text stringlengths 1 2.12k | source dict |
|---|---|
// positions multiplied by 10^6 so we can work with integers
long[] positions = new long[numFriends];
for (int i = 0; i < numFriends; i++) {
positions[i] = Long.parseLong(stp.nextToken()) * 1000000L;
}
int[] speeds = new int[numFriends];
for (int i = 0; i < numFriends; i++) {
speeds[i] = Integer.parseInt(sts.nextToken());
}
// Find min, max position
long minPosition = Long.MAX_VALUE;
long maxPosition = Long.MIN_VALUE;
for (long p : positions) {
minPosition = Math.min(minPosition, p);
maxPosition = Math.max(maxPosition, p);
}
// Binary search to find minimum time
long low = minPosition;
long high = maxPosition;
double minTime = Double.MAX_VALUE;
while (low <= high) {
long mid = (low + high) / 2;
double midTime = time(positions, speeds, mid);
double lowerTime = time(positions, speeds, mid - 1);
double higherTime = time(positions, speeds, mid + 1);
minTime = Math.min(Math.min(minTime, midTime), Math.min(lowerTime, higherTime));
if (midTime < lowerTime && midTime < higherTime) break;
else if (lowerTime > midTime || midTime > higherTime) {
low = mid + 1;
} else {
high = mid - 1;
}
}
System.out.printf("%.6f\n", minTime / 1000000);
}
// Time is maximum time of all friends to get to point
static double time(long[] positions, int[] speeds, long position) {
double maxTime = Double.MIN_VALUE;
for (int i = 0; i < speeds.length; i++) {
maxTime = Math.max(maxTime, (double) Math.abs(positions[i] - position) / speeds[i]);
}
return maxTime;
}
}
Thanks,
rayfish
Good solution as well, this is also known as ternary search, which works a little faster by cutting into 3 equal size chunks. https://en.wikipedia.org/wiki/Ternary_search
You should add this to USACO Guide as one of the user solutions . | {
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# Is the number of primes congruent to 1 mod 6 equal to the number of primes congruent to 5 mod 6?
(I know that there's an infinity of primes congruent to 5 mod 6, but I don't know if there is an infinity of primes congruent to 1 mod 6.)
But what I'd really like to know is whether or not the number of primes ≡1 mod 6 less than a given n can be said to be asymptotically equal to the number of primes ≡5 mod 6 less than that n. So the "total" number of primes ≡1 mod 6 would be equal to the number of primes ≡5 mod 6.
(I hope this is understandable. I'd also like to know how to phrase this inquiry in a more standard fashion, if someone would tell me how to fix it.)
-
You might ask like this: let $\pi_1(n)$ be the number of primes of the form $6k+1$ that are less than $n$, and $\pi_5(n)$ similarly. Then you can ask whether $\pi_1(n)/\pi_5(n)$ approaches a limit as $n$ increases without bound, and if so, whether that limit is 1. – MJD Aug 22 '12 at 1:13
That's precisely what I mean; thank you. – Annick Aug 22 '12 at 1:14
(You didn't ask for this, but for an elementary proof that there are infinitely many primes congruent to $1 \bmod 6$, consider the possible primes dividing a number of the form $k^2 - k + 1$.) – Qiaochu Yuan Aug 22 '12 at 1:28
Got it, thanks. Great to know. – Annick Aug 22 '12 at 2:35
Yes, this is true. A more general statement follows from a suitably strong form of Dirichlet's theorem on arithmetic progressions, namely that asymptotically the proportion of primes which are congruent to $a \bmod n$ (for $\gcd(a, n) = 1$) is $\frac{1}{\varphi(n)}$.
In the particular case that $n = 6$ it is possible to give a more elementary proof of a weaker result. Define the Dirichlet L-function
$$L(s, \chi_6) = \sum_{n=1}^{\infty} \frac{\chi_6(n)}{n^s}$$ | {
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$$L(s, \chi_6) = \sum_{n=1}^{\infty} \frac{\chi_6(n)}{n^s}$$
where $\chi_6$ is the unique nontrivial Dirichlet character $\bmod 6$. This takes the form $\chi_6(n) = 1$ if $n \equiv 1 \bmod 6$, $\chi_6(n) = -1$ if $n \equiv 5 \bmod 6$, and $\chi_6(n) = 0$ otherwise. The Euler product of this L-function is
$$L(s, \chi_6) = \prod_p \left( \frac{1}{1 - \chi_6(p) p^{-s}} \right) = \prod_{p \equiv 1 \bmod 6} \left( \frac{1}{1 - p^{-s}} \right) \prod_{p \equiv 5 \bmod 6} \left( \frac{1}{1 + p^{-s}} \right).$$
It is possible to explicitly evaluate $L(1, \chi_6)$ and in particular to show that it is not zero; in fact,
$$L(1, \chi_6) = \int_0^1 \frac{1 - x^5}{1 - x^6} \, dx$$
and this can be evaluated using partial fractions (but note that the integrand is always positive so this number is definitely positive). So we conclude that
$$-\log L(s, \chi_6) = \sum_{p \equiv 1 \bmod 6} \log (1 - p^{-s}) + \sum_{p \equiv 5 \bmod 6} \log (1 + p^{-s})$$
approaches a nonzero constant as $s \to 1$ (if summed in the appropriate order) even though the first and second terms separately approach $\mp \infty$. So the contributions coming from primes in each residue class cancel out asymptotically. This is not quite as strong as the desired statement, though; if you fill in all the details in what I've said you'll show that the Dirichlet density of the primes congruent to $\pm 1 \bmod 6$ are the same but this should still be true for the natural density and this requires a further argument (I am not sure how much further, though).
For more details see any book on analytic number theory, e.g. Apostol.
-
Thanks so much for this highly detailed answer and explanatory links. – Annick Aug 22 '12 at 1:27
Look up "prime races" at http://www.dms.umontreal.ca/~andrew/PDF/PrimeRace.pdf.
Their conclusion: "It does seem that “typically” qn + a has fewer primes than qn + b if a is a square modulo q while b is not." | {
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So, since 1 is a quadratic residue mod 6, while 5 is not, there will typically be more primes of the form 6n+5 than 6n+1 (though their ratio does tend to 1).
- | {
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# Does the sequence $(a_n)_{n\in\mathbb{N}}$ have a convergent subsequence?
Does the sequence $(a_n)_{n\in\mathbb{N}}$ have a convergent subsequence?
$$a_n= \begin{cases} \sin(n), & \text{if n is odd} \\ n, & \text{if n is even} \end{cases}$$
I understand that $\sin(n)$ is a bounded sequence so I can use Bolzano Weierstrass theorem to state that the sequence has a convergent subsequence. Not sure where to go from here assuming I am along the right lines? | {
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• "so I can use Bolzano Weierstrass theorem to state that the sequence has a convergent subsequence." Yes. "Not sure where to go from here". You're not sure where to go from having reached an irrefutable conclusion? You don't have to go anywhere. You were ask whether it has a convergent subsequence and you answered that it did. So ... where else is there to go? – fleablood May 30 '18 at 18:39
• @fleablood I think the question asker was missing a subtlety for this question: $\sin(n)$ has a convergent subsequence, but it is not a subsequence of $(a_n)$ - the answer is, of course, to show that $\sin(n)$ for odd $n$ has a convergent subsequence, but the work presented in the question is not complete as it stands. – Milo Brandt May 30 '18 at 18:57
• It's not that subtle is it? $a_n$ where $n$ is odd s $a_n= \sin n$ is a subsequence. That subsequence is bounded so there is a subsequence of the subsequence that is bounded. A subsequence of a subsequence is a subsequence. You have to avoid being lazy and stupid ($a_n$ itself is not bounded and $a_n; odd$ is not $\{\sin n\}$ so you have to claim that $a_n; odd$ has the subsequence; not $\{\sin n\}$) but that's not hard to do. – fleablood May 31 '18 at 0:34
• "I understand that sin(n) is a bounded sequence so I can use Bolzano Weierstrass theorem to state that the sequence has a convergent subsequence." Okay, there are two observations you must make. $\{a_n\}$ is not bounded (but the subsequence $\{a_{2k+1}\} is$. And stating $\{\sin n\}$ has a convergent subsequence isn't enough as $\{\sin n\} \not \subset \{a_n\}$. But as $\{a_{2k+1} \}\subset \{\sin n\}$ is bounded there is a convergent $\{b_j\}\subset \{a_{2k+1}\} \subset \{\sin n\}$. That's fairly obvious and easy, but enough to make a question worth 5 marks. – fleablood May 31 '18 at 0:41
Let's construct the subsequence which converges to $0$. | {
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Let's construct the subsequence which converges to $0$.
Consider convergents of $\pi$ continued fraction: $\left\{ \dfrac{n_j}{d_j},\; j\in\mathbb{N} \right\}$. And focus on its numerators: $$n_1 = 3,\\ n_2 = 22, \\ n_3 = 333, \\ n_4 = 355, \\ n_5 = 103993, \\ n_6 = 104348, \\ \vdots$$
Then $$\lim_{j\to \infty} \sin n_j =0.$$
Indeed, easy to estimate: $$|\sin n_j| = \left| \sin\left( n_j - \pi d_j \right) \right| \approx \left| n_j - \pi d_j \right|=d_j\left| \dfrac{n_j}{d_j} - \pi \right| < \dfrac{1}{d_{j+1}}.$$
And it remains to show that the sequence $\{n_j, \; j\in\mathbb{N}\}$ contains infinite number of odd $n_j$.
If continued fraction (of $\pi$) denote as $[a_0;a_1, a_2, \ldots, a_j, \ldots]$, then $$n_{j+1} = a_{j+1}n_j + n_{j-1}.$$
There are $8$ possibilities:
\begin{array}{|l|l|l|l|} \hline (n_{j-1}, n_j) & a_{j+1} & n_{j+1} & \rightarrow (n_j, n_{j+1})\\ \hline (odd, odd) & odd & even & \rightarrow (odd, even) \\ (odd, odd) & even & odd & \rightarrow (odd, odd) \\ \hline (odd, even) & odd & odd & \rightarrow (even, odd) \\ (odd, even) & even & odd & \rightarrow (even, odd) \\ \hline (even, odd) & odd & odd & \rightarrow (odd, odd) \\ (even, odd) & even & even & \rightarrow (odd, even) \\ \hline (even, even) & odd & even & \rightarrow (even, even) \\ (even, even) & even & even & \rightarrow (even, even) \\ \hline \end{array}
which show that if at least one of $n_j$ is odd, then there is infinite number of odd $n_j$.
A subsequence of a subsequence is a subsequence. The sequence of values of $a_n$ for which $n$ is odd is a subsequence, and you've already said that that has a convergent subsequence. That convergent subsequence is a subsequence of a subsequence of the original sequence. Therefore it is a subsequence of the original sequence.
You wrote (twice) “function” where you should have written “sequence”. Other than that, you are doing fine. | {
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• Edited thanks. The question is worth 5 marks, I don't think this answer justifies this. Where do I go from here? Thanks. – Ben Jones May 30 '18 at 18:36
• You know that the sequence $(a_{2n-1})_{n\in\mathbb N}$ is bounded. Therefore, it has a convergent subsequence, which will be a subsequence of the original sequence, of course. – José Carlos Santos May 30 '18 at 18:37
• "Where do I go from here?" Well, you could go to breakfast. Or Tahiti, I hear Tahiti is nice this time of year. ... you've answered the question. You don't need to go anywhere. – fleablood May 30 '18 at 18:42
• I'm off to Tahiti then. – Ben Jones May 31 '18 at 18:36 | {
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1. ## Dimensions of Kindle (Electronic pad with a screen)
I created this problem and wanted to check my answers with the MathForum.
x = length
y = width
Scenario:
There are two Kindles.
K(1) has the dimensions of 3.6 in by 4.8 in. with a surface area (SA) of $17.28in^2$ and a diagonal that measures 6 in
K(2) has the dimensions of 5.4 in by 7.9 in. with a surface area of $42.66in^2$ and a diagonal that measures 9.56 in
a. How much bigger is K(2)'s SA compared to K(1)'s SA?
b. Write an equation that represents the maximum and minimum value of
x and y when the SA of K(3) is 42.66 in.
my answer: Either $x=42.66*y$ or $y=42.66*x$
Note: For 'c' I want to know if it's possible to have a minimum and maximum value of the length and width when the diagonal is 6 in. My guess is that you can only have 1 combination of 'x' and 'y' that gives you a SA of 42.66 and a diagonal of 6 in.
c. Write an equation/system of equations that represents the maximum and minimum value of x and y if the diagonal of K(3) is 6 inches.
my answer: $x^2 + y^2 = (6)^2$ --> y = -X + 6 (If I simplified that right) and $x=42.66*y$ or $y=42.66*x$; where the two lines intersect is the value of x and y when the diagonal is 6.
2. Originally Posted by Masterthief1324
I created this problem and wanted to check my answers with the MathForum.
x = length
y = width
Scenario:
There are two Kindles.
K(1) has the dimensions of 3.6 in by 4.8 in. with a surface area (SA) of $17.28in^2$ and a diagonal that measures 6 in
K(2) has the dimensions of 5.4 in by 7.9 in. with a surface area of $42.66in^2$ and a diagonal that measures 9.56 in
a. How much bigger is K(2)'s SA compared to K(1)'s SA?
Yes, that is correct. | {
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a. How much bigger is K(2)'s SA compared to K(1)'s SA?
Yes, that is correct.
b. Write an equation that represents the maximum and minimum value of
x and y when the SA of K(3) is 42.66 in.
my answer: Either $x=42.66*y$ or $y=42.66*x$
What? You said there were two kinds of "kindle", K(1) and K(2). Where did K(3) come from? Without any information about K(3) it is impossible to answer this. Did you leave something out?
Note: For 'c' I want to know if it's possible to have a minimum and maximum value of the length and width when the diagonal is 6 in. My guess is that you can only have 1 combination of 'x' and 'y' that gives you a SA of 42.66 and a diagonal of 6 in.
c. Write an equation/system of equations that represents the maximum and minimum value of x and y if the diagonal of K(3) is 6 inches.
my answer: $x^2 + y^2 = (6)^2$ --> y = -X + 6 (If I simplified that right)
No, you didn't. " $x^2+ y^2= z^2$" does NOT reduce to "x+ y= z". For example, $4^2+ 3^2= 16+ 9= 25= 5^2$ but $4+ 3\ne 5$.
and $x=42.66*y$ or $y=42.66*x$; where the two lines intersect is the value of x and y when the diagonal is 6.
If $x^2+ y^2= 36$ and xy= 42.66, then y= 42.66/x so $x^2+ \frac{1819.8756}{x^2}= 36$. Multiplying both sides by $x^2$, $x^4+ 1819.8756= 36x^2$ or $x^4- 36x^2+ 1819.8756= 0$. If you let $u= x^2$, this becomes the quadratic equation $u^2- 36u+ 1819.8756= 0$. Solve that for u, then take the square root to find x.
But are you sure you have interpreted the question correctly? The problem said "find maximum and minimum" values, not specific values.
3. Sorry, I did make that last problem confusing.
I hope this clarifies it.
K(3) is a hypothetical Kindle device with a surface area of $42.66in^2$ and a diagonal of 6 in.
c1. Can you have more than one dimensions that satisfy the SA & the diagonal length of 6in? Why or why not?
my answer: Not if the diagonal is 6 inch -- there is only 1 value of x and y (length and width) | {
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c2. Can you represent c1. with and equation?
my answer: Since the diagonal is 6, if I was to use the Pythagorean theorem, there is only 1 value of x and y that would fit the specifications.
(specifications)
You know indefinitely that the K(3) has a surface area of 42.66 and a diagonal of 6 in:
my equation:
x^2 + y^2 = 6 (is it 6 or 6^2?) --> y=√(z^2 - x^2)
x*y = 42.66
If I was to graph the two equations, will where they intersect be the dimensions that fit the specifications?
4. b. Write an equation that represents the maximum and minimum value of
x and y when the SA of K(3) is 42.66 in.
If K(3) has a surface area of 42.66, then you know that
$xy=42.66$
since the diagonal is 6 inches, you know that
$x^2 + y^2 = 6^2$ (draw a picture to understand why it is 6^2 and not 6)
You have two equations, you can try to solve them both for y and graph them to approximate a solution
$y=42.66/x$
$y=sqrt(36-x^2)$
Once you look at this graph you will realize that there is no solution to your problem
K(3) is physically impossible
---------------------------------------
To understand this, think of the kindle with diagonal 6 that has the largest surface area, this kindle would be square. It would have x = y = sqrt (18), so its largest surface area would be xy=18
I created this problem and wanted to check my answers with the MathForum.
To fix your problem, you must either increase the size of the diagonal or reduce the surface area of K(3) to make the two curves intersect | {
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# Number of ways in which boys and girls sit alternately if six boys and six girls sit randomly?
Six boys and six girls sit in a row randomly. What is the total number of ways in which the boys and the girls sit alternately?
My attempt: Consider these six seats _ _ _ _ _ _
The number of ways to arrange 6 boys in 6 places is 6! Now 7 gaps are created between these 6 seats. So, we can select any 6 of these 7 gaps and make girls sit there. There are 7C6 * 6! ways to do that (since the girls can shuffle amongst themselves).
Hence the total number of ways to make boys and girls sit alternately should be 6! 7C1 * 6! but the answer is 2*6!*6!.
What am I missing here?
• You can't choose any $6$...the unfilled gap must be either first or last.
– lulu
Mar 18, 2017 at 11:34
• Note "alternately". Only two of the seven ways to pick six of the gaps result in an alternating pattern, the rest have two boys next to each other. Mar 18, 2017 at 11:34
• Oh right! Thank you so much. Mar 18, 2017 at 11:35
You are implicitly assuming that there are 13 seats for $6+6=12$ people. It is likely that the original question is making the implicit assumption that there are only 12 seats. In this latter case, when a boy sits first, there $6! \cdot 6!$ ways to sit the others. Multiply this by 2 to cover the case where the first seater is a girl.
You are over-counting.
You are also counting the cases where extreme 2 gaps are both occupied and out of 5 gaps in middle 4 are occupied.
For example, let's first arrange the boys: B_B_B_B_B_B = 6! ways.
We are left with 7 gaps so we select 6 among them and permute the girls so (7C6 6! ways.)
Hence according to you answer is 6! . 7C6 . 6! = 7.6!.6! | {
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Hence according to you answer is 6! . 7C6 . 6! = 7.6!.6!
But it also counts the cases like GBGBGBBGBGBG two boys are sitting together and alternate pattern is disturbed so we eliminate these unfavorable cases by bijection. So, Unfavorable ways= 6! . 2C2 . 5C4. 6! = 5.6!.6! So hence total ways of favourable permutations = 7.6!.6! - 5.6!.6! =2.6!.6! (The required answer) | {
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# How do you calculate $2^{2^{2^{2^{2}}}}$?
From information I have gathered online, this should be equivalent to $2^{16}$ but when I punch the numbers into this large number calculator, the number comes out to be over a thousand digits. Is the calculator wrong or is my method wrong?
