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# Whats the probability a subset of an $\mathbb F_2$ vector space is a spanning set?
Let $V$ be an $n$-dimensional $\mathbb F_2$ vector space. Note that $V$ has $2^n$ elements and $\mathcal P(V)$ has $2^{2^n}$.
I'm interested in the probability (under a uniform distribution) that an element of $\mathcal P(V)$ is a spanning set for $V$. Equivalently a closed form formula (or at least one who's asymptotics as $n\rightarrow \infty$ are easy to work out) for the number of spanning sets or non-spanning sets.
It's not hard to show that the probability is greater than or equal to $1/2$. Since any subset of size greater than $2^{n-1}$ must span the space. I calculated the proportion of spanning sets of size $n$ for $n$ up to $200$ which seem to be going to a number starting with $.2887$. This leads me to believe that the probability exceeds $1/2$. I couldn't nail down a formula for arbitrary sized subsets though to continue experimental calculations.
I feel like this is something that's been done before, but googling I've mostly found things concerning counting points on varieties over finite fields or counting subspaces of finite fields. Any references would be appreciated.
-
I think it would be sufficient to count how many different bases you can have in V. Then multiply that number by all possibilities of adding elements to it, since that does not perturb the spanning property of the set. – Raskolnikov Jul 17 '12 at 15:31
@Raskolnikov The issue with that is avoiding duplicates. – JSchlather Jul 17 '12 at 15:32
Indeed, you're right. – Raskolnikov Jul 17 '12 at 15:33
A random subset of $V$ is owerwhelmingly likely to span $V$. | {
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A random subset of $V$ is owerwhelmingly likely to span $V$.
Let's look at how hard it is for a random subset not to span $V$. In order not to span $V$, there must be an $(n-1)$-dimensional subspace that contains the entire subset. There are exactly $2^n-1$ such subspaces, since they are in bijective correspondence with the nontrivial linear maps $V\to\mathbb F_2$ (each subspace is the kernel of exactly one map).
For each fixed $(n-1)$ dimensional subspace, the probability for a random subset to stay within that subspace is $2^{-2^{n-1}}$, since the $2^{n-1}$ vectors outside the subspace must all randomly decide not to be in the subset.
So the probability for a random set not to span is at most $(2^n-1)2^{-2^{n-1}} < 2^{-(2^{n-1}-n)}$ (and this is slightly too high, because there are a few non-spanning subsets that have more than one proper subspace they fit into and so are counted twice here). Everything else spans.
Even for $n$ as small as $5$ the probability for a random subset to span $V$ is above 99.9%, and the number of 9's increases exponentially with larger $n$s. | {
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-
There are indeed $2^n - 1$ subspaces of dimension $n - 1$. There is a bijection with subspaces of dimension $1$, though not via the inner product. – Zhen Lin Jul 17 '12 at 15:57
I think the problem is that the probability that some set $S$ is not contained in some subspace $U$ is not independent of the probability that $S$ is not contained in some other subspace $U'$. – Zhen Lin Jul 17 '12 at 16:01
That works, and it's certainly enough to get the conclusion that the probability goes to $1$ as $n \to \infty$. – Zhen Lin Jul 17 '12 at 16:08
(+1 to Henning) @ZhenLin: A non-degenerate bilinear form gives that correspondence (as you probably noticed) between $(2^n-1)$ non-zero vectors (=1-dim. subspaces) and $(n-1)$-dimensional subspaces. IMHO the word orthogonal might still be used - with the caveat that a vector or a subspace may be orthogonal to itself. – Jyrki Lahtonen Jul 17 '12 at 16:23
@Jyrki: I think it is cleaner to speak of nonzero vectors in $V$ and $(n-1)$-dimensional subspaces in $V^{\ast}$ (their annihilators) as this correspondence is functorial and doesn't require the addition of extra structure. – Qiaochu Yuan Jul 17 '12 at 20:18
It's worth mentioning that there is an explicit formula for the probability that $N$ random vectors drawn from $\mathbb{F}_q^n$ span. (Note: I am doing sampling with replacement. If you want to require the $N$ vectors to be distinct, you can do this using inclusion-exclusion, but the result will be much messier.) Of course, the answer is $0$ if $N<n$, so assume $N \geq n$.
Write down an $n \times N$ matrix whose columns are the vectors in question. Note that the following are equivalent:
1. The columns span $\mathbb{F}_q^n$.
2. The matrix has rank $n$.
3. The rows are linearly independent.
Thinking in terms of rows, we have the new question: If we draw $n$ vectors at random from $\mathbb{F}_q^N$, what is the probability that they are linearly independent? | {
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The probability that the first vector is nonzero is $1-q^{-N}$. Given that, the probability that the second vector is not in the span of the first is $1-q^{-N+1}$. Given that, the probability that the third vector is not in the span of the first two is $1-q^{-N+2}$. Continuing in this manner, the probability that $n$ vectors in $\mathbb{F}_q^n$ are independent is $$(1-q^{-N}) (1-q^{-N+1}) \cdots (1-q^{-N+n-1}).$$ And, by the above argument, this is also the probability that $N$ random vectors will span $\mathbb{F}_q^n$.
In particular, as Henning says, once $N-n$ is of any significant size, this is extremely close to $1$.
- | {
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By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.
What is the 10 001st prime number?
At first glance, a simple iteration whilst checking if index position is prime, would suffice.
A pseudocode attempt:
sum = 0
for i = 2; i <= ?; i++ {
if(isPrime(i)){
sum++
if ( sum >= 10001) {
return i
}
}
}
Using the solution in go to check if is prime from former examples:
func isPrime(x int64) bool {
var i int64 = 2
for ; i < x; i++ {
if x%i == 0 {
return false
}
}
return true
}
The solution would scale to
$$O(n^2)$$.
This is not viable. Also, there's no upper bound to set our iteration limit, meaning, we don't know up to which number we should iterate to find the 10,001 prime.
So digging further, I found about the Erathostenes Sieve:
An ancient algorithm for finding all prime numbers up to any given limit
This is exactly what we need. Let's look at the procedure:
To find all the prime numbers less than or equal to a given integer n by Erathostenes' method:
1. Create a list of consecutive integers from 2 ... n (2,3,4,...,n)
2. Initially, let p equal 2, smallest prime.
3. Enumerate multiples of p by counting in increments of p from 2p to n, and mark them in the list (these will be 2p, 3p, 4p, ...; the p itself should not be marked)
4. Find the first number greater than p in the list that is not marked, if there is not such number, stop. Otherwise, let p now equal to this number (which is next prime), and repeat step 3
5. When algorithm terminates, the numbers remain not marked in the list, are all primes below n.
Here's the code I came up:
package main
import (
"fmt"
"math"
"time"
)
func primeSieve(n, limit int) int {
// Create and populate array of values
// lim := c
a := make([]bool, limit+1)
for i := 0; i < limit+1; i++ {
a[i] = true
} | {
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// p * p <= limit === p <= int(math.Sqrt(float64(limit)))
for p := 2; p <= int(math.Sqrt(float64(limit))); p++ {
// At first all will be true
if a[p] == true {
// i = 2 * 2, i = 3 * 2, i = 4 * 2, ..., i = i * i
// Calculate multiples of 2, then 3, then 5, then 7
for i := p * 2; i <= limit; i += p {
a[i] = false
}
}
}
// Count primes up to n
// if primes == 10001, return sievePrime
var primes, sievePrime int
for p := 2; p <= limit; p++ {
if a[p] == true {
primes++
// Sum up primes and print 10,001th prime
if primes <= n {
sievePrime = p
}
}
}
return sievePrime
}
func main() {
start := time.Now()
// Nth prime to find
var n, limit int = 10001, 105000 // Arbitrary limit find by testing
fmt.Println("Execution time: ", time.Since(start))
fmt.Println("10,001th Prime: ", primeSieve(n, limit))
} | {
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# Limit of $2^{1/n}$ as $n\to\infty$ is 1
How do I prove that:
$\lim \limits_{n\to \infty}2^{1/n}=1$
Thank you very much.
-
Using the binomial theorem for integer exponents:
Can you see that $(1+\frac 1 n)^n > 2>1$
Take the nth root, to give:
$(1+\frac 1 n) > 2^{\frac 1 n} >1$
-
oh, that's really nice. how to you prove that $(1+\frac{1}{n})^n>2$? Thanks! – Anonymous Mar 28 '12 at 20:18
$(1+x)^n = 1+nx+$ other positive terms – Mark Bennet Mar 28 '12 at 20:20
And $n$ has to be greater than 1. – Mark Bennet Mar 28 '12 at 20:21
sweet, thank you. – Anonymous Mar 28 '12 at 20:25
Now look at Sivaram Ambikasaram's answer and catch hold of how the two are related - then you'll have learned some maths. – Mark Bennet Mar 28 '12 at 20:33
Note that for $a\gt 0$, $$a^b = e^{b\ln a}.$$ So $$\lim_{n\to\infty}2^{1/n} = \lim_{n\to\infty}e^{\frac{1}{n}\ln 2}.$$ Since the exponential is continuous, we have $$\lim_{n\to\infty}e^{\frac{1}{n}\ln 2} = e^{\lim\limits_{n\to\infty}\frac{1}{n}\ln 2}.$$
Can you compute $\displaystyle\lim_{n\to\infty}\frac{\ln 2}{n}$ ?
-
Thank you for the help, however I don't want to prove this using ln and e. – Anonymous Mar 28 '12 at 20:09
@Anonymous: Thank you for your comment; perhaps next time you can state, in your post, what it is you do and do not want, so that other people may not waste their time giving you information you don't care about. – Arturo Magidin Mar 28 '12 at 20:11
I really appreciate your time and sorry for the wasted time for me, however there are probably people who would find this informative and want to know how to prove it this way. I didn't state that I don't want lan and e because I wasn't aware that it can be proven this way. Thank you very much again. – Anonymous Mar 28 '12 at 20:16
This answer deserves more than one upvote, as it not only answers the question, but also deals neatly with the ambiguity in the way the question is asked. – Mark Bennet Mar 28 '12 at 21:37
HINT: Use squeeze theorem. | {
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HINT: Use squeeze theorem.
Since $1 < 2$, we have $1 = 1^{1/n} < 2^{1/n}$ for all $n \in \mathbb{N}$.
To bound the limit from above, note that $1 + n \epsilon < \left( 1 + \epsilon \right)^n$.
Hence, given any $\epsilon >0$, $\forall n > \displaystyle 1/{\epsilon}$, we have $2 < 1 + n \epsilon < \left(1 + \epsilon \right)^n$ and hence $2^{1/n} < 1 + \epsilon$.
-
Here are a few different proofs which don't use $e$ or $\log$ and can be regarded completely elementary.
Proof 1)
We use the following theorem:
If $a_n \gt 0$ and $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = L$, then $\lim_{n\to\infty} a_n^{1/n} = L$
This is a standard theorem, and a proof can be found in almost any textbook. You can also find a proof in my answer here: Show that this limit is equal to $\liminf a_{n}^{1/n}$ for positive terms.
Apply the theorem to the sequence $a_n = 2$.
Proof 2)
We use the $\text{AM} \ge \text{GM}$ inequality on $n-1$ ones and one $2$.
$$\frac{1 + 1 + \dots + 1 + 2}{n} \ge 2^{1/n}$$
$$\frac{n+1}{n} \ge 2^{1/n}$$
Thus we have
$$1 + \frac{1}{n} \ge 2^{1/n} \ge 1$$
so by Squeeze theorem, $\lim_{n \to \infty} 2^{1/n} = 1$.
Proof 3)
We can use Bernouli's inequality (essentially similar to Sivaram's answer) to show that
$$\left(1 + \frac{1}{n}\right)^n \ge 1 + \frac{1}{n} \times n = 2$$
and we get inequalities similar to the proof in 2).
Proof 4)
The sequence $a_n = 2^{1/n}$ is bounded below (by $1$) and montonically decreasing.
Thus it is convergent, to say $L$. Since $a_{2n}$ also converges to $L$, we have that $L = \sqrt{L}$, as $2^{1/2n} = \sqrt{2^{1/n}}$. So $L = 0$ or $L = 1$. Since the limit is not less than $1$ ($2^{1/n} \ge 1$), the limit is $1$.
Proof 5)
For $n \gt 2$, we have that $1 \le 2^{1/n} \le n^{1/n}$.
Now use the fact that $\lim_{n \to \infty} n^{1/n} = 1$. | {
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Now use the fact that $\lim_{n \to \infty} n^{1/n} = 1$.
An elementary proof of that can be found here: http://math.stackexchange.com/a/115825/1102. Any proof for $n^{1/n}$ now becomes a proof for $2^{1/n}$. Proof 1) above can also be used for $n^{1/n}$.
Proof 6)
Using combinatorics.
The number of $n$ digit numbers in base-$n+1$ is $(n+1)^n$ (allowing for leading zeroes). The number of $n$ digits numbers in base-$n$ is $n^n$. We can show that $(n+1)^n \ge 2 \times n^n$: consider the base-$n$ numbers. Replace the last digit with $n$. You get a base-$n+1$ $n$ digit number. Counting the base-$n$ numbers (which are also base-$n+1$ numbers) and the "last digit modified" numbers, gives us the inequality.
This inequality implies that $1 + \frac{1}{n} \ge 2^{1/n}$ and can be used to give a proof using the squeeze theorem, similar to proofs 2 and 3.
-
Sorry, had some free time :-) I am pretty sure there are more proofs... – Aryabhata Mar 28 '12 at 21:31
Very nice list! – Pedro Tamaroff Mar 29 '12 at 23:50
@PeterT.off: Thanks! – Aryabhata Mar 29 '12 at 23:59 | {
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# Sum of series
## Homework Statement
Find the sum of the series:
$$S = 1~+~\frac{1 + 2^3}{1 + 2}~+~\frac{1 + 2^3 + 3^3}{1 + 2 + 3}~+~\frac{1 + 2^3 + 3^3 + 4^3}{1 + 2 + 3 + 4}~+~...$$
upto 11 terms
## Homework Equations
Sum of first 'n' natural numbers: $$S = \frac{n(n + 1)}{2}$$
Sum of the squares of the first 'n' natural numbers: $$S = \frac{n(n + 1)(2n + 1)}{6}$$
Sum of the cubes of the first 'n' natural numbers: $$S = \left(\frac{n(n+1)}{2}\right)^2$$
## The Attempt at a Solution
In the series, the $n^{th}$ term is given by:
$$T_n = \frac{1 + 2^3 + 3^3 + ... + n^3}{1 + 2 + 3 + ... + n}$$
$$T_n = \frac{\left(\frac{n(n+1)}{2}\right)^2}{\left(\frac{n(n + 1)}{2}\right)}$$
$$T_n = \frac{1}{2}(n^2 + n)$$
Hence,
$$S_n = \frac{1}{2}\left(\frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2}\right)$$
On substituting n = 11, I get:
$$S_n = 286$$
But this isn't one of the options I got on the test. All the options were between 300-400 and all had their last digit '9' [that's all i remember.. they took the question paper away].
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# What is $\Bbb E(X)$?
Let there be $$100$$ balls in a box out of which $$50$$ are red and $$50$$ are green. Pick $$75$$ balls at random from the box and throw them away. Now pick one ball at random from the remaining balls in the box. Let $$X$$ be the random variable which takes the value $$100$$ when the ball drawn is red in colour and takes the value $$25$$ if the ball drawn is green in colour. Find the expectation $$\Bbb E(X).$$
My attempt $$:$$ Let $$Y$$ denote the number of red balls thrown away. Then the number of green balls thrown away is $$75-Y.$$ Clearly $$Y \geq 25.$$ So \begin{align*} \Bbb P(X=100) & = \sum\limits_{n=25}^{49} \Bbb P(X=100 \mid Y=n)\ \Bbb P(Y=n) \\ & = \sum\limits_{n=25}^{49} \frac {\binom {50-n} {1}} {\binom {25} {1}} \times \frac {\binom {50} {n}} {\binom {100} {n}} \end{align*} Similarly \begin{align*} \Bbb P(X=25) & = \sum\limits_{n=25}^{49} \Bbb P(X=25 \mid (75-Y) = n)\ \Bbb P((75-Y) = n) \\ & = \sum\limits_{n=25}^{49} \Bbb P(X=25 \mid Y = 75-n)\ \Bbb P(Y = 75-n) \\ & = \sum\limits_{n=25}^{49} \frac {\binom {50-n} {1}} {\binom {25} {1}} \times \frac {\binom {50} {75-n}} {\binom {100} {75-n}} \end{align*}
Then the required expectation would be $$100\ \Bbb P(X=100) + 25\ \Bbb P(X=25).$$ But the computation is very tough. Is there any simpler way to approach the problem?
Any help will be highly appreciated. Thank you very much. | {
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Any help will be highly appreciated. Thank you very much.
• It seems to me that red and green are indistinguishable in this scenario. Whatever the probability of picking a red ball is, it must be equal to the probability of picking a green ball. So P(X=100) = P(X=25) = 1/2 – Ant Jun 13 '20 at 8:51
• But @Ant after throwing $75$ balls away it might not be the case. Obviously the green and red balls in the box are present with different proportions after throwing $75$ balls. Right? Then the red and green balls are not equally likely to be drawn. – Phi beta kappa Jun 13 '20 at 8:55
• I have posted an answer, so we can keep discussing it there. To answer your question, they will be present in a different proportion after being drawn, but that's just a single observation. Their distribution is the same; the distribution of how many red vs green balls end up in the urn after you throw some away is the same. Therefore the distribution of which ball you pick after is also the same. So probabilities are the same. Right? – Ant Jun 13 '20 at 9:02
• It is clear by symmetry that $$\mathbb P(X=25)=\mathbb P(X=100)=\frac12.$$ There is no need for elaborate computation here. – Math1000 Jun 13 '20 at 10:09
• @Math1000 can you show it mathematically without making some vague assertion? I know that you are a genius; but it's better not to think others as genius as you. – Phi beta kappa Jun 13 '20 at 10:22
It seems to me that red and green are indistinguishable in this scenario. Whatever the probability of picking a red ball is, it must be equal to the probability of picking a green ball. So $$P(X=100) = P(X=25) = 1/2$$
Which implies
$$E(X) = 100\cdot P(X=100) + 25 \cdot P(X=25) = 62.5$$
A simple script confirms that the expectation is indeed $$62.5$$. You can also double check by computing the probability. Using your terminology,
$$P(Y=n) = \frac{{75 \choose n} \cdot {25 \choose 50-n}} {100\choose 50}$$
So that | {
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$$P(Y=n) = \frac{{75 \choose n} \cdot {25 \choose 50-n}} {100\choose 50}$$
So that
$$P(X=100) = \sum_{n=25}^{50} \frac{50-n}{25} \cdot \frac{{75 \choose n} \cdot {25 \choose 50-n}} {100\choose 50}$$
And you can check numerically that this is equal to $$1/2$$
• How have you computed $\Bbb P(Y=n)$? – Phi beta kappa Jun 13 '20 at 9:43
• The probability of removing n red balls can be computed as follows: Line up all the 100 balls in a row. Count how many of the permutations end up with n balls in the first 75 spots. Divide by total number of permutations, and you're done. To compute the first number, you have 75 balls of which n are red. So the permutations are 75!/n!/(75-n)!, i.e. 75 choose n. Then you have to multiply with the number of permutations of the remaining 25 balls; it's the same, except that you now have 50-n red balls. The second number is the total number of permutations, therefore 100 choose 50. – Ant Jun 13 '20 at 9:48
• I don't think so. If according to your assumption the balls are indistinguishable then we have just $26$ ways to throw $75$ balls from the urn and each way is equally likely. If the balls are indistinguishable then it doesn't matter in which order you throw them away or which red/green balls are thrown away. In this case which really matters that how many red/green balls you throw away out of $75$ balls. Right? If I present them as ordered pairs as (number of red balls thrown away, number of green balls thrown away) then there are $26$ such pairs $(25,50),(26,50), \cdots, (50,25).$ – Phi beta kappa Jun 13 '20 at 10:15
• Each pair is equally likely to occur. So for each $25 \leq n \leq 50$ we have $$\Bbb P(Y=n) = \frac {1} {26}.$$ – Phi beta kappa Jun 13 '20 at 10:17
• No. I know it for the case of distinguishable but not for the case of indistinguishable. – Phi beta kappa Jun 13 '20 at 10:28 | {
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# The Dynamics of Contrapositive Proofs
The Wikipedia Link for contrapositive proofs states that proving if p then q is the same as proving if not q then not p. I don't completely follow why.
Is there any way to understand what's happening without involving logic tables ? If logic tables are inevitable, can anyone give me a basic explanation?
