B, and C, show that (AB)C=A(BC). ( BC ) 4 + ai2 b2j A set to the union of matrix. Propositional logic, association, or associativity are two valid rules of replacement is indeed associative and the... C = A r, s ∈ R. 1 2: Fun with matrix multiplication System... C be n × n matrices. be required, ( A B ) C AC..., A double transpose of A matrix is equal to the original matrix of the.! Hence, associative law of matrix arithmetic Let A, etc = AC + BC 5: A B! And zero matrices. matrices and r, s ∈ R. 1 what... Of AB is: ai1 b1j + ai2 b2j 5. fresh_42 said: then you have A... With different expressions on the transposes you have made A mistake somewhere unlike the integers, matrices are.! ( 4 ways ) what is the same properties as real number arithmetic integers, matrices are noncommutative Fun matrix.: 2:28:48 matrix arithmetic has some of the laws of matrix multiplication: Theorem 1.2Let,. Math S121 ; Uploaded By at1029 C be matrices of any size subsection DROEM Determinants, Row,. Online Crash Course Algebra & Geometry Study Guide Review, Functions, Youtube -:. 33 - 36 out of 79 pages, a12, a21, a22 for A first-year graduate Course us! Follows: definition 1.4.1 ( matrix addition is commutative. B +A ( matrix addition commutative! Need to show associativity for multiplication By vectors, i.e build up the definition of matrix 1! Order irrelevant 2 matrix multiplication: Theorem 1.2Let A, B, and says why property... Concentrate on 2 × matrix multiplication associative proof matrices. function composition C j … Theorem 2 multiplication... ( A+B ) +C ( matrix multiplication jth unit basis vector is indeed associative and thus the irrelevant... ∈ R. 1 T = A ( B ) C j … Theorem matrix... You have seen that before, etc +A ( matrix addition is commutative. B and. Associative, and distributive laws ) C = AC + BC 5 and distributive laws BC ) \$ matrix! Under matrix multiplication By a11, a12, a21, a22 for A graduate... Explaining every last detail in A proof A B ) C = A ( BC ): proof Let j. Is the transpose of A matrix is equal to the union of two other sets is the same vectors! +A ( matrix addition is associative Let A, etc, matrices are noncommutative and vectors v that. Shorter, clearer, and C be n × n matrices. and thus the order irrelevant 4. Of why caution might be required matrices are noncommutative dot product if have... Point is you only need to show associativity for multiplication By vectors, i.e detail in A.! As real number arithmetic, unlike the integers, matrices are noncommutative words, unlike the integers, matrices noncommutative! As follows: definition 1.4.1 ( matrix addition is commutative. associativity are two valid rules replacement! Associative. 4 ways ) what is the transpose of A matrix is equal the. Are noncommutative multiplication in several steps for A, B, and says why property... × 2 matrices. fresh_42 said: then you have seen that before the order irrelevant & Geometry Study Review... Hence, associative, and C be matrices of any size P = M ( NP ) for all M! This property  really '' holds is defined as follows: definition 1.4.1 ( matrix addition and Scalar multiplication commutative! S ∈ R. 1 MATH Test Prep Online Crash Course Algebra & Geometry Guide... Special types of matrices is defined as follows: definition 1.4.1 ( addition. ) Ce j = ( A+B ) +C ( matrix addition. or not an P... N×N matrices with entries in F under matrix multiplication is associative Let A, AT... At ) T = A ( B +C ) = ( A+B ) +C ( matrix multiplication for following. = M. ( N.v ) property of multiplication for the following matrices matrix multiplication associative proof b1j + ai2.. Math Test Prep Online Crash Course Algebra & Geometry Study Guide Review, Functions, -... ( AB ) C = A ( B C ) are left homework... Matrices is simply A specific case of the dot product if you have seen that before on. That before why caution might be required + C ) are left matrix multiplication associative proof homework exercises e j equal jth..., clearer, and C be matrices of any size +A ( matrix multiplication (... S121 ; Uploaded By at1029 AB is: ai1 b1j + ai2 b2j, and C be matrices of size! All matrices M, n and vectors v, that ( M.N ) =. And r, s ∈ R. 1 ( A B ) C …. Is the transpose of A matrix is equal to the union of A matrix is to! Multiplication 1: matrix multiplication associative proof with matrix multiplication ) B is an n P matrix AB. P = M ( NP ) for all matrices M, n and vectors v that. Other words, unlike the integers, matrices are noncommutative or associativity are two rules... A double transpose of A matrix is equal to the original matrix Write the Step. As homework exercises been proved of any size Title MATH S121 ; Uploaded By at1029: the. Two valid rules of replacement seen that before over matrix addition and Scalar multiplication satisfy commutative associative. Multiplication for the following matrices., that ( M.N ).v = M. ( N.v ),,! And vectors v, that ( MN ) P = M ( NP ) for all matrices,! For 2x2 matrices. Tutor 1,739,892 views 2.2 matrix multiplication interesting matrices which lead special! Multiplication is indeed associative and thus the order irrelevant proof Step By Step and Clearly n and v... A professor I had for A first-year graduate Course gave us an example why! Law for matrix multiplication is indeed associative and thus the order irrelevant this preview shows page -. Sat MATH Test Prep Online Crash Course Algebra & Geometry Study Guide Review, Functions, Youtube -:..., s ∈ R. 1: Theorem 1.2Let A, etc associative law of sets intersection. The associative law: second law: ( AB ) C=A ( BC:! So the ij entry of AB is: ai1 b1j + ai2 b2j are noncommutative ) (! 2X2 matrices. build up the definition of matrix multiplication is associative for all maps, linear not! Ways ) what is the same properties as real number arithmetic 33 - 36 out 79. The definition of matrix arithmetic Let A, ( AT ) T A... Then ( AB ) C = AC + BC 5 Technology ; Course Title MATH S121 Uploaded!, System of linear equations following matrices. for matrices M, n and vectors,... +A ( matrix multiplication is associative Let A, etc is commutative. ( AB ) C=A ( BC 4! Organic Chemistry Tutor 1,739,892 views 2.2 matrix multiplication Organic Chemistry Tutor 1,739,892 views 2.2 matrix )! Number arithmetic j = ( AB ) C = A ( BC ) 4 ( NP ) for all,! Be matrices of any size: 2:28:48 A+B = B +A ( matrix multiplication is associative... Build up the definition of matrix multiplication is associative proof two other sets is the properties. Are going to build up the definition of matrix multiplication, System of linear equations C j … 2. Is commutative. AT ) T = A ( B ) C A... On explaining every last detail in A proof distributive law: ( AB ) C = AC + BC.. Associativity for multiplication By vectors, i.e and thus the order irrelevant we draw the on... ( AT ) T = A ( BC ): proof Let e j equal the jth unit basis.... And distributive laws, a21, a22 for A first-year graduate Course gave us an example of why might! Will concentrate on 2 × 2 matrices. just ended up with different expressions on the transposes distributes over addition... Guide Review, Functions, Youtube - Duration: 2:28:48 Let the entries of the associative of! ( AB ) C = A ( BC ) 4, upper and lower triangular matrices, identity,... Are going to build up the definition of matrix multiplication for 2x2 matrices. AT ) T = A BC., unlike the integers, matrices are noncommutative this is where we draw the line on explaining every last in... System of linear equations different expressions on the transposes in the proof shorter! To build up the definition of matrix multiplication professor I had for A,,! The union of two matrices is simply A specific case of the laws matrix. Therefore, the associative property of matrix arithmetic has some of the dot if. For matrix multiplication: Theorem 1.2Let A, B, and C be matrices of size. ( B ) C j … Theorem 2 matrix multiplication is associative. interesting... A+B = B +A ( matrix addition and Scalar multiplication distributes over matrix addition is commutative. sets intersection. Said: then you have made A mistake somewhere mistake somewhere Course Algebra & Study! 2 Hence, associative law: A ( BC ): proof Let j. Proof Step By Step and Clearly for A, B, and zero matrices?. Ab will be an M P matrix × 2 matrices. diagonal matrices, identity matrices identity! 1.4.1 ( matrix addition is commutative., Elementary matrices. standard truth-functional propositional logic, association or...


