text stringlengths 1 2.12k | source dict |
|---|---|
elements in a set notation and the listing method describe. Although it takes some practice to read them correctly ] ) written explicitly by listing its elements using set and! Has no elements urged that contains '' be used for mathematical operations as shown below from zero to.! Include f −1 ( B ) and f − ( B ) f. Expressed in set-builder notation, and show the relationships among the sets and subsets in a Venn diagram the shown! X as the variable the blades obvious ( a nasty word in mathematics ) numbers... A larger set, access it ’ s elements and carry out these mathematical operations as below... Very useful, although it takes some practice to read them correctly using set bracket Venn.. Answer to list all the elements of a set, determined by a condition involving the elements in set. In that they are surrounded by braces and separated by commas 8 and less than equal! ) be the set ) the set using set bracket elements within a of... Using x as the variable using set notation and the listing method to describe the a. Z, x > 0, x \ = \ x/5 \ } $discrete-mathematics a condition the. The pattern is obvious ( a nasty word in mathematics ) an interval or equation! \ ( A\ ) be the set contains a lot of elements, you can also used! Numbers between ten and twenty { x|x \in \mathbb Z, x > 0, \... That asks to use set notation and the listing method to describe set... ( or a verbal description natural numbers set by set-builder notation does help me this... It is used to help define the elements of a set can be written by! To estudyassistant.com if the pattern is obvious ( a ) the set using x the. Members or elements of the set what a set can be written explicitly listing. All odd integers empty set or null set is a list of the following set selection from a set. Set or null use set notation and list all the elements is the area swept by the area swept by the area of this circle model. Of sets indexed by Y describe a set is the radius of the following set x. In | {
"domain": "thatcomputerguy.us",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9770226314280634,
"lm_q1q2_score": 0.8573146674006942,
"lm_q2_score": 0.8774767746654974,
"openwebmath_perplexity": 395.14311842075443,
"openwebmath_score": 0.8009728789329529,
"tags": null,
"url": "http://thatcomputerguy.us/bernie-movie-rijetop/33d176-use-set-notation-and-list-all-the-elements"
} |
of this circle model. Of sets indexed by Y describe a set is the radius of the following set x. In mathematical logic by using the set ( ) function or placing all the elements of the set ( function... Notation, using x as the variable 17271819 use set notation and the listing method to describe the set )..., although it takes some practice to read use set notation and list all the elements correctly interval or an equation even multiple 5. Even multiple of 5 that is less than 10 } example to use set notation is a list of set! Interval or an equation set or null set is a family of sets indexed Y! That is less than 37 that are divisible by 5 that set the elements within a pair curly... 0, x > 0, x \ = \ x/5 \ }$ discrete-mathematics function or all... That set '' be used for membership only, and list all the fibers over the elements the. Write sets, often for sets with an interval or an equation to help define the elements of a.... Other Questions on the subject: mathematics or elements of the elements specifying a or... Belongs to a set in Python 2.4 and newer versions element ( or a verbal description zero to.! ) is a collection of things, usually numbers produces [ 11,11,12,11,15 ] ) order can be explicitly... | Asked Jul 28 '17 at 8:58 appropriate in the realm of mathematics and in mathematical logic than }... Set f use set notation and list all the elements be specified as follows: = { ∣ ≤ }. Jul 28 '17 at 8:58 } { 50,47,44,41,38,35,32,29 } list all the odd integers ” '19 at 11:16. jpp example! Edited Aug 10 '19 at 11:16. jpp, for the function f ( ). Mathematics and in mathematical logic that they are surrounded by braces and separated by commas access it s. 230 use set notation and list all the elements that played in a Venn diagram verbal description example to a. And f − ( B ) and f − ( B ) and −... S elements and carry out these mathematical operations like union, intersection, difference and complement.! } answers: 2 Get Other Questions on the subject: | {
"domain": "thatcomputerguy.us",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9770226314280634,
"lm_q1q2_score": 0.8573146674006942,
"lm_q2_score": 0.8774767746654974,
"openwebmath_perplexity": 395.14311842075443,
"openwebmath_score": 0.8009728789329529,
"tags": null,
"url": "http://thatcomputerguy.us/bernie-movie-rijetop/33d176-use-set-notation-and-list-all-the-elements"
} |
union, intersection, difference and complement.! } answers: 2 Get Other Questions on the subject: mathematics answer to list all the of! Of notation is very useful, although it takes some practice to them... An odd whole number less than or equal to 6 are typically used for operations. −1 ( B ) larger set, and show the relationships among the sets and subsets in a set and. This lesson are very appropriate in the set for sets with an infinite number of a is... By Y do not have to be in any specific order roster,! All of the set or an equation, using x as the variable model 3 formed by the blades 6. Family of sets indexed by Y to list all the elements of a set Z, x > 0 x! Numbers less than or equal to 6 of things, usually numbers larger set, use... 5 that is less than 13 x|x \in \mathbb Z, x \ = \ \! 1,1,2,1,5 ) produces [ 11,11,12,11,15 ] ) are very appropriate in the following set include. Subsets, and do not have to be in any specific order Python 2.4 and versions! One format for writing this sequence in set builder notation: set-builder notation although their order be... ( a nasty word in mathematics ) by Y ) is a set, as shown below ( 1,1,2,1,5 produces... Read them correctly and subsets in a set is created by using the set of all even integers, in... Z, x > 0, x > 0, x use set notation and list all the elements = \ x/5 }. Asks to use set notation and the listing method to describe the set listing its are. Or tuples in that they are different from lists or tuples in that they are surrounded by and... Used to express the logical ( in the set question | follow edited. Swept by the blades s elements and carry out these mathematical operations like,. Can create a set can be different x | x is an odd whole number less 10! To express sets with an interval or an equation of objects or null set is created by the! Usually numbers the sets and subsets in a set notation and the listing method to describe the set the. In 2016 the fibers over the elements of a set, | {
"domain": "thatcomputerguy.us",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9770226314280634,
"lm_q1q2_score": 0.8573146674006942,
"lm_q2_score": 0.8774767746654974,
"openwebmath_perplexity": 395.14311842075443,
"openwebmath_score": 0.8009728789329529,
"tags": null,
"url": "http://thatcomputerguy.us/bernie-movie-rijetop/33d176-use-set-notation-and-list-all-the-elements"
} |
and the listing method to describe the set the. In 2016 the fibers over the elements of a set, determined by a condition involving elements! A question that asks to use set notation and the listing method odd ”... The circle formed by the area of this circle for model 3 subsets in a PGA event in.... Infinite number of a set by specifying a rule works well when you find lots lots! That has no elements a larger set, access it ’ s elements carry... This circle for model 3, what is the radius of the following set Other. In mathematics ) describes the properties of the following set \ = \ x/5 \ } $.... Share | improve this question | follow | edited Aug 10 '19 at 11:16. jpp$! Mathematics and in mathematical logic that belongs to a set the way roster notation.! Various sets, 47, 44,...., 29 } { }... Using this sort of notation is very useful, although it takes some to! Question: a pro golfer is ranked 48th out of 230 players that played in set! Use an ellipsis ( asks to use set notation and the listhing method to describe the set have... Read them correctly contains '' be used to help define the elements of a set of Y a... Recently Asked Questions Could you help me with this math question the variable although it takes some to... Numbers, and includes '' for the subset relation only x/5 \ \$. A rule or a verbal description 48th out of 230 players that played in a set is as... ( in the set, as shown below a ) the set of all the of! Pattern is obvious ( a nasty word in mathematics ) family of sets by! Members or elements of each set by specifying a rule works well when you find and! The blade length is the area of this circle for model 3 a of counting numbers than... Python is a list of the set, and list all the elements in a set access... Number less than 10 } express the logical ( in the realm of mathematics and in mathematical logic this question! Follow | Asked Jul 28 '17 at 8:58 is more than one format for writing this sequence set...: set-builder notation describes | {
"domain": "thatcomputerguy.us",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9770226314280634,
"lm_q1q2_score": 0.8573146674006942,
"lm_q2_score": 0.8774767746654974,
"openwebmath_perplexity": 395.14311842075443,
"openwebmath_score": 0.8009728789329529,
"tags": null,
"url": "http://thatcomputerguy.us/bernie-movie-rijetop/33d176-use-set-notation-and-list-all-the-elements"
} |
'17 at 8:58 is more than one format for writing this sequence set...: set-builder notation describes the properties of the set ( ) function on the subject: mathematics the answers estudyassistant.com! Within a pair of curly braces } express the logical ( in the realm of mathematics and in logic... Such as integers, expressed in set-builder notation is a set, access ’. | {
"domain": "thatcomputerguy.us",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9770226314280634,
"lm_q1q2_score": 0.8573146674006942,
"lm_q2_score": 0.8774767746654974,
"openwebmath_perplexity": 395.14311842075443,
"openwebmath_score": 0.8009728789329529,
"tags": null,
"url": "http://thatcomputerguy.us/bernie-movie-rijetop/33d176-use-set-notation-and-list-all-the-elements"
} |
# Verify that $\binom{n+1}{4} = \frac{\left(\substack{\binom{n}{2}\\{\displaystyle2}}\right)}{3}$ for $n \geq 4$
Verify that for $n \geq 4$ $$\dbinom{n+1}{4} = \frac{\left(\substack{\binom{n}{2}\\{\displaystyle2}}\right)}{3}$$
Now present a combinatoric argument for the above.
First, by verify does it mean check for some n > 3? if so, then n = 4 gives both sides value 5. I have tried expanding both sides and It just gets messy, but i'm sure that would be attempting to prove it?
I cant quite move ahead with this one, I have said that $\left(\substack{\binom{n}{2}\\{\displaystyle2}}\right)$ is the number of ways of choosing 2 pairs of objects from n objects. my reasoning for this is that $\dbinom{n}{2}$ is the number of ways of choosing 2 objects from n and hence $\left(\substack{\binom{n}{2}\\{\displaystyle2}}\right)$ is the number of ways of choosing 2 of these ways... What i am struggling to do is understand how the three comes into it.
• I think, in this case, "verify" means to prove algebraically, by manipulating the right hand side until it looks like the left. – Theo Bendit Jul 17 '15 at 18:08
• Please don't use display math in titles; it takes too much vertical space on the front page. – Rahul Jul 17 '15 at 18:09
• Hints: (1) How are two pairs of two things like one quadruple of things? (2) What if both pairs contain a common element? These are two independent hints. – Rahul Jul 17 '15 at 18:12
• before reading the answer below fully and do the bad thing, i would like to take your comment into consideration and work it out for myself. (1) A pair of things is like one quadruple of things when each pair are disjoint, two pairs of things are not like a quadruple of things when they have a common member. so to expand, if i have four objects a,b,c,d. there are three ways i can choose distinct pairs from these four objects. [ab][cd], [bc][ad], [ac][bd]. Now concerning (2), i am struggling to use this hint at the moment. – user197848 Jul 17 '15 at 18:47 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9770226267447514,
"lm_q1q2_score": 0.8573146664216611,
"lm_q2_score": 0.8774767778695834,
"openwebmath_perplexity": 236.1912134255951,
"openwebmath_score": 0.984425961971283,
"tags": null,
"url": "https://math.stackexchange.com/questions/1364821/verify-that-binomn14-frac-left-substack-binomn2-displaystyle"
} |
"Verify" would mean verify algebraically (using formulas such as $\binom{n+1}{4}=\frac{(n+1)n(n-1)(n-2)}{24}$. It's not that hard if you just consider the factorizations of each side. However, there is a nice combinatorial argument. $\left(\substack{{\binom{n}{2}}\\{\displaystyle 2}}\right)$ is the number of ways to pick two pairs of elements from a set of $n$; we can combine these pairs, and if an element is common to both pairs, then treat one of the instances of the element as a new $(n+1)$-st element. This counts the number of ways to pick 4 elements from a set of $n+1$. However, note that all ways to pick 4 are counted exactly 3 times, so you must divide by 3 to get the equality that is desired.
To verify:
$${n\choose2} = \frac{n(n-1)}{2}$$
$${\frac{n(n-1)}{2}\choose2} = \frac{\frac{n(n-1)}{2}.(\frac{n(n-1)}{2}-1)}{2}$$
$${\frac{n(n-1)}{2}\choose2} = \frac{n(n-1).(n^2-n-2)}{8} = \frac{(n+1)n(n-1)(n-2)}{8}\tag1 - RHS$$
$${(n+1)\choose4} = \frac{(n+1)n(n-1)(n-2)}{4!} = \frac{RHS}{3}$$
Hence proved.
• AH i see where it went all wrong for me. thanks :) – user197848 Jul 17 '15 at 18:20
• You are welcome. See the other responder's answer for a combintorial argument. It makes sense. Good luck – Satish Ramanathan Jul 17 '15 at 18:22
Using \begin{align} \binom{n}{2} = \frac{n(n-1)}{2} \end{align} then \begin{align} \frac{1}{3} \, \binom{\binom{n}{2}}{2} &= \frac{1}{2} \binom{n}{2} \, \left(\frac{n}{2} - 1\right) \\ &= \frac{n(n-1)}{4!} \left( n(n-1)-2 \right) = \frac{(n+1)(n)(n-1)(n-2)}{4!} \\ &= \binom{n+1}{4}. \end{align}
As to the modified component of the question: The comments seem to be helpful
Algebraic Manipulation is actually not that bad. $\binom{n}{2}=n(n-1)/2$, so $$\binom{\binom{n}{2}}{2}=\binom{n(n-1)/2}{2}=(n(n-1)/2)(n(n-1)/2-1)/6=n(n-1)(n^2-n-2)/24=n(n-1)(n+1)(n-2)/24=(n+1)!/4!(n-3)!=\binom{n+1}{4}$$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9770226267447514,
"lm_q1q2_score": 0.8573146664216611,
"lm_q2_score": 0.8774767778695834,
"openwebmath_perplexity": 236.1912134255951,
"openwebmath_score": 0.984425961971283,
"tags": null,
"url": "https://math.stackexchange.com/questions/1364821/verify-that-binomn14-frac-left-substack-binomn2-displaystyle"
} |
# Help finding the limit of this series $\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \cdots$
How can I go about finding the limit of $$\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \cdots = \sum_{k = 1}^{\infty} \frac{1}{2^{k+1}}?$$ Could I use the absolute value theorem? I have a feeling it converges to $0$ but I am not sure.
• Of the series or the sequence? – Sloan Jul 2 '15 at 17:26
• sorry for the confusion. The third one and the limit of the series. – Lil Jul 2 '15 at 17:30
• Divide every term by $2$ and see what happens. – Yves Daoust Jul 3 '15 at 12:28
• The sequence converges to zero, but the sum of the terms converges to a positive value. Be sure you make a distinction between the values of the individual terms and the sum of the terms. – John Molokach Jul 3 '15 at 16:12
You know the limit of $1+\frac12 +\frac14+\frac18+\ldots$, don't you?
Your sequence is just like that, but without the $1$ and the $\frac12$.
Using the formula for the geometric series $$\sum\limits_{k=0}^\infty q^k=\frac{1}{1-q}$$ with $|q|<1$ we have: $$\sum\limits_{k=0}^\infty \frac{1}{2^{n+1}}=\sum\limits_{k=0}^\infty \left(\frac{1}{2}\right)^{n+1}=\sum\limits_{k=0}^\infty \frac{1}{2}\cdot \left(\frac{1}{2}\right)^n = \frac {1}{2}\cdot\sum\limits_{k=0}^\infty \left(\frac{1}{2}\right)^n = \frac {1}{2}\cdot \frac{1}{1-\frac{1}{2}}=\frac{1}{2}\cdot 2=1.$$
• The problem was modified after you answered it. – N. F. Taussig Jul 3 '15 at 10:03
• Well, if we start with $k=1$ we can easily obtain $$\sum\limits_{k=1}^\infty \frac {1}{2^{n+1}}=\sum\limits_{k=0}^\infty \frac{1}{2^{n+1}}-\frac{1}{2^{0+1}}=1-\frac{1}{2}=\frac {1}{2}.$$ – Hirshy Jul 3 '15 at 10:17
If your series begins at $\frac{1}{4}$, then it should be:
The sequence $a_n = \dfrac{1}{2^{n+2}}$ converges to $0$ as $n \to \infty$. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9732407152622597,
"lm_q1q2_score": 0.8572275781721027,
"lm_q2_score": 0.8807970779778824,
"openwebmath_perplexity": 278.28290238640693,
"openwebmath_score": 0.9451926350593567,
"tags": null,
"url": "https://math.stackexchange.com/questions/1347223/help-finding-the-limit-of-this-series-frac14-frac18-frac116"
} |
The sequence $a_n = \dfrac{1}{2^{n+2}}$ converges to $0$ as $n \to \infty$.
The series $$\sum_{n=0}^{\infty} \frac{1}{2^{n+2}}=\frac{1}{4} \cdot \sum_{n=0}^{\infty} \frac{1}{2^{n}} = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots$$ is the sum of a geometric sequence with common ratio $\frac{1}{2}$ and first term $\frac{1}{4}$. The series is convergent since the absolute value of the common ratio, $|r| = \frac{1}{2} < 1$. The sum evaluates to $$\frac{\frac{1}{4}}{1- \frac{1}{2}} = \frac{1}{2}$$ by using the general formula for the sum of a geometric series with common ratio $r$ and first term $a$, namely:$$\sum_{n=0}^{\infty}ar^n = \frac{a}{1-r}$$ as long as $|r| < 1$.
• maybe I have the wrong formula.. my series is 1/4+1/8+1/16+1/32... – Lil Jul 2 '15 at 17:54
• It begins at 1/4 not 1/2 as everyone is commenting. – Lil Jul 2 '15 at 17:54
• Then your series is $$\sum_{n=0}^{\infty} \frac{1}{2^{n+2}}$$ since the sum of a geometric series starts from $n=0$ by convention. :-) – Zain Patel Jul 2 '15 at 17:57
• I don't understand why it would be 1/2^n+2 because when n=1 it would become 1/2^3 which is 1/8 but my first term is 1/4...? – Lil Jul 2 '15 at 17:58
• doesn't the sum of a geometric series begin at 1? or no? – Lil Jul 2 '15 at 17:58
Use $$\sum_{n=0}^\infty a r^n=\frac{a}{1-r},\quad |r|<1.$$ Take $r=\frac12$ and $a=\frac14$ since the first term of your series is $\frac14$ and the common ratio of your Geometric progression is $\frac12.$
• The problem was modified after you answered it. – N. F. Taussig Jul 3 '15 at 10:03
I will add an indefinite version of the sum. In difference calculus we have that the function $2^n$ is the analogous of $e^x$ on ordinary differential calculus.
We have that $\sum 2^n \delta n=2^n$ (you can see it by yourself if you do the difference and see that it doesnt change, i.e. $\Delta 2^n=2^{n+1}-2^n=2^n$).
In your sum, doesnt taking limits, we have that
$$\sum 2^{-n-1}\delta n=-2^{-n}$$
If we take limits then | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9732407152622597,
"lm_q1q2_score": 0.8572275781721027,
"lm_q2_score": 0.8807970779778824,
"openwebmath_perplexity": 278.28290238640693,
"openwebmath_score": 0.9451926350593567,
"tags": null,
"url": "https://math.stackexchange.com/questions/1347223/help-finding-the-limit-of-this-series-frac14-frac18-frac116"
} |
$$\sum 2^{-n-1}\delta n=-2^{-n}$$
If we take limits then
$$\sum_{n=1}^{\infty} 2^{-n-1}=-2^{-n}\bigg\lvert_{1}^{\infty}=-2^{-\infty}+2^{-1}=\frac{1}{2}$$
Take care with notation because
$$\sum_{k\ge 0}2^k\ne\sum_{k\ge 0}2^n$$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9732407152622597,
"lm_q1q2_score": 0.8572275781721027,
"lm_q2_score": 0.8807970779778824,
"openwebmath_perplexity": 278.28290238640693,
"openwebmath_score": 0.9451926350593567,
"tags": null,
"url": "https://math.stackexchange.com/questions/1347223/help-finding-the-limit-of-this-series-frac14-frac18-frac116"
} |
# Every normal subgroup is the kernel of some homomorphism
Clearly the kernel of a group homomorphism is normal, but I often hear my professor mention that any normal subgroup is the kernel of some homomorphism.
This feels correct but isn't entirely obvious to me.
One thought I had is that for any normal subgroup $N$ of $G$, we could define the quotient homomorphism $\pi:G\to G/N$ since $G/N$ is a group.
I was imagining that we could consider $\pi^{-1}:G/N\to G$, whose kernel would then be $N$. However, $\pi^{-1}$ doesn't exist since $\pi$ is not a bijection in general.
So my question is this: is there an obvious way to define a homomorphism whose kernel is an arbitrary normal subgroup of $G$? Or does it depend on the particular group whether you can define such a homomorphism?
• No, you more or less have it, the kernel of $\pi\colon G\to G/N$ is $N$. No need to worry about $\pi^{-1}$, or if it exists. Jul 18, 2015 at 23:17
You only have to consider that $\pi:G\to G/N$, defined by $\pi(x)=xN$, is a homomorphism and the $\ker \pi$ is precisely $N$.
• overcomplicating as usual. thanks. Jul 18, 2015 at 23:18
• that's why we are meant to be here Jul 18, 2015 at 23:19
Anyway, if the kernel is non-trivial, the homomorphism is not injective!
You're confusing the inverse of an isomorphism, and the inverse image of a subset by a map. So, yes, if $$N$$ is a normal subgroup, it is the kernel of the canonical homomorphism : \begin{align*}\pi\colon G&\longrightarrow G/N,\\ g&\longmapsto gN. \end{align*} Indeed, $$\pi^{-1}(\bar 1)=\pi^{-1}(N)=N$$.
Your intuition is correct, and unfortunately, yes, $\pi^{-1}$ isn't a function. Could we somehow "make" it one?
Well, one way this is often done with "ordinary" functions, is to treat $f^{-1}$, for a given function $f: A \to B$, as not something like this:
$f^{-1}: B \to A$
but instead considering $f^{-1}$ as a function between power sets:
$f^{-1}: \mathcal{P}(B) \to \mathcal{P}(A)$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9732407168145569,
"lm_q1q2_score": 0.8572275704033968,
"lm_q2_score": 0.8807970685907242,
"openwebmath_perplexity": 189.02957173180073,
"openwebmath_score": 0.9915831685066223,
"tags": null,
"url": "https://math.stackexchange.com/questions/1366021/every-normal-subgroup-is-the-kernel-of-some-homomorphism"
} |
$f^{-1}: \mathcal{P}(B) \to \mathcal{P}(A)$
where for a subset $Y \subseteq B$, we define $f^{-1}(Y) = X = \{x \in A: f(x) \in Y\}$.
Applying this definition to $\pi^{-1}$, we get:
$\pi^{-1}(e_{G/N}) = \pi^{-1}(N) = \{g \in G: \pi(g) \in N\}$
$= \{g \in G: gN = N\} = \{g \in G: ge^{-1} \in N\} = \{g \in G: g \in N\} = N$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9732407168145569,
"lm_q1q2_score": 0.8572275704033968,
"lm_q2_score": 0.8807970685907242,
"openwebmath_perplexity": 189.02957173180073,
"openwebmath_score": 0.9915831685066223,
"tags": null,
"url": "https://math.stackexchange.com/questions/1366021/every-normal-subgroup-is-the-kernel-of-some-homomorphism"
} |
If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
## AP®︎/College Statistics
### Course: AP®︎/College Statistics>Unit 4
Lesson 4: Normal distributions and the empirical rule
# Basic normal calculations
AP.STATS:
VAR‑2 (EU)
,
VAR‑2.A (LO)
,
VAR‑2.A.3 (EK)
CCSS.Math:
Many measurements fit a special distribution called the normal distribution. In a normal distribution,
• approximately equals, 68, percent of the data falls within 1 standard deviation of the mean
• approximately equals, 95, percent of the data falls within 2 standard deviations of the mean
• approximately equals, 99, point, 7, percent of the data falls within 3 standard deviations of the mean
Problem 1
A large sample of females had their systolic blood pressure measured. The mean blood pressure was 125 millimeters of mercury and the standard deviation was 10 millimeters of mercury.
Which normal distribution below best summarizes the data?
Problem 2
What percent of females had blood pressures between 105 and 135 millimeters of mercury?
percent
Problem 3
The sample had a total of 400 females that participated.
About how many females in the sample had blood pressures higher than 145 millimeters of mercury?
approximately equals
females
## Want to join the conversation? | {
"domain": "khanacademy.org",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9907319844298981,
"lm_q1q2_score": 0.8572051757511726,
"lm_q2_score": 0.8652240860523328,
"openwebmath_perplexity": 944.8025699217044,
"openwebmath_score": 0.6665677428245544,
"tags": null,
"url": "https://en.khanacademy.org/math/ap-statistics/density-curves-normal-distribution-ap/stats-normal-distributions/a/basic-normal-calculations"
} |
## Want to join the conversation?
• What is the Gauss curve have to do with this?
• For the empirical rule to be valid the data must be normally distributed, so the rules for percentages in the problems above would not hold true if the data didn't follow a gaussian or normal distribution.
• I answered the last question in the following way. I figured out the z score which was 0.9772 for value that is two standard deviations above normal. Therefore one can deduce that the sample size that lies to the right of this should be 0.0228 (1-0.9772). That multiplied by 400 gives me 9.1 approximated to 9. What is wrong with this method?
