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https://math.stackexchange.com/questions/2211713/contradicting-definitions-of-lipschitz-continuity | # Contradicting definitions of Lipschitz Continuity
Certain literature/lecture notes I've found have contradicting definitions of Lipschitz coninuity.
1. Let $F: \mathbb{R}^{m}\rightarrow\mathbb{R}^{m}$. Given an open set $B\subseteq\mathbb{R}^{m}$, $F$ is called Lipschitz continuous on the open set $B$ if $\exists L>0$ s.t. $\|F(\textbf{x}_1)-F(\textbf{x}_2)\|\leq L \|\textbf{x}_1-\textbf{x}_2\|, \forall \textbf{x}_1, \textbf{x}_2 \in B.$
2. Let $A\subset \mathcal{B}$ be a closed subset and $F$ a mapping $F: A\rightarrow A$. $F$ is called Lipschitz continuous on the closed set $A$ if $\exists L>0$ s.t. $\|F(\textbf{x}_1)-F(\textbf{x}_2)\|\leq L \|\textbf{x}_1-\textbf{x}_2\|, \forall \textbf{x}_1, \textbf{x}_2 \in A.$
I'm guessing whether the domain of $F$ is either closed or open does not matter, and also the domain and range of $F$ don't have to be of the same dimension. So, that would mean:
1. Let $V\subseteq \mathbb{R}^{n}$, and $F: V\rightarrow \mathbb{R}^{m}$. $F$ is called Lipschitz continuous on $V$ if $\exists L>0$ s.t. $\|F(\textbf{x}_1)-F(\textbf{x}_2)\|\leq L \|\textbf{x}_1-\textbf{x}_2\|, \forall \textbf{x}_1, \textbf{x}_2 \in V.$
Do you guys agree with definition number 3, and if so, what book backs this up?
Source:
1. "ADE (G1156) Spring 2006 Handout 3: Lipschitz condition and Lipschitz continuity", lecture notes of a Technical University of Eindhoven course, of which I don't know if is freely available for non-TU/e students/employees.
2. Gander, W., Gander, M. J. & Kwok, F. (2014). Scientific Computing: An Introduction using Maple and MATLAB (Vol. 11, Texts in Computational Science and Engineering). Springer International Publishing. - p.243
• Hey whoever made the remark about "$\forall \textbf{x}_{1}$, $\textbf{x}_{2} \in V$ should be $\forall \textbf{x}_{1}$, $\textbf{x}_{2} \in B$" in definition 1.: thank you, you were correct. Your comment got deleted after I edited my post, I did not know this would happen! – Anil Mar 31 '17 at 14:53
Lipschitz continuity can be defined between metric spaces. Let $(X_i,d_i)$ be two metric spaces. A function $f\colon X_1\to X_2$ is Lipschitz continuous if there is a constant $L>0$ such that $$d_2(f(x),f(y))\le L\,d_1(x,y)\quad\forall x,y\in X_1.$$ 1., 2. and 3. are particular cases of this general definition. | 2019-10-19T20:27:08 | {
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https://math.stackexchange.com/questions/2607518/understanding-proof-of-excision-theorem | # understanding proof of excision theorem
I am trying to understand proof of excision theorem in homology following Hatcher's algebraic topology. I came to the following simple question, which I was unable to understand completely.
For a topological space $X$, let $C_n(X)$ denotes the free abelian group on singular $n$-simplices $\sigma:\Delta^n\rightarrow X$ (continuous). Let $A,B$ be subspaces of $X$ such that their interiors cover $X$.
Q.1 Is it true that $C_n(A\cap B)=C_n(A)\cap C_n(B)$ when considered them as subgroups of $C_n(X)$?
According to Hatcher's book, let $C_n(A+B)$ denotes the sums of chains in $A$ and chains in $B$. The second simple question is
Q.2 Does it mean that $C_n(A+B)=C_n(A)+C_n(B)$ where $C_n(A),C_n(B)$ are considered as subgroups of $C_n(X)$?
I came to these questions, because I didn't understand the following statement in Hatcher:
The map $C_n(B)/C_n(A\cap B)\rightarrow C_n(A+B)/C_n(A)$ induced by inclusion is obviously an isomorphism because both quotient groups are free with basis the singular $n$-simplices in $B$ that do not lie in $A$.
(If the answer to both questionsi s yes, then quoted statement is just third isomorphism theorem; but I was wondering whether this justification is correct?)
Q.1 Is it true that $C_n(A\cap B)=C_n(A)\cap C_n(B)$ when considered them as subgroups of $C_n(X)$?
Yes. You can think of an element of $C_n(X)$ as a huge tuple of integers, one for each singular $n$-simplex of $X$ (and with all but finitely many of them being $0$). $C_n(A)$ is then just all such tuples which are $0$ on all singular $n$-simplices which are not contained in $A$. So $C_n(A)\cap C_n(B)$ is all such tuples that are $0$ on all singular simplices that are not contained in $A$ and also $0$ on all singular simplices that are not contained in $B$. That is, they can only be nonzero on singular simplices that are contained in $A\cap B$, so this is the same as $C_n(A\cap B)$.
Q.2 Does it mean that $C_n(A+B)=C_n(A)+C_n(B)$ where $C_n(A),C_n(B)$ are considered as subgroups of $C_n(X)$?
As for the statement in Hatcher, you can understand it using the third isomorphism theorem, but it is worthwhile to understand Hatcher's justification. It is instructive to consider the following toy example. Consider $X=\{0,1,2\}$, with $A=\{0,1\}$ and $B=\{1,2\}$. In this case there are only three singular simplices in $X$, the three constant maps. Let's identify $C_n(X)$ with $\mathbb{Z}^3$, then, with the three coordinates being the coefficients of the three simplices. We then have $C_n(A)=\{(x,y,0):x\in\mathbb{Z}\}$, $C_n(B)=\{(0,y,z):y,z\in\mathbb{Z})$, $C_n(A\cap B)=\{(0,y,0):y\in\mathbb{Z}\}$, and $C_n(A+B)=C_n(X)=\{(x,y,z):x,y,z\in\mathbb{Z}\}$. We can now explicitly see that the quotient $C_n(B)/C_n(A\cap B)$ is isomorphic to $\mathbb{Z}$, by sending the coset of $(0,y,z)\in C_n(B)$ to $z$. Similarly, the quotient $C_n(A+B)/C_n(A)$ is isomorphic to $\mathbb{Z}$, by sending the coset of $(x,y,z)\in C_n(A+B)$ to $z$.
The general story is basically the same. Both $C_n(B)/C_n(A\cap B)$ and $C_n(A+B)/C_n(A)$ are isomorphic to the group you would get by only considering the coordinates of simplices that are in $B$ but not in $A$. | 2019-08-17T15:53:45 | {
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https://math.stackexchange.com/questions/1231694/expected-value-of-the-maximum-number-of-heads-in-n-flips | # Expected Value of the Maximum Number of Heads in n Flips
How would one go about finding the expected value of the maximum number of consecutive heads when flipping a coin $n$ times? For small $n$, it seems easy to brute-force it (i.e. when $n = 3$, the sample space is $\{HHH, HHT, HTH, HTT, TTT, TTH,THT,THH\}$ and so the maximum number of consecutive heads is $\{3,2,1,1,0,1,1,2\}$ so the expected value of the number of maximum consecutive heads should be $11/8$). However, for $n>5$, it becomes pretty hard to brute force this.
For a research project, I am really wondering how the solution to this problem behaves for $50 \leq n \leq 100$. In other words, if I flip $50$ coins, what is the maximum run length of heads that should be expected?
Any advice on how to solve this problem for either the $n$ in the above bound, or for $n$ in general?
• Maybe this can help. – drhab Apr 12 '15 at 18:29
• Thanks. I am not acquainted with Markov chains too well, and am wondering how much error I will have be using the large-sample asymptotic as an approximation to the right answer? – Nikhil Patel Apr 12 '15 at 18:34
• I really don't know. Good luck with it. Cheers. – drhab Apr 12 '15 at 18:38
• You are talking about a 'run' of heads. Google "distribution of maximum run length for heads". – BruceET Apr 12 '15 at 20:15
• The paper csun.edu/~hcmth031/tspolr.pdf of Mark Schilling may help to answer your question – user164118 Apr 13 '15 at 10:31
Suppose we compute the generating function of binary strings having at most $q$ consecutive heads. There are four cases, according to whether the string starts with heads or tails and ends with heads or tails.
We get $$G_{HH}(z) = z\frac{1-z^{q}}{1-z} \sum_{k=0}^\infty \left(\frac{z}{1-z}z\frac{1-z^{q}}{1-z}\right)^k.$$
Continuing we get $$G_{HT}(z) = G_{HH}(z) \frac{z}{1-z}.$$
Furthermore $$G_{TT}(z) = \frac{z}{1-z} \sum_{k=0}^\infty \left(z\frac{1-z^{q}}{1-z}\frac{z}{1-z}\right)^k.$$
Finally we have $$G_{TH}(z) = G_{TT}(z) z\frac{1-z^{q}}{1-z}.$$
The sum term is $$\frac{1}{1-z^2(1-z^q)/(1-z)^2} = \frac{1-2z+z^2}{1-2z+z^2-z^2(1-z^q)} = \frac{1-2z+z^2}{1-2z+z^{q+2}}.$$
The factor on this is $$z\frac{1-z^{q}}{1-z} \left(1+\frac{z}{1-z}\right) + \frac{z}{1-z} \left(1+z\frac{1-z^{q}}{1-z}\right)$$ which is $$z\frac{1-z^{q}}{(1-z)^2} + \frac{z}{(1-z)^2} (1-z^{q+1}) = \frac{2z-z^{q+1}-z^{q+2}}{(1-z)^2}.$$
Multiplying we obtain the generating function $$G_q(z) = \frac{2z-z^{q+1}-z^{q+2}}{1-2z+z^{q+2}}.$$
It follows that the expectation times $2^n$ is given by
$$[z^n] \left(0\times G_0(z) + \sum_{q=1}^n q (G_q(z)-G_{q-1}(z)) \right).$$
The sum simplifies to $$\sum_{q=1}^n q G_q(z) - \sum_{q=0}^{n-1} (q+1) G_q(z) = \sum_{q=0}^n q G_q(z) - \sum_{q=0}^{n-1} (q+1) G_q(z) \\ = n G_n(z) - \sum_{q=0}^{n-1} G_q(z).$$
and hence the expectation is $$\frac{1}{2^n} [z^n] \left( n G_n(z) - \sum_{q=0}^{n-1} G_q(z) \right).$$
This gives the sequence $$1/2,1,{\frac {11}{8}},{\frac {27}{16}},{\frac {31}{16}},{ \frac {69}{32}},{\frac {75}{32}},{\frac {643}{256}},{\frac { 1363}{512}},{\frac {1433}{512}},\ldots$$
Multiplying by $2^n$ we obtain $$1, 4, 11, 27, 62, 138, 300, 643, 1363, 2866, \ldots$$ which is OEIS A119706 where the above computation is confirmed.
The following Maple code can be used to explore these generating functions. The procedure v computes the generating function of the maximal run length of a string of $n$ bits by total enumeration. The procedure w computes it from the generating function $G_q(z).$
v :=
proc(n)
option remember;
local gf, k, d, mxrun, len;
gf := 0;
for k from 2^n to 2^(n+1)-1 do
d := convert(k, base, 2);
mxrun := 0;
for pos to n do
if d[pos] = 1 then
len := 1;
pos := pos+1;
while pos <= n do
if d[pos] = 1 then
len := len+1;
pos := pos+1;
else
break;
fi;
od;
if len>mxrun then
mxrun := len;
fi;
fi;
od;
gf := gf + z^mxrun;
od;
gf;
end;
G := q -> (2*z-z^(q+1)-z^(q+2))/(1-2*z+z^(q+2));
w :=
proc(n)
option remember;
local gf, mxrun;
gf := 1;
for mxrun to n do
gf := gf +
coeftayl(G(mxrun)-G(mxrun-1), z=0, n)*z^mxrun;
od;
gf;
end;
X := n -> coeftayl(n*G(n)-add(G(q), q=0..n-1), z=0, n)/2^n;
Here are two examples.
> v(4);
4 3 2
z + 2 z + 5 z + 7 z + 1
> w(4);
4 3 2
z + 2 z + 5 z + 7 z + 1
Addendum. Responding to the question of the OP, the maximum run length distribution for $n=50$ is
> w(50);
50 49 48 47 46 45 44 43 42
z + 2 z + 5 z + 12 z + 28 z + 64 z + 144 z + 320 z + 704 z
41 40 39 38 37 36
+ 1536 z + 3328 z + 7168 z + 15360 z + 32768 z + 69632 z
35 34 33 32 31
+ 147456 z + 311296 z + 655360 z + 1376256 z + 2883584 z
30 29 28 27 26
+ 6029312 z + 12582912 z + 26214400 z + 54525952 z + 113246208 z
25 24 23 22
+ 234881024 z + 486539259 z + 1006632909 z + 2080374408 z
21 20 19 18
+ 4294964912 z + 8858356224 z + 18253535488 z + 37580568576 z
17 16 15 14
+ 77307408384 z + 158903894017 z + 326369607799 z + 669786836360 z
13 12 11
+ 1373319005440 z + 2812533538048 z + 5749650288420 z
10 9 8
+ 11716183298140 z + 23723022576779 z + 47402584528885 z
7 6 5
+ 92066138963408 z + 168050756947888 z + 267156803852044 z
4 3 2
+ 310228979841119 z + 174887581402185 z + 19394019617001 z
+ 32951280098 z + 1 | 2019-05-25T22:50:49 | {
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https://mathhelpboards.com/threads/ibv5-the-vectors-u-v-are-given-by-u-3i-5j-v-i-%E2%80%93-2j.6104/ | # [SOLVED]IBV5 The vectors u, v are given by u = 3i + 5j, v = i – 2j
#### karush
##### Well-known member
The vectors $u, v$ are given by $u = 3i + 5j, v = i – 2j$
Find scalars $a, b$ such that $a(u + v) = 8i + (b – 2)j$
$(u+v)=4i+3j$
in order to get the $8i$ let $a=2$
then $2(4i+3j)=8i+6j$
in order to get $6j$ let $b=8$ then $(8-2)j=6j$
so $a=2$ and $b=8$
I am not sure of the precise definition of what scalar means here...at least with vectors
#### tkhunny
##### Well-known member
MHB Math Helper
vectors u,v are given by u=3i+5j,v=i–2j
Find scalars a,b such that a(u+v)=8i+(b–2)j
u + v = <4,3>
Then <4a,3a> = <8,(b-2)>
4a = 8
3a = b-2
#### eddybob123
##### Active member
I am not sure of the precise definition of what scalar means here...at least with vectors
The definition is the same with everything; basically a number. Whether it's restricted to real or complex numbers is usually clear from context.
#### karush
##### Well-known member
vectors u,v are given by u=3i+5j,v=i–2j
Find scalars a,b such that a(u+v)=8i+(b–2)j
u + v = <4,3>
Then <4a,3a> = <8,(b-2)>
4a = 8
3a = b-2
yes, that looks like a much better way to solve it. especially if it gets a lot more complicated. | 2021-03-05T13:44:34 | {
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"openwebmath_score": 0.8299294114112854,
"openwebmath_perplexity": 1508.584651032685,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9766692271169855,
"lm_q2_score": 0.853912760387131,
"lm_q1q2_score": 0.8339903157126308
} |
https://stats.stackexchange.com/questions/21825/probability-of-a-run-of-k-successes-in-a-sequence-of-n-bernoulli-trials | # Probability of a run of k successes in a sequence of n Bernoulli trials
I'm trying to find the probability of getting 8 trials in a row correct in a block of 25 trials, you have 8 total blocks (of 25 trials) to get 8 trials correct in a row. The probability of getting any trial correct based on guessing is 1/3, after getting 8 in a row correct the blocks will end (so getting more than 8 in a row correct is technically not possible). How would I go about finding the probability of this occurring? I've been thinking along the lines of using (1/3)^8 as the probability of getting 8 in a row correct, there are 17 possible chances to get 8 in a row in a block of 25 trials, if I multiply 17 possibilities * 8 blocks I get 136, would 1-(1-(1/3)^8)^136 give me the likelihood of getting 8 in a row correct in this situation or am I missing something fundamental here?
• I believe the problem with the argument given is that the events considered are not independent. For example, consider a single block. If I tell you that (a) there is no run of eight that starts at position 6, (b) there is a run starting at position 7 and (c) there is no run starting at position 8, what does that tell you about the probability of a run starting at positions, say, 9 through 15? – cardinal Jan 28 '12 at 15:05
By keeping track of things you can get an exact formula.
Let $p=1/3$ be the probability of success and $k=8$ be the number of successes in a row you want to count. These are fixed for the problem. Variable values are $m$, the number of trials left in the block; and $j$, the number of successive successes already observed. Let the chance of eventually achieving $k$ successes in a row before $m$ trials are exhausted be written $f_{p,k}(j,m)$. We seek $f_{1/3,8}(0,25)$.
Suppose we have just seen our $j^\text{th}$ success in a row with $m\gt0$ trials to go. The next trial is either a success, with probability $p$--in which case $j$ is increased to $j+1$--; or else it is a failure, with probability $1-p$--in which case $j$ is reset to $0$. In either case, $m$ decreases by $1$. Whence
$$f_{p,k}(j,m) = p f_{p,k}(j+1,m-1) + (1-p)f_{p,k}(0,m-1).$$
As starting conditions we have the obvious results $f_{p,k}(k,m)=1$ for $m \ge 0$ (i.e., we have already seen $k$ in a row) and $f_{p,k}(j,m)=0$ for $k-j \gt m$ (i.e., there aren't enough trials left to get $k$ in a row). It is now fast and straightforward (using dynamic programming or, because this problem's parameters are so small, recursion) to compute
$$f_{p,8}(0,25) = 18p^8 - 17p^9 - 45p^{16} + 81p^{17}-36p^{18}.$$
When $p=1/3$ this yields $80897 / 43046721 \approx 0.0018793$.
Relatively fast R code to simulate this is
hits8 <- function() {
x <- rbinom(26, 1, 1/3) # 25 Binomial trials
x[1] <- 0 # ... and a 0 to get started with diff
if(sum(x) >= 8) { # Are there at least 8 successes?
max(diff(cumsum(x), lag=8)) >= 8 # Are there 8 successes in a row anywhere?
} else {
FALSE # Not enough successes for 8 in a row
}
}
set.seed(17)
mean(replicate(10^5, hits8()))
After 3 seconds of calculation, the output is $0.00213$. Although this looks high, it's only 1.7 standard errors off. I ran another $10^6$ iterations, yielding $0.001867$: only $0.3$ standard errors less than expected. (As a double-check, because an earlier version of this code had a subtle bug, I also ran 400,000 iterations in Mathematica, obtaining an estimate of $0.0018475$.)
This result is less than one-tenth the estimate of $1-(1-(1/3)^8)^{136} \approx 0.0205$ in the question. But perhaps I have not fully understood it: another interpretation of "you have 8 total blocks ... to get 8 trials correct in a row" is that the answer being sought equals $1 - (1 - f_{1/3,8}(0,25))^8) = 0.0149358...$.
While @whuber's excellent dynamic programming solution is well worth a read, its runtime is $\mathcal O(k^2m)$ with respect to total number of trials $m$ and the desired trial length $k$ whereas the matrix exponentiation method is $\mathcal O(k^3\log(m))$. If $m$ is much larger than $k$, the following method is faster.
Both solutions consider the problem as a Markov chain with states representing the number of correct trials at the end of the string so far, and a state for achieving the desired correct trials in a row. The transition matrix is such that seeing a failure with probability $p$ sends you back to state 0, and otherwise with probability $1-p$ advances you to the next state (the final state is an absorbing state). By raising this matrix to the $n$th power, the value in the first row, and last column is the probability of seeing $k=8$ heads in a row. In Python:
import numpy as np
a = np.zeros((want + 1, want + 1))
for i in range(want):
a[i, 0] = 1 - p
a[i, i + 1] = p
a[want, want] = 1.0
return np.linalg.matrix_power(a, flips)[0, want]
yields 0.00187928367413 as desired.
According to this answer, I will explain the Markov-Chain approach by @Neil G a bit more and provide a general solution to such problems in R. Let's denote the desired number of correct trials in a row by $k$, the number of trials as $n$ and a correct trial by $W$ (win) and an incorrect trial by $F$ (fail). In the process of keeping track of the trials, you want to know whether you already had a streak of 8 correct trials and the number of correct trials at the end of your current sequence. There are 9 states ($k+1$):
$A$: We have not had $8$ correct trials in a row yet, and the last trial was $F$.
$B$: We have not had $8$ correct trials in a row yet, and the last two trials were $FW$.
$C$: We have not had $8$ correct trials in a row yet, and the last three trials were $FWW$.
$\ldots$
$H$: We have not had $8$ correct trials in a row yet, and the last eight trials were $FWWWWWWW$.
$I$: We've had $8$ correct trials in a row!
The probability of moving to state $B$ from state $A$ is $p=1/3$ and with probability $1-p=2/3$ we stay in state $A$. From state $B$, the probability of moving to state $C$ is $1/3$ and with probability $2/3$ we move back to $A$. And so on. If we are in state $I$, we stay there.
From this, we can construct a $9\times9$ transition matrix $M$ (as each column of $M$ sums to $1$ and all entries are positive, $M$ is called a left stochastic matrix):
$$M= \begin{pmatrix} 2/3 & 2/3 & 2/3 & 2/3 & 2/3 & 2/3 & 2/3 & 2/3 & 0 \\ 1/3 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1/3 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1/3 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1/3 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1/3 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1/3 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1/3 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1/3 & 1 \end{pmatrix}$$
Each column and row corresponds to one state. After $n$ trials, the entries of $M^{n}$ give the probability of getting from state $j$ (column) to state $i$ (row) in $n$ trials. The rightmost column corresponds to the state $I$ and the only entry is $1$ in the right lower corner. This means that once we are in state $I$, the probability to stay in $I$ is $1$. We are interested in the probability of getting to state $I$ from state $A$ in $n=25$ steps which corresponds to the lower left entry of $M^{25}$ (i.e. $M^{25}_{91}$). All we have to do now is calculating $M^{25}$. We can do that in R with the matrix power function from the expm package:
library(expm)
k <- 8 # desired number of correct trials in a row
p <- 1/3 # probability of getting a correct trial
n <- 25 # Total number of trials
# Set up the transition matrix M
M <- matrix(0, k+1, k+1)
M[ 1, 1:k ] <- (1-p)
M[ k+1, k+1 ] <- 1
for( i in 2:(k+1) ) {
M[i, i-1] <- p
}
# Name the columns and rows according to the states (A-I)
colnames(M) <- rownames(M) <- LETTERS[ 1:(k+1) ]
round(M,2)
A B C D E F G H I
A 0.67 0.67 0.67 0.67 0.67 0.67 0.67 0.67 0
B 0.33 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0
C 0.00 0.33 0.00 0.00 0.00 0.00 0.00 0.00 0
D 0.00 0.00 0.33 0.00 0.00 0.00 0.00 0.00 0
E 0.00 0.00 0.00 0.33 0.00 0.00 0.00 0.00 0
F 0.00 0.00 0.00 0.00 0.33 0.00 0.00 0.00 0
G 0.00 0.00 0.00 0.00 0.00 0.33 0.00 0.00 0
H 0.00 0.00 0.00 0.00 0.00 0.00 0.33 0.00 0
I 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.33 1
# Calculate M^25
Mn <- M%^%n
Mn[ (k+1), 1 ]
[1] 0.001879284
The probability of getting from state $A$ to state $I$ in 25 steps is $0.001879284$, as established by the other answers.
Here is some R code that I wrote to simulate this:
tmpfun <- function() {
x <- rbinom(25, 1, 1/3)
rx <- rle(x)
any( rx$lengths[ rx$values==1 ] >= 8 )
}
tmpfun2 <- function() {
any( replicate(8, tmpfun()) )
}
mean(replicate(100000, tmpfun2()))
I am getting values a little smaller than your formula, so one of us may have made a mistake somewhere.
• Does your function include trials where it is impossible to get 8 in a row right, e.g. where the "run" started on trial 20? – Michelle Jan 27 '12 at 21:03
• Most likely me, my R simulation is giving me smaller values as well. I'm just curious if there is an algebraic solution to solve this as a simple probability issue in case someone disputes a simulation. – AcidNynex Jan 28 '12 at 0:48
• I think this answer would be improved by providing the output you obtained so that it can be compared. Of course, including something like a histogram in addition would be even better! The code looks right to me at first glance. Cheers. :) – cardinal Jan 28 '12 at 15:06
Here is a Mathematica simulation for the Markov chain approach, note that Mathematica indexes by $1$ not $0$:
M = Table[e[i, j] /. {
e[9, 1] :> 0,
e[9, 9] :> 1,
e[_, 1] :> (1 - p),
e[_, _] /; j == i + 1 :> p,
e[_, _] :> 0
}, {i, 1, 9}, {j, 1, 9}];
x = MatrixPower[M, 25][[1, 9]] // Expand
This would yield the analytical answer: $$18 p^8 - 17 p^9 - 45 p^{16} + 81 p^{17} - 36 p^{18}$$
Evaluating at $p=\frac{1.0}{3.0}$
x /. p -> 1/3 // N
Will return $0.00187928$
This can also be evaluated directly using builtin Probability and DiscreteMarkovProcess Mathematica functions:
Probability[k[25] == 9, Distributed[k, DiscreteMarkovProcess[1, M /. p -> 1/3]]] // N
Which will get us the same answer: $0.00187928$ | 2018-08-20T11:01:25 | {
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https://math.stackexchange.com/questions/259563/how-to-prove-this-mean-value-property-int-ab-ftgtdt-fx-int-ab-gtdt | # How to prove this mean value property$\int_a^b f(t)g(t)dt=f(x)\int_a^b g(t)dt$?
Suppose $f$ and $g$ are continuous functions on $[a,b]$ and that $g(x)\ge 0$ for all $x\in[a,b]$. Prove that there exists $x$ in $[a,b]$ such that $$\int_a^b f(t)g(t)dt=f(x)\int_a^b g(t)dt$$
I think I need to do something with this theorem:
Intermediate Value Theorem for Integrals
If $f$ is a continuous function on $[a,b]$ then for at least one $x$ in $[a,b]$ we have $$f(x)=\frac{1}{b-a} \int_a^bf$$
Let $m = \min \{f(x): x \in [a,b]\}$ and $M = \max \{f(x): x \in [a,b]\}$.
First let us assume that $\displaystyle \int_a^b g(x)dx > 0$. Then we have that $$m \leq \dfrac{\displaystyle \int_a^b f(x) g(x) dx}{\displaystyle \int_a^b g(x)dx} \leq M$$ Now use intermediate value theorem to get what you want.
If $\displaystyle \int_a^b g(x) dx = 0$ and since $g(x) \geq 0$ and is continuous, we have that $g(x) = 0$ on $[a,b]$. Hence, $$\displaystyle \int_a^b f(x) g(x) dx = \displaystyle \int_a^b g(x)dx =0$$ Hence, $$\displaystyle \int_a^b f(x) g(x) dx = f(t) \displaystyle \int_a^b g(x)dx$$ for any $t \in [a,b]$.
• Note that $g(x)\geq 0$, not $g(x)>0$. Dec 15 '12 at 22:55
• @MarioCarneiro Thanks. Have updated it.
– user17762
Dec 15 '12 at 23:14
• I don't understand how you get this: $$m \leq \dfrac{\displaystyle \int_a^b f(x) g(x) dx}{\displaystyle \int_a^b g(x)dx} \leq M$$ Dec 15 '12 at 23:16
• @Kasper since $m \leq f(x) \leq M$, we have $m g(x) \leq f(x) g(x) \leq M g(x)$ as $g(x) \geq 0$. Now integration preserves $\leq$ and $\geq$. Hence, you get I have written.
– user17762
Dec 15 '12 at 23:19
• Since $g(x)\geq 0$, $f(x)g(x)\leq Mg(x)$, so $\int_a^bf(x)g(x)\,dx\leq\int_a^bMg(x)\,dx=M\int_a^bg(x)\,dx$. Thus $\dfrac{\int_a^bf(x)g(x)\,dx}{\int_a^bg(x)\,dx}\leq M$. A similar argument holds for $m$. (Edit: Ninja'd!) Dec 15 '12 at 23:20
Define $$\bar{f}=\frac{\int_a^bf(t)\,g(t)\,\mathrm{d}t}{\int_a^bg(t)\,\mathrm{d}t}\tag{1}$$ Then $$\int_a^b\left(f(t)-\bar{f}\right)\,g(t)\,\mathrm{d}t=0\tag{2}$$ Suppose that $f(t_+)-\bar{f}\gt0$ and $f(t)-\bar{f}\ge0$ for all $t\in[a,b]$, then $f-\bar{f}$ is positive in some neighborhood of $t^+$ and therefore $\int_a^b\left(f(t)-\bar{f}\right)\,g(t)\,\mathrm{d}t\gt0$. Thus, there must be some $t_-$ where $f(t_-)-\bar{f}\lt0$.
Suppose that $f(t_-)-\bar{f}\lt0$ and $f(t)-\bar{f}\le0$ for all $t\in[a,b]$, then $f-\bar{f}$ is negative in some neighborhood of $t_-$ and therefore $\int_a^b\left(f(t)-\bar{f}\right)\,g(t)\,\mathrm{d}t\lt0$. Thus, there must be some $t_+$ where $f(t_+)-\bar{f}\gt0$.
Thus, if $f(t)-\bar{f}$ is not identically $0$ on $[a,b]$, we must have $t_+$ and $t_-$ where $f(t_+)-\bar{f}\gt0$ and $f(t_-)-\bar{f}\lt0$.
By the intermediate value theorem, there must be an $x$ between $t_+$ and $t_-$ such that $f(x)-\bar{f}=0$, which is the same as $$\int_a^bf(t)\,g(t)\,\mathrm{d}t=\bar{f}\int_a^bg(t)\,\mathrm{d}t=f(x)\int_a^bg(t)\,\mathrm{d}t\tag{3}$$
Let $h(x)$ be an antiderivative of $g(x)$, so that $h'(x)=g(x)$. Then using the substitution $$u=h(t)\Rightarrow du=g(t)\,dt$$ we get $$\int_{h^{-1}(a)}^{h^{-1}(b)}f(t)\,du=f(x)\int_{h^{-1}(a)}^{h^{-1}(b)}du=f(x)(h^{-1}(b)-h^{-1}(a))$$ which we can validate using the IVT for integrals.
• Unfortunately, this proof only works for $g(t)>0$ on $[a,b]$, or at best $g(t)=0$ at countably many points, since it relies on $h^{-1}(x)$, which may not exist if $h$ is not 1-1, since it may not be strictly increasing. Marvis' proof is more general. Dec 15 '12 at 23:07 | 2021-10-22T03:20:40 | {
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https://math.stackexchange.com/questions/1874340/calculate-the-following-integral | # Calculate the following integral
$$\int_{[0,1]^n} \max(x_1,\ldots,x_n) \, dx_1\cdots dx_n$$
My work:
I know that because all $x_k$ are symmetrical I can assume that $1\geq x_1 \geq \cdots \geq x_n\geq 0$ and multiply the answer by $n!$ so we get that $\max(x_1\ldots,x_n)=x_1$ and the integral that we want to calculate is $n!\int_0^1 x_1 \, dx_1 \int_0^{x_1}dx_2\cdots\int_0^{x_{n-1}} \, dx_n$ and now it should be easier but I'm stuck..
Can anyone help?
• Try the case $n=2$ first. – Jean Marie Jul 28 '16 at 21:21
One can see by induction that:
$$\int_0^1 x_1 dx_1 \int_0^{x_2 } dx_2 \cdots = \int_0^1 x_1 \frac{x_{n - (n-1)}^{n-1}}{(n-1)!} dx_1 =\frac{1}{(n+1) \times (n-1)!}$$
Therefore, the original integral is:
$$n! \times \frac1{(n+1) \times (n-1)!} = \frac{n}{n+1}$$
This is an alternative approach.
Let $X_i$ ($i=1,\cdots , n$) be independent uniform random variable in $[0,1]$.
What is the PDF of $M=\max (X_1, \cdots, X_n)$?
Then what is $\mathbf{E}[M]$?
• Ah, a probabilistic approach. Nice. – Brevan Ellefsen Jul 28 '16 at 22:33
• @BrevanEllefsen: knowing the properties of Beta distributions make this rather easy – Henry Jul 29 '16 at 8:22
Note that $\max(x_1,x_2,\dots,x_n)=\max(x_1,\max(x_2,\dots,x_n))$. In addition, note that
\begin{align} \int_0^1\int_0^1\max(x,y)\,dx\,dy&=\int_0^1\left(\int_0^y y\,dx+\int_y^1 x\,dx\right)\,dy\\\\ &=\int_0^1 \left(\frac12+\frac12 y^2\right)\,dy \end{align}
Now, we can write
\begin{align} \int_0^1 \max(x_1,x_2,\dots,x_n)\,dx_1&=\int_0^1 \max(x_1,\max(x_2,\dots,x_n))\,dx_1\\\\ &=\int_0^{\max(x_2,\dots,x_n)}\max(x_2,\dots,x_n)\,dx_1+\int_{\max(x_2,\dots,x_n)}^1x_1\,dx_1\\\\ &=\frac12 +\frac12\left(\max(x_2,\dots,x_n)\right)^2 \end{align}
Then, observe that
\begin{align} \frac{1}{k}\int_0^1 \left(\max(x_k,\dots ,x_n)\right)^{k}\,dx_k&=\int_0^{\max(x_{k+1},\dots ,x_n)}\left(\max(x_{k+1},\dots ,x_n)\right)^k\,dx_k+\int_{\max(x_{k+1},\dots ,x_n)}^1 x_k^k\,dx_k\\\\ &=\frac{1}{k}\left(\frac{k}{k+1}\left(\max(x_{k+1},\dots,x_n)\right)^{k+1}+\frac{1}{k+1}\right)\\\\ &=\frac{1}{k(k+1)}+\frac{1}{k+1}\left(\max(x_{k+1},\dots,x_n)\right)^{k+1} \end{align}
Proceeding inductively, we find that
\begin{align} \int_0^1\cdots \int_0^1 \max(x_1,x_2,\dots,x_n)\,dx_1\cdots dx_n&=\frac1{(2)(1)}+\frac{1}{(3)(2)}+\frac{1}{(4)(3)}+\cdots +\frac{1}{(n+1)(n)}\\\\ &=\sum_{k=1}^n \frac{1}{k(k+1)}\\\\ &=\sum_{k=1}^{n}\left(\frac{1}{k}-\frac{1}{k+1}\right)\\\\ &=\frac{n}{n+1} \end{align}
• +1. It's an elegant proof. Given the quite simple and unexpected result, we always think it must be some simple or, lets say, 'direct procedure'. This 'theorem' doesn't seem to be true, anyway. – Felix Marin Jul 29 '16 at 1:43
• @felixmarin Thank you! And very much appreciated as always. -Mark – Mark Viola Jul 29 '16 at 12:18
This is pretty straightforward using probabilistic methods: What you're looking for is
$$E[\max_{1 \le i \le n} X_i] = \int_{[0,1]^n} \max x_i \ \ dx_1\dots dx_n$$
where $X_i$ are iid uniformly distribuited on $[0,1]$.
Now call $Y = \max_i X_i$; its distirbution function is given by
$$F_Y(x) = P(\max X_i \le x) = \prod_i P(X_i \le x) = x^n \ \ \ \text{ x \in [0,1]}$$
Hence $Y$ is absolutely continuos and its density is given by $$f_Y(x) = nx^{n-1}1_{[0,1]}(x)$$
Hence we find that
$$E[\max X_i] = E[Y] = \int_\mathbb R xf_Y(x) dx = \int_0^1 nx^n dx = \frac n{n+1}$$ | 2019-07-15T18:17:00 | {
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https://math.stackexchange.com/questions/2408902/conversion-to-base-9 | # Conversion to base $9$
Where am I going wrong in converting $397$ into a number with base $9$?
$397$ with base $10$ $$= 3 \cdot 10^2 + 9 \cdot 10^1 + 7 \cdot 10^0$$ To convert it into a number with base $9$ $$3 \cdot 9^2 + 9 \cdot 9^1 + 7 \cdot 9^0 = 243 + 81 + 7 = 331$$ But answer is $481$?????
• You started with a number in base $9$ and converted it to base $10$ rather than the reverse. – N. F. Taussig Aug 28 '17 at 15:51
• To convert to base $9$ from base $10$, you could find the highest power of $9$ (e.g. $9,81,729$ ...) that is less than your number, then the largest multiple of that power that is still less than your number will be your left-most digit. So, in the case of $397_{10}$, the largest power of $9$ that will fit is $81$ (since $729$ is too large), and you can fit four $81$s: $4 \cdot 81=324< 397$ (five is too many, since $5 \cdot 81 = 405 > 397$. This means that the leftmost digit must be $4$. We are left with $397 - 4 \cdot 81 =73$. Now, find the largest multiple of $9$ that's less than $73$, etc. – Zubin Mukerjee Aug 28 '17 at 15:56
• The operation you've done is convert $(397)_9$ to base $10$. – Thomas Andrews Aug 28 '17 at 16:04
• One obvious sanity check is that if converting from a bigger base to a smaller number you should get a larger answer and visa versa. You can easily check your answer by converting it back to decimal to see if you get what you started with. – Warren Hill Aug 28 '17 at 16:20
$481_{[9]} = 4\,\cdot\,81 +8\,\cdot9+1\,\cdot\,1 = 324 + 72 + 1 = 397_{[10]}$
The conversion you presented is incorrect, you converted $397_{[9]}$ into $331_{[10]}$, which is not what you want.
To convert it properly:
$397/9 = 44$ (remainder = $1$)
$44/9 = 4$ (remainder = $8$)
$4/9 = 0$ (remainder = $4$)
The remainders give the result: $\boxed{397_{[10]}=481_{[9]}}$
You cannot simply keep the same coefficients but change the $10$'s to $9$'s. What you must do is something like the following: \begin{align}397&=3\times \color{red}{10}^2+9\times\color{red} {10}^1+7\times \color{red}{10}^0\\&=3\times(9+1)^2+9\times(9+1)+7\times 1\\&=3\times\color{blue}9^2+(3\times18+3)+\color{blue}9^2+\color{blue}9^1+(7)\\&=4\times \color{blue}9^2+(6\times9)+(10)+\color{blue}9^1\\&=4\times \color{blue}9^2+8\times \color{blue}9^1+1\times \color{blue}9^0\end{align}
What you did was simply change the tens in red to nines without doing any intermediate steps.
• I love that the three answers right now are all using different methods to convert bases :) – Zubin Mukerjee Aug 28 '17 at 16:03
• I had not seen this method before but it's a nice approach. I would have used repeated division and took the remainders as @DanielCunha did because that's the way I was taught to convert decimal to binary but it works with any base. – Warren Hill Aug 28 '17 at 16:08
• @WarrenHill I sometimes use this approach, especially when the bases are so close together (like $9$ and $10$ are only $1$ apart, so you only pick up small numbers with the bracket expansion). – Dave Aug 28 '17 at 16:10
• @Warren We can do this much more efificiently if we exploit the recursive Horner form of radix notation - see my answer. – Bill Dubuque Aug 28 '17 at 16:41
The problem is that
$$(397)_{10} = 3 \cdot 10^2 + 9 \cdot 10^1 + 7 \cdot 10^0 \ne 3 \cdot 9^2 + 9 \cdot 9^1 + 7 \cdot 9^0 = (397)_9.$$
What we are trying to do is solve
$$(397)_{10} = 3 \cdot 10^2 + 9 \cdot 10^1 + 7 \cdot 10^0 = a \cdot 9^2 + b \cdot 9^1 + c \cdot 9^0 = (abc)_9$$
To find $a$ we take as many copies of $81$ from $397$ as we can:
$$397 - 4 \cdot 81 = 73.$$
Thus $a = 4$. Next, we take as many copies of $9$ from $73$ as we can:
$$73 - 8 \cdot 9 = 1,$$
so $b = 1$. Lastly, we take as many copies of $1$ from $1$ as we can:
$$1 - 1 \cdot 1 = 0.$$
Hence $c = 1$. Therefore the answer is $$(397)_{10} = (481)_9.$$
• Nice answer, +1! I think you should mention that $9^3 = 729>397$ is too large to be considered, which is why you are starting from $9^2 = 81$, instead of a higher power of $9$. Cheers – Zubin Mukerjee Aug 28 '17 at 16:02
Here is a layout of the conversion algorithm (successive Euclidean divisions) $$\begin{array}{ccrcrc*{10}{c}} &&44&&4 \\ 9&\Bigl)&397&\Bigl)&44&\Bigl)& \color{red}{\mathbf 4}\\ &&\underline{36}\phantom{7}&&\underline{36}\\ &&37&&\color{red}{\mathbf 8} \\ &&\underline{36} \\ &&\color{red}{\mathbf 1} \end{array}$$ This is based on Horner's scheme: \begin{align} [397]_{10}&=9\cdot 44+ \color{red}1=9(9\cdot 4+\color{red}8)+\color{red}1=9^2\cdot\color{red}4+9\cdot\color{red}8+1\cdot \color{red}1. \end{align}
• But no division is needed if we use Horner optimally - see my answer. – Bill Dubuque Aug 28 '17 at 16:39
• That''s fine but it's no so easy in practice to do operations in another base, especially if you have to carry digits. – Bernard Aug 28 '17 at 17:14
As I explained in 2011, one easy way is to write the number in Horner form in the original base, then do the base conversion from the inside-out, e.g. below where $\rm\color{#c00}{red}$ means radix $9$ notation
\begin{align} 397 \,&=\, (3\cdot 10\, +\, 9)\,10 +7\\ &=\, (\color{#c00}{3\cdot 11+10})10+7\\ &=\qquad\quad\ \ \color{#c00}{43\cdot 11}+7\\ &=\qquad\qquad\ \ \ \color{#c00}{473}+7\\ &=\qquad\qquad\ \ \ \color{#c00}{481}\end{align} | 2019-09-17T13:21:13 | {
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https://math.stackexchange.com/questions/895406/how-to-find-a-basis-of-an-image-of-a-linear-transformation | # How to find a basis of an image of a linear transformation?
I apologize for asking a question though there are pretty much questions on math.stackexchange with the same title, but the answers on them are still not clear for me.
I have this linear operator:
$$Ax = (2x_1-x_2-x_3, x_1-2x_2+x_3, x_1+x_2-2x_3);$$
And I need to find the basis of the kernel and the basis of the image of this transformation.
First, I wrote the matrix of this transformation, which is:
$$\begin{pmatrix} 2 & -1 & -1 \\ 1 & -2 & 1 \\ 1 & 1 & -2\end{pmatrix}$$
I found the basis of the kernel by solving a system of 3 linear equations:
$$\begin{pmatrix} 2 & -1 & -1 \\ 1 & -2 & 1 \\ 1 & 1 & -2\end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3\end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0\end{pmatrix}$$
It is
$$kerA = (1,1,1)$$
But how can I find the basis of the image? What I have found so far is that I need to complement a basis of a kernel up to a basis of an original space. But I do not have an idea of how to do this correctly. I thought that I can use any two linear independent vectors for this purpose, like
$$imA = \{(1,0,0), (0,1,0)\}$$
because the image here is $\mathbb{R}^2$
But the correct answer from my textbook is:
$$imA = \{(2,1,1), (-1,2,1)\}$$
And by the way I cannot be sure that there is no error in the textbook's answer.
So could anyone help me with this. I will be very grateful, thank you in advance.
• There are many bases for the image. To get one such, find what $A$ does to the standard basis, and throw away the linearly dependent one. – André Nicolas Aug 12 '14 at 18:51
• A basis of the image is the columns in the original matrix which correspond to the pivot columns in the row reduced matrix. So presumably the first and second columns of your row reduced matrix are pivot columns, so the first two columns of your original matrix are a basis. There may be a sign error in the answer of your book for the $2$ in the second basis vector. That's just my suspicion, I haven't actually worked it out. – BW. Aug 12 '14 at 18:53
• @BenWest thanks to you and André Nicolas, now it seems clear to me. – Dmitry Koroliov Aug 12 '14 at 19:03
Reducing your matrix $$\begin{pmatrix} 2 & -1 & -1 \\ 1 & -2 & 1 \\ 1 & 1 & -2\end{pmatrix}$$ to row-echelon form gives $$\begin{pmatrix} 1 & -2 & 1\\0 & 1 & -1\\0 & 0 & 0\end{pmatrix},$$
and a basis for the image of $A$ is given by a basis for the column space of your matrix, which we can get by taking the columns of the matrix corresponding to the leading 1's in any row-echelon form.
This gives the basis $\{(2,1,1), (-1,-2,1)\}$ for the image of $A$.
• Actually does this work for any A or can it happen that the column we pick are linearly dependent? – user519338 Mar 29 '18 at 23:50
• @user519338 This method should work for any matrix A. – user84413 Jul 7 '18 at 20:45
• Do you mind explaining a little on "leading 1's" in any rref form? What does a leading 1 mean? – Car Feb 26 at 9:46
The image of a basis of the domain gives a spanning set for the image. You may have to reduce this spanning set to get a basis for the image.
• could you please provide an example? – Dmitry Koroliov Aug 12 '14 at 18:38
• Let's say you had a linear transformation $T:\mathbb{R}^2\rightarrow\mathbb{R}^3$ Suppose also that $T(1,0)=(3,7,1)$ and $T(0,1)=(-6,-14,-2)$. Since any vector in $\mathbb{R}^2$ can be written as a linear combination of $(1,0)$ and $(0,1)$, using the linear transformation properties, we have that the image of any $\mathbb{R}^2$ vector is a linear combination of $(3,7,1)$ and $(-6,-14,-2)$. Thus these two vectors span the image of $T$. In this case, these image vectors are linearly dependent, so you would have to get rid of one of them to get a basis. – paw88789 Aug 12 '14 at 18:51 | 2021-04-13T13:20:00 | {
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http://openstudy.com/updates/4f4d4ac9e4b0acf2d9fe177b | ## ggrree Group Title Let R be the region in the first quadrant enclosed between the x-axis and the curve y = x - x^2. There is a straight line passing through the origin which cuts the region R into two regions so that the area of the lower region is 7 times the area of the upper region. Find the slope of this line. 2 years ago 2 years ago
1. TuringTest Group Title
always a good idea to have a picture in front of you when you do these problems
2. TuringTest Group Title
|dw:1331329952729:dw|here is just the region R under y=x-x^2 I like to just write out the integral for R first just as a sort of starting point:$R=\int_{0}^{1}x-x^2dx$
3. TuringTest Group Title
now we want to draw in our line that will divide R into a ratio of 1:7|dw:1331330236307:dw|this is a straight line through the origin, so the equation for it is given by$y=mx$where m is the slope we will need the intersection point P of the two graphs as well for the bounds of our new integrals
4. TuringTest Group Title
to find the intersection point we set the equations for the two graphs equal$x-x^2=mx\to x=0,1-m$so we confirm that they intersect at the origin, and we get the x-coordinate of our point P, which is x=m-1 we are now ready to make our integrals
5. TuringTest Group Title
|dw:1331330760501:dw|notice the upper bound of our integral will be x=1-m since we found that top be the intersection point for the area of the top portion the integral will then be$R_u=\int_{0}^{1-m}(x-x^2)-mxdx$
6. TuringTest Group Title
*sorry, typo, that should be 1-m in the drawing
7. TuringTest Group Title
for the area of the bottom portion we will need to split the integral up accordingly|dw:1331331073448:dw|the integrals will be divided at x=1-m, and will be$R_l=\int_{0}^{1-m}mxdx+\int_{1-m}^{2}x-x^2dx$
8. TuringTest Group Title
we know that the ratio of these areas is 1:7, so we have the following relationship$R_l=7R_u$$\int_{0}^{1-m}mxdx+\int_{1-m}^{2}x-x^2dx=7\int_{0}^{1-m}(x-x^2)-mxdx$integrate and solve for m
9. ggrree Group Title
Thank you! That's exactly how far we got, but we are having trouble actually solving for m. Could you please show how you would finish the problem and find an answer?
10. TuringTest Group Title
oh man, you're making me do the messy part? ok, hold on...
11. TuringTest Group Title
ok that was painful but I think I got it...
12. TuringTest Group Title
this is all about simplifying things in a clever way$\int_{0}^{1-m}mxdx+\int_{1-m}^{2}x-x^2dx=7\int_{0}^{1-m}(x-x^2)-mxdx$$\int_{0}^{1-m}mxdx+\int_{1-m}^{2}x-x^2dx=7\int_{0}^{1-m}(1-m)x-x^2dx$from here on I will have to write the left and right hand sides on separate lines because they are so long Left side:$\frac12m(1-m)^2+\frac12(2^2)-\frac13(2^3)-\frac12(1-m)^2+\frac13(1-m)^3$Right side:$=\frac72(1-m)^3-\frac72(1-m)^3$now multiply it all by 6 to get rid of the fractions...
13. TuringTest Group Title
left:$3m(1-m)^2+3(2^2)-2(2^3)-2(2^3)-3(1-m)^2+2(1-m)^3$right:$=21(1-m)^3-14(1-m)^3$now simplify (do not expand anything!)
14. TuringTest Group Title
now I can write left and right on the same line again :)$3m(1-m)^2-3(1-m)^2-4+2(1-m)^3=7(1-m)^2$$(3m-3)(1-m)^2-4=5(1-m)^3$notice the trick that we can do on the left here:$(3m-3)(1-m)^2=-3(1-m)(1-m)^2=-3(1-m)^3$so the problem becomes$-3(1-m)^3-4=5(1-m)^3$$8(1-m)^3=-4$$1-m=-\frac1{\sqrt[3]2}$$m=1+\frac1{\sqrt[3]2}$whew!
15. ggrree Group Title
Thank you, but the answer comes up as 1/2. Earlier up, you integrated from 1-m to 2, rather than from 1-m to 1, also. 2 posts ago, when you retyped the left side and the right side again, you added an additional -2(2^3).
16. ggrree Group Title
We still can't get the answer to come up as 1/2 though. :( But everything else you've done is beautiful.
17. ggrree Group Title
Thanks for taking all this time. We really appreciate it. :)
18. TuringTest Group Title
yes I saw that typo with the -2(2^3) but I did not use it but damn I totally didn't don't know why I changed the bound to 2 now I have to try it again for my own sanity, hold on...
19. TuringTest Group Title
'didn't don't' lol I can't type for some reason..
20. ggrree Group Title
That's totally okay! Thanks!!!
21. TuringTest Group Title
oh the answer 1/2 comes easily if you just put$\frac12(1^2)-\frac13(1^3)$where I put the 2, and do everything the exact same
22. TuringTest Group Title
taking it from where I multiplied everything by 6 and just looking at the left side$3m(1-m)^2+3(2^2)+3(1^3)-2(1^3)-3(1-m)^2+2(1-m)^3$$3m(1-m)^2+1-3(1-m)^2+2(1-m)^3$$-3(1-m)^3+2(1-m)^2+1$so the -4 in my work above become a 1...
23. TuringTest Group Title
that 3(2^3) in the first line above is another typo :/
24. TuringTest Group Title
also the last line has a 2 in the exponent where it should be 3$-3(1-m)^3+2(1-m)^2+1$so yeah, follow it through the same way as before and you get to$-3(1-m)^3+1=5(1-m)^3$$8(1-m)^3=1$$1-m=\frac12$$m=\frac12\huge\checkmark$tadah...
25. TuringTest Group Title
wow I cannot seem to correct my typos the correction I meant to make above was$-3(1-m)^3+2(1-m)^3+1$but the work is correct
26. ggrree Group Title
May we ask you another integration question? $\int\limits\limits_{-1}^{1} (x^2 + x^3)/(1+x^6)$ we tried integration by partial fractions (which ended up being horribly messy) and separating the fraction into $\int\limits_{-1}^{1} x^2 /(1+x^6) + \int\limits_{-1}^{1} x^3/(1+x^6)$ which made the first integral easy to solve via trig substitution (letting x^3 = tan u), but makes the second integral unsolvable (ie. we weren't clever enough to solve it:P ) $\int\limits_{3\pi/4}^{\pi/4}3+ \int\limits\limits_{-1}^{1} x^3/(1+x^6)$
27. TuringTest Group Title
you are in luck, look at the second integrand: it is odd
28. TuringTest Group Title
also I think your first integral is$\int\frac13du$
29. TuringTest Group Title
is the fact that the second integrand is odd not meaning anything to you? look at the bounds and try to remember a theorem about this situation
30. ggrree Group Title
YOu ARE BEAUTIFUL
31. TuringTest Group Title
practice is all, lol that trick has cracked some tough nuts
32. ggrree Group Title
33. TuringTest Group Title
sure, they're good ones
34. ggrree Group Title
You are so awesome! Thanks a googleplex!
35. TuringTest Group Title
you're welcome, and thanks for asking good questions you should really post a separate thread though, long ones get slow
36. ggrree Group Title
Okay! Donne dealios!
37. TuringTest Group Title
lol
38. ggrree Group Title
Thank you TuringTest! We need to go now, but we really appreciated all your help! :)
39. TuringTest Group Title
welcome, anytime :D | 2014-08-01T16:20:57 | {
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https://www.physicsforums.com/threads/probability-theory-question.788759/ | # Probability theory question
1. Dec 22, 2014
### Astudious
1. The problem statement, all variables and given/known data
Alice attends a small college in which each class meets only once a week. She is
deciding between 30 non-overlapping classes. There are 6 classes to choose from
for each day of the week, Monday through Friday. Trusting in the benevolence
of randomness, Alice decides to register for 7 randomly selected classes out of
the 30, with all choices equally likely. What is the probability that she will have
classes every day, Monday through Friday?
2. Relevant equations
Seems like basic probability theory
3. The attempt at a solution
I'm confused here, because my thinking was that I should be able to say she can choose any course for the five days of the week (65 ways of doing this), and then any two of the remaining 25 courses (C(25,2) ways of doing this), so the total probability is 65*C(25,2) / C(30,7). But this is wrong (the answer is above 1).
I considered also that if, after she has picked one course on each day of the week (65), she can then pick another, she has 25 options, and then 24 more after that (since those are how many courses are available at each stage, and she can pick anything since she has fulfilled the every-day-class criterion). But this gives 65*25*24 / C(30,7) which is also wrong (above 1).
Why are these methods wrong?
2. Dec 22, 2014
### Ray Vickson
Since there are 5 slots and 6 chosen courses, the event E = {at least one course each day} must have one day with two courses and four days with one course each.
The first course is chosen at random. What is the probability that the second chosen course is on the same day as the first? What is the probability it is on a different day? What is the probability the third chosen course is on the same day as one of the first two (if possible)? What is the probability it is on a different day from the first two?
Keep going like that. The main issue is to keep track of the "overlaps".
I think the main problem with your method is that she does not choose "days", but courses. When she chooses a course, its day is pre-determined and so is not under her control.
3. Dec 22, 2014
### haruspex
For a start, you get to count the same selection more than once (hence the > 1 result). E.g. if the 5 one-per-day selections you start with are A, B, C, D, E, respectively and your last two are F and G, which happen to be on Wednesday and Thursday, you get the same set of 7 starting with A, B, F, G, E, then picking C and D.
Are you familiar with the principle of inclusion and exclusion? Consider how many ways there are of picking the 7 from a specific four days (or subset thereof).
4. Dec 23, 2014
### Astudious
I'm not sure I understand this. I do see the way of solving this problem by brainstorming, but not why the two methods in the OP are wrong?
It's similar, I feel, to the conundrum in this link: http://mathforum.org/library/drmath/view/62620.html. I would think the answer to the question there should be (2 * C(4,1) * C(51,4)) / C(52,5) - you pick an ace, out of the 4, and then any 4 cards (to total 5) out of the remainder (51), and there are two ways to arrange the two cards you have, so multiply by 2. What's wrong with the logic?
5. Dec 23, 2014
### haruspex
Dr Math's explanation of Matti's error at that link is wrong. Matti's error is the same as yours in this thread. As I posted above, you are counting some possibilities twice over.
In the poker hand example, Matti is counting hands with two Aces twice. E.g. suppose the hand is DA, CA, HK, HQ, HJ. Matti counts DA, plus the other four cards, then counts CA plus the other four cards. Hands with three Aces Matti counts three times, etc.
6. Dec 23, 2014
### Ray Vickson
First: sorry, my previous response was for choosing 6 courses, not 7. But my final objection stands: you need to choose courses, not days.
Let's look at the case of choosing 6 courses (which I will do now in another way); the case of choosing 7 course is a bit more involved, as it needs consideration of more cases, but I will leave that up to you to pursue.
So (for 6 courses) let p(i,j,k,l,m) = probability of choosing i courses on day 1, j courses on day 2, ..., m courses on day 5. You want
$$\text{answer} = p(2,1,1,1,1) + p(1,2,1,1,1)+p(1,1,2,1,1)+p(1,1,1,2,1) + p(1,1,1,1,2)$$
Now each of these terms is a simple (multi-class) hypergeometric probability: if we have $N$ items, $N_1$ of type 1, $N_2$ of type 2, ..., $N_r$ of type r, then the probability $p(k_1,k_2, \ldots, k_r)$ of choosing $k_1$ items of type 1, $k_2$ items of type 2, ..., $k_r$ items of type r in a sample of $n$ items selected without replacement (and with $k_1 + k_2 + \cdots + k_r = n$ is
$$p(k_1, k_2, \ldots, k_r) = \frac{C(N_1,k_1)\, C(N_2, k_2)\, \cdots \,C(N_r,k_r)}{C(N,n)}$$
Here, $N = N_1 + N_2 + \cdots + N_r$. See, eg., http://en.wikipedia.org/wiki/Hypergeometric_distribution (esp., last section) or
http://www.epixanalytics.com/modela...distributions/Multivariate_Hypergeometric.htm
So, we have $p(2,1,1,1,1) = C(6,2) C(6,1)^4/C(30,6) = 432/13195 \doteq 0.03274$. You can quite easily see that all the other terms in the answer are equal to the first one, so
$$\text{answer} = 5\, p(2,1,1,1,1) = C(5,1) \, p(2,1,1,1,1) = 432/2639 \doteq 0.16370$$
Can you see how to extend this to the case of choosing 7 courses?
BTW: you should always check your answers. Had you done so you would have seen that your first answer gave a value of $\text{ans. 1} = 432/377 \doteq 1.146$, while your second one was $\text{ans. 2} = 1512/377 \doteq 4.011$.
I think "haruspex" has already explained to you where your errors lie, so I am restricting myself to giving a positive answer (what to do), rather than a negative one (what you did wrong).
Last edited: Dec 23, 2014 | 2017-08-23T07:37:36 | {
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https://www.physicsforums.com/threads/mass-of-a-rubber-stopper-being-swung-in-a-horizontal-circle.762466/ | # Mass of a rubber stopper being swung in a horizontal circle
## Homework Statement
Homework Statement [/b]
The purpose of this lab is to determine the mass of a rubber stopper being swung in a horizontal circle.
I made an apparatus that resembles the picture attached, with a string (with a rubber stopper on one end, and a weight on the other end) put through a tube. I whirl the tube above my head so that a horizontal circle is made by the stopper.
The radius was increased from 0.25m, to 0.45m, to 0.65m, to 0.85m.
This increased the period of the stopper's circular motion, 0.330s, 0.37575s, 0.45925s, 0.50625s, respectively.
The lab requires me to draw graph of my results, with period squared in order to straighten the graph, and calculate the mass of the stopper using the slope of my graph
Fc=4(pi)2mr/T2
## The Attempt at a Solution
using the equation 4(pi)2mr/T2, I rearranged it and found that r is proportional to t2, as everything else in this experiment is supposed to be constant.
Graphing the radius and period2, the slope of the line of best fit was 0.25555555s2/m.
Now this is where I'm not sure what to do, using the slope to find the mass of the stopper.
Thanks for reading this and I hope you can help me out
#### Attachments
• ucm_app.gif
2.4 KB · Views: 2,368
Last edited:
I assume you know the value of the weight you used that is at the end of the string. The radial force, is of course, equal to this weight ##w##. So you can write $$r = \frac{w}{4\pi^2 m} T^2$$. The slope you calculated is equal to ##\frac{w}{4\pi^2 m}## where you can get the mass you're looking for.
I assume you know the value of the weight you used that is at the end of the string. The radial force, is of course, equal to this weight ##w##. So you can write $$r = \frac{w}{4\pi^2 m} T^2$$. The slope you calculated is equal to ##\frac{w}{4\pi^2 m}## where you can get the mass you're looking for.
So, in your equations, W is the force of gravity on the hanging weight, which is also the centripetal force on the stopper?
I was, in fact, given the mass of the hanging weight, 200g
In that case, 0.200kg x 9.81m/s2 = 1.962N
Now 1.962N is my Fg and also Fc, I can put it in this as W in ##\frac{w}{4\pi^2 m}##, this expression is equal to my slope of 0.255555555, and i solve from there
##0.255555555=\frac{1.962}{4\pi^2 m}##
giving me a mass of 0.194470594kg?
To be clear, the centripetal acceleration is equal to the tension on the string, the tension in the string is equal to the force due to the gravity on the hanging weight.
Does a mass of 0.1945 kg or 194.5 g for the rubber stopper makes sense? Which implies it is as massive as the weight.
1 person
To be clear, the centripetal acceleration is equal to the tension on the string, the tension in the string is equal to the force due to the gravity on the hanging weight.
Does a mass of 0.1945 kg or 194.5 g for the rubber stopper makes sense? Which implies it is as massive as the weight.
ahh, of course that mass makes no sense. The equation should be more like
##\frac{1}{0.2555555555}=\frac{1.962}{4\pi^2 m}##
I forgot that the slope is really ##\frac{0.2555555555s^2}{1m}##
Now I can manipulate the equations:
##m=\frac{(1.962)(0.2555555555)}{4\pi^2}##
so the mass of the stopper = 0.0127006104kg, or 12.7 grams
Much more reasonable, and as a matter of fact, I weighed the stopper and it was 12.67g!
Thanks so much for the help! | 2021-06-15T21:40:19 | {
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http://math.stackexchange.com/questions/68191/let-r-be-a-commutative-ring-with-1-then-why-does-a-in-nr-rightarrow-1a-in/68201 | # Let $R$ be a commutative ring with 1 then why does $a\in N(R) \Rightarrow 1+a\in U(R)$?
Let $R$ be a commutative ring with 1, we define
$$N(R):=\{ a\in R \mid \exists k\in \mathbb{N}:a^k=0\}$$
and
$$U(R):=\{ a\in R \mid a\mbox{ is invertible} \}.$$
Could anyone help me prove that if $a\in N(R) \Rightarrow 1+a\in U(R)$?
I've been trying to construst a $b$ such that $ab=1$ rather than doing it by contradiction, as I don't see how you could go about doing that.
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If $a\in N(R)$, then we can find $k\in\mathbb N^*$ such that $a^k=0$. What about $\sum_{j=0}^{k-1}(-a)^j$? – Davide Giraudo Sep 28 '11 at 10:44
@LHS You may want to change your notation: I would not use $a$ to describe $U(R)$ and then say if $a \in N(R)$ ... – user38268 Sep 28 '11 at 11:36
@DBLim Do not know him sorry, but I appear to have 5 mutual friends with him, Wadham i'm guessing? – Freeman Sep 28 '11 at 13:13
I know Adam from a few years ago when I met him at Warwick. Then he was doing his A-levels. I am no longer friends with him on the internet, but a last check reveals we have 16 mutual friends. – user38268 Sep 28 '11 at 13:50
@DBLim Small world ;) I'll keep a look out for him in my lectures! – Freeman Sep 28 '11 at 14:20
## 5 Answers
This really is a problem in disguise: How did you derive the formula for the sum of the geometric series in year (something) at school??
$(a + 1)(a^{k-1} - a^{k-2} + \ldots 1) = 1-(-a)^n$
but as $a^n = 0$, we have (it does no matter whether $n$ is even or odd) that $(a+1)$ is invertible. Prove the following analogous problem, it may strengthen your understanding:
Let $A$ be a square matrix. If $A^2 = 0$, show that $I - A$ is invertible.
If $A^3 = 0$, show that $I - A$ is invertible.
Hence in general show that if $A^n = 0$ for some positive integer $n$, then $I -A$ is invertible.
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Ah this is excellent, I don't think I would have spotted that! Need to get my game back I think ;) Thanks! – Freeman Sep 28 '11 at 13:09
@LHS Sometimes maths gives the illusion that everything is so complicated; rings, nilradicals, commutative rings, etc and the simple ideas get lost under a heap of terminology.... – user38268 Sep 28 '11 at 13:48
Indeed, just an extra thing, how'd you think this lets you deduce that is $u\in U(R)$ and $a\in N(R) \Rightarrow u+a\in U(R)$? What i'm currently working on! – Freeman Sep 28 '11 at 14:19
As U(R) forms a group with multiplication i've been trying to multiply $a+u$ by things to get $u(a+1)=au+u \in U(R)$ or as a factor.. just use that fact in some way – Freeman Sep 28 '11 at 14:23
@LHS When $n$ is odd, $a^n + b^n = (a+b)(a^{n-1} + \ldots b^{n-1})$. There might be some alternating signs in there. – user38268 Sep 28 '11 at 21:50
Note the following:
$(1+a)(1-a) = 1-a^2$.
$(1-a^2)(1+a^2) = 1-a^4$
$(1-a^4)(1+a^4) = 1-a^8$
...
Thus, continuing in this way, we may find some $b_n$ such that $(1+a)b_n = 1-a^{2^n}$
For large enough $n$, this will give us $(1+a)b_n = 1$.
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can you explain the downvote? is this incorrect? – the L Sep 28 '11 at 11:25
No, this is correct, just not obviously how so. $(1+a)$ is a factor of $(1-a^{2^n})$ for $n\geq 1$. If we pick the smallest $n$ such that $a^{2^n} = 0$, taht will yield your result. – Arthur Sep 28 '11 at 11:54
$N(R)$ is at least in some texts referred to as the nilradical of $R$. It is contained in all prime ideals (in fact, it is the intersection of all prime ideals, Atiyah, MacDonald prop. 1.8), so taking a nilpotent element $a$, since it is contained in all maximal ideals, $a+1$ is not in any maximal ideal. Then the ideal generated by $a+1$ must neccesarily be the whole ring, which means that it specifically generates 1 at some point.
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Thanks! nice to know the background – Freeman Sep 28 '11 at 13:08
Hint $\$ A nilpotent $\rm\,n\,$ lies in every prime ideal $\rm\,P,\,$ because $\rm\, n^k = 0\in P\ \Rightarrow\ n\in P.\,$ In particular, $\rm\,n\,$ lies in every maximal ideal. Hence $\rm\,n\!+\!1\,$ is a unit, since it lies in no maximal ideal $\rm\,M\,$ (else $\rm\,n\!+\!1,\,n\in M\, \Rightarrow\, (n\!+\!1)-n = 1\in M),\,$ i.e. elements coprime to every prime are units.
You may recognize a hint of this in proofs of Euclid's theorem that that are infinitely many primes. Namely, if there are only finitely many primes then their product $\rm\,n\,$ is divisible by every prime, so $\rm\,n\!+\!1\,$ is coprime to all primes, so it must be the unit $1,\,$ so $\rm\,n = 0,\,$ a contradiction.
Remark You'll meet related results later when you study the structure theory of rings. There the intersection of all maximal ideals of a ring $\rm\,R\,$ is known as the Jacobson radical $\rm\,Jac(R).\,$ The ideals $\rm\,J\,$ with $\rm\,1+J \subset U(R)=$ units of $\rm R,\,$ are precisely those ideals contained in $\rm\,Jac(R).\,$ Indeed, we have the following theorem, excerpted from my post on the fewunit ring theoretic generalization of Euclid's proof of infinitely many primes.
THEOREM $\$ TFAE in ring $\rm\,R\,$ with units $\rm\,U,\,$ ideal $\rm\,J,\,$ and Jacobson radical $\rm\,Jac(R).$
$\rm(1)\quad J \subseteq Jac(R),\quad$ i.e. $\rm\,J\,$ lies in every max ideal $\rm\,M\,$ of $\rm\,R.$
$\rm(2)\quad 1+J \subseteq U,\quad\ \$ i.e. $\rm\, 1 + j\,$ is a unit for every $\rm\, j \in J.$
$\rm(3)\quad I\neq 1\ \Rightarrow\ I+J \neq 1,\qquad\$ i.e. proper ideals survive in $\rm\,R/J.$
$\rm(4)\quad M\,$ max $\rm\,\Rightarrow M+J \ne 1,\quad$ i.e. max ideals survive in $\rm\,R/J.$
Proof $\,$ (sketch) $\$ With $\rm\,i \in I,\ j \in J,\,$ and max ideal $\rm\,M,$
$\rm(1\Rightarrow 2)\quad j \in all\ M\ \Rightarrow\ 1+j \in no\ M\ \Rightarrow\ 1+j\,$ unit.
$\rm(2\Rightarrow 3)\quad i+j = 1\ \Rightarrow\ 1-j = i\,$ unit $\rm\,\Rightarrow I = 1.$
$\rm(3\Rightarrow 4)\ \,$ Let $\rm\,I = M\,$ max.
$\rm(4\Rightarrow 1)\quad M+J \ne 1 \Rightarrow\ J \subseteq M\,$ by $\rm\,M\,$ max.
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A lot to think about there! I am definitely only scratching the surface of this area! – Freeman Sep 28 '11 at 21:21
Let $a\in N(R)$ and $k$ such that $a^k=0$. We have \begin{align*}(1+a)\left(\sum_{j=0}^{k-1}(-a)^j \right)&=\sum_{j=0}^{k-1}(-a)^j+\sum_{j=0}^{k-1}-(-a)^{j+1}\\ & =\sum_{j=0}^{k-1}(-a)^j-\sum_{j=1}^k(-a)^j\\ &=1-(-a)^k=1+(-1)^ka^k=1, \end{align*} and we have $\left(\sum_{j=0}^{k-1}(-a)^j\right)(1+a)=1$ by the same computation (it's true even if the ring is not commutative), hence $1+a$ is invertible, and it's inverse is $\left(\sum_{j=0}^{k-1}(-a)^j\right)(1+a)$.
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This seems a very complicated way of deriving a formula learnt at school... – user38268 Sep 28 '11 at 11:34
@DBLim: Indeed, yours does explain the thought process needed to come to the conclusion, I guess this is another way of expressing the same idea. Thanks though, this is helpful! – Freeman Sep 28 '11 at 13:10
@DavideGiraudo Cette question n'a rien à voir, vous étudiez à quelle université maintenant? Vous avez dit que l'anglais n'est pas votre première langue, le même que le franćais n'est pas le mien. On peut se discuter en anglais/franćais pour que... – user38268 Sep 28 '11 at 13:55
@DavideGiraudo Merci de me dire. – user38268 Sep 28 '11 at 21:50 | 2015-09-02T10:50:05 | {
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http://math.stackexchange.com/questions/170489/finding-the-derivative-of-a-function-using-the-product-rule/170503 | # Finding the derivative of a function using the Product Rule
I'm home teaching myself calculus because I'm 16 and therefore too young to take an actual class with a teacher, so I apologise if this seems simple.
I understand the definition of the Product Rule and its formula:
"If a function $h(x)=f(x)\times g(x)$ is the product of two differentiable functions $f(x)$ and $g(x)$, then $h'(x) = f(x)\times g'(x)+f'(x)\times g(x)$".
I did a question to find the derivative of $g(x) = (2x+1)(x+4)$ using the Product Rule.
Now on the solutions sheet it says I must begin by writing:
$g'(x)=(2x+1){\bf (1)}+{\bf (2)}(x+4)$
What confuses me are the terms that I have put in bold. (the terms $(1)$ and $(2)$). I believe the term $(1)$ is $g'(x)$ from the formula and the term $(2)$ is $f'(x)$ from the formula.
How am I supposed to know these 2 terms? Am I supposed to find the derivative of $(2x+1)$ and $(x+4)$ before going on to the question?
I also apologise if this is quite messy.
-
Note that using the index rule you can easily determine that the derivative of $(x+4)$ is $1$, and the derivative of $(2x+1)$ is $2$. This is where the terms come from. – Shaktal Jul 13 '12 at 21:32
I don't really understand what the stars are suppose to mean, but I can try to explain the product rules.
As you pointed out,
$h'(x) = f'(x)g(x) + f(x)g'(x)$.
In your example, the function $h(x) = (2x +1)(x + 4)$. First, you need to recognize that $h$ is the product of two functions:
$f(x) = 2x + 1$
$g(x) = x + 4$
In order to apply the product rule formula, you need to find $f'(x)$ and $g'(x)$. I assume that you know how to do this for the two functions above. Hence you should obtain
$f'(x) = 2$
$g'(x) = 1$
Sticking all of these terms into the product the rule
$h'(x) = f'(x)g(x) + f(x)g'(x)$
$h'(x) = (2)(x + 4) + (2x + 1)(1) = 2x + 8 + 2x + 1 = 4x + 9$
-
I was trying to make them bold so as to say "these are the parts I'm confused about" – Olly Price Jul 13 '12 at 21:39
@OllyPrice The product rule assumes that you know how to find $f'(x)$ and $g'(x)$. For example, the $f(x)$ and $g(x)$ are polynomial functions, so you can compute their derivative $2$ and $1$ respectively using the power rules for polynomials or just plug these function into the definition of the derivative $\lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}$. – William Jul 13 '12 at 21:43
You’re supposed to find the derivatives of $2x+1$ and $x+4$ as part of finding the derivative of their product:
\begin{align*} \Big((2x+1)(x+4)\Big)'&=(2x+1)(x+4)'+(x+4)(2x+1)'\\ &=(2x+1)(1)+(x+4)(2)\\ &=(2x+1)+(2x+8)\\ &=4x+9\;. \end{align*}
Note that you can check this by multiplying out the original function to get $2x^2+9x+4$ and taking its derivative directly.
-
So you have $(2x+1)(x+4) = h(x) = f(x)g(x)$ where $f(x) = 2x + 1$ and $g(x) = x + 4$. So you find the derivative of both $f(x)$ and $g(x)$: \begin{align} f'(x) &= 2 \\ g'(x) &= 1. \end{align} And so $$h'(x) = f'(x)g(x) + f(x)g'(x) = 2(x+4) + (2x+1)1.$$ It seems like the $**$ you have above are supposed to be unknowns - like a fill in the blank.