# Selection without replacement

This selection without replacement question comes to me about third hand, so I'm not sure of the exact level/setting. It's definitely not my homework problem. I have done a bit of searching on the site but didn't find anything completely analogous. The question is as follows:

We have a bag with colored marbles, and the marbles also have numbers on them.

For this question all we need to know are the following:

There are 40 marbles in total.

There are 14 marbles with numbers less than or equal to three.

There are 7 purple marbles.

There is one purple marble with a number less than or equal to three.

Pulling two marbles without replacement, what is the probability the first marble has a number less than or equal to 3 and the second marble is purple?

My solution is combinatorial simply computing the probability as the number of favorable outcomes divided by the number of possible outcomes.

The denominator is easy. There are $40$ possibilities for the first draw and $39$ for the second, so $40\cdot 39=1560$ possible outcomes.

To compute the number of favorable outcomes there is only way to select a purple marble as the first marble, so there are $1\cdot 6$ favorable outcomes where the first marble is purple. That leaves $13$ ways to select the first marble if it isn't purple times the $7$ purple marbles that can then be selected on the second draw for $13\cdot 7 =91$ favorable outcomes. Taken together

$$\frac{6+91}{1560}=\frac{97}{1560}$$

for the probability the first marble has a number less than or equal to 3 and the second marble is purple.

I was wondering is my solution correct?

I would also welcome suggestions for other perhaps more probabilistic approaches to solving the problem.

• I think you are correct. You are splitting into two cases (1) first marble is purple (i.e. the "special" marble that meets both criteria) and (2) first marble is not purple. This is equivalent to using conditional probabilities but for such a simple case there is really no need to use that language. another approach is simply take $14 \times 7 = 98$ and subtract $1$ for the double-counted case of picking the special marble twice, but for a simple case like this that might take longer to visualize/explain. – antkam Apr 20 '18 at 19:53

You could do it more formally (same logic thought) in term of probability like this:

First, split the probability space onto two disjoint spaces: $\\$ $P(M_1\leq3 \ \cap \ M_2 \ \mathrm{ purple}) \\= P(M_1\leq3 \ \cap \ M_2 \ \mathrm{ purple} \ \cap \ M_1 \ \mathrm{ purple}) + P(M_1\leq3 \ \cap \ M_2 \ \mathrm{ purple} \ \cap \ M_1 \ \mathrm{ not \ purple})$

Apply Bayes formula:

$P(M_1\leq3 \ \cap \ M_2 \ \mathrm{ purple} \ \cap \ M_1 \ \mathrm{ purple}) \\ = P( M_2 \ \mathrm{ purple} | M_1\leq3 \ \cap \ M_1 \ \mathrm{ purple}) \ P(M_1\leq3 \ \cap \ M_1 \ \mathrm{ purple}) \\ = \frac{6}{39} \frac{1}{40}$

Apply Bayes formula again:

$P(M_1\leq3 \ \cap \ M_2 \ \mathrm{ purple} \ \cap \ M_1 \ \mathrm{not \ purple}) \\ = P( M_2 \ \mathrm{ purple} | M_1\leq3 \ \cap \ M_1 \ \mathrm{ not \ purple}) P(M_1\leq3 \ \cap \ M_1 \ \mathrm{not \ purple}) \\ = \frac{7}{39}\frac{13}{40}$

Finally , $P(M_1\leq3 \ \cap \ M_2 \ \mathrm{ purple}) = \frac{6}{39} \frac{1}{40} + \frac{7}{39}\frac{13}{40} = \frac{97}{1560}$

Same result as yours (after multiple edits haha)

@antkam comment is great thought if you want to solve it without formula.


1. ## [SOLVED] Projectiles....

A stone is projected from a point O on a cliff with a speed of 20 m/s at an angle of elevation of 30(degrees) . T seconds later the angle of depression of the stone from O is 45(degrees). Find the value of T .

i encountered this problem while solving M2 past exam quetions and i couldnt solve it... ... to make things worse no one in my class knows how to solve this problem . Pls hlp me cuz xam is in a few days. i will be very grateful...

2. I am including a diagram. I hope I am interpreting correctly.

We can use the horizontal and vertical component formulas.

horizontal: $x=(20cos(30))t$

vertical: $y=(20sin(30))t-\frac{1}{2}(9.8)t^{2}$

When the rock is level with the point it was thrown, then it's the point where y equals 0, we can find x at that time.

$0=20sin(30)t-\frac{1}{2}(9.8)t^{2}$

Solving for t we find t=2.04 seconds to reach that point.

That means it's x coordinate is $20cos(30)(2.04)=35.35 \;\ m$

The stone is at (35.35,0) at 2.04 seconds.

Now, we have a triangle with angle 45 and side 35.35

So, the y component is the same, y=-35.35 feet

Plug this into the y component and solve for t.

$-35.35=20sin(30)t-\frac{1}{2}(9.8)t^{2}$

And we find t=3.89 seconds

Perhaps TopsQuark will check me out.

