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# What's the difference between MUTUALLY EXCLUSIVE and PAIRWISE DISJOINT?
When I study Statistical Theory, I find that these two concepts confuse me a lot.
By definition, if we say two events are PAIRWISE DISJOINT, that means the intersection of these two event is empty set. If we say that two events are MUTUALLY EXCLUSIVE, that means if one of these two events happens, the other will not. But doesn't it means that these two events are PAIRWISE DISJOINT？
If we say two events are MUTUALLY EXCLUSIVE, then they are not INDEPENDENT. Can we say that two PAIRWISE DISJOINT events are not INDEPENDENT as well？
If these two concepts are different (actually my teacher told me they are), could you please give me an example that two events are MUTUALLY EXCLUSIVE but not PAIRWISE DISJOINT, or they are PAIRWISE DISJOINT but not MUTUALLY EXCLUSIVE. | {
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"Disjoint" is a property of sets. Two sets are disjoint if there is no element in both of them, that is if $$A \cap B = \emptyset$$.
In some (but not all!) texts, "mutually exclusive" is a slightly different property of events (sets in a probability space). Two events are mutually exclusive if the probability of them both occurring is zero, that is if $$\operatorname{Pr}(A \cap B) = 0$$. With that definition, disjoint sets are necessarily mutually exclusive, but mutually exclusive events aren't necessarily disjoint.
Consider points in the square with each coordinate uniformly distributed from $$0$$ to $$1$$. Let $$A$$ be the event where the $$x$$-coordinate is $$0$$, and $$B$$ be the event that the $$y$$-coordinate is $$0$$. $$A \cap B = \{(0,0)\}$$ so $$A$$ and $$B$$ are not disjoint, but $$\operatorname{Pr}(A \cap B) = 0$$ so they are mutually exclusive.
As a second (silly, but finite) example, let the sample space be $$S = \{x, y, z\}$$ with probabilities $$\operatorname{Pr}(\{x\}) = 0$$, $$\operatorname{Pr}(\{y\}) = \frac{1}{2}$$, and $$\operatorname{Pr}(\{z\}) = \frac{1}{2}$$. If $$A = \{x, y\}$$ and $$B = \{x, z\}$$, then $$A \cap B = \{x\}$$, but $$\operatorname{Pr}(A \cap B) = \operatorname{Pr}(\{x\}) = 0$$. They are mutually exclusive but not disjoint. | {
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# Real Roots and Differentiation
Prove that the equation $$x^5 − 1102x^4 − 2015x = 0$$ has at least three real roots.
So do I sub in values of negative and positive values of $$x$$ to show that there are at least three real roots? The method to do this question is not by finding the factors of $$x$$ right? Because it will be too tedious so I want to ask whats the other solution to prove this? Help appreciated. Thank you very much.
• one root is $x=0$. So your problem is reduced to $x^4-1102x^3-2015=0$. Then, try to find two pairs $x_1, x_2$ where $x^4-1102x^3-2015$ changes sign. You can then conclude from the Intermediate value theorem, that there are at least 2 additional roots. – MrYouMath Sep 17 '15 at 18:10
• let $$f(x)=x^5-1102x^4-2015x$$ then calculate $$f(-2)$$ and $$f(-1)$$ and $$f(1000)$$ and $$f(2000)$$ – Dr. Sonnhard Graubner Sep 17 '15 at 18:13
It's clear that $x=0$ is one of the roots. Hence, if we prove there are atleast 2 zeros to $f(x) := x^4-1102x^3-2015$, we are done.
Observe, $f(0) < 0$ and $f(-2) > 0$, so from Intermediate Value Theorem there exists at least one root between $-2$ and $0$.
Now, lets say there is exactly one real root to $f$ which means that there are 3 non real complex roots to $f$. This can not be possible as complex roots occur in conjugate pairs. Hence, there are at least 2 real roots to $f=0$
It's about the real zeros of $x q(x)$ with $q(x):=x^4-1102 x^3-2015$. There is the obvious zero $x=0$. Furthermore $q(0)<0$ and $\lim_{x\to\pm\infty} q(x)=+\infty$ guarantee two more real zeros.
You can use Descartes' rule of signs to tell you the number of real roots as long as you are not interested in the value of each.
First observe that $x=0$ is a root for:
$f(x)=x^5 − 1102x^4 − 2015x$
Second, count positive real roots by counting sign changes in $f(x)$, we have: (+-)(--), that is 1 sign change indicating 1 positive root. | {
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third, count negative real roots by counting sign changes in $f(-x)$ where: $f(-x)=-x^5 - 1102x^4 + 2015x$
here we have the signs: (--)(-+), so we have 1 negative root.
From the above, we have 3 real roots for $f(x)$.
$$x^5 − 1102 \cdot x^4 − 2015 \cdot x = 0$$
This factors into,
$$\left( x^4 − 1102x^3 − 2015 \right) \cdot x = 0$$
Therefore $x=0$ is a root. Keep simplifying,
$$x^4 − 1102x^3 − 2015=0$$
Set this equal to $f(x)$,
$$f(x)=x^4 − 1102x^3 − 2015$$
$f(-1)=-912$ and $f(-2)=6817$, thus by the Intermediate Value Theorem, there is a zero in-between $-2$ and $-1$.
The same thing holds for $f(1100)$ and $f(1110)$
Now, in the spirit of fairness, let's quickly come up with the method that will allow us to "guess" where the roots of $f(x)$ lie. We'll use Newton's Method.
We get,
$$x_{n+1}=x_{n}-{{x^4 − 1102x^3 − 2015} \over {4 \cdot x^3-2204 \cdot x^2}}$$
Now, the property we'll use is the fact that the convergence for the method is oscillatory. That means if you guess too low, then the next guess will be to high $^1$. Applying this principle results in a guess of $1000$ for the root resulting in a new guess of $1170$ for the next root. Once again, the Intermediate Value Theorem applies.
$^1$ (There are subtleties about convergence and when this works and doesn't work, but generally speaking, this is the case if you pick a reasonable guess)
• must we write the statement that the equation is continuous on IR – user271716 Sep 18 '15 at 15:34
• What? I don't follow... – Zach466920 Sep 18 '15 at 15:40
The Newton polygon tells us that the dominant binomials are
• $x^5-1102x^4$ for large roots, resulting in a root close to $1102$ and
• $-1102x^4-2015x$ for small roots, resulting in roots close to $0$ and the three third roots of $-\frac{2015}{1102}\approx (-1.22282549628)^3$. | {
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Single real roots stay real under small perturbations, thus giving exactly 3 real roots and a pair of complex conjugate roots. Indeed the numerical approximations confirm this, they are (thanks to http://www.akiti.ca/PolyRootRe.html):
0
0.6111860961336238 + 1.0593896464200445 i
0.6111860961336238 - 1.0593896464200445 i
-1.2223736979388697
1102.0000015056714 | {
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# Showing Lagrange polynomials form a basis for $\Pi_3$.
Here's the question:
Let $x_0=-2, x_1=0, x_2=1, x_3=4$ and let $L_j$ be the Lagrange polynomial for $j=0,1,2,3$. Show that $L_0, L_1, L_2, L_3$ form a basis for $\Pi_3$.
So I've calculated $L_0,...,L_3$. I'll supply $L_3$ and $L_0$ just to see if I got them right. I am, on the other hand, completely lost on what I should be doing next.
$L_0= x(x-1)(x-4)/(-36)$
$L_3= x(x+2)(x-1)/(24)$
First of all, as far as I can see (if I haven't miscalculated as well ^^), $L_3$ should be $$L_3 = \frac{x(x+2)(x-1)}{(4-(-2)) \cdot 4 \cdot (4-1)} = \frac{x(x+2)(x-1)}{72}$$
Now to the question why they form a basis. The Lagrange polynomials are defined in such a way that you have $$L_j(x_i) = \delta_{ij}, i,j=0,\dots,3$$ where $\delta_{ij}$ is the Kronecker delta. We know that $\dim(\Pi_3) = 4$ (because $\{1,x,x^2,\dots,x^3\}$ is a basis). Thus it suffices to show that $L_0, \dots, L_3$ are linearly independent. So let $\lambda_0, \dots, \lambda_3$ so that $$\phi := \sum_{j=0}^{3}{\lambda_j L_j} = 0$$
We now need to show that $\lambda_j = 0\ \ \ \forall j \in \{0,\dots,3\}$. But this follows easily from $$0 = \phi(x_i) = \sum_{j=0}^{3}{\lambda_j L_j}(x_i) = \sum_{j=0}^{3}{\lambda_{j}\delta_{ij}} = \lambda_i$$
• Yes, only $L_3$ was incorrect. Thanks for the guidance sir! Mar 12 '16 at 13:20
• @user314580 you're welcome. But bubba is right. Upvoting / accepting answer is a nice way of approving efforts. Mar 12 '16 at 13:42
• Must I register to do so? Mar 12 '16 at 14:08
• @user314580 No, simply click the "up" or "down" array over/beneath the score of a post. If the post is an answer to a question you asked, you can also accept one answer (note: you can upvote multiple answers, but "accept" only one). This can be done by clicking the checkmark below the score number. Mar 12 '16 at 15:18
I assume that $\Pi_3$ means the space of all polynomials of degree 3. | {
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I assume that $\Pi_3$ means the space of all polynomials of degree 3.
This space has dimension $4$, so, if we can show that $L_0$, $L_1$, $L_2$, $L_3$ span it, then we're done.
So, given any cubic polynomial $a_0 + a_1x + a_2x^2 + a_3x3$, we have to show that there exist numbers $\lambda_0$, $\lambda_1$, $\lambda_2$, $\lambda_3$ such that $$a_0 + a_1x + a_2x^2 + a_3x^3 = \lambda_0 L_0(x) + \lambda_1 L_1(x) + \lambda_2 L_2(x) + \lambda_3 L_3(x)$$ Equating coefficients of powers of $x$, you'll get a system of four linear equations for $\lambda_0$, $\lambda_1$, $\lambda_2$, $\lambda_3$. Use your favorite technique to show that this linear system has a solution (e.g. non-zero determinant). | {
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Why does this algorithm generate a standard normal distribution?
I have this algorithm which I encountered:
(1) Generate $$U_1$$, $$U_2$$ independently from Uniform(0,1)
(2) Set $$Y_1 = -\log{U_1}, Y_2 = -\log{U_2}$$. If $$Y_2 > \frac{(1-Y_1)^2}{2}$$, accept $$(Y_1, Y_2)$$. Else reject and return to step 1.
(3) Generate $$U_3$$ from Uniform(0,1). If $$U_3 < 0.5$$, accept $$X=Y_1$$. Else, $$X=-Y_1$$
Why does X follow a standard normal distribution in the end? I know $$Y_1$$ and $$Y_2$$ here are exponential R.Vs. I probably need to understand how comparing $$Y_2$$ and $$\frac{(1-Y_1)^2}{2}$$ here does the trick and the third step is probably the result of a distribution formed after step 2 which is a folded standard normal distribution.
Edit: Source: Class Notes on Monte Carlo Methods | {
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Edit: Source: Class Notes on Monte Carlo Methods
• Step (2) draws a value of $Y_1$ conditional on $Y_2\gt(1-Y_1)^2/2.$ Find the chance that this value is less than or equal to $z$ by integrating the joint density.
– whuber
Sep 2, 2022 at 17:03
• And the answer is columbia.edu/~ks20/4703-Sigman/4703-07-Notes-ARM.pdf (Note that this is not a great algorithm to generate Normals.) Sep 2, 2022 at 18:04
• @Xi'an Right. Partially compensating for the need for almost 3 uniforms per variate is the relative simplicity of the calculations. But I understand now what you were referring to. What appeals to me about this approach is its potential to generate variates from analytically less-tractable distributions.
– whuber
Sep 2, 2022 at 18:28
• Please edit your question to credit the original source of all material written by others (e.g., your screenshot): stats.stackexchange.com/help/referencing
– D.W.
Sep 3, 2022 at 19:18
• Note that this doesn't really require 2.64 uniform samples - the last one only requires sampling one random bit, since you're only interested in the truth value of the expression (U3 < 0.5). Purely theoretically, you can take a single uniform sample X to generate U_1 = 2X (modulo 1) and U_3 = X, and you will have that the variables U_1 and (U_3 < 0.5) are independent. In practice, computationally speaking you are probably generating floating-point numbers in the range [0, 1) which involves generating 64 random bits and only using 53 of them, so the extra bit is free. Sep 4, 2022 at 13:58
I was wondering how anyone would come up with this idea.
You observe, correctly, that the $$Y_i$$ have exponential distributions. They were easy to generate from a standard uniform number generator. The question could be put like this:
Find a simple way to exploit your ability to generate $$(Y_1,Y_2)$$ to draw values $$Z$$ following any continuous distribution with positive support. | {
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Such a distribution is one with a density function proportional to $$e^{g(z)}$$ for a function $$g$$ defined on the positive numbers.
The key terms "simple" and "proportional to" suggest trying a rejection sampling method. That leads to the algorithm in the question in the following generalized form:
Generate $$(Y_1,Y_2)$$ and keep $$Y_1$$ provided $$Y_2 \gt f(Y_1)$$ for some function $$f$$ to be determined.
Although it might feel more natural to reject when $$Y_2\le f(Y_1),$$ as we will see this equivalent formulation leads to a simple calculation.
The result of this sampling procedure evidently produces values of $$Y_1$$ conditional on the event $$Y_2 \gt f(Y_1).$$ To find its distribution function, apply the (elementary) definition of conditional probability to the event $$Y_1 \le z$$ for an arbitrary positive number $$z.$$ It states
$$\Pr(Y_1\le z \mid Y_2 \gt f(Y_1)) \ \propto\ \Pr(Y_1\le z\text{ and }Y_2 \gt f(Y_1)).$$
We needn't be concerned about the constant of proportionality because we can work it out at the very end, knowing the result has to evaluate to $$1$$ as $$z\to\infty$$ (by the axiom of Total Probability).
Because $$(Y_1,Y_2)$$ is independent, their joint density is exponential. Thus, assuming $$f(z) \ge 0$$ for all $$z\gt 0,$$
\begin{aligned} \int_0^z e^{g(y_1)}\,\mathrm{d}y_1 &= \Pr(Y_1\le z\text{ and }Y_2 \gt f(Y_1)) \\ &\propto \int_0^z e^{-y_1}\int_{f(y_1)}^\infty e^{-y_2}\,\mathrm{d}y_2\mathrm{d}y_1\\ &= \int_0^z e^{-y_1 - f(y_1)}\,\mathrm{d}y_1. \end{aligned}
Equality will hold for all $$z$$ provided the two integrands are equal. Solving for $$f$$ gives
$$f(z) = -z - g(z) + C$$
where the number $$C$$ accounts for the neglected proportionality constant $$e^C.$$
Consider the Half Normal distribution where $$g(z) = -z^2/2.$$ We find
$$f(z) = -z + z^2/2 + C = (1 - z)^2/2 + C - 1/2.$$ | {
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$$f(z) = -z + z^2/2 + C = (1 - z)^2/2 + C - 1/2.$$
We will need $$C \ge 1/2$$ to assure $$f(z)$$ is nonnegative. Larger values of $$C$$ work, too, but cause more rejections and are thereby less efficient.
Clearly, step (3) in the question converts any positive variable (like a Half Normal variable) into a variable symmetrically distributed around $$0.$$
Applications
For this method to succeed, we need $$f$$ to attain a minimum that is not too negative. This implies the target distribution must not be too heavy-tailed. One example is the generalized Gamma distribution with density proportional to $$\exp(-z^3/3)$$ on the positive numbers.
Here are histograms based on a million draws of $$(Y_1,Y_2)$$ for the Half Normal and Generalized Gamma problems. The red curves plot the target densities to demonstrate the correctness of this algorithm. The (empirical) acceptance rates show how efficient it is.
This R code produced these plots.
set.seed(17)
n <- 1e6
Y <- matrix(-log(runif(2*n)), ncol = 2) # Step (1): obtain iid exponential variates
#
# The function f. The constant can be any non-negative value, with 0 being the
# most efficient.
#
Dists <- list(Half Normal = function(z, C = 0) (1 - z)^2/2 + C,
Generalized Gamma = function(z, C = 0) -z + z^3/3 + 2/3 + C)
pars <- par(mfrow = c(1, length(Dists)))
for(D in names(Dists)) {
f <- Dists[[D]]
z <- Y[Y[, 2] > f(Y[, 1]), 1] # Step (2) of the rejection sampling
rate <- length(z) / nrow(Y)
hist(z, freq = FALSE, main = D,
sub = bquote(paste("Acceptance rate is ", .(signif(rate, 2)))))
g <- function(x) exp(-x - f(x)) | {
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A <- integrate(g, 0, Inf)$value # The constant of integration curve(g(x) / A, add = TRUE, col = "Red", lwd = 2) } par(pars) • (+1) Very nice reverse engineering. Historically von Neumann started by dominating the half-Normal density with a standard Exponential density and then achieved this condensed version. Sep 2, 2022 at 18:34 • @Xi'an I have realized -- and checked in code -- that one can overcome the limitation on tail behavior by means of a suitable transformation such as a logarithm or root. For instance, to generate a Gamma$(a)$variate use$f(z) = -(1+a)z + \exp(z) + C$after finding$C$through numerical optimization. For smallish$a$the acceptance rate isn't bad; e.g. it's 66% for$a=0.3$and 39% for$a=3.$This makes the approach much more general than I had thought. – whuber Sep 2, 2022 at 18:47 • @whuber Thank you! Sep 3, 2022 at 6:54 • A more natural way to come up with it: since going from Cartesian to polar gives$$\frac{1}{2\pi}e^{-(y_1^2+y_2^2)/2}dy_1dy_2=re^{-r^2/2}dr\frac{d\theta}{2\pi},$$it's now easy to get the desired IIDs viz.$u_1=e^{-y_1^2}$(we can drop the$1-$you'd get if you wanted an order-preserving transformation),$u_2=\theta/(2\pi)$. This speaks to a deeper truth: the only way for two Cartesian coordinates to be IIDs, with the polar coordinates also independent, is for the IIDs to be$N(0,\,\sigma^2)\$.
– J.G.
Sep 4, 2022 at 20:44
To draw a comparison between this Normal generator (that I will consider as von Neumann's) and the Box-Müller polar generator,
#Box-Müller
bm=function(N){
a=sqrt(-2*log(runif(N/2)))
b=2*pi*runif(N/2)
return(c(a*sin(b),a*cos(b)))
}
#vonNeumann
vn=function(N){
u=-log(runif(2.64*N))
v=-2*log(runif(2.64*N))>(u-1)^2
w=(runif(2.64*N)<.5)-2
return((w*u)[v])
}
here are the relative computing times
> system.time(bm(1e8))
utilisateur système écoulé
7.015 0.649 7.674
> system.time(vn(1e8))
utilisateur système écoulé
42.483 5.713 48.222 | {
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• FWIW, vn does not return a Normal variate, nor does it return anywhere near N values. When it's fixed up, the relative timings are still about the same as you report. But when a better implementation of vn is used--you are welcome to borrow the one in my answer--the timing ratio is 3:1 (21:6.9). With a little optimization it drops to the expected 2.6:1, reflecting the RNG costs. vn <- function(N) { N <- ceiling(1.32 * N); y1 <- log(runif(N, 0, exp(1))); y2 <- -2 * log(runif(N)); y1 <- y1[y2 > y1^2]; sample(c(-1,1), length(y1), replace = TRUE) * (y1 - 1) }
– whuber
Sep 3, 2022 at 16:37
• @whuber: thank you, issues fixed after a busy weekend... Sep 5, 2022 at 14:17
• Why is this vn func only returning the left-hand half of a normal dist? While one could replace the output with K <- c(K,-K) , that gives an artificial perfect mirror, not what's desired. Sep 15, 2022 at 11:51
• Oh, -- you forgot the part " U3<0.5, accept X=Y1. Else, X=−Y1 " ... at least I think so. Sep 15, 2022 at 12:16
• @Xi'an then shouldn't it be w = ( (runif(2.64*N)<.5) -0.5) * 2 ? Sep 15, 2022 at 18:02 | {
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# What is the expected number of randomly generated numbers in the range [a, b] required to reach a sum $\geq X$?
We are generating random numbers (integers) in the range $$[a, b]$$. All values are equally likely.
We will continue to generate random numbers in this range, and add up successive values until their combined sum is greater than or equal to a set number $$X$$.
What is the expected number of rolls to reach at least $$X$$?
Example:
a = 1000
b = 2000
X = 5000
Value 1: 1257 (total sum so far = 1257)
Value 2: 1889 (total sum so far = 3146)
Value 3: 1902 (total sum so far = 5048; all done)
So it took $$3$$ rolls to reach $$\geq5000$$. Intuitively, we can say that it will not take more than $$5$$ rolls if each roll is $$1000$$. We can also say that it will not take less than $$3$$ rolls if each roll was $$2000$$. So it stands to reason that in the example above, the expected number of rolls lies somewhere between $$3$$ and $$5$$.
How would this be solved in the general case for arbitrary values $$[a, b]$$ and $$X$$? It's been quite a while since I last took statistics, so I've forgotten how to work with discrete random variables and expected value.