-
Just to add to and perhaps simplify the answers given above, the apostrophe in $f'(x)$ in the formula for the product rule means "take the derivative of $f(x)$." When reading $f'(x)$, we say "f prime of x." So you did in fact answer your own question and the others here have added more detail about how it works. Good luck with your studies! | 2013-12-18T12:59:40 | {
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https://nileshverma.com/midterm-ekdpfnh/page.php?ae5d50=covariance-of-beta-0-hat-and-beta-1-hat-proof | # covariance of beta 0 hat and beta 1 hat proof
It can take several seconds to load all equations. The basic idea is that the data have $n$ independent normally distributed errors. {/eq} are regression Coefficient. \be… Let Hbe a symmetric idempotent real valued matrix. Because $$\hat{\beta}_0$$ and $$\hat{\beta}_1$$ are computed from a sample, the estimators themselves are random variables with a probability distribution — the so-called sampling distribution of the estimators — which describes the values they could take on over different samples. Suppose a simple linear regression model: This post will explain how to obtain the following formulae: ①. Yes, part of what you wrote in 2008 is correct, and that is the conclusion part. For an example where the covariance is 0 but X and Y aren’t independent, let there be three outcomes, ( 1;1), (0; 2), and (1;1), all with the same probability 1 3. answer! ‘Introduction to Econometrics with R’ is an interactive companion to the well-received textbook ‘Introduction to Econometrics’ by James H. Stock and Mark W. Watson (2015). Simple Linear regression is a linear regression in which there is only one explanatory variable and one dependent variable. Beta shows how strongly one stock (or portfolio) responds to systemic volatility of the entire market. Note this sum is e0e. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … Derivation of the normal equations. Close. 1. The matrix Z0Zis symmetric, and so therefore is (Z0Z) 1. I am sorry to tell you this, but your proposition is not correct. I'm pretty stuck in this problem, bascially we are given the simple regression model: y*i* = a + bx*i* _ e*i* where e*i* ~ N(0, sigma2) i = 1,..,n. Then with xbar and ybar are sample means and ahat and bhat are the MLEs of a and b. 38 CHAPTER 3 Useful Identities for Variances and Covariances Since ¾(x;y)=¾(y;x), covariances are symmetrical. How can I derive this solution by not using matrix? … and deriving it’s variance-covariance matrix. A symmetric idempotent matrix such as H is called a perpendicular projection matrix. All rights reserved. Because $$\hat{\beta}_0$$ and $$\hat{\beta}_1$$ are computed from a sample, the estimators themselves are random variables with a probability distribution — the so-called sampling distribution of the estimators — which describes the values they could take on over different samples. If Beta >1, then the level of risk is high and highly volatile as compared to the stock market. From this table, we may conclude that: The Null model clearly does not fit. If Beta > 0 and Beta < 1, then the stock price will move with … if we were to repeatedly draw samples from the same population) the OLS estimator is on average equal to the true value β.A rather lovely property I’m sure we will agree. Properties of Least Squares Estimators Proposition: The variances of ^ 0 and ^ 1 are: V( ^ 0) = ˙2 P n i=1 x 2 P n i=1 (x i x)2 ˙2 P n i=1 x 2 S xx and V( ^ 1) = ˙2 P n i=1 (x i x)2 ˙2 S xx: Proof: V( ^ 1) = V P n Covariance Matrix of a Random Vector • The collection of variances and covariances of and between the elements of a random vector can be collection into a matrix called the covariance matrix remember so the covariance matrix is symmetric. I tried using the definition of Cov(x, y) = E[x*y] - E[x]E[y]. Average the PRE Yi =β0 +β1Xi +ui across i: β β N i 1 i N i 1 0 1 i N i 1 Yi = N + X + u (sum the PRE over the N observations) N u + N X + N N N Y N i 1 i N i 1 0 N i 1 ∑ i ∑ ∑ β= β = (divide by N) Y = β0 + β1X + u where Y =∑ iYi N, X =∑ iXi N, and u =∑ = E{ [(ybar - b1 * xbar) - (ybar - beta_1 * xbar)] [b1 - beta_1] } substituting for b0, E(b0), and E(b1) based on above, = E{[-b1 * xbar + beta_1 * xbar)] [b1 - beta_1]} simplifying, = E{[ -xbar(b1 - beta_1)] [b1 - beta_1]} factoring out -xbar, = E{-xbar(b1 - beta1)2 } simplifying a bit, = E{-xbar} * E{(b1 - beta1)2 } I split expectation to see how we get the variance, = -xbar * var(b1) definition of variance, = -xbar * [sigma2 / sum(x_i - xbar)2 ] definition for slope variance, New comments cannot be posted and votes cannot be cast, More posts from the HomeworkHelp community. © copyright 2003-2020 Study.com. ... [b1 - E(b1)]} definition of covariance. Problem Solving Using Linear Regression: Steps & Examples, Regression Analysis: Definition & Examples, Coefficient of Determination: Definition, Formula & Example, The Correlation Coefficient: Definition, Formula & Example, Factor Analysis: Confirmatory & Exploratory, Measures of Dispersion: Definition, Equations & Examples, Line of Best Fit: Definition, Equation & Examples, Type I & Type II Errors in Hypothesis Testing: Differences & Examples, Analysis Of Variance (ANOVA): Examples, Definition & Application, The Correlation Coefficient: Practice Problems, Difference between Populations & Samples in Statistics, What is Standard Deviation? They are saying that you're approximating the population's regression line from a sample of it. We'll have 1 minus 0, so you'll have a 1 times a 3 minus 4, times a negative 1. If we choose $$\lambda=0$$, we have $$p$$ parameters (since there is no penalization). It describes the influence each response value has on each fitted value. DISTRIBUTIONAL RESULTS 5 Proof. Then the eigenvalues of Hare all either 0 or 1. Answer to: Prove that variance for hat{beta}_0 is Var(hat{beta}_0) = frac{sum^n_{i=1} x^2_i}{n sum^n_{i=1}(x_i - bar{x})^2} sigma^2 . Answer to: Prove that variance for hat{beta}_0 is Var(hat{beta}_0) = frac{sum^n_{i=1} x^2_i}{n sum^n_{i=1}(x_i - bar{x})^2} sigma^2 . {/eq} and {eq}\beta_1 All other trademarks and copyrights are the property of their respective owners. It follows that the hat matrix His symmetric too. More specifically, the covariance between between the mean of Y and the estimated regression slope is not zero. {eq}\hat \beta_1=\sum_{i=1}^n k_iy_i It means the stock is volatile like the stock market. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … Earn Transferable Credit & Get your Degree, Get access to this video and our entire Q&A library. Yes, part of what you wrote in 2008 is correct, and that is the conclusion part. Finding variance-covariance of $\hat\beta$ from $\hat\beta = (X^TX)^{-1}X^Ty$ 26 The proof of shrinking coefficients using ridge regression through “spectral decomposition” Archived [University Statistics] Finding Covariance in linear regression. Haifeng (Kevin) Xie: Dear all, Given a LME model (following the notation of Pinheiro and Bates 2000) y_i = X_i*beta + Z_i*b_i + e_i, is it possible to extract the variance-covariance matrix for the estimated beta_i hat and b_i hat from the lme fitted object? A matrix formulation of the multiple regression model. Suppose that there are rones. We can find this estimate by minimizing the sum of . Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … Services, Simple Linear Regression: Definition, Formula & Examples, Working Scholars® Bringing Tuition-Free College to the Community. Make sure you can see that this is very different than ee0. Sign in to make your opinion count. Cookies help us deliver our Services. Become a Study.com member to unlock this which is equivalent to minimization of $$\sum_{i=1}^n (y_i - \sum_{j=1}^p x_{ij}\beta_j)^2$$ subject to, for some $$c>0$$, $$\sum_{j=1}^p \beta_j^2 < c$$, i.e. It's getting really weird from there and I don't know how to continue it! Covariance of beta hat times k transpose and when I … Just look at the key part of your proof: beta_0 = y^bar-beta_1*x^bar, Y^bar is the only random variable in this equation, how can you equate a unknown constant with a random variable? $\bar{y}$ refers to the average of the response (dependent variable). So you're going to have 1 times negative 1, which is negative 1. If Beta = 1, then risk in stock will be the same as a risk in the stock market. Don't like this video? constraining the sum of the squared coefficients. 5.2 Confidence Intervals for Regression Coefficients. No one is wasting your time! - Examples, Advantages & Role in Management, Confidence Interval: Definition, Formula & Example, Normal Distribution: Definition, Properties, Characteristics & Example, Production Function in Economics: Definition, Formula & Example, TExES Mathematics 7-12 (235): Practice & Study Guide, TExES Physics/Mathematics 7-12 (243): Practice & Study Guide, High School Algebra II: Homework Help Resource, Ohio Assessments for Educators - Mathematics (027): Practice & Study Guide, Saxon Math 7/6 Homeschool: Online Textbook Help, NY Regents Exam - Integrated Algebra: Help and Review, Biological and Biomedical It can take several seconds to load all equations. independent, then ¾(x;y)=0, but the converse is not true — a covariance of zero does not necessarily imply independence. This means that in repeated sampling (i.e. Simply, it is: And you might see this little hat notation in a lot of books. $\hat{\beta_1}$ refers to the estimator of the slope. Then H= P r i=1 p ip 0. Where X is explanatory variable , Y is dependent variable {eq}\beta_0 In statistics, the projection matrix (), sometimes also called the influence matrix or hat matrix (), maps the vector of response values (dependent variable values) to the vector of fitted values (or predicted values). Need help with homework? Calculation of Beta in Finance #1-Variance-Covariance Method. Well, it's telling us at least for this sample, this one time that we sampled the random variables X and Y, X was above it's expected value when Y was below its expected value. \be… We will learn the ordinary least squares (OLS) method to estimate a simple linear regression model, discuss the algebraic and statistical properties of the OLS estimator, introduce two measures of goodness of fit, and bring up three least squares assumptions for a linear regression model. Using ordinary least square and solving the Normal equation. 2.4. Est-ce donc la raison de la transposition que l'on peut faire la multiplication à l'intérieur de $E()$? This indeed holds. We use $k$ dimensions to estimate $\beta$ and the remaining $n-k$ dimensions to estimate $\sigma^2$. The basic idea is that the data have $n$ independent normally distributed errors. It describes the influence each response value has on each fitted value. By using our Services or clicking I agree, you agree to our use of cookies. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … One of the major properties of the OLS estimator ‘b’ (or beta hat) is that it is unbiased. Posted by 7 years ago. With no loss of generality, we can arrange for the ones to precede the zeros. As we already know, estimates of the regression coefficients $$\beta_0$$ and $$\beta_1$$ are subject to sampling uncertainty, see Chapter 4.Therefore, we will never exactly estimate the true value of these parameters from sample data in an empirical application. Along with y*i* hat = ahat + bhat * x*i* we are supposed to find Cov(ahat, bhat). The equation for var.matrix() is The purpose of this subreddit is to help you learn (not complete your last-minute homework), and our rules are designed to reinforce this. 1. e0e = (y −Xβˆ)0(y −Xβˆ) (3) which is quite easy to minimize using standard calculus (on matrices quadratic forms and then using chain rule). Given that S is convex, it is minimized when its gradient vector is zero (This follows by definition: if the gradient vector is not zero, there is a direction in which we can move to minimize it further – see maxima and minima. This is most easily proven in the matrix form. The covariance matrix of ^ is Cov( 0^) = ... Var(~c0 ^) Which concludes the proof. The basic idea is that it is unbiased have a 1 times a 1... Is high and highly volatile as compared to the average of the market! Where the hat over β indicates the OLS estimate of β the estimated regression slope is not.... Average of the entire market and that is the conclusion part if we choose \ ( \lambda=0\ ), may... Ols estimate of β notation in a lot of books à multiplier pourraient $... Fois m$, $n \ fois m$ the estimated regression slope is correct. And copyrights are the property of their respective owners model with one regressor a! Être $n \ not = m$ parameters ( since there is no penalization.. ˆ Y ˆ X β0 = −β1 eq } \hat \beta_1=\sum_ { i=1 } ^n {! In the linear model new picture of my solving steps University Statistics ] covariance. Get access to this shortly ; see Figure 3.3. little hat in! Assignment help/ homework help/Online Tutoring in Economics pls visit www.learnitt.com in a lot of books the matrix Z0Zis,... Fitted value pourraient être $n \ not = m$, $n not! Derive this solution by not using matrix they are saying that you approximating... ^ ) which concludes the Proof to precede the zeros, ridge regression puts further constraints on parameters! 3.3. sample, what is a Decision Tree okay, the second we... Homework help/Online Tutoring in Economics pls visit www.learnitt.com )$ simple linear regression model: post. Β indicates the OLS estimator ‘ B ’ ( or beta hat ) is that it is: the. Of Hare all Either 0 or = 1, on average be Y = X * beta + Epsilon where... Ridge regression puts further constraints on the parameters other trademarks and copyrights are the property their. Simple linear regression model with one regressor called a simple linear regression in which there is no penalization.! = ∑ = unbiasedness of βˆ 0: Start with the market, on average text use. [ /math ] independent normally distributed errors I agree, you agree our. Lot of books the average of the OLS estimator 1, which is the conclusion part m.. And hehe1223 pointed your mistake out correctly for you so you 're going to talk about let. Answer your tough homework and study questions 're going to talk about is let 's look at covariance. Symmetric idempotent matrix such as H is called a perpendicular projection matrix approximating population! Not a good fit since the value of Y [ math ] n [ /math ] independent normally errors! To learn the rest of the two estimators have [ math ] [... Learn the rest of the entire market que l'on peut faire la multiplication à de! L'Intérieur de $E ( ) 10 ), P_2 ( 3, 5 ), P_2 (,! X ; Y ) can be 0 for variables that are not inde-pendent this, but your proposition not! Βˆ 0: Start with the formula ˆ Y ˆ X β0 −β1... Or = 1, then risk in stock will be the same as risk... Remainder of this course$ \hat { \beta_1 } $refers to the average of the properties! We can find this estimate by minimizing the sum of but the B model fits significantly better than Null..., the user should use function beta_hat ( ), which is 1. They are saying that you 're approximating the population 's regression line from a sample of it the user-friendly.! – mavavilj 06 déc.. 16 2016-12-06 17:04:33 this lecture introduces a linear regression:. In tandem with the formula ˆ Y ˆ X β0 = −β1 my solving steps 3! Will be the same as a risk in the stock market, (! Economics pls visit covariance of beta 0 hat and beta 1 hat proof population 's regression line from a sample of.. But your proposition is not correct objective can be rewritten = ∑ = our entire Q a! Covariance between between the mean of Y ] Finding covariance in linear regression ( dependent variable ) are inde-pendent. Linear regression model with one regressor called a perpendicular projection matrix pointed your out... Of beta hat times k transpose and when I … Either = or. = 0 or = 1 Null model hat over β indicates the estimate... Of Xdetermines the value of Xdetermines the value of Y choose \ \beta_j\. = beta_0 and E [ b0 ] = beta_1 since these are unbiased estimators of their owners! I=1 } ^n k_iy_i { /eq } sample of it ( X ; Y ) be! Response value has on each fitted value matrix His symmetric too how strongly one stock ( or portfolio ) to. Therefore is ( Z0Z ) 1 two estimators \beta_1=\sum_ { i=1 } k_iy_i! For function betahat_mult_Sigma ( )$ hat ) is that it is 4.5. High and highly volatile as compared to the estimator of the parameters on! From a sample of it help/ homework help/Online Tutoring in Economics pls visit www.learnitt.com are that. Distributed errors model: this post will explain how to obtain the following:... Be Y = X * beta + Epsilon, where all elements of Epsilon have mean 0 variance. User-Friendly version, but your proposition is not a good fit since the goodness-of-fit chi-square is. From this table, we may conclude that: the Craft of Writing -! Needing this is because I want to have interval prediction on the predicted values ( at level = 0:1.. The Normal equation in which there is no penalization ) puisque les deux matrices à covariance of beta 0 hat and beta 1 hat proof pourraient être n. Can I derive this solution by not using matrix: ① model with one regressor called a perpendicular matrix. Not zero que l'on peut faire la multiplication à l'intérieur de $E ( b1 ) ] } definition covariance! Than ee0 X ; Y ) can be rewritten = ∑ = b0 ] = beta_0 and E b1... Concludes the Proof la transposition que l'on peut faire la multiplication à l'intérieur de$ (. Of Xdetermines the value of Y and one dependent variable ) ( dependent variable ) the! Usually write, where all elements of Epsilon have mean 0 and variance.... Agree, you agree to our use of cookies regression puts further constraints the! Specifically, the second thing we are going to talk about is let 's look at the covariance between... \Beta_1=\Sum_ { i=1 } ^n k_iy_i { /eq } that: the Craft of Writing -. Is correct, and that is the user-friendly version mean of Y and the estimated regression slope is not.. Tandem with the market, on average the Null model clearly does not fit value has on covariance of beta 0 hat and beta 1 hat proof fitted.... The rest of the keyboard shortcuts is very different than ee0 wrote in 2008 is,... \Bar { Y } $refers to the average of the two estimators use beta_hat! Z0Zis symmetric, and so therefore is ( Z0Z ) 1 } \hat \beta_1=\sum_ i=1... More specifically, the covariance between between the mean of Y and the estimated slope... Some text books use Greek letters for the unknown parameters and Roman for.: 4.5 the Sampling Distribution of the major covariance of beta 0 hat and beta 1 hat proof of the OLS estimator if we \... This table, we can arrange for the ones to precede the zeros for variables that are inde-pendent. Our entire Q & a library what you wrote in 2008 is correct, and that is the user-friendly.... À l'intérieur de$ E ( ), we can arrange for the ones to precede the zeros not.. You this, but your proposition is not a good fit since the value of Xdetermines the value of and... Degree, Get access to this shortly ; see Figure 3.3. the sum of volatile like the stock.! And so therefore is ( Z0Z ) 1 we are going to have interval prediction on parameters. Pointed your mistake out correctly for you take several seconds to load all equations: ① yes part... ’ re clearly not indepen-dent since the goodness-of-fit chi-square value is very different than ee0 can... Agree to our use of cookies is that the hat indicates that we are dealing an! { eq } \hat \beta_1=\sum_ { i=1 } ^n k_iy_i { /eq } fits... 10 ), we may conclude that: the Craft of Writing Effectively - Duration:.... Correct, and so therefore is ( Z0Z ) 1 and Roman letters for the unknown and. - E ( ) minus 4, times a 3 minus 4, times a 1. Do n't know how to obtain the following formulae: ① all elements of Epsilon have mean 0 variance... 4.5 the Sampling Distribution of the slope = beta_1 since these are unbiased estimators with one regressor a! & sample, what is a wrapper for function betahat_mult_Sigma ( ),... for ones. Response ( dependent variable ) βˆ 0: Start with the market on! Transposition que l'on peut faire la multiplication à l'intérieur de \$ E ( b1 ]. The eigenvalues of Hare all Either 0 or 1 Transferable Credit & Get Degree... Set shown below how strongly one stock ( or beta hat ) is it! The Sampling Distribution of the major properties of the entire market that the data set shown.. Normally distributed errors = 0 or = 1, then risk in the stock responds to systemic volatility of OLS!
Updated: December 5, 2020 — 2:38 PM | 2021-03-02T17:08:31 | {
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https://math.stackexchange.com/questions/983760/convergence-of-the-series-sum-limits-n-1-infty-1n-frac-lnn-sqrt | # Convergence of the series $\sum\limits_{n=1}^{\infty}(-1)^{n}\frac{\ln(n)}{\sqrt{n}}$
I would like to see whether or not $$\sum\limits_{n=1}^{\infty}(-1)^{n}\dfrac{\ln(n)}{\sqrt{n}}$$ is a convergent series.
Root test and ratio test are both inconclusive. I tried the alternating series test after altering the form of the series: $$\sum\limits_{n=1}^{\infty}(-1)^{n-1}\left[\dfrac{-\ln(n)}{\sqrt{n}}\right]\text{.}$$ After using L-Hospital, it's clear that $\lim\limits_{n \to \infty}\left[\dfrac{-\ln(n)}{\sqrt{n}}\right] = 0$. To show that it's decreasing led to me finding the derivative $\dfrac{\ln(n)}{2n^{3/2}}-\dfrac{1}{n^{3/2}}$, which I could set to be less than $0$, but a plot has shown that $n < e^{2}$ is not where $\dfrac{-\ln(n)}{\sqrt{n}}$ is decreasing.
So all that remains is a comparison test. I can't think of a clever comparison to use for this case. Any ideas?
• Use Dirichlet test. I think alternating test is also works. – Hanul Jeon Oct 21 '14 at 5:08
• Continue with your alternating series test work. You can show that after a while the function $\frac{\log t}{t^{1/2}}$ is decreasing, and that is good enough. – André Nicolas Oct 21 '14 at 5:09
• This is $-\eta'\bigg(\dfrac12\bigg).~$ See Dirichlet $\eta$ function for more information. – Lucian Oct 21 '14 at 6:03
If $f(x)=\dfrac{\ln x}{\sqrt{x}}$ then
$$f'(x) = \dfrac{2-\ln x}{2\sqrt{x^3}}.$$
Which is negative for all $x>e^2$. So $f(n)=b_n$ is decreasing for all integers $n\ge 8$.
$\dfrac{\ln(n)}{\sqrt{n}}$ is mono-tonic decreasing after $n=\lceil e^2 \rceil$ and remains bounded between $n=1$ to $\lfloor e^2 \rfloor$, so from alternating series test $\sum\limits_{n=1}^{\infty}(-1)^{n}\dfrac{\ln(n)}{\sqrt{n}}$, must converge.
You can use Leibniz test:
if you have a series like $\sum\limits_{n=1}^{\infty}(-1)^{n}{a_n}$ and ${a_n}$ is decrescent and infinitesimal when $n \rightarrow \infty$, then the series
$\sum\limits_{n=1}^{\infty}(-1)^{n}{a_n}$ converges | 2019-08-24T02:22:15 | {
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http://math.stackexchange.com/questions/119695/show-that-a-definite-integral-vanishes-for-all-values-of-a-b-c-0 | # Show that a definite integral vanishes for all values of $a,b,c > 0$.
Show $$\int_0^\infty \int_0^\infty \frac{(ax-by) {\rm e}^{-x} {\rm e}^{-y}}{(a^2 x + b^2 y + c x y)^{\frac{3}{2}}} \,dx \,dy = 0$$ for any $a,b,c > 0$.
I came upon the above double integral when simplifying an expression for a probability density function. How can I demonstrate that the integral is zero for any positive constants, $a$, $b$ and $c$?
I've verified the result numerically. At the moment it seems rather fascinating to me that the integral should always be zero given the presence of three free variables, and I would expect there to be a relatively simple derivation.
I've tried all sorts of approaches (integral substitutions including polar coordinates, differentiating under the integral, splitting the domain of integration into pieces) without seeming to make progress.
To answer Ali's comment, below I give my crude Matlab code for evaluating the double integral (with the trapezoidal rule) and one output. Regardless of what positive values of $a$,$b$,$c$ I put in, I get something close enough to zero for my liking.
Output of program:
double_int = getDoubleInt(2,3,4)
double_int =
-6.1549e-010
Program code:
function double_int = getDoubleInt(a,b,c)
x_max = 25;
y_max = 25;
NN_x = 1000;
NN_y = 1000;
x_vec = logspace(log10(x_max/10^10),log10(x_max),NN_x);
y_vec = logspace(log10(y_max/10^10),log10(y_max),NN_y);
XX = x_vec'*ones(1,NN_y);
YY = ones(NN_x,1)*y_vec;
ZZ = (a*XX-b*YY)./(a^2*XX+b^2*YY+c*XX.*YY).^(3/2).*exp(-XX).*exp(-YY);
int_1 = zeros(size(x_vec));
for i = 1:NN_x
int_1(i) = trapz(y_vec,ZZ(i,:));
end
double_int = trapz(x_vec,int_1);
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Since MathJax is down: 1.618034.com/blog_data/math/formula.17081.png – user2468 Mar 13 '12 at 15:11
I'm still incredulous this could be true and haven't yet heard from other numerical tests, but at first glance I think you could interpret it as a function of $a,b,c$ and show it's constant... – anon Mar 13 '12 at 16:09
@djws: can yu show us a numerical test of this equality? – user17090 Mar 13 '12 at 16:40
For $a=b$, I believe the result. For $a\neq b$ this looks like a miracle (and I don't believe in miracles). – Fabian Mar 13 '12 at 20:41
@Fabian: I totally agree. Now, I'm doing a few numerical experiments with Mathematica, and so far I get zero for $a=.1$, $b=1$, and values of $c$ ranging from $.01$ to $5$. Every value I've tried gives equality. The caveat is that for some values Mathematica complains that the convergence of the numerical integration is too slow. – Martin Argerami Mar 13 '12 at 20:49
## 3 Answers
The result is surprising because one would think that the combination with the exponentials would be too complicated to make the integral vanish for all $a,b,c$. However, the exponentials don't enter into it because the integral along every line with constant sum $x+y$ is zero, and the exponential factor is constant on these lines. Consider
$$x=u+v\;, \\ y=u-v\;,$$
which, up to a constant factor from the Jacobian, transforms the integral into
$$\int_0^\infty\mathrm du\mathrm e^{-2u}\int_{-u}^u\mathrm dv\frac{a(u+v)-b(u-v)}{\left(a^2(u+v)+b^2(u-v)+c(u^2-v^2)\right)^{3/2}}\;.$$
The inner integral is readily performed; Wolfram|Alpha gives
$$\int_0^\infty\mathrm du\mathrm e^{-2u}\frac2{(a-b)^2+2cu}\left[\frac{a(u+v)+b(u-v)}{\sqrt{a^2(u+v)+b^2(u-v)+c(u^2-v^2)}}\right]_{v=-u}^{v=u}\;,$$
and the result is $0$ since the antiderivative takes the same value at $u$ and $-u$.
-
very nice answer, thanks! – djws Mar 15 '12 at 21:17
@djws: You're welcome! – joriki Mar 15 '12 at 21:21
@joriki: +1 (+ bounty) thanks for helping us out ;-) – Fabian Mar 16 '12 at 8:25
@Fabian: My pleasure :-) – joriki Mar 16 '12 at 8:29
Using the substitution $$r\frac{1+t}{1-t}=\frac1r\frac{1+s}{1-s}$$ maps $[-1,1]\mapsto[-1,1]$ and has proven useful in several situations similar to this. Some formulas pertinent to this substitution are $$\frac{\mathrm{d}t}{1-t^2}=\frac{\mathrm{d}s}{1-s^2}$$ $$\frac{r(1+t)+\frac1r(1-t)}{1-t^2}=\frac{\frac1r(1+s)+r(1-s)}{1-s^2}$$ $$\frac{r(1+t)-(1-t)}{\sqrt{1-t^2}}=\frac{(1+s)-r(1-s)}{\sqrt{1-s^2}}$$ Change variables $u=x+y$ and $v=x-y$, then $v=ut$: \begin{align} &\int_0^\infty\int_0^\infty\frac{(ax-by)\,\mathrm{e}^{-x}\,\mathrm{e}^{-y}}{(a^2 x + b^2 y + c x y)^{\frac{3}{2}}}\,\mathrm{d}x\,\mathrm{d}y\\[6pt] &=2\int_0^\infty\int_{-u}^u\frac{(a-b)u+(a+b)v}{\left(cu^2-cv^2+2(a^2+b^2)u+2(a^2-b^2)v\right)^{3/2}}\,\mathrm{e}^{-u}\,\mathrm{d}v\,\mathrm{d}u\\[6pt] &=2\int_0^\infty\int_{-1}^1\frac{(a-b)+(a+b)t}{\left(cu^2(1-t^2)+2u(a^2(1+t)+b^2(1-t))\right)^{3/2}}\,\mathrm{d}t\,\mathrm{e}^{-u}u^2\,\mathrm{d}u\tag{1} \end{align} Using the substitution $\frac{a}{b}\frac{1+t}{1-t}=\frac{b}{a}\frac{1+s}{1-s}$ with $(1)$ yields $$2\int_0^\infty\int_{-1}^1\frac{(b-a)+(a+b)s}{\left(cu^2(1-s^2)+2u(b^2(1+s)+a^2(1-s))\right)^{3/2}}\,\mathrm{d}s\,\mathrm{e}^{-u}u^2\,\mathrm{d}u$$ Then substituting $w=-s$ yields $$-2\int_0^\infty\int_{-1}^1\frac{(a-b)+(a+b)w}{\left(cu^2(1-s^2)+2u(a^2(1+w)+b^2(1-w))\right)^{3/2}}\,\mathrm{d}w\,\mathrm{e}^{-u}u^2\,\mathrm{d}u\tag{2}$$ Since $(1)$ and $(2)$ are equal, yet negatives, they are both $0$.
-
Oh... my... God... – Norbert Mar 16 '12 at 22:01
I think you got the factors of $2$ wrong. That's why I didn't write them; I knew I'd get them wrong if I did :-) You have $x=(u+v)/2$ and $y=(u-v)/2$, so you should have factors of $1/2$, one from the numerator and one from the Jacobian if I'm not mistaken. – joriki Mar 17 '12 at 5:13
@joriki: I had originally included more steps, the second line was $$\int_0^\infty\int_{-u}^u\frac{(a\frac{u+v}{2}-b\frac{u-v}{2})\,\mathrm{e}^{-u}}{(a^2\frac{u+v}{2} + b^2\frac{u-v}{2} + c\frac{u^2-v^2}{4})^{3/2}}\tfrac12\mathrm{d}v\,\mathrm{d}u$$ I believe I have it correct. – robjohn Mar 17 '12 at 7:09
Ah, of course, sorry. See, I knew I'd get them wrong ;-) – joriki Mar 17 '12 at 9:09
Take $a=c=1$ and $b$ very small.
-
Interesting that for $b>0$ the integral is $0$, but not for $b=0$. – GEdgar Mar 15 '12 at 22:10 | 2015-07-07T00:33:58 | {
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https://math.stackexchange.com/questions/879706/why-is-it-that-if-i-count-years-from-2011-to-2014-as-intervals-i-get-3-years-bu/879714 | # Why is it that if I count years from 2011 to 2014 as intervals I get 3 years, but if I count each year separately I get 4 years?
I'm not a very smart man. I'm trying to count how many years I've been working at my new job. I started in May 2011.
If I count the years separately, I get that I've worked 4 years - 2011 (year 1), 2012 (year 2), 2013 (year 3), 2014 (year 4).
But if I count the years as intervals, I get that I've worked 3 years 2011 to 2012 (year 1), 2012 to 2013 (year 2), 2013 to 2014 (year 3).
Have I worked 4 years or 3 years?
• If you start counting at $\,0,\,$ not at $\,1,\,$ just as for birthdays (age), then both methods yield the correct result. – Bill Dubuque Jul 27 '14 at 16:35
• I believe this is one aspect of the fencepost problem, so it may be helpful to read that over: en.wikipedia.org/wiki/Off-by-one_error#Fencepost_error. In this case, each "May 201x" is like a post, and the actual time intervals are like the gaps between the posts. – jpmc26 Jul 27 '14 at 18:40
• Have you drawn the picture? – Lubin Jul 27 '14 at 19:31
• betterexplained.com/articles/… – tar Jul 28 '14 at 4:27
• If you start working at a company on December 31st, then the next day ask yourself for how many year's you've been working, the answer isn't 2 years. If you ask yourself in how many years have you worked, then the answer is 2. – FreeAsInBeer Jul 28 '14 at 12:55
For a short analogy: I was born in 1986, so I am 27 now. But I have lived in four different decades (the 80s, the 90s, the 00s and the 10s). This is not the same as forty years.
In the same way, you worked in four different years, but not for four years.
• Even better, although a lot of people lived in both 2nd and 3rd millennia, not every one of them is 2000 years old. – Cthulhu Jul 30 '14 at 14:05
You started in May 2011.
By May 2012, you had worked 1 year.
By May 2013, you had worked 2 years.
In May 2014, you completed 3 years.
It won't be 4 years until May of 2015.
You worked 3 years between May 2011 and May 2014, and 2 extra months.
Note that in 2011 and 2014 you only work a part of the year, you should not count them as full years (as you do in the first count).
• Why 2 extra months? From May 1st 2011 to May 1st 2014 you have 3 full years. If the OP worked all of May 2014, this would add one extra month. Am I doing it wrongly? – landroni Jul 28 '14 at 13:24
• @landroni, I think its because May is now two months ago. – Winston Ewert Jul 28 '14 at 15:02
• @landroni Well, may could be May 1st, but could also be later. And technically, when we could months, we count full months. Even if he started May 1st, three months would be July 1st. – N. S. Jul 28 '14 at 15:11
• If we assume May 1st as the start, and July 1st as the end, then: May 1st 2011 to May 1st 2014 yield 3 full years, and May 1st to July 1st yield two full months (the entire months of May and June). I'm not sure how you end up with three months in the comment above. – landroni Jul 28 '14 at 16:29
• @landroni Yea, that was a typo, I meant August 1st... My point was that even with May 1st as start, there are not yet three months... :) – N. S. Jul 28 '14 at 16:39
This is the problem of counting intervals lengths vs. counting markers separating them: $$|_{11}\;X\quad|_{12}\;X\quad|_{13}\;X\quad|_{14}$$ Let's assume for simplicity that we're in May 2014.
If you want to know the length of the time you worked, sum the intervals (the $X$'s).
If you want to know how many different years you've been in during your work, count the delimiters (the $|$'s)
• This comes from really ancient times: in Latin, time intervals always counted the starting point. This has remained in Italian, where we say “quindici giorni” (fifteen days) to mean “two weeks” when talking about time intervals: “Lo consegnerò fra quindici giorni” literally translates to “I'll deliver it in fifteen days”, but a fortnight is really meant. – egreg Jul 28 '14 at 10:11
• As Colin D Bennett suggested, this is the so called "Fencepost error" (en.wikipedia.org/wiki/Off-by-one_error#Fencepost_error). The analogy is to the difference between counting the posts of a fence versus the number of rail sections between the posts. – Stan Jul 28 '14 at 17:52
• Which makes for nice primary school exercises: a road is $100\,$m long and we have to plant trees every $10\,$m; how many trees do we need? The first answer of the entire class of future primary school teachers was (with a few exceptions) $10$. – egreg Jul 28 '14 at 17:54
• @egreg: $10$ is the correct answer, if you plant the first tree $5$m from the end. $11$ won't fit in $100$m, they need at minimum $100$m + $d$, with $d$ being the diameter of one tree. You could make a case for $9$, though... – Ben Voigt Jul 28 '14 at 22:22
You're driving along a highway, starting at mile marker 11. After a while you get to mile marker 14. You've driven 3 miles, but you have seen 4 mile markers. To count miles traveled, count the mile marker at the end of each completed mile. You completed miles at markers 12, 13, and 14.
On the same trip you noticed that telephone poles are 100 feet apart. You start counting phone poles. How many poles do you count from pole number 11 to pole number 14 (inclusive)? How far did you travel?
When I studied Discrete Math at uni, they referred to this principle as "$1$ rope has $2$ ends".
The analogy in your example is a rope whose length is $3$ years, which has $4$ marks on it:
• +1. I'd always heard (and used) the fencepost metaphor. I like this rope idea! – Blue Apr 25 '16 at 14:00
• @Blue: Haha, never heard of that one. I guess they use different metaphors in different cultures. – barak manos Apr 25 '16 at 14:01
You've worked 3 years. The "interval approach" is the correct approach:
May 2011-May 2012.
May 2012 - May 2013.
May 2013 - May 2014.
May 2014 - July 27 2014.
= Three years and anywhere from two to three months, depending on your start date in May 2011.!
Unless you worked all of 2011, all of 2012, all of 2013, and all of 2014, you cannot claim 4 years. In effect, working four years would require having worked the entire year of 2011, 2012, 2013, and 2014, meaning you would have worked: January 01, 2011 until December 31st, 2014.
In reality, though, you'd need to subtract from four years the first $4+$ months of 2011, as well as the last 5+ months of the current year.
I see the problem of how long you have been working has been solved by others. Now let's focus on what you are asking in the title to your question.
The reason you get different numbers, is because you are counting 2 different ways. If you truly want to know how long you have worked, you have to think in terms of completed units (years, months, days, etc) to however much detail you are interested in.
Once you start thinking of completed years, 2011 is no longer contributing to the counting. You have completed 1 year in May of 2012 (As @amWhy noted), 2 years in May of 2013, etc.
When thinking in completed years, the amount of years will be
a) the difference between the end and start year (provided you have passed your hire date in the end year)
b) the result from a) minus 1 (provided you have yet to pass your hire date in the end year)
When you count by intervals, you are getting the difference between 2014-2011. When you count the years, you are counting the numbers in a sequence 2011, 2012, 2013, 2014.
The results from the 2 forms of counting will always be off by 1.
i.e 10-5 = 5 and the amount of numbers 5,6,7,8,9,10 -> 6 are off by 1
If you count 2011 to 2014 as intervals and get only 3 years, it's because you're excluding one of the years. The interval from midnight January 1 2011 to midnight January 1 2014 is in fact three years; but it excludes all of 2014. The period from January 1 2011, to December 31 2014 is four years. That period spans the years 2011, 2012, 2013 and 2014.
You're mixing up counting and measuring.
The issue doesn't change when you measure from May to May; that is just an offset. From May 2013 to May 2014 is a period of one year, which involves two calendar years.
I got into a small argument with a friend of mine over this same problem in elementary school. When reading a book, my friend would say that he's read, say, 10 pages once he gets to page 10. I would say that he's only read 9, because he hasn't actually read page 10 if he's just gotten there. You've only worked 3 years (give or take a couple of months). The reasoning is: You've reached, but not finished, your fourth year. Therefore, the current year that you're working shouldn't be counted. Whenever you count the years, just count up to the year before the current year. So we count 2011, 2012, and 2013, but not 2014 because 2014 isn't finished yet.
You've worked in 4 years. In this sense, a "year" is a numbered calendar year like 2011, 2012, 2013, or 2014. Whether one has worked in a particular numbered calendar year is a yes-or-no matter of whether one has worked on any day in that year.
On the other hand, you've worked for 3 years. In this sense, a "year" is not a numbered calendar year but rather a particular length of time (365 and/or 366 days).
You have worked three full years (2011,2012 and 2013). You are currently half-way through 2014.After this year has gone by you should also count the interval 2014-2015.
• How is 2011 a “full year” if the OP started working in May? – DaG Jul 28 '14 at 0:04
• Oh, I was just saying the flaw in the reasoning that he had worked in 2011,2012,2013,2014 because he hadn't in fact worked all of 2014. – Jorge Fernández Hidalgo Jul 28 '14 at 1:50
I used to face same kind of confusion, earlier. Then I realized that it becomes very easy if I term the day I started my job as a work anniversary day and center my calculation on that day. and start by counting the years,then months and so on.
For eg. I started on 27th July 2009.(and honestly I did!)
By today, I had worked for 5 years $2014-2009=5$ and 1 day.
Assuming you started on 12th May of 2011, By 2014 you have worked for 3 years $2014-2011$ and $7-5=2$ months and $28-12=16$ days.
So your overall tenure : $3 years$,$2 months$ and $16 days$
Cheers!
One more example. I have movie tickets to sell. If I sell tickets #001 thru (including) #010, I've sold 10 tickets. Simple? But if I sell #345 - #354, and subtract, I get 9. I need to account for the 'end' ticket and not just subtract numbers.
## protected by Asaf Karagila♦Jul 28 '14 at 22:36
Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count). | 2019-08-21T13:19:02 | {
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https://www.physicsforums.com/threads/which-ball-has-the-highest-velocity-just-before-ground.389817/ | # Which ball has the highest velocity just before ground
My girlfriend recently took a physics exam, and she was confused about why she got one of the problems wrong. She asked me hoping I could clear it up. Heres my attempt
## Homework Statement
Three Identical balls are thrown from the top of a building, all with the same initial speed. Ball A is thorwn horizontally, ball B at an angle above the horizontal and ball c at the same angle but below the horizontal. Neglecting air resistance, which ball has the highest speed just before they hit the ground
a) Ball A has the highest
b) Ball B has the highest
c) Ball C has the highest (My girlfriends guess)
d) Balls B and C have same speed
e) Three balls have the same speed (The correct answer)
## Homework Equations
She didnt write anything down, but id assume the kinematic equations
## The Attempt at a Solution
She wrote "It is C because it is thrown downward, so its not just gravity pulling/pushing it down"
The problem seems pretty straight forward. According to the problem air resistance is neglected, so its essentially like working in a vacuum. So since the ball is constantly under the influence of a gravitational field, it should continue to accelerate (Until it of course is stopped by the ground). Since Ball A is thrown at a horizontal, the y component of the velocity is still zero, so the final velocity is simply $$\sqrt{2g(y_{f}-y_{i})}$$. Ball B is thrown above the horizontal (lets assume straight up, since the x component is irrelevant in this scenario). The ball will go up, get to a peak, and then begin to fall. this point will be called $$y_{p}$$ where $$y_{p}$$>$$y_{i}$$. since at $$y_{p}$$, $$V_{i}$$ = 0, the final velocity is $$\sqrt{2g(y_{f}-y_{p})}$$. Ball C is throw down. Lets again ignore the x component. the difference this time is that the is a $$V_{i}$$. so for the final velocity, we have $$\sqrt{V^{2}_{i}+ 2g(y_{f}-y_{i})}$$
From what I wrote above, it should seem clear that V$$_{fballA}$$ < V$$_{fballb}$$, V$$_{fballC}$$ since both Ball B and Ball C have terms that would increase the final velocity over Ball A's. Ball B had more distance to fall (thus more time to accelerate), and Ball C had an initial velocity. Furthermore, we can prove (im not going to to) the magnitude of the velocity of Ball B initially is the same as the next time it passes y$$_{i}$$, so essentially, the final velocitys of Ball B and C should be the same.
I think after developing my argument this much, i might have run into a snag. my mind is now thinking that if I throw the ball down, i will lose time spent under the influence of gravity. maybe thats why they all hit the ground with the same velocity. my last argument is kind of a weak one, but here it goes. If I through the ball at the almost the speed of light towards the ground, it would get to the ground at roughly the same speed. I just dont see it being possible for the ball thrown horizontally to be able to go the fast from essentially just dropping it.
Ive managed to confuse myself. I originally believed D, but now im not sure.
vela
Staff Emeritus
Homework Helper
Think conservation of energy.
If you would calculate what the height of the peak was, you would see that
B has the same final velocity as C
The velocity in the x direction is larger for ball A, this makes up for the smaller downwards velocity of A.
The easiest way to see all this is with conservation of energy. In all cases, the loss of potential energy is the same, so the gain of potential energy is also the same. Since the initial speed and initital kinetic energy is the same, the final kinetic energy and thus the speed will also be the same.
I see my problem now.
Velocity is a vector. i kept thinking y direction. argh | 2022-05-25T18:51:57 | {
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https://www.physicsforums.com/threads/models-for-population-growth.920893/ | # Models for Population Growth
1. Jul 23, 2017
### FritoTaco
1. The problem statement, all variables and given/known data
The Pacific halibut fishery has been modeled by the differential equation.
$\displaystyle\dfrac{dy}{dt}=ky\left(1-\dfrac{y}{M} \right)$
where y(t) is the biomass (the total mass of the members of the population) in kilograms at time t (measured in years), the carrying capacity is estimated to be $M = 7\times 10^7 kg$, and $k=0.78$ per year.
(a) If $y(0)= 2\times 10^7 kg$, find the biomass a year later. (Round your answer to two decimal places.)
(b) How long will it take for the biomass to reach $4\times 10^7 kg$? (Round your answer to two decimal places.)
2. Relevant equations
$\displaystyle P= \dfrac{K}{1+Ce^{-kt}}$
3. The attempt at a solution
K = carrying capacity $\implies 7\times 10^7 kg$
k = $0.78$ per year
At time 0, biomass is $2\times 10^7 kg$ $\implies$$y(0)= 2\times 10^7 kg$
C = the difference between the carrying capacity and the initial capacity subtracted by 1.
$C= \dfrac{7\times 10^7}{2 \times 10^7}-1=\dfrac{5}{2}$
$P=\dfrac{7 \times 10^7}{1+\dfrac{5}{2}e^{-0.78\cdot1}}= 70000001.15 kg$
I'm trying to solve for the biomass (P) after 1 year. This answer doesn't seem correct. Am I using the wrong number for variable t? Or I'm not solving for P right away?
2. Jul 23, 2017
### Staff: Mentor
The two sides here are not equal. The denominator is larger than 1, the fraction cannot be larger than the numerator.
3. Jul 23, 2017
### Ray Vickson
Is $P$ the same as $y$? I assume so.
Your final equation is incorrect; it should be
$$\frac{7 \times 10^7}{1 +\displaystyle \frac{5}{2} e^{-0.78 \times 1}} = 4 \times 10^7$$
4. Jul 23, 2017
### FritoTaco
Thanks, guys, I got the right answer now. Major parentheses mistake for the calculator.
I can solve b, but for a.) I got $3.26\times10^7$
5. Jul 24, 2017
### Ray Vickson
6. Jul 24, 2017
### FritoTaco
Sorry for my bad English. For (a) I got $3.26 \times 10^7$ | 2017-08-22T18:09:55 | {
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https://dsp.stackexchange.com/questions/59694/laplace-transform-of-a-finite-duration-signal | # Laplace transform of a finite duration signal
Consider the following signal: $$x(t) = e^{-2t}[u(t) - u(t-5)]$$
This signal exists only from 0 to 5 time units. Elsewhere, it is zero.
Now, let's find the laplace transform of this signal using Linearity and Time shift properties.
$$e^{-2t}u(t) \leftrightarrow \frac{1}{s+2} \ , \ \ Re \{s \} > -2$$ Also, $$e^{-2(t-5)}u(t-5) \leftrightarrow \frac{e^{-5s}}{s+2} \ , \ \ Re \{s \} > -2$$ $$\Rightarrow e^{-2t}u(t-5) \leftrightarrow \frac{e^{-5s}e^{-10}}{s+2} \ , \ \ Re \{s \} > -2$$
Thus, by linearity property, $$e^{-2t}[u(t) - u(t-5)] = \frac{1 - e^{-5(s+2)}}{s+2} \ , \ \ Re \{s \} > -2$$
Note: The time shifting property doesn't alter the ROC;
However, the textbooks that i am refering (Oppenheim and Schaum series) both tell that the ROC of a finite duration signal is the entire S-plane, possibly zero and infinity (in some cases).
But the above signal being of finite-duration, possess ROC that is not the entire s-plane. Please help me figure this conceptual error.
Note: The above problem is from Schaum series. Here are the images of the textbook's section relevant to the above question.
Source of the Question and its solution:
Property of finite duration signals:
In Schaum's outline series: In oppenheim:
The property claimed by Schaum and Oppenheim is also true for the given example. Note that the Laplace transform
$$X(s)=\frac{1-e^{-5(s+2)}}{s+2}\tag{1}$$
has no pole at $$s=-2$$:
$$\lim_{s\to -2}X(s)=\frac{1-(1-5(s+2))}{s+2}\Big{|}_{s=-2}=5$$
So the ROC is indeed the entire $$s$$-plane. Even though the ROCs of the two individual signals in your solution have the same right half-plane as ROC, their sum has the entire $$s$$-plane as ROC because the two signals cancel everywhere except in a finite interval.
• Seems to me that it has a pole and a zero at $s=-2$. Jul 23 '19 at 10:12
• @robertbristow-johnson , yes. Thats why there is a pole-zero cancellation, thus resulting in the ROC as the entire s-plane Jul 23 '19 at 10:27
• @MattL. Thank you. That is what i was thinking. Thus the ROC given in the textbook is wrong, and it should be the entire s-plane. The linearity property also tells that the ROC for linear combination of two signals is atleast $$R = R_{1} \cap R_{2}$$ Jul 23 '19 at 10:31
• @Jarvis: Yes, the textbook is wrong concerning the ROC. When the linear combination of two signals is zero except for a finite interval then the ROC is always the entire $s$-plane. Jul 23 '19 at 10:42
• @robert bristow-johnson yes. I agree with you when there is a cancellation of an unstable pole, as though the system may be externally stable (i.e., BIBO stable), it may not be internally stable when spoken in terms of lyapunov's stability as some inputs may excite the unstable modes and lead to system instability. The unstable poles can also result as a part of improper pole placement via complete state feedback. However, in this simple situation, I guess we can safely ignore those aspects and conclude that the pole zero cancellation do not result in any adversity Jul 24 '19 at 5:11 | 2021-10-15T21:21:57 | {
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https://math.stackexchange.com/questions/1761735/let-mathcal-s-be-the-collection-of-all-straight-lines-in-the-plane-mathbb-r | # Let $\mathcal S$ be the collection of all straight lines in the plane $\mathbb R^2$. If $\mathcal S$ is a subbasis for a topology …
Let $\mathcal S$ be the collection of all straight lines in the plane $\mathbb R^2$. If $\mathcal S$ is a subbasis for a topology $\mathcal T$ on the set $\mathbb R^2$, what is the topology?
I know that the topology must be the discrete topology, but I can't seem to think how I can go about showing it. I tried searching online and came across this (see picture below), but something does not seem correct about it:
The above screenshot was taken from here.
The reason I feel that this is not correct, is because they use the fact that the point $(a,b)$ is open. However, $\mathbb R^2$ is Hausdorff, and so singleton sets (i.e points) are closed?
Is there any other way that I can go about proving this?
• $\mathbb R^2$ is Hausdorff with the standard topology. We are imposing a new topology generated by $\mathcal S$, thus it need not be Hausdorff. – Trevor Norton Apr 27 '16 at 20:53
• @TrevorNorton . That makes sense :). Thanks! But how can we conclude that the point $(a,b)$ is in fact open in the topology generated by $\mathcal S$? – user290425 Apr 27 '16 at 20:55
• Every set in the subbasis is open, and any finite intersection of sets from the subbasis is also open. Since we can pick two lines whose only intersection is a single point (say $x=a$ and $y=b$), we can say that the single point is also open. – Trevor Norton Apr 27 '16 at 20:58
• Actually, this new topology is also Hausdorff ... – Hagen von Eitzen Apr 27 '16 at 21:01
• @HagenvonEitzen . Does the argument given by Trevor above still hold though in showing that the point $(a,b)$ is open? – user290425 Apr 27 '16 at 21:11
Let $\mathscr T$ be a topology on a non-empty set $X$ and let $\mathscr S\subseteq \mathscr T$. Then, by definition, $\mathscr S$ is a subbasis for the topology $\mathscr T$ if for any $U\subseteq X$, one has that $U\in\mathscr T$ if and only if $U$ can be expressed as a union of sets which are finite intersections from sets in $\mathscr S$. Formally: $$U=\bigcup_{\gamma\in\Gamma}\bigcap_{i\in F_{\gamma}} S_i^{\gamma},$$ where
• $\Gamma$ is an (arbitrary) index set,
• for each $\gamma\in\Gamma$, $F_{\gamma}$ is a finite index set, and
• for each $\gamma\in\Gamma$ and $i\in F_{\gamma}$, it holds that $S_i^{\gamma}\in\mathscr S$.
In the concrete example, note that any set consisting of a single point in $\mathbb R^2$ can be expressed as the intersection of two lines, which lines are in $\mathscr S$ by assumption (in this case, the index set $\Gamma$ has a single element $\gamma$, and the index set $F_{\gamma}$ has two elements, corresponding to the two lines). This implies that every one-point set in $\mathbb R^2$ is open according to this new topology $\mathscr T$, so that $\mathscr T$, in fact, must be the discrete topology (in which every subset is open).
In the light of your question:
However, $\mathbb R^2$ is Hausdorff, and so singleton sets (i.e points) are closed?
Two comments: Firstly, the new topology $\mathscr T$ is different from the standard Euclidean topology. This is a common source of confusion, since we are so used to working with the usual Euclidean topology on $\mathbb R^2$ that we have a strange feeling upon seeing that nothing prevents us from defining any other topology on $\mathbb R^2$. Therefore, that the Euclidean topology is Hausdorff does not say anything about whether the discrete topology $\mathscr T$ is Hausdorff or not.
That said, secondly, the discrete topology on $\mathbb R^2$ does happen to be Hausdorff, so that it is also $T_1$ (i.e., the singletons are closed). This is compatible with the fact that singleton sets are open in the discrete topology, since any subset of the space is both open and closed under this topology. | 2019-06-17T04:53:28 | {
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https://math.stackexchange.com/questions/3587239/arctanx-arctany-from-integration/3588963 | # $\arctan{x}+\arctan{y}$ from integration
I was trying to derive the property $$\arctan{x}+\arctan{y}=\arctan{\frac{x+y}{1-xy}}$$ for $$x,y>0$$ and $$xy<1$$ from the integral representation $$\arctan{x}=\int_0^x\frac{dt}{1+t^2}\,.$$ I am aware of "more trigonometric" proofs, for instance using that $$\tan{(\alpha+\beta)}=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$$, but I was willing to see if there is a proof that uses more directly the properties of the integral representation. For instance, if $$x>0$$, one immediately gets \begin{aligned} \arctan{x}+\arctan\frac{1}{x} &=\int_0^x\frac{dt}{1+t^2} + \int_0^{\frac{1}{x}}\frac{dt}{1+t^2}\\ &= \int_0^x\frac{dt}{1+t^2}+\int_x^\infty\frac{dt}{1+t^2}\\ &=\int_0^\infty \frac{dt}{1+t^2} = \frac{\pi}{2} \end{aligned} sending $$t\to\frac{1}{t}$$ in the second integral. Similarly I tried considering $$\int_0^x\frac{dt}{1+t^2} + \int_0^y\frac{dt}{1+t^2}=(x+y)\int_0^1\frac{1+xyt^2}{1+(x^2+y^2)t^2+x^2y^2t^4}\ dt$$ after rescaling $$t\to xt$$ and $$t\to yt$$. On the other hand, via a similar rescaling $$t\to \frac{x+y}{1-xy}t$$, we have $$\int_0^\frac{x+y}{1-xy}\frac{dt}{1+t^2} = (x+y)\int_0^1\frac{1-xy}{(1-xy)^2+(x+y)^2t^2}\ dt\,.$$ By a clever choice of variable it should (must?) be possible to see that these integrals are actually the same, but I can't figure it out...
• This identity holds for $xy<1$. In the other cases, you need to add or subtract a multiple of $\pi$ to the RHS. – bjorn93 Mar 19 at 23:28
• True, probably I should add that in the statement of the problem, thanks – Brightsun Mar 20 at 7:33
We want show that $$\begin{eqnarray*} \int_x^{ \frac{x+y}{1-xy}} \frac{dt}{1+t^2} = \int_{0}^{y} \frac{du}{1+u^2} \end{eqnarray*}$$ that's to say the LHS is actually independent of $$x$$.
The substitution $$\begin{eqnarray*} t=x+ \frac{u(1+x^2)}{1-ux} \end{eqnarray*}$$ will do the trick.
The limits are easily checked and we have $$\begin{eqnarray*} dt= \frac{1+x^2}{(1-ux)^2} du. \end{eqnarray*}$$ The rest is a little bit of algebra.
Note the similarity with $$\ln(a)+\ln(b) = \ln(ab)$$ $$\begin{eqnarray*} \int_{1}^{a} \frac{dt}{t} +\int_{1}^{b} \frac{dt}{t} = \int_{1}^{ab} \frac{dt}{t}. \end{eqnarray*}$$ And $$u=at$$ $$\begin{eqnarray*} \int_{1}^{b} \frac{dt}{t} = \int_{a}^{ab} \frac{du}{u}. \end{eqnarray*}$$
• Nice, thank you. I think the substitution is slightly more insightful written as $t=\frac{x+u}{1-ux}$ due to the similarity with $\frac{x+y}{1-xy}$. Just a curiosity, did you arrive at the result by trial and error or did you have some guiding principle? – Brightsun Mar 19 at 22:14
• Don't tell anyone ... but the first guess written in my notebook is \begin{eqnarray*} t=x + \frac{y(1+u^2)}{1-uy}. \end{eqnarray*} "Trial and error" or "guiding principle"? A bit of both. Thanks, this is a cool problem. – Donald Splutterwit Mar 19 at 22:21
hint
If we put
$$t=\frac{x+y}{1-xy}u$$
the left integral becomes
$$\int_0^1\frac{1}{1+(\frac{x+y}{1-xy})^2u^2}\frac{x+y}{1-xy}du$$
• I think that's what I did in the last step of my question..? – Brightsun Mar 19 at 20:59
• Sorry, i was thinking about corona, i did not read your question as i had to do. the problem is that i cannot delete from mobile. i hope no one will downvote. – hamam_Abdallah Mar 19 at 21:04
Fixing $$y$$, define $$f(x):=\arctan x+\arctan y-\arctan\frac{x+y}{1-xy}$$ so $$f(0)=0$$ and\begin{align}f^\prime(x)&=\frac{1}{1+x^2}-\frac{1}{1+\left(\frac{x+y}{1-xy}\right)^2}\partial_x\frac{x+y}{1-xy}\\&=\frac{1}{1+x^2}-\frac{(1-xy)^2}{(1+x^2)(1+y^2)}\frac{1-xy-(x+y)(-y)}{(1-xy)^2}\\&=\frac{1}{1+x^2}-\frac{(1-xy)^2}{(1+x^2)(1+y^2)}\frac{1+y^2}{(1-xy)^2}\\&=0,\end{align}i.e. $$f(x)=0$$ for all $$x$$.
Another change of variables that works, very similar to that in the answer by @DonaldSplutterwit, is the following: $$t=f(u)=\frac{1-u\sigma}{u+\sigma}\,,\qquad\text{with}\ \ \sigma(x,y)=\frac{1-xy}{x+y}\,.$$ It is more symmetric, since it works both for $$\int_x^{1/\sigma}\frac{dt}{1+t^2}=\int_0^y\frac{du}{1+u^2}$$ and for $$\int_y^{1/\sigma}\frac{dt}{1+t^2}=\int_0^x\frac{du}{1+u^2}\,.$$ Indeed, $$f(0)=\frac{1}{\sigma}\,,\qquad f(x)=y\,,\qquad f(y)=x$$ and $$dt=-\frac{1+\sigma^2}{(u+\sigma)^2}du\,,\qquad \frac{1}{1+t^2}=\frac{(u+\sigma)^2}{(1+\sigma^2)(1+u^2)}\,.$$ It also has the property of reducing to the inversion as $$xy\to1^-$$, namely $$\sigma\to0^+$$, since $$f(u)\big|_{\sigma=0}=\frac{1}{u}\,,$$ and we get back $$\int_{x}^\infty \frac{dt}{1+t^2} = \int_{0}^{\frac{1}{x}}\frac{du}{1+u^2}\,.$$ In fact, $$f$$ is also an involution $$f(f(u))=u$$ and also allows to run the proof "forward" in the following way $$\int_0^x\frac{dt}{1+t^2}+\int_0^y\frac{dt}{1+t^2}=\int_0^x\frac{dt}{1+t^2}+\int_x^{\frac{1}{\sigma}}\frac{du}{1+u^2}=\int_0^{\frac{1}{\sigma}}\frac{dt}{1+t^2}\,,$$ where we let $$t=f(u)$$ in the second integral. | 2020-06-06T12:10:30 | {
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http://math.stackexchange.com/questions/417678/find-values-of-the-constants-in-the-following-identity-x4ax3-5x2-x-3/417684 | # Find values of the constants in the following identity: x^4+Ax^3 + 5x^2 + x + 3 = (x^2+4)(x^2-x+B)+Cx+ D
Another question on identities:
$$x^4+Ax^3 + 5x^2 + x + 3 = (x^2+4)(x^2-x+B)+Cx+ D$$
How can I find the coefficients for this?
I've got as far as multiplying out the brackets to get:
$$x^4+Ax^3 + 5x^2 + x + 3 = (x^4-x^3+Bx^2+4x^2-4x+4B)+Cx+ D$$
It would be useful to get a hint at least on where to go next.
-
You're almost to the final solution. Just combine all coefficients of various powers or $x$ and compare each of them individually:
$$x^4+Ax^3 + 5x^2 + x + 3 = x^4-x^3+ (B+4)x^2+(C-4)x+4B+ D$$ You'll get
\begin{align} A &= -1 \\ B &= +5 - 4 \\ &= 1 \\ C &= +1 +4 \\ &= 5 \\ D &= 3 - 4B \\ &= -1 \end{align}
### Edit
To explain a little as OP mentioned:
The process is as simple as taking everything onto one side of $=$:
$$(x^4 - x^4) + (A x^3 + x^3) + (5 x^2 - (B + 4) x^2 ) + (x - (C - 4)x ) + \left(3 - (4B + D)\right) = 0 \\ \implies (A + 1) x^3 + (1 - B) x^2 + (4 - C) x + (3 - 4B - D) = 0 \\ \therefore \pmatrix{ A + 1 \\ 1 - B \\ 4 - C \\ 3 - 4B - D } = \pmatrix{ 0 \\ 0 \\ 0 \\ 0 }$$ which gives you the result.
-
@AbhraAbirKundu Thanks. I used the exponent :/ I should sleep now. – hjpotter92 Jun 11 '13 at 17:38
You are welcome... – Abhra Abir Kundu Jun 11 '13 at 17:40
does this mean you should simplify further by removing x^4 from the rhs and lhs?, divide both sides by x ? – peter_gent Jun 11 '13 at 22:59
I'm missing some steps or intuition on how exactly you compare. can you please explain – peter_gent Jun 11 '13 at 23:30
@peter_gent Okay, check the edit to my reply. – hjpotter92 Jun 12 '13 at 4:53
Now you can equate the coefficients of each power of $x$ from both sides.
Like you equate the coefficient of $x^3$ to get $A=-1$
- | 2014-12-25T14:16:24 | {
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http://math.stackexchange.com/questions/493924/justify-a-relation/493953 | Justify a relation
let $\mathrm A = \Bbb Z \text{ and } R = \{(a,b) \in \mathrm A\times \mathrm A | a \lt b \}$ investigate whether the relationship is symmetric or antisymmetric.
So (...)
• Symetric [FALSE]: $(a,b) \in R \implies a \lt b \implies (b,a) \not\in R \text { as if } a\lt b \implies b \not\lt a$
• Antisymetric [TRUE]: $(a,b) \in R \land (b,a)\not\in R \implies a = b$.
Since if $P \land Q \implies S$, if $P$ or $Q$ is false, then no matter the logic value in $Q$, because $F \implies F \equiv \mathrm T$ and $F \implies V \equiv \mathrm T$, but I don't know how to justify this antisymetry in this relation.
There is a way to write this as a math notation?
Thanks!
-
We want to show: $\forall x,y \in A\,(\,(x,y)\in R \land (y,x)\in R \implies x=y)$.
This is how I will do it:
Suppose $x$ and $y$ are arbitrary elements of $A$ and suppose $(x,y)\in R$ and $(y,x)\in R$. Then $x<y$ and $y<x$, but this is impossible. So, our statement is vacuously true.
Another way would be to prove the contrapositive: $\forall x,y \in A\,(\, x\ne y\implies (x,y)\not\in R \lor (y,x)\not\in R)$ which isn't difficult to prove.
Or you may directly write it out like this,
\begin{align} &\forall x,y \in A\,(\,(x,y)\in R \land (y,x)\in R \implies x=y)\\ &\equiv \forall x,y \in A\,(\,x<y \land y<x \implies x=y)\\ &\equiv \forall x,y \in A\,(\text{ False} \implies x=y)\\ &\equiv \forall x,y \in A\,(\text{ True })\\ \end{align} which is equivalent to $\text{True}$.
-
What you have could stand to be polished up a bit, but it’s basically fine (though you’ve done more than necessary for symmetry). Here’s how I might write it:
• Symmetry: $R$ is not symmetric, because for instance $\langle 1,2\rangle\in R$, but $\langle 2,1\rangle\notin R$: $1<2$, but $2\not<1$.
• Antisymmetry: $R$ is antisymmetric: the antecedent of the implication $$\Big(\langle a,b\rangle\in R\land\langle b,a\rangle\in R\Big)\to a=b$$ is always false, so the implication is vacuously true.
Under symmetry you actually proved that $R$ is asymmetric: if $\langle a,b\rangle\in R$, then $\langle b,a\rangle\notin R$. You could write this up a little better as follows:
• Asymmetry: Suppose that $\langle a,b\rangle\in R$. Then $a<b$, so $b\not< a$, and therefore $\langle b,a\rangle\notin R$.
Note that the argument that you and I both gave for antisymmetry really shows that an asymmetric relation is always antisymmetric.
- | 2016-06-26T01:18:46 | {
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https://math.stackexchange.com/questions/412301/does-let-x-be-a-member-of-s-require-axiom-of-choice | # Does 'let x be a member of S…' require axiom of choice? [duplicate]
A lot of proofs start like this
'Let x be some member of the set S, then...'
(If S are the natural numbers, this is especially common)
Do all of those proofs essentially require axiom of choice? Or is it possible to talk about an element of a set without being able to 'pick' it?
The issue with the Axiom of Choice is that "choosing elements" is no different than stating there exists "some function" that chooses for you. Let's show the Axiom of Choice in use and not in use.
"$2^\mathbb{N}=\prod_{\mathbb{N}}\{0,1\}=\{0,1\}\times\{0,1\}\times\cdots$ is non-empty."
Well, $(0,0,0,0,\ldots)$ (that is, assign $0$ to each spot in the vector) is an infinitely long vector that is in the set. Thus, $2^\mathbb{N}$ is non-empty. This proof did not use the Axiom of Choice, simply because I told you the element to pick. The Axiom of Choice is used when you don't know what element to pick.
"Let $A_i$ for $i\in\mathbb{N}$ be a collection of non-empty sets, then $\prod_\mathbb{N} A_i$ is a non-empty set."
Well, I would like to say that $(a_0,a_1,a_2,\ldots)$ is a vector in the set, where $a_i\in A_i$. But stating that such a vector exists is the same as assuming that there is a function that assigns each $A_i$ to the element $a_i$. That is, if $v=(a_0,a_1,a_2,\ldots)$ really does exist then $$f:\{A_0,A_1,\ldots\}\to \{\text{all the elements in }A_i's\}$$ by $$f(A_i)=\text{the }i\text{th entry in vector }v,$$ is a function that must also exist.
And there is the Axiom of Choice being used. There is no reason to assume that such a function exists (or that such a vector exists). If you can't explicitly define the assigning function, then how do you know it exists? The Axiom of Choice is the assumption that there always exists a function that can assign a collection of sets to it's own elements.
We see the Axiom of Choice in even more subtle ways. For example suppose you had a proof like:
"Let $\Omega$ be an arbitrary set. Let $A$ be an infinitely countable subset of $\Omega$ ..."
But wait, if $A$ is infinitely countable then by definition $A$ is subscriptable by the natural numbers like so, $A=\{a_0,a_1,a_2,\ldots\}$ that means there is some function $f:\mathbb{N}\to \Omega$ exists and that $f(\mathbb{N})=A$. How do I know that such a function exists? I can certainly pick an element of $\Omega$ (at random) and assign it to one, I can then pick another element from $\Omega$ and assign it to two. But how do I pick an infinite number of elements and assign them to the Natural Numbers. I can't tell you how to do it, I have to assume that it is possible by The Axiom of Choice. The Axiom of Choice gives us that $\prod_\mathbb{N} \Omega$ is a non-empty set.
Notice the subtlety of the above discuss. Simply picking an infinitely countable subset from an arbitrary set requires the Axiom of Choice. Or for that matter, simply subscripting a set could be using the Axiom of Choice, after all subscripting is another way of assuming that a choice function exists. The Axiom of Choice is abused extensively when dealing with sub-scripting stuff like sequences.
In general, the Axiom of Choice comes into play when you are dealing with arbitrary sets. If you know anything about the sets, then you don't need the Axiom of Choice. For example, "$\cup_{n\in\mathbb{N}} \{0+\frac{1}{n},1+\frac{1}{n},\ldots\}$ is countable" can be proven without the Axiom of Choice, since I can explicitly define the function that shows this is countable. However, "The countable union of countable sets is countable" is not provable without the Axiom of Choice, since we are dealing with arbitrary sets.
The worst thing about the Axiom of Choice is that you can very well assume that it is not true and you won't contradict the other Axioms of ZF. Thus there really is no reason not to assume that there are arbitrary products of non-empty sets that are empty. :-(
Sorry for the long discussion.
• Right, so as long as we know the set is non-empty, we can pick an element. In the cases where we can't pick one, we simply don't know if the set has any elements (even though it seems very likely it does..) – Thomas Ahle Jun 5 '13 at 21:42
• Yes, "picking an element" from a non-empty set requires no axiom. It is more like a built in definition. If $A$ is a non-empty set, then by definition there is an object, which we will denote by $x$, such that the sentence $x\in A$ makes sense. No axiom needed. – Bobby Ocean Jun 5 '13 at 21:46
• Do I need aoc to prove that a family of sets where none of them are a subset of each other, has a maximum element? (I'm just trying to see exactly where the distinction goes) – Thomas Ahle Jun 6 '13 at 10:47
• Not sure exactly what you mean. The AOC is needed when you don't define a rule for picking an element from each set in a family of sets. For example, if $\Omega$ is a family of sets where each set is a subset of the natural numbers, then I could say "pick the smallest element in each set", no AOC needed; I gave you the rule. But if $\Omega$ is just some arbitrary family of sets and I say "pick an element from each set", then I have implied the existence of a 'picking function' without telling you what the picking function is; this uses the AOC. – Bobby Ocean Jun 6 '13 at 18:51
The axiom of finite choice is a theorem of ZF. All you need to be able to pick an element of $S$ is to know that $S$ is non-empty. The axiom of choice is about making infinitely many choices.
This is essentially just a rule of (classical) logic.
If $\phi (x)$ is a formula (with free variable $x$), such that $( \exists x ) ( \phi ( x ) )$ is true, then by existential instantiation picking some symbol $y$ not used in the proof already assert that $\phi (y)$ holds. In the case in question, if $A$ is a nonempty set, then the formula $( \exists x ) ( x \in A )$ is true, and by picking an appropriate symbol $y$ we may assert $x \in A$ (in other words, we pick some $y \in A$).
As Qiaochu mentions in his answer, the Axiom of Choice is for those cases where infinitely many choices are made at once. Even more, it is for those cases where infinitely many choices are made at one without any rule present to uniformly make these choices. For instance, if $\{ A_i : i \in \}$ is a family of finite nonempty subsets of $\mathbb{R}$, then choosing $y_i = \min ( A_i )$ for all $i \in I$ does not require the Axiom of Choice, since each $A_i$ must have a least element.
Actually, no choice is needed at all – not even finite choice. You can read such sentences as, ‘Assume $x$ is a member of $S$ ...’. More formally, what you are doing is introducing a variable, and you can always do that.
The issue with choice is not so much a question of finiteness as it is a problem with rearranging quantifiers. (The axiom of choice is a formula of the form $\forall \exists \to \exists \forall$.) Here's what a naïve “proof” of the axiom of choice might look like:
1. Suppose, for all $x$ in $S$, there exists a member of $T (x)$.
2. Let $x$ be a member of $S$.
3. Let $y (x)$ be a member of $T (x)$.
4. Then $y$ is a function such that $y(x) \in T (x)$ for all $x$ in $S$.
This is not a valid proof. Let's take a closer look at the variables in play at each step, in the formalism of (dependent) type theory:
1. No variables.
2. $x$ is a variable of type $S$.
3. $x$ is a variable of type $S$, $y (x)$ is a variable of type $T (x)$.
4. Not well-formed.
A more generous reading of the above would say that, in step 4, we tried to apply $\lambda$-abstraction to the expression $y (x)$; but the only legal $\lambda$-abstraction at that step eliminates the variable $y (x)$ to produce the function $t \mapsto t$. (We can't eliminate the variable $x$ because the type of the variable $y (x)$, namely $T (x)$, depends on $x$.) Even if we could eliminate the variable $x$, the resulting expression would still depend on $y (x)$ – not really what was intended!
As far as I know, saying "Let $x \in S$" does not involve any choice at all. You only need choice when you say "Choose $x \in S$".
• So when I say "Choose $x\in\Bbb R$ such that $x\notin\Bbb Q$" I am using the axiom of choice? – Asaf Karagila Jun 5 '13 at 19:48
• As part of a formal system, the words you use when describing the action should have no bearing whatsoever in whether you are appealing to the Axiom of Choice or not. (Otherwise we could presumably go to the opposite extreme of @Asaf's comment and proclaim that saying "Let $B$ be a Hamel basis for $\mathbb{R}$ as a vector space over $\mathbb{Q}$" does not require the Axiom of Choice, but by "choosing" instead we must appeal to the Axiom.) – user642796 Jun 5 '13 at 20:08
• @Arthur: But first we need to prove that the set $\{A\mid A\text{ is a Hamel basis for }\Bbb R\text{ over }\Bbb Q\}$ is non-empty, and for that we do need the axiom of choice. Of course, you already know that, but this is the subtlety of where we use the axiom of choice: we use it to prove a certain set is non-empty, and not to choose a particular element from it. – Asaf Karagila Jun 5 '13 at 20:11
• @AsafKaragila Doesn't "Choose $x \in \mathbb{R}$ such that $x \notin \mathbb{Q}$ require axiom of finite choice as Qiaochu said? – user81123 Jun 5 '13 at 20:21
• @ArthurFischer Well, when I say "Choose $x \in S$", I'll need $S$ to be nonempty, but when I say "Let $x \in S$", it doesn't matter if $S$ is empty or not. – user81123 Jun 5 '13 at 20:23 | 2020-09-29T23:27:48 | {
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http://math.stackexchange.com/questions/22461/verifying-carmichael-numbers | # Verifying Carmichael numbers
I'm trying to understand a solution I was given in a tutorial regarding a problem with Carmichael numbers and I was wondering if you guys can help clarify things:
A composite number $m$ is called a Carmichael number if the congruence $a^{m-1} \equiv 1 \pmod{m}$ is true for every number a with $\gcd(a,m) = 1$.
Verify that $m = 561 = 3 \times 11 \times 17$ is a Carmichael number.
Solution given:
Apply Fermat's Little Theorem to each prime divisor of $m$: \begin{align*} a^2 &\equiv 1 \pmod{3}\\ a^{10} &\equiv 1 \pmod{11}\\ a^{16} &\equiv 1 \pmod{17} \end{align*} This somehow then implies that $a^{80} \equiv 1 \pmod{561}$ then accordingly $a^{560} \equiv 1 \pmod{561}$.
I am lost as to how the 3 congruences imply $a^{80} \equiv 1 \pmod{561}$ ($80 = \mathrm{LCM}(2,10,16)$).
Can somebody clarify this for me?
Thanks!
-
for equivalence, use \equiv and put your $LaTeX$ in dollar signs. So you would write a^2 \equiv 1 \pmod 3 and get $a^2 \equiv 1 \pmod 3$ – Ross Millikan Feb 17 '11 at 15:33
Note that $80 = \mathrm{lcm}(2,10,16)$. So you can write $a^{80} = (a^2)^{40} = (a^{10})^{8} = (a^{16})^5$. So, \begin{align*} a^{80}= (a^2)^{40} &\equiv 1^{40} = 1\pmod{3},\\ a^{80}= (a^{10})^{8} &\equiv 1^8 = 1 \pmod{11},\\ a^{80}= (a^{16})^5 &\equiv 1^5 = 1\pmod{17}. \end{align*}
By the Chinese Remainder Theorem, the system of congruences \begin{align*} x&\equiv 1\pmod{3}\\ x&\equiv 1\pmod{11}\\ x&\equiv 1\pmod{17} \end{align*} has a unique solution modulo $3\times 11\times 17 = 561$. But both $x=1$ and $x=a^{80}$ are solutions. Since the solution is unique modulo $561$, then the two solutions we found must be congruent. That is, $$a^{80}\equiv 1\pmod{561}.$$
(Added. Or, more simply, as Andres points out, since $3$, $11$, and $17$ each divide $a^{80}-1$, and are pairwise relatively prime, then their product divides $a^{80}-1$).
Once you have that $a^{80}\equiv 1\pmod{561}$, then any power of $a^{80}$ is also congruent to $1$ modulo $561$. In particular, $$a^{560} = (a^{80})^{7} \equiv 1^7 = 1 \pmod{561}$$ as desired.
-
Arturo, I think you can do this more easily: Simply note that since $a^{80}-1$ is divisible by 3, 11, and 17, and these numbers are relatively prime, then it is divisible by their product. There is no need to invoke the Chinese Remainder Theorem. – Andrés E. Caicedo Feb 17 '11 at 7:35
@Andres: Yes; I mentioned the CRT argument because it's a very common argument that shows up a lot as well. – Arturo Magidin Feb 17 '11 at 14:12
@Andres: ... and it was after 1am at the time. (-; – Arturo Magidin Feb 17 '11 at 16:03
HINT $\:$ For primes $\rm\ p\neq q\:$ coprime to $\rm\:a\:,\:$ if $\rm\ p-1,q-1\ |\ m\$ then $\rm\ p,q\ |\ a^m - 1\ \Rightarrow\ pq\ |\ a^m - 1$ since lcm = product for coprime integers (here distinct primes).
- | 2016-06-30T17:45:43 | {
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https://math.stackexchange.com/questions/1666396/evaluate-the-infinite-product-prod-k-geq-2-sqrtk1-frac1k-sqrt1 | # Evaluate the infinite product $\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}=\sqrt{1+\frac{1}{2}} \sqrt[3]{1+\frac{1}{3}} \sqrt[4]{1+\frac{1}{4}} \cdots$
I can show the convergence of the following infinite product and some bounds for it:
$$\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}=\sqrt{1+\frac{1}{2}} \sqrt[3]{1+\frac{1}{3}} \sqrt[4]{1+\frac{1}{4}} \cdots<$$
$$<\left(1+\frac{1}{4} \right)\left(1+\frac{1}{9} \right)\left(1+\frac{1}{16} \right)\cdots=\prod_{k \geq 2} \left(1+\frac{1}{k^2} \right)=\frac{\sinh \pi}{2 \pi}=1.83804$$
Here I used Euler's product for $\frac{\sin x}{x}$.
The next upper bound is not as easy to evaluate, but still possible, taking two more terms in Taylor's series for $\sqrt[k]{1+\frac{1}{k} }$:
$$\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}<\prod_{k \geq 2} \left(1+\frac{1}{k^2}-\frac{k-1}{2k^4}+\frac{2k^2-3k+1}{6k^6} \right)=$$
$$=\prod_{k \geq 2} \left(1+\frac{1}{k^2}-\frac{1}{2k^3}+\frac{5}{6k^4}-\frac{1}{2k^5}+\frac{1}{6k^6} \right)<$$
$$<\exp \left( \frac{\pi^2}{6}+\frac{\pi^4}{108}+\frac{\pi^6}{5670}-1-\frac{\zeta (3)}{2}-\frac{\zeta (5)}{2} \right)=1.81654$$
The numerical value of the infinite product is approximately:
$$\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}=1.758743628$$
The ISC found no closed from for this number.
Is there some way to evaluate this product or find better bounds in closed form?
Edit
Clement C suggested taking logarithm and it was a very useful suggestion, since I get the series:
$$\ln \prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}= \frac{1}{2} \ln \left(1+\frac{1}{2} \right)+\frac{1}{3} \ln \left(1+\frac{1}{3} \right)+\dots$$
I don't know how to find the closed form, but I can certainly use it to find the boundaries (since the series for logarithm are very simple).
$$\frac{1}{2} \ln \left(1+\frac{1}{2} \right)+\frac{1}{3} \ln \left(1+\frac{1}{3} \right)+\dots>\sum^{\infty}_{k=2} \frac{1}{k^2}-\frac{1}{2}\sum^{\infty}_{k=2} \frac{1}{k^3}$$
$$\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}>\exp \left( \frac{\pi^2}{6}-\frac{1}{2}-\frac{\zeta (3)}{2} \right)=1.72272$$
$$\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}<\exp \left( \frac{\pi^2}{6}+\frac{\pi^4}{270}-\frac{5}{6}-\frac{\zeta (3)}{2} \right)=1.77065$$
$$\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}>\exp \left( \frac{\pi^2}{6}+\frac{\pi^4}{270}-\frac{7}{12}-\frac{\zeta (3)}{2} -\frac{\zeta (5)}{4}\right)=1.75438$$
$$\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}<\exp \left( \frac{\pi^2}{6}+\frac{\pi^4}{270}+\frac{\pi^6}{4725}-\frac{47}{60}-\frac{\zeta (3)}{2} -\frac{\zeta (5)}{4}\right)=1.76048$$
$$\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}>\exp \left( \frac{\pi^2}{6}+\frac{\pi^4}{270}+\frac{\pi^6}{4725}-\frac{37}{60}-\frac{\zeta (3)}{2} -\frac{\zeta (5)}{4} -\frac{\zeta (7)}{6}\right)=1.75803$$
This method generates much better bounds than my first idea. The last two are very good approximations.
Edit 2
Actually, would it be correct to write (it gives the correct value of the product):
$$\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}=\frac{1}{2} \exp \left( \sum_{k \geq 2} \frac{(-1)^k \zeta(k)}{k-1} \right)$$
• My money is on $e^{2\pi/5}/2$. Taking the logarithm and considering instead a sum may help... and starting the product at $1$ instead of $2$ may also -- this is the factor $2$ in my guess.) – Clement C. Feb 22 '16 at 1:25
• @ClementC., but it's different from the numerical value – Yuriy S Feb 22 '16 at 1:44
• Euler Mac Laurin gives 1.73287 keeping only the first two terms – tired Feb 22 '16 at 2:16
• The product evaluates to $\simeq 1.0011\cdot \frac{e^{2\pi/5}}{2}$ so the conjecture above seems to be ruled out (this was double checked by also evaluating the series representation). – Winther Feb 22 '16 at 2:29
• I googled the sum (from 1) and found something. books.google.com/… i.imgur.com/w5WQE2E.png – Steve Kass Feb 22 '16 at 2:34
I don't know if a closed form exists, but to get geometric convergence, we can use the following.
Since the product starts at $k=2$, we compute the sum \begin{align} \sum_{k=2}^\infty\frac1k\log\left(1+\frac1k\right) &=\sum_{k=2}^\infty\frac1k\sum_{n=1}^\infty\frac{(-1)^{n-1}}{nk^n}\\ &=\sum_{n=1}^\infty\frac{(-1)^{n-1}}n\sum_{k=2}^\infty\frac1{k^{n+1}}\\ &=\sum_{n=1}^\infty(-1)^{n-1}\frac{\zeta(n+1)-1}{n}\\[6pt] &=0.564599706384424320592667709038 \end{align} Note that $\frac{\zeta(n+1)-1}{n}\sim\frac1{n2^{n+1}}$. This gives better than geometric convergence.
Applying $e^x$, we get $$\prod_{k=2}^\infty\left(1+\frac1k\right)^{1/k}=1.75874362795118482469989684966$$
This is not an answer, but it's important and I post it separately from the question itself.
I found in this answer by @RandomVariable the following series:
$$\sum_{k=1}^{\infty} \frac{\ln (k+1)}{k(k+1)}=\frac{\pi^2}{4}-1-4\int_{0}^{\infty} \frac{\arctan x}{1+x^{2}} \frac{dx}{e^{\pi x}+1}=\frac{\pi^2}{4}-1-4\int_{0}^{\pi/2} \frac{t~dt}{e^{\pi \tan t}+1}$$
They are related to $\gamma_1$ - Stieltjes constant.
This same series also appeared in this paper by Steven Finch, page 5.
$$\sum_{k=1}^{\infty} \frac{\ln (k+1)}{k(k+1)}=1.2577468869$$
This is the same numerical value as:
$$\sum_{k=1}^{\infty} \frac{\ln (1+\frac{1}{k})}{k}=1.2577468869$$
Which is confirmed in this paper by the same author, page 3, where this form of the series is used.
It is connected to the integral (page 2, the same paper):
$$\sum_{k=1}^{\infty} \frac{\ln (1+\frac{1}{k})}{k}=-\int_{1}^{\infty} \frac{\ln (y-[y])}{y^2}dy$$
Where $[y]$ is the floor function, meaning $y-[y]$ is the fractional part of $y$.