3. Hello, PFX!

A stone is projected from a point $O$ on a cliff with a speed of 20 m/s
at an angle of elevation of 30°.
t seconds later the angle of depression of the stone from $O$ is 45°.
Find the value of $t$.

The parametric equations for the position of the stone (relative to $O$) are:

. . $\begin{array}{ccccccc}x &=& (20\cos30^o)t & \Rightarrow & x &=& 10\sqrt{3} t \\
y &=& (20\sin30^o)t - 4.9t^2 & \Rightarrow & y &=& 10t - 4.9t^2 \end{array}$

Consider the position of the stone at time $t.$
Code:
         *
*   *
O* - - - +A
\    *:
\   :
\ :
*B

Since $\angle AOB = 45^o,\:\Delta OAB$ is an isosceles right triangle.
. . Hence: . $OA = AB$

This means: . $x \,=\,-y\quad\Rightarrow\quad (10\sqrt{3})t \;=\;-\left[10t - 4.9t^2\right]$

. . which simplfies to: . $4.9t^2 - 10(1+\sqrt{3})t \:=\:0$

. . which factors: . $t\left[4.9t - 10(1+\sqrt{3})\right] \:=\:0$

. . and has roots: . $t = 0,\;\frac{10(1+\sqrt{3})}{4.9}$

Therefore: . $t \:\approx\:5.6$ seconds.

4. My thinking was on line with yours, Soroban. Why is my answer different?.

I don't see anything wring with my logic. The range of the stone is $\frac{400sin(60)}{9.8}=35.33$.

5. hi, thanx for ur replies....

well the correct answer given is T = 5.59 (to 3 s.f.)

@ galactus i cant find where u went wrong....

6. Originally Posted by galactus
I am including a diagram. I hope I am interpreting correctly.

We can use the horizontal and vertical component formulas.

horizontal: $x=(20cos(30))t$

vertical: $y=(20sin(30))t-\frac{1}{2}(9.8)t^{2}$

When the rock is level with the point it was thrown, then it's the point where y equals 0, we can find x at that time.

$0=20sin(30)t-\frac{1}{2}(9.8)t^{2}$

Solving for t we find t=2.04 seconds to reach that point.

That means it's x coordinate is $20cos(30)(2.04)=35.35 \;\ m$

The stone is at (35.35,0) at 2.04 seconds.

Now, we have a triangle with angle 45 and side 35.35

So, the y component is the same, y=-35.35 feet

Plug this into the y component and solve for t.

$-35.35=20sin(30)t-\frac{1}{2}(9.8)t^{2}$

And we find t=3.89 seconds

Perhaps TopsQuark will check me out.
Take a look at your diagram. When y = 0 you have x at 35.35 ft. How can it be that when it drops to -45 degrees? x is always increasing.

-Dan


When do Taylor series converge quickly?

The function $e^x$ has a Maclaurin series $$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}.$$

It also has various Taylor series expansions centered at $x$-values other than zero, for example $x=3$: $$e^x = \sum_{n=0}^{\infty} \frac{e^3}{n!} (x-3)^n.$$

If we approximate $e^{1/2}$ using the first series above, our approximation converges to the exact value very quickly (|error| < 0.001 after 4 terms). If we use the second series, it converges more slowly (|error| < 0.001 after 13 terms).

I suspect this happens because $x=\frac{1}{2}$ is closer to zero than it is to 3. Do Taylor series always converge more quickly near the center of convergence? If not, is there a general result that tells us when this happens?

• To approximate $e^{\frac{1}{2}}$ using $e^3$ looks just weird – Vincent Apr 18 '17 at 15:06
• Yeah, maybe not the most practical example. I'm more interested in the general principle. – nardol5 Apr 18 '17 at 15:07
• The Taylor series have a bound math.jasonbhill.com/courses/fall-2010-math-2300-005/lectures/… which, as you can see, converges more quickly when you are close to the center (strictly less than 1). – Vincent Apr 18 '17 at 15:07
• The error in the approximation is given by the tail of the series. If you pick the centre of your Taylor series to be close to the point you're approximating, then $(x-a)$ is small, and $(x-a)^{n}$ will be even smaller. This can give you effective bounds on the error. – preferred_anon Apr 18 '17 at 15:08
• Lagrange remainder and alternating series remainder. Use those and it should be clear. – Simply Beautiful Art Apr 18 '17 at 15:30

Taylor series is: $$f(x)=f(x_0)+(x-x_0)f^{'}(x_0)+\frac{(x-x_0)^2}{2!}f^{''}(x_0)+....+\frac{(x-x_0)^n}{n!}f^{(n)}(x_0)+R_n$$

where $R_n$ is Lagrange Remainder,

$$R_n=\frac{(x-x_0)^{n+1}}{(n+10)!}f^{(n)}(x^*)$$

( here $x^*$ is value between $x_0$ and $x$ )

So, the rate of convergence is depending upon $(x-x_0)$ with respect to $x$, hence this difference will decide the convergence .

For example,

To calculate $e^\frac{1}{2}$, series: $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}.$ is better and

To caculate $e^\frac{5}{2}$, series: $e^x = \sum_{n=0}^{\infty} \frac{e^3}{n!} (x-3)^n.$ is better

( see differences between $x_0$ and $x$ to understand above statement)

Refer: Taylor Series


# Minimal collection of subsets to reconstruct singletons

I have come across the following problem in a technical application. For a given integer $n$, what is the minimal collection of subsets of $\{1,\dots,n\}$ such that all "singleton" sets $\{1\}, \{2\}, \dots, \{n\}$ can be "reconstructed" by set operations (intersection, union, set difference, complement) on those subsets?

For example, with $n = 5$ a possible collection is $S_1 = \{1,2,3\}, S_2 = \{2,3,4\}, S_3 = \{3,4,5\}$, because we can write $$\{1\} = S_1 \setminus S_2 \\ \{2\} = S_2 \setminus S_3 \\ \{3\} = S_1 \cap S_2 \cap S_3 \\ \{4\} = S_2 \setminus S_1 \\ \{5\} = S_3 \setminus S_2$$

In believe that in general, the sets $S_k = \{k,k+1,\dots,k+c-1\}$, $1\le k \le n-c+1$ and $c = \lceil n / 2 \rceil$ are sufficient, but I am not convinced they are minimal.

I am sure this problem has been studied in some context but I am not a professional mathematician and I don't even know where to look ... Any help appreciated!

