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http://mathhelpforum.com/business-math/41170-compound-interest-q-help.html | 1. Compound Interest Q Help!
Can you guys help me with this question? I';m so confused on how I should approach it:
Tom’s father paid $500 into an account on the day Tom was born. ?After that, he paid$500 into the account on Tom’s bday until Tom’s 18th bday. ?If the account accrued interest at 8%p.a compounded monthly, calculate how much Tom would receive on his 18th bday.
A:
20880.97
2. Just think it through, one payment at a time.
From the start, I'm not absolutely certain there was a payment ON the 18th birthday. let's assume that there was. If we get $500 too much, we'll discard this payment. Starting from the last payment and working backwards, we have: 18th:$500 and no accumulation for interest
17th: $500(1+i) -- 1 year's accumulation for interest 16th:$500(1+i)^2 -- 2 year's accumulation for interest
15th: $500(1+i)^3 -- 3 year's accumulation for interest ... Birth:$500(1+i)^18 -- 18 year's accumulation for interest
With any luck, one should notice this is a Geometric Sequence and we should be able to add them all up.
$500 +$500(1+i) + $500(1+i)^2 +$500(1+i)^3 + ... + $500(1+i)^18 =$500(1 + (1+i) + (1+i)^2 + (1+i)^3 + ... + (1+i)^18) =
$500\frac{(1+i)^{19}-1}{i}$
Our only remaining concern is 'i'. What is it? The formula above uses 'i' as an annual effective interest rate. We need to find one of those. We are given 8% Nominal Interest and Monthly Compounding. This gives:
$\left(1 + \frac{0.08}{12}\right)^{12} - 1 = 0.082999511 = i$
Thus,
$500\frac{(1+i)^{19}-1}{i}\;=\;21,380.97$
As can be seen, $21,380.97 -$500.00 = \$20,880.97, so I guess there was not a payment made ON the 18th Birthday Anniversary. This leaves us with a bit of a dilemma. On a written exam or a homework assignment, I would state my assumptions and provide both answers, citing the ambiguity of the word "until". Anything marked wrong would get a vigorous challenge. On a multiple-choice exam, I would be prepared to find either answer. In my view, if both appear on the multiple-choice exam, the question probably should be discarded as accepting either answer will not necessarily provide any information about a student's knowledge. One may simply have done it badly. In any case, questions should be clear. If you have discussed the word "until" in class, and it has been defined clearly to mean "NOT on the end date", then you can be expected to get the unique value. There may also be a diagram explaining the intent. It is a very hard thing to write perfectly clear questions. It is up to the student to explain any point of ambiguity. The exam writer cannot be expected to think of every possible translation, but I'm sure the exam writer tries to do that.
Well, enough of exam philosophy... | 2017-05-25T04:58:10 | {
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http://math.stackexchange.com/questions/878038/whats-the-name-of-this-algebraic-property/878272 | # What's the name of this algebraic property?
I'm looking for a name of a property of which I have a few examples:
$(1) \quad\color{green}{\text{even number}}+\color{red}{\text{odd number}}=\color{red}{\text{odd number}}$
$(2) \quad \color{green}{\text{rational number}}+\color{red}{\text{irrational number}}=\color{red}{\text{irrational number}}$
$(3) \quad\color{green}{\text{algebraic number}}+\color{red}{\text{transcendental number}}=\color{red}{\text{transcendental number}}$
$(4) \quad\color{green}{\text{real number}}+\color{red}{\text{non-real number}}=\color{red}{\text{non-real number}}$
If I were to generalise, this, I'd say that if we partition a set $X$ into two subsets $S$ and $S^c=X\setminus S$, then the sum of a member of $S$ and a member of $S^c$ is always in either $S^c$ or $S$.
My question is:
"Is there a name for this property (in these four cases) and is this property true in general?"
Also, does anyone have any more examples of this property?
-
Note that you're saying more than just set here: You're requiring that the objects of that set can be added. – Semiclassical Jul 25 at 16:50
@Semiclassical Yeah, thanks! What word should I use instead of set, then? – alexqwx Jul 25 at 16:51
This can be written as: $S$ is closed under subtraction. That is, if $a,b\in S$ and $a-b$ exists, then $a-b\in S$. – Thomas Andrews Jul 25 at 16:52
@Semiclassical I don't see how this contradicts $(2).$ – alexqwx Jul 25 at 16:54
@Semiclassical No, you misread my comment. The rationals are closed under subtraction. That is what the above says... – Thomas Andrews Jul 25 at 16:55
I think this comes from the fact that if you have a group $G$ and $H$ a subgroup of $G$ then if $h\in H$ and $x\not\in H$ we get $xh\not\in H$.
The proof is by contradiction, suppose $xh=l$ with $l\in H$. Then postmultiplying by $h^{-1}$ gives $x=lh^{-1}$ which is in $H$ since $H$ is a subgroup of $G$.
-
Why have you used $xh$ instead of $x+h$ in your generalisation (does it matter?)? – alexqwx Jul 25 at 16:58
$xh$ just means the group operator acting on the pair $(x,h)$ if the operation in the particular group is addition then $xh$ is just $x+h$. the plus sign is normally used for abelian groups. But this result generalizes to any group. – Jorge Fernández Jul 25 at 17:03
Got it. Thanks! One final thing: why did you mention post-multiplying rather than just multiplying (since addition is commutative)? – alexqwx Jul 25 at 17:05
well, in general groups don't need to be commutative, however this assumption is not required for the result to be true. – Jorge Fernández Jul 25 at 17:22
Talking about this in terms of groups seems just a bit too strict. The first example given in the question works just as well, for example, if we were to only talk about positive even integers (no inverse or zero element under addition.) – Semiclassical Jul 25 at 17:51
I call this the complementary subgroup law, because the composition law arises via the following complementary view of the Subgroup Test ($\rm\color{#c00}{ST}$), cf. below from one of my old sci.math posts.
Theorem Let $\rm\,G\,$ be a nonempty subset of an abelian group $\rm\,H,\,$ with complement set $\rm\,\bar G = H\backslash G.\,$ Then $\rm\,G\,$ is a subgroup of $\rm\,H\iff G + \bar G\, =\, \bar G.$
Proof $\$ $\rm\,G\,$ is a subgroup of $\rm\,H\!\overset{\ \large \color{#c00}{\rm ST}}\iff\! G\,$ is closed under subtraction, so, complementing
$\begin{eqnarray} & &\ \ \rm G\text{ is a subgroup of }\, H\ fails\\ &\iff&\ \rm\ G\ -\ G\ \subseteq\, G\,\ \ fails\\ &\iff&\ \rm\ g_1\, -\ g_2 =\,\ \bar g\ \ \ for\ some\ \ g_1,g_2\in G,\ \ \bar g\in \bar G\\ &\iff&\ \rm\ g_2\, +\ \bar g\ \ =\,\ g_1\ for\ some\ \ g_1,g_2\in G,\ \ \bar g\in \bar G\\ &\iff&\ \rm\ G\ +\ \bar G\ \subseteq\ \bar G\ \ fails\qquad\ {\bf QED} \end{eqnarray}$
Instances of this law are ubiquitous in concrete number systems, e.g. below. For many further examples see some of my prior posts here (and also on sci.math).
-
Nice! ByteErrant, in the answers below, has shown that prime +non-prime can be either prime or non-prime itself, which seems to contradict the above theorem. Could you explain? – alexqwx Jul 25 at 21:36
@alex The set $\,\color{#0a0}G\,$ of $\rm\color{#0a0}{primes}$ is not a subgroup of $\,H =$ integers under addition, since $\,\color{#0a0}G\,$ is not closed under subtraction, e.g. $\,\color{#0e0}{11}-\color{#0a0}7 = \color{#c00}4\,$ is not $\rm\color{#0a0}{prime,\,}$ so $\, \color{#0a0}7+\color{#c00}4 = \color{#0e0}{11}\,\Rightarrow\, \color{#0a0}G+\color{#c00}{\bar G}\subseteq \color{#c00}{\bar G}\,$ fails. That agrees with the (negation of) the theorem, i.e. $\,G+\bar G\subseteq G\,$ fails $\iff$ $\,G\,$ is not a subgroup of $\,H\,$ $\iff$ $\,G\,$ is not closed under subtraction. – Bill Dubuque Jul 25 at 23:21
This is basically just "the complement" of the statement that the sets of numbers in green are subgroups of $(\Bbb R,+)$.
Given that for a subgroup $S<\Bbb R$ $a,b\in S$ implies $a-b\in S$, you can quickly compute that if $c\notin S$, $a+c=b\in S$ implies $a-b=c\in S$, a contradiction. Therefore $a+c\notin S$ for any $a\in S$, $c\notin S$.
You could also add to your list anything like "an integer plus a noninteger is a noninteger" and "for any subring $S\subseteq \Bbb R$, an element in $S$ plus an element outside of $S$ results in an element outside of $S$." Again, you don't even really need a subring: this is true for any proper subgroup.
-
Don't prime numbers offer a counter-example to the general truth of this property?
Prime $+$ not-prime $=$ not-prime $==> 17 + 4 = 21$
Prime $+$ not-prime $=$ prime $==> 7+4=11$
-
Ah. I hadn't thought about this. Thanks! Do you (or does anyone else) have any explanation for this? – alexqwx Jul 25 at 21:34
@alexqwx Primes don't form a group under addition. Evens do, rationals do, algebraics do, and reals do. – Cory Jul 25 at 21:41
I would say that the behaviour of this property under addition is isomorphic to addition in $\mathbb{Z}_{2}$. E.g.: If you assign "even" to 0, "odd" to 1, then even + odd = odd is expressed in 1 + 0 = 1.
- | 2014-11-24T01:26:26 | {
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http://www.intmath.com/blog/learn-math/calculus-made-easy-free-book-2432 | # Calculus Made Easy (Free book)
By Murray Bourne, 25 Apr 2009
OK, it looks old and dusty, but Calculus Made Easy [PDF] is an excellent book and I strongly recommend it to those of you who are struggling with calculus concepts. It's also great for teachers, to give you ideas on how to explain calculus so it doesn't confuse the hell out of everyone. He quite rightly points out that many math text book writers are more interested in impressing the reader with sophisticated calculus techniques than explaining the basic concepts.
One of the early pages has:
BEING A VERY-SIMPLEST INTRODUCTION TO
THOSE BEAUTIFUL METHODS OF RECKONING
WHICH ARE GENERALLY CALLED BY THE
TERRIFYING NAMES OF THE
DIFFERENTIAL CALCULUS
AND THE
INTEGRAL CALCULUS.
BY
SILVANUS P. THOMPSON
In other words, this was one of the first ever "Calculus for Dummies" books. Thompson puts great effort into explaining what is going on, rather than jumping straight into the calculations. He humbly calls himself a "fool", but doesn't treat the reader as one.
He quotes from an "ancient Simian proverb":
"What one fool can do another can."
To give you an idea of how the book is written, in Chapter 1, "To Deliver You From the Preliminary Terrors", we read:
∫ which is merely a long S, and may be called (if you like) "the sum of." Thus ∫dx means the sum of all the little bits of x; or ∫dt means the sum of all the little bits of t. Ordinary mathematicians call this symbol "the integral of".
Now any fool can see that if x is considered as made up of a lot of little bits, each of which is called dx, if you add them all up together you get the sum of all the dx's, (which is the same thing as the whole of x). The word "integral" simply means "the whole".
The book is now copyright free. Grab the PDF: Calculus Made Easy.
[Thanks to Denise at LetsPlayMath for the link.]
1. solarhene says:
thanks so much for sharing this excellent book!
best regards,
an old fool
2. Murray says:
You're welcome Solarhene. I'm glad that you find it useful.
3. ayuk federick besong says:
i believe it can work and it will work
4. Noor says:
I'm so interested in re-learning calculus that I watched many different video tutorials, but I found this book summarizing Integral Calculus in its first 2 pages!
Thank you so much!
5. Sarah says:
Thanks so much for posting this. I'm so glad I clicked on your link while viewing my friend Dana's blog. My 9 year old read it and was very excited that it made so much sense. He's currently taking Physics and some of the math problems were made to be so confusing so this will help him so much.
6. lawal says:
Good DAY SIR,i want to know how i can get this text book,calculus made easy.i am mailing from Lagos Nigeria.thank u sir.
7. Murray says:
Hi Lawal
This is an e-book in PDF form (not a physical book). The link is in the article above, in the first line.
You could print it out from the PDF if you would rather hard copy.
8. D. Pal, India. says:
Its a brilliant book. Thanks for sharing the book
9. B Atchley says:
Thanks for sharing this book! It will be a great tool for my independent study students!
10. Winstone Kapanje says:
Thanks for great job well done. Do the same with other topics such as trigonometry and complex numbers
11. Judith Kanyama Chikoti says:
the way you have started culculus is very interesting but how could one get the whole PDF paper?
12. Murray says:
Hi Judith. The link to the PDF is in the article above! Do you mean that you want a hard copy? You would need to print it yourself.
13. jared okoth says:
if every mathematics topic could be introduced the way this has, then i sure no one will think of maths as the hardest of courses.
14. eduardo says:
make me realize the math is easy
15. Kaset One says:
one of the greatest books.. many thanks for sharing with us
16. Rod says:
I was contemplating purchasing a course on dvd called "Calculus Made Clear" selling for \$60.00 US. Then I decided to google "Calculus Made Easy" and found your link to this great book. I read the intro and was amazed at how easy it was to follow. I will continue to read the whole text! by the way, I was searching for books on calculus for my son so now we can read it together. Thanks!
17. Murray says:
You're welcome, Rod. Glad you found it useful.
18. fatemeh says:
i from iran and i don't english.
ineed a study book reference and i have n't book english and i request from you that a link book free download for laplas, integral ,...
thank you very much
19. Murray says:
@Fatemeh: I'll look for such materials and post it if I find any good ones.
20. Piracha, J. L. says:
Thanks
thanks! and great work which nobody cannot be appreciated for your effort for sharing such good book.
22. Dalcde says:
http://www.gutenberg.org/ebooks/33283
which is typed in LaTeX and the file size is reduced tenfold.
23. Murray says:
@Dalcde: Thanks for the tip!
24. Nayyir says:
if u r one of those looking to understand calculas at an elementary level,then u've got it here.Calculas Made Easy is indeed very helpful a book for those struggling to understand it as a concept.so do go for it , it is certainly one of the best materials available. go for it.have a nice maths day.
25. Ian Thomson says:
Hi :
Thanks for the pdf on Calculus Made Easy.
I have always been curious and terrified at the same time of calculus.
Chapter One says it all. Getting past the fancy notation, helps a huge amount.
Kind Regards Ian Thomson
26. percival says:
Hi! Ian,My name is percy and I teach Maths in grade 12. Please foreward me the calculus doc as I also struggle on calculus section.
27. Murray says:
@Percival: The link to the PDF is at the top (and again at the bottom) of the article.
28. Joe Kenyon says:
Dear Murray
I am surprised and delighted to see my old Dear Friend Sylvanus P. Thompson here. As a very discouraged high school dropout I found an original hardback version at the rubbish tip site where I used to spend my spare time. I picked it up and the proverb just grabbed me. I read the book it and loved it. I went back to school and excelled at the age of 25. Now after 40 years in senior positions in Telecommunications in Australia my copy is very much treasured and repaired and again being used as I transition into becoming a teacher myself! The bit that initially grabbed me was "What one fool can do another can" I ask that you put that wonderful saying up on this site. It was so powerfuly encouraging it got me off the bad path I was on and led to a wonderful career and a wonderful wife and family. Imagine, it hardly seems possible that an ancient seven word proverb, repeated in one book by a man long dead could do so much good.
I love this guy and would have loved to have met him.
Kind Regards
Joe Kenyon
29. Murray says:
Thanks for your inspiring story, Joe!
I have included the quote in the article.
30. sachin sharma says:
Hi,
I have been searching such work for past couple of months and at last I got here in the form of "Calculus Made Easy" May GOD Bless all who have made such efforts in preparing such an excellent book.
Thanks to all.....
Sachin Sharma
India
31. drcobol2000 says:
Murray,
Thanks for posting this...and for free! While I struggled with math from grade school to grad school, I'm now a bit of a math junkie and books with subjects like this really pique my interest. I wish I could get my hands on an actual copy of that book for myself.
One question comes to mind, though. I noticed there are two different dates in the front of the book that confuse me (I'm a sucker for old books). There is a date of 1914 in the preface and another of 1943 by itself on another page. Do you know which date was the actual date of publication? There isn't an actual copyright year specified and as I read through the text, the absence of that date combined with the style and choice of words lead me to believe this was a work from the early 20th century. I rely on a barely mediocre literary acumen to make that assessment, so take it with an enormous grain of salt.
Thanks and best wishes!
32. Murray says:
@drcobol2000 The publication date would be 1914, as that's when he wrote the Preface to this second edition. So your "early 20th century" language observation is right on the button.
I don't believe the "6-14-43" on the separate page is a date, since in the US, the date order is normally day then month. I suspect it's there to trigger thought.
33. kevin says:
Thanks you so much sir for this ebook
I am an engineering student, and i am struggling with calculus.
This book helped me a lot even though this book is quite old.
34. Pyae Phyo Aung says:
Thank you really really much sir. You help me a lot.
35. punith says:
thank u so much its very helpful
36. rahillah says:
hi Murry, thanks for sharing the brilliant book,hope you have a lovley day!
37. Murray says:
Glad you found it useful, Rahillah.
38. Nae says:
I'm not sure if it'll work but I'll give it a try. I'll write back at the end of the semester to let you know how it worked for me.
Thanks
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https://math.stackexchange.com/questions/2626576/how-do-we-define-the-domain-of-a-function-with-removable-and-nonremovable-discon | # How do we define the domain of a function with removable and nonremovable discontinuities?
For instance consider the function
$$f(x) = \frac{x-2}{(x-2)(x+2)}$$
I don't understand why some discontinuities are considered "removable". Yes, we can cancel out the $x-2$'s but isn't this invalid at $x=2$ because it's like canceling out $\frac{0}{0}$? I don't understand why we're allowed to do this.
Anyway if we "remove" that piece we get:
$$f(x) = \frac{1}{x+2}$$
Which has a discontinuity at $x=-2$ but we can't "remove" that.
So we end up with two discontinuities -- do we define the domain to include or exclude these? Is the domain all real numbers? All except $2$? All except $-2$? All except $-2, 2$?
• Because limit . – user202729 Jan 29 '18 at 14:39
The discontinuity at $x=2$ can be removed because $\lim_{x\to2}f(x)=1/4$. When 'removing' the discontinuity, we define a new function $f^\prime$ that takes on the values of $f$ at all points in the domain of $f$ and $f^\prime(2)=1/4$, thereby extending the domain.
The discontinuity at $x=-2$ cannot be removed because the limit there is improper: $\lim_{x\to-2}f(x)=\infty$.
• How do you know the limit at each point? Is there an analytic way? Is the only difference between removable and nonremovable discontinuities whether or not the limit is a number or not? – user525966 Jan 29 '18 at 14:57
• When the limit at a certain point is improper, it cannot be represented by a real number. Only proper limits can be used to remove discontinuities. – Jeroen van Riel Jan 29 '18 at 15:03
• What then would be the domain of $f$? All reals except $-2,2$? – user525966 Jan 29 '18 at 15:04
• Yes, as @57Jimmy already pointed out in his answer. – Jeroen van Riel Jan 29 '18 at 15:13
You clearly understand the algebra, and the idea behind "removable singularity". The definition of the domain is a little subtle (and usually not particularly important, given your understanding).
The expressions on the right in
$$f(x) = \frac{x-2}{(x-2)(x+2)}$$
and
$$g(x) = \frac{1}{x+2}$$
define the same function where both make sense. That function has an unremovable singularity at $x=-2$.
The domain of $g$ contains the point $x=2$; the domain of $f$ does not, so strictly speaking they are not the same function. But the limit of $f$ at $x=2$ does exist, and has value $g(2) = 1/4$. That's exactly what we mean when we say the singularity is removable.
• So a removable discontinuity is an undefined point that has a defined limit, whereas a nonremovable discontinuity has a limit that does not exist or is infinity? – user525966 Jan 29 '18 at 14:52
• @user525966 Yes. – Ethan Bolker Jan 29 '18 at 15:33
The domain of $f$ is $\mathbb{R} \backslash \{-2,2\}$, because at this two points $f$ is not defined. Having said that, some singularities are better behaved than others: $f$ goes to infinity as $x \to -2$, whereas $\underset{x \to 2}{\lim} f(x) = \frac{1}{4}$ is well-defined. This is exactly the difference between poles and removable singularities. | 2021-07-28T07:53:27 | {
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https://mathhelpboards.com/threads/cartesian-product-proof.26383/ | Cartesian Product - Proof
Yankel
Active member
Dear all,
I am trying to prove a simple thing, that if AxA = BxB then A=B.
The intuition is clear to me. If a pair (x,y) belongs to AxA it means that x is in A and y is in A. If a pair (x,y) belongs to BxB it means that x is in B and y is in B. If the sets of all pairs are equal, it means that every x in A is also in B and vice versa.
How do I prove it formally ?
Thank you !
Opalg
MHB Oldtimer
Staff member
Dear all,
I am trying to prove a simple thing, that if AxA = BxB then A=B.
The intuition is clear to me. If a pair (x,y) belongs to AxA it means that x is in A and y is in A. If a pair (x,y) belongs to BxB it means that x is in B and y is in B. If the sets of all pairs are equal, it means that every x in A is also in B and vice versa.
How do I prove it formally ?
Thank you !
This may not be as simple as you think. To start with, what do you mean by saying that two sets are equal? I think that the only way to make sense of that is to interpret "A=B" to mean that A and B have the same cardinality.