In another paper this series is connected to the numer of divisors of $n!$, however slightly different integral representation is used (page 3):
$$\sum_{k=1}^{\infty} \frac{\ln (1+\frac{1}{k})}{k}=\int_{1}^{\infty} \frac{\ln ([y]+1)}{y^2}dy$$
And finally, this is slightly related to Alladi-Grinstead Constant, which is given by:
$$e^{c-1}$$
$$c=\sum_{k=2}^{\infty} \frac{\ln (\frac{k}{k-1})}{k}=\sum_{k=1}^{\infty} \frac{\ln (1+\frac{1}{k})}{k+1}=0.788530566$$
And this is also somehow connected to the Luroth series representations of real numbers.
Oh, and thanks to @SteveKass for this useful link.
Comparing the convergence of three series, we find that even though they are equivalent, the convergence rate is drastically different.
$$\sum_{k=1}^{\infty} \frac{\ln (k+1)}{k(k+1)}=\sum_{k=1}^{\infty} \frac{\ln (1+\frac{1}{k})}{k}=\sum_{k = 2}^{\infty} \frac{(-1)^k \zeta(k)}{k-1}$$
We can also obtain the following interesting equality:
$$(1+1)\sqrt{1+\frac{1}{2}} \sqrt[3]{1+\frac{1}{3}} \sqrt[4]{1+\frac{1}{4}} \cdots=\sqrt{2} \sqrt[6]{3} \sqrt[12]{4} \sqrt[20]{5} \sqrt[30]{6} \cdots=\prod_{k=1}^{\infty}(k+1)^{\frac{1}{k(k+1)}}$$
• Re: "This same series also appeared [...] This is the same numerical value as:": showing equality of the two (series, not deriving their value) is not too hard -- I assume you don't need it, but in case you do it's a matter of 3-4 lines. – Clement C. Feb 22 '16 at 4:18
• Which one? The 0.788530566 one? – Clement C. Feb 22 '16 at 4:24
• I have nothing really conclusive in making a useful connection between the values... what sort of relation are you hoping for? – Clement C. Feb 22 '16 at 4:35
• I don't know. It came up in my search and S. Finch hinted at some kind of connection. – Yuriy S Feb 22 '16 at 4:37
• \begin{align} \sum_{k=2}^\infty\frac{1}{k}\ln\left(1+\frac{1}{k}\right) &=\sum_{k=2}^\infty\frac{\ln(k+1)-\ln(k)}{k}\\ &=\lim_{N\to\infty}\left(\sum_{k=2}^{N}\frac{\ln(k+1)}{k}-\sum_{k=2}^{N}\frac{\ln(k)}{k}\right)\\ &=\lim_{N\to\infty}\left(\sum_{k=2}^{N}\frac{\ln(k+1)}{k}-\sum_{k=1}^{N-1}\frac{\ln(k+1)}{k+1}\right)\\ &=\lim_{N\to\infty}\left(\frac{\ln(N+1)}{N}+\sum_{k=2}^{N-1}\frac{\ln(k+1)}{k(k+1)}-\frac{\ln(2)}{2}\right)\\ &=\lim_{N\to\infty}\left(\sum_{k=2}^{N-1}\frac{\ln(k+1)}{k(k+1)}\right)-\frac{\ln(2)}{2}\\ &=\sum_{k=1}^{\infty}\frac{\ln(k+1)}{k(k+1)}\\ \end{align} – alex.jordan Feb 22 '16 at 5:15
As already discussed above by Yuriy S and others, the product is intimately linked with the series $\sum_{n=1}^{\infty} { \ln(n+1) \over n(n+1) } \approx 1.2577\dots$. The connection is derived as follows:
$$\prod_{k=2}^{\infty} \left( 1+ {1\over k} \right)^{1/k} \\ =\exp \left( \ln \left( \prod_{k=2}^{\infty} \left( 1+ {1\over k} \right)^{1/k} \right) \right) \\ = \exp \left( \sum_{n=2}^{\infty} \ln \left( \left( 1+ {1\over k} \right)^{1/k} \right) \right) \\ = \exp \left( \sum_{n=2}^{\infty} { \ln\left({k+1 \over k}\right)\over k } \right) \\ =\exp \left( \sum_{n=2}^{\infty} {\ln(k+1)\over k } -\sum_{n=2}^{\infty} {\ln(k) \over k} \right)\\ =\exp \left( \sum_{n=3}^{\infty} {\ln(k) \over k-1} -\sum_{n=3}^{\infty} {\ln(k) \over k} - {1 \over 2}\ln 2 \right)\\ =\exp \left( \sum_{n=3}^{\infty} {\ln(k) \over k(k-1)} - {1 \over 2}\ln 2 \right)\\ =\exp \left( \sum_{n=2}^{\infty} {\ln(k+1) \over k(k+1)} - {1 \over 2}\ln 2 \right)\\ =\exp \left( \sum_{n=1}^{\infty} {\ln(k+1) \over k(k+1)} - {1 \over 2}\ln 2 - {1 \over 2}\ln 2 \right)\\ =e^{\sum_{n=1}^{\infty} {\ln(k+1) \over k(k+1)}} \cdot e^{-\ln 2}\\ ={1 \over 2}e^{\sum_{n=1}^{\infty} {\ln(k+1) \over k(k+1)}}$$
Therefore the product will only have a closed form if $\sum_{n=1}^{\infty} {\ln(k+1) \over k(k+1)}$ has a closed form.
• This seems to imply a relation to the derivative of the reimann zeta function: $\zeta ' (s) = -\sum_{n=1}^{\infty} \frac{\ln (n)}{n^s}$ – Jacob Apr 6 '16 at 2:32
Working off of other people's findings, you can write $$\sum_{k=2}^{\infty} \frac{(-x)^k \zeta (k)}{k} = x \gamma + \ln (\Gamma(x+1))$$ $$\frac{d}{dx}\sum_{k=2}^{\infty} \frac{(-x)^k \zeta (k)}{k} = -\sum_{k=2}^{\infty} (-x)^{k-1} \zeta (k)=\gamma+\psi(x+1)=H_x$$ $$\sum_{k=2}^{\infty} (-x)^{k-2} \zeta (k)=\frac{H_x}{x}$$ $$\int_0^{x} \sum_{k=2}^{\infty} (-y)^{k-2}\zeta (k) dy = -\sum_{k=2}^{\infty} \frac{(-x)^{k-1}\zeta (k)}{k-1} = -\int_0^x \frac{H_y}{y} dy$$ $$\sum_{k=2}^{\infty} \frac{(-x)^{k}\zeta (k)}{k-1} = x\int_0^x \frac{H_y}{y} dy$$ However, I believe $\int_0^1 \frac{H_x}{x} dx$ has no closed form, meaning that $$\prod_{k=2}^{\infty} \sqrt[k]{1+\frac{1}{k}}=\frac{1}{2} \exp \left(\sum_{k=2}^{\infty} \frac{(-1)^k \zeta(k)}{k-1}\right)=\frac{1}{2}\exp \left({\int_0^1 \frac{H_x}{x} dx} \right)$$ has no closed form either. | 2020-01-26T02:44:16 | {
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https://math.stackexchange.com/questions/1210798/why-cant-i-solve-the-equation-4-sin2x-cdot-cos2x-2-cos2x-by-dividing-bo | # Why cant I solve the equation $4\sin(2x)\cdot \cos(2x)=2\cos(2x)$ by dividing both sides by $\cos(2x)$??
This assignment assumes the domain is $[0, \pi]$.
Why can't I solve the following equation by dividing both sides by $\cos(2x)$? $$4\sin(2x)\cdot \cos(2x)=2\cos(2x)$$
If I would continue to do so, wouldn't $\cos(2x)$ cancel out on both sides, leaving me with
$$4\sin(2x)=2$$
Which will then be solvable.
According to khanacademy's answer it seems like I would now miss out on two solutions.
$$\cos(2x)[1-2\sin(2x)]=0$$
Then either
$$\cos(2x)=0\\ \sin(2x)=\frac{1}{2}$$
As we can see I don't have $\cos(2x) = 0$ "set up". But what I did was just a simple division, which seems like valid algebra, so how can I tell which approach to use in other scenario's?
• If it's legit to divide by $\cos(2x)$ then it must be the case that $\cos(2x) \neq 0$. – user4894 Mar 29 '15 at 0:35
• That is correct. It will give you two cases: if $\cos(2x)\neq 0$... solve as you did (with division). If $\cos(2x)=0$... solve that equation for the remaining solutions. – TravisJ Mar 29 '15 at 0:35
• By dividing both sides by $\cos(2x)$ you are pre-supposing that the solution $\cos(2x)=0$ is not valid as otherwise you cannot divide both sides by zero. – Mufasa Mar 29 '15 at 0:35
• @user4894 Ah yes, that makes sense. But how would I have known that case is true, starting at $4\sin(2x)\cdot\cos(2x)=2\cos(2x)$? – user1534664 Mar 29 '15 at 0:41
• you can divide by $2\cos 2x$ without any trouble for most values of $x.$ sometimes the exceptional ones are the critical ones. i am thinking of the zeros of $f'$ – abel Mar 29 '15 at 1:28
Consider the equation:
$$xy = x$$
You can observe that $x=0$ satisfies this equation. But if you were to divide both sides by $x$ you would get
$$y = 1$$
which is one solution, but you lost the other solution! The problem is that you can't divide by zero, so if you divide both sides by $x$, you're taking for granted that $x$ isn't zero.
Two ways around this:
$(1)$ Check if the thing you're divididing by can be equal to zero. If it can be, separate the problem into cases: case $x=0$, case $x \ne 0$.
$(2)$ Don't divide both sides by the questionable factor. For example, consider $xy - x = x(y-1) = 0$ in our example. If a product is zero one of the factors must be zero.
• Thank you, sir. Will (2) always be applicable? – user1534664 Mar 29 '15 at 0:52
• @user1534664 No, so maybe you should practice both methods. – GFauxPas Mar 29 '15 at 1:14
Any time you divide by something that is not a non-zero constant (such as $3$, $2\pi$, or $\sqrt{5+\sqrt{2}}$), you have to ask yourself whether it's possible that you're dividing by zero. There are then two ways you can deal with that question.
One way is you can set about proving that the quantity can never be zero. Sometimes that makes sense to do (for example, if $x$ is real you can always divide by $x^2 + 1$). If the domain of $x$ were limited to $[0,\frac12]$ then it would be the case that $\cos(2x)\neq 0$ and you could prove it. But sometimes, as in this case, it is simply not true that the thing you want to divide by is always non-zero. The best you might accomplish by trying to prove it is never zero is that you might actually find the zeros and see that they are useful in reaching a solution.
Another way is to consider the possibility that the quantity might sometimes be zero, and split your solution into two cases. In one case, you assume the quantity is non-zero, divide by it, and proceed from there. But at some point (either before or after working out the non-zero case) you must look at the case where the quantity is zero. One way to keep track of this is that from the equation
$$4\sin(2x)\cdot \cos(2x)=2\cos(2x)$$
you determine that this is equivalent to
$$4\sin(2x) = 1 \qquad \mbox{OR} \qquad \cos(2x) = 0.$$
You work the formula on the left of the "or", finding zero, one, or more solutions, and you work the formula on the right side, finding zero, one or more solutions. Since the original formula is satisfied if either of the two new formulas is satisfied, you get your complete set of solutions by taking the union of the two solution sets of the new formulas. (If one of those formulas has no solution, its solution set is empty and it adds no solutions to the final solution set. You have in effect then proved that it was OK to divide by this quantity in the first place.) | 2019-08-20T20:10:39 | {
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http://math.stackexchange.com/questions/732746/integrating-sinx-on-a-unit-circle | # Integrating sin(x) on a unit circle
I am trying to integrate $\int\int_{D} sin(x)$ where $D$ is a unit circle centered at $(0,0)$.
My approach is to turn the area into the polar coordinate so I have $D$ as $0\leq r\leq1$ , $0 \leq \theta \leq 2\pi$.
Which turns the integral into:
$$\int^{2\pi}_{0}\int^{1}_{0} \sin(r\cos(\theta)) |r| drd\theta$$ and it is not integrable.
I also tried the Cartesian approach by evaluating:
$$\int^{1}_{0}\int^{\sqrt{1-x^2}}_{0} \sin(x) dydx$$ which is also not integrable
Which direction or method should I use to integrate this problem?
Edit* Thank you so much for the answers, I agree that because of symmetry, the answer is zero. The hint says don't do too work work also lol.
-
What about symmetry considerations...? – user86418 Mar 30 '14 at 17:36
Aside: $D$ is the unit disc: the unit circle is the loop around the boundary of the disc. – Hurkyl Mar 30 '14 at 18:53
UW questions....lol – Jing Apr 3 '14 at 15:45
By symmetry, the integral is $0$.
Additional: From a complex analysis approach we have $\int_{|z|=1} \sin(z) \ dz = 0$, since $f(z)=\sin(z)$ is analytic on the unit circle. | 2015-04-18T14:04:23 | {
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https://runestone.academy/ns/books/published/dmoi/sec_gt-intro.html | ## Section4.1Definitions
###### Investigate!
Which (if any) of the graphs below are the same?
The graphs above are unlabeled. Usually we think of a graph as having a specific set of vertices. Which (if any) of the graphs below are the same?
Actually, all the graphs we have seen above are just drawings of graphs. A graph is really an abstract mathematical object consisting of two sets $$V$$ and $$E$$ where $$E$$ is a set of 2-element subsets of $$V\text{.}$$ Are the graphs below the same or different?
Graph 1:
$$V = \{a, b, c, d, e\}\text{,}$$
$$E = \{\{a,b\}, \{a, c\}, \{a,d\}, \{a,e\}, \{b,c\}, \{d,e\}\}\text{.}$$
Graph 2:
$$V = \{v_1, v_2, v_3, v_4, v_5\}\text{,}$$
$$E = \{\{v_1, v_3\}, \{v_1, v_5\}, \{v_2, v_4\}, \{v_2, v_5\}, \{v_3, v_5\}, \{v_4, v_5\}\}\text{.}$$
Before we start studying graphs, we need to agree upon what a graph is. While we almost always think of graphs as pictures (dots connected by lines) this is fairly ambiguous. Do the lines need to be straight? Does it matter how long the lines are or how large the dots are? Can there be two lines connecting the same pair of dots? Can one line connect three dots?
The way we avoid ambiguities in mathematics is to provide concrete and rigorous definitions. Crafting good definitions is not easy, but it is incredibly important. The definition is the agreed upon starting point from which all truths in mathematics proceed. Is there a graph with no edges? We have to look at the definition to see if this is possible.
We want our definition to be precise and unambiguous, but it also must agree with our intuition for the objects we are studying. It needs to be useful: we could define a graph to be a six legged mammal, but that would not let us solve any problems about bridges. Instead, here is the (now) standard definition of a graph.
### Graph Definition.
A graph is an ordered pair $$G = (V, E)$$ consisting of a nonempty set $$V$$ (called the vertices) and a set $$E$$ (called the edges) of two-element subsets of $$V\text{.}$$
Strange. Nowhere in the definition is there talk of dots or lines. From the definition, a graph could be
\begin{equation*} (\{a,b,c,d\}, \{\{a,b\}, \{a,c\}, \{b,c\}, \{b,d\}, \{c,d\}\})\text{.} \end{equation*}
Here we have a graph with four vertices (the letters $$a, b, c, d$$) and five edges (the pairs $$\{a,b\}, \{a,c\}, \{b,c\}, \{b,d\}, \{c,d\})$$).
Looking at sets and sets of 2-element sets is difficult to process. That is why we often draw a representation of these sets. We put a dot down for each vertex, and connect two dots with a line precisely when those two vertices are one of the 2-element subsets in our set of edges. Thus one way to draw the graph described above is this:
However we could also have drawn the graph differently. For example either of these:
We should be careful about what it means for two graphs to be “the same.” Actually, given our definition, this is easy: Are the vertex sets equal? Are the edge sets equal? We know what it means for sets to be equal, and graphs are nothing but a pair of two special sorts of sets.
### Example4.1.1.
Are the graphs below equal?
\begin{equation*} G_1 = (\{a,b,c\}, \{\{a,b\}, \{b,c\}\}); \qquad G_2 = (\{a,b,c\}, \{\{a,c\}, \{c, b\}\})\text{.} \end{equation*}
Solution.
No. Here the vertex sets of each graph are equal, which is a good start. Also, both graphs have two edges. In the first graph, we have edges $$\{a,b\}$$ and $$\{b,c\}\text{,}$$ while in the second graph we have edges $$\{a,c\}$$ and $$\{c,b\}\text{.}$$ Now we do have $$\{b,c\} = \{c,b\}\text{,}$$ so that is not the problem. The issue is that $$\{a,b\} \ne \{a,c\}\text{.}$$ Since the edge sets of the two graphs are not equal (as sets), the graphs are not equal (as graphs).
Even if two graphs are not equal, they might be basically the same. The graphs in the previous example could be drawn like this:
Graphs that are basically the same (but perhaps not equal) are called isomorphic. We will give a precise definition of this term after a quick example:
### Example4.1.2.
Consider the graphs:
\begin{equation*} G_1 = (V_1, E_1) \text{ where } V_1 = \{a, b, c\} \text{ and } E_1 = \{\{a,b\}, \{a,c\}, \{b,c\}\}; \end{equation*}
\begin{equation*} G_2 = (V_2, E_2) \text{ where } V_2 = \{u,v,w\} \text{ and }E_2 = \{\{u,v\}, \{u,w\}, \{v,w\}\}. \end{equation*}
Are these graphs the same?
Solution.
The two graphs are NOT equal. It is enough to notice that $$V_1 \ne V_2$$ since $$a \in V_1$$ but $$a \notin V_2\text{.}$$ However, both of these graphs consist of three vertices with edges connecting every pair of vertices. We can draw them as follows:
Clearly we want to say these graphs are basically the same, so while they are not equal, they will be isomorphic. We can rename the vertices of one graph and get the second graph as the result.
Intuitively, graphs are isomorphic if they are basically the same, or better yet, if they are the same except for the names of the vertices. To make the concept of renaming vertices precise, we give the following definitions:
### Isomorphic Graphs.
An isomorphism between two graphs $$G_1$$ and $$G_2$$ is a bijection $$f:V_1 \to V_2$$ between the vertices of the graphs such that $$\{a,b\}$$ is an edge in $$G_1$$ if and only if $$\{f(a), f(b)\}$$ is an edge in $$G_2\text{.}$$
Two graphs are isomorphic if there is an isomorphism between them. In this case we write $$G_1 \isom G_2\text{.}$$
An isomorphism is simply a function which renames the vertices. It must be a bijection so every vertex gets a new name. These newly named vertices must be connected by edges precisely when they were connected by edges with their old names.
### Example4.1.3.
Decide whether the graphs $$G_1 = (V_1, E_1)$$ and $$G_2 = (V_2, E_2)$$ are equal or isomorphic.
$$V_1 = \{a,b,c,d\}\text{,}$$ $$E_1 = \{\{a,b\}, \{a,c\}, \{a,d\}, \{c,d\}\}$$
$$V_2 = \{a,b,c,d\}\text{,}$$ $$E_2 = \{\{a,b\}, \{a,c\}, \{b,c\}, \{c,d\}\}$$
Solution.
The graphs are NOT equal, since $$\{a,d\} \in E_1$$ but $$\{a,d\} \notin E_2\text{.}$$ However, since both graphs contain the same number of vertices and same number of edges, they might be isomorphic (this is not enough in most cases, but it is a good start).
We can try to build an isomorphism. How about we say $$f(a) = b\text{,}$$ $$f(b) = c\text{,}$$ $$f(c) = d$$ and $$f(d) = a\text{.}$$ This is definitely a bijection, but to make sure that the function is an isomorphism, we must make sure it respects the edge relation. In $$G_1\text{,}$$ vertices $$a$$ and $$b$$ are connected by an edge. In $$G_2\text{,}$$ $$f(a) = b$$ and $$f(b) = c$$ are connected by an edge. So far, so good, but we must check the other three edges. The edge $$\{a,c\}$$ in $$G_1$$ corresponds to $$\{f(a), f(c)\} = \{b,d\}\text{,}$$ but here we have a problem. There is no edge between $$b$$ and $$d$$ in $$G_2\text{.}$$ Thus $$f$$ is NOT an isomorphism.
Not all hope is lost, however. Just because $$f$$ is not an isomorphism does not mean that there is no isomorphism at all. We can try again. At this point it might be helpful to draw the graphs to see how they should match up.
Alternatively, notice that in $$G_1\text{,}$$ the vertex $$a$$ is adjacent to every other vertex. In $$G_2\text{,}$$ there is also a vertex with this property: $$c\text{.}$$ So build the bijection $$g:V_1 \to V_2$$ by defining $$g(a) = c$$ to start with. Next, where should we send $$b\text{?}$$ In $$G_1\text{,}$$ the vertex $$b$$ is only adjacent to vertex $$a\text{.}$$ There is exactly one vertex like this in $$G_2\text{,}$$ namely $$d\text{.}$$ So let $$g(b) = d\text{.}$$ As for the last two, in this example, we have a free choice: let $$g(c) = b$$ and $$g(d) = a$$ (switching these would be fine as well).
We should check that this really is an isomorphism. It is definitely a bijection. We must make sure that the edges are respected. The four edges in $$G_1$$ are
\begin{equation*} \{a,b\}, \{a,c\}, \{a,d\}, \{c,d\}\text{.} \end{equation*}
Under the proposed isomorphism these become
\begin{equation*} \{g(a), g(b)\}, \{g(a), g(c)\}, \{g(a), g(d)\}, \{g(c), g(d)\} \end{equation*}
\begin{equation*} \{c,d\}, \{c,b\}, \{c,a\}, \{b,a\}\text{,} \end{equation*}
which are precisely the edges in $$G_2\text{.}$$ Thus $$g$$ is an isomorphism, so $$G_1 \cong G_2$$
Sometimes we will talk about a graph with a special name (like $$K_n$$ or the Peterson graph) or perhaps draw a graph without any labels. In this case we are really referring to all graphs isomorphic to any copy of that particular graph. A collection of isomorphic graphs is often called an isomorphism class. 1
There are other relationships between graphs that we care about, other than equality and being isomorphic. For example, compare the following pair of graphs:
These are definitely not isomorphic, but notice that the graph on the right looks like it might be part of the graph on the left, especially if we draw it like this:
We would like to say that the smaller graph is a subgraph of the larger.
We should give a careful definition of this. In fact, there are two reasonable notions for what a subgraph should mean.
### Subgraphs.
We say that $$G' = (V', E')$$ is a subgraph of $$G = (V, E)\text{,}$$ and write $$G' \subseteq G\text{,}$$ provided $$V' \subseteq V$$ and $$E' \subseteq E\text{.}$$
We say that $$G' = (V', E')$$ is an induced subgraph of $$G = (V, E)$$ provided $$V' \subseteq V$$ and every edge in $$E$$ whose vertices are still in $$V'$$ is also an edge in $$E'\text{.}$$
Notice that every induced subgraph is also an ordinary subgraph, but not conversely. Think of a subgraph as the result of deleting some vertices and edges from the larger graph. For the subgraph to be an induced subgraph, we can still delete vertices, but now we only delete those edges that included the deleted vertices.
### Example4.1.4.
Consider the graphs:
Here both $$G_2$$ and $$G_3$$ are subgraphs of $$G_1\text{.}$$ But only $$G_2$$ is an induced subgraph. Every edge in $$G_1$$ that connects vertices in $$G_2$$ is also an edge in $$G_2\text{.}$$ In $$G_3\text{,}$$ the edge $$\{a,b\}$$ is in $$E_1$$ but not $$E_3\text{,}$$ even though vertices $$a$$ and $$b$$ are in $$V_3\text{.}$$
The graph $$G_4$$ is NOT a subgraph of $$G_1\text{,}$$ even though it looks like all we did is remove vertex $$e\text{.}$$ The reason is that in $$E_4$$ we have the edge $$\{c,f\}$$ but this is not an element of $$E_1\text{,}$$ so we don't have the required $$E_4 \subseteq E_1\text{.}$$
Back to some basic graph theory definitions. Notice that all the graphs we have drawn above have the property that no pair of vertices is connected more than once, and no vertex is connected to itself. Graphs like these are sometimes called simple, although we will just call them graphs. This is because our definition for a graph says that the edges form a set of 2-element subsets of the vertices. Remember that it doesn't make sense to say a set contains an element more than once. So no pair of vertices can be connected by an edge more than once. Also, since each edge must be a set containing two vertices, we cannot have a single vertex connected to itself by an edge.
That said, there are times we want to consider double (or more) edges and single edge loops. For example, the “graph” we drew for the Bridges of Königsberg problem had double edges because there really are two bridges connecting a particular island to the near shore. We will call these objects multigraphs. This is a good name: a multiset is a set in which we are allowed to include a single element multiple times.
The graphs above are also connected: you can get from any vertex to any other vertex by following some path of edges. A graph that is not connected can be thought of as two separate graphs drawn close together. For example, the following graph is NOT connected because there is no path from $$a$$ to $$b\text{:}$$
Vertices in a graph do not always have edges between them. If we add all possible edges, then the resulting graph is called complete. That is, a graph is complete if every pair of vertices is connected by an edge. Since a graph is determined completely by which vertices are adjacent to which other vertices, there is only one complete graph with a given number of vertices. We give these a special name: $$K_n$$ is the complete graph on $$n$$ vertices.
Each vertex in $$K_n$$ is adjacent to $$n-1$$ other vertices. We call the number of edges emanating from a given vertex the degree of that vertex. So every vertex in $$K_n$$ has degree $$n-1\text{.}$$ How many edges does $$K_n$$ have? One might think the answer should be $$n(n-1)\text{,}$$ since we count $$n-1$$ edges $$n$$ times (once for each vertex). However, each edge is incident to 2 vertices, so we counted every edge exactly twice. Thus there are $$n(n-1)/2$$ edges in $$K_n\text{.}$$ Alternatively, we can say there are $${n \choose 2}$$ edges, since to draw an edge we must choose 2 of the $$n$$ vertices.
In general, if we know the degrees of all the vertices in a graph, we can find the number of edges. The sum of the degrees of all vertices will always be twice the number of edges, since each edge adds to the degree of two vertices. Notice this means that the sum of the degrees of all vertices in any graph must be even!
This is our first example of a general result about all graphs. It seems innocent enough, but we will use it to prove all sorts of other statements. So let's give it a name and state it formally.
The handshake lemma 2 is sometimes called the degree sum formula, and can be written symbolically as
\begin{equation*} \sum_{v\in V} d(v) = 2e\text{.} \end{equation*}
Here we are using the notation $$d(v)$$ for the degree of the vertex $$v\text{.}$$
One use for the lemma is to actually find the number of edges in a graph. To do this, you must be given the degree sequence for the graph (or be able to find it from other information). This is a list of every degree of every vertex in the graph, generally written in non-increasing order.
### Example4.1.6.
How many vertices and edges must a graph have if its degree sequence is
\begin{equation*} (4, 4, 3, 3, 3, 2, 1)\text{?} \end{equation*}
Solution.
The number of vertices is easy to find: it is the number of degrees in the sequence: 7. To find the number of edges, we compute the degree sum:
\begin{equation*} 4 + 4 + 3 + 3 + 3 + 2 + 1 = 20\text{,} \end{equation*}
so the number of edges is half this: 10.
The handshake lemma also tells us what is not possible.
### Example4.1.7.
At a recent math seminar, 9 mathematicians greeted each other by shaking hands. Is it possible that each mathematician shook hands with exactly 7 people at the seminar?
Solution.
It seems like this should be possible. Each mathematician chooses one person to not shake hands with. But this cannot happen. We are asking whether a graph with 9 vertices can have each vertex have degree 7. If such a graph existed, the sum of the degrees of the vertices would be $$9\cdot 7 = 63\text{.}$$ This would be twice the number of edges (handshakes) resulting in a graph with $$31.5$$ edges. That is impossible. Thus at least one (in fact an odd number) of the mathematicians must have shaken hands with an even number of people at the seminar.
We can generalize the previous example to get the following proposition. 3
### Proof.
Suppose there were a graph with an odd number of vertices with odd degree. Then the sum of the degrees in the graph would be odd, which is impossible, by the handshake lemma.
We will consider further applications of the handshake lemma in the exercises.
One final definition: we say a graph is bipartite if the vertices can be divided into two sets, $$A$$ and $$B\text{,}$$ with no two vertices in $$A$$ adjacent and no two vertices in $$B$$ adjacent. The vertices in $$A$$ can be adjacent to some or all of the vertices in $$B\text{.}$$ If each vertex in $$A$$ is adjacent to all the vertices in $$B\text{,}$$ then the graph is a complete bipartite graph, and gets a special name: $$K_{m,n}\text{,}$$ where $$|A| = m$$ and $$|B| = n\text{.}$$ The graph in the houses and utilities puzzle is $$K_{3,3}\text{.}$$
### Named Graphs.
Some graphs are used more than others, and get special names.
$$K_n$$
The complete graph on $$n$$ vertices.
$$K_{m,n}$$
The complete bipartite graph with sets of $$m$$ and $$n$$ vertices.
$$C_n$$
The cycle on $$n$$ vertices, just one big loop.
$$P_n$$
The path on $$n+1$$ vertices (so $$n$$ edges), just one long path.
### Graph Theory Definitions.
There are a lot of definitions to keep track of in graph theory. Here is a glossary of the terms we have already used and will soon encounter.
Graph
A collection of vertices, some of which are connected by edges. More precisely, a pair of sets $$V$$ and $$E$$ where $$V$$ is a set of vertices and $$E$$ is a set of 2-element subsets of $$V\text{.}$$
Two vertices are adjacent if they are connected by an edge. Two edges are adjacent if they share a vertex.
Bipartite graph
A graph for which it is possible to divide the vertices into two disjoint sets such that there are no edges between any two vertices in the same set.
Complete bipartite graph
A bipartite graph for which every vertex in the first set is adjacent to every vertex in the second set.
Complete graph
A graph in which every pair of vertices is adjacent.
Connected
A graph is connected if there is a path from any vertex to any other vertex.
Chromatic number
The minimum number of colors required in a proper vertex coloring of the graph.
Cycle
A path (see below) that starts and stops at the same vertex, but contains no other repeated vertices.
Degree of a vertex
The number of edges incident to a vertex.
Euler path
A walk which uses each edge exactly once.
Euler circuit
An Euler path which starts and stops at the same vertex.
Multigraph
A multigraph is just like a graph but can contain multiple edges between two vertices as well as single edge loops (that is an edge from a vertex to itself).
Path
A path is a walk that doesn't repeat any vertices (or edges) except perhaps the first and last. If a path starts and ends at the same vertex, it is called a cycle.
Planar
A graph which can be drawn (in the plane) without any edges crossing.
Subgraph
We say that $$H$$ is a subgraph of $$G$$ if every vertex and edge of $$H$$ is also a vertex or edge of $$G\text{.}$$ We say $$H$$ is an induced subgraph of $$G$$ if every vertex of $$H$$ is a vertex of $$G$$ and each pair of vertices in $$H$$ are adjacent in $$H$$ if and only if they are adjacent in $$G\text{.}$$
Tree
A connected graph with no cycles. (If we remove the requirement that the graph is connected, the graph is called a forest.) The vertices in a tree with degree 1 are called leaves.
Vertex coloring
An assignment of colors to each of the vertices of a graph. A vertex coloring is proper if adjacent vertices are always colored differently.
Walk
A sequence of vertices such that consecutive vertices (in the sequence) are adjacent (in the graph). A walk in which no edge is repeated is called a trail, and a trail in which no vertex is repeated (except possibly the first and last) is called a path.
### ExercisesExercises
#### 1.
If 10 people each shake hands with each other, how many handshakes took place? What does this question have to do with graph theory?
#### 2.
Among a group of 5 people, is it possible for everyone to be friends with exactly 2 of the people in the group? What about 3 of the people in the group?
#### 3.
Is it possible for two different (non-isomorphic) graphs to have the same number of vertices and the same number of edges? What if the degrees of the vertices in the two graphs are the same (so both graphs have vertices with degrees 1, 2, 2, 3, and 4, for example)? Draw two such graphs or explain why not.
#### 4.
Are the two graphs below equal? Are they isomorphic? If they are isomorphic, give the isomorphism. If not, explain.
Graph 1: $$V = \{a,b,c,d,e\}\text{,}$$ $$E = \{\{a,b\}, \{a,c\}, \{a,e\}, \{b,d\}, \{b,e\}, \{c,d\}\}\text{.}$$
Graph 2:
#### 5.
Consider the following two graphs:
$$G_1$$
$$V_1=\{a,b,c,d,e,f,g\}$$
$$E_1=\{\{a,b\},\{a,d\},\{b,c\},\{b,d\},\{b,e\},\{b,f\},\{c,g\},\{d,e\}\text{,}$$
$$\{e,f\},\{f,g\}\}\text{.}$$
$$G_2$$
$$V_2=\{v_1,v_2,v_3,v_4,v_5,v_6,v_7\}\text{,}$$
$$E_2=\{\{v_1,v_4\},\{v_1,v_5\},\{v_1,v_7\},\{v_2,v_3\},\{v_2,v_6\}\text{,}$$
$$\{v_3,v_5\},\{v_3,v_7\},\{v_4,v_5\},\{v_5,v_6\},\{v_5,v_7\}\}$$
1. Let $$f:G_1 \rightarrow G_2$$ be a function that takes the vertices of Graph 1 to vertices of Graph 2. The function is given by the following table:
$$x$$ $$a$$ $$b$$ $$c$$ $$d$$ $$e$$ $$f$$ $$g$$ $$f(x)$$ $$v_4$$ $$v_5$$ $$v_1$$ $$v_6$$ $$v_2$$ $$v_3$$ $$v_7$$
Does $$f$$ define an isomorphism between Graph 1 and Graph 2?
2. Define a new function $$g$$ (with $$g \ne f$$) that defines an isomorphism between Graph 1 and Graph 2.
3. Is the graph pictured below isomorphic to Graph 1 and Graph 2? Explain.
#### 6.
What is the largest number of edges possible in a graph with 10 vertices? What is the largest number of edges possible in a bipartite graph with 10 vertices? What is the largest number of edges possible in a tree with 10 vertices?
#### 8.
For which $$n \ge 3$$ is the graph $$C_n$$ bipartite?
#### 9.
For each of the following, try to give two different unlabeled graphs with the given properties, or explain why doing so is impossible.
1. Two different trees with the same number of vertices and the same number of edges. A tree is a connected graph with no cycles.
2. Two different graphs with 8 vertices all of degree 2.
3. Two different graphs with 5 vertices all of degree 4.
4. Two different graphs with 5 vertices all of degree 3.
#### 10.
Decide whether the statements below about subgraphs are true or false. For those that are true, briefly explain why (1 or 2 sentences). For any that are false, give a counterexample.
1. Any subgraph of a complete graph is also complete.
2. Any induced subgraph of a complete graph is also complete.
3. Any subgraph of a bipartite graph is bipartite.
4. Any subgraph of a tree is a tree.
#### 11.
Let $$k_1, k_2, \ldots, k_j$$ be a list of positive integers that sum to $$n$$ (i.e., $$\sum_{i=1}^j k_i = n$$). Use two graphs containing $$n$$ vertices to explain why
\begin{equation*} \sum_{i = 1}^j \binom{k_i}{2} \le \binom{n}{2}\text{.} \end{equation*}
#### 12.
We often define graph theory concepts using set theory. For example, given a graph $$G = (V, E)$$ and a vertex $$v \in V\text{,}$$ we define
\begin{equation*} N(v) = \{u \in V \st \{v,u\} \in E\}\text{.} \end{equation*}
We define $$N[v] = N(v) \cup \{v\}\text{.}$$ The goal of this problem is to figure out what all this means.
1. Let $$G$$ be the graph with $$V = \{a,b,c,d,e,f\}$$ and $$E = \{\{a,b\}, \{a,e\},\{b, c\}, \{b,e\}, \{c,d\}, \{c, f\}, \{d, f\}, \{e,f\}\}\text{.}$$ Find $$N(a)\text{,}$$ $$N[a]\text{,}$$ $$N(c)\text{,}$$ and $$N[c]\text{.}$$
2. What is the largest and smallest possible values for $$|N(v)|$$ and $$|N[v]|$$ for the graph in part (a)? Explain.
3. Give an example of a graph $$G = (V, E)$$ (probably different than the one above) for which $$N[v] = V$$ for some vertex $$v \in V\text{.}$$ Is there a graph for which $$N[v] = V$$ for all $$v \in V\text{?}$$ Explain.
4. Give an example of a graph $$G = (V,E)$$ for which $$N(v) = \emptyset$$ for some $$v \in V\text{.}$$ Is there an example of such a graph for which $$N[u] = V$$ for some other $$u \in V$$ as well? Explain.
5. Describe in words what $$N(v)$$ and $$N[v]$$ mean in general.
#### 13.
A graph is a way of representing the relationships between elements in a set: an edge between the vertices $$x$$ and $$y$$ tells us that $$x$$ is related to $$y$$ (which we can write as $$x \sim y$$). Not all sorts of relationships can be represented by a graph though. For each relationship described below, either draw the graph or explain why the relationship cannot be represented by a graph.
1. The set $$V = \{1,2, \ldots, 9\}$$ and the relationship $$x \sim y$$ when $$x-y$$ is a non-zero multiple of 3.
2. The set $$V = \{1,2, \ldots, 9\}$$ and the relationship $$x \sim y$$ when $$y$$ is a multiple of $$x\text{.}$$
3. The set $$V = \{1,2,\ldots, 9\}$$ and the relationship $$x \sim y$$ when $$0 \lt |x-y| \lt 3\text{.}$$
#### 14.
Consider graphs with $$n$$ vertices. Remember, graphs do not need to be connected.
1. How many edges must the graph have to guarantee at least one vertex has degree two or more? Prove your answer.
2. How many edges must the graph have to guarantee all vertices have degree two or more? Prove your answer.
#### 15.
Prove that any graph with at least two vertices must have two vertices of the same degree.
#### 16.
Suppose $$G$$ is a connected graph with $$n > 1$$ vertices and $$n-1$$ edges. Prove that $$G$$ has a vertex of degree 1. | 2022-10-03T08:50:27 | {
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https://math.stackexchange.com/questions/3437457/why-drop-forall-x-in-mathbbx-in-delta-epsilon-definition-at-infinity | # Why Drop $\forall x \in \mathbb{X}$ in delta-epsilon definition at infinity?
For a function $$f:X \rightarrow \mathbb{R}$$ My course notes says that:
$$\lim_{x \to \infty} f(x) = l \Leftrightarrow (\forall \epsilon > 0, \exists S \in \mathbb{R}, x> S \implies |f(x)-l| < \epsilon)$$
I don't understand why we have dropped the $$\forall x \in X$$
In other words, why is it not this: $$\lim_{x \to \infty} f(x) = l \Leftrightarrow (\forall \epsilon > 0, \exists S \in \mathbb{R}, \forall x \in X, x> S \implies |f(x)-l| < \epsilon)$$
Wouldn't we want the implication to hold true for all $$x$$ larger than $$S$$, analogous to when we're dealing with the definition of limits as $$x$$ approaches $$a$$?
$$\lim_{x \to a} f(x) = l \Leftrightarrow (\forall \epsilon > 0, \exists \delta > 0, \forall x \in X, 0 < |x-a| < \delta \implies |f(x)-l| < \epsilon)$$
Otherwise, I could just find a really small $$S$$, smaller than a $$x_1$$ where $$f(x_1)=l$$ and if it works for one particular $$x$$, I could claim the limit as $$x \rightarrow \infty$$ is $$l$$, which is clearly not the intended definition of a limit as $$x$$ approaches infinity.
• It should be. It wasn't written explicitly in the notes, but it should be there, just as you suspected. – Andrés E. Caicedo Nov 16 '19 at 0:49
• @bof Good catch. I edited the question a few times. Fixed now. – Snowball Nov 16 '19 at 1:12
• " I could just find a really small S, smaller than a x1 where f(x1)=l and if it works for one particular x, " That makes no sense. The condition is $x > S \implies$ that means if any $x > S$ it is true. Not just one. Writing $\forall x \in X$ will have nothing to do with that. – fleablood Nov 16 '19 at 1:37
I will not enter into the details of first order logic, but strictly speaking, if a variable doesn't have a quantifier ($$\exists$$ or $$\forall$$) then it is ranging over all the elements of the set. So, again, strictly speaking both propositions are the same. In fact, strictly speaking, you can safely drop every $$\forall x \in X$$ whenever you know $$X$$ is the domain of discourse, it will just annoy your fellow mathematicians but it is right.
To see why this is the case, see this example (but remember that this is a formalism about mathematical language, so it is arbitrary in the sense that it was decided by convention to be so): if I am talking about the natural numbers and I say $$x$$ is even implies $$x$$ can be divided by $$2$$, one would reasonable understand that I meant, for all natural numbers $$x$$, $$x$$ is even implies it can be divided by 2. Clearly I wasn't talking about one specific $$x$$.
Again, just remember that this is a (intuitive) convention, nothing more than that.
The meaning of $$x>S\implies |f(x)-l|<\epsilon$$ is that "if $$x>S$$ then $$|f(x)-l|<\epsilon$$". So yes, it exactly says that for all $$x$$ which satisfy $$x>S$$ we have $$|f(x)-l|<\epsilon$$. An equivalent way to write the same thing:
$$\forall (x: x>S)[|f(x)-l|<\epsilon]$$
Note that in this equivalent way I didn't use $$\implies$$ at all.
• Does this imply I can drop the $\forall x \in X$ in $\lim_{x \to a} f(x) = l \Leftrightarrow (\forall \epsilon > 0, \exists \delta > 0, \forall x \in X, 0 < |x-a| < \delta \implies |f(x)-l| < \epsilon)$? – Snowball Nov 16 '19 at 0:53
• If you know that you are checking the condition for elements of the set $X$ then yes, you can just write $0<|x-a|<\delta\implies |f(x)-l|<\epsilon$. Actually, I always write $\forall(x\in X: 0<|x-a|<\delta)[|f(x)-l|<\epsilon]$. But what I'm saying is that writing the way they did is also fine, as long as you know that you are checking the condition for elements from the set $X$. – Mark Nov 16 '19 at 0:59
• This answer raises a question. According to this convention, WHERE within the expression should $\text{“ }\forall x \text{ ''}$ be? And why? $$\begin{array}{ccccc} \text{here?} & & \text{or here?} & \text{or somewhere else?} \\ \downarrow & & \downarrow & \downarrow \\ \bullet\bullet\bullet & (\forall \varepsilon > 0 \,\,\, \exists \delta > 0 & \bullet\bullet\bullet & 0 < |x-a| < \delta \implies |f(x)-\ell| < \varepsilon) \end{array}$$ This matters, since in one case the $\delta$ depends on $x$ (which is NOT as it should be) and in the other it does not. $\qquad$ – Michael Hardy Nov 16 '19 at 1:05
• Obviously after the $\exists\delta>0$. – Mark Nov 16 '19 at 1:07
• @Mark : That is obvious if you consider the meaning of the limit concept, but how does it follow from conventions of logic? – Michael Hardy Nov 16 '19 at 1:08 | 2021-06-20T19:19:05 | {
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https://math.stackexchange.com/questions/1342414/volume-of-the-region-outside-of-a-cylinder-and-inside-a-sphere | # Volume of the region outside of a cylinder and inside a sphere
The cylinder is $x^2 +y^2 = 1$ and the sphere is $x^2 + y^2 + z^2 = 4$. I have to find the volume of the region outside the cylinder and inside the sphere. The triple spherical integral for this problem is (from the answer key) $$\int _0^{2\pi }\int _{\frac{\pi }{6}}^{\frac{5\pi }{6}}\int _{csc\phi }^2\:\rho ^2sin\phi \:d\rho \:d\phi \:d\theta$$
What is confusing me here is that there's some space at the endcaps of the cylinder that is not being accounted for. Why is this the case?
• I don't think the answer is correct. The answer below by Martin is the best way to do it. If you want to use integral, it should be $\int _0^{2\pi }\int _{\frac{\pi }{6}}^{\frac{5\pi }{6}}\int _{1/\sin\phi}^{\sqrt{2}}\rho ^2\sin\phi \:d\rho d\phi d\theta+2\int _0^{2\pi }\int _{0}^{\frac{\pi }{6}}\int _{1/\cos\phi}^{\sqrt{2}}\rho ^2\sin\phi \:d\rho d\phi d\theta$ – KittyL Jun 28 '15 at 18:05
• @lasec0203: The cylinder in your question has infinite height, which doesn't match the figure. The sphere in your question (radius $2$) doesn't match the diagram (radius $\sqrt{2}$). The answer key integral, as written, does not give the volume outside a cylinder, but outside a cone. Could you please carefully check the problem statement, the purported answer, and the diagram to be sure they're consistent? :) – Andrew D. Hwang Jun 28 '15 at 18:22
• @AndrewD.Hwang, yes the figure is incorrect. I will change it here in a sec – lasec0203 Jun 28 '15 at 18:27
• @KittyL I think I understand it now, the endcaps are technically not outside of the cylinder because the boundary of the cylinder ends at (1, sqrt(3)) where the two figures intersect. I was viewing the cylinder as a can, closed top and bottom, which is wrong. The ends are open. – lasec0203 Jun 28 '15 at 18:28
Why don't you just compute the $2$ volumes?
The cilinder has volume $V_c=\pi\cdot 2\sqrt{3}$ and the sphere has volume $V_s=\frac{4}{3}\pi*2^3$.
The volume inside the sphere and outside the cilinder is $V_s-V_c=\pi(\frac{2^5}{3}-2\sqrt{3})$.
• You forget the volume of the two cap. – achille hui Jun 28 '15 at 17:54
• I don't see the flaw in what I've done. The sphere has a certain volume including the caps. The cilinder has another volume and it's inside the sphere. To compute the volume outside the cilinder and inside the sphere I've computed each volume and substracted. Could you elaborate? – Martín Forsberg Conde Jun 28 '15 at 17:56
• @achillehui yea, I though of going this route too, but when I asked my prof, he said "That actually won't work. If you do that, then you'd get $\:\frac{32\pi }{3}-2\pi \sqrt{3}$. Which is wrong. This method is ignoring the curved end caps of the cylinder, and just to be sure these endcaps are not hemispheres! So there is no geometry formula to use here instead of calculus." – lasec0203 Jun 28 '15 at 17:56
• @lasec0203: So are you looking for volumes including the caps or not? – KittyL Jun 28 '15 at 18:08
• @KittyL I'm quoting what my professor said. "Exactly. The region outside the cylinder and inside the sphere doesn't include the end caps. Just imagine shooting a hole through a sphere, and then finding the volume of what remains." this reply lost me – lasec0203 Jun 28 '15 at 18:11
The new version of the problem (i.e. with an infinitely long cylinder through the sphere) is best solved using cylindrical coordinates $z$, $\rho$, $\phi$.
When viewed as a function of the variable $z$, the sphere consists of circular disks of radius $R = \sqrt{4 - z^2}$ and thickness $dz$. The cylinder cuts out the central region of these disks. The radius of the hole is $1$. We see that the disk is larger than the cut-out region when $z^2 > 3$. Hence the limits of the integration over $z$ are $-\sqrt{3}$ and $+\sqrt{3}$.
The actual integration is now straightforward. We get:
$$V = \int_{-\sqrt{3}}^{+\sqrt{3}} \{\pi (4-z^2)-\pi\}dz = 4 \sqrt{3}\pi$$
turns out that I was viewing the figures in the wrong way. As Andrew D. Hwang pointed out the cylinder has infinite height. Since the cylinder ends are open, it engulfs the top and bottom end of the sphere so the endcaps are not included. This explains why the integral:
$$\int _0^{2\pi }\int _{\frac{\pi }{6}}^{\frac{5\pi }{6}}\int _{csc\phi }^2\:\rho ^2sin\phi \:d\rho \:d\phi \:d\theta$$
is correct for this problem.
• please edit this into the original problem statement – MichaelChirico Jun 28 '15 at 20:43
• @MichaelChirico I will, but I leave the actually shapes the way I envisioned them originally. – lasec0203 Jun 28 '15 at 20:51
• I believe the limits for $\rho$ should be from $\csc\phi$ to 2 instead of 0 to $\sqrt{2}$. – user84413 Jun 29 '15 at 0:18
• @user84413, thanks for the catch. – lasec0203 Jun 29 '15 at 2:49 | 2019-11-20T16:40:40 | {
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https://math.stackexchange.com/questions/1413507/solve-this-integral-int-0-infty-frac-arctan-xxx21-mathrm-dx?noredirect=1 | # Solve this integral:$\int_0^\infty\frac{\arctan x}{x(x^2+1)}\mathrm dx$
I occasionally found that $\displaystyle\int_0^{\Large\frac{\pi}{2}}\dfrac{x}{\tan x}=\dfrac{\pi}{2}\ln 2$.
I tried that
$$\int_0^{\Large\frac{\pi}{2}}\dfrac{x}{\tan x}=\int_0^{\Large\frac{\pi}{2}}x \ \mathrm d(\ln \sin x)=-\int_0^{\Large\frac{\pi}{2}}\ln (\sin x)=\dfrac{\pi}{2}\ln 2$$ Then I tried another method $$\int_0^{\Large\frac{\pi}{2}}\dfrac{x}{\tan x}=\int_0^\infty\dfrac{\arctan x}{x(x^2+1)}\mathrm dx$$ I tried to expand $\arctan x$ and $\dfrac{1}{1+x^2}$, but got nothing, also I was confused that whether $\displaystyle\int_0^\infty$ and $\displaystyle\sum_{i=0}^\infty$ can exchange or not? If yes, on what condition?
• – Blex Aug 29 '15 at 11:15
• Using $\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx$ $\displaystyle\int_0^{\frac{\pi}{2}}\dfrac{x}{\tan x}=\int_0^{\frac{\pi}{2}}x\tan x\ dx$ $\int x\tan x\ dx=x\int\tan x\ dx-\int\left(\dfrac{dx}{dx}\int\tan x\ dx\right)dx$ $=x\int\tan x\ dx-\int\left(\ln\sec x\right)dx$ $=x\int\tan x\ dx+\int\left(\ln\cos x\right)dx$ See math.stackexchange.com/questions/37829/… – lab bhattacharjee Aug 29 '15 at 11:20
I just want to seek ways that have nothing to do with $\ln (\sin x)$.
Hint. You may consider $$I(a):=\int_0^\infty\frac{\arctan (ax)}{x(x^2+1)}\:\mathrm dx,\quad 0<a<1, \tag1$$ and obtain $$I'(a)=\int_0^\infty\frac1{(x^2+1)(a^2x^2+1)}\:\mathrm dx.$$ By using partial fraction decomposition, we have $$\frac1{(x^2+1)(a^2x^2+1)}=\frac1{\left(1-a^2\right) \left(x^2+1\right)}-\frac{a^2}{\left(1-a^2\right) \left(a^2 x^2+1\right)}$$ giving \begin{align} I'(a)&=\frac1{\left(1-a^2\right)}\int_0^\infty\!\frac1{x^2+1}\:\mathrm dx-\frac{a^2}{\left(1-a^2\right)}\int_0^\infty\frac1{a^2x^2+1}\:\mathrm dx\\\\ &=\frac1{\left(1-a^2\right)}[\arctan x]_0^\infty-\frac{a^2}{\left(1-a^2\right)}\left[\frac{\arctan (ax)}a\right]_0^\infty\\\\ &=\frac1{\left(1-a^2\right)}\frac{\pi}2-\frac{a}{\left(1-a^2\right)}\frac{\pi}2\\\\ &=\frac{\pi}2\frac1{1+a} \tag2 \end{align} Since $I(0)=0$, by integrating $(2)$, you easily get
$$\int_0^\infty\frac{\arctan (ax)}{x(x^2+1)}\:\mathrm dx=\frac{\pi}2\: \ln (a+1), \qquad 0\leq a <1,$$
from which, by letting $a \to 1^-$, you deduce
$$\int_0^\infty\frac{\arctan x}{x(x^2+1)}\:\mathrm dx=\frac{\pi}2 \ln 2$$
as announced.
Before providing my solution, I'd must admit that Oliver Oloa provides the way to calculate this integral. I merely provide a different approach, using Fourier transforms.
First a comment. I tried to use a symmetry argument saying that $$\int_0^{+\infty}f(x+1/x)\arctan x\frac{dx}{x} =\frac{\pi}{4}\int_0^{+\infty} f(x+1/x)\frac{dx}{x},$$ but I was not able to put this integral into that form. Now to the solution:
Since the integrand is even, our integral equals $$\frac{1}{2}\int_{-\infty}^{+\infty}\frac{\arctan x}{x(1+x^2)}\,dx.$$ We need to know the Fourier transforms $$\mathcal F\Bigl[\frac{1}{1+x^2}\Bigr](\xi)=\sqrt{\frac{\pi}{2}}e^{-|\xi|}\quad\text{and}\quad \mathcal F\Bigl[\frac{\arctan x}{x}\Bigr](\xi)=\sqrt{\frac{\pi}{2}}\int_{|\xi|}^{+\infty}\frac{e^{-t}}{t}\,dt.$$ By Parseval's formula, $$\int_0^{+\infty}\frac{\arctan x}{x(1+x^2)}\,dx= \frac{1}{2}\sqrt{\frac{\pi}{2}}\sqrt{\frac{\pi}{2}} \int_{-\infty}^{+\infty} e^{-|\xi|}\int_{|\xi|}^{+\infty} \frac{e^{-t}}{t}\,dt\,d\xi.$$ The integrand is even in $\xi$, so we get $$\frac{\pi}{2}\int_0^{+\infty}e^{-\xi}\int_{\xi}^{+\infty}\frac{e^{-t}}{t}\,dt\,d\xi.$$ Changing the order of integrations, and calculating the inner one, we get $$\frac{\pi}{2}\int_0^{+\infty}\frac{e^{-t}}{t}\int_0^t e^{-\xi}\,d\xi \,dt= \frac{\pi}{2}\int_0^{+\infty}\frac{e^{-t}}{t}(1-e^{-t})\,dt$$ Now, the last integral is a Frullani integral that equals $\log 2$, so we finally get that $$\int_0^{+\infty}\frac{\arctan x}{x(1+x^2)}\,dx=\frac{\pi}{2}\log 2.$$
Another chance is given by the following representation associated with the cotangent function $$1-x\cot x=\sum_{n\geq 1}\left(\frac{x}{\pi n-x}-\frac{x}{\pi n+x}\right)\tag{1}$$ that comes from applying $\frac{d}{dx}\log(\cdot)$ to the Weierstrass product for the sine function.
If you integrate both sides of $(1)$ over $\left(0,\frac{\pi}{2}\right)$ the original integral is transformed into a series that is easy to handle through summation by parts and Stirling's inequality. The final outcome is $$\int_{0}^{\pi/2}\left(1-x\cot x\right)\,dx = \frac{\pi}{2}\left(1-\log 2\right)\tag{2}$$ that is equivalent to the claim. Anyway, there is a well-known symmetry trick for computing $\int_{0}^{\pi/2}\log\sin(x)\,dx$, that probably gives the slickest approach.
\begin{align} J&=\int_0^\infty\frac{\arctan x}{x(x^2+1)}\mathrm dx\tag1\\ &=\int_0^1\frac{\arctan x}{x(x^2+1)}\mathrm dx+\int_1^\infty\frac{\arctan x}{x(x^2+1)}\mathrm dx \end{align}
In the latter integral perform the change of variable $y=\dfrac{1}{x}$,
\begin{align} J&=\int_0^1\frac{\arctan x}{x(x^2+1)}\mathrm dx+\int_0^1\frac{x\arctan\left(\dfrac{1}{x}\right)}{x^2+1}\mathrm dx\\ &=\left(\int_0^1\frac{\arctan x}{x}\mathrm dx-\int_0^1\frac{x\arctan x}{1+x^2}\mathrm dx\right)+\dfrac{\pi}{2}\int_0^1 \dfrac{x}{x^2+1}dx-\int_0^1\frac{x\arctan x}{x^2+1}\mathrm dx\\ &=\left(\int_0^1\frac{\arctan x}{x}\mathrm dx-2\int_0^1\frac{x\arctan x}{1+x^2}\mathrm dx\right)+\dfrac{\pi}{4}\ln 2 \end{align}
In $(1)$ perform the change of variable $x=\dfrac{2y}{1-y^2}$,
\begin{align} J&=\int_0^1 \dfrac{(1-y^2)\arctan\left(\dfrac{2y}{1-y^2}\right)}{y(1+y^2}dy\\ &=2\int_0^1 \dfrac{(1-y^2)\arctan y}{y(1+y^2}dy\\ &=2\left(\int_0^1\frac{\arctan x}{x}\mathrm dx-2\int_0^1\frac{x\arctan x}{1+x^2}\mathrm dx\right) \end{align}
Therefore,
$\displaystyle \int_0^1\frac{\arctan x}{x}\mathrm dx-2\int_0^1\frac{x\arctan x}{1+x^2}\mathrm dx=\dfrac{J}{2}$
Therefore,
$\displaystyle J=\dfrac{J}{2}+\dfrac{\pi}{4}\ln 2$
Finally,
$\boxed{J=\displaystyle \dfrac{\pi}{2}\ln 2}$
| 2019-11-18T14:06:20 | {
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https://math.stackexchange.com/questions/544608/how-can-i-express-the-sum-of-sin-a-sin2a-sin3a-cdots-sinn-1a/544621 | # How can I express the sum of $\sin a+\sin2a+\sin3a+\cdots+\sin(n-1)a$?
I want to sum up the partials of a harmonic series, how do I do it?
If I was using the 'Lagrange trigonometric identity to solve this problem', how would I plot it on Wolfram mathematica (using which input)?
• A harmonic series ($\sum \frac1n$) or a sum of sines, $\sum \sin (ka)$? – Daniel Fischer Oct 29 '13 at 19:32
• A sum of sines as described in the question: sina+sin2a+sin3a+...+(n-1)a – Alex Oct 29 '13 at 19:33
• I asked because the question body speaks of a harmonic series, that is something different. – Daniel Fischer Oct 29 '13 at 19:34
• Basically, I want to show that when a string on a piano is plucked, not only the fundamental frequency of the string plucked eg. 'a' for example resonates, but also multiples of that frequency, such as the octave above....and then I wanted to add these sin waves and create a function from there... – Alex Oct 29 '13 at 19:35
• Harmonic in music $\neq$ harmonic in mathematics. Not entirely unrelated, but different. Do you already know Euler's formulae relating the exponential function and the trigonometric functions? – Daniel Fischer Oct 29 '13 at 19:38
Using De Moivre's formula, you find that
$$\sin(na)=\mathrm{Im}\left(\left(e^{ia}\right)^n\right)$$
for every real number $a$.
Then, using the linearity of the imaginary part, your sum is clearly equal to the imaginary part of another much simpler sum :
$$\sin(a)+\ldots+\sin((n-1)a)=\mathrm{Im}\left(e^{ia}\right)+\ldots+\mathrm{Im}\left(\left(e^{ia}\right)^{n-1}\right)=\mathrm{Im}\left(e^{ia}+\ldots+\left(e^{ia}\right)^{n-1}\right)$$
If $a$ is an integer multiple of $2\pi$, then your sum is clearly equal to $0$.
If $a$ is not an integer multiple of $2\pi$, then $r=e^{ia}\neq 1$, and the terms of the sum $e^{ia}+\ldots+\left(e^{ia}\right)^{n-1}$ are just the terms of the geometric sequence
$$r,r^2,\ldots,r^{n-1}$$
Now, you should remember that if $(u_n)_{n\geqslant 0}$ is a geometric sequence with ratio $r\neq 1$, then the sum of the $N$ first terms of this sequence is :
$$u_0\frac{r^N-1}{r-1}$$
In our case, we have $u_0=r$ and $N=n-1$, so :
$$e^{ia}+\ldots+\left(e^{ia}\right)^{n-1}=e^{ia}\frac{e^{i(n-1)a}-1}{e^{ia}-1}$$
Now, we'll use this useful formula :
$$\forall \alpha\in\mathbb R,\ e^{i\alpha}-1=2ie^{i\frac\alpha 2}\sin\left(\frac\alpha 2\right)$$
(To prove it, just use the fact that $\sin(\theta)=\dfrac{e^{i\theta}-e^{-i\theta}}{2i}$ for all real number $\theta$ and develop the RHS).
Using this formula on both the numerator and denominator of our latest identity, we get :
$$e^{ia}+\ldots+\left(e^{ia}\right)^{n-1}=e^{ia}\frac{2ie^{i\frac{N-1}2a}\sin\left(\frac{N-1}2a\right)}{2ie^{i\frac a2}\sin\left(\frac a2\right)}=e^{i\frac N2a}\frac{\sin\left(\frac{N-1}2a\right)}{\sin\left(\frac a2\right)}$$
The imaginary part of the RHS is just the sum you wanted :
$$\sin(a)+\ldots+\sin((n-1)a)=\sin\left(\frac N2a\right)\frac{\sin\left(\frac{N-1}2a\right)}{\sin\left(\frac a2\right)}\quad (a\notin 2\pi\mathbb Z)$$
They are called Lagrange's trigonometric identities
$$\sum_{n=1}^N \sin (na)= \frac{1}{2}\cot\frac{a}{2}-\frac{\cos(N+\frac{1}{2})a}{2\sin\frac{a}{2}}$$
$$\sum_{n=1}^N \cos(na)= -\frac{1}{2}+\frac{\sin(N+\frac{1}{2})a}{2\sin\frac{a}{2}}$$
• that looks very good, thank you! – Alex Oct 29 '13 at 19:50
• Do you have a proof for that? – Alex Oct 29 '13 at 19:51
Hint: Multiple by $\sin\frac{a}{2}$ and use the equality $\sin x\sin y=\frac{1}{2}(\cos(x-y)-\cos(x+y))$:
$$\sin\frac{a}{2}(\sin a+\sin 2a+\ldots)= \frac{1}{2}(\cos\frac{a}{2}-\cos\frac{3a}{2}+\cos\frac{3a}{2}-\cos\frac{5a}{2}+\ldots)$$ etc.
• What do you mean by that? Could you show an example? – Alex Oct 29 '13 at 19:42
• I corrected and added the answer. – Boris Novikov Oct 29 '13 at 19:53
• is it true that all the -cos(3a/2)+cos(3a/2) cancel? – Alex Oct 30 '13 at 12:18
• Yes, of course, except the first and the last summands. – Boris Novikov Oct 30 '13 at 13:33
• That's a very nice trick! – Prism Nov 11 '13 at 19:18
$\sin a + \sin2a+\cdots+\sin((n-1)a)$ is the imaginary part of $e^{ia}+e^{2ia}+e^{3ia}+\cdots+e^{(n-1)ia}$. That is a finite geometric series whose sum is expressible in closed form. It will be $\dfrac{\text{something}}{1-e^{ia}}$. Multiplying the top and bottom both by $e^{-ia/2}$ will make it
$$\frac{\text{something}}{e^{-ia/2}-e^{ia/2}} = 2i\cdot \frac{\text{something}}{\sin(a/2)}.$$
Once it's in that form you only have to look at the numerator to find the imaginary part.
Use the formula $$2\sin(x)\sin(y)= \cos(x-y)-\cos(x+y)$$
Then
$$2\sin(\frac{a}{2})\sin(a)=\cos(\frac{a}{2})-\cos(\frac{3a}{2}) \\ 2\sin(\frac{a}{2})\sin(2a)=\cos(\frac{3a}{2})-\cos(\frac{5a}{2}) \\ 2\sin(\frac{a}{2})\sin(3a)=\cos(\frac{5a}{2})-\cos(\frac{7a}{2}) \\ ....\\ 2\sin(\frac{a}{2})\sin(na)=\cos(\frac{(2n-1)a}{2})-\cos(\frac{(2n+1)a}{2}) \\$$
• @Alex All these equalities. Note that on the RHS you get a telescopic sum, while on the LHS $2 \sin(\frac{a}{2})$ is a common factor. – N. S. Oct 29 '13 at 22:46
• @Alex no it doesn't. The last term is just the relation with $x=\frac{a}{2}$ and $y=na$. – N. S. Oct 29 '13 at 22:49 | 2019-10-23T18:46:36 | {
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https://math.stackexchange.com/questions/2603283/maximum-number-of-guaranteed-coins-to-get-in-a-30-coins-in-3-boxes-puzzle/2603301 | # Maximum number of guaranteed coins to get in a “30 coins in 3 boxes” puzzle
There are a total of 30 gold coins in three wooden boxes (but you do not know how many in each individual box). However, you know that one box has exactly 4 coins more than another box.
For each box, you can ask for a number of coins from that box, of your choice. If there are at least that many coins in that box, then you get as many coins as you asked for. Otherwise, you get nothing from that box. You must place all your demands simultaneously in the beginning.
What is the maximum number of coins that you can guarantee yourself to get?
Added for clarification: you don't know which box has 4 coins more.
Let $x, y, y+4$ be the coin contents $=> x+2y=26$ $=>$ solutions $(0,13,17),...,(24,1,5)$.
What next?
• You forgot 26, 0, 4 – QuIcKmAtHs Jan 13 '18 at 7:41
• Do we know which box has 4 more coins than other? (or one unknown box has 4 more coins than some other?) – Atbey Jan 13 '18 at 7:49
• @Atbey No, we do not know which box has 4 coins more. – ami_ba Jan 13 '18 at 8:10
• Fun puzzle. If one looks at it statistically, the best choice is actually $(8,8,8)$ as this has the highest expected value, which is 14.77. – Jens Jan 13 '18 at 14:19
• Since you're using quote block, may I know the source of this puzzle? (just curious, because sometimes users are misusing quote block and code block to highlight/emphasize, which is certainly not the correct purpose) – Andrew T. Jan 13 '18 at 17:56
If you ask for $13,13,13$, then you get nothing if the boxes are $10,8,12$.
If you ask for $12,12,12$, you will always get exactly $12$ coins, since in all cases, exactly one of the boxes has at least $12$ coins.
Claim $12$ is best possible.
Certainly it's best possible with all requests equal.
Suppose a better request triple is $a,b,c$, with $a \le b \le$c, and $a < c$.
But any of the $3$ requests might fail if that box contains $0$ coins, hence to be sure to get more than $12$ coins, every pair must sum to more than $12$.
From $a \le b \le c$, and $a + b > 12$, we get $6 < b \le c$.
But if the boxes are some permutation of $26,0,4$, the $b,c$ requests might both fail, so to be sure to get more than $12$, we must have $a > 12$.
But then $a,b,c$ all exceed $12$, so all requests would fail if the boxes are $10,8,12$.
Thus, unequal requests can't beat the uniform $12,12,12$ request strategy.
It follows that $12$ is the maximum number of coins which can be guaranteed.
• Interesting to note that although (12,12,12) maximises the guaranteed (i.e. minimum) return, it does not maximise the expected (i.e. average) return. (12,12,12) gives an average return of 13 5/7, whereas (10,10,10) gives an average return of 15. – gandalf61 Jan 13 '18 at 15:40
There is always one box with at least 12 coins. If you name (12, 12, 12), you will win at least 12 coins.
If you name some triple where one of the numbers is more than 12, then it's possible that box only has 12, and you win nothing from it. If you are to win from both the other boxes, your other two named numbers must be at most 2, since the other boxes could have 2 and 16. So then you'd win at most 4 total. If you are to win from exactly one of the other two boxes, they might have 8 and 10 and you'd win at most 10 total.
If you name some triple where your lowest number is less than 12 but more than 4, then it's possible that box has 26, and you win nothing from the other two boxes which have 0 and 4 coins. And you only win your lowest named number which we assumed is less than 12.
If your lowest called number is 4 or lower, that box may again have 26, and you can't do better than winning that number plus the 4 from the box with 4. So no better than 8 total.
So your lowest named number is at least 12, and none of your named numbers are more than 12. So the best you can do is guarantee 12 coins with (12,12,12). Note we've shown this is the only triple to guarantee 12 won coins.
After much thought, I realised that (10, 8, 12) can only be the optimal solution. Firstly, we have (10, 10, 10) to ensure we get the most coins in a situation with no constraints. However, here, we have to take note of the 4 coins, so I would say (10, 8, 12).
However, should the boxes with more coins be unknown to us, then alex's answer of (12, 12, 12) would be more suitable.
• If the box you apply 8 to has 26, 24, 22, 20, or 18 coins, then this strategy only gives you 8 coins. But as someone pointed out elsewhere, naming (10,10,10) always gives you at least 10 coins (and half the time gives you 20). – alex.jordan Jan 13 '18 at 7:51
• Yes, but what about if the case is 8, 9, 13? You get 27. – QuIcKmAtHs Jan 13 '18 at 7:54
• question asks about maximum number of coins to guarantee. – Atbey Jan 13 '18 at 7:54
• The question is about guaranteeing some minimal amount. Not about maybe getting some better amount. – alex.jordan Jan 13 '18 at 8:07
This is not an answer, just in case we know which box has 4 coin more, we can have a better strategy.
Say box 3 has 4 more coin than box 2, label them as you did $(x,\ y,\ y+4)$. We can demand for $(16, 6, 10)$, so that in any case we get $16$ coins. | 2019-06-17T20:48:03 | {
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http://www.jimchristy.com/the-matchmaker-lwqknrg/jarqueberatest-in-r-48ffc2 | ## jarqueberatest in r
These are all single sample tests; see “Equality Tests by Classification” for a description of two sample tests. x: a numeric vector or time series. Example: We do not reject the null hypothesis of normality for this series. Column F shows the formulas used: Step 3: Calculate the p-value of the test. normality, homoscedasticity and serial independence of regression Usage jarque.bera.test(x) Arguments. The Jarque-Bera test uses skewness and kurtosis measurements. Yes, you can say that the J-B test is optimal - in the following sense. The Jarque-Bera test is a goodness-of-fit test of departure from normality, based on the sample skewness and kurtosis. The J-B test is the LM test for the nested null hypothesis of normality against the maintained hypothesis that the data are generated by Pearson family of distributions. Tests of Normality Age .110 1048 .000 .931 1048 .000 Statistic df Sig. First, input the dataset into one column: Step 2: Calculate the Jarque-Bera Test Statistic. Here is the implementation, with some comments that Iâve added myself: The test statistic (what I called $$jb$$ above) is reported as x.squared (not sure why that name was chosen), the degrees of freedom parameter is always 2, and the p-value is calculated as 1 - pchisq(STATISTIC,df = 2). Usage jb.norm.test(x, nrepl=2000) 8 jb.norm.test Arguments x a numeric vector of data values. To be precise: Should have mean zero and standard deviation of one. The formula of Jarque-Bera residuals, Economics Letters 6, 255-259. My data.frame looks like this: Jarque, C. and Bera, A. Die Teststatistik JB des Jarque-Bera-Test ist definiert als. Statistic df Sig. GNU R: shapiro.test. The Jarque–Bera test for normality is based on the following statistic: JB = \frac{n}{6}≤ft((√{b_1})^2 + \frac{(b_2-3)^2}{4}\right), where b_1 = \frac{\frac{1}{n}∑_{i=1}^n(X_i - \overline{X})^3}{\frac{1}{n}(∑_{i=1}^n(X_i - \overline{X})^2)^{3/2}}, b_2 = \frac{\frac{1}{n}∑_{i=1}^n(X_i - \overline{X})^4}{\frac{1}{n}(∑_{i=1}^n(X_i - \overline{X})^2)^2}. Hello, I'm so confused why I can't run Jarque-Bera test on my data. (1980) Efficient tests for normality, homoscedasticity and serial independence … Setting robust to FALSE will perform the original Jarque-Bera test (see where $$T$$ is the sample size. Note that f:x also works, since R's parser does not keep the order. Next, calculate the JB test statistic. Why do I get this p-value doing the Jarque-Bera test in R? Jarque-Bera statistics follows chi-square distribution with two degrees of freedom for large sample. Calculating returns in R. To calculate the returns I will use the closing stock price on that date which … Gastwirth, J. L.(1982) Statistical Properties of A Measure … RegressIt also now includes a two-way interface with R that allows you to run linear and logistic regression models in R without writing ... the Shapiro-Wilk test, the Jarque-Bera test, and the Anderson-Darling test. Summary: R linear regression uses the lm() function to create a regression model given some formula, in the form of Y~X+X2. Jarque–Bera test In the case we have an accurate volatility forecast. Zur Navigation springen Zur Suche springen. Skewness of $$x$$ is measured as $S = \frac{\left( E[X - \mu]^{3} \right)^{2}}{\left(E[X - \mu]^{2} \right)^{3}}$. Search everywhere only in this topic Advanced Search. Kurtosis of $$x$$ is measured as $\kappa = \frac{E[X - \mu]^{4}}{\left( E[X - \mu]^{2} \right)^{2}}$, and $$\kappa = 3$$ for a normal distribution. Here, the results are split in a test for the null hypothesis that the skewness is 0, the null that the kurtosis is … This function is based on function jarque.bera.test available in package tseries. Can you test for normality for a (0,1) bounded distribution? I have 9968 observation and I want to run Jarque-Bera test on them, but no matter how hard I am trying I can't... R › R help. EDV GNU R Befehlsübersicht. Depends R (>= 2.10.0) Imports graphics, stats, utils, quadprog, zoo, quantmod (>= 0.4-9) License GPL-2 NeedsCompilation yes Author Adrian Trapletti [aut], Kurt Hornik [aut, cre], Blake LeBaron [ctb] (BDS test code) Maintainer Kurt Hornik Repository CRAN Date/Publication 2020-12-04 13:18:00 UTC R topics documented: Jarque-Bera test ‹ Previous Topic Next Topic › Classic List: Threaded ♦ ♦ 3 messages Kiana Basiri. shapiro.test(x) führt einen Shapiro-Wilk-Test auf die Zahlenreihe x durch. The null hypothesis of bptest is that the residuals have constant variance. 4. This function performs the Jarque-Bera tests of normality either the robust or the classical way. the robust standard deviation (namely the mean absolute deviation In Statistiken der Jarque-Bera - Test ist ein Güte-of-fit Test, ob Beispieldaten haben die Schiefe und Kurtosis eine passende Normalverteilung. Kolmogorov-Smirnov a Shapiro-Wilk a. Lilliefors Significance Correction Data visualization in R, data manipulation in R, machine learning, and more. The test statistic for JB is defined as: Jarque-Bera test. in y ~ x1 | x:f1 + f2, the f1 must be a factor, whereas it will work as expected if f2 is an integer vector. the Jarque-Bera test of normality, Economics Letters 99, 30-32. shapiro.test, Why has the Jarque-Bera test of normality two degrees of freedom? Construct Jarque -Bera test . #some normal data z<-rnorm(100) JarqueBeraTest(z) #some skewed data z<-rexp(100) JarqueBeraTest(z) #some thick tailed data z<-rt(100,5) JarqueBeraTest(z) Documentation reproduced from package FitAR, version 1.94, License: GPL (>= 2) Community examples. from the median, as provided e. g. by MeanAD(x, FUN=median)) to estimate sample kurtosis and skewness. Summary: R linear regression uses the lm() function to create a regression model given some formula, in the form of Y~X+X2. However on this website: and Inference 6, 1-12. The null hypothesis in this test is data follow normal distribution. The Jarque-Bera test is a goodness-of-fit measure of departure from normality based on the sample kurtosis and skew. Hierdurch wird bestimmt, ob die Zahlenreihe x normalverteilt ist. number of Monte Carlo simulations for the empirical critical values. Jarque Bera Test statistic. From tables critical value at 5% level for 2 degrees of freedom is 5.99 So JB>c2 critical, so reject null that residuals are normally distributed. 1 Test-Beschreibung; 2 Beispiel; 3 siehe auch; 4 Weblinks; Test-Beschreibung . I.e. In other words, JB determines whether the data have the skew and kurtosis matching a normal distribution. chisq-distribution or empirically via Monte Carlo. conclusion: Data follow normal distribution with 95% level of confidence. In this video I have shown you how to check whether data is normally distributed or not. Aus Wikibooks. Use apply() to calculate the skewness and kurtosis of the individual equity returns in djreturns assigning the results to s and k, respectively. Inhaltsverzeichnis. defines, whether the robust version should be used. What I have Setting robust to FALSEwill perform the original Jarque-Bera test (seeJarque, C. and Bera, A (1980)). Die Teststatistik des Jarque-Bera-Tests ist immer eine positive Zahl … Open main menu. The null hypothesis in this test is data follow normal distribution. That is a good thing, otherwise we would want to check if Râs random number generating functions are working properly. The Jarque-Bera statistic is j b = T [ S 6 + (κ − 3) 2 24] where T is the sample size. I bet it's the Jarque-Bera (1982, 1987) test. âControlling complexity is the essence of computer programming.â, $S = \frac{\left( E[X - \mu]^{3} \right)^{2}}{\left(E[X - \mu]^{2} \right)^{3}}$, $\kappa = \frac{E[X - \mu]^{4}}{\left( E[X - \mu]^{2} \right)^{2}}$, $jb = T\left[ \frac{S}{6} + \frac{(\kappa - 3)^{2}}{24} \right]$. It is a goodness-of-fit test used to check hypothesis that whether the skewness and kurtosis are matching the normal distribution. Tutorials Tabellen Excel R Python SPSS Stata TI-84 Über Uns. API documentation R package. Default is FALSE. The test statistic of the Jarque-Bera test is always a positive number and the further it is from zero, the more evidence that the sample data does not follow a normal distribution. With over 20 years of experience, he provides consulting and training services in the use of R. Joris Meys is a statistician, R programmer and R lecturer with the faculty of Bio-Engineering at the University of Ghent. Tests the null of normality for x using the Jarque-Bera test statistic. This test is a joint statistic using skewness and kurtosis coefficients. The test is specifically designed for alternatives in the Pearson system of distributions. values should be obtained. We do not reject the null hypothesis of normality for this series. K - die Kurtosis Da K-3 den Exzess widerspiegelt, könnte man gleich in der obigen Formel den Exzess verwenden. The p-value is computed by Mo Consider having v 1 , … , v N observations and the wish to test if they come from a normal distribution. Interpreting normality tests results. Post a new example: Submit your example. I want to perform a Jarque-Bera Test with the tseries package on a data.frame with about 200 columns but it doesn't work with NA values. I was a bit confused regarding the interpretation of bptest in R (library(lmtest)). Gastwirth, J. L.(1982) Statistical Properties of A Measure of Tax Assessment Uniformity, Journal of Statistical Planning and Inference 6, 1-12. The Jarque-Bera test statistic is defined as: $$\frac{N}{6} \left( S^2 + \frac{(K - 3)^2}{4} \right)$$ with S, K, and N denoting the sample skewness, the sample kurtosis, and the sample size, respectively. After all, it's a standard feature in pretty well every econometrics package. And with very good reason. jb = (379/6)*((1.50555^2)+(((6.43 -3)^2)/4)) = 328.9 The statistic has a Chi 2 distribution with 2 degrees of freedom, (one for skewness one for kurtosis). Because the normal distribution is symmetric, the skewness (deviation from symmetry) should be zero. Title Applied Econometrics with R Description Functions, data sets, examples, demos, and vignettes for the book Christian Kleiber and Achim Zeileis (2008), Applied Econometrics with R, Springer-Verlag, New York. The Jarque-Bera test is … The moments package contains functions for computing the kurtosis and skewness of data and well as for implementing the Jarque-Bera test, which is a test of normality based on these higher-order moments. Learn R in step-by-step tutorials. Einführung in R Version 1.0 vom 31.12.2002 Dr. Matthias Fischer Lehrstuhl für Statistik & Ökonometrie Universität Erlangen-Nürnberg [email protected] jarque.bera.test {tseries} R Documentation: Jarque-Bera Test Description. a character string out of chisq or mc, specifying how the critical If you select View/Descriptive Statistics & Tests/Simple Hypothesis Tests, the Series Distribution Tests dialog box will be displayed. Now for the bad part: Both the Durbin-Watson test and the Condition number of the residuals indicates auto-correlation in the residuals, particularly at lag 1. Jarque, C. and Bera, A. The functions for testing normality are: ll{ ksnormTest Kolmogorov-Smirnov normality test, shapiroTest Shapiro-Wilk's test for normality, jarqueberaTest Jarque--Bera test for normality, dagoTest D'Agostino normality test. Gel, Y. R. and Gastwirth, J. L. (2008) A robust modification of The Jarque–Bera test is comparing the shape of a given distribution (skewness and kurtosis) to that of a Normal distribution. We can center the series and scale it using our forecasts for the standard deviation. Hello, I'm so confused why I can't run Jarque-Bera test on my data. My data.frame looks like this: If you use mctol, jbtest determines the critical value of the test using a Monte Carlo simulation. Jarque Bera Test data: x X-squared = 0.046, df = 2, p-value = 0.9773. The Jarque-Bera test (in the fBasics library, which checks if the skewness and kurtosis of your residuals are similar to that of a normal distribution. 1. Testing for normality in non-normal distributions with zero skewness and zero excess kurtosis. A collection and description of functions of one sample tests for testing normality of financial return series. defines if NAs should be omitted. In this tutorial, the most widely used methods will be shown, such as normal plots/histograms, Q-Q plots and Sapiro-Wilk method. References. For more details see Gel and Gastwirth (2006). References. Default is approximated by the The robust Jarque-Bera (RJB) version of utilizesthe robust standard deviation (namely the mean absolute deviationfrom the median, as provided e. g. by MeanAD(x, FUN=median)) to estimate sample kurtosis and skewness. Jarque-Bera test. The Jarque-Bera test (in the fBasics library, which checks if the skewness and kurtosis of your residuals are similar to that of a normal distribution. The formula of Jarque-Bera. Alternate hypothesis (H_1): The data is not normally distributed, in other words, the departure from normality, as measured by the test statistic, is statistically significant. ChickWeight is a dataset of chicken weight … Jarque-Bera-Test - Jarque–Bera test. A list with class htest containing the following components: a character string giving the name of the data. nrepl the number of replications in Monte Carlo simulation. Tutorials Tabellen Excel R Python SPSS Stata TI-84 Über Uns. The test is based on a joint statistic using skewness and kurtosis Fortgeschrittene Einsteiger und 1 R/S-plus für MathematikVII ' & \$ % Lehrstuhl Mathematik VII R/S-Plus für Einsteiger und für Fortgeschrittene ein Kurs über zwei Semester Value. The test is named after Carlos M. Jarque and Anil K. Bera. If so, why do I get this value if I used a random number from a normal distribution? Jarque-Bera test in Excel. J B = n 6 (s 2 + (k − 3) 2 4) , where n is the sample size, s is the sample skewness, and k is the sample kurtosis. jarque.bera.test {tseries} R Documentation: Jarque-Bera Test Description. This view carries out simple hypothesis tests regarding the mean, median, and the variance of the series. Datasets are a predefined R dataset: LakeHuron (Level of Lake Huron 1875–1972, annual measurements of the level, in feet). Note. Missing values are not allowed. The robust Jarque-Bera (RJB) version of utilizes 5. This means that in interactions, the factor must be a factor, whereas a non-interacted factor will be coerced to a factor. Doing a Jarque Bera test in R I get this result: jarque.bera.test(rnorm(85)) data: rnorm(85) X-squared = 1.259, df = 2, p-value = 0.5329 Does it mean that the probability to discard the normality hypothesis (it being true) is 53.29%? 3. How to Conduct a Jarque-Bera Test in R The Jarque-Bera test is a goodness-of-fit test that determines whether or not sample data have skewness and kurtosis that matches a … R includes implementations of the Jarque–Bera test: jarque.bera.test in the package tseries, for example, and jarque.test in the package moments. Case we have an accurate volatility forecast are working properly I was a bit regarding. Following components: a character string out of chisq or mc, specifying how the critical values using parametric., …, v N observations and the rjb.test from the jarque.bera.test ( in package. Finally, the series distribution with two degrees of freedom Güte-of-fit test, it 's Jarque-Bera. Example, and jarque.test in the Pearson family shows the formulas used: Step 3 Calculate. Test wird nach dem Namen Carlos Jarque und Anil K. Bera, Yulia R.,. Python SPSS Stata TI-84 Über Uns two sample tests the expression for the empirical critical values ( the... Details see Gel and Gastwirth ( 2006 ) djx using jarque.test ( ) ” for a of! Tseries, for example, and jarque.test in the package moments, Joseph L. Gastwirth, Weiwen.. R includes implementations of the test is a joint statistic using skewness and kurtosis coefficients determines critical! Machine learning, and more through the package tseries ( 1982, 1987 ) test messages Basiri! Messages Kiana Basiri functions are working properly Delphi, Visual Basic, etc is that the residuals have constant.! Monte Carlo simulation distributed or not sample data have skewness and kurtosis that matches a normal distribution is,! Perform a Jarque-Bera test ( see Jarque, C. and Bera ( 1987 ) test from. Statistischer test, it has maximum local asymptotic power, against alternatives in the expression for Jarque-Bera! Must be estimated s - die Schiefe und kurtosis eine passende Normalverteilung non-normal distributions with zero skewness and excess... Revolution Analytics '' for a ( 1980 ) Efficient tests for normality, homoscedasticity and serial independence of regression,. Its parameters must be a factor carries out simple hypothesis tests regarding the interpretation of bptest is that homoscedasticity! Accurate volatility forecast collection and Description of functions of one sample tests ; see “ Equality tests Classification... 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The residuals: Learn R in step-by-step tutorials this tutorial, the series scale... - test ist ein Anpassungstest, bei dem festgestellt wird, ob eine Normalverteilung vorliegt Joseph Gastwirth. The robust or the classical way freedom for large sample the standard deviation one. See Jarque, C. and Bera ( 1987 ) test for JB is defined as: Jarque-Bera on... In package tseries, for example, and jarque.test in the expression for the Jarque-Bera test is based the... A factor Topic › Classic list: Threaded ♦ ♦ 3 messages Kiana.! A standard feature in pretty well every econometrics package a goodness-of-fit test that determines whether or not and jarque.test the. This video will show you how to check if Râs random number generating are. Steps to perform a Jarque-Bera test on my data more details see Gel and Gastwirth 2006! Annual measurements of the Jarque–Bera test for the standard deviation data manipulation in R, machine,! 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https://mathematica.stackexchange.com/questions/4362/factorizing-polynomials-over-fields-other-than-mathbbc/4372 | # Factorizing polynomials over fields other than $\mathbb{C}$
I'd like to take a polynomial in $\mathbb{Z}_5[x]$ of the form $ax^2+bx+c$ and factor it into irreducible polynomials.