• ${\cal E}(n) = {2 X \over a+b}$. – David G. Stork Jan 10 at 0:50
• Actually, one must always round up, so ${\cal E}(n) = \lceil \left({2 X \over a+b} \right) \rceil$ – David G. Stork Jan 10 at 1:44
• @DavidG.Stork: Even though the random variable $n$ is always integer valued, there's no reason why its expected value has to be... – Nate Eldredge Jan 10 at 2:18
• @DavidG.Stork Right; I claim both are incorrect. – Aaron Montgomery Jan 10 at 4:50
• @AaronMontgomery, they are indeed incorrect. – Eelvex Jan 10 at 6:24
Let $$T$$ be the number of times it takes to reach $$X$$. We compute $$E[T]$$ via $$E[T]=\sum_{t=0}^\infty P(T>t)$$. | {
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In order to have $$T>t$$, the sum of the first $$T$$ samples needs to be less than $$X$$. Let $$S_i$$ be the value of the $$i^{th}$$ sample. Then an experiment where $$T>t$$ has the first $$t$$ values satisfying $$S_1+S_2+\dots+S_t Letting $$E=X-1-(S_1+\dots+S_t)$$, we get $$S_1+\dots+S_t+E=X-1,\\ a\le S_i\le b,\\ E\ge 0$$ The number of integer solutions to the above system of equations and inequalities in the variables $$S_i$$ and $$E$$ can be computed via generating functions. The number of solutions is the coefficient of $$s^{X-1}$$ in $$(s^a+s^{a+1}+\dots+s^b)^t(1-s)^{-1}=s^{at}\cdot(1-s^{b-a+1})^t(1-s)^{-(t+1)}$$ After a bunch of work, this coefficient is equal to $$\sum_{k=0}^{\frac{X-1-ta}{b-a+1}} (-1)^k\binom{t}k\binom{t+X-1-ta-(b-a+1)k}{t}$$ Finally, we get \begin{align} E[T] &=\sum_{t=0}^{X/a} P(T>t) \\&=\boxed{\sum_{t=0}^{X/a} (b-a+1)^{-t}\sum_{k=0}^{\frac{X-1-ta}{b-a+1}}(-1)^k\binom{t}k\binom{t+X-1-ta-(b-a+1)k}{t}} \end{align} Note that when $$a=0$$, the summation in $$t$$ will go from $$t=0$$ to $$\infty$$, so it can only be computed approximately with a computer. However, this caveat can be avoided using the following observation; if $$T(a,b,X)$$ is the expected time to reach $$X$$ using sums of $$\operatorname{Unif}\{a,a+1,\dots,b\}$$ random variables, then $$E(T(0,b,X))=\frac{b+1}bE(T(1,b,X))$$.
When I write $$\sum_{k=0}^a f(k)$$ and $$a$$ is not an integer, what I mean is $$\sum_{k=0}^{\lfloor a \rfloor}f(k)$$. | {
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The expected number is dominated by the term based on the mean $$\frac{2X}{a+b}$$ - but we can do better than that. As $$X\to\infty$$, $$E(n) = \frac{2X}{a+b}+\frac{2(a^2+ab+b^2)}{3(a^2+2ab+b^2)}+o(1)$$ Why? Consider the excess. In the limit $$X\to\infty$$, the frequency of hitting a particular point approaches a uniform density of $$\frac{2}{b+a}$$. For points in $$[X,X+a)$$, this is certain to be the first time we've exceeded $$X$$, so the density function for that first time exceeding $$X$$ is approximately $$\frac{2}{b+a}$$ on that range. For $$t\in[X+a,X+b)$$, the probability that it's the first time exceeding $$X$$ is $$\frac{X+b-t}{b-a}$$; we need the last term added to be at least $$t-X$$. Multiply by $$\frac{2}{b+a}$$, and the density function for the first time exceeding $$X$$ is approximately $$\frac{2(X+b-t)}{b^2-a^2}$$ on $$[X+a,X+b)$$. For $$t or $$t\ge X+b$$, of course, the density is zero. Graphically, this density is a rectangular piece followed by a triangle sloping to zero. Now, the expected value of the first sum to be greater than $$X$$ is approximately $$\int_{X}^{X+a}\frac{2t}{b+a}\,dt+\int_{X+a}^{X+b}\frac{2t(X+b-t)}{b^2-a^2}\,dt$$ $$=\frac{2a(2X+a)}{2(b+a)}+\frac{2(b-a)(2X+a+b)(X+b)}{2(b^2-a^2)}-\frac{2(b-a)((X+a)^2+(X+a)(X+b)+(X+b)^2)}{3(b^2-a^2)}$$ $$=\frac{6aX+3a^2+6X^2+3(a+3b)X+3ab+3b^2-6X^2-6(a+b)X-2(a^2+ab+b^2)}{3(b+a)}$$ $$=\frac{3(a+b)X+a^2+ab+b^2}{3(b+a)}=X+\frac{a^2+ab+b^2}{3(a+b)}$$ Divide by the mean $$\frac{a+b}{2}$$ of each term in the sum, and we have an asymptotic expression for $$E(n)$$.
This is not a rigorous argument. If I wanted to do that, I'd bring out something like the moment-generating function.
All right, now that rigorous argument I alluded to. Credit to the late Kent Merryfield for laying out the argument cleanly, although I'm sure I contributed in the development stage. AoPS link | {
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Suppose we add independent copies of a reasonably nice nonnegative continuous random variable with density $$f$$, mean $$\mu$$, and variance $$\sigma^2$$ until we exceed some fixed $$y$$. As $$y\to\infty$$, the expected number $$g(y)$$ of these variables needed to exceed $$y$$ is $$g(y)=\boxed{\frac{y}{\mu}+\frac12+\frac{\sigma^2}{2\mu^2}+o(1)}$$ In the case of the uniform $$[a,b]$$ distribution of this problem, that is $$g(y)=\frac{2y}{a+b}+\frac12+\frac{(b-a)^2}{6(b+a)^2}+o(1)$$.
To prove this, condition on the first copy of the variable. If it has value $$x$$, the expected number of additional terms needed is $$g(y-x)$$. Add the one for the term we conditioned on, integrate against the density, and $$g(y)=1+\int_0^y f(x)g(y-x)\,dx$$ This is a convolution. To resolve it into something we can work with more easily, apply a transform method. Specifically, in this case with a convolution over $$[0,\infty]$$, the Laplace transform is most appropriate. If $$F(s)=\int_0^{\infty}e^{-sx}f(x)\,dx$$ and $$G(s)=\int_0^{\infty}e^{-sy}g(y)\,dy$$, then, transforming that integral equation for $$g$$, we get $$G(s)=\frac1s+F(s)G(s)$$ Solve for $$G$$, and $$G(s)=\frac1{s(1-F(s))}$$ Now, note that $$F$$ is essentially the moment-generating function for the variable with density $$f$$; there's a sign change, but it's otherwise the same. As such, it has the power series expansion $$F(s)=1-\mu s+\frac{\mu^2+\sigma^2}{2\mu^2}s^2 + O(s^3)$$ near zero. Then $$G(s)=\frac1{\mu s^2-\frac{\mu^2+\sigma^2}{2}s^3+O(s^4)} = \frac1{\mu}s^{-2}+\frac{\mu^2+\sigma^2}{2\mu^2}s^{-1}+Q(s)$$ where $$Q(s)$$ has a removable singularity at zero. Also, from the "reasonably nice" condition, $$Q$$ decays at $$\infty$$ How fast? Well, as long as the density of $$f$$ is bounded above by some exponential, its transform $$F$$ goes to zero as $$s\to\infty$$. Then $$G(s)=\frac1{s(1-F(s)}=O(s^{-1})$$ as $$s\to\infty$$; subtracting off the two other terms, $$Q(s)=O(s^{-1})$$ as $$s\to\infty$$. | {
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Now, we take the inverse Laplace transform of $$G$$. The inverse transform of $$s^{-2}$$ is $$y$$ and the inverse transform of $$s^{-1}$$ is $$1$$, so there's the $$\frac{y}{\mu}+\frac{\sigma^2+\mu^2}{2\sigma^2}$$ terms of the expansion of $$g$$. For the remainder $$Q$$, we bring out the integral formula $$q(y)=\frac1{2\pi}\int_{\infty}^{\infty}Q(it)e^{ity}\,dt$$. Again for nice enough $$f$$ - a square-integrable density will do - this decays to zero as $$y\to\infty$$ (Riemann-Lebesgue lemma).
And now, it's time to get more concrete. Our original example was a uniform distribution on $$[a,b]$$, which has mean $$\mu=\frac{a+b}{2}$$ and variance $$\sigma^2=\frac{(b-a)^2}{12}$$. That gives $$g(y)=\boxed{\frac{2y}{a+b}+\frac12+\frac{(b-a)^2}{6(a+b)^2}+o(1)}$$ Also, we can be more explicit; the transforms are $$F(s)=\frac1{(b-a)s}(e^{-as}-e^{-bs})$$ and $$G(s)=\frac{1}{s-\frac1{b-a}e^{-as}+\frac1{b-a}e^{-bs}}$$ This $$G$$ is analytic in the right half-plane except for the pole at zero, and is comparable to $$\frac1s$$ at $$\infty$$ in all directions in that half-plane. The remainder term $$Q$$ is thus smooth and decays like $$\frac1s$$ along the imaginary axis, so its inverse transform $$q$$ will have a jump discontinuity (at zero, where it goes from zero to $$1$$) and decay rapidly. | {
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If we want an explicit form for $$g$$ - it's a mess. Three different exponentials in that denominator pushes this far off the standard tables of transforms, and splitting it up as a geometric series introduces way too many terms. I'll try to write something down anyway: \begin{align*}G(s) &= \frac{1}{s-\frac1{b-a}e^{-as}+\frac1{b-a}e^{-bs}}\\ &= s^{-1} +\frac{s^{-2}}{b-a}\left(e^{-as}-e^{-bs}\right)+\frac{s^{-3}}{(b-a)^2}\left(e^{-as}-e^{-bs}\right)^2+\cdots\\ &= \sum_{n=0}^{\infty} \frac{s^{-n-1}}{(b-a)^n}\left(e^{-as}-e^{-bs}\right)^n\\ &= \sum_{n=0}^{\infty}\sum_{k=0}^n\frac{s^{-n-1}}{(b-a)^n}\binom{n}{k}(-1)^k e^{-s(kb+(n-k)a)}\end{align*} Now, in the table of Laplace transforms I'm using, there's an entry "delayed $$n$$th power with frequency shift"; the transform of $$(y-\tau)^n e^{-\alpha(y-\tau)}u(y-\tau)$$ (where $$u$$ is the unit step function) is $$n!e^{-\tau s}(s+\alpha)^{-n-1}$$. From this and the linearity of the transform, we get $$g(y) = \sum_{n=0}^{\infty}\sum_{k=0}^n[y-(kb+(n-k)a)]^n\frac{(-1)^k}{k!(n-k)!}u[t-(kb+(n-k)a)]$$ $$g(y) = \sum_{j=0}^{\infty}\sum_{k=0}^{\infty} (-1)^k\frac{[y-(ja+kb)]^{j+k}}{j!k!}u[t-(kb+(n-k)a)]$$ For any given $$y$$, this is a finite sum; we only sum those terms for which $$ja+kb \le y$$. It is horrible to use in practice, since it involves adding and subtracting large numbers to get something relatively small.
The $$a=0$$ case allows us to split things up differently, and comes out with something that's actually reasonably practical. Follow the link at the start of the post if you want to see it. | {
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Given $$m$$ discrete variables in the range $$[a,b]$$, the number of ways to get a sum $$s$$ out of them corresponds to \eqalign{ & N_b (s - ma,d,m) = \cr & = {\rm No}{\rm .}\,{\rm of}\,{\rm solutions}\,{\rm to}\;\left\{ \matrix{ a \le {\rm integer}\;y_{\,j} \le b \hfill \cr y_{\,1} + y_{\,2} + \; \cdots \; + y_{\,m} = s \hfill \cr} \right.\quad = \cr & = {\rm No}{\rm .}\,{\rm of}\,{\rm solutions}\,{\rm to}\;\left\{ \matrix{ {\rm 0} \le {\rm integer}\;x_{\,j} \le b - a = d \hfill \cr x_{\,1} + x_{\,2} + \; \cdots \; + x_{\,m} = s - ma \hfill \cr} \right. \cr} where $$N_b$$ is given by $$\bbox[lightyellow] { N_b (s-ma,d,m)\quad \left| {\;0 \leqslant \text{integers }s,m,d} \right.\quad = \sum\limits_{\left( {0\, \leqslant } \right)\,\,k\,\,\left( { \leqslant \,\frac{s-ma}{d+1}\, \leqslant \,m} \right)} {\left( { - 1} \right)^k \binom{m}{k} \binom { s -ma+ m - 1 - k\left( {d + 1} \right) } { s-ma - k\left( {d + 1} \right)}\ } } \tag{1}$$ as widely explained in this post.
Let's also notice the symmetry property of $$N_b$$ $$N_b (s - ma,d,m) = N_b (md - \left( {s - ma} \right),d,m) = N_b (mb - s,d,m)$$
The probability of obtaining exactly the sum $$s$$ in $$m$$ rolls is therefore \bbox[lightyellow] { \eqalign{ & p(s\;;\,m,a,b) = {{N_b (s - ma,d,m)} \over {\left( {d + 1} \right)^{\,m} }} = {{N_b (mb - s,d,m)} \over {\left( {d + 1} \right)^{\,m} }} = \cr & = {1 \over {\left( {d + 1} \right)^{\,m} }} \sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( { \le \,{{s - ma} \over {d + 1}}\, \le \,m} \right)} { \left( { - 1} \right)^k \binom{m}{k} \binom{ s -ma+ m - 1 - k\left( {d + 1} \right) }{ s-ma - k\left( {d + 1} \right)} } = \cr & = {1 \over {\left( {d + 1} \right)^{\,m} }} \sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( { \le \,{{mb - s} \over {d + 1}}\, \le \,m} \right)} { \left( { - 1} \right)^k \binom{m}{k} \binom{ mb-s+ m - 1 - k\left( {d + 1} \right) }{ mb-s - k\left( {d + 1} \right)} } \cr} } \tag{2} and the sum of $$p$$ over $$s$$ is in fact one. | {
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The probability $$p$$ quickly converges, by Central Limit Theorem, to the Gaussian in the variable $$s$$ $$\cal N\left( {\mu ,\sigma ^{\,2} } \right)\quad \left| \matrix{ \;\mu = m\left( {{d \over 2} + a} \right) = m\left( {{{a + b} \over 2}} \right) \hfill \cr \;\sigma ^{\,2} = m{{\left( {d + 1} \right)^{\,2} - 1} \over {12}} \hfill \cr} \right.$$ see for instance this post.
It is easy to demonstrate (by the "double convoluton" of binomials) that the Cumulative version, i.e. the probability to obtain a sum $$\le S$$, is \bbox[lightyellow] { \eqalign{ & P(S\;;\,m,a,b) = {1 \over {\left( {d + 1} \right)^{\,m} }}\sum\limits_{0\, \le \,\,s\,\, \le \,S} {N_b (s - ma,d,m)} = \cr & = {1 \over {\left( {d + 1} \right)^{\,m} }}\sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( { \le \,{{s - ma} \over {d + 1}}\, \le \,m} \right)} { \left( { - 1} \right)^k \binom{m}{k} \binom{S - ma + m - k\left( {d + 1} \right) }{S - ma - k\left( {d + 1} \right) } } \cr & \approx {1 \over 2}\left( {1 + {\rm erf}\left( {\sqrt 6 {{S + 1/2 - m\left( {a + b} \right)/2} \over {\left( {b - a + 1} \right)\sqrt m \;}}} \right)} \right) \cr} } \tag{3}
The following plot allows to appreciate the good approximation provided by the asymptotics even at relatively small values of the parameters involved.
That premised,
the probability that we reach or exceed a predefined sum $$X$$ at the $$m$$-th roll and not earlier
is given by the probability that
we get $$S as the sum of the first $$m-1$$ rolls, and then that the $$m$$-th one has a value $$s$$ such that $$X \le S+s$$,
or, what is demonstrably the same, by
the probability of getting $$X \le S$$ in $$m$$ rolls minus the probability of getting the same in $$m-1$$ rolls. | {
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That is, in conclusion \bbox[lightyellow] { \eqalign{ & Q(m\;;\,X,a,b)\quad \left| \matrix{ \;1 \le m \hfill \cr \;0 \le X \hfill \cr \;0 \le a \le b \hfill \cr} \right.\quad = \cr & = 1 - P(X - 1\;;\,m,a,b) - 1 + P(X - 1\;;\,m - 1,a,b) = \cr & = P(X - 1\;;\,m - 1,a,b) - P(X - 1\;;\,m,a,b) \cr} } \tag{4} and the sum of $$Q$$ over $$m$$ correctly checks to one.
Example
a) For small values of $$X,a,b$$ the exact expression for $$Q$$ in (4) gives the results shown (e.g. $$a=1, \; b=4$$) which look to be correct and the row sums correctly check to be $$1$$.
b) For large values of the parameters, like in your example $$a=1000,\; b=2000, \; X=5000$$ we shall use the asymptotic version of $$Q$$ which gives the results plotted below | {
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# 9-Digit Combination Lock Chances And Probability
• June 25th 2011, 11:46 AM
FatalSylence
9-Digit Combination Lock Chances And Probability
I have been thinking on combination locks recently, and would like some help with this particular problem.
Problem:
I have a lock with digits 1-9, and the length of the combination is 4 digits. You cannot press the digits 1-9 more than once, and the combination does not have to be typed in any particular order.
Questions:
1. Is it true, then, that the possible amount of combinations is 3,024, because 9*8*7*6=3,024?
2. Is it also true that 24 of the combinations are correct, since you do not have to type the combination in any particular order? Therefore 1*2*3*4=24.
3. Is it true that the chances of guessing it right are 1 in 125 since 3,000/24=125?
1. If you would change the lock to have a 5 digit combination, would the chance to guess it right be the same as if it were a 4 digit combination? Since the possible combination amount is 15,120 (9*8*7*6*5), the amount of right combinations is 120 (1*2*3*4*5), therefore 15,000/120=125. The same amount of chance as if it were a 4-digit combination?
2. If all this is true, then wouldn't a 6-digit combination actually be less secure? I.e., you have a greater chance of guessing the right combination? (1 in 84, if I am doing it right.)
Help is greatly appreciated. I have not done probability in a long time and am wondering if I am even doing it right.
Thanks, guys.
Fatal Sylence
• June 25th 2011, 12:01 PM
Plato
Re: 9-Digit Combination Lock Chances And Probability
Quote:
Originally Posted by FatalSylence
[/B]I have a lock with digits 1-9, and the length of the combination is 4 digits. You cannot press the digits 1-9 more than once, and the combination does not have to be typed in any particular order.
Questions: | {
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Questions:
1. Is it true, then, that the possible amount of combinations is 3,024, because 9*8*7*6=3,024?
2. Is it also true that 24 of the combinations are correct, since you do not have to type the combination in any particular order? Therefore 1*2*3*4=24.
3. Is it true that the chances of guessing it right are 1 in 125 since 3,000/24=125?
If order makes a difference then the answer is $9\cdot 8 \cdot 7\cdot 6=3024$.
If order makes no difference then $\binom{9}{4}=\frac{9!}{4!\cdot 5!}=126$
• June 25th 2011, 12:07 PM
FatalSylence
Re: 9-Digit Combination Lock Chances And Probability
Quote:
Originally Posted by Plato
If order makes a difference then the answer is $9\cdot 8 \cdot 7\cdot 6=3024$.
If order makes no difference then $\binom{9}{4}=\frac{9!}{4!\cdot 5!}=126$
I'm not sure I understand your answer. 126 as in a 1 in 126 chance of guessing it right?
• June 25th 2011, 12:12 PM
Plato
Re: 9-Digit Combination Lock Chances And Probability
Quote:
Originally Posted by FatalSylence
I'm not sure I understand your answer. 126 as in a 1 in 126 chance of guessing it right?
That is the combination of 9 things taken 4 at a time.
By the way that is $\frac{3024}{24}$.
• June 25th 2011, 12:24 PM
FatalSylence
Re: 9-Digit Combination Lock Chances And Probability
Quote:
Originally Posted by Plato
That is the combination of 9 things taken 4 at a time.
By the way that is $\frac{3024}{24}$.
Hmm, okay. Can you direct me towards a resource that explains the "n things taken r at a time"?
Also, were my other questions correct?
Finally, would a 5 digit combination be more secure? A 4 digit combination has the same chance to guess right as a a 5-digit combination, but don't you have a greater chance to guess it wrongwith 5-digits?
• June 25th 2011, 12:28 PM
Plato
Re: 9-Digit Combination Lock Chances And Probability
Quote:
Originally Posted by FatalSylence
Hmm, okay. Can you direct me towards a resource that explains the "n things taken r at a time"? | {
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Also, were my other questions correct?
Finally, would a 5 digit combination be more secure? A 4 digit combination has the same chance to guess right as a a 5-digit combination, but don't you have a greater chance to guess it wrongwith 5-digits?
Also look at this page. You change those numbers and hit $\boxed{=}$.
• June 25th 2011, 12:37 PM
FatalSylence
Re: 9-Digit Combination Lock Chances And Probability
Quote:
Originally Posted by Plato
Also look at this page. You change those numbers and hit $\boxed{=}$.
Can you explain to me if in a 4-digit combination, the amount of correct combinations is 126, or 24 like I originally thought? I'm just trying to understand this here.
• June 25th 2011, 12:44 PM
Plato
Re: 9-Digit Combination Lock Chances And Probability
Quote:
Originally Posted by FatalSylence
Can you explain to me if in a 4-digit combination, the amount of correct combinations are 126, or 24 like I originally thought? I'm just trying to understand this here.
I think that we are talking about two different ideas here.
There are 24 ways to arrange the string $9,4,2,6$.
But there are 126 ways to select a set of four from $\{1,2,3,4,5,6,7,8,9\}$.
One of those 126 sets is $\{2,4,6,9\}$.
Then the numbers in that set can be arranged in 24 ways.
• June 25th 2011, 12:56 PM
FatalSylence
Re: 9-Digit Combination Lock Chances And Probability
Quote:
Originally Posted by Plato
I think that we are talking about two different ideas here.