• The number is equal to $2^r$ where $r=2^{16}$. – Aravind Mar 16 '17 at 17:17
• this is quiet not clear use parentesis – Dr. Sonnhard Graubner Mar 16 '17 at 17:17
• If parenthesis are not used, it is assumed that exponents are evaluated from top-down as opposed to bottom-up. $(a^b)^c=a^{bc}\neq a^{(b~^c)}$. Exponentiation is not associative. The answer of $2^{16}$ is for if it were evaluated bottom-up as (((2^2)^2)^2)^2 instead of top-down which is 2^(2^(2^(2^2))) which is much larger – JMoravitz Mar 16 '17 at 17:22
• Possible duplicate of What is the order when doing $x^{y^z}$ and why? – Simply Beautiful Art Mar 16 '17 at 21:11
• My TI calculator has an inline option of showing this as 2^2^2^2^2, which evaluates as 65536 (left to right evaluation). But when I switch to math print mode it shows the tower of powers, which it tries, but fails, to evaluate top down...overflow. I'm not saying that a TI calculator is the final arbiter of math truth, but it is what I get. – paw88789 Mar 17 '17 at 2:22
## 6 Answers
$$2^{2^{2^{2^2}}}=2^{2^{2^4}}=2^{2^{16}}=2^{65536}\tag1$$
The number of digits:
$$\mathcal{A}=1+\lfloor\log_{10}\left(2^{65536}\right)\rfloor=19729\tag2$$ | {
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The number of digits:
$$\mathcal{A}=1+\lfloor\log_{10}\left(2^{65536}\right)\rfloor=19729\tag2$$
• This is all true enough, but it doesn't really answer the question as actually asked. – hBy2Py Mar 16 '17 at 18:59
• It currently does answer the question in the title. Equation 1 shows how to evaluate the expression. It also implies, though doesn't explicitly states that the answer the asker's method (or at least the answer he got using the method) was wrong. – Fluidized Pigeon Reactor Mar 16 '17 at 21:30
• The title is a reference for searching and tracking. The question is the text beneath that. NAA. – Nij Mar 17 '17 at 8:43
What you have is a power tower or "tetration" (defined as iterated exponentiation). From the latter link, you would most benefit from this brief excerpt on the difference between iterated powers and iterated exponentials.
The comment by JMoravitz really gets to the heart of the matter, namely that exponential towers must be evaluated from top to bottom (or right to left). There actually is a notation for your particular question: ${}^52=2^{2^{2^{2^{2}}}}$. You really need to look at ${}^42$ before you get something meaningful because, unfortunately, $${}^32=2^{2^{2}}=2^4=16=4^2=(2^2)^2;$$ however, $${}^42=2^{2^{2^{2}}}=2^{2^{4}}=2^{16}\neq2^8=(4^2)^2=((2^2)^2)^2.$$ Hence, your method is wrong, but everything in those links should provide more than enough for you to become comfortable with tetration.
• It seems that the Knuth's notation $2\uparrow\uparrow n$ has gained popularity over Rucker's one $^n2$ nowadays. – zwim Mar 16 '17 at 17:47
• @Daniel W. Farlow look like you beat me to posting. Got to love it when that happens. – Sentinel135 Mar 16 '17 at 17:53
• @zwim I actually prefer Rucker's notation, but I do see the appeal of Knuth's very unambiguous notation. – Daniel W. Farlow Mar 16 '17 at 17:55 | {
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By convention, the meaning of things written $\displaystyle a^{b^{c^d}}$ without brackets is $\displaystyle a^{\left(b^{\left(c^d\right)}\right)}$ and not $\left(\left(a^b\right)^c\right)^d$.
This is because $\left(\left(a^b\right)^c\right)^d$ equals $a^{b\cdot c\cdot d}$ anyway, so it makes pragmatic sense to reserve the raw power-tower notation $\displaystyle a^{b^{c^d}}$ for the case that doesn't have an alternative notation without parentheses.
As others have explained, $\displaystyle 2^{2^{2^{2^2}}}$ interpreted with this convention is $2^{65536}$, a horribly huge number, whereas $(((2^2)^2)^2)^2$ is $2^{16}=65536$, as you compute.
• "Horribly huge number" rubs me the wrong way in this context. Huge numbers are horrible when they count something that you don't want. The number's not horrible when it's your bank balance! (Actually it might be. If you put together that many pennies it'd probably collapse into a black hole the size of the Milky Way.) – Matt Samuel Mar 16 '17 at 22:44
• @MattSamuel: Milky Way? This number exceeds the number of Planck volumes in the observable universe ... to the hundredth power! – Henning Makholm Mar 17 '17 at 1:24
• Do you really expect me to do coordinate transformations in my head while strapped to a centrifuge??? – Matt Samuel Mar 17 '17 at 1:25
• 2^65536 really isn't a horribly huge number when you consider that its binary representation fits in a mere 8 KB of memory. I mean, we are already using RSA moduli around 2^4096 already, so this number is only around 10× longer. – Nayuki Mar 17 '17 at 3:28
• Actually all finite numbers are pretty small because all but a finite number of the rest are bigger. Come to that all transfinite numbers are pretty small as well. – Martin Rattigan Apr 2 '17 at 21:55
I would calculate it using Maxima (which evaluates repeat exponentiation correctly, right-to-left), since there is no point wasting brain cells on something that a machine can do: | {
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$maxima Maxima branch_5_39_base_2_gc9edaee http://maxima.sourceforge.net using Lisp GNU Common Lisp (GCL) GCL 2.6.12 Distributed under the GNU Public License. See the file COPYING. Dedicated to the memory of William Schelter. The function bug_report() provides bug reporting information. (%i1) 2^2^2^2^2; (%o1) 200352993040684646497907235156025575044782547556975141926501697371089405\ 955631145308950613088093334810103823434290726318182294938211881266886950636476\ 154702916504187191635158796634721944293092798208430910485599057015931895963952\ 486337236720300291696959215610876494888925409080591145703767520850020667156370\ 236612635974714480711177481588091413574272096719015183628256061809145885269982\ 614142503012339110827360384376787644904320596037912449090570756031403507616256\ 247603186379312648470374378295497561377098160461441330869211810248595915238019\ 533103029216280016056867010565164675056803874152946384224484529253736144253361\ 437372908830379460127472495841486491593064725201515569392262818069165079638106\ 413227530726714399815850881129262890113423778270556742108007006528396332215507\ 783121428855167555407334510721311242739956298271976915005488390522380435704584\ 819795639315785351001899200002414196370681355984046403947219401606951769015611\ 972698233789001764151719005113346630689814021938348143542638730653955296969138\ 802415816185956110064036211979610185953480278716720012260464249238511139340046\ 435162386756707874525946467090388654774348321789701276445552940909202195958575\ 162297333357615955239488529757995402847194352991354376370598692891375715374000\ 198639433246489005254310662966916524341917469138963247656028941519977547770313\ 806478134230959619096065459130089018888758808473362595606544488850144733570605\ 881709016210849971452956834406197969056546981363116205357936979140323632849623\ 304642106613620022017578785185740916205048971178182040018728293994344618622432\ 800983732376493181478984811945271300744022076568091037620399920349202390662626\ | {
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438917541526750091501119885687270884819553167693168127289214303137681801644547\ 736751835349785792427646335416243360112596025210950161226411034608346564823559\ 793427405686884922445874549377675212032470380303549115754483129527589193989368\ 087632768543876955769488142284431199859570072752139317683783177033913042306095\ 899913731468456901042209516196707050642025673387344611565527617599272715187766\ 001023894476053978951694570880272873622512107622409181006670088347473760515628\ 553394356584375627124124445765166306408593950794755092046393224520253546363444\ 479175566172596218719927918657549085785295001284022903506151493731010700944615\ 101161371242376142672254173205595920278212932572594714641722497732131638184532\ 655527960427054187149623658525245864893325414506264233788565146467060429856478\ 196846159366328895429978072254226479040061601975197500746054515006029180663827\ 149701611098795133663377137843441619405312144529185518013657555866761501937302\ 969193207612000925506508158327550849934076879725236998702356793102680413674571\ 895664143185267905471716996299036301554564509004480278905570196832831363071899\ 769915316667920895876857229060091547291963638167359667395997571032601557192023\ 734858052112811745861006515259888384311451189488055212914577569914657753004138\ 4717124577965048175856395072895337539755822087777506072339445587895905719156736 (%i2) bfloat(%); (%o2) 2.003529930406846b19728 (%i3) Of course if I just wanted to estimate the magnitude of the number without resorting to the use of arbitrary precision computer software, I'd note that the exponent is$2^{2^{2^2}}=2^{(2^{(2^2)})}=2^{(2^4)}=2^{16}=65536$; multiplying it by$\log_{10} 2\sim 0.30103$gives$19728.302$, so the result is approximately$10^{0.302}\times 10^{19728}\sim 2\times 10^{19728}$. • Wow. Someone actually posted the digits. – Matt Samuel Mar 16 '17 at 23:36 • There is a mistake in the computed result. The 2 in the middle should be a 3. Probably a typo. – augustin Mar 17 '17 at 3:43 • "2 in the | {
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The 2 in the middle should be a 3. Probably a typo. – augustin Mar 17 '17 at 3:43 • "2 in the middle"... can you be a bit more specific? – Viktor Toth Mar 17 '17 at 3:48 • @augustin: The digit in the middle is an 8. And yes, I checked it; with the help of the computer, of course. Given that the total number of digits (19729) is odd, the digit in the middle is well defined (it is the digit which is preceded and followed by the same number of digits). – celtschk Mar 17 '17 at 7:59 • @celtschk now you sir are fun at parties! – Pierre Arlaud Mar 17 '17 at 8:44 This looks an awfully close to what is known as a tetration (a.k.a. power tower). This is$^{(k)}a=a^{^{(k-1)}a}$where$^1a=a$. For numbers greater than one, these usually get really big really fast, and faster than exponents do. So in your case, you have$^52=2^{2^{16}}$. Now if you want to see an interesting one look at$\lim_{k\to \infty}\;^{(k)}(\sqrt{2})$. • Wouldn't that work for any$\sqrt[n]{n}$? – Random832 Mar 16 '17 at 20:04 • Wouldn't what work for any$\sqrt[n]{n}\$? I never said the answer. And yes, but it depends on what you think the answer is. ;D – Sentinel135 Mar 16 '17 at 20:13 | {
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• Looks like that limit goes to infinity really quickly. – Joshua Mar 16 '17 at 20:36
• @Joshua are you sure? can you try and prove it? – Sentinel135 Mar 16 '17 at 20:56
• Darn. I hate accumulated roundoff. – Joshua Mar 16 '17 at 21:24 | {
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Your equation can be simplified using Knuth's up arrow notation. \begin{equation*} 2^{2^{2^{2^2}}} = 2 \uparrow\uparrow 5 \end{equation*}
(because we can calculate tetration with Knuth's up arrow notation)
By definition of Knuth's up arrow notation, You can get this result.
\begin{equation*} 2\uparrow\uparrow5 = 2^{(2^{(2^{(2^2)})})} \end{equation*}
And according to web2.0calc,
\begin{equation*} 2^{(2^{(2^{2})})} = 65536 \end{equation*}
Finally, the answer would be:
\begin{equation*} 2^{65536} \end{equation*}
(correct me if I'm wrong, this was my first answer on Mathematics SE) | {
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# block matrix multiplication
If $A,B$ are $2 \times 2$ matrices of real or complex numbers, then
$$AB = \left[ \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right]\cdot \left[ \begin{array}{cc} b_{11} & b_{12} \\ b_{21} & b_{22} \end{array} \right] = \left[ \begin{array}{cc} a_{11}b_{11}+a_{12}b_{21} & a_{11}b_{12}+a_{12}b_{22} \\ a_{21}b_{11}+a_{22}b_{21} & a_{22}b_{12}+a_{22}b_{22} \end{array} \right]$$
What if the entries $a_{ij}, b_{ij}$ are themselves $2 \times 2$ matrices? Does matrix multiplication hold in some sort of "block" form ?
$$AB = \left[ \begin{array}{c|c} A_{11} & A_{12} \\\hline A_{21} & A_{22} \end{array} \right]\cdot \left[ \begin{array}{c|c} B_{11} & B_{12} \\\hline B_{21} & B_{22} \end{array} \right] = \left[ \begin{array}{c|c} A_{11}B_{11}+A_{12}B_{21} & A_{11}B_{12}+A_{12}B_{22} \\\hline A_{21}B_{11}+A_{22}B_{21} & A_{22}B_{12}+A_{22}B_{22} \end{array} \right]$$ This identity would be very useful in my research.
• Yes it does if the "blocking" is "conforming". May 9, 2014 at 14:12
• Yes. (The blocking is "confirming" in the situation you have given.) Discussed in detail in §6.12 of cip.ifi.lmu.de/~grinberg/primes2015/sols.pdf (specifically Exercise 38 and Remark 6.73; search for "block-matrix notation" if these numbers change). Dec 4, 2015 at 10:16
It depends on how you partition it, not all partitions work. For example, if you partition these two matrices
$$\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}, \begin{bmatrix} a' & b' & c' \\ d' & e' & f' \\ g' & h' & i' \end{bmatrix}$$
in this way
$$\left[\begin{array}{c|cc}a&b&c\\ d&e&f\\ \hline g&h&i \end{array}\right], \left[\begin{array}{c|cc}a'&b'&c'\\ d'&e'&f'\\ \hline g'&h'&i' \end{array}\right]$$
and then multiply them, it won't work. But this would
$$\left[\begin{array}{c|cc}a&b&c\\ \hline d&e&f\\ g&h&i \end{array}\right] ,\left[\begin{array}{c|cc}a'&b'&c'\\ \hline d'&e'&f'\\ g'&h'&i' \end{array}\right]$$ | {
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What's the difference? Well, in the first case, all submatrix products are not defined, like $\begin{bmatrix} a \\ d \\ \end{bmatrix}$ cannot be multiplied with $\begin{bmatrix} a' \\ d' \\ \end{bmatrix}$
So, what is the general rule? (Taken entirely from the Wiki page on Block matrix)
Given, an $(m \times p)$ matrix $\mathbf{A}$ with $q$ row partitions and $s$ column partitions $$\begin{bmatrix} \mathbf{A}_{11} & \mathbf{A}_{12} & \cdots &\mathbf{A}_{1s}\\ \mathbf{A}_{21} & \mathbf{A}_{22} & \cdots &\mathbf{A}_{2s}\\ \vdots & \vdots & \ddots &\vdots \\ \mathbf{A}_{q1} & \mathbf{A}_{q2} & \cdots &\mathbf{A}_{qs}\end{bmatrix}$$
and a $(p \times n)$ matrix $\mathbf{B}$ with $s$ row partitions and $r$ column parttions
$$\begin{bmatrix} \mathbf{B}_{11} & \mathbf{B}_{12} & \cdots &\mathbf{B}_{1r}\\ \mathbf{B}_{21} & \mathbf{B}_{22} & \cdots &\mathbf{B}_{2r}\\ \vdots & \vdots & \ddots &\vdots \\ \mathbf{B}_{s1} & \mathbf{B}_{s2} & \cdots &\mathbf{B}_{sr}\end{bmatrix}$$
that are compatible with the partitions of $\mathbf{A}$, the matrix product
$\mathbf{C}=\mathbf{A}\mathbf{B}$
can be formed blockwise, yielding $\mathbf{C}$ as an $(m\times n)$ matrix with $q$ row partitions and $r$ column partitions.
• Use \pmatrix{a&b&c\\d&e&f\\g&h&i} May 9, 2014 at 14:37
• @Berci I know that. But getting those horizontal and vertical lines is the difficult part. May 9, 2014 at 14:38
• @PandaBear You can use $\color{blue}{\text{colors}}$ :) May 9, 2014 at 14:45
• @PavelJiranek Your comment is funny looking. May 9, 2014 at 14:46
• I think this is wrong. You can't partition both of them same way. If you partition after x rows in first matrix , you've to partition after x columns (not rows ) in the second matrix. Otherwise while multiplying you'll have to multiply mn block with another mn block which is not possible. (you need np block) Try it with your example. Oct 1, 2015 at 18:08 | {
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It is always suspect with a very late answer to a popular question, but I came here looking for what a compatible block partitioning is and did not find it:
For $$\mathbf{AB}$$ to work by blocks the important part is that the partition along the columns of $$\mathbf A$$ must match the partition along the rows of $$\mathbf{B}$$. This is analogous to how, when doing $$\mathbf{AB}$$ without blocks—which is of course just a partitioning into $$1\times 1$$ blocks—the number of columns in $$\mathbf A$$ must match the number of rows in $$\mathbf B$$.
Example: Let $$\mathbf{M}_{mn}$$ denote any matrix of $$m$$ rows and $$n$$ columns irrespective of contents. We know that $$\mathbf{M}_{mn}\mathbf{M}_{nq}$$ works and yields a matrix $$\mathbf{M}_{mq}$$. Split $$\mathbf A$$ by columns into a block of size $$a$$ and a block of size $$b$$, and do the same with $$\mathbf B$$ by rows. Then split $$\mathbf A$$ however you wish along its rows, same for $$\mathbf B$$ along its columns. Now we have $$A = \begin{bmatrix} \mathbf{M}_{ra} & \mathbf{M}_{rb} \\ \mathbf{M}_{sa} & \mathbf{M}_{sb} \end{bmatrix}, B = \begin{bmatrix} \mathbf{M}_{at} & \mathbf{M}_{au} \\ \mathbf{M}_{bt} & \mathbf{M}_{bu} \end{bmatrix},$$
and $$AB = \begin{bmatrix} \mathbf{M}_{ra}\mathbf{M}_{at} + \mathbf{M}_{rb}\mathbf{M}_{bt} & \mathbf{M}_{ra}\mathbf{M}_{au} + \mathbf{M}_{rb}\mathbf{M}_{bu} \\ \mathbf{M}_{sa}\mathbf{M}_{at} + \mathbf{M}_{sb}\mathbf{M}_{bt} & \mathbf{M}_{sa}\mathbf{M}_{au} + \mathbf{M}_{sb}\mathbf{M}_{bu} \end{bmatrix} = \begin{bmatrix} \mathbf{M}_{rt} & \mathbf{M}_{ru} \\ \mathbf{M}_{st} & \mathbf{M}_{su} \end{bmatrix}.$$
All multiplications conform, all sums work out, and the resulting matrix is the size you'd expect. There is nothing special about splitting in two so long as you match any column split of $$\mathbf A$$ with a row split in $$\mathbf B$$ (try removing a block row from $$\mathbf A$$ or further splitting a block column of $$\mathbf B$$). | {
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The nonworking example from the accepted answer is nonworking because the columns of $$\mathbf A$$ are split into $$(1, 2)$$ while the rows of $$\mathbf B$$ are split into $$(2, 1)$$.