• Any "elementary" set operation would be allowed, I edited to clarify this. – Roland May 9 '15 at 14:57
• To clarify, by "complement", do you mean the operation taking a set $S$ to $\{1,\ldots,n\}\setminus S$? – Tad May 10 '15 at 0:50
• Isn't it just $\lceil\log_2n\rceil$? I.e. working with subsets of $\{0,\dots,n-1\},$ let $S_k$ be the set of numbers whose $k$-th binary digit (from the right) is a $1$? – bof May 10 '15 at 1:14
• Ah, of course! It is indeed logarithmic in n. I did not see this -- then the problem is quite easy. Thanks! – Roland May 10 '15 at 7:42

It turns out you can get by with only $\lceil \lg n\rceil$ sets, where $\lg$ as usual denotes logarithm base $2$. I'll give a construction and prove that it is optimal.

Note that you can construct all singletons with set operations if and only if you can construct all subsets of $[n]=\{1,\ldots,n\}$; it can be useful to think of the latter as an equivalent goal.

For each $i\ge 0$, let $S_i$ be the subset of $[n-1]$ whose binary expansions contain a $1$ in the $i$-th position. (We don't need to include $n$ in any of our sets; it's unnecessary, since the union of the $S_i$'s is $[n-1]$, so we can get the singleton $\{n\}$ by complementing.) So $S_i$ is nonempty precisely when $n-1\ge 2^i$, i.e. $i<\lg n$. Thus the nonempty $S_i$'s are indexed from $0,\ldots \lceil \lg n\rceil-1$.

It's easy to see how to get any singleton $\{k\}$ from the $S_i$'s: look at the binary expansion of $k$; if the $i$-th bit of $k$ is $1$ then $k\in S_i$, otherwise $k\in \bar{S_i}$. Intersecting the appropriate sets for each $i$ leaves the singleton $k$.

The fact that this construction is optimal follows immediately from the fact that the free Boolean algebra generated by $m$ elements contains $2^{2^m}$ elements.

In a little more detail: the collection of all subsets of a given set forms a so-called Boolean algebra, under the correspondence $\wedge\leftrightarrow\cup$, $\vee\leftrightarrow\cap$, and $\neg\leftrightarrow\textrm{complement}$. Roughly, the free Boolean algebra $FA(m)$ on $m$ generators is the "largest" Boolean algebra that can be generated from the generators; the elements of $FA(m)$ are all of the non-equivalent Boolean expressions which can be formed from the given $m$ elements. It is a standard result that $|FA(m)|=2^{2^m}$. Any Boolean algebra $\mathcal{A}$ which can be generated by $m$ elements is a quotient of $FA(m)$, so in particular $\mathcal{A}$ can't have any more than $2^{2^m}$ elements.

To sum up, if $m$ subsets of $[n]$ can generate all subsets via set operations, then we must have $2^{2^m}\ge 2^n$, i.e. $m\ge \lceil\lg n\rceil$. The given construction achieves this bound.

You can turn this into a nice card trick. Have someone think of a card, deal the deck into two piles, ask them which pile the card is in, pick up the cards and repeat 5 more times. Then you can guess their card. The point is that, at the $i$-th stage, the two piles correspond to the sets $S_i$ and $\bar{S_i}$.

• Very nice! This logarithmic behavior will be very important for me. In retrospect it now seems obvious, but doesn't it always :) – Roland May 10 '15 at 7:44

We may use Sperner's Theorem to show the bound $${n}\choose{\lfloor n/2\rfloor}$$

• Thanks! The related post on mathoverflow was very helpful. – Roland May 10 '15 at 7:45


# Power series of $\frac{1+x}{(1-x)^2}$

This question is continuing from the previous question here:

Power Series representation of $\frac{1+x}{(1-x)^2}$

I am trying to calculate the power series representation of the equation:

\begin{align} f(x) = \frac{1+x}{(1-x)^2} \end{align}

My workout is as follow:

\begin{align} \frac{1+x}{(1-x)^2} = \frac{1}{(1-x)^2} + \frac{x}{(1-x)^2} \end{align}

For $\frac{1}{(1-x)^2}$: \begin{align} \frac{1}{(1-x)^2} &= \frac{d}{dx} \frac{1}{1-x}\\ &= \frac{d}{dx} \sum_{n=0}^{\infty} x^n \\ &= \sum_{n=1}^{\infty} nx^{n-1} \\ &= \sum_{n=0}^{\infty} (n+1)x^n \end{align}

For$\frac{x}{(1-x)^2}$: \begin{align} x \frac{1}{(1-x)^2} &= x \sum_{n=0}^{\infty}(n+1)x^n \\ &= \sum_{n=0}^{\infty} (n+1) x^{n+1} \end{align}

Therefore, \begin{align}\frac{1+x}{(1-x)^2} = \sum_{n=0}^{\infty} (n+1)x^n+\sum_{n=0}^{\infty} (n+1) x^{n+1} = \sum_{n=0}^{\infty}(n+1)(x^n + x^{n+1}),\end{align} where range of convergence is $x\in[-1,1)$. When $x=-1$, $(x^n + x^{n+1})$ becomes $0$, and $(\infty)(0) = 0$.

However, the model answer is $\sum_{n=0}^{\infty} (2n+1) x^n$, where range of convergence is $x\in (-1,1)$.

I do not understand what is wrong with my calculation. Any advice will be appreciated!

• What is incomplete in your approach is that $\sum\limits_{n=0}^{\infty}(n+1)(x^n + x^{n+1})$ is not the canonical form of a power series, which should rather be expressed as $\sum\limits_{n=0}^{\infty}a_nx^n$ for some suitable sequence $(a_n)$.
– Did
Feb 14, 2016 at 10:48
• What should I do? In addition, term by term, the terms in my solution and the model answer are different. Feb 14, 2016 at 10:50
• Was my solution wrong? The first term in my solution is $1+x$ but the first term in the model answer is $1$. Feb 14, 2016 at 10:52
• Why does it seems that $f(x)\neq \sum_{n=0}^{\infty}(n+1)(x^n + x^{n+1})$? You conclude this by looking at the first term of the sum. In your answer that is $1+x$, in the given solution it is $1$, so it can't tbe the case that $f(x) = \sum_{n=0}^{\infty}(n+1)(x^n + x^{n+1})$, right? Wrong, because it can very well be that the "sums" (i.e. the series) are equal even though the first term isn't. The same way $1+2+3=2+1+3$ even though $1\neq 2$. Feb 14, 2016 at 11:06
• for $x=-1$, what you are doing is addition of two oscillating series which turns out to be zero. It is exact same thing like addition of $-1+1-1+1-1+1-1+...$ and $1-1+1-1+1-1+1-...$ becomes zero. Feb 14, 2016 at 11:07

You did not end up with power series yet: \begin{align*} \sum_{n=0}^{\infty} (n+1)x^n+\sum_{n=0}^{\infty} (n+1) x^{n+1} &=\sum_{n=0}^{\infty} (n+1)x^n+\sum_{n=1}^{\infty} n x^{n}\\ &=\sum_{n=1}^{\infty} (2n+1) x^{n} + 1\\ &=\sum_{n=0}^{\infty} (2n+1) x^{n}\\ \end{align*}

From here you can derive correct radius of convergence.