I can't seem to follow how if p then q gets simplified to not p or q I understand that the truth tables of the two are same. But, that's that.
-
It could help to think the about contrapositive of some simple theorems, like the Pythagorean theorem. – Ragib Zaman Jul 8 '12 at 16:43
Semiformally speaking: Suppose $P\implies Q$. Further suppose $Q$ is false; if $P$ were true then $Q$ would also be true and we'd have a contradiction - impossible - so $P\implies Q$ and $\lnot Q$ implies $\lnot P$, i.e. $P\implies Q$ implies $\lnot Q\implies\lnot P$, ie an implication entails its contrapositive. Of course this means the contrapositive entails the contrapositive's contrapositive, ie the original statement, hence we have an equivalence, $(P\implies Q)\iff (\lnot Q\implies \lnot P)$. – anon Jul 8 '12 at 16:54
The sentence "if $p$ then $q$" asserts that in every single situation (model) in which $p$ is true, $q$ must be true. The only way this can be incorrect is if we have a situation in which $p$ is true but $q$ is false. We can show this cannot be the case by showing that whenever $q$ fails, $p$ must fail, that is, by showing that if "not $q$" then "not $p$."
Conversely, suppose that in every situation in which $p$ is true, then $q$ must be true. Then there cannot be a situation in which $q$ fails and $p$ is true. So if "not $q$" holds, then "not $p$" must hold.
Thus the two assertions "if $p$ then $q$" and "if not $q$ then not $p$" hold in precisely the same cases. | {
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Remark: The answer above is much too abstract. To understand what's going on, it is useful to examine a number of concrete cases. Suppose that we want to show that if $x^2=2$ then $x$ is not an ordinary fraction (integer divided by integer). So $p$ is the assertion $x^2=2$, and $q$ is the assertion that $x$ is not a fraction.
We prove the result by showing that if $x$ is a fraction, then $x^2$ cannot be equal to $2$, that is, by showing that "not $q$" implies "not $p$." That shows that if $x^2=2$, then $x$ could not possibly be a fraction, which is exactly what we wanted.
-
Much Clearer now. Thanks André. – Inquest Jul 8 '12 at 17:23
"If $P$ then $Q$" tells you that whenever you know that $P$ is true, you may automatically conclude that $Q$ is true.
"If not $Q$, then not $P$" tells you that whenever you know that $Q$ is not present, you may conclude that $P$ is not present.
If ($P$ implies $Q$) is true, then knowledge of the presence of $P$ always gives us that $Q$ is present. If $Q$ isn't there, then there must be no $P$ to have put it there. Likewise, if the absence of $Q$ automatically gives us the absence of $P$, and if $P$ is present, there can be no not-$Q$ to have given us a not-$P$.
Similarly, either we have $P$ or we don't have $P$. In the case that we have $P$, we are guaranteed to have $Q$. The other case is that we don't have $P$. So we either don't have $P$, or we have $Q$ (which we got by having $P$): not-$P$ or $Q$.
On the other hand, let's say all we know is that either $P$ isn't true or $Q$ is true. If $P$ isn't true, then $P\Rightarrow Q$ is trivially true. Let's deal with the case where $P$ is true. So we know that $P$ is true and also that either $P$ isn't true or $Q$ is true. That is, since $P$ and not-$P$ can't both be true, if $P$ is true, then $Q$ is true. | {
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-
+1. Took me a few reads to understand. The last paragraph's overall flow and structure is why most of my non-math friends don't speak to me when I'm doing math. – Inquest Jul 8 '12 at 17:22
Sorry. I'll take a look to revise it. – Neal Jul 8 '12 at 17:26 | {
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# Is every correlation matrix positive definite?
I'm talking here about matrices of Pearson correlations.
I've often heard it said that all correlation matrices must be positive semidefinite. My understanding is that positive definite matrices must have eigenvalues $> 0$, while positive semidefinite matrices must have eigenvalues $\ge 0$. This makes me think that my question can be rephrased as "Is it possible for correlation matrices to have an eigenvalue $= 0$?"
Is it possible for a correlation matrix (generated from empirical data, with no missing data) to have an eigenvalue $= 0$, or an eigenvalue $< 0$? What if it was a population correlation matrix instead?
Consider three variables, $X$, $Y$ and $Z = X+Y$. Their covariance matrix, $M$, is not positive definite, since there's a vector $z$ ($= (1, 1, -1)'$) for which $z'Mz$ is not positive.
However, if instead of a covariance matrix I do those calculations on a correlation matrix then $z'Mz$ comes out as positive. Thus I think that maybe the situation is different for correlation and covariance matrices.
My reason for asking is that I got asked over on stackoverflow, in relation to a question I asked there.
• If, for example, two attributes are one thing, only having different names, the matrix is singular. If two attributes add to a constant, it is again singular, et cetera. – ttnphns Nov 21 '15 at 16:54
• If a covariance matrix is singular correlation matrix is singular as well. – ttnphns Nov 21 '15 at 17:05
• Near-duplicates: Is every correlation matrix positive semi-definite? which has less focus on the definite versus semi-definite angle, and Is every covariance matrix positive definite? which is relevant because a covariance is essentially a rescaled correlation. – Silverfish Nov 21 '15 at 21:20
Correlation matrices need not be positive definite. | {
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Correlation matrices need not be positive definite.
Consider a scalar random variable X having non-zero variance. Then the correlation matrix of X with itself is the matrix of all ones, which is positive semi-definite, but not positive definite.
As for sample correlation, consider sample data for the above, having first observation 1 and 1, and second observation 2 and 2. This results in sample correlation being the matrix of all ones, so not positive definite.
A sample correlation matrix, if computed in exact arithmetic (i.e., with no roundoff error) can not have negative eigenvalues.
• May be worth mentioning the possible effects of missing values on the sample correlation matrix. Numerical fuzz isn't the only reason to get a negative eigenvalue in a sample correlation/covariance matrix. – Silverfish Nov 21 '15 at 21:22
• Yes, I didn't make it explicit, but I was assuming, per the question statement, "with no missing data". Once you get into the wild, wacky world of missing data and adjustments therefor, anything goes. – Mark L. Stone Nov 21 '15 at 21:35
• Yes, sorry, you're quite right the question said "no missing data" - just thought it worth mentioning somewhere since future searchers might be interested even if the OP's appetite is sated! – Silverfish Nov 21 '15 at 21:40
The answers by @yoki and @MarkLStone (+1 to both) both point out that a population correlation matrix can have zero eigenvalues if variables are linearly related (such as e.g. $X_1 = X_2$ in the example of @MarkLStone and $X_1 = 2X_2$ in the example of @yoki). | {
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In addition to that, a sample correlation matrix will necessarily have zero eigenvalues if $n<p$, i.e. if the sample size is smaller than the number of variables. In this case covariance and correlation matrices will both be at most of rank $n-1$, so there will be at least $p-n+1$ zero eigenvalues. See Why is a sample covariance matrix singular when sample size is less than number of variables? and Why is the rank of covariance matrix at most $n-1$?
• True 'dat. I suppose I could have and should have provided this info as well, but my goal was to produce a counterexample to refute the OP's hypothesis, thereby showing its invalidity Nevertheless, you should adjust your second sentence to be "In this case covariance and correlation matrices will be at most rank n−1, so there will be at least (p−n+1) zero eigenvalues." – Mark L. Stone Nov 21 '15 at 15:21
Consider $X$ to be an r.v. with mean 0 and variance of 1. Let $Y=2X$, and calculate the covariance matrix of $(X,Y)$. Since $2X=Y$, $E[Y^2]=4E[X^2]=\sigma_Y^2$, and $E[XY]=2E[X^2]$. Due to the zero mean configuration, the second moments are equal to the suitable covariances, for instance: $\mbox{Cov}(X,Y)=E[XY]-EXEY=E[XY]$.
So the covariance matrix will be: $$\Lambda = \left(\array{1 & 2 \\ 2 & 4 }\right),$$ having a zero eigenvalue. The correlation matrix will be: $$\Lambda = \left(\array{1 & 1 \\ 1 & 1 }\right),$$ having a zero eigenvalue as well. Due to the linear correspondence between $X$ and $Y$ it is easy to see why we get this correlation matrix - the diagonal will always be 1, and the off-diagonal is 1 because of the linear relationship. | {
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• Just for math-challenged readers like myself, let me point out that the 2 in $\Lambda$ is the $cov(X,Y)= \mathbb{E}(XY)-\mathbb{E}(X)\mathbb{E}(Y)= 2\mathbb{E}[X^2]=2\left(\sigma_X^2+[E(X)]^2\right)$ this last equality resulting from: $E(X^2)=\text{Var}(X)+[E(X)]^2$. – Antoni Parellada Nov 21 '15 at 19:45
• +1 for your post. I wanted to make it easy to follow for everyone by including the $diag \Lambda^{-1/2}\,\Lambda\, diag \Lambda^{1/2}$ formula, but the comments format won't allow it. Do you think there is any valid point in including it in your post? – Antoni Parellada Nov 21 '15 at 20:44
• @AntoniParellada , I'm not exactly sure what you mean - the covariance here is a direct calculation. But I will edit and make that clearer. Thanks. – yoki Nov 22 '15 at 14:59 | {
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Toronto Math Forum
APM346-2022S => APM346--Lectures & Home Assignments => Chapter 2 => Topic started by: Xiangmin.Z on January 17, 2022, 04:40:37 PM
Title: Week 2 Lec 1 (Chapter 2) question
Post by: Xiangmin.Z on January 17, 2022, 04:40:37 PM
Hello, I have a question about example 4 from W2 L1:
We are given :$u_{t}+xu_{x}=xt$
after calculation we get:
$x=Ce^t$
$du=xt dt=Cte^t dt$, so $u=C(t-1)e^t+D=x(t-1)+D$,
but how did we get $D=\phi({xe^{-t}} )$? we know it is a constant, but why is D depended on $xe^{-t}$?
Also, why would the initial condition $u|_{t=0} =0$ implies that $\phi({x}) =x$ ?
Thanks.
Title: Re: Week 2 Lec 1 (Chapter 2) question
Post by: Xiangmin.Z on January 17, 2022, 05:27:38 PM
I checked the calculation and I think the answer for x is correct, and does anyone know why D is depended on $xe^{-t}$?
Title: Re: Week 2 Lec 1 (Chapter 2) question
Post by: Victor Ivrii on January 17, 2022, 07:48:38 PM
Now it is correct $x=Ce^{t}$ and then $C=?$
Title: Re: Week 2 Lec 1 (Chapter 2) question
Post by: Xiangmin.Z on January 17, 2022, 08:19:53 PM
$C=xe^{-t}$, so D is a function of $\phi$, therefore $D=\phi({xe^{-t}} )$ ? Yes, but "$D$ is a function of it" | {
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# Prove that $f(x) = x^3 -x$ is NOT Injective
Okay, so I know there are plenty of injective/surjective (and thus, bijective) questions out there but I'm still not happy with the rigor of what I have done. So I'd really appreciate some help!
So the question actually asks me to do two things:
(a) give an example of a cubic function that is bijective. Explain why it is bijective.
(b) give an example of a cubic function that is not bijective. Explain why it is not bijective.
So for (a) I'm fairly happy with what I've done (I think):
$$f: \mathbb R \rightarrow \mathbb R , f(x) = x^3$$
So we know that to prove if a function is bijective, we must prove it is both injective and surjective.
Proof: $f$ is injective
Let: $$x,y \in \mathbb R : f(x) = f(y)$$ $$x^3 = y^3$$ (take cube root of both sides) $$x=y$$
Proof: $f$ is surjective
Let: $$y \in \mathbb R$$
$$x = \sqrt[3]{y}$$
$$f(x) = (\sqrt[3]{y})^3 = y$$
So I believe that is enough to prove bijectivity for $f(x) = x^3$. Keep in mind I have cut out some of the formalities i.e. invoking definitions and sentences explaining steps to save readers time. This is just 'bare essentials'.
So for (b)
$$f: \mathbb R \rightarrow \mathbb R , f(x) = x^3 - x$$
Now I'm just going to try and prove it is NOT injective, as that should be sufficient to prove it is NOT bijective.
Let:
$$x,y \in \mathbb R : f(x) = f(y)$$ $$x^3 - x = y^3 - y$$
This is about as far as I get. Send help. Thanks everyone.
• What happens if $x=0$ or $x=1$? – Cameron Williams Mar 28 '16 at 2:59
• To prove that a function $f$ is not injective, you just need to find two different numbers $x$ and $y$ such that $f(x)\not=f(y)$. – Christopher Carl Heckman Mar 28 '16 at 3:02
• @Carl you mean two numbers $x,y$ such that $x\neq y$ but $f(x)=f(y)$ – JMoravitz Mar 28 '16 at 3:03
• @CarlHeckman as JMoravitz said, you may confuse op... – YoTengoUnLCD Mar 28 '16 at 3:07
• Thanks for the help guys, +1's given. – Rubicon Mar 28 '16 at 3:53
Injectivity: | {
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Injectivity:
A function $f$ from $X\to Y$ is said to be injective iff the following statement holds true:
for every $x_1,x_2\in X$ if $x_1\neq x_2$ then $f(x_1)\neq f(x_2)$
The negation of this then yields:
A function $f$ from $X\to Y$ is not injective iff there exists $x_1,x_2\in X$ such that $x_1\neq x_2$ but $f(x_1)=f(x_2)$
In the case of the cubic in question, it is an easily factorable polynomial and we can find multiple distinct roots. $f(x)=x^3-x=x(x^2-1)=x(x+1)(x-1)$
We know that a root of a polynomial is a number $\alpha$ such that $f(\alpha)=0$
So, $f(1)=f(0)=f(-1)=0$ despite $1,0,-1$ all being distinct unequal numbers in the domain. This shows that it is not injective, and thus not bijective.
As an aside, one can prove that any odd degree polynomial from $\Bbb R\to \Bbb R$ must be surjective by the fact that polynomials are continuous and the intermediate value theorem.
• This is great. I think it has helped my understanding significantly. So what you're sort of saying is because f(1) and f(-1) equal 0, there is more than one-to-one mapping to the elements in the co-domain (hence not injective)? So in this case the set might even look like this: {(1,0) , (-1,0)} (just as a way to visualise it)? – Rubicon Mar 28 '16 at 3:53
Regarding (a), when you say "take cube root of both sides" you are (at least implicitly) assuming that the function is injective -- if it were not, the cube root would not be a well-defined operation. Depending on what you are / are not allowed to assume, that could be regarded as a circular argument.
However, there is another approach. Beginning with the assumption that $x^3=y^3$, you can manipulate the expression as follows: $$x^3-y^3=0$$ $$(x-y)(x^2+xy+y^2)=0$$ Now if you can find a convincing argument that the second factor, $x^2+xy+y^2$, is always positive for any choice of $x$ and $y$, then it follows that the only way $x^3=y^3$ can be true is if $x=y$. | {
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The factorization $x^3-y^3=(x-y)(x^2+xy+y^2)$, by the way, could also give you some insight into how to handle (b).
• $$x^2+xy+y^2 = {x^2+y^2\over 2} + {1\over 2}(x^2 + 2xy+y^2) = {x^2+y^2\over 2} + {1\over2}(x+y)^2 \ge 0.$$ The only way to get equality is to have $x=0$ and $y=0$, but then $x=y$ in that case as well. – Christopher Carl Heckman Mar 29 '16 at 6:30
• @CarlHeckman Yes, I know. Also $x^2+xy+y^2 = (x + \frac{1}{2}y)^2 + \frac{3}{4}y^2$. I was intentionally leaving that out to leave the OP something to finish. – mweiss Mar 29 '16 at 14:16 | {
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# Can an eigenvalue $\lambda_i=0$?
I was doing some work with diagonalization of a matrix $A$ in order to find a matrix $P$ such that $\,P^{-1}AP\,$ was diagonal. In order to that I set $\;\lambda I_{n}=0\;$ and found the characteristic polynomial and its roots.
When I factored my characteristic polynomial I obtained $\;\lambda^2(\lambda-2),\,$ so $\,\lambda=0,\,2$.
I was taught that the eignenvalues$\,\lambda_{i}\,$ I found become the entries of the diagonal matrix $\,P^{-1}AP.\,$ If this is indeed true, then two of the diagonal entries would be $\,0.\,$ Is this allowed, or must a diagonal matrix strictly have non-zero diagonal entries?
• Zero is allowed. You may be thinking of eigenvectors --- the zero vector can't be an eigenvector. But for eigenvalues, no problem. – Gerry Myerson Jun 21 '13 at 23:44
Absolutely yes: It is very possible $\lambda = 0$. Zero is allowed.
You may be mixing up what you know about eigenvectors --- the zero vector cannot be an eigenvector.
But for an eigenvalue $\lambda$, it is certainly possible and admissible that $\lambda = 0$.
With respect to your last question:
"($\lambda = 0$): Is this allowed, or does a diagonal matrix strictly have to have the diagonal entries as non-zero?"
Yes, it is allowed for zero's to be on the diagonal. No, the diagonal entries need not be non-zero.
• Dang! Beat me to the punch with the eigenvalue vs. eigenvector observation. +1 – Cameron Buie Jun 21 '13 at 23:58
• +1 what that Dang! means Amy? Dang Dang... Is that like a sound of a big Bell? – mrs Jun 22 '13 at 0:09
• +1 it's a non-profane way of saying something like "Damn": meaning, if I hadn't posted what I posted, Cameron intended to. – amWhy Jun 22 '13 at 0:12
• @amWhy Thanks; makes perfect sense now. I think I was confusing eigenvalues with eigenvectors. – Sujaan Kunalan Jun 22 '13 at 0:36
• @amWhy: Spot on +1 – Amzoti Jun 22 '13 at 0:38 | {
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A matrix is called diagonal if its off-diagonal entries ($a_{ij}$ for $i \not= j$) are all zero. This does not require the diagonal entries ($a_{jj}$) to be nonzero. For instance, the zero matrix is diagonal. | {
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# Solving problem using related rates yields incorrect result
I've been trying for a while to figure out what I did wrong on this problem, help would be appreciated.
"A 25-ft ladder is leaning against a wall. If we push the ladder toward the wall at a rate of 1 ft/sec, and the bottom of the ladder is initially 20 ft away from the wall, how fast does the ladder move up the wall 5 sec after we start pushing?"
$$x$$ = distance from wall, $$y$$ = height of ladder on wall, $$h$$ = length of ladder $$\frac {dx}{dt} = -1, t = 5$$ $$x^2 + y^2 = h^2$$ $$x^2 +y^2 = 25^2$$ $$2x\frac{dx}{dt}+2y\frac{dy}{dt} = 0$$ $$\frac{dy}{dt} = \frac{-x\frac{dx}{dt}}{y}$$ Since the ladder is moving toward the wall at 1 ft/sec for 5 sec, $$x(t = 5) =20-5 = 15$$
$$y(t=5) = \sqrt{25^2 - 15^2}=20$$ Substitute variables in equation: $$\frac{dy}{dt} = \frac{-(15)(-1)}{20}$$ $$\frac{dy}{dt} = \frac{3}{4}$$ The issue is that if $$\frac{dy}{dt} = \frac{3}{4},$$ the ladder would have risen 3.75 ft. in 5 seconds instead of 5 feet it SHOULD have risen, which is given by: $$\Delta y = y(t=5) - y(t=0)$$ $$(\sqrt{25^2 - 15^2}) - (\sqrt{25^2 - 20^2})$$ $$15-20$$ $$5$$ Thank you for reading and any answers.
At the start you have $$y(0)=\sqrt {25^2-20^2}=15$$. You are correct that the top has risen $$5$$ feet in $$5$$ seconds, but that does not mean that it has risen steadily at $$1$$ ft/sec for the whole time. Because $$\frac {dx}{dt}=-1$$ you found $$\frac {dy}{dt}=\frac xy$$ At $$t=0$$ that means that $$\frac {dy}{dt}=\frac {20}{15}=\frac 43$$ and at $$t=5$$ that means $$\frac {dy}{dt}=\frac {15}{20}=\frac 34$$. There is no contradiction here.
• Oh, I assumed that the rate of change had to be constant for some reason. Thank you for your help! – Kebab Boy Mar 14 at 20:07
Keeping your notation, you have that $$x(t) = 20 - t, \quad y(t)=\sqrt{25^2-x(t)^2},$$
and so, $$y'(t) = \frac{-2 x'(t)x(t)}{2 \sqrt{25^2-x(t)^2}},$$
which means that $$y'(5)=\frac{15}{\sqrt{25^2-15^2}}=\frac{3}{4}$$ | {
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which means that $$y'(5)=\frac{15}{\sqrt{25^2-15^2}}=\frac{3}{4}$$
The vertical displacement does not have to match the horizontal one. They are just connected through the expression $$x(t)^2 + y(t)^2=25^2$$. | {
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Solving uartics
BloodyFrozen
Is there a method to finding the roots of quartics besides Ferrari's formula?