• So basically the formula is valid whenever it makes sense? Apr 3, 2021 at 16:10
• @Filippo pretty much, though it did not make sense to me at the time which is why I had to go looking for it Apr 5, 2021 at 11:33
The recipe for multiplication of (scalar) matrices $$(AB)_{i,j}=\sum_k A_{i,k}B_{k,j}\tag1$$ is saying: to obtain the $$(i,j)$$ element of $$AB$$, form the dot product as you walk along row $$i$$ of $$A$$ while simultaneously walking down column $$j$$ of $$B$$. In other words,
The element at row $$i$$, column $$j$$ of $$AB$$ is the product of row $$i$$ of $$A$$ with column $$j$$ of $$B$$.
Using the notation $$A_{i,\ast}$$ to denote row $$i$$ of $$A$$ and $$B_{\ast,j}$$ to denote column $$j$$ of $$B$$, this can be restated symbolically as $$(AB)_{i,j}=A_{i,\ast}B_{\ast, j}\ .\tag2$$
This all works provided you can match up any row of $$A$$ with any column of $$B$$, which means the number of columns of $$A$$ must equal the number of rows of $$B$$. (This is what it means for $$A$$ and $$B$$ to be conformable.)
You can generalize (1) to block matrices, i.e., the formula still works when every $$A_{i,k}$$ and $$B_{k,j}$$ is itself a matrix [and the subscripts enumerate rows and columns of blocks], provided every product on the RHS of (1) makes sense. This means the number of columns of $$A_{i,k}$$ must equal the number of rows of $$B_{k,j}$$. | {
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Example: Partition matrix $$A$$ into $$A:=\pmatrix{A_1 &A_2}$$ and $$B$$ into $$B:=\pmatrix{B_1\\ B_2}$$, where #columns($$A_1$$) = #rows($$B_1$$), and #columns($$A_2$$) = #rows($$B_2$$). Then $$AB=\pmatrix{A_1 & A_2}\pmatrix{B_1\\ B_2}=A_1B_1 + A_2 B_2\ .\tag3$$ To see this, consider the element at row $$i$$, column $$j$$ of $$AB$$. This element is computed as the product of row $$i$$ of $$A$$ with column $$j$$ of $$B$$. But row $$i$$ of $$A$$ is just row $$i$$ of $$A_1$$ followed by row $$i$$ of $$A_2$$. Similarly, column $$j$$ of $$B$$ is column $$j$$ of $$B_1$$ atop column $$j$$ of $$B_2$$. Assuming conformability, the product of row $$i$$ of $$A$$ with column $$j$$ of $$B$$ is the sum of two pieces: one piece is the product of row $$i$$ of $$A_1$$ with column $$j$$ of $$B_1$$, and the other is the product of row $$i$$ of $$A_2$$ with column $$j$$ of $$B_2$$. In symbols we've argued that $$(AB)_{i,j}=A_{i,\ast}B_{\ast,j}=(A_1)_{i,\ast}(B_1)_{\ast,j}+(A_2)_{i,\ast}(B_2)_{\ast,j}=(A_1B_1)_{i,j} + (A_2B_2)_{i,j},\tag4$$ and we're done.
Example: Consider conformable partitioned matrices $$C:=\pmatrix{C_1\\ C_2}$$ and $$D:=\pmatrix{D_1 & D_2}$$. Then $$CD=\pmatrix{C_1\\ C_2}\pmatrix{D_1 & D_2} =\pmatrix{C_1D_1 & C_1D_2\\C_2D_1&C_2 D_2}.\tag5$$ Again, this can be seen by considering what happens when you multiply row $$i$$ of $$C$$ with column $$j$$ of $$D$$. | {
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# Math Help - Challenging Integral
1. ## Challenging Integral
I was tutoring someone in Calculus today, and they had an extra credit assignment to do. These integrals appeared on past PUTNAM exams, and they can get help on these. However, I've been trying to work this one out and I (and all the other tutors I work with) couldn't figure it out.
So for the sake of me giving a hint or two to this person, I would like some guidance on how to tackle this beast:
$\int_2^4\frac{\sqrt{\ln\!\left(9-x\right)}\,dx}{\sqrt{\ln\!\left(9-x\right)}+\sqrt{\ln\!\left(x+3\right)}}$
All I know is that this equals 1 but I don't know how to get it. XD
Any input or suggestions would be appreciated!!
2. Originally Posted by Chris L T521
I was tutoring someone in Calculus today, and they had an extra credit assignment to do. These integrals appeared on past PUTNAM exams, and they can get help on these. However, I've been trying to work this one out and I (and all the other tutors I work with) couldn't figure it out.
So for the sake of me giving a hint or two to this person, I would like some guidance on how to tackle this beast:
$\int_2^4\frac{\sqrt{\ln\!\left(9-x\right)}\,dx}{\sqrt{\ln\!\left(9-x\right)}+\sqrt{\ln\!\left(x+3\right)}}$
All I know is that this equals 1 but I don't know how to get it. XD
Any input or suggestions would be appreciated!!
Consider
$
\int_2^4 \frac{\sqrt{\ln 9-x} + \sqrt{\ln x+3} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx = \int_2^4 1 = 2
$
Expanding gives
$
\int_2^4 \frac{\sqrt{\ln 9-x} + \sqrt{\ln x+3} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx = \int_2^4 \frac{\sqrt{\ln 9-x} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx$
$+ \int_2^4 \frac{ \sqrt{\ln x+3} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx$
Let $x = 6-u$ in the second integral
$\int_2^4 \frac{\sqrt{\ln 9-x} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx$ $- \int_4^2 \frac{ \sqrt{\ln 9-u} }{\sqrt{\ln 3+u} + \sqrt{\ln 9-u}}\,dx$
$= 2 \int_2^4 \frac{\sqrt{\ln 9-x} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx = 2$ | {
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$= 2 \int_2^4 \frac{\sqrt{\ln 9-x} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx = 2$
$\int_2^4 \frac{\sqrt{\ln 9-x} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx = 1$
3. Originally Posted by danny arrigo
Consider
$
\int_2^4 \frac{\sqrt{\ln 9-x} + \sqrt{\ln x+3} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx = \int_2^4 1 = 2
$
Expanding gives
$
\int_2^4 \frac{\sqrt{\ln 9-x} + \sqrt{\ln x+3} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx = \int_2^4 \frac{\sqrt{\ln 9-x} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx$
$+ \int_2^4 \frac{ \sqrt{\ln x+3} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx$
Let $x = 6-u$ in the second integral
$\int_2^4 \frac{\sqrt{\ln 9-x} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx$ $- \int_4^2 \frac{ \sqrt{\ln 9-u} }{\sqrt{\ln 3+u} + \sqrt{\ln 9-u}}\,dx$
$= 2 \int_2^4 \frac{\sqrt{\ln 9-x} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx = 2$
$\int_2^4 \frac{\sqrt{\ln 9-x} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx = 1$
I would have never thought of that. Thank you very much!
4. It is very simple actually.Look at this standard property in definite integral
$
\int_a^bf(x)dx=\int_a^bf(a+b-x)dx
$
and the rest is obvious
Interesting corollary to above
$\int_a^b\frac{f(x)}{f(x)+f(a+b-x)}dx=\frac{b-a}{2}$
5. Originally Posted by pankaj
It is very simple actually.Look at this standard property in definite integral
$
\int_a^bf(x)dx=\int_a^bf(a+b-x)dx
$
and the rest is obvious
Interesting corollary to above
$\int_a^b\frac{f(x)}{f(x)+f(a+b-x)}dx=\frac{b-a}{2}$
The first part of this is in Stewart's Calculus book (problems plus). | {
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# Probability of drawing exactly $1$ ace upon drawing $2$ cards from a deck
I got this question and answered it incorrectly. I haven't yet seen the correct answer. The possible answers were:
1. $\frac{4}{52}$
2. $\frac{16}{221}$ (my answer)
3. $\frac{2}{52}$
4. $\frac{32}{221}$
My reasoning is the following:
Event A: Card is not an ace.
Event B: Card is an ace.
$$P(A)\times P(B|A)=\frac{4}{52}\times \frac{48}{51}=\frac{16}{221}$$
This is assuming the cards are drawn sequentially. Drawing exactly one ace from a single draw is $4/52$ but to ensure that only a single ace was drawn one should consider the probability of not getting a second ace.
How am I wrong?
• What if the first card is not an ace but the second is? – Ethan Bolker Dec 25 '17 at 16:21
• If you are Lemmy, then in probability 1 you will draw the ace of spades. – Asaf Karagila Dec 26 '17 at 8:18
• You can also do it in another way (as a check). The probability of two aces is $\frac{4}{52}\cdot\frac{3}{51}$. The probability of zero aces (that is two non-aces), is $\frac{48}{52}\cdot\frac{47}{51}$. So the answer should be $$1-\frac{4}{52}\cdot\frac{3}{51}-\frac{48}{52}\cdot\frac{47}{51}$$ – Jeppe Stig Nielsen Dec 26 '17 at 13:44
You forgot to assume $P(B)\times P(A|B)=\frac{48}{52}\times\frac4{51}=\frac{16}{221}$.
So $P(A)\times P(B|A)+P(B)\times P(A|B)= \frac{16}{221}+ \frac{16}{221}= \frac{32}{221}$.
The correct answer is $4$ , i.e. $\frac{32}{221}$ | {
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The correct answer is $4$ , i.e. $\frac{32}{221}$
• Why? Isn't it the same thing, just reversed? – Rivasa Dec 25 '17 at 21:02
• @Annabelle, there are two possible ways you can get it, so you account for both. Thinking about it intuitively, because there aren’t many aces, you would expect your chances to be better by taking two cards than by taking one. Otherwise, if you kept taking more and more, your chances wouldn’t change, which would be strange when you got to 50 cards and are guaranteed to get two aces... – Jeff Dec 25 '17 at 21:12
• @Annabelle: Consider that you have a deck of 3 cards, 2 aces. You draw two cards. What's the probability of drawing exactly one ace? Well, the event "drawing one" is equivalent to "an ace remains in the deck after drawing two cards". What is the probability of that? Clearly 2/3 as there are two aces in a deck of three if you imagine randomly picking a card to remain in the deck after drawing. But computing it as the OP did here you would end with P=1/3 because of mistakenly not considering that both drawing an ace first and drawing an ace second must be taken into account. – Jon Dec 25 '17 at 21:58
• I think I get it now... @Jon – Rivasa Dec 26 '17 at 2:46
We have: $$P_{\text{ only one ace }} = P_{\text{ ace }1} P_{\text{ non-ace }2} + P_{\text{ non-ace }1} P_{\text{ ace }2} = \frac{4}{52}\times \frac{48}{51} + \frac{48}{52}\times \frac{4}{51} = \frac{32}{221}$$
• Why the downvote? – user371838 Dec 26 '17 at 13:40
Answer 4 seems correct to me because:
$$P=\frac {\binom4 1\binom {48} 1}{\binom {52} 2}=\frac {32}{221}$$
You can figure this out without any knowledge of conditional probabilities:
There are 52 choices for the first pick and 51 choice for the second pick. So, in total, there are $52 \times 51 = 2652$ possible "hands" that consist of two cards.
Now, the question is, how many of these 2652 contain a single ace. Let's do some counting of the possible one-ace hands: | {
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There are $4 \times 48 = 192$ hands that have an ace as their first pick and a non-ace as the second pick (because there are 4 aces and 48 non-aces). Similarly, there are $48 \times 4 = 192$ hands that have a non-ace as their first pick and an ace as the second pick. So the total number of hands that have a single ace is $192+192=384$.
So, the probability of a one-ace hand is $384/2652$. Dividing top and bottom by $12$ gives us $384/2652 = 32/221$.
• Does the downvoter care to explain his or her reasoning? – bubba Dec 27 '17 at 2:51 | {
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# Number of different permutations
Consider some text $T$
How many different permutations of this text can we achieve ?
The easiest case is when every letter appears only once in the text, so the answer is $|T|!$
But when we have for example $T=aab$ the number of different permutations is 3, because $aab$ $aba$ $baa$
How can we solve this problem without actually generating all possible permutations of this text ?
-
It is given by the multinomial theorem:
$$\frac{N!}{m_1! m_2! m_3! \ldots m_k!} = \text{choose}(n, (m_1,m_2, \ldots, m_k))$$
Here $N$ is the number of elements in $T$, $|T|=N$. If there are $k$ unique elements in $T$, $m_i$ is the number of times that element shows up. From your example, $N=3$, $m_1 = 2$, $m_2 = 1$, so we get
$$\frac{3!}{2! \cdot 1!} = \frac{6}{2} = 3$$
-
Thank you, i didn't know it's that easy :) – Chris Feb 1 '12 at 16:45
An example might help. Consider the text $probability$. If we thought of all letters as distinct: $p\ r\ o\ b_1\ a\ b_2\ i_1\ l\ i_2\ t\ y$, then there would be $11!$ permutations.
But this overcounts, as not all letters are distinct. In particular, amongst the $11!$ factorial arrangements above, we have both \eqalign{ &r\ p\ o\ b_1\ a\ b_2\ \color{maroon}{i_1}\ l\ \color{maroon}{i_2}\ t\ y\cr &r\ p\ o\ b_1\ a\ b_2\ \color{maroon}{i_2}\ l\ \color{maroon}{i_i}\ t\ y\cr} For any particular arrangement amongst the $11!$ arrangements of the letters, we can find one other (by permuting the $i$'s) that is the same when we regard the $i$'s as indistinguishable.
So we've over counted by a factor of 2. Similarly, because there are two $b$'s, we overcounted by another factor of 2.
So, the correct number of arrangements is $11!\over 2\cdot2$. | {
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So, the correct number of arrangements is $11!\over 2\cdot2$.
A similar argument can be made to establish the formula that Hooked gives in his answer. In particular if one letter was repeated thrice, then if we found the number of arrangements where we thought of all letters as distinct, we would have over counted by a factor of $3!$; since there are $3!$ ways to permute the three repititions of the letter.
-
The general formula for a permutation of a text with length $n$ with $n_1$ indistinguishable objects of one type, $n_2$ of the next type is:
$$\frac{n!}{n_1!n_2!\dots n_k!}$$
For your text: aab n=3, $n_1$(the number of times a appears)=2 and $n2=1$(Since b appears once). This formula simply divides out all of the objects that $n!$ overcounts.
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# Math Help - Simplifying fractions
1. ## Simplifying fractions
This is the problem
Simplify the following equation:
(5x/(2x^2+7x+3))-((x-3)/(2x^2-3x-2))+((2x+1)/(x^2+x-6))
The way i did it was factoring the denominators then adding them, therefore getting common terms where the denominator would be (2x+1)(x+3)(x-2). After that, I don't know what to do next, I'm stumped XD
Would appreciate the help..
2. Hello, mcroldan08!
Simplify the following expression:
. . $\frac{5x}{2x^2+7x+3} - \frac{x-3}{2x^2-3x-2} + \frac{2x+1}{x^2+x-6}$
It sounds like your intentions were correct.
But since you didn't show us what you did, we can't tell if you made any errors.
We have: . $\frac{5x}{(x+3)(2x+1)} - \frac{x-3}{(x-2)(2x+1)} + \frac{2x+1}{(x-2)(x+3)}$
You are correct . . . the LCD is: . $(x+3)(2x+1)(x-2)$
And we must convert the fractions to the LCD.
$\frac{5x}{(x+3)(2x+1)}\cdot{\color{blue}\frac{x-2}{x-2}} - \frac{x-3}{(x-2)(2x+1)}\cdot{\color{blue}\frac{x+3}{x+3}} + \frac{2x+1}{(x-2)(x+3)}\cdot{\color{blue}\frac{2x+1}{2x+1}}$
. . $= \;\frac{5x(x-2)}{(x+3)(2x +1)(x-2)} - \frac{(x-3)(x+3)}{(x-2)(2x+1)(x+3)} + \frac{(2x+1)(2x+1)}{(x-2)(x+3)(2x+1)}$
. . $= \;\frac{5x(x-2) - (x-3)(x+3) + (2x+1)(2x+1)}{(x+3)(2x+1)(x-2)}$
. . $= \;\frac{5x^2 - 10x - x^2 + 9 + 4x^2 + 4x + 1}{(x+3)(2x+1)(x-2)}$
. . $= \;\frac{8x^2 - 6x + 10}{(x+3)(2x+1)(x-2)}$
. . $= \;\frac{2(4x^2 - 3x + 5)}{(x+3)(2x+1)(x-2)}$
3. That is actually what I got... however, I was wondering if this can be simplified further. This is actually a multiple choice question and the choices are:
a.) 4/(x+3)
b.) 4/(x-3)
c.) 2/(x-3)
d.) 2/(x+3)
Could it be that the none of the choices are correct? | {
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# Math Help - Rebound Superball Problem
1. ## Rebound Superball Problem
The rebound ratio of a speckled green superball is 75%. It is dropped from a height of 16 feet. Consider the instant when the ball strikes the ground for the fiftieth time.
a. How far downward has the ball traveled at this instant?
b. How far ( upward and downward has the ball traveled at this instant?
c. How far would the ball travel if you just let it bounce...?
I have a feeling for the first two it would be something along the lines of "16(0.75)^0+16(0.75)^1...+16(0.75)^50", and for the second 2(" "), but there must be something that would allow me to do this calculation quicker than just adding 50 numbers together. Would it have something to do with ! ?
Thanks.
2. Originally Posted by bkap
The rebound ratio of a speckled green superball is 75%. It is dropped from a height of 16 feet. Consider the instant when the ball strikes the ground for the fiftieth time.
a. How far downward has the ball traveled at this instant?
b. How far ( upward and downward has the ball traveled at this instant?
c. How far would the ball travel if you just let it bounce...?
I have a feeling for the first two it would be something along the lines of "16(0.75)^0+16(0.75)^1...+16(0.75)^50", and for the second 2(" "), but there must be something that would allow me to do this calculation quicker than just adding 50 numbers together. Would it have something to do with ! ?
Thanks.
Yes your approach to the first problem is correct, if you factorize things out a bit, it looks something like:
$16*((3/4)^0+(3/4)^1+...+(3/4)^{49})$
it ends with the 49th power because 0-49 gives a total of 50 drops
now this is just a standard progression, whose sum is:
$a(r^{n+1}-1)/(r-1)$
where a is the first term, in this case 1, r is the successive ratio, in this case 3/4, and n is how far the last term goes, in this case 49. | {
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The second part is trivial once you got the first part, however you have to keep in mind that the first upward bounce is the same distance as the second downward bounce.
Finally for the third part, you simply let n tend to infinity and calculate. If you haven't done limit, just make n=1000 on the calculator, this will give you a pretty good approximation
3. Originally Posted by bkap
The rebound ratio of a speckled green superball is 75%. It is dropped from a height of 16 feet. Consider the instant when the ball strikes the ground for the fiftieth time.
a. How far downward has the ball traveled at this instant?
b. How far ( upward and downward has the ball traveled at this instant?
c. How far would the ball travel if you just let it bounce...?
I have a feeling for the first two it would be something along the lines of "16(0.75)^0+16(0.75)^1...+16(0.75)^50", and for the second 2(" "), but there must be something that would allow me to do this calculation quicker than just adding 50 numbers together. Would it have something to do with ! ?
Thanks.