If a set $A$ is finite then its cardinality is just the number of elements it contains, denoted by $|A|$. If $|A| = m$ then $|A\times A| = m^2.$ So if $|B| = n$ and $|A\times A| = |B\times B|$ then $m^2 = n^2$, from which it follows that $m=n$. This proves that if "$A\times A = B\times B$" then "$A=B$" in the case of finite sets.
For infinite sets the situation is more complicated. There is a theorem of Zermelo that if $A$ is an infinite set then $|A\times A| = |A|$. From that it follows immediately that if $|A\times A| = |B\times B|$ then $|A| = |B|$. However, the proof of Zermelo's theorem requires the Axiom of Choice. In models of set theory that do not satisfy this axiom, it may be that your result does not hold.
HallsofIvy
Well-known member
MHB Math Helper
Opalg said "
I think that the only way to make sense of that is to interpret "A=B" to mean that A and B have the same
cardinality
."
I disagree. To say that sets A and B are equal means "$$x\in A$$ if and only if $$x\in B$$". If two sets are equal they have the same cardinality but the converse is not true. The sets A= {1, 2, 3} and B= {a, b, c} have the same cardinality but are not equal.
Opalg
MHB Oldtimer
Staff member
Opalg said "
I think that the only way to make sense of that is to interpret "A=B" to mean that A and B have the same
cardinality
."
I disagree. To say that sets A and B are equal means "$$x\in A$$ if and only if $$x\in B$$". If two sets are equal they have the same cardinality but the converse is not true. The sets A= {1, 2, 3} and B= {a, b, c} have the same cardinality but are not equal.
In that case, the result becomes trivially true. If $A\times A$ and $B\times B$ are just two different names for the same set, then the diagonal elements of $A\times A$ (those of the form $(a,a):a\in A$) are duplicates of the elements of $A$. The same holds for the diagonal elements of $B\times B$. If those diagonals are the same, it follows that the elements of $A$ are the same as the elements of $B$, so $A=B$.
HallsofIvy
Well-known member
MHB Math Helper
Yes, it is. Saying that "A= B", where A and B are sets, means that if x is in A then it is also in B and if y is in B then it is also in A.
If x is a member of A. then (x, x) is in AxA= BxB so x is in B. If y is a member of B then (y, y) is in BxB= AxA so y is in A. Therefore A= B.
It is trivial but that is the question asked.
Olinguito
Well-known member
I am trying to prove a simple thing, that if AxA = BxB then A=B.
The intuition is clear to me. If a pair (x,y) belongs to AxA it means that x is in A and y is in A. If a pair (x,y) belongs to BxB it means that x is in B and y is in B. If the sets of all pairs are equal, it means that every x in A is also in B and vice versa.
How do I prove it formally ?
Thank you !
Let $a\in A$. Then $(a,a)\in A\times A$. Since we’re assuming $A\times A=B\times B$, this means $(a,a)\in B\times B$ and thus $a\in B$. Therefore $A\subseteq B$. The same argument with $A$ and $B$ interchanged shows that $B\subseteq A$. Hence $A=B$. | 2020-08-12T09:05:49 | {
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https://www.physicsforums.com/threads/pair-in-a-poker-again.641732/ | # Pair in a poker, again
1. Oct 6, 2012
### Astr
Im New on this forum, so i hope this is the right place to ask this question. I've tried to solve the problem of finding how many hands are in a poker game with exactly one pair, this way:
I can choose the first card from 52 ways. for the second card I can choose it from 3 ways (to match the pair). For the third card I can choose it from 48 possibilities (for ensure I´ve got no trios in the hand). For the fourth and fifth i can choose them in 44 and 40 different ways, respectively. And because order is not important I must divide for 5! (ways of permuting 5 cards). My (wrong) answer is:
(52x3x48x44x40)/5!=109 824
This is a tenth of the correct answer. Please can you tell whats wrong in the procedure. (English isn't my first language, sorry if Ive got some mistakes)
2. Oct 7, 2012
### CWatters
That gives you too many because a QS + QD is the same pair as QD+SQ.
There are 6 ways to get a pair of (for example) Queens..
QS + QH
QS + QD
QS + QC
QH + QC
QH + QD
QD + QC
but there is a choice of 13 denomination (values) so for pair it's
6 * 13
then the other cards must not have the same value but can be of any suit.
3. Oct 7, 2012
### Astr
Thank you Cwatters. But I think i avoid the problem QS+QD=QD+QS, dividing by 5!, becasuse that ensures that the order of the 5 cards does not matter.
4. Oct 10, 2012
### CWatters
I can't immediatly see why your method is wrond but continuing from above..
Then the next three cards can have one of 12 other values (3 from 12 = 220) and be from any of 4 suits (4*4*4 = 64)
220 * 16 = 14080
Now that is also 48*44*40/3! so your last part seems ok?
5. Oct 10, 2012
### Astr
Thank you again, CWatters. Well I know how to solve the problem, and this is out of discussion:
For the pair:
52x3/2!= 78 ways
For the three different cards:
48x44x40/3!=14 080 ways
then: 78x14 080= 1098240 ways to get exactly one pair.
but my problem is: Whats wrong in the procedure i post first?
6. Oct 10, 2012
### Ray Vickson
You should not divide by 5!; instead, you should divide by 2! × 3!, because the order of the cards within the pair does not matter and the order within the three non-pairs does not matter. That would give the number of hands as 1098240, as you want.
There are some subtle points here, so let me expand. When counting the hands containing exactly one pair, the person posing the question is distinguishing between hands such as PPNNN, PNNNP, PNNPN, NNNPP, etc., even though the actual cards involved are exactly the same (and so all would be regarded as a *single* hand by a poker player). Basically the question is asking for the number of different ways you could be *dealt* a hand containing a single pair, even though after being dealt the cards you would re-arrange them to your own, personal liking.
Another way to see this is to look at P{pair}, the probability of getting dealt a pair. Suppose, first that we have 2 aces and one each of 2 3 4. The probability, p, of getting that is p = C(4,2)*C(4,1)^3/C(52,5), where C(a,b) = "a choose b" = a!/[b!(a-b)!]. This holds because there are C(4,2) ways to pick 2 aces from 4 and for each of the 2 3 and 4 there are C(4,1) ways to choose the particular card, and because there are C(52,5) says of being dealt a 5-card hand althogether. Now, of course, the pair may not be aces and the non-pairs may not be 2 3 and 4. However, for any choice of ranks, the probability of the pair and the three non-pairs are all the same as the p above. Therefore, P{pair} is p times the number of ways to choose the rank of the pair (13) times the number of ways of choosing the three different ranks from the remaining 12 (which is C(12,3)). Thus, we have
$$P\{ \text{pair}\} = \frac{13 \, C(12,3) \, C(4,2) \, C(4,1)^3}{C(52,5)} = \frac{N_p}{C(52,5)},$$
where Np = number of hands containing one pair. Thus, we have
$$N_p = 13 \, C(12,3) \, C(4,2) \, C(4,1)^3 = 1098240.$$
RGV
Last edited: Oct 10, 2012
7. Oct 10, 2012
### Astr
Ray, Thank you for your answer. Can you explain me why i should not divide by 5!, because this is my whole problem. Thanks. | 2018-02-19T14:29:19 | {
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http://math.stackexchange.com/questions/181239/solving-a-b | # Solving $|a| < |b|$
I apologize if this question in general, but I've been having trouble finding solutions as Google discards absolute value signs and inequality symbols.
I am looking for a way to eliminate absolute value functions in $|a| < |b|$.
I can solve $|a| < b$ and $|a| > b$, but I am unsure what method / combination of methods to use to eliminate absolute value signs from both sides.
Thank you!
An example problem:
$$|x + 2| < |x - 4|$$
-
perhaps you could give some specific problems that trouble you. – Will Jagy Aug 11 '12 at 5:19
Good suggestion. I've added one above. – Mark V. Aug 11 '12 at 5:41
It's useful to review the definition of the absolute value function: $$\text{abs}(a) = \begin{cases} a & 0 \leq a \\ -a & a < 0 \end{cases}$$ In order to eliminate one absolute value function from an expression, you will need to consider two cases: the case where its argument is non-negative, and the case where its argument is negative. This will give you an expression with two cases with one less absolute value function. Iterating this process will give an expression without the absolute value sign. – danportin Aug 11 '12 at 7:13
Since you mentioned trying Google, maybe you should try Wolfram Alpha instead – Ben Millwood Aug 11 '12 at 11:45
We have $|a| \lt |b|\,$ if any of these is true:
(i) $\,a$ and $b$ are $\gt 0$ and $a \lt b$
(ii) $a\lt 0$ and $b\ge 0$ and $-a \lt b$
(iii) $b \lt 0$ and $a \gt 0$ and $a \lt -b\,$
(iv) $\,a\lt 0$ and $b\lt 0$ and $-a\lt -b$. We can rewrite this as $b \lt a$.
Four cases! Not surprising, since eliminating a single absolute value sign often involves breaking up the problem into $2$ cases.
Sometimes, one can exploit the simpler $|a| \lt| b|\,$ iff $\,a^2\lt b^2$. But squaring expressions generally makes them substantially messier.
Added: With your new sample problem, squaring happens to work nicely. We have $|x+2| \lt |x-4|$ iff $(x+2)^2 \lt (x-4)^2$. Expand. We are looking at the inequality
$$x^2+4x+4 \lt x^2-8x+16.$$
The $x^2$ cancel, and after minor algebra we get the equivalent inequality $12x \lt 12$, or equivalently $x\lt 1$. The squaring strategy works well for any inequality of the form $|ax+b| \lt |cx+d|$.
But the best approach for this particular problem is geometric. Draw a number line, with $-2$ and $4$ on it. Our inequality says that we are closer to $-2$ than we are to $4$. The number $1$ is halfway between $-2$ and $4$, so we must be to the left of $1$.
-
Ha! I never would of thought of this. This is brilliant! Thank you so much for your help! – Mark V. Aug 11 '12 at 18:10
There are different approaches; one is to look at the zeroes of the expressions inside the absolute values, and split up $\mathbb R$ into intervals accordingly. In your example $|x + 2| < |x - 4|$, the points of interest are at $x=-2$ and $x=4$. You can therefore consider three cases:
1. If $x \in (-\infty,-2)$, then $x+2 < 0$ and $x-4 < 0$, so the absolute values will reverse the signs of both. This gives: \begin{align} -(x+2) &< -(x-4) \\ x+2 &> x-4 \\ 2 &> -4 \end{align} This is true for all $x$ in the interval.
2. If $x \in [-2,4)$, then $x+2 \geq 0$ so its sign is unaffected by the absolute value, but $x-4 <0$ so its sign will be reversed: \begin{align} x+2 &< -(x-4) \\ 2x+2 &< 4 \\ x &< 1 \end{align} Combining this last inequality with the assumption that $x \in [-2,4)$, we see that any $x$ in $[-2,1)$ is valid.
3. Finally, if $x \in [4,\infty)$, neither expression's sign is reversed: \begin{align} x+2 &< x-4 \\ 2 &< -4 \end{align} This is false for all $x$ in the interval.
Putting all the information together from the above three cases, we have $x \in (-\infty, 1)$.
Note: as some other contributors have mentioned, there are simpler ways to deal with your problem, such as viewing it geometrically. The method that I have shown above is more useful when the expressions are more complicated or when you have several absolute values; for example, an inequality like $3 |x^2-1|+|x-2|+|x^2-3x| > 5$.
-
Pedantry: you probably want $x\in[-2,4)$ for step 2 and $x\in[4,\infty)$ for step 3 (or change where you say $x-4<0$, but I think my way is more natural/consistent) – Ben Millwood Aug 11 '12 at 11:03
@BenMillwood: Fair enough! I've changed it. – Théophile Aug 11 '12 at 18:13
We can see $|x-a|$ as a distance point $x$ from point $a$.
Now, the question with the above "definition" would be like:
For which $x$ values distance point $x$ from $-2$ be less than distance point $x$ from $4$?
Clearly, by drawing it maybe, you can observe that the answer is for all $x<1$.
-
We can divide by $|b|$ to get $|a/b|<1$. Let $x=a/b$ then $|x|<1$ so $-1<x<1$.
Now multiply through by $b$.
If $b>0$ then $-b<a<b$.
If $b<0$ then $-b>a>b$
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http://mathhelpforum.com/differential-geometry/108094-problem-riemann-integrable.html | # Math Help - A problem of Riemann Integrable
1. ## A problem of Riemann Integrable
If $f(x)$ is Riemann Integrable in $[a,b]$,and for any integer $n\geq0$, we have $\int_{a}^{b}x^{n}f(x)dx = 0$. can we get the conclusion: $f(x)=0$ a.e.?
(a.e. stands for almost everywhere)
2. Yes. Note that if $\int_{a} ^{b} \ x^nf(x)dx =0$ for all $n \in \mathbb{N}$ then $\int_{a} ^{b} \ p(x)f(x)dx =0$ for any polynomial $p(x)$. By Wierstrass aproximation theorem, this set (polynomials) is dense in $C^0 ([a,b])$ where this last is considered with the uniform norm. The functional $I: C^0([a,b]) \rightarrow \mathbb{R}$ $g \mapsto \int_{a} ^{b} \ g(x)f(x)dx$ is linear and continous and is zero on a dense subset of the domain so it must be zero everywhere. So $\int_{a} ^{b} \ g(x)f(x)dx =0$ for all $g(x) \in C^0([a,b])$. It is a well known result that under this conditions $f \equiv 0$.
Edit: Actually, we would need f to be continous, but since it's R-I it's continous a.e. so there is no real problem.
3. Originally Posted by Xingyuan
If $f(x)$ is Riemann Integrable in $[a,b]$,and for any integer $n\geq0$, we have $\int_{a}^{b}x^{n}f(x)dx = 0$. can we get the conclusion: $f(x)=0$ a.e.?
(a.e. stands for almost everywhere)
Well you made your question almost trivial allowing n >= 0 ==> taking n = 0 we get that INT{1*f(x)} dx = 0 ==> f(x) = 0 a.e.
Tonio
4. Originally Posted by tonio
Well you made your question almost trivial allowing n >= 0 ==> taking n = 0 we get that INT{1*f(x)} dx = 0 ==> f(x) = 0 a.e.
Tonio
No,In the problem,I am not assuming $f(x)\geq0$
5. Originally Posted by Xingyuan
No,In the problem,I am not assuming $f(x)\geq0$
I didn't say anything about f(x) >= 0. I took n = 0, since you wrote n >= 0, and then x^0 =1 and INT{x^0*f(x)} dx = INT fx dx = 0 ==> f(x) = 0 a.e.
Tonio
6. Originally Posted by tonio
Well you made your question almost trivial allowing n >= 0 ==> taking n = 0 we get that INT{1*f(x)} dx = 0 ==> f(x) = 0 a.e.
Tonio
Huh? That doesn't follow.
7. Originally Posted by tonio
I didn't say anything about f(x) >= 0. I took n = 0, since you wrote n >= 0, and then x^0 =1 and INT{x^0*f(x)} dx = INT fx dx = 0 ==> f(x) = 0 a.e.
Tonio
Consider $\int_{-\pi}^{\pi}\sin(x)\,dx$
It's only $0$ at the two endpoints and $x=0$, but the whole integral is $0$.
Or, for that matter, consider $\int_{-a}^a f(x)\,dx$ where $f(x)$ is any odd function.
8. Originally Posted by rn443
Huh? That doesn't follow.
Yes it does
Tonio
9. Originally Posted by tonio
Yes it does
Tonio
Are you assuming f is nonnegative?
10. Originally Posted by rn443
Are you assuming f is nonnegative?
Ok, I thought you gave f(x) >= 0 but you actually didn't, but then the claim is false as redsoxfan showed already.
Tonio
11. Does the theorem still hold if we instead assume $\int_{-\infty}^\infty x^n f(x) \mathrm dx = 0$ for all n >= 0, provided that $\int_{-\infty}^\infty |x^n f(x)| \mathrm dx$ exists and is finite?
12. Lets suppose, without loss of generality, that is $a=-\pi$ and $b=\pi$. Since $f(*)$ is Riemann integrable in $[-\pi,\pi]$ is can be expanded in Fourier series ...
$f(x)= \frac{a_0}{2} + \sum_{k=1}^{\infty} a_{k}\cdot \cos kx + b_{k}\cdot \sin kx$ (1)
... where...
$a_{k}= \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)\cdot \cos kx\cdot dx$
$b_{k}= \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)\cdot \sin kx\cdot dx$ (2)
Now is...
$\cos kx = \sum_{n=0}^{\infty} \frac{(-1)^{n}\cdot (kx)^{2n}}{(2n)!}$
$\sin kx = \sum_{n=0}^{\infty} \frac{(-1)^{n}\cdot (kx)^{2n+1}}{(2n+1)!}$ (3)
... and $\forall n\ge 0$ is...
$\int_{-\pi}^{\pi} x^{n}\cdot f(x)\cdot dx =0$ (4)
The conclusion is that all the integrals in (2) are null and the same is for the $a_{k}$ and $b_{k}$ ...
Kind regards
$\chi$ $\sigma$
13. Originally Posted by Jose27
Yes. Note that if $\int_{a} ^{b} \ x^nf(x)dx =0$ for all $n \in \mathbb{N}$ then $\int_{a} ^{b} \ p(x)f(x)dx =0$ for any polynomial $p(x)$. By Wierstrass aproximation theorem, this set (polynomials) is dense in $C^0 ([a,b])$ where this last is considered with the uniform norm. The functional $I: C^0([a,b]) \rightarrow \mathbb{R}$ $g \mapsto \int_{a} ^{b} \ g(x)f(x)dx$ is linear and continous and is zero on a dense subset of the domain so it must be zero everywhere. So $\int_{a} ^{b} \ g(x)f(x)dx =0$ for all $g(x) \in C^0([a,b])$. It is a well known result that under this conditions $f \equiv 0$.
Edit: Actually, we would need f to be continous, but since it's R-I it's continous a.e. so there is no real problem.
Yes,Riemann Integrable function is continous a.e.
But Weierstrass' approximation theorem is:
Every function defined and continuous on the finite interval $[a,b]$can be approximated uniformly on $[a,b]$ by polynomials to any degree of accuracy.
Because Weierstrass' approximation theorem must be use on interval ,so we must make another function $g(x)$
If we denote the set $E$ of all discontinuity point,then define:
$g(x)=\{ \begin{array}{cc}f(x), &\mbox{if x is a continuous point of} f(x) \\ \lim_{t \rightarrow x}f(t) &\mbox{if x is a discontinuity point of} f(x) \end{array}$
then $g(x)$ is a continuous function on $[a,b]$,right?
then use the conclusion above,we get the conclusion $g(x)\equiv0$.
so $f(x)=0$a.e.
Is this proof right?
14. Originally Posted by Xingyuan
Yes,Riemann Integrable function is continous a.e.
But Weierstrass' approximation theorem is:
Every function defined and continuous on the finite interval $[a,b]$can be approximated uniformly on $[a,b]$ by polynomials to any degree of accuracy.
Because Weierstrass' approximation theorem must be use on interval ,so we must make another function $g(x)$
If we denote the set $E$ of all discontinuity point,then define:
$g(x)=\{ \begin{array}{cc}f(x), &\mbox{if x is a continuous point of} f(x) \\ \lim_{t \rightarrow x}f(t) &\mbox{if x is a discontinuity point of} f(x) \end{array}$
then $g(x)$ is a continuous function on $[a,b]$,right?
then use the conclusion above,we get the conclusion $g(x)\equiv0$.
so $f(x)=0$a.e.
Is this proof right?
Notice that I'm using Weierstrass to expand the set of functions $g$ such that $\int_{a} ^{b} g(x)f(x)dx =0$, I'm not saying anything about $f$.
Another thing, the $g$ you defined could be not well defined if f has a jump discontinuity.
Ok, I thought you gave f(x) >= 0 but you actually didn't, but then the claim is false as redsoxfan showed already.
This doesn't contradict the claim since $\int_{-\pi } ^{\pi } x\sin(x) dx=2\int_{0 } ^{ \pi } x\sin(x) dx >0$
15. Originally Posted by Jose27
This doesn't contradict the claim since $\int_{-\pi } ^{\pi } x\sin(x) dx=2\int_{0 } ^{ \pi } x\sin(x) dx >0$
I was only contradicting the case $n=0$.
Page 1 of 2 12 Last | 2015-11-30T06:14:42 | {
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https://math.stackexchange.com/questions/3303103/how-to-choose-a-control-input-so-that-the-nonlinear-system-stays-at-a-fixed-poin | # How to choose a control input so that the nonlinear system stays at a fixed point?
The dynamics of a magnetically suspended steel ball can be described by
$$m \ddot{h} = mg - c \frac{u^2}{h^2}$$
where $$m$$ is the mass of the ball, $$g$$ is gravitational acceleration, $$c$$ is a positive constant, and $$h$$ represents the vertical position of the ball. The input $$u$$ is the current supplied to the electromagnet.
(a) Write down a nonlinear state space model using $$x_1 = h$$ and $$x_2 = \dot{h}$$.
(b) Determine the equilibrium control input $$u_e$$ that has to be applied to suspend the ball at some position $$h=h_0 > 0$$.