I have the equation
$$x^{4}+5x^{2}+4x+5=0$$
I know one of the factor is something like $$x^{2}+x+1$$ and the other one can be found using sythetic division, but how can I find the factors without knowing one of them in the first place?
Thanks.
Homework Helper
If the quartic has "nice" quadratic factors, then they will be of the form:
$$x^4+5x^2+4x+5\equiv (x^2+ax\pm 1)(x^2+bx\pm 5)$$
Can you see why this is true? If we expand the right-hand side, we just need to compare coefficients to find the values of a and b and then we should be able to determine the nature of the $\pm$ operator.
BloodyFrozen
If the quartic has "nice" quadratic factors, then they will be of the form:
$$x^4+5x^2+4x+5\equiv (x^2+ax\pm 1)(x^2+bx\pm 5)$$
Can you see why this is true? If we expand the right-hand side, we just need to compare coefficients to find the values of a and b and then we should be able to determine the nature of the $\pm$ operator.
Oh, I see, but how do we know when the factors are "nice"?
Homework Helper
Oh, I see, but how do we know when the factors are "nice"?
We don't. We can always check to see if they are, just like we check to see if there are any rational roots.
With the quartic $x^4+5x^2+4x+5$ we end up solving a system of equations for a,b that works out nicely, but if we changed the quartic around to say, $x^4+6x^2+4x+5$ then we find no solutions for a,b in the system of equations.
BloodyFrozen
Ok, thanks!
BloodyFrozen
Ah, I believe I've seen this before. The method is nice, but it seems a little lengthy. Thanks for the link
BloodyFrozen
We don't. We can always check to see if they are, just like we check to see if there are any rational roots. | {
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With the quartic $x^4+5x^2+4x+5$ we end up solving a system of equations for a,b that works out nicely, but if we changed the quartic around to say, $x^4+6x^2+4x+5$ then we find no solutions for a,b in the system of equations.
I end up getting two systems but one of them has no solution. Did I proceed correctly?
Homework Helper
I end up getting two systems but one of them has no solution. Did I proceed correctly?
I don't know, what did you get?
BloodyFrozen
Sorry for not having LATEX since I'm on my phone, but I got:
System 1:
a+b=0
5+1+ab=5
5a+b=4
System 2:
a+b=0
-5-1+ab=5
-5a-b=4
Homework Helper
Sorry for not having LATEX since I'm on my phone, but I got:
System 1:
a+b=0
5+1+ab=5
5a+b=4
Right, so simplifying this system, we have
a+b=0
ab=-1
5a+b=4
Which we can then deduce,
a=-b
therefore,
a2=1 -> a=$\pm1$
4a=4 -> a=1, b=-1
System 2:
a+b=0
-5-1+ab=5
-5a-b=4
For this system we have no real solution, so what does that tell you?
BloodyFrozen
For the second system, I don't get any solutions. Therefore, that can't be the right system.
Homework Helper
For the second system, I don't get any solutions. Therefore, that can't be the right system.
Exactly, so what must your factors be?
BloodyFrozen
Ah, that's what I thought. Thanks! | {
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# The idea behind the sum of powers of 2
I know that the sum of power of $2$ is $2^{n+1}-1$, and I know the mathematical induction proof. But does anyone know how $2^{n+1}-1$ comes up in the first place.
For example, sum of n numbers is $\frac{n(n+1)}{2}$. The idea is that we replicate the set and put it in a rectangle, hence we can do the trick. What is the logic behind the sum of power of $2$ formula?
• This addition of powers of two, and many other examples, are all geometric series. There's a lot of visualisations of this out there. – Parcly Taxel Oct 29 '16 at 10:58
• Write the numbers in base 2: The powers of $2$ starting from $1=2^0$ will be in binary, $1+10+100+1000$ will always be a number that will be a n with all binary digits 1. This is the largest number having that many digits. SO it is of the form $2^{n+1}-1$ – P Vanchinathan Oct 29 '16 at 11:01
• I think there's a counting/probabilistic way to approach this something about having $n$ binary choices – BCLC Oct 31 '16 at 11:17
## 8 Answers
The binary expansion of $$\sum_{k=0}^n2^k$$ is a string of $$n+1$$ 1's: $$\underbrace{111\dots111}_{n+1}$$ If I add a 1 to this number, what do I get? $$1\underbrace{000\dots000}_{n+1}$$ 1 followed by $$n+1$$ 0's, hence $$2^{n+1}$$. Therefore $$\sum_{k=0}^n2^k=2^{n+1}-1$$
• Very clever proof! – user3932000 Nov 28 '20 at 4:16
This works for any partial sum of geometric series.
Let $$S = 1 + x + x^2+\ldots +x^n$$. Then $$xS = x + x^2 + \ldots +x^n + x^{n+1} = S - 1 + x^{n+1}$$.
All you have to do now is solve for $$S$$ (assuming $$x\neq 1$$).
• Yes, but the OP said that he already knew this. What he is asking for is a visual representation of the proof. – Alex M. Oct 29 '16 at 11:01
• Actually, they said that they knew proof by induction, but didn't know how one would come up with the formula in the first place. At least that is how I understood the question. @AlexM. – Ennar Oct 29 '16 at 11:02
There is a geometrical explanation. | {
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There is a geometrical explanation.
Take a box long enough to put twice the amount of items equal to the last term of the sum, and try to pack it.
Let's take a box of length $$2 * 2^n$$.
Let's put the items from the first term ($$2^n$$) into the box. Now exactly one half of the space is left for the other terms, from $$2^{n-1}$$ down to $$1$$, so we repeat this process, starting from the next largest term.
As we put the items from each term, we notice that they take exactly one half of the empty space left in the box, because both the largest term left and the space left decreases in half on each step.
At some point we reach the first term, which is equal to 1, and there are two empty places left in the box for just one item, so after we put the last item into the box, there is still space for one more item.
The length of the box is $$2*2^n = 2^{n+1}$$, but it could be shorter by one, which is $$2^{n+1} - 1$$, and this is our formula.
For example, let's pack: $$\sum_{i=0}^3 2^i$$
Box length: $$2 * 2^3 = 16$$
2^3 |● ● ● ● ● ● ● ●| |
2^2 |○ ○ ○ ○ ○ ○ ○ ○|● ● ● ●| |
2^1 |○ ○ ○ ○ ○ ○ ○ ○|○ ○ ○ ○|● ●| |
2^0 |○ ○ ○ ○ ○ ○ ○ ○|○ ○ ○ ○|○ ○|●| |
It could be shorter by one: $$2^{3+1}-1 = 15$$
By the way, similar geometrical explanation can be used for the ever decreasing geometric progression $$\sum_{i=0}^\infty 2^{-i}$$ with the only difference that we do not need to account for the last piece of empty space, because it tends to 0. Instead, we take some continuous medium as an example, such as a ribbon or a thread, which can be divided virtually infinitely so it sums to exactly a length of 2 units. | {
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There is a combinatorial interpretation. Consider the collection of all binary sequences of length $$n+1$$ with at least one $$1$$ (call this set $$E$$). There are $$2^{n+1}-1$$ such sequences because only $$0...0$$ isn't. Now let $$E_j$$ be the set of binary sequences of length $$n+1$$ such that the first $$1$$ is in the $$j$$ th component for $$j=1,\dotsc, n+1$$. Then $$|E_j|=2^{n+1-j}$$. Then the $$(E_j)$$ partition $$E$$ and we have that $$1+2+2^2+\dotsb+2^n=\sum_{j=1}^{n+1}2^{n+1-j}=2^{n+1}-1.$$ Alternatively consider the telescoping sum: $$\sum_{k=0}^n 2^{k}=\sum_{k=0}^n (2^{k+1}-2^k)=2^{n+1}-1.$$
Another famous approach is to count the $$2$$-player games to determine a winner among $$2^{n+1}$$ people. It's $$2^{n+1}-1$$, the number of people to eliminate. It's also $$2^n$$ people eliminated in the first round, halving each round until we reach the final.
• Very beautiful and efficient proof ! – projetmbc Oct 23 '20 at 9:57
Another pictorial proof, similar to the box packing reply, is to count the number of nodes in a complete binary tree with n+1 levels.
$$\sum_{i = 0}^{n} 2^i$$
is the number of nodes in the complete binary tree for the levels 0 (the root) through n.
If you draw an example and number the nodes like this
1
/ \
2 3
/ \ / \
4 5 6 7
/ \
8 ...
etc. you will notice that the left-most node of each level is a power of 2 (1, 2, 4, 8, 16, ...).
so the number of nodes in the tree from root to level n is 2^{n+1} - 1.
$$\sum_{i = 0}^{n} 2^i = 2^{n+1} - 1 \,.$$
Another geometric interpretation.
Start with a line segment, AB. Extend this into a line. Use compassto mark off a point C by setting center at B and passing radius through A. Then lengths, AB=BC. Create point D in a similar way. We have thus constructed a length double that of the original line segment.
This doubling can continue AS MUCH S you like.
Now ignore the first line segment, AB. | {
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This doubling can continue AS MUCH S you like.
Now ignore the first line segment, AB.
The series of segments now starts with BD which has double the length of AB.
In general the nth segment in the first case is the n-1th segment upon removal of AB. Further, each is double the length of its corresponding section.
So By subtracting a part we've doubled what remains, but with fewer sections to account for.
This gives the formula.
Not sure if my answer has been posted yet, but here's how I understand it.
You are trying to understand why
$$2^{n+1} - 1 = 2^n + 2^{n-1} + 2^{n-2} . . . + 2^1 + 2^0$$
Suppose we take 2^n in the sum. We know since these are powers of two, that the previous term will be half of 2^n, and the term before that a quarter of 2^n.
Let n in 2^n be 1, or 2^1 = 2. The term before in the sum will be half of 2, so we can also write the entire sum as:
$$2^1 + \frac{1}{2}(2^1)$$
If you do this but for different values of n for 2^n you will find you can rewrite the sums as:
$$2^n + \frac{ 2^n - 1}{2^n} ( 2^n)$$
You can simplify this because you're dividing (2^n) - 1 by 2^n, and then multiplying it by 2^n which cancel each other out. The new simplified version with the cancellation will look like:
$$2^n + 2^n - 1$$
This can also be written as: $$2( 2^n) - 1$$
Further simplified into: $$2^{n + 1} - 1$$
And that is why that formula shows up! Hope this helps ( I'm not sure if this was the mathematical induction proof you already knew, I'm young and discovered this on my own so I don't know what this is). | {
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×
# Elementary Techniques used in the IMO (International Mathematical Olympiad) - Cauchy Schwarz Inequality
If people asked you, what is the most elementary inequality you know? I bet your answer would be AM-GM. But in this series of posts I will try to show you the power that Cauchy Schwarz has over that of AM-GM. But as an introduction, let us first state and prove the theorem.
Cauchy Schwarz Inequality: Let $$(a_1, a_2, \ldots , a_n)$$ and $$(b_1, b_2, \ldots, b_n)$$ be two sequences of real numbers, then we have:
$$(a_1^2 + a_2^2 + \ldots + a_n^2)(b_1^2 + b_2^2 + \ldots + b_n^2) \geq (a_1b_1 + a_2b_2 + \ldots + a_nb_n)^2$$
In particular, equality holds iff there exists $$k \in \mathbb{R}$$ for which $$a_i = k b_i$$ for $$i = {1, \ldots, n}$$.
Proof: We will present $$2$$ proofs, one originating from analysis on the equality case, the other by wishful thinking on small cases of $$n = 2,3$$.
(i) Consider defining the following function $$f$$:
$$f(x) = (a_1x - b_1)^2 + (a_2x - b_2)^2 + \ldots + (a_nx - b_n)^2$$.
We will expand this to get:
$$f(x) = (a_1^2 + a_2^2 + \ldots + a_n^2)x^2 - 2(a_1b_1 + a_2b_2 + \ldots a_nb_n)x + (b_1^2 + b_2^2 + \ldots + b_n^2)$$
From our first way of representing $$f(x)$$, we can conclude that $$f(j) ≥ 0 \leftrightarrow ∆_f \leq 0$$ or
$$(a_1^2 + a_2^2 + \ldots + a_n^2)(b_1^2 + b_2^2 + \ldots + b_n^2) \geq (a_1b_1 + a_2b_2 + \ldots + a_nb_n)^2$$
Equality holds if the equation $$f(x) = 0$$ has one root.■
(ii) Just remark that:
$$(a_1^2 + a_2^2 + \ldots + a_n^2)(b_1^2 + b_2^2 + \ldots + b_n^2) - (a_1b_1 + a_2b_2 + \ldots + a_nb_n)^2 = \sum_{i,j=1}^n (a_ib_j + a_jb_i)^2 ≥ 0$$■
Let us note the following positive things regarding Cauchy-Schwarz:
• it is effective in proving symmetric inequalities
• Try to form squares
• Helps to clear up square roots
Look forward to the next few posts to see applications of this extremely elegant inequality!
Note by Anqi Li
3 years, 3 months ago
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Note by Anqi Li
3 years, 3 months ago
Sort by:
Hi everyone, I'm part of #TorqueGroup too. My assignment was to post olympiad level problems for the community that I found interesting. Since Anqi is posting these great topics about math things useful in such math, it seems like we're a perfect fit. So, here are some examples:
Show that for real numbers $$a_1, a_2, \cdots$$ and $$b_1, b_2, \cdots$$
$$\displaystyle \sum_{i = 1}^{k} \frac {a_i^2}{b_i} \geq \frac{(\sum_{i = 1}^k a_i)^2}{\sum_{i = 1}^k b_i}$$
IMO 1995
Let $$a, b, c \in \mathbb{R}^+$$ such that $$abc = 1$$. Prove that
$$\frac{1}{a^3(b+c)} + \frac{1}{b^3(a+c)} + \frac{1}{c^3(a+b)} \geq \frac{3}{2}$$.
..
Hint: It may help to make a symmetric substitution for $$a = \alpha, b = \beta, c = \gamma$$ such that the condition $$\alpha \beta \gamma = 1$$ still holds.
USAMO 2009
For $$n \geq 2$$ let $$a_1, a_2, \cdots, a_n$$ be positive reals such that
$$(a_1 + a_2 + \cdots + a_n)(\frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}) \leq (n + \frac{1}{2})^2$$
Prove that $$max (a_1, a_2, \cdots, a_n) \leq 4 min (a_1, a_2, \cdots, a_n)$$.
..
Hint: The LHS should scream cauchy. Choose which $$a_i$$ multiplies which $$b_k$$ to isolate a desired max and min, and then do some algebra. · 3 years, 3 months ago
I will be happy to post my solutions if requested (as long as you try them first!) · 3 years, 3 months ago
Could you add these examples to the Applications of Cauchy Schwarz Inequality Wiki page? Staff · 2 years, 5 months ago
For the second one, following your hint, let $$a = \frac{1}{x}, b = \frac{1}{y}, c = \frac{1}{z}$$. The expression is then $$\sum_{cyc} \frac{x^3yz}{y+z}$$. Since $$xyz = 1$$, this is $$\sum_{cyc} \frac{x^2}{y+z}$$. Using Cauchy, this is greater than $$\frac{(x+y+z)^2}{2(x+y+z)}$$. So now it is left to prove that $$x+y+z \geq 3$$. This is simple using AM-GM, so $$\frac{x+y+z}{3} \geq \sqrt[3]{xyz}$$, and we are done. $$\blacksquare$$ | {
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For the second one, WLOG assume $$a_1$$ is the maximum and $$a_n$$ is the minimum. Using Cauchy,
$$(a_1 + a_2 + \cdots + a_n)(\frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}) \geq (\sqrt{\frac{a_1}{a_n}} + \sqrt{\frac{a_n}{a_1}} + n - 2)$$
$$(n + \frac{1}{2})^2 \geq (\sqrt{\frac{a_1}{a_n}} + \sqrt{\frac{a_n}{a_1}} + n - 2)^2$$
Taking square roots of both sides, we get
$$n + \frac{1}{2} \geq \sqrt{\frac{a_1}{a_n}} + \sqrt{\frac{a_n}{a_1}} + n - 2$$
$$\frac{5}{2} \geq \sqrt{\frac{a_1}{a_n}} + \sqrt{\frac{a_n}{a_1}}$$
Now, squaring, we get
$$\frac{25}{4} \geq \frac{a_1}{a_n} + 2 + \frac{a_n}{a_1}$$
Multiplying both sides by $$a_1 a_n$$ we get
$$a_1^2 - \frac{17}{4} a_1 a_n + a_n^2 \leq 0$$
Factoring...
$$(a_1 - 4a_n)(a_1 - \frac{1}{4}a_n) \leq 0$$
Since $$a_1$$ and $$a_n$$ are positive reals, $$(a_1 - 4a_n)$$ must be nonpositive. Thus,
$$(a_1 - 4a_n \leq 0 \rightarrow a_1 \leq 4a_n$$, and we are done. · 3 years, 3 months ago
Could you explain the second one as in how did you use Cauchy Schwartz in it. · 2 years, 1 month ago
What are symmetric inequalities? · 2 years, 8 months ago
Hope you share many more such techniques!!thnx a lot for this. · 3 years, 3 months ago
Yup I will. In fact, I posted some on invariants and I also wrote some worked examples in my newest post. :) · 3 years, 3 months ago
@Anqi Li Can you add this to the Cauchy Schwarz Inequality Wiki page? Staff · 2 years, 5 months ago
One can easily prove Cauchy-Schwarz using vectors too. It is similar to the proof above, but is much cleaner. The interesting this is that almost every other inequality can be derived from this inequality. · 3 years, 3 months ago | {
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# Calculate limit without L'Hopital's rule
I have to calculate this limit whitout using L'Hopital's rule or Taylor polynomials:
$$\lim_{ x\to \pi/4 } \frac{1 - \tan(x)}{x-\frac{\pi}{4}}$$
I know how to make it using L'Hopital and that the result is $-2$ ,but I'm getting nowhere when I try without it. Any advice?
Hint:
Let $t=x-\frac{\pi}{4}$, then $$\frac{1-\tan x}{x-\frac{\pi}{4}}=\frac{1-\tan\left(t+\frac{\pi}{4}\right)}{t}=\frac{1}{t}\cdot\left(1-\frac{\tan t+\tan\frac{\pi}{4}}{1-\tan \frac{\pi}{4}\tan t}\right)=\frac{1}{t}\left(1-\frac{\tan t+1}{1-\tan t}\right)=\frac{-2\tan t}{t(1-\tan t)}$$ Now, take the limit as $t\to 0$: \begin{align} \lim_{x\to \frac{\pi}{4}}\frac{1-\tan x}{x-\frac{\pi}{4}}&=\lim_{t\to 0}\frac{-2\tan t}{t(1-\tan t)}\\ &=\lim_{t\to 0}\frac{-2\tan t\cos t}{t(1-\tan t)\cos t}\\ &=-2\lim_{t\to 0}\frac{\sin t}{t(\cos t-\sin t)}\\ &=-2\left(\lim_{t\to 0}\frac{\sin t}{t}\right)\left(\lim_{t\to 0}\frac{1}{\cos t-\sin t}\right)\\ &=-2(1)(1)\\ &=\color{blue}{-2} \end{align}
• While all the answers are valid and helped me,I chose this because it doesn't imply the use of more advanced concepts like derivatives.Thanks! – Der Rosenkavalier Oct 4 '15 at 20:48
Hint: What is the definition of the derivative of $\tan(x)$ at $x=\pi/4$? | {
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Hint: What is the definition of the derivative of $\tan(x)$ at $x=\pi/4$?