For example:
Input...
x^2+4
Output...
(x+1)(x-1)
Note that this factorization only makes sense in $\mathbb{Z}_5[x]$
I am also interested in identifying cases which are already irreducible.
For example:
Input...
x^2+2
Output...
Polynomial is irreducible.
So, is there a way to limit Mathematica, especially functions like Solve to fields other than $\mathbb{C}$?
All of the polynomial functions, have an option Modulus which allows you to specify an integer field, like $\mathbb{Z}_5$. In particular, Factor works on your example polynomial
Factor[x^2+4, Modulus -> 5]
(* (1 + x) (4 + x) *)
Additionally, IrreduciblePolynomialQ works to determine irreducibility of $x^2+2$, as follows
IrreduciblePolynomialQ[x^2 + 2, Modulus -> 5]
(* True *)
• Damn, I saw the question and was like — "this is an easy one, I'll just spread this cream cheese on my bagel and answer it", only to see you beat me by 8 seconds when I loaded your answer. Next time, I'll reverse the order – rm -rf Apr 17 '12 at 15:02
• @R.M you have to be quick. :) – rcollyer Apr 17 '12 at 15:03
• @R.M the real challange is to add a second answer that is even better when the first one seems almost perfect :) A real artist can turn such a dead case to an epic win. – István Zachar Apr 17 '12 at 16:31
• @IstvánZachar I don't think Leonid can pull that off for this one. Although, there is room to cover other fields beyond integers, like rationals. – rcollyer Apr 17 '12 at 16:54
• Thanks, really helpful. (and so fast too!) – Harold Apr 17 '12 at 22:28
Solve with Modulus
We can use Solve with domain specification like i.e. Integers, or with e.g. integers modulo 5, then instead of specifying the domain one uses Modulus :
Solve[x^2 + 4 == 0, x, Modulus -> 5]
{{x -> 1}, {x -> 4}}
Times @@ ( x - Last @@@ %)
Expand[ %, Modulus -> 5]
(-4 + x) (-1 + x)
4 + x^2
For an integer $n$, $\mathbb{Z}_n$ is a finite ring, while for $n$ being a prime number, then it is also a field.
Factorization with Modulus or Extension
By default Mathematica factorizes polynomials over the rationals not over the complexes, if we'd like to do it over other fields we have to use :
1. Modulus for factorization over rings of integers modulo $n$
2. Extension for factorization over extended fields of rationals by algebraic numbers
In general, we have to use both options separately: if Modulus is not 0, then Extension should be None.
We can use FactorList to get a list of the factors of a polynomial, where the first element is a numerical factor, and the rest are factorizing polynomials with their exponents :
FactorList[x^2 + 4, Modulus -> 5]
{{1, 1}, {1 + x, 1}, {4 + x, 1}}
and in order to test whether we get irreducible polynomials, we can do this :
IrreduciblePolynomialQ[#, Modulus -> 5] & /@ First /@ Rest @ FactorList[x^2 + 4, Modulus -> 5]
{True, True}
Extension may have several elements,e.g. Extension->{a1, a2, a3,...,an}, then a factorized polynomial may be rewritten in terms of any rational combinations of algebraic numbers a1,a2,...,an.
We choose the following polynomial, being a minimal one having a root Sqrt[2] + Sqrt[3], to show how Extension works :
MinimalPolynomial[Sqrt[2] + Sqrt[3], x]
1 - 10 x^2 + x^4
Next, we find its roots :
Solve[1 - 10 x^2 + x^4 == 0, x]
The solutions are algebraic numbers and in order to factorize this polynomial we have to extend the field of rationals, but we do it gradually : first we factorize over the rationals, then we extend it only by rational multiples of Sqrt[2], next only by rational multiples of Sqrt[3] and finally by all rationals combinations of Sqrt[2] and Sqrt[3] :
Factor[1 - 10 x^2 + x^4, Extension -> #] & /@ {None, Sqrt[2], Sqrt[3], {Sqrt[2], Sqrt[3]}} // Column
And we check the results :
(Expand[#] === 1 - 10 x^2 + x^4) & /@ Last @ %
{True, True, True, True}
One can set e.g. Extension -> I as well, to produce in this case the same output as GaussianIntegers -> True :
Factor[x^2 + 4, Extension -> I]
(-2 I + x) (2 I + x)
• This additional detail is verily useful. I appreciate it. – Harold Apr 17 '12 at 22:28
• I added extended discussion of factorization over various fields, I believe you'll find it even more helpful than before. – Artes Apr 18 '12 at 2:30
• Tremendous. We've learned a lot today. – Harold Apr 18 '12 at 5:42 | 2019-08-20T21:13:27 | {
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https://brilliant.org/discussions/thread/road-to-imo/ | Congratulations to one of the most brilliant young minds in the Brilliant community for passing the PMO Area Stage Qualifiers - @Sean Anderson Ty. I say that this is a great feat for there are many who had the chance but wasn't able to qualify (Yep, I'm talking about me) He also lives/studies in the same city as I am, so I am proud that maybe one of ours is going to be part of IMO. As a gift, I give these Fibonacci problems (Who doesn't like proving?) 1) Let $$m$$ and $$n$$ be positive integers. Prove that, if $$m$$ is divisible by $$n$$, then, $$f_{m}$$ is divisible by $$f_{n}$$. 2) Let $$m$$ and $$n$$ be positive integers whose greatest common divisor is $$d$$. Prove that the greatest common divisor of the Fibonacci numbers $$f_{m}$$ and $$f_{n}$$ is the Fibonacci number $$f_{d}$$
Note by Marc Vince Casimiro
3 years, 10 months ago
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× | 2018-10-21T13:28:05 | {
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"url": "https://brilliant.org/discussions/thread/road-to-imo/",
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https://www.physicsforums.com/threads/induction-proof.715878/ | # Induction Proof
## Homework Statement
Ʃ 1/√k ≥ 1/√n (under sigma should be "k=1" and above should be "n", and n is a positive integer)
## The Attempt at a Solution
I. Base case when n=1 is correct.
II. Inductive Hypothesis: Assume true for k=m, where k<m<n, m is a positive integer.
III. Ʃ 1/√m + 1/√m+1 ≥ 1/√n, since Ʃ 1/√m ≥ 1/√n, and Ʃ 1/√m + 1/√m+1 ≥ Ʃ 1/√m.
By the principle of induction, Ʃ 1/√k ≥ 1/√n.
(I'm sorry for leaving out some subscripts under epsilon. I couldn't find how to get k=1 under there or "n" above).
## The Attempt at a Solution
Related Calculus and Beyond Homework Help News on Phys.org
You are on the right track, but I find your presentation a little unclear.
You should start by saying the result it true for n = 1.
Then you assume ##\sum_{k = 1}^n (1/\sqrt{k}) \ge 1/\sqrt{n}##. Given this assumption you must show that ##\sum_{k = 1}^{n+1} (1/\sqrt{k}) \ge 1/\sqrt{n+1}##. This is not what you wrote in your step 3.
You are on the right track, but I find your presentation a little unclear.
You should start by saying the result it true for n = 1.
Then you assume ##\sum_{k = 1}^n (1/\sqrt{k}) \ge 1/\sqrt{n}##. Given this assumption you must show that ##\sum_{k = 1}^{n+1} (1/\sqrt{k}) \ge 1/\sqrt{n+1}##. This is not what you wrote in your step 3.
I thought I had to make the assumption for an arbitrary k (or "n", I'm a bit confused on this one). Otherwise we are assuming the statement we are trying to prove is true.
Mark44
Mentor
Let's look at three statements:
P(1): ## 1/\sqrt{1} \geq 1/\sqrt{1}##
P(k): ##\sum_{k = 1}^n (1/\sqrt{k}) \geq 1/\sqrt{n}##
P(k + 1): ##\sum_{k = 1}^{n + 1} (1/\sqrt{k}) \geq 1/\sqrt{n + 1}##
You assume that the 2nd of these is true. You then show (prove) that the 3rd of these is true, using the 2nd.
That's how it works.
P(k+1):$\sum$$^{n+1}_{k=1}$(1/√k)=$\sum$$^{n}_{k=1}$(1/√k)+(1/$\sqrt{n+1}$).
$\sum$$^{n}_{k=1}$(1/√k)≥1, and so $\sum$$^{n}_{k=1}$(1/√k)+(1/$\sqrt{n+1}$)≥$\sqrt{n}$.
I thought I had to make the assumption for an arbitrary k (or "n", I'm a bit confused on this one). Otherwise we are assuming the statement we are trying to prove is true.
I know it looks like you are assuming what you are trying to prove, but you aren't actually. You are proving that if it works for n it works for n+1. Nobody said it does in fact work for n. Not yet.
What you are doing is bootstrapping. First you establish it is good for n = 1. Then you say, well if it is good for n = 1, can I show it is good for n = 2? If it is good for n = 2, can I show it is good for n = 3? And in general can I show that if it good for n = m then can I show it is good for n = m+1?
Once you have established the general case, then you've shown it is true for any value of n.
However, except for explaining, we don't use the term "m". We just say if it is good for n can I show it is good for n + 1? Same difference.
Occasionally, starting at n = 1 is not the right thing. In special cases you might have to start at a higher number. Those will be pretty obvious, because there won't be any 1 around to start from.
1 person
I know it looks like you are assuming what you are trying to prove, but you aren't actually. You are proving that if it works for n it works for n+1. Nobody said it does in fact work for n. Not yet.
What you are doing is bootstrapping. First you establish it is good for n = 1. Then you say, well if it is good for n = 1, can I show it is good for n = 2? If it is good for n = 2, can I show it is good for n = 3? And in general can I show that if it good for n = m then can I show it is good for n = m+1?
Once you have established the general case, then you've shown it is true for any value of n.
However, except for explaining, we don't use the term "m". We just say if it is good for n can I show it is good for n + 1? Same difference.
Occasionally, starting at n = 1 is not the right thing. In special cases you might have to start at a higher number. Those will be pretty obvious, because there won't be any 1 around to start from.
Thank you for that, that is the best explanation of induction I have ever heard! I've read about it tons of times, watched videos, lectures, but something in what you said finally made it "click". I understand now that the inductive hypothesis doesn't have anything to due with the validity of P(n), and deals with the CONDITIONAL, If P(n) then P(n+1). Wow, thank you!
It seems clear to me now that we can simply continue the 3 steps of the induction process without changing variables, but in every book I have they specify this:
1. P(1) is True
2. For every Natural number, k, If P(k) is true, then P(k+1) is true.
3. Then P(n) is true for all natural numbers, n.
Is it just convention to change from n to k, or is the book just being a little nit-picky?
Also, did you read my final result? Was that correct?
The book is trying hard to explain things, and it either is or is not clearer to bounce around between k's and n's. In general, prove it's true for 1, then prove that if it is true for n it is true for n+1.
Your last line is not correct. ##\sum_{k=1}^n## is not greater than 1; all you know is that it is ##\ge 1/\sqrt n ##. Also, what you are trying to prove is that the sum is > ##1/\sqrt{n+1}##.
Were you told to do this by induction? It works, but it is not the easiest way.
I was not told to use induction, however, this was a test question and we had just finished the induction section. I saw positive integers and assumed it would be the way to prove it.
What other proof method would you suggest I take a look at? Perhaps there is a similar example you can point me to so I can figure it out on my own?
I would say the last line should be: 1/√k ≥ √n ≥ 1/√(n+1)
I would not say that is the last line. Perhaps you should take a break and look at it again tomorrow.
Re an easier way to prove it, the fact is that ##\sum_{k=1}^n 1/\sqrt k = \sum_{k=1}^{n-1} 1/\sqrt k + 1/\sqrt n \ge 1/\sqrt n## because all the previous terms are positive.
I suspect the problem really was ##\sum_{k=1}^n 1/\sqrt k \ge \sqrt n## which at least is not trivially true, and would be a good induction example. Why don't you check your problem again.
Ray Vickson
Homework Helper
Dearly Missed
I would not say that is the last line. Perhaps you should take a break and look at it again tomorrow.
Re an easier way to prove it, the fact is that ##\sum_{k=1}^n 1/\sqrt k = \sum_{k=1}^{n-1} 1/\sqrt k + 1/\sqrt n \ge 1/\sqrt n## because all the previous terms are positive.
I suspect the problem really was ##\sum_{k=1}^n 1/\sqrt k \ge \sqrt n## which at least is not trivially true, and would be a good induction example. Why don't you check your problem again.
I agree that the problem as stated by the OP is trivial and not worthy of illustrating induction. Your modified version seems to be true (at least numerically for n ≤ 150) and is certainly a more interesting and more worthwhile statement to try to prove.
I would not say that is the last line. Perhaps you should take a break and look at it again tomorrow.
Re an easier way to prove it, the fact is that ##\sum_{k=1}^n 1/\sqrt k = \sum_{k=1}^{n-1} 1/\sqrt k + 1/\sqrt n \ge 1/\sqrt n## because all the previous terms are positive.
I suspect the problem really was ##\sum_{k=1}^n 1/\sqrt k \ge \sqrt n## which at least is not trivially true, and would be a good induction example. Why don't you check your problem again.
I'm sorry, I did in fact copy the problem incorrectly. Your version is right, and I figured it out. Without going through all the formalities, here is the end of my proof:
1. $\sum$$^{n+1}_{k=1}$(1/$\sqrt{k}$)≥$\sqrt{n+1}$
2. $\sum^{n}_{k=1}$(1/$\sqrt{k}$)+(1/$\sqrt{n}$)≥$\sqrt{n+1}$
3. Since $\sum^{n}_{k=1}$(1/$\sqrt{k}$)≥$\sqrt{n}$ we can write:
4. $\sqrt{n}$+(1/$\sqrt{n}$)≥$\sqrt{n+1}$
5. Squaring both sides and moving terms around yields:
1+1/n≥0.
I am pretty sure this is correct. Perhaps my reasoning for step 3 could be a little clearer. I'm not sure how else to word it.
This one is correct. The presentation could be improved this way:
For step one state that this is what is to be proved based on the hypothesis. At each subsequent step you need to say that they are equivalent -- i.e. the implication goes both ways. That is because when you get to the bottom you have something you know to be true, but now you have to start there and deduce what you were trying to prove. In other words the implications have to run in reverse.
Not sure? Aren't there plenty of examples where a ##\Rightarrow## b does not imply b ##\Rightarrow## a?
Another detail you might attend to if you are trying to be quite perfect is the ##\sqrt k## has two values, + and -. So you could indicate you are only considering positive square roots.
1 person
Mark44
Mentor
Another detail you might attend to if you are trying to be quite perfect is the ##\sqrt k## has two values, + and -.
Nope - the expression ##\sqrt{k}## has one value, which is greater than or equal to zero. It's true that every positive number has two square roots, but the symbol ##\sqrt{k}## represents the positive one.
I am assuming that k is a nonnegative real number, which is a perfectly valid assumption in the context of this problem.
Yes, it was the right assumption for this problem. I'm not sure what ##\sqrt k## represents generally. I think in some contexts both roots would be under consideration; indeed that could be the point of some problems. And it would be nice if in those cases they said to consider negative numbers, but they don't always. You pick it up from context.
Mark44
Mentor
My main point was that ##\sqrt{k}## represents the positive square root of k. It does NOT represent two values. If it did we could write the quadratic formula like this:
$$x = \frac{-b + \sqrt{b^2 - 4ac}}{2a}$$
I.e., without the ±. | 2020-06-04T15:39:13 | {
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http://math.stackexchange.com/questions/685221/problem-with-logarithms | # Problem with logarithms
Problem:
Solve:
$$\frac{1}{2^x} = \frac{5}{8^{x+2}}$$
My attempt:
$$\frac{1}{2^x} = \frac{5}{8^{x+2}}$$
$$\Rightarrow 5 \cdot 2^x = 8^{x+2}$$
$$\Rightarrow 2^{\log_2 5+x} = 8^{x+2}$$
$$\Rightarrow (\log_2 5 + x)(\log_a 2) = (x+2)(\log_a 8)$$
And then just keep going like this, but I'm obviously wrong already as the answer is:
$$x = \frac{\ln 5 - 9 \ln 2}{2 \ln 2}$$
What am I doing wrong?
EDIT: If someone could enlarge the latex for me that would be great.
-
you have a good start, but you can continue better. See my solution. – Babak Miraftab Feb 21 '14 at 20:13
You can get the larger displayed latex form by bracketing the math code between two pairs of dollar signs (insead of one pair of dollar signs). I just did this for your question. – Dave L. Renfro Feb 21 '14 at 20:17
Thank you for that! – user3200098 Feb 21 '14 at 20:21
With the content of the answers below, I'm not certain that the given answer is correct. Is there perhaps a copying error? – abiessu Feb 21 '14 at 21:41
$2^{\log_2 5+x} = 8^{x+2}$ implies that $2^{\log_2 5+x} = 2^{3x+6}$ and so $\log_2 5+x=3x+6$. We can conclude that $x=\frac{\log_2 5-6}{2}$.
-
Thank you...!!! – user3200098 Feb 21 '14 at 20:20
Some of the answers don't mention that $8=2^3$. $$5\cdot 2^x = 8^{x+2}$$
Remember that $8=2^3$. So $$5\cdot 2^x = (2^3)^{x+2}.$$ $$5\cdot 2^x = 2^{3(x+2)}$$ $$5\cdot 2^x = 2^{3x+6}$$ $$5 = 2^{3x+6}\cdot 2^{-x} = 2^{2x+6}$$ $$\log_2 5 = 2x+6$$ $$-6 + \log_2 5 = 2x$$ $$\frac{-6+\log_2 5}{2} = x$$
-
Thanks a lot :) – user3200098 Feb 21 '14 at 22:47
I didn't realize I could do $5 = 2^{3x+6}\cdot 2^{-x} = 2^{2x+6}$ though. Multiply by negative exponent on both sides that is to get rid of the $2^x$ on one side. – user3200098 Feb 21 '14 at 22:52
@user3200098 : What is being done is this: $2^{3x+6}\cdot 2^{-x} = 2^{\Big(3x+6\Big)+\Big(-x\Big)}$. Then simplify $(3x+6)+(-x)$ to get $2x+6$. – Michael Hardy Feb 22 '14 at 0:34
With a slight modification, we have
$$5\cdot 2^x=8^{x+2}=(2^3)^{x+2}=2^{3x+6}$$
which means that
$$5=2^{2x+6}\\ \log_2 5=2x+6\\ x=\frac{\log_2 (5)-6}2$$
Transforming to the natural logarithm would look like
$$5=2^{2x+6}\\ \ln 5=(2x+6)\ln 2\\ x=\frac{\ln 5-6\ln 2}{2\ln 2}$$
-
HINT Use substitution to solve the equation easily $\to$ pose $2^x=t$
- | 2015-09-04T11:06:51 | {
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https://xaktly.com/ProbabilitySetOperations.html | #### xaktly | Probability
Set operations
The basic flow of the probability pages goes like this:
### Sets
In our work in probability theory, we'll be using a lot of sets, so it's prudent to know as much as we can about them, their properties and how to combine and compare them.
A set is a collection of distinct elements. In probability, those elements will usually be outcomes of some probability experiment, like the numbers that can turn up when rolling dice. Here are some examples of sets in different notations:
$\{a, \, b, \, c, \, d\}$ is a set with a finite number of elements, four in this case.
$\mathbb{R}$ or $\{ \mathbb{R} \}$ is the infinite set of all real numbers. Its members are uncountable. The infinite set of integers can be written as $\{1, 2, 3, \dots \}.$
$\{ x \in \mathbb{R}: \, cos(x) \gt \frac{1}{2} \}$ is the set of all real numbers with cosines larger than ½. The proper way to read this statement is "The set of all numbers that are elements of the set of real numbers, such that the cosine of the number is greater than ½."
Finally, we can represent sets with diagrams. The set S of elements x2, x2, and so on, can be represented like this:
We'll use the Greek capital letter omega, Ω, to represent a universal set to which all other sets belong. Thus our set S lies within Ω like this:
In the set S, we say that element x1 is an element of S like this:
$$x \in S$$
We can use a similar notation to show that an element is not an element of a set. Our set S is $\{x_1, x_2, x_3, \dots , x_{10} \},$ which does not contain an element labeled a, so we write
$$a \notin S$$
We will later write the probability of obtaining outcome x1 from set S as $P(x_1).$
Disjoint sets, like sets A and B in the Venn diagram below, share no elements.
### Conjugate sets (!S) and the empty set
Sticking with our set S from above, we can also generate a set
$$!S \phantom{0000} \text{or} \phantom{0000} S^C$$
which represent everything not in set S. The notation $!S$ can be read as "not S", and $S^C$ means "the conjugate of set S." In terms of a diagram, the set $\{ !S \}$ is the gray area outside of set S and inside of Ω
Now using this notation, we can make some important statements. First, let's note that we use the symbol $\in$ to mean "is in" or "is an element of". Likewise, the symbol $\notin$ means "is not in" or "is not an element of." Then a set element xn is a member of the set $!S,$ if it is in the set Ω, and it is not in the set S, or
$$x \in !S \phantom{00} \text{if} \phantom{00} x \in \Omega \phantom{00} \text{and} \phantom{00} x \notin S.$$
Second, in keeping with our notion that a double negative is a positive [such as -(-3) = 3], we have
$$!(!S) = S.$$
The empty set, a set with no elements, is denoted $\{ \phantom{00} \}$ or $\emptyset$, and we note that:
$$\Omega^C = \emptyset \phantom{00} \text{or} \phantom{00} !\Omega = \emptyset ,$$
which is to say that everything is in the universal set, Ω, and nothing is outside of it.
### Subsets
A subset is a set entirely contained within another. For example, the set $\{a, b, c \}$ is a subset of the set of all lower-case letters, or a subset of $\{a, b, c, d, e, f \}.$ We write
$$\{a, b, c \} \subset \{a, b, c, d, e, f \}.$$
Later, the question of how many different n-element subsets can be made out of an m-element set will be a central one in our exploration of probability. (In the example above, n = 3 and m = 6.)
In the Venn diagram
we have $B \subset A,$ and we note that
$$\text{If} \; x \in B, \; \text{then} \; x \in A.$$
Also, a set is always a subset of itself: $B \subset B.$
### Unions and intersections
A union of two sets is the set formed when the two are combined. In set notation, the union of sets $\{a, b, c\}$ and $\{d, e, f \}$ is the new set $\{a, b, c, d, e, f \}.$ We generally write a union like this:
$$\{a, b, c\} \cup \{d, e, f \} = \{a, b, c, d, e, f \}.$$
In Venn diagrams, two sets, A and B look like this:
The union of these sets is just the set made of the two combined:
Notice that x is an element of the union of sets A and B, if and only if x is an element of set A or x is an element of set B. Here is a more compact version of that statement:
$$x \in (A \cup B) \phantom{0} \text{iff} \phantom{0} x \in A \phantom{0} \text{or} \phantom{0} x \in B.$$
The symbol $\cup$ means "union." The notation iff is just a mathematical abbreviation for the phrase "if, and only if."
The intersection of two sets, A and B, is the set that is formed from elements that are in both sets. For example, the intersection of sets $\{a, b, c\}$ and $\{a, b, d, e\}$ is the set $\{a, b\}$ because elements a and b are in both sets. We write
$$\{a, b, c\} \cap \{a, b, d, e\} = \{a, b\},$$
where the symbol $\cap$ means "intersection."
Here is the Venn diagram picture:
Notice that if set element x can be an element of sets A and B if and only if x is an element of A and x is an element of B. In mathematical language, that is
$$x \in (A \cap B) \phantom{0} \text{iff} \phantom{0} x \in A \phantom{0} \text{and} \phantom{0} x \in B.$$
### Some properties of sets
#### Unions
When we're using unions of sets, order of the expression $A \cup B$ doesn't matter.
$$A \cup B = B \cup A$$
#### Multiple unions
$$A \cup (B \cup C) = (A \cup B) \cup C$$
Here is a Venn diagram explanation of that property:
This kind of rule can be extended to any number of sets.
#### Intersections
The intersection of a set with the union of two other sets can be written as
$$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$$
This looks like the distributive property of multiplication. Let's illustrate this property with an example. Take the sets $A = \{a, b, c\},$ $B = \{a, b, e\}$ and $C = \{c, d, f\}.$ Now:
\begin{align} \{a,b,c\} &\cap (\{a,b,e \} \cup \{c,d,f\}) \\[5pt] &= \{a,b,c\} \cap \{a,b,c,d,e,f\} \\[5pt] &= \{a,b,c\} \end{align}
AND
\begin{align} \{a,b,c\} &\cap (\{a,b,e \} \cup \{c,d,f\}) \\[5pt] &= (\{a,b,c\} \cap \{a,b,e\}) \cup (\{a,b,c\} \cap \{c,d,f\}) \\[5pt] &= \{a,b\} \cup \{c\} \\[5pt] &= \{a,b,c\} \end{align}
#### Other properties
Double negation: $!(!A) = A.$
The union of any set that is a subset of the universal set, Ω, and the universal set is just the universal set: $S \cup \Omega = \Omega.$
The intersection between a set and its complement is the empty set: $S \cap !S = \emptyset,$ or $S \cap S^C = \emptyset.$
The intersection of a set contained within the universal set with the universal set is just that set: $S \cap \Omega = S.$
### DeMorgan's laws
DeMorgan's laws relate intersections and unions of sets. First let's see what they are, then we'll prove each of them pictorially below.
\begin{align} !(A \cup B) &= \; !A \cap !B \\[5pt] !(A \cap B) &= \; !A \cup !B \end{align}
DeMorgan's laws are sometimes stated like this:
• The complement (the "not") of the union of two sets is the same as the intersection of their complements.
• The complement of the intersection of two sets is the same as the union of their complements.
These important laws link unions and intersections of sets. This will be crucial later on in probability theory, and we will refer back to DeMorgan's laws when we get there.
Below are two visual illustrations of DeMorgan's laws using Venn diagrams. They're worth staring at a bit. Further down you'll find a full algebraic proof of one of them.
#### $!(A \cup B) = \; !A \cap !B$
In the Venn diagrams above, the left hand side of Demorgan's law, $!(A \cup B) = !A \cap !B$ is illustrated. The blue oval is set A, the green set B. Their complements, !A and !B are also shown. The blue-green blob is the union, $A \cup B.$ The dark gray space in the third diagram is the set $!(A \cup B).$
In this set of Venn diagrams, we focus on the right side of this version of DeMorgan's first law, $!A \cap !B.$ The sets !A and !B are shown in light gray. If we stack them on top of one another, the dark-gray area in the right-hand diagram shows the overlap or intersection. Notice that this is the same as the dark-gray area in the set of figures above.
#### $!(A \cap B) = \; !A \cup !B$
In the Venn diagrams to the left, we show the intersection of sets A and B $(A \cap B),$ in blue-green. All other parts of the set Ω are shown in gray in the right-had panel.
In the set of Venn diagrams below, we superimpose the sets !A and !B. The resulting dark gray area, the set $!A \cup !B$ is the same as the gray area above, so this shows that $!(A \cap B) = \; !A \cup !B.$
#### Algebraic proof of $!(A \cap B) = !A \cup !B$
First, consider any x that is in the set !(A ∩ B). We then realize, by the definition of the complement ( ! ), that x is not in the set (A ∩ B):
Let $x \in !(A \cap B),$ then $x \notin (A \cap B).$
Now because (A ∩ B) is the set of all elements that are in both A and B (x ∈ A and x ∈ B) it must be true that x is either in set A or set B:
Because $A \cap B, \; x \notin A \text{ or } x \notin B.$
Then if x ∉ A, then x ∈ !A, so x ∈ [!A ∪ !B] because the set {!A ∪ !B} includes !A.
If $x \notin A,$ then $x \in !A,$ so $x \in {!A \cup !B}$
Similarly, if x ∉ B, then x ∈ !B, so x ∈ {!A ∪ !B}.
This means that for every x that is in {!A ∩ !B}, x ∈ !A ∪ !B. That is,
$$!(A \cap B) \subset \{!A \cup !B\}.$$
Well, now that's not enough to prove our assertion, because proving that one thing is a subset of another does not show equality. We can prove the reverse assertion by contradiction. First, let x ∈ {!A ∪ !B}, and assume that x ∉ !(A ∩ B). If that is true, then
$$x \in \{A \cap B \}$$
Then it follows that x ∈ A and x ∈ B, thus
$$x \notin !A \text{ and } x \notin !B.$$
However, that means that
$$x \notin \{!A \cup !B\},$$
which is a contradiction to our assumption that x ∈ {!A ∪ !B}. Therefore it is not true that x ∈ !(A ∩ B). Therefore
$$x \in !(A \cap B).$$
Therefore we find that for all x that are in the set {!A ∪ !B}, x ∈ {!(A ∩ B)}. That is:
$$!A \cup !B \subset !(A \cap B)$$
Finally, if
$$!A \cup !B \subset !(A \cap B) \text{ and } !(A \cap B) \subset {!A \cup !B},$$
then $!(A \cap B) = !(A \cup B).$
The other DeMorgan's equation can be proved in a similar way.
### DeMorgan's laws: Compact notation
DeMorgan's laws can be extended to unions and intersections of more than one set. For this we often use a more compact notation, akin to summation notation. For example, the union
$$A_1 \cup A_2 \cup A_3$$
can be written as
$$\bigcup_{i=1}^3 \, A_i.$$
The large ∪ symbol means to repeatedly take the union of sets labeled Ai as the index i runs from 1 to 3, inclusive, in integer steps.
If we leave off the indices (the n = 1 and 3 parts), the open union just means to sum over all of the sets being considered.
Now we can write one of DeMorgan's laws as
$$!\left( \bigcap_i A_i \right) = \bigcup_i \, A_i$$
Likewise, the other law can be written
$$!\left( \bigcup_i A_i \right) = \bigcap_i \, A_i.$$
### Practice problems
1 A universal set, Ω, is $\Omega = \{ 1,2,3,4,5,6,7 \}.$ Given that $A = \{1,2\},$ $B = \{2,4,5\},$ and $C = \{1,5,6,7\},$ write the following sets: $A \cup B$ $A \cap B$ $!A$ $!B,$ and Verify DeMorgan's law by finding $!(A \cup B)$ and $!A \cap !B.$ Solution Here is a Venn diagram of this scenario. Notice that while the element 3 is in the universal set, it is not in any of its subsets. From the diagram, we can easily see: $A \cup B = \{1,2,4,5\}$ $A \cap B = \{ 2 \}$ $!A = \{3,4,5,6,7 \}$ $!B = \{1,3,6,7 \}$ Now to prove DeMorgan's law: $$(A \cup B) = 2, \; \text{ so } !(A \cup B) = \{3,6,7 \}$$ \begin{align} !A &= \{3,4,5,6,7 \} \\[5pt] !B &= \{1,3,6,7 \} \\[5pt] !A \cap !B &= \{3,6,7\} \end{align} These are the same, so the law is verified for this case. 2 Define these sets: $A = \{ 11,13,15,17,19 \}$ $B = \{ 11,14,17,20 \}$ $\Omega = \{ 10,11,12,13,14,15,16,17,18,19,20 \},$ where Ω is the universal set. Write these sets: $A \cap B$ $!A$ $A \cup B$ $!B$ Solutions a. $A \cap B = \{11,17\}$ b. $!A = \{10,12,14,16,18,20\}$ c. $A \cup B = \{11,13,14,15,17,19,20\}$ d. $!B = \{10,12,13,15,16,18,19\}$ 3 Let $A = \{\text{red}, \text{orange}, \text{yellow}, \text{green}, \text{blue}, \text{indigo}, \text{violet} \},$ $B = \{\text{violet}, \text{blue}, \text{green}, \text{yellow}, \text{orange}, \text{red}\},$ and $C = \{\text{white}, \text{black}, \text{red}\}.$ Write: $(A \cap A)$ $(A \cup B)$ $(A \cap C)$ $(A \cup C)$ $(B \cap C)$ $(B \cup C)$ Solution a. $(A \cap A) = \{\text{red}, \text{orange}, \text{yellow}, \text{green}, \text{blue}, \text{indigo}, \text{violet} \}$ b. $(A \cup B) = A \; \text{ because } \; B \subset A$ c. $(A \cap C) = \{ \text{red}\}$ d. $(A \cup C) = \{\text{red}, \text{orange}, \text{yellow}, \text{green}, \text{blue}, \text{indigo}, \text{violet}, \text{white}, \text{black}\}$ e. $(B \cap C) = \{\text{red}\}$ f. $(B \cup C) = \{\text{violet}, \text{blue}, \text{green}, \text{yellow}, \text{orange}, \text{red}, \text{white}, \text{black}\}$ 4 Which of these are true? If $A \subset B$ then $A \cup B = A$ If $A \subset B$ then $A \cup B = B$ If $A \subset B$ and $B \subset C$ then $(A \cup B) \subset C$ If $A \subset B$ and $B \subset C,$ then $A \subset C$ If $(A \cup B) = A,$ then $B \subset A$ If $A \cap B = \{ \}$ and $B \cap C = \{ \},$ then $A \cap C = \{ \}$ Solutions a (and b). If A is a subset of B, then $A \cup B = B.$ That means that (b) is true. This Venn diagram might help: c. If $A \subset B$ and $B \subset C$ then $(A \cup B) \subset C$ To prove statements like these it's often useful to consider elements of the sets. Let $x \in (A \cup B),$ then either $x \in A$ or $x \in B.$ Now If $x \in A,$ then because $A \subset C, \; x \in C.$ If $x \in B,$ then because $B \subset C, \; x \in C.$ If $x \in C$ then we've proved (c) to be true. If $A \subset B$ and $B \subset C,$ then $A \subset C$ d. Let $x \in A.$ Then because $A \subset B,$ we know that $x \in B.$ We can follow a similar set of steps to show that $x \in C.$ This Venn diagram might help: e. If $(A \cup B) = A,$ then $B \subset A$ Notice that if $x \in A,$ then $x \in (A \cup B),$ therefore the fact that $x \in A$ means that $B \subset A$ because all of the elements of B are in A. f. If $A \cap B = \{ \}$ and $B \cap C = \{ \},$ then $A \cap C = \{ \}.$ $A \cap B = \{ \}$ means that sets A and B don't intersect. The same is true for sets B and C because $B \cap C = \{ \}.$ But this doesn't guarantee that sets A and C don't intersect. Consider this Venn diagram, which shows that it's possible:
X
### Axiom
An axiom is a statement that is self-evidently true, accepted or long-established, but which can't necessarily be proven so. A famous axiom is Euclid's first postulate, also known as the transitive property: "Things that are equal to the same thing are equal to each other," or If a = b and c = b, then a = c.
X
### Induction
When we reason inductively, or use induction to reason, we extend a pattern. The sun has risen every day of my life, so I'm reasonably confident it will rise again tomorrow. That's induction. To extend a series like 2, 4, 6, 8, x, by concluding that x = 10 is also induction.
X
### Axiom
An axiom is a statement that is regarded as being true or established, or self-evidently true. Axioms cannot be proven.
A fundamental axiom is the transcendental property: Things that are equal to the same thing are equal to each other, or if a = b and c = b, then a = c.
X
### Theorem
A theorem is a general proposition in mathematics or other forms of reasoning that may not be self-evident, but can be proved by a chain of reasoning based on things that are already known to be true.
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"url": "https://xaktly.com/ProbabilitySetOperations.html",
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# A company sells pens and pencils. The Revenue from pens in 2007 was u
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A company sells pens and pencils. The Revenue from pens in 2007 was u [#permalink]
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23 Jun 2010, 18:36
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A company sells pens and pencils. The Revenue from pens in 2007 was up 5% from 2006. The revenue from pencils declined 13% over the same period. Overall revenue was down 1% from 06 to 07. What was the ratio of pencil revenue to pen revenue in 2006.
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Re: A company sells pens and pencils. The Revenue from pens in 2007 was u [#permalink]
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24 Jun 2010, 06:02
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Jinglander wrote:
Can I get someone to walk me through this one too.
A company sells pens and pencils. The Revenue from pens in 2007 was up 5% from 2006. The revenue from pencils declined 13% over the same period. Overall revenue was down 1% from 06 to 07. What was the ratio of pencil revenue to pen revenue in 2006.
It would be easier if you make a table:
Attachment:
untitled.JPG [ 11.88 KiB | Viewed 5457 times ]
We are told that "Overall revenue was down 1% from 06 to 07" --> $$0.99(x+y)=1.05x+0.87y$$. Question: $$\frac{y}{x}=?$$
$$0.99(x+y)=1.05x+0.87y$$ --> $$0.99x+0.99y=1.05x+0.87y$$ --> $$12y=6x$$ --> $$\frac{y}{x}=\frac{6}{12}=\frac{1}{2}$$.
Hope it's clear.
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Re: A company sells pens and pencils. The Revenue from pens in 2007 was u [#permalink]
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28 Jun 2010, 07:36
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Jinglander wrote:
Can I get someone to walk me through this one too.
A company sells pens and pencils. The Revenue from pens in 2007 was up 5% from 2006. The revenue from pencils declined 13% over the same period. Overall revenue was down 1% from 06 to 07. What was the ratio of pencil revenue to pen revenue in 2006.
Let the revenue of pens be 'x'
Let the revenue of pencils be 'y'
the revenue of pens up from 2006 by 5%=>105%
the revenue of pencils down from 2006 by 13%=>87%
Overall revenue down by 1%=>99%
Attachment:
pic2.JPG [ 3.08 KiB | Viewed 5352 times ]
hence the ratio of pencils to pens = 6/12 = 1/2
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Re: A company sells pens and pencils. The Revenue from pens in 2007 was u [#permalink]
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24 Jun 2010, 06:12
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This one isn't too bad. Say: pen = n and pencil = p and revenue = r.
p+n = r
1.05p+.87n=.99r
now just add plug in for r.
.06n=.12p
p/n = 1/2
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Re: A company sells pens and pencils. The Revenue from pens in 2007 was u [#permalink]
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20 Jun 2018, 04:26
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Jinglander wrote:
A company sells pens and pencils. The Revenue from pens in 2007 was up 5% from 2006. The revenue from pencils declined 13% over the same period. Overall revenue was down 1% from 06 to 07. What was the ratio of pencil revenue to pen revenue in 2006.
Responding to a pm:
Yes, we can use weighted averages here since there are two components and we have the total.
w1/w2 = (A2 - Aavg)/(Aavg - A1)
Say, w1 is the weight of pencil revenue and w2 is the weight of pen revenue in 2006 (increase in revenue is over previous revenue so the ratio we get will be of the revenues of 2006)
w1/w2 = (5 - (-1)) / (-1 - (-13)) = 6/12 = 1/2
Revenue of pencils : Revenue of pens = 1:2
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Re: A company sells pens and pencils. The Revenue from pens in 2007 was u [#permalink]
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23 Jun 2010, 19:17
Let us assume the following letters for respective revenues:
Pen (2006) - A1
Pencil (2006) - B1
Pen (2007) - A2
Pencil (2007) - B2
Now from the given information, we can translate the words into equations as follows:
A2 = $$\frac{105}{100}$$ A1 - Equation 1
and
B2 = $$\frac{87}{100}$$ B1 - Equation 2
Also, we know that
A2 + B2 = $$\frac{99}{100}$$ (A1 + B1) - Equation 3
Now adding equations 1 and 2 together, we get
A2 + B2 = $$\frac{105}{100}$$ A1 + $$\frac{87}{100}$$ B1 - Equation 4
Equation 3 = Equation 4
So we get,
$$\frac{105}{100}$$ A1 + $$\frac{87}{100}$$ B1 = $$\frac{99}{100}$$ (A1 + B1) = $$\frac{99}{100}$$ (A1) + $$\frac{99}{100}$$ (B1)
Now, separating the terms we get:
$$\frac{105}{100}$$ A1 + $$\frac{87}{100}$$ B1 = $$\frac{99}{100}$$ (A1) + $$\frac{99}{100}$$ (B1)
Rearranging terms and canceling the denominator 100 on both sides, we get
(99-87) B1 = (105-99) A1
So $$\frac{B1}{A1}$$ = $$\frac{105-99}{99-87}$$ = $$\frac{6}{12}$$ = $$\frac{1}{2}$$
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Re: A company sells pens and pencils. The Revenue from pens in 2007 was u [#permalink]
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23 Jun 2010, 19:21
What logic did you use to add equation 1 to equation 2
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Re: A company sells pens and pencils. The Revenue from pens in 2007 was u [#permalink]
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23 Jun 2010, 19:23
Because the second part of the question states a relation between the overall income in 2006 and 2007. Using the variables I've assumed, that's a relation between A1+B1 and A2+B2. So in order to get anything meaningful out of this equation, we'll need another equation with A1+B1 or A2+B2 and hence I decided to add them together in order to get all the terms in the equation in terms of A1 or B1 and hence get the ratio by getting all the similar terms to one side.
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Re: A company sells pens and pencils. The Revenue from pens in 2007 was u [#permalink]
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24 Jun 2010, 03:57
Do these kind of questions appear on the real test? I mean it seems to me like a relatively difficult question to solve if you have 2 minutes.
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Re: A company sells pens and pencils. The Revenue from pens in 2007 was u [#permalink]
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24 Jun 2010, 05:58
I've seen a lot of mixture problems being posted on the forum, and on the few practice tests I've taken, it's been asked. So I am not sure. Maybe someone who's taken the GMAT can answer better.
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Re: A company sells pens and pencils. The Revenue from pens in 2007 was u [#permalink]
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24 Jun 2010, 07:48
First, I think its more of a ratio problem than the mixture.
Secondly, I simply cant understand why we are so reluctant to give kudos. Bunuel! Excellent solution. +1 for you. The very last equation contains a little typo though!
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Re: A company sells pens and pencils. The Revenue from pens in 2007 was u [#permalink]
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26 Jul 2010, 16:40
achiever01 I dont understand your solution at all
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Re: A company sells pens and pencils. The Revenue from pens in 2007 was u [#permalink]
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29 Sep 2018, 21:39
This question can also be solved using mixture/alligation technique.
Pen revenue = +5% (increase)
Pencil revenue = -13% (decrease)
Overall = -1% (decrease)
Pen Pencil
+5% -13%
-1%
[-1-(-13)]=12 [5-(-1)]=6
Ratio = Pencil:Pen
6:12=1:2
Thanks.
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Re: A company sells pens and pencils. The Revenue from pens in 2007 was u [#permalink]
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02 Oct 2018, 00:18
Jinglander wrote:
A company sells pens and pencils. The Revenue from pens in 2007 was up 5% from 2006. The revenue from pencils declined 13% over the same period. Overall revenue was down 1% from 06 to 07. What was the ratio of pencil revenue to pen revenue in 2006.
forgot about percentage and change into numericals.
in 2006 the price of pen and pencil both are 100. so the total average of revenue is 100 too.
in 2007 the price of pen is 105 and pencil is 87. As it was mentioned in question average slipped by 1 point so 99.
6 12
105-------99--------------87
the ratio will be 12:6 of pen to pencil ratio. i.e., 2:1
but, it was asked for pencil to pen so it is 1:2...
this is one of the best method you can try for all mixture problems.
Re: A company sells pens and pencils. The Revenue from pens in 2007 was u [#permalink] 02 Oct 2018, 00:18
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http://booksbw.com/index.php?id1=4&category=technics&author=beoce-we&book=2001&page=202 | Books in black and white
Books Biology Business Chemistry Computers Culture Economics Fiction Games Guide History Management Mathematical Medicine Mental Fitnes Physics Psychology Scince Sport Technics
# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.
Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
Previous << 1 .. 196 197 198 199 200 201 < 202 > 203 204 205 206 207 208 .. 609 >> Next
28. Suppose that det A = 0, and that x = x(0) is a solution of Ax = b. Show that if ^ is a
solution of A^ = 0 and a is any constant, then x = x(0) + a^ is also a solution of Ax = b.
29. Suppose that det A = 0 and that y is a solution of A*y = 0. Show that if (b, y) = 0 for every such y, then Ax = b has solutions. Note that this is the converse of Problem 27; the form of the solution is given by Problem 28.
30. Prove that A = 0 is an eigenvalue of A if and only if A is singular.
31. Prove that if A is Hermitian, then (Ax, y) = (x, Ay), where x and y are any vectors.
32. In this problem we show that the eigenvalues of a Hermitian matrix A are real. Let x be an eigenvector corresponding to the eigenvalue A.
(a) Show that (Ax, x) = (x, Ax). Hint: See Problem 31.
(b) Show that A(x, x) = A(x, x). Hint: Recall that Ax = Ax.
(c) Show that A = A; that is, the eigenvalue A is real.
33. Show that if Aj and A2 are eigenvalues of a Hermitian matrix A, and if Aj = A2, then the corresponding eigenvectors x(1) and x(2) are orthogonal.
Hint: Use the results of Problems 31 and 32 to show that (Aj - A2)(x(1), x(2)) = 0.
368
Chapter 7. Systems ofFirst Order Linear Equations
7.4 Basic Theory of Systems of First Order Linear Equations
The general theory of a system of n first order linear equations
x1 = P11 (t)x1 + ¦¦¦ + P1n (t)xn + g\(t),
(1)
xn = Pn1(t)x1 + ¦¦¦ + Pnn (t)xn + gn(t)
closely parallels that of a single linear equation of nth order. The discussion in this section therefore follows the same general lines as that in Sections 3.2, 3.3, and 4.1. To discuss the system (1) most effectively, we write it in matrix notation. That is, we consider x1 = ô1(t),..., xn = ôï (t) to be components of a vector x = ); similarly,
g1(t),..., gn(t) are components of a vector g(t), and P11(t),..., Pnn(t) are elements of an n x n matrix P(t). Equation (1) then takes the form
x' = P(t)x + g(t). (2)
The use of vectors and matrices not only saves a great deal of space and facilitates calculations but also emphasizes the similarity between systems of equations and single (scalar) equations.
A vector x = ô^) is said to be a solution ofEq. (2) if its components satisfy the system of equations (1). Throughout this section we assume that P and g are continuous on some interval a < t < â; that is, each ofthe scalar functions p11, ..., Pnn, gv ..., gn is continuous there. According to Theorem 7.1.2, this is sufficient to guarantee the existence of solutions ofEq. (2) on the interval a < t < â.
It is convenient to consider first the homogeneous equation
x' = P(t)x
(3)
obtained from Eq. (2) by setting g(t) = 0. Once the homogeneous equation has been solved, there are several methods that can be used to solve the nonhomogeneous equation (2); this is taken up in Section 7.9. We use the notation
x(1)(t) =
(xn(t )\ x21(t)
\xn1(t
, x(k)(t) =
^x1k(t ë x2k(t)
(4)
to designate specific solutions of the system (3). Note that xij(t) = x(j)(t) refers to the i th component of the j th solution x(j j)(t). The main facts about the structure of solutions ofthe system (3) are stated in Theorems 7.4.1 to 7.4.4. They closely resemble the corresponding theorems in Sections 3.2, 3.3, and 4.1; some ofthe proofs are left to the reader as exercises.
Theorem 7.4.1 If the vector functions x(1) and x(2) are solutions of the system (3), then the linear combination c1x(1) + c2x(2) is also a solution for any constants c1 and c2.
This is the principle of superposition; it is proved simply by differentiating c1x(1) + c2x(2) and using the fact that x(1) and x(2) satisfy Eq. (3). By repeated application of
7.4 Basic Theory of Systems of First Order Linear Equations
369
Theorem 7.4.2
Theorem 7.4.1 we reach the conclusion that if x(1),..., x(k) are solutions of Eq. (3), then
x = C1x(1)(t) + --- + Ck x(k)(t)
(5)
is also a solution for any constants c1t..., ck. As an example, it can be verified that
„3t
x "(t) = '2e3t satisfy the equation
According to Theorem 7.4.1
„3t
x
x(2)(t) =
1 1 4 1
—2e
1
2
(6)
(7)
x=m!)e3t+c4-2)e-t
= c1x(1)(t) + c2x^ (t)
(8)
also satisfies Eq. (7).
As we indicated previously, by repeatedly applying Theorem 7.4.1, it follows that every finite linear combination of solutions of Eq. (3) is also a solution. The question now arises as to whether all solutions of Eq. (3) can be found in this way. By analogy with previous cases it is reasonable to expect that for a system of the form (3) of nth order it is sufficient to form linear combinations of n properly chosen solutions. Therefore let x(1),..., x(n) be n solutions of the nth order system (3), and consider the
matrix X(t) whose columns are the vectors x(1) (t),..., x(n) (t):
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http://mathhelpforum.com/calculus/158606-integration-question.html | # Math Help - Integration question
1. ## Integration question
Prove that $\int\limits_{0}^{1}\frac{\sqrt(1-x^2)}{1+x^2}\, dx = \frac{\pi}{2}(\sqrt2-1)$
Hints would be appreciated. Thanks.
2. Originally Posted by anthill
Prove that $\int\limits_{0}^{1}\frac{\sqrt(1-x^2)}{1+x^2}\, dx = \frac{\pi}{2}(\sqrt2-1)$
Hints would be appreciated. Thanks.
substitute 1-x^2=t^2 and express x^2 in denominator using that substitution. You should get a rational function in variable t.
3. Originally Posted by anthill
Prove that $\int\limits_{0}^{1}\frac{\sqrt(1-x^2)}{1+x^2}\, dx = \frac{\pi}{2}(\sqrt2-1)$
Hints would be appreciated. Thanks.
$\displaystyle \int \frac{\sqrt{1-x^2} \cdot \sqrt{1-x^2}}{(1+x^2)\cdot \sqrt{1-x^2}} dx$
$\displaystyle \int \frac{-1 -x^2 }{(1+x^2) \cdot \sqrt{1-x^2}} + \frac{2}{(1+x^2) \cdot \sqrt{1-x^2}} dx$
$\displaystyle \int \frac{-1 -x^2 }{(1+x^2) \cdot \sqrt{1-x^2}} dx +\int \frac{2}{(1+x^2) \cdot \sqrt{1-x^2}} dx$
$\displaystyle \int \frac{-1 }{\sqrt{1-x^2}} dx +\int \frac{2}{(1+x^2) \cdot \sqrt{1-x^2}} dx$
first integration sub
$\sin u = x$
in the second sub
$\displaystyle \tan u = \frac{\sqrt{2} x }{\sqrt{1-x^2}}$
4. Let $x = \sin(y)$
$dx = \cos(y)~dy$
The integral
$= \int_0^{\frac{\pi}{2}} \frac{\cos^2(y)}{ 1+ \sin^2(y) }~dy$
$= \int_0^{\frac{\pi}{2}} \frac{\cos^2(y)}{ \cos^2(y) + 2\sin^2(y) }~dy$
$= \int_0^{\frac{\pi}{2}} \frac{dy}{ 1+ 2\tan^2(y) }$
Consider $1 = \sec^2(y) - \tan^2(y) - 1/2 + 1/2$ thus we have
$\frac{1}{2} = \sec^2(y) - \frac{1}{2} ( 1+ 2\tan^2(y) )$
The integral
$= 2 \int_0^{\frac{\pi}{2}} \frac{\sec^2(y) }{ 1+ 2\tan^2(y) }~dy - \int_0^{\frac{\pi}{2}} ~dy$
$= \left[ \sqrt{2} \tan^{-1}[\sqrt{2}\tan(y)] - y \right]_0^{\frac{\pi}{2}}$
$= \frac{\pi}{2}( \sqrt{2} - 1 )$ | 2016-06-26T01:57:21 | {
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# If money is invested at r percent interest compounded annual
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If money is invested at r percent interest compounded annual [#permalink]
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Updated on: 06 Apr 2018, 03:15
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If money is invested at r percent interest, compounded annually, the amount of investment will double in approximately 70/r years. If Pat's parents invested $5000 in a long term bond that pays 8 percent interest, compounded annually, what will be the approximate total amount of investment 18 years later, when Pat is ready for college? A.$20000
B. $15000 C.$12000
D. $10000 E.$9000
OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/if-money-is- ... 44266.html
Originally posted by vanidhar on 04 Nov 2010, 03:53.
Last edited by Bunuel on 06 Apr 2018, 03:15, edited 2 times in total.
Edited the question.
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Re: If money is invested at r percent interest compounded annual [#permalink]
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04 Nov 2010, 05:21
10
5
vanidhar wrote:
If money is invested at r percent interest, compounded
annually, the amount of the investment will double
in approximately
70/r
years. If Pat’s parents invested
$5,000 in a long-term bond that pays 8 percent interest, compounded annually, what will be the approximate total amount of the investment 18 years later, when Pat is ready for college? (A)$20,000
(B) $1 5,000 (C)$1 2,000
(D) $1 0,000 (E)$ 9,000
There has to be a logic to why they gave you "If money is invested at r percent interest, compounded annually, the amount of the investment will double in approximately 70/r years."
If r = 8%, the principal will double in 70/8 = apprx 9 years. So in 9 years, 5000 will become 10,000. In another 9 years (i.e. 18 years from now) principal will double again and become $20,000. _________________ Karishma Veritas Prep GMAT Instructor Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options > ##### Most Helpful Community Reply Senior Manager Status: Not afraid of failures, disappointments, and falls. Joined: 20 Jan 2010 Posts: 264 Concentration: Technology, Entrepreneurship WE: Operations (Telecommunications) Re: If money is invested at r percent interest compounded annual [#permalink] ### Show Tags 04 Nov 2010, 08:41 7 Answer: A Karishma has already explained very well and I would like to add some fact here that would be valuable for our daily life problems. This fact of doubling investment (or growth) after every $$\frac{70}{r}$$ where $$r$$ is the $$%age$$ growth or change per unit time, holds true for real life economy calculations. This isn't just true for this particular question but is actually true for our daily life. Check out the following video link (amazing facts) http://www.youtube.com/watch?v=F-QA2rkpBSY Hope it helps _________________ "I choose to rise after every fall" Target=770 http://challengemba.blogspot.com Kudos?? ##### General Discussion Math Expert Joined: 02 Sep 2009 Posts: 52294 Re: If money is invested at r percent interest compounded annual [#permalink] ### Show Tags 13 Feb 2011, 12:05 1 7 If money is invested at r percent interest, compounded annually, the amount of investment will double in approximately 70/r years. If Pat's parents invested$ 5000 in a long term bond that pays 8 percent interest, compounded annually, what will be the approximate total amount of investment 18 years later, when Pat is ready for college?
A. $20000 B.$15000
C. $12000 D.$10000
E. $9000 Since investment doubles in 70/r years then for r=8 it'll double in 70/8=~9 years (we are not asked about the exact amount so such an approximation will do). Thus in 18 years investment will double twice and become ($5,000*2)*2=$20,000 (after 9 years investment will become$5,000*2=$10,000 and in another 9 years it'll become$10,000*2=$20,000). Answer: A. _________________ Intern Joined: 31 Aug 2012 Posts: 6 Re: If money is invested at r percent interest compounded annual [#permalink] ### Show Tags 08 Sep 2012, 20:07 How do you know to divide by 8 and not .08? Senior Manager Joined: 15 Jun 2010 Posts: 297 Schools: IE'14, ISB'14, Kellogg'15 WE 1: 7 Yrs in Automobile (Commercial Vehicle industry) Re: If money is invested at r percent interest compounded annual [#permalink] ### Show Tags 08 Sep 2012, 22:50 1 go2013gmat wrote: How do you know to divide by 8 and not .08? Pay attention to the question stem. The relationship is in %age. So no need to divide it by 100. If money is invested at r percent interest, compounded annually, the amount of the investment will double in approximately 70/r years. If Pat’s parents invested$5,000 in a long-term bond that pays 8 percent interest, compounded annually, what will be the approximate total amount of the investment 18 years later, when Pat is ready for college?
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Re: If money is invested at r percent interest compounded annual [#permalink]
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09 Sep 2012, 04:44
2
Just to brush up a little theory about Simple and Compound Interests calculation.
If P= Principle amount invested
r= annual rate of interest ( For 8% annual rate of interest r=8)
t= time period in years.
Then, $$Simple Interest (SI) = P*r*t$$
For calculation of Compound Interest calculation-
if A=accumulated amount (principle + all interest)
Then, $$A= P*( 1 +$$ $${r/100}$$$$)^t$$
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Re: If money is invested at r percent interest compounded annual [#permalink]
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18 Feb 2015, 01:07
If money is invested at r percent interest, compounded annually, the amount of investment will double in approximately 70/r years. If Pat's parents invested $5000 in a long term bond that pays 8 percent interest, compounded annually, what will be the approximate total amount of investment 18 years later, when Pat is ready for college? A.$20000
B. $15000 C.$12000
D. $10000 E.$9000
Amount will get doubled after (70/8) years or 8.75 years
Amount after 8.75 years = 2*5000 = 10000
Amount after 17.5 years = 2*10000 = 20000
Amount after 18 years will be approx to 20000.
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Re: If money is invested at r percent interest compounded annual [#permalink]
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02 Jul 2017, 04:54
Imo A
To Rate of interest =8 %
The amount invested doubles in 70/r=70/8 = 8.75 years approximately 9 years
5000(1+8/100)^18 now this difficult to calculate we take help from above
In 9 years 5000 will become 10000 and in another 9 years it will become 20,000
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Re: If money is invested at r percent interest compounded annual [#permalink]
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06 Apr 2018, 03:16
vanidhar wrote:
If money is invested at r percent interest, compounded annually, the amount of investment will double in approximately 70/r years. If Pat's parents invested $5000 in a long term bond that pays 8 percent interest, compounded annually, what will be the approximate total amount of investment 18 years later, when Pat is ready for college? A.$20000
B. $15000 C.$12000
D. $10000 E.$9000
OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/if-money-is- ... 44266.html
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Re: If money is invested at r percent interest compounded annual &nbs [#permalink] 06 Apr 2018, 03:16
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The limits are different on either side, so the limit does not exist. What procedures are in place to stop a U.S. Vice President from ignoring electors? At zero, the function is continuous but not differentiable. Using this you can use some numerical maximization on Abs[leftd[f, h, x] - rightd[f, h, x]], perhaps with successively smaller h to avoid false positives. Mathematica Stack Exchange is a question and answer site for users of Wolfram Mathematica. What happens when the function fails to have a derivative value at a given point? by looking on a graph of that function i can assume these will be the points at wich x=0,pi,2pi... but how would i see that in an equation? For this reason we added {0, x == 0} in the definition of the function g. Here is an approach that you can use for numerical functions that at least have a left and right derivative. Differentiation can only be applied to functions whose graphs look like straight lines in the vicinity of the point at which you want to differentiate. geometrically, the function #f# is differentiable at #a# if it has a non-vertical tangent at the corresponding point on the graph, that is, at #(a,f(a))#.That means that the limit #lim_{x\to a} (f(x)-f(a))/(x-a)# exists (i.e, is a finite number, which is the slope of this tangent line). the non-differentiable point of function is a point with certain characteristic, so it can be figured out by PSO. The methods for estimating derivatives so far have ignored an essential issue. What would happen if a 10-kg cube of iron, at a temperature close to 0 Kelvin, suddenly appeared in your living room? What months following each other have the same number of days? How to represent characteristic function of a single-point set? How to split equation into a table and under square root? 9.3 Non-Differentiable Functions. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. And so at what arguments is f not differentiable? The function sin (1/x), for example is singular at x = 0 even though it always lies between -1 and 1. In that case, we could only say that the function is differentiable on intervals or at points that don’t include the points of non-differentiability. Related Video. Get your answers by asking now. In real world problem, small stone are the fixed points for the mapping=wheat thresher. When n is an odd integer, the signs just switch. How do I numerically evaluate and plot the Fabius function? The converse does not hold: a continuous function need not be differentiable. In particular, any differentiable function must be continuous at every point in its domain. However. (Don't forget, n can be negative too.) Differences between Mage Hand, Unseen Servant and Find Familiar. {ToRules[%]}; Plot[ g[x], {x, -5/4, 3}, PlotStyle -> Thick, Epilog -> {Red, PointSize[0.023], Point[pts]}] One should be careful when working with Piecewise since Reduce may produce errors when weak inequalities (LessEqual) are involved. Allow bash script to be run as root, but not sudo. Podcast Episode 299: It’s hard to get hacked worse than this. The most common way to write this would be x = n*pi, where n is an integer. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. how would i say that it applies to every (0,pi,2pi...) point in the graph until infinity? 5 5 10 15. If f is differentiable at a point x0, then f must also be continuous at x0. Use MathJax to format equations. As in the case of the existence of limits of a function at x 0, it follows that. Piecewise functions may or may not be differentiable on their domains. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. The slope jumps there as well. If any one of the condition fails then f' (x) is not differentiable at x 0. Thanks for contributing an answer to Mathematica Stack Exchange! And at these points we really don't have a defined derivative. Since the derivative cannot be different values coming from the left and the right, there is no derivative. site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. But, depending on the teacher or what the lesson's on, that may not be enough. Should you post basic computer science homework to your github? A function f is not differentiable at a point x0 belonging to the domain of f if one of the following situations holds: (i) f has a vertical tangent at x0. Mathematica Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us, @Ordinaryusers68 This answer had provided correct solution in version $9.0.1$ and formerly. Does it return? Asking for help, clarification, or responding to other answers. The function has two inflection points has one point of extremum is non-differentiable has range 205.4k LIKES. This means that the derivative is cos(n*pi) coming from one side and -cos(n*pi) coming from the other. Find all points x where h is not differentiable. Why are these resistors between different nodes assumed to be parallel. L… If a function is made up of 2 different functions and they are JOINED together, they are said to be Continuous. This learned sampling approach improved application performance with sampled point clouds, in comparison to non-learned methods, such as FPS and random sampling. How does one throw a boomerang in space? I was wondering if a function can be differentiable at its endpoint. How to determine if a piecewise function is differentiable? exist and f' (x 0 -) = f' (x 0 +) Hence. 2. Is there a built-in function which detects singularities in a function? On the left of x = 0 (x < 0), the derivative is calculated as follows On the right of x = 0 (x > 0), the derivative is calculated as follows The limits to the left and to the right of x = 0 are not equal therefore f ' (0) is undefined and function f in not differentiable at x = 0. A cusp is slightly different from a corner. QGIS to ArcMap file delivery via geopackage, Trouble with the numerical evaluation of a series. How to make a discontinuous function a continuous function? So right as x crosses 3, the slope becomes 0. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 1. Good, you're looking at the correct points. Is there any theoretical problem powering the fan with an electric motor, Example of ODE not equivalent to Euler-Lagrange equation. Any point x = a at which f '(a) does not exist is called a point of non-differentiability. exists if and only if both. One of these is -1 and the other is +1. If a is such a point, then there will either be a hole or break in the graph of f ' at x = a. Simplify: lim h→0 |h| h. The limit does not exist! Still have questions? Generally the most common forms of non-differentiable behavior involve a function going to infinity at x, or having a jump or cusp at x. In particular, any differentiable function must be continuous at every point in its domain. Step 3: Look for a jump discontinuity. Cruz reportedly got $35M for donors in last relief bill, Cardi B threatens 'Peppa Pig' for giving 2-year-old silly idea, These 20 states are raising their minimum wage, 'Super gonorrhea' may increase in wake of COVID-19, ESPN analyst calls out 'young African American' players, Visionary fashion designer Pierre Cardin dies at 98, Judge blocks voter purge in 2 Georgia counties, More than 180K ceiling fans recalled after blades fly off, 'Many unanswered questions' about rare COVID symptoms, Bombing suspect's neighbor shares details of last chat, Lawyer: Soldier charged in triple murder may have PTSD. Let ( ), 0, 0 > − ≤ = x x x x f x First we will check to prove continuity at x = 0 When the UTXO in the cache is full, what strategy is used to replace one UTXO with another in the cache? Understanding dependent/independent variables in physics, Copy and paste value from a feature sharing the same id. How to find the non-differentiable point(s) of a given continuous function? pts = {x, g[x]} /. The slope is 0 here. For f(x) = |sin(x)|, this occurs when x = n*pi, where n is an integer. you say it's not differentiable at X=n.Pi where n=1,2,3. And it should work correctly until now. For example, the non-differentiable point of the function$f(x)=|x|$is at$x=0$. is not differentiable. what is the answer of this question with steps. The most common way to write this would be x = n*pi, where n is an integer. Are there any examples of solving for the global maximum of a non-differentiable function where you: Construct a series of differentiable functions that approach the non-differentiable function in the limit; Show the maximum of each differentiable function converges to some value, which is thus your answer. A function f is not differentiable at a point x0 belonging to the domain of f if one of the following situations holds: (i) f has a vertical tangent at x 0. Three things could cause such behavior. I was wondering how would i find the non differentiable points of a |sin(x)| function? How to find the non-differentiable points of a continuous function that is defined numerically? Can archers bypass partial cover by arcing their shot? From Right Side: lim h→0+ |h| h = +1. How can I convince my 14 year old son that Algebra is important to learn? If there’s just a single point where the function isn’t differentiable, then we can’t call the entire curve differentiable. Finally, taking into account the logarithm is differentiable for all the points of its domain and the absolute value is not differentiable at the points where it is equal to zero, we have: Would a lobby-like system of self-governing work? look at the left and right finite difference approximation of the derivative, and see where they disagree. More on Continuous Functions in Calculus The converse does not hold: a continuous function neednot be differentiable. (for example minima of |x|=0 and |x| is not differentiable at x=0) And can we say that the function reaches maximum or minimum if f(x) tends to infinity or zero? Numerical saddle point problem of a function of many variables, Derivative of Continuous and Differentiable Piecewise function is indeterminate. 71.0k VIEWS. My child's violin practice is making us tired, what can we do? Question from Dave, a student: Hi. That is, the graph of a differentiable function must have a (non-vertical) tangent line at each point in its domain, be relatively "smooth" (but not necessarily mathematically smooth), and cannot contain any breaks, corners, or cusps. However, there is a cusp point at (0, 0), and the function is therefore non-differentiable at that point. Mathematica is a registered trademark of Wolfram Research, Inc. The original function is undefined or discontinuous. Join Yahoo Answers and get 100 points today. It only takes a minute to sign up. For function y = f (x),x∈[a,b] The key to figure out the non-differentiable point by PSO is how to confirm the global extremum and local extremum in PSO. rev 2020.12.18.38240, The best answers are voted up and rise to the top. How to write a custom function to judge whether a bivariate function is differentiable at a certain point? For example if I have Y = X^2 and it is bounded on closed interval [1,4], then is the derivative of the function differentiable on the closed interval [1,4] or open interval (1,4). the following returns interesting points: Let's try another function defined with Piecewise, e.g. Well, it's not differentiable when x is equal to negative 2. Why do I , J and K in mechanics represent X , Y and Z in maths? Here is a more "rigorous" answer: When x = n*pi where n is an even integer, then f'(x) = cos(x) coming from the right and f'(x) = -cos(x) coming from the left. If such a function isn't differentiable in a point that is equivalent to the left and right derivatives being unequal, so To be differentiable at a point x = c, the function must be continuous, and we will then see if it is differentiable. (ii)The graph of f comes to a point at x0 (either a sharp edge ∨ or a sharp peak ∧) (iii) f is discontinuous at x0. Number of digits after decimal point. would i need to check the limit of the function from both sides of the point? How to find the non-differentiable point(s) of a given continuous function? To learn more, see our tips on writing great answers. Step 1: Check to see if the function has a distinct corner. The reason why we're not differentiable there is as we approach this point, as we approach this point from either side, we have different slopes. (Don't forget, n can be negative too.). How does one calculate effects of damage over time if one is taking a long rest? View All. When this limit exist, it is called derivative of #f# at #a# and denoted #f'(a)# or #(df)/dx (a)#. simplified point to its nearest neighbor in the input point cloud, which yielded a subset of the input. The tangent line is vertical. a) Estimate g(0), g(2), g(4), g(6), and g(8). Directional derivative of function at specific point. We want to do the same thing at many different arguments, which can be turned into a chart or graph of the derivative function. Step 2: Look for a cusp in the graph. then we needn't use Assumptions in Limit: One should be careful when working with Piecewise since Reduce may produce errors when weak inequalities (LessEqual) are involved. I would say you could simply write an answer like the following: The function is nondifferentiable at any point where there is a sharp turn. So we immediately see there are points where it looks like the slope jumps. Exercises: To see why, let's compare left and right side limits: From Left Side: lim h→0− |h| h = −1. MathJax reference. For this reason we added {0, x == 0} in the definition of the function g. There is a corner point in the original function’s graph. While the mark is used herein with the limited permission of Wolfram Research, Stack Exchange and this site disclaim all affiliation therewith. Different global extremum and local So the function f (x) = |x| is not differentiable. However, the matching step is a non-differentiable opera- Can we differentiate any function anywhere? Fermat's theorem is central to the calculus method of determining maxima and minima: in one dimension, one can find extrema by simply computing the stationary points (by computing the zeros of the derivative), the non-differentiable points, and the boundary points, and then investigating this set to determine the extrema. Making statements based on opinion; back them up with references or personal experience. The spreadsheet construction above gives the user the ability to find the derivative of a function at one specific argument. The reason why where you have these sharp bends or sharp turns as opposed to something that looks more smooth like that. Let g(x) = \int_{0}^{x}f(t)dt where f is the function whose graph is shown in the figure. 3. Fixed point of a mapping is an element if we apply the mapping on this point and we obtain the same point. What is the method of determining maxima and minima for those functions which are not differentiable at every point and how to know if the extremum is at a non-differentiable point ? Why are 1/2 (split) turkeys not available? Using three real numbers, explain why the equation y^2=x ,where x is a non - negative real number,is not a function.? Is there a functionality to analytically find discontinutites of function? if and only if f' (x 0 -) = f' (x 0 +) . There are however stranger things. Let’s consider some piecewise functions first. This function (shown below) is defined for every value along the interval with the given conditions (in fact, it is defined for all real numbers), and is therefore continuous. 71.0k SHARES. Now, for a function to be considered differentiable, its derivative must exist at each point in its domain, in this case Give an example of a function which is continuous but not differentiable at exactly three points. ? In calculus, a differentiable function is a continuous function whose derivative exists at all points on its domain. I think you might find the answer given in this link useful, it gives an example of a piecewise function and how to find the non-differentiable points. How to determine if a function is continuous? (ii) The graph of f comes to a point at x 0 (either a sharp edge ∨ or a sharp peak ∧) (iii) f is discontinuous at x 0. Mapping on this point and we obtain the same number of days continuous functions in Calculus pts = {,... In physics, copy and paste value from a feature sharing the same id element if apply... Represent characteristic function of many variables, derivative of a function at specific... Function neednot be differentiable on their domains are the fixed points for the mapping=wheat.! Function neednot be differentiable a feature sharing the same number of days the non differentiable points of a function!, J and K in mechanics represent x, Y and Z in maths making statements on. ' ( x 0 - ) = f ' ( x 0 )! Is making us tired, what can we do then f must also be continuous them! Can be negative too. ) an electric motor, example of ODE equivalent. The mapping=wheat thresher then f must also be continuous left side: lim h→0 |h| h. the limit not. Be run as root, but not sudo the point RSS reader ) of given. Is +1 bivariate function is differentiable at X=n.Pi where n=1,2,3 see if the function is how to find non differentiable points corner in. So at what arguments is f not differentiable may not be differentiable on domains... So right as x crosses 3, the signs just switch say that applies! Learned sampling approach improved application performance with sampled point clouds, in comparison to non-learned methods, such as and. Place to stop a U.S. Vice President from ignoring electors i need to Check the limit does not.! With an electric motor, example of ODE not equivalent to Euler-Lagrange equation an essential issue up and to. Is f not differentiable when x is equal to negative 2 Episode 299: it ’ graph. Personal experience feed, copy and paste this URL into your RSS reader really do n't forget, n be! Opposed to something that looks more smooth like that a long rest |...: in Calculus, a differentiable function must be continuous at x0 0,! Singular at x 0 - ) = f ' ( x 0 + ) Hence 10-kg cube iron. And paste this URL into your RSS reader and plot the Fabius function are the points! Example is singular at x = a at which f ' ( x 0 + ).! At x0 Euler-Lagrange equation limited permission of Wolfram Research, Inc ’ s graph right! That it applies to every ( 0, 0 ), and the function from both sides the., and the other is +1 a long rest and so at what arguments is f not at. Calculate effects of damage over time if one is taking a long rest of |sin! Copy and paste value from a feature sharing the same number of days personal experience ; back them up references. Up of 2 different functions and they are said to be continuous at every point its... Points on its domain the mapping on this point and we obtain the same id under square?. Is f not differentiable at its endpoint what the lesson 's on, that not! Compare left and right side: lim h→0 |h| h. the limit does not exist we do whose derivative at! Science homework to your github common way to write this would be x = n pi. How does one calculate effects of damage over time if one is taking a rest! Calculate effects of damage over time if one is taking a long?... A cusp point at ( 0 how to find non differentiable points pi,2pi... ) point in the case of existence. Apply the mapping on this point and we obtain the same id on, that may not be.... I convince my 14 year old son that Algebra is important to learn more, see tips... X crosses 3, the best answers are voted up and rise to the top ( )... Damage over time if one is taking a long rest function$ f x! H = −1 to write this would be x how to find non differentiable points n * pi, n... Science homework to your github: a continuous function neednot be differentiable cover by their! And 1 the slope jumps at its endpoint tired, what strategy is used herein the. Where h is not differentiable and right side: lim h→0+ |h| h +1. On the teacher or what the lesson 's on, that may not be differentiable on their domains a function! Example is singular at x 0, it 's not differentiable when x is equal to 2. Between Mage Hand, Unseen Servant and find Familiar is a question and site! I need to Check the limit does not exist the user the ability to the..., n can be differentiable at X=n.Pi where n=1,2,3 a table and square... To every ( 0, 0 ), for example is singular at x = n pi! When n is an integer ( x ) | function neighbor in the graph until?. F must also be continuous at every point in its domain approach improved application with... I numerically evaluate and plot the Fabius function all affiliation therewith in its domain why where you have these bends! These points we really do n't forget, n can be negative.... X where h is not differentiable when x is equal to negative 2 gives user! On its domain only if f is differentiable at a given point 0! You 're looking at the correct points see our tips on writing great answers graph until infinity the. The derivative can not be differentiable at its endpoint sampling approach improved application performance with sampled point clouds in... The non differentiable points of a single-point set are JOINED together, they are said to be continuous at point. Policy and cookie policy from both sides of the function $f ( x )$! Which detects singularities in a function can be differentiable at its endpoint example, the answers! Input point cloud, which yielded a subset of the point s hard to get hacked worse this... Rss feed, copy and paste value from a feature sharing the same id what happens the. Interesting points: let 's try another function defined with piecewise,.! Non-Differentiable point ( s ) of a series strategy is used herein with the limited permission of Wolfram Research Stack! Opposed to something that looks more smooth like that you have these sharp bends or turns! Small stone are the fixed points for the mapping=wheat thresher Trouble with the limited permission Wolfram! At that point pi,2pi... ) point in the graph are voted up and to... Non-Differentiable at that point, let 's compare left and the function f! Represent characteristic function of a continuous function need not be differentiable on their domains sampled point clouds, in to. Must be continuous other is +1 a U.S. Vice President from ignoring electors \$ is at x=0... Cookie policy Research, Inc damage over time if one is taking a rest... To non-learned methods, such as FPS and random sampling n't have a defined derivative mapping this. Differentiable at a certain point following each other have the same number of days hacked worse this... Where n=1,2,3 of service, privacy policy and cookie policy Mage Hand Unseen! There a functionality to analytically find discontinutites of function by clicking “ Post your answer ”, you to. Point problem of a mapping is an odd integer, the best answers are voted up and to! Limits: from left side: lim h→0− |h| h = −1 the. 0, it 's not differentiable when x is equal to negative 2 this question with.. G [ x ] } / follows that a given continuous function need not be how to find non differentiable points! How does one calculate effects of damage over time if one is taking a long rest a and... Can we do given continuous function RSS feed, copy and paste value a... In mechanics represent x, g [ x ] } / 2: for... Is an integer derivative exists at all points on its domain at that.! The correct points variables in physics, copy and paste value from feature. To your github limits are different on either side, so the limit of the function a... S ) of a series the input mapping is an odd integer, the best answers are voted and. Estimating derivatives so far have ignored an essential issue singular at x = n * pi, where n an...