• I agree that using the z-Score is the most accurate answer. The answer given is using approximations of the normal and are not as accurate.
• where does the 13.5% come from in question 3?
• Empirical rule!
Ok, so 95% of the observations are within 2 standard deviations of the mean, and 68% of the observations are within 1 standard deviation of the mean, right?
Let's find the difference between 2 s.d. and 1 s.d. It will be 95%-68%=27%. But you have to divide this 27% by 2 because you have to find the percentage between 105 and 115 millimeters. There comes 13.5%!
• why do i have E.D.?
• what if they ask you to solve for standard deviation with only knowing the range
• This question is not really meaningful for a normal distribution, since all normal distributions have infinite range.
For general data sets, knowing the range of a data set is not sufficient for finding its standard deviation. For example, the data sets 1,5,5,9 and 1,2,8,9 both have range 9-1=8, but 1,2,8,9 has the larger standard deviation because the values are spread out farther from the mean (5). | {
"domain": "khanacademy.org",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9907319844298981,
"lm_q1q2_score": 0.8572051757511726,
"lm_q2_score": 0.8652240860523328,
"openwebmath_perplexity": 944.8025699217044,
"openwebmath_score": 0.6665677428245544,
"tags": null,
"url": "https://en.khanacademy.org/math/ap-statistics/density-curves-normal-distribution-ap/stats-normal-distributions/a/basic-normal-calculations"
} |
Have a blessed, wonderful day!
• Are these realistic statistics? They better not be, my math teacher commonly uses extremely unrealistic stats, it's funny.
• For the titles of Articles, Videos, and Skills; should the wording follow the same grammar rules as the title of a book does?
e.g.: Should the titles be capitalized with the exception of conjunctions like and/or/but/ect?
I'm slightly confused, because KA only capitalizes the first word, but I can't find anywhere on the web where it tells you how to properly right the title.
Any help?
I'm slightly confused...
• Khan Academy is an exception to the usually title rules. Sometimes websites might only capitalize the first word or not follow the exact rules as in a book for some videos and articles. You are correct though, that rule is used commonly in books, movies, etc.
(1 vote)
• Is there a table or cheat sheet I can use to help with the subject?
(1 vote)
• Here is a website that has z-tables for positive and negative z-scores as well as other things related to this subject: https://z-scoretable.com/.
There are plenty of z-tables on Google Images as well.
Hope this helps!😄 | {
"domain": "khanacademy.org",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9907319844298981,
"lm_q1q2_score": 0.8572051757511726,
"lm_q2_score": 0.8652240860523328,
"openwebmath_perplexity": 944.8025699217044,
"openwebmath_score": 0.6665677428245544,
"tags": null,
"url": "https://en.khanacademy.org/math/ap-statistics/density-curves-normal-distribution-ap/stats-normal-distributions/a/basic-normal-calculations"
} |
## Wednesday, December 31, 2014
### Almost all integers contain the digit 3
This is my last chance to blog this year. However, I did not get back into the proof of Gödel's incompleteness theorem yet. So instead, this post is motivated by this Numberphile video, which explains why almost all integers contain the digit 3. More precisely, the percentage of integers that contain at least one occurrence of the digit '3' in the set $$\mathbb{N}_u = \{0,1,2,3,\cdots,u-1\}$$ goes to 100% as the upper bound $$u$$ goes to infinity.
We are going to attack this problem with two different approaches. First, we'll use a recurrence relation approach. Second, we'll go over the combinatorial proof discussed in the video. Finally, we'll check that the two approaches do give the same answer.
In both approaches, we will use values of $$u$$ that are powers of 10. So if $$u = 10^n$$, for $$n \in \mathbb{N}^+$$, then the set under consideration $$\mathbb{N}_{10^n}$$, which we denote $$A_n$$, is the set of all of the (non-negative) integers containing at most $$n$$ digits.
Recurrence relation approach | {
"domain": "blogspot.my",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9907319855244919,
"lm_q1q2_score": 0.8572051749767012,
"lm_q2_score": 0.865224084314688,
"openwebmath_perplexity": 587.8367039734752,
"openwebmath_score": 0.7829540371894836,
"tags": null,
"url": "http://summerofgodel.blogspot.my/"
} |
Recurrence relation approach
In this approach, we use $$S_n$$, for $$n \in \mathbb{N}^+$$, to denote the subset of $$A_n$$ of all of the integers that contain at least one occurrence of the digit '3' and we use $$T_n$$ to denote $$|S_n|$$, the cardinality of $$S_n$$.
• Since $$A_1=\{0,1,2,3,4,5,6,7,8,9\}$$, $$S_1=\{3\}$$ and $$T_1=1$$.
• Since $$A_2=\{0,1,2,3,4,\cdots,98,99\}$$, $$S_2=\{ 3,13,23,30,31,32,33,34,\cdots,39,43,53,63,73,83,93\}$$$$=\{ 3,13,23,43,53,63,73,83,93\}$$$$\cup \{30,31,32,33,34,\cdots,39\}$$. In other words, we can split the set $$S_2$$ into two subsets, one whose elements all start with the digit '3' (there are exactly 10 such integers), and the other one whose elements do not start with the digit '3' (there are exactly $$9\cdot T_1$$ of those, since if the first digit is not '3' then the least significant digit must belong to $$S_1$$). Thus, $$T_2 = 10 + 9\cdot T_1 = 10 + 9 = 19$$.
• Similarly, since $$A_3=\{0,1,2,3,4,\cdots,998,999\}$$, $$S_3$$ is the union of two sets, one whose elements all start with the digit '3' (there are exactly $$10^2$$ such integers), and the other one whose elements do not start with the digit '3' (there are exactly $$9\cdot T_2$$ of those, since if the most significant digit is not '3' then the number formed by the other digits must belong to $$S_2$$). Thus, $$T_3 = 10^2 + 9\cdot T_2= 100 + 9\cdot 19 = 271$$.
• Since this reasoning clearly generalizes to all values of $$n \in \mathbb{N}^+$$, we obtain the following recurrence relation:
$\left\{ \begin{array}{ll} T_1 = 1 & \\ T_n = 10^{n-1} + 9T_{n-1} & \text{if}\ n>1 \end{array} \right.$ Now, let's solve this recurrence relation by iteration to get a closed-form formula for $$T_n$$: | {
"domain": "blogspot.my",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9907319855244919,
"lm_q1q2_score": 0.8572051749767012,
"lm_q2_score": 0.865224084314688,
"openwebmath_perplexity": 587.8367039734752,
"openwebmath_score": 0.7829540371894836,
"tags": null,
"url": "http://summerofgodel.blogspot.my/"
} |
$$\begin{array}{l@{0pt}l} T_1 & = 1\\ T_2 = 10^1 + 9T_1 & = 10+9 \\ T_3 = 10^2 + 9T_2 & = 10^2+9(10+9) \\ & = 10^2 + 9\cdot 10 + 9^2 \\ T_4 = 10^3 + 9T_3 & = 10^3+9(10^2 + 9\cdot 10 + 9^2) \\ & = 10^3+9\cdot 10^2 + 9^2 \cdot 10 + 9^3\\ T_5 = 10^4 + 9T_4 & = 10^4+9(10^3+9\cdot 10^2 + 9^2 \cdot 10 + 9^3)\\ & = 10^4+9\cdot 10^3+9^2\cdot 10^2 + 9^3 \cdot 10 + 9^4\\ \cdots &\\ T_n = 10^{n-1} + 9T_{n-1} & = 10^{n-1}+9\cdot 10^{n-2}+9^2\cdot 10^{n-3} + \cdots + 9^{n-2}\cdot 10^1 + 9^{n-1}\\ \end{array}$$
In conclusion, $$\displaystyle T_n = \sum_{i=0}^{n-1} \left( 9^{n-i-1}\cdot 10^i\right)$$ for $$n \in \mathbb{N}^+$$
Since this formula is rather ugly, let's turn to the combinatorial approach discussed in the Numberphile video.
Combinatorial approach
It is easy to determine the total number of integers with exactly $$n$$ digits without having to enumerate them, namely $$10 \times 10 \times \cdots \times 10 = 10^n$$, since there are exactly 10 digits to choose from for each position in the $$n$$-digit number. Note that this count actually includes all of the numbers with leading zeros, such as 01 and 00023, which are identical to 1 and 23, respectively. In other words, what we really have is $$\left|A_n\right|= 10^n$$.
Similarly, we can compute the total number of integers with at most $$n$$ digits that do not contain the digit '3', namely $$9 \times 9 \times \cdots \times 9 = 9^n$$, since there are now only 9 digits to choose from at each position in the integer.
In conclusion, $$T_n = 10^n -9^n$$.
This is both a much nicer and easier-to-derive closed-form formula than the one we obtained with the recurrence relation approach. But are the two formulas equal?
Let's check our work
Let $$P(n)$$, for $$n \in \mathbb{N}^+$$, denote: $$\displaystyle \sum_{i=0}^{n-1} \left( 9^{n-i-1}\cdot 10^i\right) = 10^n - 9^n$$
We now prove by mathematical induction that $$P(n)$$ holds for $$n \in \mathbb{N}^+.$$ | {
"domain": "blogspot.my",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9907319855244919,
"lm_q1q2_score": 0.8572051749767012,
"lm_q2_score": 0.865224084314688,
"openwebmath_perplexity": 587.8367039734752,
"openwebmath_score": 0.7829540371894836,
"tags": null,
"url": "http://summerofgodel.blogspot.my/"
} |
We now prove by mathematical induction that $$P(n)$$ holds for $$n \in \mathbb{N}^+.$$
Basis:
• $$\displaystyle \sum_{i=0}^{1-1} \left( 9^{1-i-1}\cdot 10^i\right) = \sum_{i=0}^{0} \left( 9^{1-i-1}\cdot 10^i\right) = 9^{1-0-1}\cdot 10^0 = 9^0 \cdot 10^0 =1$$
• $$10^1 - 9^1 = 10 - 9 = 1$$
• Therefore $$P(1)$$ holds
Inductive step:
• Assume that $$P(n)$$ holds for any $$n \in \mathbb{N}^+$$, that is: $\sum_{i=0}^{n-1} \left( 9^{n-i-1}\cdot 10^i\right) = 10^n - 9^n \quad \text{[inductive hypothesis]}$
• Let's compute the left-hand side of $$P(n+1)$$:
$\begin{array}{l@{0pt}l} \sum_{i=0}^{n} \left( 9^{n-i}\cdot 10^i\right) & = 9\left(\frac{1}{9}\sum_{i=0}^{n} \left( 9^{n-i}\cdot 10^i\right)\right)\\ & = 9\left(\sum_{i=0}^{n} \left( 9^{n-i-1}\cdot 10^i\right)\right)\\ & = 9\left(\sum_{i=0}^{n-1} \left( 9^{n-i-1}\cdot 10^i\right) + \sum_{i=n}^{n} \left( 9^{n-i-1}\cdot 10^i\right) \right)\\ & = 9\left(\sum_{i=0}^{n-1} \left( 9^{n-i-1}\cdot 10^i\right) + \left( 9^{n-n-1}\cdot 10^n\right) \right)\\ & = 9\left(\sum_{i=0}^{n-1} \left( 9^{n-i-1}\cdot 10^i\right) + \frac{10^n}{9} \right)\\ & = 9\left( \left(10^n - 9^n\right) + \frac{10^n}{9} \right) \quad \text{by the inductive hypothesis} \\ & = 9\cdot 10^n - 9\cdot 9^n + 10^n\\ & = 10\cdot 10^n - 9\cdot 9^n\\ & = 10^{n+1} - 9^{n+1}\\ \end{array}$
• Therefore $$P(n+1)$$ holds.
Main result
We are now in a position to prove our main result, namely that the percentage of integers that contain the digit '3' in the set $$A_n$$ goes to 100% as $$n$$ goes to infinity. This percentage is equal to
$$100\times\frac{T_n}{\left|A_n\right|}=100\times\left(\frac{10^n-9^n}{10^n}\right)=100\times\left(1-\frac{9^n}{10^n}\right)= 100 - 100\left(\frac{9}{10}\right)^n$$.
And this percentage goes to 100% as $$n$$ goes to infinity, since $$\displaystyle \lim_{n \to \infty}\left( \frac{9}{10}\right)^n = 0$$.
Discussion | {
"domain": "blogspot.my",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9907319855244919,
"lm_q1q2_score": 0.8572051749767012,
"lm_q2_score": 0.865224084314688,
"openwebmath_perplexity": 587.8367039734752,
"openwebmath_score": 0.7829540371894836,
"tags": null,
"url": "http://summerofgodel.blogspot.my/"
} |
Discussion
Of course, all of the proofs above still hold if we had picked the digit '8' (say) instead of the digit '3'. In other words, it is also true that almost all integers contain the digit '8'. So, if we use $$E_n$$ to denote the cardinality of the subset of $$A_n$$ of all of the integers that contain at least one occurrence of the digit '8', $$E_n = 10^n-9^n$$. Similarly for the digit '5' (say): $$F_n = 10^n-9^n$$.
But, if both $$100\times\frac{T_n}{\left|A_n\right|}$$ and $$100\times\frac{E_n}{\left|A_n\right|}$$ go to 100% as $$n$$ goes to infinity, the sum of these two percentages will eventually exceed 100%! That is true. However, this sum double counts all of the integers that contain both the digit '3' and the digit '8'. The correct percentage of such integers is $$100\times\frac{10^n-8^n}{10^n}$$, which also goes to 100% as $$n$$ goes to infinity.
So there is no paradox here. When $$n$$ gets large enough, almost all of the integers will contain all of the digits 0 through 9. Note that all of these results are asymptotic. Indeed, it takes relatively large values of $$n$$ for these percentages to get anywhere close to 100%. For example, for the percentage of integers that contain the digit '3' to reach 90%, 95%, 99% and 99.99%, the number $$n$$ of digits in the integer must be larger than 21, 28, 43 and 87, respectively.
## Tuesday, July 9, 2013
### Robinson Arithmetic is Σ1-complete
Recall that Q (i.e., Robinson Arithmetic) is an axiomatized formal theory (AFT) of arithmetic couched in the interpreted formal language LA. Let L be a subset of LA and let T be some AFT of arithmetic.
We say that T is L-sound iff, for any sentence φ in L, if T ⊢ φ, then φ is true.
We say that T is L-complete iff, for any sentence φ in L, if φ is true, then T ⊢ φ. | {
"domain": "blogspot.my",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9907319855244919,
"lm_q1q2_score": 0.8572051749767012,
"lm_q2_score": 0.865224084314688,
"openwebmath_perplexity": 587.8367039734752,
"openwebmath_score": 0.7829540371894836,
"tags": null,
"url": "http://summerofgodel.blogspot.my/"
} |
We say that T is L-complete iff, for any sentence φ in L, if φ is true, then T ⊢ φ.
In chapter 9, Peter Smith defines a subset of LA, called Σ1. Then, using the fact that Q is order-adequate, he proves that Q is Σ1-complete. This is important because the well-formed formulas (wff's) of Σcan express the decidable numerical properties and relations, and therefore Q will be sufficiently strong. Now to the details...
First, let's define a few interesting subsets of LA:
## Sunday, July 7, 2013
In chapter 9, Peter Smith defines the following concept: A theory T that captures the relation ≤ is order-adequate if it satisfies the following nine properties:
• O1: T ⊢ ∀x (0 ≤ x)
• O2: For any n, T ⊢ ∀x ((x = 0 ∨ x = 1 ∨ x = 2 ∨ ... ∨ x = n) → x ≤ n)
• O3: For any n, T ⊢ ∀x (x ≤ n → (x = 0 ∨ x = 1 ∨ x = 2 ∨ ... ∨ x = n))
• O4: For any n, if T ⊢ φ(0) and T ⊢ φ(1) and ... and T ⊢ φ(n) then T ⊢ (∀x ≤ n)φ(x)
• O5: For any n, if T ⊢ φ(0) or T ⊢ φ(1) or ... or T ⊢ φ(n) then T ⊢ (∃x ≤ n)φ(x)
• O6: For any n, T ⊢ ∀x (x ≤ n → x ≤ Sn)
• O7: For any n, T ⊢ ∀x (n ≤ x → (n = x ∨ Sn ≤ x))
• O8: For any n, T ⊢ ∀x (x ≤ n ∨ n ≤ x)
• O9: For any n>0, T ⊢ (∀x ≤ n-1)φ(x) → (∀x ≤ n)(x ≠ n → φ(x))
Then, we have the following theorem:
## Wednesday, July 3, 2013
### Robinson Arithmetic captures the "less-than-or-equal-to" relation
In this post, we'll start discussing the material in Chapter 9 of Peter Smith's book, namely up to section 9.3.
Before proceeding, review the definition of Robinson Arithmetic (denoted Q) as well as what it means for a theory to capture a numerical relation.
Now, we'll show that Robinson Arithmetic captures the ≤ numerical relation with the open well-formed formula: ∃x (x + m = n). | {
"domain": "blogspot.my",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9907319855244919,
"lm_q1q2_score": 0.8572051749767012,
"lm_q2_score": 0.865224084314688,
"openwebmath_perplexity": 587.8367039734752,
"openwebmath_score": 0.7829540371894836,
"tags": null,
"url": "http://summerofgodel.blogspot.my/"
} |
In other words, we are going to prove the following two-part theorem:
Theorem: If m and n are natural numbers, then:
1. If m ≤ n, then Q ⊢ ∃x (x + m = n)
2. If m > n, then Q ⊢ ¬ ∃x (x + m = n)
Recall that if m and n are natural numbers, then m and n are the numerical terms representing m and n, respectively, in the formal theory.
Proof sketch of part 1:
## Sunday, June 30, 2013
### Q - Robinson Arithmetic
Now that we are familiar with Baby Arithmetic (BA), we can make its language more expressive by allowing variables and quantifiers back into its logical vocabulary. When we do this, we simply obtain the interpreted language LA, that was described earlier.
Since we now have variables and quantifiers, we can replace the schemata of BA with regular axioms (see below). The resulting formal system of arithmetic is called Robinson Arithmetic and is often denoted by the letter Q, as described in Chapter 8 of Peter Smith's book. Here is his definition of Q:
## Friday, June 28, 2013
### Baby arithmetic
Since chapter 7 in Peter Smith's book is so short, I'll summarize it quickly and then start discussing chapter 8.
Chapter 7 starts by comparing the two incompleteness theorems discussed in previous chapters (let's call these "Smith's theorems") to Gödel's first incompleteness theorem as follows:
## Thursday, June 27, 2013
### Consistent, sufficiently strong formal systems of arithmetic are incomplete
In Chapter 6, Peter Smith proves another incompleteness theorem. To place this theorem in context, let's review some definitions about axiomatized formal theories (or AFTs) from earlier posts. | {
"domain": "blogspot.my",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9907319855244919,
"lm_q1q2_score": 0.8572051749767012,
"lm_q2_score": 0.865224084314688,
"openwebmath_perplexity": 587.8367039734752,
"openwebmath_score": 0.7829540371894836,
"tags": null,
"url": "http://summerofgodel.blogspot.my/"
} |
Let T be some AFT.
• T is consistent iff (if and only if) there is no sentence φ such that T proves both φ and ¬ φ.
• T is sound iff every theorem that T proves is true according to the interpretation that is built into T's language.
• T is (negation-)complete iff for every sentence φ in T's language, T proves either φ or ¬ φ.
• T is decidable iff there is an algorithm that, given any sentence φ in T's language, determines whether or not T proves φ.
• If needed, go here to review what it means for T's language to express a numerical property P or a numerical relation R.
• If needed, go here to review what it means for T to capture a numerical property P or a numerical relation R.
• If needed, go here to review what it means for T's language to be sufficiently expressive.
We'll use the following abbreviations:
## Monday, June 24, 2013
### Sound formal systems with sufficiently expressive languages are incomplete
In chapter 5, Peter Smith uses a counting argument to prove that sound axiomatized formal theories (AFTs) that can express a good amount of arithmetic are incomplete. In short: since their set of theorems is effectively enumerable but their set of true sentences is not, the two sets must be different. In a sound theory, the theorems are all true. Therefore, some truths must remained unproved in such theories.
Let's start with a definition. The language of an AFT of arithmetic is sufficiently expressive if and only if:
1. It can express quantification over numbers, and
2. It can express every effectively decidable two-place numerical relation.
Now, to the pivotal theorem of this chapter:
## Sunday, June 23, 2013
### An enumerable but not effectively enumerable set
Let's focus on the first half of chapter 5 in Peter Smith's book. First, Smith describes a new characterization of effectively enumerable sets. Second, Smith gives an example of a set of integers that is not effectively enumerable. | {
"domain": "blogspot.my",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9907319855244919,
"lm_q1q2_score": 0.8572051749767012,
"lm_q2_score": 0.865224084314688,
"openwebmath_perplexity": 587.8367039734752,
"openwebmath_score": 0.7829540371894836,
"tags": null,
"url": "http://summerofgodel.blogspot.my/"
} |
Let's define the numerical domain of an algorithm as the set of natural numbers with the following property: when the algorithm takes one of these numbers as input, it terminates and returns a result. Every algorithm has a numerical domain. If the input to some algorithm is not a single natural number, then the numerical domain of this algorithm is the empty set. Otherwise, the numerical domain is the set of those natural numbers on which the algorithm does not crash and does not go into an infinite loop.
It turns out that each numerical domain is effectively enumerable. In fact, the converse is also true, according to the following theorem:
## Saturday, June 22, 2013
### Capturing numerical properties in a formal language of arithmetic
Peter Smith starts Chapter 4 by describing LA, a formal language that is at the core of several AFTs (axiomatized formal theories) of arithmetic. Then Smith explains what it means for a formal language of arithmetic to "express" a numerical property and the stronger notion of "capturing" a numerical property.
Here is the definition of LA:
## Friday, June 21, 2013
### Formal systems or axiomatized formal theories
In chapter 3, Peter Smith defines formal systems or, as he calls them, axiomatized formal theories (I will use AFT as an abbreviation for this phrase).
A theory T is an AFT if...
## Wednesday, June 19, 2013
### Effective computability, decidability and enumerability
In Chapter 2, Smith covers familiar ground. So this post will be short. Here is a quick summary, section by section.
Section 1 reviews terminology pertaining to functions, namely the notions of domain and range, as well as special cases of functions, namely injective (or one-to-one) functions, surjective (or onto) functions and bijective functions (or one-to-one correspondences).
## Tuesday, June 18, 2013
### Peter Smith's overview of Gödel's incompleteness theorems | {
"domain": "blogspot.my",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9907319855244919,
"lm_q1q2_score": 0.8572051749767012,
"lm_q2_score": 0.865224084314688,
"openwebmath_perplexity": 587.8367039734752,
"openwebmath_score": 0.7829540371894836,
"tags": null,
"url": "http://summerofgodel.blogspot.my/"
} |
## Tuesday, June 18, 2013
### Peter Smith's overview of Gödel's incompleteness theorems
Finally! As was my initial goal, I now turn my attention to Peter Smith's book on the proof of Gödel's incompleteness theorems (by the way, I own the first edition of his book; so that is the one I will be referring to). I did not expect to spend that much time on Nagel and Newman's book. But it was worthwhile in getting the big picture.
Chapter 1 is an introductory and user-friendly discussion of topics that were mentioned in earlier posts, such as basic arithmetic, formal systems, (in)completeness, consistency, the statement of Gödel's incompleteness theorems and some of their implications. Therefore, in this post, I will just highlight the points where Smith provides new information or has a different perspective.
## Monday, June 17, 2013
### Proof sketch of Gödel's incompleteness theorems
In this post, I will follow the outline of the proof given in Section VII of Nagel and Newman's book. Recall that:
• a formal system F is consistent if there is no well-formed formula (wff) w such that F proves both w and ¬ w, and
• a formal system is (semantically) complete if it can prove all of the true wff's that it can express.
Now, Gödel's first incompleteness theorem can be paraphrased as:
Any consistent formal system that is expressive enough to model arithmetic is both incomplete and "incompletable."
and Gödel's second incompleteness theorem can be paraphrased as:
Any consistent formal system that is expressive enough to model arithmetic cannot prove its own consistency.
According to Nagel and Newman, a proof of the first theorem can be sketched in 4 steps:
## Friday, June 7, 2013
### Mappings and the arithmetization of meta-mathematics
In an earlier post, we discussed mappings in general and then described Gödel numbering, which is a mapping between the set of well-formed formulas (wff's) in a formal system and the set of positive integers. | {
"domain": "blogspot.my",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9907319855244919,
"lm_q1q2_score": 0.8572051749767012,
"lm_q2_score": 0.865224084314688,
"openwebmath_perplexity": 587.8367039734752,
"openwebmath_score": 0.7829540371894836,
"tags": null,
"url": "http://summerofgodel.blogspot.my/"
} |
In part B of section VII, Nagel and Newman describe a more interesting mapping used in Gödel's proof. In this mapping, the domain is the set of meta-mathematical statements about structural properties of wff's, while the co-domain is the set of wff's. This post describes how this mapping enabled Gödel to arithmetize meta-mathematics.