• $\frac{1}{\cos^2(\frac{\pi}{4})}$ But what's the point? I can't use L'Hopital's rule because the exercise ask it, not because I don't how to use it.I don't understand how can I use the derivative of $\tan(x)$ at $\frac{\pi}{4}$ in this exercise. – Der Rosenkavalier Sep 30 '15 at 21:47
• @DerRosenkavalier By definition: $$\tan'(\pi/4)=\displaystyle\lim_{x \to \pi/4}\frac{\tan(x)-1}{x-\pi/4}$$. How does this compare to your limit? – Reveillark Sep 30 '15 at 21:49
• L'Hopital isn't being used here; just the definition of the derivative. It's important to know the difference. – zhw. Sep 30 '15 at 22:00
• @zhw I agree with your comment. This type of question sometimes rasises debate as to "allowed ways forward. For example, is the use of asymptotic analysis permitted here? IMHO, yes absolutely. Yet others would argue that that approach is tantamount to the use of LHR. ;-) – Mark Viola Sep 30 '15 at 22:21
• @DerRosenkavalier : My answer posted here explicitly explains how you can use the value of the derivative of $\tan x$ at $\pi/4$. ${}\qquad{}$ – Michael Hardy Oct 1 '15 at 4:32
Recall that $$f'(a) = \lim_{x\to a}\frac{f(x) - f(a)}{x-a}.$$ Apply it to the case where $f(x) = \tan x$ and $a=\pi/4$. Then $f(a) = 1$ and $f'(a) = \sec^2 a = \sec^2(\pi/4) = 2$. Therefore $$2 = \lim_{x\to\pi/4}\frac{\tan x - \tan(\pi/4)}{x-\pi/4}.$$
So $-2$ is the answer to the question as you've posed it.
We have the identities
\begin{align}\frac{1-\tan x}{x-\pi/4}&=-\frac{\tan(x-\pi/4)}{x-\pi/4}(1+\tan x)\\\\ &=-\frac{\sin(x-\pi/4)}{x-\pi/4}\frac{1+\tan x}{\cos(x-\pi/4)} \end{align}
Can you finish from here?
• Yes,I could use that $\lim_{ x\to 0 }\frac{\sin(x)}{x}=1$ . Thanks for the help! – Der Rosenkavalier Oct 4 '15 at 20:58
• You're welcome. My pleasure. This is a very efficient approach that circumvents use of L'Hospital's Rule, as you requested. – Mark Viola Oct 4 '15 at 21:07 | {
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you can use the following hint ,
try to write $\tan(x)$ in terms of $\sin(x)$ and $\cos(x)$. A little bit of rearranging and then use the following
$\lim_{x\rightarrow 0} \frac{\sin(x)}{x}=1$ A little bit of work yields the result . | {
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How to formalize the fact that $f(i)=\lceil i/2\rceil$ is surjective but not one-to-one from $\Bbb N$ to itself?
If the set $S$ is countably infinite, prove or disprove that if $f$ maps $S$ onto $S$ (i.e. $f\colon S\to S$ is a surjective function), then $f$ is one-to-one mapping.
Please give a formal mathematical proof for this statement.
I have a counter-example: Suppose the mapping is from $\Bbb N$ (the natural numbers, a countably infinite set) to $\Bbb N$. And $f(i) = \lceil i/2\rceil$. It is onto function but not one-one.
But I am not getting that how this mapping is onto. Intuitively, since $1$ and $2$ will be mapped to $1$; $3$ and $4$ will be mapped to $2$; $5$ and $6$ will be mapped to $3$; similarly, $n-1$ and $n$ will be mapped to $n/2$. So in the codomain, there will be some elements that have no pre-image for this countably infinite set. So, please correct me where I am wrong. | {
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• Welcome to MSE. Please use MathJax. – José Carlos Santos May 15 '18 at 16:35
• "So in the codomain, there will be some elements that have no pre-image for this countably infinite set." Why do you say that? You just seemed to give an argument that it would. Anyway for any $n$ then both $2n$ and $2n+1$ will map to $n$. So $f$ is onto. – fleablood May 15 '18 at 16:47
• sir , since , cardinality of both sets are same and every 2 elements of domain are mapped to one element of codomain . so intuitively , it looks that some elements of codomain will not have any pre-image. please correct me – ankit1729 May 15 '18 at 16:51
• You just should that $1,2,3$ and $\frac n2$ (assuming $n$ is even) all have pre-images. What elements don't have pre-images? I honestly don't understand why you said that. – fleablood May 15 '18 at 16:51
• Are you thinking because you will run out of elements? But the set is infinite. You will never run out. So it's perfectly fine to have 2, 3 or even an infinite number of elements all map to the same element and still be onto. In fact that is exactly what you are proving. It is possible to be surjective and not 1-1. – fleablood May 15 '18 at 16:54
2 Answers
To prove that a function $f\colon A\to B$ is surjective, you need to be able to prove that given $b\in B$, there is some $a\in A$ such that $f(a)=b$.
In your case, $A=B=\Bbb N$, and $f(x)=\lceil \frac x2\rceil$.
So you need to prove that given any $m\in\Bbb N$, there is some $n\in\Bbb N$ such that $\lceil\frac n2\rceil=m$. Even though this $n$ is not unique, I am sure that you can come up with a way to find such $n$, from a given $m$.
Maybe I'm missing something. For every n in codomain, 2n and 2n-1 are in preimage, so the mapping is onto but not 1-1
• sir , it may or may not be one-one – ankit1729 May 15 '18 at 17:11 | {
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# what is the magnitude of the difference vector?
I read this question in Sears and Zemansky's University Physics book:
"Two displacement vectors, S and T, have magnitudes S = 3 m and T = 4 m. Which of the following could be the magnitude of the difference vector S - T ?(There may be more than one answer.) i. 9 m; ii. 7 m; iii. 5 m; iv. 1 m; v. 0m; vi. -1m "
My question is: shouldn't the direction of both vectors be specified in order for me to solve it? Or do I just assume that S and T are positive and negative respectively as per the equation S - T
• The question is asking you what the possible values of $\mathbf S - \mathbf T$ could be for all possible directions. So for example if one of the options was $666$m you need consider if there is any possible arrangement of the two vectors that could give $|\mathbf S - \mathbf T| = 666$. This freedom to orient the vectors in any direction means more than one of the answers can be correct. – John Rennie Jan 12 '17 at 16:20
• That's what I thought initially, but I had doubts. Thanks for confirming! – DigiNin Gravy Jan 12 '17 at 16:23
• Re Answer (vi): can the magnitude of a vector be negative? – DJohnM Jan 12 '17 at 17:18
• @DJohnM I know that the magnitude can never be negative, but the vector can. – DigiNin Gravy Jan 12 '17 at 18:20
• @DigiNinGravy A vector which is anti-parallel to, say, the $x$-axis of a coordinate system is sometimes called "a vector in the negative-$x$ direction." But it is an incorrect oversimplification to call such an object "a negative vector." – rob Jan 13 '17 at 7:26
The magnitude of the difference vectors depends on the orientation of $$\bf\vec{S}$$ and $$\bf \vec{T}$$. If they are parallel then $$|\bf \vec{S}-\bf \vec{T}|=|\,|\bf \vec{S}|-|\bf \vec{T}|\,|$$ and if they are anti-parallel then $$|\bf \vec{S}-\bf \vec{T}|=|\bf \vec{S}|+|\bf \vec{T}|$$.
Thus the possible values of $$|\bf \vec{S}-\bf \vec{T}\|$$ lie in the range: | {
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Thus the possible values of $$|\bf \vec{S}-\bf \vec{T}\|$$ lie in the range:
$$|\,|\bf \vec{S}|-|\bf \vec{T}|\,| \leq |\bf \vec{S}-\bf \vec{T}| \leq |\bf \vec{S}|+|\bf \vec{T}|$$
I'll let you work out which answers comply with this.
• |S - T| = 5 or | |S| - |T| | = -1 or |S| + |T| = 7. I just don't understand how |S - T| = |S| + |T|. I mean to have |S| + |T| both vectors must be parallel - having the same direction - so | |S| - |T| | seems off, unless we consider T as -T? – DigiNin Gravy Jan 12 '17 at 17:09
• @DigiNinGravy I think you gave the aswer yourself both $\bf T and$-\bf T$have se same magnitude, and the magnitude does not give the direction of the vector. – Mikael Fremling Jan 12 '17 at 18:10 In above Figure move the end point$\:\mathrm{B}\:$of the vector$\:\mathbf{S}\:$around a circle of radius$\:|\mathbf{S }|=3\:$. Try to find the length$\:|\mathbf{S}-\mathbf{T}|\:$of the vector$\:\mathbf{S}-\mathbf{T}\:$when$\:\mathrm{B}\:$is on points$\:\mathrm{P_{ii}},\mathrm{P_{iii}},\mathrm{P_{iv}}\:\$.
• Thanks, but you reversed the magnitudes of S and T. Can you tell me the name of the program you used? – DigiNin Gravy Jan 12 '17 at 18:17
• @DigiNin Gravy : GeoGebra – Frobenius Jan 12 '17 at 18:35 | {
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# Finding $n$ permutations $r$ with repetitions
I have given $n$ items and I have to find ${}^nP_r$ and $\sum_{r=0}^n{}^nP_r$ with repetitions allowed. Is there any closed formula for this?
For $n=3$ and $r=1$, possible permutations are:
$\{1\},\{2\},\{3\}$
Total: 3
For $n=3$ and $r=2$, possible permutations are:
$\{1,1\},\{2,2\},\{3,3\},\{1,2\},\{2,1\},\{1,3\},\{3,1\},\{2,3\},\{3,2\}$
Total: 9
For $n=3$ and $r=3$, possible permutations are:
$\{1,1,1\},\{2,2,2\},\{3,3,3\},$
$\{1,1,2\},\{1,2,1\},\{2,1,1\},$
$\{1,2,2\},\{2,1,2\},\{2,2,1\},$
$\{1,1,3\},\{1,3,1\},\{3,1,1\},$
$\{1,3,3\},\{3,1,3\},\{3,3,1\},$
$\{2,2,3\},\{2,3,2\},\{3,2,2\},$
$\{2,3,3\},\{3,2,3\},\{3,3,2\}$,
$\{1,2,3\},\{1,3,2\}$,
$\{2,1,3\},\{2,3,1\}$,
$\{3,1,2\},\{3,2,1\}$,
Total: 27
Can we come up with any closed formula for individual ${}^nP_r$ with repetitions and also for their sum i.e. here $3+9+27=39$
I understand that I cannot call this exactly the permutation, since ${}^3P_3$ is strictly $6$, while above its $27$, since I allow repetitions, but then whatever it is, how do I get the count?
Note that permutations with repetition is usually the well known case corresponding to $\frac{n!}{n_1!n_2!...n_i!}$, which is not what I am asking here. Is what am asking also some well know case, and I am stupidly not able to guess it? My primary guess is that, there cannot be any closed formula. Is it right? | {
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• What exactly is the condition you're looking for? I notice that the permutation $1,2,3$ is excluded - is this intentional? – platty Aug 20 '17 at 15:31
• In fact, it looks like you left out all the permutations of $1,2,3$ - is this intentional? – platty Aug 20 '17 at 15:41
• Nope, I missed it. Sorry. – anir Aug 20 '17 at 15:43
• I dont know it feels fuzzy. Is it just ${}^nP_r \text{with repetition} = n^r$?. Feels like so. Let me read question and answers again. – anir Aug 20 '17 at 15:47
• Your formula is not for permutations with replacements. It's for combinations with replacements. Are you actually looking for the expression for multinominal coefficients? – Vim Aug 20 '17 at 15:48
If you just want unrestricted strings consisting of $r$ letters, chosen with replacement from $n$, you can just use the multiplication rule to get $n^r$; there's $n$ choices for the first one, $n$ choices for the second, etc.
Extending this, we can use this to find the number of strings with length up to $r$ by summing the intermediate results: $$1+n+n^2+\dots +n^r$$ This is the sum of a geometric series, which means we can apply the formula to get $\frac{n^{r+1}-1}{n-1}$.
If you mean "permutations with replacements" then the answer is just $n^r$. But this doesn't match your result for $n=r=3$ which should be $27$? If you actually mean combinations with replacements, then see the below hint.
Hint: you are basically putting $r$ identical balls into $n$ different jars, or equivalently, finding the number of non-negative integer solutions to $$x_1+\cdots+x_n=r$$ To solve this, first let $y_i:=x_i+1$ (which are bijections) then try finding the number of positive integer solutions to the equation $$y_1+\cdots+y_n=r+n.$$ Try Stars & Bars technique. | {
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• I don't think this is correct - OP has order mattering in the listed examples. – platty Aug 20 '17 at 15:39
• @platty this is really confusing, because in this case the third result should be 27. Clearly there is some misphrasing in the question. – Vim Aug 20 '17 at 15:41
Note that in your last paragraph, you say that $\frac{n!}{n_1!n_2! \dots n_i!}$ is the formula for permutations with repetition, but perhaps a better way to word it would be that this is a generalized formula for partitioning a set of $n$ objects into $i$ cells with each "cell" being distinguishable, but elements within their respective cells are not.
As noted before, permutations with repetition allowed can be represented with $n^r$, which @platty already explained. This is the formula you would use to solve the second example I gave, as well as what you seem to be looking for in your example.
the number of words, from alphabet $\{1,\cdots,n\}$, of length $r$ and max repetition $r$,
i.e. simply the number of words, from alphabet $\{1,\cdots,n\}$, of length $r$
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Math Help - 2, 5, 8, 11, 14...
1. 2, 5, 8, 11, 14...
How can I prove that no square number can be in the sequence?
2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, 62, 65, 68, 71, 74, 77, 80, 83, 86, 89, 92, 95, 98, 101, 104, 107, 110, 113, 116, 119, 122, 125, 128, 131, 134, 137, 140, 143, 146, 149, 152, 155, 158, 161, 164, 167, 170, 173, 176, 179, ...
Is it to do with mod 5? And if so how would I do it
2. Originally Posted by Natasha
How can I prove that no square number can be in the sequence?
2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, 62, 65, 68, 71, 74, 77, 80, 83, 86, 89, 92, 95, 98, 101, 104, 107, 110, 113, 116, 119, 122, 125, 128, 131, 134, 137, 140, 143, 146, 149, 152, 155, 158, 161, 164, 167, 170, 173, 176, 179, ...
Is it to do with mod 5? And if so how would I do it
First you need find the general term of your sequence. Now it looks to me
as though the sequence is:
$a_n=2+3n, n=0,1,2,..$
Which suggests you consider the squares modulo 3. Now any natural number
may be written:
$N=3k+\rho$,
for some natural number $k$, and $\rho= 0, 1\ \mbox{or}\ 2$. So
$N^2=9k^2+6k \rho + \rho^2$.
Therefore $N^2(mod\ 3)$ is $\rho^2(mod\ 3)$, but
$
\rho^2(mod\ 3)=0, 1, \mbox{or}\ 1(mod\ 3) \mbox{as}\ \rho=0,1 \mbox{or}\ 2
$
.
So as $a_n, n=0,1,2,..$ is congruent to $2(mod\ 3)$ it cannot be a square for any
natural number $n$
RonL
3. Originally Posted by CaptainBlack
First you need find the general term of your sequence. Now it looks to me
as though the sequence is:
$a_n=2+3n, n=0,1,2,..$
Which suggests you consider the squares modulo 3. Now any natural number
may be written:
$N=3k+\rho$,
for some natural number $k$, and $\rho= 0, 1\ \mbox{or}\ 2$. So
$N^2=9k^2+6k \rho + \rho^2$.
Therefore $N^2(mod\ 3)$ is $\rho^2(mod\ 3)$, but
$
\rho^2(mod\ 3)=0, 1, \mbox{or}\ 1(mod\ 3) \mbox{as}\ \rho=0,1 \mbox{or}\ 2
$
. | {
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$
\rho^2(mod\ 3)=0, 1, \mbox{or}\ 1(mod\ 3) \mbox{as}\ \rho=0,1 \mbox{or}\ 2
$
.
So as $a_n, n=1,2,..$ is congruent to $2(mod\ 3)$ it cannot be a square for any
natural number $n$
RonL
Captain is it the same to say that the general form 3n - 1 is the same as 3n + 2??
4. Originally Posted by Natasha
Captain is it the same to say that the general form 3n - 1 is the same as 3n + 2??
Yes, but the +2 form is slightly more convienient when dealing with modular
arithmetic.
RonL
5. Originally Posted by CaptainBlack
Yes, but the +2 form is slightly more convienient when dealing with modular
arithmetic.
RonL
I see. Thanks! | {
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# Find the Ratio in Which the Point P(X, 2) Divides the Line Segment Joining the Points A(12, 5) and B(4, −3). Also, Find the Value of X. - Mathematics
Find the ratio in which the point P(x, 2) divides the line segment joining the points A(12, 5) and B(4, −3). Also, find the value of x.
#### Solution 1
Let the point P (x, 2) divide the line segment joining the points A (12, 5) and B (4, −3) in the ratio k:1.
Then, the coordinates of P are ((4k+12)/(k+1),(-3k+5)/(k+1))
Now, the coordinates of P are (x, 2).
therefore (4k+12)/(k+1)=x and (-3k+5)/(k+1)=2
(-3k+5)/(k+1)=2
-3k+5=2k+2
5k=3
k=3/5
Substituting k=3/5 " in" (4k+12)/(k+1)=x
we get
x=(4xx3/5+12)/(3/5+1)
x=(12+60)/(3+5)
x=72/8
x=9
Thus, the value of x is 9.
Also, the point P divides the line segment joining the points A(12, 5) and (4, −3) in the ratio 3/5:1 i.e. 3:5.
#### Solution 2
Let k be the ratio in which the point P(x,2) divides the line joining the points
A(x_1 =12, y_1=5) and B(x_2 = 4, y_2 = -3 ) . Then
x= (kxx4+12)/(k+1) and 2 = (kxx (-3)+5) /(k+1)
Now,
2 = (kxx (-3)+5)/(k+1) ⇒ 2k+2 = -3k +5 ⇒ k=3/5
Hence, the required ratio is3:5 .
Concept: Section Formula
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Class 10 Maths
Chapter 6 Co-Ordinate Geometry
Exercise 6.3 | Q 20 | Page 29
RS Aggarwal Secondary School Class 10 Maths
Chapter 16 Coordinate Geomentry
Q 5 | {
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# How to evaluate this definite double integral?
$$\int\int_D 6x\sqrt{y^2-x^2}dA, D=\{(x,y)|0\leq y\leq 2, 0 \leq x \leq y\}$$
I tried: $$\int_0^2 \int_0^y 6x\sqrt{y^2-x^2}dxdy$$
But that is incorrect.
• Why do you say that's incorrect? Looks fine to me... – Eli Berkowitz Dec 14 '15 at 3:32
• Your integral looks perfect! Now just evaluate it, try a u-substitution for the inside integral. – Alex Dec 14 '15 at 3:37
• I got -8, I must have made an arithmetic error or something! – d0rmLife Dec 14 '15 at 3:39
Setting $y^2-x^2 = t$, we obtain $-2xdx = dt$. Hence, we obtain $$I = \int_0^2 \int_0^y 6x\sqrt{y^2-x^2}dxdy = \int_0^2 \int_{y^2}^0(-3\sqrt{t}dt)dy = \int_0^2 \int_0^{y^2} 3\sqrt{t}dtdy = \int_0^22y^3 dy = 8$$
• Why does the range of the inner integral change to $y^2,0$ instead of $y^2,y^2-y$ - I thought you put the original range of integration into the u substitution equation? – d0rmLife Dec 14 '15 at 3:47
• @d0rmLife because the first inner integral goes from $x=0$ to $x=y$, and when replacing $y^2-x^2=t$ in the first case you get $y^2-0^2=y^2=t$ and in the second $y^2-y^2=0=t$. When you u-substitute you must change the range accordingly or else it is "wrong": you can get away with it if you evaluate the result in terms of x instead of u in the end, but still it is best to change it. – Fede Poncio Dec 14 '15 at 4:13
Notice, $$\int_{0}^2\int_0^y 6x\sqrt{y^2-x^2} \ dxdy=\int_{0}^2\left(\int_0^y 6x\sqrt{y^2-x^2} \ dx\right)dy$$ $$=\int_{0}^2\left(-3\int_0^y (y^2-x^2)^{1/2} \ d(y^2-x^2)\right)dy$$ $$=\int_{0}^2dy\left(-3\frac{2(y^2-x^2)^{3/2}}{3}\right)_{0}^{y}$$ $$=\int_{0}^2\left(2y^3\right)\ dy$$ $$=\left(2\frac{y^4}{4}\right)_{0}^{2}=\left(\frac{2^4}{2}-0\right)=\color{red}{8}$$ | {
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# Recognizing the sequence 1/16, 1/8, 3/16, 1/4, 5/16, …
What is the missing number? $$\frac{1}{16}, \frac{1}{8}, \frac{3}{16}, \frac{1}{4}, \frac{5}{16}, \ \ \ [?]$$
$$A. \frac{5}{4}\quad B. \frac{3}{4}\quad C. \frac{5}{8}\quad D. \frac{3}{8}$$
Spoiler: Answer is $D$, but I don't know why.