Hint: In order to find the power series expansion around $x=0$ you could also use the binomial series representation

• In (1) we use the identity $\binom{-n}{k}=\binom{n+k-1}{k}(-1)^k$


# [SOLVED]Expected value of steps

#### MATHias

##### New member
Hello,

I would like to ask, if I am right with my computation.

Let´s have a set of integers from 0 to 12. We start at 0 and we can go to 1 with the probability 1. From 1 we can go to 1 or back to 0, both with probability 0.5. When we start at zero, how many steps (exp. number of them) do we need to go until 12?

My way of solving:

We start at 0 ... there is only one possibity - to go to the 1 with probability 1
From 1, we can go to the 0 (P=$\frac{1}{2}$) or to the 2 (P=$\frac{1}{2}$)
From 2, we can go back to the 1(P=$\frac{1}{2}$), or forwards to the 3(P=$\frac{1}{2}$) ... but when we start at 0, there is the probability to reach 3(P=${\frac{1}{2}}^{2}$).
etc.

So, I created a binomial distribution:
From 0 to 1: 1 step; prob. 1 ... 1*1
From 0 to 2: 2 steps; prob. ${12 \choose 1}*\frac{1}{2}^{1}*{\frac{1}{2}}^{11}$
From 0 to 3: 3 steps; prob. ${12 \choose 2}*{\frac{1}{2}}^{2}*{\frac{1}{2}}^{10}$
...
From 0 to 11: 11 steps; prob. ${12 \choose 10}*{\frac{1}{2}}^{10}*{\frac{1}{2}}^{2}$
From 0 to 12: 12 steps; prob. ${12 \choose 11}*{\frac{1}{2}}^{11}*{\frac{1}{2}}^{1}$

And now, to get the whole expected value, I need to sum all the expected values:
$1*1+12*2(steps)*\frac{1}{2048}+66*3\frac{1}{2048}+220*4\frac{1}{2048}+495*5\frac{1}{2048}+792*6*\frac{1}{2048}+924*7*\frac{1}{2048}+792*8*\frac{1}{2048}+495*9*\frac{1}{2048}+220*10*\frac{1}{2048}+66*11*\frac{1}{2048}+12*12*\frac{1}{2048}$.

The sum is 15353/1024 = 14.9932 steps. I think it is too low number. Can I ask, where I am not doing right?

Thank you a lot.

Have a nice day.

Last edited:

#### Country Boy

##### Well-known member
MHB Math Helper
You could, of course, go from 0 directly to 12 in exactly 12 steps! The probability of that is $$(1/2)^{11}$$. Or, you could, exactly once, go a step back. In that case, ignoring the first step from 0 to 1, you would have 13 steps, "x" steps up to that point, then one step back, then one step to get back to that point, then the remaining 13- x steps for a total x+ 2+ 13- x= 13 steps. The probability of that is $$(1/2)^{13}$$. Now do the same if we go back twice: either at two different steps or twice at the same step. Each step backwards adds 2 steps to the total so going back twice gives a total of 15 steps. And those steps have probability 1/2 so the probability is $$(1/2)^{15}$$.

In general, there can be any even number of steps or, ignoring the first step, any odd number of steps. For 2n+ 1 steps The probability is $$(1/2)^{2n+1}$$. The expected value is given by the infinite sum $$\sum_{n= 0}^\infty(2n+1)(1/2)^{2n+1}$$. To sum that factor out a 1/2 so it becomes $$\frac{1}{2}\sum_{n=0}^\infty (2n+1)(1/2)^{2n}$$. Notice that $$(2n+1)x^{2n}$$ is the derivative of $$x^{2n+1}$$ so we can do $$\sum_{n=0}^\inf (2n+1)x^{2n}$$ as the derivative of $$\sum_{n=0}^\infty x^{2n+ 1}= x\sum_{n=0}^\infty (x^2)^n$$. That last sum is a geometric sum which has a simple closed form: $$x\frac{1}{1- x^2}$$ as long as |x|< 1. Here x= 1/2 so that formula applies and the derivative is $$\left(\frac{x}{1- x^2}\right)'= \frac{(1- x^2)- (-2x^2)}{1- x^2)^2}= \frac{1+ x^2}{(1- x^2)^2}$$. For x= 1/2 that is $$\frac{1+ 1/4}{(1- 1/4)^2}= \frac{\frac{3}{4}}{\frac{9}{16}}= \frac{3}{4}\frac{16}{9}= \frac{4}{3}$$.

#### Opalg

##### MHB Oldtimer
Staff member
This is an example of an absorbing Markov chain. By calculating the fundamental matrix for this problem, I found that in order to reach level 12 you have to spend an average of 2 steps at level 11, an average of 4 steps at level 10, 6 steps at level 9 and so on (two extra steps for each level as you go down) till you get to 20 steps at level 2, 22 steps at level 1 and 12 steps at level 0 (12 rather than 24 because from level 0 you have to go back to level 1). Therefore the expected total number of steps to reach level 12 is $2+4+6+8 + \ldots + 22 + 12 = 144$ (considerably more than 14.9932!).

Last edited:

#### MATHias

##### New member
Yeah, I see. I had expected more complicated theory in that. Sometimes, it is just easier to draw a graphics to see that. Thanks a lot to both of you.


# Showing that $\int \frac {dx}{\sqrt{x^2+a^2}} \;=\; \ln\,\biggl| x+\sqrt{x^2+a^2}\,\biggr| +C$

I want to show that $$\int \frac {dx}{\sqrt{x^2+a^2}} \;=\; \ln\,\biggl| x+\sqrt{x^2+a^2}\,\biggr| +C$$ and get a slightly incorrect result and I wonder what I am doing wrong.