Nah, Factorials only really occur in probability - you'll be better off with a sequence. Let s be the distance it bounces back up
$s_0: 16 = 16 \times 0.75^0$
$s_1: 16 \times 0.75^1 = 0.75s_0$
$s_2: (16 \times 0.75) \times 0.75 = 0.75s_1$
...
$s_n = 16 \times 0.75^n = 0.75s_{n-1}$
From this we can see it's a geometric sequence so use the sum of a geometric sequence to solve how far it's bounced up. To find down and up multiply by 2
4. Hello, bkap!
Everyone has given you excellent advice . . .
The rebound ratio of a speckled green superball is 75%.
It is dropped from a height of 16 feet.
Consider the instant when the ball strikes the ground for the 50th time.
a. How far downward has the ball traveled at this instant?
Total downward distance is given by:
. . $D \;=\;16 + 16(.075) + 16(.075)^2 + 16(0.75)^3 + \hdots + 16(0.75)^{49}$ | {
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. . $D \;=\;16 + 16(.075) + 16(.075)^2 + 16(0.75)^3 + \hdots + 16(0.75)^{49}$
. . . . . $=\;16\underbrace{\bigg[1 + (0.75) + (0.75)^2 + (0.75)^3 + \hdots + (0.75)^{49}\bigg]}_{\text{geometric series}}$
. . The sum of the geometric series is: . $\frac{1-(0.75)^{50}}{1-0.75} \:=\:3.999997735$
Therefore: . $D \;=\;16(3.999997736) \;=\;63.99996376$ feet.
b. How far ( upward and downward) has the ball traveled at this instant?
We must be very careful . . .
The balls falls 16 feet.
Then bounces up $16(0,75)$ feet and falls $16(0.75)$ feet.
Then bounces up $16(0.75)^2$ feet and falls $16(0.75)^2$ feet . . . and so on.
The total distance is:
. . $T \;=\;16 + 2\!\cdot\!16(0.75) + 2\!\cdot\!16(0.75)^2 + 2\!\cdot\!16(0.75)^3 + \hdots + 2\!\cdot\!16(0.75)^{49}$
. . . . $=\; 16 + 32(0.75)\bigg[1 + (0.75) + (0.75)^2 + \hdots + (0.75)^{49}\bigg]$
. . . . $=\; 16 + 24(3.999997735) \;=\;111.9999275$ feet.
c. How far would the ball travel if you just let it bounce?
If we let it bounce (forever), the total distance is:
. . $T \;=\;16 + 24\underbrace{\bigg[1 + (0.75) + (0.75)^2 + (0.75)^3 + \hdots\bigg]}_{\text{infinite geomtric series}}$
. . The sum of the infinite series is: . $\frac{1}{1-0.75} \:=\:4$
Therefore: . $T \;=\;16 + 24(4) \;=\;112$ feet. | {
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# Solving the infinite series $1-\frac{2^3}{1!}+\frac{3^3}{2!}-\frac{4^3}{3!}+\cdots$
I have the following question:
Evaluate the infinite series: $$S=1-\frac{2^3}{1!}+\frac{3^3}{2!}-\frac{4^3}{3!}+\cdots$$ (a) $\displaystyle\frac1e$ (b) $\displaystyle\frac{-1}e$ (c) $\displaystyle\frac{2}e$ (d) $\displaystyle\frac{-2}e$
Now in the book they have given a strange explanation using $$(n+1)^3=[n(n-1)(n-2)+6n(n-1)+7n+1].$$ I don't understand how did they get this. So I have two doubts:
1. How did they get this "trick"?
2. If you would have got this question in a competitive exam, how would you have solved it (in the minimum amount of time), either directly or with numerical methods?
• The trick seems to be typical when you want to see $An(n-1)(n-2)+Bn(n-1)+Cn+D$. Just identify. Sep 5, 2015 at 14:54
You want to evaluate $\sum_{n=0}^{\infty} \frac{(-1)^n(n+1)^3}{n!}$.
Now the idea (at the latest if you see some $e$ as a possible answer) should be to relate this to the well-known formula $$\sum_{n=0}^{\infty} \frac{x^n}{n!}=e^x$$ In this particular case (given the $(-1)^n$ included) it is useful to plug in $x=-1$ and obtain $$\sum_{n=0}^{\infty} \frac{(-1)^n}{n!}=e^{-1}=\frac{1}{e}$$ But you still have a $(n+1)^3$ in the numerator where you wish to have $1$.
So you might want to split this into several parts such that you get one (or several) related sums.
As an example of this technique, if you want to know $\sum_{n=0}^{\infty} \frac{n+1}{n!}$ you might try: $$\sum_{n=0}^{\infty} \frac{n+1}{n!}=\sum_{n=0}^{\infty} \frac{n}{n!}+\sum_{n=0}^{\infty} \frac{1}{n!}$$ Now, the second sum is just $e$ and the first one is (after cancelling) $$\sum_{n=1}^{\infty} \frac{1}{(n-1)!}=\sum_{n=0}^{\infty} \frac{1}{n!}=e$$ and so the total sum is $2e$. Can you now understand the "trick" and tackle your problem by yourself?
Concerning the second question: | {
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Concerning the second question:
If one happens to know this "trick", it is not hard to get the result by just a few computations so it would be my preferential choice.
But if you sit in the exam and have absolutely no idea of how to solve this algebraically (and especially if you don't have to give a proof of your result!) it might be a good idea to estimate the sum by computing a few terms. At least the sign of the result ($\pm$) should be approachable...
• Yes thanks a lot, Now I understood the trick. But can you please answer the 2nd part of the question? Sep 5, 2015 at 14:59
• @Kartik by learning the trick now.Its a very common question :-)!! I felt the same as you the first time...!
– user220382
Sep 5, 2015 at 15:29
• I changed "nominator" into "numerator Sep 5, 2015 at 15:37
In general, if $p_i(x)$ is a sequence of polynomials with $\deg p_i=i$, then it is relatively easy to prove that every polynomial can be written as a linear combination of a finite number of $p_i$. When the $p_i$ are monic, you can do this easily. Let $P(x)=a_kx^k + ... a_0$. Then $P(x)-a_kp_k(x)$ is a polynomial of lower degree. Find it's first coefficient of that, and proceed inductively.
The standard basis for polynomials is $e_i(x)=x^i$.
So in your case; $p_i(n)=x(x-1)\cdots(x-(i-1))$ with $p_0(x)=1$. We chose this particular basis because $$\frac{p_i(n)}{n!} =\begin{cases}\frac{1}{(n-i)!} & n\geq i\\0&0\leq n<i\end{cases}$$ So:$$\sum_{n=0}^\infty \frac{p_i(n)z^n}{n!} = z^ie^{z}$$
In your case, $z=-1$.
With these polynomials, we get: \begin{align}(n+1)^3 &= n^3+3n^2+3n+1\\ (n+1)^3-p_3(n) &= n^3 +3n^2+3n+1 - (n^3-3n^2+2n)\\& = 6n^2+n+1\\ (n+1)^3-p_3(n)-6p_2(n) &= 6n^2+n+1 - 6n(n-1)\\&= 7n+1\\ \end{align}
So you get: $$(n+1)^3=p_3(n) + 6p_2(n) + 7p_1(n) +p_0(n)$$ | {
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So you get: $$(n+1)^3=p_3(n) + 6p_2(n) + 7p_1(n) +p_0(n)$$
The key is that the leading coefficient of the cubic $(n+1)^3$ is $1$, so we subtract $1\cdot p_3(n)$. Then the leading coefficient of the remaining quadratic is $6$, so we subtract $6p_2(n)$. The leading coefficient of the remaining linear polynomial is $7$, so we subtract $7p_1(x)$. You are left with a constant.
The relationship between the standard basis of the polynomials and the basis $p_i(x)=x(x-1)\cdots(x-(i-1))$ is actually fairly interesting and comes up again and again. The $p_i(x)$ are often called "falling factorials."
My way to solve it:
If the numerators were all $1$, we would have the well-known series for $e^{-1}$, obtained from the entire series
$$e^{-x}=\sum_{k=0}^\infty(-1)^k\frac{x^k}{k!}.$$
Now we want to make numerators $(k+1)^3$ appear. Multiply by $x$ and derive, to get
$$(xe^{-x})'=\left(\sum_{k=0}^\infty(-1)^k\frac{x^{k+1}}{k!}\right)'=\sum_{k=0}^\infty(-1)^k\frac{(k+1)x^k}{k!}.$$
Repeating this three times,
$$(x(x(xe^{-x})')')'=\sum_{k=0}^\infty(-1)^k\frac{(k+1)^3x^k}{k!}.$$
Then evaluate at $x=1$.
$$e^{-x}\to (1-x)e^{-x}\to (1-3x+x^2)e^{-x}\to (1-7x+6x^2-x^3)e^{-x}\to -\frac1e.$$
For such problems, you can often start from a known Taylor series (in particular $e^x$ and $1/(1-x)$)and perform the following operations
• multiply or divide by a power of $x$ (shifts the exponents),
• derive on $x$ (multiplies by the exponents),
• integrate on $x$ (divides by the exponents),
to modify the coefficients.
Since $(n+1)^3 = \color{red}{1}\cdot n(n-1)(n-2)+\color{red}{6}\cdot n(n-1) +\color{red}{7}\cdot n+\color{red}{1}$ we have: | {
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$$\begin{eqnarray*}\sum_{n=0}^{+\infty}\frac{(-1)^n (n+1)^3}{n!} &=& \frac{13}{2}+\sum_{n\geq 3}\frac{(-1)^n (n+1)^3}{n!}\\&=&\frac{13}{2}+\color{red}{1}\cdot\sum_{n\geq 3}\frac{(-1)^n}{(n-3)!}+\color{red}{6}\cdot\sum_{n\geq 3}\frac{(-1)^n}{(n-2)!}+\color{red}{7}\cdot\sum_{n\geq 3}\frac{(-1)^n}{(n-1)!}+\color{red}{1}\cdot\sum_{n\geq 3}\frac{(-1)^n}{n!}\\&=&\frac{13}{2}-\sum_{n\geq 0}\frac{(-1)^n}{n!}+6\cdot\sum_{n\geq 1}\frac{(-1)^n}{n!}-7\cdot\sum_{n\geq 2}\frac{(-1)^n}{n!}+\sum_{n\geq 3}\frac{(-1)^n}{n!}\\&=&\sum_{n\geq 0}(-1+6-7+1)\frac{(-1)^n}{n!}\\&=&-\sum_{n\geq 0}\frac{(-1)^n}{n!}=\color{red}{-\frac{1}{e}}.\end{eqnarray*}$$
We simply get the initial decomposition by induction. Obviously $(n+1)^3-n(n-1)(n-2)$ is a quadratic polynomial in $n$, whose leading term is $6n^2$. So we substract $6n(n-1)$ and we are left with a linear polynomial and so on. Another chance is to tackle our problem this way: $$\begin{eqnarray*} \sum_{n\geq 0}\frac{(-1)^n(n+1)^3}{n!}&=&\sum_{n\geq 1}\frac{(-1)^n n(n+1)^2}{n!}+\sum_{n\geq 0}\frac{(-1)^n(n+1)^2}{n!}\\&=&\sum_{n\geq 0}\frac{(-1)^n(n+1)^2}{n!}-\sum_{n\geq 0}\frac{(-1)^n(n+2)^2}{n!}\\&=&-\sum_{n\geq 0}\frac{(-1)^n(2n+3)}{n!}\\&=&-3\sum_{n\geq 0}\frac{(-1)^n}{n!}+2\sum_{n\geq 0}\frac{(-1)^n}{n!}=\frac{2-3}{e}.\end{eqnarray*}$$
Looking at the given identity: \begin{align} (n+1)^3 &= [n(n-1)(n-2)+6n(n-1)+7n+1] \\ &= n(n^2 - 3n + 2) + 6 n^2 + n + 1 \\ &= n^3 + 3 n^2 + 3 n + 1 = (n+1)^3 \end{align}
Now, consider differentiation of the exponential series: \begin{align} D \, e^{t} &= \sum_{n=0}^{\infty} \frac{1}{n!} \, \frac{d}{dt} \, t^{n} = \sum_{n=0}^{\infty} \frac{n \, t^{n-1}}{n!} \end{align} Further differentiation can be applied. | {
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Now, consider the series in question: \begin{align} S &= 1 - \frac{2^{3}}{1!} + \frac{3^{3}}{2!} - \cdots \\ &= \sum_{n=0}^{\infty} \frac{(-1)^{n} \, (n+1)^{3}}{n!} \\ &= \left. \sum_{n=0}^{\infty} \frac{(-1)^{n} \, (n+1)^{3} \, t^{n}}{n!} \, \right|_{t=1} \\ &= \left. t^{3} \, \sum_{n=0}^{\infty} \frac{(-1)^{n} \, n(n-1)(n-2) \, t^{n-3}}{n!} + 6 t^{2} \, \sum_{n=0}^{\infty} \frac{(-1)^{n} \, n(n-1) \, t^{n-2}}{n!} + 7 t \, \sum_{n=0}^{\infty} \frac{(-1)^{n} \, n \, t^{n-1}}{n!} + e^{-t} \, \right|_{t=1} \\ &= \left. t^3 \, D^{3} e^{-t} + 6 \, t^{2} \, D^{2} e^{-t} + 7 \, t \, D e^{-t} + e^{-t} \right|_{t=1} \\ &= \left. (-t^3 + 6 \, t^2 - 7 \, t + 1) \, e^{-t} \right|_{t=1} \\ &= - \frac{1}{e} \end{align}
• Thanks for the answer but this explanation is already given in the book which I am solving. I just wanted the explanation of the "Trick" in the first identity, Sep 5, 2015 at 15:16 | {
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## Calculus: Early Transcendentals 8th Edition
Published by Cengage Learning
# Chapter 2 - Section 2.2 - The Limit of a Function - 2.2 Exercises: 1
#### Answer
As $x$ approaches 2, $f(x)$ approaches 5. Yes, it is possible for this to be true and for $f(2)=3$.
#### Work Step by Step
It means that $f(x)$ becomes arbitrarily close to $3$ as we take $x$ close enough to $2$, but $x\ne2$. It is possible for $f(2)$ to equal $3$ because the limit only relies on when $f(x)$ is near $2$; there can be a hole at $x=2$.
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# Should I use a binomial cdf or a normal cdf when flipping coins?
A coin needs to be tested for fairness. 30 heads come up after 50 flips. Assuming the coin is fair, what is the probability that you would get at least 30 heads in 50 flips?
The right way to do this problem, according to my teacher, is to do
normalcdf(min = .6, max = ∞, p = .5, σ = sqrt(.5 * .5 / 50) = 0.0786
However, I took a binomial cumulative distribution function like this
1 - binomcdf(n = 50, p = .5, x = 29) = 0.1013
I believe the criteria for a binomial distribution are satisfied: the individual events are independent, there are only two possible outcomes (heads vs. tails), the probability is constant for the question (0.5), and the number of trials is fixed at 50. Yet obviously, the two methods give different answers, and a simulation supports my answer (at least the few times I ran it; obviously, I can't guarantee that you'd get the same results).
Is my teacher wrong in assuming that a Normal distribution curve would also be a valid way to do this problem (at no point is it said that the distribution is Normal, but n*p and n*(1-p) are both greater than 10), or have I misunderstood something about binomial distributions?
• A person with experience in using Normal approximations to the Binomial would proceed a little differently: they would apply the (usual) continuity correction, as in 1 - pnorm((30-0.5)/50, mean=0.5, sd=sqrt(0.5*(1-0.5)/50)) (this is an R expression), whose value is 0.1015, in quite close agreement with the Binomial cdf. – whuber Jan 18 '12 at 21:23
Here is an illustration of the answers of whuber and onestop.
In red the binomial distribution $\mathcal Bin(50,0.5)$, in black the density of the normal approximation $\mathcal N(25, 12.5)$, and in blue the surface corresponding to $\mathbb P(Y > 29.5)$ for $Y \sim \mathcal N(25, 12.5)$. | {
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The height of a red bar corresponding to $\mathbb P(X=k)$ for $X\sim\mathcal Bin(50,0.5)$ is well approximated by $\mathbb P\left( k -{1\over 2} < Y < k + {1\over 2}\right)$. To get a good approximation of $\mathbb P(X \ge 30)$, you need to use $\mathbb P(Y>29.5)$.
(edit) This is $$\mathbb P(Y>29.5) \simeq 0.1015459,$$ (obtained in R by 1-pnorm(29.5,25,sqrt(12.5))) whereas $$\mathbb P(X \ge 30) \simeq 0.1013194:$$ the approximation is correct.
This is called continuity correction. It allows you to compute even "point probabilities" like $\mathbb P(X=22)$ : \begin{align*} \mathbb P(X=22) &= {50 \choose 22} 0.5^{22} \cdot 0.5^{28} \simeq 0.07882567, \\ \mathbb P(21.5 < Y < 22.5) & \simeq 0.2397501 - 0.1610994 \simeq 0.07865066.\end{align*}
The normal distribution gives a closer approximation to the binomial if you use a continuity correction. Using this for your example, I get 0.1015. As this is homework, I'll leave it to you to fill in the details.
Consider this. In the discrete binomial distribution you have actual probabilities for individual numbers. In the continuous normal that isn't the case, you need a range of values. So... if you were going to approximate the probability of an individual value, let's say X, from the binomial with the normal how would you do that? Look at a probability histogram of the binomial distribution with the normal curve laid over it. You would need to actually select from X ± 0.5 to capture something similar to what the binomial probability of X is with the normal approximation.
Now extend that to when you're selecting a tail of the distribution. When you use the binomial method you're selecting your entire value's probability (30 in your case) plus everything higher. Therefore, when you do the continuous you have to make sure you capture that and select 0.5 less as well, so the cutoff on the continuous distribution is 29.5. | {
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• Actually, the question exhibits a thoughtful understanding of the problem and does not appear to be looking for an answer to a routine homework question. Although it's tagged homework, consider making an exception here. In particular, a good discussion of using the Normal distribution to approximate discrete distributions (such as Binomials and Poissons with large N's) would be appropriate and most welcome here. – whuber Jan 18 '12 at 23:37 | {
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# Prove that $(l^\infty,\|.\|_\infty)$ is a Banach space.
$(l^\infty,\|.\|_\infty)$ is a Banach space. In the proof, $\mathbb{F}$ is either the field of complex numbers or the field of real numbers.
Proof: Let $(x^n)_{n\in\mathbb{N}}$ be a Cauchy sequence in $l^\infty$, where $x^n=(x_k^n)_{k\in\mathbb{N}}$. Consider the sequence $(x^0_k,x^1_k,\cdots,x^n_k,\cdots)$ of $k$th coordinates of the sequence $(x^n)_{n\in\mathbb{N}}$.
Let $\epsilon>0$. Since $(x^n)_{n\in\mathbb{N}}$ is Cauchy, there exists $n_0\in\mathbb{N}$ such that $$\forall m,n>n_0,\|x^m-x^m\|_\infty<\epsilon.$$ Therefore for each $m,n>n_0$: $$|x^m_k-x^n_k|<\epsilon.$$
So the sequence $(x^0_k,x^1_k,\cdots,x^n_k,\cdots)$ is Cauchy. Therefore it converges to some $y_k\in\mathbb{F}\$ ($\mathbb{F}$ is complete).