There are 24 ways to arrange the string $9,4,2,6$.
But there are 126 ways to select a set of four from $\{1,2,3,4,5,6,7,8,9\}$.
One of those 126 sets is $\{2,4,6,9\}$.
Then the numbers in that set can be arranged in 24 ways.
OH!!! Gosh! I get it! It does not matter how many ways you can arrange a set of four since order does not matter. I understand. One of these 126 sets encompasses all the ways you can arrange it since order does not matter. | {
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So, if the person was good at math, they would realize that since order does not matter, there are only 126 possible combinations. Wow, I get it. Thank you.
EDIT:
Is there a formula or calculator that can display all the possible sets for a 1-9, 4 digit combination? Or any other numbers for that matter?
• June 25th 2011, 03:39 PM
Plato
Re: 9-Digit Combination Lock Chances And Probability
Quote:
Originally Posted by FatalSylence
EDIT:[/B] Is there a formula or calculator that can display all the possible sets for a 1-9, 4 digit combination? Or any other numbers for that matter?
The good news is that this is a simple programming problem.
But the bad news I doubt that you can easily find it on the web.
I suggest that you discuss this with a friend who does programming. | {
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# A compact locally connected metric space is “uniformly locally connected”
A compact locally connected metric space is "uniformly locally connected"\ That is, for any $\epsilon > 0$, there is some $\delta > 0$ such that whenever $\rho(x, y) < \delta$, then $x$ and $y$ both lie in some connected subset of $X$ of diameter $<\epsilon$.
proof:-
Since $X$ is locally connected metric space then each $x\in X$ has a nhood base of open connected sets\ Given $\epsilon > 0$, let $x \in X$ and $U_x=\rho(x, \epsilon)$ be a nhood of $x$\ There exist an open connected basic nhood $V_x$ with diameter $<\epsilon$, Now $$X=\bigcup_{x\in X }{V_x}$$, hence cover $X$ by open connected nhoods of diameter $<\epsilon$.\ Since $X$ is compact, reduce this to a finite subcover $\{V_{x1},. . . , V_{xn}\}$ and let $\delta$ be a Lebesgue number (22.5) for this cover.\ Then if $\rho(x, y) < \delta$, both $x$ and $у$ belong to some $V_{xi}$. \
{Theorem 22.5} (Lebesgue covering lemma). If $\{U_1..., U_n\}$ is a finite open cover of a compact metric space X, there is some $\delta > 0$ such that if A is any subset of $X$ of diameter $< \delta$, then $A \subset U_i$ for some i.
I try to write the proof better than this.
I would like to confirm this proof
If acceptable, I would like to clarify and improve it (Language and Mathematical)as much as possible
• Do not use \ or , or .\ at the end of a sentence. Just use . $\quad$ And if a sentence ends in a displayed line, include the . in that line. Write in the same style as for an an essay about something else. What you have is not so bad. I've seen far worse. And your proof is correct . – DanielWainfleet Dec 12 '17 at 9:25
The proof is correct. | {
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The proof is correct.
As for writing it up. That depends on your audience. For a general audience, the first rule is - avoid abbreviations ("neighborhood", not "nhood"). And I assume that the "/" scattered through the text was some sort of misguided attempt to show where you put new lines? Just let Latex have it's way here. If a new line is needed, let it form a paragraph break.
Also "and $U_x=\rho(x,\epsilon)$ be a nhood of $x$" doesn't make sense. $\rho$ is a function that takes points in your space in both arguments, and produces real numbers. You have it with a real number as the second argument, and producing a set. Now it is evident that you mean $U_x$ to be a ball of radius $\epsilon$ about $x$. But if someone has told you this is a good way to denote the ball, do NOT trust them about notations in the future! Commonly used ways of denoting balls include $B(x, \epsilon)$ and $B_x(\epsilon)$. If the metric to be used is not obvious, then usually $B_\rho(x, \epsilon)$ is preferred. (Of course, it you had intended $B(x, \epsilon)$ and only accidently used $\rho$, just be careful in your final write-up.)
But you don't do anything with "$U_x$", which makes introducing a notation for it pointless. You don't actually need to name the balls, since you don't plan on using that name.
There are also some phrasing differences I would suggest. Your phrasing is clear enough, but sounds awkward sometimes to a native English speaker (at least - to this native English speaker). There is nothing particularly bad, though.
I would write it up like this: | {
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I would write it up like this:
Since $X$ is locally connected, for $\epsilon > 0$, each point $x \in X$ has a open connected neighborhood $V_x$ contained within the ball of radius $\epsilon$ about $x$. The collection $\{V_x \mid x \in X\}$ forms an open cover of $X$ by sets of diameter $< \epsilon$. Since $X$ is compact, it has a finite subcover $\{V_1, V_2, ..., V_m\}$. By the Lesbegue covering lemma (Theorem 22.5), there is a $\delta > 0$ such that if $\rho(x, y) < \delta$, then there is some $V_i$ with $\{x, y\} \subset V_i$, which completes the proof. | {
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My Math Forum Progression .
Algebra Pre-Algebra and Basic Algebra Math Forum
April 15th, 2014, 02:49 AM #1 Senior Member Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 Progression . For the same job specifications, two companies offer a different salary scale: Company A :Starting monthly pay = RM900.00 Monthly increment = RM50.00 Company B:Starting monthly pay = RM750.00 Monthly increment = RM60.00 (a)In January 2014, Ali starts to work in Company A and Ahmad in Company B. When will the monthly salary of Ali and Ahmad be the same? Use three methods. Include the use of ICT. (b)Which salary scale is the better deal? Justify. How to solve this problem ? Especially problem b.
April 15th, 2014, 10:20 PM #2
Math Team
Joined: Dec 2006
From: Lexington, MA
Posts: 3,267
Thanks: 408
Hello, jiasyuen!
Quote:
For the same job, two companies offer a different salary scale: Company A: Starting monthly pay = \$900.00 Monthly increment = \$50.00 Company B: Starting monthly pay = \$750.00 Monthly increment = \$60.00 (a) In January 2014, Ali starts to work in Company A and Ahmad in Company B. When will the monthly salary of Ali and Ahmad be the same? Use three methods. Include the use of ICT.
You need to construct algebraic expressions for Ali's and Ahmad's monthly pay.
Ali starts at \$900 per month and gets a \$50 raise each month.
When he has worked for $\displaystyle m$ months,
$\displaystyle \quad$his monthly pay will be: $\displaystyle \:900 + 50m$ dollars.
Ahmad starts at \$750 per month and gets a \$60 raise each month.
When he has worked for $\displaystyle m$ months,
$\displaystyle \quad$his monthly pay will be $\displaystyle \,750 +60m$ dollars.
When are these equal?
$\displaystyle \quad 750 + 60m \:=\:900 + 50m \quad\Rightarrow\quad 10m \:=\:150 \quad\Rightarrow\quad m \,=\,15$
Their salaries will be equal in 15 months (March 2015).
Quote:
(b)Which salary scale is the better deal? Justify. | {
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Quote:
(b)Which salary scale is the better deal? Justify.
If Ali considers this a temporary job (less than 15 months),
$\displaystyle \quad$ he has the better deal.
If both are planning to work more than 15 months,
$\displaystyle \quad$ then Ahmad has the better deal.$\displaystyle \;\;{\color{red}{**}}$
$\displaystyle \begin{array}{c|ccccccccc} \hline \text{Month} & 0 & 1 & 2 & 3 & \cdots & 15 & 16 & 17 & \cdots \\ \hline \text{Ali} & 900 & 950 & 1000 & 1050 & \cdots & 1650 & 1700 & 1750 & \cdots \\ \hline \text{Ahmad} & 750 & 810 & 870 & 930 & \cdots & 1650 & 1710 & 1770 & \cdots \\ \hline \end{array}$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
$\displaystyle {\color{red}{**}}$
I gave this problem some extra thought . . .
In 15 months their monthly pay is the same.
But Ali has made \$18,750 while Ahmad has made only \$17,550.
It will take 31 months for Ahmad's total earnings to equal that of Ali.
After that, Ahmad's total earnings will exceed Ali's.
April 16th, 2014, 03:43 AM #3 Senior Member Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 The question stated there have to use three methods to find. Including of ICT. What it means ? How to do ?
April 16th, 2014, 09:08 AM #4
Math Team
Joined: Oct 2011
Posts: 14,597
Thanks: 1038
Quote:
Originally Posted by jiasyuen The question stated there have to use three methods to find. Including of ICT. What it means ? How to do ?
Well, Soroban gave you one; seems fair that YOU find two more;
it is YOUR homework, not ours. Do you agree?
April 16th, 2014, 09:12 AM #5 Senior Member Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 I don't know what's another two methods ? Can you give me some guidance ?
April 23rd, 2014, 04:36 AM #6
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Joined: Sep 2013
From: Earth
Posts: 827
Thanks: 36
Quote:
Originally Posted by soroban Hello, jiasyuen! You need to construct algebraic expressions for Ali's and Ahmad's monthly pay. Ali starts at \$900 per month and gets a \$50 raise each month. When he has worked for $\displaystyle m$ months, $\displaystyle \quad$his monthly pay will be: $\displaystyle \:900 + 50m$ dollars. Ahmad starts at \$750 per month and gets a \$60 raise each month. When he has worked for $\displaystyle m$ months, $\displaystyle \quad$his monthly pay will be $\displaystyle \,750 +60m$ dollars. When are these equal? $\displaystyle \quad 750 + 60m \:=\:900 + 50m \quad\Rightarrow\quad 10m \:=\:150 \quad\Rightarrow\quad m \,=\,15$ Their salaries will be equal in 15 months (March 2015). If Ali considers this a temporary job (less than 15 months), $\displaystyle \quad$ he has the better deal. If both are planning to work more than 15 months, $\displaystyle \quad$ then Ahmad has the better deal.$\displaystyle \;\;{\color{red}{**}}$ $\displaystyle \begin{array}{c|ccccccccc} \hline \text{Month} & 0 & 1 & 2 & 3 & \cdots & 15 & 16 & 17 & \cdots \\ \hline \text{Ali} & 900 & 950 & 1000 & 1050 & \cdots & 1650 & 1700 & 1750 & \cdots \\ \hline \text{Ahmad} & 750 & 810 & 870 & 930 & \cdots & 1650 & 1710 & 1770 & \cdots \\ \hline \end{array}$ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ $\displaystyle {\color{red}{**}}$ I gave this problem some extra thought . . . In 15 months their monthly pay is the same. But Ali has made \$18,750 while Ahmad has made only \$17,550. It will take 31 months for Ahmad's total earnings to equal that of Ali. After that, Ahmad's total earnings will exceed Ali's.
Not 16 month?
Ali in Company A:
Tn =a+(n-1)d
=900+(n-1)(50)
=900+50n-50
=850+50n
Tn = a+(n-1)d
=750+(n-1)(60)
=750+60n-60
=690+60n
When the monthly salary of Ali and Ahmad the same,
850+50n=690+60n
160n=10n
n=16 (i.e. in the 16th month)
Tags progression
, | {
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Tags progression
,
### in january 2014 ali starts to work in company A and ahmad company b when will the monthly salary of ali and ahmad be the same
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# Math Help - [SOLVED] integration with sinxcosx
1. ## [SOLVED] integration with sinxcosx
Hey guys i kind of have a weird little problem which i can't figure out.
I'm integrating sinxcosx, and depending on what i use as my substution you get two answers.
1) sin^2x/2 + C and 2) -cos^2x/2 + C
Now if we set these two answers equal to eachother we get
sin^2x + cos^2x = 0 but we know there exists a trig identity that states sin^2x + cos^2x = 1
So i'm confused here. How can this be possible? I know that 0 cannot equal 1. Or at least i hope not hehe. Any help would be great cheers
2. Originally Posted by block
Hey guys i kind of have a weird little problem which i can't figure out.
I'm integrating sinxcosx, and depending on what i use as my substution you get two answers.
1) sin^2x/2 + C and 2) -cos^2x/2 + C
Now if we set these two answers equal to eachother we get
sin^2x + cos^2x = 0 but we know there exists a trig identity that states sin^2x + cos^2x = 1
So i'm confused here. How can this be possible? I know that 0 cannot equal 1. Or at least i hope not hehe. Any help would be great cheers
Two solutions of an integral may be of different form if they are off by a constant. You have two solutions, but who says the constant of integration is the same? What this is telling you is that $C_{cos} = C_{sin} + 1$
-Dan
3. Originally Posted by block
Hey guys i kind of have a weird little problem which i can't figure out.
I'm integrating sinxcosx, and depending on what i use as my substution you get two answers.
You have,
$\int \sin x \cos x dx$
And I think the problem you have you think you get two distinct answers when you use $u=\sin x$ and when you switch to $u=\cos x$.
1) $u=\sin x$ then $u'=\cos x$
Thus,
$\int \sin x \cos x dx= \int u du=\frac{1}{2}u^2 +C=\frac{1}{2}\sin^2 x+C$
2) $u=\cos x$ then $u'=-\sin x$
Thus,
$\int \sin x \cos x dx= -\int u du=\frac{1}{2}u^2 +C=-\frac{1}{2}\cos^2 x+C$ | {
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But in reality these are the same (the set of all anti-derivates) because,
$\frac{1}{2}\sin^2 x+C=\frac{1}{2}(1-\cos^2 x)+C=-\frac{1}{2}\cos^2 x+\left( \frac{1}{2}+C\right)=-\frac{1}{2}\cos^2 x+C_1$
I just renamed the constant function,
$\frac{1}{2} +C=C_1$
Thus, you get the same thing.
And the last thing you could have done is,
$\frac{1}{2}\int 2\sin x \cos x dx$
I just multiplied and divide by 2, unchanged expression.
From trignometry you know that,
$\frac{1}{2}\int \sin 2x dx=-\frac{1}{4}\cos 2x+C$
But again though it looks different the space of solutions is still the same. You can use some trignometry to confirm this.
4. Hello, block!
This is a classic puzzler.
I ran across it across it in Calculus I.
There are three ways (at least) to integrate it.
We have: . $\int\sin x\cos x\,dx$
[1] If we look at it as: . $\int \underbrace{\sin x}_{u}\underbrace{(\cos x\,dx)}_{du}$
. . . we get: . $\boxed{\frac{1}{2}\sin^2\!x + C_1}$
[2] If we write it as: . $\int \cos x(\sin x\,dx)$
. . . we let: $u = \cos x\quad\Rightarrow\quad du = -\sin x\,dx\quad\Rightarrow\quad dx = -\frac{du}{\sin x}$
Substitute: . $-\int u\,du \;=\;-\frac{1}{2}u^2 + C\;=\;\boxed{-\frac{1}{2}\cos^2\!x + C_2}$
Since these two answers must be equal, we have:
. . $\frac{1}{2}\sin^2\!x + C_1 \;=\;-\frac{1}{2}\cos^2\!x + C_2\quad\Rightarrow\quad \frac{1}{2}\sin^2\!x + \frac{1}{2}\cos^2\!x\;=\;C_2-C_1$
Multiply by 2: . $\sin^2\!x + \cos^2\!x \;=\;2(C_2-C_1)$
The time the constants $(C_1,\:C_2)$ aren't completely arbitrary.
. . They must satisfy: . $2(C_2-C_1) \:=\:1$
[3] We can use the identity: . $2\sin\theta\cos\theta \,=\,\sin2\theta$
So: . $\int\sin x\cos x\,dx\;=\;\frac{1}{2}\int2\sin x\cos x\,dx \;=\;\frac{1}{2}\int\sin2x\,dx$ $=\;\boxed{-\frac{1}{4}\cos2x + C_3}$
. . I'll let you justify this one . . .
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
A similar problem: . $\int\sec^2\!x\tan x\,dx$ | {
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A similar problem: . $\int\sec^2\!x\tan x\,dx$
$[1]\;\;\int \underbrace{\sec x}_{u}\underbrace{(\sec x\tan x\,dx)}_{du} \;=\;\frac{1}{2}\sec^2\!2x + C$
$[2]\;\;\int\underbrace{\tan x}_{u}\underbrace{(\sec^2x\,dx)}_{du}\;=\;\frac{1} {2}\tan^2x + C$ | {
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## anonymous 2 years ago how would I solve for x if 2x-3/x+1<1
1. kc_kennylau
$\frac{2x-3}{x+1}<1$
2. kc_kennylau
I would split it into two cases
3. anonymous
how would you do that
4. myininaya
I would first write as f(x)<0 Then find where f(x) is zero and undefined Then draw a number line with the numbers that make f zero or undefined Then test the intervals around those numbers to see if f is less than 0 or not
5. kc_kennylau
Since we are to multiply (x-1) on both sides, I would split it into x-1<0 and x-1>0
6. myininaya
you mean x+1?
7. kc_kennylau
Yes sorry
8. anonymous
SO i take x+ 1 and get $2x-3<x+1 and 2x-3>x+1$
9. kc_kennylau
Yes
10. kc_kennylau
Well actually I believe we should do it case by case
11. kc_kennylau
When x+1<0: 2x-3>x+1 ... When x+1>0: 2x-3<x+1 ...
12. anonymous
so then my next step would to add in the 3 on both sides corect? so it would turn into $2x>x+4$ and $2x<x+4$
13. kc_kennylau
Yep
14. kc_kennylau
And for presentation purpose I would write it as 2x>x+4 where x+1<0 and 2x<x+4 where x+1>0
15. anonymous
then do I divide by 2 on both of the equations or divide by 2x?
16. kc_kennylau
You subtract x from both sides
17. anonymous
so It would be $x-2x>4 where x+1<0$ and just the oposit sign on the otherone?
18. kc_kennylau
No, it should be 2x-x>4 where x+1<0
19. anonymous
ok now im starting to understand so then the next step would be to hmove the 4 over
20. kc_kennylau
Nah, the next step is subtracting it
21. anonymous
ok so it would then be 2x-x-4> 0 and where x+1<0 and 2x-x-4< 0 and where x+1>0 and t
22. kc_kennylau
I mean you can write 2x-x as x
23. kc_kennylau
When x+1<0: 2x-3>x+1 2x>x+4 2x-x>4 x>4 When x+1>0: 2x-3<x+1 2x<x+4 2x-x<4 x<4
24. kc_kennylau
However, when x+1<0, x can't be >4, and the same goes to the second case
25. kc_kennylau
Therefore there is no solution
26. kc_kennylau
Wait
27. aum
The inequality is valid for x in the interval (-1, 4).
28. kc_kennylau | {
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Wait
27. aum
The inequality is valid for x in the interval (-1, 4).
28. kc_kennylau
@aum tell me what is wrong in my solution, I have no time to double-check
29. kc_kennylau
bye
30. anonymous
@aum I thought so. I see how to get the 4 but not the -1 could you help me solve this?
31. aum
My method is what @myininaya suggested. (2x - 3) / (x + 1) < 1 (2x - 3) / (x + 1) - 1 < 0 (2x-3 - x - 1) / (x + 1) < 0 (x - 4) / (x + 1) < 0 The x values that are of interest to us here are the ones that will make f(x) = (x-4)/(x+1) zero or undefined. f(x) = 0 when x = 4 f(x) is undefined when x = -1. So the number line is split into three intervals: (-infinity, -1); (-1, 4), (4, infinity). Pick a convenient number in each interval and see if the inequality is valid. f(x) = (x - 4) / (x + 1) < 0 when x = -2, f(x) = (-6) / (-1) = 6 which is greater than 0. So not a solution. when x = 0, f(x) = -4/1 = -4 which is less than 0. This is a solution. when x = 5, f(x) = 1 / 6 which is greater than 0. Not a solution. So the solution is x in the interval (-1, 4).
32. anonymous
thank you so much
33. aum
you are welcome.
34. aum
To follow @kc_kennylau 's method: (2x - 3) / (x + 1) < 1 Case 1: x + 1 > 0 Since (x+1) is positive, we can multiply both sides by (x+1) and the "less than" inequality will remain intact. 2x - 3 < x + 1 x < 4 Case 2: x + 1 < 0 Since (x+1) is negative, multiplying both sides by (x+1) would change the "less than" inequality to a "greater than" inequality. 2x - 3 > x + 1 x > 4. But for the second case we assumed x + 1 < 0 which implies x < -1 But we get the solution as x > 4. x cannot be less than -1 and also be greater than 4 simultaneously. Thus, case 2 yields NO solution. Therefore, there is no solution in the domain x < -1. Combining the two cases we get x should be less than 4 but greater than -1.
35. kc_kennylau
I incorrectly rejected case 1
36. aum
yes. | {
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# Expected no. of coin flips to win
Two players A and B play a game in which they alternately flips a coin. Player A starts the game. If a player gets T and another player got H before, he/she is the winner. What is the expected no. of flips for A to win ?
Ex:-
Game-1: HT --> B wins
Game-2: HHT --> A wins
• You must share your thoughts on the problem. – StubbornAtom Aug 26 '17 at 16:34
• What I can think is to find probability for A to win in n flips and then use the formula of expectation to find the expected no. of flips. But this method is quite lengthy. Is there any better solution? – Sumit Kumar Aug 26 '17 at 16:38
• So the person who wins is the first person to get tails after someone has gotten heads, is that right? – Arthur Aug 26 '17 at 16:41
• If B wins, presumably A never wins – Henry Aug 26 '17 at 16:43
• @Arthur A player who get tails wins if and only if another player has gotten heads in the previous flip i.e. a sequence HT occurs – Sumit Kumar Aug 26 '17 at 16:43
We expect four coin flips. That's because we first expect two flips to get the first heads, and then we expect two flips to get the first tails.