For part (a), the system is \begin{align} \dot{x}_1 &= x_2 \tag{1} \\ \dot{x}_2 &= g - \frac{c}{m} \frac{u^2}{x^2_1} \tag{2} \end{align}
For part (b), the equilibrium point $$x_e=(h_0,0)$$ which is not at the origin, so we need first to make sure the equilibrium point $$x_e$$ is transformed to the origin by introducing new variables: let $$y=x-x_e$$, we get:
\begin{align} y_1 &= x_1 - h_0 &\implies \dot{x}_1 = \dot{y}_1 \tag{3}\\ y_2 &= x_2 - 0 &\implies \dot{x}_2 = \dot{y}_2 \tag{4} \end{align} From (1),(2),(3), and (4), the new system is \begin{align} \dot{y}_1 &= y_2 \\ \dot{y}_2 &= g - \frac{c}{m} \frac{u^2}{(y_1 + h_0)^2} \end{align} The equilibrium point $$y_e = (\sqrt{\frac{c}{gm}} u - h_0, 0)$$. The control input $$u_e$$ that makes $$y_e$$ is zero is $$u_e = \frac{ h_0}{\sqrt{\frac{c}{gm}}}$$, therefore, the equilibrium control input $$u_e$$ that has to be applied to suspend the ball at some position $$h=h_0 > 0$$ is $$u_e = \frac{ h_0}{\sqrt{\frac{c}{gm}}}$$
Is this correct? if not, any suggestions how to tackle this problem.
Looks correct, but you can always check this yourself by plugging your $$u_e$$ into the expression for either $$\dot{x}_2$$ or $$\dot{y}_2$$. Also note that the input only appears as $$u^2$$, so using $$-u$$ should have the same effect as $$u$$.
Lastly it is also possible to solve for $$u_e$$ by setting $$\dot{x}_2=0$$ while using $$x_1=h_0$$. However, you might also have to linearize the system around this equilibrium point, so doing this coordinate translation would probably have to be done anyway in a later stage.
• I've fixed that, thanks. But when I run the simulation via Simulink Matlab, it doesn't seem the the state $x_1 \rightarrow h_0$. I'm not sure if the input choice is correct. – CroCo Jul 25 at 0:42
• @CroCo Having a desired equilibrium does not mean that the system converges to it. For that you also need that that equilibrium is stable. This probably isn't the case. In order to achieve this you need to let $u$ deviate from $u_e$ using some sort of feedback. – Kwin van der Veen Jul 25 at 0:47
• The Jacobain matrix $A$ is $$A =\frac{\partial f}{ \partial x} = \begin{bmatrix} 0 & 1 \\ 2gh^2_0x^{-3}_1 & 0 \end{bmatrix}_{x_e=(h_0,0),u_e=\sqrt{\frac{gm}{c}} h_0} = \begin{bmatrix} 0 & 1 \\ 2gh^{-1}_0 & 0 \end{bmatrix}$$ The eigenvalues of $A$ are $\lambda_{1,2}=\pm \sqrt{\frac{2g}{h_0}}$. They are real and sign opposite, hence, the system is unstable for this input. Is this correct? – CroCo Jul 25 at 2:55 | 2019-12-14T03:02:14 | {
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https://ekiim.me/books/linear-algebra-strang/exercises/01-2.htm | ekiim's blog
## Exercises
1. For the eeauations $x + y = 4, 2x - 2y = 4$, draw the row picture (two intersecting lines) and the column picture (a combination of two columns, equatl to the column vector $(4, 4)$ on the right side).
2. Solve the non singular triangular system: \begin{aligned} u + v + w &= b_1 \\ v + w &= b_2 \\ w &= b_3 \end{aligned} Show that your solution gives a combination of columns that equal the column on the write
3. Describe the intersection of three planes $u + v + w + z = 6$ and $u + w + z = 4$ and $u + w = 2$ (all in 4-dimensional space). Is it a line or a point or an empty set? What is the intersection of the fourth plane u = -1 is excluded?
4. Sketch the lines \begin{aligned} x + 2y &= 2\\ x - y &= 2\\ y &= 1 \\ \end{aligned} Can the three equations be solved simultaneously? What happens to the ficutre if all right hand sides zero? Is there any nonzero choice of right hand sides which allows the three lines to intersect at the same point and the three equations to have a solution?
5. Find two points on the line intersection of three plances $t=0$, and $z=0$ and $x+y+z+y=1$ in four-dimensional space.
6. When $b=(2,5,7)$, find a solution $(u, v w)$ to equation (4) other than the solutions $(1, 0 1)$ and $(4, -1, -1)$, metnioned in the text.
• Equation 4: $u \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} v \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} w \begin{bmatrix} 1 \\ 3 \\ 4 \end{bmatrix} = b$
7. Give two more righthand sides in addition to $b=(2,5,7)$ for which equation (4) can be solved. Give two more right hand sides in addition to $b=(2,5,6)$ for which it cannot be solved.
8. Explain why the system: \begin{aligned} u + w + w &= 2\\ u + 2v + 3w &= 1\\ v + 2w &= 0 \\ \end{aligned} if singular, by finding a combination of three equations that adds up $0 = 1$. What value should replace the last zero on the rightside to allow the equation to have solutions — and what is one of the solutions?
9. The column picture for the prevous exercises is $u \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} v \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} w \begin{bmatrix} 1 \\ 3 \\ 2 \end{bmatrix} = b$ Show that the three columns on the left lie in the same plane by expressing the third column as a combination of the first two. What are the solutions $(u, v, w)$, if b is the zero vector $(0, 0, 0)$?
10. Under what condiction on $y_1, y_2, y_3$, do the points $(0, y_1), $(1, y_2)$, (2, y_3)$ lie on a straight line?
11. The equations \begin{aligned} ax + 2y &= 0\\ 2x + ay &= 0 \\ \end{aligned} are certain to have the solution $x=y=0$. For which values of $a$, is there a whole line of solutions?
12. Sketch the plane $x+y+z=1$, or the part of the plane that is in the positivie octant where $x\geq0, y\geq0, z\geq0$. Do the same for $x+y+z=2$ in the same figure. What vector is Perpendicular to does planes?
13. Starting with the line $x+4y=7$, fin d the equation for the parallel line thruough $x=0, y=0$. Find the equation of another line that meets the first at $x=3, y=1$ | 2019-09-17T10:16:46 | {
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https://byjus.com/question-answer/a-triangle-abc-is-drawn-to-circumscribe-a-circle-of-radius-3-cm-such-that-1/ | Question
# A $$\triangle{ABC}$$ is drawn to circumscribe a circle of radius $$3$$ cm. such that the segments $$BD$$ and $$DC$$ are respectively of length $$6$$ cm and $$9$$ cm. If the area of $$\triangle{ABC}$$ is $$54{cm}^{2}$$, then find the lengths of sides $$AB$$ and $$AC$$
Solution
## Area of $$\triangle{ABC}=$$area of $$\triangle{OBC}+$$area of $$\triangle{OAC}+$$area of $$\triangle{OAB}$$We have $$BD=6$$cm, $$BE=6$$cm(equal tangents)$$DC=9$$cm, $$CF=9$$cm(equal tangents)Area of $$\triangle{OBC}=\dfrac{1}{2}\times b\times h=\dfrac{1}{2}\times 15\times 3=\dfrac{45}{2}{cm}^{2}$$Area of $$\triangle{OAC}=\dfrac{1}{2}\times \left(x+9\right)\times 3=\dfrac{3\left(x+9\right)}{2}{cm}^{2}$$Area of $$\triangle{OAB}=\dfrac{1}{2}\times \left(x+6\right)\times 3=\dfrac{3\left(x+6\right)}{2}{cm}^{2}$$By Heron's formula,area of $$\triangle{ABC}=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}{cm}^{2}$$$$s=\dfrac{x+9+x+6+15}{2}=\dfrac{2x+30}{2}=x+15$$$$\therefore \Delta=\sqrt{\left(x+15\right)\left(x+15-15\right)\left(x-15-x-4\right)\left(x+15-x-6\right)}=\sqrt{x\left(x+15\right)\times 6\times 9}$$$$\Rightarrow \sqrt{54x\left(x+15\right)}=\dfrac{3\left(x+9\right)}{2}+\dfrac{3\left(x+16\right)}{2}+\dfrac{45}{2}$$$$\Rightarrow \sqrt{54x\left(x+15\right)}=\dfrac{3}{2}\left[x+9+x+6+15\right]$$$$\Rightarrow \sqrt{54x\left(x+15\right)}=\dfrac{3}{2}\left(2x+30\right)$$$$\Rightarrow \sqrt{54x\left(x+15\right)}=3x+15$$Squaring on both sides we get$$\Rightarrow 54x\left(x+15\right)=9{\left(x+15\right)}^{2}$$$$\Rightarrow 6x=x+15$$$$\Rightarrow 6x-x=15$$$$\Rightarrow 5x=15$$$$\therefore x=\dfrac{15}{5}=3$$cmHence the sides are $$15$$cm, $$12$$cm and $$9$$cmMathematics
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View More | 2022-01-17T23:24:04 | {
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https://math.stackexchange.com/questions/2311346/show-that-s-n2 | # Show that $s_n<2$
If $s_1=\sqrt{2}$ and $s_{n+1}=\sqrt {2+s_n}$ for $n\geq 1$ show that $s_n<2$ $\forall, n\geq1$ and $s_n$ is convergent.
The second part I supposed that is immediate from Cauchy sequence definition that $s_n$ is convergent.
The first part I think to use induction
For $n=1$ we have that $s_1=\sqrt{2}<2$
Suppose that is true for $n=k$ then we need to show for $n=k+1$. I wrote a few terms of this recursive sequence
$$s_1=\sqrt{2},\quad s_2=\sqrt{2+\sqrt{2}},\quad s_3=\sqrt{2+s_2}=\sqrt{2+\sqrt{2+\sqrt{2}}}$$ Then I think that $s_{n+1}$ can written as $$s_{n+1}=\sum_{i=1}^n2^{\frac{1}{2n}}$$
I do not know how to argue that the above term is less than $2$.
• $$s_{n+1} = \sqrt{2+s_n} < \sqrt{2 + 2} = 2$$ – peterwhy Jun 5 '17 at 23:25
• Your general form is wrong. – Simply Beautiful Art Jun 5 '17 at 23:25
• And it is not "obvious" that the sequence is Cauchy – Luiz Cordeiro Jun 5 '17 at 23:28
As peterwhy mentions,
$$s_n<2\implies s_{n+1}=\sqrt{2+s_n}<\sqrt{2+2}=2\tag{\color{green}\checkmark}$$
To show that it is convergent, you want to show that it is monotone increasing, which combined with the fact that it is bounded above will mean it converges:
$$s_n>s_{n-1}\implies s_{n+1}=\sqrt{2+s_n}>\sqrt{2+s_{n-1}}=s_n\tag{\color{green}\checkmark}$$
and thus you are done.
By the induction assumption, for some $k\in\mathbb N$, $$s_k < 2$$
Consider $n = k+1$,
$$s_{k+1} = \sqrt{2+s_k} < \sqrt{2 + 2} = 2$$
Together with the fact that $s_1 = \sqrt 2 < \sqrt 4 = 2$, by induction, all $s_n < 2$.
For $0<x<2$, $x^2 < 2x$ and $2x < 2 + x$.
To prove that the sequence $s_n$ is increasing,
\begin{align*}s_{n+1} &= \sqrt{2+s_n}\\ &> \sqrt{s_n + s_n}\\ &= \sqrt{2s_n}\\ &> \sqrt{s_n^2}\\ &= s_n \end{align*}
Since the sequence $s_n$ is increasing and bounded above, limit exists.
Hint: $\,s_{n+1}-2=(\sqrt {2+s_n}-2) \cdot \cfrac{\sqrt {2+s_n}+2}{\sqrt {2+s_n}+2} = \cfrac{2+s_n-4}{\sqrt {2+s_n}+2} = \cfrac{s_n-2}{\sqrt {2+s_n}+2}\,$, therefore $s_{n+1}-2$ has the same sign as $s_n -2\,$ and, by induction/telescoping, the same sign as $s_1-2\,$.
Using the fact that the square root function is increasing, and the induction hypothesis that $s_k<2$, we have: $$s_{k+1}=\sqrt{2+s_k}<\sqrt{2+2}=2$$
Assuming that the sequence does converge (you prove that later) write the limit as "S". Letting n go to infinity in $s_{n+1}= \sqrt{2+ s_n}$ we get $S= \sqrt{2+ S}$. Squaring both sides, $S^2= 2+ S$ so S must satisfy $S^2- S- 2= (S- 2)(S+ 1)= 0$ so S is either 2 or -1. Since all terms are positive, the limit (again, if there is a limit) must be 2.
Two show that this is convergent (so that the above is correct) you can show this sequence is increasing and bounded above. Since we got 2 as the limit above, it makes sense to show that $S_n< 2$ for all n. Yes, I would do that using induction on n. When n= 1, $S_n= \sqrt{2}< 2$. Assume that, for some k, $S_k< 2$. Then $S_{k+1}= \sqrt{2+ S_k}< \sqrt{2+ 2}= \sqrt{4}= 2$. | 2019-12-10T08:28:36 | {
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https://math.stackexchange.com/questions/2588234/are-positive-definite-matrices-robust-to-small-changes/2588243 | # Are positive definite matrices robust to “small changes”?
Let $A$ be a positive-definite matrix and let $B$ be some other symmetric matrix. Consider the matrix $$C=A+\varepsilon B.$$ for some $\varepsilon>0$. Is it true that for $\varepsilon$ small enough $C$ is also positive definite?
• Is $B$ required to be symmetric? – kimchi lover Jan 2 '18 at 1:02
• I guess, since "positive definite" usually refer to symmetric matrices. I have updated the question. – user_lambda Jan 2 '18 at 1:02
• Eigenvalues of $A+\epsilon B$ depends continuously on $\epsilon$. If they are all positive for $\epsilon=0$ then they are positive for all small $\epsilon$. – A.Γ. Jan 2 '18 at 1:37
Positive definite means that $\langle v, Av \rangle > 0$ for all nonzero vectors $v$; actually it suffices to check this condition for unit vectors. We have
$$\langle v, Cv \rangle = \langle v, Av \rangle + \epsilon \langle v, Bv \rangle.$$
Now, by the compactness of the unit sphere, $\langle v, Av \rangle$ takes on a minimum nonzero value $m$ on unit vectors (the smallest eigenvalue of $A$, although we don't need this), and $| \langle v, Bv \rangle |$ takes on a maximum nonzero value $M$ on unit vectors (the largest eigenvalue of $B$ in absolute value).
So we can take $\epsilon < \frac{m}{M}$, which gives
$$\langle v, Cv \rangle \ge m - \epsilon M > 0$$
for all unit vectors $v$. So $C$ is positive definite as desired.
• Why did you use the spectral radius of $B$ instead of $| \lambda_{\min} (B) |$? – Rodrigo de Azevedo Jan 2 '18 at 10:13
• That would also work. I just needed anything that would clearly give a lower bound. – Qiaochu Yuan Jan 2 '18 at 10:36
If by "positive definite" you mean "strictly positive definite", the answer is "yes". The set of strictly positive definite matrices is an open set in the space of symmetric matrices.
For the following reason. The positive definite matrices are the ones which satisfy a certain finite set of determinental inequalites (the principal minor determinants must all be strictly positive), each one of of which cuts out an open set in the space of matrices. Alternatively, from first principles, let $X$ be the closed unit sphere in vector space, let $Y$ be the symmetric matrices. The function $(v,A)\mapsto \|Av\|$ is continuous, so the set $S=\{(v,A): \|Av\|\le0\}\subset X\times Y$ is closed. Since $X$ is compact, the map $\pi:(v,A)\mapsto A$ is a closed map, so $\pi(S)$ is closed in $Y$, so the complement of $\pi(S)$ is open in $Y$. But that complement is the set of all $A$ for which $\|Av\|>0$ for all $v\in X$, that is to say, the set of all (strictly) positive definite matrices.
• And for positive semi-definite, it's strictly false: $\pmatrix{1 & 0 \\ 0 & 0 } + \epsilon \pmatrix{0 & 0 \\ 0 & -1}$ has eigenvalues $1, -\epsilon$ for every $\epsilon$. – John Hughes Jan 2 '18 at 1:09
• Is "strictly positive definite" different than "positive definite"? I'm unaware of that definition. – user_lambda Jan 2 '18 at 1:14
• @user_lambda: some authors use "positive definite" to mean "positive semidefinite," in the same way that some authors use "positive" to mean "nonnegative." – Qiaochu Yuan Jan 2 '18 at 1:17
• @QiaochuYuan: For what it's worth, I haven't heard of anyone using the word "positive" to mean "non-negative", but what I have seen is the notation of $\mathbb{R}_+$ to mean the non-negative reals. – Mehrdad Jan 2 '18 at 5:42
• @Mehrdad How about Bourbaki? – Daniel Fischer Jan 2 '18 at 15:16
To complement Qiaochu's answer, we have
$$\rm v^\top C \, v = v^\top A \, v + \varepsilon \, v^\top B \, v$$
where $\| \rm v \|_2 = 1$ and $\varepsilon > 0$. Note that
$$\{ \rm v^\top A \, v : \| v \|_2 = 1 \} = [ \lambda_{\min} (\mathrm A), \lambda_{\max} (\mathrm A) ]$$
$$\{ \rm v^\top B \, v : \| v \|_2 = 1 \} = [ \lambda_{\min} (\mathrm B), \lambda_{\max} (\mathrm B) ]$$
and, thus,
$$\{ \rm v^\top C \, v : \| v \|_2 = 1 \} = \left[ \color{blue}{\lambda_{\min} (\mathrm A) + \varepsilon \, \lambda_{\min} (\mathrm B)}, \lambda_{\max} (\mathrm A) + \varepsilon \, \lambda_{\max} (\mathrm B) \right]$$
We know that $\rm A \succ 0$, i.e., $\lambda_{\min} (\mathrm A) > 0$. If $\rm B$ is positive semidefinite, then $\lambda_{\min} (\mathrm A) + \varepsilon \, \lambda_{\min} (\mathrm B) > 0$, i.e., matrix $\rm C$ is positive definite for all values of $\varepsilon > 0$. After all, the conic combination of positive semidefinite matrices is also positive semidefinite. If $\rm B$ is not positive semidefinite, then
$$\lambda_{\min} (\mathrm A) + \varepsilon \, \lambda_{\min} (\mathrm B) = \lambda_{\min} (\mathrm A) - \varepsilon \, |\lambda_{\min} (\mathrm B)| > 0$$
yields the following upper bound on $\varepsilon$
$$\varepsilon < \color{blue}{\frac{\lambda_{\min} (\mathrm A)}{|\lambda_{\min} (\mathrm B)|}}$$ | 2019-10-17T19:00:50 | {
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https://www.physicsforums.com/threads/composition-of-functions.295719/ | # Composition of functions
1. ### mnb96
638
This might be a silly question:
given a function $$g$$ is it possible to find a function $$f$$ such that $$f = f \circ g$$?
321
f=const.
3. ### csprof2000
287
That is the trivial solution. Perhaps there are others?
f(x) = f(g(x)) if...
f(x) = c, c constant.
g(x) = x, f any function.
If these two conditions don't hold, though...
Assume f has an inverse function. For instance, if f(x) = 2x, then (inv f)(x) = x/2. Then
f = f o g
<=>
f(x) = f(g(x))
<=>
x = g(x).
So if f has an inverse, g must equal x if f = f o g.
So you'd only be looking for functions which don't have an inverse.
But what we have is a little stronger than that, no? Since my argument makes no reference to intervals, it must be true on any interval. So g = x if you want a function f which is invertible over any interval.
The only function which is not invertible over any interval is - you guessed it - constant functions.
So, in summary:
if g(x) = x, then any function f(x) will do.
Otherwise, f(x) = c , c constant, is the only solution.
What about functions of more variables? Or generalized operations like differentiation? No idea.
4. ### John Creighto
811
If g(x)=x Then f(x) can be anything.
Let
g(n)=2n
Then f(n) can map even numbers to one constant and odd numbers to a different constant. (At least it works if n is discrete). Not sure if it works in the continuous case but I think it might.
Last edited: Feb 27, 2009
5. ### yyat
321
a) f(x)=abs(x), g(x)=-x
b) f(x)=cos(x), g(x)=x+2pi
6. ### csprof2000
287
Good point.
I guess then that my argument only works for functions which don't have an inverse where Domain(inv f) = Range(f). This, naturally, precludes functions such as abs(x) and cos(x)...
So I guess more though will have to be put into functions which are not bijections.
7. ### mnb96
638
Thanks a lot. You all made very good observations that helped me a lot.
BTW, it seems that if the function f admits an inverse there are not many choices, while if the function f is not invertible, many solutions exist but the problem is non-trivial, and it is difficult to say what kind of functions f and g would have to be, in order to satisfy f(x)=f(g(x)).
At the moment I am trying to solve the following (similar) problem:
$$f = (f \circ g) g'$$
where g is invertible, and g' denotes its derivative
If you find it interesting, suggestions are always welcome.
Thanks!
8. ### csprof2000
287
Well, similar suggestions - cases - are possible.
Assume f(x) = c, c constant. Then c = cg', g' = 1, and g = x + k, k constant.
Assume f(x) = x. Then x = gg', gdg = xdx, and (g^2)/2 = (x^2)/2 + k, k constant.
Assume f(x) = x^n. Then x^n = g^n g', g^n dg = x^n dx, [1/(n+1)]g^(n+1) = [1/(n+1)]x^(n+1) + k, k constant
Wow, that's an alright result. So for x to any power at all, it's possible. Is is true for any polynomial? Yes, it seems like it should be. So... for any polynomial, I believe you can use the above formula to reduce it.
f(x) = cos x, cos x = (cos g)g', (cos x)dx = (cos g)dg, sin(x) = sin(g) + k.
It seems like, unless I'm mistaken, this is the same thing every time:
f(x) = f(g(x))g'(x) is the same as solving the differential equation f(x)dx = f(g(x))dg.
So, as long as f is integrable, the problem is actually quite easy. Maybe I'm wrong. Thoughts?
9. ### mnb96
638
csproof2000: I think you just made it!