Thanks
-
OMG, of course. Thank you for the heads up. – user1495024 Aug 2 at 1:22
Am I the only one who thought $\frac{1}{2}$ before reading the possible options? :) – Thomas Aug 2 at 4:15
No, the next two numbers should definitely be 1/2 and 7/16 :-) – CompuChip Aug 2 at 8:23
@Thomas I did consider it, but then thought it wasn't just right. First of all, I'd be generalizing from just two data points. Secondly I'd be ignoring the other three data points, which supposedly were there for a reason. – kasperd Aug 2 at 9:37
$$\frac{1}{16}, \frac{1}{8}=\frac{2}{16}, \frac{3}{16}, \frac{1}{4}=\frac{4}{16}, \frac{5}{16}$$
So the $i$th term is of the form $$\frac{i}{16}$$ Therefore, the next term is $$\frac{6}{16}=\frac{3}{8}$$
-
$$\frac{1}{16}, \frac{1}{8}, \frac{3}{16}, \frac{1}{4}, \frac{5}{16}$$ The above is the same as $\displaystyle\frac1{16},\frac2{16},\frac3{16},\frac4{16},\frac5{16}$.
-
Another sequence-recognizing technique is to look at the difference between consecutive terms.
In this case, $\frac{1}{8}-\frac{1}{16} = \frac{1}{16}$, $\frac{3}{16}-\frac{1}{8} = \frac{1}{16}$, $\frac{1}{4}-\frac{3}{16} = \frac{1}{16}$, and $\frac{5}{16}-\frac{1}{4} = \frac{1}{16}$.
Since the difference between consecutive terms is $\frac{1}{16}$, the next term should be $\frac{5}{16}+\frac{1}{16} =\frac{6}{16} =\frac{3}{8}$.
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1. ## relations help (2)
hey guys, question:
Determine whether the following relation is a reflexive, symmetric or transitive. If any is an equivalence relation, describe its equivalence classes.
xRy iff x has the same integer part as y, on <----- I need help on what this relation is asking? thanks
2. Originally Posted by jvignacio
hey guys, question:
Determine whether the following relation is a reflexive, symmetric or transitive. If any is an equivalence relation, describe its equivalence classes.
xRy iff x has the same integer part as y, on <----- I need help on what this relation is asking? thanks
The integer part of x is the integer n such that $n\le x < n+1$. (There is exactly one such integer for each real x.) It is usually denoted by [x].
For example:
[3.14]=3
[0.52]=0
[1]=1
[-1.26]=-2
So 2 and 2.5 are in relation, since [2]=2=[2.5], but 1.5 and 2.5 are not.
Hope this helps.
3. Originally Posted by kompik
the integer part of x is the integer n such that $n\le x < n+1$. (there is exactly one such integer for each real x.) it is usually denoted by [x].
For example:
[3.14]=3
[0.52]=0
[1]=1
[-1.26]=-2
so 2 and 2.5 are in relation, since [2]=2=[2.5], but 1.5 and 2.5 are not.
Hope this helps.
Is that like saying the next full integer down?
4. Originally Posted by jvignacio
Is that like saying the next full integer down?
Exactly.
5. Originally Posted by kompik
Exactly.
So in this case,
- It IS reflexive since x has the same integer part as x.
- It IS symmetric since if x has the same integer part as y then y has the same integer part as x.
- It IS transitive since if x has the same integer part as y and if y has the same integer part as z then x has the same integer part as z.
correct?
6. Originally Posted by jvignacio
So in this case, | {
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correct?
6. Originally Posted by jvignacio
So in this case,
- It IS reflexive since x has the same integer part as x.
- It IS symmetric since if x has the same integer part as y then y has the same integer part as x.
- It IS transitive since if x has the same integer part as y and if y has the same integer part as z then x has the same integer part as z.
correct?
Yes, it is an equivalence relation.
Another way to see this is to notice that you're in fact given a decomposition of R and every decomposition gives you an equivalence relation. (But if you haven't heard much about the correspondence between equivalences and decompositions at your lessons, you should perhaps ignore this comment.)
7. Originally Posted by kompik
Yes, it is an equivalence relation.
Another way to see this is to notice that you're in fact given a decomposition of R and every decomposition gives you an equivalence relation. (But if you haven't heard much about the correspondence between equivalences and decompositions at your lessons, you should perhaps ignore this comment.)
Yeah ive never herd of that... Not yet anyway. What would the equivalence class be in this case now that its an equivalence relation? All numbers?
8. Originally Posted by jvignacio
Yeah ive never herd of that... Not yet anyway. What would the equivalence class be in this case now that its an equivalence relation? All numbers?
No.
Take, for instance, the equivalent class of 3.14. It contains all numbers such that x R 3.14.
This is equivalent to
[x]=[3.14]=3.
And what are the numbers such that [x]=3? Precisely the numbers from the interval $\langle 3,4)$, right?
Can you find the remaining classes? | {
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9. Originally Posted by kompik
No.
Take, for instance, the equivalent class of 3.14. It contains all numbers such that x R 3.14.
This is equivalent to
[x]=[3.14]=3.
And what are the numbers such that [x]=3? Precisely the numbers from the interval $\lange 3,4)$, right?
Can you find the remaining classes?
Sorry whats in the interval for [x]=3? theres a latex error..
10. Originally Posted by jvignacio
Sorry whats in the interval for [x]=3? theres a latex error..
Sorry, did not notice that. I've edited my post.
11. Originally Posted by kompik
No.
Take, for instance, the equivalent class of 3.14. It contains all numbers such that x R 3.14.
This is equivalent to
[x]=[3.14]=3.
And what are the numbers such that [x]=3? Precisely the numbers from the interval $\langle 3,4)$, right?
Can you find the remaining classes?
Ok I understand the equivalent class of 3.14 and the numbers such that [x]=3 are all numbers between 3 and 4 but Which remaining classes are you referring too? Since my question is a x and y question, how can I write this in a general form. If you get what I mean...
12. Originally Posted by jvignacio
Ok I understand the equivalent class of 3.14 and the numbers such that [x]=3 are all numbers between 3 and 4 but Which remaining classes are you referring too? Since my question is a x and y question, how can I write this in a general form. If you get what I mean...
If I were your teacher, I would expect answer to be something like:
The equivalent classes are intervals .... (fill in the dots) for $n\in\mathbb{N}$.
(I guess the example with 3.14 might help you to see what to fill in.)
13. Originally Posted by kompik
If I were your teacher, I would expect answer to be something like:
The equivalent classes are intervals .... (fill in the dots) for $n\in\mathbb{N}$.
(I guess the example with 3.14 might help you to see what to fill in.)
intervals $\langle n,n+1)$ , $n \in \mathbb{N}$ ? | {
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14. Originally Posted by jvignacio
intervals $\langle n,n+1)$ , $n \in \mathbb{N}$ ?
Exactly.
15. Originally Posted by kompik
Exactly.
mate thanks alot for the help. Really appreciate your time.
Page 1 of 2 12 Last | {
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# Use proof by induction to prove $\frac{1}{n!}<\frac{1}{2^n-1}$ for all $n\geq 4$
Use proof by induction to prove that that $\frac{1}{n!}<\frac{1}{2^n-1}$ for all $n\geq 4$, .\Base case: $$\frac{1}{4}=\frac{1}{24}\leq \frac{1}{2^4-1}$$ Inductive hypothesis: Assume there exists $k\in \mathbb{N}$ s.t.
$$\frac{1}{k!}\leq\frac{1}{2^k-1}$$ Inductive step: Show that:$$\frac{1}{(k+1)!}\leq\frac{1}{2^{k+1}-1}$$ Now, $$\frac{1}{(k+1)!}=\frac{1}{k!}\cdot\frac{1}{k+1}$$ Using the hypothesis $$\frac{1}{(k+1)!}\leq\frac{1}{2^k-1}\cdot \frac{1}{k+1}$$ Because $n\geq4, \frac{1}{k+1}<\frac{1}{2}$ $$\frac{1}{(k+1)!}\leq\frac{1}{2^k-1}\cdot \frac{1}{2}=\frac{1}{2^{k+1}-2}\leq\frac{1}{2^{k+1}-1}$$ Hence by mathematical induction we have proved that $\frac{1}{n!}<\frac{1}{2^n-1}$ for all $n\geq 4$
Firstly I need to know if the proof is correct, secondly it has to be as concise as possible hence I would like to know if there are any lines I can change/delete
And lastly can anyone explain to me why every sentence starts with "\" It looks perfectly fine in www.sharelatex.com ;(
• You must start with $n=4$. – User3101 Feb 18 '15 at 16:11
• I see and I am lost know because it is a question copied from the coursework given by a professor, I just blindly followed so lets assume $n\geq4$ – Scavenger23 Feb 18 '15 at 16:12
• So I should say Assume that $\frac{1}{k!}\leq\frac{1}{2^k-1}$ holds for some $k\geq4$ and then proceed ? – Scavenger23 Feb 18 '15 at 16:21
• One error I notice in your logic - in the last statement you state that $\frac 1{2^{k+1}-2} \leq \frac 1{2^{k+1}-1}$. This is incorrect. The latter (-1) fraction has a larger denominator than the -2 fraction, which means it's a smaller fraction. – Duncan Feb 18 '15 at 16:34
All in all it looks like you have the idea down on how induction works. It looks like you are correctly making the induction step to get your result. I have a couple things to point out, however. | {
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$(1)$ During your induction step you introduce the "$\geq$" symbol, when initially you are trying to write a proof about a strict inequality (which should only use the "$>$" symbol). If the strict inequality holds, it is not incorrect to use "$\geq$", but for the sake of consistency you should just use one.
$(2)$ Your induction hypothesis is not stated quite right. It is not enough to assume there exists $k \in \Bbb{N}$ such that the inequality holds. You want to make the stronger assumption that you can find this $k$, and the inequality holds for all $n \leq k$. It is in doing this that you will be allowed to conclude at the end of the proof that the inequality holds for all $n \geq 4$, instead of just $n=4$, and one arbitrary $k$.
$(3)$. If you are looking for a more concise proof, I would instead prove the equivalent statement that $2^n-1<n!$ for all $n \geq 4$. It is clear for $n \geq 4$ that $$2^n-1<2^n = 2 \cdot 2\cdot 2\cdot 2\cdot \ldots < 1 \cdot 2\cdot 3 \cdot 4 \cdot \ldots = n!$$ This is easier than dealing with fractions. But, upon proving this result one need merely flip the inequality around and put the quantities on each side under a numerator of $1$. | {
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• Never mind (4). in the preview mode every sentence had "\" in front of it while in here as you see Everything is fine. I see the mistake $< and \leq$ thank you for pointing it out. Thank you. I started induction only 2 weeks ago and it is a difficult topic for me. – Scavenger23 Feb 18 '15 at 16:37
• You see I didn't have the "flipping the fractions" tool in the toolbox so I didn't even think about what you have done in (3). The grammar unfortunately will remain an issue, since English is not my native language. – Scavenger23 Feb 18 '15 at 16:42
• You know what now I realised the whole mistake. I can't copy correctly. The question is not asking to prove $\frac{1}{n!}<\frac{1}{2^n-1}$ but $\frac{1}{n!}<\frac{1}{2^{n-1}}$ – Scavenger23 Feb 18 '15 at 16:50
• I don't want to use you but if I finish the right question this time, would you be able to have a one last look on it? I will stop spamming now since the website tells me to avoid extended discussions in comments. – Scavenger23 Feb 18 '15 at 16:54
• Use proof by induction to prove that that $\frac{1}{n!}<\frac{1}{2^n-1}$ for all $n\geq 3$\\I will use the fact that $k!>2^{k-1}$ is an equivelant statement to $\frac{1}{k!}<\frac{1}{2^k-1}$ .\\Base case $3!>2^{2}$ Inductive step: Assume $k!>2^{k-1}$ holds for all $n\leq k$ for some $k \in \bbb{N}$\\\ Inductive hypothesis\\Show that $(k+1)!>2^k$\\It is clear that when $k\geq 3$\\$2^{k-1}<2^k=2*2*2*2...<1*2*3*4...=k!<(k+1)!$ – Scavenger23 Feb 18 '15 at 17:18
It's much easier proving that $2^n<1+n!$, for $n\ge4$, which is completely equivalent to your assignment. The base step is obvious. Suppose it holds for $n$; then $$2^{n+1}=2\cdot2^n<2\cdot(1+n!)=2+2\cdot n!<1+(n-1)\cdot n!+2\cdot n!=1+(n+1)!$$ because $(n-1)n!>1$.
You can, if you want, transform this into a proof of your assigned inequality, but it's not necessary.
Rewrite as$$n!\ge2^n.$$ Then $$4!\ge2^4$$ and $$n!\ge2^n\land n+1\ge2\implies(n+1)!\ge2^{n+1}.$$ | {
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# How do I prove that $\lim_{x→0} x⋅\ln x=0$
How do I prove (without L'Hôpital's rule) that $$\lim_{x→0} x⋅\ln x = 0$$.
I'm trying to get some intuitive sense for this, but it's quite hard. It's like trying to prove that $x$ goes faster to $\ 0 \$ then $\ln x$ goes to $-∞$, right ?
I tried this, for $x∈(0,1)$ $$x \ln x=x⋅-∫_x^1 \frac{1}{t}dt>x⋅\frac{(1-x)}{-x}=-1+x$$
So when $x→0$, I conclude that $x⋅\ln x≥-1$.
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I don't know how to implement this, but how about squeeze theorem? It was the first thing that came to mind. – 000 Dec 13 '12 at 22:35
Since $\ln(\frac{1}{x})=-\ln x$, $$\lim_{x\to 0^{+}} x \cdot \ln x = \lim_{x\to \infty} \frac{1}{x} \cdot\ln \frac{1}{x} = - \lim_{x\to \infty} \frac{\ln x}{x}$$
But $\ln x$ is "slower" than $x$ (since $e^x$ is faster than $x$), so this limit is zero.
(Rigorously: for positive $x$, $2 e^x > x^2$ (by comparing power series of both sides). Plug $\sqrt{x}$ instead of $x$ and take logarithm of both sides: $\ln x < \ln 2 + \sqrt{x} =o(x)$.) | {
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Is it always true that you can rewrite $\lim_{x→0^+}f(x) = \lim_{x→∞}f(1/x)$? – Kasper Dec 13 '12 at 22:48
@Kasper, that's a great question. My intuition is no, and the reason is that there are tricky situations that arise when moving from a one-sided limit to a two-sided limit. I would say, however, that it may be true that $$\lim_{x \to 0}f(x)=\lim_{x \to \infty}f\left(\frac{1}{x}\right).$$ However, I am uncertain. – 000 Dec 13 '12 at 22:55
@Kasper - The answer is "Yes". Try to prove it using the definition of one-sided limit at 0 and limit at infinity. Show that if one limit exists, the other must equal it. – Ofir Dec 13 '12 at 23:03
@Limitless - There's no problem because the limit at infinity is not two-sided. Actually, your proposed equality is false, because in the LHS you evaluate $f$ on both negative and positive values, and at the RHS you evaluate $f$ only on positive values. Take $f(x)=\frac{1}{x}$ to reach a contradiction. – Ofir Dec 13 '12 at 23:05
@Ofir Let $s_n>0$ and $\lim_{n→∞} s_n = 0$. By definition $$\lim_{x→0^+}f(x)=\lim_{n→∞}f(s_n)$$. Let $t_n=1/s_n$. Then $\lim_{n→∞}t_n=+∞$. So $$\lim_{x→0^+}f(x)=\lim_{n→∞}f(s_n)=\lim_{n→∞}f(1/t_n)=\lim_{x→∞}f(1/x)$$ Do you mean something like this ? – Kasper Dec 13 '12 at 23:21
$$x \ln x=-x⋅\int_x^1 \frac{1}{t}dt$$
Now, let $0<a$ be arbitrary. Then, for all $x \in (0,1)$ we have
$$\frac{1}{t^{1-a}}\leq \frac{1}{t} \leq \frac{1}{t^{1+a}}$$
Thus
$$-x⋅\int_x^1 \frac{1}{t^{1+a}}dt \leq -x \ln(x) \leq -x⋅\int_x^1 \frac{1}{t^{1-a}}dt$$ $$-x⋅\frac{x^a-1}{a} \leq -x \ln(x) \leq -x⋅\frac{x^{-a}-1}{-a}⋅$$
now let $x \to 0$.
P.S. The argument works directly with any $0<a<1$, so picking $a=\frac{1}{2}$ makes the proof much cleaner....
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Make the change $x = e^{-y}$: $$\lim_{x \to 0} x \ln x = - \lim_{y \to \infty} y e^{-y}$$
Show that for $y > 0$: $$e^y > \frac{y^2}{2!}$$
And use the squeeze theorem.
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# What am I doing wrong with this second order ODE?
I have an ODE: $$\ddot v=-kt^{-1}\dot v+kt^{-2}v$$ I want to solve it by reducing it to a first order ODE, by defining $u(t)=t^{m}v(t)$ If I rewrite the ODE in terms of $u$, this gives me: $$\ddot u =(-k+2m)t^{-1}\dot u +(km+k-1-m^2)t^{-2}u$$ Setting the coefficient of $u$ to zero gives $2m=k+\sqrt{k^2-4+4k}$, which I will write as $k+\alpha$.
This gives me the single order ODE $$\ddot u=\alpha t^{-1}\dot u$$ Solving it gives $$\dot u=2c\alpha t$$ $$u(t)=c\alpha t^2$$ Plugging this back into $v$, gives $$v(t)=c\alpha t^{2-m}$$ $$\dot v = (2-m)c\alpha t^{1-m}$$ $$\ddot v = (1-m)(2-m)c\alpha t ^{-m}$$
These equations are supposed to hold for any $k$. If we plug this into the original ODE, however, this reduces to the equation $$3k=4+\sqrt{k^2-4+4k}$$
Which obviously does not hold for all $k$.
What am I doing wrong?
• Hint: Since this is a Euler-Cauchy type, let $v = t^m$. Find $v''$ and $v'$ and substitute in, solve for $m$ and solve.
– Moo
Mar 30 '17 at 17:39
• most likely your solution changes between (over)damped and oscillatory depending on $k$ Mar 30 '17 at 18:00
• Do you need to reduce it to a first order ODE? Or are you fine solving it with any method? Mar 30 '17 at 18:06
• I want to learn as much as possible about these things, so I was working on reducing it to a first order ODE. So I'd like to learn about other methods as well, but also I'd like to make sure I can do this method Mar 30 '17 at 18:09
• Please check again all signs and coefficients, the one before $u$ should reduce to $-(m+k)(m-1)$. Mar 30 '17 at 18:53
What you should realize is that you don't need to do a change of variables here but recognize that your solution is a Euler-Cauchy problem and therefore can be solved by:
Let $v=t^m$ then subbing the values into your DE you get:
$$m(m-1)t^{m-2} = -kmt^{m-2} + kt^{m-2} \\ \implies m(m-1) +km -k =0 \\ \implies m=1,m=-k \\$$
therefore giving: | {
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therefore giving:
$$v(t) = At^1 + Bt^{-k} \implies v(t) = At + Bt^{-k}$$
Which does hold $\forall k \in \mathbb{R}$ but you should also note that it does hold $\forall t \in (-\infty,0) \cup (0,+\infty)$ because $t=0$ is a case basis in that if $k>0$ then $t \neq 0$ but if $k<0$ then $t\in (-\infty,\infty)$
Since you did say you want to learn about the reduction to first order just note then when you do that you typically let u be something like $u= t^m \dot{v}$ so then $\dot{u}$ would be equal to a second derivative of v.
In your first integration, you should get from $$\ddot u=αt^{-1}\dot u$$ to $$\ln|\dot u|=α\ln|t|+c$$ or $$\dot u = C·t^α$$ leading to $$u=\frac{C}{α+1}·t^{α+1}+D$$ | {
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# Proving ⊢ A∨(A→B) using Hilbert system
I'm self-studying mathematical logic from "Introduction to Mathematical Logic" by Detlovs and Podnieks (available free here under CC license). Unfortunately, it doesn't come with any solutions. I'm stuck trying to prove ⊢ A∨(A→B), from Exercise 1.4.2 (i) on page 43. The text appears to be using the Hilbert-style deduction method.