More detailed: as a player is about to throw, the game is in one of two states: either there hasn't been any heads yet, or there has been at least one head (and since the game isn't over, the last throw must've been a head). Let's call the expected number of throws left until the game is over if we're in the first state $E_1$, and the expected number of throws left if we're in the second state $E_2$. | {
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First we calculate $E_2$. If a player is about to throw, and the last throw was a heads, then there's a probability of $0.5$ of the game ending in one more throw, and a probability of $0.5$ of the game continuing to the next player still in the same state, which means that after that we expect another $E_2$ throws. This implies $$E_2=0.5\cdot1+0.5(1+E_2)\\ E_2=2$$ Now we get to $E_1$. If a player is about to throw, and there hasn't been any heads yet, then there is a probability of $0.5$ that the player throws heads, and the game continues in state two, which means that we expect another $E_2=2$ throws, and there is a probability of $0.5$ of the player throwing tails, which means that the game continues in the first state, and we expect another $E_1$ throws. This gives $$E_1=0.5(1+E_2)+0.5(1+E_1)\\ E_1=4$$ which is our answer.
Given that $A$ wins, we have an odd number of flips.
Last flip is $T$ and others consist of a possible run of tails, followed by a definite run of heads.
Let $N$ be the number of flips total.
$$P(N=n)=\frac{1}{2}\sum_{k=1}^{n-1}\left(\frac{1}{2}\right)^{n-1}=\frac{n-1}{2^n}$$ $$P(A)=P(N\equiv 1\mod 2)=\sum_{k=1}^\infty P(N=2k+1)=\sum_{k=0}^\infty\frac{k}{4^k}=\frac{4}{9}$$.
$$E(N|A)=\frac{9}{4}\sum_{k=1}^\infty\frac{k(2k+1)}{4^k}=\frac{13}{3}$$
The $E(N)=4$ result above is wrong since it is the total expectation $E(N)$ of coin flips until endgame. The total expectation $E(N)$ is, however, an interesting quantity as well, since it is equivalent to a special case of the following theorem involving a typing monkey:
Let a monkey type letters on a typewriter that has some alphabet $\Sigma$ and let $w\in\Sigma^*$. Also let $\{w_1,\cdots,w_n\}$ be the set of those strings that are both a prefix and suffix of $w$.
If the monkey types until the last letters typed form $w$ and then stops typing, let $X$ be the total number of letters typed. Then $$E(X)=\sum_{k=1}^n|\Sigma|^{|w_k|}$$
The theorem is provable with Martingale Theory.
Example: | {
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The theorem is provable with Martingale Theory.
Example:
Typewriter alphabet: $\{A,B,\cdots,Z\}$ (Standard English).
Monkey's Goal: $ABRACADABRA$.
Prefix-Suffix Strings: $\{A,ABRA,ABRACADABRA\}$ of sizes $1,4,11$.
So the monkey will on average type $26+26^4+26^{11}=3670344487444778$ letters.
• I have done exactly the same except that i have not multiplied 9/4 as you did in the last equation. Can you please explain why have you done that? – Sumit Kumar Aug 27 '17 at 6:11
• We are only concerned with expected value $E(N|A)$ of coin flips given that $A$ wins. $A$ wins if $N$ is odd, and loses if $N$ is even. We know how to easily get the total expectation $E(N)$ and the probability of $A$ winning. It is known that if $N$ is a random variable and $A$ is any event, then $E(N|A)$ is given as the weighted arithmetic mean of $E(N|X=x)$ with weights being probabilities of those outcomes $X=x$ that make $A$ happen. Recall that in the discrete case, the weighted mean is $\frac{w_1x_1+\cdots+w_nx_n}{w_1+\cdots+w_n}$ – Roman Chokler Aug 27 '17 at 13:40 | {
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# Floor function repeated addition
I've come across this problem: $$\left\lfloor r + \frac{19}{100}\right\rfloor + \left\lfloor r + \frac{20}{100} \right\rfloor + \left\lfloor r + \frac{21}{100}\right\rfloor + \dots + \left\lfloor r + \frac{91}{100}\right\rfloor = 546$$ $$\text{Find} \ \lfloor 100r\rfloor. \textit{(Source: AIME)}$$ Here is my work:
$$\lfloor r + \frac{19}{100}\rfloor = r + \frac{19}{100} - \{r + \frac{19}{100}\}$$, so the figure can be restated as $$\{r + \frac{19}{100}\} = r + \frac{19}{100} + a - 546$$, where $$a = \lfloor r + \frac{20}{100}\rfloor + \lfloor r + \frac{21}{100}\rfloor + \dots + \lfloor r + \frac{91}{100}\rfloor$$.
Because $$\{r + \frac{19}{100}\}$$ is the fractional part, $$0 \le r + \frac{19}{100} + a - 546 < 1$$, so after some more maniuplation, $$545 + \frac{81}{100} \le r + a < 546 + \frac{81}{100}$$. $$a$$ is an integer, so the fractional part of $$r$$ must be $$\frac{81}{100}$$.
$$r = \lfloor r\rfloor + \{r\}$$, so $$\lfloor r + \frac{19}{100}\rfloor = \lfloor \lfloor r\rfloor + \frac{81}{100} + \frac{19}{100}\rfloor$$ = $$\lfloor \lfloor r\rfloor + 1\rfloor$$. Because both of the terms inside that floor function are integers, it must equal $$\lfloor r\rfloor + 1$$. This same reasoning can be applied to each of the individual floor functions of the given figure's LHS, and they each turn out to be $$\lfloor r\rfloor + 1$$.
Therefore, $$73 \lfloor r\rfloor + 73 = 546$$, so $$73\lfloor r\rfloor = 473$$. However, this cannot be true unless there is no answer (which I assume is not the case), or unless I did something wrong in my process, because then $$\lfloor r\rfloor$$ is not an integer.
If you do see the solution, it would be really nice if you did not give the answer in your response! Instead, maybe some helpful hints or partial solutions would be preferred. | {
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• The line "...so the fractional part of $r$ must be $\frac {81}{100}$" is where I think I see an issue. Couldn't that inequality equally have any of the fractions in the interval $[\frac 9{100},\frac{81}{100}]$ based on applying that logic in the same manner to any of the other terms from the original? If you apply the correct term to your next line of reasoning, or else find the next-smallest multiple of $73$ below $473$ I think you'll have your answer... Mar 25 '20 at 17:40
• Sorry, "next-largest" instead of "next-smallest"... Really, just begin with the "$73\lfloor r\rfloor+k=546$" logic, and deduce what $k$ has to be. Mar 25 '20 at 17:52
• The average of the terms is about $7.5$ so each term must be $7$ or $8$. How many of each? Mar 25 '20 at 17:58
• Don't worry about the fractional part. And don't worry about $[r]$. Find the precise value where $[r + \frac k{100}]\ne [r+\frac {k+1}{100}]$. That is where $r + \frac {k}{100} < n \le r + \frac {k+1}{100}$ for some integer $n$. Mar 25 '20 at 18:23
## 2 Answers
Don't worry about the fractional part. And don't worry about $$[r]$$. Find the precise value of $$k$$ where $$[r + \frac k{100}]\ne [r+\frac {k+1}{100}]$$. That is where $$r + \frac {k}{100} < m \le r + \frac {k+1}{100}$$ for some integer $$m$$.
====== my answer below ====
Well, what jumps at me is $$0 < \frac k{100} < 1$$ so all the $$[r +\frac {k}{100}]$$ are either one integer, call it $$n$$ or the next, $$n+1$$.
.
So if $$b$$ of them equal $$n+1$$ and $$(73 -b)$$ of them equal $$n$$ we have $$(73-b)n + b(n+1) = 73n + b =546$$ where $$0\le b < 73$$.
.
So as $$546\equiv 35 \pmod {73}$$ and $$546= 7*73 + 35$$ so $$b=35$$ and $$n=7$$.
.
So we have (I'll have to be careful not to do a fencepost error... $$91-35=56$$ so....) $$[r+\frac{19}{100}],....,[r+\frac{56}{100}] = 7$$ and $$[r+\frac {57}{100}],...,[r+\frac {91}{100}] = 8$$.
.
So $$r + \frac {56}{100} < 8\le r+\frac {57}{100}$$ so
.
$$100r + 56 < 800 \le 100r + 57$$
. | {
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.
So $$r + \frac {56}{100} < 8\le r+\frac {57}{100}$$ so
.
$$100r + 56 < 800 \le 100r + 57$$
.
$$100r < 744 \le 100r + 1$$
.
And $$100r -1 < 743 \le 100r$$ so $$743 \le 100r < 744$$
.
So $$[100r] = 743$$.
• I solved the problem after user abieussu helped me realize that $\{r\}$ isn't necessarily equal to $\frac{81}{100}$, and your answer is wrong but very close. In your second to last step, you state this: "And 100r−<741≤100r so 741≤100r<742", which is the mistake, I believe.
– mpnm
Mar 25 '20 at 19:00
• If the third to last step is correct then my second to last ought to be. But arithmetic error can abound anywhere. Mar 25 '20 at 19:21
• Oh, for gosh sake. I worried so much about making a fencepost error and figure $57$ was the correct fencepost marker.... and then immediately replaced it with $58$. So everything from then on is off by one\$. Mar 25 '20 at 19:25
• So .... corrected it but... might have made and error. Mar 25 '20 at 19:31
• Argh... a fencepost error off by two... that's pretty annoying. Mar 25 '20 at 19:34
$$\left\lfloor r + \frac{19}{100}\right\rfloor + \left\lfloor r + \frac{20}{100} \right\rfloor + \left\lfloor r + \frac{21}{100}\right\rfloor + \dots + \left\lfloor r + \frac{91}{100}\right\rfloor = 546$$
First, let $$r = s - \dfrac{19}{100}$$. Then the sum becomes
$$\left\lfloor s + \frac{0}{100}\right\rfloor + \left\lfloor s + \frac{1}{100} \right\rfloor + \left\lfloor s + \frac{2}{100}\right\rfloor + \dots + \left\lfloor s + \frac{72}{100}\right\rfloor = 546$$
If $$s$$ were an integer, then you would get $$73s = 546$$, which has solution $$s = 7 \dfrac{35}{73}$$.
Noting that $$7 \cdot 73 = 511$$ and $$8 \cdot 73 = 584$$, we see that $$7 < s < 8$$. Now $$546-511 = 35$$. So where are you going to pick up that extra $$35$$? | {
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# Resistor factory and Bayes rule - which solution is correct?
Problem Statement
At a certain electronics factory, in a typical day’s output 10% percent of resistors are bad, and the rest are good. Good resistors have an 80% chance of passing the factory’s test, and bad resistors have a 30% chance of passing the test. Suppose the factory tests each resistor three times. If a particular resistor passes the test 2 times out of 3, what are the chances it is good?
Solution 1
First, let's define some events:
$B :=$ resistor is bad, $G :=$ resistor is good
$P :=$ resistor passes one test, $F :=$ resistor fails one test
$Q :=$ resistor passes 2 out of 3 tests
From the problem statement, we can say that $P[B] = 0.10, P[G] = 0.90, P[P|G] = 0.80, P[P|B] = 0.30$. Since the tests are independent, we can also say that $P[Q|G] = {3 \choose 2}(0.8)^2(0.2)$ and that $P[Q|B] = {3 \choose 2}(0.3)^2(0.7)$.
The quantity we seek is $P[G|Q]$. We proceed using Bayes' theorem:
\begin{align} P[G|Q] &= \frac{P[Q|G]P[G]}{P[Q]}\\ &= \frac{P[Q|G]P[G]}{P[Q|G]P[G] + P[Q|B]P[B]}\\ &= \frac{{3 \choose 2}(0.8)^2(0.2)(0.9)}{{3 \choose 2}(0.8)^2(0.2){0.9} + {3 \choose 2}(0.3)^2(0.7)(0.1)}\\ &\approx 0.95 \end{align}
Solution 2
This solution is the same as solution 1, but we calculate $P[Q]$ differently. We can say that $P[P] = P[P|G]P[G] + P[P|B]P[B] = (0.8)(0.9) + (0.3)(0.1) = 0.75$. Since the tests are independent, $P[Q] = {3 \choose 2}(0.75)^2 (0.25)$.
We proceed using Bayes' theorem just like we did in solution 1:
\begin{align} P[G|Q] &= \frac{P[Q|G]P[G]}{P[Q]}\\ &= \frac{{3 \choose 2}(0.8)^2(0.2)(0.9)}{{3 \choose 2}(0.75)^2 (0.25)}\\ &\approx 0.82 \end{align}
My Question
To me, both of these approaches seem right, but the answers are different. I've done the math several times, so I don't think the difference is due to an arithmetic error. Which solution is correct? Why? | {
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• The choice of using $P$ with two distinct meanings ("probability" and "passes the test") is confusing. – mlc Mar 13 '17 at 16:41
• P(Q) the probability we pass the test 2 out of 3 times - the first part is what type of resistor is it. then the second part is get the probability it passes or fails based on that fact - you can't treat each test as independent, once it passes a test it is more likely to be good than it was before the test – Cato Mar 13 '17 at 16:53
• if 1% of boys always pass a test and 99% of girls always pass a test - I can't calculate the probability of a person passing two tests as 0.5^2 = .25 - it is actually (1/2)(.99^ 2 + .01^2) = .49 That is to say the girls pass both the tests usually - just to dream up a quick sexist example – Cato Mar 13 '17 at 16:58
The difference in the two approaches is in how you compute $P(Q)$. The first approach is the correct one.
(Let me write $Y$ instead of $P$ for "passing the test one time".) The second approach computes $P(Y)$ and then feeds it in the binomial distribution to derive the probability of passing the test twice out of three attempts. The flaw is in the assumption that the three trials are independent: the resistor is either good or bad (and the outcomes of the trials depend on its state).
Your second approach may be re-examined as follows. Let $Q$ be the event "passing the test two times out of three". Then $P[Q] = P[Q|G]P[G] + P[Q|B]P[B]$ as in your application of the Bayes' rule.
• So the the trials are not independent if you don't know whether the resistor is good or bad? How can that be? – SplitInfinity Mar 13 '17 at 16:59
• See third comment to the question with an example with boys and girls. Independence has nothing to do with what you know. If the resistor is good (but you do not know it), this affects the odds for $Q$. If the first trial is a success, this increases the probability that the resistor is good and hence the odds for $Q$. – mlc Mar 13 '17 at 17:02 | {
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P(GOOD | Passes 2/3 times ) = P(Passes 2/3 times AND Good) / P(passes 2/3 times)
= 0.9 x 0.8 x 0.8 x 0.2 x 3 / (.9 x 0.8 x 0.8 x 0.2 x 3 + 0.1 x 0.3 x 0.3 x 0.7 x 3) = (216/625)/(216/625 + 189/10000) = 128/135 = .95
the first one | {
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Probability that linear combination of normal random variables exceeds a value
I'm new to statistics, and I'm wondering how can I compute $$P(2Y_1 + 4Y_2 - 3Y_3 \geq 40)$$ given the following information?
• $$Y$$ follows a multivariate normal distribution.
• $$Y_1, Y_2, Y_3$$ follow a normal distribution.
• $$Y_1, Y_2, Y_3$$ have means $$10, 12, 14$$.
• $$Y_1, Y_2, Y_3$$ have variances $$2, 2, 8$$.
• The covariance between $$Y_1, Y_2$$ is $$0.50$$, and the covariance between $$Y_1, Y_3$$ is $$-0.75$$. The variables $$Y_2$$ and $$Y_3$$ are independent.
Here's the covariance matrix I found for $$Y_1, Y_2, Y_3$$:
$$\begin{bmatrix} 2 & .50 & -.75 \\ .5 & 2 & 0 \\ -.75 & 0 & 8 \end{bmatrix}$$
I know how to do it when $$Y_1, Y_2, Y_3$$ are all independent by using the fact that linear combinations of independent normal random variables are normal. However, I'm really not sure about how to do it in this case. I can easily find the covariance matrix, but I'm not sure how to proceed from there.
• Just knowing their means, variances and covariances along with the fact that each variables is (marginally) normal is insufficient. Sep 18, 2020 at 3:32
If you're looking for an analytic solution, this is what you should do. Start by defining a new variable $$X = 2Y_1 + 4Y_2 - 3Y_3$$ A sum of jointly normal random variables is also normal, even if the variables are not independent (see Wikipedia article here). All that remains is to compute the mean and variance.
The mean of $$X$$ can be determined as \begin{align*} E[X] &= 2 E[Y_1] + 4 E[Y_2] - 3 E[Y_3] \\ &= (2 \times 10) + (4 \times 12)-(3 \times 14) \\ &= 26 \end{align*} The variance of $$X$$ is slightly easier to write in matrix form. Let $$Y \equiv (Y_1,Y_2,Y_3)$$ and $$\omega \equiv (2,4,-3)$$ be vectors of random variables and weights, respectively. The variance is given by
$$Var(X) = Var(\omega'Y) = \omega'Var(Y)\omega$$ | {
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$$Var(X) = Var(\omega'Y) = \omega'Var(Y)\omega$$
Numerically this is equal to $$Var(X) = \begin{bmatrix} 2 & 4 & -3 \end{bmatrix} \begin{bmatrix} 2& .5 & -.75 \\ .5 & 2 & 0 \\ -.75 & 0 & 8 \end{bmatrix} \begin{bmatrix} 2 \\ 4 \\ -3 \end{bmatrix}= 129$$ If $$\Phi$$ is the CDF of a standard normal, \begin{align*} P(X > 40) &= 1-P\left( X \le 40 \right) \\ &= 1-P\left( \frac{X - E[X]}{\sqrt{Var(X)}} \le \frac{40 - E[X]}{\sqrt{Var(X)}} \right) \\ &= 1-P\left( Z \le \frac{40 - E[X]}{\sqrt{Var(X)}} \right) \\ &= 1-\Phi\left( \frac{40 - E[X]}{\sqrt{Var(X)}} \right) \\ &= 0.108867 \end{align*}
• "A sum of normal random variables is also normal, even if the variables are not independent" - I don't think that's true. Do you have a proof?
– Paul
Sep 17, 2020 at 7:31
• @Paul you are correct, the necessary assumption is that the random variables are jointly gaussian (which they are in this case). The easy counterexample is $X \sim \mathcal N (0,1)$ and $Y = -X$. Then $X + Y = 0$ is not gaussian. Sep 17, 2020 at 7:55
• Another nice counterexample is here math.stackexchange.com/questions/563364/… Sep 17, 2020 at 8:05
• Great counter examples! I have amended the statement to emphasize that "joint" normality is important. Your comments emphasize the pitfalls of focusing on the marginal distributions only. Sep 18, 2020 at 2:56
You can obtain the mean and variance of the resulting linear combination or you can also do a simulation and obtain the results.
Via simulation.
> covmat=matrix(c(2,0.5,-0.75,0.5,2,0,-0.75,0,8),nrow=3)
> is.positive.definite(x, tol=1e-8)
[1] TRUE
> means=c(10,12,14)
> weights=c(2,4,-3)
> mat=mvrnorm(10^7,means,covmat)
> mat=mat %*% weights
> result=sum(mat>40)/10^7
> result
[1] 0.1088587
Non simulated
The mean is a linear combination of means
> new_mean= means %*% weights
> new_mean
[1] 26
The variance is obtained multiplying
> weights %*% covmat %*% weights
[,1]
[1,] 129
> 1-pnorm(40,26,sqrt(129)) | {
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# Definition of limit as $x\rightarrow \infty$
Every time i get confused with the definition of $\lim_{x\rightarrow \infty}f(x)=L$. I could not find a reference that will give the definition.
I am trying to write what i understood. See if this is correct.
• By $\lim_{x\rightarrow \infty}f(x)=L$ we mean the following : Given $\epsilon >0$ there exists $R>0$ such that $|f(x)-L|<\epsilon$ for all $x>R$.
• By $\lim_{x\rightarrow -\infty}f(x)=L$ we mean the following : Given $\epsilon >0$ there exists $R<0$ such that $|f(x)-L|<\epsilon$ for all $x<R$.
• By $\lim_{x\rightarrow \infty}f(x)=\infty$ we mean the following: Given $R>0$ there exists $L>0$ such that $|f(x)|>R$ for all $x>L$
• By $\lim_{x\rightarrow \infty}f(x)=-\infty$ we mean the following: Given $R<0$ there exists $L>0$ such that $f(x)<R$ for all $x>L$
• By $\lim_{x\rightarrow -\infty}f(x)=\infty$ we mean the following: Given $R>0$ there exists $L<0$ such that $|f(x)|>R$ for all $x<L$
• By $\lim_{x\rightarrow \infty}f(x)=-\infty$ we mean the following: Given $R<0$ there exists $L<0$ such that $f(x)<R$ for all $x<L$.
Let me know if i understood somethings wrongly.
• It's quite fine for me. The signs for $R$ and $L$ are unncessary (they're implicit). – Bernard Apr 15 '16 at 14:25
• You don't, technically, need $R<0//R>0$. for the first two. It can be any $R$. It's the "for all $x>R$" and "for all $x<R$" that makes the difference, not the sign of $R$. – Thomas Andrews Apr 15 '16 at 14:25
• If there is no need for sign then $x\rightarrow \infty$ and $x\rightarrow -\infty$ would mean the same thing? @Bernard – user311526 Apr 15 '16 at 14:27
• If there is no need for sign then $x\rightarrow \infty$ and $x\rightarrow -\infty$ would mean the same thing? @ThomasAndrews – user311526 Apr 15 '16 at 14:27
• @topgeomj No, because one definition uses $x>R$ and the other $x<R$. That is the difference, not the sign of $R$. – Thomas Andrews Apr 15 '16 at 14:28
I will use $+\infty$ in this answer to avoid ambiguity. | {
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I will use $+\infty$ in this answer to avoid ambiguity.
• $\lim_{x\to +\infty} f(x) = L$ means for all $\epsilon>0$ there exists $M$ such that for all $x>M,$ $|f(x)-L|<\epsilon$ .