Let $y=(y_k)_{k\in\mathbb{N}}$. Since $(x^0_k,x^1_k,\cdots,x^n_k,\cdots)$ is Cauchy it is bounded. Choose $M>0$ such that for each $n\in\mathbb{N}$, $|x^n_k|<M$. But since $$y_k=\lim_{n\to\infty}x_k^n,$$ we have $|y_k|\leq M$ for each $k$. Therefore, $y\in l^\infty$.
Fix $m>n_0$. Then we have $\|x^m-x^n\|_\infty<\epsilon.$ Therefore $\|x^m-y\|_\infty<\epsilon$ as $n\to\infty$.
Therefore for each $m>n_0,\ \|x^m-y\|_\infty<\epsilon$. Hence $(x^n)_{n\in\mathbb{N}}$ converges in $l^\infty$ and the proof is complete.
Could someone please tell me if the above proof is alright? Thanks. | {
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Could someone please tell me if the above proof is alright? Thanks.
• Looks alright to me. The readability was fine even before it was over-pedantically edited. – uniquesolution Aug 23 '17 at 7:35
• The general procedure seems right, however near the end $\|x^m - x^m\|_\infty$ is something you should fix. – mlk Aug 23 '17 at 7:36
• @uniquesolution Then you are just more skilled than I am in reading wall-of text style questions. But I hope even you will agree that the readability is much better now. – 5xum Aug 23 '17 at 7:37
• When I first looked at it it was properly Latex formatted. If it was text-style and you converted to Latex, then thank you. – uniquesolution Aug 23 '17 at 7:39
• @mlk fixed it, and thanks everyone. – Janitha357 Aug 23 '17 at 7:40
Your proof is very rigorous and very detailed all the way up to the point where you say
Fix $m>n_0$. Then we have $\|x^m-x^n\|_\infty<\epsilon.$ Therefore $\|x^m-y\|_\infty<\epsilon$ as $n\to\infty$.
Now I know the inequality stands, but as you were very thorough with all your other inequalities, I think it would be nice if you wrote a little more justification for this one as well - it is not entirely obvious how the right inequality follows from the left one.
Other than that, the proof is very well written and easy to follow.
Rather a comment, but my reputation does not allow me to comment yet. You write
choose $M>0$ that for each $n\in\mathbb{N}, |x^n_k|<M$
I am curios whether you have to write $M_k$, since hypothetically $M$ can depend on the choice of the sequence $(x^0_k,x^1_k,\cdots,x^n_k,\cdots)$. But then we have difficulties with proving that $y\in l^\infty$. | {
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EDIT: the proof is more subtle than I initially thought. First we have to settle the issue 5xum mentioned. We start with $$\forall k\in\mathbb{N}\ \forall \epsilon > 0 \ \exists n_0:\forall n,p>n_0:|x^n_k-x^{n+p}_k|<\epsilon.$$ This is inequality in $\mathbb{F}$, so we can let $p\to\infty$. Thus we obtain $$\forall k\in\mathbb{N}\ \forall \epsilon > 0 \ \exists n_0:\forall n>n_0:|x^n_k-y_k|<\epsilon.$$ That means $\|x^n-y\|_\infty<\epsilon$, thus $(x^n)_{n\in\mathbb{N}}$ is converging to $y$.
We still have to prove that $y\in l^\infty$. But $$\|y\|_\infty \le \|x^n-y\|_\infty + \|x^n\|_\infty \le \epsilon + \|x^n\|_\infty < \infty$$
• Could you explain the last bit please? – Janitha357 Aug 23 '17 at 9:53
• I am using triangle inequality and addition of zero to prove that $y\in l^\infty$: $\|y\|_\infty \le \|x^n - x^n+y\|_\infty \le \|x^n-y\|_\infty + \|x^n\|_\infty \le \epsilon + \|x^n\|_\infty < \infty$ – EugenR Aug 23 '17 at 11:10 | {
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# An interesting problem of polynomials
In the polynomial $$(x-1)(x^2-2)(x^3-3) \ldots (x^{11}-11)$$ what is the coefficient of $$x^{60}$$?
I've been trying to solve this question since a long time but I couldn't. I don't know whether opening the brackets would help because that is really a mess. I have run out of ideas. Would someone please help me to solve this question?
• Think of how many ways the terms can form a term $x^{60}$ ... It could be $x^1 \cdot x^{59}$, or $x^2 \cdot x^{58}$, or even $x^3 x^8 x^{49}$ ... Can you think of a systematic way to create $x^{60}$ like this? Nov 5 '19 at 10:38
• Old question, see here. Also on this site there are several such question. Please have a look, say here. Next time, please search before posting. Nov 5 '19 at 10:39
• @DietrichBurde it's not easy to search for mathematical expressions (unless you know to use approach 0 or something like it), and there isn't a lot of text to go off of for these problems. I don't think that there's much reason to assume that the asker didn't at least try searching. Nov 5 '19 at 10:44
• @Omnomnomnom I agree, that it is not always immediate to see these duplicates, but honestly my experience on this site is that several users are not even aware that it is possible to find the same question with several beautiful answers. This is true in particular for these "standard questions". One of the goals for students also is to learn how to search. Nov 5 '19 at 10:47
• @Aryan The previous edit was better actually. Nov 5 '19 at 10:48
The highest exponent possible is $$1+2+ \cdots + 11 = 66$$.
Now, to create the exponent $$60$$, you can only leave out the factors containing $$(1,2,3),(2,4),(1,5)$$ and $$6$$. Hence,
• $$1+2+3 \Rightarrow$$ gives coefficient $$(-1)(-2)(-3) = -6$$
• $$2+4 \Rightarrow$$ gives coefficient $$(-2)(-4) = 8$$
• $$1+5 \Rightarrow$$ gives coefficient $$5$$
• $$6 \Rightarrow$$ gives coefficient $$-6$$
Summing up gives $$13-12 = \boxed{1}$$. | {
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Summing up gives $$13-12 = \boxed{1}$$.
Hint : $$1+2+3 +...+11= \frac {11×12}{2} =66$$ so we must find how we can construct number $$6=6+0=5+1=4+2=3+3=1+2+3$$ and note that $$3+3$$ impossible.
Expanding the brackets is definitely not the way to go, but thinking about what would happen if you did is helpful. Every term in the expanded polynomial will come from multiplying one term from each bracket (e.g. the highest degree term will come from multiplying $$\,x \cdot x^2 \cdot x^3 \cdot x^4 \dots \cdot x^{11}\,$$), and since each bracket is a binomial that doesn't leave many options.
The first thing to notice is that the aforementioned highest-degree term will be $$\,x^{66};\,$$ thus, to get a term with $$\,x^{60},\,$$ we will need to use MOST BUT NOT ALL of the $$\,x$$'s in the factors. We need to reduce the maximum degree by $$\,6\,$$; one obvious way to do that is to not use the $$\,x^6\,$$ in the sixth factor, so instead we use the $$\,-6\,$$ from that bracket. However, there are other ways, e.g. not using the $$\,x^2\,$$ or the $$\,x^4\,$$ from the second and fourth factors, respectively. Hopefully by now you're getting the sense that what we need to do is come up with all the different ways to make $$\,x^6\,$$ with any of the powers available to us, and then for each way find the corresponding coefficient.
There are not that many ways of making $$\,x^6;\,$$ note that the most we can use is three powers of $$\,x\,$$, as the minimum power would then be $$\,x\cdot x^2\cdot x^3=x^6.\,$$ The list of ways is: $$\,x^6, x\cdot x^5, x^2 \cdot x^4,\,$$ and $$\,x\cdot x^2\cdot x^3.\,$$
Now we just need to find the coefficient of each term that EXCLUDES one of the above combinations. As an example, let's consider the term that excludes $$\,x^2\,$$ and $$\,x^4.\,$$ Written out in detail, this term would be
$$(x)(-2)(x^3)(-4)(x^5)(x^6)(x^7)(x^8)(x^9)(x^{10})(x^{11}) = 8x^{60}$$ | {
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$$(x)(-2)(x^3)(-4)(x^5)(x^6)(x^7)(x^8)(x^9)(x^{10})(x^{11}) = 8x^{60}$$
Similarly, it should be easy to see that the coefficients for the terms excluding the other combinations are:
• For $$\,x^6,\,$$, the coefficient will just be $$\,-6\,$$;
• For $$\,x\cdot x^5,\,$$ the coefficient will be $$\,(-1)(-5)=5\,$$;
• For $$\,\,x\cdot x^2\cdot x^3,\,$$ the coefficient will be $$\,(-1)(-2)(-3)=-6.\,$$
Thus the four terms that will have $$\,x^{60}\,$$ are $$\,-6x^{60}, 5x^{60}, 8x^{60}\,$$ and $$\,-6x^{60}.\,$$ The sum of these will be just $$\,x^{60},\,$$ so the answer to the question is $$\,\boxed{1}.\,$$
• Thanks a lot A.J. Nov 6 '19 at 9:12 | {
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# Sum in a sudoku with sum total of 15 and 45
I have created a sudoku puzzle with the following restrictions:
• Each row and column sum to $$45$$.
• Each row and column in the nine $$3$$ by $$3$$ sub-grids sum to $$15$$.
Is such a sudoku unique?
• "...where each row and column adds to a total of 45..." is redundant, since it is in the nature of a sudoku puzzle that its rows and columns sum to $45$. This is because each row contains the numbers $1$ through $9$, and $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$. – Hugh Feb 11 at 5:04
• "where each (3*3) 9 individual squares of rows and columns total a sum of 15" >>> do you mean 45? – Kryesec Feb 11 at 5:33
• @Kryesec I've assumed that that means that each of the 3 by 3 sub-grids is a magic square with "magic constant" 15. In other words, each row/column in each 3 by 3 sub-grid sums to 15. – Hugh Feb 11 at 5:47
No.
Of course, there's rotation and reflection, but even beyond that you can swap any two rows within its group of three rows (eg swap row 1 and 2 or swap 4 and 6), and you can swap a group of three with one of the other groups of three (eg swap 1,2,3 with 4,5,6), and the same applies to the columns.
• I wonder if there's any kinda of transformation that would work here that doesn't work on sudoku in general. – Rawling Feb 11 at 12:00
I was beaten to a very similar answer, but you can create such a square based on the magic square below:
abc
def
ghi
Turn it into this:
abcdefghi
defghiabc
ghiabcdef
bcaefdhig
efdhigbca
higbcaefd
cabfdeigh
fdeighcab
ighcabfde
Since there are multiple possible magic squares, it's not unique. When you change the number at the very center (i), which you can, you get a different sudoku (not even a rotation or reflection).
One such arrangement of the 3x3 grid could be:
1 5 9
8 3 4
6 7 2
This would then be repeated along to the right, but rotating the order of the rows to create:
1 5 9 8 3 4 6 7 2
8 3 4 6 7 2 1 5 9
6 7 2 1 5 9 8 3 4 | {
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1 5 9 8 3 4 6 7 2
8 3 4 6 7 2 1 5 9
6 7 2 1 5 9 8 3 4
The same could be done to repeat downwards but rotating the columns this time:
1 5 9 8 3 4 6 7 2
8 3 4 6 7 2 1 5 9
6 7 2 1 5 9 8 3 4
5 9 1 3 4 8 7 2 6
3 4 8 7 2 6 5 9 1
7 2 6 5 9 1 3 4 8
9 1 5 4 8 3 2 6 7
4 8 3 2 6 7 9 1 5
2 6 7 9 1 5 4 8 3
However:
This creates one such possibility of your given solution, as each of the 3x3 sections could be placed in the top left and the rotations occur from there.
The answer is no. A Sudoku satisfying the puzzle requirement (i.e. each 3x3 sub-grid is a semi-magic square) is not unique even if we do not make a distinction between Sudokus differing just by rotations or reflections. And the simplest reason, as pointed out by @DR Xorile, is that there are natural Sudoku transformations (namely certain rows and/or columns swaps) which are not rotations and reflections but which preserve the integrity of the Sudoku and the semi-magic property of the sub-grids. This brings up a question alluded to by @Rawling: what if we extend a rotations and reflections group of transformations with row and column permutations suggested by @DR Xorile (let's call this group RRRCP for brevity) and not make distinctions between Sudokus if they can be transformed into each other via any RRRCP transformations. Will it make a Sudoku built out of 3x3 sub-grid of semi-magic squares "unique"? At first I thought it might. On a smaller scale, a 3x3 semi-magic square is "unique" in a sense that all 72 semi-magic squares can be generated by applying transformations from a similar extended group (rotations, reflections, columns/rows permutations) to just one semi-magic square. But it turns out that for a full Sudoku the answer is still no - a Sudoku built out of 3x3 sub-grid of semi-magic squares is not "unique" even modulo RRRCP transformations. Here is why. | {
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For the sake of precision here is the description of an RRRCP group. Let's call each set of three columns (1, 2, 3), (4, 5, 6), and (7, 8, 9) - a "macro" column and each set of three rows (1, 2, 3), (4, 5, 6), and (7, 8, 9) - a "macro" row. "Macro" columns and "macro" rows constitute a 3x3 "macro" grid. A RRRCP group consists of compositions of the following Sudoku transformations:
• rotations and reflections;
• permutations of the columns within a "macro" column (ex. columns 4, 5, 6)
• permutations of the rows within a "macro" row (ex. rows 7, 8, 9)
• permutations of whole "macro" columns within a 3x3 "macro" grid (ex swap columns 1 2 3 with columns 7, 8, 9)
• permutations of whole "macro" rows within a 3x3 "macro" grid (ex swap rows 1 2 3 with rows 4, 5, 6).
The following two Sudokus S1 and S2 are built out of semi-magic squares but cannot be transformed into each other by RRRCP transformations.
2 7 6 1 9 5 3 8 4
9 5 1 8 4 3 7 6 2
4 3 8 6 2 7 5 1 9
3 8 4 2 7 6 1 9 5
7 6 2 9 5 1 8 4 3
5 1 9 4 3 8 6 2 7
1 9 5 3 8 4 2 7 6
8 4 3 7 6 2 9 5 1
6 2 7 5 1 9 4 3 8
-------------------------------
2 7 6 5 1 9 8 4 3
9 5 1 3 8 4 6 2 7
4 3 8 7 6 2 1 9 5
8 4 3 2 7 6 5 1 9
6 2 7 9 5 1 3 8 4
1 9 5 4 3 8 7 6 2
5 1 9 8 4 3 2 7 6
3 8 4 6 2 7 9 5 1
7 6 2 1 9 5 4 3 8
S1 and S2 cannot be transformed into each other by rotations or reflections because they have an identical central 3x3 sub-grid.
S1 and S2 cannot be transformed into each other by permutations within a "macro" row/column because they have an identical "macro" diagonal.
S1 and S2 cannot be transformed into each other by permutations of "macro" rows/columns because they are built out of different semi-magic squares. | {
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# What is the sum of number of digits of the numbers $2^{2001}$ and $5^{2001}$
What is the sum of number of digits of the numbers $$2^{2001}$$ and $$5^{2001}$$? (Singapore 1970)
I attempted to solve this question by working out what each digit must be, and maybe find some pattern, but I couldn't find any, apart from the fact that $$2^{2001}\mod{10}\equiv 4$$ and $$2^{2001}\pmod{10}\equiv 5$$. Could you please explain to me how to solve this question? This question is multiple choice with options $$1999, 2003, 4002, 6003, 2002$$
• @samerivertwice yes base 10 Mar 15 at 17:15
• Not hard to do with a computer program, but perhaps that's not allowed. Mar 15 at 17:18
• @Sil I am certain that that is what the question asks Mar 15 at 17:36
• @Sil I'll ask the author for an explanation, as soon as I get a response, I'll post it here Mar 15 at 18:05
• @sil I worked it out given the multi choice 😉 Mar 15 at 18:08
It's $$2002$$. It's asking for the sum of the number of digits of $$2^{2001}$$ and $$5^{2001}$$ in base $$10$$, so just take the log base $$10$$ of each, take the ceiling function and hey presto:
$$603+1399=2002$$
• remark: without taking the ceiling function, $$2001\log_{10}(2)+2001\log_{10}(5) = 2001(\log_{10}(2)+\log_{10}(5)) = 2001\log_{10}(10) = 2001$$ so no calculator is needed to know the answer is in $[2001,2003)$ Mar 15 at 18:14
• "the sum of the number of digits" Well, that's just nasty..... Mar 15 at 18:16
• @hgmath nice. I figured there was something like that although I just used the calculator ;) Mar 15 at 18:17
• @samerivertwice your solution is brilliant, thank you very much Mar 15 at 18:18
• @MichaelBlane for the logarithm I just used a calculator. If needed to do without a calculator this can be done, using an infinite series for the log function but the real magic is in hgmath's comment - if you want to learn something, that comment is the thing to think about. Mar 15 at 18:20 | {
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It looks like you don't need logarithms or any calculator to solve this problem. Let's start.
First, observe that the following inequalities hold:
$$10^m<\underbrace {2^{2001}}_{m+1 ~ \text{digits}}<10^{m+1}$$
$$10^n<\underbrace{5^{2001}}_{n+1 ~ \text{digits}}<10^{n+1}$$
You get,
$$10^{m+n}<10^{2001}<10^{m+n+2}$$
$$2001=m+n+1$$
$$m+n=2000$$
Finally, the sum of digits of $$2^{2001}$$ and $$5^{2001}$$ is equal :
\begin{align}\color {gold}{\boxed {\color{black}{m+1+n+1=m+n+2\\ \qquad \qquad \qquad\thinspace=2000+2 \\\qquad \qquad \qquad \thinspace=2002.}}}\end{align}
The answer is $$\overbrace{\lfloor2001\log_{10}(2)\rfloor+1}^\text{digits in 2^{2001}}+\overbrace{\lfloor2001\log_{10}(5)\rfloor+1}^\text{digits in 5^{2001}}$$ However, we also have, using Iverson Brackets, $$\lfloor x\rfloor+\lfloor y\rfloor=\lfloor x+y\rfloor-[\{x\}+\{y\}\ge1]$$ So we need to know $$\{2001\log_{10}(2)\}+\{2001\log_{10}(5)\}$$, but since $$2001\log_{10}(2)+2001\log_{10}(5)=2001$$, we know that the sum of their fractional parts is exactly $$0$$ or exactly $$1$$. Since the fractional parts are both positive, we must have exactly $$1$$.
Therefore, \begin{align} \lfloor2001\log_{10}(2)\rfloor+1+\lfloor2001\log_{10}(5)\rfloor+1 &=\lfloor2001\log_{10}(2)+2001\log_{10}(5)\rfloor+1\\ &=2002 \end{align}
• I see that I answered the question after it was edited to be correct. It was probably that edit which brought this question to the top of the front page just as I was looking.