You gave the solution to the problem, and for some reason I didn't immediately spot that what I was trying to do is actually solving a differential equation. | 2015-03-02T03:39:13 | {
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http://math.stackexchange.com/questions/171274/iint-ayx-da-evaluating | # $-\iint_{A}(y+x)\,dA$ Evaluating
I am bit unsure about the following problem:
Evaluate the double integral:
$$-\iint_{A}(y+x)\,dA$$
over the triangle with vertices $(0,0), (1,1), (2,0)$
OK, so I figured here that I would do this by first evaluating the integral over the region bounded by the vertices $(0,0), (1,1), (1,0)$ and then evaluate the integral over the region bounded by the vertices $(1,0), (1,1), (2,0)$ before adding the two answers together, and then reversing the sign of this answer (since there is a minus sign in front of the original double integral). Thus, I begin by finding:
$$\int_{0}^{1}dx \int_{0}^{x}(y+x)\,dy$$
When solved this gives me the answer $\frac{1}{2}$.
Next I solve:
$$\int_{1}^{2}dx \int_{1}^{2-x}(y+x)\,dy$$
When solved this gives me the answer $-\frac{7}{6}$.
I have verified both the integrals in Wolframalpha, and they give me the same answer. I would therefore believe that the final answer should be:
$$-(\frac{1}{2} - \frac{7}{6}) = \frac{2}{3}$$
However, the final answer should, according to the book, be $-\frac{4}{3}$.
Thus, obviously I do something wrong here. If anyone can help me out, I would greatly appreciate it. Is it perhaps that it is not allowed to "split up" this into two separate integrals? I couldn't find a way to solve this without doing this.
-
I see one problem at least: The inner ($y$) integral in the second part should start at $0$. Anyway, you can avoid the splitting by doing the integrals with the $x$ integral on the inside. Try it! – Harald Hanche-Olsen Jul 15 '12 at 20:32
Thank you for your answer. But why should the second integral start at 0? The y-values being at 1 and end up at 0 (hence I chose y = 2-x for the upper value of the integral). – Kristian Jul 15 '12 at 20:39
Draw a picture. The whole triangle has its base at the $x$-axis, and so does each of the two pieces resulting from the split. – Harald Hanche-Olsen Jul 15 '12 at 20:41
You can also solve this with the Gauss integral theorem on $F(x,y) = xy$ and evaluate the integral on the boundary instead. Maybe this is easier, since the boundary is made of line segments. – Cocopuffs Jul 15 '12 at 20:42
It seems you may be confusing the two bounds. The bound that begins at $1$ and ends up at $0$ is the upper bound; the lower one is $0$ throughout. Also, you can tell from the sign that $-\frac76$ must be wrong, since the integrand is non-negative throughout the triangle. – joriki Jul 15 '12 at 20:45
Your second integral should be $$\int_{1}^{2}dx \int_{0}^{2-x}(y+x)dy.$$ Your lower $y$ limit was 1 instead of 0. | 2016-05-26T20:38:22 | {
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https://math.stackexchange.com/questions/4057952/drawing-the-grid-without-lifting-the-pen/4058111 | # Drawing the grid without lifting the pen
The below shape consist of $$24$$ segments with unit length. if we want to draw this shape without lifting the pen, what is the minimum length of the line we should draw? $$1)25\quad\quad\quad\quad\quad\quad2)26\quad\quad\quad\quad\quad\quad3)27\quad\quad\quad\quad\quad\quad4)28\quad\quad\quad\quad\quad\quad5)29\quad\quad\quad\quad\quad\quad$$
I think we can do it somehow by segments with the length $$28$$ (by passing twice the four middle segments on each side of outer big square).
I'm not sure how to solve this problem. is it possible to solve it mathematically or I should just try different ways to draw this?
Notice that if you draw it without lifting the pen, except for the starting and ending point you must draw the same number of lines going into each point as going out. This means that you must draw an even number of segments meeting every point except possibly two. How many extra segments does this imply you need at a minimum? Can you achieve this?
• Thank you for the answer. but unfortunately, I can't figure out how to continue from your hint. can you please elaborate more on that? Mar 11, 2021 at 17:04
You will be able to draw all the edges of the graph without lifting the pen and drawing every edge exactly once (starting and ending at the same node), iff there exists an Euler circuit in the graph.
We shall use the following theorem: a connected graph has an Euler circuit iff all the nodes of the graph has even degree.
The given grid graph does not have all nodes with even degree. Let's first find the nodes with odd degrees, as shown in the next figure. Notice that the nodes B,C, E,I, H,L and N,O have odd degrees (namely 3).
Let's add 4 additional (red) edges to the grid graph as shown in the next figure to make all the nodes have even degrees.
Now, in the augmented graph above, by the theorem, it must have an Euler circuit, so we shall be able to draw all the edges of the graph exactly once without lifting the pen.
We shall use Flury's algorithm to find an Euler circuit in the augmented graph first. The key idea is that when you have a choice between a bridge and a non-bridge, always choose the non-bridge (don't burn the bridges).
Since the augmented graph has all nodes with even degrees, we can start at any node, let's start at B, a node with degree 4 in the augmented graph.
Using the algorithm, the following animation shows how to construct an Euler circuit in the augmented graph.
Now, let's find where the additional edges (ones not present in the original graph) were used in the circuit found in the augmented graph.
Let's go back to our original graph by replacing those edges by the ones in the original graph, we had exactly 4 such additional edges needed to make all nodes in the original graph to have even degrees.
It implies that in order to have the Euler circuit in the original graph we need 4 additional edges (requiring 24+4=28 edges). It also means that in the original graph we need to traverse the corresponding 4 edges twice, in order to draw all the edges without lifting the pen, as shown below:
[EDIT]
If we don't need to start and end at the same node, it's sufficient to have an Euler trail instead of a circuit, needing one less edge: $$28-1=27$$ edges (only 3 edge segments, namely, EI, ON and HL are needed to be traversed twice, leading to minimum length of the line as $$24+3=27$$).
In this case the trail shown above starts at $$B$$ but ends at $$C$$ (instead of $$B$$ as before), since the edge $$BC$$ has already been visited once, we don't need to visit it again (as shown in the next figure).
• Wow! thank you very much for the answer and beautiful animation you used. I really appreciate that. Mar 11, 2021 at 19:15
• Great answer, except that you don't need an Euler circuit, but merely an Euler path. Which means we can save one line segment. Mar 12, 2021 at 1:04
The figure has 4 nodes with 2 lines connected to node. 8 nodes with 3 lines connected, and 4 nodes with 4 lines connected.
If you want to complete the circuit without lifting your pen, you can start at a node with an odd number of connections, and you can end at a node with an odd number of connections, for all the rest you must have an even number of connections. That is, if you get to the node on one path you leave the node on a different path.
This means that you must double up on some of the lines where it appears that there is a odd number of lines to make it an even number. What is the minimum number of lines you need to double up such that there are at most two nodes with an odd number of connections?
• Thank you very much. So for the start we connect two of the points with odd degrees and we left with three more segments passing through other odd degree points on other sides of outer big square. so we must pass them twice and the answer is $24+3=27$. am I right? Mar 11, 2021 at 17:42
• It must be at least 27. It is probably worth the effort to show that it can be done with 27 segments. Mar 11, 2021 at 17:46
Every node must have the same number of edges entering and leaving, except for the end points. There are 8 nodes on the graph with an odd degree, but we can make two of them the endpoints, so we have 6 nodes that need an "extra" edge. Finally, we can note that the odd-degree nodes come in adjacent pairs, so the "extra" edge leaving one node can be the "extra" edge entering another, so we only need half as many "extra" edges as we have non-endpoint, odd-degree nodes. | 2022-12-03T21:22:00 | {
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http://math.stackexchange.com/questions/60261/polynomial-px-0-for-all-x-implies-coefficients-of-polynomial-are-zero | # Polynomial $p(x) = 0$ for all $x$ implies coefficients of polynomial are zero
I am curious why the following is true. The text I am reading is "An Introduction to Numerical Analysis" by Atkinson, 2nd edition, page 133, line 4.
$p(x)$ is a polynomial of the form:
$$p(x) = b_0 + b_1 x + \cdots + b_n x^n$$
If $p(x) = 0$ for all $x$, then $b_i = 0$ for $i=0,1,\ldots,n$.
Why is this true? For example, for $n=2$, I can first prove $b_0=0$, then set $x=2$ to get a linear system of two equations. Then I can prove $b_1=b_2 = 0$. Similarly, for $n=3$, I first prove $b_0=0$, then I calculate the rank of the resulting linear system of equations. That shows that $b_1=b_2=b_3=0$. But if $n$ is very large, I cannot keep manually solving systems of equations. Is there some other argument to show all the coefficients must be zero when the polynomial is always zero for all $x$?
Thanks.
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Are you familiar with the fact that if $p(a) = 0$, then $p(x)$ is divisible by $x - a$? Use this $n+1$ times. – Qiaochu Yuan Aug 28 '11 at 3:36
As explained in the answers, we can solve the problem by showing that a non-zero polynomial of degree $d$ has no more than $d$ roots. However, your approach can be made to work. To carry it out, one uses properties of the Vandermonde matrix. The details are somewhat more messy than the approach through counting roots, but accessible. – André Nicolas Aug 28 '11 at 3:55
Hi Qiaochu Yuan, can you give me some more hints? I am aware of the Fundamental Theorem of Algebra, which can be found here: tutorial.math.lamar.edu/Classes/Alg/ZeroesOfPolynomials.aspx. If I divide a polynomial of degree $n$ by (x-a), and a is guaranteed to be a root (since the polynomial is a horizontal line at $y=0$), then I get as a result, another polynomial with degree $n-1$. After $n-1$ divisions, I get as a result, something like: $d + ex = 0$. $d$ and $e$ might not be the original polynomial coefficients. What would be the next step after this? – jrand Aug 28 '11 at 4:35
For purposes of argumentation: false over finite fields... – GEdgar Aug 28 '11 at 13:31
As you are interested in numerical methods the following is probably not of interest to you, but for the sake of completeness I feel compelled to point out that the conclusion is false, if your domain is finite. A non-zero polynomial can easily vanish at all points of, e.g. $\mathbf{Z}_m$ without being the zero polynomial. The easiest example is the polynomial $p(x)=x^2-x\in \mathbf{Z}_2[x]$ that vanishes at all the elements of $\mathbf{Z}_2$. If you don't study algebra, you can safely ignore this comment for now. – Jyrki Lahtonen Aug 28 '11 at 13:37
HINT $\$ A nonzero polynomial over a field (or domain) has no more roots than its degree, as is easily proved by induction and the Factor Theorem. In fact if every natural was a root then the polynomial would be divisible by $\rm\:(x-1)\:(x-2)\:\cdots\:(x-n)\:$ for all $\rm\:n\in \mathbb N\:,\:$ which yields a contradiction for $\rm\:n\:$ greater than the degree of the polynomial.
Note that the proof of said statement depends crucially on the hypothesis that coefficient ring is an integral domain, i.e. a ring satisfying $\rm\:ab = 0\iff a=0\ \ or\ \ b=0\:.\:$ Over non-domains such as the integers modulo $\rm\:m\:$ not prime, polynomials can have more roots than their degree. In fact if this is true then one can use such roots to factor $\rm\:m\:,\:$ see here.
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Following reading Bill Dubuque's, Shawn's, and Qiaochu Yuan's answer, I believe I must agree to the following fact. If a polynomial is equal to zero, then it must be of degree 0, and the coefficient must be set such that $b_0=0$. Since it is degree 0, the other coefficients must also be set to 0. – jrand Aug 28 '11 at 12:42
@jrand: The degree of the polynomial zero is not zero. – Did Aug 28 '11 at 13:02
Note that $p(x)=0$ implies derivatives of all orders are also $0$. Let $x=0$ into $p(x) = b_0 + b_1 x + \cdots + b_n x^n$ to obtain $b_0 = 0$.
Now differentiate both sides: $p'(x) = b_1 + 2b_2 x + 3b_3 x^2 + \cdots + nb_n x^{n-1}$. Let $x=0$ again, and we get that $b_1=0$.
If we continue to differentiate and substitution $x=0$ we will get that $b_k=0$ for $k=0,1,2,\cdots, n$. This idea can easily be made into a rigorous induction proof.
A nice corollary of this result is that two polynomials are equal if and only if they have the same degree and coefficients.
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If you are willing to accept $\displaystyle \lim_{x \rightarrow \infty}\frac{1}{x^{m}}=0$ and $\displaystyle \lim_{x \rightarrow \infty}x^m=\infty$ for all $m>0$, then you can argue as follows. Suppose you have a polynomial $f(x)=b_nx^n+ \ldots b_0$ with $b_n \neq 0$ and $n\geq 1$, then $$\lim_{x\rightarrow \infty}f(x)=\lim_{x \rightarrow \infty} x^n(b_n+ \ldots \frac{a_0}{x^n})=\lim_{x \rightarrow \infty} x^nb_n=\infty.$$
Since the constant function $0$ does not have this property, it cannot be equal to a polynomial with degree greater or equal to $1$.
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+1. Nice analytical solution with no derivatives. – Did Aug 28 '11 at 22:22
A polynomial can be uniquely fitted by knowing its value at $d+1$ points, where $d$ is the degree of the polynomial. If it's 0 at all points, that's clearly more than $d+1$ points.
But we also know that a polynomial can be written as $(x-r_0)(x-r_1)...(x-r_k)$, where the $r_i$ are the roots of polynomial (possibly complex). Again, if there are an infinite number of roots...
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Let $n\in \mathbb{N}$, and $$p(x)=b_0+b_1x+\ldots +b_nx^n$$ a polynomial of degree at most $n$. If $p(x)=0$ for every $x$, then $b_i=0$ for $i=0,\, 1,\ldots\, n$.
Proof:
By induction on $n$:
If $\deg(p)\leq 1$, then $p(x)=b_0+b_1x$. Since $p(0)=p(1)=0$ you get $b_0=b_1=0$.
Suppose that for any $$r(x)=c_0+c_1x+\ldots +c_nx^n$$ of degree at most $n$, if $r(x)=0$ for every $x$, then $c_i=0$ for $i=0,\, 1,\ldots\, n$. Let $$p(x)=b_0+b_1x+\ldots +b_{n+1}x^{n+1}.$$ Suppose that $p\equiv 0$. Since $p(0)=0$, you get that $b_0=0$. From here, I have two possible arguments. The first, that was my original one, is as follows:
We have $$p(x)=b_1x+b_2x^2+\ldots+b_{n+1}x^{n+1}=x(b_1+b_2x+\ldots+b_{n+1}x^{n}).$$ Then for all $x$, \begin{align*} 0&= p(x)\\ 0&= x(b_1+b_2x+\ldots+b_{n+1}x^{n}).\end{align*} Then for any $x\neq 0$, $$q(x):=b_1+b_2x+\ldots+b_{n+1}x^{n}=0$$ If $q\not\equiv 0$, then we have a polynomial of degree at most $n$ with infinitely many roots. This can't be, therefore $q\equiv 0$. Now, $\deg(q)\leq n$ and $q\equiv 0$, therefore by the induction hypothesis, $b_i=0$ for $i=1,\,\ldots ,\, n+1$.
The second argument was inspired by @Ragib Zaman's answer. Just differentiate both sides of $p(x)=0$, then $$b_1+2b_2x+\ldots+(n+1)b_{n+1}x^n=0$$ for all $x$. By the induction hypothesis, $kb_k=0$ for $k=1,\,\ldots,\,n+1$, and this implies that $b_k=0$ for $k=1,\,\ldots,\,n+1$
Since $b_0=0$, this proves that $b_i=0$ for $i=0,\, 1,\ldots,\, n+1$.
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In general, by the fundamental theorem of algebra we can say that the number of different roots of a polynomial of degree $n$ is at most $n$. I the argument above $q$ must be identically $0$ because if not we have a polynomial with infinite different roots. – leo Aug 28 '11 at 6:46
My comment is the hint of @Bill – leo Aug 28 '11 at 7:12
This induction is flawed because during your nth inductive step you use as a fact a property (for every q of degree n, if q(x)=0 for every nonzero x, then q=0) that the hypothesis at the beginning of the step (for every polynomial p of degree n, if p(x)=0 for every x, then p=0) does not imply. – Did Aug 28 '11 at 9:03
@Didier Piau: I'm trying to prove a statement about the coefficients of polynomials, not about the polynomials. If I understand correct your comment, you say me that the hypothesis of induction does not implies that the polynomial $q$ is identically $0$. Yes that is true, but i don't pretend it. The fact that $q\equiv 0$ is something that I want to apply my hypothesis of induction. – leo Aug 28 '11 at 10:06
I think that the answer is now more clear. – leo Aug 28 '11 at 10:14 | 2014-07-26T14:00:02 | {
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https://math.stackexchange.com/questions/1970304/2-different-methods-of-calculating-possibilities-of-people-sitting-at-round-tabl | # 2 different methods of calculating possibilities of people sitting at round table
I have 2 different problems that seem identical:
8 men and 8 women are going to sit at a round table where there are 16 seats. They take their seats randomly. How many ways can the 16 seats be taken so that no 2 women are sitting next to each other?
I solved this by doing $$2 * 8!8!$$
The method I used was that $8*8*7*7*6*6*5*5*4*4*3*3*2*2*1*1= 8!8!$ is all the ways you can sit these 8 women alternating them with the 8 men.
I then multiplied by two because you have 2 possibilities: the women can be sitting at the odd numbered seats, or at the even numbered ones.
Then I have this other problem:
In how many ways can a party of 4 men and 4 women be seated at a circular table so that no two women are adjacent?
(I took the problem from here: http://www.imsc.res.in/~kamalakshya/cupboard/comb_mag.pdf)
And the solution, according to the website, is:
Answer: The 4 men can be seated at the circular table such that there is a vacant seat between every pair of men in (4-1)! =3! Ways. Now 4 vacant seats can be occupied by 4 women in 4! Ways. Hence the required number of seating arrangements = 3!4! = 144
However, I can't seem to use the method from the 1st problem on this one. All the possible combinations of men and women, sitting alternated with men, would be $4!4!$ Then, multiplying by 2 the solution would be $2 * 4!4! = 1152$
If I use the method from the answer of the 2nd problem with the 1st one, I get $7!8!$, which is different from $2 * 8!8!$
My question is, how can this be? Is there any difference between these 2 problems that I am not seeing?
EDIT: I believe the 1st solution is correct. Check here Probability of men and women sitting at a table alternately
• You are assuming that the seats are labeled. The answer to the second problem assumes the seats are not labeled and takes rotational invariance into account. – N. F. Taussig Oct 16 '16 at 1:29
• Take into account circular permutations. The 1st solution is wring 2nd is correct. – Navin Oct 16 '16 at 1:30
• @navinstudent math.stackexchange.com/questions/1969572/… – SilenceOnTheWire Oct 16 '16 at 1:32
• @N.F.Taussig That makes sense, but how do you infer which method to use from reading the problems, without looking at their solutions? Neither problem mentions anything about the seats being labeled. – SilenceOnTheWire Oct 16 '16 at 1:52
• If the seats are not described as labeled, I would assume that they are unlabeled and that rotational invariance applies, as in the second problem. – N. F. Taussig Oct 16 '16 at 2:57 | 2019-07-21T19:44:30 | {
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http://math.stackexchange.com/questions/318362/conditional-probability-with-balls-in-an-urn | # Conditional Probability with balls in an urn
Two balls, each equally likely to be colored either red or blue, are put in an urn. At each stage one of the balls is randomly chosen, its color is noted, and it is then returned to the urn. If the first two balls chosen are colored red, what is the probability that (a) both balls in the urn are colored red; (b) the next ball chosen will be red?
I'm wondering if my method for part (a) is correct:
Let $P(R)$ be the probability of picking a red ball: $P(R)=\frac{1}{2}$
Let $P(B)$ be the probability of picking a blue ball: $P(B)=\frac{1}{2}$
Let $P(C)$ be the probability of the condition, i.e., picking two red balls consecutively: $P(C)=P(C|H_1)P(H_1)+P(C|H_2)P(H_2)+P(C|H_3)P(H_3)$
where $P(H_1)$ is the probability both balls in the urn are red: $P(H_1)=(P(R))^2=(\frac{1}{2})^2=\frac{1}4$
$P(H_2)$ is the probability that both balls in the urn are blue: $P(H_2)=(P(B))^2=(\frac{1}{2})^2=\frac{1}4$
$P(H_3)$ is the probability that one ball is red and one is blue inside the urn: $P(H_3)=1-(P(H_1)+P(H_2))=1-\frac{1}2=\frac{1}2$ since the sum of the mutually exclusive hypotheses or events must sum to 1.
Thus $P(C)=1\times\frac{1}4+\frac{1}4\times0+\frac{1}2\times\frac{1}4=\frac{3}8$
and $P(H_1|C)= \frac{P(C \bigcap H_1)}{P(C)}=\frac{P(C|H_1)P(H_1)}{P(C)}=\frac{1\times\frac{1}4}{\frac{3}8}=\frac{2}3$
For part (b), I know that $P(R|C)$, the probability of picking a red ball given that the first two balls picked were red is to be found
$P(R|C)= \frac{P(C \bigcap R)}{P(C)}=\frac{P(C|R)P(R)}{P(C)}$
How can I find $P(C|R)$?
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There are three possible urns that you might have: both balls red (RR), both balls blue (BB), one of each color (RB) which occur with probabilities $1/4, 1/4, 1/2$ respectively. Now calculate $P(\text{red,red}\mid \text{RR})$, $P(\text{red,red}\mid \text{BB})$, $P(\text{red,red}\mid \text{RB})$, and combine them via the law of total probability to get $P(\text{red,red})$. Then, use Bayes' formula to obtain $P(\text{BB}\mid \text{red,red})$ – Dilip Sarwate Mar 2 '13 at 3:04
The probability that the contents of the urn are two red is indeed $\frac{1}{4}$, as is the probability of two blue, and the probability of mixed is therefore $\frac{1}{2}$. The derivation could have been done more quickly.