• $\lim_{x\to+\infty} f(x)=+\infty$ means for all $R$ there exists $M$ such that for all $x>M$, $f(x)>R$.
Those two definitions let you define the other limits by symmetry:
\begin{align} \lim_{x\to+\infty} f(x) = -\infty&\iff \lim_{x\to+\infty} -f(x)=+\infty\\ \lim_{x\to-\infty} f(x) = M &\iff \lim_{x\to+\infty} f(-x)=M \end{align}
Where the $M$ in the second case can be any of either a real value, or $+\infty,-\infty$.
So $\lim_{x\to-\infty} f(x)= -\infty$ means $\lim_{x\to +\infty} -f(-x)=+\infty$, which means:
For any $R$ there is an $M$ so that for all $x>M$, $-f(-x)>R$.
Now, given any $R'$, you can set $R=-R'$ and find $M$ with this condition, and set $M'=-M$. Then if $x<M'$, $-x>M$, and thus $-f(x)>R$ or $f(x)<R'=-R$. So we get back the definition that we want.
The reason to distinguish $+\infty$ from $\infty$ is that some books use $\infty$ means a single point at infinity, in both directions - essentially, merging the two values $+\infty,-\infty$ into a single point at infinity.
Then:
$$\lim_{x\to\infty} f(x)= L \iff \lim_{x\to+\infty} f(x)=\lim_{x\to-\infty} f(x)=L$$
where $L$ can be any real or $+\infty,-\infty$.
$$\lim_{x\to W} f(x) = \infty\iff \lim_{x\to W} |f(x)|=+\infty$$
Where $W$ can be any real, or $+\infty$ or $-\infty$, or $\infty$.
Your question is somehwat confused, because you seem to distingush $\infty$ from $+\infty$ with absolute values when $\infty$ is the limit, but not when $x\to\infty$.
• Thanks... I understand this clearly... :) – user311526 Apr 15 '16 at 15:04
One also does not need the absolute values on $|f(x)| > R$. If f is going to infinity, then $f(x) > R$ is accurate. | {
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• For $f(x) = (-1)^{[x]}e^x$, $f(x) \to \infty | x \to +\infty$, but not $f(x) \to +\infty | x \to +\infty$. – Abstraction Apr 15 '16 at 14:34
• Ok.. Can you say something about other definitions as well.. – user311526 Apr 15 '16 at 14:34
• @Abstraction Some people distinguish between $+\infty$ and $\infty$, but quite often they are used interchangably. – Thomas Andrews Apr 15 '16 at 14:59
• @ThomasAndrews Seeing as third and fifth examples in the initial post clearly distinguish between $f(x) \to +\infty$ and $f(x) \to \infty$ (using absolute value of $f(x)$), I think here the distinction is important. – Abstraction Apr 15 '16 at 15:08
• Well, the main indication that the OP is using $\infty$ to mean a single point at infinity is the absolute values, but in the $x\to\infty$ examples, the definitions do not use $|x|>L$. So the clarity is not there in the question. @Abstraction – Thomas Andrews Apr 15 '16 at 15:11
Let me know if i understood somethings wrongly.
You did. When you write $x \to \infty$ (as opposed to $x \to +\infty$), for $x \in \mathbb{R}$ it roughly means "when absolute value of $x$ is arbitrarily large" ($|x| > R$, not $x > R$)
There is a universal definition of limit using filters. A filter $\mathcal{F}$ is a collection of sets such that it doesn't contain an empty set and $\forall f_1, f_2 \in \mathcal{F}, \exists f_3 \in \mathcal{F} : f_3 \subseteq f_1 \cap f_2$. For example, a set of neighbourhoods of a real point $x$ is a filter ($\{(x-\epsilon, x+\epsilon) | \epsilon \in \mathbb{R}^+\}$, called neighbourhoods filter); a set of intervals with infinite endpoint ($\{(a,+\infty) | a \in \mathbb{R}\}$) is a filter; a set of segment complements ($\{(-\infty,a)\cup(b,+\infty) | a,b \in \mathbb{R}\}$) is a filter. | {
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Now, let's take function $h \in \{X \to Y\}$ and there is a filter $\mathcal{F}$ on $X$ and a concept of "neighbourhoods" on $Y$ (called topology). Then let's take this sentence: "there is such $y \in Y$ that for any $y$ neighbourhood $O_y$, there is a filter element $f \in \mathcal{F}$ so that $h(f) \subseteq O_y$" and write it for short as $\lim_{\mathcal{F}}h(x) = y$.
Lets write filter $\{(a,+\infty) | a \in \mathbb{R}\}$ as $x \to +\infty$, filter $\{(-\infty,a) | a \in \mathbb{R}\}$ as $x \to -\infty$ and filter $\{(-\infty,a) \cup (a,+\infty) | a \in \mathbb{R}\}$ as $x \to \infty$.
Finally, let's call set $\{(L-\epsilon,L+\epsilon) | \epsilon \in \mathbb{R}\}$ a set of neighbourhoods of $L$, set $\{(b,+\infty) | b \in \mathbb{R}\}$ a set of neighbourhoods of $+\infty$, set $\{(-\infty,b) | b \in \mathbb{R}\}$ a set of neighbourhoods of $-\infty$ and set $\{(-\infty,b) \cup (b,+\infty) | b \in \mathbb{R}\}$ a set of neighbourhoods of $\infty$.
Now from these rules we can derive any definition. Take $\lim_{x \to -\infty}f(x) = +\infty$. It comes to "for any $+\infty$ neighbourhood $(b,+\infty), b \in \mathbb{R}$, there is a filter element $(-\infty,a), a \in \mathbb{R}$ so that $f((-\infty,a)) \subseteq (b,+\infty)$". Or, in more traditional notation, $$\forall b \in \mathbb{R}, \exists a \in \mathbb{R} : \forall x \in (-\infty,a), f(x) \in (b,+\infty)$$
• May be you are complication the question by saying more advanced things as filter, topology... – user312648 Apr 15 '16 at 16:09
• @cello Maybe. But memorizing definitions for all cases of $x \to \pm\infty$ is actually hard, filters help to see some logic in all this. "Topology" here is simply a proper naming for a system of neighbourhoods; as you can see, it's properties aren't used. – Abstraction Apr 15 '16 at 16:21 | {
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# Help with matrix derivative
In my studies of applied mathematics, specifically optimization and applied linear algebra, I have come across the following expression which I need help differentiating:
$z(B,C) = \lVert Y-B \phi(CX) \rVert_F ^2 = \text{trace} \left( (Y-B \phi(CX))(Y-B \phi(CX))^T \right)$
where $Y \in \mathbb{R}^{m\times s},X\in \mathbb{R}^{n\times s}$ are two constant real matrices, $B \in \mathbb{R} ^{m\times k}$ and $C\in\mathbb{R}^{k \times n}$ are the two variable matrices of the function $z(B,C)$ defined above, $\lVert \bullet \rVert_F$ is the Frobenius norm and all matrix dimensions involved are constant and non-variable.
The function $\phi : \mathbb{R} \to \mathbb{R}$ is a nonlinear function defined for real numbers as follows:
$\phi(x) = \left\{ \begin{array}{ll} 0 & x \leq 0 \\ x & x > 0 \\ \end{array} \right.$
and we extend its definition to matrices element-wise, that is $(\phi(A))_{i,j} = \phi((A)_{i,j})$.
I would like to find the matrix derivative of the function $z$, as stated above, with respect to $C$, that is $\frac{\partial z}{\partial C}$.
Differentiating with respect to $B$ is easy enough as $C$ doesn't impede applying standard formulas, but my problem is with differentiating with respect to $C$ as it is the argument of a nonlinear function and I do not know how to derive it with respect to matrices.
I was hoping someone could please come to the rescue and help me differentiate the function $z$ with respect to $C$. I thank all helpers.
• In the first derivative, is A supposed to be C? – John Polcari Jul 8 '18 at 16:31
• Does this book have copy right? This not allowed on this website – Cloud JR K Jul 8 '18 at 16:46
• @JohnPolcari : yes fixed it now – kroner Jul 8 '18 at 16:49
• @CloudJR : no, it is OK – kroner Jul 8 '18 at 16:50
• @JohnPolcari : thanks for pointing out – kroner Jul 8 '18 at 16:51 | {
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Let's use a convention where uppercase Latin letters represent matrices, lowercase Latin vectors, and Greek letters are scalars.
The function you've denoted by $\phi$ is the ReLU function, $r(\alpha)$, whose derivative is the Heaviside step function $$h(\alpha) = \frac{dr(\alpha)}{d\alpha} \implies dr = h\,d\alpha$$ Applying these scalar functions element-wise on a matrix argument $A=CX,$ produces matrix results, which we'll denote as \eqalign{ R &= r(A) \cr H &= h(A) \implies dR = H\odot dA = H\odot(dC\,X) \cr } where $\odot$ is the elementwise/Hadamard product.
Define a new matrix variable $$M=BR-Y \implies dM = B\,dR + dB\,R$$ Write the function in terms of this new variable, then find its differential. \eqalign{ \lambda &= \|M\|^2_F = M:M \cr d\lambda &= 2M:dM \cr &= 2M:B\,dR + 2M:dB\,R \cr &= 2B^TM:dR + 2MR^T:dB \cr &= 2B^TM:H\odot(dC\,X) + 2MR^T:dB \cr &= 2(B^TM)\odot H:(dC\,X) + 2MR^T:dB \cr &= 2((B^TM)\odot H)X^T:dC + 2MR^T:dB \cr } Setting $dB=0$ yields the gradient wrt $C$ $$\frac{\partial\lambda}{\partial C} = 2((B^TM)\odot H)X^T$$ And setting $dC=0$ yields the gradient wrt $B$ $$\frac{\partial\lambda}{\partial B} = 2MR^T$$
NB: Depending on your preferred layout convention, you may need to transpose these results.
Also, the colon notation used above (called the Frobenius product) is just a convenient way of writing the trace function, i.e. $\,\,A:B={\rm tr}(A^TB)$.
The cyclic property of the trace leads to several ways to rearrange the terms in a Frobenius product. For example, all of the following expressions are equivalent \eqalign{ A:BC &= A^T:(BC)^T \cr &= BC:A \cr &= AC^T:B \cr &= B^TA:C \cr }
• Thanks for pointing out the name of $\phi$ as I was not aware of this. – kroner Jul 8 '18 at 17:26
• Thanks for your masterpiece answer, I appreciate it. – kroner Jul 8 '18 at 17:27 | {
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# I Help with set theory notation
1. Jan 27, 2017
### Jehannum
I bought a maths book and have discovered it's somewhat above my level.
In particular I'm confused about one bit of notation. I understand the "is a member of" operator when it takes a set as argument (e.g. n ∈ ℝ) but not when the book uses it with functions (e.g. n ∈ f)
Does n ∈ f mean that the number n can be returned by function f?
2. Jan 27, 2017
### PeroK
You need to give more context. My guess is that a function can be seen as a set. For example, the function $f(x) = x^2$ can be seen as the set of points $\lbrace (x, x^2): x \in \mathbb{R} \rbrace$, also commonly known as the graph of $f$.
In this case, you could use the notation: $f = \lbrace (x, x^2): x \in \mathbb{R} \rbrace$.
And, here, $n \in f$ would mean that $n = (x, x^2)$ for some real $x$.
3. Jan 27, 2017
### Staff: Mentor
This is strange, not to say wrong. A function $f\, : \,X \rightarrow Y$ can be written as a subset $f=\{(x,y)\,\vert \,y=f(x)\} \subseteq X \times Y$ of the Cartesian product, but this cannot explain the notation $n \in f$ unless $n$ is a pair. Can you give some more context?
4. Jan 27, 2017
### Staff: Mentor
5. Jan 29, 2017
### Jehannum
I've copied part of the book below:
Theorem 2.2 There are infinite sets that are not enumerable.
Proof Consider the powerset of N, in other words the collection P whose members are all the sets of numbers (so X ∈ P iff X ⊆ N).
Suppose for reductio that there is a function f : N→P which enumerates P, and consider what we’ll call the diagonal set D ⊆ N such that n ∈ D iff n /∈ f(n).
Since D ∈ P and f by hypothesis enumerates all the members of P, there must be some number d such that f(d) = D.
So we have, for all numbers n, n ∈ f(d) iff n /∈ f(n). Hence in particular d ∈ f(d) iff d /∈ f(d). Contradiction!
6. Jan 29, 2017
### pasmith
That's not what the text is saying. | {
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6. Jan 29, 2017
### pasmith
That's not what the text is saying.
This is defining $f$ as a function from the natural numbers $\mathbb{N}$ to the collection $P$ of subsets of natural numbers. Thus if $n \in \mathbb{N}$ then $f(n) \subset \mathbb{N}$. The statement $n \in f(n)$ is therefore perfectly sensible, since it always makes sense to ask if a particular natural number is or is not a member of a particular subset of the natural numbers.
7. Jan 29, 2017
### Staff: Mentor
O.k., but this is not what you wrote. $n \in f$ is something different to $n \in f(n)$ where $f: \mathbb{N}\rightarrow \mathcal{P}(\mathbb{N})$ is a function which values are sets $X$, so $n \in X=f(n) \in \mathcal{P}(\mathbb{N})$ makes sense. This is not the same as $n \in f \in \mathcal{P}( \mathbb{N} \times \mathbb{N})$ which we explained in the previous posts.
The theorem is proven by the so called diagonal argument. It is a formal notation of the following principle:
We want to show, that the set of all subsets of $\mathbb{N}$ isn't enumerable. | {
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We want to show, that the set of all subsets of $\mathbb{N}$ isn't enumerable.
If we have an element $X \in \mathcal{P}(\mathbb{N})$, i.e. a subset $X \subseteq \mathbb{N}$, then we can write it down number by number: $X= n_1n_2n_3n_4 \ldots$
Now let's assume $\mathcal{P}(\mathbb{N})$ is enumerable. Then there would be a complete list like
$$\begin{matrix} 1: &X_1 & = & n_{11}n_{12}n_{13} \ldots \\ 2: &X_2 & = & n_{21}n_{22}n_{23} \ldots \\ 3: & X_3 & = & n_{31}n_{32}n_{33} \ldots \\ &\vdots &&\\ \end{matrix}$$
which contains all elements of $\mathcal{P}(\mathbb{N})$, or subsets of $\mathbb{N}$.
The diagonal $d$ is now the number $d=n_{11}n_{22}n_{33} \ldots$ We next add $+1$ to all of the numbers $n_{ii}\,$.
Since by assumption we have enumerated all, this new number $d'$ (or subset $d'=\{n_{11}+1,n_{22}+1,n_{33}+1, \ldots\}$ if you like) has to occur somewhere, say $d'=X_k$. However the $k-$th position of $d'$ is $n_{kk}$ by enumeration of the $X_i$ and $n_{kk}+1$ by construction of $d'$. This cannot both be true, which means there is no way to enumerate $\mathcal{P}(\mathbb{N})$.
You might want to read the proof in the book again and figure out, why both are basically the same. It is also the method used to show why $[0,1]$ or $\mathbb{R}$ are uncountable. Whether $|\mathcal{P}(\mathbb{N})| = |\mathbb{R}|$ holds is called the continuum hypothesis. It cannot be deduced from the usual axioms of set theory, i.e. by the Zermelo-Fraenkel axioms (ZF). If you read ZFC somewhere, then Zermelo-Fraenkel + Contnuum hypothesis is meant. (Usually it is also assumed, if nothing is said.)
8. Jan 29, 2017
### micromass
Staff Emeritus
That is not the continuum hypothesis and it can be proven in ZFC.
9. Jan 29, 2017
### micromass
Staff Emeritus
ZFC means Zermelo-Fraenkel with choice. Not the continuum hypothesis.
10. Jan 29, 2017
### Jehannum
Thanks. It's becoming clear now. | {
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10. Jan 29, 2017
### Jehannum
Thanks. It's becoming clear now.
Boggled by some of the unfamiliar notation and concepts, I was not appreciating the fact that the function f returns a set as its result, so the ∈ operator can of course be used on the outputs of f such as f (n).
The book itself moves onto such diagonal proofs.
11. Jan 29, 2017
### Staff: Mentor
Thanks for clarification.
Oops, I just had a quick look on Wiki to be sure, for I know I tend to forget where the "gap" really is, and only saw $\mathfrak{c}=\aleph_1$ and next $2^{\aleph_0}=\aleph_1$. I should have read it with more care ...
Thanks for the corrections. | {
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# Math Help - 3*2 Matrix with complex elements
1. ## 3*2 Matrix with complex elements
Folks,
Given that the set of all 3*2 matrices with complex elements is a complex vector space, with the usual definitions of addition and scalr multiplication of matrices, what is its dimension?
Are the following subsets subspaces?
1) The set of 3*2 with real entries
2) The set of matrices with first row (0,0)
3) The set of matrices with first row (1,1)
I understand the dimension of a vector space is the number of vectros in any basis for the space, ie the number of coordinates necessary to specify any vector. How do I find out what its dimension is?
Thanks
2. ## Re: 3*2 Matrix with complex elements
the "naive" way is to ask yourself: how many complex numbers do i need to specify to identify my "vector"? for 3x2 matrices, that number is 6 (one for each entry).
however, mathematics professors being what they are, will usually insist you display a basis (a linearly independent spanning set). can you think of a set of six matrices,
each of which captures the idea of "a single matrix coordinate"? once you have done this, show linear independence and spanning.
again, with any vector space V, there are 3 things you need to show for any subset W:
1) W is non-empty (preferrably by showing the 0-element is a member. if the 0-element (0-vector) is not in W, you will not obtain a vector space).
2) if u,v are in W then their vector sum u+v must also be.
3) if c is any scalar in your underlying field, and u is in W, then cu must also be in W. be careful with this one. if you are working with a complex vector space, it is not sufficient to check this property for real scalars only.
attempt to show these properties (or give a counter-example) for each of the sets in your post. that's how it's done.
3. ## Re: 3*2 Matrix with complex elements | {
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3. ## Re: 3*2 Matrix with complex elements
Originally Posted by Deveno
the "naive" way is to ask yourself: how many complex numbers do i need to specify to identify my "vector"? for 3x2 matrices, that number is 6 (one for each entry).
can you think of a set of six matrices, each of which captures the idea of "a single matrix coordinate"? once you have done this, show linear independence and spanning.
Sorry, I am still struggling to get a start...I dont know how to approach this? If its a 3*2 matrix then would it have 3 solutions in ${\mathbb{R}^3}$?
4. ## Re: 3*2 Matrix with complex elements
If I write a sample 3*2 matrix as
$\begin{bmatrix}2 & 5\\ 1 & 4 \\ 6 & 8 \end{bmatrix}$
THis matrix contains 6 real entries, but I am unbale to write expression since I believe I need it to be a 3*3 or a 2*3 matrix. How would I proceed to show its closed under vector addition and scalar multiplication?
Thanks
5. ## Re: 3*2 Matrix with complex elements
you're not being asked to "solve" a system of linear equations, you're being asked to display a basis for a vector space.
here is a similar problem:
find the dimension of the vector space of all 2x2 complex matrices.
i claim:
$\left\{\begin{bmatrix}1&0\\0&0\end{bmatrix},\begin {bmatrix}0&1\\0&0\end{bmatrix}, \begin{bmatrix}0&0\\ 1&0 \end{bmatrix},\begin{bmatrix}0&0\\0&1\end{bmatrix} \right\}$
is a basis for this vector space. clearly this is a spaning set, since:
$\begin{bmatrix}a&b\\c&d\end{bmatrix} = a\begin{bmatrix}1&0\\0&0\end{bmatrix} + b\begin{bmatrix}0&1\\0&0\end{bmatrix} + c\begin{bmatrix}0&0\\1&0\end{bmatrix} + d\begin{bmatrix}0&0\\0&1\end{bmatrix}$
now, to show linear independence, suppose that:
$\begin{bmatrix}0&0\\0&0\end{bmatrix} = c_1\begin{bmatrix}1&0\\0&0\end{bmatrix} + c_2\begin{bmatrix}0&1\\0&0\end{bmatrix} + c_3\begin{bmatrix}0&0\\1&0\end{bmatrix} + c_4\begin{bmatrix}0&0\\0&1\end{bmatrix}$
then:
$\begin{bmatrix}0&0\\0&0\end{bmatrix} = \begin{bmatrix}c_1&c_2\\c_3&c_4\end{bmatrix}$ | {
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then:
$\begin{bmatrix}0&0\\0&0\end{bmatrix} = \begin{bmatrix}c_1&c_2\\c_3&c_4\end{bmatrix}$
so $c_1 = c_2 = c_3 = c_4 = 0$ so our set is linearly independent, and is thus a basis.
since our basis has 4 elements, the dimension of the vector space is 4.
6. ## Re: 3*2 Matrix with complex elements
Originally Posted by Deveno
you're not being asked to "solve" a system of linear equations, you're being asked to display a basis for a vector space.
here is a similar problem:
find the dimension of the vector space of all 2x2 complex matrices.
i claim:
$\left\{\begin{bmatrix}1&0\\0&0\end{bmatrix},\begin {bmatrix}0&1\\0&0\end{bmatrix}, \begin{bmatrix}0&0\\ 1&0 \end{bmatrix},\begin{bmatrix}0&0\\0&1\end{bmatrix} \right\}$
is a basis for this vector space. clearly this is a spaning set, since:
$\begin{bmatrix}a&b\\c&d\end{bmatrix} = a\begin{bmatrix}1&0\\0&0\end{bmatrix} + b\begin{bmatrix}0&1\\0&0\end{bmatrix} + c\begin{bmatrix}0&0\\1&0\end{bmatrix} + d\begin{bmatrix}0&0\\0&1\end{bmatrix}$
now, to show linear independence, suppose that:
$\begin{bmatrix}0&0\\0&0\end{bmatrix} = c_1\begin{bmatrix}1&0\\0&0\end{bmatrix} + c_2\begin{bmatrix}0&1\\0&0\end{bmatrix} + c_3\begin{bmatrix}0&0\\1&0\end{bmatrix} + c_4\begin{bmatrix}0&0\\0&1\end{bmatrix}$
then:
$\begin{bmatrix}0&0\\0&0\end{bmatrix} = \begin{bmatrix}c_1&c_2\\c_3&c_4\end{bmatrix}$
so $c_1 = c_2 = c_3 = c_4 = 0$ so our set is linearly independent, and is thus a basis.
since our basis has 4 elements, the dimension of the vector space is 4.