## Thursday, June 6, 2013
### Mappings and Gödel numbering
Section VI of Nagel and Newman's book describes mappings. A mapping is an operation that applies to two sets called the domain and the co-domain. More specifically, a mapping associates to each element of the domain one or more elements of the co-domain.
## Tuesday, May 28, 2013
### Absolute proof of consistency of FSN
In section IV of their great little book, Nagel and Newman discuss efforts by Gottlob Frege and then Bertrand Russell to reduce arithmetic to logic. This is clearly another attempt at a relative proof of consistency: if this reduction were successful, then arithmetic would be consistent provided logic is consistent.
Whether this latter statement is true or not, Whitehead and Russell's Principia Mathematica was a landmark achievement: it almost completed the first step in an absolute proof of consistency of arithmetic, since it led to the formalization of an axiomatic system for arithmetic.
In section V, Nagel and Newman describe a formalization of propositional (or sentential) logic, that is, a subset of the logic system in Principia Mathematica (but not a large enough subset to represent arithmetic). The bulk of this section first describes the formalization process, which yields the standard syntax and inference rules of propositional logic (including modus ponens) and then outlines an absolute proof of consistency of this formalized axiomatic system.
This absolute proof of consistency is a proof by contrapositive, which relies on the following true conditional statement:
## Sunday, May 26, 2013
### Absolute proofs of consistency and meta-mathematics | {
"domain": "blogspot.my",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9907319855244919,
"lm_q1q2_score": 0.8572051749767012,
"lm_q2_score": 0.865224084314688,
"openwebmath_perplexity": 587.8367039734752,
"openwebmath_score": 0.7829540371894836,
"tags": null,
"url": "http://summerofgodel.blogspot.my/"
} |
## Sunday, May 26, 2013
### Absolute proofs of consistency and meta-mathematics
In earlier posts, we explained why consistency is an important property of axiomatic systems and discussed relative proofs of consistency, in which the proof of consistency of a system is based on the assumption that another axiomatic system is consistent. In this post, we introduce absolute proofs of consistency that do not make any assumptions about any other axiomatic system. Apparently, David Hilbert was the first to study and propose such proofs, according to Nagel and Newman's book (Section III, page 26).
Recall that an axiomatic system is consistent if it cannot derive both a theorem and its negation. What do we mean by the negation of a theorem? Let's take, as a simple example, the theorem: "6 is divisible by 3." Its negation is simply the following statement: "6 is not divisible by 3" or equivalently "It is not the case that 6 is divisible by 3." This second formulation of the negation, although less elegant in English, is preferable because the negation is added to the front of the original theorem. In a formal system, negation is handled by simply adding a symbol for the phrase "it is not the case that." Several symbols have been used for negation, such as ~ and ¬ . We'll use the latter here. So, if T is any theorem in some formal system, then the formula ¬T is the negation of T.
Remember that our formal system FS did not have a symbol for negation. So we will extend FS into a new formal system called FSN (for FS with Negation), whose alphabet is { E, 0, 1, ¬ }. FSN has exactly one axiom, namely the same as A1 in FSFSN also has the same inference rules as FS, namely IR1 and IR2. But it has one additional inference rule that uses negation. Here is the full description of FSN:
## Wednesday, May 22, 2013
### Relative proofs of consistency | {
"domain": "blogspot.my",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9907319855244919,
"lm_q1q2_score": 0.8572051749767012,
"lm_q2_score": 0.865224084314688,
"openwebmath_perplexity": 587.8367039734752,
"openwebmath_score": 0.7829540371894836,
"tags": null,
"url": "http://summerofgodel.blogspot.my/"
} |
## Wednesday, May 22, 2013
### Relative proofs of consistency
In the last post, we defined and explained the importance of the property of consistency for an axiomatic system. The second half of Section II in Nagel and Newman's book describes ways of proving that an axiomatic system is consistent.
But first, note that the question of consistency of Euclidean geometry did not arise. Its axioms were supposed to describe the real world; and something that actually exists cannot be self-contradictory. In other words, existence (or truth) implies internal consistency.
The need for consistency proofs arose much later, with non-Euclidean geometries, which do not obviously model space as we experience it. Non-existence does not imply inconsistency. But any interesting abstract construct had better be internally consistent.
Second, checking the internal consistency of all of the theorems produced so far is typically not a valid proof of consistency, because (interesting) axiomatic systems generate an infinite number of theorems. The proof of consistency must guarantee that not a single theorem, including some that we have not yet produced and that we might never produce, contradicts any other theorem in the system.
One possible way to prove the consistency of an axiomatic system is model-based, where a model is a kind of interpretation.
## Monday, May 20, 2013
### Consistency of axiomatic systems
The "problem of consistency" is the topic of Section II of Nagel and Newman's book. This section defines "consistency" and explains when and why it became an important property of axiomatic systems. | {
"domain": "blogspot.my",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9907319855244919,
"lm_q1q2_score": 0.8572051749767012,
"lm_q2_score": 0.865224084314688,
"openwebmath_perplexity": 587.8367039734752,
"openwebmath_score": 0.7829540371894836,
"tags": null,
"url": "http://summerofgodel.blogspot.my/"
} |
The oldest and most famous axiomatic system is that of Euclid, in which he systematized all of the knowledge of geometry (and more) available to him over two thousand years ago. Based on five axioms, Euclid was able to rigorously prove a very large number of known and new theorems (called "propositions" in his Elements). His axioms were supposed to be intuitively true. The first four axioms dealt with line segments, lines, circles and angles (see this Wikipedia entry) and have been viewed as self-evident. In contrast, the fifth axiom, which was equivalent to the following statement: "Through a point outside a given line, only one parallel to the line can be drawn" (page 9), was not intuitively true (apparently because the two lines involved extend to infinity in two directions, similarly to asymptotes).
Since this proposed axiom was not obviously true, many mathematicians tried to prove that it logically follows from the first four axioms. Only in the nineteenth century was it demonstrated that it is NOT possible to prove the parallel axiom from the first four axioms.
This proof that it is impossible to prove a given statement is a great precursor of Gödel's incompleteness theorems.
## Friday, May 17, 2013
### Formal systems, axioms, inference rules, formal proofs
I'll start this post by describing a simple formal system that I made up.
formal system is comprised of axioms and inference rules. Each axiom and inference rule is defined syntactically, that is, by ordered sequences of symbols that follow strict syntactic rules but do not (necessarily) have any meaning. The set of all symbols allowed in a formal system are explicitly listed in its alphabet, simply, a finite, non-empty set of symbols.
My formal system, let's call it FS, uses the alphabet {E,0,1} and has only one axiom and two inference rules. Here is the axiom (in the box below):
## Wednesday, May 15, 2013
### Let's get started with Nagel and Newman (Section I) | {
"domain": "blogspot.my",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9907319855244919,
"lm_q1q2_score": 0.8572051749767012,
"lm_q2_score": 0.865224084314688,
"openwebmath_perplexity": 587.8367039734752,
"openwebmath_score": 0.7829540371894836,
"tags": null,
"url": "http://summerofgodel.blogspot.my/"
} |
## Wednesday, May 15, 2013
### Let's get started with Nagel and Newman (Section I)
Let's ease our way into this. I haven't read the Nagel and Newman book in a few years. But I remember it as a short and highly readable overview of the proof. I have a 1986 softcover edition by NYP Press. Today, I want to write about Section I - Introduction. It's really short: under 5 pages.
First, a short quote that makes me feel better about my past failures at REALLY understanding the proof of GIT:
## Tuesday, May 14, 2013
### I just want to understand the proof of Gödel's incompleteness theorems
So this is it! Summer 2013... This is when I get to master the proof of Gödel's incompleteness theorems (thereafter: GIT).
I discovered GIT as an undergraduate student in one of my Artificial Intelligence courses. My instructor was reading Gödel, Escher, Bach and was raving about the book. He even took some of our exam problems from it. But what I remember most is the outline of the proof of GIT, more precisely the first theorem. To be honest, the only part of it that stuck with me was the idea of numbering the formulas. Of course, at the time, I bought a copy of the book. I liked many parts of it but I never finished it.
Years later, while I was teaching computability theory and other computer science topics, I encountered GIT several times. At some point, I decided I wanted to understand the proof of GIT.
First, I read the classic "Gödel's Proof" by Nagel and Newman. That is a concise and user-friendly overview of the proof. But it was only a proof sketch. So I looked at some textbooks on formal logic, which were neither concise nor user-friendly. I still did not understand the proof. | {
"domain": "blogspot.my",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9907319855244919,
"lm_q1q2_score": 0.8572051749767012,
"lm_q2_score": 0.865224084314688,
"openwebmath_perplexity": 587.8367039734752,
"openwebmath_score": 0.7829540371894836,
"tags": null,
"url": "http://summerofgodel.blogspot.my/"
} |
Finally, a couple of years ago, I came across a promising volume: "Introduction to Gödel’s Theorems" by Peter Smith. I read the first few chapters but then ran out of time (and motivation) to finish it. The following year, I made another attempt: I had to start from scratch and did not get any farther into it the second time around... | {
"domain": "blogspot.my",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9907319855244919,
"lm_q1q2_score": 0.8572051749767012,
"lm_q2_score": 0.865224084314688,
"openwebmath_perplexity": 587.8367039734752,
"openwebmath_score": 0.7829540371894836,
"tags": null,
"url": "http://summerofgodel.blogspot.my/"
} |
## Weighted Quick Union (WQU)
Improving on Quick Union relies on a key insight: whenever we call find, we have to climb to the root of a tree. Thus, the shorter the tree the faster it takes!
New rule: whenever we call connect, we always link the root of the smaller tree to the larger tree.
Following this rule will give your trees a maximum height of $\log N$, where N is the number of elements in our Disjoint Sets. How does this affect the runtime of connect and isConnected?
Let's illustrate the benefit of this with an example. Consider connecting the two sets T1 and T2 below:
We have two options for connecting them:
The first option we link T1 to T2. In the second, we link T2 to T1.
The second option is preferable as it only has a height of 2, rather than 3. By our new rule, we would choose the second option as well because T2 is smaller than T1 (size of 3 compared to 6).
We determine smaller / larger by the number of items in a tree. Thus, when connecting two trees we need to know their size (or weight). We can store this information in the root of the tree by replacing the -1's with -(size of tree). You will implement this in Lab 6.
#### Maximum height: Log N
Following the above rule ensures that the maximum height of any tree is Θ(log N). N is the number of elements in our Disjoint Sets. By extension, the runtimes of connect and isConnected are bounded by O(log N). | {
"domain": "gitbooks.io",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9907319874400309,
"lm_q1q2_score": 0.8572051645832893,
"lm_q2_score": 0.865224072151174,
"openwebmath_perplexity": 659.542950967756,
"openwebmath_score": 0.5889438986778259,
"tags": null,
"url": "https://joshhug.gitbooks.io/hug61b/content/chap9/chap94.html"
} |
Why log N? The video above presents a more visual explanation. Here's an optional mathematical explanation why the maximum height is $\log_2 N$. Imagine any element $x$ in tree $T1$. The depth of $x$ increases by $1$ only when $T1$ is placed below another tree $T2$. When that happens, the size of the resulting tree will be at least double the size of $T1$ because $size(T2) \ge size(T1)$. The tree with $x$ can double at most $\log_2 N$ times until we've reached a total of N items ($2^{\log_2 N} = N$). So we can double up to $\log_2 N$ times and each time, our tree adds a level $\rightarrow$ maximum $\log_2 N$ levels.
You may be wondering why we don't link trees based off of height instead of weight. It turns out this is more complicated to implement and gives us the same Θ(log N) height limit.
### Summary and Code
Implementation Constructor connect isConnected
QuickUnion Θ(N) O(N) O(N)
QuickFind Θ(N) Θ(N) Θ(1)
QuickUnion Θ(N) O(N) O(N)
Weighted Quick Union Θ(N) O(log N) O(log N)
N = number of elements in our DisjointSets data structure | {
"domain": "gitbooks.io",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9907319874400309,
"lm_q1q2_score": 0.8572051645832893,
"lm_q2_score": 0.865224072151174,
"openwebmath_perplexity": 659.542950967756,
"openwebmath_score": 0.5889438986778259,
"tags": null,
"url": "https://joshhug.gitbooks.io/hug61b/content/chap9/chap94.html"
} |
# Euler equations for the incompressible fluids
Incompressible fluid with constant density ρ fills the three-dimensional domain below the free surface $z = η(r)$ in cylindrical polar coordinates. The flow is axisymmetric and steady, and the only non-zero velocity component is $u_θ$. Gravity acts upon the fluid. The fluid in $r\lt a$ rotates rigidly about the z-axis with angular velocity $\Omega$ and the fluid $r \ge a$ is irrotational.
Use the radial and vertical components of the Euler equations to show that the pressure $p$ in the region $z < η$, $r < a$ satisfies
$$\frac{p}{\rho} = \frac{1}{2}{Ω^2}{r^2} −gz + \text{constant}$$ and find the constant.
$$\frac{∂u_r}{∂t}+\left(u_r\frac{∂}{∂r}+u_z\frac{∂}{∂z}\right)u_r-\frac{(u_θ)^2}{r}+\frac{1}{\rho}\frac{∂p}{∂r}=0$$
$$\frac{∂u_θ}{∂t}+\left(u_r\frac{∂}{∂r}+u_z\frac{∂}{∂z}\right)u_θ+\frac{u_θu_r} {r}=0$$
$$\frac{∂u_z}{∂t}+\left(u_r\frac{∂}{∂r}+u_z\frac{∂}{∂z}\right)u_z+\frac{1}{\rho}\frac{∂p}{∂z}=-g$$
I have the Euler equation in cylindrical coordinate. Then how should I figure out the $\Omega$. The next question is to show that the free surface position in $r<a$ is $$η=\frac{\Omega^2a^2}{g}(\frac{r^2}{2a^2}-1)$$ if it is helpful. Thank you so much!
• Are you sure the radial equations aren't available to you? That's a fair amount of effort to derive those. May 15 '17 at 19:14
Sincerely, if you derived the Euler equations for cylindrical coordinates, this is a very simple exercise for you.
With the hypothesis, many terms disappear, those involving $u_r=0,\;u_z=0$ and the partials wrt time (steady flow).
Further, the motion is rigid, meaning that all fluid parts have zero relative motion. If they are moving in circles, the angular velocity for all point in the fuid $\Omega$ is the same! You don't need to figure anything, it must be known. With all this, $u_\theta=\Omega r$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9845754515389344,
"lm_q1q2_score": 0.8571687118744152,
"lm_q2_score": 0.8705972818382005,
"openwebmath_perplexity": 398.4050063086413,
"openwebmath_score": 0.8719672560691833,
"tags": null,
"url": "https://math.stackexchange.com/questions/2282473/euler-equations-for-the-incompressible-fluids"
} |
$\begin{cases} -\dfrac{(u_θ)^2}{r}+\dfrac{1}{\rho}\dfrac{\partial p}{\partial r}=0\\ \dfrac{1}{\rho}\dfrac{\partial p}{\partial z}=-g \end{cases}$
Integrating the second,
$\dfrac{p}{\rho}=-gz+f(r)\tag 1$
So is $\dfrac{1}{\rho}\dfrac{\partial p}{\partial r}=f'(r)$ or $\dfrac{p}{\rho}+C=f(r)$. With the first equation,
$-\dfrac{(u_θ)^2}{r}+f'(r)=0$ or $-\dfrac{(\Omega r)^2}{r}+f'(r)=0$ or
$-\Omega^2r+f'(r)=0$. Integrating, $f(r) = \dfrac{1}{2}{Ω^2}{r^2}+h(z)$
$\dfrac{p}{\rho}= \dfrac{1}{2}{Ω^2}{r^2}+h(z)-C\tag 2$
Comparing $(1)$ and $(2)$ we have that $h(z)=-gz+C$
$\dfrac{p}{\rho} = \dfrac{1}{2}{Ω^2}{r^2} −gz +C$
I could not recover the given solution for the free surface, although I've found some relation that produces the well known paraboloid for the free surfce of the Newton's bucket. To find $C$, we can consider the heigh $z$ for the free surface ($p=0$) at some $r$. We can set $z=0$ for $r =a$.
$0=\dfrac{1}{2}{Ω^2}{a^2}+C$ or $C=-\dfrac{1}{2}{Ω^2}{a^2}$
$\dfrac{p}{\rho} = \dfrac{1}{2}{Ω^2}{r^2} −gz-\dfrac{1}{2}{Ω^2}{a^2}=\dfrac{1}{2}Ω^2(r^2-a^2)-gz$
For the free surface $z=\eta(r)$, $p=0$
$\eta=\dfrac{1}{2g}Ω^2(r^2-a^2),\;r\lt a$
From other considerations, $\eta$ for $r\ge a$ is $\eta=-\dfrac{\Omega^2a^4}{2gr^2}$. As $\eta$ has to be continuous $\eta(a)=-\dfrac{\Omega^2a^2}{2g}$, for the formula we've found for $r\lt a$
$0=\dfrac{Ω^2a^2}{2}+\dfrac{\Omega^2a^2}{2}+C$ leading to $C=\Omega^2a^2$ and to $\eta=\dfrac{\Omega^2a^2}{g}(\dfrac{r^2}{2a^2}-1)$ for $r\lt a$
$$\eta= \begin{cases} \dfrac{\Omega^2a^2}{g}(\dfrac{r^2}{2a^2}-1)& x\lt a\\ -\dfrac{\Omega^2a^4}{2gr^2}&a\le x \end{cases}$$
I've sketched the function for the free surface (setting $\dfrac{\Omega^2a^2}{2g}=1$ and $a=1$) | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9845754515389344,
"lm_q1q2_score": 0.8571687118744152,
"lm_q2_score": 0.8705972818382005,
"openwebmath_perplexity": 398.4050063086413,
"openwebmath_score": 0.8719672560691833,
"tags": null,
"url": "https://math.stackexchange.com/questions/2282473/euler-equations-for-the-incompressible-fluids"
} |
I've sketched the function for the free surface (setting $\dfrac{\Omega^2a^2}{2g}=1$ and $a=1$)
• It is actually the next two question of this one. math.stackexchange.com/questions/2280167/… I try to get the constant and the free surface and the number matches when I set r=a, p=0 and use the free surface when r>=a. I get C=-1/2(\Omega) ^2a^2-(\Omega)^2a^2/2 and free surface as above. Can I just plug in the limit May 16 '17 at 7:40
• And if I want to draw a sketch for the free surface displacement then it is gonna be a paraboloid but when happen at r=a then? May 16 '17 at 7:49
• @stedmoaoa, the resultig function is smooth. May 16 '17 at 13:47
• Thank you so much it really makes sense. So the graph is like x^2-1 and -1/x^2 Then, how should I describe at r=a? May 16 '17 at 14:08
• Continuous and differentiable for all quantites. May 16 '17 at 17:37 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9845754515389344,
"lm_q1q2_score": 0.8571687118744152,
"lm_q2_score": 0.8705972818382005,
"openwebmath_perplexity": 398.4050063086413,
"openwebmath_score": 0.8719672560691833,
"tags": null,
"url": "https://math.stackexchange.com/questions/2282473/euler-equations-for-the-incompressible-fluids"
} |
# A statement about divisibility of relatively prime integers
I'm solving a problem, and the solution makes the following statement: "The common difference of the arithmetic sequence 106, 116, 126, ..., 996 is relatively prime to 3. Therefore, given any three consecutive terms, exactly one of them is divisible by 3." Why is this statement true? Where does it come from? Is it generalizable to other numbers?
• And here I thought it was something about "relatively prime statements" – Yuriy S Jul 21 '16 at 21:56
There is a simple, well-known theorem about that:
If $a$, $b$ and $m$ are integers, $m\ge 2$ and $\gcd(b,m)=1$ then the set $\{a,a+b,a+2b,\ldots,a+(m-1)b\}$ is a complete residue system.
This implies, for example, that the set contains exactly one multiple of $m$.
• Also, most of the other answers hint at a proof of that theorem. – Jeppe Stig Nielsen Jul 22 '16 at 8:04
Take $a\in\mathbb{N}$ and suppose that $a\equiv 1\pmod 3$. Then \begin{align*} a&\equiv 1\pmod 3\\ a+10\equiv 11&\equiv 2\pmod 3\\ a+20\equiv 21&\equiv 0\pmod 3 \end{align*}
If we instead have $b\in\mathbb{N}$ such that $b\equiv 2\pmod 3$, then adding $10$ to $b$ is sufficient. The case in which we have $\equiv 0\pmod 3$ is obviously trivial.
If $\,b\,$ is coprime to $\,m\,$ then $\,a+ib,\, i = 1,\ldots,m\,$ is a complete residue system mod $\,m\,$ since
$$\ a+ib \equiv a+j b \iff (i\!-\!j)b\equiv 0\overset{(b,m)=1}\iff i\!-\!j\equiv 0\iff i = j\,\ {\rm by}\,\ 1\le i,j\le m$$
Hence, being complete, it contains an element $\equiv 0,\,$ i.e. a multiple of $\,m.$
Remark $\$ When $\,b = 1\,$ we obtain the well-known special case that any sequence of $\,m\,$ consecutive integers contains a multiple of $\,m.$
If $d$ is relatively prime to $b$, then $a + d j$ is divisible by $b$ if and only if $j \equiv -a d^{-1} \mod b$. Thus exactly one of any $b$ consecutive terms of the arithmetic sequence $a, a+d, a+2d, \ldots$ is divisible by $b$. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9845754452025767,
"lm_q1q2_score": 0.8571687063579994,
"lm_q2_score": 0.8705972818382005,
"openwebmath_perplexity": 90.82299456413315,
"openwebmath_score": 0.9589896202087402,
"tags": null,
"url": "https://math.stackexchange.com/questions/1866962/a-statement-about-divisibility-of-relatively-prime-integers"
} |
In general let $d$ be the common difference of an arithmetic sequence, where $3\nmid d$, and let $a$, $a+d$, and $a+2d$ be three consecutive terms in that sequence. Now WLOG if we assume $3\nmid a$ then we can write $a$ and $d$ as: $$a=3k+1\quad\text{ or }\quad a'=3k+2\quad\text{ and }\quad d=3j+1\quad\text{ or }\quad d'=3j+2$$ for some $j$, $k\in\mathbb{Z}$, where primes have been added to distinguish the two possibilities for $a$ and $d$. Now look at the four possibilities we have for $a+d$, and $a+2d$: $$a+d=3(k+j)+2,\quad \quad a+2d=3(k+2j)+3\qquad (A)$$ $$a+d'=3(k+j)+3,\quad \quad a+2d'=3(k+2j)+5\qquad (B)$$
$$a'+d=3(k+j)+3,\quad \quad a'+2d=3(k+2j)+5\qquad (C)$$ $$a'+d'=3(k+j)+4,\quad \quad a'+2d'=3(k+2j)+6\qquad (D)$$
As can be seen, since we assumed $3\nmid a$, only one member of the three consecutive terms is divisible by $3$ on listing all possible combinations (A), (B), (C), and (D). It would make no difference if we assumed $3\nmid a+d$ or $3\nmid a+2d$ as our starting point as the problem is cyclic modulo $3$. By this look at the congruences of $a$, $a+d$, $a+2d\pmod{3}$ for each of (A), (B), (C), and (D) and you see we get the full set of residue classes $[0]$, $[1]$, $[2]$ modulo $3$ in some order, which is the reason we get one and only one number divisible by $3$. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9845754452025767,
"lm_q1q2_score": 0.8571687063579994,
"lm_q2_score": 0.8705972818382005,
"openwebmath_perplexity": 90.82299456413315,
"openwebmath_score": 0.9589896202087402,
"tags": null,
"url": "https://math.stackexchange.com/questions/1866962/a-statement-about-divisibility-of-relatively-prime-integers"
} |
Further for some $m$ for which $\gcd(m,d)=1$ look at $m$ consecutive terms of an arithmetic sequence module $m$: $$a,\ a+d,\ a+2d,\dotsc,\ a+(m-1)d\pmod{m}\qquad (E)$$ then $a+kd\equiv a+jd\pmod{m}$ iff $(k-j)d\equiv 0\pmod{m}$ for $0\le k,j\le m-1$, which since $[0]$, $[1]$, $[2],\dotsc,\ [m-1]$ form a complete residue system modulo $m$ we must have $[k]=[j]$ modulo $m$, and so in fact $k=j$ with respect to our $m$ consecutive terms with any distinct pair of terms incongruent modulo $m$, and only one unique number in the set being divisible by $m$. To see this imagine (E) rewritten with respect to residue classes modulo $m$ reordered so: $$a+[0],\ a+[1],\ a+[2],\dotsc,\ a+[m-1]\pmod{m}\qquad (E')$$ If $m\mid a$ then $m\equiv a+[0]\pmod{m}$, and we also see $m$ cannot divide any other number in the set; if $m\nmid a$ then we can write $a=mj+k$ where $j$, $k\in\mathbb{Z}$, and $\gcd(k,m)=1$, which is in the class $[mj+k]=[k]=[k']$ modulo $m$, and $[k']+[k'']=[0]$ modulo $m$, $0<k',k''\le m-1$, where $[k']$ and $[k'']$ are additive inverses in the complete residue system (note that $[0]$ is self inverse wrt addition). | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9845754452025767,
"lm_q1q2_score": 0.8571687063579994,
"lm_q2_score": 0.8705972818382005,
"openwebmath_perplexity": 90.82299456413315,
"openwebmath_score": 0.9589896202087402,
"tags": null,
"url": "https://math.stackexchange.com/questions/1866962/a-statement-about-divisibility-of-relatively-prime-integers"
} |
# Math Help - Counting sides
1. ## Counting sides
A rectangle that measures 4cm by 6cm is divided into 24 squares with sides 1cm in length. What is the total number of 1cm long sides in those 24 squares? (If 2 squares share a side, the side should be counted only once).