– robjohn
Mar 15 at 22:09
Generalization of the problem:
• What is the sum of number of digits of the numbers $$2^N$$ and $$5^N$$?
$$10^m<\underbrace {2^{N}}_{m+1 ~ \text{digits}}<10^{m+1}$$
$$10^n<\underbrace{5^{N}}_{n+1 ~ \text{digits}}<10^{n+1}$$
$$10^{m+n}<10^{N}<10^{m+n+2}$$
$$N= m+n+1$$
$$m+n=N-1$$
The sum of digits of the numbers $$2^{N}$$ and $$5^{N}$$ will be equal : | {
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$$N= m+n+1$$
$$m+n=N-1$$
The sum of digits of the numbers $$2^{N}$$ and $$5^{N}$$ will be equal :
\begin{align}\color {gold}{\boxed {\color{black}{m+1+n+1=m+n+2\\ \qquad \qquad \qquad\thinspace=N-1+2 \\\qquad \qquad \qquad \thinspace=N+1.}}}\end{align}
• Short answer: $$N+1$$ digits. | {
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# help factoring
#### slowle4rner
##### New member
I'm having trouble factoring the following binomial... can someone try to point me in the right direction please?
4y^3+4
It has been 7 years since I took algebra and I am trying to get my review done. This one just does not make sense to me right now...
Thanks!
#### MarkFL
Staff member
First factor the 4 out, to get:
$$\displaystyle 4y^3+4=4\left(y^3+1 \right)=4\left(y^3+1^3 \right)$$
Now apply the sum of cubes formula:
$$\displaystyle a^3+b^3=(a+b)\left(a^2-ab+b^2 \right)$$
What do you get?
#### slowle4rner
##### New member
So...A^3+b^3 = (a+b)(a^2-ab+b^2)
4y^3+4 = 4(y^3+1) = 4(y^3+1^3) = 4(y+1)(y^2-y+1)
That's basically what I came up with. I guess I am confused because I 4 is not a perfect cube and I didn't realize you could factor out the 4 and work the problem from there. But I suppose this makes sense assuming that I can multiply in the four and then multiply (4y+4)(y^2-y+1) or multiply 4(y+1)(y^2-y+1) and distribute the 4.
Would I be correct to presume that 4(y+1)(y^2-y+1) is the most factored and appropriate answer? That would seem to make sense to me...
Thanks for the help. I wasn't seeing it. Gratitude.
#### SuperSonic4
##### Well-known member
MHB Math Helper
So...$$\displaystyle a^3+b^3 = (a+b)(a^2-ab+b^2)$$
$$\displaystyle 4y^3+4 = 4(y^3+1) = 4(y^3+1^3) = 4(y+1)(y^2-y+1)$$ | {
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$$\displaystyle 4y^3+4 = 4(y^3+1) = 4(y^3+1^3) = 4(y+1)(y^2-y+1)$$
That's basically what I came up with. I guess I am confused because I 4 is not a perfect cube and I didn't realize you could factor out the 4 and work the problem from there. But I suppose this makes sense assuming that I can multiply in the four and then multiply $$\displaystyle (4y+4)(y^2-y+1)$$ or multiply $$\displaystyle 4(y+1)(y^2-y+1)$$ and distribute the 4.
The 4 is a bit of a red herring because, as you said, it's not a perfect cube but you can leave that alone out the front. You could say $$\displaystyle (4y+4)(y^2-y+1)$$ (as well as $$\displaystyle (y+1)(4y^2-4y+4)$$) but it's good form to leave it fully factored which means not distributing the four. If you ever do exams they usually want it fully factored.
Would I be correct to presume that 4(y+1)(y^2-y+1) is the most factored and appropriate answer? That would seem to make sense to me...
Thanks for the help. I wasn't seeing it. Gratitude.
Yes, that is the correct answer.
#### HallsofIvy
##### Well-known member
MHB Math Helper
Even if you were not able to factor that "4" out, it just a number! If it had been, say, $$x^3+ 4$$ you could write it as $$x^3+(\sqrt[3]{4})^3$$ and use that same fornula. | {
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# Common logarithm question
I'm studying logarithms and am doing an exercise where you're supposed to evaluate the solutions of common logarithms without using a calculator. I'm very stuck on this one particular question. I know the answer because I used my calculator, but I'd like to know how to solve it without one. The question is $$\log\left(\frac{10}{\sqrt[\large3]{10}}\right)$$
How do I solve this without a calculator? (Please provide a step-by-step solution, this has really confused me.)
• Can you write $10/\root3\of {10}$ as $10^x$ for some $x$? If so, what would $\log 10^x$ be? – David Mitra Jul 29 '14 at 10:39
• @DavidMitra I got as far as $10(10^{-\frac13})$ but don't know where to go from there – imulsion Jul 29 '14 at 10:40
Hint
$$\frac{10}{\sqrt[3]{10}} = \frac{10}{10^{1/3}} = 10^{1-1/3} = 10^{2/3}$$
Now, what would the logarithm (assuming base 10) of that final expression be?
• Thanks for your answer, but I don't understand why $\frac{10}{10^{1/3}} = 10^{1-1/3}$. Probably me being a moron, but I don't understand how you got there. – imulsion Jul 29 '14 at 10:46
• @imulsion The following rule is very good to know: $\frac{x^a}{x^b} = x^{a-b}$. Let $x=10$, $a=1$ and $b=1/3$. (Remember that $10 = 10^1$.) – naslundx Jul 29 '14 at 10:48
• I can't believe that I forgot that :facepalm:. Thanks very much for your help though, I will accept your answer when I can – imulsion Jul 29 '14 at 10:49
• @imulsion No need to feel stupid, we are all here to learn. :) – naslundx Jul 29 '14 at 10:50
Remember that the logarithm of a quotient is the difference of logarithms: $$log\left(\frac{10}{\sqrt[3]{10}}\right)=log(10)-log(\sqrt[3]{10})=1-log(10^{1/3})=1-\frac {1}{3}\cdot log(10)=1-\frac {1}{3}=\frac {2}{3}$$
To get a good understanding of logarithms it is good to realize that the following two questions are equivalent:
1) What is the logarithm of $a$ on base of $g$? I.e. $\log$$_{g}\left(a\right)=?$. | {
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1) What is the logarithm of $a$ on base of $g$? I.e. $\log$$_{g}\left(a\right)=?$.
2) To what power must $g$ be raised to get $a$ as outcome? I.e. $g^{?}=a$.
Here $a>0$, $g>0$ and $g\neq1$.
So $10^{\frac{2}{3}}=\frac{10}{\sqrt[3]{10}}$ is the same information as $\log_{10}\frac{10}{\sqrt[3]{10}}=\frac{2}{3}$ | {
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# Is the action of $G$ on $H_1(T^n, \mathbb{Z}) = \mathbb{Z}^n$ faithful?
Let $G$ be a finite group of diffeomorphisms of the torus $T^n$ fixing some point $p$, i.e. $p$ is fixed by every element of $G$. I have two questions.
1. Is the action of $G$ on $H_1(T^n, \mathbb{Z}) = \mathbb{Z}^n$ faithful?
2. What if $G$ is not assumed to fix a point $p$?
• 2. If $G$ acts as a finite group of translations, then the action on $H_1$ is trivial (while if $G$ is nontrivial it fixes no point).
– YCor
Feb 26, 2016 at 9:14
• This question is not really research level and should be asked at math stack exchange. Nevertheless, the techniques used to answer such a question may not be well-known. Nov 25, 2016 at 15:50
• @user99720 but my comment is really immediate, there are already nice answers on the less trivial part of the question.
– YCor
Nov 26, 2016 at 22:10
From Edmonds' notes on Transformation groups: http://www.indiana.edu/~jfdavis/seminar/transformationgroupsb.pdf
Problem 9. [p.28] Show that if a [finite] group $G$ acts on the torus $T^n$ with a fixed point x and induces the trivial action on $\pi_1(T^n, x)$, then the action is trivial.
So the answer to question 1 is yes. Question 2 is answered by @YCor's comment and also discussed in the above notes.
This follows from a 1931 theorem of M. H. A. Newman. Theorem 1 of the paper says that for a connected manifold $M$ with a metric and an integer $m>1$, there is a constant $d$ so that any (uniformly continuous) periodic transformation of $M$ of order $m$ must move a point of $M$ at least a distance $d$. | {
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Given a homeomorphism $f$ of a torus $T^n$ which acts trivially on $\pi_1(T^n)$, has order $m$, and fixes a point, there is a lift $\tilde{f}$ of $f$ to the universal cover $\mathbb{R}^n$ which fixes a point and has order $m$ and is uniformly continuous. Moreover, $f$ is homotopic to the identity, since $T^n$ is a $K(\mathbb{Z}^n,1)$. Then in some Euclidean metric on $T^n$, every point will be homotopic to its image under $f$ by a path of length at most $D$ (the tracks of the homotopy have bounded length). The homotopy of $f$ to the identity lifts to a homotopy of $\tilde{f}$ to the identity, so $\tilde{f}$ also has the property that it moves points at most $D$ and has order $m$. Let $d$ be the constant from Newman's theorem for the manifold $\mathbb{R}^n$ with Euclidean metric and $m$, then rescale the metric by $d/D$, we see that $\tilde{f}$ is periodic of order $m$ and moves points at most distance $d$, hence is the identity. But then $f$ is also the identity.
Hence for a finite group $G$ acting (faithfully) on a torus $T^n$, every non-trivial element $f$ of $G$ must act non-trivially on $\pi_1(T^n)$.
We can expand the answer from user83633, following hints in the notes by Edmonds, as follows.
Let $G$ be a finite group acting faithfully on an $n$-dimensional torus $T$ and fixing a point $p$. We can write $T$ as $V/L$ for some vector space $V$ and lattice $L<V$. As $T$ is homogeneous, we can assume that this identifies $p$ with the zero element. Let $\pi\colon V\to T$ be the obvious projection, which is a universal covering. Covering theory tells us that there is a unique way to let $G$ act on $V$ such that $0$ is fixed and $\pi$ is equivariant. This action preserves $\pi^{-1}\{0\}=L$, and we can identify $L$ with $H_1(T)$ equivariantly. We want to show that the action of $G$ on $L$ is faithful. If not, we can choose a subgroup $C\leq G$ of prime order $p$ that acts trivially on $L$. We will show that this leads to a contradiction. | {
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First note that the free action map $L\times V\to V$ is $G$-equivariant, and $C$ acts trivially on $L$, so we get an induced free action of $L$ on $V^C$. The map $\pi$ gives an injection $V^C/L\to T^C$; we claim that this is a homeomorphism. As everything is compact and hausdorff, we need only check that it is surjective. Suppose that $x\in T^C$, and choose $v\in V$ with $\pi(v)=x$. If $g$ is a generator of $C$, we must then have $g.v=a+v$ for some $a\in L$. As $g$ acts trivially on $L$ this gives $g^k.v=ka+v$ for all $k$, so in particular $v=g^p.v=pa+v$ so $pa=0$ so $a=0$ so $g.v=v$ as required.
For the rest of the argument we will use various cohomology groups; these are always taken with coefficients $\mathbb{Z}/p$.
Next, recall that $H^*(BC)$ is the tensor product of a polynomial algebra on a class $x$ of degree $2$ with an exterior algebra on a class $a$ of degree $1$. For any $C$-space $X$ we have a map $EC\times_CX\to EC/C=BC$, which makes the Borel cohomology group $H_C^*(X)=H^*(EC\times_CX)$ into an algebra over $H^*(BC)$. We write $\widehat{H}^*_C(X)$ for the ring obtained by inverting $x$ in $H_C^*(X)$. If $Y$ is a $C$-subspace of $X$ then we have relative groups $\widehat{H}_C^*(X,Y)$ defined in a similar way. If $C$ acts freely on $X\setminus Y$ then these relative groups are trivial, as one can prove by cellular induction starting from the case where $X=C\times B^d$ and $Y=C\times S^{d-1}$. On the other hand, if $C$ acts trivially on $X$ then $\widehat{H}^*_C(X)=H^*(X)\otimes R^*$, where $$R^*=\widehat{H}^*_C(\text{point})=\mathbb{Z}/p[x,x^{-1}]\otimes E[a].$$ Because $C$ has prime order, it must act freely on $X\setminus X^C$, so $\widehat{H}^*_C(X,X^C)=0$, so $$\widehat{H}^*_C(X) \simeq \widehat{H}^*_C(X^C) \simeq H^*(X^C)\otimes R^*.$$ This is essentially what is called Smith theory. | {
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Now take $X$ to be our vector space $V$, so $V$ is contractible, and the usual spectral sequence $H^*(G;H^*(X))\Longrightarrow H^*_G(X)$ gives $H^*_G(V)=H^*(BG)$ and therefore $\widehat{H}^*_G(V)=R^*$. By comparison with Smith theory, we get $H^*(V^C)=\mathbb{Z}/p$. As $L$ acts freely on $V^C$ with $V^C/L=T^C$ we get a spectral sequence $$H^*(L;H^*(V^C)) \Longrightarrow H^*(T^C)$$ This admits a map from the similar spectral sequence $$H^*(L;H^*(V)) \Longrightarrow H^*(T).$$ As $H^*(V^C)=H^*(V)=\mathbb{Z}/p$, the map is an isomorphism on the initial page, and there is no room for any differentials anyway, so the map $H^*(T)\to H^*(T^C)$ is an isomorphism. In particular, the map $H^n(T)\to H^n(T^C)$ is an isomorphism.
On the other hand, if $X$ is any proper subset of $T$ then the restriction $H^n(T)\to H^n(X)$ factors through the group $H^n(T\setminus \{x\})=0$ for some $x\in T$. Thus, we must have $T^C=T$, which means that $C$ acts trivially on $T$. This contradicts our assumption that the action of $G$ is faithful.
For a simpler and more geometric proof: It suffices to show the only element which acts as the identity is the identity. Suppose $g$ is our element which acts as the identity on $H_1$ and thus on $\pi_1$. Lift $g$ to act on $\mathbb{R}^2$. Define a homotopy $g_t(p) = (1 - t) p + t g(p)$ between $g$ and the identity. The homotopy descends to $\mathbb{T}^2$. By surface theory we can upgrade this to an isotopy between $g$ and the identity on $\mathbb{T}^2$. Then, and I'm not sure how to prove this (and in some sense this seems to be the crux of the issue) one can show that maps isotopic to the identity have infinite order. | {
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# Mistake in solving $-\int \frac{1}{x\sqrt{x^2-1}}$
I have this function $$f:(-\infty ,-1)\rightarrow \mathbb{R}, f(x)=\frac{1}{x\sqrt{x^{2}-1}}$$ and I need to find the primitives of $$f(x)$$.So because $$x<-1$$ I need to calculate $$-\int \frac{1}{x\sqrt{x^2-1}}\,dx=-\arctan(\sqrt{x^2-1})+C$$ but in my book the correct answer is $$\arcsin\left(\frac{1}{x}\right)+C$$
Where is my mistake?I solved the integral using $$u=\sqrt{x^{2}-1}$$
It is known that $$\arctan(x)+\arctan(1/x)=C$$ where $$C$$ is a constant. So your solution can be written as $$\arctan\left(\frac1{\sqrt{x^2-1}}\right):=u$$Then using some trig identities, $$\frac1{\sqrt{x^2-1}}=\tan u\\\sec^2 u=1+\tan^2 u=\frac{x^2}{x^2-1}\\\sin^2 u=1-\cos^2 u=1-\left(\frac{x^2-1}{x^2}\right)=\frac1{x^2}\\u=\arcsin\left(\frac1x\right)$$
Hint:
You might want to plot both $$\arcsin(\frac{1}{x})$$ and $$-\arctan(\sqrt{x^2 - 1})$$, from, say, $$-3$$ to $$-1$$. You could do this using Desmos (https://www.desmos.com/). You might even insert an extra minus sign here or there...
$$\arcsin(\frac{1}{x}) +C = \operatorname{arccot}(\sqrt{x^2 - 1}) + C$$
$$=\frac{\pi}{2} - \arctan(\sqrt{x^2 - 1})+ C$$
You can form another Constant $$C_1 = \pi/2 + C$$.
So you solved correctly, the book answer and your answer are equivalent, you just had to manipulate it a little to get there. The difference is just of a constant which doesn't matter when calculating indefinite integrals. | {
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• Yes, I understand what you're saying but my book gives both results as anwers and I just need to pick one answer.For example, at exam how to know which is good ? – DaniVaja May 1 '19 at 18:48
• @Vizag: I think that the relationship you've asserted in the first line is only correct for $x > 0$; you might want to check that by plotting both sides when $x < 0$. – John Hughes May 1 '19 at 18:50
• @DaniVaja Both are fine, since they are both correct. So you shouldn't worry too much. You could use these answers to see how to show that both forms are equivalent (up to constants). – John Doe May 1 '19 at 18:50
• @DaniVaja: Both the answers look correct to me. The expressions are equivalent up to constants. Are you sure it's not a multiple answers correct kind of question? – Vizag May 1 '19 at 18:53
• Here is the original exercise: imgur.com/hhQUzoi (i don't know the code for posting images).The book says that one answer from 5 is correct and in this exercise I have both answers, which are both good like you said.But for example in an exam, how to know which answer to pick ?In this case they chose arcsin(1/x) + c as correct answer and not -arctan... – DaniVaja May 1 '19 at 18:56 | {
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# [SOLVED]Laplace Transform
#### nacho
##### Active member
How do you find the laplace transform of this without expanding it?
$L(t^2+1)^2$
Last edited:
#### chisigma
##### Well-known member
How do you find the laplace transform of this without expanding it?
$L(t^2+1)^2$
You can use the fact that for n positive integer is...
$\displaystyle \int_{0}^{\infty} t^{n}\ e^{- s\ t}\ dt = \frac{n!}{s^{n+1}}\ (1)$
... expanding $\displaystyle (1 + t^{2})^{2}$ in powers of t...
Kind regards
$\chi$ $\sigma$
I didn't realize immediately that it was requested 'without expanding' ... very sorry!...
#### chisigma
##### Well-known member
How do you find the laplace transform of this without expanding it?
$L(t^2+1)^2$
The possible solution is the use of a 'forgotten formula' ['forgotten' in the sense that it is neglected from most of the Complex Analysis 'Holybooks'...] according to which if You have two L-transformable functions $f_{1}(t)$ and $f_{2} (t)$ with L-transforms $F_{1} (s)$ and $F_{2} (s)$ and abscissas of convergence $\sigma_{1}$ and $\sigma_{2}$, then the L-transform of the product is given by the integral...
$\displaystyle \mathcal {L} \{f_{1}(t)\ f_{2}(t)\} = \frac{1}{2\ \pi\ i}\ \int_{\gamma - i\ \infty}^{\gamma + i\ \infty} F_{1} (z)\ F_{2}(s - z)\ dz\ (1)$
... where $\displaystyle \sigma_{1} < \gamma < \sigma - \sigma_{2},\ \sigma> \sigma_{1} + \sigma_{2}$. In Your case is $\displaystyle f_{1}(t) = f_{2} (t) = 1 + t^{2} \implies F_{1}(s) = F_{2}(s) = \frac{1}{s} + \frac{2}{s^{3}},\ \sigma_{1}=\sigma_{2}=0$. May be it exists a more comfortable way but till now I didnn't succed to find it!...