I can only write 4 of the required 6 matrices...
$\left[ \begin{array}{ccc} 1 & 0 \\ 0 & 1\\ 1 & 1\end{array},\begin{array}{ccc} 0 & 1\\ 1 & 0\\ 0 & 0\end{array},\begin{array}{ccc} 1 & 1 \\ 0 & 0\\ 0 & 1\end{array},\begin{array}{ccc} 0 & 0 \\ 1 &1\\ 1 & 0\end{array}, \right]$
There is a lack of symmetry for me to complete....
7. ## Re: 3*2 Matrix with complex elements | {
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There is a lack of symmetry for me to complete....
7. ## Re: 3*2 Matrix with complex elements
One of the things you need to clarify is whether you are thinking of this as a vector space over the real numbers or over the complex numbers. For example, with the set of pairs of complex numbers, with the usual addition and scalar multiplication, over the real numbers, we can take (a+ bi, c+ di)= a(1, 0)+ b(i, 0)+ c(0, 1)+ d(0, i) so it has dimension 4. The same set over the complex numbers has (a+ bi, c+di)= (a+ bi)(1, 0)+ (c+di)(0, 1) and so the dimension is 2.
8. ## Re: 3*2 Matrix with complex elements
Originally Posted by HallsofIvy
One of the things you need to clarify is whether you are thinking of this as a vector space over the real numbers or over the complex numbers. For example, with the set of pairs of complex numbers, with the usual addition and scalar multiplication, over the real numbers, we can take (a+ bi, c+ di)= a(1, 0)+ b(i, 0)+ c(0, 1)+ d(0, i) so it has dimension 4. The same set over the complex numbers has (a+ bi, c+di)= (a+ bi)(1, 0)+ (c+di)(0, 1) and so the dimension is 2.
1) What is the significance of saying 'vector space over the real numbers' or 'vector space over the complex numbers'? You have written to different ways (RHS) of expressing the same LHS. I dont understand the difference.
2) Is the above an example for a 2*2 matrix?
3) How does one do the above for a 3*2?
9. ## Re: 3*2 Matrix with complex elements
the "underlying field" of a vector space can make a difference when talking about dimension.
i don't understand why you came up with those particular 4 3x2 matrices. why not the 6 matrices who have one entry 1, the rest 0?
10. ## Re: 3*2 Matrix with complex elements
Originally Posted by Deveno
you're not being asked to "solve" a system of linear equations, you're being asked to display a basis for a vector space.
here is a similar problem:
find the dimension of the vector space of all 2x2 complex matrices.
i claim: | {
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find the dimension of the vector space of all 2x2 complex matrices.
i claim:
$\left\{\begin{bmatrix}1&0\\0&0\end{bmatrix},\begin {bmatrix}0&1\\0&0\end{bmatrix}, \begin{bmatrix}0&0\\ 1&0 \end{bmatrix},\begin{bmatrix}0&0\\0&1\end{bmatrix} \right\}$
is a basis for this vector space. clearly this is a spaning set, since:
$\begin{bmatrix}a&b\\c&d\end{bmatrix} = a\begin{bmatrix}1&0\\0&0\end{bmatrix} + b\begin{bmatrix}0&1\\0&0\end{bmatrix} + c\begin{bmatrix}0&0\\1&0\end{bmatrix} + d\begin{bmatrix}0&0\\0&1\end{bmatrix}$
now, to show linear independence, suppose that:
$\begin{bmatrix}0&0\\0&0\end{bmatrix} = c_1\begin{bmatrix}1&0\\0&0\end{bmatrix} + c_2\begin{bmatrix}0&1\\0&0\end{bmatrix} + c_3\begin{bmatrix}0&0\\1&0\end{bmatrix} + c_4\begin{bmatrix}0&0\\0&1\end{bmatrix}$
then:
$\begin{bmatrix}0&0\\0&0\end{bmatrix} = \begin{bmatrix}c_1&c_2\\c_3&c_4\end{bmatrix}$
so $c_1 = c_2 = c_3 = c_4 = 0$ so our set is linearly independent, and is thus a basis.
since our basis has 4 elements, the dimension of the vector space is 4.
$\left[\begin{array}{ccc} 0 & 0 \\ 0 & 0\\0 & 0\end{array}\right]=c_1\left[ \begin{array}{ccc} 1 & 0 \\ 0 & 0\\0 & 0\end{array}\right]+c_2\left[ \begin{array}{ccc} 0 & 1\\ 0 & 0\\ 0 & 0\end{array}\right]+c_3\left[ \begin{array}{ccc} 0 & 0 \\ 1 & 0\\ 0 & 0\end{array}\right]+c_4\left[ \begin{array}{ccc} 0 & 0 \\ 0 & 1\\ 0 & 0\end{array}\right]+c_5\left[ \begin{array}{ccc} 0 & 0 \\ 0 & 0\\ 1 & 0\end{array}\right]+c_6\left[ \begin{array}{ccc} 0 & 0 \\ 0 & 0\\ 0& 1\end{array}\right]$
implies $c_1+c_2+c_3+c_4+c_5+c_6=0$
This implies the set is linearly independant with a dimension of 6? What is next?
My attempt based on HallsofIvy post
3*2 matrix over the real numbers
$\left[\begin{array}{ccc} a+bi & c+di \\ e+fi & g+hi\\j+ki & l+mi\end{array}\right]=a(1,0)+b(i,0)+c(1,0)+d(i,0)........$ Hence 12 dimensions
3*2 over the complex numbers | {
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3*2 over the complex numbers
$\left[\begin{array}{ccc} a+bi & c+di \\ e+fi & g+hi\\j+ki & l+mi\end{array}\right]=(a+bi)(1,0)+(c+di)(0,1)+........$ Hence 6 dimensions.........?
11. ## Re: 3*2 Matrix with complex elements
Every element of the matrix is an ordered pair. If the matrix has only one non-zero element, all matrices with only this one element are a linear combination of two matrices, one with (0,1) as the sole element, and the other with (1,0) as the sole element. So for each element in the matrix you need two matrices with (1,0) and (0,1) as the sole elements. You need a total of 12 matrices, two for each element. The dimension of the vector space is 12.
12. ## Re: 3*2 Matrix with complex elements
Originally Posted by Hartlw
Every element of the matrix is an ordered pair. If the matrix has only one non-zero element, all matrices with only this one element are a linear combination of two matrices, one with (0,1) as the sole element, and the other with (1,0) as the sole element. So for each element in the matrix you need two matrices with (1,0) and (0,1) as the sole elements. You need a total of 12 matrices, two for each element. The dimension of the vector space is 12.
For both the real case and the complex case? Is this not conflicting with post # 7
13. ## Re: 3*2 Matrix with complex elements
Originally Posted by bugatti79
For both the real case and the complex case? Is this not conflicting with post # 7
For the real case, every base matrix has a 1 in one element and every other element a 0, for a total of six, the dimension is six. For complex case, see my previous post. | {
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EDIT: I think post seven is discussing a complex vector space consisting of complex vectors a + bi. Not the same thing as a complex vector space each element of which is a matrix. You can probably think of each matrix as the sum of a real (a,0) at each location and imaginary, (,0,b) at each location, component . In any event, any complex matrix can be expressed as a linear combination of the 12 base matrices I described in my previous post.
EDIT: The matrices themselves satisfy requirements of linear vector space with the usual definition of matrix addition and multiplication by a scalar.
The base matrices are:
(0,1) (0,0)
(0,0) (0,0)
(0,0) (0,0)
(1,0) (0,0)
(0,0) (0,0)
(0,0) (0,0)
(0,0) (1,0)
(0,0) (0,0)
(0,0) (0,0)
..... for 12 base matrices (dim 12).
14. ## Re: 3*2 Matrix with complex elements
Originally Posted by Hartlw
I think post seven is discussing a complex vector space consisting of complex vectors a + bi.
So based on your post is this the answer to the following original question?
Given that the set of all 3*2 matrices with complex elements is a complex vector space, with the usual definitions of addition and scalr multiplication of matrices, what is its dimension?
3*2 matrix over the complex numbers
$\left[\begin{array}{ccc} a+bi & c+di \\ e+fi & g+hi\\j+ki & l+mi\end{array}\right]=a(1,0)+b(i,0)+c(1,0)+d(i,0)........$
15. ## Re: 3*2 Matrix with complex elements
No
If you denote the matrices in my previous post by E1, E2, E3 , E4, ...
Then an arbitrary complex matrix is expressed as a linear combination of E1, E2, E3,...,E12
(1,3) (5,3)
(3,5) (5,7) = 1xE1 + 3xE2 + 5xE3 + ..........
(3,3) (4,3)
Above is a matrix. The ordered pair (1,3) is just a std shortcut way of expressing 1+i3.
A complex matrix has to be expressed as a sum of complex matrices. Not as a sum of complex numbers. The addition of two complex matrices is a complex matrix by term by term addition, added just like complex numbers.
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# Empty intersection and empty union
If $A_\alpha$ are subsets of a set $S$ then
$\bigcup_{\alpha \in I}A_\alpha$ = "all $x \in S$ so that $x$ is in at least one $A_\alpha$"
$\bigcap_{\alpha \in I} A_\alpha$ = "all $x \in S$ so that $x$ is in all $A_\alpha$"
It is the convention that $\bigcup_{\alpha \in \emptyset}A_\alpha = \emptyset$ and $\bigcap_{\alpha \in \emptyset} A_\alpha = S$.
But if $x$ is in $\bigcap_{\alpha \in \emptyset} A_\alpha = S$ then $x$ is in all $A_\alpha$ with $\alpha \in \emptyset$ and therefore $x$ is certainly in at least one $A_\alpha$ with $\alpha \in \emptyset$. But then $x \in \bigcup_{\alpha \in I}A_\alpha$.
Can someone help me and tell me what is wrong with this? Thank you.
-
This might help. math.stackexchange.com/q/309986/35983 – Gautam Shenoy Apr 23 at 9:13
Element in the empty set is the problem. – simplicity Apr 23 at 9:14
– Martin Sleziak Apr 23 at 11:55
Some texts consider it a convention, but it is in fact a computation!
For the fixed set $S$, we are looking at the set $\mathcal P(S)$, the set of all subsets of $S$. With the operations of intersection and unions (of arbitrary many subsets) the set $\mathcal P(S)$ is what is known as a complete lattice (don't worry if you don't know what that means).
First, one can argue intuitively: the more subsets of $S$ you intersect, the smaller the intersection is. Or, the fewer subsets you intersect, the larger the intersection is. So, the intersection of no subsets at all, the least amount of sets you can intersect, should be the largest subset possible. Namely, $\bigcap_{i\in \emptyset}A_i=S$. Similarly, the fewer subsets you take the union of, the smaller the union. So, the union of no subsets at all should be the smallest set possible. Namely, $\bigcup _{i\in \emptyset }A_i=\emptyset$. | {
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Now, to make things more formal, lets define the intersection and union in $\mathcal P(S)$. The definition will be equivalent to the set-theoretic definitions but will only make use of the partial order relation of inclusion. Given a collection $\{A_i\}_{i\in I}$ of subsets of $S$, their intersection is the largest subset of $S$ that is contained in each $A_i$ (notice that this is saying that the intersection is a greatest lower bound). Similarly, the union of the family of subsets is the smallest subset of $S$ that contains each $A_i$ (notice that this says that the union is a least upper bound). Incidentally, this point of view very clearly points to a duality between union and intersection.
So now, the intersection of no subsets is the largest subset of $S$ that is contained in each one of the given subsets. There are no given subsets at all, so (vacuously) any subset $B\subseteq S$ contains each of the non-existent $A_i$. The largest of those is $S$, proving that $\bigcap_{i\in \emptyset}A_i=S$. Similarly, the union of no subsets is the smallest subset of S that contains each of the given subsets. No subsets are given, so any subsets $B\subseteq S$ contains each of the non-existent $A_i$. The smallest of these is $\emptyset$, thus proving that $\bigcup_{i\in \emptyset}A_i=\emptyset$.
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I realized, thank you. – Anna Apr 23 at 9:37
You are welcome @Anna – Ittay Weiss Apr 23 at 9:38
In the last paragraph: did you mean "...is contained in each of..."? – Anna Apr 23 at 10:02
+1, excellent explanation. – Andreas Caranti Apr 23 at 10:08
Please note that the fact that
$$\bigcap_{\alpha \in \emptyset} A_\alpha = S$$
is not a convention - it follows from the definition
$$\bigcap_{\alpha \in I} A_\alpha = \{ x \in S : \text{x \in A_\alpha, for all \alpha \in I}\}.$$ | {
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$$\bigcap_{\alpha \in I} A_\alpha = \{ x \in S : \text{x \in A_\alpha, for all \alpha \in I}\}.$$
This is best understood by asking when is it that for a given $x \in S$ we have $x \notin \bigcap_{\alpha \in \emptyset} A_\alpha$. This can only happen if there exists $\alpha \in \emptyset$ such that $x \notin A_{\alpha}$. Since $\emptyset$ has no elements, there cannot be such an $\alpha$.
PS One can avoid introducing a (somewhat arbitrary) index set in an intersection, say. Instead, fix a set $S$, take a subset $\mathfrak{S}$ of $\mathcal{P}(S)$, and define $$\bigcap \mathfrak{S} = \{ x \in S : \text{x \in A for all A \in \mathfrak{S}} \}.$$ Then the argument above becomes $$\bigcap \emptyset = \{ x \in S : \text{x \in A for all A \in \emptyset} \} = \{ x \in S : \ \} = S,$$ since there are no $A$ to consider here.
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It should be pointed out that this intersection is $S$ only if it is given that the computation is performed in $\mathcal P(S)$. Otherwise, if the computation is performed just in the ambient universe of sets, then the intersection does not exist (since it's trying to be a set that contains every other set). In contrast, the empty union is the empty set no matter where it is computed. – Ittay Weiss Apr 23 at 9:47
@IttayWeiss, an excellent point. – Andreas Caranti Apr 23 at 9:52
What is the ambient universe of sets? – Anna Apr 23 at 10:06
– Andreas Caranti Apr 23 at 10:07
Thank you Andreas. I think that my point of confusion is why you can write $\left\{ x \in S : x \in A \text{ for all } A \in \emptyset \right\} = \left\{ x \in S : \text{} \right\}$, whereas you cannot write (considering the case of the union of the empty collection) $\left\{ x \in S : x \in A \text{ for at least one } A \in \emptyset \right\} = \left\{ x \in S : \text{} \right\}$. Would you mind helping me understand the distinction? – Dan Douglas Dec 12 at 15:59
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Those conventions comes from the fact that $(\mathcal P (S), \cup, \emptyset)$ and $(\mathcal P (S), \cap, S)$ are monoids.
Hence, those conventions are just the usual $\sum_{k \in \emptyset} k = 0$ or $\prod_{k \in \emptyset} k = 1$ that you certainly know for $(\mathbb N, +)$ or $(\mathbb Z, \times)$ for example. | {
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-
This is like saying "it's so because it is so". In fact, it is not a convention at all. It can be proven so by following the definition. In a monoid though, the unit is (by definition if you like) the result of applying the operation to no elements. But for $\mathcal P(S)$ it follows from a deeper reason than just saying it's so. – Ittay Weiss Apr 23 at 9:34
@IttayWeiss The question was not "Is it a convention or not ?" but "Can anyone enlighten me about those conventions/properties ?". I was just pointing out that the OP was actually familiar with those but did not recognize it at first sight. Plus, I could argue that your proof can actually show that $\bigcap_\emptyset A_i = \emptyset$ : indeed, taking $B$ a subset of $S$, $\forall A_i \in \emptyset, A_i \not\subseteq B$ is as true as $\forall A_i \in \emptyset, A_i \subseteq B$, making $\emptyset$ the intersection over the emptyset. – Pece Apr 23 at 19:16
I see your your points with the monoid analogy. If you'll think about it though, you'll see that your criticism of my proof is false (or are you claiming that you just proved a contradiction?) – Ittay Weiss Apr 23 at 19:45
Sorry, I mixed up a little. What I was saying is : as $A_i \in \emptyset$ is universally false, your claim $\forall A_i, A_i \in \emptyset \implies B \subseteq A_i$ (and so $S$, the greatest of all, is the intersection) is a valid as my claim $\exists A_i, A_i \in \emptyset \implies B \not\subseteq A_i$ (and so no subset of $S$ is a lower bound of the $A_i$ except $\emptyset$). So in my opinion, you're making a convention on the complete lattices (the meet of nothing is the top) while I was making conventions about monoid. – Pece Apr 24 at 8:34
I think you are still confused. I did not make use of any convention. I followed the definition. – Ittay Weiss Apr 24 at 8:56
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# How to prove that $\mathbb{Q}$ ( the rationals) is a countable set
I want to prove that $\mathbb{Q}$ is countable. So basically, I could find a bijection from $\mathbb{Q}$ to $\mathbb{N}$. But I have also recently proved that $\mathbb{Z}$ is countable, so is it equivalent to find a bijection from $\mathbb{Q}$ to $\mathbb{Z}$?
• Sure. Better yet, it's sufficient to find an injection from $\mathbb Q$ to $\mathbb Z$. Or for that matter a surjection from $\mathbb Z$ to $\mathbb Q$, or from $\mathbb N$ to $\mathbb Q$. – bof Feb 1 '14 at 8:25
• is just an injection sufficient to prove countability, though? – furashu Feb 1 '14 at 8:33
• I guess it depends on how you define "countable". Are finite sets countable? If not, then besides finding an injection from $\mathbb Q$ to $\mathbb Z$, you also have to prove that $\mathbb Q$ is infinite. But I you already know that. – bof Feb 1 '14 at 8:36
• i would say finite sets are not "countable" but "finite" .. seems weird, but countable is a term I would reserve specifically for distinction between cardinality of infinite sets. – furashu Feb 1 '14 at 8:40
• Fine. $\mathbb Q$ is an infinite set. An infinite set is countable if it has an injection into $\mathbb N$. Or into any countable set, such as $\mathbb Z$, which you already know is countable. – bof Feb 1 '14 at 8:46
Clearly $\mathbb{Z}$ injects into $\mathbb{Q}$.
Let $p_i$ enumerate all the prime numbers.
If $q \neq 0, 1, -1$, let $q = \pm \frac{p_{i_0}^{n_0} ... p_{i_k}^{n_k}}{p_{j_0}^{m_0} ... p_{j_p}^{m_p}}$ be the prime decomposition the numerator and denominator of $q$ written in simplest form. Define | {
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$\Phi(q) = \begin{cases} 0 & \quad q = 0 \\ 1 & \quad q = 1 \\ -1 & \quad q = -1 \\ p_{2 i_0}^{n_0} ... p_{2 i_k}^{n_k} p_{2 j_0 + 1}^{m_0} ... p_{2 j_p + 1}^{m_p} & \quad q = \frac{p_{i_0}^{n_0} ... p_{i_k}^{n_k}}{p_{j_0}^{m_0} ... p_{j_p}^{m_p}} \\ - p_{2 i_0}^{n_0} ... p_{2 i_k}^{n_k} p_{2 j_0 + 1}^{m_0} ... p_{2 j_p + 1}^{m_p} & \quad q = - \frac{p_{i_0}^{n_0} ... p_{i_k}^{n_k}}{p_{j_0}^{m_0} ... p_{j_p}^{m_p}} \end{cases}$
$\Phi$ is an injection of $\mathbb{Q}$ into $\mathbb{Z}$.
By the Cantor Schroder Theorem, there is a bijection between $\mathbb{Z}$ and $\mathbb{Q}$.
As bof mentioned, a nicer injection would be
$\Phi(q) = \begin{cases} 0 & \quad q = 0 \\ 1 & \quad q = 1 \\ -1 & \quad q = -1 \\ 2^m (2n + 1) & \quad q = \frac{m}{n} \text{ simplest form } \\ - 2^m(2n + 1) & \quad q = - \frac{m}{n} \text{ simplest form} \end{cases}$
• this is one of the most impressive and elegant answers i've seen on here, well done! – furashu Feb 1 '14 at 9:18
• Instead of using prime factorizations, you could just map the reduced fraction $\frac mn$ to the integer $2^m(2n+1)$ – bof Feb 1 '14 at 9:23
• @bof True. This would have saved me like ten minutes of typing. – William Feb 1 '14 at 9:27
• @William can you explain what bof said? what's the significance of that integer? – furashu Feb 1 '14 at 9:39
• @William oh i am checking it now, I just like to know where things coem from, to get a glimpse on peoples intuition and knowledge. – furashu Feb 1 '14 at 9:46
If you know that $\mathbb{Z}$ is countable, you know there is a bijection $\chi:\mathbb{N} \rightarrow \mathbb{Z}$. Hence, it is sufficient to find a bijection $\nu:\mathbb{Z} \rightarrow \mathbb{Q}$ since then $\chi \circ \nu$ is a bijection between $\mathbb{N}$ and $\mathbb{Q}$.
In any case, the following figure illustrates a bijection between $\mathbb{Z}$ and $\mathbb{Q}$.