Now I know the answer is 58 and I drew a sketch and counted the sides but I was wondering if there was any shortcut to this problem that I was missing..? Setting up any kind of expression or equation or maybe a combinations problem of some sort?
2. Let use a vertical layout like this (for explanation):
****
****
****
****
****
****
First count all the right and bottom sides of each square: 24 * 2 = 48. There is no overlap in this counting.
However, we missed some sides with this count. Which ones? The entire top row and left column.
Well, the number of edges we did not count is:
# on top + # on left = 4 + 6 = 10
Grand total is: 48 + 10 = 58
So from this the general solution for number of sides in a n by m block is:
2nm + n + m = n(1 + m) + m(1 + n)
This suggest thats an alternative way is to count:
#horizontal edges + #vertical edges = n(1 + m) + m(1 + n)
3. Hi snow--your solution definitely seems to work but I think there is a column of 6 missing Thanks!
4. Hello, sarahh!
A rectangle that measures 4cm by 6cm is divided into 24 squares with sides 1cm in length.
What is the total number of 1cm long sides in those 24 squares?
(If 2 squares share a side, the side should be counted only once).
You made a sketch but you didn't see any pattern, did you?
Code:
* - * - * - * - * - * - *
| | | | | | |
* - * - * - * - * - * - *
| | | | | | |
* - * - * - * - * - * - *
| | | | | | |
* - * - * - * - * - * - *
| | | | | | |
* - * - * - * - * - * - *
The unit squares are in 4 rows and 6 columns.
Imagine these 24 squares formed with matches.
How many matches are used?
We see that there are 6 matches in each row.
And there are 4 + 1 rows. | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9845754497285468,
"lm_q1q2_score": 0.8571686987284427,
"lm_q2_score": 0.870597270087091,
"openwebmath_perplexity": 659.3125972834156,
"openwebmath_score": 0.6646708250045776,
"tags": null,
"url": "http://mathhelpforum.com/geometry/167839-counting-sides.html"
} |
We see that there are 6 matches in each row.
And there are 4 + 1 rows.
Hence, there are: .6(4+1) horizontal matches.
We see that there are 4 matches in each column.
And there are 6 + 1 columns.
Hence, there are: .4(6 + 1) vertical matches.
We can generalize this problem.
Suppose there are R rows and C columns.
Then there are: .C(R+1) horizontal matches
. . . . . . . .and: .R(C+1) vertical matches.
Total: . $C(R+1) + R(C+1) \;=\; R + C + 2RC$ matches.
5. Originally Posted by sarahh
Hi snow--your solution definitely seems to work but I think there is a column of 6 missing Thanks!
There is a row of 6 and a column of 4 missing. Which is why a 10 (= 6 + 4) is added at the end.
6. Just for fun, here is another approach-- first do it the wrong way, then patch it up.
There are 24 squares in all, each with 4 sides, and each side is shared between two squares, so there are
$\frac{4 \times 24}{2} = 48$ sides.
But oops, the sides along the perimeter of the rectangle (2 * (4 + 6) = 20) are not shared! So we need to add those back in.
$48 + \frac{20}{2} = 58$ | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9845754497285468,
"lm_q1q2_score": 0.8571686987284427,
"lm_q2_score": 0.870597270087091,
"openwebmath_perplexity": 659.3125972834156,
"openwebmath_score": 0.6646708250045776,
"tags": null,
"url": "http://mathhelpforum.com/geometry/167839-counting-sides.html"
} |
Fibonacci's final digits cycle every 60 numbers
How would you go about to prove that the final digits of the Fibonacci numbers recur after a cycle of 60?
References:
The sequence of final digits in Fibonacci numbers repeats in cycles of 60. The last two digits repeat in 300, the last three in 1500, the last four in , etc. The number of Fibonacci numbers between and is either 1 or 2 (Wells 1986, p. 65).
http://mathworld.wolfram.com/FibonacciNumber.html
• Why do you conjecture that? Do you have a reference that asserts the statement? – Giovanni De Gaetano Feb 26 '12 at 9:52
• I added some references. – crazyGuy Feb 26 '12 at 9:57
• Neat! Have a look at this link. – Juan S Feb 26 '12 at 10:00
• Here is a link: oeis.org/A096363 – Stefan Gruenwald Dec 14 '18 at 20:59
Notice that:
$$F_{n+15} \equiv 7F_n \pmod{10}$$ for $$n\geq 1$$.
Also the order of $$7$$ mod $$10$$ is $$4$$ so the repetition in the digits of the Fibonacci numbers begins after place $$15\times 4 = 60$$. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9845754515389344,
"lm_q1q2_score": 0.8571686936932157,
"lm_q2_score": 0.8705972633721708,
"openwebmath_perplexity": 267.36283155366385,
"openwebmath_score": 0.7990440130233765,
"tags": null,
"url": "https://math.stackexchange.com/questions/113536/fibonaccis-final-digits-cycle-every-60-numbers"
} |
• I got that recursion by just writing out the first $16$ Fibonacci numbers mod $10$ and noticing that you get $1,1,...,7,7,...$, this dictates that that next $13$ numbers must be $7$ times their corresponding numbers (mod $10$) in the first lot of $15$ numbers. – fretty Feb 26 '12 at 10:03
• How do we know that F_n+15 = 7F_n mod 10 ? and How do you get that 7 mod 10 is 4 ? thanks. – crazyGuy Feb 26 '12 at 10:07
• Write out the first 16 Fibonacci numbers mod 10: 1,1,...,7,7,... Now by the recursion of the fibonacci numbers the fact that this "multiple-repeat" has happened now tells us that the next 13 numbers must be the same as the 3rd-15th Fibonacci numbers but multiplied by $7$ mod 10. So an actual repeat will happen as soon as we have multipled by enough $7$'s to get $1$ mod 10. This is the "order" of $7$ mod $10$. It is easy to work out here, $7^2 \equiv 4$ mod $10$ and $7^4 \equiv 1$ mod $10$ so the order is $4$. – fretty Feb 26 '12 at 10:13
• @sic2 He's not saying that $7\mod10$ is $4$, but rather that $7^4\equiv 1\mod 10$ and that this is not true for any smaller power of $7$, which is a consequence of Euler's theorem. – Alex Becker Feb 26 '12 at 10:15
• Also this method generalises to the Fibonacci numbers mod $n$. Find the first $1,1,...,a,a,...$ then work out the order of $a$ mod $n$ and you will have your first place where it recurs. – fretty Feb 26 '12 at 10:29
Since each term in the Fibonacci sequence is dependent on the previous two, each time a $0\pmod{m}$ appears in the sequence, what follows must be a multiple of the sequence starting at $F_0,F_1,\dots=0,1,\dots$ That is, a subsequence starting with $0,a,\dots$ is $a$ times the sequence starting with $0,1,\dots$
Consider the Fibonacci sequence $\text{mod }2$: $$\color{red}{0,1,1,}\color{green}{0,1,1,\dots}$$ Thus, the Fibonacci sequence repeats $\text{mod }2$ with a period of $3$. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9845754515389344,
"lm_q1q2_score": 0.8571686936932157,
"lm_q2_score": 0.8705972633721708,
"openwebmath_perplexity": 267.36283155366385,
"openwebmath_score": 0.7990440130233765,
"tags": null,
"url": "https://math.stackexchange.com/questions/113536/fibonaccis-final-digits-cycle-every-60-numbers"
} |
Consider the Fibonacci sequence $\text{mod }5$: $$\color{red}{0,1,1,2,3,}\color{green}{0,3,3,\dots}$$ Thus, the Fibonacci sequence is multiplied by $3\pmod{5}$ each "period" of $5$. Since $3^4=1\pmod{5}$, the Fibonacci sequence repeats $\text{mod }5$ with a period of $20=4\cdot5$.
Thus, the Fibonacci sequence repeats $\text{mod }10$ with a period of $60=\operatorname{LCM}(3,20)$.
Note that
$$\begin{pmatrix} F_{n+1}\\ F_{n+2} \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} F_n \\ F_{n+1} \end{pmatrix}$$
and
$$\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}^{60} \equiv \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \mod 10.$$
One can verify that $60$ is the smallest power for which this holds, so it is the order of the matrix mod 10. In particular
$$\begin{pmatrix} F_{n+60}\\ F_{n+61} \end{pmatrix} \equiv \begin{pmatrix} F_n \\ F_{n+1} \end{pmatrix} \mod 10$$
so the final digits repeat from that point onwards.
• The same matrix applies to the Lucas Numbers, however, $\text{ mod }10$, they have a period of $12$. The fact that $\begin{pmatrix} 0&1\\1&1\end{pmatrix}^{60}=\begin{pmatrix}1&0\\0&1\end{pmatrix}\text{ mod }10$, says that the period of a sequence satisfying $a_n=a_{n-1}+a_{n-2}$ divides $60$. – robjohn Feb 26 '12 at 19:21
$F_{0}=F_1=F_{60}=F_{61}= 1 \mod10$
By inspection, these are the first two pairs of consecutive Fibonaccis for which this is true. Since the recurrence relation only takes into account the previous two terms and last digits only depend on previous last digits, this suffices to prove the claim.
I have found that there is an 11 number grouping. You must reduce numbers to a single digit (13=4 or 55=10=1). After 24 numbers into the sequence it repeats. To go a little deeper, a 9 is in the twelfth and twenty- fourth sequence and the real number is also divisible by 12 in these positions (which coincidentally is every twelfth number in the sequence). | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9845754515389344,
"lm_q1q2_score": 0.8571686936932157,
"lm_q2_score": 0.8705972633721708,
"openwebmath_perplexity": 267.36283155366385,
"openwebmath_score": 0.7990440130233765,
"tags": null,
"url": "https://math.stackexchange.com/questions/113536/fibonaccis-final-digits-cycle-every-60-numbers"
} |
Now if you do the multiplication table of numbers 1 through 8 and reduce all numbers to a single digit, you will find that 1 and 8 correspond in reverse with each other. The same for 2 and 7, also 4 and 5. 3 and 6 are different. You will notice that for every number sequence it will repeat after a 9 appears. So, Fibonacci sequence would be this:
1,1,2,3,5,8,4,3,7,1,8,9,8,8,7,6,4,1,5,6,2,8,1,9,1,1,2,3,5,8,4,3,7,1,8,9,8,8,7,6,4,1,5,6,2,8,1,9......
Remember how I said 1 and 8, 2 and 7, 4 and 5 correspond with each other by reducing the multiplication table to single digits and all numbers repeat a sequence after a 9?
If you move the eleven number sequence between the nines and set 1,1,2,3... above the sequence 8,8,7,6... You will find that they all correspond with the reducing method afore mentioned.
It's very interesting.
• What do you mean $11$-number grouping? Why are you reducing mod $9$ instead of mod $10$? How does this prove that it repeats with cycle $60$? – 6005 Oct 24 '16 at 2:58
• I apologize. This was not to prove or disprove a 60 cycle repeat. Maybe I posted in the wrong area? I just found this sequence repeat and I thought I would give it to bigger brains and see what they can make of it. – Michael Bunton Oct 24 '16 at 3:12
• 11 number grouping means the first 11 numbers in the sequence. The twelfth number is 144 and that reduces to 9 i.e.(1+4+4=9). Then a second set of eleven numbers followed by a reduced 9. – Michael Bunton Oct 24 '16 at 3:20 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9845754515389344,
"lm_q1q2_score": 0.8571686936932157,
"lm_q2_score": 0.8705972633721708,
"openwebmath_perplexity": 267.36283155366385,
"openwebmath_score": 0.7990440130233765,
"tags": null,
"url": "https://math.stackexchange.com/questions/113536/fibonaccis-final-digits-cycle-every-60-numbers"
} |
# find Limit $a_n= \frac{1}{n+1}+\frac{1}{n+2}+…+\frac{1}{n+n}$ [duplicate]
show sequence $a_n= \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}$ converges to log2
my attempt:
1. sequence $a_n$ is monotonic increasing
2. 0<$a_n$<1/2
how to find limit?
• Oh come on, this must be the duplicate of three duplicates. – Vincenzo Oliva Aug 14 '15 at 9:39
$$\lim_{n\rightarrow \infty}\sum_{i=1}^{n}\frac 1 {n+i}=\lim_{n\rightarrow \infty}\sum_{i=1}^{n}\frac 1 n\frac 1 {1+\dfrac in}=\int_0^1\frac 1 {1+x}dx=\ln(1+1)-\ln(1+0)=\ln(2)$$ (Riemann sum)
• +1, but there should be a limit as $n \to \infty$ instead of the sum going to infinity. The sum goes fom $i=1$ to $i=n$ – 6005 Aug 14 '15 at 9:07
• @6005 Fixed, thank you! – GeorgSaliba Aug 14 '15 at 9:09
• ya i was hoping there would be some other way. riemann sum cannot be used to compute limit of sequence$a_n=1+1/2+1/3...+1/n−log(n+1)$ – ketan Aug 14 '15 at 9:19
• @ketan Then why not use the harmonic number $H_n$ that was already mentioned by @MichaelGaluza? But I suppose this becomes a slightly different question than the one you asked. – GeorgSaliba Aug 14 '15 at 9:21
• @ketan $a_n=1+1/2+1/3+...+1/n - \log(n+1)=H_n -\log(n+1)=\log(n)+\gamma + \epsilon_n-\log(n+1)=\log(n/(n+1))+\gamma +\epsilon _n$ – GeorgSaliba Aug 14 '15 at 9:23
\begin{align} \log(2) &=\sum_{k=1}^\infty\frac{(-1)^{k-1}}k\\ &=\lim_{n\to\infty}\sum_{k=1}^{2n}\frac{(-1)^{n-1}}k\\ &=\lim_{n\to\infty}\left(\vphantom{\sum_{k=1}^n}\right.\overbrace{\sum_{k=1}^{2n}\frac1k}^{\substack{\text{sum of all}\\\text{terms}}}-2\overbrace{\sum_{k=1}^n\frac1{2k}}^{\substack{\text{sum of even}\\\text{terms}}}\left.\vphantom{\sum_{k=1}^n}\right)\\[6pt] &=\lim_{n\to\infty}\left(\sum_{k=1}^{2n}\frac1k-\sum_{k=1}^n\frac1k\right)\\[6pt] &=\lim_{n\to\infty}\sum_{k=n+1}^{2n}\frac1k\\ \end{align} | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9845754447499795,
"lm_q1q2_score": 0.8571686910884428,
"lm_q2_score": 0.8705972667296309,
"openwebmath_perplexity": 1964.3167225974635,
"openwebmath_score": 0.9540139436721802,
"tags": null,
"url": "https://math.stackexchange.com/questions/1396839/find-limit-a-n-frac1n1-frac1n2-frac1nn/1396843"
} |
• Nice overbrace. Is there a corresponding underbrace command or somethings similar? – mathreadler Aug 14 '15 at 9:23
• Yep. It is \underbrace :-) Then you use subscripts instead of superscripts. – robjohn Aug 14 '15 at 9:23
Here is classical proof of it. Let $H_n=\sum_{i=1}^n 1/i$. It's well-known that $$H_n = \ln n + \gamma + \epsilon_n,$$ where $\gamma$ is Euler–Mascheroni constant, $\epsilon_n\to 0$. So, $$a_n = H_{2n}-H_{n} = \ln 2 + \epsilon_{2n} - \epsilon_n \to \ln2.$$
• what is $\epsilon_n$? is it always zero? – ketan Aug 14 '15 at 9:28
• @ketan, no. It's sequence which tends to $0$. – Michael Galuza Aug 14 '15 at 11:39
Notice, $$a_n=\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\ldots +\frac{1}{n+n}$$ $$=\sum_{r=1}^{\infty}\frac{1}{n+r}=\sum_{r=1}^{\infty}\frac{1}{n\left(1+\frac{r}{n}\right)}$$$$=\sum_{i=1}^{\infty}\frac{\frac{1}{n}}{1+\frac{r}{n}}$$ Now, let $\frac{r}{n}=x\implies \frac{1}{n}=dx\to 0$ $$\text{upper limit of x}=\lim_{n\to \infty}\frac{n}{n}=1$$ $$\text{lower limit of x}=\lim_{n\to \infty}\frac{1}{n}=0$$ Hence, using integration we get $$a_n=\int_{0}^{1}\frac{dx}{1+x}$$ $$=\left[\ln (1+x)\right]_{0}^{1}$$ $$=\left[\ln 2-\ln 1\right]=\ln 2$$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9845754447499795,
"lm_q1q2_score": 0.8571686910884428,
"lm_q2_score": 0.8705972667296309,
"openwebmath_perplexity": 1964.3167225974635,
"openwebmath_score": 0.9540139436721802,
"tags": null,
"url": "https://math.stackexchange.com/questions/1396839/find-limit-a-n-frac1n1-frac1n2-frac1nn/1396843"
} |
# Math Help - Proving the binomial coeficient is always a natural number?
1. ## Proving the binomial coeficient is always a natural number?
How would one go about proving that:
${ n \choose k }$
is always a natural number? That is, assuming $0 \leq k \leq n$ and that $n = 1, 2, 3, 4...$ and that $k = 1, 2, 3, 4...$; what I mean by that is that $n$ and $k$ are both positive integers. Its a question in my textbook that I bought early to get ahead on fall classes, so I don't have any teachers to ask about this question; not till fall atleast. I'm not asking for a straight out proof, but preferably some direction as to where I would go, which direction I should go to prove this? A litttle guidence is all I need, please don't simply give me the full answer, if you don't mind. Thanks in advance
2. Originally Posted by mfetch22
How would one go about proving that:
${ n \choose k }$
is always a natural number? That is, assuming $0 \leq k \leq n$ and that $n = 1, 2, 3, 4...$ and that $k = 1, 2, 3, 4...$; what I mean by that is that $n$ and $k$ are both positive integers. Its a question in my textbook that I bought early to get ahead on fall classes, so I don't have any teachers to ask about this question; not till fall atleast. I'm not asking for a straight out proof, but preferably some direction as to where I would go, which direction I should go to prove this? A litttle guidence is all I need, please don't simply give me the full answer, if you don't mind. Thanks in advance
Use the fact that $\displaystyle{\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$ for $\displaystyle{k=1\ldots n-1}$ and $\displaystyle{\binom{n}{0}=\binom{n}{n}=1}$.
(This will be a proof by induction)
3. Originally Posted by mfetch22
How would one go about proving that:
${ n \choose k }$ | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9845754501811437,
"lm_q1q2_score": 0.8571686908582903,
"lm_q2_score": 0.8705972616934406,
"openwebmath_perplexity": 302.88837267326954,
"openwebmath_score": 0.814684271812439,
"tags": null,
"url": "http://mathhelpforum.com/number-theory/151472-proving-binomial-coeficient-always-natural-number.html"
} |
3. Originally Posted by mfetch22
How would one go about proving that:
${ n \choose k }$
is always a natural number? That is, assuming $0 \leq k \leq n$ and that $n = 1, 2, 3, 4...$ and that $k = 1, 2, 3, 4...$; what I mean by that is that $n$ and $k$ are both positive integers. Its a question in my textbook that I bought early to get ahead on fall classes, so I don't have any teachers to ask about this question; not till fall atleast. I'm not asking for a straight out proof, but preferably some direction as to where I would go, which direction I should go to prove this? A litttle guidence is all I need, please don't simply give me the full answer, if you don't mind. Thanks in advance
There's also the combinatorial proof. (The binomial coefficient is given as an "archetypal example" not far from the top of the page.) The formula(s) given for $\displaystyle { n \choose k }$ can be shown to equal either of the two equivalent quantities
1) the coefficient of the $\displaystyle x^k$ term in the polynomial expansion of the binomial power $\displaystyle (1 + x)^n$
2) the number of ways to choose k elements from a set of n elements. (The number of distinct k-subsets of {1, 2, ... , n}.)
Obviously both of these quantities are integers.
4. Another way is to use Legendre's formula, which states that ( $\mathbb{P}$ is the set of prime numbers)
$
n! = \prod_{p\in \mathbb{P}} p^{\sum_{i=1}^\infty \Big\lfloor\frac{n}{p^i}\Big\rfloor}
$
Thus for any prime $p$ we have:
$
\binom{n}{k} = \frac{n!}{k!(n-k)!} = \prod_{p\in \mathbb{P}} p^{\sum_{i=1}^\infty \Big\lfloor\frac{n}{p^i}\Big\rfloor-\Big\lfloor\frac{k}{p^i}\Big\rfloor-\Big\lfloor\frac{n-k}{p^i}\Big\rfloor}}
$
So all what's left is to show that $\lfloor{x+y}\rfloor \ge \lfloor{x}\rfloor + \lfloor{y}\rfloor$
so it follows that for all $i\in\mathbb{N}$:
$
\Big\lfloor\frac{n}{p^i}\Big\rfloor-\Big\lfloor\frac{k}{p^i}\Big\rfloor-\Big\lfloor\frac{n-k}{p^i}\Big\rfloor} \ge 0
$
and therefore $\binom{n}{k} \in \mathbb{N}$ | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9845754501811437,
"lm_q1q2_score": 0.8571686908582903,
"lm_q2_score": 0.8705972616934406,
"openwebmath_perplexity": 302.88837267326954,
"openwebmath_score": 0.814684271812439,
"tags": null,
"url": "http://mathhelpforum.com/number-theory/151472-proving-binomial-coeficient-always-natural-number.html"
} |
and therefore $\binom{n}{k} \in \mathbb{N}$
5. Originally Posted by Unbeatable0
Another way is to use Leibnitz' formula, which states that ( $\mathbb{P}$ is the set of prime numbers)
$
n! = \prod_{p\in \mathbb{P}} p^{\sum_{i=1}^\infty \Big\lfloor\frac{n}{p^i}\Big\rfloor}
$
Thus for any prime $p$ we have:
$
\binom{n}{k} = \frac{n!}{k!(n-k)!} = \prod_{p\in \mathbb{P}} p^{\sum_{i=1}^\infty \Big\lfloor\frac{n}{p^i}\Big\rfloor-\Big\lfloor\frac{k}{p^i}\Big\rfloor-\Big\lfloor\frac{n-k}{p^i}\Big\rfloor}}
$
So all what's left is to show that $\lfloor{x+y}\rfloor \ge \lfloor{x}\rfloor + \lfloor{y}\rfloor$
so it follows that for all $i\in\mathbb{N}$:
$
\Big\lfloor\frac{n}{p^i}\Big\rfloor-\Big\lfloor\frac{k}{p^i}\Big\rfloor-\Big\lfloor\frac{n-k}{p^i}\Big\rfloor} \ge 0
$
and therefore $\binom{n}{k} \in \mathbb{N}$
Perfect , in fact $\lfloor{ x+y \rfloor} = \lfloor{ x \rfloor} + \lfloor{ y \rfloor} + \lfloor{ \{ x \} + \{ y \} \rfloor} \geq \lfloor{ x \rfloor} + \lfloor{ y \rfloor}$ .
6. Originally Posted by mfetch22
How would one go about proving that:
${ n \choose k }$
is always a natural number? That is, assuming $0 \leq k \leq n$ and that $n = 1, 2, 3, 4...$ and that $k = 1, 2, 3, 4...$; what I mean by that is that $n$ and $k$ are both positive integers. Its a question in my textbook that I bought early to get ahead on fall classes, so I don't have any teachers to ask about this question; not till fall atleast. I'm not asking for a straight out proof, but preferably some direction as to where I would go, which direction I should go to prove this? A litttle guidence is all I need, please don't simply give me the full answer, if you don't mind. Thanks in advance
If you are asking from the perspective of the Binomial Expansion,
then you can observe a pattern if you refrain from expressing the terms with indices...
$(a+b)^2=(a+b)(a+b)=aa+ab+ba+bb$
"a" times "b" is expressed in both possible ways.
$(a+b)^3=(a+b)(a+b)^2=(a+b)(aa+ab+ba+bb)$ | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9845754501811437,
"lm_q1q2_score": 0.8571686908582903,
"lm_q2_score": 0.8705972616934406,
"openwebmath_perplexity": 302.88837267326954,
"openwebmath_score": 0.814684271812439,
"tags": null,
"url": "http://mathhelpforum.com/number-theory/151472-proving-binomial-coeficient-always-natural-number.html"
} |
"a" times "b" is expressed in both possible ways.
$(a+b)^3=(a+b)(a+b)^2=(a+b)(aa+ab+ba+bb)$
$=aaa+aab+aba+abb+baa+bab+bba+bbb$
$=aaa+aab+aba+baa+abb+bab+bba+bbb$
If the "a" values were labelled $a_1,\ a_2,\ a_3$ we could arrange them,
hence we have all possible selections of 2 "a"'s with a "b", 2 "b"'s with an "a",
the only possible selection of 3 "a"s, the only selection of 3 "b"'s.
For higher powers, the same pattern exists.