Relevant axioms (classical logic)
• $$L_1: B \to (C \to B)$$
• $$L_2: (B \to (C \to D)) \to ((B \to C) \to (B \to D))$$
• $$L_3: B \to B \lor C$$
• $$L_4: C \to B \lor C$$
• $$L_8: (B \to D) \to ((C \to D) \to (B \lor C \to D))$$
• $$L_9: (B \to C) \to ((B \to \lnot C) \to ¬B)$$
• $$L_{10}: \lnot B \to (B \lor C)$$
• $$L_{11}: B \lor \lnot B$$
Relevant inference rules
• Modus Ponens
Attempt
1. $$(A \to A \lor (A \to B)) \to ((\lnot A \to A \lor (A \to B)) \to (A \lor \lnot A \to A \lor (A \to B)))$$ --- $$L_8$$
2. $$A \to A \lor (A \to B)$$ --- $$L_6$$
3. $$A \lor \lnot A$$ --- $$L_{11}$$
4. $$(\lnot A \to A \lor (A \to B)) \to (A \lor \lnot A \to A \lor (A \to B))$$ --- from (1) and (2) by MP
5. $$(\lnot A \to A \lor (A \to B)) \to A \lor (A \to B)$$ --- from (3) and (4) by MP
6. $$(\lnot A \to (A \to A \lor (A \to B))) \to ((\lnot A \to A) \to (\lnot A \to A \lor (A \to B)))$$ --- $$L_2$$
7. $$(\lnot A \to (A \to A \lor (A \to B)))$$ --- $$L_{10}$$
8. $$(\lnot A \to A) \to (\lnot A \to A \lor (A \to B))$$ --- from (6) and (7) by MP
I'm stuck at this point because $$\lnot A \to A$$ appears to be a contradiction. I'm also not sure whether this is the right approach. It looks alright at formula 5, but I'm not sure how to prove $$\lnot A \to A \lor (A \to B)$$, since it requires proving that $$A \lor (A \to B)$$ is always true, which is what we're trying to prove in the first place.
The text states that it can be solved using 14 formulas, 13 being the shortest yet. | {
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The text states that it can be solved using 14 formulas, 13 being the shortest yet.
• The "proof strategy" is to use L11: $A \lor \lnot A$ and derive $A \lor (A \to B)$ from $A$ with L6 and from $\lnot A$ with L10. Then use L8 – Mauro ALLEGRANZA Jul 5 '20 at 16:48
• @MauroALLEGRANZA I seem to have done the first three steps in my attempt (if I understood correctly); substituting B=A, C=¬A for L8 seems to yield (A→D)→((¬A→D)→D) after simplifying. The only straightforward substitution for D that I haven't tried seems to be D=(¬A→A∨(A→B)) but that leads to a double negation case with L10, which hasn't been proved yet. – user383527 Jul 5 '20 at 20:36
## 1 Answer
I assume that you are forced not tu use the Deduction Theorem.
But you can use the so-called Law of Syllogism (transitivity of $$\to$$).
If so, here is a sketch of a derivation:
1. $$\vdash A \to (A \lor (A \to B))$$ --- L6
2. $$\vdash \lnot A \to (A \to B)$$ --- L10
3. $$\vdash (A \to B) \to (A \lor (A \to B))$$ --- L7
4. $$\vdash \lnot A \to (A \lor (A \to B))$$ --- from 2. and 3. by Syllogism.
Now we can "cook them" together using L8:
1. $$\vdash (A \to (A \lor (A \to B))) \to [(\lnot A \to (A \lor (A \to B))) \to ((A \lor \lnot A) \to (A \lor (A \to B)))]$$
Now, from 5., 1. and 4. by MP twice:
1. $$\vdash (A \lor \lnot A) \to (A \lor (A \to B))$$
Finally, using L11: $$\vdash A \lor \lnot A$$, by MP:
$$\vdash A \lor (A \to B)$$.
• Smart use of the law of syllogism! The text has proved it as a theorem, so it is straightforward to show it from (2) and (3). In total, this gives 13 formulas (verbosely written), which the text says is the shortest proof. – user383527 Jul 6 '20 at 22:35 | {
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# Sums of rising factorial powers
Doodling in wolfram, I found that $$\sum^{k}_{n=1}1=k$$ The formula is pretty obvious, but then you get $$\sum^{k}_{n=1}n=\frac{k(k+1)}{2}$$ That is a very well known formula, but then it gets interesting when you calculate $$\sum^{k}_{n=1}n(n+1)=\frac{k(k+1)(k+2)}{3}\\ \sum^{k}_{n=1}n(n+1)(n+2)=\frac{k(k+1)(k+2)(k+3)}{4}$$ And so on. There is an obvious pattern that I really doubt is a coincidence, but I have no idea how to prove it in the general case. Any ideas?
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Pascal's triangle comes to mind here – Omnomnomnom Sep 4 '13 at 3:47
@Omnomnomnom I thought about it but I was unable to introduce binomials there. However, you can generalize the hypothesis using factorials, but usually working with factorials is harder than not doing so(At least I would have a very hard time trying to do so). – chubakueno Sep 4 '13 at 3:51
These are essentially the sums for the general $d$ dimensional simplex. – Jaycob Coleman Sep 4 '13 at 4:02
It's overkill, but you could expand and apply en.wikipedia.org/wiki/Faulhaber%27s_formula – dls Sep 4 '13 at 4:11
@dls I think that in that case the interesting part would go backwards: That expanding and a applying Faulhaber´s yields such a regular and simple result :) – chubakueno Sep 4 '13 at 4:22
You can argue any given case by induction. I will take your last,$$\sum^{k}_{n=1}n(n+1)(n+2)=\frac{k(k+1)(k+2)(k+3)}{4}$$ for the example, but I think it is easy to see how it gets carried forward. The base case is simply $1\cdot 2\cdot 3=1\cdot 2\cdot 3\cdot \frac 44$ If it is true up to $k$, then $$\sum^{k+1}_{n=1}n(n+1)(n+2)\\=\sum^{k}_{n=1}n(n+1)(n+2)+(k+1)(k+2)(k+3)\\=\frac{k(k+1)(k+2)(k+3)}{4}+(k+1)(k+2)(k+3)\frac {k+4-k}4\\=\frac{(k+1)(k+2)(k+3)(k+4)}{4}$$
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Thanks! Just changing the $2$, $3$ and $4$ by $x$,$x+1$ and $x+2$ does the work of generalizing it very nicely. – chubakueno Sep 4 '13 at 3:59 | {
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The easy way to deal, for example, with $\sum_{i=1}^n i(i+1)(i+2)(i+3)$ is to let $F(i)=i(i+1)(i+2)(i+3)(i+4)$. We calculate $F(i)-F(i-1)$. We get $$i(i+1)(i+2)(i+3)(i+4)-(i-1)(i)(i+1)(i+2)(i+3).$$ There is a common factor of $i(i+1)(i+2)(i+3)$. When we "take it out" we are left with $(i+4)-(i-1)=5$.
Let $G(i)=\frac{F(i)}{5}$. Then by our calculation $i(i+1)(i+2)(i+3)=G(i)-G(i-1)$.
Now consider the sum $\sum_{i=1}^n i(i+1)(i+2)(i+3)$. This is $$(G(1)-G(0))+(G(2)-G(1))+G(3)-G(2)) +\cdots+(G(n)-G(n-1)).$$ Observe the telescoping. Since $G(0)=0$, the above sum is equal to $G(n)$. Thus $$\sum_{i=1}^n i(i+1)(i+2)(i+3)=G(n)=\frac{n(n+1)(n+2)(n+3)(n+4)}{5}.$$
Exactly the same idea works in general.
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The hypothesized equality can be written as follows: for any $m$, we conjecture $$\sum^{k}_{n=1}\frac{(n+m)!}{(n-1)!}=\frac{(k+m+1)!}{(m+2)(k-1)!}$$ Dividing both sides by $(m+1)!$, we have $$\sum^{k}_{n=1}\frac{(n+m)!}{(n-1)!(m+1)!}=\frac{(k+m+1)!}{(m+2)!(k-1)!}$$ Or, in other words $$\sum^{k}_{n=1}\binom{n+m}{m+1}=\binom{k+m+1}{m+2}$$ I'm not sure how to prove this (yet), but it seems very likely that there's a neat trick for all this.
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Partial sums of sequences $a_n$ that can be expressed as polynomials in $n$ are easily found using discrete calculus.
We start with the discrete version of the Fundamental Theorem of Integral Calculus:
\begin{align*} \sum\limits_{n=1}^k a_n &= \sum\limits_{n=1}^k (\Delta b)_n \\ &= (b_2 - b_1)+(b_3 - b_2)+ \ldots + (b_k - b_{k-1}) + (b_{k+1} - b_k) \\ &= b_{k+1} - b_1 \end{align*}
where $(\Delta b)_n = b_{n+1} - b_n$ is the forward difference. Finding the partial sum has now been reduced to finding a sequence $b_n$ such that $(\Delta b)_n = a_n$.
We will find $b$, the antiderivative of $a$, using falling powers, which are defined by
$$n^{\underline{k}} = n(n-1)(n-2)\ldots (n-k+1)$$
where $k$ is an integer and, by a second definition, $n^{\underline{0}}=1$. For example
$$n^{\underline{3}} = n(n-1)(n-2).$$ | {
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$$n^{\underline{3}} = n(n-1)(n-2).$$
We now need one more result: the discrete derivative of $n^{\underline{k}}$ is given by
\begin{align*} \Delta n^{\underline{k}} &= (n+1)^{\underline{k}} - n^{\underline{k}} \\ &= (n+1)n^{\underline{k-1}} - n^{\underline{k-1}}(n-k+1) \\ &= kn^{\underline{k-1}} \end{align*}
Let's now find the partial sum for a particular case:
\begin{align} \sum^{k}_{n=1}n(n+1)(n+2) &= \sum^{k}_{n=1} (n+2)^{\underline{3}} \\ &= \sum^{k}_{n=1} \Delta \left[\frac{1}{4} (n+2)^{\underline{4}}\right] \\ &= \frac{(k+3)(k+2)(k+1)k}{4} - \require{cancel}\cancelto{0}{\frac{(1+2)(1+1)(1-0)(1-1)}{4}} \end{align}
The general case:
\begin{align} \sum^{k}_{n=1} (n+p)^{\underline{p+1}} &= \sum^{k}_{n=1} \Delta \left[\frac{1}{p+2} (n+p)^{\underline{p+2}}\right] \\ &= \frac{(k+1+p)(k+p)\ldots [(k+1+p)-(p+2)+1)]}{p+2} \\ &= \frac{(k+1+p)(k+p)\ldots k}{p+2} \end{align}
where $p>0$ is an integer.
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[I haven't finished yet. Will be gradually improved]
If we do it this way, then probably, it will become more insightful.
$$\sum^{k}_{n=1}1=k$$
then we preserve everything from the right side exactly as it is. $$\sum^{k}_{n=1}n=\frac{k(k+1)}{1 \cdot 2}$$ [there will be a picture of the right triangle]
Now again, preserve everything from the right side: $$\sum^{k}_{n=1}\frac{n(n+1)}{1 \cdot 2}=\frac{k(k+1)(k+2)}{1 \cdot 2 \cdot 3}$$
[there will be a picture of the 6 pyramids composed into rectangular parallelepiped]
$$\sum^{k}_{n=1}\frac{k(k+1)(k+2)}{1 \cdot 2 \cdot 3}=\frac{k(k+1)(k+2)(k + 3)}{1 \cdot 2 \cdot 3 \cdot 4}$$
[There should be a description of the connection between simple combinations and combinations with repetition]
$${n \choose k} = \frac{n \cdot (n - 1) \cdot \ldots \cdot (n - k + 1)}{k \cdot (k - 1) \cdot \ldots \cdot 1} = \frac{n^{\underline{k}}}{k!}$$ $$\left(\!\middle(\genfrac{}{}{0pt}{}{n}{k}\middle)\!\right) = \frac{n \cdot (n + 1) \cdot \ldots \cdot (n + k - 1)}{k \cdot (k - 1) \cdot \ldots \cdot 1} = \frac{n^{\overline{k}}}{k!}$$ | {
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You should use the sandbox if you are part way through an answer. – Michael Albanese 2 days ago
@Michael: Sorry, I don't understand how to use it. Probably, I just don't have the rights to use that post or I need to do something beforehand, but I don't see [Add answer] button there and I cannot edit them neither. – Pixar 2 days ago
Ah, you might not have enough reputation yet. Oh well, keep it in mind for future use. – Michael Albanese 2 days ago | {
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0 First we highlight the important concepts introduced in the previous post on permutations and combinations. , n ) 6 k − ) ⋅ {\displaystyle n} n y . Multiset coefficients may be expressed in terms of binomial coefficients by the rule, One possible alternative characterization of this identity is as follows: 1 Binomial coefficients have been known for centuries, but they're best known from Blaise Pascal's work circa 1640. n n … = n + $\binom{n+k}k$ divides $\frac{\text{lcm}(n,n+1,\ldots,n+k)}n$. We can use the formula (nk)=n!k!(n−k)! 1 This shows in particular that For instance, if k is a positive integer and n is arbitrary, then. z x {\displaystyle 1={\tbinom {6}{6}}={\tbinom {6}{0}}={\tbinom {43}{0}}} k − → 1 {\displaystyle k=0} 6 n which explains the name "binomial coefficient". ) [/math], ${n\choose k_1,k_2,\ldots,k_r} ={n-1\choose k_1-1,k_2,\ldots,k_r}+{n-1\choose k_1,k_2-1,\ldots,k_r}+\ldots+{n-1\choose k_1,k_2,\ldots,k_r-1}$, ${n\choose k_1,k_2,\ldots,k_r} ={n\choose k_{\sigma_1},k_{\sigma_2},\ldots,k_{\sigma_r}}$, \begin{align} {z \choose k} = \frac{1}{k! n {\displaystyle {\tbinom {n}{k}}} Is it possible to have perfect pitch but zero sense of relative pitch? □. weißen irgendwie aufgereihten Elementen. {\displaystyle {\tbinom {n}{k}}} &=\left(\!\!\binom{-7}{7}\!\!\right)\left(\!\!\binom{4}{7}\!\!\right)=\binom{-1}{7}\binom{10}{7}.\end{align}, ${x \choose y}= \frac{\Gamma(x+1)}{\Gamma(y+1) \Gamma(x-y+1)}= \frac{1}{(x+1) \Beta(x-y+1,y+1)}. An equivalent question: how many permutations are there of the letters A, B, C, D, E, F and G? For example, the 2nd2^{\text{nd}}2nd number in the 4th4^{\text{th}}4th row is represented as (31)=3\binom{3}{1} = 3(13)=3. □. In Counting Principles, we studied combinations.In the shortcut to finding$\,{\left(x+y\right)}^{n},\,$we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. {\displaystyle {\frac {{\text{lcm}}(n,n+1,\ldots ,n+k)}{n\cdot {\text{lcm}}({\binom | {
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of the binomial. {\displaystyle {\frac {{\text{lcm}}(n,n+1,\ldots ,n+k)}{n\cdot {\text{lcm}}({\binom {k}{0}},{\binom {k}{1}},\ldots ,{\binom {k}{k}})}}} + = 0 when either k > n or k < 0.$ It is the coefficient of the xk term in the polynomial expansion of the binomial power (1 + x)n, and it is given by the formula, For example, the fourth power of 1 + x is. An integer n ≥ 2 is prime if and only if n ) {\displaystyle \alpha -z} α {\displaystyle m,n\in \mathbb {N} ,}. (2003). Für die Anzahl der möglichen Ziehungen oder Tippscheine beim deutschen Lotto 6 aus 49 (ohne Zusatzzahl oder Superzahl) gilt: Es gibt hier offensichtlich genau eine Möglichkeit, 6 Richtige zu tippen. A direct implementation of the multiplicative formula works well: (In Python, range(k) produces a list from 0 to k–1.). The coefficient of the middle term in the binomial expansion in powers of x of (1 + αx)^4 and of (1 – αx)^6 is the same if α equals asked Nov 5 in Binomial Theorem by Maahi01 ( 19.5k points) binomial theorem + Binomial coefficient formula reduction. 4 α -elementigen Teilmenge. A related combinatorial problem is to count multisets of prescribed size with elements drawn from a given set, that is, to count the number of ways to select a certain number of elements from a given set with the possibility of selecting the same element repeatedly. {\displaystyle k} {\displaystyle \psi (n)} ) m Randomly select a ball. z ( {\displaystyle n} n ln , sondern den Bruch &=(-1)^7\;\frac{4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10} k When the exponent is 1, we get the original value, unchanged: (a+b) 1 = a+b. k , dann ist. ) for all positive integers r and s such that s < pr. , ( k ≥ The right side counts the same parameter, because there are ways of choosing a set of q marks and they occur in all subsets that additionally contain some subset of the remaining elements, of which there are. -elementigen Menge ist, ergibt sich durch die Summation die Anzahl aller ihrer Teilmengen, also ) ! Other notations for | {
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ist, ergibt sich durch die Summation die Anzahl aller ihrer Teilmengen, also ) ! Other notations for the binomial coefficient are , and . ( ) The binomial coefficients form the entries of Pascal's triangle.. für das dritte usw., bis hin zu □, Or, in other terms, the sum of two adjacent binomial coefficients is equal to the one "below" it, assuming Pascal's triangle is produced in the equilateral-triangle-like shape shown above. r 2 {\displaystyle p} https://de.wikipedia.org/w/index.php?title=Binomialkoeffizient&oldid=205380223, „Creative Commons Attribution/Share Alike“. ∈ − ≤ ) More precisely, fix an integer d and let f(N) denote the number of binomial coefficients with n < N such that d divides . One of the more visually striking properties is the Sierpiński sieve, which is obtained by taking mod 2 of every binomial coefficient. 7!} where $\ln$ $\Gamma(n)$ denotes the natural logarithm of the gamma function at $n$. 5 ( = k ) , n n 5 When n is composite, let p be the smallest prime factor of n and let k = n/p. und r }{k!\big((n-k)!\big)} + \frac{n!}{(k+1)!\big(n-(k+1)\big)!} {\displaystyle 43={\tbinom {43}{1}}} wie auch ) ( ! k binomial coefficients: These can be proved by using Euler's formula to convert trigonometric functions to complex exponentials, expanding using the binomial theorem, and integrating term by term. 6 {\displaystyle x\in X} α m We can verify this visually by looking at Pascal''s triangle and using our guideline for construction: (nk)+(nk+1)=(n+1k+1).\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}.(kn)+(k+1n)=(k+1n+1). 0 The Chu–Vandermonde identity, which holds for any complex-values m and n and any non-negative integer k, is, $\sum_{j=0}^k \binom m j \binom{n-m}{k-j} = \binom n k$, and can be found by examination of the coefficient of $x^k$ in the expansion of (1 + x)m (1 + x)n − m = (1 + x)n using equation (2). und 0 ( X erhält man die Beziehung. will remain the same. k α ∈ \binom{n+1}{2} = T_{n} &= \dfrac{(n+1)(n)}{2}, p Grinshpan, A. | {
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will remain the same. k α ∈ \binom{n+1}{2} = T_{n} &= \dfrac{(n+1)(n)}{2}, p Grinshpan, A. ( , can be calculated by logarithmic differentiation: Over any field of characteristic 0 (that is, any field that contains the rational numbers), each polynomial p(t) of degree at most d is uniquely expressible as a linear combination ( ⋅ }{k!\cdot l! ) Ein Binomialkoeffizient hängt von zwei natürlichen Zahlen ) l How many permutations are there of 7 objects? This article incorporates material from the following PlanetMath articles, which are licensed under the Creative Commons Attribution/Share-Alike License: Binomial Coefficient, Upper and lower bounds to binomial coefficient, Binomial coefficient is an integer, Generalized binomial coefficients. k The binomial theorem is a formula for deriving the power of a binomial, i.e. Certain trigonometric integrals have values expressible in terms of n Binomial coefficients are a family of positive integers that occur as coefficients in the binomial theorem. However this is not true of higher powers of p: for example 9 does not divide + , Alternative notations include C(n, k), nCk, nCk, Ckn, Cnk, and Cn,k in all of which the C stands for combinations or choices. 5 z Exercise 2 Binomial coefficients can be generalized to multinomial coefficients defined to be the number: While the binomial coefficients represent the coefficients of (x+y)n, the multinomial coefficients is convergent for k ≥ 2. = ( over For example, for nonnegative integers ${n} \geq {q}$, the identity. t ) This is done by interpreting as the number of ways to partition the set into two subsets of size k and n-k. Pascal's rule is the important recurrence relation. Die Anzahl aller so zusammengestellten 1 \end{cases}[/math], $\tbinom n0,\tbinom n1,\tbinom n2,\ldots$, $\sum_{k=0}^\infty {n\choose k} x^k = (1+x)^n. {\displaystyle \alpha } Differentiating (2) k times and setting x = −1 yields this for Identifying Binomial Coefficients. The notation [math] {n \choose k}$ is | {