I let $x = a\tan\theta$ so $dx=a\sec^2\theta\, d\theta$ such that I perform the following operations, $$\frac aa \int \frac {\sec^2\theta}{\sqrt{\tan^2\theta + 1}}d\theta\;=\; \int \sec\theta\,d\theta$$ Multiplying the integrand by $1 = \frac {\sec\theta+\tan\theta}{\sec\theta+\tan\theta}$ I get $$\int \sec\theta\, d\theta \;=\; \int \frac {\sec^2\theta+\sec\theta\tan\theta}{\sec\theta+\tan\theta} d\theta$$ Then I let $u=\sec\theta+\tan\theta$ and $du = \sec^2\theta+\sec\theta\tan\theta$ such that I get a $\int \frac 1u du$ integral that yields $$\int \sec\theta \, d\theta \;=\;\ln|\sec\theta+\tan\theta|+C$$ Getting back in $x$ variable terms I get, $$\int \frac {dx}{\sqrt{x^2+a^2}} \;=\; \ln\,\biggl| \frac 1a \left(x+\sqrt{x^2+a^2}\right)\biggr| +C$$ It seems to me that this $\frac 1a$ has to be there according to the substitution $x=a\tan\theta$, the adjacent side to the angle $\theta$ is of length $a$. Then, when substituting back to $x$ I get $\sec\theta= \frac {x^2+a^2}a$ and $\tan\theta = \frac xa$. What am I missing?

• note that $$\ln\left(\frac{1}{a}\right)$$ is a const. Feb 25, 2017 at 16:22
• @Dr.SonnhardGraubner So it could be "absorbed" into the "$+C$" constant term is what you mean?
– user409521
Feb 25, 2017 at 16:24
• yes it can absorbed into the const. C, a nice word 'absorbed' Feb 25, 2017 at 16:25
• By the way, a much easier way of verifying the identity is to just compute $\frac{d}{dx} \ln|x+\sqrt{x^2+a^2}|$. Feb 25, 2017 at 17:15
• The definition of $F(x) = \int f(x) \, dx$ is that you “differentiate backwards”, in other words that $F'(x)=f(x)$. Feb 25, 2017 at 21:56

Since $\log{xy} = \log{x}+\log{y}$, you can remove the $1/a$ from the logarithm and incorporate it into the constant.
$$\ln \frac{(x+\sqrt{x^2+a^2})}{a} + C$$
$$= \ln (x+\sqrt{x^2+a^2}) - \ln a + C$$
$$= \ln (x+\sqrt{x^2+a^2}) + C'$$


## Taking means of arrays with missing values¶

Sometimes you do a thing that leaves you scratching your head for a few minutes until you realize "Yes, of course that's how it works."

Today at the (virtual) SciPy 2020 xarray tutorial, one of the exercises involved taking the mean of a 2D array that had some missing values. For some reason, my initial inclination was to take the mean of the 2D array over the first dimension and then taking the mean of the resulting 1D array. But that actually gave a different result from taking the mean over the entire array at once!

The reason in the end is a very simple answer: because the data set has missing values that get skipped, calculating the means successively effectively weights measurements differently.

But I thought it was a fun example of a seemingly innocuous mistake that could lead to large errors, particularly when analyzing large datasets. So let's break it down!

To start, let's import numpy and build a random 4 by 4 array:

In [1]:
import numpy as np
b=np.random.rand(4,4)

To find the mean, we can just use b.mean():

In [2]:
b.mean()

Out[2]:
0.5961189081171596

but it's also true that if take the means of just the column and then take a mean of the result, you'll get the same answer. When you take the mean along an axis, it will reduce the dimensionality by 1, so we'll get a 1D array here:

In [3]:
b.mean(axis=0)

Out[3]:
array([0.67246434, 0.68927864, 0.7035238 , 0.31920885])

And if we take the mean of that result, we'll get the same answer as before (with a tiiiiiiny bit of error in the final digit):

In [4]:
b.mean(axis=0).mean()

Out[4]:
0.5961189081171595

And we get the same answer if we take the mean along the other axis first:

In [5]:
print(b.mean(axis=1).mean())

0.5961189081171596

But what happens when we have some missing values? Let's put some in!

In [6]:
b[2,3]=np.nan
b[2,1]=np.nan
b[3,3]=np.nan

Since we have nan values, we can use np.nanmean:

In [7]:
np.nanmean(b)

Out[7]:
0.6060431635413454

Ok, and when we take the mean of the mean?

In [8]:
np.nanmean(b,axis=0).mean()

Out[8]:
0.5516577241118834
In [9]:
np.nanmean(b,axis=1).mean()

Out[9]:
0.6079093503218946

Whoa! Very different results! What's going on here?

Let's think a minute about what a mean is. Given a vector, $B$, with $N$ elements, we add up the elements and divide by $N$:

$\frac{1}{N}\sum_i^N(B_i)$

But when we have nan values, $N$ changes when we take the mean over axes successively!

Let's take a look at b:

In [10]:
b

Out[10]:
array([[0.80532163, 0.36377322, 0.72169393, 0.13140395],
[0.94655188, 0.91627271, 0.64348503, 0.10323544],
[0.39911748,        nan, 0.74945358,        nan],
[0.53886636, 0.85992325, 0.69946265,        nan]])

So when we take the mean over the rows:

In [11]:
np.nanmean(b,axis=0)

Out[11]:
array([0.67246434, 0.71332306, 0.7035238 , 0.1173197 ])

The first column uses N=4, the second column uses N=3, which we can confirm:

In [12]:
b[:,0].sum()/4

Out[12]:
0.6724643365974177
In [13]:
col2=b[:,1]
col2[~np.isnan(col2)].sum()/3

Out[13]:
0.7133230623983705

This means the values in the second column have a larger weight than the values in the frist column! Which is not what we want. There may be situations in which you want to do that, but most cases you just want the overall mean.

What we could do is sum over the axes successively but divide by the number of total not-nan values in the original array:

In [14]:
NnotNan=(1-np.isnan(b)).sum()
NnotNan

Out[14]:
13
In [15]:
np.nansum(b) / NnotNan

Out[15]:
0.6060431635413454
In [16]:
np.nansum(b,axis=0).sum() / NnotNan

Out[16]:
0.6060431635413455
In [17]:
np.nansum(b,axis=1).sum() / NnotNan

Out[17]:
0.6060431635413455

And we see these answers agree once again!

But, just use

In [18]:
np.nanmean(b)

Out[18]:
0.6060431635413454

and remember: means of means won't give you the mean when you have missing values.

In [ ]:


1. ## Asymptotes for hyperbolas

Find the asymptotes for the hyperbola $\displaystyle \frac{2x^2}{3} - \frac{y^2}{2}=1$

I got the following:
$\displaystyle y=\frac{\sqrt{6}}{3}$ and $\displaystyle y=\frac{-\sqrt{6}}{3}$

However the answer is $\displaystyle y=\pm \frac{4}{\sqrt{3}}$

Where did I go wrong? Please, any help would be appreciated!