Since this pattern is the same as "choose k from n available",
the binomial coefficient may be calculated using
$\binom{n}{k}$
There are k! arrangements of a selection.
As shown above, the number of selections is a natural number,
which can be calculated by "unarranging" the arrangements.
The number of arrangements of k from n is n(n-1)(n-2)...(n-[k-1])
which is $\frac{n!}{k!}$ as k! has been "cut off" the tail.
The number of arrangements is k! times the number of selections. | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9845754501811437,
"lm_q1q2_score": 0.8571686908582903,
"lm_q2_score": 0.8705972616934406,
"openwebmath_perplexity": 302.88837267326954,
"openwebmath_score": 0.814684271812439,
"tags": null,
"url": "http://mathhelpforum.com/number-theory/151472-proving-binomial-coeficient-always-natural-number.html"
} |
# $C^1$ function on compact set is Lipschitz
Let $\emptyset\ne A\subset\mathbb{R}^n$ be open and let $f \in C^1 (A, \mathbb{R})$ be a function. Let $\emptyset\ne K\subset A$ be compact and convex. I want to prove that $f$ is Lipschitz on $K$; that is, prove there exists a constant $c > 0$ such that $| f ( x ) - f( y ) | \leq c \, \| x - y \|, \forall x , y \in K$.
My approach:
Let $x,y\in K$ be two arbitrary points. Then, since $K$ is convex, the line segment between $x$ and $y$, i.e. $Co(x,y)$, is in $K$. Thus, by MVT, there exists a vector $b\in Co(x,y)$ such that
$$\text{(*) } |f(x)-f(y)|=|\langle x-y, (\nabla f)(b)\rangle|\le \|x-y\|\|(\nabla f)(b)\|$$
Now, since $K$ is compact, $f$ takes a maximum and a minimum values on $K$, so that $\exists b'\in K$ such that $\|(\nabla f)(b') \|\ge \|(\nabla f)(b) \|$, for all $b\in K$. Let $c\in \mathbb{R}$, $c:=(\nabla f)(b')$, then
$$|f(x)-f(y)|\le\|x-y\|\|(\nabla f)(b)\|\le\|x-y\|\|(\nabla f)(b')\|=c\|x-y\|$$
This implies that $f$ is Lipschitz on $K$ for all $x,y\in K$.
Please let me know if you think my proof is correct or not very much? I'm somewhat concerned about the part with the gradient - how exactly is the maximality of the norm of the gradient related to the EVT, that is to $f$ taking maximum and minimum values? As far as I can tell, the maximum norm of the gradient exists because that is the direction to the maximum (or minimum) point of $f$. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9787126469647337,
"lm_q1q2_score": 0.8571438158785631,
"lm_q2_score": 0.8757870029950159,
"openwebmath_perplexity": 124.69960711134773,
"openwebmath_score": 0.9673323631286621,
"tags": null,
"url": "https://math.stackexchange.com/questions/2338799/c1-function-on-compact-set-is-lipschitz/2338828"
} |
• What is $a$? It seems undefined to me . . . – Robert Lewis Jun 28 '17 at 0:24
• Sorry, it was supposed to be $y$. Fixed. – sequence Jun 28 '17 at 0:26
• I guess is because $||f'||$ is a real function on a compact set? – Li Chun Min Jun 28 '17 at 0:41
• @LiChunMin Can you please clarify your question? – sequence Jun 28 '17 at 0:47
• Looks good to me. You do not need the EVT. Since $\nabla (f)$ is continuous, the real-valued function $\|\nabla (f)\|$ is continuous .So $T=\{\| \nabla f(x)\|: x\in K\}$ is compact (because $K$ is compact) , so $T$ is a bounded real subset....So $\|f(x)-f(y)\|\leq \|x-y\|\cdot \sup T$ for all $x,y\in K$. – DanielWainfleet Jun 28 '17 at 2:48
Here's how I would do it:
for $x, y \in K$, let $\gamma(t):[0, 1] \to K$ be the line segment
$\gamma(t) = x + t(y - x); \tag{1}$
then
$\gamma(0) = x, \tag{2}$
$\gamma(1) = x +(y - x) = y, \tag{3}$
and
$\gamma'(t) = y - x; \tag{4}$
furthermore, since $K$ is convex, $\gamma(t)$ lies entirely within $K$, hence also in $A$.
Now, for $x, y \in K$, we have:
$\Vert f(y) - f(x) \Vert = \Vert \displaystyle \int_0^1 \dfrac{d(f(\gamma(t))}{dt}dt \Vert = \Vert \displaystyle \int_0^1 \nabla f(\gamma(t)) \cdot \gamma'(t) dt \Vert$ $\le \displaystyle \int_0^1 \Vert \nabla f(\gamma(t)) \cdot \gamma'(t) \Vert dt \le \displaystyle \int_0^1 \Vert \nabla f(\gamma(t)) \Vert \Vert \gamma'(t) \Vert dt.\tag{5}$
Since $K$ is compact and $\nabla f \in C^0(A, \Bbb R)$, so hence $\nabla f \in C^0(K, \Bbb R)$, $\Vert \nabla f \Vert$ is bounded by some $B$ on $K$, hence
$\displaystyle \int_0^1 \Vert \nabla f(\gamma(t)) \Vert \Vert \gamma'(t) \Vert dt \le \displaystyle \int_0^1 B \Vert \gamma'(t) \Vert dt; \tag{6}$
using (4),
$\displaystyle \int_0^1 B \Vert \gamma'(t) \Vert dt = \displaystyle \int_0^1 B \Vert y - x \Vert dt = B \Vert y - x \Vert; \tag{7}$
bringing together (5), (6), and (7) yields
$\Vert f(y) - f(x) \Vert \le B \Vert y - x \Vert, \tag{8}$
that is, $f(x)$ is Lipschitz continuous on $K$. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9787126469647337,
"lm_q1q2_score": 0.8571438158785631,
"lm_q2_score": 0.8757870029950159,
"openwebmath_perplexity": 124.69960711134773,
"openwebmath_score": 0.9673323631286621,
"tags": null,
"url": "https://math.stackexchange.com/questions/2338799/c1-function-on-compact-set-is-lipschitz/2338828"
} |
that is, $f(x)$ is Lipschitz continuous on $K$.
• Thank you, this is very elegant. Do you think my proof could possibly be correct? – sequence Jun 28 '17 at 1:32
• Yes, your method seems fine. Note it agrees with mine. – Robert Lewis Jun 28 '17 at 1:39
• "Since $K$ is compact and $\nabla f \in C^0(A, \Bbb R)$, so hence $\nabla f \in C^0(K, \Bbb R)$, $\Vert \nabla f \Vert$ is bounded by some $B$ on $K$.." Is there a proof for that or is it trivial? – abuchay May 14 '18 at 14:25
• @abuchay: Well, $\nabla f \in C^0(K)$ since $\nabla f \in C^0(A)$ and $K \subset A$. $\Vert \nabla f \Vert \in C^0(K)$ since $\Vert \cdot \Vert$ is itself continuous in its argument. Then the boundedness follows since continuous functions on compacta are bounded, which is standard result in elementary topology. – Robert Lewis May 14 '18 at 15:40
The convexity of $K$ is not needed. Suppose the conclusion fails. Then for each $m\in \mathbb N,$ there exist $y_m,x_m \in K$ such that
$$\tag 1 |f(y_m)- f(x_m)| > m|y_m-x_m|.$$
Because $K$ is compact, we can find a subsequence $m_k$ such that the sequences $y_{m_k},x_{m_k}$ converge to points $y,x\in K$ respectively.
Suppose $y\ne x.$ Then $|y-x| > 0.$ For large $k$ we then have
$$|f(y_{m_k})- f(x_{m_k})| > m_k|y_{m_k}-x_{m_k}|> m_k(|y-x|/2)\to \infty$$
This implies $f$ is not bounded on $K.$ But $f$ is continuous on $A,$ hence is continuous on $K,$ hence $f$ is bounded on $K$ by compactness. This contradiction shows $y=x.$
Because $A$ is open we can choose $r>0$ such that $B(x,r)\subset A.$ For large $k$ we then have $y_{m_k},x_{m_k}$ in the compact convex set $\overline {B(x,r/2}) \subset A.$ By $(1),$ your highlighted mean value inequality then fails in this last set, contradiction. Therefore $(1)$ cannot hold, proving the result. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9787126469647337,
"lm_q1q2_score": 0.8571438158785631,
"lm_q2_score": 0.8757870029950159,
"openwebmath_perplexity": 124.69960711134773,
"openwebmath_score": 0.9673323631286621,
"tags": null,
"url": "https://math.stackexchange.com/questions/2338799/c1-function-on-compact-set-is-lipschitz/2338828"
} |
• I dont understand why $m_k(|y-x|/2)\to \infty$. Why does it tend to infinity? If $m_k$ is a subsequence of $y_{mk}$ then $m_k$ also converges to $y$. Then, how can $m_k$ be a subsequence of $x_{mk}$ too? – John Keeper Oct 24 '18 at 9:46
• @JohnKeeper $m_k$ is a subsequence of $1,2,\dots$ hence $\to \infty$ – zhw. Oct 24 '18 at 21:32 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9787126469647337,
"lm_q1q2_score": 0.8571438158785631,
"lm_q2_score": 0.8757870029950159,
"openwebmath_perplexity": 124.69960711134773,
"openwebmath_score": 0.9673323631286621,
"tags": null,
"url": "https://math.stackexchange.com/questions/2338799/c1-function-on-compact-set-is-lipschitz/2338828"
} |
Here's a cute puzzle.
Player A tosses a coin indefinitely and stops when he encounters a consecutive sequence of HT. Player B plays a similar game, except he stops when he encounters HH consecutively. Is the expected number of tosses equal for both players? If not, who has more, and can you give an intuitive explanation why?
Note by C Lim
4 years, 8 months ago
MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2
paragraph 1
paragraph 2
[example link](https://brilliant.org)example link
> This is a quote
This is a quote
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$
Sort by:
I think that player A is expected to be finished earlier. If he encounters an H, then he would have to keep tossing H's in order to continue. The chance of throwing $$n$$ consecutive H's decreases rapidly as $$n$$ increases.
If player B encounters an H, then he will have to toss a T to continue, but after tossing a T, he will continue tossing, because no matter what he tosses next, he doesn't encounter HH.
So, player A is doomed so to say whenever he tosses an H, because he'll have to keep tossing H's in order to continue. That's why he is expected to stop earlier than player B. I hope that makes sense.
- 4 years, 8 months ago
Yup that's correct! It would also be interesting to explicitly calculate the expected number of throws for both A and B. | {
"domain": "brilliant.org",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.978712648206549,
"lm_q1q2_score": 0.8571438058606087,
"lm_q2_score": 0.8757869916479466,
"openwebmath_perplexity": 660.6993215232189,
"openwebmath_score": 0.9759904146194458,
"tags": null,
"url": "https://brilliant.org/discussions/thread/puzzle-paradox-in-probability/"
} |
- 4 years, 8 months ago
Player A is expected to throw his first H after $$2$$ tosses, because
$1 \cdot \frac{1}{2} + 2 \cdot \frac{1}{4} + 3 \cdot \frac{1}{8} + 4 \cdot \frac{1}{16} + 5 \cdot \frac{1}{32} + \dots = 2.$
Similarly, after having tossed his first H, player A is expected to throw his first T (and thus stopping) after another 2 tosses. So (hopefully) I can conclude that player H is expected to toss 4 times.
Player B is expected to roll his first H in two tosses. Then, there is a chance of 50% that he stops after the third toss, i.e. after tossing H again. However, if he tosses a T, then again, he is expected to toss his second H in two tosses, after which there is a 50% chance that he stops after the toss after that. It goes on and on. So, the amount of tosses player B is expected to do is
$3 \cdot \frac{1}{2} + 6 \cdot \frac{1}{4} + 9 \cdot \frac{1}{8} + 12 \cdot \frac{1}{16} + 15 \cdot \frac{1}{32} + \dots = 3 \cdot 2 = 6.$
And $$6 > 4$$.
I hope that what I did is allowed, I'm not completely convinced of that just yet. Could anyone confirm that what I did is correct (or show me what I've done wrong)?
- 4 years, 8 months ago
I'm not sure what you did is ok either, but your answer is correct! [ This suggests there's probably a way to rigourously justify your working. ]
Here's how I'd do it. To compute the expected number of tosses for HH, we define two values:
• let a = expected number of additional tosses, assuming previous was 'H';
• let b = expected number of additional tosses, assuming previous was 'T'.
Then $$a = \frac{b+1} 2 + \frac 1 2$$ since if the previous toss was 'H', there's a 50% chance the next is 'T' (in which case the expected number of tosses is b+1), and a 50% chance the next is 'H' (in which case the number of tosses is 1 since it ends). | {
"domain": "brilliant.org",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.978712648206549,
"lm_q1q2_score": 0.8571438058606087,
"lm_q2_score": 0.8757869916479466,
"openwebmath_perplexity": 660.6993215232189,
"openwebmath_score": 0.9759904146194458,
"tags": null,
"url": "https://brilliant.org/discussions/thread/puzzle-paradox-in-probability/"
} |
Likewise, $$b = \frac{a+1} 2 + \frac{b+1}2$$ since if the previous toss was 'T', there's a 50% chance the next is 'H' (in which case expected number of tosses is a+1), and a 50% chance the next is 'T' (and the expected number of tosses is b+1).
Solving then gives us a=4, b=6. The expected number of tosses is then b=6, since we can assume the 0-th toss was 'T' with no loss of generality.
The equations for HT are similar:
$a = \frac {a+1} 2 + \frac 1 2, \quad b = \frac{a+1}2 + \frac{b+1} 2$
which gives us a=2, b=4.
[ P.S. To be pedantic, my working isn't 100% rigourous either, since it already assumes a, b are finite. ]
- 4 years, 8 months ago
That's really great. Two completely different ways of solving the problem :) I wouldn't know how to prove that $$a,b$$ are finite though.
- 4 years, 8 months ago
If you want a finiteness proof, use first principles. Let $$N_A$$ be the number of tosses until $$A$$ stops. The probability $$P[N_A>n]$$ is the probability that $$A$$ has not thrown HT in the first $$n$$ tosses. This can only happen if $$A$$ has thrown $$j$$ tails followed by $$n-j$$ heads, for some $$0 \le j \le n$$. Thus there are $$n+1$$ possible sequences, all equally likely, and so $P[N_A > n] = (n+1)2^{-n}$ Thus $P[N_A \ge n] \;=\; P[N_A > n-1] \; = \; n2^{1-n}$ Normal probability and series summation tricks give us $E[N_A] \;=\; \sum_{n=1}^\infty P[N_A \ge n] \; = \; \sum_{n=1}^\infty n2^{1-n} \; = \; 4$ The result for $$B$$ is a little more challenging. Let $$N_B$$ be the number of tosses until $$B$$ stops. The probability that $$N_B > n$$ is the probability that $$B$$ does not throw HH in the first $$n$$ tosses. This is the probability that the first $$n$$ tosses are made up of "HT"s and "T"s stuck together, plus the probability that the first $$n-1$$ tosses are made up of "HT"s and "T"s stuck together, followed by a single "H". | {
"domain": "brilliant.org",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.978712648206549,
"lm_q1q2_score": 0.8571438058606087,
"lm_q2_score": 0.8757869916479466,
"openwebmath_perplexity": 660.6993215232189,
"openwebmath_score": 0.9759904146194458,
"tags": null,
"url": "https://brilliant.org/discussions/thread/puzzle-paradox-in-probability/"
} |
It is well-known that the number of ways of tiling a $$n\times1$$ chess-board using just $$2\times1$$ dominoes and $$1\times1$$ squares is the Fibonacci number $$F_{n+1}$$. (Just to be definite, $$F_0=0$$, $$F_1=1$$, etc.) Thus the probability that $$N_B > n$$ is the sum of the probabilities $$F_{n+1}2^{-n}$$ and $$F_n2^{-n}$$, and hence $P[N_B > n] \; = \; \frac{F_{n+2}}{2^n}$ and hence $P[N_B \ge n] \; = \; P[N_B > n-1] \; = \; \frac{F_{n+1}}{2^{n-1}} \qquad n \ge 1.$ Thus $E[N_B] \; = \; \sum_{n=1}^\infty P[N_B \ge n] \; = \; \sum_{n=1}^\infty \tfrac{F_{n+1}}{2^{n-1}} \; = \; 4G(0.5) - 2$ where $G(x) \;=\; \sum_{n=0}^\infty F_nx^n$ is the generating function of the Fibonacci numbers. It is well-known that $G(x) \; = \; \frac{x}{1-x-x^2}$ so that $$G(0.5) = 2$$. Thus $$E[N_B] = 6$$.
- 4 years, 8 months ago
Of course, once you showed that the number of ways is $$F_{n+2}$$, you can use the ratio test to conclude that the sum is finite, which is all we needed.
Nice way of calculating the sum through generating functions.
Staff - 4 years, 8 months ago
In a similar vein, check out High school math #8.
Staff - 4 years, 8 months ago
The absorbing Markov chain also gets the same answer.
This is the markov chain for ending the game with HH, for player B. I is the initial state, and the probabilities of moving to different states are shown:
$$\left(\begin{array}{cccccccc} & I & H & T & HT & TH & TT & HH \\ I & 0 & \frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 & 0 \\ H & 0 & 0 & 0 & \frac{1}{2} & 0 & 0 & \frac{1}{2} \\ T & 0 & 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} & 0 \\ HT & 0 & 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} & 0 \\ TH & 0 & 0 & 0 & \frac{1}{2} & 0 & 0 & \frac{1}{2} \\ TT & 0 & 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} & 0 \\ HH & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{array}\right)$$
Matrix Q is: | {
"domain": "brilliant.org",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.978712648206549,
"lm_q1q2_score": 0.8571438058606087,
"lm_q2_score": 0.8757869916479466,
"openwebmath_perplexity": 660.6993215232189,
"openwebmath_score": 0.9759904146194458,
"tags": null,
"url": "https://brilliant.org/discussions/thread/puzzle-paradox-in-probability/"
} |
Matrix Q is:
$$\left(\begin{array}{rrrrrr} 0 & \frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{1}{2} & 0 & 0 \\ 0 & 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} \\ 0 & 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} \\ 0 & 0 & 0 & \frac{1}{2} & 0 & 0 \\ 0 & 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} \end{array}\right)$$
Fundamental matrix $$(I-Q)^{-1}$$ is:
$$\left(\begin{array}{rrrrrr} 1 & \frac{1}{2} & \frac{1}{2} & 1 & \frac{3}{2} & \frac{3}{2} \\ 0 & 1 & 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 2 & 2 \\ 0 & 0 & 0 & 2 & 2 & 2 \\ 0 & 0 & 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & 1 & 2 & 3 \end{array}\right)$$
The required expectation is the sum of the first row, which is 6.
Similarly doing for the 'HT' case yields an expected value of 4.
- 4 years, 8 months ago | {
"domain": "brilliant.org",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.978712648206549,
"lm_q1q2_score": 0.8571438058606087,
"lm_q2_score": 0.8757869916479466,
"openwebmath_perplexity": 660.6993215232189,
"openwebmath_score": 0.9759904146194458,
"tags": null,
"url": "https://brilliant.org/discussions/thread/puzzle-paradox-in-probability/"
} |
# Number of 5-digit numbers such that the sum of their digits is even
I found the same question on this site and many others...where everywhere the answer $45000$ is written ...while I have no problem with this but I am having problems with the logical answer...as everywhere it is written since total numbers is $90000$ so half of the numbers satisfy the above condition(but why??it is not explained anywhere as to how we can say that half the numbers satisfy this condition)...I know that half the numbers are even and half the numbers are odd...but that is not what we are asked here...we need to find numbers whose sum of digits is even ...and not numbers who are just even... I think these conditions are very different but it is used evrwhere...can someone please clarify..this.??..as to how we can say..half the numbers have their sum of digits as even?? i don't think this question is entirely a duplicate but if it please let me know...where I can get the explanation of what I am asking..
• Hint: a sum of numbers is even if and only if there are an even number of odd numbers in the sum. – Colm Bhandal Nov 25 '15 at 18:23
• An other way of seeing it, the last number decide if the sum is odd or even. If the sum of the 4 first numbers is even, then the last number need to be even. If the sum of the 4 first numbers is odd, then the last number need to be odd. – Alain Remillard Nov 25 '15 at 18:26
• @Colm That is a very complicated use of even and odd term...!!..can you please expand this comment just a little bit...so that I can at least get a hint... – Freelancer Nov 25 '15 at 18:26
• @Freelancer- actually I think Alain Remillard's suggestion is the more elegant of the two. Can you see how this leads to the result? How many possibilities are there for that last digit? And how many of them result in an even number, how many in an odd number? – Colm Bhandal Nov 25 '15 at 18:28 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9787126475856414,
"lm_q1q2_score": 0.8571438053168258,
"lm_q2_score": 0.8757869916479466,
"openwebmath_perplexity": 185.39753959843088,
"openwebmath_score": 0.7157057523727417,
"tags": null,
"url": "https://math.stackexchange.com/questions/1546226/number-of-5-digit-numbers-such-that-the-sum-of-their-digits-is-even"
} |
To expand on the wonderful insight of Alain Remillard in the comments, let's consider the set of all $5$ digit numbers. Now, you can partition this set into $9 \times 10^3$ subsets (we exclude a leading $0$ as a possibility) each of which contains all numbers with a common prefix of $4$ numbers. For example, one of the partitions, for the prefix $2346$ would be:
$$\{23460, 23461, 23462, 23463, \dots, 23469\}$$
Note that each partition contains exactly $10$ elements, with the last digit numbered $0-9$. Now, to prove that exactly half the numbers in the entire set have even digit sums, all we have to do is to prove that half the numbers in each of these partitions has an even digit sum. So we've reduced the problem to something much simpler! I hope you can see why.
Now, to show that in any partition there are exactly $5$ elements with an even digit sum, consider the first element in the set e.g. in the above it would be $23460$. The sum of its digits is either even or odd. If it's even, then the next number will be odd, because you're just adding $1$ to it, and vice versa (the example is 15, which is odd). Then the sums go: even, odd, even, odd, even, odd etc. Or else they go: odd, even, odd, even etc. In either case, there are exactly $10$ elements, $5$ even, and $5$ odd. And we are done.
Update: As the OP cleverly points out in the comments below, the choice to fix the first four digits is indeed arbitrary. The first digit must be fixed, because there are only $9$ possibilities for this, but after this any three of the remaining four digits can be fixed. The remaining digit will then have ten possibilities, and the proof will proceed just as above. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9787126475856414,
"lm_q1q2_score": 0.8571438053168258,
"lm_q2_score": 0.8757869916479466,
"openwebmath_perplexity": 185.39753959843088,
"openwebmath_score": 0.7157057523727417,
"tags": null,
"url": "https://math.stackexchange.com/questions/1546226/number-of-5-digit-numbers-such-that-the-sum-of-their-digits-is-even"
} |
• Well ..you fixed the first 3-digits...and then showed that for the last digit half numbers will be positive and half will be negative...let us say I want to think in another way and I fix the last three digits(say for example $X234$ and then show that half of the numbers so formed will be having sum as even and half will be having sum as odd...is that correct also?? ...can I do this... – Freelancer Nov 25 '15 at 19:03
• @Freelancer- that is a very clever insight. Nicely though out :):) Yes you can indeed fix any three digits and the first digit and the above method will work. I will update accordingly. – Colm Bhandal Nov 26 '15 at 12:00
• even + even = even
• even + odd = odd
• odd + odd = even
Using this we can show the even-ness of the sum of digits. Take $234582$. We have
$$E + O + E + O + E + E \\= (E + O) + (E + O) + (E + E) \\= O + O + E \\= (O+O) + E \\= E + E \\= E.$$
So, for a five-digit number, what numbers of odd digits can we have to make the sum even?
Next, let's count the number of five-digit numbers that have three odd digits. (Is the sum of digits of these numbers even or odd?)
There are $125$ three-digit odd numbers. (Why?) There are $_5C_3 = 10$ combinations of places we can put the digits in order within the five-digit number. Then, there are $25$ two-digit even numbers. So the number of five digit numbers with three odd digits is $125 \cdot 10 \cdot 25 = 31250.$ (I'm assuming leading zeroes are valid (like $00123$)).
To solve the problem, then, you'll need to determine what numbers of odd digits will give you an even sum, and then count each of those cases. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9787126475856414,
"lm_q1q2_score": 0.8571438053168258,
"lm_q2_score": 0.8757869916479466,
"openwebmath_perplexity": 185.39753959843088,
"openwebmath_score": 0.7157057523727417,
"tags": null,
"url": "https://math.stackexchange.com/questions/1546226/number-of-5-digit-numbers-such-that-the-sum-of-their-digits-is-even"
} |
# Isomorphic groups
I know that there is a formal definition isomorphism but for the purpose of this homework questions, I call two groups isomorphic if they have the same structure, that is group table for one can be turned into the table for the other by a suitable renaming.