Kind regards
$\chi$ $\sigma$
#### Chris L T521
##### Well-known member
Staff member
How do you find the laplace transform of this without expanding it?
$L(t^2+1)^2$
To compute this without expanding requires a little trick (especially if you're allowed to do this without using the integral definition of $\mathcal{L}\{f(t)\}$). | {
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Let $f(t) = (t^2+1)^2$. We note that $f(0)=1$.
Thus, $f^{\prime}(t) = 4t(t^2+1)$ and hence $f^{\prime}(0) = 0$.
Finally, we see that $f^{\prime\prime}(t) = 4(t^2+1) + 8t^2 = 12t^2+4$.
Why did we compute all these derivatives? Well, we know that $\mathcal{L}\{f^{\prime\prime}(t)\} = s^2F(s) - sf(0) - f^{\prime}(0)$, which in our case would be $\mathcal{L}\{f^{\prime\prime}(t)\}=s^2F(s)-s$.
Therefore,
$\mathcal{L}\{12t^2+4\} = \mathcal{L}\{f^{\prime\prime}(t)\} = s^2F(s)-s$
We now compute the appropriate Laplace transforms and solve for $F(s)$:
$s^2F(s) - s = \mathcal{L}\{12t^2+4\} = 12\cdot\frac{2!}{s^3}+4\cdot\frac{1}{s}=\frac{24}{s^3}+\frac{4}{s} \implies F(s) = \frac{24}{s^5} + \frac{4}{s^3}+\frac{1}{s}$
Therefore, $\mathcal{L}\{(t^2+1)^2\} = \dfrac{24}{s^5}+\dfrac{4}{s^3}+\dfrac{1}{s}$.
I hope this makes sense!
#### Jester
##### Well-known member
MHB Math Helper
Question for the OP. Why would you want to compute this transform without expanding? I mean expanding would be the easiest approach.
BTW - Nice work Chris!
Following Chris's lead, you could keep going
\begin{align}
f(t) &= (t^2+1)^2,\;\;\; &f(0)&=1\\
f'(t) &= 4t(t^2+1)\;\;\; &f'(0)&=0\\
f''(t) &=4(t^2+1) + 8t^2,\;\;\;&f''(0)&=4\\
f'''(t) &= 8t + 16t,\;\;\;&f'''(0)&=0\\
f^{(4)}(t)&=24,\;\;\;&f^{(4)}(0)&=24\\
f^{(5)}(t)& = 0,
\end{align}
Then
$\mathcal{L}\{f^{(5)}(t)\} = s^5F(s) - f(0) s^4 - f'(0) s^3 - f''(0) s^2 - f'''(0) s - f^{(4)}(0) = 0.$
Substitute $f$ and its derivative at zero and solve for $F$ (same answer)
Last edited:
#### nacho
##### Active member
Thanks for the responses.
@Jester, really no other reason than for the sake of it. I've just started to learn laplace transforms and am finding a little out of the box thinking helps a lot.
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# How do I prove that two empty sets are equivalent?
I need to prove that two empty sets have the same cardinality, I know that proving equivalence is done by defining an onto and one-to-one function from one group to the other, but how can I define a function from an empty set to an empty set?
-
Do you mean define a function from an empty set to an empty set? – Git Gud Feb 6 '13 at 22:36
yes of course, sorry. – Georgey Feb 6 '13 at 22:37
@amWhy That is correct. I will will delete my comment. – Git Gud Feb 6 '13 at 23:49
Let $A=B=\varnothing$. A function from $A$ to $B$ is a subset $f$ of $A\times B$ with the property that if $\langle a,b_1\rangle$ and $\langle a,b_2\rangle\in f$, then $b_1=b_2$. The only subset of $A\times B=\varnothing\times\varnothing=\varnothing$ is $\varnothing$. Does it have the function property? Yes: it contains no ordered pairs at all, so the hypothesis
if $\langle a,b_1\rangle$ and $\langle a,b_2\rangle\in f$
is never satisfied, and the property
if $\langle a,b_1\rangle$ and $\langle a,b_2\rangle\in f$, then $b_1=b_2$
never actually requires anything to be true $-$ it holds vacuously.
Thus, $\varnothing$ is a function from $\varnothing$ to $\varnothing$. Is it a bijection? Is it true that for each $a\in A=\varnothing$ there is a unique $b\in B=\varnothing$ such that $\langle a,b\rangle\in f$? Again the answer is yes, vacuously: there’s no $a\in A=\varnothing$ at all, so the requirement is never actually invoked.
However, all of that work is completely unnecessary, because it’s always true that if $A=B$, then $|A|=|B|$. And if $A$ and $B$ are both empty, then they have exactly the same members, so they are the same set, a fact that I implicitly used above when I wrote $A=B=\varnothing$.
-
Nice answer, Brian. +1 – Rick Decker Feb 7 '13 at 2:00
Thanks, @Rick. ${}$ – Brian M. Scott Feb 7 '13 at 2:03 | {
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Hint: Recall that there are no elements in an empty set. Show that if $A$ and $B$ are empty sets then for every $x\in A$ we have $x\in B$, and vice versa. Therefore $A\subseteq B$ and $B\subseteq A$ and so $A=B$.
If two sets are equal then certainly they have the same cardinality.
-
Hint:
Use the definition of the (an) empty set - a set with no elements - to show, it is vacuously true that for all $x$ in the domain, $\;x \in \emptyset_1 \implies x \in \emptyset_2,\;$ so $$\emptyset_1 \subseteq \emptyset_2$$ and do similarly to show that it is vacuously true that $$\emptyset_2 \subseteq \emptyset_1$$
Hence, $\emptyset_1 = \emptyset_2$.
Hence, $|\emptyset_1| = |\emptyset_2|$
-
The extension and specification axioms implies that there is only one set with no elements.
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# Partial Fractions Decomposition $\frac{6x^2 - 29x - 29}{(x+1)(x-3)^2}$ explanation repeated factors
I am trying to solve the fraction $$\frac{6x^2 - 29x - 29}{(x+1)(x-3)^2}$$ into partial fractions.
Now, I thought it could be solved into the following
$$\frac{6x^2 - 29x - 29}{(x+1)(x-3)^2} = \frac{A}{x+1} + \frac{B}{(x-3)^2}$$
but this is apparently incorrect.
According to the text, the decomposition is
$$\frac{6x^2 - 29x - 29}{(x+1)(x-3)^2} = \frac{A}{x+1} + \frac{B}{x-3} + \frac{C}{(x-3)^2}$$
I discussed this with my friend that the fraction first decomposes into
$$\frac{6x^2 - 29x - 29}{(x+1)(x-3)^2} = \frac{A}{x+1} + \frac{Bx +C}{(x-3)^2}$$
but I can't see how he derived this.
I don't understand how he is correct.
• Anything of your textbook's form can easily be expressed in your friend's form and vice versa. – Lord Shark the Unknown Sep 1 '17 at 20:08
Both decompositions used by your friend and the text are analogous since $$\frac{Bx+C}{(x-3)^2}=\frac{Bx}{(x-3)^2}+\frac{C}{(x-3)^2}=\frac{B(x-3)+3B}{(x-3)^2}+\frac{C}{(x-3)^2}$$ $$=\frac{B}{x-3}+\frac{3B+C}{(x-3)^2}=\frac{B'}{x-3}+\frac{C'}{(x-3)^2}$$
When you decompose, the degree of the numerator will be less than the degree of the denominator. How much less is yet to be determined. Assume that it is one degree less, and then if you get a zero coefficient, so be it.
If you had something like...$\frac {P(x)}{(x^2 +x + 1)(x-1)}$ your first step would be $\frac {Ax + B}{x^2+ x+ 1} + \frac {C}{(x-1)}$
With each numerator of degree 1 less than each denominator.
And the you might want to break down the first term to make it easier to integrate.
e.g. $\frac {A(x+\frac12)}{x^2 + x + 1} + \frac {B}{(x+\frac 12)^2 + \frac 34}$
With this problem.
$(x-3)^2$ makes for a degree 2 denominator, and you need a degree 1 numerator.
So $\frac {A}{x+1} + \frac {Bx+C}{(x-3)^2}$ would be a good place to start. | {
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So $\frac {A}{x+1} + \frac {Bx+C}{(x-3)^2}$ would be a good place to start.
but $\frac {B}{(x-3)} + \frac {C}{(x-3)^2}$ is easier to integrate, so you might want to skip the intermediate step. | {
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# Let $f(x)$ be continous in $x_0$ and $f(x_0)\ne 0$. Prove that $\exists C>0$ and neighbouhood of $x_0$ such that $|f(x)| \ge C$
Let a function $$f(x)$$ be continuous in some point $$x_0$$ and $$f(x_0) \ne 0$$. Prove that there exists a number $$C > 0$$ and neighbourhood of $$x_0$$ such that forall $$x$$ in that neighbourhood the following inequality holds: $$|f(x)| \ge C$$
The problem statement says that $$f(x_0) \ne 0$$, so it is either greater than $$0$$ or less than $$0$$. Let's consider the case for $$f(x_0) > 0$$. In such case by continuity of $$f(x)$$ we know: $$\lim_{x\to x_0} f(x) = f(x_0) > 0$$
Or in other words: $$\forall \epsilon > 0\ \exists \delta_\epsilon > 0\ \forall x: |x-x_0| < \delta_\epsilon \implies |f(x) - f(x_0)| < \epsilon$$
Since $$f(x_0) \ne 0$$ we may let $$\epsilon = {f(x_0)\over 2}$$. In such case: $$|f(x) - f(x_0)| < \epsilon \\ |f(x) - f(x_0)| < { f(x_0)\over 2 } \\ -{ f(x_0)\over 2 } < f(x) - f(x_0) < { f(x_0)\over 2 }\\ 0<{f(x_0)\over 2} < f(x) < {3f(x_0)\over 2}$$
So we may now choose any $$C \in \left(0; {f(x_0)\over 2}\right)$$, which would imply $$f(x) \ge C$$.
Similar reasoning is applied to the case when $$f(x_0) < 0$$.
I'm not sure my reasoning above is valid, so I would like to kindly ask for verification. Or for a correct proof if the above makes no sense. Thank you!
• Yes, this absolutely fine ;) The neighbourhood would be $(-f(x_0)/2,f(x_0)/2)$ when $f(x_0)>0$. – weirdo May 24 at 16:24
• This looks correct. You have $f(x) \geq C$ for all $|x-x_0|\leq\delta_\epsilon$. The reason this works is because of continuity, i.e. the statement is for all $\epsilon >0$..., the the particular choice of $\epsilon = f(x_0)/2$ works. – Dayton May 24 at 16:24 | {
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Yep, the above proof is valid. But for a couple suggestions in terms of proof writing, in the last line you said that $$f(x_0) \geq C$$, but this is only true for all $$x$$ such that $$|x-x_0|<\delta$$. Just make sure you include that in your last line so that people don't think you are saying that you are proving it for every $$x$$.
Have you been introduced to the neighbourhood notation? It is basically $$B_r(x_0) := \{|x-x_0| < r \quad| \quad \forall x \text{ in your domain.}\}$$ This way you can say $$|f(x) - f(x_0)|<\epsilon$$ as $$B_\epsilon (x_0)$$ | {
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# Show convergence and find the limit of the sequence given by $a_1=1$ and $a_{n+1}=\frac{1}{3+a_n}$
I've been trying to solve this exam question on an exam in real analysis. Thus, only such methods may be used. The problem is as follows.
Show that the sequence $a_n$ defined by $a_1=1$ and $a_{n+1}=(3+a_n)^{-1}$ for $n=1,2,\dots$ converges and determine the limit.
I'm not really able to find a closed form for the sequence. I guess that computing that and then showing that it's a Cauchy sequence works for showing convergence.
I just have to find that closed form first, which I'm unable to do. If there's a general strategy I'd love to see how that works as I've been unable to find any.
• Hint: monotonicity. Maybe monotonicity for subsequences. – xbh Aug 19 '18 at 19:56
Recursively $a_n>0$, $f(x)={1\over{x+3}}$, $f'(x)=-{1\over{(x+3)^2}}$, implies that $|a_{n+2}-a_{n+1}|=|f'(c_n)||a_{n+1}-a_n|\leq {1\over 3^2}|a_{n+1}-a_n|$, $c_n$ is an element of $(a_n,a_{n+1})$ or $(a_{n+1},a_n)$. You deduce that the sequence is a Cauchy sequence. (recursively you can show that $|a_{n+1}-a_n|\leq {1\over 3^{2(n-2)}}|a_2-a_1|$.
so it converges, the limit verifies $f(l)=l$ which implies that $l^2+3l-1=0$ ans is the positive root of this quadratic equation.
There need not be such a closed form for the sequence. If a limit $l$ exists it must be that$$l=\dfrac{1}{l+3}$$which yields to $$l=-1.5\pm\sqrt{3.25}$$where $l=-1.5-\sqrt{3.25}$ is not possible. To show that the sequence converges to $\sqrt{3.25}-1.5$ lets define $$e_n=a_n-(\sqrt{3.25}-1.5)$$therefore $$e_{n+1}=a_{n+1}-(\sqrt{3.25}-1.5)=\dfrac{1}{a_n+3}-(\sqrt{3.25}-1.5)=\dfrac{1-(e_n+1.5+\sqrt{3.25})(\sqrt{3.25}-1.5)}{a_n+3}=\dfrac{-e_n(\sqrt{3.25}-1.5)}{a_n+3}$$therefore $$|e_{n+1}|<\dfrac{|e_n|(\sqrt{3.25}-1.5)}{3}$$which shows that $e_n\to 0$ and $$a_n\to \sqrt{3.25}-1.5$$ | {
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• +1. I had thought of the same approach. Do you have something in mind to show that the convergence happens without having to assume that the limit exists first? – Vizag Aug 19 '18 at 20:18
• Well! Consider $$a_{n+2}=\dfrac{1}{3+a_{n+1}}=\dfrac{a_n+3}{3a_n+10}<\dfrac{4}{10}a_n$$ therefore we can conclude what we want – Mostafa Ayaz Aug 19 '18 at 20:23
Monotonicity method.
Rewrite the recursive relation as $$a_{n+2} = \frac 1 {3+ \dfrac 1 {3 + a_n}} = \frac {3+a_n} {3(3+a_n) + 1} = \frac {a_n + 3} {3 a_n + 10},$$ or $$a_{n+2} = \frac 13 - \frac 1{9a_n + 30} \leqslant \frac 13.$$ Since $$a_{n+4}-a_{n+2} = \frac 1 {9a_n + 30} - \frac 1{9a_{n+2} +30} = \frac {9(a_{n+2}-a_n )} {(9a_n +30) (9a_{n+2}+30)},$$ and clearly $a_n \geqslant 0$, we conclude that subsequences $a_{2k}, a_{2k+1}$ are monotonic [not necessarily the same type]. Since $a_1 = 1, a_2 = 1/4, a_3 = 4/13, a_4=13/34, a_1 > a_3, a_2 < a_4$, we know that $a_{2k} < a_{2k+2} \leqslant 1/3, 1 \geqslant a_{2k+1} > a_{2k+3}$. Hence $a_{2k}, a_{2k+1}$ both converge [since both are monotonic and bounded]. By the recursive relation, they converge to the same limit $L > 0$.
Let $n \to \infty$ in the first formula, $$L = \frac {L + 3}{3L + 10},$$ or simply, $$L(L+3) =1.$$
Here is a somewhat systematic approach. The crux of the method is to separate the numerator and denominator. Write $a_n = p_n / q_n$. Can we work out $p_{n+1}, q_{n+1}$? We have $$x_{n+1} = \frac{1}{3 + p_n/q_n} = \frac{q_n}{3q_n + p_n}$$ So $p_{n+1} = q_n$ and $q_{n+1} = 3q_n + p_n = 3q_n + q_{n-1}$. Now we can see that the important value is the denominator. We are lucky because our second equation is a second order recurring relation for $q_n$, so we can solve for the general form using the following theorem: | {
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Let $(x_n)_{n \ge 0}$ be a sequence such that there exists $(a,b)\in \mathbb{C}^2$ $x_{n+2} = ax_{n+1} + bx_n$. Then the sequence $(u_n)_{n \ge 0}$, $u_n = r^n$ is a solution iff $r$ satisfies the characteristic equation: $$r^2 - ar - b = 0$$ Furthermore if the characteristic equation admits two distinct roots then any solution is of the form $$\alpha r_1^n + \beta r_2 ^n$$
We don't mention the case of repeated roots. In our case the characteristic equation is $$r^2 - 3r - 1 = 0$$
The two roots are $j_1 = \frac{3 + \sqrt{13}}{2}, j_2 = \frac{3-\sqrt{13}}{2} =$. The best part is that we don't need to solve for $\alpha$ and $\beta$ (though you do need to check $\alpha \ne 0$, which is easily done); notice that $|j_2| < 1$. Now we can write $$x_n = \frac{p_n}{q_n} = \frac{q_{n-1}}{q_n} = \frac{\alpha j_1^{n-1} + \beta j_2^{n-1}}{\alpha j_1^{n} + \beta j_2^{n}}$$ Since $|j_2|<1$, as $n\to \infty$, this goes to $1/j_1$. You can check that $1/j_1$ satisfies $L = 1/(3 + L)$ | {
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Q&A
# difference between quotient rule and product rule
+2
−1
Product rule :
$$\frac{d}{dx} f(x)g(x)=f'(x)g(x)+f(x)g' (x)$$
Quotient rule :
$$\frac{d}{dx} \frac{f(x)}{g(x)}=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}$$
Suppose, the following is given in question. $$y=\frac{2x^3+4x^2+2}{3x^2+2x^3}$$
Simply, this is looking like Quotient rule. But, if I follow arrange the equation following way
$$y=(2x^3+4x^2+2)(3x^2+2x^3)^{-1}$$
Then, we can solve it using Product rule. As I was solving earlier problems in a pdf book using Product rule. I think both answers are correct. But, my question is, How does a Physicist and Mathematician solve this type question? Even, is it OK to use Product rule instead of Quotient rule in University and Real Life?
Why does this post require moderator attention?
Why should this post be closed?
+1
−0
How does a Physicist and Mathematician solve this type question? Even, is it OK to use Product rule instead of Quotient rule in University and Real Life?
is that any experienced scientist knows several methods to solve problems and uses those that are most convenient for them at that particular time.
I would look at that derivative and use the quotient rule. But if there was something in the source of the problem that suggested that it made more sense to write the denominator as $(3x^2+2x^3)^{-1}$ then the product rule would be more appropriate.
Why does this post require moderator attention?
+1
−0
is it OK to use Product rule instead of Quotient rule in University and Real Life?
Sure. For whatever reason, I long had a hard time remembering the quotient rule, and instead used the process you describe.
Why does this post require moderator attention?
+1
−0 | {
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Why does this post require moderator attention?