Question (a) asks for the probability both are red given that the two drawn balls are red. Let $R$ be the event both are red, and $D$ be the event both drawn balls are red. We want $\Pr(R|D)$. By the usual formula this is $\frac{\Pr(R\cap D)}{\Pr(D)}$.
To find $\Pr(D)$, note that if both balls are red (probability $\frac{1}{2}$), then the probability of $D$ is $1$, while if one ball is red and the other is not (probability $\frac{1}{2}$) then the probability of $D$ is $\frac{1}{4}$. Thus the probability of $D$ is $\left(\frac{1}{4}\right)(1)+\left(\frac{1}{2}\right)\left(\frac{1}{4}\right)$. This is $\frac{3}{8}$.
The probability of $R\cap D$ is $\frac{1}{4}$. So the ratio is indeed $\frac{2}{3}$.
For (b), you can use the calculation of (a). With probability $\frac{2}{3}$ we are drawing from a double red, and we will get red with probability $1$. With probability $\frac{1}{3}$ the urn is a mixed one, and the probability of drawing a red is $\frac{1}{2}$, for a total of $\frac{2}{3}\cdot 1+\frac{1}{3}\cdot\frac{1}{2}$.
One can also solve (b) without using the result of (a). With I hope self-explanatoru notation, we want $\Pr(RRR|RR)$. The probabilities needed in the conditional probability formula are easily computed. We have $\Pr(RRR\cap RR)=\Pr(RRR)=\frac{1}{4}\cdot 1+\frac{1}{2}\cdot\frac{1}{8}=\frac{5}{16}$. Divide by $\frac{3}{8}$.
- | 2016-07-28T18:42:28 | {
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https://math.stackexchange.com/questions/497584/finding-a-curve-that-intersects-any-line-on-the-plane/497588 | # Finding a curve that intersects any line on the plane
Question Is there a curve on plane such that any line on the plane meets it (a non zero ) finite times ?
What are the bounds on the number of such intersections.
My question was itself inspired by this "Can you draw circles on the plane so that every line intersects at least one of them but no more than 100 of them?"
• Clearly if your curve is the graph of an odd polynomial, the number of intersections is at most the degree of the polynomial. – Cameron Williams Sep 18 '13 at 15:08
• One possibility is a pair of parabolas, say one concave up and one concave down, situated so that they touch tangentially (or intersect) and therefore cannot be separated by a hyperplane. – hardmath Sep 18 '13 at 15:08
• Please try to make titles of questions more informative. E.g., Why does $a\le b$ imply $a+c\le b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. – Lord_Farin Sep 18 '13 at 15:08
• @CameronWilliams Correct, but just being an odd degree polynomial; does not guarantee that every line in the plane will intersect it – ARi Sep 18 '13 at 15:10
• @ARi: Actually it does imply that, except degree 1. – hardmath Sep 18 '13 at 15:13
Cubic parabola $$y=x^3$$ has this property. The max number of such intersections is given by the Fundamental theorem of algebra:
$$x^3=ax+b$$ can have at most 3 solutions.
• Yes but will all of the solutions be real for all real (a,b) pairs – ARi Sep 18 '13 at 15:14
• For any $a$, there will be only one real solution for all sufficiently large $|b|$. – hardmath Sep 18 '13 at 15:17
• I think the important fact here (i.e., relevant to the question), is that $\,x^3-ax-b=0\;$ has always at least one real solution and, of course, at most three different ones. Nice answer. +1 – DonAntonio Sep 18 '13 at 15:22
• So the upper bound for intersections will be three and the lower one 0. – ARi Sep 18 '13 at 15:29
• Well, a tighter and always attainable lower bound is one, in fact...and this fulfills the question's conditions. – DonAntonio Sep 18 '13 at 15:31
As already shown, any cubic polynomial (and indeed, any odd-degree polynomial) has the requisite property by the fundamental theorem of algebra.
What's more, a simple perturbation argument should be enough to show that any (sufficiently) smooth curve that meets every line in at least one point will meet some lines in at least three points. Consider a point tangent to the curve where the second derivative 'with respect to the tangent line' is non-zero; that is, a non-reflex tangent point, or locally extremal point. (Such points must exist if the curve is non-trivial). Now, consider pencils of lines 'near' this intersection point; displaced infinitesimally one way from the tangent, they must have another point of intersection with the curve, and this point can be made 'generic' so that it doesn't vanish under small perturbations. Then displace infinitesimally the other direction; the 'generic' point of intersection is still a point of intersection, but the tangent turns into two points of intersection. | 2019-11-15T09:54:44 | {
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http://mathhelpforum.com/algebra/136341-arithmetic-series-help.html | 1. ## Arithmetic Series Help
First of all I'm not sure if this is the right section, so I apologise if it's not.
I was hoping that someone could help me out with the following question:
From a piece of wire 5000mm long, n pieces are cut, each piece being 10mm longer than the previous piece. Ui is the length of the ith piece and Si is the cumulative length of i pieces.
Express Un and Sn in terms of U1 and n.
2. $U_n = u_1 + 10(n-1)$
$S_n = \sum_{i=1}^n U_i = \sum_{i=1}^n [u_1 + 10(i-1)]$ $= u_1 + 10\sum_{i=1}^n i -\sum_{i=1}^n 1 = u_1 + 10\frac{n(n+1)}{2} - n = u_1 +n(5n + 4)$
Also, $S_n = 5000 \Rightarrow u_1 = 5000 - 5n^2 - 4n.$
3. Your answer for Un looks right so thank you for that.
However the others don't seem to add up, I had to find U1 as the second part of the question and I've calculated it as being 80 which I've checked and it is right. If you put 80 into your formulae it doesn't equal 5000.
Thank you anyway though bud.
4. Hello, JP103!
Anonymous1 made some small errors. . .
$S_n \;=\; \sum_{i=1}^n U_i$
. . $=\; \sum_{i=1}^n \bigg[U_1 + 10(i-1)\bigg]$
. . $=\; \sum_{i=n}^n\bigg[U_1 + 10i - {\color{red}10}\bigg]$
. . $= \;\sum_{n=1}^nU_1 + 10\sum_{n=1}^ni - 10\sum_{n-1}^n1$
. . $=\;{\color{red}n}U_1 + 10\frac{n(n+1)}{2} - 10n$
$\boxed{S_n \;=\;nU_1 + 5n(n-1)}$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Using the sum-formula: . $S_n \:=\:\frac{n}{2}\bigg[2a + (n-1)d\bigg]$
. . we have: . $S_n \;=\;\frac{n}{2}\bigg[U_1 + (n-1)10\bigg] \quad\Rightarrow\quad S_n \;=\;nU_1 + 5n(n-1)$ | 2017-01-17T21:33:13 | {
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http://math.stackexchange.com/questions/165434/how-to-prove-if-a-function-is-bijective | How to prove if a function is bijective?
I am having problems being able to formally demonstrate when a function is bijective (and therefore, surjective and injective). Here's an example:
How do I prove that $g(x)$ is bijective?
\begin{align} f &: \mathbb R \to\mathbb R \\ g &: \mathbb R \to\mathbb R \\ g(x) &= 2f(x) + 3 \end{align}
However, I fear I don't really know how to do such. I realize that the above example implies a composition (which makes things slighty harder?). In any case, I don't understand how to prove such (be it a composition or not).
For injective, I believe I need to prove that different elements of the codomain have different preimages in the domain. Alright, but, well, how?
As for surjective, I think I have to prove that all the elements of the codomain have one, and only one preimage in the domain, right? I don't know how to prove that either!
EDIT
f is a bijection. Sorry I forgot to say that.
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Normally one distinguishes between the two different arrows $\mapsto$ and $\to$. One writes $f:\mathbb{R}\to\mathbb{R}$ to mean $f$ is a function from $\mathbb{R}$ into $\mathbb{R}$. The notation $x\mapsto x^3$ means the function that maps every input value to its cube. – Michael Hardy Jul 1 '12 at 20:39
Wouldn't you have to know something about $f$? Is $f$ a bijection? – Dylan Moreland Jul 1 '12 at 20:40
My apologies! Yes, f is a bijection. – Omega Jul 1 '12 at 20:40
Both of your deinitions are wrong. Maybe all you need in order to finish the problem is to straighten those out and go from there. I've posted the definitions as an answer below. – Michael Hardy Jul 1 '12 at 20:45
The way to verify something like that is to check the definitions one by one and see if $g(x)$ satisfies the needed properties.
Recall that $F\colon A\to B$ is a bijection if and only if $F$ is:
1. injective: $F(x)=F(y)\implies x=y$, and
2. surjective: for all $b\in B$ there is some $a\in A$ such that $F(a)=b$.
Assuming that $R$ stands for the real numbers, we check.
Is $g$ injective?
Take $x,y\in R$ and assume that $g(x)=g(y)$. Therefore $2f(x)+3=2f(y)+3$. We can cancel out the $3$ and divide by $2$, then we get $f(x)=f(y)$. Since $f$ is a bijection, then it is injective, and we have that $x=y$.
Is $g$ surjective?
Take some $y\in R$, we want to show that $y=g(x)$ that is, $y=2f(x)+3$. Subtract $3$ and divide by $2$, again we have $\frac{y-3}2=f(x)$. As before, if $f$ was surjective then we are about done, simply denote $w=\frac{y-3}2$, since $f$ is surjective there is some $x$ such that $f(x)=w$. Show now that $g(x)=y$ as wanted.
Alternatively, you can use theorems. What sort of theorems? The composition of bijections is a bijection. If $f$ is a bijection, show that $h_1(x)=2x$ is a bijection, and show that $h_2(x)=x+2$ is also a bijection. Now we have that $g=h_2\circ h_1\circ f$ and is therefore a bijection.
Of course this is again under the assumption that $f$ is a bijection.
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"Subtract 3 and divide by 2, again we have (y−3)/2=f(x). As before, if f was surjective then we are about done" I'm not sure, but why would we be done there? – Omega Jul 1 '12 at 21:18
@Omega: If $f$ was surjective, then there is some $x$ such that $f(x)=\frac{y-3}2$, show now that $g(x)=y$. – Asaf Karagila Jul 1 '12 at 21:20
You know, it had me thinking: according to your method to find out if it is injective, no matter what function I test it with, I always manage to get the final equality (x = y). How would a function ever be not-injective? – Omega Jul 1 '12 at 22:34
@Omega: No, assume that $f(x)=0$ for all $x$, suppose that $x,y$ are any two real numbers (perhaps different and perhaps not), does $f(x)=f(y)$ tell you something about $x=y$ or $x\neq y$? No, because taking $x=1$ and $y=2$ gives $f(1)=0=f(2)$, but $1\neq 2$. Note that my answer relies critically on the fact that $f$ itself is injective. – Asaf Karagila Jul 1 '12 at 22:37
To prove a function is bijective, you need to prove that it is injective and also surjective.
"Injective" means no two elements in the domain of the function gets mapped to the same image.
"Surjective" means that any element in the range of the function is hit by the function.
Let us first prove that $g(x)$ is injective. If $g(x_1) = g(x_2)$, then we get that $2f(x_1) + 3 = 2f(x_2) +3 \implies f(x_1) = f(x_2)$. Since $f(x)$ is bijective, it is also injective and hence we get that $x_1 = x_2$.
Now let us prove that $g(x)$ is surjective. Consider $y \in \mathbb{R}$ and look at the number $\dfrac{y-3}2$. Since $f(x)$ is surjective, there exists $\hat{x}$ such that $f(\hat{x}) = \dfrac{y-3}2$. This means that $g(\hat{x}) = 2f(\hat{x}) +3 = y$. Hence, given any $y \in \mathbb{R}$, there exists $\hat{x} \in \mathbb{R}$ such that $g(\hat{x}) = y$. Hence, $g$ is also surjective.
Hence, $g(x)$ is bijective.
In general, if $g(x) = h(f(x))$ and if $f(x) : A \to B$ and $h(x): B \to C$ are both bijective then $g(x): A \to C$ is also bijective.
In your case, $f(x)$ was bijective from $\mathbb{R} \to \mathbb{R}$ and $h(x) = 2x+3$ is also bijective from $\mathbb{R} \to \mathbb{R}$.
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You haven't said enough about the function $f$ to say whether $g$ is bijective.
"Injective" means different elements of the domain always map to different elements of the codomain.
"Surjective" means every element of the codomain has at least one preimage in the domain.
Later edit: What you've now added---that $f$ is a bijection---bring us to the point where we can answer the question. However, maybe you should look at what I wrote above. Since both definitions that I gave contradict what you wrote, that might be enough to get you there.
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Sorry, yes, f is bijective. Thank you for clarifying that :) – Omega Jul 1 '12 at 20:43
First show that $g$ is injective ($1$-$1$) by showing that if $g(x)=g(y)$, then $x=y$. This isn’t hard: if $g(x)=g(y)$, then $2f(x)+3=2f(y)+3$, so by elementary algebra $f(x)=f(y)$. By hypothesis $f$ is a bijection and therefore injective, so $x=y$.
Now show that $g$ is surjective. To do this, you must show that for each $y\in\Bbb R$ there is some $x\in\Bbb R$ such that $g(x)=y$. That requires finding an $x\in\Bbb R$ such that $2f(x)+3=y$ or, equivalently, such that $f(x)=\frac{y-3}2$. But $f$ is known to be a bijection and hence a surjection, so you know that there is such an $x\in\Bbb R$.
In general this is one of the two natural ways to show that a function is bijective: show directly that it’s both injective and surjective. The other is to construct its inverse explicitly, thereby showing that it has an inverse and hence that it must be a bijection. You could take that approach to this problem as well:
$$g^{-1}(y)=f^{-1}\left(\frac{y-3}2\right)\;,$$
since
\begin{align*} g\left(f^{-1}\left(\frac{y-3}2\right)\right)&=2f\left(f^{-1}\left(\frac{y-3}2\right)\right)+3\\ &=2\left(\frac{y-3}2\right)+3\\ &=y\;, \end{align*}
and since $f$ is a bijection, $f^{-1}\left(\frac{y-3}2\right)$ exists for every $y\in\Bbb R$.
Added: As Marc reminds me, this is only half the job: if you take this approach, you must either show directly that $g$ is injective, as I did above, or verify that the function that I called $g^{-1}$ above is a two-sided inverse, i.e., that $g^{-1}\big(g(x)\big)=x$ for $x\in\Bbb R$. This is not particularly difficult in this case:
\begin{align*} g^{-1}\big(g(x)\big)&=g^{-1}\big(2f(x)+3\big)\\ &=f^{-1}\left(\frac{\big(2f(x)+3\big)-3}2\right)\\ &=f^{-1}\big(f(x)\big)\\ &=x\;, \end{align*}
since $f$ is a bijection.
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When using the "inverse" criterion, you should be careful in really checking that a purported inverse is an inverse, both ways. With $g^{-1}$ denoting your purported inverse, your final argument checked that $g(g^{-1}(y))=y$ for all $y\in\mathbb R$; this only shows that $g$ is surjective (it has a right inverse, also called a section). To show that $g$ is also injective you need to separately check that $g^{-1}(g(x))=x$ for all $x\in\mathbb R$. – Marc van Leeuwen Jul 1 '12 at 21:22
@Marc: Yes, I should probably say as much; I hadn’t originally intended to mention this approach at all, and did so only as an afterthought. I was implicitly assuming that the obvious injectivity had already been checked, but that’s not clear from what I wrote. – Brian M. Scott Jul 1 '12 at 21:26 | 2014-04-17T07:30:29 | {
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http://math.stackexchange.com/questions/154087/angle-of-a-triangle-inscribed-in-a-square | Angle of a triangle inscribed in a square
Say we have a square $ABCD$. Put points $E$ and $F$ on sides $AB$ and $BC$ respectively, so that $BE = BF$. Let $BN$ be the altitude in triangle $BCE$. What is $\angle DNF$?
I'm inclined to say that it's a right angle because that's what it looks like from what I've drawn, but I have no idea how to proceed.
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I have given an argument why your conjecture is true. I am more interested in why you looked at this problem and how you came up with this neat little conjecture. – user17762 Jun 5 '12 at 5:29
If you want a purely Euclidean-geometry proof, you might start by noting the similarity of triangles $EBC$, $ENB$, and $BNC$, and then trying to prove that triangles $BNF$ and $CND$ are similar. I'm not sure exactly how the proof will go, but it feels doable. – Rahul Jun 5 '12 at 5:32
@Marvis I'm in a 2nd year geometry course, and this is part of my homework due on Friday. I'm being a keener and getting it done early :) I just guessed on the right angle part. – Charlie Jun 5 '12 at 5:46
$\angle DNF = 90^\circ \Longleftrightarrow \angle BNF = \angle CND$. It suffics to prove that $\triangle BNF \sim \triangle CND$. Well, it's trivial: $\angle NBF = \angle BEC = \angle NCD$ and $\displaystyle \frac{NB}{BF} = \frac{NB}{BE} = \sin\angle NEB = \cos\angle NCB = \frac{NC}{CB} = \frac{NC}{CD}$. Q.E.D
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That $NB/BE = NC/CB$ follows from the similarity of right triangles $ENB$ and $BNC$ without need for trigonometry. Nice proof! – Rahul Jun 5 '12 at 7:16
@RahulNarain Trigonometry is an abbreviation of similiar triangles here because we don't use angle transformation formulas such as $\sin(\alpha+\beta)=\cdots$. – Frank Science Jun 5 '12 at 7:20
Indeed; what I meant to say was that the use of trigonometric notation is not needed. It suggests that more powerful machinery is being employed than actually is. – Rahul Jun 5 '12 at 9:12
Firstly We can write
(1) Oklid relation from $\triangle CNB$
$h^2=m(k+x)$
(2) Pisagor relation from $\triangle DPN$
$a^2=(x+k)^2+(x+k+m-h)^2$
(3) Pisagor relation from $\triangle FRN$
$b^2=h^2+k^2$
(4) Pisagor relation from $\triangle DCF$
$c^2=x^2+(x+k+m)^2$
if DNF triange is a right triangle,It must satisfy $a^2+b^2=c^2$.
$(x+k)^2+(x+k+m-h)^2+h^2+k^2=x^2+(x+k+m)^2$
$(x+k)^2+(x+k+m)^2-2h(x+k+m)+h^2+h^2+k^2=x^2+(x+k+m)^2$
$(x+k)^2-2h(x+k+m)+2h^2+k^2=x^2$
$xk+k^2-h(x+k+m)+h^2=0$
$xk+k^2-h(x+k+m)+m(k+x)=0$
$-h(x+k+m)+(k+m)(k+x)=0$
$\frac{x+k}{x+k+m}=\frac{h}{k+m}$
This result is equal to the rates of thales formula for similar triangles $\triangle CRN \sim \triangle CBE$
Thus $a^2+b^2=c^2$ is correct for $\triangle DNF$ .
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Let us do it through coordinate geometry. Let $B$ be the origin. Lets fix the coordinates first. $$A = (0,a)\\ B = (0,0) \\ C = (a,0)\\ D = (a,a)$$ Since $E$ and $F$ are equidistant from $B$, lets say $$F = (b,0)\\ E = (0,b)$$ where $0 \leq b \leq a$. The equation of the line $CE$ is $\dfrac{x}{a} + \dfrac{y}{b} = 1$. The equation of $BN$ is $y = \dfrac{a}{b}x$. This gives us the coordinate of $N$ as $\left( \dfrac{ab^2}{a^2 + b^2},\dfrac{a^2b}{a^2 + b^2} \right)$.
The slope of $FN$ is $m_1 = \dfrac{a^2b/(a^2+b^2) - 0}{ab^2/(a^2+b^2)-b} = \dfrac{a^2}{ab - a^2 - b^2}$.
The slope of $DN$ is $m_2 = \dfrac{a^2b/(a^2+b^2) - a}{ab^2/(a^2+b^2)-a} = \dfrac{ab - a^2 - b^2}{b^2 - a^2 - b^2} = - \left(\dfrac{ab-a^2-b^2}{a^2} \right)$.
Hence, the product of the slopes is $m_1m_2 = -1$. Hence your conjecture is indeed correct. I am waiting for someone to post a nicer geometric argument.
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HINT: Try to resort to Homothetic transformation and you're done immediately. Another way is to connect $N$ to the middle of the $DF$ (let's call that point $M$) and prove that you have there $NM$=$DM$=$MF$ (here you may resort to Apollonius' theorem). You also could resort to the cyclic quadrilaterals as another approaching way.
- | 2015-07-07T22:35:09 | {
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https://math.stackexchange.com/questions/2002785/density-of-rational-and-irrational-numbers | # Density of rational and irrational numbers
Since $\mathbb{Q}$ is countable and $\mathbb{R} \backslash \mathbb {Q}$ is not, what does this tell us about the density of rational and irrational numbers along the real number line? Saying that there exists more irrational numbers than rational numbers seems rather vague becuase we're comparing infinites. How do we even define density here?
• There is a mathematical notion of dense, and the rationals and irrationals are each dense in the real numbers. However, it seems you are thinking of a different notion of denseness, perhaps more related to cardinality... – angryavian Nov 7 '16 at 0:31
• "what does this tell us about the density of rational and irrational numbers along the real number line?" Nothing: there are countable non-dense sets, countable dense sets, uncountable dense sets and uncountable non-dense sets. Countability and density are totally unrelated notions. – Crostul Nov 7 '16 at 0:50
• @Alephnull The Cantor set is such an example. – Crostul Nov 7 '16 at 8:09
• @Crostul But there are not co-countable nowhere dense sets (sets which are nowhere dense by have countable complement). So because $\mathbb{Q}$ is countable, we can at least conclude that $\mathbb{R}\setminus\mathbb{Q}$ is somewhere dense. – Reese Nov 7 '16 at 9:13
• @Alephnull It's pretty straightforward - a nowhere dense set by definition must exclude an entire interval. That interval cannot be countable. – Reese Nov 7 '16 at 11:03
One can have any combination of (countably infinite, uncountable) and (dense in the line, not dense in the line). (Unless, as was suggested in the comments, you have some notion of density other than the one defined by Captain Falcon.)