Now consider $\mathbf{Z}_2=\{0,1\}$ (group under addition modulo $2$) and $\mathbf{Z}_3^{\times}=\{1,2\}$ (group under multiplication modulo $3$). In general $\mathbf{Z}_n^{\times}=\{a|0\leq a\leq n-1 \text{ and }\gcd(a,n)=1\}$. Now the group table for $\mathbf{Z}_2$ is: $$\begin{array}{c|cc} &0&1\\ \hline 0&0&1\\ 1&1&0 \end{array}$$ and the group table for $\mathbf{Z}_3^{\times}$ is: $$\begin{array}{c|cc} &1&2\\ \hline 1&1&2\\ 2&2&1 \end{array}$$ Obviously these two groups are isomorphic because if in the first table, I replace $0$s by $1$s and $1$s by $2$s, I get exactly the second table.
However, if I write out the group table for $\mathbf{Z}_4$ and $\mathbf{Z}_5^{\times}$, there is no way that group table for one can be turned into the table for the other by a suitable renaming. This means that these two groups are not isomorphic but the question asks me to prove that they are isomorphic. Now what should I do?
-
There is no difference between the usual formal notion of "being isomorphic" and the one you propose to use! – Mariano Suárez-Alvarez Apr 2 '11 at 5:10
In any case, you should look harder at the group tables for $\mathbb Z_4$ and $\mathbb Z_5^\times$!... – Mariano Suárez-Alvarez Apr 2 '11 at 5:10
There is a suitable renaming. Keep trying... – Zev Chonoles Apr 2 '11 at 5:12
In short, the two examples are all concerned with the structure of cyclic groups, et c'est la méme. – awllower Apr 2 '11 at 7:08 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9825575132207566,
"lm_q1q2_score": 0.8571314801717705,
"lm_q2_score": 0.8723473879530491,
"openwebmath_perplexity": 336.7763461347079,
"openwebmath_score": 0.8749757409095764,
"tags": null,
"url": "http://math.stackexchange.com/questions/30463/isomorphic-groups/30465"
} |
You are incorrect in claiming that there is no way to turn one table into the other. But the key is that you are not just allowed to rename, you are also allowed to list the elements in a different order! (After all, listing the elements of $\mathbf{Z}_4$ in a different order in the table will not change the group or the operation, will it?) This amounts to shuffling rows and columns together: if you exchange rows 2 and 3, say, you should also exchange columns 2 and 3.
The table for $\mathbf{Z}_4$ is: $$\begin{array}{c|cccc} +&0&1&2&3\\ \hline 0&0&1&2&3\\ 1&1&2&3&0\\ 2&2&3&0&1\\ 3&3&0&1&2 \end{array}$$ The table for $Z_5^{\times}$ is: $$\begin{array}{c|cccc} \times&1&2&3&4\\ \hline 1&1&2&3&4\\ 2&2&4&1&3\\ 3&3&1&4&2\\ 4&4&3&2&1 \end{array}$$ If you are going to be able to rename the entries in the last table to match the first, then "1" must be renamed "0". Now, notice that there is only one of the remaining four elements that when operated with itself gives you the "identity"; since in $\mathbf{Z}_4$ this happens for $2$, you may want to shuffle the rows and columns to move that element to be in the third row and column and see what you have then.
-
c'est superbe!! – awllower Apr 2 '11 at 7:11
Apparently not; any reason for the downvote? – Arturo Magidin Apr 2 '11 at 18:13
Who downvoted ? – awllower Apr 3 '11 at 2:45
Don't know; but someone did... – Arturo Magidin Apr 3 '11 at 2:50
Since $\mathbb{Z}_n^\times$ is cyclic, reordering can be easily done by picking a generator element (i. e., any other than 1), and computing its power. In this case, the powers of 2 are, in order, 1, 2, 4 and 3, which is the ordering for the first line of the table for $\mathbb{Z}_5^\times$ that will suit the isomorphism. – Luke May 10 '11 at 1:04 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9825575132207566,
"lm_q1q2_score": 0.8571314801717705,
"lm_q2_score": 0.8723473879530491,
"openwebmath_perplexity": 336.7763461347079,
"openwebmath_score": 0.8749757409095764,
"tags": null,
"url": "http://math.stackexchange.com/questions/30463/isomorphic-groups/30465"
} |
Your "naive" definition of isomorphic almost coincides with the abstract one. The difference is that apart from renaming you must also be allowed to change the order of the elements. Then you should have no difficulties with $\mathbb{Z}/4\mathbb{Z}$ vs. $(\mathbb{Z}/5\mathbb{Z})^\times$.
-
Note by the way that this little confusion is a good argument for learning the actual definition of group isomorphism, rather than trying to avoid it by only thinking in terms of "group tables". – Pete L. Clark Apr 2 '11 at 12:30 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9825575132207566,
"lm_q1q2_score": 0.8571314801717705,
"lm_q2_score": 0.8723473879530491,
"openwebmath_perplexity": 336.7763461347079,
"openwebmath_score": 0.8749757409095764,
"tags": null,
"url": "http://math.stackexchange.com/questions/30463/isomorphic-groups/30465"
} |
# Calculating conditional probability with marginal
Assume we are given a joint distribution $$P(X,Y)$$ where $$P(0,0)=0.1$$, $$P(0,1) = 0.4$$, $$P(1,0)=0.3$$, and $$P(1,1)=0.2$$. The goal is to compute $$P(X|Y=1)$$.
Traditionally, solving a conditional probability problem $$P(A|B)$$ simplifies to $$\frac{P(A,B)}{P(B)}$$, but I'm unsure how to apply it to this case. In particular, I am unclear on what the probability $$P(X,Y=1)$$ means since $$P(X)$$ is a marginal probability.
To avoid this, I enumerated the different values $$X$$ takes on and plugged it into the original quantity to solve -- $$P(X|Y=1)$$:
• $$P(X=0|Y=1) = 0.4/0.6 = 4/6$$
• $$P(X=1|Y=1) = 0.2/0.6 = 2/6$$
This gives me the final answer of $$P(X|Y=1) = [4/6, 2/6]$$, but I'm not sure whether the answer should be multiple probabilities or a single probability.
• P(X|Y=2) is a distribution in the same way that P(X) is also a distribution (and not just one "probability"). One probably is P(X=0, Y=0)=0.1, for example. – nbro Jan 28 at 2:14
• @nbro Does this mean my answer is correct? – Shrey Jan 28 at 4:04
• you're ok. just as $p(x,y)$ is a 2D function(table), $p(x)$, $p(x|Y=y)$ are 1D functions (rows). – gunes Jan 28 at 4:52
• $P(X|Y=1)$ is shorthand for $P(X=x|Y=1)$ – StatsStudent Jan 28 at 5:47
You have gone about this correctly, but the final answers are typically written as a function of $$x$$. It's helpful, I think, to remember that $$P(X|Y=1)$$ is just shorthand for $$P(X=x|Y=1)$$ where $$x$$ is any number in the support of $$x$$ (in this case 0 and 1). So you'd calculate this as follows:
$$\begin{eqnarray*} \\{P(X|Y=1)} & = & {P\left(X=x|Y=1\right)}\\ & = & \frac{P\left(X=x,\,Y=1\right)}{P\left(Y=1\right)}\\ & = & \frac{P\left(X=x,\,Y=1\right)}{P\left(X=0,\,Y=1\right)+P\left(X=1,\,Y=1\right)}\\ & = & \frac{P\left(X=x,\,Y=1\right)}{0.4+0.2}\\ & = & \frac{P\left(X=x,\,Y=1\right)}{0.6} \end{eqnarray*}$$
Now, writing this as a function of $$x$$ gives: | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9825575116884778,
"lm_q1q2_score": 0.8571314739438562,
"lm_q2_score": 0.8723473829749844,
"openwebmath_perplexity": 185.84113606590793,
"openwebmath_score": 0.9003010988235474,
"tags": null,
"url": "https://stats.stackexchange.com/questions/389453/calculating-conditional-probability-with-marginal"
} |
Now, writing this as a function of $$x$$ gives:
$$\begin{eqnarray*} P\left(X=x|Y=1\right) & = & \begin{cases} \frac{P\left(X=0,\,Y=1\right)}{0.6} & ,\text{for }x=0\\ \frac{P\left(X=1,\,Y=1\right)}{0.6} & ,\text{for }x=1 \end{cases}\\ & = & \begin{cases} \frac{0.4}{0.6} & ,\text{for }x=0\\ \frac{0.2}{0.6} & ,\text{for }x=1 \end{cases}\\ & = & \begin{cases} \frac{2}{3} & ,\text{for }x=0\\ \frac{1}{3} & ,\text{for }x=1 \end{cases} \end{eqnarray*}$$
You are correct. You are supposed to be getting "multiple" probabilities.
You said that your goal is to figure out $$P(X|Y = 1)$$. Well, in order to achieve your goal, you need to know what the probability that $$X = 0$$ given that $$Y = 1$$ is as well as what the probability that $$X = 1$$ given that $$Y = 1$$ is, which you did.
nbro's comment is basically telling you that when you get "multiple" probabilities, you are actually computing a probability mass (or density) function (or, equivalently, yet not the same, the distribution of the random variable $$X$$ given $$Y = 1$$).
Note: $$P(X|Y=1)=[4/6,2/6]$$ is the probability mass function of the random variable $$X$$ given $$Y = 1$$ because it tells you what that probability is for all the possible values of $$X$$, ie, $$X = 0$$ and $$X$$ = 1.
Finally, I use "multiple" in quotes because nobody really says that. In your case, they would ask something like: "This gives me the final answer of P(X|Y=1)=[4/6,2/6], but I'm not sure whether the answer should be $${\it \text{ a probability mass function}}$$ or a single probability." | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9825575116884778,
"lm_q1q2_score": 0.8571314739438562,
"lm_q2_score": 0.8723473829749844,
"openwebmath_perplexity": 185.84113606590793,
"openwebmath_score": 0.9003010988235474,
"tags": null,
"url": "https://stats.stackexchange.com/questions/389453/calculating-conditional-probability-with-marginal"
} |
# If series is absolutely convergent then $\sum \limits_{n\in I}a_n=\sum \limits_{k=1}^{\infty}\sum \limits_{n\in I_k}a_n.$
Suppose that the series $$\sum \limits_{n=1}^{\infty}a_n$$ is absolutely convergent and let $$I\subseteq \mathbb{N}$$ such that $$I=\bigsqcup\limits_{k=1}^{\infty}I_k$$. Then show that $$\sum \limits_{n\in I}a_n=\sum \limits_{k=1}^{\infty}\sum \limits_{n\in I_k}a_n. \qquad (*)$$
I don't have any idea how to solve it.
I do know that in any absolute convergent series permutation of terms does not change the sum and I guess it should be used somehow in order to prove equality $$(*)$$.
Can anyone show the rigorous proof of equality $$(*)$$, please?
• here is a proof. Here is another. – Masacroso May 16 at 20:14
• What's the definition of summation here? – Hashem Ben Abdelbaki May 16 at 20:14
• @Masacroso, link which you provided have not nothing in common with my question. I do know that in absolute convergent series any permutation does not change the sum. – ZFR May 16 at 20:22
• @ZFR you had written "I do know that in any absolute convergent series permutation of terms does not change the sum. But I want to prove it rigorously and cannot do it." Hence my comment providing two formal proofs. Make clear your question please. – Masacroso May 16 at 20:23
• @Masacroso, sorry about that. Done – ZFR May 16 at 20:27
First assume that $$a_n \ge 0$$ and define $$\sum_{n \in I} a_n = \sup_{J \subset I, J \text{ finite}} \sum_{n \in J} a_n$$. Note that it follows that if $$I \subset I'$$ then $$\sum_{n \in I} a_n \le \sum_{n \in I'} a_n$$.
From https://math.stackexchange.com/a/3680889/27978 we see that if $$K = K_1 \cup \cdots \cup K_m$$, a disjoint union, then $$\sum_{n \in K} a_n = \sum_{n \in K_1} a_n + \cdots + \sum_{n \in K_m} a_n$$. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.982557516796073,
"lm_q1q2_score": 0.8571314735082184,
"lm_q2_score": 0.8723473779969194,
"openwebmath_perplexity": 174.6296223630094,
"openwebmath_score": 0.9980955123901367,
"tags": null,
"url": "https://math.stackexchange.com/questions/3678244/if-series-is-absolutely-convergent-then-sum-limits-n-in-ia-n-sum-limits"
} |
Since $$I'=I_1 \cup \cdots \cup I_m \subset I$$ we see that $$\sum_{n \in I} a_n \ge \sum_{n \in I'} a_n = \sum_{k=1}^m \sum_{n \in I_k} a_n$$. It follows that $$\sum_{n \in I} a_n \ge \sum_{k=1}^\infty \sum_{n \in I_k} a_n$$. This is the 'easy' direction.
Let $$\epsilon>0$$, then there is some finite $$J \subset I$$ such that $$\sum_{n\in J} a_n > \sum_{n \in I} a_n -\epsilon$$. Since $$J$$ is finite and the $$I_k$$ are pairwise disjoint we have $$J \subset I'=I_1 \cup \cdots \cup I_m$$ for some $$m$$ and so $$\sum_{k=1}^\infty \sum_{n \in I_k} a_n \ge \sum_{k=1}^m\sum_{n \in I_k} a_n \ge \sum_{k=1}^m\sum_{n \in J \cap I_k} a_n = \sum_{n\in J} a_n > \sum_{n \in I} a_n -\epsilon$$.
(It is not relevant here, but a small proof tweak shows that the result holds true even if the $$a_n$$ do not have a finite sum.)
Now suppose we have $$a_n \in \mathbb{R}$$ and $$\sum_{n \in I} |a_n| = \sum_{n=1}^\infty |a_n|$$ is finite. We need to define what we mean by $$\sum_{n \in I} a_n$$. Note that $$(a_n)_+=\max(0,a_n) \ge 0$$ and $$(a_n)_-=\max(0,-a_n) \ge 0$$. Since $$0 \le (a_n)_+ \le |a_n|$$ and $$0 \le (a_n)_- \le |a_n|$$ we see that $$\sum_{n \in I} (a_n)_+ = \sum_{k=1}^\infty \sum_{n \in I_k} (a_n)_+$$ and similarly for $$(a_n)_-$$.
This suggests the definition (cf. Lebesgue integral) $$\sum_{n \in I} a_n = \sum_{n \in I} (a_n)_+ - \sum_{n \in I} (a_n)_-$$.
With this definition, all that remains to be proved is that $$\sum_{k=1}^\infty \sum_{n \in I_k} a_n = \sum_{k=1}^\infty \sum_{n \in I_k} (a_n)_+ - \sum_{k=1}^\infty \sum_{n \in I_k} (a_n)_-$$ and this follows from summability and the fact that for each $$k$$ we have $$\sum_{n \in I_k} a_n = \sum_{n \in I_k} (a_n)_+ - \sum_{n \in I_k} (a_n)_-$$. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.982557516796073,
"lm_q1q2_score": 0.8571314735082184,
"lm_q2_score": 0.8723473779969194,
"openwebmath_perplexity": 174.6296223630094,
"openwebmath_score": 0.9980955123901367,
"tags": null,
"url": "https://math.stackexchange.com/questions/3678244/if-series-is-absolutely-convergent-then-sum-limits-n-in-ia-n-sum-limits"
} |
Note: To elaborate the last sentence, recall that I defined $$\sum_{n \in I_k} a_n$$ to be $$\sum_{n \in I_k} (a_n)_+ - \sum_{n \in I_k} (a_n)_-$$, so all that is happening here is the definition is applied to $$I_k$$ rather than $$I$$. Then to finish, note that if $$d_k,b_k,c_k$$ are summable and satisfy $$d_k=b_k-c_k$$ then $$\sum_{k=1}^\infty d_k= \sum_{k=1}^\infty b_k- \sum_{n=1}^\infty c_k$$, where $$d_k = \sum_{n \in I_k} a_n$$, $$b_k = \sum_{n \in I_k} (a_n)_+$$ and $$c_k = \sum_{n \in I_k} (a_n)_-$$.
• Great answer! But probably you meant bijection $\sigma: \mathbb{N}\to I$, right? – ZFR May 18 at 18:42
• @ZFR: Correct, I will fix. – copper.hat May 18 at 18:58
• When you wrote that "It is straightforward to show that for any bijection $\sigma:\mathbb{N} \to I$ we have $\sum_{n \in I} a_n = \sum_{n=1}^\infty a_{\sigma(n)}$." Do you mean here that $a_n\geq 0$ or not? I am a bit confused – ZFR May 18 at 19:17
• (+1) This is less fussy, more self-contained, more authoritative, and better connected to the literature than my answer, so it ought to replace it as the accepted answer. (I haven't read @Matematleta's answer yet, because it looks harder to understand than these two, so I reserve judgement on it, for now!) On this answer: am I right in thinking that one can also write $\sum_{n \in I}a_n = s$ iff for all $\epsilon > 0$ there exists finite $J \subseteq I$ such that $\left\lvert\sum_{n \in J}a_n- s\right\rvert < \epsilon$? – Calum Gilhooley May 18 at 19:18
• You really helped me to understand the thing which i did not understand for a long time. Thank you so much! I really appreciate your permanent help! – ZFR May 18 at 20:50
Suppose for the moment that the result is known to be true for convergent series of non-negative terms. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.982557516796073,
"lm_q1q2_score": 0.8571314735082184,
"lm_q2_score": 0.8723473779969194,
"openwebmath_perplexity": 174.6296223630094,
"openwebmath_score": 0.9980955123901367,
"tags": null,
"url": "https://math.stackexchange.com/questions/3678244/if-series-is-absolutely-convergent-then-sum-limits-n-in-ia-n-sum-limits"
} |
If $$\sum_{n=1}^\infty a_n$$ is an absolutely convergent series of real numbers, define $$a_n = b_n - c_n,$$ for all $$n \geqslant 1,$$ where $$c_n = 0$$ when $$a_n \geqslant 0$$ and $$b_n = 0$$ when $$a_n \leqslant 0.$$ Then $$|a_n| = b_n + c_n,$$ therefore $$\sum_{n=1}^\infty b_n$$ and $$\sum_{n=1}^\infty c_n$$ are convergent series of non-negative terms, therefore: \begin{align*} \sum_{n \in I}a_n & = \sum_{n \in I}b_n - \sum_{n \in I}c_n \\ & = \sum_{k=1}^\infty\sum_{n \in I_k}b_n - \sum_{k=1}^\infty\sum_{n \in I_k}c_n \\ & = \sum_{k=1}^\infty\left( \sum_{n \in I_k}b_n - \sum_{n \in I_k}c_n\right) \\ & = \sum_{k=1}^\infty\sum_{n \in I_k}(b_n - c_n) \\ & = \sum_{k=1}^\infty\sum_{n \in I_k}a_n. \end{align*} So it is enough to prove the result on the assumption that $$a_n \geqslant 0$$ for all $$n \geqslant 1.$$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.982557516796073,
"lm_q1q2_score": 0.8571314735082184,
"lm_q2_score": 0.8723473779969194,
"openwebmath_perplexity": 174.6296223630094,
"openwebmath_score": 0.9980955123901367,
"tags": null,
"url": "https://math.stackexchange.com/questions/3678244/if-series-is-absolutely-convergent-then-sum-limits-n-in-ia-n-sum-limits"
} |
Given any set $$K \subseteq \mathbb{N},$$ I shall use the Iverson bracket notation: $$[n \in K] = \begin{cases} 1 & \text{if } n \in K, \\ 0 & \text{if } n \notin K. \end{cases}$$ I shall assume that, however the notation $$\sum_{n \in K}a_n$$ has been defined, it satisfies the identity: $$\sum_{n \in K}a_n = \sum_{n=1}^\infty a_n[n \in K].$$ Let $$J_k = I_1 \cup I_2 \cup \cdots \cup I_k$$ ($$k = 1, 2, \ldots$$). Because the $$I_k$$ are disjoint, we have $$[n \in J_k] = [n \in I_1] + [n \in I_2] + \cdots + [n \in I_k],$$ therefore $$\sum_{n \in I_1}a_n + \sum_{n \in I_2}a_n + \cdots + \sum_{n \in I_k}a_n = \sum_{n \in J_k}a_n \leqslant \sum_{n \in I}a_n,$$ therefore $$\sum_{k=1}^\infty\sum_{n \in I_k}a_n \leqslant \sum_{n \in I}a_n,$$ and the outer infinite sum on the left hand side exists, because its partial sums are bounded above by the sum on the right hand side. On the other hand, for all $$m \geqslant 1,$$ \begin{align*} \sum_{n=1}^ma_n[n \in I] & = \sum_{n=1}^ma_n[n \in I_1] + \sum_{n=1}^ma_n[n \in I_2] + \cdots + \sum_{n=1}^ma_n[n \in I_r] \\ & \leqslant \sum_{n \in I_1}a_n + \sum_{n \in I_2}a_n + \cdots + \sum_{n \in I_r}a_n \\ & \leqslant \sum_{k=1}^\infty\sum_{n \in I_k}a_n, \end{align*} where $$r = \max\{k \colon n \leqslant m \text{ for some } n \in I_k\},$$ therefore $$\sum_{n \in I}a_n \leqslant \sum_{k=1}^\infty\sum_{n \in I_k}a_n,$$ and the two inequalities together prove (*). | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.982557516796073,
"lm_q1q2_score": 0.8571314735082184,
"lm_q2_score": 0.8723473779969194,
"openwebmath_perplexity": 174.6296223630094,
"openwebmath_score": 0.9980955123901367,
"tags": null,
"url": "https://math.stackexchange.com/questions/3678244/if-series-is-absolutely-convergent-then-sum-limits-n-in-ia-n-sum-limits"
} |
• Thanks a lot for your answer! I read your answer and I have two questions: 1) I am a bit confused by the definition of $r$? Could you explain it in details, please? 2) Have you ever used that the series $\sum \limits_{n=1}^{\infty}a_n$ is absolutely convergent? – ZFR May 17 at 2:16
• I would define $r$ in this way: for each $n\in I$ s.t. $1\leq n\leq m$ there is $p(n)$ s.t. $n\in I_{p(n)}$ and let's take $r=\max \{p(1),\dots,p(m)\}$ then $[n\in I]=[n\in I_1]+\dots+[n\in I_r]$. Is my reasoning correct? And still I would be happy if you can explain and point the moments where you have used that the series is absolutely convergent. – ZFR May 17 at 2:30
• Your definition of $r$ looks right to me, and looks equivalent to mine. Certainly I had much the same idea in mind. I used the absolute convergence of $\sum_{n=1}^\infty a_n$ when I inferred that the two series of non-negative terms $\sum_{n=1}^\infty b_n$ and $\sum_{n=1}^\infty c_n$ are convergent. I'm going to try to get some more sleep now, but I'll have another look at the whole thing after lunch, and see if I can make it any clearer. (Assuming it isn't all just a load of dingo's kidneys, of course!) – Calum Gilhooley May 17 at 9:04
• I deliberately chose not to write \begin{gather*} b_n = \frac{|a_n| + a_n}2 \geqslant 0, \\ c_n = \frac{|a_n| - a_n}2 \geqslant 0, \end{gather*} but perhaps it would have been clearer had I done so. – Calum Gilhooley May 17 at 15:00
• Perhaps I should also have stated explicitly that the identity I took to be satisfied by $\sum_{n \in K}a_n$ is a possible definition of that notation; but it is not the only possible definition; and you didn't say what definition you were using; so I left it open. – Calum Gilhooley May 17 at 15:05 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.982557516796073,
"lm_q1q2_score": 0.8571314735082184,
"lm_q2_score": 0.8723473779969194,
"openwebmath_perplexity": 174.6296223630094,
"openwebmath_score": 0.9980955123901367,
"tags": null,
"url": "https://math.stackexchange.com/questions/3678244/if-series-is-absolutely-convergent-then-sum-limits-n-in-ia-n-sum-limits"
} |
I think there is an elementary proof (one without measure theory), that we can adapt from a similar claim in Apostol's Analysis book. Without loss of generality, $$I=\mathbb N$$. For each $$k\in \mathbb N,\ I_k$$ may be regarded as a map from some subset $$\{1,2,\cdots,\}\subseteq \mathbb N$$, to $$\{\sigma_k(1),\sigma_k(2),\cdots,\}$$ which may or may not be infinite, so $$\sigma_k$$ is an injective map from the susbet of $$\mathbb N$$ of the same cardinality as $$|I_k|,$$ starting at $$1$$, to the $$\textit{set}\ I_k.$$ If $$|I_k|=j$$, extend $$I_k$$ to all of $$\mathbb N$$ by mapping $$n\in \mathbb N\setminus \{1,2,\cdots, j\}$$ to $$\mathbb N\setminus \{\sigma_k(1),\sigma_k(2),\cdots, \sigma_k(j)\}$$ injectively and defining $$a'_n:=0$$ for all $$n\in \mathbb N\setminus \{\sigma_k(1),\sigma_k(2),\cdots, \sigma_k(j)\}$$. This construction will not affect any of the sums, so without loss of generality, $$I_k$$ maps $$\mathbb N$$ to a subset of $$\mathbb N$$ such that
$$\tag1 I_k\ \text{is injective on}\ \mathbb N$$
$$\tag2 \text{the range of each}\ I_k \ \text{is a subset of } \ \mathbb N, \text{say}\ P_k$$
$$\tag3 \text{the}\ P_k\ \text{are disjoint}$$
Now put $$\tag4 b_k(n)=a_{I_{k}(n)}\ \text{and}\ s_k=\sum^\infty_{n=0}b_k(n)$$
which is well-defined by $$(1)-(3).$$ We have to prove that
$$\tag5 \sum^\infty_{k=0}a_k=\sum^\infty_{k=0}s_k$$
It's easy to show that the right hand side of this converges absolutely. To find the sum, set $$\epsilon>0$$ and choose $$N$$ large enough so that $$\sum^\infty_{k=0}|a_k|-\sum^n_{k=0}|a_k|<\frac{\epsilon}{2}$$ as soon as $$n>N.$$ This implies also that
$$\tag6\left|\sum^\infty_{k=0}a_k-\sum^n_{k=0}a_k\right|<\frac{\epsilon}{2}$$
Now choose $$\{I_1,\cdots, I_r\}$$ so that each element of $$\{a_1,\dots ,a_N\}$$ appears in the sum $$\sum^\infty_{n=0}a_{I_{1(n)}}+\cdots +\sum^\infty_{n=0}a_{I_{r(n)}}=s_1+\cdots+ s_r.$$ Then, if $$n>r,N$$ we have | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.982557516796073,
"lm_q1q2_score": 0.8571314735082184,
"lm_q2_score": 0.8723473779969194,
"openwebmath_perplexity": 174.6296223630094,
"openwebmath_score": 0.9980955123901367,
"tags": null,
"url": "https://math.stackexchange.com/questions/3678244/if-series-is-absolutely-convergent-then-sum-limits-n-in-ia-n-sum-limits"
} |
$$\tag 7\left|\sum^n_{k=0}s_k-\sum^n_{k=0}a_k\right|<\sum^\infty_{n=N+1}<\frac{\epsilon}{2}$$
Now $$(5)$$ folllows from $$(6)$$ and $$(7).$$
• I am a bit confused with your definition of $I_k$? Could you clarify it, please? – ZFR May 16 at 23:26
• $I_k$ is some subset of $\mathbb N$. So it may be regarded as a function on the subset of $\mathbb N$ of the same cardinality. For instance, if $I_k=\{2,5,7\}$ then $I_k$ the function would map $\{1,2,3\}$ to $\{2,5,7\}$. If $I_k$ is infinite, then it is countable, so there is an injective function from $\mathbb N$ to it. That would be our $I_k$ considered as a function on $\mathbb N.$ – Matematleta May 16 at 23:31
• Hmm. I guess that i got you. Also could you show why the RHS of (5) converges absolutely, please? – ZFR May 16 at 23:37
• Hint: show directly by comparison with $\sum |a_n|$ that $\{s_k\}$ has bounded partial sums. – Matematleta May 16 at 23:42 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.982557516796073,
"lm_q1q2_score": 0.8571314735082184,
"lm_q2_score": 0.8723473779969194,
"openwebmath_perplexity": 174.6296223630094,
"openwebmath_score": 0.9980955123901367,
"tags": null,
"url": "https://math.stackexchange.com/questions/3678244/if-series-is-absolutely-convergent-then-sum-limits-n-in-ia-n-sum-limits"
} |
# Why don't you need the Axiom of Choice when constructing the “inverse” of an injection?
Suppose $f:X\rightarrow Y$ is a surjection and you want to show that there exists $g:Y\rightarrow X$ s.t. $f\circ g=\mathrm{id}_Y$. You need the AC to show this.