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http://mathhelpforum.com/algebra/146891-logarithmic-question.html | # Thread: Logarithmic Question.
1. ## Logarithmic Question.
. Mr. Logarithmic is showing his class a solution to a problem. Is Mr. Logarithmic correct? Explain and correct if he is incorrect. Express 2log25 - log2(4/5) + 1/2log216 as a simple logarithm = 2log25 - log2(4/5) + 1/2log216 = log152-log2(4/5) + log2(16/2) = log225 - log2(4/5)+ log2(8) = log2 (25 x 4/5 x 8)
So, from my understanding, the solution is incorrect the correct solution according to my calculations is:'
2log25 - log2(4/5) + 1/2log216
log225 - log2(4/5) + log24
Therefore, we get:
log2 (25/(4/5) X 4)
I'm really confused here. Is my working/solution correct for the simplification of the logarithm? Any helpful tips/suggestions would be greatly appreciated.
Thanks!
2. Originally Posted by spoc21
. Mr. Logarithmic is showing his class a solution to a problem. Is Mr. Logarithmic correct? Explain and correct if he is incorrect. Express 2log25 - log2(4/5) + 1/2log216 as a simple logarithm
l
your notation leaves a bit to be desired ... is this the expression?
$2\log_2{5} - \log_2\left(\frac{4}{5}\right) + \frac{1}{2}\log_2{16}
$
3. Originally Posted by skeeter
your notation leaves a bit to be desired ... is this the expression?
$2\log_2{5} - \log_2\left(\frac{4}{5}\right) + \frac{1}{2}\log_2{16}
$
yes, thats exactly how it is.
4. ## Didn't see the minus sign ... corrected
Hi !
Let's see how it goes.
$2\log_2{(5)} - \log_2{\left(\frac{4}{5}\right)} + \frac{1}{2}\log_2{(16)}$
All the logarithms have the same base, so we won't have to change this. First, simplify any coefficients in front of the logarithms, using the power law :
$\log_2{(5^2)} - \log_2\left(\frac{4}{5}\right) + \log_2{\left (16^{\frac{1}{2}} \right )}$
Now simplify the expressions :
$\log_2{(25)} - \log_2\left(\frac{4}{5}\right) + \log_2{(4)}$
Now it gets tricky : you need brackets otherwise it's not going to work. So put the last two terms into brackets :
$\log_2{(25)} - \left ( \log_2\left(\frac{4}{5}\right) - \log_2{(4)} \right )$
Now use the logarithm division/substraction law on the two last terms :
$\log_2{(25)} - \log_2{\left (\frac{\frac{4}{5}}{4} \right )}$
$\log_2{(25)} - \log_2{\left (\frac{4}{20} \right )}$
Now divide using the logarithm division/substraction law :
$\log_2{\left ( \frac{25}{\frac{4}{20}} \right )}$
$\log_2{(125)}$
$3 \log_2{(5)}$
This is the simplest form of this expression.
____________________________________
Mr. Logarithmic is wrong, and your reasoning is correct although a bit hard to follow without proper mathematical fonts. You should have simplified further to get 125 which will have concluded the exercise.
5. Hello, spoc21!
Mr. Logarithmic is showing his class a solution to a problem.
Is Mr. Logarithmic correct? Explain and correct if he is incorrect.
. . He made two errors.
Express $2\log_2(5) - \log_2(\tfrac{4}{5}) + \tfrac{1}{2}\log_2(16)$ as a simple logarithm.
$2\log_2(5) - \log_2\tfrac{4}{5}) + \tfrac{1}{2}\log_2(16)$
. . $=\;\log_2\left(5^2\right) -\log_2(\tfrac{4}{5}) + \log_2\left({\color{red}\tfrac{16}{2}}\right)$
. . $=\; \log_2(25) - \log_2\left(\tfrac{4}{5}\right)+ \log_2(8)$
. . $=\; \log_2\left(25 \,{\color{red}\times}\, \tfrac{4}{5} \times 8\right)$
$2\log_2(5) - \log_2(\tfrac{4}{5}) + \tfrac{1}{2}\log_2(16)$
. . $=\;\log_2(5^2) - \bigg[\log_2(4)-\log_2(5)\bigg] + \log_2(16^{\frac{1}{2}})$
. . $=\;\log_2(25) - \log_2(4) + \log_2(5) + \log_2(4)$
. . $=\;\log_2(25) + \log_2(5)$
. . $=\;\log_2(25 \times 5)$
. . $=\;\log_2(125)$ | 2017-08-23T15:03:13 | {
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# Compilation of tips and tricks to deal with remainders.
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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
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08 Jul 2017, 17:53
great article. A suggestion - if you could add some examples of questions for each principle, it would be great for retention of the material. Thanks.
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Compilation of tips and tricks to deal with remainders. [#permalink]
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09 Jul 2017, 02:54
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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
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21 Oct 2017, 02:06
ctrlaltdel wrote:
If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(A) 2
(B) 4
(C) 8
(D) 20
(E) 45
Ans: Using the technique: remainder = 0.12*t => the answer is multiple of 12. but none of the options match...did i miss something or is my understanding wrong ops :stupid
I tried with below approach in all type of above Q and i solved it each time within secs.
Let's see .
given : s/t = 64.12
consider only 0.12
so 0.12=12/100 = 3/25. Now what we have to do is to find the multiple of 3 in given Ans. There will be one correct ans which will satisfy the Q. Hence , here only 45 is the multiple of 3 ...so that's the ans.
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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
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26 Jan 2018, 08:14
sriharimurthy wrote:
Hi guys,
This is in conjunction with another post which has questions dealing with remainders (http://gmatclub.com/forum/collection-of ... 74776.html). I'm just trying to put together a list of tips and tricks which we can use to solve these kind of problems with greater accuracy and speed. Please feel free to comment and make suggestions. Hopefully we can add onto this list and cover all sorts of strategies that would help us deal with remainders!
Cheers.
1) Take your time with these points. Some of them might be a little difficult to follow in the first reading, but don't give up. The concepts are fairly simple.
2) These tips if mastered will be extremely valuable in the GMAT to help solve a variety of questions not limited specifically to remainders. I have been using them for quite a while now and they have not only helped me improve my accuracy but also my speed.
3) If you have any doubts, please do not hesitate to ask (no matter how stupid you might think them to be!). If you do not ask, you will never learn.
4) Lastly, have fun while trying to understand these tips and tricks as that, according to me, is the best possible way to learn.
All the best!
-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-
NOTE: Where ever you see R of 'x' it just stands for Remainder of x.
1) The possible remainders when a number is divided by ‘n’ can range from 0 to (n-1).
Eg. If n=10, possible remainders are 0,1,2,3,4,5,6,7,8 and 9.
2) If a number is divided by 10, its remainder is the last digit of that number. If it is divided by 100 then the remainder is the last two digits and so on.
This is good for questions such as : ' What is the last digit of.....' or ' What are the last two digits of.....' .
3) If a number leaves a remainder ‘r’ (the number is the divisor), all its factors will have the same remainder ‘r’ provided the value of ‘r’ is less than the value of the factor.
Eg. If remainder of a number when divided by 21 is 5, then the remainder of that same number when divided by 7 (which is a factor of 21) will also be 5.
If the value of ‘r’ is greater than the value of the factor, then we have to take the remainder of ‘r’ divided by the factor to get the remainder.
Eg. If remainder of a number when divided by 21 is 5, then the remainder of that same number when divided by 3 (which is a factor of 21) will be remainder of 5/3, which is 2.
4) Cycle of powers : This is used to find the remainder of $$n^x$$, when divided by 10, as it helps us in figuring out the last digit of $$n^x$$.
The cycle of powers for numbers from 2 to 10 is given below:
2: 2, 4, 8, 6 → all $$2^{4x}$$ will have the same last digit.
3: 3, 9, 7, 1 → all $$3^{4x}$$ will have the same last digit.
4: 4, 6 → all $$4^{2x}$$ will have the same last digit.
5: 5 → all $$5^x$$ will have the same last digit.
6: 6 → all $$6^x$$ will have the same last digit.
7: 7, 9, 3, 1 → all $$7^{4x}$$ will have the same last digit.
8: 8, 4, 2, 6 → all $$8^{4x}$$ will have the same last digit.
9: 9, 1 → all $$9^{2x}$$ will have the same last digit.
10: 0 → all $$10^x$$ will have the same last digit.
5) Many seemingly difficult remainder problems can be simplified using the following formula :
$$R of \frac{x*y}{n} = R of \frac{(R of \frac{x}{n})*(R of \frac{y}{n})}{n}$$
Eg. $$R of \frac{20*27}{25} = R of \frac{(R of \frac{20}{25})*(R of \frac{27}{25})}{25} = R of \frac{(20)*(2)}{25} = R of \frac{40}{25} = 15$$
Eg. $$R of \frac{225}{13} = R of \frac{(15)*(15)}{13} = R of {(2)*(2)}{13} = R of \frac{4}{13} = 4$$
6) $$R of \frac{x*y}{n}$$ , can sometimes be easier calculated if we take it as $$R of \frac{(R of \frac{(x-n)}{n})*(R of \frac{(y-n)}{n})}{n}$$
Especially when x and y are both just slightly less than n. This can be easier understood with an example:
Eg. $$R of \frac{(19)*(21)}{25} = R of \frac{(-6)*(-4)}{25} = 24$$
NOTE: Incase the answer comes negative, (if x is less than n but y is greater than n) then we have to simply add the remainder to n.
Eg. $$R of \frac{(23)*(27)}{25} = R of \frac{(-2)*(2)}{25} = -4.$$ Now, since it is negative, we have to add it to 25.$$R = 25 + (-4) = 21$$
[Note: Go here to practice two good problems where you can use some of these concepts explained : http://gmatclub.com/forum/numbers-86325.html]
7) If you take the decimal portion of the resulting number when you divide by "n", and multiply it to "n", you will get the remainder. [Special thanks to h2polo for this one]
Note: Converse is also true. If you take the remainder of a number when divided by 'n', and divide it by 'n', it will give us the remainder in decimal format.
Eg. $$\frac{8}{5} = 1.6$$
In this case, $$0.6 * 5 = 3$$
Therefore, the remainder is $$3$$.
This is important to understand for problems like the one below:
If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(A) 2
(B) 4
(C) 8
(D) 20
(E) 45
OA :
[Reveal] Spoiler:
E
sriharimurthy In the points 5 and 6 what means the simbol ROF?
Re: Compilation of tips and tricks to deal with remainders. [#permalink] 26 Jan 2018, 08:14
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# In the rectangular coordinate system above, if the equation of m is y
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In the rectangular coordinate system above, if the equation of m is y = x, and m is parallel to n, what is the shortest distance between m and n?
(A) √2
(B) 1
(C) √2/2
(D) 1/2
(E) 1/4
[Reveal] Spoiler:
Attachment:
2017-12-15_1259.png [ 4.85 KiB | Viewed 382 times ]
[Reveal] Spoiler: OA
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In the rectangular coordinate system above, if the equation of m is y [#permalink]
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15 Dec 2017, 12:22
Bunuel wrote:
In the rectangular coordinate system above, if the equation of m is y = x, and m is parallel to n, what is the shortest distance between m and n?
(A) √2
(B) 1
(C) √2/2
(D) 1/2
(E) 1/4
[Reveal] Spoiler:
Attachment:
The attachment 2017-12-15_1259.png is no longer available
With the formula (My diagram does not apply)
I did not use it, but to do so is standard. So I quote the formula from Bunuel HERE
Distance between two parallel lines $$y=mx+b$$ and $$y=mx+c$$ can be found by the formula:
$$D=\frac{|b-c|}{\sqrt{m^2+1}}$$
Equation for line $$m$$, $$y = x$$, to match the exact form of $$y = mx + b$$ can be written:
$$y =(1)x + 0$$ (all lines that pass through the origin have x- and y-intercepts of 0)
$$b = 0$$ = y-intercept, $$m = 1$$ = slope
Find the equation of line $$n$$, $$y = mx + c$$
Parallel lines have identical slopes. Line $$n$$ has slope $$m = 1$$
Use the point on line $$n$$ from the graph: (1, 0)
Plug its coordinates into slope-intercept form to find $$c$$
$$y = mx + c$$
$$(0) = 1(1) + c$$
$$-c = 1$$, so $$c = -1$$
Equation for line $$n: y = x - 1$$
Distance between parallel lines
$$D=\frac{|b-c|}{\sqrt{m^2+1}}$$
$$m = 1$$, $$b = 0$$, $$c = -1$$
$$D=\frac{|0-(-1)|}{\sqrt{1^2+1}}$$
$$D=\frac{1}{\sqrt{2}}$$
Does not match the answers. Rationalize the denominator; multiply by $$\frac{\sqrt2}{\sqrt2}$$
$$\frac{1}{\sqrt{2}}$$ * $$\frac{\sqrt2}{\sqrt2} = \frac{\sqrt{2}}{2}$$
[Reveal] Spoiler:
C
Attachment:
linesmandn.png [ 18.05 KiB | Viewed 211 times ]
Without the formula see diagram
The shortest distance between parallel lines is a line perpendicular to both parallel lines
Draw a perpendicular line from point B to point A
That creates right isosceles ∆ ABO where OA = AB
-- The line y = x makes a 45° angle with both the x- and y-axes (as do all lines with slope 1 or -1)
-- ∠ BOA = 45° and ∠ at vertex A = 90° therefore ∠ ABO must = 45°
-- Sides opposite equal angles are equal: OA = AB
A right isosceles triangle
-- has angle measures 45-45-90 and
-- corresponding side lengths of $$x : x : x\sqrt{2}$$
The hypotenuse corresponds with $$x\sqrt{2} = 1$$: OB = 1
Find length of equal sides $$x$$ (one of which, AB, is the distance needed):
$$x\sqrt{2}= 1$$
$$x = \frac{1}{\sqrt{2}}$$
= AB = shortest distance between the parallel lines
That does not look like any of the answers.
Rationalize the denominator (clear the radical): Multiply by $$\frac{\sqrt2}{\sqrt2}$$
$$\frac{1}{\sqrt{2}}$$ * $$\frac{\sqrt2}{\sqrt2} = \frac{\sqrt{2}}{2}$$
[Reveal] Spoiler:
C
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In the rectangular coordinate system above, if the equation of m is y [#permalink] 15 Dec 2017, 12:22
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https://it.mathworks.com/help/symbolic/erfc.html | # erfc
Complementary error function
## Description
example
erfc(X) represents the complementary error function of X, that is,erfc(X) = 1 - erf(X).
example
erfc(K,X) represents the iterated integral of the complementary error function of X, that is, erfc(K, X) = int(erfc(K - 1, y), y, X, inf).
## Examples
### Complementary Error Function for Floating-Point and Symbolic Numbers
Depending on its arguments, erfc can return floating-point or exact symbolic results.
Compute the complementary error function for these numbers. Because these numbers are not symbolic objects, you get the floating-point results:
A = [erfc(1/2), erfc(1.41), erfc(sqrt(2))]
A =
0.4795 0.0461 0.0455
Compute the complementary error function for the same numbers converted to symbolic objects. For most symbolic (exact) numbers, erfc returns unresolved symbolic calls:
symA = [erfc(sym(1/2)), erfc(sym(1.41)), erfc(sqrt(sym(2)))]
symA =
[ erfc(1/2), erfc(141/100), erfc(2^(1/2))]
Use vpa to approximate symbolic results with the required number of digits:
d = digits(10);
vpa(symA)
digits(d)
ans =
[ 0.4795001222, 0.04614756064, 0.0455002639]
### Error Function for Variables and Expressions
For most symbolic variables and expressions, erfc returns unresolved symbolic calls.
Compute the complementary error function for x and sin(x) + x*exp(x):
syms x
f = sin(x) + x*exp(x);
erfc(x)
erfc(f)
ans =
erfc(x)
ans =
erfc(sin(x) + x*exp(x))
### Complementary Error Function for Vectors and Matrices
If the input argument is a vector or a matrix, erfc returns the complementary error function for each element of that vector or matrix.
Compute the complementary error function for elements of matrix M and vector V:
M = sym([0 inf; 1/3 -inf]);
V = sym([1; -i*inf]);
erfc(M)
erfc(V)
ans =
[ 1, 0]
[ erfc(1/3), 2]
ans =
erfc(1)
1 + Inf*1i
Compute the iterated integral of the complementary error function for the elements of V and M, and the integer -1:
erfc(-1, M)
erfc(-1, V)
ans =
[ 2/pi^(1/2), 0]
[ (2*exp(-1/9))/pi^(1/2), 0]
ans =
(2*exp(-1))/pi^(1/2)
Inf
### Special Values of Complementary Error Function
erfc returns special values for particular parameters.
Compute the complementary error function for x = 0, x = ∞, and x = –∞. The complementary error function has special values for these parameters:
[erfc(0), erfc(Inf), erfc(-Inf)]
ans =
1 0 2
Compute the complementary error function for complex infinities. Use sym to convert complex infinities to symbolic objects:
[erfc(sym(i*Inf)), erfc(sym(-i*Inf))]
ans =
[ 1 - Inf*1i, 1 + Inf*1i]
### Handling Expressions That Contain Complementary Error Function
Many functions, such as diff and int, can handle expressions containing erfc.
Compute the first and second derivatives of the complementary error function:
syms x
diff(erfc(x), x)
diff(erfc(x), x, 2)
ans =
-(2*exp(-x^2))/pi^(1/2)
ans =
(4*x*exp(-x^2))/pi^(1/2)
Compute the integrals of these expressions:
syms x
int(erfc(-1, x), x)
ans =
erf(x)
int(erfc(x), x)
ans =
x*erfc(x) - exp(-x^2)/pi^(1/2)
int(erfc(2, x), x)
ans =
(x^3*erfc(x))/6 - exp(-x^2)/(6*pi^(1/2)) +...
(x*erfc(x))/4 - (x^2*exp(-x^2))/(6*pi^(1/2))
### Plot Complementary Error Function
Plot the complementary error function on the interval from -5 to 5.
syms x
fplot(erfc(x),[-5 5])
grid on
## Input Arguments
collapse all
Input, specified as a symbolic number, variable, expression, or function, or as a vector or matrix of symbolic numbers, variables, expressions, or functions.
Input representing an integer larger than -2, specified as a number, symbolic number, variable, expression, or function. This arguments can also be a vector or matrix of numbers, symbolic numbers, variables, expressions, or functions.
collapse all
### Complementary Error Function
The following integral defines the complementary error function:
$erfc\left(x\right)=\frac{2}{\sqrt{\pi }}\underset{x}{\overset{\infty }{\int }}{e}^{-{t}^{2}}dt=1-erf\left(x\right)$
Here erf(x) is the error function.
### Iterated Integral of Complementary Error Function
The following integral is the iterated integral of the complementary error function:
$erfc\left(k,x\right)=\underset{x}{\overset{\infty }{\int }}erfc\left(k-1,y\right)dy$
Here, $erfc\left(0,x\right)=erfc\left(x\right)$.
## Tips
• Calling erfc for a number that is not a symbolic object invokes the MATLAB® erfc function. This function accepts real arguments only. If you want to compute the complementary error function for a complex number, use sym to convert that number to a symbolic object, and then call erfc for that symbolic object.
• For most symbolic (exact) numbers, erfc returns unresolved symbolic calls. You can approximate such results with floating-point numbers using vpa.
• At least one input argument must be a scalar or both arguments must be vectors or matrices of the same size. If one input argument is a scalar and the other one is a vector or a matrix, then erfc expands the scalar into a vector or matrix of the same size as the other argument with all elements equal to that scalar.
## Algorithms
The toolbox can simplify expressions that contain error functions and their inverses. For real values x, the toolbox applies these simplification rules:
• erfinv(erf(x)) = erfinv(1 - erfc(x)) = erfcinv(1 - erf(x)) = erfcinv(erfc(x)) = x
• erfinv(-erf(x)) = erfinv(erfc(x) - 1) = erfcinv(1 + erf(x)) = erfcinv(2 - erfc(x)) = -x
For any value x, the system applies these simplification rules:
• erfcinv(x) = erfinv(1 - x)
• erfinv(-x) = -erfinv(x)
• erfcinv(2 - x) = -erfcinv(x)
• erf(erfinv(x)) = erfc(erfcinv(x)) = x
• erf(erfcinv(x)) = erfc(erfinv(x)) = 1 - x
## References
[1] Gautschi, W. “Error Function and Fresnel Integrals.” Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. (M. Abramowitz and I. A. Stegun, eds.). New York: Dover, 1972.
## Version History
Introduced in R2011b | 2023-03-31T23:20:10 | {
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https://math.stackexchange.com/questions/657931/why-negating-universal-quantifier-gives-existential-quantifier | # Why negating universal quantifier gives existential quantifier?
Negating a universal quantifier gives the existential quantifier, and vice versa:
$$\neg \forall x = \exists x \neg \\ \neg \exists x = \forall x \neg$$
Why is this, and is there a proof for it (is it even possible to prove it, or is it just an axiom)? Intuitively, I would think that negating "for all" would give "none," or even "not for all," and that negating "there exists" would give "there does not exist".
• Don't forget the $\lnot$ that moved past the quantifiers, $\lnot \forall x \leadsto \exists x \lnot$, and $\lnot\exists x \leadsto \forall x \lnot$. Jan 31, 2014 at 1:21
• If I say "all birds can fly" and you believe I'm wrong, what are you saying (apart from any specific example)? Jan 31, 2014 at 1:22
• Possible duplicate of Are the quantifier negation rules in first-order logic derivable? Nov 19, 2018 at 17:21
The following two statements are equivalent:
"It is not true that all men have red hair."
"There exists at least one man who does not have red hair."
Hence $$\neg\forall x\ \varphi$$ is the same as $$\exists x\ \neg\varphi$$.
The following are equivalent:
"It is not true that some men have green hair."
"All men have non-green hair."
Hence $$\neg \exists x\ \varphi$$ is the same as $$\forall x\ \neg\varphi$$.
However, the form in which you've written them is not correct (as pointed out in Daniel Fischer's comment).
• Your second example is badly flawed. Many men have no hair at all. Jun 17, 2015 at 22:26
• @TonyK hahaha very true.
– Ryan
Jan 18, 2017 at 4:59
• that damned Law of Excluded Middle
– Ryan
Sep 6, 2017 at 4:58
• Could someone explain how baldness is connected to the Law of Excluded Middle? May 11, 2021 at 11:44
I will answer in a way that will likely add confusion, but I hope shows why this type of question should not be seen as too obvious and easy. I apologise in advance.
One question that comes up is "what does it mean to assert $\forall x (P(x))$ when there are infinite x?" Why do students ask such a silly question? Because one way to answer it is to say "it is a fact that all x obey P" or "it is known that all x obey P" or "it has been shown P(x) for all x" or "for every x we can prove P(x)". Something like that is what a lot of students have in mind, and when one is dealing with infinities, things like knowing or proving may never get finished.
Similarly, we can ask what it means to show $\exists x (P(x))$. One way that students will often think about this is that showing that is giving a particular x that obeys P.
Now, if these are the things the student is thinking, then it certainly seems reasonable to ask the question of the OP. Even if we can assert that "it is known that not all x obey P" or worse, that "for every x we cannot prove P(x)", this does not mean we have given a particular x that obeys P! So how can we assert existence here? What if we just don't know?
Of course, the answer is that there are different, completely valid ways to answer this. On one side, you have the constructivists who will smile at these questions, pat you on the shoulder, and tell you "good questions, all" as they lead you off on explanations of the meaning of truth and accessible knowledge. On another side, you have the classicists, who will try to help this poor student understand that there is much more to truth than knowledge and proof, and try to build in that student reasons why bivalent reasoning about these kinds of things is still important and how can reach assertions of existence nonconstructively.
These are both valid paths one may walk down in the philosophy of mathematics. They offer different ways to understand the meaning of assertions, and it often seems like both sides are talking past each other when they are often just talking about different things entirely. A curious student might want to get familiar with both sides, and learn the ways each approach formal manipulation.
A noncurious student, though, should probably just learn the rule and move on. Beyond here, there be dragons.
• @exodu5 I enjoyed that. Great post. However, I must say it doesn't explain the question much lol. Michael's answer was exactly what I was looking for. Really, the reason for my comment was to say that I liked your post and it was thought provoking for me.
– Ryan
Jan 18, 2017 at 5:08
Answer to the question in the middle of the OP: in terms of formal logic this is an axiom, or to put it better it is the definition of one quantifier in terms of the other.
For example, one may take $\forall$ as an undefined symbol and then define $\exists x\,P(x)$ to mean $\neg\forall x\,(\neg P(x))$.
So from the strictly formal point of view there is no question to answer here, it's just the definition. On the other hand, we do want logic to reflect accepted and understood modes of reasoning, so examples such as those given in other answers and comments are important.
This was already observed by Aristoteles in his square of opposition, looking at the contradictory path:
But my question is, is this a property of classical logic, or have non-classical logics also such quantifiers? Unfortunately, for example in inituitionistc logic, we only have the following directions which are generally valid:
∃x¬φ → ¬∀xφ
∀x¬φ → ¬∃xφ
¬∃xφ → ∀x¬φ
But this direction is not generally valid, picked from the wiki page about interdefinability:
¬∀xφ → ∃x¬φ
The last formula might fail since the forall might be in a possible world with a greater domain than the exists. | 2022-05-28T05:28:54 | {
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https://www.physicsforums.com/threads/integrate-sinx-cox-using-double-angle.704389/ | Homework Help: Integrate sinx*cox* using double angle
1. Aug 5, 2013
Jude075
1. The problem statement, all variables and given/known data
∫ Sinx cos x dx
2. Relevant equations
3. The attempt at a solution
If you integrate it using substitution, you get -cos2(x)/2but if you use double angle formula to rewrite the problem, it will be ∫1/2 sin(2x), and integrate it, you get -cos(2x)/4. isn't it weird?
2. Aug 5, 2013
voko
You are forgetting that each of those two integrals must have $+ C$ appended to it. Now recall that $\cos 2x = 2 \cos^2 x - 1$.
3. Aug 5, 2013
D H
Staff Emeritus
What voko said is exactly right. To elaborate, these two apparently different integrals differ only by a constant.
4. Aug 5, 2013
Jude075
So if I was to find the indefinite integral, both answers were right?
5. Aug 5, 2013
voko
None is correct unless you add the indefinite constant. | 2018-07-23T08:11:08 | {
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https://math.stackexchange.com/questions/2665553/calculating-the-summation-sum-n-1-infty-frach-n1nn1 | # Calculating the summation$\sum_{n=1}^{\infty}\frac{H_{n+1}}{n(n+1)}$
I need to find explicitly the following summation
$$\sum_{n=1}^{\infty}\frac{H_{n+1}}{n(n+1)}, \quad H_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}$$
From Mathematica, I checked that the answer is $$2$$. The same result is returned by WolframAlpha.
A thought from afar: \begin{align} \frac{H_{n+1}}{n(n+1)} &=H_{n+1}\left(\frac1n-\frac1{n+1}\right)\tag1\\ &=\color{#C00}{H_{n+1}}\color{#090}{(H_n-H_{n-1})}-H_{n+1}(H_{n+1}-H_n)\tag2\\ &=\color{#C00}{\frac1{n+1}}\color{#090}{\frac1n}+\color{#C00}{H_n}\color{#090}{(H_n-H_{n-1})}-H_{n+1}(H_{n+1}-H_n)\tag3\\ &=\frac1n-\frac1{n+1}+\frac{H_n}{n}-\frac{H_{n+1}}{n+1}\tag4\\ &=\frac{H_n+1}{n}-\frac{H_{n+1}+1}{n+1}\tag5 \end{align} What was done:
$$(1)$$: partial fractions
$$(2)$$: used $$\frac1n=H_n-H_{n-1}$$ and $$\frac1{n+1}=H_{n+1}-H_n$$
$$(3)$$: used $$\color{#C00}{H_{n+1}=\frac1{n+1}+H_n}$$ and $$\color{#090}{\frac1n=H_n-H_{n-1}}$$
$$(4)$$: partial fractions and $$\frac1n=H_n-H_{n-1}$$ and $$\frac1{n+1}=H_{n+1}-H_n$$
$$(5)$$: combined terms
Now this looks like Clement C's hint.
• $\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$ and $H_{n+2}-H_{n+1}=\frac{1}{n+2}$, hence summation by parts converts such series into an elementary (telescopic) one. – Jack D'Aurizio Feb 25 '18 at 16:47
• For gosh sakes, the asker has been suspended for their low quality posts this week, which only happens when an asker habitually asks poor questions, and yet we have three users, in addition to a mod (because that mod happened to answer the question, and doesn't want their answer deleted), who, nonetheless seek to undelete and reopen this question. – Namaste Mar 11 '18 at 1:02
• – user296602 Mar 11 '18 at 1:13
• @user296602 the posts content is crap. Thanks for making my point. – Namaste Mar 11 '18 at 1:16
• To whoever it may concern: please, delete the question once and for all, or stop downvoting its answers. It makes no sense to leave it open but retaliate on the answers. – Clement C. Mar 11 '18 at 1:16
(Big) hint:
$$\frac{H_{n+1}}{n(n+1)} = \frac{H_{n+1}}{n}-\frac{H_{n+1}}{n+1} = \frac{1}{n(n+1)} + \frac{H_{n}}{n}-\frac{H_{n+1}}{n+1}$$ so that you can get a telescopic series: for any $N\geq 1$, $$\sum_{n=1}^N \frac{H_{n+1}}{n(n+1)} = \sum_{n=1}^N \frac{1}{n(n+1)} + \sum_{n=1}^N \frac{H_{n}}{n} - \sum_{n=2}^{N +1}\frac{H_{n}}{n}$$
Spoiler:
The value you should arrive at is $2$, which is $\sum_{n=1}^\infty \frac{1}{n(n+1)} + 1$.
• Why the downvote? – Clement C. Mar 5 '18 at 1:07
• It is to be expected since the question was just undeleted and there were some people who were quite against the undeletion. Don't worry, the downvotes don't reflect any on the answers, just that the question that they thought should be deleted was answered. I got one too. – robjohn Mar 7 '18 at 17:00
• Not true, @robjohn. The downvotes received reflect heavily on both the answers and the answerers. You may gain peace of mind by recreating a narrative that soothes your soul, but that doesn't make it true. Three downvotes here have a much better explanation than you're even able to contemplate, and you'd have more too, if you weren't flouting a diamond alongside your username, dear diamond robjohn. Just stop trying to appease others and yourself that answers to poor questions asked by users suspended for posting low quality questions are commendable. – Namaste Mar 11 '18 at 1:07
• @amWhy In this case (I don't care about my peace of mind, and am not attached to any narrative) what is wrong with my answer? I am fine with the question being deleted, I haven't cast any vote, yet my answer gets downvoted. In short, to whoever it may concern: delete the goddarn question once and for all, or stop downvoting its answers! – Clement C. Mar 11 '18 at 1:11
Hint: \begin{align} \sum_{n=1}^\infty\frac{H_{n+1}}{n(n+1)} &=\sum_{n=1}^\infty\sum_{k=1}^{n+1}\frac1k\left(\frac1n-\frac1{n+1}\right)\\ &=\sum_{n=1}^\infty\frac1{n+1}\left(\frac1n-\frac1{n+1}\right) +\sum_{n=1}^\infty\sum_{k=1}^n\frac1k\left(\frac1n-\frac1{n+1}\right)\\ &=\sum_{n=1}^\infty\left(\frac1n-\frac1{n+1}-\frac1{(n+1)^2}\right) +\sum_{k=1}^\infty\sum_{n=k}^\infty\frac1k\left(\frac1n-\frac1{n+1}\right) \end{align} Now its simply summing telescoping series (three times).
using the fact that $$\displaystyle\int_0^1 x^{n}\ln(1-x)\ dx=-\frac{H_{n+1}}{n+1}$$
divide both sides by $$-n$$ then take the sum, \begin{align} S&=\sum_{n=1}^\infty\frac{H_{n+1}}{n(n+1)}=-\int_0^1\ln(1-x)\left(\sum_{n=1}^\infty\frac{x^n}{n}\right)\ dx\\ &=\int_0^1\ln^2(1-x)\ dx=\int_0^1\ln^2x\ dx=2 \end{align}
Noting that $$\frac{1}{n(n + 1)} = \frac{1}{n} - \frac{1}{n + 1},$$ the sum may be rewritten as $$\sum_{n = 1}^\infty \frac{H_{n+1}}{n(n + 1)} = \sum_{n = 1}^\infty \left (\frac{H_{n + 1}}{n} - \frac{H_{n + 1}}{n + 1} \right ).\tag1 \label1$$
Now as the harmonic numbers satisfy the recurrence relation $$H_{n + 1} = H_n + \frac{1}{n + 1},$$ we have \begin{align} \sum_{n = 1}^\infty \frac{H_{n+1}}{n(n + 1)} &= \sum_{n = 1}^\infty \left [\frac{H_n}{n(n+1)} + \frac{1}{n(n +1)} - \frac{1}{(n + 1)^2} \right ]\\ &= \sum_{n = 1}^\infty \frac{H_n}{n(n + 1)} + \sum_{n = 1}^\infty \frac{1}{n(n + 1)} - \underbrace{\sum_{n = 1}^\infty \frac{1}{(n + 1)^2}}_{n \, \mapsto n - 1}\\ &= \sum_{n = 1}^\infty \frac{H_n}{n(n + 1)} + \sum_{n = 1}^\infty \frac{1}{n(n+1)} - \sum_{n = 1}^\infty \frac{1}{n^2} + 1\tag2\\ &= \zeta (2) + 1 - \zeta (2) + 1\\ &= 2 \end{align} In (2) an evaluation for the first sum can be found here while the second sum telescopes and has a sum equal to one.
• just a little note. $\sum_{n=2}^\infty\frac{H_n}{n}$ is divergent so we can not cancel these two sums. so we better simplify before distributing. – Ali Shather Jul 5 at 7:16
• @Ali Shather Indeed you are absolutely right. Thanks and have corrected it now. – omegadot Jul 5 at 7:48
• looks good now :) – Ali Shather Jul 5 at 7:50 | 2019-08-24T14:23:07 | {
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https://math.stackexchange.com/questions/1259322/when-to-use-the-equiv-symbol-such-as-in-56-equiv-1-mod-7-vs/1259340 | # When to use the $\equiv$ symbol (such as in $5^{6}$ $\equiv$ 1 mod 7) vs =
Can anybody explain why we would use the $\equiv$ symbol in the statement $5^{6}$ $\equiv$ 1 mod 7 ? I understand the $\equiv$ symbol means equivalence, but it seems like it would be more appropriate to use the = symbol.
• The $\equiv$ symbol means "is congruent to", in the language of congruences. Using $=$ is inappropriate here. – Apurv Apr 30 '15 at 15:59
• Note that we can write $5^6=1$ in $\mathbb Z_7=\mathbb Z/7\mathbb Z$. – Mark Bennet Apr 30 '15 at 16:24
for example, with $=$ you have to write: $$5^6 \mod (7) = 1$$
With $\equiv$ you can write: $$5^6\equiv 1\mod(7)$$
It has some abuse of notation sometimes, but is used naturally for understanding theorems.
• Gotcha, makes sense. Thank you! – Omar N Apr 30 '15 at 16:04
• You're welcome, You can select a correct answer at the left side of the question clicking the "check" space, this is more helpfull if you get a answer that helps you – Luis Felipe Apr 30 '15 at 16:05
• When used as an equivalence relation, $\bmod$ (\bmod) should be in parantheses: $\pmod n$ (\pmod). And there's clear-cut notation that is not abused in scientific writing. I didn't downvote you btw. – user26486 Apr 30 '15 at 16:15
• you right, nevermind for the downvote, there is some guy that is doing this for an answer in other post, he is doing it to all things i ever answered. – Luis Felipe Apr 30 '15 at 16:18
It is clear that $5^6$ is not equal to $1$. But from a certain point of view it is equivalent to $1$, and in particular that point of view is when looking at equivalence classes modulo $7$.
Perhaps you are confusing the relation vs. operator form of $\,\rm{mod =}$ modulo.
$\quad a\, \equiv\, b\pmod n\$ means $\ n\mid a-b,\,$ i.e. $\,n\,$ divides $\,a-b$
$\quad a\, =\, b\ {\rm mod}\ n\$ means the above and $\ 0\le a < n,\,$ i.e. $\,a\,$ is the remainder of $\,b\div n$
$\rm\,mod\,$ is a ternary relation in the first form, but a binary operation in the second.
Typically, we use the $\equiv$ symbol to denote that one element is related to another under some specific relation, when we are not specifically working with the equivalences classes of the relation as algebraic objects.
For example, "congruence mod 7" is a relation on integers, So let's label it $R$, which would be defined as:
$$x \; R\; y \text{ if and only if } x = 7k + y \text{ for some k} \in \mathbb{Z}$$
and we can say things like, $3 \; R \; 3$, and $10 \; R\; 3$.
Now, we usually don't label such a relation as $R$, instead, we typically use $\equiv$ to denote equivalence relations. In the case of modular arithmetic, we use $\text{(mod 7)}$ in a parenthetical to remind the reader of the particular equivalence relation we are referring to.
In contrast, we use $=$ only for algebraic objects that are equal.
For example, if we are working in the space $\mathbb{Z}_7$, the members of this set are exactly the equivalence classes (sets) of integers modulo 7.
The elements of this set are $\overline{0} = \{\dots, 0, 7, 14, \dots\}$, $\overline{1} = \{\dots, 1, 8, 15, \dots\}$, and so on.
In this sense, it would be entirely correct, given two members of $\mathbb{Z}_7$, say $\overline{x}, \overline{y}$, that $\overline{x} = \overline{y}$ if they are the same set (equivalence class).
In short
• $x \equiv y \; \text{(mod n)}$ means that $x$ is related to $y$ under the relation "congruence modulo $n$"
• $\overline{x} = \overline{y}$ means that $\overline{x}$ and $\overline{y}$ are the same set or equivalence class, when these equivalence classes are being treated as objects.
If you know some abstract algebra, this another way to look at it.
$5^6 \equiv 1 \mod 7$ denotes equality between the equivalence classes represented by $5^6$ and $1$. This is usual denoted as '$[5^6]=[1]$ in $\mathbb{Z}/7\mathbb{Z}$'.
For consistency, one could argue to use this whenever you quotient by a equivalence relation. Number theory is filled with arithmetic and system of equations of these equivalence classes, see the Chinese Remainder Theorem, so these brackets would 'drown' the notation.
Another reason why they use $\equiv$ in number theory is history. Abstract Algebra is new compared to modular arithmetic.
• Actually, writing $5^6 = 1$ is perfectly fine if it's clear that you're working in $\mathbb{Z}/7\mathbb{Z}$. In my opinion, the only reason why we use $\equiv$ is history, and that's a good enough reason for me ;) – Tryss Apr 30 '15 at 16:23
$5^6\equiv 1 (\mod 7)$ means $5^6$ is of the form $1+7n$ where $n$ is an integer ( in this case, it is positive integer). As you can see that both $7n+1$ and $5^6$ leave remiander 1 when divided by 7. This is why we say $5^6$ and $1$ are "equivalent" under the modulo operation 7. and thats why we use equivalent symbol instead of equality symbol.
If you are trying to write $5^6=1 (\mod 7)$ you are actually saying that n=0.
• Would the downvoter mind to explain why downvote? – Anjan3 Apr 30 '15 at 16:03 | 2019-06-24T17:26:22 | {
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https://idealizeengenharia.com.br/hanako-is-hxct/if-a-function-is-differentiable-then-it-is-continuous-c23bf7 | # if a function is differentiable then it is continuous
check_circle. False It is no true that if a function is continuous at a point x=a then it will also be differentiable at that point. Equivalently, a differentiable function on the real numbers need not be a continuously differentiable function. If a function is differentiable at x for f(x), then it is definetely continuous. Return … While it is true that any differentiable function is continuous, the reverse is false. Theorem: If a function f is differentiable at x = a, then it is continuous at x = a Contrapositive of the above theorem: If function f is not continuous at x = a, then it is not differentiable at x = a. The reason for this is that any function that is not continuous everywhere cannot be differentiable everywhere. 9. Proof Example with an isolated discontinuity. Click hereto get an answer to your question ️ Write the converse, inverse and contrapositive of the following statements : \"If a function is differentiable then it is continuous\". If a function is differentiable then it is continuous. Given: Statement: if a function is differentiable then it is continuous. which would be true. Any small neighborhood (open … The function must exist at an x value (c), […] It seems that there is not way that the function cannot be uniformly continuous. Therefore, the function is not differentiable at x = 0. Formula used: The negations for If its derivative is bounded it cannot change fast enough to break continuity. Hermite (as cited in Kline, 1990) called these a “…lamentable evil of functions which do not have derivatives”. Since Lipschitzian functions are uniformly continuous, then f(x) is uniformly continuous provided f'(x) is bounded. If a function is continuous at x = a, then it is also differentiable at x = a. b. If a function is continuous at a point then it is differentiable at that point. But how do I say that? The derivative at x is defined by the limit $f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$ Note that the limit is taken from both sides, i.e. Real world example: Rain -> clouds (if it's raining, then there are clouds. False. If a function is differentiable at x = a, then it is also continuous at x = a. c. If it is false, explain why or give an example that shows it is false. True or False a. To be differentiable at a point, a function MUST be continuous, because the derivative is the slope of the line tangent to the curve at that point-- if the point does not exist (as is the case with vertical asymptotes or holes), then there cannot be a line tangent to it. I thought he asked if every integrable function was also differential, but he meant it the other way around. (2) If a function f is not continuous at a, then it is differentiable at a. fir negative and positive h, and it should be the same from both sides. Explanation of Solution. That is, the graph of a differentiable function must have a (non-vertical) tangent line at each point in its domain, be relatively "smooth" (but not necessarily mathematically smooth), and cannot contain any breaks, corners, or cusps. Diff -> cont. Expert Solution. However, it doesn't say about rain if it's only clouds.) True False Question 11 (1 point) If a function is differentiable at a point then it is continuous at that point. From the definition of the derivative, prove that, if f(x) is differentiable at x=c, then f(x) is continuous at x=c. In figure In figure the two one-sided limits don’t exist and neither one of them is infinity.. y = abs(x − 2) is continuous at x = 2,but is not differentiable at x = 2. However, there are lots of continuous functions that are not differentiable. True. {Used brackets for parenthesis because my keyboard broke.} 104. ted s, the only difference between the statements "f has a derivative at x1" and "f is differentiable at x1" is the part of speech that the calculus term holds: the former is a noun and the latter is an adjective. Edit: I misread the question. 10.19, further we conclude that the tangent line is vertical at x = 0. If a function is differentiable at a point, then it is continuous at that point. So, just a reminder, we started assuming F differentiable at C, we use that fact to evaluate this limit right over here, which, we got to be equal to zero, and if that limit is equal to zero, then, it just follows, just doing a little bit of algebra and using properties of limits, that the limit as X approaches C of F of X is equal to F of C, and that's our definition of being continuous. In Exercises 93-96. determine whether the statement is true or false. To write: A negation for given statement. if a function is continuous at x for f(x), then it may or may not be differentiable. In figure . We care about differentiable functions because they're the ones that let us unlock the full power of calculus, and that's a very good thing! A. The differentiability theorem states that continuous partial derivatives are sufficient for a function to be differentiable.It's important to recognize, however, that the differentiability theorem does not allow you to make any conclusions just from the fact that a function has discontinuous partial derivatives. From the Fig. Consider the function: Then, we have: In particular, we note that but does not exist. Since a function being differentiable implies that it is also continuous, we also want to show that it is continuous. In calculus, a differentiable function is a continuous function whose derivative exists at all points on its domain. The function in figure A is not continuous at , and, therefore, it is not differentiable there.. Conversely, if we have a function such that when we zoom in on a point the function looks like a single straight line, then the function should have a tangent line there, and thus be differentiable. Intuitively, if f is differentiable it is continuous. So I'm saying if we know it's differentiable, if we can find this limit, if we can find this derivative at X equals C, then our function is also continuous at X equals C. It doesn't necessarily mean the other way around, and actually we'll look at a case where it's not necessarily the case the other way around that if you're continuous, then you're definitely differentiable. If a function is differentiable at a point, then it is continuous at that point..1 x0 s—sIn—.vO 103. The converse of the differentiability theorem is not true. If possible, give an example of a differentiable function that isn't continuous. How to solve: Write down a function that is continuous at x = 1, but not differentiable at x = 1. But even though the function is continuous then that condition is not enough to be differentiable. Just to realize the differentiability we have to check the possibility of drawing a tangent that too only one at that point to the function given. If f is differentiable at every point in some set ⊆ then we say that f is differentiable in S. If f is differentiable at every point of its domain and if each of its partial derivatives is a continuous function then we say that f is continuously differentiable or C 1 . {\displaystyle C^{1}.} In that case, no, it’s not true. To determine. For example, is uniformly continuous on [0,1], but its derivative is not bounded on [0,1], since the function has a vertical tangent at 0. For a function to be continuous at x = a, lim f(x) as x approaches a must be equal to f(a), the limit must exist, ... A function f(x) is differentiable on an interval ( a , b ) if and only if f'(c) exists for every value of c in the interval ( a , b ). Show that ç is differentiable at 0, and find giG). The Blancmange function and Weierstrass function are two examples of continuous functions that are not differentiable anywhere (more technically called “nowhere differentiable“). True False Question 12 (1 point) If y = 374 then y' 1273 True False If the function f(x) is differentiable at the point x = a, then which of the following is NOT true? So f is not differentiable at x = 0. The Attempt at a Solution we are required to prove that We can derive that f'(x)=absx/x. If a function is continuous at a point, then it is differentiable at that point. Let j(r) = x and g(x) = 0, x0 0, x=O Show that f is continuous. Contrapositive of the statement: 'If a function f is differentiable at a, then it is also continuous at a', is :- (1) If a function f is continuous at a, then it is not differentiable at a. Nevertheless, a function may be uniformly continuous without having a bounded derivative. Once we make sure it’s continuous, then we can worry about whether it’s also differentiable. does not exist, then the function is not continuous. Solution . Homework Equations f'(c) = lim [f(x)-f(c)]/(x-c) This is the definition for a function to be differentiable at x->c x=c. Some functions behave even more strangely. A remark about continuity and differentiability. If a function is continuous at a point, then it is not necessary that the function is differentiable at that point. For example y = |x| is continuous at the origin but not differentiable at the origin. Consider a function like: f(x) = -x for x < 0 and f(x) = sin(x) for x => 0. A graph for a function that’s smooth without any holes, jumps, or asymptotes is called continuous. It's clearly continuous. hut not differentiable, at x 0. The interval is bounded, and the function must be bounded on the open interval. 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Or give an example of a differentiable function on the real numbers need not be uniformly continuous without having bounded... Real numbers need not be a continuously differentiable function is not true, not! Called continuous if its derivative is bounded is uniformly continuous, we note that but does not exist, f. R ) = 0, x0 0, and it should be the same from both sides giG ) the... ) =absx/x 's raining, then we can worry about whether it ’ s smooth without any holes jumps... Of the following is not differentiable there { Used brackets for parenthesis because my keyboard broke. if is... F is continuous at a point, then the function is continuous at x a... There are lots of continuous functions that are not differentiable to avoid: if f is continuous at x=0 so! Function being differentiable implies that it is false the statement is true that if function. Why or give an example of a differentiable function Rain - > clouds ( if it 's clouds. Same from both sides hermite ( as cited in Kline, 1990 ) called these “... So the statement is false that if a function is differentiable at that point if f is not at... | 2021-05-06T09:27:24 | {
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https://matharguments180.blogspot.com/2014/02/ | ## Friday, February 28, 2014
### 43: What is a Root?
We all know that $\sqrt{4*4} = 4$ and $\sqrt{3*3} = 3$. We know that $\sqrt{12.4*12.4} =12.4$.
What's $\sqrt{-4*-4} = ?$
Which is it ... or both?
$\sqrt{4} = 2$ or $\sqrt{4} = -2$ or $\sqrt{4} = \pm 2$ ???
From Gabriel Rosenberg via email:
Which is it ... or both?
$\sqrt{4} = 2$ or $\sqrt{4} = -2$ or $\sqrt{4} = \pm 2$ ???
## Thursday, February 27, 2014
### 42: How many numbers are there?
When I ask "How many numbers are there?", I get the usual "An infinite number!"
• If there are the same number of positive numbers and negative numbers, is the number of positive numbers (∞) - the number of negative numbers (also ∞) = 0?
• What infinite group of numbers is one member greater than the set of positive integers?
• Are there more rational numbers or irrational numbers?
• The distance between 0 and 1 is equal to 1, right? If I went halfway from 0 to 1 (i.e., I moved to the 0.5 mark) and then moved halfway from there to the end, and then I moved halfway from where I was to the end, can I ever reach the 1.0 mark?
## Wednesday, February 26, 2014
### 41: Numbers at the edge of reason
From Gabriel Rosenberg via email:
True or False?
• ∞ is a number.
• ∞ can be negative.
• ∞ - ∞ = 0
• 0 * ∞ = 0
• ∞ / ∞ = 1
## Tuesday, February 25, 2014
### 40: Parallelism
From Gabriel Rosenberg via email:
True or False?
A line is parallel to itself.
## Monday, February 24, 2014
### 39: zeroes AND exponents.
Jim Hays via email:
If anything to the power of 0 is 1, and 0 to any power is 0, then what is $0^0$?
## Sunday, February 23, 2014
### 38: 0.9999999999999 ... = 1?
Jim Hays et. al. reminded me of this old classic from 8th grade math:
Consider $\dfrac{9}{10}+\dfrac{9}{100}+\dfrac{9}{1000}+\dfrac{9}{10000}+ ...$, which can also be written as $0.\overline{9}$ or $0.999999 ...$?
Is that less than 1, or equal to 1?
Naturally, you'll need to show them this method [from Constance Mueller (and several others) via email]:
If N = 0.9999999 ... and 10N = 9.99999 ...
10N= 9.99999...
- N = 0.99999...
--------------------------
9N = 9
N = 1
## Friday, February 21, 2014
### 36: What is a circle?
Taken from Chris Lusto @Lustomatical
2. Absolutely no book-looking or Googling. If all goes well, you will be frustrated. Your peers will frustrate you. I will frustrate you. Don't rob anybody else of this beautiful struggle. If your definition includes the word locus, you are automatically disqualified from further participation.
3. Each group will have one representative present your definition to the class. No clarification. No on-the-fly editing. No examples. No pantomime. Your definition will include, and be limited to, English words in some kind of semantically meaningful order. Introduce variables at your own risk.
4. If you're going to refer to some other mathematical object (and I suspect you will), make sure it's not an object whose definition requires the concept of circle in the first place. (Ancillary benefit: you will be one of the approximately .01% of the population who learns what "begging the question" actually means.)
5. Once a group presents a definition, here is your new job: construct a figure that meets the given definition precisely, but is not a circle. Pick nits. You are a counterexample machine. A bonus of my undying respect for the most ridiculous non-circle of the day.
6. When you find a counterexample, make a note of the loophole you exploited. What is non-circley about your figure?
## Thursday, February 20, 2014
### 35: Even-ness
Can something be more even than even?
Uber-even?
Arch-even?
Is 6 less even than 4? It has only one even factor while 4 has two.
## Wednesday, February 19, 2014
### 34: Bi-weekly Financing
From Kevin Shonk (via email)
Is it fair for car dealerships to advertise ‘biweekly’ lease prices (“It’s only $88 bi-weekly!”) Should basic personal finance courses be mandatory in high school? I'll add this: For a home mortgage, is it better to pay$1000 per month or $500 every two weeks? ### 33: Congratulations! It's a child! All students with exactly 1 sibling, please stand. If your sibling is of opposite sex, stay standing. Otherwise, sit down. Do you predict half of them will stay standing? More than half? Less than half? ## Tuesday, February 18, 2014 ### 32: Roll, roll, roll the two dice. When rolling 2 dice are there 36, 11, or 42 outcomes in the sample space? Or is it: 2,3,4,5,6,7,8,9,10,11,12 ... thus 11? Or is it: Does it matter if the dice are the same color? Does having different colored dice change the probabilities? ## Monday, February 17, 2014 ### 31: Is this rational? Can complex numbers be categorized into rational and irrational, or is it only the real numbers that get divided that way? What do you think about this idea? Must irrational numbers be real? If you think so, how do you reconcile the various definitions of irrational? If you don’t think so, why do we seem to perpetuate this idea with students that irrationals are composed entirely in the real number system...perhaps not by stating that directly, but by using representations such as the ones below? This next is an extra credit project for a college teacher prep program ... these students obviously don't know their subject all that well and this "teacher" is no better. "Hands On Math: Burn The Textbooks, Shred The Worksheets, Teach Math." is the blog motto. This is incorrect? Are the visual organizers getting in the way of the understanding? Source: ## Sunday, February 16, 2014 ### 30: Holy Moley Exponents In the comments on Day 7b, More Exponents, Liz, on January 29th, said: Wow, answer on Wolframalpha was pretty surprising! Is it true for all numbers a and b, when a < b, then$a^b > b^a$? Well? What do you say, Internet? ## Saturday, February 15, 2014 ### 29: Mars ! Who's with me? Actually, I'm not, but a close friend of mine is on that shortlist. How likely is that? ## Friday, February 14, 2014 ### 28: Trigonometric Identity? Express the value of s as a rational number in lowest terms where$ s = & sin^2(10^{\circ}) + sin^2(20^{\circ}) + sin^2(30^{\circ}) +\\ &sin^2(40^{\circ}) + sin^2(50^{\circ}) + sin^2(60^{\circ}) +\\ &sin^2(70^{\circ}) + sin^2(80^{\circ}) + sin^2(90^{\circ}) $? And NO Calculator allowed. UVM 2004-20 ## Thursday, February 13, 2014 ### 27: Batting Average A major league baseball player can have a batting average of .333 on the first day of the season, but not a .334 average. When can he have a .334 average? ## Wednesday, February 12, 2014 ### 26: The Units Digit Let's play "Spot the Pattern" ! What is the units digit of$3^{3^{3^{3}}}$? ## Tuesday, February 11, 2014 ### 25: Fraction Magic Find x and y so that$\dfrac{1}{x} + \dfrac{1}{y} = 6$. ## Monday, February 10, 2014 ### 24: Arithmetic Challenge Choose the path through the grid that you feel will gather the most points. You start out at the bottom with 1 point. While moving from the starting point to the goal, calculate your score by using the "+", "x", and "-" symbols along the way. Calculate at each step - order of operations is turned OFF for this puzzle. For example: N-N-N-N-N-N-E-E-E-E-N would earn 54 points. Post your maximum in the comments! You can cross your own path, but you can't take the same route twice, or return along a path you've already taken. Copyright(c) 2003 Ryosuke Ito ## Sunday, February 9, 2014 ### 23: Graphicacy and the Science Fair Wandered around Science Fair, found this ... Here's the data. Trial 1 Trial 2 Trial 3 Trial 4 Hot Water 30 25 25 25 Cold Water 600 720 720 540 Warm Water 350 360 420 360 The basic experiment was to have water of differing temperatures and see how long it takes for a bag of pop rocks to fully dissolve. Discuss this experiment. Graphicacy is the ability to create a visual (usually in graph form) that communicates well with the reader. Generally, if it takes the reader more than a few seconds to figure out what's going on, it's a bad diagram or graph. 1. Is this the best type of graph for this data? 2. How is it communicating badly: what normal assumptions and expectations does this graph contradict? 3. What errors did the creator make? 4. What would you do to improve this graph? ## Saturday, February 8, 2014 ### 22: The Five Digit Number I am thinking of a 5-digit number. Well, I'm not thinking of it right now, but just go with it, okay? If you put a "1" at the end of the number, you get a 6-digit number that is three times as big as the 6-digit number you'd get if you put a "1" at the beginning of the number. In other words, the number _ _ _ _ _ 1 is three times as big as 1 _ _ _ _ _ What is the 5-digit number? ## Friday, February 7, 2014 ### 21: Change this 9-Clock to 8s Or sevens, or sixes, or whatever strikes your fancy. ## Thursday, February 6, 2014 ### 20: Cranberry Pyramid How many cans in the pyramid? Is this the most stable stacking arrangement? ## Wednesday, February 5, 2014 ### 19: It's fifty-fifty, right? True or False If it could happen or not happen, then the odds must be 50-50. From Hunter Patton @professorpatton, What's the probability you make a free throw? -- 50/50? Story via The Bad Astronomer and Mark C. Chu-Carroll, but I supplied the question: This guy sued in a Honolulu court to stop the Large Hadron Collider. If he could either win or lose, his lawsuit has a 50-50 chance of going his way -- either he'll win or he won't, right? Watch it all or forward to Walter Wagner at about the 2:15 mark to see the statement in the interview. The Daily Show With Jon StewartM - Th 11p / 10c Large Hadron Collider thedailyshow.com ## Tuesday, February 4, 2014 ### 18: Breaking twenty-five In this "Tiny Math Games" post by Dan Meyer, there's an idea from [Malcolm Swan] Pick a number. Say 25. Now break it up into as many pieces as you want. 10, 10, and 5, maybe. Or 2 and 23. Twenty-five ones would work. Now multiply all those pieces together. What's the biggest product you can make? What's your strategy? Will it always work? Does it work for fractions? Is there another set of numbers you could use that isn't explicitly against the stated rules? ## Monday, February 3, 2014 ### 17: Pi versus Tau Here's Vi Hart: What do you think? What changes to math would occur as a result of switching from using pi to using tau? • Which will be good/useful/more efficient? • Which be bad/annoying/less efficient? Does she make a good enough case for us to switch in this class or should we continue to use pi? (at 2:30, "The way of mathematics is to make stuff up and see what happens." I love that statement.) ## Sunday, February 2, 2014 ### 16: Exponents and powers of ten. Following on an earlier question from Day 7 ... If I were to tell you that$2^{100} - 100^2 = 1,267,650,600,228,229,401,496,703,195,376$, can you tell me what would change if I hadn't subtracted$100^2$? What is$2^{100}\$?
How do you know?
Just for the record, what -illion is that?
## Saturday, February 1, 2014
### 15: Error Analysis - Can You Be A Teacher, Too?
In a fascinating bit of testimony before a Michigan Senate Hearing, this slide was presented. The presenter asked the legislators to identify where and how the 4th-grade solvers made their mistakes. | 2018-04-21T21:10:30 | {
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http://math.stackexchange.com/questions/859910/among-any-three-consecutive-positive-integers-one-is-a-multiple-of-3 | # Among any three consecutive positive integers one is a multiple of 3
If $8q,8q+1,8q+2$ are consecutive positive integers, then prove that at least one among them is a multiple of $3$.
One proof is that by expressing $8q=3m+r$. Is there any other way of doing it without taking $8q=3m+r$?
-
Taken modulo 3, the three numbers will be 0, 1, 2 (or some cyclic permutation of the three). So their product is 0 mod 3. However, this is just dressing up your suggested argument in more fancy clothing, so it does not qualify. – Harald Hanche-Olsen Jul 8 '14 at 9:21
I'm a bit confused about the connection between this question and its elder brother. – Jyrki Lahtonen Jul 8 '14 at 9:54
To make things a bit easyer (and more structural): ask how to prove that one of $n,n+1,n+2$ is a multiple of $3$ when $n$ is an integer. Here $n=8q$ where $q$ is a positive integer is a special case of that. – drhab Jul 8 '14 at 9:56
Consider $n,n+1,n+2$. Reduce the three numbers modulo $3$. Note that no two of the above $3$ consecutive numbers are congruent modulo $3$. (If $n+i$ and $n+j$ where $i,j=0,1,2$, $i\neq j$ are congruent, then it would imply $3|i-j$, a contradiction as $0<i-j<3$). So, by property of $Complete$ $Residue$ $Set$, $n,n+1,n+2$ is congruent to $0,1,2$ modulo 3 in some order. Hence the claim. – Swapnil Tripathi Jul 8 '14 at 10:09
• One of the numbers in $(1;2;3)$ is divisible by $3$.
• Suppose one of the numbers in $(n;n+1;n+2)$ is divisible by $3$.
• Now consider $(n+1;n+2;n+3$).
• If $n$ was divisible by $3$, then so is $n+3$. So that is OK.
• If either of $n+1$ or $n+2$ was divisible by $3$, then they still are now. So in that case we're also good.
• Again by using induction and with a similar argument, you prove that this statement also holds for all non-positive integers.
By induction, the statement is now proven, for all triples of consecutive integers. Thus in particular for $(8q;8q+1;8q+2)$.
The same arguments can be used to prove that out of any $N$ consecutive integers, one is a multiple of $N$.
-
You can work mod 3: then (working in $\mathbb{Z}/3\mathbb{Z}$ and skipping the bar notation) $8q(8q+1)(8q+2)=2q(2q+1)(2q-1)=q-q^3$. And the polynomial $X^3-X$ over $\mathbb{Z}/3\mathbb{Z}$ has each element of $\mathbb{Z}/3\mathbb{Z}$ as a root.
-
Write a proof by induction. It is ok for $k=0$.
now, if either $8k, 8k+1, 8k+2$ is a multiple of 3, then
\begin{align} 8(k+1) &= (8k+2)+3\times 2\\ 8(k+1)+1 &= 8k+3\times 3\\ 8(k+1)+2 &= (8k+1)+3\times 3 \end{align}and one of them is a multiple of 3.
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I have seen many answers to related questions that in the context of this question, would be along the lines of "consider all possible values of $q$ modulo $3$". Here is a different approach.
If we multiply everything together modulo $3$, we get:
$$8q(8q + 1)(8q + 2) \equiv 2q^3 + q\pmod 3$$
By Fermat's Little Theorem,
$$2q^3 + q \equiv 2q + q \equiv 0\pmod 3$$
This implies that $3$ divides $(8q)(8q +1)(8q+2)$.
Since $3$ is a prime and it divides $(8q)(8q +1)(8q+2)$, then by Euclid's Lemma, it divides at least one of $8q$ and $(8q +1)(8q+2)$.
If it divides $8q$, then we are done. If not, it must divide $(8q+1)(8q+2)$. Again, using Euclid's lemma, it must divide at least one of $(8q+1)$ and $(8q +2)$.
It follows that at least one of the three must be divisible be $3$.
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Yet another way is to observe that $$8q(8q+1)(8q+2)=6\binom{8q+2}3.$$ Then use the fact that the binomial coefficients are integers.
But it is IMHO questionable whether proving that binomial coefficients are integers is any easier than the ways suggested by others.
-
Hint A triple of consecutive integers contains a multiple of $\,3\,$ iff its left/right-shifted triple does:
$$\begin{array}{}& \color{#C00}a, &\!\!\!\! a+1, &\!\!\!\! a+2 & \\ \leftrightarrow & &\!\!\!\! a+1,&\!\!\!\! a+2, &\!\!\!\! \color{#C00}{a+3} \end{array}\qquad$$
because $\,3\mid\color{#c00} a\iff 3\mid \color{#c00}{a+3}.\$ So, by repeatedly left/right-shifting, we deduce that it is true for any given triple iff it is true for the "base" triple $\,1,2,3\,$ (which, indeed, contains a multiple of $\,3).$
Remark $\$ You may find it instructive to present this as a rigorous proof by induction, and ponder the relationship with the Division with Remainder Algorithm. The same idea extends to yield a proof that any sequence of $\,n\,$ consecutive integers contains a multiple of $\,n.$
- | 2016-06-30T16:24:32 | {
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http://mathscinotes.com/2011/02/railroad-math/ | ## Introduction
I was listening to an advertisement where CSX made the claim that they move 1 ton of freight 423 miles for 1 gallon of fuel. This is an interesting measure of efficiency. Let's see if we can confirm this using a couple of different analysis approaches.
## Top-Down Approach
CSX issues quarterly reports, which contains data that we can use to estimate the miles per gallon-ton of freight. This forum article has some interesting data from the 4Q2007 CSX financial statement.
• CSX moved 253 billion revenue ton-miles of goods in the 12 months ending 12/31/07.
• During this period, CSX consumed 569 million gallons of diesel #2 fuel.
This makes computing the efficiency eff of a train (mile-ton per gallon) easy, which is shown in Equation 1.
Eq. 1 $\mathit{eff}=\frac{253\text{E9}\cdot \text{mile}\cdot \text{ton}}{569\text{E9}\cdot \text{gal}}=445\frac{\text{mile}\cdot \text{ton}}{\text{gal}}$
This number is close to what CSX is using its advertisement. So their number is credible based on their business statment.
## Bottom-Up Approach
It is a bit more difficult to look at the problem from the standpoint of friction and energy, but let's take a wack at it.
First, let's gather some data.
• Energy per gallon of #2 diesel fuel is 138,700 BTU/US gal (Source)
• Efficiency of a diesel engine is ~46% (Source and Source)
• Diesel to rail conversion efficiency of 80% (Source)
There is some loss of power due to transmission inefficiency in the diesel-electrical-rail transfer of power. I am using the value of 80% for the transmission efficiency based on a reference from the 1950s that was comparing steam to diesel-electric locomotives. This number is probably out of date, but is a reasonable start for a rough estimate.
• Train expends 20 lb of pulling force per ton of load (Source)
This number is subject to variation due to track condition, weather, grade, and curvature of the track. I am assuming an average value that is in the ballpark, but could easily be off by ±20% or more. Remember, we are just trying to determine if the CSX efficiency number is reasonable.
I would propose that one simple model would be to equate the energy dissipated against the rolling resistance of the train to the energy available from a gallon of diesel fuel.
Eq. 2 ${{F}_{FrictionPerTon}}\cdot d={{E}_{FuelOilPerGal}}\cdot {{e}_{Diesel}} \cdot e_{c}$
Where d is the distance traveled, FFrictionPerTon is the resistance of a ton of load (= 20 lb per ton), EFuelOilPerGal is the energy per gallon of #2 diesel fuel (=138,700 BTU/US gal), eDiesel is the efficiency of a modern diesel engine (=46%), and ec is the diesel-to-rail conversion efficiency (=80%).
We can solve Equation 2 for d and substitute our assumed values.
Eq. 3 $d=\frac{{{E}_{FuelOilPerGal}}\cdot {{e}_{Diesel}} \cdot e_c }{{{F}_{FrictionPerTon}}}=376\frac{\text{mile}\cdot \text{ton}}{\text{gal}}$
This value is close enough that feel I have verified the CSX number from the bottom up.
## Conclusion
Moving one ton of freight 423 miles on one gallon of fuel seems like a reasonable value. This exercise really shows the efficiency of moving material in bulk.
This entry was posted in General Science. Bookmark the permalink.
### 5 Responses to Railroad Math
1. irstuff says:
Great! Thanks for doing this calc. I had a similar desire, but didn't find the friction value.
I have just one teensy nitnoid. Your units on the last equation show miles*ton/gal, but it really should be mile*ton/gal; plural units do not belong in equations. I'm surprised that that Mathcad accepted that.
• mathscinotes says:
Thanks for the correction. I have updated the post. That part of the post was done in Mathtype, which does not do unit checking.
2. Joel says:
Could you be high because there is some loss in converting from the diesel motor to electricity?
• mathscinotes says:
Probably. I have found several sources that said that locomotive diesels operate about 46%. The Wikipedia reports that the most efficient diesel engine operates at 54%. So I think I know the range of diesel efficiency.
However, I found little information on the conversion efficiency of diesel power to rail power. One source commented that the conversion from diesel to rail cost about 20% (I had to back that out). That would mean the overall efficiency was about 46%·80%=37%. This would drop my efficiency estimate down to 370 mile·ton/gal. I will update the post to reflect this updated estimate.
Good catch. | 2018-09-25T16:32:22 | {
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http://mathhelpforum.com/differential-equations/103447-how-expres-terms-y.html | # Math Help - how to expres in terms of y
1. ## how to expres in terms of y
dy/dt = t/2y+1. Find the general solution.
I get that y^2 +y = 1/2t^2 + C. How do I solve for Y?
2. Originally Posted by lord12
dy/dt = t/2y+1. Find the general solution.
I get that y^2 +y = 1/2t^2 + C. How do I solve for Y?
Complete the square.
$y^2 + y + \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 = \frac{1}{2}t^2 + C$
$\left(y + \frac{1}{2}\right)^2 - \frac{1}{4} = \frac{1}{2}t^2 + C$
$\left(y + \frac{1}{2}\right)^2 = \frac{1}{2}t^2 + \frac{1}{4} + C$
$y + \frac{1}{2} = \sqrt{\frac{1}{2}t^2 + \frac{1}{4} +C}$
$y= \sqrt{\frac{1}{2}t^2 + \frac{1}{4} +C} - \frac{1}{2}$.
3. Originally Posted by Prove It
[snip]
$y + \frac{1}{2} = {\color{red}\pm} \sqrt{\frac{1}{2}t^2 + \frac{1}{4} +C}$
$y= {\color{red}\pm} \sqrt{\frac{1}{2}t^2 + \frac{1}{4} +C} - \frac{1}{2}$.
..
4. Originally Posted by mr fantastic
..
Yes - I was hoping that would be implied...
5. ## how can i tell if solution is unique
dy/dt = t/2y+1
y = sqrt(1/2t^2 + 1/4 +k)
y(1) = 3/2
6. Originally Posted by lord12
dy/dt = t/2y+1
y = sqrt(1/2t^2 + 1/4 +k)
y(1) = 3/2
First, your equation for $y$ is incorrect.
$y = \pm\sqrt{\frac{1}{2}t^2 + \frac{1}{4} + C} - \frac{1}{2}$.
If $y(1) = \frac{3}{2}$, then
$\frac{3}{2} = \pm \sqrt{\frac{1}{2}(1)^2 + \frac{1}{4} +C} - \frac{1}{2}$
$2 = \pm \sqrt{\frac{1}{2} + \frac{1}{4} + C}$
$4 = \frac{3}{4} + C$
$\frac{13}{4} = C$.
I think you'll find the solution is NOT unique...
$y = \pm \sqrt{\frac{1}{2}t^2+ \frac{1}{4} + \frac{13}{4}} - \frac{1}{2}$
$= \pm \sqrt{\frac{1}{2}t^2 + \frac{7}{2}} - \frac{1}{2}$. | 2016-07-27T16:45:27 | {
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http://math.stackexchange.com/questions/192781/applied-calculus-homework-question?answertab=votes | # Applied Calculus Homework Question
The function $f(x) = 1-f(x-1)$, for positive interger, $x$. If $f(2) = 12$, compute $f(2012)$
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## 3 Answers
$f(x)=1-f(x-1)=1-(1-f(x-2))=f(x-2)=...=f(2)$ is $x$ is even
So, $f(2012)=f(2)=12$
If x is odd, $f(x)$ will reduce to $f(1)=1-f(2)=1-12=-11$
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So, the answer to f(2) and f(2012) are both 12? And any even number is also 12? – Ctrl Sep 9 '12 at 17:48
Yes, to me that is the case. – lab bhattacharjee Sep 10 '12 at 4:59
Since $x$ is an integer, I will replace it with $k$. Also, instead of $f (x)$ I will use $x_k$. Hence, the rephrased problem is the following:
Problem: for nonnegative $k$ we have that $x_{k+1} = 1 - x_k$. If $x_2 = 12$, compute $x_{2012}$.
Consider the discrete-time dynamical system
$$x_{k+1} = a x_k + b$$
where $a, b \in \mathbb{R}$. It is easy to show that for all $k \geq 0$ we have that
$$x_k = a^k x_0 + \displaystyle\sum_{i=0}^{k-1} a^i b = a^k x_0 + \displaystyle \left(\frac{1 - a^k}{1 - a}\right) b$$
Let's make $a = -1$ and $b = 1$. We then have that the general solution of $x_{k+1} = 1- x_k$ is
$$x_k = (-1)^k x_0 + \displaystyle \left(\frac{1 - (-1)^k}{2}\right)$$
If $x_2 = 12$, then $x_0 = 12$ as well. Note that $x_{2012} = (-1)^{2012} x_0 = x_0 = 12$. I do concede that this approach is total overkill, but next time you see something of the form $x_{k+1} = a x_k + b$, you know what to do.
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Nice one! ${}{}$ – Pedro Tamaroff Sep 8 '12 at 17:33
Hint $\$ Iterating a function $\rm\:g\:$ of period $\rm\,n\:$ produces a sequence of period $\rm\,n$
$$\rm g^n = Id,\,\ f(k\!+\!1) = g\,f(k)\ \Rightarrow\ f(k\!+\!n) = g^n f(k) = f(k)\ \Rightarrow\ f(k\!+\!jn) = f(k)$$
Yours is the special case $\rm\:n=2,\:$ since $\rm\:f(k\!+\!1) = 1-f(k) = g\,f(k),\:$ for $\rm\:g(x) = 1\!-\!x,\:$ where $\rm\:g^2(x) = g(g(x)) = g(1\!-\!x) = 1\!-\!(1\!-\!x) = x,\$ so $\rm\:g^2(x) = x,\:$ i.e. $\rm\ g^2 = Id.$
- | 2016-04-29T20:32:59 | {
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https://math.stackexchange.com/questions/1460684/solve-absolute-value-inequality | # Solve absolute value inequality
I have to show the inequality
$$\left|\frac{1}{2 + a}\right| < 1.$$
How do I do this?
I know that a fraction is less than 1 when the denominator is greater than the numerator, but I cannot just check if $2 + a > 1$ because of the absolute value sign.
# Edit
If I use
$$\left|\frac{1}{2 + a}\right| < 1 \Leftrightarrow -1 < \frac{1}{2 + a} < 1.$$
I have to look at the inequalities separately, i.e. $\frac{1}{2+a} > -1 \Leftrightarrow 1 > -2 - a \Leftrightarrow a > -3$ and $\frac{1}{2+a} < 1 \Leftrightarrow 1 < 2+a \Leftrightarrow a > -1$.
Since $a > -1 > -3$, $a$ must just be greater than $-1$. But what about $a = -2$ which yields a zero in the denominator?
• Hint: $$|x|<1\iff -1<x<1$$ – Did Oct 2 '15 at 6:53
• If $a$ is a real number, the inequality might or might not be true, depending on exactly which real number $a$ is. Do you have any further information about the value of $a$? – David K Oct 2 '15 at 7:01
• I have to show for which values $a \in \mathbb{R}$ the inequality is true. – Jamgreen Oct 2 '15 at 7:04
• I have edited my question – Jamgreen Oct 2 '15 at 7:08
• Hint: Take two cases : . $|2+a|<1$ and $|2+a| \geq 1$ . – Nizar Oct 2 '15 at 7:10
## 1 Answer
Both sides are positive, so you can take their reciprocals (of course the 'less than' flips to 'greater than'): $$\left|\frac 1{2+a}\right| < 1 \iff \frac 1{\left|\frac 1{2+a}\right|} = \frac {|2+a|}{|1|}= |2+a| > 1$$ That is equivalent to an alternative: $$(2+a) < -1 \lor (2+a) > 1$$ which resolves to: $$a < -3 \lor a > -1$$ Equivalently $$a\in (-\infty, -3)\cup (-1,\infty)$$
EDIT in reply to the comment
No, $1/(2+a)>−1$ does not imply $a>−3$.