Countably infinite and dense in the reals: Rationals
Countably infinite and not dense in the reals: Integers
Uncountable and dense in the reals: Irrationals
Uncountable and not dense in the reals: Unit interval
• Ah ah, I appreciate the reference to Captain Falcon. However, "C." stands for "Cyril" ; Falcon being my name. :) – C. Falcon Nov 7 '16 at 0:58
• Another example, a bit worse, is the Cantor set. It is uncountable and not dense, in fact it equals its closure which has (Lebesgue) length zero. – Jeppe Stig Nielsen Nov 7 '16 at 10:11
A subset $A$ of $\mathbb{R}$ is said dense if and only if any elements of $\mathbb{R}$ is a limit of a sequence of elements of $A$.
Indeed, both $\mathbb{Q}$ and $\mathbb{R}\setminus\mathbb{Q}$ are dense in $\mathbb{R}$ in that sense.
As you pointed out, $\mathbb{Q}$ is countable, whereas $\mathbb{R}\setminus\mathbb{Q}$ is not, but the notion of density as nothing to do with countability.
Remark. To see that $\mathbb{Q}$ is dense in $\mathbb{R}$, let us pick $x\in\mathbb{R}$ and for all $n\in\mathbb{N}$ let us define: $$x_n:=\frac{\lfloor 10^nx\rfloor}{10^n}.$$ It is a sequence of $\mathbb{Q}$ which converges towards $x$ using squeeze theorem. For the density of $\mathbb{R}\setminus\mathbb{Q}$ consider: $$y_n:=x_n+\frac{\sqrt{2}}{n+1}.$$
• Nice answer, Captain! (salute) – Todd Wilcox Nov 7 '16 at 5:35
As clarified in other answers, cardinality by itself does not answer density questions.
Regarding the definition of density, there are two definitions. One is topological, saying that a set $A$ is dense if it intersects every non-empty open set. For the real line this is equivalent to saying that $A$ is dense (in the real line $\Bbb R$) if it intersects every open interval $(x,y)$ with $x<y$. In other words, for all choices of $x$ and $y$ with $x<y$ there is $a\in A$ with $x<a<y$. The latter condition usually is termed "order-dense" (in $\Bbb R$) and could be defined without reference to topology, for any linear order in place of $\Bbb R$. That is, $A$ is order-dense in a linear order $(L,<)$ is for all $x,y\in L$ with $x<y$ there is $a$ in $(x,y)$, that is, $x<a<y$ with $a\in A$. Finally a set $A$ is order-dense in itself if $A$ is order-dense in the linear order $(A,<)$. The set $\Bbb Q$ of all rational numbers is order-dense in the set $\Bbb R.$ The set $\Bbb Q$ is also order-dense in itself. The set $\Bbb Z$ of all integers is not order-dense in itself, and it is not order-dense in $\Bbb R$. The middle-third Cantor set $C$ is not order-dense in itself as it has "gaps", e.g. the interval $(\frac13,\frac23)$ contains no elements of $C$. (But, $C$ is (topologically) dense in itself, as it has no isolated points.) $C$ is no-where dense (in $\Bbb R$), meaning that the complement of its (topological) closure in $\Bbb R$ is (topologically) dense (and, as it happens, order-dense).
If $X$ is a topological space and $A$ is a subset then we say that $A$ is dense in $X$ provided that the closure of $A$ contains $X$. When $X=\Bbb R$ this means either of the following two conditions: (1) the set $A$ intersects every non-empty open interval, or (2) for every real number $x$ there is a sequence of points of $A$ converging to $x$. On the other hand, if $A=\Bbb Z$ then $A$ is dense in itself (topologically), but not order-dense in itself. Indeed, for every $m\in\Bbb Z$ the constant sequence $\langle m,m,...\rangle$ converges to $m$. On the other hand, the interval $(m,m+1)$ contains no integers.
In case you interpreted "dense" not in the topological definition, but as saying "many more irrationals than rationals", there's another field of math that provides an alternative way of seeing this.
In Measure Theory there's the concept of the measure of a set, which is an alternative characterization of its size. We want a way to define a function on sets, so let $X\subseteq 2^\mathbb{R}$ be a collection of sets (I have a specific one in mind here, namely the Lebesgue $\sigma$-algebra, but don't worry about that). Now, define the function: $$\mu:X\to\mathbb{R}_{\geq 0}\cup\lbrace\infty\rbrace$$ that must fulfill three axioms:
1. Non-negativity. For any subset $E\in X$, we have that $\mu(E)\geq 0$. Intuitively, things can't have negative size.
2. $\mu(\emptyset) = 0$. This should also seem intuitive, "the lack of a set" has no measure (there will be other sets with no measure though!).
3. Countable additivity on disjoint subsets. If $E_1,E_2$ are disjoint subsets of $\mathbb R$ (so $E_1\cap E_2 = \emptyset$), then: $$\mu(E_1\cup E_2) = \sum_{i = 1}^2 \mu(E_i)$$ This will actually be true for countably many disjoint subsets, so if $\lbrace E_i\rbrace$ are all pairwise disjoint, then: $$\mu(\cup_{i= 1}^\infty E_i)= \sum_{i = 1}^\infty \mu(E_i)$$ For finite sums, the intuition for this should be clear (think about each set as a circle, like how you make a Venn diagram. If there's no overlap between circles, the total "area" should be just all the individual areas added together).
Unfortunately, the above identity working for countable sums is less clearly true to our intuition. If we accept it though, we get some interesting results. The most applicable one to this is that we can calculate: $$\mu(\mathbb R\setminus\mathbb Q) = \infty,\quad\mu(\mathbb Q) = 0$$
So, in terms of measure theory, there are essentially no rational numbers, but infinitely many irrationals. | 2019-06-18T05:30:00 | {
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http://math.stackexchange.com/questions/66996/what-is-the-probability-that-xy | What is the probability that $X<Y$?
If two dice are rolled repeatedly, and $X$ is the number of tosses until $3$ is rolled, and $Y$ is the number of tosses until a $5$ is rolled, what is the probability$(X < Y)$? Also, if $Z$ is the number of rolls until a $10$ is rolled, what the probability$(X < Y < Z)$?
Is $P(X < Y) = \sum_{n=2}^{\infty}\sum_{k=1}^{\infty}((32/36)^{(n-1)}(4/36)-(34/36)^{(k-1)}(2/36)]$?
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Is it assumed that $X$, $Y$ and $Z$ are independent ? – Sasha Sep 23 '11 at 16:50
Yes X Y and Z are independent as the dice are balanced – lord12 Sep 23 '11 at 16:50
No, $X$, $Y$ and $Z$ are not independent. For example, if $X = 1$ then $Y \ne 1$ and $Z \ne 1$. – Robert Israel Sep 23 '11 at 18:40
It's not clear whether "a 3 is rolled" is referring to a 3 coming up on at least one of the two dice, or the sum of the two being 3. But since "a 10 is rolled" can only refer to the sum, I guess we can assume the 3 and 5 are also referring to the sum. – Robert Israel Sep 23 '11 at 18:43
@Robert: Those were my thoughts too. But also notice that the first line says "until 3 is rolled" and not "until a 3 is rolled". So I suppose this indeed refers to the sum being 3, rather than one of them being 3. – TMM Sep 23 '11 at 18:47
(a) $P(X<Y)$
Suppose $q = P(3 \text{ before } 5)$. When will we see a $3$ before we see a $5$? That is if we first see a $3$ on the first roll (we are done), or if we don't see a $3$ or a $5$ on the first roll and then see a $3$ before a $5$ after that. In other words:
$$q = P(3) \cdot 1 + P(5) \cdot 0 + P(\text{neither } 3 \text{ nor } 5) \cdot q$$
Since we know that $P(3) = \frac{2}{36}$, $P(5) = \frac{4}{36}$ we get $P(\text{neither } 3 \text{ nor } 5) = 1 - \frac{2}{36} - \frac{4}{36} = \frac{30}{36}$, so
$$q = \frac{2}{36} + \frac{30}{36} q \quad \Longleftrightarrow \quad q = \frac{1}{3}$$
Note that we can also rewrite the first equation as
$$q = P(3) \cdot 1 + P(5) \cdot 0 + P(\text{neither } 3 \text{ nor } 5) \cdot q \quad \Longleftrightarrow \quad q = \frac{P(3)}{P(3) + P(5)} = \frac{P(3)}{P(3 \text{ or } 5)} = P(3\ |\ 3 \text{ or } 5)$$
This is in accordance with a comment made by Dilip: We can simply ignore all throws which are neither a $3$ or a $5$, as they are irrelevant. Then the probability of seeing a $3$ first is simply the conditional probability of throwing a $3$, given that it's either a $3$ or a $5$.
(b) $P(X<Y<Z)$
We can use the same approach as above. First, let us ignore all throws which are not $3$, $5$ or $10$. We now need that the first event is a $3$, which happens with probability
$$P(X<Y,X<Z) = \frac{P(X)}{P(X \text{ or } Y \text{ or } Z)} = \frac{P(X)}{P(X) + P(Y) + P(Z)} = \frac{2}{2+4+3} = \frac{2}{9}$$
Now if this happens, we only need that in all successive events of $3,5,10$ we first see some number of $3$s, and then see a $5$ (and not a $10$ yet). But that means that after this first event, we can again ignore all events $X$, and look at the first throw resulting in either a $5$ or a $10$. Then
$$P(Y<Z | X<Y,X<Z) = \frac{P(Y)}{P(Y \text{ or } Z)} = \frac{P(Y)}{P(Y)+P(Z)} = \frac{4}{4+3} = \frac{4}{7}$$
So the combined probability is then
$$P(X<Y<Z) = P(X<Y,X<Z) \cdot P(Y<Z | X<Y, X<Z) = \frac{2}{9} \cdot \frac{4}{7} = \frac{8}{63}$$
-
Following on from @Thijs analysis, which is actually a tool called "first step analysis", we can get the probability that $P(X<Y<Z)$. So we have to wait until one of these events happens. $P(X)=\frac{2}{36}$ $P(Y)=\frac{4}{36}$ $P(Z)=\frac{3}{36}$. So we have probability of $\frac{27}{36}$ of achieving "nothing" each trial, and effectively going back to the first step.
So at effective trial one, if $X$ occurs we have success, if $Y$ or $Z$ occurs, we have failure.
$$P(X<Y<Z)=\frac{2}{36}P(X<Y<Z|X<Y,Z)+\frac{3}{36}P(X<Y<Z|Z<X,Y)$$ $$+\frac{4}{36}P(X<Y<Z|Y<X,Z)+\frac{27}{36}P(X<Y<Z)$$
The middle two probabilities are zero, and the first is simply $P(Y<Z)$, and applying first step analysis again we have
$$P(Y<Z)=\frac{4}{7}$$
Plugging back into the previous equation and solving we get
$$P(X<Y<Z)=\frac{8}{36(7)}+\frac{27}{36}P(X<Y<Z)=\frac{8}{63}$$
-
Nice, your follow-up is more in line with my previous analysis than my own follow-up! :) – TMM Sep 23 '11 at 18:37
Here's a simpler way than one involving infinite series (such as that posted by Sasha). Roll a die until you get either a 3 or a 5. What's the probability that it's a 3? It's just the conditional probability of getting a 3, given that you've got either a 3 or a 5: $$\Pr(W=3\mid W=3\text{ or }W=5) = \frac{\Pr(W=3)}{\Pr(W=3\text{ or }W=5)} = \frac{2/36}{6/36}= \frac13.$$
- | 2014-11-24T02:52:01 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/66996/what-is-the-probability-that-xy",
"openwebmath_score": 0.8715810775756836,
"openwebmath_perplexity": 189.553806010762,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9838471618625457,
"lm_q2_score": 0.865224073888819,
"lm_q1q2_score": 0.8512482494706641
} |
https://perfectionatic.org/?tag=julia | 11/9/19
# Evaluating Arbitrary Precision Integer Expressions in Julia using Metaprogramming
While watching the Mathologer masterclass on power sums
I came across a challenge to evaluate the following sum
$$1^{10}+2^{10}+\cdots+1000^{10}$$
This can be easily evaluated via brute force in Julia to yield
function power_cal(n)
a=big(0)
for i=1:n
a+=big(i)^10
end
a
end
julia> power_cal(1000)
91409924241424243424241924242500
Note the I had to use big to makes sure we are using BigInt in the computation. Without that, we would be quickly running in into an overflow issue and we will get a very wrong number.
In the comment section of the video I found a very elegant solution to the above sum, expressed as
(1/11) * 1000^11 + (1/2) * 1000^10 + (5/6) * 1000^9 – 1000^7 + 1000^5-(1/2) * 1000^3 + (5/66) * 1000 = 91409924241424243424241924242500
If I try to plug this into the Julia, I get
julia> (1/11) * 1000^11 + (1/2) * 1000^10 + (5/6) * 1000^9 - 1000^7 + 1000^5-(1/2) * 1000^3 + (5/66) * 1000
-6.740310541071357e18
This negative answer is not surprising at all, because we obviously ran into an overflow. We can, of course, go through that expression and modify all instances of Int64 with BigInt by wrapping it in the big function. But that would be cumbersome to do by hand.
## The power of Metaprogramming
In Julia, metaprogramming allows you to write code that creates code, the idea here to manipulate the abstract syntax tree (AST) of a Julia expression. We start to by “quoting” our original mathematical expressing into a Julia expression. In the at form it is not evaluated yet, however we can always evaluate it via eval.
julia> ex1=:((1/11) * 1000^11 + (1/2) * 1000^10 + (5/6) * 1000^9 - 1000^7 + 1000^5-(1/2) * 1000^3 + (5/66) * 1000)
:((((((1 / 11) * 1000 ^ 11 + (1 / 2) * 1000 ^ 10 + (5 / 6) * 1000 ^ 9) - 1000 ^ 7) + 1000 ^ 5) - (1 / 2) * 1000 ^ 3) + (5 / 66) * 1000)
julia> dump(ex1)
Expr
args: Array{Any}((3,))
1: Symbol +
2: Expr
args: Array{Any}((3,))
1: Symbol -
2: Expr
args: Array{Any}((3,))
1: Symbol +
2: Expr
args: Array{Any}((3,))
1: Symbol -
2: Expr
3: Expr
3: Expr
args: Array{Any}((3,))
1: Symbol ^
2: Int64 1000
3: Int64 5
julia> eval(ex1)
-6.740310541071357e18
The output of dump show the full AST in all its glory (…well almost the depth is a bit truncated). Notice that here all our numbers are interpreted as Int64.
Now we walk through the AST and change all occurrences of Int64 with BigInt by using the big function.
function makeIntBig!(ex::Expr)
args=ex.args
for i in eachindex(args)
if args[i] isa Int64
args[i]=big(args[i])
end
if args[i] isa Expr
makeIntBig!(args[i])
end
end
end
julia> ex2=copy(ex1)
:((((((1 / 11) * 1000 ^ 11 + (1 / 2) * 1000 ^ 10 + (5 / 6) * 1000 ^ 9) - 1000 ^ 7) + 1000 ^ 5) - (1 / 2) * 1000 ^ 3) + (5 / 66) * 1000)
julia> makeIntBig!(ex2)
julia> eval(ex2)
9.14099242414242434242419242425000000000000000000000000000000000000000000000014e+31
We see an improvement here, but the results are not very satisfactory. The divisions yield BigFloat results, which had a tiny bit of floating point errors. Can we do better?
Julia has support for Rational expressions baked in. We can use that improve the results. We just need to search for call expressions the / symbol and replace it by the // symbol. For safety we just have to makes sure the operands are as subtype of Integer.
function makeIntBig!(ex::Expr)
args=ex.args
if ex.head == :call && args[1]==:/ &&
length(args)==3 &&
all(x->typeof(x) <: Integer,args[2:end])
args[1]=://
args[2]=big(args[2])
args[3]=big(args[3])
else
for i in eachindex(args)
if args[i] isa Int64
args[i]=big(args[i])
end
if args[i] isa Expr
makeIntBig!(args[i])
end
end
end
end
julia> ex2=copy(ex1);
julia> makeIntBig!(ex2)
julia> eval(ex2)
91409924241424243424241924242500//1
Now that is much better! We have not lost any precision and we ended us with a Rational expression.
Finally, we can build a macro so the if we run into such expressions in the future and we want to evaluate them, we could just conveniently call it.
macro eval_bigInt(ex)
makeIntBig!(ex)
quote # Removing the denominator if it is redundant
local x=$ex (x isa Rational && x.den==1) ? x.num : x end end and we can now simply evaluate our original expression as julia> @eval_bigInt (1/11) * 1000^11 + (1/2) * 1000^10 + (5/6) * 1000^9 - 1000^7 + 1000^5-(1/2) * 1000^3 + (5/66) * 1000 91409924241424243424241924242500 07/28/18 # Exploring left truncatable primes Recently I came across a fascinating Numberphile video on truncatable primes I immediately thought it would be cool to whip a quick Julia code to get the full enumeration of all left truncatable primes, count the number of branches and also get the largest left truncatable prime. using Primes function get_left_primes(s::String) p_arr=Array{String,1}() for i=1:9 number_s="$i$s" if isprime(parse(BigInt, number_s)) push!(p_arr,number_s) end end p_arr end function get_all_left_primes(l) r_l= Array{String,1}() n_end_points=0 for i in l new_l=get_left_primes(i) isempty(new_l) && (n_end_points+=1) append!(r_l,new_l) next_new_l,new_n=get_all_left_primes(new_l) n_end_points+=new_n # counting the chains append!(r_l,next_new_l) end r_l, n end The first function just prepends a number (expressed in String for convenience) and checks for it possible primes that can emerge from a single digit prepending. For example: julia> get_left_primes("17") 2-element Array{String,1}: "317" "617" The second function, just makes extensive use of the first to get all left truncatable primes and also count the number of branches. julia> all_left_primes, n_branches=get_all_left_primes([""]) (String["2", "3", "5", "7", "13", "23", "43", "53", "73", "83" … "6435616333396997", "6633396997", "76633396997", "963396997", "16396997", "96396997", "616396997", "916396997", "396396997", "4396396997"], 1442) julia> n_branches 1442 julia> all_left_primes 4260-element Array{String,1}: "2" "3" "5" "7" "13" "23" ⋮ "96396997" "616396997" "916396997" "396396997" "4396396997" So we the full list of possible left truncatable primes with a length 4260. Also the total number of branches came to 1442. We now get the largest left truncatable primes with the following one liner: julia> largest_left_prime=length.(all_left_primes)|>indmax|> x->all_left_primes[x] "357686312646216567629137" After this fun exploration, I found an implementation in Julia for just getting the largest left truncatable prime for any base in Rosseta Code. 04/1/18 # Iterating with Dates and Time in Julia Julia has good documentation on dealing with Dates and Time, however that is often in the context constructing and Date and Time objects. In this post, I am focus on the ability to iterate over Dates and Times. This is very useful in countless application. We start of by capturing this moment and moving ahead into the future julia> this_moment=now() 2018-04-01T23:13:33.437 In one hour that will be julia> this_moment+Dates.Hour(1) 2018-04-02T00:13:33.437 Notice that Julia was clever enough properly interpret that we will be on the in another day after exactly one hour. Thanks to it multiple dispatch of the DateTime type to be able to do TimeType period arithmatic. You can then write a nice for loop that does something every four hours for the next two days. julia> for t=this_moment:Dates.Hour(4):this_moment+Dates.Day(2) println(t) #or somethings special with that time end 2018-04-01T23:13:33.437 2018-04-02T03:13:33.437 2018-04-02T07:13:33.437 2018-04-02T11:13:33.437 2018-04-02T15:13:33.437 2018-04-02T19:13:33.437 2018-04-02T23:13:33.437 2018-04-03T03:13:33.437 2018-04-03T07:13:33.437 2018-04-03T11:13:33.437 2018-04-03T15:13:33.437 2018-04-03T19:13:33.437 2018-04-03T23:13:33.437 Often we are not so interested in the full dates. For example if we are reading a video file and we want to get a frame every 5 seconds while using VideoIO.jl. We can deal here with the simpler Time type. julia> video_start=Dates.Time(0,5,20) 00:05:20 Here we are interested in starting 5 minutes and 20 seconds into the video. Now we can make a nice loop from the start to finish for t=video_start:Dates.Second(5):video_start+Dates.Hour(2) h=Dates.Hour(t).value m=Dates.Minute(t).value s=Dates.Second(t).value ms=Dates.Millisecond(t).value # Do something interesting with ffmpeg seek on the video end 02/9/18 # When Julia is faster than C, digging deeper In my earlier post I showed an example where Julia is significantly faster than c. I got this insightful response So I decided to dig deeper. Basically the standard c rand() is not that good. So instead I searched for the fastest Mersenne Twister there is. I downloaded the latest code and compiled it in the fastest way for my architecture. /* eurler2.c */ #include <stdio.h> /* printf, NULL */ #include <stdlib.h> /* srand, rand */ #include "SFMT.h" /* fast Mersenne Twister */ sfmt_t sfmt; double r2() { return sfmt_genrand_res53(&sfmt); } double euler(long int n) { long int m=0; long int i; for(i=0; i<n; i++){ double the_sum=0; while(1) { m++; the_sum+=r2(); if(the_sum>1.0) break; } } return (double)m/(double)n; } int main () { sfmt_init_gen_rand(&sfmt,123456); printf ("Euler : %2.5f\n", euler(1000000000)); return 0; } I had to compile with a whole bunch of flags which I induced from SFMT‘s Makefile to get faster performance. gcc -O3 -finline-functions -fomit-frame-pointer -DNDEBUG -fno-strict-aliasing --param max-inline-insns-single=1800 -Wall -std=c99 -msse2 -DHAVE_SSE2 -DSFMT_MEXP=1279 -ISFMT-src-1.5.1 -o eulerfast SFMT.c euler2.c And after all that trouble we got the performance down to 18 seconds. Still slower that Julia‘s 16 seconds. $ time ./eulerfast
Euler : 2.71824
real 0m18.075s
user 0m18.085s
sys 0m0.001s
Probably, we could do a bit better with more tweaks, and probably exceed Julia‘s performance with some effort. But at that point, I got tired of pushing this further. The thing I love about Julia is how well it is engineered and hassle free. It is quite phenomenal the performance you get out of it, with so little effort. And for basic technical computing things, like random number generation, you don’t have to dig hard for a better library. The “batteries included” choices in the Julia‘s standard library are pretty good.