+1
−0
If you work out deriving the quotient rule yourself using the exact trick you're highlighting, you can see that the quotient rule is nothing more than the product rule and the chain rule used together. If you rearrange a quotient to use the product rule, then you'll very likely be using the chain rule shortly thereafter on the $(\cdots)^{-1}$ part, and you will inevitably arrive at exactly the same result as if you had used the quotient rule but with more steps. The only potential difference is whether your result looks like $\frac{a}{b} + \frac{c}{b^2}$ or $\frac{ab + c}{b^2}$—but of course, as I'm sure you can see, that's no difference at all.
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# Invertible Matrix and Linearly Independent Vectors Proof
Trying to do this one:
Suppose $A$ is an invertible $n$ x $n$ matrix and the vectors $v_1$, $v_2$, ..., $v_n$ are linearly independent. Show that the vectors $Av_1$, $Av_2$, ..., $Av_n$ are linearly independent.
I know that A's column vectors are linearly independent since A is invertible. I also know there is no relation amongst the $v_i$ because they're linearly independent.
My idea is to write the product of A and a given $v_i$ in terms of the columns of A. Not sure if this is right, any guidance much appreciated!
Thanks, Mariogs
-
Maybe you can do a proof by contradiction? If they are lin dependent, then the zero vector can be written as a lin combination (with nonzero scalars) of your $Av's$ That means that a particular $Av$ vector can be written as a lin combination of the remaining $Av's$. Now what does that mean? – imranfat Feb 27 '14 at 16:33
Let $w_i=Av_i$ for $i=1,\dots,n$. To show linear independence, we must show that $$c_1w_1+\cdots+c_nw_n=0 \Longrightarrow c_i=0,\quad i=1,\dots n.$$ Since $A$ is invertible, we can left-multiply $c_1w_1+\dots=0$ by $A^{-1}$ to get...
Can you take it from there? Remember that the vectors $v_1,\dots v_n$ are linearly independent.
-
Ah this makes sense to me, thanks! – bclayman Feb 27 '14 at 16:56
To finish it: We multiply on the left by $A^-1$ and get $c_1 v_1$ + ... + $c1 v_n$ = 0. Since $v_1, ..., v_n$ are linearly independent, we must have $c_i$ = 0 for all i. Yes? – bclayman Feb 27 '14 at 17:00
That's right, yes. – user12477 Mar 3 '14 at 12:05
Notice that since $A$ is invertible then $$Ax=0\iff x=0$$ Now let $a_1,a_2,\ldots,a_n\in \Bbb R$ such that $$\sum_{k=1}^na_k Av_k=0\iff A\left(\sum_{k=1}^na_k v_k\right)=0\iff \sum_{k=1}^na_k v_k=0\Rightarrow a_i=0\;\forall i$$ since $v_1$, $v_2$, ..., $v_n$ are linearly independent. Conclude.
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Whoa! You've been busy with answers! – amWhy Feb 28 '14 at 13:34 | {
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-
Whoa! You've been busy with answers! – amWhy Feb 28 '14 at 13:34
Why not use the definition directly? Justify/explain the following:
$$\text{For scalars}\;\;c_1,...,c_n\;:\;\;0=\sum_{k=1}^nc_kAv_k=\sum_{k=1}^nA(c_kv_k)=A\left(\sum_{k=1}^nc_kv_k\right)\iff$$
$$\iff\sum_{k=1}^nc_kv_k=0\iff c_1=c_2=\ldots=c_n=0$$
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Miscellaneous questions: Part I: tutorial practice for preRMO and RMO
Problem 1:
The sixty four squares of a chess board are filled with positive integers one on each in such a way that each integer is the average of the of the integers on the neighbouring squares. (Two squares are neighbours if they share a common edge or vertex. Thus, a square can have 8,5 or 3 neighbours depending on its position). Show that all sixty four entries are in fact equal.
Problem 2:
Let T be the set of all triples (a,b,c) of integers such that $1 \leq a < b < c \leq 6$. For each triple (a,b,c) in T, take the product abc. Add all these products corresponding to all triples in I. Prove that the sum is divisible by 7.
Problem 3:
In a class of 25 students, there are 17 cyclists, 13 swimmers, and 8 weight lifters and no one in all the three. In a certain mathematics examination, 6 students got grades D or E. If the cyclists, swimmers and weight lifters all got grade B or C, determine the number of students who got grade A. Also, find the number of cyclists, who are swimmers.
Problem 4:
Five men A, B, C, D, E are wearing caps of black or white colour without each knowing the colour of his cap. It is known that a man wearing a black cap always speaks the truth while a man wearing a white cap always lies. If they make the following statements, find the colour of the cap worn by each of them:
A: I see three black and one white cap.
B: I see four white caps.
C: I see one black and three white caps.
D: I see four black caps.
Problem 5:
Let f be a bijective (one-one and onto) function from the set $A=\{ 1,2,3,\ldots,n\}$ to itself. Show that there is a positive integer $M>1$ such that $f^{M}(i)=f(i)$ for each $i \in A$. Note that $f^{M}$ denotes the composite function $f \circ f \circ f \ldots \circ f$ repeated M times.
Problem 6:
Show that there exists a convex hexagon in the plane such that:
a) all its interior angles are equal
b) its sides are 1,2,3,4,5,6 in some order.
Problem 7: | {
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Problem 7:
There are ten objects with total weights 20, each of the weights being a positive integer. Given that none of the weights exceed 10, prove that the ten objects can be divided into two groups that balance each other when placed on the pans of a balance.
Problem 8:
In each of the eight corners of a cube, write +1 or -1 arbitrarily. Then, on each of the six faces of the cube write the product of the numbers written at the four corners of that face. Add all the fourteen numbers so writtein down. Is it possible to arrange the numbers +1 and -1 at the corners initially so that this final sum is zero?
Problem 9:
Given the seven element set $A = \{ a,b,c,d,e,f,g\}$ find a collection T of 3-element subsets of A such that each pair of elements from A occurs exactly in one of the subsets of T.
Try these !!
Regards,
Nalin Pithwa
Towards Baby Analysis: Part I: INMO, IMO and CMI Entrance
$\bf{Reference: \hspace{0.1in}Introductory \hspace{0.1in} Real Analysis: \hspace{0.1in} Kolmogorov \hspace{0.1in} and \hspace{0.1in} Fomin; \hspace{0.1in}Dover \hspace{0.1in }Publications}$
$\bf{Equivalence \hspace{0.1in} of \hspace{0.1in} Sets \hspace{0.1in} The \hspace{0.1in}Power \hspace{0.1in }of \hspace{0.1in }a \hspace{0.1in}Set}$
$\bf{Section 1}$:
$\bf{Finite \hspace{0.1in} and \hspace{0.1in} infinite \hspace{0.1in} sets}$ | {
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$\bf{Section 1}$:
$\bf{Finite \hspace{0.1in} and \hspace{0.1in} infinite \hspace{0.1in} sets}$
The set of all vertices of a given polyhedron, the set of all prime numbers less than a given number, and the set of all residents of NYC (at a given time) have a certain property in common, namely, each set has a definite number of elements which can be found in principle, if not in practice. Accordingly, these sets are all said to be $\it{finite}$.$\it{Clearly \hspace{0.1in} we \hspace{0.1in}can \hspace{0.1in} be \hspace{0.1in} sure \hspace{0.1in} that \hspace{0.1in} a \hspace{0.1in} set \hspace{0.1in}is \hspace{0.1in}finite \hspace{0.1in} without \hspace{0.1in} knowing \hspace{0.1in} the \hspace{0.1in} number \hspace{0.1in} of elements \hspace{0.1in}in \hspace{0.1in}it.}$
On the other hand, the set of all positive integers, the set of all points on the line, the set of all circles in the plane, and the set of all polynomials with rational coefficients have a different property in common, namely, $\it{if \hspace{0.1in } we \hspace{0.1in}remove \hspace{0.1in} one \hspace{0.1in} element \hspace{0.1in}from \hspace{0.1in}each \hspace{0.1in}set, \hspace{0.1in}then \hspace{0.1in}remove \hspace{0.1in}two \hspace{0.1in}elements, \hspace{0.1in}three \hspace{0.1in}elements, \hspace{0.1in}and \hspace{0.1in}so \hspace{0.1in}on, \hspace{0.1in}there \hspace{0.1in}will \hspace{0.1in}still \hspace{0.1in}be \hspace{0.1in}elements \hspace{0.1in}left \hspace{0.1in}in \hspace{0.1in}the \hspace{0.1in}set \hspace{0.1in}in \hspace{0.1in}each \hspace{0.1in}stage}$. Accordingly, sets of these kind are called $\it{infinite}$ sets. | {
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Given two finite sets, we can always decide whether or not they have the same number of elements, and if not, we can always determine which set has more elements than the other. It is natural to ask whether the same is true of infinite sets. In other words, does it make sense to ask, for example, whether there are more circles in the plane than rational points on the line, or more functions defined in the interval [0,1] than lines in space? As will soon be apparent, questions of this kind can indeed be answered.
To compare two finite sets A and B, we can count the number of elements in each set and then compare the two numbers, but alternatively, we can try to establish a $\it{one-\hspace{0.1in}to-\hspace{0.1in}one \hspace{0.1in}correspondence}$ between the elements of set A and set B, that is, a correspondence such that each element in A corresponds to one and only element in B, and vice-versa. It is clear that a one-to-one correspondence between two finite sets can be set up if and only if the two sets have the same number of elements. For example, to ascertain if or not the number of students in an assembly is the same as the number of seats in the auditorium, there is no need to count the number of students and the number of seats. We need merely observe whether or not there are empty seats or students with no place to sit down. If the students can all be seated with no empty seats left, that is, if there is a one-to-one correspondence between the set of students and the set of seats, then these two sets obviously have the same number of elements. The important point here is that the first method(counting elements) works only for finite sets, while the second method(setting up a one-to-one correspondence) works for infinite sets as well as for finite sets.
$\bf{Section 2}$:
$\bf{Countable \hspace{0.1in} Sets}$. | {
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$\bf{Section 2}$:
$\bf{Countable \hspace{0.1in} Sets}$.
The simplest infinite set is the set $\mathscr{Z^{+}}$ of all positive integers. An infinite set is called $\bf{countable}$ if its elements can be put into one-to-one correspondence with those of $\mathscr{Z^{+}}$. In other words, a countable set is a set whose elements can be numbered $a_{1}, a_{2}, a_{3}, \ldots a_{n}, \ldots$. By an $\bf{uncountable}$ set we mean, of course, an infinite set which is not countable.
We now give some examples of countable sets:
$\bf{Example 1}$:
The set $\mathscr{Z}$ of all integers, positive, negative, or zero is countable. In fact, we can set up the following one-to-one correspondence between $\mathscr{Z}$ and $\mathscr{Z^{+}}$ of all positive integers: (0,1), (-1,2), (1,3), (-2,4), (2,5), and so on. More explicitly, we associate the non-negative integer $n \geq 0$ with the odd number $2n+1$, and the negative integer $n<0$ with the even number $2|n|$, that is,
$n \leftrightarrow (2n+1)$, if $n \geq 0$, and $n \in \mathscr{Z}$
$n \leftrightarrow 2|n|$, if $n<0$, and $n \in \mathscr{Z}$
$\bf{Example 2}$:
The set of all positive even numbers is countable, as shown by the obvious correspondence $n \leftrightarrow 2n$.
$\bf{Example 3}$:
The set 2,4,8,$\ldots 2^{n}$ is countable as shown by the obvious correspondence $n \leftrightarrow 2^{n}$. | {
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$\bf{Example 4}: The set$latex \mathscr{Q}$of rational numbers is countable. To see this, we first note that every rational number $\alpha$ can be written as a fraction $\frac{p}{q}$, with $q>0$ with a positive denominator. (Of course, p and q are integers). Call the sum $|p|+q$ as the “height” of the rational number $\alpha$. For example, $\frac{0}{1}=0$ is the only rational number of height zero, $\frac{-1}{1}$, $\frac{1}{1}$ are the only rational numbers of height 2, $\frac{-2}{1}$, $\frac{-1}{2}$, $\frac{1}{2}$, $\frac{2}{1}$ are the only rational numbers of height 3, and so on. We can now arrange all rational numbers in order of increasing “height” (with the numerators increasing in each set of rational numbers of the same height). In other words, we first count the rational numbers of height 1, then those of height 2 (suitably arranged), then those of height 3(suitably arranged), and so on. In this way, we assign every rational number a unique positive integer, that is, we set up a one-to-one correspondence between the set Q of all rational numbers and the set $\mathscr{Z^{+}}$ of all positive integers. $\it{Next \hspace{0.1in}we \hspace{0.1in} prove \hspace{0.1in}some \hspace{0.1in}elementary \hspace{0.1in}theorems \hspace{0.1in}involving \hspace{0.1in}countable \hspace{0.1in}sets}$ $\bf{Theorem1}$. $\bf{Every \hspace{0.1in} subset \hspace{0.1in}of \hspace{0.1in}a \hspace{0.1in}countable \hspace{0.1in}set \hspace{0.1in}is \hspace{0.1in}countable}$. $\bf{Proof}$ Let set A be countable, with elements $a_{1}, a_{2}, a_{3}, \ldots$, and let set B be a subset of A. Among the elements $a_{1}, a_{2}, a_{3}, \ldots$, let $a_{n_{1}}, a_{n_{2}}, a_{n_{3}}, \ldots$ be those in the set B. If the set of numbers $n_{1}, n_{2}, n_{3}, \ldots$ has a largest number, then B is finite. Otherwise, B is countable (consider the one-to-one correspondence $i \leftrightarrow a_{n_{i}}$). $\bf{QED.}$ $\bf{Theorem2}$ $\bf{The \hspace{0.1in}union \hspace{0.1in}of \hspace{0.1in}a | {
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$\bf{QED.}$ $\bf{Theorem2}$ $\bf{The \hspace{0.1in}union \hspace{0.1in}of \hspace{0.1in}a \hspace{0.1in}finite \hspace{0.1in}or \hspace{0.1in}countable \hspace{0.1in}number \hspace{0.1in}of \hspace{0.1in}countable \hspace{0.1in}sets \hspace{0.1in}A_{1}, A_{2}, A_{3}, \ldots \hspace{0.1in}is \hspace{0.1in}itself \hspace{0.1in}countable.}$ $\bf{Proof}$ We can assume that no two of the sets $A_{1}, A_{2}, A_{3}, \ldots$ have any elements in common, since otherwise we could consider the sets $A_{1}$, $A_{2}-A_{1}$, $A_{3}-(A_{1}\bigcup A_{2})$, $\ldots$, instead, which are countable by Theorem 1, and have the same union as the original sets. Suppose we write the elements of $A_{1}, A_{2}, A_{3}, \ldots$ in the form of an infinite table $\begin{array}{ccccc} a_{11} & a_{12} & a_{13} & a_{14} &\ldots \\ a_{21} &a_{22} & a_{23} & a_{24} & \ldots \\ a_{31} & a_{32} & a_{33} & a_{34} & \ldots \\ a_{41} & a_{42} & a_{43} & a_{44} & \ldots \\ \ldots & \ldots & \ldots & \ldots & \ldots \end{array}$ where the elements of the set $A_{1}$ appear in the first row, the elements of the set $A_{2}$ appear in the second row, and so on. We now count all the elements in the above array “diagonally”; that is, first we choose $a_{11}$, then $a_{12}$, then move downwards, diagonally to “left”, picking $a_{21}$, then move down vertically picking up $a_{31}$, then move across towards right picking up $a_{22}$, next pick up $a_{13}$ and so on ($a_{14}, a_{23}, a_{32}, a_{41}$)as per the pattern shown: $\begin{array}{cccccccc} a_{11} & \rightarrow & a_{12} &\hspace{0.1in} & a_{13} & \rightarrow a_{14} & \ldots \\ \hspace{0.1in} & \swarrow & \hspace{0.1in} & \nearrow & \hspace{0.01in} & \swarrow & \hspace{0.1in} & \hspace{0.1in}\\ a_{21} & \hspace{0.1in} & a_{22} & \hspace{0.1in} & a_{23} \hspace{0.1in} & a_{24} & \ldots \\ \downarrow & \nearrow & \hspace{0.1in} & \swarrow & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in}\\ a_{31} & \hspace{0.1in} & a_{32} & \hspace{0.1in} & | {
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"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9883127433812521,
"lm_q1q2_score": 0.8783716472247783,
"lm_q2_score": 0.8887587993853654,
"openwebmath_perplexity": 294.34835768822796,
"openwebmath_score": 0.9011602997779846,
"tags": null,
"url": "https://madhavamathcompetition.com/category/chennai-mathematical-institute-entrance/"
} |
& \hspace{0.1in} & \hspace{0.1in}\\ a_{31} & \hspace{0.1in} & a_{32} & \hspace{0.1in} & a_{33} & \hspace{0.1in} & a_{34} & \ldots \\ \hspace{0.1in} & \swarrow & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in}\\ a_{41} & \hspace{0.1in} & a_{42} &\hspace{0.1in} & a_{43} &\hspace{0.1in} &a_{44} &\ldots\\ \ldots & \hspace{0.1in} & \ldots & \hspace{0.1in} & \ldots & \hspace{0.1in} & \ldots & \hspace{0.1in} \end{array}$ It is clear that this procedure associates a unique number to each element in each of the sets $A_{1}, A_{2}, \ldots$ thereby establishing a one-to-one correspondence between the union of the sets $A_{1}, A_{2}, \ldots$ and the set $\mathscr{Z^{+}}$ of all positive integers. $\bf{QED.}$ $\bf{Theorem3}$ $\bf{Every \hspace{0.1in}infinite \hspace{0.1in}subset \hspace{0.1in}has \hspace{0.1in}a \hspace{0.1in}countable \hspace{0.1in}subset.}$ $\bf{Proof}$ Let M be an infinite set and $a_{1}$ any element of M. Being infinite, M contains an element $a_{2}$ distinct from $a_{1}$, an element $a_{3}$ distinct from both $a_{2}$ and $a_{1}$, and so on. Continuing this process, (which can never terminate due to “shortage” of elements, since M is infinite), we get a countable subset $A= \{ a_{1}, a_{2}, a_{3}, \ldots, a_{n}, \ldots\}$ of the set $M$. $\bf{QED.}$ $\bf{Remark}$ Theorem 3 shows that countable sets are the “smallest” infinite sets. The question of whether there exist uncountable (infinite) sets will be considered below. $\bf{Section3}$ $\bf{Equivalence \hspace{0.1in} of \hspace{0.1in} sets}$ We arrived at the notion of a countable set M by considering one-to-one correspondences between set M and the set $\mathscr{Z^{+}}$ of all positive integers. More generally, we can consider one-to-one correspondences between any two sets M and N. $\bf{Definition}$ Two sets M and N are said to be $\bf{equivalent}$ (written $M \sim N$) if there is a one-to-one correspondence between the elements of M and the elements of | {
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"lm_name": "Qwen/Qwen-72B",
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"openwebmath_perplexity": 294.34835768822796,
"openwebmath_score": 0.9011602997779846,
"tags": null,
"url": "https://madhavamathcompetition.com/category/chennai-mathematical-institute-entrance/"
} |
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