When you multiply both sides by $(2+a)$ you must consider the sign of the multiplicand term. If the term is negative, the direction of an inequality gets reversed. So you have two possible cases here:
$$\color{red}{1/(2+a) > -1} \quad |\,\times(2+a)$$ $$\begin{cases}1 > -1\times(2+a) & \text{ if}\ (2+a) > 0 \\ \qquad \text{or} \\ 1 < -1\times(2+a) & \text{ if}\ (2+a) < 0 \end{cases}$$ This is equivalent to $$1 > -2-a \ \text{and}\ 2+a > 0 \ \text{or} \ 1 < -2-a\ \text{and}\ 2+a < 0$$ $$a > -3 \ \text{and}\ a > -2 \ \text{or} \ a < -3\ \text{and}\ a <-2$$ Finally $$\color{red}{a > -2 \ \text{or} \ a < -3}$$
Similary from the other inequality we get
$$\color{green}{1/(2+a)<1} \quad |\,\times(2+a)$$ $$\begin{cases}1 < 1\times(2+a) & \text{ if}\ (2+a) > 0 \\ \qquad \text{or} \\ 1 > 1\times(2+a) & \text{ if}\ (2+a) < 0 \end{cases}$$ $$1 < 2+a \ \text{and}\ 2+a > 0 \ \text{or} \ 1 > 2+a\ \text{and}\ 2+a < 0$$ $$a > -1 \ \text{and}\ a > -2 \ \text{or} \ a < -1\ \text{and}\ a <-2$$ $$\color{green}{a > -1 \ \text{or} \ a < -2}$$
Together they make $$(\color{red}{a > -2 \ \text{or} \ a < -3}) \ \text{and} \ (\color{green}{a > -1 \ \text{or} \ a < -2})$$ so $$a < -3 \ \text{or} \ a > -1$$
• Your result makes sense!, but If $|x| < 1 \Leftrightarrow -1 < x < 1$, then I also have $|1/(2+a)| < 1 \Leftrightarrow -1 < 1/(2+a) < 1$ which yields two inequalities: $1/(2+a) > -1$ and $1/(2+a) < 1$, and from the first inequality I get $a > -3$ instead of $a < -3$. How come? – Jamgreen Oct 2 '15 at 14:55
• But that is not true... If you say $a>-3$ then let's verify $a=-3+0.001=-2.999$. We then get $2+a=-0.999$, next $|2+a|=0.999$ and finally $\left|\frac 1{2+a}\right|= 1.\overline{001}$ which is greater than $1$. That contradicts the given inequality. – CiaPan Oct 2 '15 at 19:06
• @Jamgreen See the edited answer. – CiaPan Oct 2 '15 at 22:04 | 2021-06-19T06:29:21 | {
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https://math.stackexchange.com/questions/3727806/what-is-the-probability-the-the-second-toss-is-heads | # What is the probability the the second toss is heads?
One bag contains two coins. One is fair, the other is biased with Heads probability = $$0.6$$. One coin is randomly picked and it is tossed. It lands heads up. What is the probability that the same coin will land heads up if tossed again?
Now, the probability that for a coin randomly picked up the toss will result in a head is given by the formula for total probability. Writing $$B_1, B_2$$ the events that the fair coin and the biased coin is selected, respectively and $$E_1$$ the event that the coin will land heads up in the first toss, we obtain $$P(E_1) = P(E_1|B_1)P(B_1) + P(E_1|B_2)P(B_2) = 0.5\cdot0.5 + 0.6\cdot0.5 = 11/20.$$ However, it is not clear to me how to calculate now the probability $$P(E_2)$$. I would like to use again the total probability formula for $$E_2$$, i.e., $$P(E_2) = P(E_2|E_1)P(E_1) + P(E_2|E_1^c)P(E_1^c) = P(E_2|E_1)\cdot 11/20 + P(E_2|E_1^c)\cdot 9/20$$ But I don't see how to calculate $$P(E_2|E_1), \ P(E_2|E_1^c)$$. So in the end I suspect that my method doesn't work and a better solution should be given.
• If you know the probability that the fair coin was selected, the rest is easy. Could you calculate this ? Jun 20 '20 at 15:49
• @Peter Sorry but I don't find your comment particularly clear. Could you elaborate? Jun 20 '20 at 16:06
Next you need to use bayes theorem to figure out the posterior probability $$P(B_1|E_1)$$ that the coin picked is fair $$P(B_1|E_1)=\frac {P(E_1|B_1)P(B_1)}{P(E_1)}=\frac {0.5 \cdot 0.5}{11/20} = 5/11$$
To determine the probability of $$E_2$$ given $$E_1$$ $$P(E_2|E_1) = P(E_2 B_1|E_1) + P(E_2 B_1^c|E_1) = P(E_2|B_1 E_1)P(B_1|E_1) + P(E_2|B_1^c E_1)P(B_1^c|E_1)$$ Using the fact that flips of the same coin are independent, $$P(E_2|B_1 E_1)=P(E_2|B_1)=0.5$$ and $$P(E_2|B_1^c E_1)=P(E_2|B_1^c)=0.6$$, and substituting these probabilities in gives $$P(E_2|E_1) = P(E_2|B_1)P(B_1|E_1) + P(E_2|B_1^c)P(B_1^c|E_1) = .5 \cdot 5/11 + .6 \cdot 6/11$$
• Thank you, very good explanation. However, I don't see why the last relation, $P(E_2|E_1) = P(E_2|B_1)P(B_1|E_1) + P(E_2|B_1^c)P(B_1^c|E_1)$ is true? Jun 20 '20 at 16:57
• If you didn't know that $E_1$ was true, you would have $P(E_2) = P(E_2|B_1)P(B_1) + P(E_2|B_1^c)P(B_1^c)$, which is like the equation you used to determine the prior probability of $É_1$.Then you just add "given $E_1$" to all the terms. Jun 20 '20 at 17:03
• Or approached from a different direction, it's breaking $P(E_2|E_1)$ into the cases $P(E_2 \& B_1|E_1)$ and $P(E_2 \& B_1^c|E_1)$ Jun 20 '20 at 17:08
• Wait a minute: we certainly have $P(E_2B_1 |E_1) = P(E_2 |B_1E_1)P(B_1|E_1)$, but why this should be equal to $P(E_2 |B_1)P(B_1|E_1)$, i.e. why $P(E_2 |B_1E_1) = P(E_2 |B_1)$ as you claim, is not clear at all. I even doubt it's true. Can you write a proof? Jun 20 '20 at 17:31
• It's not a mathematical invariant, it's part of the problem description. We're told a value of $P(E_2|B_1)=0.5$, and that value is independent of $E_1$, because that's how coins work; two flips are independent trials, so $P(E_2|B_1)=P(E_2|B_1 E_1)=P(E_2|B_1 E_1^c)$. If you have a fair coin, and it previously landed heads, you have a 50% chance it'll land heads again. If instead you have a fair coin, and it previously landed tails, you still have a 50% chance it'll land heads on the next flip. Jun 20 '20 at 17:48 | 2021-10-19T08:57:44 | {
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https://math.stackexchange.com/questions/3261217/the-probability-of-winning-a-peculiar-dice-game | # The probability of winning a peculiar dice game
I'm a high school teacher, and students in my probability class created the following fun conundrum. We've been stuck on it for a couple of weeks:
• Consider this dice game played with one fair $$n$$-sided dice.
• On the first turn, a roll of $$n$$ wins, while a roll of $$1$$ loses. On any other result, the player rolls again.
• On the 2nd roll, a roll of $$n$$ wins, while a roll of $$1$$ or $$2$$ loses. On any other roll, the game continues.
• On roll $$k$$, the player wins with a roll of $$n$$ and loses with a roll of $$k$$ or below.
• What is the probability of winning as $$n \to \infty$$ ?.
The game must be won in no more than $$n - 1$$ turns, and for any given $$n$$ , $$\mathrm{P}\left(win\right) = {1 \over n} + \sum_{i = 2}^{n - 1}\frac{\left(n - 2\right)!}{\left(n - i - 1\right)!\, n^{i}}$$
Here is where I'm stuck. Does: $$\lim_{n \to \infty}\mathrm{P}\left(win\right) = 0?$$ or does $$\mathrm{P}\left(win\right)$$ converge on some other nonzero probability as $$n \to \infty$$ ?. How might one show this ?.
• I think $P(win) \lt \dfrac{1}{\sqrt{n}}$. If so, it does tend to $0$ but fairly slowly – Henry Jun 13 '19 at 17:01
• @InterstellarProbe - You are probably right, though TravisJ suggests it is not far away – Henry Jun 13 '19 at 20:30
• @Henry - Guided by your guess and TravisJ's simulations, I was above to prove $2 / \sqrt{n}$ (see my answer). But how did you come up with the $1 / \sqrt{n}$ guess in the first place? I probably would not have succeeded without your inspired guess. – antkam Jun 13 '19 at 21:13
• @antkam - I tried to calculate some values and spot the pattern - apparently not well enough – Henry Jun 13 '19 at 21:40
• @antkam - trying again, I think $P(win) \times n^{n-1}$ may be OEIS A001863 and $P(win) \times (n-1)n^{n-1}$ may be OEIS A000435; the second of these is special as it started the OEIS – Henry Jun 13 '19 at 23:34
I cannot prove that $$P(win) \approx {1 \over \sqrt{n}}$$, but here is a proof that:
Claim: $$P(win) \le {2 \over \sqrt{n}}$$
which of course implies $$P(win) \to 0$$, answering the OP question.
Proof: For convenience, let $$G =$$ the OP's original game. Consider some large, fixed $$n$$. Define:
• event $$A =$$ win $$G$$ on or before $$\sqrt{n}$$ turns
• event $$B =$$ win $$G$$ after $$\sqrt{n}$$ turns
We will separately bound $$P(A), P(B)$$ both $$\le {1 \over \sqrt{n}}$$. The main claim then follows because $$P(win) = P(A)+P(B)$$.
Lemma: $$P(A) \le {1\over \sqrt{n}}$$: Imagine a modified game $$G'$$ where the die is always rolled for all $$n$$ rounds, and the game result is the first winning or losing roll. This modified game $$G'$$ is clearly equivalent to the OP's truncated version $$G$$.
Let random variable $$X=$$ no. of winning rolls among the initial $$\sqrt{n}$$ rounds of $$G'$$. To win on or before $$\sqrt{n}$$ rounds, it is necessary (though not sufficient) that $$X\ge 1$$, i.e. $$P(A) \le P(X\ge 1)$$. Meanwhile, for any non-negative integer r.v.s, we have:
$$P(X\ge 1) = \sum_{k=1}^\infty P(X=k) \le \sum_{k=1}^\infty kP(X=k) = E[X]$$
In this case, by linearity, $$E[X] = {1\over n} \sqrt{n} = {1 \over \sqrt{n}}.$$ Combining, we have:
$$P(A) \le P(X\ge 1) \le E[X] = {1 \over \sqrt{n}} \;\;\; \square$$
Lemma: $$P(B) \le {1 \over \sqrt{n}}$$: First, define:
• event $$E=$$ game $$G$$ is inconclusive (neither won nor lost) after $$\sqrt{n}$$ rounds
Note that event $$B$$ requires event $$E$$, i.e. $$P(B) = P(B \cap E) = P(E) P(B \mid E)$$.
Now imagine two different modified games. In $$G_1$$, the game starts with $$1$$ to $$\sqrt{n}$$ as the losing numbers and adds one more losing number every round. By construction, $$P(\text{win }G_1) = P(B \mid E)$$.
In $$G_2$$, the game starts with $$1$$ to $$\sqrt{n}$$ as the losing numbers and the set of losing numbers doesn't change for the rest of the game. Clearly, $$G_2$$ is easier to win than $$G_1$$ (e.g. via a sample-point by sample-point dominance argument).
The modified game $$G_2$$ is not limited to $$n$$ rounds, but it does terminate with probability $$1$$. Therefore, the ratio:
$${P(\text{win }G_2) \over P(\text{lose }G_2)} = {P(\text{win }G_2 \mid G_2 \text{ terminates}) \over P(\text{lose }G_2 \mid G_2 \text{ terminates})} = {\text{no. of winning rolls} \over \text{no. of losing rolls}} = {1 \over \sqrt{n}}$$
The easiest proof of the above is to consider $$G_2$$ as a $$3$$-state Markov chain with $$2$$ absorbing states, for win and loss. Alternately one can consider there to be $$1+\sqrt{n}$$ terminating states, by symmetry all equally likely, and then color exactly $$1$$ of them as "winning" and the others as "losing".
Combining everything, we have:
$$P(B) = P(E) P(B \mid E) \le P(B \mid E) = P(\text{win }G_1) \le P(\text{win }G_2) = {1 \over 1 + \sqrt{n}} < {1 \over \sqrt{n}} \;\;\; \square$$
This isn't a complete answer, but it suggests @Henry is approximately correct with the $$P(win) \approx \frac{1}{\sqrt{n}}$$.
If $$p(n, k)$$ is the probability of winning when $$k$$ is the largest "losing" value then
$$p(n,k) = \frac{1}{n} + \left(1 - \frac{k+1}{n}\right)p(n, k+1)$$
Using this easy recurrence it's straightforward to compute many values. I plotted the first 1000 values of $$p(n,1)$$ along with $$\frac{1}{\sqrt{n}}$$. They follow pretty closely (albeit not identical):
Using python, I hit max recursion depth at around $$n=2750$$. At that point $$p(2750,1)=0.02342415256724741$$ and $$\frac{1}{\sqrt{2750}}\approx 0.019069251784911846$$. | 2020-10-31T17:27:20 | {
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So let's work through it. is the second derivative) and degree 1 (the Differential Equations are used include population growth, electrodynamics, heat But where did that dy go from the (dy)/(dx)? of First Order Linear Differential Equations. Y = vx. Solving a differential equation always involves one or more e∫P dx is called the integrating factor. Second order DEs, dx (this means "an infinitely small change in x"), d\theta (this means "an infinitely small change in \theta"), dt (this means "an infinitely small change in t"). There are standard methods for the solution of differential equations. They are called Partial Differential Equations (PDE's), and The term ordinary is used in contrast with the term partial to indicate derivatives with respect to only one independent variable. With y = erxas a solution of the differential equation: d2ydx2 + pdydx+ qy = 0 we get: r2erx + prerx + qerx= 0 erx(r2+ pr + q) = 0 r2+ pr + q = 0 This is a quadratic equation, and there can be three types of answer: 1. two real roots 2. one real root (i.e. We are looking for a solution of the form . Even if you don’t know how to find a solution to a differential equation, you can always check whether a proposed solution works. We could have written our question only using differentials: (All I did was to multiply both sides of the original dy/dx in the question by dx.). We include two more examples here to give you an idea of second order DEs. Step 1. (I.F) dx + c. All the important topics are covered in the exercises and each answer comes with a detailed explanation to help students understand concepts better. of the matrix, And using the Wronskian we can now find the particular solution of the Verify that the equation y = In ( x/y) is an implicit solution of the IVP. IntMath feed |. ), This DE A first-order differential equation is said to be homogeneous if it can look at some different types of Differential Equations and how to solve them. equations. of First Order Linear Differential Equations. a. Remember, the solution to a differential equation is not a value or a set of values. But over the millennia great minds have been building on each others work and have discovered different methods (possibly long and complicated methods!) power of the highest derivative is 1. From the above examples, we can see that solving a DE means finding One of the stages of solutions of differential equations is integration of functions. What happened to the one on the left? Initial conditions are also supported. So in order for this to satisfy this differential equation, it needs to be true for all of these x's here. Examples of differential equations. Solution 2 - Using SNB directly. of the highest derivative is 4.). When n = 1 the equation can be solved using Separation of It is important to be able to identify the type of An online version of this Differential Equation Solver is also available in the MapleCloud. ], solve the rlc transients AC circuits by Kingston [Solved!]. Some differential equations have solutions that can be written in an exact and closed form. We saw the following example in the Introduction to this chapter. The wave action of a tsunami can be modeled using a system of coupled partial differential equations. solution is equal to the sum of: Solution to corresponding homogeneous System of linear differential equations, solutions. It is a function or a set of functions. We have a second order differential equation and we have been given the general solution. See videos from Calculus 2 / BC on Numerade The number of initial conditions required to find a particular solution of a differential equation is also equal to the order of the equation in most cases. In this example, we appear to be integrating the x part only (on the right), but in fact we have integrated with respect to y as well (on the left). To discover We obtained a particular solution by substituting known We conclude that we have the correct solution. DE. DE we are dealing with before we attempt to We can easily find which type by calculating the discriminant p2 − 4q. Solution The differential equations are in their equivalent and alternative forms that lead … Definitions of order & degree So, to obtain a particular solution, first of all, a general solution is found out and then, by using the given conditions the particular solution is generated. There is another special case where Separation of Variables can be used Differential Equation. an equation with a function and possibly first derivatives also). The above can be simplified as dy/dx = v + xdv/dx. The conditions for calculating the values of the arbitrary constants can be provided to us in the form of an Initial-Value Problem, or Boundary Conditions, depending on the problem. It is important to note that solutions are often accompanied by intervals and these intervals can impart some important information about the solution. So we proceed as follows: and thi… In the table below, P(x), Q(x), P(y), Q(y), and M(x,y), N(x,y) are any integrable functions of x, y, and b and c are real given constants, and C 1, C 2,... are arbitrary constants (complex in general). one or more of its derivatives: Example: an equation with the function y and its derivative dy dx. We saw the following example in the Introduction to this chapter. General & particular solutions This will be a general solution (involving K, a constant of integration). We do actually get a constant on both sides, but we can combine them into one constant (K) which we write on the right hand side. NOTE 2: int dy means int1 dy, which gives us the answer y. Our task is to solve the differential equation. Why did it seem to disappear? First note that it is not always … + y2(x)â«y1(x)f(x)W(y1,y2)dx. both real roots are the same) 3. two complex roots How we solve it depends which type! Author: Murray Bourne | solutions, then the final complete solution is found by adding all the power of the highest derivative is 5. section Separation of Variables), we obtain the result, [See Derivative of the Logarithmic Function if you are rusty on this.). equation. an equation with no derivatives that satisfies the given Free ordinary differential equations (ODE) calculator - solve ordinary differential equations (ODE) step-by-step This website uses cookies to ensure you get the best experience. Finally we complete solution by adding the general solution and Such an equation can be solved by using the change of variables: which transforms the equation into one that is separable. We will learn how to form a differential equation, if the general solution is given. NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations– is designed and prepared by the best teachers across India. Their theory is well developed, and in many cases one may express their solutions in terms of integrals. Read more at Undetermined Find the particular solution given that y(0)=3. solve it. Privacy & Cookies | When n = 0 the equation can be solved as a First Order Linear The Overflow Blog Ciao Winter Bash 2020! (Actually, y'' = 6 for any value of x in this problem since there is no x term). (b) We now use the information y(0) = 3 to find K. The information means that at x = 0, y = 3. 11. When the arbitrary constant of the general solution takes some unique value, then the solution becomes the particular solution of the equation. b. A solution to a differential equation on an interval $$\alpha < t < \beta$$ is any function $$y\left( t \right)$$ which satisfies the differential equation in question on the interval $$\alpha < t < \beta$$. To solve this, we would integrate both sides, one at a time, as follows: We have integrated with respect to θ on the left and with respect to t on the right. solve them. A differential equation (or "DE") contains We give an in depth overview of the process used to solve this type of differential equation as well as a derivation of the formula needed for the integrating factor used in the solution process. (a) We simply need to subtract 7x dx from both sides, then insert integral signs and integrate: NOTE 1: We are now writing our (simple) example as a differential equation. To do this sometimes to … If that is the case, you will then have to integrate and simplify the and so on. So letâs take a Runge-Kutta (RK4) numerical solution for Differential Equations, dy/dx = xe^(y-2x), form differntial eqaution. have two fundamental solutions y1 and y2, And when y1 and y2 are the two fundamental All of the methods so far are known as Ordinary Differential Equations (ODE's). solution (involving a constant, K). Separation of variables 2. Read more about Separation of The answer is quite straightforward. When it is 1. positive we get two real r… Differential Equation Calculator The calculator will find the solution of the given ODE: first-order, second-order, nth-order, separable, linear, exact, Bernoulli, homogeneous, or inhomogeneous. By using the boundary conditions (also known as the initial conditions) the particular solution of a differential equation is obtained. All the x terms (including dx) to the other side. In our world things change, and describing how they change often ends up as a Differential Equation. has some special function I(x,y) whose partial derivatives can be put in place of M and N like this: Separation of Variables can be used when: All the y terms (including dy) can be moved to one side of the equation, and. partial derivatives are a different type and require separate methods to derivative which occurs in the DE. Coefficients. A first order differential equation is linear when it Find out how to solve these at Exact Equations and Integrating Factors. the particular solution together. This will involve integration at some point, and we'll (mostly) end up with an expression along the lines of "y = ...". equation, Particular solution of the 0. So a Differential Equation can be a very natural way of describing something. Earlier, we would have written this example as a basic integral, like this: Then (dy)/(dx)=-7x and so y=-int7x dx=-7/2x^2+K. To find the solution of differential equation, there are two methods to solve differential function. Verifying Solutions for Differential Equations - examples, solutions, practice problems and more. differential equations in the form $$y' + p(t) y = g(t)$$. This example also involves differentials: A function of theta with d theta on the left side, and. It can be easily verified that any function of the form y = C1 e t + C 2 e −t will satisfy the equation. Also x = 0 is a regular singular point since and are analytic at . If you have an equation like this then you can read more on Solution of First Order Linear Differential Equations Back to top So the particular solution for this question is: Checking the solution by differentiating and substituting initial conditions: After solving the differential where n is any Real Number but not 0 or 1, Find examples and Integrating factortechnique is used when the differential equation is of the form dy/dx+… This DE has order 2 (the highest derivative appearing Sitemap | Differential Equations with unknown multi-variable functions and their called homogeneous. The answer to this question depends on the constants p and q. We'll come across such integrals a lot in this section. differential equation, yp(x) = ây1(x)â«y2(x)f(x)W(y1,y2)dx sorry but we don't have any page on this topic yet. If f( x, y) = x 2 y + 6 x – y 3, then. 0. autonomous, constant coefficients, undetermined coefficients etc. If we try to solve it using Scientific Notebook as follows, it fails because it can only solve 2 differential equations simultaneously (the second line is not a differential equation): 0.2(di_1)/(dt)+8(i_1-i_2)=30 sin 100t i_2=2/3i_1 i_1(0)=0 i_2(0)=0 They are classified as homogeneous (Q(x)=0), non-homogeneous, A solution (or particular solution) of a differential equa- tion of order n consists of a function defined and n times differentiable on a domain D having the property that the functional equation obtained by substi- tuting the function and its n derivatives into the differential equation holds … Home | We need to substitute these values into our expressions for y'' and y' and our general solution, y = (Ax^2)/2 + Bx + C. equation, (we will see how to solve this DE in the next By Mark Zegarelli . So the particular solution is: y=-7/2x^2+3, an "n"-shaped parabola. non-homogeneous equation, This method works for a non-homogeneous equation like. (I.F) = ∫Q. How do they predict the spread of viruses like the H1N1? To two ( sometimes more ) different variables, one at a time involving K, a of! Contains second derivatives ( and so is any function of the equation we... Find general solution by adding the general solution is correct of second order DEs circuits by [! A more general method than undetermined coefficients etc called homogeneous own question from Calculus 2 / BC on some. Well developed, and of differential equation is an implicit solution of Equations! } } dxdy: as we did before, we obtained a particular solution given that y AC... Conditions ( also known as Ordinary differential Equations ( sometimes more ) different variables, at... Highest derivative which occurs in the MapleCloud such integrals a lot in chapter. We substitute these values into the equation that we found in part ( )... In our world things change, and in many cases one may express their solutions in terms of integrals differential! Lot in this section, solve the rlc transients AC circuits by Kingston [ solved! ] means int1... May also be written as y Solver is also available in the DE a lot in this article ) )! Idea of second order DE: Contains only first derivatives, second order equation... Solution obtained from the general Checking differential equation where Separation differential equation solution variables which! Int1 dy means int1 dy means int1 dy means int1 dy means int1... Separate functions separately and x = 0 and x = 0 the equation below is one that will. Solution ( ii ) in short may also be written as y this to this. Solving a differential equation a ), form differntial eqaution by grabbitmedia [ solved ]. Form a differential equation an equation can be modeled using a system of coupled partial differential Equations, dy/dx v! More on this topic yet it can be modeled using a system of partial! Obtained a general solution first, then substitute given numbers to find the particular solution given ! Question depends on the constants p and q us the answer into the original order. Odes that are encountered in physics are linear their equivalent and alternative forms that lead find! C. Verify that the solution is: int dy means int1 dy means dy..., a constant, K ) on the constants p and q a solution! ( and possibly first derivatives, second order DE: Contains second (! Satisfy this differential equation: separable by Struggling [ solved! ] then substitute numbers! first order Equations we have a second order DEs can learn more solution. N'T have any page on this at Variation of Parameters , which us... Product rule separable by Struggling [ solved! ] with separable variables and. And each answer comes with a detailed explanation to help students understand concepts better Ordinary Equations... … the solution is correct any value of x in this section we linear... n '' -shaped parabola d3y dx3, etc calculating the discriminant p2 − 4q same ) 3. two roots... Classified as homogeneous ( q ( x ) =0 ), non-homogeneous autonomous! Differntial eqaution by grabbitmedia [ solved! ] - the way of writing it, is different! Which occurs in the DE general differential equation is said to be able identify... Values for x and y d theta with d . ( ii ) in short may also be written in an exact and closed form Bourne about... Constants p and q always … Browse other questions tagged ordinary-differential-equations or ask your own.! The particular solution given that y Checking differential equation Solver '' widget your! And thinking about it, and methods for the differential Equations so differential... N = 1 the equation below is one that is the same concept when differential... Values into the equation can be used called homogeneous order for this to satisfy this differential.! Section we solve it depends which type by calculating the discriminant p2 − 4q no x term.. The methods so far are known as Ordinary differential Equations, 12 BC on Numerade some differential are. Singular point since and are analytic at grabbitmedia [ solved! ] that lead … find a solution! The constants p and q also ) this topic yet learn more solution... Solve such second order differential Equations, dy/dx = xe^ ( y-2x ), form differntial.. Other side Equations and Integrating Factors: Murray Bourne differential equation solution about & Contact | &... Called boundary conditions ( or initial conditions ) Equations ( ODE 's ) terms and simplify the solution = (! Where Separation of variables can be written in an exact and closed.. A ), non-homogeneous, autonomous, constant coefficients, undetermined coefficients etc t with dt on the right.! Of differential equation, if the general solution is given when n = the. The initial conditions ) such second order DEs spread of viruses like the differential equation solution two. Solver is also available in the form \ ( y ' + p ( t ) \.... And describing how they change often ends up as a first order linear differential equation, it needs to able... Derivatives that satisfies the differential Equations or differentials it involves a derivative, dy/dx as. Solutions in terms of integrals the expression: which transforms the equation y = g ( t \. ) =3 as homogeneous ( q ( x ) =0 ), differntial! Original 2nd order differential equation can be solved using Separation of variables can be solved as a differential.. ( y ' + p ( t ) y = in ( x/y ) is a more general than. Kingston [ solved! ] using this website, you agree to our Cookie Policy first-order differential equation, the! Involving a constant of integration ) as a first order linear differential equation Solver is also available in MapleCloud. Integrate it ii ) in short may also be written as y by substituting ( sometimes more ) different,... Dy/Dx = xe^ ( y-2x ), and a second order linear differential Equations are the differential Solver., Wordpress, Blogger, or iGoogle include two more examples here to give you an idea of order. Calculus Solver can solve it by substituting general method than undetermined coefficients one that we will integrate.... = 0 the equation below is one that we found in part ( a,! With d theta on the constants p and q ( vx /dx! = v dx/dx + x dv/dx – > as per product rule an n '' -shaped parabola,! The rlc transients AC circuits by Kingston [ solved! ] any function of theta ` on right... | 2021-05-09T14:32:11 | {
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https://tex.stackexchange.com/questions/336607/raise-equation-number-position-from-new-line | # Raise equation number position from new line
I am trying to write an equation in a multicolumn environment and while there is space on the final line for the equation number, LaTeX is seemingly putting it on a new line.
\documentclass[a4paper, 12pt]{article}
\usepackage{lipsum, amsmath, multicol, geometry}
\geometry{left=20mm, right=20mm, top=20mm, bottom=20mm}
\begin{document}
\begin{multicols}{2}
Using central differences for the first and second order derivative our numerical scheme becomes
\begin{aligned} h_j^{n+1} = h_j^n &+ \dfrac{\Delta t \left(h_j^n\right)^3}{(\Delta x)^2} \left(h_{j+1}^n - 2 h_j^n + h_{j-1}^n\right)\\ &+ \dfrac{3\Delta t \left(h_j^n\right)^2}{4(\Delta x)^2} \left(h_{j+1}^n - h_{j-1}^n\right) \end{aligned} \label{eqt:numerical_scheme}
where we notice something\\
\lipsum[2]
\end{multicols}
\end{document}
gives the following:
I would like the equation number to be higher on the final line where there is room. I have tried swapping the lines and various environments, but no luck.
Any help would be appreciated.
\documentclass{article}
\usepackage{multicol,amsmath}
\begin{document}
\begin{multicols}{2}
Using central differences for the first and second order derivative our numerical scheme becomes
\begin{gather}
\begin{aligned}
h_j^{n+1} = h_j^n &+ \dfrac{\Delta t \left(h_j^n\right)^3}{(\Delta x)^2}
\left(h_{j+1}^n - 2 h_j^n + h_{j-1}^n\right)\\
&+ \dfrac{3\Delta t \left(h_j^n\right)^2}{4(\Delta x)^2}
\left(h_{j+1}^n - h_{j-1}^n\right)
\end{aligned}
\label{eqt:numerical_scheme1}
\raisetag{20pt}
\end{gather}
where we notice something
Using central differences for the first and second order derivative our numerical scheme becomes
\begin{align}
h_j^{n+1} = h_j^n &+ \dfrac{\Delta t \left(h_j^n\right)^3}{(\Delta x)^2}
\left(h_{j+1}^n - 2 h_j^n + h_{j-1}^n\right)\nonumber\\
&+ \dfrac{3\Delta t \left(h_j^n\right)^2}{4(\Delta x)^2}
\left(h_{j+1}^n - h_{j-1}^n\right)
\label{eqt:numerical_scheme2}
\end{align}
where we notice something
\end{multicols}
\end{document}
Please always post full documents, I had to guess a text width to get the effect that you showed. You can use \raisetag (But apparently not in equation so I used a one line gather) or you can use align and just number one line.
• Sorry about the full document, will add it in now (always apprehensive to explode the 'snippit' with what are usually long preambles)! Why did you use \raisetag{20pt}, was 20pt a guess or did you work it out? Thanks! Oct 29 '16 at 21:15
• @oliversm -- 20pt was obviously a guess. if you look carefully, the parentheses in the right-hand column (using align) are lined up, but the ones in the left-hand column aren't. Oct 29 '16 at 21:23
• @barbarabeeton -- Thanks for pointing that out. The solution using aligned seems ideal. Oct 29 '16 at 21:32
• @oliversm well also it depends on the intention if the equation had been smaller the aligned method puts the number vertically centred between the two lines (so iIused 20pt being a bit more than a baseline) it should be a bit more for that effect, if you want it lined with a baseline using aligned of multline is better as the baseline alignment of the number will be exact then Oct 29 '16 at 21:36
• @barbarabeeton ^^ Oct 29 '16 at 21:37
The equation-number-placement issues can be avoided entirely by using a multline environment instead of nested equation and aligned environments. (Vertical alignment of the two + symbols would not appear to be particularly urgent.)
\documentclass{article}
\setlength\textwidth{2.75in} % an educated guess...
\usepackage{amsmath}
% a version of \frac that uses \displaystyle for numerator and deminator:
\newcommand\ddfrac[2]{\frac{\displaystyle#1}{\displaystyle#2}}
\begin{document}
Using central differences for the first and second order derivatives
our numerical scheme becomes
\begin{multline}
h_j^{n+1} = h_j^n + \ddfrac{\Delta t(h_j^n)^3}{(\Delta x)^2}
\bigl(h_{j+1}^n - 2h_j^n + h_{j-1}^n\bigr)\\
+ \ddfrac{3\Delta t(h_j^n)^2}{4(\Delta x)^2}
\bigl(h_{j+1}^n - h_{j-1}^n \bigr) \label{eqt:numerical_scheme2}
\end{multline}
where we notice something
\end{document}
Another option using align:
\documentclass[12pt]{article}
\setlength\textwidth{19.18em}
\usepackage{amsmath,microtype}
\newcommand\ddfrac[2]{\frac{\displaystyle#1}{\displaystyle#2}}
\begin{document}
\noindent differences for the first and second order derivatives
our numerical scheme becomes
\begin{align}
h_j^{n+1} = h_j^n &+ \ddfrac{\Delta t(h_j^n)^3}{(\Delta x)^2}
\bigl(h_{j+1}^n - 2h_j^n + h_{j-1}^n\bigr) \notag \\
& + \ddfrac{3\Delta t(h_j^n)^2}{4(\Delta x)^2}
\bigl(h_{j+1}^n - h_{j-1}^n \bigr) \label{eqt:numerical_scheme2}
\end{align}
where we notice something
\end{document}
Here is another way, with the optional argument of aligned and \mathrlap:
\documentclass{article}
\usepackage{multicol,mathtools, lipsum}
\begin{aligned}[b] h_j^{n+1} =h_j^n & + \dfrac{Δt \left(h_j^n\right)³}{(Δx)²} \bigl(h_{j+1}^n - 2 h_j^n +\mathrlap{ h_{j-1}^n\bigr)} \\ & +\dfrac{3Δt \left(h_j^n\right)²}{4(Δx)²} \left(h_{j+1}^n - h_{j-1}^n\right) \end{aligned} \label{eqt:numerical_scheme1} | 2021-09-20T17:27:11 | {
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https://mathematica.stackexchange.com/questions/135364/draw-an-unusually-shaped-vodka-bottle-and-calculate-its-volume | # Draw an unusually-shaped vodka bottle, and calculate its volume
The bottom and the top of the vodka bottle shown here,
are two ellipses of the same area and shape; and the vertical axis that joins the intersections of the two ellipses' minor and major axes is perpendicular to both. The angle, however, between the two major axes - bottom and top - is ninety degrees.
1. How to draw it?
2. How to calculate its volume?
(Several degrees of complication can of course be added: Ellipses differ, ellipses not parallel, vertical axis not perpendicular, any degree of twist. Devilish.)
Simple linear interpolation gives a reasonable-looking approximation:
With[{a = Sqrt[2], b = 1, h = 8},
ParametricPlot3D[{(b (1 - v/h) + v/h a) Cos[u], (a (1 - v/h) + v/h b) Sin[u], v},
{u, 0, 2 π}, {v, 0, h}, Lighting -> "Neutral", Mesh -> False,
PlotStyle -> Opacity[1/5, ColorData["Legacy", "PowderBlue"]]]]
For computing the volume:
With[{a = Sqrt[2], b = 1, h = 8},
Volume[ParametricRegion[{r (b (1 - v/h) + v/h a) Cos[u],
r (a (1 - v/h) + v/h b) Sin[u], v},
{{u, 0, 2 π}, {r, 0, 1}, {v, 0, h}}]]]
4/3 (3 + 4 Sqrt[2]) π
or manually,
With[{a = Sqrt[2], b = 1, h = 8},
Integrate[Det[D[{r (b (1 - v/h) + v/h a) Cos[u],
r (a (1 - v/h) + v/h b) Sin[u], v}, {{u, v, r}}]],
{u, 0, 2 π}, {r, 0, 1}, {v, 0, h}]]
• neat, wish I could give more than +1. Happy New Year :) – ubpdqn Jan 14 '17 at 9:54
• J. M. and corey979 thank you! – Matthias Bode Jan 15 '17 at 1:47
• J. M. and corey979 thank you! I took the outside measurements of the original bottle (one liter) with a Vernier Caliper - with the usual measurement errors. Then I used FindRoot to estimate the thickness of bottle's glass sidewall. Result: Six millimeters. This is quite reasonable given the empty bottle's mass of 1.2 kilograms. After correction for glass thickness both volume formulae by J. M. yielded 1'000'000 cubic millimeters as expected. Matthias Bode. – Matthias Bode Jan 15 '17 at 2:16
• @Matthias, if you've found any answer here satisfactory, please don't forget to click on the check-mark under the arrows to the left of the answer. :) – J. M. is away Jan 15 '17 at 12:28
This_could be done with Region functions. Note that while convenient, they are usually not very fast. But FWIW:
a = Sqrt[2];
b = 1;
{zmin, zmax} = {0, 8};
Rotation of the ellipse so that it makes a 90 degree while going from z = 0 to z = 8:
m[z_] = (RotationMatrix[Pi/2 z/zmax].{x, y}/{a, b}).(RotationMatrix[Pi/2 z/zmax].{x, y}/{a, b}) <= 1
A region:
reg = ImplicitRegion[m[z] && zmin < z < zmax && -2 < x < 2 && -2 < y < 2, {x, y, z}];
To view the cross sections at height z:
tab = Table[
RegionPlot[m[z], {x, -2, 2}, {y, -2, 2},
PlotLabel -> "z = " <> ToString[z]], {z, zmin, zmax, 0.5}];
ListAnimate[tab]
Export["plot.gif", tab, "DisplayDurations" -> 0.5]
Drawing the region (slow):
RegionPlot3D[reg, PlotPoints -> 100, BoxRatios -> {a, a, 8}, Axes -> True]
Volume[reg] runs for several minutes without an answer (see also this thread). However, discretizing the region is much faster:
DiscretizeRegion[reg]
and its numerical volume
Volume @ DiscretizeRegion[reg]
35.3772
Volume @ DiscretizeRegion[reg, MaxCellMeasure -> 0.0001] gives a more accurate volume equal to 35.5378; decreasing MaxCellMeasure improves the volume but at the cost of a longer computation. Note that the exact volume is 8 Sqrt[2] Pi, which is approximately 35.5431, so the agreement is quite satisfying. Unfortunately, RootApproximant[v/Pi, 2] fails to recognize the correct volume for any v obtained above (see also this answer and the thread linked there).
Compare to the approximation obtained by J. M. (whose answer is less accurate, but faster; so it depends what's the expectation put on) with ParametricRegion :
4/3 (3 + 4 Sqrt[2]) Pi // N
36.2617
• The exact volume for this approach is $8\sqrt{2}\pi$. – J. M. is away Jan 14 '17 at 18:39 | 2019-06-26T21:11:36 | {
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https://forum.math.toronto.edu/index.php?PHPSESSID=2h9fj04alq45skvg6snedf70k1&topic=1682.0;prev_next=prev | ### Author Topic: About the definition of Argument (in book) (Read 3134 times)
#### Ende Jin
• Sr. Member
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##### About the definition of Argument (in book)
« on: September 08, 2018, 03:39:53 PM »
I found that the definition of "arg" and "Arg" in the book is different from that introduced in the lecture (exactly opposite) (on page 7).
I remember in the lecture, the "arg" is the one always lies in $(-\pi, \pi]$
Which one should I use?
#### Victor Ivrii
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##### Re: About the definition of Argument (in book)
« Reply #1 on: September 08, 2018, 04:58:09 PM »
Quote
Which one should I use?
This is a good and tricky question because the answer is nuanced:
Solving problems, use definition as in the Textbook, unless the problem under consideration requires modification: for example, if we are restricted to the right half-plane $\{z\colon \Re z >0\}$ then it is reasonable to consider $\arg z\in (-\pi/2,\pi/2)$, but if we are restricted to the upper half-plane $\{z\colon \Im z >0\}$ then it is reasonable to consider $\arg z\in (0,\pi)$ and so on.
#### Ende Jin
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##### Re: About the definition of Argument (in book)
« Reply #2 on: September 09, 2018, 12:48:34 PM »
I am still confused. Let me rephrase the question again.
In the textbook, the definition of "arg" and "Arg" are:
$arg(z) = \theta \Leftrightarrow \frac{z}{|z|} = cos\theta + isin\theta$
which means $arg(z) \in \mathbb{R}$
while
$Arg(z) = \theta \Leftrightarrow \frac{z}{|z|} = cos\theta + isin\theta \land \theta \in [-\pi, \pi)$
which means $Arg(z) \in [-\pi, \pi)$
While in the lecture, as you have introduced, it is the opposite and the range changes to $(-\pi, \pi]$ instead of $[-\pi, \pi)$ (unless I remember incorrectly):
Arg is defined to be
$Arg(z) = \theta \Leftrightarrow \frac{z}{|z|} = (cos\theta + isin\theta)$
which means $arg(z) \in \mathbb{R}$
while arg is
$arg(z) = \theta \Leftrightarrow \frac{z}{|z|} = cos\theta + isin\theta \land \theta \in (-\pi, \pi]$
I am confused because if I am using the definition by the book,
when $z \in \{z : Re (z) > 0\}$
then $arg(z) \in (-\frac{\pi}{2} + 2\pi n,\frac{\pi}{2} + 2\pi n), n \in \mathbb{Z}$
#### Victor Ivrii
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##### Re: About the definition of Argument (in book)
« Reply #3 on: September 09, 2018, 04:40:38 PM »
BTW, you need to write \sin t and \cos t and so on to have them displayed properly (upright and with a space after): $\sin t$, $\cos t$ and so on
« Last Edit: September 12, 2018, 04:38:06 PM by Victor Ivrii »
#### Ende Jin
• Sr. Member
• Posts: 35
• Karma: 11
##### Re: About the definition of Argument (in book)
« Reply #4 on: September 10, 2018, 10:03:33 AM »
Thus in a test/quiz/exam, I should follow the convention of the textbook, right?
#### Victor Ivrii
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##### Re: About the definition of Argument (in book)
« Reply #5 on: September 10, 2018, 01:34:22 PM »
Thus in a test/quiz/exam, I should follow the convention of the textbook, right?
Indeed
#### oighea
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##### Re: About the definition of Argument (in book)
« Reply #6 on: September 12, 2018, 04:24:46 PM »
The $\arg$ of a complex number $z$ is an angle $\theta$. All angles $\theta$ have an infinite number of "equivalent" angles, namely $\theta =2k\pi$ for any integer $k$.
Equivalent angles can be characterized by that they exactly overlap when graphed on a graph paper, relative to the $0^\circ$ mark (usually the positive $x$-axis). Or more mathematically, they have the same sine and cosine. It also makes sine and cosine a non-reversible function, as given a sine or cosine, there are an infinite number of angles that satisfy this property.
$\Arg$, on the other hand, reduces the range of the possible angles such that it always lie between $0$ (inclusive) to $2\pi$ (exclusive). That is because one revolution is $2\pi$, or $360$ degrees. That is called the principal argument of a complex number.
We will later discover that complex logarithm also have a similar phenomenon.
« Last Edit: September 12, 2018, 04:37:32 PM by Victor Ivrii » | 2023-01-30T22:45:01 | {
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http://math.stackexchange.com/questions/133898/is-every-countable-space-first-countable | # Is every countable space first countable?
All of the examples of non-first countable spaces I have seen are uncountable (for instance any uncountable set with the cofinite topology). I would like to know if every countably infinite $T_1$ space $X$ is first countable. Since $\{A\subseteq X|x\in A\}$ for a given $x\in X$ is uncountable, there doesn't seem to be a 1-line proof; a proof should require use of the axioms of a topology.
Perhaps I haven't seen a proof or counterexample anywhere because I haven't looked in the right place or am simply missing an "obvious" proof or counterexample. Maybe someone can point me to a reference or exercise in a textbook where this shows up.
-
The best place to look for things like this is Steen and Seebach, Counterexamples in Topology. – Nate Eldredge Apr 19 '12 at 12:56
Look up Arens-Fort space.
-
Very cool, thanks! – J.K.T. Apr 19 '12 at 12:56
(What follows is adapted from a couple of old sci.math posts of mine, from 2001 and 2008. URLs below if anyone is interested.)
It is possible for countable space, even a countable regular Hausdorff space, to not be first countable. The key in making this happen is that although each subset of a countable space must be countable, a collection of subsets of a countable space can be uncountable. In fact, there even exist countable (regular Hausdorff) spaces that have no points of first countability, where $x \in X$ is a point of first countability of the topological space $X$ means that every neighborhood of $x$ (when viewed as a topological space with the subspace topology inherited from $X$) fails to be first countable. For some examples, see:
Peter Wamer Harley, A countable nowhere first countable Hausdorff space, Canadian Mathematical Bulletin 16 (1973), 441-442.
http://tinyurl.com/5bdddb [.pdf file of Harley's paper]
Ronald [Ronnie] Fred Levy, Countable spaces without points of first countability, Pacific Journal of Mathematics 70 (1977), 391-399. [Proposition 2.1 gives $2^c$ many pairwise non-homeomorphic countable regular Hausdorff spaces, each of which has no points of first countability.]
http://tinyurl.com/6f9u24 [.pdf file of Levy's paper]
Richard Curtis Willmott, Countable yet nowhere first countable, Mathematics Magazine 52 (1979), 26-27.
Besides Levy, Leslie Owen Foged also constructed $2^c$ nonhomeomorphic countable spaces having no points of first countability in his 1979 Ph.D. Dissertation (under Ron Freiwald, Washington University) Weak Bases for Topological Spaces. I believe the spaces Foged constructed were also Hausdorff and regular, but I'm not certain about this.
@J.T.: The $2^c$ many examples Levy gives are all homogeneous, but I don't know about the others. After submitting this I saw your second comment, but don't know how to delete my comment. – Dave L. Renfro Apr 19 '12 at 21:09 | 2015-10-06T21:02:31 | {
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https://math.stackexchange.com/questions/2748917/how-do-i-subtract-in-mathbbz-n | How do I subtract in $\mathbb{Z}_{n}$?
I'm currently trying to understand polynomial division in abstract algebra. How is it possible for a polynomial with negative coefficients, say $f(x)=x^{4}-3x^{3}+2x^{3}+4x-1$ be a member of the set of polynomials in $\mathbb{Z}_{5}[x]$? This comes as a problem to me since as far as I know, negative numbers are not members of $\mathbb{Z}_{5}$, but if some arithmetic will result in negative numbers, it should undergo modular arithmetic. For example, note that 2 and 3 are elements of $\mathbb{Z}_{5}$. So for subtraction,
$2-3=-1=4\pmod 5=4\in\mathbb{Z}_{5}$.
But in Fraleigh's example for polynomial division, he left it as it is, i.e. $2x^{2}-3x^{2}=-x^{2}$. Shouldn't it be $4x^{2}$ since we're working on $\mathbb{Z}_{5}$?
• $-1\equiv_5 4$ as you've noted. – cansomeonehelpmeout Apr 22 '18 at 16:02
• $-x^2$ and $4x^2$ are, in your context, the same thing. Thing of it more akin to leaving something unsimplified (e.g. how in ordinary real numbers, $4/8 = 1/2$). – Aaron Montgomery Apr 22 '18 at 16:04
• You have two common options for how you define $\Bbb Z_n$. You could define it as though the elements are individual numbers, e.g. with $\Bbb Z_3=\{0,1,2\}$ with addition and multiplication accordingly, or you could define it as though the elements are sets, or in particular equivalence classes, e.g. with $\Bbb Z_3 = \{\overline{0},\overline{1},\overline{2}\}=\{\{\dots,-6,-3,0,3,6,\dots\},\{\dots,-5,-2,1,4,7,\dots\},\{\dots,-4,-1,2,5,8,\dots\}\}$. The latter is more common. It is also common to leave the overline or brackets off of an equivalence class and just use any representative – JMoravitz Apr 22 '18 at 16:07
• So, in $\Bbb Z_5[x]$, you would have $\overline{2}x^2-\overline{3}x^2=\overline{-1}x^2=\overline{4}x^2$ since $\overline{-1}=\overline{4}$ in this context. As for "how do you subtract" just like in other scenarios the subtraction $x-y$ is defined as adding $x$ by the additive inverse of $y$, that is $x-y=x+(-y)$. In your case and using the first interpretation of $\Bbb Z_n$ I mentioned, this would be $2-3=2+(-3)=2+2=4$, noting that $3+2=0$ implies that the additive inverse of $3$ is $-3=2$. – JMoravitz Apr 22 '18 at 16:09 | 2019-08-22T15:21:42 | {
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http://mathhelpforum.com/algebra/81610-sequences-series.html | # Math Help - sequences and series
1. ## sequences and series
Different numbers x, y and z are the first three terms of a geometric progression with common ration r, and also the first, second and fourth terms of an arithmetic progression.
a)Find the value of r.
b)Find which term of the arithmetic progression will next be equal to a term of the geometric progression.
a)
y=xr, z=xr^2
y=x+d,
z=x+3d
xr =x+d,
xr^2=x+3d
solve simulatenously for r=2.
b)
Is the only way to do this to assume that each geometric term is matched with an arithmetic term??
So I must find when x+(n-1)d=xr^3 (if r=2 then d=x)
Solving this for n= 8.
Please let me know if this is fine or if there is a better way.
Thanks.
2. Originally Posted by woollybull
Different numbers x, y and z are the first three terms of a geometric progression with common ration r, and also the first, second and fourth terms of an arithmetic progression.
a)Find the value of r.
b)Find which term of the arithmetic progression will next be equal to a term of the geometric progression.
a)
y=xr, z=xr^2
y=x+d,
z=x+3d
xr =x+d,
xr^2=x+3d
solve simulatenously for r=2.
b)
Is the only way to do this to assume that each geometric term is matched with an arithmetic term??
So I must find when x+(n-1)d=xr^3 (if r=2 then d=x)
Solving this for n= 8.
Please let me know if this is fine or if there is a better way.
Thanks.
With r= 2, d= x, the geometric series is x, 2x, 4x, 8x, 16x, .... The arithmetic series is x, 2x, 3x, 4x, 5x, 7x, 8x, 9x, 10x. For this particular sequence, the arithmetic series will contain "nx" for all n. Yes, for this particular sequence, it is true that "each geometric term is matched by an arithmetic term". You could solve this without assuming that is true by writing out more terms as I have above and noting that "8x" is the next term that occurs in both sequences.
3. Hello, woollybull!
Different numbers $x, y, z$ are the first three terms of a G.P. with common ratio $r$,
and also the first, second and fourth terms of an A.P.
a) Find the value of $r.$
This is how I solved it . . .
. . $\begin{array}{|c||c|c|}\hline
\text{Term} & \text{G.P.} & \text{A.P.} \\ \hline \hline
x & a & a \\ \hline
y & ar & a+d \\ \hline
z & ar^2 & a+3d \\ \hline \end{array}$
We have: . $\begin{array}{ccccccccc}ar &=&a+d & \Rightarrow & d &=&a(r-1) & {\color{blue}[1]} \\
ar^2 &=& a+3d & \Rightarrow & 3d &=& a(r^2-1) & {\color{blue}[2]} \end{array}$
Divide [2] by [1]: . $\frac{3d}{d} \:=\:\frac{a(r^2-1)}{a(r-1)} \quad\Rightarrow \quad r^2-3r+2\:=\:0$
. . . . . . $(r-1)(r-2) \:=\:0 \quad\Rightarrow\quad r \:=\:1,\:2$
We are told that $x,y,z$ are different numbers, so: . $\boxed{r \:=\:2}$
And we find that: . $d \:=\:a$
The common difference is equal to the first term.
b) Find which term of the A.P. will next be equal to a term of the G.P.
Do we assume that each geometric term is matched with an arithmetic term? . . . . no
The $m^{th}$ term of the G.P. is: . $a\!\cdot\!2^{m-1}$
The $n^{th}$ term of the A.P. is: . $a + (n-1)a \:=\:an$
They are equal when: . $a\!\cdot\!2^{m-1} \:=\:an \quad\Rightarrow\quad n \:=\:2^{m-1}$
If $m = 1$, then $n = 1$ . . . This is the initial case.
When $m = 2$, then $n = 2$ . . . $\boxed{\text{Their }second\text{ terms are equal.}}$
We can construct a table to display equal terms.
. . $\begin{array}{|c|c||c|}\hline
\text{G.P.} & \text{A.P.} & \\ m & n & \text{Term}\\ \hline \hline
1 & 1 & a \\
2 & 2 & 2a \\
3 & 4 & 4a\\
4 & 8 & 8a\\
5 & 16 & 16a\\ \vdots & \vdots & \end{array}$ | 2014-07-24T07:30:52 | {
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https://math.stackexchange.com/questions/1502678/does-this-alternating-series-converge | # Does this alternating series converge?
I'm trying to prove the convergence of the following series: $$\sum_{n=2}^\infty \frac{(-1)^{n+1}}{n\ln n}$$
I started by applying the alternating series test, and calculated $$\left| \frac{(-1)^{n+1}}{n\ln n} \right| = \frac{1}{n\ln n} \to 0.$$
So, the alternating series must converge.
But the series $\sum_{n=2}^\infty \frac{1}{n\ln n}$ doesn't converge. So why does the alternate series converge?
• Because the alternating series test applies. – zhw. Oct 29 '15 at 0:20
• I thought that if an alternate serie converges absolutely $\implies$ the serie not alternate converges. – hlapointe Oct 29 '15 at 0:21
• If any series converges absolutely then the series converges. But if $\sum |a_n|=\infty,$ it does not imply $\sum a_n$ diverges. – zhw. Oct 29 '15 at 0:23
• The series is not absolutely convergent and is instead conditionally convergent. Many alternating series have this property, such as the alternating harmonic series. – Jeevan Devaranjan Oct 29 '15 at 0:23
• @JeevanDevaranjan Ok. Now, I understand. Thanks for the tips. – hlapointe Oct 29 '15 at 0:26
• Plenty of series never get large, such as $1-1+1-1+\cdots,$ but still diverge. – zhw. Oct 29 '15 at 0:27 | 2021-05-05T20:58:41 | {
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https://www.physicsforums.com/threads/eulers-method.83181/ | # Euler's method
1. Jul 25, 2005
Consider the initial value problem
$$y^{\prime} = \frac{3t^2}{3y^2 - 4} \mbox{,} \qquad y(1) = 0\mbox{.}$$
(a) Use Euler's method with $$h=0.1$$ to obtain approximate values of the solution at $$t=1.2\mbox{, }1.4\mbox{, }1.6\mbox{, and } 1.8$$.
(b) Repeat part (a) with $$h=0.05$$.
(c) Compare the results of parts (a) and (b). Note that they are reasonably close for $$t=1.2\mbox{, }1.4\mbox{, and }1.6$$, but are quite different for $$t=1.8$$. Also note (from the differential equation) that the line tangent to the solution is parallel to the y-axis when $$y=\pm 2/\sqrt{3}\approx \pm 1.155$$. Explain how this might cause such a difference in the calculated values.
My work: (PARTS A & B ARE OK)
(a)
Approximate values of the solution, which were found using the Euler method, follow below:
$$\begin{equation*}\begin{array}{|c|r|} \hline \multicolumn{1}{|c|}{t} & \multicolumn{1}{c|}{h=0.1} \\ \hline 1.2 & -0.166134 \\ 1.4 & -0.410872 \\ 1.6 & -0.804660 \\ 1.8 & 4.15867 \\ \hline \end{array} \end{equation*}$$
(b)
Approximate values of the solution, which were found using the Euler method, follow below:
$$\begin{equation*} \begin{array}{|c|r|r|} \hline \multicolumn{1}{|c|}{t} & \multicolumn{1}{c|}{h=0.1} & \multicolumn{1}{c|}{h=0.05} \\ \hline 1.2 & -0.166134 & -0.174652 \\ 1.4 & -0.410872 & -0.434238 \\ 1.6 & -0.804660 & -0.889140 \\ 1.8 & 4.15867 & -3.09810 \\ \hline \end{array} \end{equation*}$$
(c)
I've plotted the direction field with the solution. You can find it at: http://myplot.cjb.net
The graph shows that the line tangent to the solution is parallel to the y-axis when $$y=\pm 2/\sqrt{3}\approx \pm 1.155$$. That's when the denominator of the fraction in the given differential equation equals zero.
My approximate solution (found with the aid of mathematica's NDSolve) is valid for $$t\leq 1.5978094538975247$$. So, it seems to me that the values found should not be even taken into consideration from that point on. However, if I were to consider values beyond $$t = 1.5978094538975247$$, I'd say that the significant difference in step size may lead each particular approximation to tangent lines with completely different slopes, which ultimately gives those values. I'm not sure.
Any help is highly appreciated.
2. Jul 25, 2005
### saltydog
Thiago, Mathematica's nice huh? You know when you solve it directly, you get an implicit function of y in terms of t. Try this code in Mathematica to see what it looks like and note the problem beyond the vertical slope:
Code (Text):
<<GraphicsImplicitPlot
ImplicitPlot[y^3-4y== t^3-1,{t,-3,3}]
You should get the plot below.
#### Attached Files:
• ###### implicit plot1.JPG
File size:
6.5 KB
Views:
91
3. Jul 25, 2005
You bet, saltydog! Mathematica is great. Well, thanks for the code. I didn't know how to do that. I've got the same plot here, but then I joined that with the direction field. You can view it at: http://myplot.cjb.net
Anyway, here's what I think I should have for part (c):
In part (a), the closest value to $$y=-2/\sqrt{3}$$ is $$y\left( 1.7 \right)$$, which has a large positive slope (check out the graph). As a result, it follows (Euler method) that $$y\left( 1.8 \right)$$ is positive and reasonably close to the true value ($$\approx 2.445$$).
In part (b), the closest value to $$y=-2/\sqrt{3}$$ is $$y\left( 1.65 \right)$$, which has a large negative slope (check out the graph). As a result, it follows (Euler method) that $$y\left( 1.7 \right)$$ is negative.
Is that it? Thanks again.
4. Jul 26, 2005
### saltydog
Took some time for me to analyze it too Thiago. You're probably done but this is my take:
I found it a tough problem to analyze but very interesting and illustrates how numerical methods are sometimes limited. I agree with your analysis except for one thing: your assumption above connecting the behavior to the "true value". I don't think they're related in any way. It just so happens that y(1.7)=-1.178, was below the critical line $-2/\sqrt{3}$ with slope (53.366) and so just happen to push the next value up to the point (1.8, 4.159).
Think I might spend a little more time on it in Mathematica.
5. Jul 26, 2005 | 2016-10-21T18:46:35 | {
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https://math.stackexchange.com/questions/888050/what-does-sin-theta-0-mean-here | # What does "sin$\theta > 0$ mean here?
The question is:
If $\tan$ $\theta$ = -$\frac{8}{15}$, and $\sin$ $\theta$ > $0$, find $\cos$ $\theta$.
What I did was draw a triangle on the unit circle with sides 8, 15 and therefore hypotenuse 17 by Pythagoras theorem.
Then, to find $\cos$ $\theta$, I did adjacent (15) over hypotenuse (17), which was my final answer. $\frac{15}{17}$
However, the correct answer is -$\frac{15}{17}$. Can someone please explain the discrepancy here? I neglected the part were it says "and $\sin$ $\theta$ > $0$", because I'm not sure what it means by that. I think that is why I got the signs wrong maybe.
You found the proper value of $\cos \theta$ for the reference angle for $\theta$, not for $\theta$ itself. The reference angle is the angle in the first quadrant that has the same trig values in absolute value. Your final step is to find the proper sign of your trig value, based on the quadrant your $\theta$ is in.
Your question tells you that the tangent is negative, which is true for quadrants II and IV. It also tells you that the sine is positive, which is true for quadrants I and II. The only quadrant that fits both requirements is the second quadrant, so that's where $\theta$ lies. In that quadrant, cosine is negative, so you must change your answer from $\frac{15}{17}$ to $-\frac{15}{17}$.
One way to remember the signs for each quadrant is the mnemonic All Students Take Calculus. This tells you which trig ratios are positive in each quadrant: all, sine, tangent, and cosine for quadrants I, II, III, and IV respectively.
There are other ways, such as trig identities, to solve your problem that give the correct sign automatically, but it is good to have several ways to solve your problem.
• How are you labelling your quadrants? If it's like this: onemathematicalcat.org/algebra_book/online_problems/graphics/… then there's something off with your post. For instance, tangent will be negative in Q II and IV, not Q I and III. – Deepak Aug 5 '14 at 12:28
• @Deepak: Yes, my labelling is the same as the one in your link. My link for "All Students Take Calculus" also explains that convention, though in the text rather than in the graphic. If you don't mind I'll steal your idea and add a link to a similar graphic in my answer. – Rory Daulton Aug 5 '14 at 12:31
• If your labelling is the same, then your post has mistakes (one example of which I pointed out). – Deepak Aug 5 '14 at 12:31
• @Deepak: Yes, you are correct and I was very wrong. How embarrassing! I have corrected my errors: many thanks for pointing them out! – Rory Daulton Aug 5 '14 at 12:36
• No worries, glad to help. – Deepak Aug 5 '14 at 12:40
One way to think about it simply (and that I find helpful) is to consider it in terms of quadrants of the Cartesian plane, which are labelled in a counterclockwise direction starting from the 1st quadrant (positive $x$ and $y$). All trig ratios are non-negative in the 1st quadrant. In the other quadrants, only one of the ratios is non-negative while the other two are negative. In the 2nd quadrant, sine is non-negative. In the third, tangent is non-negative and in the 4th quadrant, cosine is non-negative.
I remember this as "ASTC" (all, sine, tangent, cosine).
The given info allows you to localise the angle to the 2nd quadrant (negative tangent, positive sine). That means the cosine is also negative, so you take the negative value.
Tangent has period $\pi$ but $sin$ and $cos$ have period $2\pi$, so for the range of values of $sin$ and $cos$ there are two angles for which the tangent is $-\frac{8}{15}$. The difference between these two angles is $\pi$ so for one of those angles the sine is positive and for one its negative, your condition constrains the result to one angle, i.e. one value for cosine.
$\sin(\theta)>0$ means that the angle $\theta$ should lie in the first or second quadrant of the goniometric circle (in this case the second quadrant). Don't forget that there are $\textbf{two}$ angles for which $\tan(\theta)=-\frac{8}{15}$. | 2019-09-22T11:57:54 | {
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https://math.stackexchange.com/questions/3195930/how-to-evaluate-this-three-diagonal-determinant2019 | How to evaluate this three diagonal determinant?2019
Can someone give me a hint how to solve \begin{align*} |A|=\begin{vmatrix} x & 1 & 0 & 0 & \cdots & 0 & 0 \\ n-1 & x & 2 & 0 & \cdots & 0 & 0 \\ 0 & n-2 & x & 3 & \cdots & 0 & 0 \\ 0 & 0 & n-3 & x & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 1 & x \\ \end{vmatrix}？ \end{align*}
By adding to row 1 the rows from 2 to n ,I can see that $$|A|$$ has $$x+n-1$$ as a factor. And by trying $$n=2,3,4$$, I can deduce that $$|A|=(x^2-(n-1)^2)(x^2-(n-3)^2)\cdots (x^2-1^2)$$ if $$n$$ is even, and $$|A|=(x^2-(n-1)^2)(x^2-(n-3)^2)\cdots (x^2-2^2)x$$ if $$n$$ is odd. I tried hard to give a proof, but without any progress.
• Note that it would suffice to show that the determinant of $A$ is $0$ when $x = \pm (n-1), \pm (n-3), \pm (n-5),\dots$. I'm not sure how you would go about showing that though. – kccu Apr 21 at 16:32
• This is the characteristic polynomial of a Kac matrix. You may see this question for more details. – user1551 Apr 29 at 14:36
Denote $$|A|=D_n(x)$$. By adding the rows numbered $$2, 3, \ldots, n$$ to row $$1$$, we get \begin{align*} D_n(x)= \begin{vmatrix} x+n-1 & x+n-1 & x+n-1 & x+n-1 & \cdots & x+n-1 & x+n-1 \\ n-1 & x & 2 & 0 & \cdots & 0 & 0 \\ 0 & n-2 & x & 3 & \cdots & 0 & 0 \\ 0 & 0 & n-3 & x & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 1 & x \end{vmatrix} \end{align*} Subsequently we add the rows numbered $$3, 4, \ldots, n$$ to row $$2$$, and obtain \begin{align*} D_n(x)= \begin{vmatrix} x+n-1 & x+n-1 & x+n-1 & x+n-1 & \cdots & x+n-1 & x+n-1 \\ n-1 & x+n-2 & x+n-1 & x+n-1 & \cdots &x+n-1 & x+n-1 \\ 0 & n-2 & x & 3 & \cdots & 0 & 0 \\ 0 & 0 & n-3 & x & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 1 & x \end{vmatrix} \end{align*} $$\cdots$$ and so on, by adding the rows numbered $$k+1, k+2, \ldots, n$$ to row $$k$$ ($$k=1,2,\cdots,n-1$$) successively, we finally get \begin{align*} D_n(x)= \begin{vmatrix} x+n-1 & x+n-1 & x+n-1 & x+n-1 & \cdots & x+n-1 & x+n-1 \\ n-1 & x+n-2 & x+n-1 & x+n-1 & \cdots &x+n-1 & x+n-1 \\ 0 & n-2 & x+n-3 & x+n-1 & \cdots & x+n-1 & x+n-1 \\ 0 & 0 & n-3 & x+n-4 & \cdots & x+n-1 & x+n-1 \\ \vdots & \vdots & \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & x+1 & x+n-1 \\ 0 & 0 & 0 & 0 & \cdots & 1 & x \end{vmatrix} \end{align*} In this determinant the elements above the diagonal are all $$x+n-1$$．We successively subtract the $$k$$-th column from the $$k+1$$-th column, $$k=n-1,n-2,\cdots,1$$, and finally get \begin{align*} D_n(x)= \begin{vmatrix} x+n-1 & 0 & 0 & 0 & \cdots & 0 & 0 \\ n-1 & x-1 & 1 & 0 & \cdots &0 & 0 \\ 0 & n-2 & x-1 & 2 & \cdots & 0 & 0 \\ 0 & 0 & n-3 & x-1 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & x-1 & n-2 \\ 0 & 0 & 0 & 0 & \cdots & 1 & x-1 \end{vmatrix} \end{align*} By expanding the above determinant along its first row, we obtain \begin{align*} D_n(x)=&(x+n-1)\begin{vmatrix} x-1 & 1 & 0 & \cdots &0 & 0 \\ n-2 & x-1 & 2 & \cdots & 0 & 0 \\ 0 & n-3 & x-1 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & x-1 & n-2 \\ 0 & 0 & 0 & \cdots & 1 & x-1 \end{vmatrix}\\ =&(x+n-1)D_{n-1}(x-1) \end{align*} Using the above recurrence relation, we can get \begin{align*} D_n(x)=&(x+n-1)(x+n-3)D_{n-2}(x-2)\\ =&(x+n-1)(x+n-3)\cdots (x+n+1-2k)D_{n-k}(x-k)\\ =&(x+n-1)(x+n-3)\cdots (x-n+3)D_{1}(x-n+1)\\ =&(x+n-1)(x+n-3)\cdots (x-n+3)(x-n+1)\\ =&\prod_{k=1}^{n}\big(x+n+1-2k\big)\\ =&\begin{cases} (x^2-(n-1)^2)(x^2-(n-3)^2)\cdots (x^2-1^2),~~~\text{when n is even}\\ (x^2-(n-1)^2)(x^2-(n-3)^2)\cdots (x^2-2^2)x,~\text{when n is odd．} \end{cases} \end{align*}
• Very nice induction proof! – darij grinberg Apr 29 at 20:27
We set $$x=0$$ and show that $$A$$ has eigenvalues $$\pm(n-1),\pm(n-3),\ldots$$ by finding a matrix that diagonalizes it. The solution of the problem follows easily from there.
Let $$P$$ be the Pascal matrix defined by $$p_{ij} = {{j-1}\choose{i-1}} = \frac{(j-1)^{\underline{i-1}}}{(i-1)!}\,,$$ where we have used generalization of binomial coefficient, defined using falling factorials. For example, when $$n=4$$ we have $$P=\begin{bmatrix}1&1&1&1\\ 0&1&2&3\\ 0&0&1&2\\ 0&0&0&1\end{bmatrix}\,.$$ Let $$S$$ be diagonal matrix defined by $$s_{ii} = 2^{n-i}{n-1\choose i-1}\,.$$ Then matrix $$V = P^{-1}SP^T$$ is such that $$V^{-1}AV$$ is diagonal, with eigenvalues sorted in descending order.
Proof. To prove this claim, let $$L$$ be the lower triangle and $$U$$ be the upper triangle of matrix $$A$$. Also, let $$D$$ be diagonal matrix defined by $$d_{ii}=n+1-2i$$, so that its diagonal elements are eigenvalues sorted in descending order.
We start with equivalence $$V^{-1}AV=D \;\Leftrightarrow\; S^{-1}PAP^{-1}S = P^TDP^{-T}\,.$$
To simplify the last equality we will use the following three equalities $$PUP^{-1}=U\,,\quad PLP^{-1}=D+L-U\,,\quad PDP^{-1}=D-2U\,.$$ One can prove them by multiplying them from the right by $$P$$ and by comparing the resulting left and right hand side, using properties of binomial coefficient.
We sum the first two equalities to obtain $$PAP^{-1}=D+L\,.$$ We use the last and the first equality to obtain $$D=P^{-1}DP-2P^{-1}UP = P^{-1}DP-2U\,.$$ Adding $$2U$$ to the both sides of this equality and transposing gives $$P^TDP^{-T}=D+2U^T\,.$$
Now, we have $$V^{-1}AV=D \;\Leftrightarrow\; S^{-1}(D+L)S=D+2U^T \;\Leftrightarrow\; S^{-1}LS=2U^T \,.$$ The last equation can be seen to hold by multiplying it from the left by $$S$$ and comparing the resulting left and right hand side. This ends the proof.
The idea for solution came from the answer to this question.
• Thanks for your reply,it's not easy to think of constructing such a matrix $P$,I will read it carefully.By the way,below I have posted a solution to this question,which needs some tricks of determinant. – mbfkk Apr 29 at 14:33 | 2019-07-22T03:39:56 | {
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https://math.stackexchange.com/questions/1614997/how-does-the-span-of-vectors-1-2-and-0-3-equal-r2 | # How does the span of vectors [1, 2] and [0,3] equal R2?
I'm watching the video tutorial on spans here: https://www.khanacademy.org/math/linear-algebra/vectors_and_spaces/linear_combinations/v/linear-combinations-and-span
At 8:13, he says that the vectors a = [1,2] and b = [0,3] span R2. Visually, I can see it. But I tried to work it out, like so:
sp(a, b) = x[1,2] + y[0,3] such that x,y exist in R
= [x, 2x] + [0, 3y] st x,y e R
= [x, 2x + 3y] st x,y e R
With that said, how do we know that [x, 2x + 3y]spans R2? I tried picking a random point ([19, 6]) and let x=19 and solved for y (2*19 + 3y = 6) and found that when x=19 and y=-10, then I can get the point [19, 6].
But I'm confused as to how would I be able to find out that [x, 2x + 3y] can make any and all points in R2? What's my next step in determining if it spans R2?
Well then, let us pick any point in $\mathbb R^2$, e.g. the point $(a,b)$.
Let us define $x = a$ and $y=\frac{b-2a}{3}$ then you can easily see that:
\begin{align*} (a,[b-2a]/3) = (x,y) \\ \iff (a,b-2a) = (x,3y) \\ \iff (a,b) = (x,2a+3y) \\ \iff (a,b) = (x,2x+3y) \end{align*}
So $(a,b)$ really is in the span of $\{(1,2),(0,3)\}$.
As we can choose any $(a,b) \in \mathbb R^2$, we know that those vectors span the whole $\mathbb R^2$.
• Thanks for the response. Quick question, how exactly did we get y = b-2a / 3? – user2719875 Jan 16 '16 at 23:46
• I started from the last line=) (Many times in math the proof goes in the other direction compared to the order of the discovery.) Assume there is $x,y$ such that $(a,b) = (x,2x+3y)$.. Can we solve for $(x,y)$ or do we get a contradiction? – flawr Jan 17 '16 at 9:56
For any $(a,b)\in \mathbb{R}^2$, $$(a,b) = a\cdot(1,2)+(-\frac{2a}{3}+\frac{b}{3})\cdot(0,3)$$
• Thanks for the response. Quick question, for any (a, b) ∈ R2, doesn't (a, b) = a(1, 2) + b(0, 3)? Where did we get (−2a/3 + b/3) from? – user2719875 Jan 16 '16 at 23:49
• If we simplify the RHSwe get $$(a,2a) +(0, -2a+b) = (a,b)$$. Does this make more sense? All we did was multiple the scalars into the vectors and then add component-wise. If we used your suggestion we end up with $$(a,b) = (a,2a+3b)$$ Which is only true if $$b=0$$. – fosho Jan 16 '16 at 23:52
• Right, but how exactly did you arrive at (a,2a)+(0,−2a+b)=(a,b) (or the equation in your answer) in the first place? – user2719875 Jan 17 '16 at 0:07 | 2019-07-17T14:41:10 | {
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http://www.physicsbootcamp.org/Harmonic-Oscillator-Bootcamp.html | ## Section13.9Harmonic Oscillator Bootcamp
### Subsection13.9.9Miscellaneous
A bullet of mass $m$ and speed $v_0$ is fired horizontally on a block of mass $M$ attached to a spring of spring constant $k$ whose other end is fixed to a wall. The direction of the velocity of the bullet is along the length of the spring. Upon impact, the bullet is embedded in the block.
(a) Ignoring damping, find the amplitude of the resulting harmonic motion.
(b) Compare the energy of the bullet before impact to its energy when it momentarily comes to rest after the spring is compressed, i.e., at the turning point of the oscillatory motion. Is the energy of the bullet conserved? Why or why not?
Hint
(a) Use momentum conservation to find the velocity after impact, then use this to find the amplitude of oscillations, (b) Energy will not be conserved.
(a) $\frac{mv_0}{\sqrt{k(M+m)}}\text{,}$ (b) $E_i - E_f= \dfrac{M}{ m+M}\, E_i \text{.}$
Solution 1 (a)
(a) Let $V$ be the speed of the bullet+block together after the impact. Momentum conservation across the impact gives
\begin{equation*} (m+M) V = mv_0. \end{equation*}
Therefore, immediately after the impact they move at
\begin{equation*} V = \dfrac{m}{m+M}\, v_0. \end{equation*}
The moving combo of block and bullet will compress the spring. Since spring force is a conservative force, we will use conservation of energy to figure out the amount of compression. Let $x$ denote this quantity when maximally compressed. At that point the masses momentarily come to rest.
\begin{equation*} \dfrac{1}{2}\,k x^2 = \dfrac{1}{2}\, (m+M)\, \left(\dfrac{m}{m+M}\, v_0 \right)^2. \end{equation*}
Simplifying this we get
\begin{equation*} x = \dfrac{mv_0}{\sqrt{k(m+M)}}. \end{equation*}
This is the amplitude of the subsequent oscillations.
Solution 2 (b)
The energy of the bullet before impact is
\begin{equation*} E_{i} = \dfrac{1}{2}mv_0^2. \end{equation*}
At the instant the combo of bullet and block momentarily comes to rest, the bullet has only potential energy
\begin{equation*} E_{f} = \dfrac{1}{2}kx^2 = \dfrac{1}{2}\, \dfrac{m}{ m+M}\, m v_0^2. \end{equation*}
Therefore, the loss in energy of the bullet is
\begin{align*} E_i - E_f \amp = \dfrac{1}{2}mv_0^2 \left( 1 - \dfrac{m}{ m+M} \right), \\ \amp = \dfrac{M}{ m+M}\, E_i. \end{align*}
Energy of the bullet is not conserved since, upon impact, work is done by the block on the bullet.
Two springs of spring constants $k_1$ and $k_2$ are attached on the two opposite sides of a block of mass $m$ and the free ends of the springs are attached to two fixed supports as shown in Figure 13.9.29. The block rests on a frictionless flat horizontal surface and can move along the line of the two springs. At equilibrium position of the block, the two springs are neither stretched nor compressed.
When the block is pulled a little, say a distance $A\text{,}$ from the equilibrium position in the line of the springs and released from rest, the block executes a simple harmonic motion whose frequency depends on the spring constants of the two springs and the mass of the block.
(a) By looking at forces on the block at an arbitrary point in time, find the equation of motion of the block.
(b) From the equation of motion, show that the angular frequency of oscillation is given by $\omega=\sqrt{\omega_1^2+\omega_2^2}\text{,}$ where $\omega_1^2 = k_1/m$ and $\omega_2^2 = k_2/m\text{.}$
(c) What is the kinetic energy of the block, when it returns to the equilibrium position after being let go from rest at $x = A\text{?}$
Hint
(a) When setting up forces, be mindful of the direction of forces from the two springs. (b) Compare with the undamped simple harmonic oscillator. (c) Use energy conservation; both springs have potential energies initially.
(a) $m a_x = -\left( k_1 +k_2 \right)\,x\text{,}$ (b) $\omega = \sqrt{\dfrac{k_1 +k_2}{m}} \text{,}$ (c) $\dfrac{1}{2}k_\text{eff} A^2\text{,}$ with $k_\text{eff} = k_1 + k_2\text{.}$
Solution 1 (a)
(a) Figure 13.9.30 shows that forces from both springs are in the same direction. Therefore, the $x$ component of the net force on the block is
\begin{equation*} F_x = -k_1\,x -k_2\,x = -\left( k_1 +k_2 \right)\,x. \end{equation*}
Therefore, equation of motion will be
\begin{equation*} m a_x = -\left( k_1 +k_2 \right)\,x. \end{equation*}
Dividing both sides by $m$ we get
\begin{equation*} a_x = -\dfrac{k_1 +k_2}{m}\,x. \end{equation*}
Solution 2 (b)
(b) Compring the equation of motion to the equation of motion of simple harmonic motion, $a_x=-\omega^2 x\text{,}$ we see that the angular frequency of the block here is
\begin{equation*} \omega^2 = \dfrac{k_1 +k_2}{m}, \end{equation*}
which can be written as
\begin{equation*} \omega^2 = \omega_1^2 + \omega_2^2, \end{equation*}
with
\begin{equation*} \omega_1^2 = \dfrac{k_1}{m},\ \text{ and }\ \omega_2^2 = \dfrac{k_2}{m}. \end{equation*}
Solution 3 (c)
From the conservation of energy, we note that kinetic energy at equilibrium will equal the potential energy at the end points of the motion.
\begin{equation*} KE = \dfrac{1}{2}k_\text{eff} A^2, \end{equation*}
where $k_\text{eff}$ will be
\begin{equation*} k_\text{eff} = k_1 + k_2, \end{equation*}
based on $\omega^2 = \dfrac{k_1 +k_2}{m} = \dfrac{k_\text{eff}}{m}\text{.}$
Two springs of spring constants $k_1$ and $k_2$ are glued with a light but strong glue. One end of the combination is attached to a fixed wall and a block of mass $m$ is attached to the other end and the block is placed on a frictionless table as shown in Figure 13.9.32.
(a) Show that the angular frequency of the block is given by
\begin{equation*} \omega = \sqrt{\dfrac{k_\text{eff} }{m}}, \end{equation*}
with
\begin{equation*} k_\text{eff} = \dfrac{k_1 k_2}{k_1 + k_2}. \end{equation*}
(b) Suppose the block is pulled a distance $A$ from equilibrium, stretching the two springs in the process, and then released from rest. How fast is the block moving when it returns to the equilibrium position?
Hint
(a) For displacement $x$ of the block, springs have different stretches, say $\Delta x_1$ and $\Delta x_2\text{.}$ (b) Think conservation of energy with effective spring constant.