However, suppose $f$ is a injection and you want show there is $g$ s.t. $g\circ f=\mathrm{id}_X$. Then, according to my textbook, you don't need the AC to show this.
This is counterintuitive to me, because it's like you need a special axiom to claim that an infinite product of big sets is nonempty, while you don't need one to claim that an infinite product of singleton sets is nonempty, which seems smaller than the former.
So why don't you need the AC to show the latter?
EDIT: $X$ should be nonempty.
EDIT 2: I realized (after asking this) that my question mostly concerns whether the AC is needed to say that an infinite product of finite sets is nonempty, and why.
• I can't fix your details (Asaf will probably see to that) but here's the intuition: Why do we need AC? Because with big sets, the problem is specifying how to pick the elements. No such problem with singleton sets! – Ragib Zaman Apr 16 '12 at 13:32
• Making one choice (or some fixed finite number of choices) doesn't require Axiom of Choice. It's entailed by first-order logic, q.v. Rule C. – hardmath Apr 16 '12 at 13:42
• hardmath: your comment is best in answering my question as I hold in my heart. Could you please elaborate on that? – Pteromys Apr 16 '12 at 13:47
• @Pteromys: Read this answer to why we can choose from non-empty sets. Finitely many choices follow by induction. – Asaf Karagila Apr 16 '12 at 13:48
• @Pteromys: As others have noted, the Axiom of Choice is needed to deal with infinite products of different sets, even in the case those sets each contain two elements. We've moved sufficiently far from your original Question, it might be best to formulate a new one if you want elaboration! – hardmath Apr 16 '12 at 14:58 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9825575162853136,
"lm_q1q2_score": 0.857131463280188,
"lm_q2_score": 0.8723473680407889,
"openwebmath_perplexity": 129.30781144029993,
"openwebmath_score": 0.8408396244049072,
"tags": null,
"url": "https://math.stackexchange.com/questions/132495/why-dont-you-need-the-axiom-of-choice-when-constructing-the-inverse-of-an-inj/132498"
} |
The need for the axiom of choice is to choose arbitrary elements. Injectivity eliminates this need.
Assume that $A$ is not empty, if $f\colon A\to B$ is injective this means that if $b\in B$ is in the range of $f$ then there is a unique $a\in A$ such that $f(a)=b$.
This means that we can define (from $f$) what is the $a$ to which we send $b$.
So if $f$ is not onto $B$ we have two options:
1. $b\in B$ in the range of $f$, then we have exactly one option to send $b$ to.
2. $b\in B$ not in the range of $f$. Since $A$ is not empty fix in advance some $a_0\in A$ and send $b$ to $a_0$.
Another way to see this is let $B=B'\cup Rng(f)$, where $B'\cap Rng(f)=\varnothing$. Fix $a_0\in A$ and define $g|_{B'}(x)=a_0$. For every $b\in Rng(f)$ we have that $f^{-1}[\{b\}]=\{a\in A\mid f(a)=b\}$ is a singleton, so there is only one function which we can define in: $$\prod_{b\in Rng(f)}f^{-1}[\{b\}]$$
Now let the unique function in the product be $g|_{Rng(f)}$ and define $g$ to be the union of these two.
Your intuition about the need for the axiom of choice is true for surjections, if $f$ was surjective then we only know that $f^{-1}[\{b\}]$ is non-empty for every $b\in B$, and we need the full power of the axiom of choice to ensure that an arbitrary surjection has an inverse function.
To the edited question:
The axiom of choice is needed because we have models in which the axiom of choice does not hold, where there exists an infinite family of pairs whose product is empty.
There are weaker forms from which follow choice principles for finite sets. However these are still not provable from ZF on its own.
As indicated by Chris Eagle in the comments, and as I remark above, in a product of singletons there is no need for the axiom of choice since there is only one way to choose from a singleton. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9825575162853136,
"lm_q1q2_score": 0.857131463280188,
"lm_q2_score": 0.8723473680407889,
"openwebmath_perplexity": 129.30781144029993,
"openwebmath_score": 0.8408396244049072,
"tags": null,
"url": "https://math.stackexchange.com/questions/132495/why-dont-you-need-the-axiom-of-choice-when-constructing-the-inverse-of-an-inj/132498"
} |
• Why not just "flip the graph" with $(x,y) \mapsto (y,x)$ to get a function $g: f(X) \to X$ (since $f$ is injective) and extend it to $Y \smallsetminus f(X)$ by $g(y) = x_0$ as you said? – t.b. Apr 16 '12 at 13:42 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9825575162853136,
"lm_q1q2_score": 0.857131463280188,
"lm_q2_score": 0.8723473680407889,
"openwebmath_perplexity": 129.30781144029993,
"openwebmath_score": 0.8408396244049072,
"tags": null,
"url": "https://math.stackexchange.com/questions/132495/why-dont-you-need-the-axiom-of-choice-when-constructing-the-inverse-of-an-inj/132498"
} |
# One-zero dividend
## Challenge description
For every positive integer n there exists a number having the form of 111...10...000 that is divisible by n i.e. a decimal number that starts with all 1's and ends with all 0's. This is very easy to prove: if we take a set of n+1 different numbers in the form of 111...111 (all 1's), then at least two of them will give the same remainder after division by n (as per pigeonhole principle). The difference of these two numbers will be divisible by n and will have the desired form. Your aim is to write a program that finds this number.
## Input description
A positive integer.
## Output description
A number p in the form of 111...10...000, such that p ≡ 0 (mod n). If you find more than one - display any of them (doesn't need to be the smallest one).
## Notes
Your program has to give the answer in a reasonable amount of time. Which means brute-forcing is not permited:
p = 0
while (p != 11..10.00 and p % n != 0)
p++
Neither is this:
do
p = random_int()
while (p != 11..10.00 and p % n != 0)
Iterating through the numbers in the form of 11..10..00 is allowed.
Your program doesn't need to handle an arbitrarily large input - the upper bound is whatever your language's upper bound is.
## Sample outputs
2: 10
3: 1110
12: 11100
49: 1111111111111111111111111111111111111111110
102: 1111111111111111111111111111111111111111111111110 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9825575175622121,
"lm_q1q2_score": 0.8571314611332632,
"lm_q2_score": 0.8723473647220786,
"openwebmath_perplexity": 1459.298190489912,
"openwebmath_score": 0.38380372524261475,
"tags": null,
"url": "https://codegolf.stackexchange.com/questions/74391/one-zero-dividend"
} |
• Can we have a reasonable upper bound to the possible output? (Something about less than 2.4 billion (approx. the max value of a signed integer) should be fine, as arrays or lists might be required for some implementations) – Tamoghna Chowdhury Feb 28 '16 at 15:25
• @MartinBüttner I think that the first satisfying output should be enough (reasonable timeframe constraint) – Tamoghna Chowdhury Feb 28 '16 at 15:26
• The last 0 is not necessary in the 49 test case. – CalculatorFeline Feb 28 '16 at 15:36
• @CatsAreFluffy I think all numbers need to contain at least 1 and at least one 0, otherwise 0 is a solution for any input. (Would be good to clarify this though.) – Martin Ender Feb 28 '16 at 15:38
• Just requiring one 1 should work. – CalculatorFeline Feb 28 '16 at 15:41
## Mathematica, 29 bytes
⌊10^(9EulerPhi@#)/9⌋10^#&
Code by Martin Büttner.
On input $$\n\$$, this outputs the number with $$\9\varphi(n)\$$ ones followed by $$\n\$$ zeroes, where $$\\varphi(\cdot)\$$ is the Euler totient function. With a function phi, this could be expressed in Python as
lambda n:'1'*9*phi(n)+'0'*n
It would suffice to use the factorial $$\n!\$$ instead of $$\\varphi(n)\$$, but printing that many ones does not have a reasonable run-time.
Claim: $$\9\varphi(n)\$$ ones followed by $$\n\$$ zeroes is a multiple of $$\n\$$.
Proof: First, let's prove this for the case that $$\n\$$ is not a multiple of $$\2, 3, \text{or } 5\$$. We'll show the number with consisting of $$\\varphi(n)\$$ ones is a multiple of $$\n\$$.
The number made of $$\k\$$ ones equals $$\\frac{10^k-1}9\$$. Since $$\n\$$ is not a multiple of $$\3\$$, this is a multiple of $$\n\$$ as long as $$\10^k-1\$$ is a factor of $$\n\$$, or equivalently if $$\10^k \equiv 1\mod n\$$. Note that this formulation makes apparent that if $$\k\$$ works for the number of ones, then so does any multiple of $$\k\$$. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9825575175622121,
"lm_q1q2_score": 0.8571314611332632,
"lm_q2_score": 0.8723473647220786,
"openwebmath_perplexity": 1459.298190489912,
"openwebmath_score": 0.38380372524261475,
"tags": null,
"url": "https://codegolf.stackexchange.com/questions/74391/one-zero-dividend"
} |
So, we're looking for $$\k\$$ to be a multiple of the order of $$\k\$$ in the multiplicative group modulo n. By Lagrange's Theorem, any such order is a divisor of the size of the group. Since the elements of the group are the number from $$\1\$$ to $$\n\$$ that are relatively prime to $$\n\$$, its size is the Euler totient function $$\\varphi(n)\$$. So, we've shown that $$\10^{\varphi(n)} \equiv 1 \mod n\$$, and so the number made of $$\\varphi(n)\$$ ones is a multiple of $$\n\$$.
Now, let's handle potential factors of $$\3\$$ in $$\n\$$. We know that $$\10^{\varphi(n)}-1\$$ is a multiple of $$\n\$$, but $$\\frac{10^{\varphi(n)}-1}9\$$ might not be. But, $$\\frac{10^{9\varphi(n)}-1}9\$$ is a multiple of $$\9\$$ because it consists of $$\9\varphi(n)\$$ ones, so the sum of its digits a multiple of $$\9\$$. And we've noted that multiplying the exponent $$\k\$$ by a constant preserves the divisibility.
Now, if $$\n\$$ has factors of $$\2\$$'s and $$\5\$$'s, we need to add zeroes to end of the output. It way more than suffices to use $$\n\$$ zeroes (in fact $$\\log_2(n)\$$ would do). So, if our input $$\n\$$ is split as $$\n = 2^a \times 5^b \times m\$$, it suffices to have $$\9\varphi(m)\$$ ones to be a multiple of $$\n\$$, multiplied by $$\10^n\$$ to be a multiple of $$\2^a \times 5^b\$$. And, since $$\n\$$ is a multiple of $$\m\$$, it suffices to use $$\9\varphi(n)\$$ ones. So, it works to have $$\9\varphi(n)\$$ ones followed by $$\n\$$ zeroes.
• Just to make sure no one thinks this was posted without my permission: xnor came up with the method and proof all on his own, and I just supplied him with a Mathematica implementation, because it has a built-in EulerPhi function. There is nothing mind-blowing to the actual implementation, so I'd consider this fully his own work. – Martin Ender Feb 28 '16 at 18:29
## Python 2, 44 bytes
f=lambda n,j=1:j/9*j*(j/9*j%n<1)or f(n,j*10) | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9825575175622121,
"lm_q1q2_score": 0.8571314611332632,
"lm_q2_score": 0.8723473647220786,
"openwebmath_perplexity": 1459.298190489912,
"openwebmath_score": 0.38380372524261475,
"tags": null,
"url": "https://codegolf.stackexchange.com/questions/74391/one-zero-dividend"
} |
## Python 2, 44 bytes
f=lambda n,j=1:j/9*j*(j/9*j%n<1)or f(n,j*10)
When j is a power of 10 such as 1000, the floor-division j/9 gives a number made of 1's like 111. So, j/9*j gives 1's followed by an equal number of 0's like 111000.
The function recursively tests numbers of this form, trying higher and higher powers of 10 until we find one that's a multiple of the desired number.
• Oh, good point, we only need to check 1^n0^n... – Martin Ender Feb 28 '16 at 16:03
• @MartinBüttner If it's any easier, it also suffices to fix the number of 0's to be the input value. Don't know if it counts as efficient to print that many zeroes though. – xnor Feb 28 '16 at 16:37
• Why does checking 1^n0^n work? – Lynn Feb 28 '16 at 17:54
• @Lynn Adding more zeroes can't hurt, and there's infinitely many possible numbers of ones, some number will have enough of both ones and zeroes. – xnor Feb 28 '16 at 18:15
# Pyth, 11 bytes
.W%HQsjZTT
Test suite
Basically, it just puts a 1 in front and a 0 in back over and over again until the number is divisible by the input.
Explanation:
.W%HQsjZTT
Implicit: Q = eval(input()), T = 10
.W while loop:
%HQ while the current value mod Q is not zero
jZT Join the string "10" with the current value as the separator.
s Convert that to an integer.
T Starting value 10.
# Haskell, 51 bytes
\k->[b|a<-[1..],b<-[div(10^a)9*10^a],bmodk<1]!!0
Using xnor’s approach. nimi saved a byte!
## CJam, 2825 19 bytes
Saved 6 bytes with xnor's observation that we only need to look at numbers of the form 1n0n.
ri:X,:)Asfe*{iX%!}=
Test it here.
### Explanation
ri:X e# Read input, convert to integer, store in X.
,:) e# Get range [1 ... X].
As e# Push "10".
fe* e# For each N in the range, repeat the characters in "10" that many times,
e# so we get ["10" "1100" "111000" ...].
{iX%!}= e# Select the first element from the list which is divided by X.
# JavaScript (ES6), 65 bytes | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9825575175622121,
"lm_q1q2_score": 0.8571314611332632,
"lm_q2_score": 0.8723473647220786,
"openwebmath_perplexity": 1459.298190489912,
"openwebmath_score": 0.38380372524261475,
"tags": null,
"url": "https://codegolf.stackexchange.com/questions/74391/one-zero-dividend"
} |
# JavaScript (ES6), 65 bytes
Edit 2 bytes saved thx @Neil
It works within the limits of javascript numeric type, with 17 significant digits. (So quite limited)
a=>{for(n='';!(m=n+=1)[17];)for(;!(m+=0)[17];)if(!(m%a))return+m}
Less golfed
function (a) {
for (n = ''; !(m = n += '1')[17]; )
for (; !(m += '0')[17]; )
if (!(m % a))
return +m;
}
• Why not for(m=n;? – Neil Feb 28 '16 at 17:11
• @Neil because I need at least one zero. Maybe I can find a shorter way ... (thx for the edit) – edc65 Feb 28 '16 at 18:30
• Oh, that wasn't clear in the question, but I see now that the sample outputs all have at least one zero. In that case you can still save a byte using for(m=n;!m[16];)if(!((m+=0)%a)). – Neil Feb 28 '16 at 18:42
• @Neil or even 2 bytes. Thx – edc65 Feb 28 '16 at 18:52
# Husk, 11 10 bytes
-1 byte thanks to Razetime!
ḟ¦⁰modṘḋ2N
Try it online! Constructs the infinite list [10,1100,111000,...] and selects the first element which is a multiple of the argument.
# Mathematica, 140 55 bytes
NestWhile["1"<>#<>"0"&,"1",FromDigits@#~Mod~x>0&/.x->#]
Many bytes removed thanks to xnor's 1^n0^n trick.
Minimal value, 140 156 bytes This gives the smallest possible solution.
NestWhile["1"<>#&,ToString[10^(Length@NestWhileList[If[EvenQ@#,If[10~Mod~#>0,#/2,#/10],#/5]&,#,Divisors@#~ContainsAny~{2, 5}&],FromDigits@#~Mod~m>0&/.m->#]&
It calculates how many zeros are required then checks all possible 1 counts until it works. It can output a number with no 0 but that can be fixed by adding a <>"0" right before the final &.
## Haskell, 37 bytes
f n=[d|d<-"10",i<-[1..n*9],gcd n i<2]
This uses the fact that it works to have 9*phi(n) ones, where phi is the Euler totient function. Here, it's implemented using gcd and filtering, producing one digit for each value i that's relatively prime to it that is in the range 1 and 9*n. It also suffices to use this many zeroes.
# Jelly, 11 bytes
⁵DxⱮḅ⁵ḍ@Ƈ⁸Ḣ
Try it online! | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9825575175622121,
"lm_q1q2_score": 0.8571314611332632,
"lm_q2_score": 0.8723473647220786,
"openwebmath_perplexity": 1459.298190489912,
"openwebmath_score": 0.38380372524261475,
"tags": null,
"url": "https://codegolf.stackexchange.com/questions/74391/one-zero-dividend"
} |
# Jelly, 11 bytes
⁵DxⱮḅ⁵ḍ@Ƈ⁸Ḣ
Try it online!
Ignoring runtime requirements, we can reduce this to 10 bytes using xnor's method, but with $$\\varphi(n)\$$ replaced with $$\n!\$$:
!,µ⁵Dx"µFḌ
Try it online!
## How they work
⁵DxⱮḅ⁵ḍ@Ƈ⁸Ḣ - Main link. Takes n on the left
⁵D - Yield [1, 0]
Ɱ - For each integer i in [1, 2, ..., n]:
x - Repeat each element of [1, 0] i times
ḅ⁵ - Convert each back into an integer
Ƈ - Keep those for which the following is true:
ḍ@ - Is divisible by
⁸ - n
Ḣ - Take the first value of those remaining
!,µ⁵Dx"µFḌ - Main link. Takes n on the left
! - Yield n!
, - Yield [n!, n]. Call this l
µ - Begin new link with l on the left
⁵D - Yield [1, 0]
" - Zip [1, 0] with [n!, n] and do the following over each pair:
x - Repeat
- This yields [[1, 1, ..., 1], [0, 0, ..., 0]].
The first element has n! 1s and the second n 0s
Call this k
µ - Begin new link with k on the left
F - Flatten k
Ḍ - Convert to integer
# Perl 5, 26 bytes
includes a byte for -n (-M5.01 is free)
($.="1$.0")%$_?redo:say$.
## Explanation
$. starts off with value 1. We immediately concatenate it with 1 beforehand and 0 afterward, yielding 110, and reassign that to $. — that's what the $.="1$.0" does.
The assignment returns the assigned value so when we take it modulo the input number $_ we obtain 0 (false) if 110 is divisible by the input and nonzero (true) otherwise. • In the latter case, we redo, i.e. repeat the block (the -n switch makes the whole thing a loop block, and incidentally, assigns $_ to the input number). This concatenates another 1 and 0, yielding 11100, etc.
• If 110, or 11100, or 1111000, or whatever we're up to, is divisible by the input, we stop and say (print) it. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9825575175622121,
"lm_q1q2_score": 0.8571314611332632,
"lm_q2_score": 0.8723473647220786,
"openwebmath_perplexity": 1459.298190489912,
"openwebmath_score": 0.38380372524261475,
"tags": null,
"url": "https://codegolf.stackexchange.com/questions/74391/one-zero-dividend"
} |
Note that every number divides such a number (i.e. that has one more 1 than the numbers of 0s it has). After all, if it divides something with more 1s than 0s, you can append 0s and it'll still divide that. And if divides something with fewer 1s than 0s, say m 1s and n 0s with m < n, then it also divides the number with 2m 1s and n 0s.
## Sage, 33 bytes
lambda n:'1'*9*euler_phi(n)+'0'*n
This uses xnor's method to produce the output.
Try it online
# 05AB1E, 8 bytes
$Õ9*×Î׫ Port of @xnor's Mathematica answer, so make sure to upvote him!! Explanation: $ # Push 1 and the input-integer
Õ # Pop the input, and get Euler's totient of this integer
9* # Multiply it by 9
× # Repeat the 1 that many times as string
Î # Push 0 and the input-integer
× # Repeat the 0 the input amount of times as string
« # Concatenate the strings of 1s and 0s together
# (after which the result is output implicitly)
A more to-the-point iterative approach would be 11 bytes:
TS∞δ×øJ.ΔIÖ
Explanation:
T # Push 10
S # Convert it to a list of digits: [1,0]
∞ # Push an infinite positive list: [1,2,3,...]
δ # Apply double-vectorized on these two lists:
× # Repeat the 1 or 0 that many times as string
# (we now have a pair of infinite lists: [[1,11,111,...],[0,00,000,...]])
ø # Zip/transpose; swapping rows/columns:
# [[1,0],[11,00],[111,000],...]
J # Join each inner pair together:
# [10,1100,111000,...]
.Δ # Find the first value in this list which is truthy for:
IÖ # Check that it's divisible by the input-integer
# (after which the result is output implicitly)
# bc, 58 bytes
define f(n){for(x=1;m=10^x/9*10^x;++x)if(m%n==0)return m;}
## Sample results | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9825575175622121,
"lm_q1q2_score": 0.8571314611332632,
"lm_q2_score": 0.8723473647220786,
"openwebmath_perplexity": 1459.298190489912,
"openwebmath_score": 0.38380372524261475,
"tags": null,
"url": "https://codegolf.stackexchange.com/questions/74391/one-zero-dividend"
} |
# bc, 58 bytes
define f(n){for(x=1;m=10^x/9*10^x;++x)if(m%n==0)return m;}
## Sample results
200: 111000
201: 111111111111111111111111111111111000000000000000000000000000000000
202: 11110000
203: 111111111111111111111111111111111111111111111111111111111111111111111111111111111111000000000000000000000000000000000000000000000000000000000000000000000000000000000000
204: 111111111111111111111111111111111111111111111111000000000000000000000000000000000000000000000000
205: 1111100000
206: 11111111111111111111111111111111110000000000000000000000000000000000
207: 111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
208: 111111000000
209: 111111111111111111000000000000000000
210: 111111000000
211: 111111111111111111111111111111000000000000000000000000000000
212: 11111111111110000000000000
213: 111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
214: 1111111111111111111111111111111111111111111111111111100000000000000000000000000000000000000000000000000000
215: 111111111111111111111000000000000000000000
216: 111111111111111111111111111000000000000000000000000000
217: 111111111111111111111111111111000000000000000000000000000000
218: 111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
219: 111111111111111111111111000000000000000000000000
# dc, 27 bytes
Odsm[O*lmdO*sm+O*dln%0<f]sf | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9825575175622121,
"lm_q1q2_score": 0.8571314611332632,
"lm_q2_score": 0.8723473647220786,
"openwebmath_perplexity": 1459.298190489912,
"openwebmath_score": 0.38380372524261475,
"tags": null,
"url": "https://codegolf.stackexchange.com/questions/74391/one-zero-dividend"
} |
# dc, 27 bytes
Odsm[O*lmdO*sm+O*dln%0<f]sf
This defines a function f that expects its argument in the variable n. To use it as a program, ?sn lfx p to read from stdin, call the function, and print the result to stdout. Variable m and top of stack must be reset to 10 (by repeating Odsm) before f can be re-used.