02/8/18
# When Julia is faster than C
On e-day, I came across this cool tweet from Fermat’s library
So I spend a few minutes coding this into Julia
function euler(n)
m=0
for i=1:n
the_sum=0.0
while true
m+=1
the_sum+=rand()
(the_sum>1.0) && break;
end
end
m/n
end
Timing this on my machine, I got
julia> @time euler(1000000000)
15.959913 seconds (5 allocations: 176 bytes)
2.718219862
Gave a little under 16 seconds.
Tried a c implementation
#include <stdio.h> /* printf, NULL */
#include <stdlib.h> /* srand, rand */
#include <time.h> /* time */
double r2()
{
return (double)rand() / (double)((unsigned)RAND_MAX + 1);
}
double euler(long int n)
{
long int m=0;
long int i;
for(i=0; i<n; i++){
double the_sum=0;
while(1) {
m++;
the_sum+=r2();
if(the_sum>1.0) break;
}
}
return (double)m/(double)n;
}
int main ()
{
printf ("Euler : %2.5f\n", euler(1000000000));
return 0;
}
and compiling with either gcc
gcc -Ofast euler.c
or clang
clang -Ofast euler.c
gave a timing twice as long
$time ./a.out Euler : 2.71829 real 0m36.213s user 0m36.238s sys 0m0.004s For the curios, I am using this version of Julia julia> versioninfo() Julia Version 0.6.3-pre.0 Commit 93168a6 (2017-12-18 07:11 UTC) Platform Info: OS: Linux (x86_64-linux-gnu) CPU: Intel(R) Core(TM) i7-4770HQ CPU @ 2.20GHz WORD_SIZE: 64 BLAS: libopenblas (USE64BITINT DYNAMIC_ARCH NO_AFFINITY Haswell) LAPACK: libopenblas64_ LIBM: libopenlibm LLVM: libLLVM-3.9.1 (ORCJIT, haswell) Now one should not put too much emphasis on such micro benchmarks. However, I found this a very curious examples when a high level language like Julia could be twice as fast a c. The Julia language authors must be doing some amazing mojo. 01/2/18 # Visualizing the Inscribed Circle and Square Puzzle Recently, I watched a cool mind your decsions video on an inscribed circle and rectangle puzzle. In the video they showed a diagram that was not scale. I wanted to get a sense of how these differently shaped areas will match. There was a cool ratio between the outer and inner circle radii that is expressed as $$\frac{R}{r}=\sqrt{\frac{\pi-2}{4-\pi}}$$. I used Compose.jl to rapidly do that. using Compose set_default_graphic_size(20cm, 20cm) ϕ=sqrt((pi -2)/(4-pi)) R=10 r=R/ϕ ctx=context(units=UnitBox(-10, -10, 20, 20)) composition = compose(ctx, (ctx, rectangle(-r/√2,-r/√2,r*√2,r*√2),fill("white")), (ctx,circle(0,0,r),fill("blue")), (ctx,circle(0,0,R),fill("white")), (ctx,rectangle(-10,-10,20,20),fill("red"))) composition |> SVG("inscribed.svg") 08/6/17 # Solving the code lock riddle with Julia I came across a neat math puzzle involving counting the number of unique combinations in a hypothetical lock where digit order does not count. Before you continue, please watch at least the first minute of following video: The rest of the video describes two related approaches for carrying out the counting. Often when I run into complex counting problems, I like to do a sanity check using brute force computation to make sure I have not missed anything. Julia is fantastic choice for doing such computation. It has C like speed, and with an expressiveness that rivals many other high level languages. Without further ado, here is the Julia code I used to verify my solution the problem. 1. function unique_combs(n=4) 2. pat_lookup=Dict{String,Bool}() 3. for i=0:10^n-1 4. d=digits(i,10,n) # The digits on an integer in an array with padding 5. ds=d |> sort |> join # putting the digits in a string after sorting 6. get(pat_lookup,ds,false) || (pat_lookup[ds]=true) 7. end 8. println("The number of unique digits is$(length(pat_lookup))")
9. end
In line 2 we create a dictionary that we will be using to check if the number fits a previously seen pattern. The loop starting in line 3, examines all possible ordered combinations. The digits function in line 4 takes any integer and generate an array of its constituent digits. We generate the unique digit string in line 5 using pipes, by first sorting the integer array of digits and then combining them in a string. In line 6 we check if the pattern of digits was seen before and make use of quick short short-circuit evaluation to avoid an if-then statement.
07/14/17
# Julia calling C: A more minimal example
Earlier I presented a minimal example of Julia calling C. It mimics how one would go about writing C code, wrapping it a library and then calling it from Julia. Today I came across and even more minimal way of doing that while reading an excellent blog on Julia’s syntactic loop fusion. Associated with the blog was notebook that explores the matter further.
Basically, you an write you C in a string and pass it directly to the compiler. It goes something like
using Libdl
C_code= raw"""
double mean(double a, double b) {
return (a+b) / 2;
}
"""
const Clib=tempname()
open(gcc -fPIC -O3 -xc -shared -o \$(Clib * "." * Libdl.dlext) -, "w") do f
print(f, C_code)
end
The tempname function generate a unique temporary file path. On my Linux system Clib will be string like "/tmp/juliaivzRkT". That path is used to generate a library name "/tmp/juliaivzRkT.so" which will then used in the ccall:
meanc(a,b)=ccall((:mean,Clib),Float64,(Float64,Float64),a,b)
julia> meanc(3,4)
3.5
This approach would not be recommended if you are writing anything sophisticated in C. However, it is fun to experiment with for short bits of C code that you might like to call from Julia. Saves you the hassle of creating a Makefile, compiling, etc…
06/29/17
# Solving the Fish Riddle with JuMP
Recently I came across a nice Ted-Ed video presenting a Fish Riddle.
I thought it would be fun to try solving it using Julia’s award winning JuMP package. Before we get started, please watch the above video-you might want to pause at 2:24 if you want to solve it yourself.
To attempt this problem in Julia, you will have to install the JuMP package.
julia> Pkg.add("JuMP")
JuMP provides an algebraic modeling language for dealing with mathematical optimization problems. Basically, that allows you to focus on describing your problem in a simple syntax and it would then take care of transforming that description in a form that can be handled by any number of solvers. Those solvers can deal with several types of optimization problems, and some solvers are more generic than others. It is important to pick the right solver for the problem that you are attempting.
The problem premises are:
1. There are 50 creatures in total. That includes sharks outside the tanks and fish
2. Each SECTOR has anywhere from 1 to 7 sharks, with no two sectors having the same number of sharks.
3. Each tank has an equal number of fish
4. In total, there are 13 or fewer tanks
5. SECTOR ALPHA has 2 sharks and 4 tanks
6. SECTOR BETA has 4 sharsk and 2 tanks
We want to find the number of tanks in sector GAMMA!
Here we identify the problem as mixed integer non-linear program (MINLP). We know that because the problem involves an integer number of fish tanks, sharks, and number of fish inside each tank. It also non-linear (quadratic to be exact) because it involves multiplying two two of the problem variables to get the total number or creatures. Looking at the table of solvers in the JuMP manual. pick the Bonmin solver from AmplNLWriter package. This is an open source solver, so installation should be hassle free.
julia> Pkg.add("AmplNLWriter")
We are now ready to write some code.
using JuMP, AmplNLWriter
# Solve model
m = Model(solver=BonminNLSolver())
# Number of fish in each tank
@variable(m, n>=1, Int)
# Number of sharks in each sector
@variable(m, s[i=1:3], Int)
# Number of tanks in each sector
@variable(m, nt[i=1:3]>=0, Int)
@constraints m begin
# Constraint 2
sharks[i=1:3], 1 <= s[i] <= 7
numfish[i=1:3], 1 <= nt[i]
# Missing uniqueness in restriction
# Constraint 4
sum(nt) <= 13
# Constraint 5
s[1] == 2
nt[1] == 4
# Constraint 6
s[2] == 4
nt[2] == 2
end
# Constraints 1 & 3
@NLconstraint(m, s[1]+s[2]+s[3]+n*(nt[1]+nt[2]+nt[3]) == 50)
# Solve it
status = solve(m)
sharks_in_each_sector=getvalue(s)
fish_in_each_tank=getvalue(n)
tanks_in_each_sector=getvalue(nt)
@printf("We have %d fishes in each tank.\n", fish_in_each_tank)
@printf("We have %d tanks in sector Gamma.\n",tanks_in_each_sector[3])
@printf("We have %d sharks in sector Gamma.\n",sharks_in_each_sector[3])
In that representation we could not capture the restriction that “no two sectors having the same number of sharks”. We end up with the following output:
We have 4 fishes in each tank.
We have 4 tanks in sector Gamma.
We have 4 sharks in sector Gamma.
Since the problem domain is limited, we can possible fix that by adding a constrain that force the number of sharks in sector Gamma to be greater than 4.
@constraint(m,s[3]>=5)
This will result in an answer that that does not violate any of the stated constraints.
We have 3 fishes in each tank.
We have 7 tanks in sector Gamma.
We have 5 sharks in sector Gamma.
However, this seems like a bit of kludge. The proper way go about it is represent the number of sharks in the each sector as binary array, with only one value set to 1.
# Number of sharks in each sector
@variable(m, s[i=1:3,j=1:7], Bin)
We will have to modify our constraint block accordingly
@constraints m begin
# Constraint 2
sharks[i=1:3], sum(s[i,:]) == 1
u_sharks[j=1:7], sum(s[:,j]) <=1 # uniquness
# Constraint 4
sum(nt) <= 13
# Constraint 5
s[1,2] == 1
nt[1] == 4
# Constraint 6
s[2,4] == 1
nt[2] == 2
end
We invent a new variable array st to capture the number of sharks in each sector. This simply obtained by multiplying the binary array by the vector $$[1,2,\ldots,7]^\top$$
@variable(m,st[i=1:3],Int)
@constraint(m, st.==s*collect(1:7))
We rewrite our last constraint as
# Constraints 1 & 3
@NLconstraint(m, st[1]+st[2]+st[3]+n*(nt[1]+nt[2]+nt[3]) == 50)
After the model has been solved, we extract our output for the number of sharks.
sharks_in_each_sector=getvalue(st)
…and we get the correct output.
This problem might have been an overkill for using a full blown mixed integer non-linear optimizer. It can be solved by a simple table as shown in the video. However, we might not alway find ourselves in such a fortunate position. We could have also use mixed integer quadratic programming solver such as Gurobi which would be more efficient for that sort of problem. Given the small problem size, efficiency hardly matters here.
06/12/17
# Reading DataFrames with non-UTF8 encoding in Julia
Recently I ran into problem where I was trying to read a CSV files from a Scandinavian friend into a DataFrame. I was getting errors it could not properly parse the latin1 encoded names.
I tried running
using DataFrames
dataT=readtable("example.csv", encoding=:latin1)
but the got this error
ArgumentError: Argument 'encoding' only supports ':utf8' currently.
The solution make use of (StringEncodings.jl)[https://github.com/nalimilan/StringEncodings.jl] to wrap the file data stream before presenting it to the readtable function.
f=open("example.csv","r")
s=StringDecoder(f,"LATIN1", "UTF-8")
close(f)
The StringDecoder generates an IO stream that appears to be utf8 for the readtable function. | 2021-11-29T17:24:34 | {
"domain": "perfectionatic.org",
"url": "https://perfectionatic.org/?tag=julia",
"openwebmath_score": 0.4335174560546875,
"openwebmath_perplexity": 3075.453846992643,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9664104904802131,
"lm_q2_score": 0.8807970858005139,
"lm_q1q2_score": 0.8512115437020169
} |
https://math.stackexchange.com/questions/3145781/can-a-sequence-be-neither-decreasing-nor-increasing/3145807 | Can a sequence be neither decreasing nor increasing?
Given the definition of a increasing sequence: "A sequence $$(a_n)_{n\in\mathbb{N}}$$ is increasing if for all $$n\in \mathbb{N}$$, $$a_n\le a_{n+1}$$."
My question is: by this definition isn't the sequence $$(1,1,1,1,1,\dotsc)$$ an increasing sequence then?
• Yes, of course. Some people will say the sequence is strictly increasing if $a_n<a_{n+1}$ for all $n$. ... The answer to your title question is of course — just take a sequence like $1,2,1,2,1,2,\dots$. – Ted Shifrin Mar 12 at 22:21
• If you are upset by the language that "increasing" should in your opinion be reserved only for strict inequality between terms, then you might prefer to instead use the term "weakly increasing" instead. Also, before you ask, yes, the definitions work out then that a constant sequence like $1,1,1,\dots$ is simultaneously an increasing sequence and a decreasing sequence and it is easy to prove that any sequence which is simultaneously both must be a constant sequence. – JMoravitz Mar 12 at 22:27
Definition: A sequence $$(a_n)$$ is increasing if $$a_n \le a_{n+1}$$ for all $$n$$,
(or something similar), then the sequence $$(1,1,1,\dotsc)$$ is an increasing sequence. It is increasing precisely because it satisfies the property which defines an increasing sequence (per the definition given).
This may not feel right, as our intuition from natural language is that this sequence is constant, and therefore not increasing. There are two bits of advice that I would give (the first when you have to deal with other people's exposition, the second when you are doing your own work):
1. Just get used to it. There are many terms in mathematics which come from natural language, but which don't mean the same thing in mathematics as they do in vernacular English. The terms "open" and "closed" are a big bugaboo for intro to topology students, for example. As a mathematician, you should learn to get used to words that are given technical definitions which contradict plain English interpretation (and, indeed, may contradict each other, as different authors may define the same term differently.
2. Find another term. In his series on analysis, Simon comments early on that weak vs strict inequalities are possibly ambiguous. He therefore declares that, in his text, "increasing" always means "nondecreasing" (and, if I recall correctly, he also declares that "positive" means "nonnegative"; the point is that he doesn't want to have to say things like "nonnegative nondecreasing"). A similar strategy may help you: if you don't want to call a constant sequence "increasing", use the words "nondecreasing" (for a sequence where $$a_n \le a_{n+1}$$), and use "increasing" (or even "strictly increasing") for sequences which satisfy $$a_n < a_{n+1}$$.
Simon, Barry, Real analysis. A comprehensive course in analysis, part 1, Providence, RI: American Mathematical Society (AMS) (ISBN 978-1-4704-1099-5/hbk). xx, 789 p. (2015). ZBL1332.00003. | 2019-11-20T19:24:33 | {
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https://physics.stackexchange.com/questions/359583/unitary-time-condition | # Unitary time condition
I have a confusion with regards to the principle of QM that states that time evolution must be unitary. In particular, given that states transform through time as $|\Psi(t)\rangle = U(t)|\Psi(0)\rangle$; does the condition: $$\langle\Phi(0)|\Psi(0)\rangle=0 \ \implies \ \langle\Phi(t)|\Psi(t)\rangle=0$$
imply that $U$ must be unitary, or is it imposed on $U$?
I realize that with the before mentioned condition it is posible to prove that for two orthonormal memebers of the basis, $i,j$: $\langle i| U^{\dagger}U | j \rangle = 0$. Is it posible to prove that for the same member the product is 1?
• If $U$ is unitary then $U^{\dagger}$ is the inverse of $U$ by definition. Sep 27 '17 at 14:10
• Yes, but the question is about how to show that, that is the case Sep 27 '17 at 14:18
• Since $\vert \Psi(t)\rangle = U\vert\Psi(0)\rangle$ then clearly $$\langle \Phi(t)\vert \Psi(t)\rangle = \langle \Phi(0)\vert U^\dagger\,U\vert \Psi(0)\rangle = \langle\Phi(0)\vert\Psi(0)\rangle$$ valid $\forall t$ implies $U^\dagger U=\hat 1\, \forall t$ as well, provided your kets are arbitrary. (Hopefully this should be enough.) Sep 27 '17 at 14:32
• The condition of orthogonality preservation you stated alone would also allow for operators that satisfy $U^\dagger U = c \cdot 1$ with any constant $c$. Usually one says that unitarity comes from norm preservation, i.e. $< \psi(t) ,\psi(t)> = <\psi(0),\psi(0)>$.
– Luke
Sep 27 '17 at 15:23
It is an interesting elementary problem. After I while I proved the following proposition where I use $A^*$ for the adjoint of $A$.
Proposition. Let $U: H \to H$ be a bounded operator over a Hilbert space $H$. The following conditions are equivalent.
(a) For $x,y \in H$,$\quad$ $\langle x|y\rangle =0$ implies $\langle Ux|Uy\rangle =0$
(b) $U^*U = cI$ for some real $c\geq 0$.
Before proving the statement I observe that even if $c=1$, $U$ is not necessarily unitary, because unitarity is $U^*U=UU^*=I$. And here $UU^*=I$ generally fails when $H$ is infinite dimensional (otherwise it is trivially true as a consequence of $U^*U=I$). For $c=1$, $U$ is an isometry not necessarily surjective.
Proof. It is obvious that (b) implies (a), so we prove that (a) implies (b). Condition (a) can be rephrased as $y \perp x$ implies $y \perp U^*Ux$. As a consequence $U^*Ux \in \{\{x\}^\perp\}^\perp$ which is the linear span of $x$. In other words $U^*U x = \lambda_x x$ for some $\lambda_x\in \mathbb C$. My goal is now proving that $\lambda_x$ does not depend on $x$.
To this end, consider a couple of vectors $x \perp y$ with $x, y \neq 0$. Using the argument above we have $$U^*U x = \lambda_x x\:,\quad U^*U y = \lambda_y y\:, \quad U^*U (x+y) = \lambda_{x+y} (x+y)\:.\tag{1}$$ Linearity of $U^*U$ applied to the last identity leads to $$U^*Ux + U^*Uy = \lambda_{x+y}x + \lambda_{x+y}y\:,$$ namely $$U^*Ux- \lambda_{x+y}x = -(U^*Uy- \lambda_{x+y}y)\:\:.$$ Exploiting the first two identities in (1) we get $$(\lambda_x- \lambda_{x+y})x = -(\lambda_y- \lambda_{x+y})y\:.$$ Since $x \perp y$ and $x,y \neq 0$, the only possibility is that $$\lambda_x = \lambda_{x+y} = \lambda_y\:.$$ So a couple of orthogonal non-vanishing vectors has the same $\lambda_x$.
To conclude consider a Hilbert basis $\{x_n\}$ of $H$ so that, if $z\in H$, $$z = \sum_n c_n x_n \tag{2}$$ for complex numbers $c_n$. Since $U^*U$ is continuous ($U$ is bounded), $$U^*Uz = \sum_n c_n U^*Ux_n = \sum_n c_n \lambda_{x_n}x_n\tag{3}$$ But we know from the previous argument that $\lambda_{x_n} = \lambda_{x_m}$ so that, indicating with $c$ the common value of the $\lambda_{x_n}$, (3) can be rewritten as $$U^*Uz = \sum_n c_n cx_n = c\sum_n c_n x_n = cz\:.$$ Since $z\in H$ was arbitrary, we have found that $$U^*U=c I\:.$$ Taking the adjoint of both sides we obtain $c=\overline{c}$ so that $c$ is real. Finally, $$0 \leq \langle Ux | Ux\rangle = \langle x| U^*U x\rangle = c \langle x| x \rangle$$ so that $c\geq 0$. QED
• This is nice but does the part "Is it posible to prove that for the same member the product is 1?" not imply $c=\lambda_x=1$? Sep 27 '17 at 19:32
• I do not understand, we already know that $U$ such thst $U^*U= cI$ satisfies the initial hypotheses also for $c\neq 1$. If it were possible to prove that $c=1$ we would exclude this case that we know exists. Sep 27 '17 at 19:40
• Something's clearly throwing me off here. I'll come bac k to in a couple of days. Sep 27 '17 at 19:57 | 2022-01-18T04:08:50 | {
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http://math.stackexchange.com/questions/421760/combinations-question-why-is-my-approach-wrong | # Combinations question - why is my approach wrong?
I'm learning Permutations and Combinations and while trying to solve this simple question, I stuck on the solution:-
From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?
The solution was to consider all three possibilities one by one- 3 men and 2 women, 4 men and 1 women, all 5 men. Calculate the combinations for each case and add them.
But I was trying it from a different approach- Firstly select 3 men from 7 (gives 35 combinations) and multiply it with the combinations of selecting 2 more committee members from 10 remaining members. So my answer came out to be 35 x 45 = 1575. But the answer is 756 which is not even a multiple of 35. So is my approach wrong? Why so?