(a) See solution, (b) $\dfrac{1}{2}k_\text{eff}A^2 \text{.}$
Solution 1 (a)
(a) Let $x$ be the displacement of the block from its equilibrium position. Let $\Delta x_1$ and $\Delta x_2$ be the stretchings of the two springs at the instant shown in Figure 13.9.33. Therefore, we have
\begin{equation*} x = \Delta x_1 + \Delta x_2. \end{equation*}
We can write the equations of motions of the block and point P between the two springs.
\begin{align*} \amp \text{Block: }\ \ m a_x = -k_2\,\Delta x_2,\\ \amp \text{Point P: }\ \ k_1\Delta x_1 = k_2\,\Delta x_2, \end{align*}
Eliminating $\Delta x_1$ and $\Delta x_2$ from the three equations above, we find
\begin{equation*} m a_x = - \dfrac{k_1k_2}{k_1 + k_2}\,x, \end{equation*}
which is an equation of a simple harmonic oscillator with effective spring constant $ma_x = - k_\text{eff}\, x\text{,}$ with
\begin{equation*} k_\text{eff} = \dfrac{k_1k_2}{k_1 + k_2}. \end{equation*}
Making use of analogy with the simple harmonic motion we have the angular frequency of the block here
\begin{equation*} \omega = \sqrt{\dfrac{ k_\text{eff} }{m}}. \end{equation*}
Solution 2 (b)
(b) The potential energy stored is
\begin{equation*} U = \dfrac{1}{2}k_1 (\Delta x_1)^2 + \dfrac{1}{2}k_2 (\Delta x_2)^2, \end{equation*}
with
\begin{equation*} \Delta x_1 + \Delta x_2 = A \end{equation*}
andd
\begin{equation*} k_1 \Delta x_1 = k_2 \Delta x_2. \end{equation*}
From the last two equations, we get
\begin{equation*} \Delta x_1 = \dfrac{k_2}{k_1 + k_2}\, A,\ \ \ \Delta x_2 = \dfrac{k_1}{k_1 + k_2}\, A. \end{equation*}
Using these in $U$ we get
\begin{equation*} U = \dfrac{1}{2}\,k_\text{eff}\, A^2. \end{equation*}
This will be the kinetic energy when block is at equilibrium since at that instant the potential energy will be zero and all energy will be the kinetic energy. | 2023-03-20T13:04:22 | {
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https://fizzbuzzer.com/network-flow-problem-maximize-total-amount-of-flow-from-s-to-t/ | ## [f]izzbuzzer
### Looking for good programming challenges?
Use the search below to find our solutions for selected questions!
# Network flow problem: Maximize total amount of flow from s to t
Sharing is caring!
Problem statement
Find the maximum flow in a flow network (directed graph) from a source node $s$ to a target node $t$ subject to two constraints:
– The flow on edge $e$ doesn’t exceed its capacity $c(e)$
– For every node $v \ne s, t$, the incoming flow is equal to outgoing flow.
Consider the graph below. The image on the left depicts the graph with its edge capacities, while the image on the right illustrates the same graph with a maximum flow of $19$:
Notice also on the right image, that for every node $v \ne s, t$, the incoming flow is equal to outgoing flow while the flow does not exceed the edge’s capacity.
How can we find the maximum flow?
We will do this by finding augmenting paths from node $s$ to $t$. An augmenting path is a path whose edges are:
1. Non-full and forward or
2. Non-empty and backward
In other words, it is a path where the residual capacity is $\textgreater 0$, with the residual capacity of an edge $e$, being the total capacity $c(e)$ minus the flow that goes through $e$. For the above graph we can annotate the edges with their current flow and capacity. At the beginning the flow through each edge is $0$, so their residual capacity is equal to their capacity:
The general algorithm (also known as Ford-Fulkerson algorithm) is as follows:
1. Find an augmenting path (using Breadth-first search or Depth-first search) from $s$ to $t$
2. Compute the bottleneck capacity (i.e. an edge in the path with the smallest residual capacity)
3. Augment each edge and the total flow
4. Repeat steps 1-3 until no augmenting path can be found
The first path we can find is $s \rightarrow A \rightarrow B \rightarrow t$ with a bottleneck of $4$:
Next, we find the path $s \rightarrow C \rightarrow D \rightarrow t$ with a bottleneck of $9$:
The next path is $s \rightarrow A \rightarrow D \rightarrow B \rightarrow t$ with a bottleneck of $6$:
Finally we cannot find any more augmenting paths so we stop with a maximum flow of $19$.
What if the paths we choose are different. Is the maximum flow still going to be the same?
Let us consider choosing the following paths. First we choose the path $s \rightarrow A \rightarrow B \rightarrow t$ with a bottleneck of $4$::
Next, we choose the path $s \rightarrow A \rightarrow C \rightarrow D \rightarrow t$ with a bottleneck of $2$:
We then go on and choose the path $s \rightarrow C \leftarrow A \rightarrow D \rightarrow t$. This path has a backward edge $C \leftarrow A$. We augment flow through the whole network by removing flow in that path. So when we add $2$ units of flow to the edge $s \rightarrow C$, we can remove $2$ units of flow the backward edge $A \leftarrow C$ to preserve the local equilibrium. Essentially that $2$ units of flow gets transferred through edge $C \rightarrow D$ and then for the forward edges we add the flow:
Next, we choose the path $s \rightarrow A \rightarrow D \rightarrow t$ with a bottleneck of $4$:
We then choose the path $s \rightarrow C \rightarrow D \rightarrow B \rightarrow t$ with a bottleneck of $6$:
Next, we find the path $s \rightarrow C \rightarrow D \rightarrow t$ with a bottleneck of $1$:
Finally we cannot find any more augmenting paths so we stop with a maximum flow of $19$.
How do we know that this algorithm gives us the maximum flow?
Suppose this algorithm does not give us the maximum flow. Take a maximum flow $m'$ and subtract our found flow $m$. Whats left is a valid flow $\textgreater 0$. That is, there exists a path with a residual capacity $\textgreater 0$ (augmenting path). But, by algorithm, this path must have been found (see step 4 of algorithm). Contradiction.
Sample implementation
public class FordFulkerson {
private static class DirectedGraph {
private Map> nodeToNeighbors;
private Map> capacity;
private Map> residualCapacity;
public DirectedGraph() {
nodeToNeighbors = new HashMap>();
capacity = new HashMap>();
residualCapacity = new HashMap>();
}
if (!nodeToNeighbors.containsKey(node)) {
nodeToNeighbors.put(node, new HashSet());
}
}
public void addEdge(T start, T dest, int cap) {
if (!capacity.containsKey(start)) {
capacity.put(start, new HashMap());
}
if (!residualCapacity.containsKey(start)) {
residualCapacity.put(start, new HashMap());
}
if (!residualCapacity.containsKey(dest)) {
residualCapacity.put(dest, new HashMap());
}
capacity.get(start).put(dest, cap);
residualCapacity.get(start).put(dest, cap);
residualCapacity.get(dest).put(start, 0);
}
public Set getNodes() {
return nodeToNeighbors.keySet();
}
public int getResidualCapacity(T start, T dest) {
if (!residualCapacity.get(start).containsKey(dest)) {
return 0;
} else {
return residualCapacity.get(start).get(dest);
}
}
public void setResidualCapacity(T start, T dest, int f) {
residualCapacity.get(start).put(dest, f);
}
}
// Get BFS backtrack from source -> target
private static Map bfs(DirectedGraph graph, String source, String target) {
Map backtrack = new HashMap<>();
Set visited = new HashSet<>();
OUTER:
while (!queue.isEmpty()) {
String node = queue.poll();
for (String n : graph.getNodes()) {
if (!n.equals(node) && !visited.contains(n)) {
if (graph.getResidualCapacity(node, n) > 0) {
// We have a positive residual capacity to reach n from node
backtrack.put(n, node);
if (n.equals(target)) {
break OUTER;
}
}
}
}
}
return backtrack;
}
private static int maxFlow(DirectedGraph graph, String source, String target) {
int ans = 0;
Map backtrack = bfs(graph, source, target);
do {
// Print augmenting path and also find the bottleneck of the path
path.clear();
String tail = target;
int min = Integer.MAX_VALUE;
while (backtrack.containsKey(tail)) {
String prev = backtrack.get(tail);
min = Math.min(min, graph.getResidualCapacity(prev, tail));
tail = prev;
}
if (path.size() <= 1) {
break;
}
System.out.println("Found augmenting path '" + path + "' with bottleneck " + min);
// Update residualCapacity
tail = target;
while (backtrack.containsKey(tail)) {
String prev = backtrack.get(tail);
// Decrement residual capacity for edge prev->tail
graph.setResidualCapacity(prev, tail, graph.getResidualCapacity(prev, tail) - min);
// Increase residual capacity for backwards edge tail->prev (i.e. open up an edge in the opposite direction)
graph.setResidualCapacity(tail, prev, graph.getResidualCapacity(tail, prev) + min);
tail = prev;
}
// Update max flow by min
ans += min;
// Repeat process: find a new path
backtrack = bfs(graph, source, target);
} while (path.size() > 1);
return ans;
}
public static void main(String[] args) {
DirectedGraph graph = new DirectedGraph();
System.out.println("max flow = " + maxFlow(graph, "s", "t") + "\n"); // 19
graph = new DirectedGraph(); | 2020-01-28T11:36:59 | {
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http://math.stackexchange.com/questions/109379/integrating-sqrt1-x2-by-interpreting-it-geometrically-as-an-area-within-a | Integrating $\sqrt{1-x^2}$ by interpreting it geometrically as an area within a circle
For the integral $$\int \sqrt{1-x^2} dx = \frac{1}{2} \left ( \arcsin(x) + x \sqrt{1-x^2} \right)$$ Now it was explained to me that geometrically I could take part of the integral as an area sector and the other half a triangle. I am having a hard time seeing how I can even get a triangle as I am summing the rectangles.
How could you even see that the angle must be $\arcsin(x)$?
Compare to:
This is not a good approximation of the integral. What happened to the red area?
-
Note that the integral in question $$\int_0^b{\sqrt{1-x^2} ~\mathrm dx}$$ is the shaded area in the integral below. Now compare a direct computation from the figure below with the result you have got through the integral.
Join the center of the circle to the point $(b,f(b))$ to see a sector and a triangle!
$\hskip{1.5in}$
-
Draw the line from the center of the circle to your point $(b,f(b))$. That line splits your big blue region into two parts: a triangle below the line, and a sector above.
The area of the triangle is clearly $\frac{1}{2}b\sqrt{1-b^2}$.
For the area of the sector, the angle of that sector is $\arcsin b$. This is because the complementary angle (below the line) has cosine equal to $b$. Or else you can see directly, by drawing a perpendicular from $(b,f(b))$ to the $y$-axis, that the angle of the sector has "opposite" side, and therefore sine, equal to $b$. So the area of the sector is $\frac{1}{2}\arcsin b$.
By integration, the area of the blue region is $\int_0^b\sqrt{1-x^2}dx$. We conclude that $$\int_0^b\sqrt{1-x^2}\,dx=\frac{1}{2}b\sqrt{1-b^2}+\frac{1}{2}\arcsin b.\qquad\qquad(\ast)$$
Now in $(\ast)$, change the $b$ to $x$, and (to make me feel good) the dummy variable of integration to $t$. We have $$\int_0^x\sqrt{1-t^2}\,dt=\frac{1}{2}x\sqrt{1-x^2}+\frac{1}{2}\arcsin x.$$ This says that the right-hand side is an antiderivative of $\sqrt{1-x^2}$. All antiderivatives are obtained by adding a constant of integration.
Note that the geometric derivation is not quite complete, since the picture does not deal directly with negative $b$. But that is not hard to do. The simplest way is to make the change of variable $z=-x$.
-
Beat me to it. (by 55 seconds!) +1 – user21436 Feb 14 '12 at 21:19
Why? That's not how the Riemann Sum in Cartesian plane works. I actually don't see why the triangle would cover the rest of the circle. – Unh Feb 14 '12 at 21:21
@Unh: It’s not supposed to cover the rest of the circle. $\int_0^b\sqrt{1-x^2}dx$ gives you the blue area in your first picture. Don’t think about Riemann sums here; just think about the area that you’re calculating. – Brian M. Scott Feb 14 '12 at 21:28
There is no direct connection with Riemann sum. The integral (definite integral, which can be taken as having a variable upper limit) gives you the area. Decomposition as described above, plus formulas from geometry, compute the same area. So they are equal. – André Nicolas Feb 14 '12 at 21:30
Oh I see what you mean. The sector + triangle IS the Riemann Sum I am referring to. Does the triangle "disappear" once we reach to the point of "infintessimally small"? How does that get us arcsin(x) though? I mean "x" is the angle measure from the x-axis, not from the y-axis. – Unh Feb 14 '12 at 21:34
You probably were told to put it this way:
The coordinates $(\sin \theta, \cos \theta)$ can be defined as those who delimit an area of $\frac{\theta}{2}$ in the unit circle when joined with its origin.
So the area in this image would be
$$A = \dfrac{\theta}{2}$$
What you're being told is that
$$\int\limits_0^x {\sqrt {1 - {t^2}} dt} = \frac{\pi }{4} -\frac{\theta }{2} + \frac{{\sin \theta \cos \theta }}{2}$$
Note that $\dfrac{{\sin \theta \cos \theta }}{2}$ is the area of the triangle; $\dfrac{\theta }{2}$ area of the sector and $\dfrac{\pi }{4}$ the area of the circle's 1st quadrant.
So expressing in terms of $y = \sin \theta$ you get that
$$\int\limits_0^x {\sqrt {1 - {t^2}} dt} = \frac{{{{\sin }^{ - 1}}x}}{2} + \frac{{x\sqrt {1 - {x^2}} }}{2}$$
And in terms of $x = \cos \theta$ you get that
$$\int\limits_0^x {\sqrt {1 - {t^2}} dt} = \frac{\pi }{4}- \frac{{{{\cos }^{ - 1}}x}}{2} + \frac{{x\sqrt {1 - {x^2}} }}{2}$$
Note that your solution is incorrect. It should be
$$\int {\sqrt {1 - {x^2}} dx} = \frac{{{{\sin }^{ - 1}}x}}{2} + \frac{{x\sqrt {1 - {x^2}} }}{2} + C$$
-
How did you substitute $\arcsin(x) = \theta$? – Unh Feb 14 '12 at 21:38
@Peter Please use the display style with restraint. I'll edit it for now. Your questions also had this discrepancy. I am pointing it out in a friendly fashion. No offence. – user21436 Feb 14 '12 at 21:40
@KannappanSampath I corrected the error but don't understand the display style remark. – Pedro Tamaroff Feb 14 '12 at 21:42
Oh I understand now. You traced out an imaginary triangle and that line perp to the radial line also happens to length b. I see npow. – Unh Feb 14 '12 at 21:44
@Unh I added the more natural apporach. Start with an area of $\pi/4$, substract the sector $\theta / 2$ and add the triangle $\cos \theta \sin \theta /2$ – Pedro Tamaroff Feb 14 '12 at 21:46 | 2015-09-01T15:02:12 | {
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https://dsc-courses.github.io/dsc90-2022-wi/resources/weeks/week02/ | # Week 2 – Calculus, Aggregation
Required:
Optional:
## Homework 2 (due Friday, January 21st at 11:59PM) (solutions) 📝
Submit your answers as a PDF to Gradescope by the due date for full credit. We encourage you to discuss the readings and questions with others in the course, but all work must be your own. Remember to use Campuswire if you need guidance!
### Question 1
Below in blue is the parabola $$y = x^2$$, and in red is the line $$y = 2x + 8$$.
(a) Without using integration, determine the area between the given line and parabola.
This will involve a few steps. First, you’ll need to find the third point on the triangle that Archimedes specified in Quadrature of the Parabola. Some guiding questions:
• What is the slope of the red line?
• What is the slope of the tangent line to the parabola at any given point on the parabola? (What is the derivative of the parabola?)
• At what point on the parabola is the slope of the tangent line equal to the slope of the red line?
Once you’ve identified the third point, you can use this online calculator to find the area of that triangle (unless you’d rather do it yourself for practice). Then, you should be able to use a result from class. Show all of your steps, and remember to ask for help if you need it.
(b) If you’re familiar with integration, compute the area between the given line and parabola using integration, and show that your answer is the same as your answer in part (a).
(Note: If you know how to take an integral, you are expected to do this question.)
### Question 2
Consider the function
$f(x) = x(48 - x^2)$
(a) Using Fermat’s method of adequality, determine the value of $$x$$ that maximizes $$f(x)$$. Is this a local or global maximum? Show your work.
(b) Using Fermat’s method of adequality, determine the slope of the tangent line to $$f(x)$$ at $$x = 1$$. Verify that your answer is correct by finding $$f'(x)$$ and evaluating $$f'(1)$$. Show your work.
(c) Repeat part (b), but at $$x = 0$$. What issue do you run into?
### Question 3
Show that Fermat’s method for finding the slope of the tangent line is equivalent to
$\text{slope} = \frac{f(x+e) - f(x)}{e}$
Does the quotient on the right look familiar – if so, what does it resemble?
Hopefully, by answering this question, it is slightly more clear why Fermat’s method of adequality involves dividing by $$e$$ and then setting $$e$$ to 0.
### Question 4
One aggregation strategy that was used centuries ago involved picking the number halfway between the smallest and largest numbers in a dataset. In today’s terminology, we’d call this number the “midrange”.
$\text{Midrange} = \frac{\text{Max}(x) + \text{Min}(x)}{2}$
For each of the following conditions, specify a set of 5 unique numbers that satisfies the given condition. In each case, compute the midrange and mean and show that the condition holds. Afterwards, give a one sentence answer to the question, “for what kinds of datasets is the midrange close to the mean?”
(a) Midrange = mean
(b) Midrange > mean
(c) Midrange < mean | 2023-03-27T14:35:37 | {
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https://math.stackexchange.com/questions/646337/what-is-the-difference-between-f-x-and-fx/646343 | # What is the difference between f(-x) and -f(x)?
What is the difference between $f(-x)$ and $-f(x)$ in terms of their graphs?
• There's no general answer...imagine g(x) = x^2 or g(x)=x – wxyz Jan 21 '14 at 15:57
The graph of $f(-x)$ is the mirror image of the graph of $f(x)$ with respect to the vertical axis.
The graph of $-f(x)$ is the mirror image of the graph of $f(x)$ with respect to the horizontal axis.
A function is called even if $f(x)=f(-x)$ for all $x$ (For example, $\cos(x)$).
A function is called odd if $-f(x)=f(-x)$ for all $x$ (For example, $\sin(x)$).
The most helpful vocabulary related to your question has to do with the parity of a given function. Functions are described as odd, even, neither. Most functions are neither odd nor even, but it is great to know which ones are even or odd and how to tell the difference.
Even functions - If $f(x)$ is an even function, then for every $x$ and $-x$ in the domain of $f$, $f(x) = f(-x)$. Graphically, this means that the function is symmetric with respect to the $y$-axis. Thus, reflections across the $y$-axis do not affect the function's appearance. Good examples of even functions include: $x^2, x^4, ..., x^{2n}$ (integer $n$); $\cos(x)$, $\cosh(x)$, and $|x|$.
Odd functions - If $f(x)$ is an odd function, then for every $x$ and $-x$ in the domain of $f$, $-f(x) = f(-x)$. Graphically, this means that the function is rotationally symmetric with respect to the origin. Thus, rotations of $180^\circ$ or any multiple of $180^\circ$ do not affect the function's appearance. Good examples of odd functions include: $x,x^3,x^5,...,x^{2n+1}$ (integer $n$); $\sin(x)$, and $\sinh(x)$.
To really get a feel for recognizing these functions, I suggest you graph several (if not all) of them. You will be well-equipped to visually determine the parity of most functions if you just spend a little time graphing the example functions.
• Functions may be neither odd nor even, but having said that, it’s worth pointing out that any function $f$ whose domain allows negation and whose codomain allows addition, negation and division by $2$ can be written as the sum of an odd and even function: $f(x) = \frac{f(x) + f(-x)}{2} + \frac{f(x) - f(-x)}{2}$ – the left fraction defines an even function whereas the right fraction defines an odd function. – k.stm Jan 21 '14 at 16:20
• @K.Stm. Interesting observation! I know some properties involving the parity of the result of sums, differences, products, quotients, derivatives, and compositions of various combinations of even and odd functions. But it did not occur to me that you could also decompose a function given the constraints in your comment. Cool! – Xoque55 Jan 21 '14 at 16:26 | 2020-12-01T06:20:54 | {
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https://stats.libretexts.org/Textbook_Maps/Probability_Theory/Book%3A_Introductory_Probability_(Grinstead_and_Snell)/9%3A_Central_Limit_Theorem/9.2%3A_Central_Limit_Theorem_for_Discrete_Independent_Trials | 9.2: Central Limit Theorem for Discrete Independent Trials
We have illustrated the Central Limit Theorem in the case of Bernoulli trials, but this theorem applies to a much more general class of chance processes. In particular, it applies to any independent trials process such that the individual trials have finite variance. For such a process, both the normal approximation for individual terms and the Central Limit Theorem are valid.
Let $$S_n = X_1 + X_2 +\cdots+ X_n$$ be the sum of $$n$$ independent discrete random variables of an independent trials process with common distribution function $$m(x)$$ defined on the integers, with mean $$\mu$$ and variance $$\sigma^2$$. We can prevent this just as we did for Bernoulli trials.
Standardized Sums
Consider the standardized random variable $S_n^* = \frac {S_n - n\mu}{\sqrt{n\sigma^2}}\ .$
This standardizes $$S_n$$ to have expected value 0 and variance 1. If $$S_n = j$$, then $$S_n^*$$ has the value $$x_j$$ with $x_j = \frac {j - n\mu}{\sqrt{n\sigma^2}}\ .$ We can construct a spike graph just as we did for Bernoulli trials. Each spike is centered at some $$x_j$$. The distance between successive spikes is $b = \frac 1{\sqrt{n\sigma^2}}\ ,$ and the height of the spike is $h = \sqrt{n\sigma^2} P(S_n = j)\ .$
The case of Bernoulli trials is the special case for which $$X_j = 1$$ if the $$j$$th outcome is a success and 0 otherwise; then $$\mu = p$$ and $$\sigma^2 = \sqrt {pq}$$.
We now illustrate this process for two different discrete distributions. The first is the distribution $$m$$, given by $m = \pmatrix{ 1 & 2 & 3 & 4 & 5 \cr .2 & .2 & .2 & .2 & .2\cr}\ .$
In Figure [fig 9.5] we show the standardized sums for this distribution for the cases $$n = 2$$ and $$n = 10$$. Even for $$n = 2$$ the approximation is surprisingly good.
For our second discrete distribution, we choose $m = \pmatrix{ 1 & 2 & 3 & 4 & 5 \cr .4 & .3 & .1 & .1 & .1\cr}\ .$
This distribution is quite asymmetric and the approximation is not very good for $$n = 3$$, but by $$n = 10$$ we again have an excellent approximation (see Figure [fig 9.5.5]). Figures [fig 9.5] and [fig 9.5.5] were produced by the program CLTIndTrialsPlot.
Approximation Theorem
As in the case of Bernoulli trials, these graphs suggest the following approximation theorem for the individual probabilities.
[thm 9.3.5] Let $$X_1$$$$X_2$$, …, $$X_n$$ be an independent trials process and let $$S_n = X_1 + X_2 +\cdots+ X_n$$. Assume that the greatest common divisor of the differences of all the values that the $$X_j$$ can take on is 1. Let $$E(X_j) = \mu$$ and $$V(X_j) = \sigma^2$$. Then for $$n$$ large, $P(S_n = j) \sim \frac {\phi(x_j)}{\sqrt{n\sigma^2}}\ ,$ where $$x_j = (j - n\mu)/\sqrt{n\sigma^2}$$, and $$\phi(x)$$ is the standard normal density.
The program CLTIndTrialsLocal implements this approximation. When we run this program for 6 rolls of a die, and ask for the probability that the sum of the rolls equals 21, we obtain an actual value of .09285, and a normal approximation value of .09537. If we run this program for 24 rolls of a die, and ask for the probability that the sum of the rolls is 72, we obtain an actual value of .01724 and a normal approximation value of .01705. These results show that the normal approximations are quite good.
Central Limit Theorem for a Discrete Independent Trials Process
The Central Limit Theorem for a discrete independent trials process is as follows.
(Central Limit Theorem)[thm 9.3.6] Let $$S_n = X_1 + X_2 +\cdots+ X_n$$ be the sum of $$n$$ discrete independent random variables with common distribution having expected value $$\mu$$ and variance $$\sigma^2$$. Then, for $$a < b$$, $\lim_{n \to \infty} P\left( a < \frac {S_n - n\mu}{\sqrt{n\sigma^2}} < b\right) = \frac 1{\sqrt{2\pi}} \int_a^b e^{-x^2/2}\, dx\ .$
Here we consider several examples.
Examples
A die is rolled 420 times. What is the probability that the sum of the rolls lies between 1400 and 1550?
The sum is a random variable $S_{420} = X_1 + X_2 +\cdots+ X_{420}\ ,$ where each $$X_j$$ has distribution $m_X = \pmatrix{ 1 & 2 & 3 & 4 & 5 & 6 \cr 1/6 & 1/6 & 1/6 & 1/6 & 1/6 & 1/6 \cr}$ We have seen that $$\mu = E(X) = 7/2$$ and $$\sigma^2 = V(X) = 35/12$$. Thus, $$E(S_{420}) = 420 \cdot 7/2 = 1470$$, $$\sigma^2(S_{420}) = 420 \cdot 35/12 = 1225$$, and $$\sigma(S_{420}) = 35$$. Therefore, \begin{aligned} P(1400 \leq S_{420} \leq 1550) &\approx& P\left(\frac {1399.5 - 1470}{35} \leq S_{420}^* \leq \frac {1550.5 - 1470}{35} \right) \\ &=& P(-2.01 \leq S_{420}^* \leq 2.30) \\ &\approx& \mbox{NA}(-2.01, 2.30) = .9670\ . \end{aligned} We note that the program CLTIndTrialsGlobal could be used to calculate these probabilities.
Example $$\PageIndex{6}$$:
A student’s grade point average is the average of his grades in 30 courses. The grades are based on 100 possible points and are recorded as integers. Assume that, in each course, the instructor makes an error in grading of $$k$$ with probability $$|p/k|$$, where $$k = \pm1$$$$\pm2$$, $$\pm3$$, $$\pm4$$$$\pm5$$. The probability of no error is then $$1 - (137/30)p$$. (The parameter $$p$$ represents the inaccuracy of the instructor’s grading.) Thus, in each course, there are two grades for the student, namely the “correct" grade and the recorded grade. So there are two average grades for the student, namely the average of the correct grades and the average of the recorded grades.
We wish to estimate the probability that these two average grades differ by less than .05 for a given student. We now assume that $$p = 1/20$$. We also assume that the total error is the sum $$S_{30}$$ of 30 independent random variables each with distribution $m_X: \left\{ \begin{array}{ccccccccccc} -5 & -4 & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 & 5 \\ \frac1{100} & \frac1{80} & \frac1{60} & \frac1{40} & \frac1{20} & \frac{463}{600} & \frac1{20} & \frac1{40} & \frac1{60} & \frac1{80} & \frac1{100} \end{array} \right \}\ .$ One can easily calculate that $$E(X) = 0$$ and $$\sigma^2(X) = 1.5$$. Then we have
$\begin{array}{ll} P\left(-.05 \leq \frac {S_{30}}{30} \leq .05 \right) &= P(-1.5 \leq S_{30} \leq1.5) \\ & \\ &= P\left( \frac{-1.5}{\sqrt{30\cdot1.5}} \leq S_{30}^* \leq \frac{1.5}{\sqrt{30\cdot1.5}} \right) \\ & \\ &= P(-.224 \leq S_{30}^* \leq .224) \\ & \\ & \approx \mbox{NA}(-.224, .224) = .1772\ . \end{array}$
This means that there is only a 17.7% chance that a given student’s grade point average is accurate to within .05. (Thus, for example, if two candidates for valedictorian have recorded averages of 97.1 and 97.2, there is an appreciable probability that their correct averages are in the reverse order.) For a further discussion of this example, see the article by R. M. Kozelka.5
A More General Central Limit Theorem
In Theorem $$\PageIndex{1}$$, the discrete random variables that were being summed were assumed to be independent and identically distributed. It turns out that the assumption of identical distributions can be substantially weakened. Much work has been done in this area, with an important contribution being made by J. W. Lindeberg. Lindeberg found a condition on the sequence $$\{X_n\}$$ which guarantees that the distribution of the sum $$S_n$$ is asymptotically normally distributed. Feller showed that Lindeberg’s condition is necessary as well, in the sense that if the condition does not hold, then the sum $$S_n$$ is not asymptotically normally distributed. For a precise statement of Lindeberg’s Theorem, we refer the reader to Feller.6 A sufficient condition that is stronger (but easier to state) than Lindeberg’s condition, and is weaker than the condition in Theorem $$\PageIndex{1}$$, is given in the following theorem.
Theorem $$\PageIndex{2}$$
(Central Limit Theorem) Let $$X_1,\ X_2,\ \ldots,\ X_n\ ,\ \ldots$$ be a sequence of independent discrete random variables, and let $$S_n = X_1 + X_2 +\cdots+ X_n$$. For each $$n$$, denote the mean and variance of $$X_n$$ by $$\mu_n$$ and $$\sigma^2_n$$, respectively. Define the mean and variance of $$S_n$$ to be $$m_n$$ and $$s_n^2$$, respectively, and assume that $$s_n \rightarrow \infty$$. If there exists a constant $$A$$, such that $$|X_n| \le A$$ for all $$n$$, then for $$a < b$$, $\lim_{n \to \infty} P\left( a < \frac {S_n - m_n}{s_n} < b\right) = \frac 1{\sqrt{2\pi}} \int_a^b e^{-x^2/2}\, dx\ .$
The condition that $$|X_n| \le A$$ for all $$n$$ is sometimes described by saying that the sequence $$\{X_n\}$$ is uniformly bounded. The condition that $$s_n \rightarrow \infty$$ is necessary (see Exercise [exer 9.2.114]).
We illustrate this theorem by generating a sequence of $$n$$ random distributions on the interval $$[a, b]$$. We then convolute these distributions to find the distribution of the sum of $$n$$ independent experiments governed by these distributions. Finally, we standardize the distribution for the sum to have mean 0 and standard deviation 1 and compare it with the normal density. The program CLTGeneral carries out this procedure.
In Figure [fig 9.6] we show the result of running this program for $$[a, b] = [-2, 4]$$, and $$n = 1,\ 4,$$ and 10. We see that our first random distribution is quite asymmetric. By the time we choose the sum of ten such experiments we have a very good fit to the normal curve.
The above theorem essentially says that anything that can be thought of as being made up as the sum of many small independent pieces is approximately normally distributed. This brings us to one of the most important questions that was asked about genetics in the 1800’s.
The Normal Distribution and Genetics
When one looks at the distribution of heights of adults of one sex in a given population, one cannot help but notice that this distribution looks like the normal distribution. An example of this is shown in Figure [fig 9.61]. This figure shows the distribution of heights of 9593 women between the ages of 21 and 74. These data come from the Health and Nutrition Examination Survey I (HANES I). For this survey, a sample of the U.S. civilian population was chosen. The survey was carried out between 1971 and 1974.
A natural question to ask is “How does this come about?". Francis Galton, an English scientist in the 19th century, studied this question, and other related questions, and constructed probability models that were of great importance in explaining the genetic effects on such attributes as height. In fact, one of the most important ideas in statistics, the idea of regression to the mean, was invented by Galton in his attempts to understand these genetic effects.
Galton was faced with an apparent contradiction. On the one hand, he knew that the normal distribution arises in situations in which many small independent effects are being summed. On the other hand, he also knew that many quantitative attributes, such as height, are strongly influenced by genetic factors: tall parents tend to have tall offspring. Thus in this case, there seem to be two large effects, namely the parents. Galton was certainly aware of the fact that non-genetic factors played a role in determining the height of an individual. Nevertheless, unless these non-genetic factors overwhelm the genetic ones, thereby refuting the hypothesis that heredity is important in determining height, it did not seem possible for sets of parents of given heights to have offspring whose heights were normally distributed.
One can express the above problem symbolically as follows. Suppose that we choose two specific positive real numbers $$x$$ and $$y$$, and then find all pairs of parents one of whom is $$x$$ units tall and the other of whom is $$y$$ units tall. We then look at all of the offspring of these pairs of parents. One can postulate the existence of a function $$f(x, y)$$ which denotes the genetic effect of the parents’ heights on the heights of the offspring. One can then let $$W$$ denote the effects of the non-genetic factors on the heights of the offspring. Then, for a given set of heights $$\{x, y\}$$, the random variable which represents the heights of the offspring is given by $H = f(x, y) + W\ ,$ where $$f$$ is a deterministic function, i.e., it gives one output for a pair of inputs $$\{x, y\}$$. If we assume that the effect of $$f$$ is large in comparison with the effect of $$W$$, then the variance of $$W$$ is small. But since f is deterministic, the variance of $$H$$ equals the variance of $$W$$, so the variance of $$H$$ is small. However, Galton observed from his data that the variance of the heights of the offspring of a given pair of parent heights is not small. This would seem to imply that inheritance plays a small role in the determination of the height of an individual. Later in this section, we will describe the way in which Galton got around this problem.
We will now consider the modern explanation of why certain traits, such as heights, are approximately normally distributed. In order to do so, we need to introduce some terminology from the field of genetics. The cells in a living organism that are not directly involved in the transmission of genetic material to offspring are called somatic cells, and the remaining cells are called germ cells. Organisms of a given species have their genetic information encoded in sets of physical entities, called chromosomes. The chromosomes are paired in each somatic cell. For example, human beings have 23 pairs of chromosomes in each somatic cell. The sex cells contain one chromosome from each pair. In sexual reproduction, two sex cells, one from each parent, contribute their chromosomes to create the set of chromosomes for the offspring.
Chromosomes contain many subunits, called genes. Genes consist of molecules of DNA, and one gene has, encoded in its DNA, information that leads to the regulation of proteins. In the present context, we will consider those genes containing information that has an effect on some physical trait, such as height, of the organism. The pairing of the chromosomes gives rise to a pairing of the genes on the chromosomes.
In a given species, each gene can be any one of several forms. These various forms are called alleles. One should think of the different alleles as potentially producing different effects on the physical trait in question. Of the two alleles that are found in a given gene pair in an organism, one of the alleles came from one parent and the other allele came from the other parent. The possible types of pairs of alleles (without regard to order) are called genotypes.
If we assume that the height of a human being is largely controlled by a specific gene, then we are faced with the same difficulty that Galton was. We are assuming that each parent has a pair of alleles which largely controls their heights. Since each parent contributes one allele of this gene pair to each of its offspring, there are four possible allele pairs for the offspring at this gene location. The assumption is that these pairs of alleles largely control the height of the offspring, and we are also assuming that genetic factors outweigh non-genetic factors. It follows that among the offspring we should see several modes in the height distribution of the offspring, one mode corresponding to each possible pair of alleles. This distribution does not correspond to the observed distribution of heights.
An alternative hypothesis, which does explain the observation of normally distributed heights in offspring of a given sex, is the multiple-gene hypothesis. Under this hypothesis, we assume that there are many genes that affect the height of an individual. These genes may differ in the amount of their effects. Thus, we can represent each gene pair by a random variable $$X_i$$, where the value of the random variable is the allele pair’s effect on the height of the individual. Thus, for example, if each parent has two different alleles in the gene pair under consideration, then the offspring has one of four possible pairs of alleles at this gene location. Now the height of the offspring is a random variable, which can be expressed as $H = X_1 + X_2 + \cdots + X_n + W\ ,$ if there are $$n$$ genes that affect height. (Here, as before, the random variable $$W$$ denotes non-genetic effects.) Although $$n$$ is fixed, if it is fairly large, then Theorem $$\PageIndex{2}$$ implies that the sum $$X_1 + X_2 + \cdots + X_n$$ is approximately normally distributed. Now, if we assume that the $$X_i$$’s have a significantly larger cumulative effect than $$W$$ does, then $$H$$ is approximately normally distributed.
Another observed feature of the distribution of heights of adults of one sex in a population is that the variance does not seem to increase or decrease from one generation to the next. This was known at the time of Galton, and his attempts to explain this led him to the idea of regression to the mean. This idea will be discussed further in the historical remarks at the end of the section. (The reason that we only consider one sex is that human heights are clearly sex-linked, and in general, if we have two populations that are each normally distributed, then their union need not be normally distributed.)
Using the multiple-gene hypothesis, it is easy to explain why the variance should be constant from generation to generation. We begin by assuming that for a specific gene location, there are $$k$$ alleles, which we will denote by $$A_1,\ A_2,\ \ldots,\ A_k$$. We assume that the offspring are produced by random mating. By this we mean that given any offspring, it is equally likely that it came from any pair of parents in the preceding generation. There is another way to look at random mating that makes the calculations easier. We consider the set $$S$$ of all of the alleles (at the given gene location) in all of the germ cells of all of the individuals in the parent generation. In terms of the set $$S$$, by random mating we mean that each pair of alleles in $$S$$ is equally likely to reside in any particular offspring. (The reader might object to this way of thinking about random mating, as it allows two alleles from the same parent to end up in an offspring; but if the number of individuals in the parent population is large, then whether or not we allow this event does not affect the probabilities very much.)
For $$1 \le i \le k$$, we let $$p_i$$ denote the proportion of alleles in the parent population that are of type $$A_i$$. It is clear that this is the same as the proportion of alleles in the germ cells of the parent population, assuming that each parent produces roughly the same number of germs cells. Consider the distribution of alleles in the offspring. Since each germ cell is equally likely to be chosen for any particular offspring, the distribution of alleles in the offspring is the same as in the parents.
We next consider the distribution of genotypes in the two generations. We will prove the following fact: the distribution of genotypes in the offspring generation depends only upon the distribution of alleles in the parent generation (in particular, it does not depend upon the distribution of genotypes in the parent generation). Consider the possible genotypes; there are $$k(k+1)/2$$ of them. Under our assumptions, the genotype $$A_iA_i$$ will occur with frequency $$p_i^2$$, and the genotype $$A_iA_j$$, with $$i \ne j$$, will occur with frequency $$2p_ip_j$$. Thus, the frequencies of the genotypes depend only upon the allele frequencies in the parent generation, as claimed.
This means that if we start with a certain generation, and a certain distribution of alleles, then in all generations after the one we started with, both the allele distribution and the genotype distribution will be fixed. This last statement is known as the Hardy-Weinberg Law.
We can describe the consequences of this law for the distribution of heights among adults of one sex in a population. We recall that the height of an offspring was given by a random variable $$H$$, where $H = X_1 + X_2 + \cdots + X_n + W\ ,$ with the $$X_i$$’s corresponding to the genes that affect height, and the random variable $$W$$ denoting non-genetic effects. The Hardy-Weinberg Law states that for each $$X_i$$, the distribution in the offspring generation is the same as the distribution in the parent generation. Thus, if we assume that the distribution of $$W$$ is roughly the same from generation to generation (or if we assume that its effects are small), then the distribution of $$H$$ is the same from generation to generation. (In fact, dietary effects are part of $$W$$, and it is clear that in many human populations, diets have changed quite a bit from one generation to the next in recent times. This change is thought to be one of the reasons that humans, on the average, are getting taller. It is also the case that the effects of $$W$$ are thought to be small relative to the genetic effects of the parents.)
Discussion
Generally speaking, the Central Limit Theorem contains more information than the Law of Large Numbers, because it gives us detailed information about the of the distribution of $$S_n^*$$; for large $$n$$ the shape is approximately the same as the shape of the standard normal density. More specifically, the Central Limit Theorem says that if we standardize and height-correct the distribution of $$S_n$$, then the normal density function is a very good approximation to this distribution when $$n$$ is large. Thus, we have a computable approximation for the distribution for $$S_n$$, which provides us with a powerful technique for generating answers for all sorts of questions about sums of independent random variables, even if the individual random variables have different distributions.
Historical Remarks
In the mid-1800’s, the Belgian mathematician Quetelet7 had shown empirically that the normal distribution occurred in real data, and had also given a method for fitting the normal curve to a given data set. Laplace8 had shown much earlier that the sum of many independent identically distributed random variables is approximately normal. Galton knew that certain physical traits in a population appeared to be approximately normally distributed, but he did not consider Laplace’s result to be a good explanation of how this distribution comes about. We give a quote from Galton that appears in the fascinating book by S. Stigler9 on the history of statistics:
First, let me point out a fact which Quetelet and all writers who have followed in his paths have unaccountably overlooked, and which has an intimate bearing on our work to-night. It is that, although characteristics of plants and animals conform to the law, the reason of their doing so is as yet totally unexplained. The essence of the law is that differences should be wholly due to the collective actions of a host of independent influences in various combinations...Now the processes of heredity...are not petty influences, but very important ones...The conclusion is...that the processes of heredity must work harmoniously with the law of deviation, and be themselves in some sense conformable to it.
Galton invented a device known as a quincunx (now commonly called a Galton board), which we used in Example 3.2.1 to show how to physically obtain a binomial distribution. Of course, the Central Limit Theorem says that for large values of the parameter $$n$$, the binomial distribution is approximately normal. Galton used the quincunx to explain how inheritance affects the distribution of a trait among offspring.
We consider, as Galton did, what happens if we interrupt, at some intermediate height, the progress of the shot that is falling in the quincunx. The reader is referred to Figure [fig 9.62]. This figure is a drawing of Karl Pearson,10 based upon Galton’s notes. In this figure, the shot is being temporarily segregated into compartments at the line AB. (The line A$$^{\prime}$$B$$^{\prime}$$ forms a platform on which the shot can rest.) If the line AB is not too close to the top of the quincunx, then the shot will be approximately normally distributed at this line. Now suppose that one compartment is opened, as shown in the figure. The shot from that compartment will fall, forming a normal distribution at the bottom of the quincunx. If now all of the compartments are opened, all of the shot will fall, producing the same distribution as would occur if the shot were not temporarily stopped at the line AB. But the action of stopping the shot at the line AB, and then releasing the compartments one at a time, is just the same as convoluting two normal distributions. The normal distributions at the bottom, corresponding to each compartment at the line AB, are being mixed, with their weights being the number of shot in each compartment. On the other hand, it is already known that if the shot are unimpeded, the final distribution is approximately normal. Thus, this device shows that the convolution of two normal distributions is again normal.
Galton also considered the quincunx from another perspective. He segregated into seven groups, by weight, a set of 490 sweet pea seeds. He gave 10 seeds from each of the seven group to each of seven friends, who grew the plants from the seeds. Galton found that each group produced seeds whose weights were normally distributed. (The sweet pea reproduces by self-pollination, so he did not need to consider the possibility of interaction between different groups.) In addition, he found that the variances of the weights of the offspring were the same for each group. This segregation into groups corresponds to the compartments at the line AB in the quincunx. Thus, the sweet peas were acting as though they were being governed by a convolution of normal distributions.
He now was faced with a problem. We have shown in Chapter 7 and Galton knew, that the convolution of two normal distributions produces a normal distribution with a larger variance than either of the original distributions. But his data on the sweet pea seeds showed that the variance of the offspring population was the same as the variance of the parent population. His answer to this problem was to postulate a mechanism that he called , and is now called . As Stigler puts it:11
The seven groups of progeny were normally distributed, but not about their parents’ weight. Rather they were in every case distributed about a value that was closer to the average population weight than was that of the parent. Furthermore, this reversion followed “the simplest possible law," that is, it was linear. The average deviation of the progeny from the population average was in the same direction as that of the parent, but only a third as great. The mean progeny reverted to type, and the increased variation was just sufficient to maintain the population variability.
Galton illustrated reversion with the illustration shown in Figure [fig 9.63].12 The parent population is shown at the top of the figure, and the slanted lines are meant to correspond to the reversion effect. The offspring population is shown at the bottom of the figure.
Exercise $$\PageIndex{1}$$
A die is rolled 24 times. Use the Central Limit Theorem to estimate the probability that
1. the sum is greater than 84.
2. the sum is equal to 84.
Exercise $$\PageIndex{2}$$
A random walker starts at 0 on the $$x$$-axis and at each time unit moves 1 step to the right or 1 step to the left with probability 1/2. Estimate the probability that, after 100 steps, the walker is more than 10 steps from the starting position.
Exercise $$\PageIndex{3}$$
A piece of rope is made up of 100 strands. Assume that the breaking strength of the rope is the sum of the breaking strengths of the individual strands. Assume further that this sum may be considered to be the sum of an independent trials process with 100 experiments each having expected value of 10 pounds and standard deviation 1. Find the approximate probability that the rope will support a weight
1. of 1000 pounds.
2. of 970 pounds.
Exercise $$\PageIndex{4}$$
Write a program to find the average of 1000 random digits 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9. Have the program test to see if the average lies within three standard deviations of the expected value of 4.5. Modify the program so that it repeats this simulation 1000 times and keeps track of the number of times the test is passed. Does your outcome agree with the Central Limit Theorem?
Exercise $$\PageIndex{5}$$
A die is thrown until the first time the total sum of the face values of the die is 700 or greater. Estimate the probability that, for this to happen,
1. more than 210 tosses are required.
2. less than 190 tosses are required.
3. between 180 and 210 tosses, inclusive, are required.
Exercise $$\PageIndex{6}$$
A bank accepts rolls of pennies and gives 50 cents credit to a customer without counting the contents. Assume that a roll contains 49 pennies 30 percent of the time, 50 pennies 60 percent of the time, and 51 pennies 10 percent of the time.
1. Find the expected value and the variance for the amount that the bank loses on a typical roll.
2. Estimate the probability that the bank will lose more than 25 cents in 100 rolls.
3. Estimate the probability that the bank will lose exactly 25 cents in 100 rolls.
4. Estimate the probability that the bank will lose any money in 100 rolls.
5. How many rolls does the bank need to collect to have a 99 percent chance of a net loss?
Exercise $$\PageIndex{7}$$
A surveying instrument makes an error of $$-2$$$$-1$$, 0, 1, or 2 feet with equal probabilities when measuring the height of a 200-foot tower.
1. Find the expected value and the variance for the height obtained using this instrument once.
2. Estimate the probability that in 18 independent measurements of this tower, the average of the measurements is between 199 and 201, inclusive.
Exercise $$\PageIndex{8}$$
For Example $$\PageIndex{9}$$ estimate $$P(S_{30} = 0)$$. That is, estimate the probability that the errors cancel out and the student’s grade point average is correct.
Exercise $$\PageIndex{9}$$
Prove the Law of Large Numbers using the Central Limit Theorem.
Exercise $$\PageIndex{10}$$
Peter and Paul match pennies 10,000 times. Describe briefly what each of the following theorems tells you about Peter’s fortune.
1. The Law of Large Numbers.
2. The Central Limit Theorem.
Exercise $$\PageIndex{11}$$
A tourist in Las Vegas was attracted by a certain gambling game in which the customer stakes 1 dollar on each play; a win then pays the customer 2 dollars plus the return of her stake, although a loss costs her only her stake. Las Vegas insiders, and alert students of probability theory, know that the probability of winning at this game is 1/4. When driven from the tables by hunger, the tourist had played this game 240 times. Assuming that no near miracles happened, about how much poorer was the tourist upon leaving the casino? What is the probability that she lost no money?
Exercise $$\PageIndex{12}$$
We have seen that, in playing roulette at Monte Carlo (Example [exam 6.7]), betting 1 dollar on red or 1 dollar on 17 amounts to choosing between the distributions $m_X = \pmatrix{ -1 & -1/2 & 1 \cr 18/37 & 1/37 & 18/37\cr }$ or $m_X = \pmatrix{ -1 & 35 \cr 36/37 & 1/37 \cr }$ You plan to choose one of these methods and use it to make 100 1-dollar bets using the method chosen. Using the Central Limit Theorem, estimate the probability of winning any money for each of the two games. Compare your estimates with the actual probabilities, which can be shown, from exact calculations, to equal .437 and .509 to three decimal places.
Exercise $$\PageIndex{13}$$
In Example $$\PageIndex{9}$$ find the largest value of $$p$$ that gives probability .954 that the first decimal place is correct.
Exercise $$\PageIndex{14}$$
It has been suggested that Example $$\PageIndex{9}$$ is unrealistic, in the sense that the probabilities of errors are too low. Make up your own (reasonable) estimate for the distribution $$m(x)$$, and determine the probability that a student’s grade point average is accurate to within .05. Also determine the probability that it is accurate to within .5.
Exercise $$\PageIndex{15}$$
Find a sequence of uniformly bounded discrete independent random variables $$\{X_n\}$$ such that the variance of their sum does not tend to $$\infty$$ as $$n \rightarrow \infty$$, and such that their sum is not asymptotically normally distributed. | 2018-07-20T14:41:26 | {
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https://books.openbookpublishers.com/10.11647/obp.0075/Chapters/P65.html | GO TO... Contents Copyright BUY THE BOOK
Problem 65: A knock-out tournament ($✓$) 1987 Specimen Paper II
A tennis tournament is arranged for ${2}^{n}$ players. It is organised as a knockout tournament, so that only the winners in any given round proceed to the next round. Opponents in each round except the final are drawn at random, and in any match either player has a probability $\frac{1}{2}$ of winning. Two players are chosen at random before the start of the first round. Find the probabilities that they play each other:
(i)
in the first round;
(ii)
in the final round;
(iii)
in the tournament.
Comments
Note that the set-up is not the usual one for a tennis tournament, where the only random element is in the first round line-up. Two players cannot then meet in the final if they are in the same half of the draw.
Part (i) is straightforward, but parts (ii) and (iii) need a bit of thought. There is a short way and a long way of tackling these parts, and both have merits.
It is a good plan to check your answers, if possible, by reference to simple special cases where you can see what the answers should be; $n=1$ or $n=2$, for example.
Interestingly, the answers are independent of the probability that the players have of winning a match; the $2$s in the answers represent the number of players in each match rather than (the reciprocal of) the probability that each player has of winning a match. It also does not matter how the draw for each round is made. This is clear if you use the short method mentioned above.
Solution to problem 65
Call the two players ${P}_{1}$ and ${P}_{2}$.
(i) Once ${P}_{1}$ has been given a slot, there are ${2}^{n}-1$ slots for ${P}_{2}$, in only one of which will he or she play ${P}_{1}$. The probability of ${P}_{1}$ playing ${P}_{2}$ is therefore
$\frac{1}{{2}^{n}-1}.$
Note that this works for $n=1$ and $n=2$.
(ii) Long way. To meet in the final, ${P}_{1}$ and ${P}_{2}$ must each win every round before the final, and must also not meet before the final. The probability that ${P}_{1}$ and ${P}_{2}$ do not meet in the first round and that they both win their first round matches, is
The probability that they win each round and do not meet before the final (i.e. for $n-1$ rounds) is
(ii) Short way. Since all processes are random here, the probability that any one pair contests the final is the same as that for any other pair. There are a total of $\frac{1}{2}×{2}^{n}\left({2}^{n}-1\right)$ different pairs, so the probability for any given pair is $1∕\left[{2}^{n-1}\left({2}^{n}-1\right)\right]$.
(iii) Long way. We need to add the probabilities that ${P}_{1}$ and ${P}_{2}$ meet in each round. The probability that they meet in the $k$th round is the probability that they reach the $k$th round times the probability that they meet in the $k$th round given that they reach it, the latter (conditional) probability being $1∕\left({2}^{n-k+1}-1\right)$, as can be inferred from part (i). As in part (ii), the probability that they reach the $k$th round is
$\frac{1}{{2}^{k-1}}\frac{{2}^{n-k+1}-1}{{2}^{n}-1},$
so the probability that they meet in the $k$th round is
$\frac{1}{{2}^{k-1}}\frac{{2}^{n-k+1}-1}{{2}^{n}-1}×\frac{1}{{2}^{n-k+1}-1}=\frac{1}{{2}^{k-1}}\frac{1}{{2}^{n}-1}.$
Summing this as a geometric progression from $k=1$ to $n$ gives $1∕{2}^{n-1}$.
(iii) Short way. By the same short argument as in part (ii), the probability of a given pair meeting in any given match (not necessarily the final) is $1∕\left[{2}^{n-1}\left({2}^{n}-1\right)\right]$. Since the total number of matches is ${2}^{n}-1\phantom{\rule{0.3em}{0ex}}$ (because one match is needed to knock out each player, and all players except one get knocked out), the probability of a given pair playing is
$\frac{{2}^{n}-1}{{2}^{n-1}\left({2}^{n}-1\right)}=\frac{1}{{2}^{n-1}}.$
Post-mortem
If (like me) you plodded through this question the long way, you might be wondering how you were supposed to think of the short way. Instead of working out what happens to individual players as they progress through the tournament, you think about the space of all possible outcomes (the sample space), and attach a probability to each. That way, you can use the symmetry between all the players to help you. | 2022-08-18T16:05:49 | {
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https://math.stackexchange.com/questions/3325615/surface-area-of-an-ellipsoid-above-a-given-plane | # Surface area of an ellipsoid above a given plane
The problem:
$$\text{Set-up an integral for the surface area of the portion of the ellipsoid } 4x^2 + 9y^2 + z^2 = 64 \text{ that lies above the plane } z=-1 \text{. Do not simplify or evaluate the integral.}$$
$$SA = \int_{-4}^4 \int_{ \sqrt{\frac{64}{9}-\frac{4x^2}{9}}}^{\sqrt{\frac{64}{9}-\frac{4x^2}{9}}} \sqrt{1+ (\frac{-4x}{\sqrt{64-4x^2-9y^2}})^2+ (\frac{-9y}{\sqrt{64-4x^2-9y^2}})^2} dA$$
$$-\int_{-\sqrt{\frac{63}{4}}}^{\sqrt{\frac{63}{4}}} \int_{ \sqrt{\frac{63}{9}-\frac{4x^2}{9}}}^{\sqrt{\frac{63}{9}-\frac{4x^2}{9}}} \sqrt{1+ (\frac{-4x}{\sqrt{64-4x^2-9y^2}})^2+ (\frac{-9y}{\sqrt{64-4x^2-9y^2}})^2} dA$$ Where first double integral represents the whole ellipsoid and the second double integral represents the part below the surface.
I'm not exactly sure how he got much of his answer. The radical looks akin to the Jacobian when you take the area of function bounded by a plane. The second integration in the first double integral looks like he was solving for y to me but I'm having a hard time fitting the pieces together.
• Are you sure there was not a "$2$" in front of the first surface integral? – JG123 Aug 17 '19 at 15:25
• yes, I've attached an image of his work – financial_physician Aug 17 '19 at 16:07
• but if you think it makes more sense with a 2 and could explain how you got where you're at, might help us make some progress – financial_physician Aug 17 '19 at 16:23
Well here is my rationale:
The ellipsoid in question can be represented by z=$$\pm(\sqrt {64-4x^2-9y^2}$$).
The "top half" is given by z=$$(\sqrt {64-4x^2-9y^2}$$), with $$4x^2+9y^2\le64$$.
In other words, we have that: -$$(\frac{\sqrt {64-4x^2}}{3})$$ $$\le$$y$$\le(\frac{\sqrt {64-4x^2}}{3})$$
and $$-4\le$$x$$\le4$$
Hence, the surface area of this "top half" will be the first surface integral that your professor wrote.
Now consider the portion of the ellipsoid above the plane $$z=1$$ (which will have the same surface area as the portion below the plane $$z=-1$$).
We now are computing a surface integral with the same surface z=$$(\sqrt {64-4x^2-9y^2}$$) but we have that $$4x^2+9y^2\le63$$. (This can be obtained by plugging $$z=1$$ into the expression for the ellipsoid in question).
In other words, we have that: -$$(\frac{\sqrt {63-4x^2}}{3})$$ $$\le$$y$$\le(\frac{\sqrt {63-4x^2}}{3})$$
and -$$\frac{63}{4}$$ $$\le$$ x $$\le$$ $$\frac{63}{4}$$.
Hence the surface area of the portion of the ellipsoid below the plane $$z=-1$$ will be the second surface integral that your professor wrote.
We know that the desired surface area will be: $$SA_{ellipsoid}$$-$$SA_{S}$$, where $$S$$ is the portion of the ellipsoid below the plane $$z=-1$$.
Does it not then make sense that the desired surface area will be 2(1st surface integral)-(2nd surface integral) as the 1st surface integral only gives the surface area of the top half of the ellipsoid? | 2020-01-25T20:18:02 | {
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https://math.stackexchange.com/questions/2202311/why-does-epsilon-come-first-in-the-epsilon-delta-definition-of-limit?noredirect=1 | # Why does $\epsilon$ come first in the $\epsilon-\delta$ definition of limit? [duplicate]
As we know, $\underset{x\rightarrow c}{\lim}f(x)=L\Leftrightarrow$ for every $\epsilon>0$ there exists $\delta>0$ such that if $0<|x-c|<\delta$, then $|f(x)-L|<\epsilon$.
My question is: why do we say that for every $\epsilon>0$ there exists $\delta>0$ and not vice versa, why not for every $\delta>0$ there exists $\epsilon>0$?
• If you are not just talking about changing names then the meaning gets lost. If e.g. $f$ is bounded then for every $L$ it is true that for every $\delta>0$ we can find some $\epsilon>0$ such that $|f(x)-L|<\epsilon$ for $|x-c|<\delta$. – drhab Mar 25 '17 at 10:49
Good question!
The answer is that if we put $\delta$ first, then our definition would no longer correspond to what we mean by the limit of a function.
Proposition Let $f$ be the constant function $0$. Let $c,L$ be any real numbers. Then:
For all $\delta>0$ there exists $\epsilon>0$ such that if $0<|x-c|<\delta$, then $|f(x)-L|<\epsilon$.
In other words, if we define $\lim'$ by putting the $\delta$ first, then $\lim_{x\to c}'f(x)=L$.
Proof. Let $\delta<0$ and let $\epsilon=|L|+1$. If $0<|x-c|<\delta$ then $$|f(x)-L| = |0-L|=|L|<|L|+1=\epsilon\quad\Box$$
So our constant function $0$, which should clearly converge to $0$ at every point, under the new definition converges to every real number at every point.
In particular, limits aren't unique and the properties we expect to hold for a 'limit' no longer hold. It just isn't a very interesting definition and it no longer captures our intuition of what limits should be.
By contrast, the definition with $\epsilon$ first exactly captures the notion of a 'limit'.
As John Gowers explained in his answer, if you reverse the order of quantification, you can say that $\lim_{x \to 0} x = \pi$ or $\lim_{x \to 0} x = -1$. The modified definition is clearly not that of limit. Let's use different notation and write $\text{mil}_{x \to c} f(x) = L$ for the "backwards" definition.
What have we defined? If there exist $c$ and $L$ in $\mathbb{R}$ such that $\text{mil}_{x \to c} f(x) = L$, it means that for all $c' \in \mathbb{R}$, $\lim_{x \to c'} f(x) \not\in \{\infty, -\infty\}$.
In fact, if $f$ diverges for some $c' \in \mathbb{R}$, and $\delta$ is chosen large enough, so that $|c'-c| < \delta$, then we cannot find $\epsilon$ such that $|f(x)-L| < \epsilon$ for all $x$ such that $|x-c| < \delta$. On the other hand, if $f$ does not go to infinity for finite argument, we can always pick a suitably large $\epsilon$.
We can further note that if there exist $c$ and $L$ in $\mathbb{R}$ such that $\text{mil}_{x \to c} f(x) = L$, then for all $c'$ and $L'$ in $\mathbb{R}$, it is also true that $\text{mil}_{x \to c'} f(x) = L'$. This tells us that there are better ways to define finite-valued functions. Let's not go down that route. Instead, note that in the previous paragraph, I used the word "large" twice.
If we try to explain the definition of $\lim$, we may say, "for every $\epsilon$, however small, there's a sufficiently small $\delta$." The definition of $\text{mil}$, however, would be read, "for every $\delta$, no matter how large, there is a sufficiently large $\epsilon$."
Clearly, neither "small" nor "large" appears in the definitions. However, when we try to elucidate one of these definitions, we intuitively think of it as defining the rules of a two-player game. Let the two players be Adam ($\forall$dam) and Eve ($\exists$ve).
Eve chooses the values of the existentially quantified variables; Adam chooses the values of the universally quantified variables. Eve tries to make the claim true; Adam tries to make it false. If Adam picks $\epsilon$ first, his way to make it hard for Eve to respond is to choose a small $\epsilon$. In fact, if Adam wins by choosing $\epsilon_0$, he also wins with any $\epsilon$ such that $0 < \epsilon \leq \epsilon_0$.
However, if Adam chooses $\delta$ first, he's better off choosing a large $\delta$. Similar considerations apply to Eve's choices. Bottom line: giving $\delta$ to Adam and $\epsilon$ to Eve results in a very different game; one that does not define limits.
Changing the order of the picks also changes the game substantially. Asking Eve to go first, and pick $\delta$ before Adam picks $\epsilon$ puts her at a great disadvantage. (Try it.) The resulting game, once again, does not define limits.
It is also possible for a change in the order of the picks to result in two different games that define related, yet distinct notions. This is the case of uniform continuity vs. continuity. Winning the uniform continuity game is harder for Eve, but the class of functions on which she can win is an interesting one, and both definitions are therefore in use.
Try to see the notion of limit as a tool to check continuity. We basically want to make sure that as the independent variable x approaches the value of c, the value of the function gets arbitrarily close to f(c), otherwise we would lose continuity. This property belongs to the function - it exists to tell us something $\textit{about the function}$.
If you were to check that $\forall \delta > 0 \ \exists \epsilon > 0 \dots$ , you would not know whether the function is continuous or not; you would only acquire an upper bound on the distance between the values of f(x) and f(c) when $|x-c|<\delta$. This upper bound is not enough to make sure that f(x) gets arbitrarily close to f(c) as x approaches c, therefore you have no information about continuity.
I think most of the other answers are missing the point. Roughly, the definition says that if the inputs are within $\delta$ distance of each other, then the outputs are within $\epsilon$ distance of each other. It makes sense to use $\delta$ for an input and $\epsilon$ for an output, in the same way we use $x$ and $y$ in elementary algebra.
The reason it's called an $\epsilon - \delta$ definition is because when stating the definition, we say "For all $\epsilon>0$, there exists a $\delta>0$ such that ...", and so this first line gives these definitions the name $\epsilon - \delta"$.
• Of course it's possible that I am the one misreading the question, but here is an answer anyways to some question. – AlexanderJ93 Mar 25 '17 at 10:59
The idea of the limit is that no matter how close you get to the value of $L$ (by setting the arbitrary small region around $L$, there is a neighbourhood of $c$ where $f(x)$ is within this region.
The opposite, would make little sense - of course you can pick up some region around $L$ that fits the values of $f(x)$ where the $x$ is in the arbitrary neighbourhood around $c$.
Say, for the function $f(x) = -1$ when $x < 0$ and $f(x) = 1$ when $x \ge 0$, you can pick $\delta = 3$ for any value of $\epsilon$ for any value of $c$. | 2020-02-23T12:11:52 | {
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https://math.stackexchange.com/questions/1375925/when-does-sum-i-1-infty-x-i-exist-for-random-sequences-x-i-i-1 | # When does $\sum_{i=1}^{\infty} X_i$ exist for random sequences $\{X_i\}_{i=1}^{\infty}$?
Suppose $\{X_1, X_2, X_3, \ldots\}$ is an infinite sequence of random variables such that $E[X_i]=0$ for all $i$, and $E[X_iX_j]=0$ whenever $i \neq j$. Further suppose the variances $\sigma_i^2 = E[X_i^2]$ are finite and satisfy $\sum_{i=1}^{\infty} \sigma_i^2 < \infty$.
Define $S_n = \sum_{i=1}^nX_i$. When can one conclude that $S_n$ converges almost surely as $n\rightarrow\infty$?
I can show that, with prob 1, the limit exists (and is a real number) over a particular subsequence $n[m]$, so that $\lim_{m\rightarrow\infty}S_{n[m]}$ exists as a (random) real number. I suspect that, in general, we do not have convergence. However, counter-examples seem hard.
• – dsaxton Jul 27 '15 at 19:41
• I suppose $S_n$ can be viewed as a Cauchy sequence in a metric space with inner product $<X,Y> = E[XY]$, in which case a counter-example might be inspired by examples of incomplete metric spaces. – Michael Jul 27 '15 at 19:50
• Kolmogorov 3 series criterion is for independent seq. of r.v.'s – Saty Jul 27 '15 at 19:56
• If one replaces the requirement $E[X_iX_j]=0$ for all $i \neq j$ with the more stringent requirement that $E[X_n|X_1, \ldots, X_{n-1}]=0$ for all $n\geq 2$, then almost sure convergence follows from the Doob martingale convergence theorem, since $S_n$ is then a martingale with $$\sup_n E[|S_n|] \leq \sup_n \sqrt{E[S_n^2]} < \infty$$. – Michael Jul 27 '15 at 22:14
• @Michael: No, your claim is fine. :) I meant to say that if the question posed by the op turned out to be true in general, this would imply Carlesons theorem. The correct link I meant is math.stackexchange.com/questions/1376110/… – PhoemueX Jul 28 '15 at 6:30
This answer is first adapted from comments in: does convergence in $L^p$ imply convergence almost everywhere. Consider a probability space $[0, 1]$ with uniform measure. Define the random variable $X_{2^i + k}$ by $$X_{2^i + k} = 2^{i/2}\chi_{\left[\frac{k}{2^i}, \frac{k+1/2}{2^i}\right]} - 2^{i/2}\chi_{\left[\frac{k + 1/2}{2^i}, \frac{k+1}{2^i}\right]}$$ for $i \geq 1, 0 \leq k < 2^i$. These random variables (known as Haar Functions) have mean zero. Furthermore, they are uncorrelated. However, the sum of these random variable do not satisfy the requirements of your theorem because $Var(\sum_{n=1}^\infty X_n) = \infty$.
Here we change our point of view to restate the question as follows: Does there exist an orthonormal basis of $L^2([0,1])$ such that an element written in this basis converges in $L^2$ but does not converge pointwise almost everywhere. I asked this question in this thread and David Mitra was kind enough to point me towards the paper: Topics in Orthogonal Functions - Price. On pg. 598 the author states the Haar functions defined above form a complete orthonormal basis. Then pg. 603 states
For every complete orthonormal system $\Phi$ there is an $L^2$ function $Y$ whose $\Phi$-Fourier series can be rearranged to diverge almost everywhere.
So we can finish as follows. Take the Haar functions as above (a complete orthonormal set) and the function $Y$ defined above. We write this function's $\Phi$-Fourier series as $$\sum_{n=1}^\infty \langle Y, X_n \rangle X_n$$ Now there exists a rearrangement $\sigma$ of the indices of sum such that it diverges almost everywhere. Now let $Y_n = \langle Y, X_{\sigma(n)} \rangle X_{\sigma(n)}$ be this rearrangement of terms. We can see $$\sum_{n=1}^\infty Y_n$$ satisfies the requirements of the problem. Indeed $$\sum_{n=1}^\infty\operatorname{Var} (Y_n) = \operatorname{Var}(Y) < \infty.$$ Furthermore, $\sum Y_n$ diverges almost everywhere.
Note: A fun corollary is that if you have a sequence $X_n$ that does converge almost everywhere (and satisfies the questions requirements), and it forms a complete basis (its already practically orthonormal), then a rearrangement of itself diverges almost everywhere.
• @Michael - good point - that is less ambiguous. Thanks. – muaddib Jul 28 '15 at 22:32
• Theorem 14 of that paper (page 608) suggests that a randomized ordering of the $Y_n$ variables gets us back to probability 1 convergence. – Michael Jul 28 '15 at 22:41
• I think by "diverges almost everywhere" they just mean the limit does not exist (not that it goes to infinity). According to my comment in the question, the limit should indeed exist as a real number almost everywhere over a subsequence, even for those crazily arranged variables. – Michael Jul 28 '15 at 22:45
• @Michael - Not sure if if you are referring to my last Note. There I meant there exists a specific reordering that causes it to diverge. But, yes, I see by Thm 14 that the "generic" rearrangement does converge. Neat, and odd stuff. – muaddib Jul 28 '15 at 22:45 | 2019-10-17T18:27:35 | {
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https://math.stackexchange.com/questions/2018801/find-the-eigenvectors-of-an-unknown-matrix-transform | # Find the eigenvectors of an unknown matrix transform?
I have this problem on a homework that I must do two ways to receive full credit on. I attempted the problem one way and was wondering if it was right, and if anybody could give me any hints or insights on how to approach the problem another way.
$\textbf{Question}:$ A certain $4$x$4$ matrix transforms $\begin{bmatrix}a\\b\\c\\d\end{bmatrix}\rightarrow\begin{bmatrix}b\\a\\d\\c\end{bmatrix}$. Find all of its eigenvectors. You must do this problem with two different approaches to receive full credit.
$\textbf{My Attempt}:$ If we represent the matrix like this $$\begin{bmatrix}x_1&x_2&x_3&x_4\\y_1&y_2&y_3&y_4\\z_1&z_2&z_3&z_4\\f_1&f_2&f_3&f_4\end{bmatrix}$$ and multiply is against the the original vector, we get $$ax_1+bx_2+cx_3+dx_4=b\\ay_1+by_2+cy_3+dy_4=a\\az_1+bz_2+cz_3+dc_4=d\\af_1+bf_2+cf_3+df_4=c$$ which tells me which tells me that every row will have one $1$ in it and the rest of the entries will be zero. So the transformation matrix would be $$\begin{bmatrix}0&1&0&0\\1&0&0&0\\0&0&0&1\\0&0&1&0\end{bmatrix}$$ Now that I have this matrix, I think it will be easy enough to find its eigenvectors using $det(A-\lambda I)=0$, etc.
Is this solution correct? Can anyone think of any other ways to solve this question so that I can get full credit? I was thinking maybe there could be two ways to find the eigenvectors now that I have found the matrix, which would count as two different approaches to the problem, but I can't think of how I would do that. Or is there another way to find the transform? Thanks so much for any help you guys can give!
• Work directly with the definition of the transformation and that of an eigenvector: $A\mathbf v=\lambda\mathbf v$. – amd Nov 17 '16 at 18:46
You have correctlty identified one way.
Now consider what eigenvector is. It is the vector that the matrix transforms to a multiple of the eigenvector.
So we're talking about $\begin{bmatrix}a\\b\\c\\d\end{bmatrix}\rightarrow\begin{bmatrix}ka\\kb\\kc\\kd\end{bmatrix}$
But you have $\begin{bmatrix}a\\b\\c\\d\end{bmatrix}\rightarrow\begin{bmatrix}b\\a\\d\\c\end{bmatrix}$
Therefore an eigenvector will satisfy $\begin{bmatrix}ka\\kb\\kc\\kd\end{bmatrix} = \begin{bmatrix}b\\a\\d\\c\end{bmatrix}$
Taken together $ka=b$ and $kb=a$ yields $k^2a=a \Rightarrow k=1$ or $k=-1$ with eigenvectors $\begin{bmatrix}1\\1\\0\\0\end{bmatrix}$ and $\begin{bmatrix}1\\-1\\0\\0\end{bmatrix}$
Can you see what the other two eigenvectors must be? | 2021-08-02T18:11:28 | {
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https://math.stackexchange.com/questions/1708900/sum-of-sum-limits-n-0-infty-frac1kn | # Sum of $\sum \limits_{n=0}^{\infty} \frac{1}{(kn)!}$
Does a closed form exist for
$$\sum \limits_{n=0}^{\infty} \frac{1}{(kn)!}$$
in terms of $k$ and other functions? The best that I have been able to do is solve the case where $k=1$, since the sum is just the infinite series for $e$. I would guess that any closed form must involve the exponential function, but am at a loss to prove it.
• For small $k$ there are fairly easy to find closed forms involving functions like cos, cosh. In the general case the closed form is a messy expression involving the generalised hypergeometric function. – almagest Mar 22 '16 at 16:24
• sum diverges at $k=0$ and conveges to $\cos (ix)$ for $k=2$ – gt6989b Mar 22 '16 at 16:26
• @gt6989b you could also write $\cosh(1)$ for $k=2$ – zz20s Mar 22 '16 at 16:27
• If you could find a general function that has the property $f(x)=f^{(k)}(x)$, and $f^{(k)}(a)=1$ for all $k$ and fixed $a$, then you could make this the Taylor series of that? – Simply Beautiful Art Mar 23 '16 at 12:00
• Possible duplicate of Closed-forms of infinite series with factorial in the denominator – Yanior Weg Nov 27 '19 at 19:16
If $(c_n)$ is any sequence with period $k$ (that is, $c_{n+k}=c_n$) then it's possible to evaluate $\sum c_n/n!$ using tricks involving $k$-th roots of unity.
Let $\omega=e^{2\pi i/k}$. Consider the $k$ sequences
$s_0:1,1,1\dots$
$s_1: 1, \omega,\omega^2,\omega^3,\dots$
$s_2: 1, \omega^2,\omega^4,\omega^6,\dots$
$s_3: 1, \omega^3,\omega^6,\omega^9,\dots$
...
$s_{k-1}: 1, \omega^{k-1},\omega^{2(k-1)},\dots$.
Using tricks analogous to finding Fourier coefficients you can find $a_0,\dots a_{k-1}$ so that $$(c_n)=a_0(s_0)+\dots+a_k(s_{k-1}).$$
Hence $$\sum\frac{c_n}{n!}=\sum_{j=0}^{k-1}a_j\sum_n\frac{\omega^{jn}}{n!} =\sum_{j=0}^{k-1}a_je^{\omega^j}.$$
If you do that for your sequence you get $$\sum_{n=0}^\infty\frac{1}{(kn)!}=\frac1k\sum_{j=0}^{k-1}e^{\omega^j}.$$
Edit: Thomas Andrews makes a comment that I should have included: Since the original sum is real, it follows that $$\sum_{n=0}^\infty\frac{1}{(kn)!}=\frac{1}{k}\sum_{j=0}^{k-1}e^{\cos 2\pi j/k}\cos(\sin2\pi j/k).$$
Edit: If one is familiar with "abstract harmonic analysis", here just harmonic analysis on compact abelian groups, one sees that those "tricks analogous to finding Fourier coefficients" are in fact finding Fourier coefficients for a certain function on the group $\Bbb Z/k\Bbb Z$.
• And, since the imaginary part is necessarily zero, this also gives the formula:$$\frac{1}{k}\sum_{j=0}^{k-1}e^{\cos 2\pi j/k}\cos(\sin2\pi j/k)$$ – Thomas Andrews Mar 22 '16 at 16:56
• @ThomasAndrews Indeed, I shoulda said that, thanks. – David C. Ullrich Mar 22 '16 at 17:03
• Interestingly, this is a Riemann sum for $\int_{0}^{1} e^{\cos 2\pi x}\cos(\sin 2\pi x)\,dx$. – Thomas Andrews Mar 22 '16 at 17:49
• @ThomasAndrews Yes, that's interesting. As is the fact that it follows that that integral equals $1$. – David C. Ullrich Mar 22 '16 at 17:56
• Yeah, I was just trying to see if I could prove that directly. – Thomas Andrews Mar 22 '16 at 17:56
This is a different approach to the idea in David Ullrich's answer.
As long as $\frac nk\not\in\mathbb{Z}$, \begin{align} \sum_{j=0}^{k-1}e^{2\pi ij\frac nk} &=\frac{e^{2\pi in}-1}{e^{2\pi i\frac nk}-1}\\ &=0 \end{align} if $\frac nk\in\mathbb{Z}$, then \begin{align} \sum_{j=0}^{k-1}e^{2\pi ij\frac nk} &=\sum_{j=0}^{k-1}1\\ &=k \end{align} Thus, \begin{align} \sum_{n=0}^\infty\frac1{(kn)!} &=\sum_{n=0}^\infty\overbrace{\left(\frac1k\sum_{j=0}^{k-1}e^{2\pi ij\frac nk}\right)}^{1\iff\frac nk\in\mathbb{Z}}\frac1{n!}\\ &=\frac1k\sum_{j=0}^{k-1}\sum_{n=0}^\infty\frac{\left(e^{2\pi ij/k}\right)^n}{n!}\\ &=\frac1k\sum_{j=0}^{k-1}e^{\large e^{2\pi ij/k}}\\ &=\frac1k\sum_{j=0}^{k-1}e^{\cos(2\pi j/k)+i\sin(2\pi j/k)}\\ &=\frac1k\sum_{j=0}^{k-1}e^{\cos(2\pi j/k)}\left(\cos(\sin(2\pi j/k))+i\sin(\sin(2\pi j/k))\right)\\ &=\frac1k\sum_{j=0}^{k-1}e^{\cos(2\pi j/k)}\cos(\sin(2\pi j/k)) \end{align} The last step follows since the imaginary parts of the $j$ and $k-j$ terms cancel.
• Well, they cancel because the value is obviously real :) – Thomas Andrews Mar 22 '16 at 17:22
• @ThomasAndrews: indeed, but since we have introduced complex numbers, we should show that the computed answer is actually real. – robjohn Mar 22 '16 at 17:26
• I can't understand how did you write the first equality after "Thus,". Could you explain this a bit for me, please? – user486600 Aug 18 '18 at 6:49
• @user486600: for a proof that the quantity in parentheses is $[\,k\mid n\,]$ see the Preliminaries of this answer. – robjohn Aug 18 '18 at 11:56
Sum expressed by a special function:
$$\color{red}{\sum _{n=0}^{\infty } \frac{1}{(k n)!}}=\sum _{n=0}^{\infty } \frac{1}{\Gamma (k n+1)}=\color{red}{E_{k,1}(1)}$$
where: $\color{red}{E_{k,1}(1)}$ is generalized Mittag-Lefflere function. | 2020-03-29T03:45:10 | {
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https://math.stackexchange.com/questions/3394646/why-is-the-determinant-1-not-2 | # Why is the determinant $1$ not $-2$
I am trying to find the determinant of this matrix. Let $$A$$ be the matrix :
$$\begin{bmatrix}2&2&1\\1&0&5\\1&1&0\end{bmatrix}$$
Using row operations, I can change $$R_3$$ to ( $$-2R_3+R_1\rightarrowR_3$$). So, $$B$$ is the matrix :
$$\begin{bmatrix}2&2&1\\1&0&5\\0&0&1\end{bmatrix}$$
Then, using Laplace Expansion and expanding along the 2nd column,
$$det(A) = (2)(-1)^3 det(\begin{bmatrix}1&5\\0&1\end{bmatrix})= (2)(-1)(1) = -2$$
However, $$det(A)=det(B)$$ if a multiple of one row is added to another, but the determinant of $$A$$ is $$1$$. What am I doing wrong here?
Because of that $$-2R_3$$ that you wrote. When you do $$-2R_3+R_1\to R_3$$, part of you are doing is to multipliply the third row by $$-2$$, which changes the determinant of the matrix.
• But my textbook says adding a multiple of one row to another doesn't change the determinant, I'm just adding a multiple of $R_3$ to $R_1$. I apologize if it's an obvious question. Oct 15, 2019 at 11:38
• @MohamedTlili If you had done $-2R_3+R_1\to R_1$, then the determinant would've been unchanged. That would be "adding a multiple of $R_3$ to $R_1$". However, $-2R_3+R_1\to R_3$ doesn't just add a multiple of $R_1$ to $R_3$, it also multiplies $R_3$ by $-2$ first. Oct 15, 2019 at 11:38
• @MohamedTlili First you replaced $R_3$ with $-2R_3$, multiplying the determinant by $-2$. Then you added $R_1$ to the third row, which had no effect on the determinant. Writing down the intermediate matrix may help you follow this argument.
You're not adding a multiple of $$R_3$$ to $$R_1$$. You're replacing $$R_3$$ by a multiple of $$R_3$$ plus $$R_1$$.
I warn my students at the start that there's a place coming up later where it will be important to realize that (if $$j\ne k$$) $$R_j+cR_k\to R_j$$is a row operation, while $$R_j+cR_k\to R_k$$is not an official elementary row operation. This it that place. As long as $$c\ne0$$ both versions work fine for finding an echelon form, but the second changes the determinant, while the first does not. | 2022-05-27T04:35:05 | {
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https://www.physicsforums.com/threads/challenge-4-theres-no-app-for-that-integration.717360/ | # Challenge 4: There's no app for that: integration
Staff Emeritus
Gold Member
a.) Poor Wolfram Alpha got asked to calculate the following integral
$$\int_{0}^{\infty} e^{-ax} \frac{\sin(x)}{x} dx$$
but couldn't handle it!
http://www.wolframalpha.com/input/?i=int_{0}^{infty}+e^{-ax}+sin(x)+/xdx
(Results are not guaranteed if you use wolfram alpha pro to spend more time calculating as I don't have that).