We follow the worm back and forth "counting" the rational numbers, skipping any numbers that are not simplified fractions. | {
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Hint: There is a natural map $$\left\{ \begin{array}{ccc} \mathbb{Z} \times \mathbb{Z}_{>0} & \to & \mathbb{Q} \\ (a,b) & \mapsto & \frac{a}{b} \end{array} \right.$$
• Is mapping from Z x Z the same as mapping from Z, though? – furashu Feb 1 '14 at 8:44
• Lemma : If $A$ is countable, then $A \times A$ is countable. – Euler....IS_ALIVE Feb 1 '14 at 8:53
• @Euler....IS_ALIVE How does one prove that? – Arjang Jan 19 '17 at 12:23
I read this proof in Amer. Math. Monthly. Suffices to find out an injective function from the set of positive rationals to positive integers. Consider the representation of numbers in base 11, where the 11 digits used are $$0,1,2,3, ..., 9, /$$ (yes, it is a slash as digit for the number ten). Now the rational number $$7/83$$ represents the 4-digit base-11 positive integer: $$(7\times 11^3) + (10\times 11^2) + (8\times 11) + 3$$.
• I didn't understand this. Can you explain why we can give a completely different meaning to the slash (from its established definition as the division symbol to a numeral in base-11) so as to create an injection. – Mr Reality Jan 12 at 20:21
• Ok. Write out a rational number which has a division symbol slash, say for example 3/4. Now replace slash by x. (Remember x is roman numeral for ten). So this becomes 3x4 which is a 3-digit number in base 11 system. The value of this is three eleven squared plus ten elevens and 4 which in base 10 is $(3\times 121 + 10 \times 11 +4= 363+110+4=477$. SO the function sends $3/4$ to 477. – P Vanchinathan Jan 13 at 2:08
Consider for $n=2,3,4,\ldots$ the sets
$$A_n=\left\{\frac pq :(p,q)=1,p,q>0,p+q=n\right\}$$
Claim Each $A_n$ is finite, and $\Bbb Q^+=A_2\cup A_3\cup A_4\cup\dots$
A002487 Stern's diatomic series (or Stern-Brocot sequence) Also called fusc(n) [Dijkstra]. a(n)/a(n+1) runs through all the reduced nonnegative rationals exactly once [Stern; Calkin and Wilf]. https://www.cs.utexas.edu/users/EWD/transcriptions/EWD05xx/EWD570.html | {
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fusc(1) = 1
fusc(2n) = fusc(n)
fusc(2n+1) = fusc(n) + fusc(n+1)
And it seems that fusc(k)/fusc(k+1) gives the positive rational numbers. There was also a contrieved proof recently that fusc does what it does here: https://www.isa-afp.org/browser_info/current/AFP/Stern_Brocot/document.pdf
Using the sign of Z we could create a mapping to the positive and negative rational numbers. | {
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# Number of edges of a non-planar graph with fixed number of vertex
Consider a connected undirected simple non-planar graph $$G$$ with $$15$$ vertices. If removing any edge from $$G$$ results in a planar graph, how many edges does $$G$$ have?
It is obvious that if the number of edges greater than $$3*15-6=39$$, then it is not planar. But, how to discuss "removing any edge make it planar"? Thanks
The number of edges of $$G$$ is either $$18$$ or $$20$$.
Since graph $$G$$ is non-planar, by Kuratowski's theorem, it contains a subdivision of either $$K_5$$ (the complete graph of $$5$$ vertices) or $$K_{3,3}$$ (the complete bipartite graph of $$6$$ vertices) as a subgraph.
Since removing any edge from $$G$$ make it planar, above subgraph exhaust all edges of $$G$$. This means $$G$$ is a subdivision of either $$K_5$$ or $$K_{3,3}$$.
Notice subdividing a graph increases the number of vertices and edges by same amount.
If we start from $$K_5$$ which has $$5$$ vertices and $$10$$ edges, we need to subdivide $$10$$ times to get $$15$$ vertices. In this case, the resulting graph will have $$10 + 10 = 20$$ edges.
If we start from $$K_{3,3}$$ which has $$6$$ vertices and $$9$$ edges, we need to subdivide $$9$$ times to get $$15$$ vertices. In that case, the resulting graph will have $$9+9 = 18$$ edges.
• Got it! Thank you so much! Rarely use subdivision to count~~ – Andy Feb 25 '19 at 11:31
I am not sure if this is the answer you are looking for, but a discussion on this topic can conclude upper and lower bounds for the number of edges in such a graph. | {
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As you have already stated, we can determine the upper limit using the Euler characteristic, $$v-e+f-c=1$$ which we can reduce to $$v-e+f=2$$ since our graph is connected. Because we know that our original graph $$G$$ is non-planar before removing an edge, we can say that $$\left|E\right|=e+1$$. We can use the maximum possible faces of a planer graph with $$v=15$$ to find $$e$$ using $$f_{max}=2v-4 = 26$$ $$15-e+26=2 \implies \left|E\right| \leq 40 \leftarrow\textrm{upper bound}$$
To find the lower bound, we can use Kuratowski's Theorem (see here for a nice explanation) to state a lower bound, $$\left|E\right| \geq v+3 = 18$$.
So, we can say that for a non-planar graph $$G$$, with $$\left|V\right|=15$$ and the property that removing an edge must make the result planar, has $$18\leq\left|E\right|\leq 40$$ edges.
Essentially, if the number of edges is less than the lower bound, we know that the graph must be planar. If the number of edges is greater than the upper bound, we would have to remove more than one edge to make it planar.
• Got it! Thank you so much. You gave a very detailed description. – Andy Feb 25 '19 at 11:35 | {
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# Expected number of steps in a 1D random walk with reflecting edges
Assume there is a row of $$k$$ tiles. A creature (monkey in some situations, ant in others, frog in others) lies on tile $$a$$. There is a 50% probability that the creature jumps to tile $$a-1$$ and a 50% probability that the creature jumps to tile $$a+1$$, unless it is on an edge tile. If it is on an edge tile, it must jump inwards, so it can't escape the system (i.e. tile 2 from tile 1 and tile $$k-1$$ from tile $$k$$). What is the expected number of steps for it to first reach tile $$b$$? $$1<=a, b<=k$$ is assumed. I feel like Markov chains might be used to get the answer, but I have a very limited understanding of them. If there is a closed form for the answer as well as a derivation for understanding, that would be perfect.
• Markov chains are the way! – Lord Shark the Unknown Jan 19 at 5:20
• @LordSharktheUnknown Can you elaborate in an answer how to use them in order to solve this problem? Like I said, I have a limited understanding of them. – automaticallyGenerated Jan 19 at 5:31
• I'have taken the liberty to modifiy your title and tags (3 tags are a good average) in order more readers are directed towards this interesting question and its interesting answer – Jean Marie Feb 15 at 12:27
Amazingly (to me) there happens to be a very simple expression for the expected number of steps to reach $$\ b\$$ from $$\ a\$$. For $$\ a < b\$$, it is: $$\left(\,b + a -2\,\right)\,\left(\,b-a\,\right)\ .$$ Although a Markov chain is the obvious way to model the process, and I'm sure it could used to derive the above result, there turns out to be a less cumbersome way of doing it.
For each $$\ i\$$ between $$\ 1\$$ and $$\ b\$$ inclusive, let $$\ e_i\$$ be the expected number of steps the creature takes to go from $$\ i\$$ to $$\ b\$$. Obviously, $$\ e_b\ = 0\$$. | {
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If the creature starts from $$\ 1\$$, then it has to take one step to $$\ 2\$$, from which the expected number of steps to reach $$\ b\$$ is $$\ e_2\$$. Thus, the expected number of steps, $$\ e_1\$$, to reach $$\ b\$$ from $$\ 1\$$ is $$\ e_2 + 1\$$.
If the creature starts from $$\ b-1\$$, then with probability $$\ \frac{1}{2}\$$ it reaches $$\ b\$$ on the very next step—that is, in just a single step—, and with probability $$\ \frac{1}{2}\$$ it jumps to $$\ b-2\$$, from which the expected number of steps to reach $$\ b\$$ is $$\ e_{b-2}\$$. Thus $$\ e_{b-1} = \frac{1}{2}\left(e_{b-2} +1\right) + \frac{1}{2}\,1=\frac{1}{2}\,e_{b-2}+1\$$.
If the creature starts from any other point $$\ i\$$, with $$\ 2\le i\le b-2\$$, then with probability $$\ \frac{1}{2}\$$ it jumps to $$\ i-1\$$, from which the expected number of steps to reach $$\ b\$$ is $$\ e_{i-1}\$$, and with probability $$\ \frac{1}{2}\$$ it jumps to $$\ i+1\$$, from which the expected number of steps to reach $$\ b\$$ is $$\ e_{i+1}\$$. Therefore, $$\ e_i = \frac{1}{2}\left(e_{i-1} +1\right) + \frac{1}{2}\left(e_{i+1} +1\right)= \frac{1}{2}\,e_{i-1} + \frac{1}{2}\,e_{i+1} +1\$$.
Putting this all together, we have
$$\begin{eqnarray} e_1 &=& e_2 + 1\\ e_i &=& \frac{1}{2}\,e_{i-1} + \frac{1}{2}\,e_{i+1} +1, \ \ \mbox{for } i=2,3, \dots, b-2\ \mbox{, and}\\ e_{b-1} &=& \frac{1}{2}\,e_{b-2}+1\ , \end{eqnarray}$$ or, equivalently,
$$\begin{eqnarray} e_1 - e_2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ &=& 1\\ \ \ \ \ \ \ \ \ \ \ \ \ \ -\frac{1}{2}\,e_{i-1} + e_i -\frac{1}{2}\,e_{i+1} &=& 1, \ \ \mbox{for } i=2,3, \dots, b-2\ \mbox{, and}\\ -\frac{1}{2}\,e_{b-2}+e_{b-1} &=& 1\ . \end{eqnarray}$$ | {
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These equations can be written as: $$M\,e = \mathbb 1\ ,$$ where $$\ M\$$ is the $$\ \left(\,b-1\,\right)\times\left(\,b-1\,\right)\$$ matrix, and $$\ \mathbb 1\$$ the $$\ \left(\,b-1\,\right)\times\,1\$$ column vector, whose entries are given by: $$\begin{eqnarray} M_{1,2} &=& -1\\ M_{i,i} &=& 1\ \ \mbox{for } i=1,2,\dots, b-1\\ M_{i,i-1} &=& -\frac{1}{2}\ \ \mbox{for } i=2,3,\dots, b-1\\ M_{i,i+1} &=& -\frac{1}{2}\ \ \mbox{for } i=2,3,\dots, b-2\\ M_{i,\,j} &=& 0 \ \ \mbox{for all other }\ i, j\\ \mathbb 1_i &=& 1\ \ \mbox{for } i=1,2,\dots, b-1\ . \end{eqnarray}$$ For $$\ b=6\$$, the matrix $$\ M\$$ looks like this: $$\left(\begin{matrix}1&-1&0&0&0 \\ -\frac{1}{2}&1&-\frac{1}{2}&0&0\\ 0&-\frac{1}{2}&1&-\frac{1}{2}&0\\ 0&0&-\frac{1}{2}&1&-\frac{1}{2}\\ 0&0&0&-\frac{1}{2}&1& \end{matrix}\right)\ ,$$ and has the following inverse: $$M^{-1} = \left(\begin{matrix} 5&8&6&4&2\\ 4&8&6&4&2\\ 3&6&6&4&2\\ 2&4&4&4&2\\ 1&2&2&2&2\\ \end{matrix}\right)\ .$$ From this, we can conjecture that the entries of the inverse of the $$\ \left(\,b-1\,\right)\times\left(\,b-1\,\right)\$$ matrix $$\ M\$$, defined above, should be the matrix $$\ L\$$ whose entries are given by: $$\begin{eqnarray} L_{i,1} &=& b-i\ \ \mbox{for } i=1,2,\dots, b-1\\ L_{1,\,j} &=& 2\,\left(b-j\right) \ \mbox{for } j=2,3,\dots, b-1\\ L_{i,\,j} &=& 2\,\min\left(b-i,b-j\right) \ \mbox{for } 2\le i\le b-1\ \ \mbox{and }\ 2\le j\le b-1\ , \end{eqnarray}$$ and on checking the product $$\ M\,L\$$, we find that it is indeed the $$\ \left(\,b-1\,\right)\times\left(\,b-1\,\right)\$$ identity matrix. So, finally, we have: $$e = M^{-1}\,\mathbb 1 = L\,\mathbb 1\ ,$$ and $$\ e_a\$$, the expected number of steps to get to $$\ b\$$ from $$\ a\$$ is the sum of the entries in the $$\ a^\mbox{th}\$$ row of $$\ L\$$: $$\begin{eqnarray} e_a &=& \left(b-a\right) + 2\,\left(\,a-1\,\right)\,\left(\,b-a\,\right) + 2\,\sum_{j=1}^{b-a-1} j\\ &=& \left(\,b + a -2\,\right)\,\left(\,b-a\,\right)\ , \end{eqnarray}$$ as stated above. | {
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• Thanks for the clear answer. I think that this is just a disguised Markov chain, as your "M" is the identity matrix minus the transition matrix. – automaticallyGenerated Jan 21 at 14:30
• Not quite. The identity matrix minus the transition matrix is a singular $\ b\times b\$ matrix. $\ M\$ is the submatrix obtained from it by chopping off its last row and column. But you're right that there is indeed a Markov chain lurking in the shadows. – lonza leggiera Jan 21 at 23:27 | {
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Question
# $$\displaystyle \int_{0}^{1}\frac{dx}{(1+x^{2})\sqrt{(2+x^{2})}}=\frac{\pi }{k}$$. Find the value of $$k$$.
Solution
## Let $$\displaystyle I=\int { \frac { 1 }{ \left( { x }^{ 2 }+1 \right) \sqrt { { x }^{ 2 }+2 } } dx }$$Put $$x=\sqrt { 2 } \tan { t } \Rightarrow dx=\sqrt { 2 } \sec ^{ 2 }{ t } dt$$$$\displaystyle I=\sqrt { 2 } \int { \frac { sec{ t } }{ \sqrt { 2 } \left( 2\tan ^{ 2 }{ t } +1 \right) } dt } =\int { \frac { sec{ t } }{ 2\tan ^{ 2 }{ t } +1 } dt }$$Multiplying numerator and denominator by $$\cos ^{ 2 }{ t }$$, we get$$\displaystyle I=\int { \frac { \cos { t } }{ 2\sin ^{ 2 }{ t } +\cos ^{ 2 }{ t } } dt } =\int { \frac { \cos { t } }{ \sin ^{ 2 }{ t } +1 } dt }$$Now put $$u=\sin { t } \Rightarrow du=\cos { t } dt$$Therefore$$\displaystyle I=\int { \frac { 1 }{ { u }^{ 2 }+1 } du } =\tan ^{ -1 }{ u } =\tan ^{ -1 }{ \sin { t } }$$$$\displaystyle =\tan ^{ -1 }{ \sin { \left( \tan ^{ -1 }{ \left( \frac { x }{ \sqrt { 2 } } \right) } \right) } } =\tan ^{ -1 }{ \left( \frac { x }{ \sqrt { { x }^{ 2 }+2 } } \right) }$$Hence$$\displaystyle \int _{ 0 }^{ 1 }{ \frac { 1 }{ \left( { x }^{ 2 }+1 \right) \sqrt { { x }^{ 2 }+2 } } dx } ={ \left[ \tan ^{ -1 }{ \left( \frac { x }{ \sqrt { { x }^{ 2 }+2 } } \right) } \right] }_{ 0 }^{ 1 }$$$$\displaystyle =\frac { \pi }{ 6 } -0=\frac { \pi }{ 6 }$$$$\Rightarrow k=6$$Mathematics
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# how many cubes have at least $1,2,3$ colors on them
I have a painted cube, which is cut into $n^3$ smaller cubes. I now want to find the number of cubes which have $1$,$2$,$3$ sides painted. I know the long way round of taking each cube and putting them into different categories... but is there a short way or formula to do it?
$1$ face painted - You have to consider all the faces. There are $6$ faces, each with $(n-2)^2$ cubes with $1$ face painted . That gives you the formula $$6*(n-2)^2$$
$2$ faces painted - You have to take the edges. There are $12$ edges, each edge having $n-2$ cubes with $2$ faces painted . That gives you the formula $$12*(n-2)$$
$3$ faces painted - You have to take the corners alone. That gives you the formula $$8$$
Extra - $0$ faces painted - You have to consider the interior alone which gives $$(n-2)^3$$
• I think that you want $8$ corners. ;-) – Sammy Black Feb 23 '16 at 7:20
• @SammyBlack Thank you for that. Edited. – Win Vineeth Feb 23 '16 at 7:20
• saying side is not corrrect it is face which is painted – Bhaskara-III Feb 23 '16 at 13:17
• @Bhaskara-III The question asked for side, hence, I used side. – Win Vineeth Feb 23 '16 at 13:20
• oops you should have correctly mentioned that it should be a painted face not a painted side in your answer because it's very confusing term – Bhaskara-III Feb 23 '16 at 13:22
This is an addendum to Win Vineeth's excellent answer: the particular counts fall out of the algebraic expansion (using the Binomial Theorem) of $n^3$, where we write $n = (n-2) + 2$: $$\bigl( (n-2) + 2 \bigr)^3 = (n-2)^3 + 6(n-2)^2 + 12(n-2) + 8$$ | {
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This approach generalizes by letting $t$ be the "thickness" of the boundary (the part that gets painted). The original question uses $t=1$. The expansion now looks like: $$\bigl( (n-2t) + 2t \bigr)^3 = (n-2t)^3 + 6(n-2t)^2t + 12(n-2t)t^2 + 8t^3$$ Now, the power of $t$ serves as a placeholder for the number of sides that got painted. For example, if you want to know how many subcubes got painted on $2$ sides, then look at the coefficient of $t^2$, namely $12(n-2t)$, which is $12(n-2)$ once you set $t=1$. This is an example of a generating function, where a sequence is encoded as the list of coefficients of a polynomial (or often infinite series).
This approach also has the advantage of counting the number of boundary subhypercubes of a hypercube of any dimension: just expand $n^d$, where $d$ is the dimension. | {
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Question
Let $$A$$ be a set containing $$n$$ elements. A subset $$P$$ of the set $$A$$ is chosen at random.The set $$A$$ is reconstructed by replacing the element of $$P$$, and another subsets $$Q$$ of $$A$$ is chosen at random. The probability that $$\left( P\cap Q \right)$$ contains exactly $$m(m<n)$$ elements is
A
3nm4n
B
nCm3m4n
C
nCm3nm4n
D
none of these
Solution
The correct option is C $$\displaystyle \frac { _{ }^{ n }{ { C }_{ m } }{ 3 }^{ n-m } }{ { 4 }^{ n } }$$We know that the number of subsets of a set containing $$n$$ elements is $${ 2 }^{ n }$$.Therefore, the number of ways of choosing $$P$$ and $$Q$$ is $$^{ { 2 }^{ n } }{ { C }_{ 1 } }\times _{ }^{ { 2 }^{ n } }{ { C }_{ 1 } }={ 2 }^{ n }\times { 2 }^{ n }={ 4 }^{ n }$$.Out of $$n$$ elements, $$m$$ elements can be chosen in $$^{ n }{ { C }_{ m } }$$ ways.If $$\left( P\cap Q \right)$$ contains exactly in elements, then from the remaining $$n-m$$ elements either an element belongs to $$P$$ or $$Q$$ but not both $$P$$ and $$Q$$.Suppose $$P$$ contains $$r$$ elements from the remaining $$n-m$$ elements.Then $$Q$$ may contain any number of elements from the remaining $$(n-m)-r$$ elements. Therefore, $$P$$ and $$Q$$ can be chosen in $$^{ n-m }{ { C }_{ r } }{ 2 }^{ \left( n-m \right) -r }$$But $$r$$ can vary from $$0$$ to $$(n-m)$$. So, $$P$$ and $$Q$$ can be chosen in general in$$\displaystyle \left( \sum _{ r=0 }^{ n-m }{ ^{ n-m }{ { C }_{ r } }{ 2 }^{ \left( n-m \right) -r } } \right) _{ }^{ n }{ { C }_{ m } }={ \left( 1+2 \right) }^{ n-m }\times _{ }^{ n }{ { C }_{ m } }=_{ }^{ n }{ { C }_{ m } }\times { 3 }^{ n-m }$$Hence, required probability $$\displaystyle =\dfrac { ^{ n }{ { C }_{ m } }\times { 3 }^{ n-m } }{ { 4 }^{ n } }$$Maths
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# Two Pumps, One Tank
• Jun 10th 2007, 04:04 AM
blueridge
Two Pumps, One Tank
Two pumps of different sizes working together can empty a fuel tank is 5 hours. The larger pump can empty this tank in 4 hours less than the smaller one. If the larger one is out of order, how long will it take the smaller one to do the job alone?
• Jun 10th 2007, 04:10 AM
Quick
Quote:
Originally Posted by blueridge
Two pumps of different sizes working together can empty a fuel tank is 5 hours. The larger pump can empty this tank in 4 hours less than the smaller one. If the larger one is out of order, how long will it take the smaller one to do the job alone?
I'll start you off...
Define your variables: I'd call the bigger pump $b$ and the smaller pump $s$
$b=$the number of tanks per hour the bigger pump can empty
$s=$the number of tanks per hour the smaller pump can empty
Both of those numbers will be fractions.
Now, since you only wanted a hint, I'm only going to give you the first equation and you'll have to find the rest...
Quote:
Two pumps of different sizes working together can empty a fuel tank is 5 hours.
$b+s=\frac{1\text{tank}}{5\text{hours}}$
Do you need any more help?
• Jun 10th 2007, 04:12 AM
blueridge
Yes
If you can set up the other equation, I can take it from there.