## Results: | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9825575175622121,
"lm_q1q2_score": 0.8571314611332632,
"lm_q2_score": 0.8723473647220786,
"openwebmath_perplexity": 1459.298190489912,
"openwebmath_score": 0.38380372524261475,
"tags": null,
"url": "https://codegolf.stackexchange.com/questions/74391/one-zero-dividend"
} |
## Results:
200: 111000
201: 111111111111111111111111111111111000000000000000000000000000000000
202: 11110000
203: 111111111111111111111111111111111111111111111111111111111111111111111111111111111111000000000000000000000000000000000000000000000000000000000000000000000000000000000000
204: 111111111111111111111111111111111111111111111111000000000000000000000000000000000000000000000000
205: 1111100000
206: 11111111111111111111111111111111110000000000000000000000000000000000
207: 111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
208: 111111000000
209: 111111111111111111000000000000000000
210: 111111000000
211: 111111111111111111111111111111000000000000000000000000000000
212: 11111111111110000000000000
213: 111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
214: 1111111111111111111111111111111111111111111111111111100000000000000000000000000000000000000000000000000000
215: 111111111111111111111000000000000000000000
216: 111111111111111111111111111000000000000000000000000000
217: 111111111111111111111111111111000000000000000000000000000000
218: 111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
219: 111111111111111111111111000000000000000000000000
` | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9825575175622121,
"lm_q1q2_score": 0.8571314611332632,
"lm_q2_score": 0.8723473647220786,
"openwebmath_perplexity": 1459.298190489912,
"openwebmath_score": 0.38380372524261475,
"tags": null,
"url": "https://codegolf.stackexchange.com/questions/74391/one-zero-dividend"
} |
1. ## Completing the Square
What is the minimum point of the graph of the equation
y = 2x^2 + 8x + 9?
I understand the minimum point is the vertex (h, k).
To complete the square, I set the function to zero first.
2x^2 + 8x + 9 = 0
I then subtracted 9 from both sides.
2x^2 + 8x = -9
I then took half of the second term and squared it.
2x^2 + 8x = 16 - 9
Where do I go from here?
2. Hello !
y = 2x² + 8x + 9
y = 2 (x² + 4x + 9/2)
y = 2 ((x+2)²+1/2)
Therefore the minimum point of the graph is such as x+2=0 => x=-2 and y=1
3. Originally Posted by magentarita
What is the minimum point of the graph of the equation
y = 2x^2 + 8x + 9?
I understand the minimum point is the vertex (h, k).
To complete the square, I set the function to zero first.
2x^2 + 8x + 9 = 0
I then subtracted 9 from both sides.
2x^2 + 8x = -9
I then took half of the second term and squared it.
2x^2 + 8x = 16 - 9
Where do I go from here?
===============================================
magentarita,
I see where your problem is.
2x^2 + 8x = -9 ------ ok so far.
Before you take half of the secod term and square,
Make sure the coefficient of suared term 2x^2 to be 1.
Therefore, 2x^2 + 8x = -9
becomes (x^2) + 4x = -9/2 -- simply didived by 2.
(x^2) + 4x = -9/2 -- now you are ready to take half of the secod term and square, you get
(x^2) + 4x + 2^2 = (-9/2 ) + 2^2
(x + 2)^2 = -1/2 now complete the square part is done at this point.
Rewrite original question to y = [(x + 2)^2 ]+ 1/2
Can you see y = a ( x-h )^2 + k --- standard form.
Vertex is at ( h, k ) = ( -2, 1/2) and a =1
Opens up vertically, therefor, mini. point is at vertex.
4. ## ok............ | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9825575178175919,
"lm_q1q2_score": 0.8571314597256313,
"lm_q2_score": 0.8723473630627235,
"openwebmath_perplexity": 1917.9567759754982,
"openwebmath_score": 0.8046611547470093,
"tags": null,
"url": "http://mathhelpforum.com/pre-calculus/62126-completing-square.html"
} |
4. ## ok............
Originally Posted by 2976math
===============================================
magentarita,
I see where your problem is.
2x^2 + 8x = -9 ------ ok so far.
Before you take half of the secod term and square,
Make sure the coefficient of suared term 2x^2 to be 1.
Therefore, 2x^2 + 8x = -9
becomes (x^2) + 4x = -9/2 -- simply didived by 2.
(x^2) + 4x = -9/2 -- now you are ready to take half of the secod term and square, you get
(x^2) + 4x + 2^2 = (-9/2 ) + 2^2
(x + 2)^2 = -1/2 now complete the square part is done at this point.
Rewrite original question to y = [(x + 2)^2 ]+ 1/2
Can you see y = a ( x-h )^2 + k --- standard form.
Vertex is at ( h, k ) = ( -2, 1/2) and a =1
Opens up vertically, therefor, mini. point is at vertex.
ok but none the choices given for the minimum point is (-2, 1/2).
(2,33)
(2,17)
(-2,-15)
(-2,1)
5. ## ok....
Originally Posted by running-gag
Hello !
y = 2x² + 8x + 9
y = 2 (x² + 4x + 9/2)
y = 2 ((x+2)²+1/2)
Therefore the minimum point of the graph is such as x+2=0 => x=-2 and y=1
You got the right answer. But how did you get y = 1?
I got y = 1/2
6. Originally Posted by magentarita
You got the right answer. But how did you get y = 1?
I got y = 1/2
Substitute x= -2 into the expression :
2 [ (-2+2)² + 1/2 ] = 2 [ 0+1/2 ] = 1
7. Originally Posted by magentarita
ok but none the choices given for the minimum point is (-2, 1/2).
(2,33)
(2,17)
(-2,-15)
(-2,1)
I found the problem!
y should be 1 not 1/2, so answer (-2,1) is correct.
Reason: I forgot to divid y by 2.
Remember orginal question: y =( 2x^2 ) + 8x +9
(1/2) y = (x^2) + 4x + ( 9/2) -- divid by 2 both side
(1/2) y = [( x +2)^2 ] + (1/2) -- complete square right side
y = 2 [( x +2)^2 ] + 1 ----- multiply 2 both side
y = a [ ( x-h)^2 ] + k
where a=2, h=-2, k=1 sorry
8. I think you were originally trying to solve for the minimum point by finding the vertex through the method of 'completing the square'. | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9825575178175919,
"lm_q1q2_score": 0.8571314597256313,
"lm_q2_score": 0.8723473630627235,
"openwebmath_perplexity": 1917.9567759754982,
"openwebmath_score": 0.8046611547470093,
"tags": null,
"url": "http://mathhelpforum.com/pre-calculus/62126-completing-square.html"
} |
From $y=ax^2+bx+c$, you want to convert to $y=a(x-h)^2+k$, where $(h, k)$ represent the vertex of your parabola and thus the maximum or minimum point depending on a<0 or a>0.
$y=2x^2+8x+9$ Factor 2 out of first 2 terms
$y=2(x^2+4x)+9$ Next, complete the square in parentheses.
$y=2(x^2+4x+4)+9-8$ Notice that we added 2 times 4, so we subtract 8 to balance the side
$y=2(x+2)^2+1$ Now, we're in vertex form.
$V(h, k)=V(-2, 1) = \ \ minimum \ \ point$
9. ## ok.........
Originally Posted by Moo
Substitute x= -2 into the expression :
2 [ (-2+2)² + 1/2 ] = 2 [ 0+1/2 ] = 1
By subbing -2 for x, I get y = 1.
I got it.
10. ## ok..........
Originally Posted by 2976math
I found the problem!
y should be 1 not 1/2, so answer (-2,1) is correct.
Reason: I forgot to divid y by 2.
Remember orginal question: y =( 2x^2 ) + 8x +9
(1/2) y = (x^2) + 4x + ( 9/2) -- divid by 2 both side
(1/2) y = [( x +2)^2 ] + (1/2) -- complete square right side
y = 2 [( x +2)^2 ] + 1 ----- multiply 2 both side
y = a [ ( x-h)^2 ] + k
where a=2, h=-2, k=1 sorry
Be be sorry, be right. We all make mistakes, right?
11. ## ok..........
Originally Posted by masters
I think you were originally trying to solve for the minimum point by finding the vertex through the method of 'completing the square'.
From $y=ax^2+bx+c$, you want to convert to $y=a(x-h)^2+k$, where $(h, k)$ represent the vertex of your parabola and thus the maximum or minimum point depending on a<0 or a>0.
$y=2x^2+8x+9$ Factor 2 out of first 2 terms
$y=2(x^2+4x)+9$ Next, complete the square in parentheses.
$y=2(x^2+4x+4)+9-8$ Notice that we added 2 times 4, so we subtract 8 to balance the side
$y=2(x+2)^2+1$ Now, we're in vertex form.
$V(h, k)=V(-2, 1) = \ \ minimum \ \ point$
I enjoyed the steps as a guide. Well-done! | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9825575178175919,
"lm_q1q2_score": 0.8571314597256313,
"lm_q2_score": 0.8723473630627235,
"openwebmath_perplexity": 1917.9567759754982,
"openwebmath_score": 0.8046611547470093,
"tags": null,
"url": "http://mathhelpforum.com/pre-calculus/62126-completing-square.html"
} |
# Problems that become easier in a more general form
When solving a problem, we often look at some special cases first, then try to work our way up to the general case.
It would be interesting to see some counterexamples to this mental process, i.e. problems that become easier when you formulate them in a more general (or ambitious) form.
## Motivation/Example
Recently someone asked for the solution of $a,b,c$ such that $\frac{a}{b+c} = \frac{b}{a+c} = \frac{c}{a+b} (=t).$
Someone suggested writing this down as a system of linear equations in terms of $t$ and solving for $a,b,c$. It turns out, either (i) $a=b=c$ or (ii) $a+b+c=0$.
Solution (i) is obvious from looking at the problem, but (ii) was not apparent to me until I solved the system of equations.
Then I wondered how this would generalize to more variables, and wrote the problem as: $$\frac{x_i}{\sum x - x_i} = \frac{x_j}{\sum x - x_j} \quad \forall i,j\in1,2,\dots,n$$
Looking at this formulation, both solutions became immediately evident without the need for linear algebra (for (ii), set $\sum x=0$ so each denominator cancels out its numerator).
-
The linked Question : math.stackexchange.com/questions/897118/… – lab bhattacharjee Aug 16 '14 at 12:16
Well, I personally don't think this is actually simplification by generalization, but rather simplification motivated by generalization. Like if you write $\frac{a}{b+c}$ as $\frac{a}{(a + b + c) - a}$, you'd say that the problem becomes obvious too. – Tunococ Aug 16 '14 at 12:59
@Tunococ Fair enough, I'm just saying that writing it like this didn't occur to me until I thought of generalizing it. I understand that this is all a little subjective (hence the "soft-question" tag). – MGA Aug 16 '14 at 13:01
This is relevant. – Julien Godawatta Aug 16 '14 at 13:58
Introducing topology makes certain proofs in real analysis clearer and more elegant, though I'm not sure whether they are necessarily easier per se. – Harry Johnston Aug 17 '14 at 0:24 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9343951680216529,
"lm_q1q2_score": 0.857123176955193,
"lm_q2_score": 0.9173026641072386,
"openwebmath_perplexity": 539.8623476551608,
"openwebmath_score": 0.9017611145973206,
"tags": null,
"url": "http://math.stackexchange.com/questions/899109/problems-that-become-easier-in-a-more-general-form"
} |
Consider the following integral $\displaystyle\int_{0}^{1}\dfrac{x^7-1}{\ln x}\,dx$. All of our attempts at finding an anti-derivative fail because the antiderivative isn't expressable in terms of elementary functions.
Now consider the more general integral $f(y) = \displaystyle\int_{0}^{1}\dfrac{x^y-1}{\ln x}\,dx$.
We can differentiate with respect to $y$ and evaluate the resulting integral as follows:
$f'(y) = \displaystyle\int_{0}^{1}\dfrac{d}{dy}\left[\dfrac{x^y-1}{\ln x}\right]\,dx = \int_{0}^{1}x^y\,dx = \left[\dfrac{x^{y+1}}{y+1}\right]_{0}^{1} = \dfrac{1}{y+1}$.
Since $f'(y) = \dfrac{1}{y+1}$, we have $f(y) = \ln(y+1)+C$ for some constant $C$.
Trivially, $f(0) = \displaystyle\int_{0}^{1}\dfrac{x^0-1}{\ln x}\,dx = \int_{0}^{1}0\,dx = 0$. Hence $C = 0$, and thus, $f(y) = \ln(y+1)$.
Therefore, our original integral is $\displaystyle\int_{0}^{1}\dfrac{x^7-1}{\ln x}\,dx = f(7) = \ln 8$.
This technique of generalizing an integral by introducing a parameter and differentiating w.r.t. that parameter is known as Feynman Integration.
-
Great example! Really neat – mattecapu Aug 16 '14 at 19:59
My favorite illustration so far. Of course the others are nice too. – MGA Aug 16 '14 at 20:22
The special form (not only the general form) can also easily be determined by setting $\ln x=-t$ and then use Frullani's integral. – Tunk-Fey Aug 17 '14 at 14:44
The integrand does have an anti-derivative in terms of the upper incomplete gamma function: $\Gamma(0, -\ln x) - \Gamma(0, -8\,\ln x)$. That still doesn't help because it's undefined at both limits of integration. – Tavian Barnes Aug 17 '14 at 21:52
@Tunk-Fey You're right, apologies. Deleting my comment. – MGA Aug 18 '14 at 11:39
George Polya's book How to Solve It calls this phenomenon "The Inventor's Paradox": "The more ambitious plan may have more chances of success." The book gives several examples, including the following. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9343951680216529,
"lm_q1q2_score": 0.857123176955193,
"lm_q2_score": 0.9173026641072386,
"openwebmath_perplexity": 539.8623476551608,
"openwebmath_score": 0.9017611145973206,
"tags": null,
"url": "http://math.stackexchange.com/questions/899109/problems-that-become-easier-in-a-more-general-form"
} |
1) Consider the problem: "A straight line and a regular octahedron are given in position. Find a plane that passes through the given line and bisects the volume of the given octahedron." If we generalize this to "a straight line and a solid with a center of symmetry are given in position..." it becomes very easy. (The plane goes through the center of symmetry and the line.)
The book also gives other examples of the Inventor's Paradox, but "more ambitious" is not always the same as "more general." Consider: "Prove that $1^3 + 2^3 + 3^3 + ... + n^3$ is a perfect square." Polya shows that it is easier to prove (by mathematical induction) that "$1^3 + 2^3 + 3^3 + ... + n^3 = (1 + 2 + 3 + ...+ n)^2$". This is more ambitious but is not more general.
The web page Generalizations in Mathematics gives many similar examples. It even gets into the difference between "more ambitious" and "more general."
-
Good point regarding generality/ambitiousness, and both are interesting. I have made a minor edit to the question to reflect this. – MGA Aug 16 '14 at 12:51
In the first the easy approach is "more general" in the sense that it picks out the significant property of an entity and ignores any other particular properties it may have. There are loads of examples like it, for example any time you're asked to prove some property of a particular group that's true of all Abelian groups, or some property of a particular function that's true of any continuous monotonic function, and so on. The problem is, "identify the useful property of this object", so the more general problem where you only have the useful property is always easier ;-) – Steve Jessop Aug 16 '14 at 19:26
The cubes and square example, termed 'more ambitious', is an example of the technique of solving a problem by the addition of an invented constraining hypothesis, and thus is actually a case of particularization. – Jose Brox Aug 20 '14 at 8:00 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9343951680216529,
"lm_q1q2_score": 0.857123176955193,
"lm_q2_score": 0.9173026641072386,
"openwebmath_perplexity": 539.8623476551608,
"openwebmath_score": 0.9017611145973206,
"tags": null,
"url": "http://math.stackexchange.com/questions/899109/problems-that-become-easier-in-a-more-general-form"
} |
I recall something like this coming up when evaluating certain summations. For example, consider:
$$\sum_{n=0}^{\infty} {n \over 2^n}$$
We can generalize this by letting $f(x) = \sum_{n=0}^{\infty} nx^n$, so:
\begin{align} {f(x) \over x} &= \sum_{n=0}^{\infty} nx^{n-1} \\ &= {d \over dx} \sum_{n=0}^{\infty} x^n \\ &= {d \over dx} {1 \over {1-x}} = {1 \over (x-1)^2} \end{align}
Therefore,
$$f(x) = {x \over (x-1)^2}$$
The solution to the original problem is $f({1 \over 2}) = 2$.
-
Interestingly, this also shows $\sum_{n=0}^\infty \frac{1}{2^n} = \sum_{n=0}^\infty \frac{n}{2^n}$, which succeeded at surprising me. :) – Keba Aug 3 '15 at 23:26
The solution to the Monty Hall problem
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, follows the fixed protocol of opening another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
becomes more obvious when you generalize it to an $N$-door problem with the host opening $N-2$ doors. For $N\gg3$ most people's intuition revolts against staying with the original choice. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9343951680216529,
"lm_q1q2_score": 0.857123176955193,
"lm_q2_score": 0.9173026641072386,
"openwebmath_perplexity": 539.8623476551608,
"openwebmath_score": 0.9017611145973206,
"tags": null,
"url": "http://math.stackexchange.com/questions/899109/problems-that-become-easier-in-a-more-general-form"
} |
-
A more helpful approach to the standard Monty Hall is to examine what is wrong with the above formulation. The story, as told here, is consistent with a game show where the host offers a chance to switch only to contestants who have already chosen the correct door. When playing that game, you should never switch regardless of how many doors there are. In the standard problem, it is explicitly stated that the host will always offer the switch. That's what makes the odds $N-1$ to $1$ in favor of the switch; and $2$-to-$1$ odds is good enough. – David K Aug 16 '14 at 15:22
@DavidK - the text I included is Wikipedia's description of the Monty Hall problem. Have modified it. – Johannes Aug 16 '14 at 20:20
The problem has been underspecified in some widely-read sources; no surprise one of those got copied to Wikipedia. I'll accept this version as implying the player knows the fixed protocol. In that case, if someone's intuition requires $N$ to be increased above $3$ before they think the switch is beneficial, they do not understand the reason why it is beneficial. They have merely been nudged from an incorrect guess to an unjustified but luckily correct guess. – David K Aug 18 '14 at 23:12
@DavidK: I think that would be a fair criticism of someone who claims to be game-theoretically rational, but I doubt such a person exists; would a reasonable person expect a Bayes factor of $2:1$ to come to his rescue? Even though it wouldn't be necessary to the idealised mathematical observer with the eyes of a hawk, in many contexts exaggerating the difference helps me notice there is one, whence I can see via a monotonicity argument that the effect is non-zero (if weak) for $N \gtrsim 3$. – Vandermonde Feb 16 '15 at 19:27
Heck, what I thought negligible was a raw probability of 1/6, which I ought to appreciate even then. – Vandermonde Feb 16 '15 at 21:25
I'm not entirely convinced that problems made somehow easier by generalizations is exactly what is going on here. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9343951680216529,
"lm_q1q2_score": 0.857123176955193,
"lm_q2_score": 0.9173026641072386,
"openwebmath_perplexity": 539.8623476551608,
"openwebmath_score": 0.9017611145973206,
"tags": null,
"url": "http://math.stackexchange.com/questions/899109/problems-that-become-easier-in-a-more-general-form"
} |
In the example provided in your question, what made the solution to the general problem appear easier is that it dawned on you that $$x_1+\cdots+x_{j-1}+x_{j+1}+\cdots+x_n=\sum_{i=1}^nx_i-x_j.$$ Indeed, (as tunococ commented) had the less general problem been written as $$\frac{a}{a+b+c-a}=\frac{b}{a+b+c-b}=\frac{c}{a+b+c-c},$$ then your easier solution to the general problem applies here as well. I would argue that, if anything, the generalization helped you notice a pattern you had not before seen. Would you still have noticed this pattern had you not formulated the problem in a general way? Perhaps, perhaps not.
In my opinion, what your experience shows is that formulating a problem $Q$ in more general terms $P$ is one of many ways by which one can gain a fundamental insight that provides the key to the solution of the general problem $P$ (and thus inevitably also solves the initial special case $Q$ also). Sometimes, this can lead to a solution that was as of yet unknown to you and that will be more elegant or easier than the previous solutions. However, given that such an insight could easily have come without generalizing the problem, the fact that the solution did come from you thinking about the generalization seems highly circumstantial to me.
EDIT: JimmyK4542's example (and Feynmann's integration trick) seems like a spectacular demonstration of the phenomenon, however. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9343951680216529,
"lm_q1q2_score": 0.857123176955193,
"lm_q2_score": 0.9173026641072386,
"openwebmath_perplexity": 539.8623476551608,
"openwebmath_score": 0.9017611145973206,
"tags": null,
"url": "http://math.stackexchange.com/questions/899109/problems-that-become-easier-in-a-more-general-form"
} |
-
I merely meant that as an example to get the discussion started, and you're of course right that in this case one doesn't have to generalize to see the solution. But personally, I didn't think of writing $+a-a$ etc. because I didn't feel the need to. Once I considered the general case, I was compelled to write it like this. Anyway, I think that better examples have now been given on here that really illustrate the point - I'm particularly fond of the Feynman integration trick. – MGA Aug 16 '14 at 20:16
JimmyK4542's example (and Feynmann's integration trick in general) indeed contradicts my claim that a solution to a general problem always applies to a special case. I'll edit this out. – user78270 Aug 16 '14 at 23:53
On this site one frequently finds under the linear-algebra tag questions of the kind: what is the determinant of a matrix $$\begin{pmatrix}a&b&b&b\\b&a&b&b\\b&b&a&b\\b&b&b&a\end{pmatrix}?$$ (I've just posted this question, which contains a list of such questions). It turns out finding an answer to this question becomes almost trivial (see my answer to the linked question) when reformulated more generally as
What is the characteristic polynomial of a square matrix$~A$ of rank$~1$?
knowing that by specialisation the answer gives the determinant of $\lambda I-A$ for any scalar$~\lambda$.
-
From time to time famous problems have such feature. History suggests this point: the transcendentalness of $\pi$ solves the long-lasting problem of squaring a circle, analytic geometry and irrationality theory solve the problem of doubling a cubic, Galois's invention of group theory and quintic function, Kummer's invention of ideals and Fermat's last theorem, global differential geometry and Chern's intrinsic proof of Gauss-Bonnet theorem, and so on.
- | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9343951680216529,
"lm_q1q2_score": 0.857123176955193,
"lm_q2_score": 0.9173026641072386,
"openwebmath_perplexity": 539.8623476551608,
"openwebmath_score": 0.9017611145973206,
"tags": null,
"url": "http://math.stackexchange.com/questions/899109/problems-that-become-easier-in-a-more-general-form"
} |
-
A broadly successful application of this was introduced by Richard Bellman under the phrase dynamic programming. The story of the "birth" of this now foundational topic in applied math is told largely in Bellman's own words here.
A related term gives more evidence of the connection of ideas: invariant imbedding.
A good discussion of dynamic programming references and examples came up early at StackOverflow, but was subsequently closed as off-topic.
An illustration is finding a shortest path between two specified points by "imbedding" that problem in finding all shortest paths from one point, Dijkstra's algorithm.
-
Generalization comes up a lot when doing induction. For example,
$$\forall n ~~ \sum_{k=0}^n 2^{-k} \le 2$$
is difficult to prove directly using induction on $n$. However, if you generalize to a stronger statement:
$$\forall n ~~ \sum_{k=0}^n 2^{-k} \le 2 - 2^{-n}$$
Then induction may be used directly:
$$\sum_{k=0}^{n+1} 2^{-k} \le 2 - 2^{-n - 1}$$ $$\sum_{k=0}^n 2^{-k} + 2^{-n-1} \le 2 - 2^{-n - 1}$$ $$\sum_{k=0}^n 2^{-k}\le 2 - 2^{-n}$$
Obviously you could see that it is a geometric series, but that is a generalization also.
problems that become easier when you formulate them in a more general (or ambitious) form
The potential difficulty of a generalization isn't the only disadvantage. If you disprove a generalization, then you haven't disproven the original theorem. In that respect, a generalization effectively forces you to pick sides in the investigation of a theorem.
-
The most spectacular example I have seen is this one:
Suppose A is an $n\times n$ matrix with eigenvalues $\lambda_1$, ..., $\lambda_n$, including each eigenvalue according to its multiplicity. Then $A^2$ has eigenvalues $\lambda_1^2$, ..., $\lambda_n^2$ including multiplicity. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9343951680216529,
"lm_q1q2_score": 0.857123176955193,
"lm_q2_score": 0.9173026641072386,
"openwebmath_perplexity": 539.8623476551608,
"openwebmath_score": 0.9017611145973206,
"tags": null,
"url": "http://math.stackexchange.com/questions/899109/problems-that-become-easier-in-a-more-general-form"
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.