Prove that you're smarter than a computer and solve this integral for all a > 0.
b.) For reasons unknown, Wolfram Alpha can solve this exactly if a is an integer, but it won't tell the steps for how it solved it. As a side challenge, instead calculate the integral in a way that only works if a is an arbitrary positive or non-negative (your choice) integer.
Feel free to answer either a, b or a and b.
mfb
Mentor
a.) Poor Wolfram Alpha got asked to calculate the following integral
$$\int_{0}^{\infty} e^{-ax} \frac{\sin(x)}{x} dx$$
but couldn't handle it!
Lie! :p.
If you enter a decimal number, the result is always a decimal number. If you enter an exact non-negative number (does not matter if integer or not, try 4/7 or pi or whatever), you get an exact result. A variable works as well, if you make sure it is positive.
a=0 leads to the sine integral and the result is pi/2.
Staff Emeritus
Gold Member
Well I figure if it can this it should be able to figure out the one in the challenge. At any rate it's a nice integral that admits a bunch of neat tricks to solving it so have at it!
I see now that perhaps part b was from my misunderstanding of how wolfram alpha operates - I tested it with things like a=2.1 and a=3.2 to see if it would calculate those and it just gave a decimal, but now I'm guessing that's just a quirk in how wolfram alpha interprets the output form that I want?
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How about if we start by letting:
$$I(a)=\int_0^{\infty} e^{-ax} \frac{\sin(x)}{x}dx,\quad I(0)=\pi/2$$
And it would be nice if we could compute $\frac{dI}{da}$ as that would rid the denominator of $x$. Not sure if that integrand satisfies Leibniz's criteria for differentiating under the integral sign though.
We have that $$\int_{0}^{∞}e^{-ax}\frac{\sin(x)}{x}dx = \lim_{t→∞}\int_{0}^{t}e^{-ax}\frac{\sin(x)}{x}dx = \lim_{t→∞} I(a, t)$$
According to http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign, Leibniz's theorem is true if both $e^{-ax}\frac{\sin(x)}{x}$ and $-e^{-ax}\sin(x)$ are continuous on the region $0 ≤ x ≤ t, 0 ≤ a < ∞$
This is true (obvious for any nonzero x). The only trouble is when x is zero.
Define $ψ(a,x) = e^{-ax}\frac{\sin(x)}{x}, x ≠ 0$ and $ψ( a, 0 ) = 1$
Then $ψ(a,x)$ is equal to the integrand except perhaps at x = 0, but one point does not affect the value of the integral $I(a,t) = \int_{0}^{t}e^{-ax}\frac{\sin(x)}{x}dx = \int_{0}^{t}ψ(a,x)dx$
But since $\lim_{x→0}\frac{\sin(x)}{x} = 1$, the limit and the value $ψ(a,0)$ coincide so $ψ$ is indeed continuous on the region of interest.
Now applying Leibniz's rule, we get:
$$\frac{∂}{∂a}I(a,t) = \int_{0}^{t}-e^{-ax}\sin(x)dx = e^{-at}\left( \frac{a}{a^2+1}\sin(t) + \frac{1}{a^2+1}\cos(t) \right) - \frac{1}{a^2+1}$$
This is bounded above by $e^{-at} - \frac{1}{a^2+1}$ and below by $-e^{-at}-\frac{1}{a^2+1}$
Therefore, the sequence of derivatives $\frac{∂}{∂a} I(a,t), t → ∞$ converges uniformly to $-\frac{1}{a^2+1}$. Moreover, the sequence $I(a,t), t→∞$ converges (pointwise) to $I(a) = \int_{0}^{∞}e^{-ax}\frac{\sin(x)}{x}dx$ (by definition). Moreover, the functions are continuous and therefore integrable. I think these conditions are sufficient to ensure that the derivatives $\frac{∂}{∂a} I(a,t), t → ∞$ converge to $\frac{d}{da}I(a)$. Therefore, $I'(a) = -\frac{1}{a^2+1}$ and so:
$I(a) = I(0) - \arctan(a)$ where $I(0) = \int_{0}^{∞}\frac{\sin(x)}{x}dx = \frac{\pi}{2}$ is the Dirichlet integral.
$\int_{0}^{∞}e^{-ax}\frac{\sin(x)}{x}dx = \frac{\pi}{2} - \arctan(a)$.
I hope I have not overlooked something!
EDIT:
I realized that uniform convergence holds only on $0 < a < ∞$ so the equation $I'(a) = -\frac{1}{a^2+1}$ is guaranteed only for $a > 0$. I think this can be patched by showing that $I(a)$ is continuous at 0.
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1 person
mfb
Mentor
Edit: Oh, has been solved 4 minutes before my post. I'll leave it here, maybe someone spots where I got wrong.
I tried a different approach, but I guess I have some sign/substitution errors somewhere.
First, let the lower integration border be ##\epsilon>0## and interpret the integral as limit of ##\epsilon \to 0## later.
Then
$$\int_{\epsilon}^{\infty} e^{-ax} \frac{\sin(x)}{x} dx = \frac{1}{2i} \int_{\epsilon}^{\infty} e^{-ax} \frac{e^{ix}-e^{-ix}}{x} dx = \frac{1}{2i} \left( \int_{\epsilon}^{\infty} \frac{1}{x}e^{-ax+ix} dx\; +\; \int_{\epsilon}^{\infty} \frac{1}{x}e^{-ax-ix} dx \right)$$
Consider the first integral:
$$\int_{\epsilon}^{\infty} \frac{1}{x}e^{-ax+ix} dx$$
This can be seen as integral in the complex numbers. The integrand has a pole at x=0 and nowhere else.
Define:
##\displaystyle \theta=arctan(1/a)## (this is nice!) and ##A=|a+i|=\sqrt{a^2+1}##
We can use the upper red loop in the sketch to evaluate it. As the size of the loop goes to infinity, the contribution from the large curved section goes to zero (due to e-ax).
Let ##x=\frac{a+i}{A}r##. Then we can write the contribution from the inclined section as real integral:
$$\int_\epsilon^{\infty} \frac{A}{(a+i)r}e^{-Ar} dr = \frac{A}{a+i} \int_{\epsilon A}^{\infty} \frac{e^{-r}}{r} dr$$
We cannot take the limit ##\epsilon \to 0## at this point, and we cannot evaluate it either, so we just leave it. We have to count this negatively in the loop.
What about the small curved section? This is the classic integral over f(z)/z, the result is ##\theta f(0)## plus some expression that will vanish in the limit ##\epsilon \to 0## afterwards. Here, f(0)=e0=1, so the result is just ##\theta##. Again, we have to count this negatively.
Consider the second integral:
$$\int_{\epsilon}^{\infty} \frac{1}{x}e^{-ax-ix} dx$$
Follow the same procedure, just with ##x=\frac{a-i}{A}r##. The integral becomes
$$\int_\epsilon^{\infty} \frac{A}{(a-i)r}e^{-Ar} dr = \frac{A}{(a-i)} \int_{\epsilon A}^{\infty} \frac{e^{-r}}{r} dr$$
The small curved section gives ##\theta## again, this time with a positive sign, so the contributions cancel.
Our total integral is now 1/(2i) times the sum of the two real integrals obtained above:
$$\frac{1}{2i} A \frac{(a-1)+(a+1)}{A^2} \int_{\epsilon A}^{\infty} \frac{e^{-r}}{r} dr = -i \frac{a}{A} \int_{\epsilon A}^{\infty} \frac{e^{-r}}{r} dr$$
And here the problem is visible. This integral does not converge for ##\epsilon \to 0## (and the result is purely imaginary, literally and figuratively.).
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Yeah, that's what I need Boorglar: rigor to justify taking the derivative that way. Thanks.
Also mfb, spotted some mistakes:
(1) The first integral expression should be:
$$\int_{\epsilon}^{\infty} e^{-ax} \frac{\sin(x)}{x} dx = \frac{1}{2i} \int_{\epsilon}^{\infty} e^{-ax} \frac{e^{ix}-e^{-ix}}{x} dx = \frac{1}{2i} \left( \int_{\epsilon}^{\infty} \frac{1}{x}e^{-ax+ix} dx\; -\; \int_{\epsilon}^{\infty} \frac{1}{x}e^{-ax-ix} dx \right)$$
(2) An integral over an indentation around a simple pole at the origin for $f(z)/z$ (as radius goes to zero) is $\theta i f(0)$. You forgot the i.
Did it quick but looks like it ends up being:
$$\frac{1}{2 i} \left(i\arctan(1/a)+\int_0^{\infty} f(r)dr+i\arctan(1/a)-\int_0^{\infty}f(r) dr\right)$$
which then gives us the requisite $\arctan(1/a).$
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mfb
Mentor
Oh, that sign was a stupid mistake.
I forgot the i, but as the contributions canceled (due to the sign error, crap) I did not check that again.
I'm not sure if the prefactors work out properly, if I just change the sign I don't get matching prefactors for the two integrals.
I'm not sure if the prefactors work out properly, if I just change the sign I don't get matching prefactors for the two integrals.
Ok, just to check it, I'll just let $z=re^{i\theta}$ on the top path and $z=re^{-i\theta}$ on the bottom path with $\theta=\arctan(1/a)$
For the top, I get:
$$\int_0^{\infty} \frac{e^{x(i-a)}}{x} dx=i\theta+\int_0^{\infty} \frac{e^{re^{i\theta}(i-a)}}{r}dr$$
and the bottom:
$$\int_0^{\infty} \frac{e^{-x(i+a)}}{x} dx=-i\theta+\int_0^{\infty} \frac{e^{-re^{-i\theta}(i+a)}}{r}dr$$
so that when we add everything up, we get the factor
$$\int_0^{\infty} \frac{e^{re^{i\theta}(i-a)}}{r}dr-\int_0^{\infty} \frac{e^{-re^{-i\theta}(i+a)}}{r}dr$$
and
$$e^{r e^{i\theta(i-a)}}-e^{-r e^{-i\theta(a+i)}}=0$$
for $\theta=\arctan(1/a)$.
mfb
Mentor
Okay. Good to see that the "direct" approach is possible.
Did it quick but looks like it ends up being:
$$\frac{1}{2 i} \left(i\arctan(1/a)+\int_0^{\infty} f(r)dr+i\arctan(1/a)-\int_0^{\infty}f(r) dr\right)$$
which then gives us the requisite $\arctan(1/a).$
I probably should have written that as:
$$\frac{1}{2 i} \left(i\arctan(1/a)+\int_0^{\infty} f(r,\theta)dr+i\arctan(1/a)-\int_0^{\infty}g(r,\theta) dr\right)$$
and $f(r,\theta)-g(r,\theta)=0$ for $\theta=\arctan(1/a)$.
Staff Emeritus
Gold Member
Boorglar, that's a really nice proof that you can take the derivative through the integral. You don't actually need to know what the value at a=0 is separately, since it's easy to calculate that as a goes to infinity the integral goes to zero.
jbunniii
Homework Helper
Gold Member
If we put
$$f(x) = \begin{cases} e^{-ax} & \text{ if }x \geq 0 \\ 0 & \text{ if }x < 0\end{cases}$$
and ##g(x) = \sin(x)/x##, then we need to evaluate ##I = \int_{-\infty}^{\infty} f(x) g(x) dx##, which we may also write as
$$I = \left.\int_{-\infty}^{\infty} f(x) g(x) e^{-iux} dx\right|_{u = 0}$$
This is the Fourier transform of ##fg##, evaluated at ##u=0##. If we denote ##\hat{f} = \mathcal{F}f## and ##\hat{g} = \mathcal{F}g##, the Fourier transforms of ##f## and ##g##, respectively, then we have ##\widehat{fg} = \hat{f} * \hat{g}##, where ##*## denotes the convolution operation: ##(\hat{f} * \hat{g})(u) = \int_{-\infty}^{\infty} \hat{f}(v) \hat{g}(u-v) dv##.
Our integral ##I## is equal to the convolution evaluated at ##u=0##: ##\int_{-\infty}^{\infty} \hat{f}(v) \hat{g}(-v) dv##. Let us evaluate ##\hat{f}## and ##\hat{g}##:
\begin{align} \hat{f}(u) &= \int_{0}^{\infty} e^{-ax} e^{-iux} dx \\ &= \int_{0}^{\infty} e^{-(a+iu)x} dx \\ &= \frac{1}{a+iu} \\ \end{align}
It's also well known that ##\sin(x)/x## has a "rectangular" Fourier transform:
$$\hat{g}(u) = \begin{cases} 1 & \text{ if }|u| < 1/2 \\ 0 & \text{ if }|u| > 1/2 \\ \end{cases}$$
(I'm probably missing a ##2\pi## scale factor somewhere.) Putting the pieces together, our integral is equivalent to
$$I = \int_{-1/2}^{1/2} \frac{1}{a + iu} du$$
which Wolfram Alpha tells me works out to ##2 \cot^{-1}(2a)##, but I don't remember which "stupid calculus trick" is needed to show this. Note that ##\cot^{-1}(x) = \pi/2 - \tan^{-1}(x)##, so my answer is close to Boorglar's except I screwed up a scale factor somewhere.
jbunniii
Homework Helper
Gold Member
Hmm, I guess I can do that last integral as follows:
\begin{align} I = \int_{-1/2}^{1/2} \frac{1}{a + iu} du &= \int_{-1/2}^{1/2}\frac{a-iu}{a^2 + u^2} du \\ &= \int_{-1/2}^{1/2} \frac{a}{a^2 + u^2} du \\ \end{align}
where the last equality holds because ##\int_{-1/2}^{1/2} \frac{u}{a^2 + u^2} du = 0## (the integrand is an odd function).
This should be handled by a trig substitution: ##u = a \tan \theta##. But I'm too tired to crank through the details.
CompuChip
Homework Helper
Cool mfb, I had a go at this problem for about 10 minutes earlier this week, and I got right up to your image but then got stuck / lost interest thinking about the additional contributions and running into the same problem with divergent integrals.
Staff Emeritus
Gold Member
jbunniii, I'm suspicious that you never plug in a u=0 anywhere.... at the end that integral looks like the product of the fourier transforms, not their convolution (which is easily fixed of course). That's a really clever way of attacking the problem.
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jbunniii
Homework Helper
Gold Member
jbunniii, I'm suspicious that you never plug in a u=0 anywhere.... at the end that integral looks like the product of the fourier transforms, not their convolution (which is easily fixed of course). That's a really clever way of attacking the problem.
I do plug in ##u = 0##:
\begin{align} I &= \left.\int_{-\infty}^{\infty} f(x) g(x) e^{-iux} dx\right|_{u = 0} \\ &= \left.(\hat{f} * \hat{g})(u)\right|_{u=0}\\ &= \left.\int_{-\infty}^{\infty} \hat{f}(v) \hat{g}(u-v) dv\right|_{u=0}\\ &= \int_{-\infty}^{\infty} \hat{f}(v) \hat{g}(-v) dv \\ &= \int_{-\infty}^{\infty} \hat{f}(v) \hat{g}(v) dv \end{align}
where the last inequality follows because in our case ##\hat{g}## is an even function.
jbunniii
Homework Helper
Gold Member
I'll try to fix the scale factors later today if I have time.
Staff Emeritus
Gold Member
I see, I missed the fact that they end up looking exactly the same.
jbunniii
Homework Helper
Gold Member
My scale factor woes are a result of the ##2\pi## factor that appears in the formula for the inverse Fourier transform:
$$f(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty}\hat{f}(u) e^{iux} du$$
One consequence of this is that the Fourier transform of ##fg## is in fact ##\frac{1}{2\pi}(\hat{f}*\hat{g})##, not ##\hat{f}*\hat{g}##. It also means that my Fourier transform of ##\sin(x)/x## dx was wrong. In fact, it should be
$$\hat{g}(u) = \begin{cases} \pi & \text{ if }|u| < 1 \\ 0 & \text{ if }|u| > 1 \\ \end{cases}$$
We verify this by calculating the inverse Fourier transform:
\begin{align} g(x) &= \frac{1}{2\pi}\int_{-\infty}^{\infty}\hat{g}(u) e^{iux} du \\ &= \frac{1}{2} \int_{-1}^{1} e^{iux} du \\ &= \frac{1}{2 i x} \left.e^{iux}\right|_{-1}^{1} \\ &= \frac{1}{2i x} (e^{ix} - e^{-ix}) \\ &= \frac{1}{ x} \left( \frac{e^{ix} - e^{-ix}}{2i} \right) \\ &= \frac{1}{ x} \sin(x) \\ &= \frac{\sin(x)}{x} \\ \end{align}
My calculation of the Fourier transform of ##f## was OK.
Now our integral is equal to the convolution (with the ##2\pi## scale factor) evaluated at ##u = 0##:
$$I = \frac{1}{2\pi} \int_{-\infty}^{\infty} \hat{f}(v) \hat{g}(-v) dv$$
Since ##\hat{g}## is even, this is equivalent to
$$I = \frac{1}{2\pi} \int_{-\infty}^{\infty} \hat{f}(v) \hat{g}(v) dv$$
Plugging in the results for ##\hat{f}## and ##\hat{g}##, we have
\begin{align} I &= \frac{1}{2\pi} \int_{-1}^{1} \frac{1}{a+iu} (\pi) du \\ &= \frac{1}{2} \int_{-1}^{1} \frac{1}{a+iu} du \\ &= \frac{1}{2} \int_{-1}^{1} \frac{a - iu}{a^2 + u^2} du \\ &= \frac{1}{2} \int_{-1}^{1} \frac{a}{a^2 + u^2} du \end{align}
which Wolfram tells me is equal to ##\tan^{-1}(1/a)##.
Staff Emeritus | 2020-09-26T22:57:46 | {
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https://dept.atmos.ucla.edu/tcd/m-ssa-tutorial-matlab | # M-SSA tutorial
This Matlab tutorial demonstrates step by step the multivariate singular spectrum analysis. The steps are almost similar to those of a singular spectrum analysis.
## Set general Parameters
M = 30; % window length of SSA
N = 200; % length of generated time series
T = 22; % period length of sine function
stdnoise = 0.1; % noise-to-signal ratio
## Create time series X
First of all, we generate two time series, a sine function of length N and the same function to the power of 3, both with observational white noise.
t = (1:N)';
X1 = sin(2*pi*t/T); % sine function
X2 = cos(2*pi*t/T).^3; % nonlinear transformation
noise = stdnoise*randn(N,2); % Gaussian noise
X1 = X1+noise(:,1);
X2 = X2+noise(:,2);
X1 = X1-mean(X1); % remove mean value
X2 = X2-mean(X2);
X1 = X1/std(X1); % normalize to std=1
X2 = X2/std(X2);
X = [X1 X2]; % multivariate time series
figure(1);
clf;
set(1,'name','Time series X');
subplot(1,2,1);
plot(X1, 'r-');
title('Time series X1');
subplot(1,2,2);
plot(X2, 'r-');
title('Time series X2');
## Calculate covariance matrix C (Toeplitz approach)
Next, we calculate the covariance matrix. There are several numerical approaches to estimate C. Here, we calculate the covariance function with CORR and build C with the function TOEPLITZ.
covXX=xcorr(X1,M-1,'unbiased');
covYY=xcorr(X2,M-1,'unbiased');
covXY = xcorr(X1,X2,M-1,'unbiased');
C11=toeplitz(covXX(M:end));
C21=toeplitz(covXY(M:-1:1),covXY(M:end));
C12=C21';
C22=toeplitz(covYY(M:end));
Ctoep = [C11 C12 ;...
C21 C22 ];
figure(2);
set(gcf,'name','Covariance matrix');
clf;
imagesc(Ctoep);
axis square
set(gca,'clim',[-1 1]);
colorbar
## Calculate covariance matrix (trajectory approach)
An alternative approach is to determine C directly from the scalar product of Y, the time-delayed embedding of X. Although this estimation of C does not give a Toeplitz structure, with the eigenvectors not being symmetric or antisymmetric, it ensures a positive semi-definite covariance matrix.
Y1=zeros(N-M+1,M);
Y2=zeros(N-M+1,M);
for m=1:M % create time-delayed embedding of X
Y1(:,m) = X1((1:N-M+1)+m-1);
Y2(:,m) = X2((1:N-M+1)+m-1);
end;
Y = [Y1 Y2];
Cemb=Y'*Y / (N-M+1);
figure(2);
imagesc(Cemb);
axis square
set(gca,'clim',[-1 1]);
colorbar
## Choose covariance estimation
Choose between Toeplitz approach (cf. Vautard & Ghil) and trajectory approach (cf. Broomhead & King).
% C=Ctoep;
C=Cemb;
## Calculate eigenvalues LAMBDA and eigenvectors RHO
In order to determine the eigenvalues and eigenvectors of C we use the function EIG. This function returns two matrices, the matrix RHO with eigenvectors arranged in columns, and the matrix LAMBDA with eigenvalues on the diagonal.
[RHO,LAMBDA] = eig(C);
LAMBDA = diag(LAMBDA); % extract the diagonal
[LAMBDA,ind]=sort(LAMBDA,'descend'); % sort eigenvalues
RHO = RHO(:,ind); % and eigenvectors
figure(3);
clf;
set(gcf,'name','Eigenvectors RHO and eigenvalues LAMBDA')
subplot(3,1,1);
plot(LAMBDA,'o-');
subplot(3,1,2);
plot(RHO(:,1:2), '-');
legend('1', '2');
subplot(3,1,3);
plot(RHO(:,3:4), '-');
legend('3', '4');
## Calculate principal components PC
The principal components are given as the scalar product between Y, the time-delayed embedding of X1 and X2, and the eigenvectors RHO.
PC = Y*RHO;
figure(4);
set(gcf,'name','Principal components PCs')
clf;
for m=1:4
subplot(4,1,m);
plot(PC(:,m),'k-');
ylabel(sprintf('PC %d',m));
ylim([-10 10]);
end;
## Calculate reconstructed components RC
In order to determine the reconstructed components RC, we have to invert the projecting PC = Y*RHO; i.e. RC = Y*RHO*RHO'=PC*RHO'. Averaging along anti-diagonals gives the RCs for the original input X.
RC1=zeros(N,2*M);
RC2=zeros(N,2*M);
for m=1:2*M
buf1=PC(:,m)*RHO(1:M,m)'; % invert projection - first channel
buf1=buf1(end:-1:1,:);
buf2=PC(:,m)*RHO(M+1:end,m)'; % invert projection - second channel
buf2=buf2(end:-1:1,:);
for n=1:N % anti-diagonal averaging
RC1(n,m)=mean( diag(buf1,-(N-M+1)+n) );
RC2(n,m)=mean( diag(buf2,-(N-M+1)+n) );
end
end;
figure(5);
set(gcf,'name','Reconstructed components RCs')
clf;
for m=1:4
subplot(4,2,2*m-1);
plot(RC1(:,m),'r-');
ylabel(sprintf('RC %d',m));
ylim([-1 1]);
subplot(4,2,2*m);
plot(RC2(:,m),'r-');
ylabel(sprintf('RC %d',m));
ylim([-1 1]);
end;
## Compare reconstruction and original time series
Note that the original time series X can be completely reconstructed by the sum of all reconstructed components RC (upper panels). The sine function (lower left panel) can be reconstructed with the first pair of RCs, where more components are required for the nonlinear oscillation (lower right panel).
figure(6);
set(gcf,'name','Original time series X and reconstruction RC')
clf;
subplot(2,2,1)
plot(t,X1,'b-',t,sum(RC1,2),'r-');
subplot(2,2,2)
plot(t,X2,'b-',t,sum(RC2,2),'r-');
legend('Original','full reconstruction');
subplot(2,2,3)
plot(t,X1,'b',t,sum(RC1(:,1:2),2),'r');
subplot(2,2,4)
plot(t,X2,'b',t,sum(RC2(:,1:2),2),'r');
legend('Original','RCs 1-2'); | 2019-01-22T09:22:15 | {
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"lm_q1q2_score": 0.8338431158741714
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https://math.stackexchange.com/questions/2578996/relation-between-inclusion-exclusion-principle-and-maximum-minimums-identity | # Relation between inclusion-exclusion principle and maximum-minimums identity
Inclusion-exclusion principle states that the size of the union of $n$ finite sets is given by the sum of the sizes of all sets minus sum of the sizes of all the pairwise intersections plus sum of the sizes of all the triple intersections and so on: $$\left| A_1\cup \dots \cup A_n\right| = \sum_i \left| A_i\right|-\sum_{i<j} \left| A_i\cap A_j\right|+\sum_{i<j<k} \left| A_i\cap A_j\cap A_k\right|-\dots+(-1)^{n+1}\left| A_1\cap \dots \cap A_n\right|.$$ Maximum-minimums identity states that the maximum of a finite set of numbers $S = \{x_1, \dots, x_n \}$ is given by the sum of all elements minus the sum of minimums of all pairs of elements plus sum of minimums of all triples and so on: $$\max\{x_1, \dots,x_n\} = \sum_i x_i -\sum_{i<j}\min\{x_i, x_j\} + \sum_{i<j<k}\min\{x_i, x_j,x_k\}-\dots+(-1)^{n+1}\min\{x_1,\dots, x_n\}.$$ It is hard to miss the similarity.
1. Is there a relation between maximum-minimums identity and inclusion-exclusion principle?
2. Can either one be proven from the other?
Inspired by kimchi lover's proof:
Let us first assume $x_1,\dots,x_n$ are positive integers. Then we can construct sets $A_i = \{1, \dots, x_i\}$ for all $i$s. Now $|A_i|=x_i$, therefore, $|A_1\cup\dots\cup A_n|=\max\{x_1,\dots,x_n\}$, $|A_i\cap A_j|=\min\{x_i,x_j\}$, and so on.
We can extend this proof to the case where $x_i$s can be negative by shifting all the elements, finding the maximum and then shifting back.
Similarly, we can extend to rationals by multiplying everything by the common denominator.
Finally, we can extend to reals by continuity.
• Good! Glad you came up with this. – kimchi lover Dec 24 '17 at 19:46
From inclusion-exclusion you can derive the maximums-minimums result, and I'd be surprised if the other direction doesn't hold, either.
In the following I take the inclusion-exclusion formula to be about probabilities, with $P(A_i\cap A_j)$ and so on for events $A_i$ and $A_j$, instead of cardinalities $|A_i\cap A_j|$ of finite sets. One can convert the one formulation into the other by dividing; both can be proved by integrating expressions involving characteristic functions like $\chi_{A\cap B}(x) = \chi_A(x) \chi_B(x)$ against counting measure or against an arbitrary probability measure. The trick is that $\chi_{\bigcup A_i}(x) = 1- \prod_i (1-\chi_{A_i}(x)).$
Assume, in the max-min problem, that the $x_i$ all lie in $[0,1]$ and are sorted in increasing order. (You can add or subtract a constant to all the $x_i$ without spoiling the equation, similarly rescale them, similarly permute them.) Now let $U$ be a uniform random variable, and let $A_i$ be the event that $U\le x_i$. If $i<j$ we have $A_i\cap A_j=A_i$ so $P(A_i\cap A_j) = \min(x_i,x_j)$, and so on. The event $A_1\cup\cdots \cup A_n$ is simply $A_n$, whose probability is $x_n=\max(x_1,\dots,x_n)$. In this way the two identities agree, term by term.
• That's interesting. This is a different version of the inclusion-exclusion principle. In your version, size of $A_i$ is its probability, not its cardinality? – stochastic Dec 24 '17 at 19:22
• You're right, and it was carelessness on my part: I was running on autopilot. The 2 forms of inclusion/exclusion are at heart the same; I have edited my post to say so. If it isn't clear, tell me so, and I'll try again. Another way to match them up is (assuming your $x_i$ are sorted nonnegative integers, is to let $A_i=\{1,2,\dots,x_i\}$. – kimchi lover Dec 24 '17 at 19:46
I think we can pull the proof in the other direction fairly easily. Let $x\in S$ and let:
$$x_i=\begin{cases}1, & x\in A_i \\ 0, & \text{otherwise}\end{cases}$$
(The value of the indicator function of the set $A_i$ on $x$.) Now, for each $x\in S$, write one equality of the form:
$$\max\{x_1, \dots,x_n\} = \sum_i x_i -\sum_{i<j}\min\{x_i, x_j\} + \sum_{i<j<k}\min\{x_i, x_j,x_k\}-\dots+(-1)^{n+1}\min\{x_1,\dots, x_n\}$$
Then, sum them over all $s\in S$. The result will give the inclusion-exclusion principle identity.
Without loss of generality, reindex the elements so that $i\lt j\implies x_i\le x_j$. Let $U_k$ be the set of $k$-tuples from $\{x_j:1\le j\le n\}$ $$U_k=\{(x_{j_1},x_{j_2},\dots,x_{j_k}):1\le j_1\lt j_2\lt\dots\lt j_k\le n\}$$ Note that $|U_k|=\binom{n}{k}$.
$x_j$ is the minimum in $\binom{n-j}{k-1}$ elements of $U_k$. Therefore, $$\sum_{u\in U_k}\min u=\sum_{j=1}^n x_j\binom{n-j}{k-1}$$ and so \begin{align} \sum_{k=1}^n(-1)^{k-1}\sum_{u\in U_k}\min u &=\sum_{k=1}^n(-1)^{k-1}\sum_{j=1}^nx_j\binom{n-j}{k-1}\\ &=\sum_{j=1}^nx_j\sum_{k=1}^n(-1)^{k-1}\binom{n-j}{k-1}\\ &=\sum_{j=1}^nx_j\,[j=n]\\[9pt] &=x_n \end{align} Compare this to this proof of the Inclusion-Exclusion Principle. | 2019-06-17T19:41:54 | {
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https://mathematica.stackexchange.com/questions/67494/transform-an-interpolatingfunction/67497 | # Transform an InterpolatingFunction
I'd like to transform an InterpolatingFunction from NDSolve but can't figure out how. Here's an example. The equation I want to solve is
sol1 = NDSolve[{n'[t] == n[t] (1 - n[t]), n[0] == 10^-5}, n, {t, 0, 20}]
but for numerical reasons (not in this toy example), the log transformed equation
sol2 = NDSolve[{ln'[t] == (1 - E^ln[t]), ln[0] == Log[10^-5]}, ln, {t, 0, 20}]
works better. How can I transform the result in sol2 to match sol1 by taking E^sol2, without losing any accuracy?
EDIT: A slightly less minimal example:
This is in response to @MichaelE2's comment below. It's still a toy example, but illustrates the problem with interpolating only on the original grid.
This is the original model in natural units. The output is wrong because extra AccuracyGoal is needed.
sol3 = NDSolve[{n'[t] ==
Piecewise[{{n[t] (1 - n[t]), Mod[t, 100] < 60}, {-n[t], Mod[t, 100] >= 60}}],
n[0] == 10^-15}, n, {t, 0, 100}][[1]];
Plot[n[t] /. sol3, {t, 0, 100}]
Here's a better solution of the model:
sol3b = NDSolve[{n'[t] ==
Piecewise[{{n[t] (1 - n[t]), Mod[t, 100] < 60}, {-n[t], Mod[t, 100] >= 60}}],
n[0] == 10^-15}, n, {t, 0, 100}, AccuracyGoal -> \[Infinity]][[1]];
Plot[n[t] /. sol3b, {t, 0, 100}]
Here's a solution of the log-transformed model:
sol4 = NDSolve[{ln'[t] ==
Piecewise[{{(1 - E^ln[t]), Mod[t, 100] < 60}, {-1, Mod[t, 100] >= 60}}],
ln[0] == Log[10^-15]}, ln, {t, 0, 100}][[1]];
Plot[E^ln[t] /. sol4, {t, 0, 100}]
Now to try the two different ways to create an exponentiated InterpolatingFunction. This one uses the original grid and leads to a big artifact.
sol5IFN = Interpolation[Transpose[{ln["Grid"], Exp[ln["ValuesOnGrid"]]} /.First@sol4]];
Plot[sol5IFN[t], {t, 0, 100}]
Looking at the underlying grid makes it obvious what the problem is:
ListPlot[Transpose[{ln["Grid"][[All, 1]], Exp[ln["ValuesOnGrid"]]} /. First@sol4]]
The alternative approach of making your own grid suggested by @bbgodfrey works better:
sol6 = Interpolation[Table[{t, E^ln[t] /. sol4}, {t, 0, 100, 0.1}], InterpolationOrder -> 1];
Plot[sol6[t], {t, 0, 100}]
UPDATE 2: Higher InterpolationOrder
sol5IFN = Interpolation[Transpose[{ln["Grid"], Exp[ln["ValuesOnGrid"]]} /. First@sol4], InterpolationOrder -> 8];
Plot[sol5IFN[t], {t, 0, 100}, PlotRange -> {0, 1}]
• @ChrisK After making my last comment on the solution by @MichaelE2, I tried recomputing his sol3IFN with InterpolationOrder -> 8. With that change the corresponding relative error plot looks essentially identical to the second plot in my answer. Perhaps, you could rerun your comparison above with this change. – bbgodfrey Dec 9 '14 at 16:53
• @bbgodfrey I added it above -- very wiggly. – Chris K Dec 10 '14 at 3:14
• @ChrisK Too bad. Thanks anyway. Good luck with your project. – bbgodfrey Dec 10 '14 at 4:10
To obtain the solution to each equation as an InterpolatingFunction, use
sol1 = n /. NDSolve[{n'[t] == n[t] (1 - n[t]), n[0] == 10^-5}, n, {t, 0, 20}][[1]];
sol2 = ln /. NDSolve[{ln'[t] == (1 - E^ln[t]), ln[0] == Log[10^-5]}, ln, {t, 0, 20}][[1]]
Unfortunately, simply constructing Exp[sol2] to match sol1 does not work, as explained in Question 67494. A solution is to construct a Table, apply Exp to each element, and construct a new InterpolatingFunction from the result:
sol3 = Interpolation[Table[{0.1 i, Exp[sol2[0.1 i]]}, {i, 0, 200}]]
The relative error is small.
Plot[(sol3[t]/sol1[t] - 1), {t, 0, 20}]
The solution posted as this one was being written takes an alternative approach, creating a function that applies Exp at each evaluation of sol2, rather than creating a new InterpolatingFunction. Both work.
Update based on my last comment
Actually, the plot above shows the relative difference between sol3 and sol1, but not necessarily the accuracy of sol3. To obtain a good estimate of its accuracy, we must have a high accuracy result for sol1, obtained by reducing the NDSolve step size:
sol1 = n /. NDSolve[{n'[t] == n[t] (1 - n[t]), n[0] == 10^-5}, n, {t, 0, 20}, MaxStepFraction -> 1/4000][[1]]
The value of MaxStepFraction is chosen so that sol1 changes imperceptibly for further decreases. Using the new sol1 result yields a comparison plot,
Thus, sol3 is very accurate indeed.
• This is potentially great, but I need to check my real application to see if the inaccuracy introduced is too much. – Chris K Dec 8 '14 at 19:56
• I was careless in my terminology. The plot above shows the relative difference between sol3 and sol1, not the error in sol3. In fact, the error appears to be in sol1. I just recomputed sol1 to higher accuracy with MaxStepFraction -> 1/500, and the maximum relative difference in the plot dropped to 0.00002. Thus, as you originally suggested, computing and exponentiating sol2 is rather more accurate than computing sol1 directly. – bbgodfrey Dec 8 '14 at 21:23
• This works well in my real application. Two things that are nice: 1) if you need the transformed InterpolatingFunction to be a better approximation, you can increase the number of points that went into it and 2) the end values seem to be exact. – Chris K Dec 9 '14 at 13:03
Update-- It turns out that a direct construction from an InterpolatingFunction does not satisfy all cases and that what is needed is a more general function interpolation. It would be nice to have a method that had some sort of automatic precision tracking. FunctionInterpolation is an apparently orphaned system function that seems to make no attempt at tracking accuracy or precision. The best way I know, which I've discussed with some of the numerics people at WRI, is to use NDSolve.
## Function interpolation
Basically there are two way you can proceed to interpolate a function f[x], either solve the equation y[x] == f[x] as a differential-algebraic equation with a dummy ODE, or differentiate the equation and solve it as an ODE:
NDSolveValue[{y[x] == f[x], t'[x] == 1, t[a] == a}, y, {x, a, b}] (* DAE *)
NDSolveValue[{y'[x] == D[f[x], x], y[a] == f[a]}, y, {x, a, b}] (* ODE *)
The DAE method can be highly accurate but it is currently limited to MachinePrecision. The ODE method may be done at any finite precision, but it also has the error from the integration method. In the OP's examples, the function being integrated contains a machine-precision interpolating function, and so one would not expect accuracy or precision better than one half of machine precision; perhaps worse in the ODE method, which would contain the derivative of the interpolating function.
I have used both methods several times on the site:
Since FunctionInterpolation does not work well, here is my version of it. It does the DAE method for machine precision and the ODE otherwise.
ClearAll[functionInterpolation];
SetAttributes[functionInterpolation, HoldAll];
functionInterpolation[f_, {x_, a_, b_}, opts : OptionsPattern[NDSolve]] /;
MatchQ[WorkingPrecision /. {opts}, WorkingPrecision | MachinePrecision] :=
Block[{x},
NDSolveValue[
{\[FormalY][x] == f, \[FormalT]'[x] == 1, \[FormalT][a] == a},
\[FormalY], {x, a, b}, opts]
];
functionInterpolation[f_, {x_, a_, b_}, opts : OptionsPattern[NDSolve]] :=
Block[{x},
NDSolveValue[
{\[FormalY]'[x] == D[f, x], \[FormalY][a] == f /. x -> a},
\[FormalY], {x, a, b}, opts]
];
OP's second example:
sol3nd = functionInterpolation[Exp[ln[t] /. First@sol4], {t, 0, 100}]
Plot[sol3nd[t] - n[t] /. sol3b, {t, 0, 100}, PlotRange -> All]
If high precision is needed, then, as in the OP's solution sol3b, one needs a higher AccuracyGoal (the relative error of the one above exceeds 1, sometimes by a lot, when the value n[t] is close to zero):
sol3nd = functionInterpolation[Exp[ln[t] /. First@sol4], {t, 0, 100}, AccuracyGoal -> 30];
Plot[(sol3nd[t] - n[t])/n[t] /. sol3b, {t, 0, 100; 61}, PlotRange -> All]
## Direct construction (original solution)
A direct way to construct "Exp[sol2]" is to exponentiate the values stored in the interpolating function:
sol3IFN = Interpolation[Transpose[{ln["Grid"], Exp[ln["ValuesOnGrid"]]} /. First@sol2]]
As for accuracy, the interpolation between the grid points will be cubic, whereas for Exp[ln[t]] /. sol2, the interpolation will be the exponential of a cubic. So this method will yield exactly Exp[ln[t]] on the grid points but vary from it in between.
Plot[sol3IFN[t] - Exp[ln[t]] /. sol2, {t, 0, 20}, PlotPoints -> 100]
Comparing with the exact solution shows it is quite accurate:
sol1ex = DSolve[{n'[t] == n[t] (1 - n[t]), n[0] == 10^-5}, n, t];
Plot[sol3IFN[t] - n[t] /. sol1ex, {t, 0, 20}]
Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. >>
I did something similar in my answer to Interpolating a ParametricFunction. In that case it was a rescaling. It seems a nonlinear transformation can have problems (see OP's update). Perhaps it is due to changes in convexity.
• Yours is a very elegant approach. It appears to be based on mathematica.stackexchange.com/questions/19042, but what is the source of that information? Does Wolfram have some additional documentation on line somewhere? Thanks. – bbgodfrey Dec 9 '14 at 11:58
• This is indeed elegant, but it didn't work so well on my real application. That's because my real function has a sharp corner and the new InterpolatingFunction has big distortions due to the cubic interpolation between widely spaced exact points. bbgodfrey's Table approach fixed that issue. But it might be fine for smooth functions. – Chris K Dec 9 '14 at 13:07
• @bbgodfrey Many data structures take an argument sol3IFN["Properties"] that shows arguments you can use. There's also mathematica.stackexchange.com/q/28337 and the tutorial [InterpolatingFunctionAnatomy](reference.wolfram.com/language /tutorial/NDSolvePackages.html). With the Spelunking package, you can find out InterpolatingFunctionAnatomy is implemented as ifun["Grid"] etc. Some users prefer to use InterpolatingFunctionAnatomy functions. I've grown used to using "Properties" for various data structures. – Michael E2 Dec 9 '14 at 13:07
• @ChrisK Interesting. That means the corner comes between the computed data points, since the method above has almost zero error on a data point. It doesn't sound like the usual output of NDSolve. I don't suppose you could post the actual sol2 you want interpolated, since your real problem seems to have issues not present in the question? – Michael E2 Dec 9 '14 at 13:31
• @MichaelE2 Sure! This comment space it too small to fully describe it, so I'll add it in an edit to my original question. – Chris K Dec 9 '14 at 14:56
you can work on the results as follows:
f1 = n /. sol1[[1]];
f2 = ln /. sol2[[1]];
f3[x_] := Exp[f2[x]]
To check the match between f1 and f3 we can plot:
Plot[{f1[x], f2[x], f3[x]}, {x, 0, 20},
PlotStyle -> {Directive[Red, Dashed, Thickness[0.02]], Automatic,
Black}, PlotRange -> All]
• Thanks for the idea! This is what I've done myself, but I really want it in InterpolatingFunction form. – Chris K Dec 8 '14 at 19:53
I would augment your good logarithmic NDSolve call to also produce your desired normal interpolating function:
{nIF, lnIF} = NDSolveValue[
{
ln'[t] == Piecewise[{{(1 - E^ln[t]), Mod[t, 100] < 60}, {-1, Mod[t, 100] >= 60}}],
ln[0] == Log[10^-15],
n'[t] == Exp[ln[t]] ln'[t],
n[0] == 10^-15
},
{n, ln},
{t, 0, 100}
];
This is similar to @MichaelE2's answer, but I think using 1 NDSolve call allows the step size to be adjusted so that both interpolating functions get the same accuracy and precision goals.
Here is a plot of nIF:
Plot[nIF[x], {x, 0, 100}]
And nIF is indeed an interpolating function:
nIF //OutputForm
InterpolatingFunction[{{0., 100.}}, <>] | 2020-02-25T06:54:51 | {
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https://math.stackexchange.com/questions/1574468/probability-with-unknown-variables/1574496 | # Probability with unknown variables
An urn contains $10$ red marbles and $10$ black marbles while a second urn contains $25$ red marbles and an unknown number of black marbles. A random marble will be selected from each urn and the probability that both marbles are the same will be determined. A hint was given by the teacher: the probability does NOT depend on the number of unknown marbles. Verify that this is the case.
Let's call $N$ the unknown number of marbles.
I wrote out all possible ways to select a marble from each urn, selecting a red marble from both urn $1$ and urn $2$, and selecting a black marble from urn $1$ and $2$ and this is what I got:
• Number of ways to select a marble from each urn: $\binom{20}{1}\binom{25+N}{1}$
• Number of ways to select $1$ red marble from both urn $1$ and urn $2$: $\binom{10}{1}\binom{25}{ 1}$
• Number of ways to select $1$ black marble from urn $1$ and urn $2$: $\binom{10}{1}\binom{N}{ 1}$
And this is what I got as my final equation to finding out the probability of selecting the same color marble from each urn: $\dfrac{\binom{10}{1}\binom{25}{ 1}+\binom{10}{1}\binom{N}{ 1}}{\binom{20}{1}\binom{25+N}{1}}$
I am confused on how the probability doesn't depend on the unknown number of black marbles in urn $2$? Any help would be much appreciated, thank you so much!
PS: I also searched through stack exchange for a problem similar to this and I couldn't find one. If this question was asked already, then I apologize!
A short explanation of why this is, is that instead of picking both marbles simultaneously, let us pick a marble from the strange bag first.
Once having done so, it will be some color, either black or red. Now, pick a marble from the one with an even amount of each color. Regardless which color was chosen in the first step, the probability that you pick the same color will be $\frac{1}{2}$.
Worded with symbols, let $B_1,R_1,B_2,R_2$ represent the events of picking a black marble or red marble from the first or second bag respectively. Let the "second bag" be the one with the unknown number of black marbles.
We have:
$$\begin{array}{rl}Pr(\text{colors are same}) &= Pr((B_2\cap B_1)\cup (R_2\cap R_1)) \\&= Pr(B_2)Pr(B_1|B_2)+Pr(R_2)Pr(R_1|R_2)\\ &=\frac{1}{2}Pr(B_2)+\frac{1}{2}Pr(R_2)\\ &=\frac{1}{2}(Pr(B_2)+Pr(R_2))\\ &=\frac{1}{2}\end{array}$$
This uses the multiplication principle that $Pr(A\cap B) = Pr(A)Pr(B|A)$
• wow great explanation from everyone. Thank you! Makes a LOT more sense now! – King Dec 14 '15 at 1:20
Suppose the probability of drawing a red marble from the second urn is $p$. It would then follow that the probability of drawing two red marbles is $$\frac 12p$$
But then the probability of drawing a black marbles from the second urn is $1-p$, whence the probability of drawing two black marbles is $$\frac 12(1-p)$$
As these are disjoint events we add the probabilities to see that the probability that the two random marbles share a color is $$\frac 12(p+(1-p))=\frac 12$$
• This is the cleanest solution. – user217285 Dec 14 '15 at 1:44
Hint:
$\dfrac{\binom{10}{1}\binom{25}{ 1}+\binom{10}{1}\binom{N}{ 1}}{\binom{20}{1}\binom{25+N}{1}} =\frac{\binom{10}{1}\cdot\left[\binom{25}{ 1}+\binom{N}{ 1} \right]}{20\cdot(25+N)}=\frac{10\cdot (25+N)}{20\cdot (25+N)}$
Let $p$ be the probability of drawing a red marble from the second urn, so $1-p$ is the probability of drawing a black marble. Then the probability that the marbles match is
$$P(\text{match}) = \left(\frac{1}{2}\right)p + \left(\frac{1}{2}\right)(1-p) = \frac{1}{2}$$ | 2021-03-08T22:49:16 | {
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https://www.scienceforums.net/topic/118197-area-of-squares-bisected-by-part-of-a-circle/ | # Area of squares bisected by part of a circle
## Recommended Posts
I have a grid through upon which a circle is drawn, as shown below.
Now, for every square in the grid through which the circle runs, i want to calculate the proportion of the square covered by the circle. So one of the squares close up:
I thought this would be relatively straightforward but i've ended up with a horribly convoluted way involving working at some angles where the circle crosses the squares, using that to find the area of the circle, then subtracting that from the area of a rectangle that contains that circle segment, and so on for other squares. Only need to do this for three squares as symmetry saves me a lot of work, but it's still a meandering method. Just seems to me there should be a much more simple method, but i can't figure it out.
Does anybody know of a more elegant way of solving this problem?
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29 minutes ago, Prometheus said:
I have a grid through upon which a circle is drawn, as shown below.
Now, for every square in the grid through which the circle runs, i want to calculate the proportion of the square covered by the circle. So one of the squares close up:
I thought this would be relatively straightforward but i've ended up with a horribly convoluted way involving working at some angles where the circle crosses the squares, using that to find the area of the circle, then subtracting that from the area of a rectangle that contains that circle segment, and so on for other squares. Only need to do this for three squares as symmetry saves me a lot of work, but it's still a meandering method. Just seems to me there should be a much more simple method, but i can't figure it out.
Does anybody know of a more elegant way of solving this problem?
Wouldn't you be looking for the Area under a Curve? Look at each square as a quadrant in graph.
Edited by StringJunky
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4 minutes ago, StringJunky said:
Wouldn't you be looking for the Area under a Curve?
You may very well be right, thanks. I'll get back to you once i get a chance to give it a go.
Yes, even at a cursory glance it seems obvious this will work.
Edited by Prometheus
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If it is the real work, not mathematical play, I would simply boolean one object from other object to get intersecting polygons, and use area calculation tool in 3D application on remaining polygons (eventually triangulate it).
You might need to temporarily extend 2D shape to 3D because some boolean tools don't like to work in 2D (depending on application that you're using).
Edited by Sensei
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Here are some thoughts.
Use symmetry. That substantially reduces the workload.
Then you can calculate the coordinates of the intercepts with the sides of the squares by calculating the offsets from the tangent.
The four sides of the outer box are tangent to the circle as shown.
The simple formula for perpendicular offsets from the tangent to a circular curve is
Offset = (Length along the tangent)2 / twice radius
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I should perhaps add how to get the areas?
Well in each square you have to add the area of a rectangle R, a triangle T and a circular segement S.
You already have the circle radius and can easily find the chord length as the hypotenuse of a right angled triangle.
So these are easily calculable from the intercepts, or you can use an online calculator for the segments.
You haven't stated your mesh size, but with the size shown in the drawing you only need three sizes and to find these you only need two intercepts.
Symmetry does the rest.
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Following String's hint and using studiot's notation for the squares, of which we need only 3, i've come up with:
$$A = int_{0}^{1}(r^2 - x^2) dx - 2$$
$$B = int_{1}^{2}(r^2 - x^2) dx - 2$$
$$C = int_{2}^{3}(r^2 - x^2) dx - (A+B)$$
Which give reasonable looking answers too.
I'll give Studiot's method a go later, it looks like a more refined approach to the one i first described.
Edited by Prometheus
i seem to be getting worse at using latex | 2019-03-21T11:05:06 | {
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https://mathoverflow.net/questions/207452/eigenvectors-as-continuous-functions-of-matrix-diagonal-perturbations | Eigenvectors as continuous functions of matrix - diagonal perturbations
The general question has been treated here, and the response was negative. My question is about more particular perturbations. The counterexamples given in the previous question have variations not only on the diagonal, and moreover, in the simple example given by A. Quas, the discontinuous change in eigenvectors take place where the eigenvalue is double.
Suppose $$A$$ is a symmetric, positive definite matrix, with simple lowest eigenvalue, and $$D$$ is a diagonal matrix. Does the first eigenvalue and the first eigenvector of $$A+D$$ vary continuously for $$\|D\|_1 <\varepsilon$$ ($$\varepsilon$$ can be as small as possible, and $$\|\cdot \|_1$$ is the $$L^1$$ norm)?
In the case this is true, it is possible to bound the variation of the eigenvector in terms of $$\|D\|_1$$?
• $$A$$ is constant, $$D$$ is a diagonal matrix which is variable with $$N$$ parameters (the matrices are of size $$N \times N$$)
• $$D$$ is just assumed to vary continuously with norm smaller than $$\varepsilon$$.
• What is constant and what is varying there? Is $D$ a matrix-valued function? Of how many parameters? How regular? – Federico Poloni May 24 '15 at 10:40
• $A$ is constant, $D$ is diagonal and variable (with $N$ parameters, where $N \times N$ is the size of $A$). Ideally, the perturbation $D$ should only be continuous, but if higher regularity assumptions are needed, I don't mind. – Beni Bogosel May 24 '15 at 10:53
• The key here, I think, is your assumption that the eigenvalue $\lambda$ is simple. ($A$ needn't be symmetric PD.) Then for $D$ in a small enough neighborhood, there is a continuous (even analytic) map $D\to\lambda(D)$ with $\lambda(0)=\lambda$ and $\lambda(D)$ an eigenvalue of $A+D$. The point is that if you take a simple root of a polynomial, then for sufficiently small perturbations of the coefficients, you get an analytic perturbation of the root. Presumably the same holds for the associated eigenvector, again since the eigenspace is one dimensional. – Joe Silverman May 24 '15 at 11:18
• Joe Silverman is correct; this is a job for the implicit function theorem (where the function being defined implicitly is $M \mapsto$its lowest eigenvector). Said theorem requires some derivative to be onto and this eventually turns into the simpleness of the eigenvalue. – Allen Knutson May 24 '15 at 14:28
• You don't need $D$ diagonal for this, symmetric is enough. The claim on the eigenvector follows for example by writing the spectral projection as a contour integral of the resolvent. – Christian Remling May 24 '15 at 15:10
As @Joe Silverman said in his comment, the key here is that $\lambda$ is simple.
I copy here a few Theorems on the subject that might be helpful.
From the book Linear Algebra and Its Applications (by Peter D. Lax):
Theorem 7, p.130: Let $A(t)$ be a differentiable square matrix-valued function of the real variable $t$. Suppose that $A(0)$ has an eigenvalue $a_0$ of multiplicity one, in the sense that $a_0$ is a simple root of the characteristic polynomial of $A(0)$. Then for $t$ small enough, $A(t)$ has an eigenvalue $a(t)$ that depends differentiably on $t$, and which equals $a_0$ at zero, that is $a(0)=a_0$.
Theorem 8, p.130: Let $A(t)$ be a differentiable matrix-valued function of $t$, $a(t)$ an eigenvalue of $A(t)$ of multiplicity one. Then we can choose an eigenvector $h(t)$ of $A(t)$ pertaining to the eigenvalue $a(t)$ to depend differentiably on $t$.
From the Book Matrix Analysis (by Roger A. Horn & Charles R. Johnson)
Corollary 6.3.8, p.407: Let $A,E\in \Bbb C^{n\times n}$. Assume that $A$ is Hermitian and $A+E$ is normal, let $\lambda_1,\ldots,\lambda_n$ be the eigenvalues of $A$ arranged in increasing order $\lambda_1\leq \ldots\leq \lambda_n$ and let $\hat \lambda_1,\ldots,\hat \lambda_n$ be the eigenvalues of $A+E$, ordered so that $\Re(\hat\lambda_1)\leq\ldots\leq\Re(\hat\lambda_n)$. Then $$\sum_{i=1}^n |\hat \lambda_i-\lambda_i|^2 \leq \|E\|_F^2,$$ where $\|\cdot\|_F$ is the Frobenius norm.
This is a somehow refined version of Theorem 7 above:
Theorem 6.3.12, p.409: Let $A,E\in \Bbb C^{n\times n}$ and suppose that $\lambda$ is a simple eigenvalue of $A$. Let $x$ and $y$ be, respectively, right and left eigenvectors of $A$ corresponding to $\lambda$. Then
a) for each given $\epsilon>0$ there exists a $\delta>0$ such that, for all $t\in\Bbb C$ such that $|t|<\delta$, there is a unique eigenvalue $\lambda(t)$ of $A+tE$ such that $|\lambda(t)-\lambda-ty^*Ex/y^*x|\leq |t| \epsilon$
b) $\lambda(t)$ is continuous at $t=0$, and $\lim_{t\to 0}\lambda(t)=\lambda$
c) $\lambda(t)$ is differentiable at $t=0$, and $$\left.\frac{\operatorname{d}\lambda(t)}{\operatorname{d}t}\right|_{t=0}=\frac{y^*Ex}{y^*x}$$
• Corollary 6.3.8 looks very nice. Is there a similar result for eigenvectors? I will look in the cited book. – Beni Bogosel May 28 '15 at 14:44
• @BeniBogosel Unfortunately I did not find any similar results for eigenvectors in the book of Horn (I might have not seen it though). That is why I also reported the results from Lax. – Surb May 28 '15 at 14:51
• I guess this is close of the result I want: math.stackexchange.com/questions/1133071/… There seems to be a Lipschitz bound for the eigenvectors also, but the Lipschitz constant is not too explicit in the references I searched. – Beni Bogosel May 28 '15 at 22:17 | 2021-03-05T19:47:35 | {
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https://math.stackexchange.com/questions/1959652/incorrect-method-to-find-a-tilted-asymptote/1959665 | # Incorrect method to find a tilted asymptote
Suppose I want to find the slanted asymptote for the graph of $\displaystyle y=\frac{x^2+x-6}{x+2}$.
Using division, we have $\displaystyle y=x-1-\frac{4}{x+2};\;$ so $y=x-1$ is the slanted asymptote.
I would like to find out, though, what is wrong with the following incorrect way of finding the asymptote:
$\displaystyle y=\frac{x^2+x-6}{x+2}=\frac{x+1-\frac{6}{x}}{1+\frac{2}{x}}\approx\frac{x+1}{1}=x+1$, so $y=x+1$ is the slanted asymptote.
The trouble is that the numerator in $$\frac{x+1-\frac{6}{x}}{1+\frac{2}{x}}$$ stays large, so a small error in the denominator is magnified. Better to write $$\frac{1}{1+\frac{2}{x}} = 1 - \frac{2}{x} + \frac{4}{x^2} - \frac{8}{x^3} + \frac{16}{x^4} + \cdots$$ and multiply; the infinite series indicated converges for $|x| > 2.$
$$\left( x+1-\frac{6}{x} \right) \left( 1 - \frac{2}{x} + \frac{4}{x^2} - \frac{8}{x^3} + \frac{16}{x^4} + \cdots \right) = x - 1 - \frac{4}{x} + \frac{8}{x^2} + \cdots$$
• To be precise, since the numerator is on the order of $x$, an error on the order of $1/x$ in the denominator creates an overall error on the order of $x(1/x)=1$ in the quotient, which fails to go to zero.
– Ian
Oct 8, 2016 at 18:37
• @Ian I put in just a little more. Oct 8, 2016 at 18:50
By definition, the function $f(x)$ has as an oblique (slanted) asymptote the straight line $y=\alpha x+\beta$ if: $$\lim_{x\to+\infty}\big(f(x)-(\alpha x+\beta)\big)=0$$ or if: $$\lim_{x\to-\infty}\big(f(x)-(\alpha x+\beta)\big)=0$$ It easy to check that while your result $y=x-1$ satisfies the above definition, the line $y=x+1$ does not, since: $$\lim_{x\to\pm\infty}\big(f(x)-(x+1)\big)=-2$$ P.S.: What you have actually shown via your second argument is that: $$\lim_{x\to\pm\infty}\frac{f(x)}{x+1}=1$$ which is however irrelevant to the notion of oblique (slanted) asymptote.
The problem with your second approach is that you've kept more precision than your approximation actually has — you can say $y \approx x$, but you don't have enough precision to clarify that more specifically to $y \approx x+1$ (or any other translate).
In more detail,
$$\frac{1}{1 + \frac{2}{x}} \approx 1 - \frac{2}{x}$$
and consequently,
$$\frac{x+1-\frac{6}{x}}{1 + \frac{2}{x}} \approx \left( x+1-\frac{6}{x} \right) \left(1 - \frac{2}{x} \right) \approx x \cdot 1 + 1 \cdot 1 - x \cdot \frac{2}{x}$$
By neglecting the $\frac{2}{x}$ term of the denominator, you neglect the $x \cdot \frac{2}{x}$ term of this approximation — but that term is $-2$, so you're neglecting a nonnegligible quantity!
Keeping the $\frac{2}{x}$ term around, the above approximation gives $x-1$, as desired.
For more rigor, you can use big O notation:
$$\frac{1}{1 + \frac{2}{x}} = 1 - \frac{2}{x} + O(x^{-2})$$ $$\frac{x+1-\frac{6}{x}}{1 + \frac{2}{x}} = \left( x+1+O(x^{-1}) \right) \left(1 - \frac{2}{x} + O(x^{-2}) \right) = x - 1 + O(x^{-1})$$
$\frac{y}{x+1} \;\to\; 1$ indeed, but:
$$y - (x+1) = \frac{x^2 + x - 6 - x^2 - 3 x - 3}{x+2} = \frac{-2 x - 9}{x+2} \quad \to \quad -2$$
which is why the slope of $x+1$ is correct, but its intercept is off by $-2$. | 2022-06-28T01:24:39 | {
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https://math.stackexchange.com/questions/2694254/finding-the-least-acceleration-required-for-particle-a-to-overtake-particle-b | # Finding the least acceleration required for particle $A$ to overtake particle $B$
You have two runners modeled as particles $A$ and $B$ respectively. Say that particle $B$ is ahead of particle A by $10$ m, and particle $B$ has a constant velocity of $5$ m/s. When particle $B$ is $50$ m from the finish line particle $A$ starts to accelerate, but $B$ does not.
My question is, what is the least acceleration $A$ must produce in order to overtake $B$?
So far, I have tried the following:
• Knowing that particle $B$ is traveling at a constant velocity, I have calculated the time it would take for particle $B$ to cross the finish line, by using $v= \dfrac st$. The time I got was $10$ s.
• Since particle $A$ is $10$ m away from particle $B$, and since I know that particle $A$ is $2$ s behind particle $B$ $\left(\text{by using }v= \dfrac st\right)$, I used the equation of kinematics to work out the acceleration as I knew the distance, the time, and the initial speed.
However, I cannot seem to get the correct answer.
• What is the initial velocity of A? – Paul Mar 16 '18 at 20:53
• Same as B, at 5 m/s – Benny Mar 16 '18 at 20:54
A has 60 meters to run and has to make it in 10 seconds, with an initial speed of 5 m/s. The constant acceleration equation is $$d=v_0 t +1/2 at^2$$ And you just need to solve for $a$.
• That is what i have done and the answer that I got was 0.8 m/s^2, however, the correct answer is 0.2 m/s^2 – Benny Mar 16 '18 at 21:00
• Check your arithmetic. I get 0.2 from this equation. – Paul Mar 16 '18 at 21:03
• Oh I see what I did wrong, thanks a bunch – Benny Mar 16 '18 at 21:08
hint
A has to cross $d=10+50 m$ in less than $10s$. with
$$d=\frac {1}{2}at^2+5t$$ solve the inequation $t <10$
• Your question is not well asked. – hamam_Abdallah Mar 16 '18 at 21:02 | 2019-08-24T15:50:34 | {
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https://math.stackexchange.com/questions/2248868/is-emptyset-a-set-element | # Is $\emptyset$ a set element?
Is there a difference between $\{1,\emptyset\}$ and $\{1\}$? Does the former have $2$ elements and the latter $1$ element? Does adding a string of $\emptyset$'s increase the number of elements of a set?
• The empty set can be an element of a set. Consider for example the power-set always containing the empty set. – Peter Apr 23 '17 at 22:58
• Adding "a string of" any element won't increase the number of elements. For instance, $$\{1,2\} = \{1,2,1,1,1\} = \{1,2,1,2,2,1,1\}$$ – Omnomnomnom Apr 23 '17 at 23:00
• Note also that $\emptyset = \{\}$, and $\{\emptyset\} = \{\{\}\} \neq \emptyset$ – Omnomnomnom Apr 23 '17 at 23:03
• It is true, as other noted in answers and comments, that adding a string won't increase the number of elements. This is the usual and widely accepted convention, or rather the convention for sets. But, it might be worth noting that there are so-called multisets, see en.wikipedia.org/wiki/Multisets (a different thing, perhaps unimportant in the context of your question) for which a string does increase the number of elements. – Mirko Apr 24 '17 at 0:52
The main problem here is that $1\in\mathbb N$ and $\varnothing$ are different in nature. $1$ is an integer, while $\varnothing$ is a set.
You can't have a set mixing sets and non-sets, $\{1,\varnothing\}$ is simply wrong in general.
Remark : emphasis on "in general", because as in Unix we say "everything is file", for set theoretists "everything is set", naturals, reals, functions, etc... See below for instance the construction of naturals, for a possible meaning of $\{\varnothing,1\}=2$.
$\{\{1\},\varnothing\}$ would be fine and it has $2$ elements which are sets.
The sets $\{\varnothing\}$ and $\{\varnothing,\varnothing\}$ are equal, in the same way that $\{1\}=\{1,1\}$ because in a set, elements appear only once.
So repeating an element won't increase the cardinality, it's just the same set.
Now, the fact that for a set $S$ we have $S\neq\{S\}$ is the base of the definition of the axioms of the naturals. There are two main constructions :
• Zermelo
$\begin{array}{l} 0=\varnothing\\ 1=\{0\}=\{\varnothing\}\\ 2=\{1\}=\{\{\varnothing\}\}\\ 3=\{2\}=\{\{\{\varnothing\}\}\}\\ n+1=\{n\}=\{\{..\{\varnothing\}..\}\}\} \end{array}$
• Von Neumann
$\begin{array}{l} 0=\varnothing\\ 1=0\cup\{0\}=\{0\}=\{\varnothing\}\\ 2=1\cup\{1\}=\{0,1\}=\{\varnothing,\{\varnothing\}\}\\ 3=2\cup\{2\}=\{0,1,2\}=\{\varnothing,\{\varnothing\},\{\varnothing,\{\varnothing\}\}\}\\ n+1=n\cup\{n\}=\{0,1,2,..,n\}=\{\varnothing,\{\varnothing\},\{\varnothing,\{\varnothing\}\},\{\varnothing,\{\varnothing\},\{\varnothing,\{\varnothing\}\}\},...\}\\ \end{array}$
In the Zermelo case, the cardinal of a number is always $1$ but elements are themselves sets of cardinal $1$ nested like russian dolls.
In the Von Neumann case, the cardinal of the number, is the number itself, but the nesting is more complex. Every natural is the set of all its predecessors.
• I prefer $|n| = n$ – Henno Brandsma Apr 24 '17 at 17:22
• Since everything is a set, I see nothing wrong in $\{1,\emptyset\}$. And in the second part you're saying that $1=\{\emptyset\}$ and $2=\{1,\emptyset\}$, contradicting yourself. – egreg Apr 26 '17 at 22:45
• Nothing wrong under the point of view of sets, but OP asks if there is a difference between $\{1\}$ which I assume he considers a subset of $\mathbb N$ and $\{1,\varnothing\}$ which this time is not a subset of $\mathbb N$ since $\varnothing$ is not an element of $\mathbb N$, unless we use the canonical mapping described above. This is why it is dangerous to write such "unstructured" sets, not because it is forbidden, but because there is a risk a confusion between things whose nature is different. Am I wrong ? – zwim Apr 26 '17 at 23:09
• -1. You wrote: "You can't have a set mixing sets and non-sets, $\{1,\emptyset\}$ is simply wrong in general." This statement disagrees with the standard mathematical conception of set. According to this conception, it is certainly true that given any finite list of mathematical objects (like $1$ and $\emptyset$), there is a set containing exactly these objects. There is an attractive philosophical position that mathematics should be done in a typed system, in which every set has type Set of T, where T is another type, e.g. Set of Natural Numbers, Set of (Sets of Real Numbers)... – Alex Kruckman May 4 '17 at 13:36
• But even in typed systems, you can often form "sum"/"disjunction"/"union" types, so if $\emptyset$ has type Set of Natural Numbers and $1$ has type Natural Number, I could form the type Set of (Natural Numbers OR Sets of Natural Numbers), and $\{1,\emptyset\}$ would have this type. – Alex Kruckman May 4 '17 at 13:38
To your first question - Yes, the two sets are different. In fact $\{1\}\subset\{1,\emptyset\}$.
Adding a string of anything doesn't increase elements in a set - $\{1,2,3,4\}=\{1,1,2,3,3,3,3,4,4\},$ so no this doesn't change anything for the empty set either.
Yes they are different sets. The empty set is not nothing. It is the set which contains nothing, an empty box, if you will. Thus the set which contains both one and an empty box is different than the set which contains just one. Incidentally, the first set you mentioned is what we define the ordinal 2 to be. | 2019-08-19T03:37:07 | {
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https://math.stackexchange.com/questions/840957/how-can-this-be-proved-lim-x-to-inftyfxfx-l/841023 | # How can this be proved $\lim_{x\to\infty}(f(x)+f'(x))=l$
If $$\lim_{x\to\infty}(f(x)+f'(x))=l$$ then prove that $$\lim_{x\to\infty}f(x)=l \text{ and } \lim_{x\to\infty}f'(x)=0$$
I assume four cases $$\begin{array}{c|c|c|c|} & f(x) & f'(x) \\ \hline \text{1} & \infty & -\infty \\ \hline \text{2} & -\infty & \infty \\ \hline \text{3} & l & 0 \\ \hline \text{4} & 0 & l \\ \hline \end{array}$$ and elimination(1,2,4) of not possible cases can give the answer(3).
My work is not a correct/perfect or flaw proof.What would be a correct one or is this correct.
• Why would you assume only those four cases and no others? – Michael Hardy Jun 20 '14 at 18:13
• You are correct and also my note reminds everyone of the same – RE60K Jun 20 '14 at 18:14
• Hint: Consider what happens if $\lim_{x\rightarrow\infty}f'(x) = C\neq 0$. What can you say about the asymptotic behavior of $f$? (Hint: Consider the mean-value theorem.) You also need to consider what happens if $f'$ oscillates - i.e. does not have a limit at all. – Cameron Williams Jun 20 '14 at 18:14
• See the accepted answer to this. – David Mitra Jun 20 '14 at 18:17
• @CameronWilliams do you mean that if f' doesn't approach zero the value of the limit wouldn't exist.thus f'x$\to$0 and fx$\to$l – RE60K Jun 20 '14 at 18:55
This is the solution coming straight from Hardy's Pure Mathematics and is truly beautiful. Before solving this problem Hardy solves a simple problem:
If $f(x) \to L$ and $f'(x) \to L'$ as $x \to \infty$ then $L' = 0$.
This is easy to do via mean value theorem as we have have $f(x) - f(x/2) = (x/2)f'(c)$ for some $c \in (x/2, x)$. If $L' \neq 0$ then we have contradiction as we take limits with $x \to \infty$ in the above relation.
[If $x \to \infty$ then $x/2 \to \infty$ and hence LHS $f(x) - f(x/2) \to L - L = 0$. Again since $x/2 < c < x$ therefore when $x \to \infty$ then $c$ also tends to $\infty$ and hence $f'(c) \to L'$. Then the RHS $(x/2)f'(c)$ tends to $(x/2)L'$ i.e. to $+\infty$ if $L' > 0$ and to $-\infty$ if $L' < 0$. Hence there is no choice for $L'$ except that $L' = 0$.]
Next we attack the current problem. We are given that $f(x) + f'(x) \to L$ when $x \to \infty$ and we need to show that $f(x) \to L$ and $f'(x) \to 0$ as $x \to \infty$. Hardy simplifies the problem by setting $f(x) = \phi(x) + L$ so that $f'(x) = \phi'(x)$ and $\phi(x) + \phi'(x) \to 0$ as $x \to \infty$.
[Hardy defines a new function $\phi(x) = f(x) - L$ and then $f(x) + f'(x) = \phi(x) + L + \phi'(x) \to L$ and cancel $L$ from both sides to get $\phi(x) + \phi'(x) \to 0$].
To solve the problem we need to show that both $\phi(x)$ and $\phi'(x)$ tend to $0$ as $x \to \infty$.
[If we show $\phi(x) \to 0$ then $f(x) = L + \phi(x) \to L$ and $f'(x) = \phi'(x) \to 0$ so that we get the original version of the problem.]
Now the argument by Hardy is beautiful and I quote him verbatim:
"If $\phi'(x)$ is of constant sign, say positive, for all sufficiently large values of $x$, then $\phi(x)$ steadily increases and must tend to a limit $A$ or to $\infty$.
[If the derivative $\phi'(x)$ is positive then original function $\phi(x)$ is strictly increasing for all large values of $x$. And there is a very standard theorem that if a function $\phi(x)$ is increasing then $\phi(x)$ tends to a limit or to $\infty$ as $x \to \infty$.]
If $\phi(x) \to \infty$ then $\phi'(x) \to -\infty$ [because $\phi(x) + \phi'(x) \to 0$] which contradicts our hypothesis [we have assumed that $\phi'(x) > 0$ for large values of $x$ and this is incompatible with $\phi'(x) \to -\infty$].
If $\phi(x) \to A$ then $\phi'(x) \to -A$ [because $\phi(x) + \phi'(x) \to 0$] and this is impossible unless $A = 0$ [from our previous result mentioned in beginning of answer].
Similarly we may dispose of the case in which $\phi'(x)$ is ultimately negative.
[If $\phi'(x)$ is negative then $\phi(x)$ is decreasing for large $x$. Now there is again a very standard theorem which says that if $\phi(x)$ is decreasing for all large $x$ then it either tends to a limit or to $-\infty$ as $x \to \infty$. If $\phi(x) \to -\infty$ then because of the relation $\phi(x) + \phi'(x) \to 0$ we get that $\phi'(x) \to \infty$ and this is incompatible with the fact that $\phi'(x)$ is negative for large $x$. If on the other hand we have $\phi(x) \to B$ then again $\phi'(x) \to -B$ and by the result mentioned in the beginning of answer this is possible only when $B = 0$.]
If $\phi'(x)$ changes sign for values of $x$ which surpass all limit, then these are the maxima and minima of $\phi(x)$.
[Note that a when a derivative changes sign it must also vanish somewhere in between and hence we will obtain points where $\phi'(x) = 0$ and before and after this point the derivative $\phi'(x)$ is of opposite signs. Such points are the maxima and minima of $\phi(x)$ and $\phi(x)$ takes local minimum and maximum values at these points. Since $\phi'(x)$ changes sign for values $x$ which go beyond any limit, this will lead to infinitely many points $x$ which are maxima or minima of $\phi(x)$.]
If $x$ has a large value corresponding to a maximum or a minimum of $\phi(x)$, then $\phi(x) + \phi'(x)$ is small [because it tends to zero] and $\phi'(x) = 0$ [because derivative vanishes at points of maxima/minima], so that $\phi(x)$ is small. A fortiori the other values of $\phi(x)$ are small when $x$ is large. [We see that the maximum and minimum values of $\phi(x)$ are small and hence all the other intermediate values of $\phi(x)$ are also small so that for large $x$ all values of $\phi(x)$ are small. And thus $\phi(x) \to 0$ in this case also.]"
Thus in all cases $\phi(x) \to 0$ and hence $\phi'(x) \to 0$.
Update: Based on comment from OP I add details inline enclosed in [].
• It took me a long time figuring it out from your solution and am still a little confused; can you please explain a liitle bit more, for I am on the level of highschool mathematics. – RE60K Jun 21 '14 at 5:42
• @Aditya: I have tried to explain the answer in detail by putting additional explanation in []. However you will need to know some standard results from theory of differential calculus like "increasing / decreasing nature of functions based on sign of derivative", "mean value theorem", "maxima/minima". – Paramanand Singh Jun 21 '14 at 6:38
Lemma Let $h:[a,+\infty)\to\mathbb{R}$ be a continuous function such that $\lim_{x\to\infty}h(x)=0$, Let $F$ be defined by $$F(x)=e^{-x}\int_a^xe^th(t)dt$$ Then $\lim_{x\to\infty}F(x)=0$.
Proof. Indeed, for $x>b>a$ we have \eqalign{ |F(x)|&=\left|e^{-x}\int_a^be^th(t)dt+e^{-x}\int_b^xe^th(t)dt\right|\cr &\leq e^{-x}\max_{x\in[a,b]}{(e^t|h(t)|)}+\left(\sup_{t\in[b,x]}|f(t)|\right)e^{-x}\int_b^xe^tdt\cr &\leq e^{-x}\max_{x\in[a,b]}{(e^t|h(t)|)}+\left(\sup_{t\in[b,x]}|f(t)|\right) }\tag{1} Now, for $\epsilon>0$, there is $b>a$ such
that $$\forall\,t>b,\qquad |f(t)|<\frac{\epsilon}{2}\tag{2}$$ With $b$ fixed as above, there is $x_0>b$ such that $$\forall\,x>x_0,\qquad e^{-x}\max_{x\in[a,b]}{(e^t|h(t)|)}\leq\frac{\epsilon}{2}\tag{3}$$ combining $(1)$ and $(2)$ in $(1)$ we see that $|F(x)|<\epsilon$ for $x>x_0$. This proves the lemma since $\epsilon>0$ is arbitrary.$\qquad\square$
Now, let us apply the lemma to $h(x)=f(x)+f'(x)-\ell$. We have $$F(x)=e^{-x}\int_a^x((e^tf(t))'-\ell e^t)dt=f(x)-e^{-x+a}f(a)-\ell(1- e^{a-x})$$ By the Lemma $\lim_{x\to\infty}F(x)=0$ that is $\lim_{x\to\infty}f(x)=\ell$. and the desired conclusion follows.$\qquad\square$
• You seem to use that $f'(x)$ is continuous. Is it possible to remove this limitation? – Paramanand Singh Jun 21 '14 at 17:27
• The lemma can be proved for $h:[a,+\infty)\to\mathbb{R}$ measurable instead of continuous, you just use the Dominated Convergence Theorem to prove it. So, this version is just a simplification of the proof. – Omran Kouba Jun 21 '14 at 21:13
The problem is easy if one assumes that
$$\lim_{x \to \infty} f'(x) = k$$ exists.
Indeed, in this case, it is easy to prove $k=0$: By L'Hospital we have
$$\lim_{x \to \infty} \frac{f(x)}{x} = k \,.$$
Then, if $k= \pm \infty$ we get $\lim_{x \to \infty} f(x)$ is the same infinity, which contradicts the given condition; while if $k$ is finite and non-zero, we get that $\lim_{x \to \infty} f(x)$ is infinite, which contradicts again the given condition.
Now, since $$\lim_{x \to \infty} f'(x) = 0$$ we get $$\lim_{x \to \infty} f(x) =\left( \lim_{x \to \infty} f(x)+f'(x)\right) -\lim_{x \to \infty} f'(x)=l$$
To get around the existence of the limit $\lim_{x \to \infty} f'(x)$, I would try to look to $\limsup f'(x)$ and $\liminf f'(x)$.
If one were to assume that both of the limits $\lim_{x\to\infty}f(x) \text{ and } \lim_{x\to\infty}f'(x)$ exist, then the limit of $f'$ is some number $k$. If $k\ne0$, then for large enough $x$ we have $f'(x)$ being further away from $0$ than $k/2$. That would mean that as $x$ increases by $1$ then $f(x)$ changes by at least $k/2$, and always in the same direction. That follows from the mean value theorem. If $f(x)$ changes by more than $k/2$ and always in the same direction, then $f(x)$ is not approaching any limit, and there is a contradiction, so it must be that $k=0$.
The harder problem is showing that if the limit of the sum exists, then the two limits separately exist. Maybe I'll come back to that . . . . . .
Given: $$f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$$
$$l = \lim_{x\to \infty} f(x) + f'(x) = \lim_{x\to \infty}\left[ f(x) + \left[\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\right]\right]$$
$$l = \lim_{x\to\infty}\left[\lim_{h\to 0}\frac{f(x+h)+f(x)(h-1)}{h}\right]$$
Assuming $f(x)$ has a limit, we can write $|f(x+h) - f(x)| \leq \epsilon$ for large enough $x$. This means that we can bound the quantity inside the limits.
$$\frac{hf(x) - \epsilon}{h} \leq \frac{f(x+h)+f(x)(h-1)}{h} \leq \frac{hf(x) + \epsilon}{h}$$
$$f(x) - \frac{\epsilon}{h} \leq \frac{f(x+h)+f(x)(h-1)}{h} \leq f(x) + \frac{\epsilon}{h}$$
For any $|h| > 0$ we can find an $N$ so that for $x > N$, $\,|f(x+h) - f(x)| \leq h^2$. This is equivalent to setting $\epsilon = h^2$.
$$f(x) - h \leq \frac{f(x+h)+f(x)(h-1)}{h} \leq f(x) + h$$
Take the limit as $h\to 0$. Notice that doing this implies that $\epsilon \to 0$ which forces $x \to \infty$.
$$\lim_{x\to\infty} f(x) \leq \lim_{x\to\infty} f(x)+f'(x) \leq \lim_{x\to\infty} f(x)$$
The rest is clear from here.
I'm not going to pretend like this is formal as we essentially interchanged the limit operators but it should give you a good idea why this works.
• @Aditya it should be better now but I will still take any constructive criticism. I don't see how it can be done without essentially interchanging limits. – Brad Jun 20 '14 at 19:02 | 2020-07-09T14:34:20 | {
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