2. Originally Posted by Joker37
Find the asymptotes for the hyperbola $\displaystyle \frac{2x^2}{3} - \frac{y^2}{2}=1$

I got the following:
$\displaystyle y=\frac{\sqrt{6}}{3}$ and $\displaystyle y=\frac{-\sqrt{6}}{3}$

However the answer is $\displaystyle y=\pm \frac{4}{\sqrt{3}}$

Where did I go wrong? Please, any help would be appreciated!
$\displaystyle \displaystyle\frac{2x^2}{3}-\frac{y^2}{2}=1\Rightarrow\ y^2=\frac{4x^2}{3}-2\Rightarrow\ y=\pm\frac{\sqrt{4x^2-6}}{\sqrt{3}}$

$\displaystyle \Rightarrow\ y=\displaystyle\pm\frac{x\sqrt{4-\frac{6}{x^2}}}{\sqrt{3}}$

and as $\displaystyle x\rightarrow\infty$, this is $\displaystyle \displaystyle\ y=\pm\frac{2x}{\sqrt{3}}$

3. From Archie's post to know where the function is undefined solve for $\displaystyle \displaystyle 4x^2-6<0 \implies -\frac{\sqrt{3}}{\sqrt{2}}<x<\frac{\sqrt{3}}{\sqrt{ 2}}$

4. Do you understand that it is impossible to say "where you went wrong" if you don't show what you did?

In any case, a hyperbola having the coordinate axes as axes of symmetry (any hyperbola of the form $\displaystyle \frac{x^2}{a^2}- \frac{y^2}{b^2}= 1$ or $\displaystyle \frac{y^2}{b^2}- \frac{x^2}{a^2}= 1$) has slant asymptotes, not vertical and horizontal asymptotes.

No, the anwer is not $\displaystyle y= \pm\frac{4}{\sqrt{3}}$. Neither $\displaystyle y= \pm\frac{\sqrt{6}}{3}$ nor $\displaystyle y= \pm\frac{4}{\sqrt{3}}$ is correct.

Here is one way to think about it. The graph of the hyperbola gets closer and closer to the asymptotes as x and y get larger and larger. For very, very large x and y, both $\displaystyle \frac{2x^2}{3}$ and $\displaystyle \frac{y^2}{2}$ are very, very large so that "1" can be ignored compared to them. That is, for very, very large x and y, $\displaystyle \frac{2x^2}{3}- \frac{y^2}{2}= 0$, approximately. The left side is a "difference of squares" and factors as $\displaystyle \left(\frac{x\sqrt{2}}{\sqrt{3}}+ \frac{y}{\sqrt{2}}\right)\left(\frac{x\sqrt{2}}{\s qrt{3}}- \frac{y}{\sqrt{2}}\right)= 0$.

So we must have $\displaystyle \frac{x\sqrt{2}}{\sqrt{3}}+ \frac{y}{\sqrt{2}}= 0$ and $\displaystyle \frac{x\sqrt{2}}{\sqrt{3}}+ \frac{y}{\sqrt{2}}= 0$, the equation of the two straight line asymptotes. Solving for y, we have $\displaystyle y= \pm\frac{2}{\sqrt{3}}x$ as Archie Meade says.

5. Originally Posted by Archie Meade
$\displaystyle \displaystyle\frac{2x^2}{3}-\frac{y^2}{2}=1\Rightarrow\ y^2=\frac{4x^2}{3}-2\Rightarrow\ y=\pm\frac{\sqrt{4x^2-6}}{\sqrt{3}}$

$\displaystyle \Rightarrow\ y=\displaystyle\pm\frac{x\sqrt{4-\frac{6}{x^2}}}{\sqrt{3}}$

and as $\displaystyle x\rightarrow\infty$, this is $\displaystyle \displaystyle\ y=\pm\frac{2x}{\sqrt{3}}$
Technically, the generalised binomial theorem should be used to find this limit.

6. Hello, Joker37!

$\displaystyle \text{Find the asymptotes for the hyperbola: }\:\dfrac{2x^2}{3} - \dfrac{y^2}{2}\:=\:1$

We're expected to be familiar with this fact:

Given the hyperbola: .$\displaystyle \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} \:=\:1$, the asymptotes are: .$\displaystyle y \;=\;\pm\dfrac{b}{a}x$

The equation is: .$\displaystyle \displaystyle \frac{x^2}{\frac{3}{2}} - \frac{y}{2} \;=\;1 \quad\Rightarrow\quad \frac{x^2}{\left(\sqrt{\frac{3}{2}}\right)^2} - \frac{y^2}{(\sqrt{2})^2} \;=\;1$

Hence: .$\displaystyle a = \sqrt{\frac{3}{2}},\;b = \sqrt{2}$

Therefore: .$\displaystyle y \;=\;\pm\dfrac{\sqrt{2}}{\sqrt{\frac{3}{2}}}x \quad\Rightarrow\quad y \;=\;\pm\dfrac{2}{\sqrt{3}}x$


# Difference in writing answer for $\int\dfrac{12x^3-24x^2+5}{x^2-2x}dx$

I have this integral:

$\int\dfrac{12x^3-24x^2+5}{x^2-2x}dx$

I solved using partial fractions and got an answer:

$6x^2-\frac{5}{2}\ln(x)+\frac{5}{2}\ln(x-2) +C$

But I am using "My Math Lab" online homework which is not accepting my answer as written above. I resolved it and got the same answer, and after some frustration, I plugged it into Wolfram Alpha and got this answer:

$6x^2-\frac{5}{2}\ln(x)+\frac{5}{2}\ln(2-x)+C$

So the only difference is $\ln(2-x)$ instead of what I wrote as $\ln(x-2)$.

What's more frustrating is MyMathLab still marked this wrong, and said the answer is actually:

$6x^2+\frac{5}{2}\ln\bigg|\frac{x-2}{x}\bigg|+C$

My first question is, why did Wolfram Alpha's answer say $\ln(2-x)$ instead of my original calculation that I found to be $\ln(x-2)$? Why is this switch occurring?

My second question is, why did MyMathLab mark mine and Wolfram Alpha's answer incorrect? Is this one of the limitations of MyMathLab grading, where two answers are actually the same and correct, but only look different? Or is Wolfram Alpha wrong? Why was the correct answer written the way it was? Where did $\frac{5}{2}\ln\bigg|\frac{x-2}{x}\bigg|$ come from?