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x = −1 yields this for Identifying Binomial Coefficients. The notation [math] {n \choose k}$ is convenient in handwriting but inconvenient for typewriters and computer terminals. > ( Then 0 < p < n and. k {\displaystyle 6={\tbinom {6}{5}}} Damit gilt für jede endliche Menge {\displaystyle {\tbinom {2n}{n}}} }{k_1!k_2!\cdots k_r! − n ψ {\displaystyle l!} n n How many different pizza can the customer create? Another form of the Chu–Vandermonde identity, which applies for any integers j, k, and n satisfying 0 ≤ j ≤ k ≤ n, is, The proof is similar, but uses the binomial series expansion (2) with negative integer exponents. Sie kann anschaulich etwa so gedeutet werden: Zunächst zählt man alle 1 k M {\displaystyle \sum _{k=0}^{d}a_{k}{\binom {t}{k}}} Compute (92)+(83)\dbinom{9}{2} + \dbinom{8}{3}(29)+(38). ) ) 0 {\displaystyle k\to \infty } Create a free website or blog at WordPress.com. This definition inherits these following additional properties from { 1 It also follows from tracing the contributions to Xk in (1 + X)n−1(1 + X). Differentiating (2) k times and setting x = −1 yields this for ( Asymptotic of binomial coefficients near the center. {\displaystyle \{1,2\}{\text{, }}\{1,3\}{\text{, }}\{1,4\}{\text{, }}\{2,3\}{\text{, }}\{2,4\}{\text{,}}} The overflow can be avoided by dividing first and fixing the result using the remainder: Another way to compute the binomial coefficient when using large numbers is to recognize that. ) in a language with fixed-length integers, the multiplication by n^{\underline{k}}/k! ( 1 . Hot Network Questions Should I respond to an "ethical hacker" who's requesting a bounty? ) ) , = 6[/math] is the coefficient of the x2 term. ! There are several ways to come up with the answer. m eigentlich einfache Binomialkoeffizienten sind. x {\displaystyle {\tfrac {(k+l)! ⋅ Newton's binomial series, named after Sir Isaac Newton, is a generalization of the binomial theorem to infinite series: The identity can be obtained by showing that both sides | {
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of the binomial theorem to infinite series: The identity can be obtained by showing that both sides satisfy the differential equation (1 + z) f'(z) = α f(z). + Thinking of the binomial coefficient as the number of ways to making a series of two-outcome decisions is crucial to the understanding of binomial distribution. + The answer can also be obtained by the multiplication principle. ) The multiplicative formula allows the definition of binomial coefficients to be extended[3] by replacing n by an arbitrary number α (negative, real, complex) or even an element of any commutative ring in which all positive integers are invertible: With this definition one has a generalization of the binomial formula (with one of the variables set to 1), which justifies still calling the n {\displaystyle k!} ) 7 C k 6 Each polynomial $\tbinom{t}{k}$ is integer-valued: it has an integer value at all integer inputs $t$. ( l A simple and rough upper bound for the sum of binomial coefficients can be obtained using the binomial theorem: which is valid by for all integers = Rekursive Darstellung und Pascalsches Dreieck, Der Binomialkoeffizient in der Kombinatorik, Summen mit alternierenden Binomialkoeffizienten, Summen von Binomialkoeffizienten mit geraden bzw. {\displaystyle {\tbinom {n}{k}}=0} k setzt. ) 0 Um unnötigen Rechenaufwand zu vermeiden, berechnet man im Fall Then it follows from our earlier definition for the sum of binomial coefficients that. ( 1\quad 4 \quad 6 \quad 4 \quad 1\\ k -Binomialsymbolen: ( , ) ⋅ Addition obiger Gleichungen ) n 1 n und {\displaystyle \operatorname {Re} z>0} . M Die wörtliche Übersetzung von „ Suppose that the customer is a meat lover. The following is one specific path that the person might take. The reasoning is based on the multiplication principle (see here). , 1 Some properties make use of symmetry, some deal with expansion, but they all can be proved rather intuitively. \binom MN = \frac{M!}{N!(M-N)!} n For each k, the polynomial {\displaystyle | {
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rather intuitively. \binom MN = \frac{M!}{N!(M-N)!} n For each k, the polynomial {\displaystyle z} {\displaystyle {\tbinom {t}{k}}} ( The Chu–Vandermonde identity, which holds for any complex-values m and n and any non-negative integer k, is, and can be found by examination of the coefficient of )^3}[/math], $\sum_{k=-a}^a(-1)^k{a+b\choose a+k} {b+c\choose b+k}{c+a\choose c+k} = \frac{(a+b+c)!}{a!\,b!\,c! One way to solve this problem is to count the number of paths in a “brute force” approach by tracing all possible paths in the diagram. , n} with the right hand side first grouping them into those which contain element n and those which don’t. über m {\displaystyle n} {\displaystyle {\tbinom {\alpha }{k}}} [math]\sum_{k=0}^n k \binom n k = n 2^{n-1}$. = k k x k ( is convergent for k ≥ 2. α ) , ϵ The left side counts the number of ways of selecting a subset of [n] = {1, 2, ..., n} with at least q elements, and marking q elements among those selected. 1 Template:Planetmath 1 ) gilt: Beweis: ( bezeichnet. (5−3)!3!=10.\binom{5}{3} = \frac{5!}{(5-3)! − | {\displaystyle M} , , [/math], that is clear since the RHS is a term of the exponential series [math] e^k=\sum_{j=0}^\infty k^j/j! of binomial coefficients. ∞ Möglichkeiten der Wahl des ersten Tupel-Elements. ∞ In ordering a pizza, a customer can choose from a list of 10 toppings: mushroom, onion, olive, bell pepper, pineapple, spinach, extra cheese, sausage, ham, and pepperoni. {\displaystyle -z,-s\notin \mathbb {N} } − = 5040. {\displaystyle m\leq n} }\\ , 5 k This formula is used in the analysis of the German tank problem. ) 1 {\displaystyle \sum _{k=0}^{[{\frac {n-1}{2}}]}{\binom {n}{2k+1}}=2^{n-1}} Assume that all the paths from any point to any point in the above diagram are available for walking. This gives. roten und 2 terms in this product is 0 {\color{blue}1 \qquad 3 \qquad 3 \qquad 1} \\ is a natural number for any natural numbers n and k. There are many other combinatorial interpretations of binomial | {
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for any natural numbers n and k. There are many other combinatorial interpretations of binomial coefficients (counting problems for which the answer is given by a binomial coefficient expression), for instance the number of words formed of n bits (digits 0 or 1) whose sum is k is given by + This article incorporates material from the following PlanetMath articles, which are licensed under the Creative Commons Attribution/Share-Alike License: Binomial Coefficient, Upper and lower bounds to binomial coefficient, Binomial coefficient is an integer, Generalized binomial coefficients. The point 1 is known to be in each -subset of the first type and the other points must be chosen from objects. ) = {\displaystyle n} A combinatorial interpretation of this formula is as follows: when forming a subset of elements (from a set of size ), it is equivalent to consider the number of ways you can pick elements and the number of ways you can exclude elements. Now, if you know your stuff about triangular numbers, you can say that. □\displaystyle { y }^{ 3 }{ a \choose 3 } = { 3 }^{ 3 }{ 10 \choose 3 } = 3240.\ _\squarey3(3a)=33(310)=3240. {\displaystyle k=m=n} und This is the statement that each row of Pascal's triangle is symmetric in either one of two ways. { (x+y) }^{ a }=\sum _{ i=0 }^{ a }{ a \choose i } { x }^{ a-i }{ y }^{ i }.(x+y)a=i=0∑a(ia)xa−iyi. n | {
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# How many bit strings?
How many bit strings of length $8$ have either exactly two $1$-bit among the first $4$ bits or exactly two $1$-bit among the last $4$ bits?
My solution:
A bit only contains $0$ and $1$, so $2$ different numbers, i.e., $0$ and $1$. For the first part we have $2^6=64$ ways. Similar for the other way. Hence there exists $2^4=16$ bit strings. Is my answer true?
Update: I mean $2^6+2^6-2^4=112$ bit strings
I lost your logic. By symmetry, amount of strings with exactly 2 ones in the first four (call this group $F$) is identical to the ones with exactly 2 ones in the last four (call this group $L$).
Then your desired amount is $|F| + |L| - |F \cap L|$.
To compute $F$, note that you have exactly $\binom{4}{2} = 6$ ways to pick the location of the ones in the first four, which fixes your picks to be ones and the other two of the first four to be zeros. In other words, this fixes the first four bits, and the rest can be manipulated in $2^4=16$ ways.
Can you finish computing $|F|$? We already said $|L|=|F|$ and can you compute $|F \cap L|$ by a similar technique? | {
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• I still belive that it should be $2^6+2^6-2^4=112$ ways.
– user530832
Feb 14, 2018 at 16:40
• @M.Rasmussen your mistake is not automatically setting the other 2 bits to zero. The way you compute, consider, for example, the first group, with exactly 2 ones in the first four bits. $2^6$ you are getting counts the number of bit strings with 6 places, but (a) you did not count that there are multiple ways in spreading the ones among the first four and (b) you are including the string 11111111 as legal, whereas you must have exactly 2 ones in the first four, and this string has 4. Feb 14, 2018 at 17:25
• Ohh okay. I see, so all other "empty" spaces must be 0 if you know what I mean.
– user530832
Feb 14, 2018 at 17:49
• Update: I get 156 now. But to make sure I understand it, what if I change the problem to this: "How many bit strings of length $12$ have either exactly four $1$-bit among the first $6$ bits or exactly four $1$-bit among the last $6$ bits?" Then I get the answer: 122655 is that correct? I did the same proces like the original question.
– user530832
Feb 14, 2018 at 17:57
• @M.Rasmussen Then both elementary groups are $|F| = |L| = \binom{6}{4} \cdot 2^6$ and their intersection is $|F \cap L| = \binom{6}{4} \binom{6}{4}$ so you end up with $$|F| + |L| - |F\cap L| = 2 \cdot \binom{6}{4} \cdot 2^6 - \binom{6}{4}^2 = 960 - 225 = 715.$$ Feb 14, 2018 at 21:45
Let $A$ be the set of bit strings with exactly two $1$-bit among the first $4$ bits, and $B$ be the set of bit strings with exactly two $1$-bit among the last $4$ bits.
\begin{align} \#A &= \binom{4}{2} 2^4 = 6\cdot2^4 \\ \#B &= 2^4 \binom{4}{2} = 6\cdot2^4 \\ \#A\cap B &= \binom{4}{2}^2 = 6^2 \\ \#A\cup B &= \#A + \#B - \# A \cap B \\ &= 6 (2^4 \cdot 2 - 6) \\ &= 6 \cdot 26 = 156 \end{align}
Among the first $4$ bits, choose $2$ to set them to one and the other two would be set to $0$ and there are $4$ of them with absolute freedom.
$$\binom{4}{2}\cdot 2^4$$
Similar when we focus on the last $4$ bits. | {
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$$\binom{4}{2}\cdot 2^4$$
Similar when we focus on the last $4$ bits.
When we focus on intersection. We would pick $2$ from the first $4$ and pick $2$ from the last $4$.
So my overall answer would be
$$2\binom42 \cdot 2^4 - \binom42^2$$
Remark: I think your mistake is thinking that you can set arbitary $6$ bits to anything.
• I see many of you get 156, but I just think on Inclusion–exclusion principle and the way I did it here was drawing the problem: i.stack.imgur.com/jYOXA.png
– user530832
Feb 14, 2018 at 16:38
• The keyword is exactly $2$ ones in the first $4$ bits, which means exactly $2$ ones and exactly $2$ zeros in the first $4$ bits. Feb 14, 2018 at 16:59
• Update: I get 156 now. But to make sure I understand it, what if I change the problem to this: "How many bit strings of length 12 have either exactly four 1-bit among the first 6 bits or exactly four 1-bit among the last 6 bits?" Then I get the answer: 122655 is that correct? I did the same proces like the original question
– user530832
Feb 14, 2018 at 17:58
• @M.Rasmussen: There are not even that many 12 bit strings period. There are only 4096 strings of length 12 so how could there possibly be 122655 of them that have some property? You need to get in the habit of checking your answers for reasonableness. Feb 14, 2018 at 20:55
• @M.Rasmussen would you like to show the working of how you obtain that number? Feb 14, 2018 at 21:58
How many bit strings of length 8 have either exactly two 1-bit among the first 4 bits or exactly two 1-bit among the last 4 bits?
The simplest solution is the best. There are only 256 possibilities to check so just list them! | {
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The simplest solution is the best. There are only 256 possibilities to check so just list them!
00000011 00000101 00000110 00001001 00001010 00001100 00010011 00010101
00010110 00011001 00011010 00011100 00100011 00100101 00100110 00101001
00101010 00101100 00110000 00110001 00110010 00110011 00110100 00110101
00110110 00110111 00111000 00111001 00111010 00111011 00111100 00111101
00111110 00111111 01000011 01000101 01000110 01001001 01001010 01001100
01010000 01010001 01010010 01010011 01010100 01010101 01010110 01010111
01011000 01011001 01011010 01011011 01011100 01011101 01011110 01011111
01100000 01100001 01100010 01100011 01100100 01100101 01100110 01100111
01101000 01101001 01101010 01101011 01101100 01101101 01101110 01101111
01110011 01110101 01110110 01111001 01111010 01111100 10000011 10000101
10000110 10001001 10001010 10001100 10010000 10010001 10010010 10010011
10010100 10010101 10010110 10010111 10011000 10011001 10011010 10011011
10011100 10011101 10011110 10011111 10100000 10100001 10100010 10100011
10100100 10100101 10100110 10100111 10101000 10101001 10101010 10101011
10101100 10101101 10101110 10101111 10110011 10110101 10110110 10111001
10111010 10111100 11000000 11000001 11000010 11000011 11000100 11000101
11000110 11000111 11001000 11001001 11001010 11001011 11001100 11001101
11001110 11001111 11010011 11010101 11010110 11011001 11011010 11011100
11100011 11100101 11100110 11101001 11101010 11101100 11110011 11110101
11110110 11111001 11111010 11111100
There are 156 such strings.
Do you still believe there are only 112? If so, then which ones did I either list twice, or list in error?
• I am convinced, thank you
– user530832
Feb 15, 2018 at 13:13
As a programmer (and a relatively naive mathematician) I immediately thought of reducing the 8-bit string to a hexadecimal string. Two digits, each made of four bits exactly as the problem is divided.
And, out 16 digits (0-F), 6 of them are valid (exactly 2 1-bits) digits. | {
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And, out 16 digits (0-F), 6 of them are valid (exactly 2 1-bits) digits.
So, we have 16×6 valid numbers (the first digit can be anything) and out of the the remaining 16×10 numbers, 6×10 of those will also be valid.
Now we have 16×6 + 6×10 = 156 | {
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# GCF of 28 and 32
The gcf of 28 and 32 is the largest positive integer that divides the numbers 28 and 32 without a remainder. Spelled out, it is the greatest common factor of 28 and 32. Here you can find the gcf of 28 and 32, along with a total of three methods for computing it. In addition, we have a calculator you should check out. Not only can it determine the gcf of 28 and 32, but also that of three or more integers including twenty-eight and thirty-two for example. Keep reading to learn everything about the gcf (28,32) and the terms related to it.
## What is the GCF of 28 and 32
If you just want to know what is the greatest common factor of 28 and 32, it is 4. Usually, this is written as
gcf(28,32) = 4
The gcf of 28 and 32 can be obtained like this:
• The factors of 28 are 28, 14, 7, 4, 2, 1.
• The factors of 32 are 32, 16, 8, 4, 2, 1.
• The common factors of 28 and 32 are 4, 2, 1, intersecting the two sets above.
• In the intersection factors of 28 ∩ factors of 32 the greatest element is 4.
• Therefore, the greatest common factor of 28 and 32 is 4.
Taking the above into account you also know how to find all the common factors of 28 and 32, not just the greatest. In the next section we show you how to calculate the gcf of twenty-eight and thirty-two by means of two more methods.
## How to find the GCF of 28 and 32
The greatest common factor of 28 and 32 can be computed by using the least common multiple aka lcm of 28 and 32. This is the easiest approach:
gcf (28,32) = $\frac{28 \times 32}{lcm(28,32)} = \frac{896}{224}$ = 4
Alternatively, the gcf of 28 and 32 can be found using the prime factorization of 28 and 32:
• The prime factorization of 28 is: 2 x 2 x 7
• The prime factorization of 32 is: 2 x 2 x 2 x 2 x 2
• The prime factors and multiplicities 28 and 32 have in common are: 2 x 2
• 2 x 2 is the gcf of 28 and 32
• gcf(28,32) = 4 | {
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In any case, the easiest way to compute the gcf of two numbers like 28 and 32 is by using our calculator below. Note that it can also compute the gcf of more than two numbers, separated by a comma. For example, enter 28,32. The calculation is conducted automatically.
The gcf is...
Frequently searched terms on our site also include:
## Use of GCF of 28 and 32
What is the greatest common factor of 28 and 32 used for? Answer: It is helpful for reducing fractions like 28 / 32. Just divide the nominator as well as the denominator by the gcf (28,32) to reduce the fraction to lowest terms.
$\frac{28}{32} = \frac{\frac{28}{4}}{\frac{32}{4}} = \frac{7}{8}$.
## Properties of GCF of 28 and 32
The most important properties of the gcf(28,32) are:
• Commutative property: gcf(28,32) = gcf(32,28)
• Associative property: gcf(28,32,n) = gcf(gcf(32,28),n) $\hspace{10px}n\hspace{3px}\epsilon\hspace{3px}\mathbb{Z}$
The associativity is particularly useful to get the gcf of three or more numbers; our calculator makes use of it.
To sum up, the gcf of 28 and 32 is 4. In common notation: gcf (28,32) = 4.
If you have been searching for gcf 28 and 32 or gcf 28 32 then you have come to the correct page, too. The same is the true if you typed gcf for 28 and 32 in your favorite search engine.
Note that you can find the greatest common factor of many integer pairs including twenty-eight / thirty-two by using the the search form in the sidebar of this page.
Questions and comments related to the gcf of 28 and 32 are really appreciated. Use the form below or send us a mail to get in touch.
Please hit the sharing buttons if our article about the greatest common factor of 28 and 32 has been useful to you, and make sure to bookmark our site.
Posted in Greatest Common Factor
###### 2 comments on “GCF of 28 and 32”
1. POOFA says:
WOW! THIS IS A GREAT ANSWER! THANK YOU! 🙂
2. POOFA says:
WOW! THIS IS A GREAT ANSWER! THANK YOU! 🙂
## Related pages | {
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144 a prime numberprime factorization for 34prime factors of 252the prime factorization of 245prime factorization 110is 63 a prime or composite numberprime factorization of 297greatest common factor of 42 and 63prime factorization of 111common denominator of 5 and 10 | {
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# Seeking methods to solve $\int_{0}^{\frac{\pi}{2}} \ln\left|2 + \tan^2(x) \right| \:dx$
As part of going through a set of definite integrals that are solvable using the Feynman Trick, I am now solving the following:
$$\int_{0}^{\frac{\pi}{2}} \ln\left|2 + \tan^2(x) \right| \:dx$$
I'm seeking methods using the Feynman Trick (or any method for that matter) that can be used to solve this definite integral.