• Jun 10th 2007, 04:22 AM
Quick
Quote:
Originally Posted by blueridge
If you can set up the other equation, I can take it from there.
Quote:
Originally Posted by blueridge
Two pumps of different sizes working together can empty a fuel tank is 5 hours. The larger pump can empty this tank in 4 hours less than the smaller one. If the larger one is out of order, how long will it take the smaller one to do the job alone?
The next equation is somewhat weird.
If $b=\frac{\text{tanks}}{\text{hours}}$
than $\frac{1}{b}=\frac{\text{hours}}{\text{tanks}}$
So in fact:
$\frac{1}{b}=$the number of hours to empty a tank | {
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So in fact:
$\frac{1}{b}=$the number of hours to empty a tank
So we know that: $\frac{1}{b}-\frac{1}{s}=4$
• Jun 10th 2007, 05:54 AM
blueridge
tell me...
I am dealing with two equations in two unknowns?
• Jun 10th 2007, 06:59 AM
CaptainBlack
Quote:
Originally Posted by blueridge
I am dealing with two equations in two unknowns?
Yes, solve for $b$ and $s$, and the required answer is $s$
RonL
• Jun 10th 2007, 07:39 AM
Soroban
Hello, blueridge!
Here's another approach . . .
Quote:
Two pumps of different sizes working together can empty a fuel tank is 5 hours.
The larger pump can empty this tank in 4 hours less than the smaller one.
If the larger one is out of order, how long will it take the smaller one to do the job alone?
Together, they can do the job in 5 hours.
. . In one hour, they can do $\frac{1}{5}$ of the job. .[1]
The smaller pump can do the job in $x$ hours. .[Note that: . $x > 4$.]
. . In one hour, it can do $\frac{1}{x}$ of the job.
The larger pump takes 4 hours less; it takes $x - 4$ hours.
. . In one hour, it can do $\frac{1}{x-4}$ of the job.
Together, in one hour, they can do: . $\frac{1}{x} + \frac{1}{x-4}$ of the job. .[2]
But [1] and [2] describe the same thing:
. . the fraction of the job done in one hour.
There is our equaton! . . . . $\boxed{\frac{1}{x} + \frac{1}{x-4} \:=\:\frac{1}{5}}$
Multiply by the common denominator: $5x(x - 4)$
. . $5(x - 4) + 5x \:=\:x(x-4)$
. . which simplifies to the quadratic: . $x^2 - 14x + 20 \:=\:0$
The Quadratic Formula gives us: . $x \;=\;\frac{14\pm\sqrt{116}}{2} \;=\;7 \pm\sqrt{29} \;\approx\;\{1.6,\:12.4\}$
Since $x > 4$. the solution is: . $x = 12.4$
Therefore, the smaller pump will take about 12.4 hours working alone.
• Jun 10th 2007, 11:14 AM
blueridge
tell me...
Soroban,
I thank you for sharing yet a more simplistic avenue to understanding this question. | {
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# Math Help - Simple expression for trig matrix cubed
1. ## Simple expression for trig matrix cubed
Does anyone know how to determine a simple expression for $A^3$, where A is the following matrix?
cos@ sin@
-sin@ cos@
*I used @ as theta
I think I understand how to cube it (perform the multiplication twice?) but I have no idea how to simplify it. Thanks!
2. Originally Posted by jennifer1004
Does anyone know how to determine a simple expression for $A^3$, where A is the following matrix?
cos@ sin@
-sin@ cos@
*I used @ as theta
I think I understand how to cube it (perform the multiplication twice?) but I have no idea how to simplify it. Thanks!
A is the standard rotation matrix by angle @. A^3 = A.A.A => rotate by 3@. Therefore A^3 = .....
Originally Posted by jennifer1004
Does anyone know how to determine a simple expression for $A^3$, where A is the following matrix?
cos@ sin@
-sin@ cos@
*I used @ as theta
I think I understand how to cube it (perform the multiplication twice?) but I have no idea how to simplify it. Thanks!
Please see attached file
4. ## Thank you!
You have helped me so much. There is nothing in my text for linear algebra that looks anything like this matrix. I wasn't sure how to simplify the sines and cosines. Thanks again. Your work is very clear!
5. Hello, jennifer1004!
We are expected to know some basic identities:
. . $\begin{array}{cccccc}\cos^2\!x-\sin^2\!x \:=\:\cos2x & & \cos x\cos y - \sin x\sin y \:=\: \cos(x+y) \\
2\sin x\cos x \:=\: \sin2x & & \sin x\cos y + \cos x\sin y \:=\:\sin(x+y) \end{array}$
Given: . $A \:=\:\begin{bmatrix}\cos\theta & \sin\theta \\ \text{-}\sin\theta & \cos\theta \end{bmatrix}$
Find $A^3$ | {
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Find $A^3$
$A^2 \:=\:\begin{bmatrix}\cos\theta & \sin\theta \\ \text{-}\sin\theta & \cos\theta\end{bmatrix}\begin{bmatrix}\cos\theta & \sin\theta \\ \text{-}\sin\theta & \cos\theta \end{bmatrix}$ . $= \;\begin{bmatrix}\cos^2\!\theta-\sin^2\!\theta & 2\sin\theta\cos\theta \\ \text{-}2\sin\theta\cos\theta & \cos^2\!\theta -\sin^2\!\theta \end{bmatrix}$ . $= \;\begin{bmatrix}\cos2\theta & \sin2\theta \\ \text{-}\sin2\theta & \cos2\theta\end{bmatrix}$
$A^3 \;=\;\begin{bmatrix}\cos\theta & \sin\theta \\ \text{-}\sin\theta & \cos\theta \end{bmatrix}\begin{bmatrix}\cos2\theta & \sin2\theta \\ \text{-}\sin2\theta & \cos2\theta \end{bmatrix}$
. . $= \;\begin{bmatrix}\cos\theta\cos2\theta - \sin\theta\sin2\theta & \sin2\theta\cos\theta + \sin\theta\cos2\theta \\ \text{-}\sin\theta\cos2\theta - \sin2\theta\cos\theta & \cos\theta\cos2\theta - \sin\theta\sin2\theta \end{bmatrix}$
. . $= \;\begin{bmatrix}\cos3\theta & \sin3\theta \\ \text{-}\sin3\theta & \cos3\theta \end{bmatrix}$
6. ## Thanks Soroban!
Those identities are what I was looking for. I wasn't sure how to simplify. Thanks for sharing! | {
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# Calculating Eigenvectors: Is my book wrong?
I have a covariance matrix: $$S= \begin{pmatrix} 16 & 10 \\ 10 & 25 \end{pmatrix}$$
I calculate my eigenvalues correctly (the same as what the book finds);
$\lambda_1 = 31.47$ , $\lambda_2 = 9.53$
But now it comes to calculating eigenvectors: I do everything as I was taught way back in Elementary Linear Algebra.
1. $S X = \lambda v$ {where v is the eigenvector}
2. $(S - I \lambda)v$
3. Get Row-Echelon Form
But when I do this I get the following reduced matrix:
$$\begin{pmatrix} 1 & -.646412 & 0 \\ 0 & 0 &0 \end{pmatrix}$$
But this result doesn't seem consistent with my textbook which says that the eigenvectors are;
$(0.54 , 0.84)^T$ and $(0.84 , -0.54)$
I looked online for calculators and found one consistent with the book and a few consistent with my result:
Consistent with Book: http://comnuan.com/cmnn01002/
Consistent with Me: http://www.arndt-bruenner.de/mathe/scripts/engl_eigenwert2.htm
Any ideas?
Additional Information:
• This problem stems from Principal Component Analysis
• $(0.54,0.84) = (9,14) \cdot 0.06,$ and $9/14 \approx 0.642857\ldots,$ so I wonder if you get $0.646412$ by rounding too early, before dividing. Maybe your reduced matrix should have been $\left[ \begin{array}{ccc} 1 & -0.642857\ldots & 0 \\ 0 & 0 & 0 \end{array} \right],$ which would be consistent with the answer being $(0.54,0.84). \qquad$ – Michael Hardy Nov 24 '17 at 19:10
• Specifically, how did you get $-0.646412\text{ ?} \qquad$ – Michael Hardy Nov 24 '17 at 19:12
• I answered your question in the answers section. You were right that I rounded early. – Nicklovn Nov 24 '17 at 19:15
• The textbook has normalized the eigenvectors so that the change-of-basis matrix is orthogonal. Remember, there’s no such thing as the eigenvector(s): any non-zero scalar multiple of an eigenvector is also an eigenvector. – amd Nov 25 '17 at 0:08
## 4 Answers
TLDR: The answers are the same. | {
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## 4 Answers
TLDR: The answers are the same.
The vectors $(0.646586,1)$ and $(0.54,0.84)$ go in (almost) the same direction (the only differences due to rounding and the magnitude of the vector). The first has the benefit of one of the entries equalling one. The second has the benefit that its magnitude is (almost) $1$, but they both give essentially the same information.
Remember that an eigenvector for a specific eigenvalue $\lambda$ is any vector such that $Av=\lambda v$ and these vectors collectively make up an entire subspace of your vector space, referred to as the eigenspace for the eigenvector $\lambda$. In the problem of determining eigenvalues and corresponding eigenvectors, you need only find some collection of eigenvectors such that they form a basis for each corresponding eigenspace. There are infinitely many correct choices for such eigenvectors.
Eigenvector is not unique.
Notice that non-zero scalar multiple of an eigenvector is still an eigenvector.
Both answers are correct.
Both answers are correct. The eigen vector you computed does not have unity norm. If you normalize your eigen vector then you will get the text-book answer.
$(0.54,0.84) = (9,14) \cdot 0.06,$ and $9/14 \approx 0.642857\ldots,$ so I wonder if you get $0.646412$ by rounding too early, before dividing. Maybe your reduced matrix should have been $\left[ \begin{array}{ccc} 1 & -0.642857\ldots & 0 \\ 0 & 0 & 0 \end{array} \right],$ which would be consistent with the answer being $(0.54,0.84).$ | {
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• Yes I admit that I rounded early. What happened is that I calculated the eigenvalues a few days ago but I lost that paper and just went with the eigenvalues listed in the book (theirs being rounded already). But I was informed that I had only to normalize the vector I had calculated to get the textbook answer. – Nicklovn Nov 24 '17 at 19:14
• @Nicklovn : This is one respect in which elementary arithmetic gets far less respect than it deserves. People round without knowing to what extent or in what way the bottom line will be affected, and then they write six or eight or ten digits in their bottom-line answer when the early rounding has the effect that only perhaps the first one or two are truthful and the rest of meaningless noise. – Michael Hardy Nov 24 '17 at 19:22
• I agree. I was doing well with it at first but when I got frustrated and lost my sheet I just relied on what the book gave me which was naturally rounded :( Thanks for your help! – Nicklovn Nov 25 '17 at 21:30
• @Nicklovn : I'm glad it helped. – Michael Hardy Nov 25 '17 at 23:19 | {
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# n distinguishable balls into n boxes
We have n distinguishable balls (say they have different labels or colours). If these balls are dropped at random in n boxes, what is the probability that:
1- No box is empty? 2- Exactly one box is empty?
For 1, I figured that we have $n^n$ ways to put the n balls into the n boxes. And I figured there are $n!$ to sort the balls so there is one ball for each box.
So is the answer to question 1 $n!/(n^n)$?
For 2, there are $n-1$ ways for the boxes to be empty. This is because you can have box 1 be empty (and just that), box 2 be empty and just that, so ultimately you can have at most $n-1$ variations of empty boxes.
So I figured the solution to part 2 was $\frac{n-1}{n^n}$
Is any of this right?
• I think your part two is incorrect, because for each box $n$ that is empty, there are $n-1$ boxes that could have the extra ball. Jan 26 '15 at 20:07
• That sounds right...I'll edit my answer. Jan 26 '15 at 20:12
For part 2, you must place the balls so that there is one empty box, one box with two balls, and the remaining balls will have one ball each. There are $n$ ways to pick the empty box, and $n-1$ ways to then pick the box with two balls.
We can now fill the $n$ spaces for the balls with the $n$ balls in any order you wish. There are of course $n!$ ways to do this. However, there is a caveat. If the box with two spaces is filled with ball $a$ and then ball $b$, that is the same as if we put ball $b$ and then ball $a$. So we are double counting, and the number of ways to fill in the boxes is $\frac{n!}{2}$.
Thus the total number of ways to have one empty box is $\frac{n(n-1)(n!)}{2}$, so the probability of having one empty box is $\frac{\frac{n(n-1)(n!)}{2}}{n^n} = \frac{(n-1)(n!)}{2n^{n-1}}$. | {
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• It seems like order matters, so why would we divide by 1/2? Jan 26 '15 at 22:00
• Of the $n^n$ possibilities, all that matters is which box each ball goes into. When we consider the $n!$ ways to put $n$ boxes into the $n$ spaces, we have two spaces in one box. So by that counting, if we put ball $a$ into space 1 of that box, and ball $b$ into space 2 of that box, that's counted separately from putting ball $b$ into space 1 and ball $a$ into space 2. But in the $n^n$ counting it's just counted once, since both balls go into the same box. So we have to divide by 2 to match up the counting. Jan 27 '15 at 0:55
• See the answer I posted above from the textbook. Jan 27 '15 at 17:56
• Yes, the answer is the same: $\frac{n!}{2} = {n \choose 2} (n-2)!$. The book solution just picks the two balls in the one box first, then the remaining $n-2$ balls, whereas I picked all $n$ at once. Jan 27 '15 at 19:10
Here is the answer to the question, based on the answer manual (it was worded slightly differently in my textbook, hence why I did not find it). My part 1 was correct.
The textbook question is as follows -- I'm not sure the cells versus boxes thing is supposed to make it be any different.
>If $n$ balls are placed at random into $n$ cells, find the probability that exactly one cell remains empty.
First, we have $n$ ways of picking the empty box. Then we have $n-1$ ways of picking the box with two balls.
Now which two balls will go into the one box that will have two balls? There are ${n \choose 2}$ ways to pick the two distinct balls to go into this cell.
Lastly we need to arrange the remaining $n-2$ cells, and there are $(n-2)!$ ways to permutate all of these.
So we have:
$n(n-1){n \choose 2}(n-2)! = n!{n \choose 2}$
So the answer to part 2 is:
$\frac{n! {n \choose 2 }}{n^n}$ | {
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So the answer to part 2 is:
$\frac{n! {n \choose 2 }}{n^n}$
The argument in 1. is quite elegant. Another way to see this is that the probability to put the first ball in an empty box is $1$, for the second it is $(1 - 1/n)$, for the third it is $1 - 2/n$. The searched for probability is thus the product and this is exactly what you found in a different way that I actually like better.
For 2. you need to consider again all possibilities to place the balls. There are a lot more than you said. It is not just which box is empty (also there are $n$ ways for this) but also how the balls are distributed over the remaining ones. | {
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# How to produce ListPlot that samples a given distribution?
I wanted to get a binned plot that a uniform background distribution with a spike at some bin. My first instinct was to use RandomVariate, and then ListPlot. What i get is this
This makes sense as the RandomVariate function is assigning a value to Y based on the distribution. But I was wondering how I would produce a plot that has a spike at some particular bin (some specified x). I would want to produce something that looks more like this, for example.
I know I could just manually enter the data, but is there a way to have a binned plot where there is a uniform background distribution and a gaussian peak at some specific x?
You are on the right track. Notice that in your mixture, the uniform distribution spans [0,2], but the Normal distribution has a mean of 3. Here is another mixture. Just adjust the distribution parameters as needed:
dist = MixtureDistribution[{1, 1},
{
UniformDistribution[{0, 100}],
NormalDistribution[50, 2]
}
];
Histogram[RandomVariate[dist, 1000], {1}]
Edit: If you want data, you can use some variant of HistogramList:
{binBoundaries, counts} =
HistogramList[RandomVariate[dist, 1000], {1}];
ListPlot[Transpose[{Most[binBoundaries], counts}], PlotRange -> All,
Filling -> Axis]
• Yes, but this is a Histogram. The actual data will still have values between 0 and a 100. I wanted a way where the data itself is spread that way. – shivangi Apr 10 at 3:38
• @shivangi, I understand. I edited my answer with a possibility. – David Keith Apr 10 at 4:17
• Thank you so much! This is exactly what I was trying to do. – shivangi Apr 13 at 0:51 | {
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Repeated Eigenvalues | {
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Proof Because x is an eigenvector of A, you know that and can write In cases for which the power method generates a good approximation of a dominant eigenvector, the Rayleigh quotient provides a correspondingly good approximation of the dominant eigenvalue. 7 Today's handouts Lecture 21 notes Tutorial 7 questions 2. Let us focus on the behavior of the solutions when (meaning the future). † Think of repeated eigenvalue case as a bifurcation between 2 distinct real eigenvalue case (2 straight-line solutions) and complex conjugate eigenvalue case (no straight-line solutions. Find more Mathematics widgets in Wolfram|Alpha. Interesting eigenvectors of the Fourier transform Berthold K. ) FINDING EIGENVECTORS • Once the eigenvaluesof a matrix (A) have been found, we can find the eigenvectors by Gaussian Elimination. A symmetric minor of A is a submatrix B obtained by deleting some rows and the corresponding columns. In particular, undamped vibration is governed by. Example Determine if the following matrices are diagonalizable. To find an eigenvector corresponding to an eigenvalue , we write. For repeated diagonal elements, it might not tell you much about the location of the eigenvalues. However, if a matrix has repeated eigenvalues, it is not similar to a diagonal matrix unless it has a full (independent) set of eigenvectors. And the lambda, the multiple that it becomes-- this is the eigenvalue associated with that eigenvector. By looking at these eigenvalues it is possible to get information about a graph that might otherwise be di cult to obtain. Recall the basic result that the roots of a polynomial depend continuously on the coefficients of the polynomial. Since we are going to be working with systems in which $$A$$ is a $$2 \times 2$$ matrix we will make that assumption from the start. The complete case. If x= a+ ibis a complex number, then we let x = a ibdenote its conjugate. The characteristic polynomial P( ) = j I Aj. • What the Hautus Keymann Theorem says is that it | {
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The characteristic polynomial P( ) = j I Aj. • What the Hautus Keymann Theorem says is that it is possible after preliminary state feedback using a matrix F. For very high or very low correlation in DVs, it is not suitable: if DVs are too. When a matrix has no repeated eigenvalues, the eigenvectors are always independent and the eigenvector matrix V diagonalizes the original matrix A if applied as a similarity transformation. This might introduce extra solutions. We can’t find it by elimination. complex eigenvalues. Recall that given a symmetric, positive de nite matrix A we de ne R(x) = xTAx xTx: Here, the numerator and denominator are1 by 1matrices, which we interpret as numbers. Repeated eigenvalues - Duration: 7:30. It is a \repeated eigenvalue," in the sense that the characteristic polynomial (T 1)2 has 1 as a repeated root. Since x ≠ 0, this equation implies λ = 1(Eigenvalue); then, from x = 1 x, every (nonzero) vector is an eigenvector of I. So lambda is an eigenvalue of A. The reason is that the eigenvalues of a given matrix, given by the roots of the characteristic polynomial of the matrix, need not be distinct nor must they necessarily be real. However, ker(B I 2) = ker 0 2 0 0 = span( 1 0 ): Motivated by this example, de ne the geometric multiplicity of an eigenvalue. An eigenvector of Ais a nonzero vector v such that Av = v for some number. MATRIX EXPONENTIALS, and REPEATED EIGENVALUES Systems and Matrix Exponentials : solve x0= Ax, for n nreal A. Repeated Eigenvalues Occasionally when we have repeated eigenvalues, we are still able to nd the correct number of linearly independent eigenvectors. On this site one can calculate the Characteristic Polynomial, the Eigenvalues, and the Eigenvectors for a given matrix. † Think of repeated eigenvalue case as a bifurcation between 2 distinct real eigenvalue case (2 straight-line solutions) and complex conjugate eigenvalue case (no straight-line solutions. Putting all these bases together gives us a list of | {
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eigenvalue case (no straight-line solutions. Putting all these bases together gives us a list of vectors: ~v 1, ~v 2, Lucky Fact 2: The geometric multiplicity of , meaning the dimension of this kernel, is equal to the number of times occurs as a root of f. Free Matrix Eigenvalues calculator - calculate matrix eigenvalues step-by-step. The linearization has a center at the origin, but the. component is being recorded, and then "removed". The study of the relations between eigenvalues and structures in graphs is. This might introduce extra solutions. An eigenvalue of Ais a number such that Av = v for some nonzero vector v. EIGENVALUES OF THE LAPLACIAN AND THEIR RELATIONSHIP TO THE CONNECTEDNESS OF A GRAPH3 (2. When a matrix has no repeated eigenvalues, the eigenvectors are always independent and the eigenvector matrix V diagonalizes the original matrix A if applied as a similarity transformation. 2 λhas a single eigenvector Kassociated to it. The reason is that the eigenvalues of a given matrix, given by the roots of the characteristic polynomial of the matrix, need not be distinct nor must they necessarily be real. To tackle the issue of non-smoothness of repeated eigenvalues, we propose an estimator constructed by averaging all repeated eigenvalues. The coefficients for the principal components are unique (except for a change in sign) if the eigenvalues are distinct and not zero. We solve a problem about eigenvalues of an upper triangular matrix and the square of a matrix. Differential Equations; Slope field; System of Linear DEs Real Repeated Eigenvalues #2; System of Linear DEs Imaginary Eigenvalues;. So that is the end of our lecturer on complex eigenvalues and next lecture were to talk about what to do when you have repeated real eigenvalues. Since our last example and that wraps up our lecture on repeated eigenvalues so, this is the systems of differential equations where we had repeated eigenvalues. (2018) A new method for computation of eigenvector | {
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