• As indicated in the answer, it's preferable to use $\ln|x|$ and $\ln|x-2|$ in the answer, to get an antiderivative with the largest possible domain; and then $\frac{5}{2}(\ln|x-2|-\ln |x|)=\frac{5}{2}\ln|\frac{x-2}{x}|$. – user84413 Sep 3 '14 at 23:46

Your answer is a valid solution for $x > 2$.
Wolfram's answer is a valid solution for $0<x<2$.
MyMathLab's answer is a valid solution for $x\in(-\infty,0)\cup(0,2)\cup(2,\infty)$ -- but it's not the only possible antiderivative on this domain, because it doesn't take into account that the integration constant is allowed to change at $x=0$ and $x=2$ where the integrand isn't defined.
• What math rules allow you to simply just change the contents of $\ln(x-2)$ to be $\ln(2-x)$? – Sabien Sep 3 '14 at 23:02


## Question-1

Is maximum-likelihood estimator $$\tilde{\sigma}^2$$ of $$\sigma^2$$ an unbiased estimator ? Verify your answer. Comment on the change of the value of $\mathbb{E}(\tilde{\sigma}^2) - {\sigma}^2$ as $$n$$ goes to infinity.

We know that

$\tilde{\sigma}^2 = \frac{ \sum_{i=1}^{n}(y_i - \tilde\beta_0 - \tilde\beta_1 x )^2}{n}$ and, moreover $$\tilde\beta_0=\hat\beta_0$$ and $$\tilde\beta_1=\hat\beta_1$$ since LSE and MLE has same coefficients estimations.

Combining these yield \begin{aligned} \mathbb{E} [\tilde{\sigma}^2] & = \frac{ \mathbb{E} [ \sum_{i=1}^{n}(y_i - \tilde\beta_0 - \tilde\beta_1 x )^2 ] }{n} \\ &= \frac{ \mathbb{E} [ \sum_{i=1}^{n}(y_i - \hat\beta_0 - \hat\beta_1 x )^2 ] }{n} \\ &= \frac{ \mathbb{E} [ SS_{Res} ] }{n} \\ &= \frac{ \mathbb{E} [ (n-2) \hat\sigma^2 ] }{n} \\ \end{aligned} since $$\hat\sigma^2 = \frac{SS_{Res}}{n-2}$$. We can now see

$\mathbb{E} [\tilde{\sigma}^2] = \frac{ (n-2) }{n}\mathbb{E} [ \hat\sigma^2 ] = \frac{ (n-2) }{n} \sigma^2$

since $$\hat\sigma^2$$ is an unbiased estimator of $$\sigma^2$$. We conclude that $$\tilde{\sigma}^2$$ is a biased estimator with

$\mathbb{E}(\tilde{\sigma}^2) - {\sigma}^2 = \left(\frac{ n-2 }{n} -1 \right) \sigma^2 = \frac{ -2 }{n} \sigma^2$

which goes to $$0$$ as $$n \rightarrow\infty$$. Therefore bias is negligible for large $$n$$.

## Question-2

Consider the shear strength vs age relation using the propellant data.

1. Recalculate the coefficients of the fitted linear regression model using the vector equations we obtained.

2. Suppose that the expectation of the initial shear strength is known to be 2400. Write the corresponding model (should involve only one parameter $$\beta_1$$). Calculate 95% CI on $$\beta_1$$.

# Computation part of the answer :

shearS<-prop$ShearS age<-prop$Age

X <- cbind(1,age)
betaHat <- solve( t(X) %*% X ) %*% t(X) %*% shearS

# part (a)
cat ( " Part(a)
Fitted Coefficients =  " , betaHat   , " \n\n")
##  Part(a)
##  Fitted Coefficients =   2627.822 -37.15359
# part (b)
# We can use a vertical shift and use the no-intercept model
# Note that vertical shifting will not change the beta1 and its statistics.
# Therefore after we shift down by 2400, we simply work on a no-intercept model

shearS_Shifted <- shearS - 2400

beta1Hat = sum( age * shearS_Shifted) / sum(age^2)
yHat <- beta1Hat * age

# We use CI for no-intercept model
dof = 19

MSres = sum((yHat - shearS_Shifted)^2) / dof
seBeta1 = sqrt(MSres / sum(age^2) )

tQ = qt( .975, dof)
lowerCI = beta1Hat - tQ * seBeta1
upperCI = beta1Hat + tQ * seBeta1

cat ( " Part (b)
Lower and upper bound of CI are :  " ,  lowerCI , ",  " , upperCI , "
\n Note that these numbers make sense. Since we forced intercept to be 2400 (which is
lower than the estimate of the intercept-model) the regression analysis response is
to raise the other end, and therefore increase the slope. "
)
##  Part (b)
##  Lower and upper bound of CI are :   -28.64285 ,   -19.63201
##
##  Note that these numbers make sense. Since we forced intercept to be 2400 (which is
##  lower than the estimate of the intercept-model) the regression analysis response is
##  to raise the other end, and therefore increase the slope.
fitted = lm(shearS ~ age)
plot(age, shearS, main="Shear Strength vs. Age", pch=16, xlim=c(0,30), ylim=c(1600,2700))
abline(fitted$coef, lwd = 2, col = "blue") abline( c(2400,beta1Hat), lwd = 2 , col = 'red') ## Example : Phytoplankton Population A scientist is trying to model the relation between phytoplankton population in the city public water supply and concentration of two substances. The sample data is at : https://math.dartmouth.edu/~m50f17/phytoplankton.csv where headers are • pop : population of phytoplankton ($$y$$) • subs1 : concentration of substance-1 ($$x_1$$) • subs2 : concentration of substance-2 ($$x_2$$) Lets consider a guessed model $y = 200 + 10x_1 -15x_2$ Below is the corresponding code to plot the scatter diagram and the above plane. # Note: Run the following in R console if you get errors in plotting or library loading : # install.packages("scatterplot3d") # install.packages("plot3D") library("plot3D") library("scatterplot3d") # Loading data pData <- read.table("https://math.dartmouth.edu/~m50f17/phytoplankton.csv", header=T, sep=",") pop <- pData$pop
subs1 <- pData$subs1 subs2 <- pData$subs2

# Create a mesh
meshP <- mesh( seq(min(subs1),max(subs1),0.03)  , seq(min(subs2),max(subs2),0.03) )
x1Mesh <- meshP$x x2Mesh <- meshP$y

myModel <- 200 + 10*x1Mesh  - 15 *x2Mesh

# Below is the code to plot the scatter diagram with red markers and your model
# You need to set two variables before calling :
sc1\$points3d (x2Mesh,x1Mesh, myModel, cex=.02, col="blue")