If one wishes to use "Feynman's Trick," then begin by defining a function $$I(a)$$, $$a>1$$ as given by
$$I(a)=\int_0^{\pi/2}\log(a+\tan^2(x))\,dx \tag1$$
Differentiation of $$(1)$$ reveals
\begin{align} I'(a)&=\int_0^{\pi/2} \frac{1}{a+\tan^2(x)}\,dx\\\\ &=\frac{\pi/2}{a-1}-\frac{\pi/2}{\sqrt a (a-1)}\tag2 \end{align}
Integration of $$(2)$$ yields
\begin{align} I(a)&=\frac\pi2\left(\log(a-1)+\log\left(\frac{\sqrt a+1}{\sqrt{a}-1}\right) \right)\\\\ &=\pi \log(\sqrt a+1)\tag3 \end{align}
Finally, setting $$a=2$$ in $$(3)$$, we obtain the coveted result
$$\int_0^{\pi/2}\log(2+\tan^2(x))\,dx=\pi \log(\sqrt 2+1)$$
• How did you resolve the constant of Integration in Step (3)? – user150203 Nov 21 '18 at 4:38
• @davidg Apology for the sign error. Note $I(0)=0$. – Mark Viola Nov 21 '18 at 5:08
• Hi Mark ! Long time no speak. Nice solution $\to +1$. Cheers. – Claude Leibovici Nov 21 '18 at 6:22
• @MarkViola do you know of any other ‘tricks that would work? – user150203 Nov 21 '18 at 6:51
• Hi David. You could try writing the logarithm as $\log(1+\cos^2(x))-\log(\sin^2(x))$ and expanding the first term as $\sum_{n=1}^\infty \frac{(-1)^{n-1}\cos^{2n}(x)}{n}$ and proceeding. I haven't tried this, but it might be worth pursuing. – Mark Viola Nov 21 '18 at 15:47
My approach
Let | {
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My approach
Let
\begin{align} \int_{0}^{\frac{\pi}{2}} \ln\left|2 + \tan^2(x) \right| \:dx &= \int_{0}^{\frac{\pi}{2}} \ln\left|1 + \left(1 + \tan^2(x)\right) \right| \:dx \\ &= \int_{0}^{\frac{\pi}{2}} \ln\left|1 + \sec^2(x) \right| \:dx \\ &= \int_{0}^{\frac{\pi}{2}} \ln\left|\frac{\cos^2(x) + 1}{\cos^2(x)} \right| \:dx \\ &= \int_{0}^{\frac{\pi}{2}} \left[ \ln\left|\cos^2(x) + 1 \right| - \ln\left|\cos^2(x)\right| \right]\:dx \\ &= \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x) + 1 \right|\:dx - \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x)\right|\:dx \end{align}
Now
$$\int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x)\right|\:dx = 2\int_{0}^{\frac{\pi}{2}} \ln\left|\cos(x)\right|\:dx = 2\cdot-\frac{\pi}{2}\ln(2) = -\pi \ln(2)$$
For detail on this definite integral, see guidance here
We now need to solve
$$\int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x) + 1 \right|\:dx$$
Here, Let
$$I(t) = \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x) + t \right|\:dx$$
Thus,
$$\frac{dI}{dt} = \int_{0}^{\frac{\pi}{2}} \frac{1}{\cos^2(x) + t}\:dx = \int_{0}^{\frac{\pi}{2}} \frac{1}{\frac{\cos(2x) + 1}{2} + t}\:dx = 2\int_{0}^{\frac{\pi}{2}} \frac{1}{\cos(2x) + 2t + 1}\:dx$$
Employ a change of variable $$u = 2x$$:
$$\frac{dI}{dt} = \int_{0}^{\pi} \frac{1}{\cos(u) + 2t + 1}\:du$$
Employ the Weierstrass substitution $$\omega = \tan\left(\frac{u}{2} \right)$$:
\begin{align} \frac{dI}{dt} &= \int_{0}^{\infty} \frac{1}{\frac{1 - \omega^2}{1 + \omega^2} + 2t + 1}\:\frac{2}{1 + \omega^2}\cdot d\omega \\ &= \int_{0}^{\infty} \frac{1}{t\omega^2 + t + 1} \:d\omega \\ &= \frac{1}{t}\int_{0}^{\infty} \frac{1}{\omega^2 + \frac{t + 1}{t}} \:d\omega \\ &= \frac{1}{t}\left[\frac{1}{\sqrt{\frac{t+1}{t}}}\arctan\left( \frac{\omega}{\sqrt{\frac{t+1}{t}}}\right)\right]_{0}^{\infty} \\ &= \frac{1}{t}\frac{1}{\sqrt{\frac{t+1}{t}}}\frac{\pi}{2} \\ &= \frac{1}{\sqrt{t}\sqrt{t + 1}}\frac{\pi}{2} \end{align}
And so, | {
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And so,
$$I(t) = \int \frac{1}{\sqrt{t}\sqrt{t + 1}}\frac{\pi}{2}\:dt = \pi\ln\left| \sqrt{t} + \sqrt{t + 1}\right| + C$$
Now
$$I(0) = \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x) + 0 \right|\:dx = -\pi \ln(2) = \pi\ln\left|\sqrt{0} + \sqrt{0 + 1} \right| + C \rightarrow C = -\pi \ln(2)$$
And so,
$$I(t) = \pi\ln\left| \sqrt{t} + \sqrt{t + 1}\right| -\pi \ln(2) = \pi\ln\left|\frac{\sqrt{t} + \sqrt{t + 1}}{2} \right|$$
Thus,
$$I = I(1) = \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x) + 1 \right|\:dx = \pi\ln\left|\frac{\sqrt{1} + \sqrt{1 + 1}}{2} \right| = \pi\ln\left|\frac{1 + \sqrt{2}}{2} \right|$$
And Finally
\begin{align} \int_{0}^{\frac{\pi}{2}} \ln\left|2 + \tan^2(x) \right| \:dx &= \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x) + 1 \right|\:dx - \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x)\right|\:dx \\ &= \pi\ln\left|\frac{1 + \sqrt{2}}{2} \right| - \left(-\pi \ln(2)\right) \\ &= \pi\ln\left|1 + \sqrt{2} \right| \end{align}
• Your question and your answer have a gap of just 1 minute. If you did solve the question earlier then you must've included your approach in the question itself. What is the need to post it as an answer? – Rohan Shinde Nov 21 '18 at 3:56
• @Digamma - Sorry, although I've been a member of Math StackExchange for 4 years, I've only become regularly active recently. When I asked a friend who is a member as to what is the best way of presenting a solution whilst asking for other solutions I was told this is the approach taken within the community. If this violate any of the rules, please advise and I will reposition accordingly. Also, I never assign my solution as 'the solution' to any post of this nature. – user150203 Nov 21 '18 at 4:00
• @Digamma The site encourages people to share the questions with answers. – Tianlalu Nov 21 '18 at 4:03
• @Tianlalu - Sorry, just to be clear - Have I employed the correct process here? – user150203 Nov 21 '18 at 4:10
• Thanks @mick - much appreciated :-) – user150203 Dec 3 '18 at 3:42 | {
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# If f(x)= |4x - 1| + |x-3| + |x + 1| (Question of the Day-II)
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If f(x)= |4x - 1| + |x-3| + |x + 1| (Question of the Day-II) [#permalink]
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If f(x) = |4x - 1| + |x-3| + |x + 1|, what is the minimum value of f(x)?
(A) 3
(B) 4
(C) 5
(D) 21/4
(E) 7
(Still high on mods! Next week, will make questions on some other topic.)
[Reveal] Spoiler: OA
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Last edited by Bunuel on 06 Jul 2013, 02:33, edited 1 time in total. Added the OA. CEO Status: Nothing comes easy: neither do I want. Joined: 12 Oct 2009 Posts: 2783 Location: Malaysia Concentration: Technology, Entrepreneurship Schools: ISB '15 (M) GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35 Re: Question of the Day - II [#permalink] ### Show Tags 29 Oct 2010, 18:51 7 This post received KUDOS VeritasPrepKarishma wrote: Q. If f(x) = |4x - 1| + |x-3| + |x + 1|, what is the minimum value of f(x)? (A) 3 (B) 4 (C) 5 (D) 21/4 (E) 7 (Still high on mods! Next week, will make questions on some other topic.) put x=0 we get f(x) = 5..rule out D and E put x=1/4 we get f(x) = 0 + 11/4 + 5/4 = 16/4 = 4.. rule out C D and E Now the answer is either 3 or 4. Reason for above checking of values: for every value of x> 3 the f(x) is quite big because of 4x-1 for every value of x < -1 the f(x) if bigger than 3 and 4. for x >1/4 and x< 3 f(x) is bigger than 4. because f(x) = 4x-1 + 3-x + x+1 = 4x+3 > 3 => only 4 is the probable answer. Thus we only need to check x>=-1 and x<= 1/4 f(x) in this domain is = 1-4x + 3-x + x+1 = 5-4x => for f(x) to be minimum the x should be +ve => for x = 1/4 , f(x) = 4. This can be solved using graph as well by plotting the f(x) in different domains. _________________ Fight for your dreams :For all those who fear from Verbal- lets give it a fight Money Saved is the Money Earned Jo Bole So Nihaal , Sat Shri Akaal Support GMAT Club by putting a GMAT Club badge on your blog/Facebook GMAT Club Premium Membership - big benefits and savings Gmat test review : http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html Retired Moderator Joined: 02 Sep 2010 Posts: 803 Location: London Re: Question of the Day - II [#permalink] ### Show Tags 30 Oct 2010, 01:59 10 This post received KUDOS 1 This post was BOOKMARKED Let g(x) = |4x - 1| & h(x)=|x-3| + |x + 1| ... So | {
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post received KUDOS 1 This post was BOOKMARKED Let g(x) = |4x - 1| & h(x)=|x-3| + |x + 1| ... So f(x)=g(x)+h(x) Now h(x) is equal to 4 between -1 and 3 and is higher everywhere else, so it minimizes between -1 & 3 g(x) is minimum at x=(1/4) where it is equal to 0. Since -1<(1/4)<3 & f(x)=g(x)+h(x) ... It is straight forward to imply that f(x) will be minimum at 1/4, since both g(x) & h(x) are minimum at that point f(1/4)=g(1/4)+h(1/4)=0+4 Hence answer is 4 _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7450 Location: Pune, India Re: Question of the Day - II [#permalink] ### Show Tags 30 Oct 2010, 08:53 45 This post received KUDOS Expert's post 13 This post was BOOKMARKED That's right! I can count on you guys to always give the correct answer and some innovative ways of solving... Let me give you the method I would use to solve it too... Where mods are concerned, the fact that mod signifies the distance from 0 on the number line is my best friend... So |x-3| is the distance from 3, |x + 1| is the distance from -1 and |4x - 1| is 4 times the distance from 1/4 So |4x - 1| + |x-3| + |x + 1| is the sum of distances from 3, -1 and four times the distance from 1/4 e.g. if x takes the value at point A, f(x) will be sum of length of red line, green line and blue line. the question here is, what is the minimum such sum possible? Attachment: Ques.jpg [ 4.56 KiB | Viewed 14954 times ] Can I say that minimum total distance will be covered from point 1/4? Attachment: Ques1.jpg [ 3.43 KiB | Viewed 14937 times ] The logic being that the distance from 3 to -1, which is 4 units has to be covered. Why to cover any distance from 1/4 at all? _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199 | {
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Re: Question of the Day - II [#permalink]
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In other words, as my mentor says, assume there is one guy on point 3, one on point -1 and 4 guys on point 1/4. If they have to meet up, but cover minimum distance, they should meet at point 1/4. What happens when there are 4 such points? Lets add another term (2x - 3) to f(x)...
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 11 Jul 2010 Posts: 224 Re: Question of the Day - II [#permalink] ### Show Tags 30 Oct 2010, 09:29 would the answer still remain 1/4? based on your 'distance' method? The 3 guys standing at the new 3/2 post + the existing persons at -1 and 3 can again meet at 1/4 as that would be the shortest distance Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7450 Location: Pune, India Re: Question of the Day - II [#permalink] ### Show Tags 30 Oct 2010, 16:17 3 This post received KUDOS Expert's post 3 This post was BOOKMARKED gmat1011 wrote: would the answer still remain 1/4? based on your 'distance' method? The 3 guys standing at the new 3/2 post + the existing persons at -1 and 3 can again meet at 1/4 as that would be the shortest distance Yes, that is right! The answer still remains 1/4. Since |2x - 3| = 2|x - 3/2| it is twice the distance from 3/2 (or we can say, there are 2 guys are 3/2). The guy at -1 and 3 still need to cover 4 units together. If the 2 guys at 3/2 come down to 1/4, they would have covered less distance than if 4 guys were made to travel anywhere from 1/4. Attachment: Ques.jpg [ 5.2 KiB | Viewed 14965 times ] Try some other combinations. e.g. f(x) = |x - 1| + |x-3| + |x + 1| + |x + 6| f(x) = |2x - 3| + |4x + 7| etc _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199
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Re: Question of the Day - II [#permalink]
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09 Nov 2010, 23:23
Great Method Karishma !!
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Re: Question of the Day - II [#permalink]
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14 Jun 2011, 02:14
for x = 0, f(x) = 5. POE options C,D and E.
for x = 1/4, f(x) = 4.
B it is.
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Re: Question of the Day - II [#permalink]
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Very good question thanks for posting
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Re: Question of the Day - II [#permalink]
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15 Mar 2012, 11:07
VeritasPrepKarishma wrote:
gmat1011 wrote:
The 3 guys standing at the new 3/2 post + the existing persons at -1 and 3 can again meet at 1/4 as that would be the shortest distance
Yes, that is right! The answer still remains 1/4.
Since |2x - 3| = 2|x - 3/2| it is twice the distance from 3/2 (or we can say, there are 2 guys are 3/2). The guy at -1 and 3 still need to cover 4 units together. If the 2 guys at 3/2 come down to 1/4, they would have covered less distance than if 4 guys were made to travel anywhere from 1/4.
Attachment:
Ques.jpg
Try some other combinations. e.g. f(x) = |x - 1| + |x-3| + |x + 1| + |x + 6|
f(x) = |2x - 3| + |4x + 7| etc
Karishman, how are you deciding that distance for 2 guys to travel from 3/2 would be less than distance by 4 guys to move from 1/4 ?
In example : |x - 1| + |x-3| + |x + 1| + |x + 6| .. the posts on the number line are : -6, -1 , 1 and 3. In order to minimize value of this expression to 9, how to chose the x ? | {
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Please correct me if I am wrong.
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Re: Question of the Day - II [#permalink]
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Karishman, how are you deciding that distance for 2 guys to travel from 3/2 would be less than distance by 4 guys to move from 1/4 ?
In example : |x - 1| + |x-3| + |x + 1| + |x + 6| .. the posts on the number line are : -6, -1 , 1 and 3. In order to minimize value of this expression to 9, how to chose the x ?
Please correct me if I am wrong.
Forget these numbers. Think logically.
My house is 10 miles away from your house. If we have to meet up, how much distance do we need to cover together? In any case, we need to cover 10 miles together at least, right? Either you come down to my place (you cover 10 miles) or I come down to yours (I cover 10 miles) or we meet mid way (10 miles covered together) or we meet up at a nice coffee place 2 miles further down from my house in the opposite direction in which case we will need to cover more than 10 miles (i.e. we cover 2 + 12 = 14 miles)
Now say, another friend is at my place. In which case will people cover minimum distance together? If two of us come down to your place, we cover 10+10 = 20 miles together but if you come down to our place, you cover only 10 miles. If instead, we meet midway, we cover 5+5 and you cover 5 miles so in all 15 miles. So less number of people should travel the entire distance.
If there are 4 people at point A and 2 at point B, minimum distance will be covered if people at point B travel to point A. So people at 3/2 should come down to 1/4.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7450 Location: Pune, India Re: Question of the Day - II [#permalink] ### Show Tags 15 Mar 2012, 11:33 5 This post received KUDOS Expert's post 2 This post was BOOKMARKED ficklehead wrote: In example : |x - 1| + |x-3| + |x + 1| + |x + 6| .. the posts on the number line are : -6, -1 , 1 and 3. In order to minimize value of this expression to 9, how to chose the x ? Please correct me if I am wrong. x is that point on the number line whose sum of distances from -6, -1, 1 and 3 is minimum. So basically there is a person each at points -6, -1, 1 and 3. You need to make them all meet by covering minimum distance. Distance between -6 and 3 is 9 which must be covered by these 2 people to meet. These 2 can meet at any point: -6, -1, 0, 1 or 3 etc they will cover a distance of 9 together. If -1 and 1 have to meet too, they need to cover a distance of 2 together. Say, if person at -1 travels down to 1 and -6 and 3 also meet at 1, the minimum distance covered will be 9+2 = 11 and they will all be able to meet. If they instead meet at -1, the situation will be the same and total distance covered will be 11 again. In fact, they can meet at any point between -1 and 1, the total distance covered will be 11. To check, put x = 1. you get |x - 1| + |x-3| + |x + 1| + |x + 6| = 11 put x = -1, you get |x - 1| + |x-3| + |x + 1| + |x + 6| = 11 put x = 0, you get |x - 1| + |x-3| + |x + 1| + |x + 6| = 11 _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199
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Re: Question of the Day - II [#permalink]
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15 Mar 2012, 14:48
VeritasPrepKarishma wrote:
In example : |x - 1| + |x-3| + |x + 1| + |x + 6| .. the posts on the number line are : -6, -1 , 1 and 3. In order to minimize value of this expression to 9, how to chose the x ?
Please correct me if I am wrong.
x is that point on the number line whose sum of distances from -6, -1, 1 and 3 is minimum. So basically there is a person each at points -6, -1, 1 and 3. You need to make them all meet by covering minimum distance.
Distance between -6 and 3 is 9 which must be covered by these 2 people to meet. These 2 can meet at any point: -6, -1, 0, 1 or 3 etc they will cover a distance of 9 together.
If -1 and 1 have to meet too, they need to cover a distance of 2 together. Say, if person at -1 travels down to 1 and -6 and 3 also meet at 1, the minimum distance covered will be 9+2 = 11 and they will all be able to meet.
If they instead meet at -1, the situation will be the same and total distance covered will be 11 again. In fact, they can meet at any point between -1 and 1, the total distance covered will be 11.
To check, put x = 1. you get |x - 1| + |x-3| + |x + 1| + |x + 6| = 11
put x = -1, you get |x - 1| + |x-3| + |x + 1| + |x + 6| = 11
put x = 0, you get |x - 1| + |x-3| + |x + 1| + |x + 6| = 11
Thanks Karishma for this detailed explanation.
I got it now.
I was getting 11 as a distance but was not sure, if I should try other values of x to check if there could be a lower value than 11.
To seal the concept, for |2x-3|+|4x+7|, minimum distance be : 3/2+7/4=13/4 ?
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Re: Question of the Day - II [#permalink]
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I was getting 11 as a distance but was not sure, if I should try other values of x to check if there could be a lower value than 11. | {
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To seal the concept, for |2x-3|+|4x+7|, minimum distance be : 3/2+7/4=13/4 ?
Look at the diagram below. Make a number line in such questions.
Attachment:
Ques3.jpg [ 5.1 KiB | Viewed 13919 times ]
For x and y to meet, they have to cover a distance of 9 together. For p and q to meet, they have to cover a distance of 2 together. They can meet anywhere between -1 and 1 and they will cover a total distance of 11 only. So x can take any value -1 < x < 1 and the value of the expression will be 11.
|2x-3|+|4x+7| = 2|x-3/2| + 4|x+7/4|
Attachment:
Ques4.jpg [ 5.35 KiB | Viewed 13934 times ]
There are 4 people at -7/4 and 2 people at 3/2. Distance between the two points is 7/4 + 3/2 = 13/4
For these people to meet covering the minimum distance, the 2 people X and Y should travel to point -7/4. (Make minimum people travel). So minimum distance that needs to be covered = 2*13/4 = 13/2 (because 2 people travel 13/4 each) which is the minimum value of the expression. The value of x when the expression takes minimum value is -7/4.
Check by putting x = -7/4. You get |2x-3|+|4x+7| = 13/2
Also see that when you put x = 0 or 3/2 etc, the value of the expression is higher.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 24 Jul 2011 Posts: 13 Re: Question of the Day - II [#permalink] ### Show Tags 16 Mar 2012, 09:01 Thanks a lot, Karishma. Intern Joined: 24 Jul 2011 Posts: 13 Re: Question of the Day - II [#permalink] ### Show Tags 16 Mar 2012, 19:48 I am wondering how can this method be used in questions where there are negative between terms : Ex: minimum value of : |x+6|-|x-1| ? Manager Joined: 28 Jul 2011 Posts: 238 Re: Question of the Day - II [#permalink] ### Show Tags 20 Mar 2012, 14:33 I got C = 5 f(x) = |4x-1| + |x-3| + |x+1| =(4x-1) + (x-3) + (x+1) or = - (4x-1) - (x-3) - (x+1) so, x= 1/2 then i used the value of x=1/2 f(x) = |4x-1| + |x-3| + |x+1| f(1/2) = |4(1/2)-1| + |(1/2)-3| + |(1/2)+1| = |1| + |-5/2| + |3/2| = 1 + 5/2 + 3/2 = 5 Is that correct? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7450 Location: Pune, India Re: Question of the Day - II [#permalink] ### Show Tags 20 Mar 2012, 21:07 kuttingchai wrote: I got C = 5 f(x) = |4x-1| + |x-3| + |x+1| =(4x-1) + (x-3) + (x+1) or = - (4x-1) - (x-3) - (x+1) so, x= 1/2 then i used the value of x=1/2 f(x) = |4x-1| + |x-3| + |x+1| f(1/2) = |4(1/2)-1| + |(1/2)-3| + |(1/2)+1| = |1| + |-5/2| + |3/2| = 1 + 5/2 + 3/2 = 5 Is that correct? Put x = 1/4 and the minimum value you will get is 4. How did you get x = 1/2? I would suggest you to check out one of the approaches mentioned above. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199
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I am wondering how can this method be used in questions where there are negative between terms :
Ex: minimum value of : |x+6|-|x-1| ? | {
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