-
If you first select 3 men and then fill up with two arbitrary people, then you count each combination with exactly three men once, but others are counted repeatedly. For example, you count all-male committees 10 times, once for each way to choose the three first men in retrospect from the final five men.
Smaller numbers example: Given John, Jack and Jill, form a committee of two people with thge constraint that at least one is a man. Actually, there are three such ocmmittees possible (becaus ethe constraint is no constraint at all). But your method would give four: First select John as male, then select either Jack or Jill as second. Or first select Jack as male, then select John or Jill as second. The committee "Jack and John" is counted twice by you.
-
I don't understand your first paragraph, but your second paragraph, with the simpler example, was very helpful. Thanks. – a.real.human.being Oct 5 '14 at 1:00
You’re counting every committee with $4$ men $\binom43=4$ times: if the men are A, B, C, and D, and the woman is F, you’re counting it once with A, B, and C as the first $3$ and D and F as the extra $2$, once with A, B, and D as the first $3$ and C and F as the extra $2$, once with A, C, and D as the first $3$ and B and F as the extra $2$, and once with B, C, and D as the first $3$ and A and F as the extra $2$.
Similarly, you’re counting every committee with $5$ men $\binom53=10$ times: if the men are A, B, C, D, and E, you’re counting the committee once with A, B, and C as the first $3$ and D and E as the extra $2$, and so one for every possible $3$-$2$ split of the five men.
Only the committees of $3$ men and $2$ women are counted correctly, exactly once each.
Since some committees are counted once, some four times, and some ten times, you don’t even get an answer that’s related to the correct one in some very simple way, as you would if, for example, you counted every committee twice. From the correct answer you know that there are $525$ committees with $3$ men, $210$ with $4$ men, and $21$ with $5$ men, and sure enough,
$$525+4\cdot210+10\cdot21=1575\;,$$
$$\binom 73\cdot\binom 62+\binom 74\cdot\binom 61+\binom 75\cdot\binom 60=35\cdot15+35\cdot6+21\cdot1=756$$ | 2015-01-25T16:52:50 | {
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http://pziq.umood.it/exponential-functions-game.html | # Exponential Functions Game
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A simple way to know differentiate between the two is to look at the output values when plugging in a number for an unknown variable. a is any value greater than 0. x -2 -1 0 1 2 y -6 -6 -4 0 6. Show students how to simplify exponential functions, including those involving multiplying (x²*x³) and dividing terms (x³/x²) with exponents or apply exponents to a term that already. If you think about it, having a negative number (such as –2) as the base wouldn't be very useful, since the even powers would give you positive answers (such as "(–2) 2 = 4") and the odd powers would give you negative answers (such as "(–2) 3 = –8"), and what would you even do with the powers that aren't. It's the exclusive-or (XOR) operator (see here). 21; Section B and Throwback 32 & 33 was due 03. The two types of exponential functions are exponential growth and exponential decay. Finding the equation of an exponential function from the graph Worked example 17: Finding the equation of an exponential function from the graph Use the given graph of $$y = -2 \times 3^{(x + p)} + q$$ to determine the values of $$p$$ and $$q$$. Thanks for contributing an answer to Signal Processing Stack Exchange! Please be sure to answer the question. Scroll down the page for more examples and solutions on the graphs of exponential. Try Remote Buzzer-Mode for even more fun!. Other examples of exponential functions include: $$y=3^x$$ $$f(x)=4. Exponential functions. Play this game to review Algebra I. To determine how the given function grows over an interval of length 3, determine the value of f at each endpoint of that interval. 90)^t Find the initial value of the car and the value after 10 years. 27: Derivatives of Exponential Functions: Section 5. This allows you to make an unlimited number of printable math worksheets to your specifications instantly. Syntax : exp(x), where x is a number. Parent Function for Exponential Decay Functions The function f ()xb= x, where 0 1,< 0. Finding an Exponential Function through 2 points. Then the following properties are true: 1. , screenshot, dump, ads, commercial, instruction. x –2 –1 0 1 2 y –6 –6 –4 0 6. Intro Lesson to Exponential functions. 25 and the exponential component is 10 taken to the negative 2 power. A comprehensive database of more than 11 exponential quizzes online, test your knowledge with exponential quiz questions. A logical value that indicates which form of the exponential function to provide. The function cards include x-axis reflections and/or vertical translations. Classwork Estimation 180 - Day 16 (Andrew Stadel) Warm Up Card Sort: Exponentials (Desmos) Practice Answers Link to Answers Standards Common Core HSF. c ccsc welc I sout I o Digit Mail MYF Knov Play Skip Answers 1046 AM 4/22/2020 A exponential growth exponential decay kahoot. Exponential. Syntax : exp(x), where x is a number. Exponential definition, of or relating to an exponent or exponents. Here is the calculator itself. In mathematics, the exponential function is the function whose derivative is equal to itself. It is a nice way to introduce the concept of Exponential Functions and start the class differently than the norm. For example, build a function that models the temperature of a cooling body by adding a constant function to a decaying exponential, and relate these functions to the model. js can automatically use backoff strategies to retry requests with the autoRetry parameter. The exponential function is one of the most important functions in mathematics (though it would have to admit that the linear function ranks even higher in importance). I have one standard for linear equations and one for exponential equations. Please help me on this Algebra 1 problem involving exponential functions (exponential growth and decay). Exponential & Logarithmic Functions. This section contains the following sections. Although exponential growth is always ultimately limited it is a good approximation to many physical processes in the Earth system for finite time intervals. In the first example, the decimal component is 6. Linear Quadratic Exponential y mx " by a x 2 "bx " cy a b x Use the data in the table to describe how the software’s cost is changing. I can model real world situations with exponential functions. Top synonym for exponentially (another word for exponentially) is rapidly. Graph exponential functions. 2 Create and graph equations in two variables to represent linear, exponential, and quadratic relationships between quantities. Herb Gross introduces the exponential function, discussing the rate of growth of the output for each unit change in input and motivates the definition of the meaning of fractional exponents. Play this game to review Algebra I. ˆ˙˝ ˆ˚ ˛ ˘ ˇ ˘ Solving Exponential and Logarithmic Equations 1. Exponential Functions - Application Problem - National Debt. EXPONENTIAL REGRESSION. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Now it was time to see how the economic concepts. An exponential function is a function which takes a point x and returns the value a to the power x. Exponential definition, of or relating to an exponent or exponents. In this example: 8 2 = 8 × 8 = 64. Creating Exponential Functions to Match Given Behaviors-Part 2 5. Determine whether each table or rule represents a linear or an exponential function. Graph of Exponentials 1 - Cool Math has free online cool math lessons, cool math games and fun math activities. Making statements based on opinion; back them up with references or personal experience. Concept Summary Exponential Function Family You can apply the four types of transformations—stretches, compressions, reflections, and translations—to exponential functions. graph of increasing exponential function going through point 0, 1. The #1 Jeopardy-style classroom review game now supports remote learning online. You don't write a function for this (unless you're insane, of course). , population growth). This exercise practices graphing exponential functions and find the appropriate graph given the function. Graphing Exponential Functions The graph of a function y = abx is a vertical stretch or shrink by a factor of ∣ a ∣ of the graph of the parent function y = bx. 5% per year. Nothing like a good criminal investigation to liven up geometry! In this project, students will work in teams to investigate the culprit of six fictional thefts. It began at a length of 6 in and grew at a rate of 14% a week. Writing Equations – Key. Exponential Growth/Decay Graph Demonstration. Click below for lesson resources. Connect Math Support Subject: (Choose one) I forgot my login information I am having a technical problem with Connect Math I just need help with Connect Math I have a suggestion or comment regarding Connect Math I would like to get more information about Connect Math Other. Learn math anywhere on your mobile device or tablet. I have one standard for linear equations and one for exponential equations. In exponential decay, the total value decreases but the proportion that leaves remains constant over time. Priming for the Laws of Exponents 3. Then we explore additional applications by looking at Half Life, Compounding Interest and Logistic Functions. & & Exponential)Ellie&beginswith100inhersavingsaccount,butearnsannual. Exponential functions always have some positive number other than 1 as the base. EXPONENTS MADE VERY SIMPLE TO UNDERSTAND. About this webmix : Glencoe Sequences Quiz Glencoe Exponential Function Write answers on a sep page Exponents Game - Otter. The Mayfair Exponential Game System or MEGS is a rules system developed for role-playing games. If f is a function, x is the input (an element of the domain), and f(x) is the output (an element of the range). The parameter value. - A 'positive constant base raised to a variable exponent' = (constant) (variable) is an 'Exponential function'. ˆ˙˝ ˆ˚ ˛ ˘ ˇ ˘ Solving Exponential and Logarithmic Equations 1. An exponential function is defined for every real number x. This repeated multiplication can be expressed using exponential functions. Modeling with exponential and power law functions in Geosciences. Engaging math & science practice! Improve your skills with free problems in 'Write and Apply Exponential and Power Functions' and thousands of other practice lessons. Exponential Growth/Decay Graph Demonstration. Date: Topic / Lesson and Handouts: Homework, and Other Resources: Mar. Students play a generalized version of connect four, gaining the chance to place a piece on the board by solving an algebraic equation. I am having a hard time researching how to handle summations of functions with exponential growth or decay. Simplify the expression:log (base 2) 8 , Write the equation log (base 3) 729 = 6 in exponential form. This calculus video tutorial explains how to find the derivative of exponential functions using a simple formula. 1 Warm-up: Math Talk: Exponents (5 minutes) 4. Categories with all finite products and exponential objects are called cartesian closed categories. Next Chapter: EXPONENTIAL AND LOGARITHMIC FUNCTIONS. Question 1109486: The dollar value v(t) of a certain car model that is t years old is given by the following exponential function. Logarithm property. LG 1 – Exponential Functions – Blank Copy. 87 which is already approximately 3 as you want. Simplify the expression: log (base 2) 8 , Write the equation log (base 3) 729 = 6 in exponential form. How to Compare Linear, Exponential, and Quadratic Functions Comparing Functions By its Intercepts x-intercept: where y=0 Locate on the graph or table OR Substitute 0 in for y and solve for x y-intercept: where x=0 Locate on the graph or table OR Substitute 0 in for x and solve for y By Rate of Change "Which one grows the fastest?" ' *Remember. Student Instructions Students will keep track of all information for the direct instruction in notability with color -coded examples of each part of the exponential function. A comprehensive database of more than 11 exponential quizzes online, test your knowledge with exponential quiz questions. Finding an exponential function given its graph. ) Exponential Functions. The library includes a great number of useful mathematical functions for manipulating floating point numbers. Computer Software Year 0 1 2 3. First Name:. Really clear math lessons (pre-algebra, algebra, precalculus), cool math games, online graphing calculators, geometry art, fractals, polyhedra, parents and teachers areas too. 01) to the 12𝘵 power, 𝘺 = (1. To help us grasp it better let us use an ancient Indian chess legend as an example. Function Machine Division: If you think the numbers are being divided by 2, simply enter ÷2. a) Multiply b) Add c) Subtract 500 Answer from Exponential Graphs and Equations :. Logarithm Worksheets Logarithms, the inverse of the exponential function, are used in many areas of science, such as biology, chemistry, geology, and physics. KEYWORDS: Course materials, Linear functions, Least Squares Line, Linear Modeling, Statistics and probability, Non-linear functions, Managing money with the exponential function, The logistic function – constrained growth and decay. If you think about it, having a negative number (such as -2) as the base wouldn't be very useful, since the even powers would give you positive answers (such as "(-2) 2 = 4") and the odd powers would give you negative answers (such as "(-2) 3 = -8"), and what would you even do with the powers that aren't. Inverse Trig Functions 21. Characteristics of Graphs of Exponential Functions Before we begin graphing, it is helpful to review the behavior of exponential growth. Exponential functions always have some positive number other than 1 as the base. Exponential Functions. An introduction page sets out some basic information about exponential functions leading to a definition of the exponential function. You don't write a function for this (unless you're insane, of course). In this case, f(x) is called an exponential growth function. I added a little more detail as I am making it a major grade and needed a clear rubric and wanted to try and add a little more detail for the students. The chip level is the domain of Moore's Law, as noted. I know that simple summations can be calculated as follows:$$\sum_{n=1}^{50} n = \frac{n(n+1)}{2}$$How do you approach problems of exponential decay or growth? Consider the following example:$$\sum_{n=1}^{50} e^{-0. Classroom Work for Exponential Functions. Introduction to Exponential Functions. If they find a match, YAY! If not, they have to turn them back over and lose. Without introducing a factor to suppress it, exponential growth is an infectious disease doctor's. Farina explained that an exponential function exhibits a rapid growth of a given variable, and in her view, the growth of prices in the economy could show exponential growth, if analyzed through a long period of time. An exponential _____ function is a function of the form f(x) = ab x where a > 0 and 0 < b < 1. Fractional exponent. Play this game to review Algebra I. About this webmix : IXL-Evaluate an exponential Graphs of Exponential Function Comparing growth rates. l 9 2A nl4lg rji 8g yh3t LsS tr RelsCeUr kv YeDd5. , Write the equation log 1000 = 3 in exponential form. Exponential functions are closely related to geometric sequences. Students will explore and interpret the characteristics of functions, using graphs, tables, and simple algebraic techniques. Concept Summary Exponential Function Family You can apply the four types of transformations—stretches, compressions, reflections, and translations—to exponential functions. Hence the range of function f is given by y > 0 or the interval (0 , +∞). We will begin by drawing up a curve for y = 10 x. As functions of a real variable, exponential functions are uniquely characterized by the fact that the growth rate of such a function (that is, its derivative) is directly proportional to. Here is the entire project for FREE if you are interested.$500 Question from Exponential Graphs and Equations In tables of exponential functions, you _____ by the same amount each time to get to the next number. By definition: log b y = x means b x = y. If you've ever earned interest in the bank (or even if you haven't), you've probably heard of "compounding", "appreciation", or "depreciation"; these have to do with exponential functions. Returns the exponential distribution. alg_1_chapter_6_review_game. This image shows. The following list outlines some basic rules that apply to exponential functions: The parent exponential function f(x) = bx always has a horizontal asymptote at y = 0, except when […]. Welcome to IXL's year 11 maths page. Click below for lesson resources. Online tutoring available for math help. Implicit in this definition is the fact that, no matter when you start measuring, the population will always take the same amount of time to double. Derivatives Of Exponential, Trigonometric, And Logarithmic Functions Exponential, trigonometric, and logarithmic functions are types of transcendental functions; that is, they are non-algebraic and do not follow the typical rules used for differentiation. js can automatically use backoff strategies to retry requests with the autoRetry parameter. Solving exponential equations using exponent rules. ˘ Inverse Properties of Exponents and Logarithms Base a Natural Base e 1. Exponential Function Quizizz - Game Code: 158847. Examples : exp(0), returns 1. Doubling time. Chapter 8 : Exponents and Exponential Functions 8. Graphing Program That Teaches a Thing or Two If you want to know anything about math, statistics, use a grapher, or just simply amuse yourself by strange information about everything, check out Wolfram Alpha. Identify functions using differences or ratios EXAMPLE 2 Use differences or ratios to tell whether the table of values represents a linear function, an exponential function, or a quadratic function. 02) to the 𝘵 power, 𝘺 = (0. Is the graph linear, exponential or neither?. It is used to represent exponential growth, which has uses in virtually all science subjects and it is also prominent in Finance. True Note that the t values are equally spaced and the ratios 3 0 0 1 4 7 = 1 4 7 7 2. The two types of exponential functions are exponential growth and exponential decay. Links to the data sets are included in the file. Recall the table of values for a function of the form $f\left(x\right)={b}^{x}$ whose base is greater than one. The function exp calculates online the exponential of a number. A geometric sequence is a list of numbers in which each number is obtained by multiplying the previous number by a fixed factor m. The exponential function is one of the most important functions in mathematics (though it would have to admit that the linear function ranks even higher in importance). Played 98 times. Exponential functions follow all the rules of functions. Exponential Function Intro – Key. In other words, it is possible to have n An matrices A and B such that eA+B 6= e eB. Implicit in this definition is the fact that, no matter when you start measuring, the population will always take the same amount of time to double. It is noteworthy for its use of an exponential system for measuring nearly everything in the game. See Also: Equations, Inequalities & Functions, Exponents, Linear Equations, Quadratic. plug x=0 into given function. I have one standard for linear equations and one for exponential equations. The Atmega8 chip, which is now dated, but still supported, does not have enough memory to be able to use the math. f(x) Kahc play Skip Answers 1047 AM 4/22/2020. This algebra activity focuses on exponential functions. Translated by the Desmos localization team into: French: https://teacher. Step 3 Make a table like the one below and record the number of sheets of paper you have in the stack after one cut. alg_1_chapter_6_review_game. A Guide to Algebraic Functions Teaching Approach Functions focus on laying a solid foundation for work to come in Grade 11 and Grade 12. If the base, b=1 then the function is constant. You can also write it verbally. First Name:. This allows you to make an unlimited number of printable math worksheets to your specifications instantly. Students have fun making a poster and using multiple representations. A decreasing exponential function has a base, b<1, while an increasing exponential function has a base b>1 as you can verify on the graph above. Examples include: population growth, economic growth, growth of carbon emissions, Other References. The quantity 1 + r in the exponential growth model y = a(1 + r) t where a is the initial amount and r is the percent increase expressed as a decimal. Student Instructions Students will keep track of all information for the direct instruction in notability with color -coded examples of each part of the exponential function. So this is new. Classroom Work for Exponential Functions. Algebra Worksheets, Quizzes and Activities. Exponential growth is a pattern of data that shows greater increases with passing time, creating the curve of an exponential function. Exponential Function REVIEW - Millionaire Game. These are functions of the form: y = a b x, where x is in an exponent (not in the base as was the case for power functions) and a and b are constants. Example 1. ! Graph!each!exponential!function. (1) Exponential Function Poster Activity: This is my very first exponential function activity that I ever created. Teacher Name: undefined undefined Student and Login Information:: Select. On the same day, Checkersville has a population of 70,000 people. Logarithmic Functions Learn the definitions of logarithmic functions and their properties, and how to graph them. It is noteworthy for its use of an exponential system for measuring nearly everything in the game. Graph of Exponentials 1 - Cool Math has free online cool math lessons, cool math games and fun math activities. Compute and plot the binomial cumulative distribution function for the specified range of integer values, number of trials, and probability of success for each trial. Com provides free math worksheets for teachers, parents, students, and home schoolers. Notice that 'a' cannot be '1' ,. 4 - Solving Log Equations EVEN only (omit 28) Day 5 Exponential. 3 Integrals of trigonometric functions and applications: Online: Proofs of some trigonometric limits: Online: New functions from old: Scaled and shifted functions: Case study: Predicting airline empty seat volume: Download data. First Name:. Practice analyzing the end behavior of two functions that model similar real-world relationship, where one function is exponential and the other is polynomial. Try Remote Buzzer-Mode for even more fun!. However this is often not true for exponentials of matrices. The exponential function is one of the most important functions in mathematics (though it would have to admit that the linear function ranks even higher in importance). Properties depend on value of "a". Use the properties of exponents to interpret expressions for exponential functions. write and solve exponential and logarithmic equations. But before you take a look at the worked examples, I suggest that you review the suggested steps below first in order to have a good grasp of the general procedure. 3 - Exponential Functions Click here to review the definition of a function. A function in which an independent variable appears as an exponent. Can I convert this exponential function in QGIS with Raster Calculator ? Stack Exchange Network Stack Exchange network consists of 177 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Students can play Memory, Go Fish, Old Made or just use them as flash cards. x -2 -1 0 1 2 y -6 -6 -4 0 6. Exponential functions follow all the rules of functions. 7182818…) is the base of the natural system. js can automatically use backoff strategies to retry requests with the autoRetry parameter. ANSWER The table of values represents a quadratic function. Probably the most important of the exponential functions is y = e x , sometimes written y = exp ( x ), in which e (2. The chip level is the domain of Moore's Law, as noted. Students have fun making a poster and using multiple representations. Com provides free math worksheets for teachers, parents, students, and home schoolers. There follows an interactive graphical page showing graphically the definition of the exponential function. In this case, f(x) is called an exponential growth function. Its population increases 7% each year. Four variables - percent change, time, the amount at the beginning of the time period, and the amount at the end of the time period - play roles in exponential functions. Exponential Function • A function in the form y = ax - Where a > 0 and a ≠ 1 - Another form is: y = abx + c • In this case, a is the coefficient • To graph exponential function, make a table • Initial Value - - The value of the function when x = 0 - Also the y-intercept. Exponential functions always have some positive number other than 1 as the base. We will discuss in this lesson three of the most common applications: population growth , exponential decay , and compound interest. activities Tarsia Puzzles - With this software you will easily be able to create, print out, save and exchange customised jigsaws, domino activities and a variety of rectangular card sort activities. Writing Numbers Using Scientific Notation. But the effect is still the same. 7182818…) is the base of the natural system. In , the lifetime of a certain computer part has the exponential distribution with a mean of ten years (X ~ Exp(0. This paper analyzes the supply decision of agricultural products based on negative exponential utility function and game analysis. Exponential/Logarithm Test Review Game Jeopardy Template. Exponential functions tell the stories of explosive change. y = y 0 · m x. As the exponent is varied, the function output will change. Concept Summary Exponential Function Family You can apply the four types of transformations—stretches, compressions, reflections, and translations—to exponential functions. We hope your visit has been a productive one. Here, y represents the number of people infected and x represents the number of days that have passed since day zero. See more ideas about Exponential functions, Exponential, Algebra. Exponential Growth is a critically important aspect of Finance, Demographics, Biology, Economics, Resources, Electronics and many other areas. I could do exponential functions all year. One Grain of Rice Activity Give them a calendar page and have them fill in the first eight days with the number of grains of rice they would receive if they had made the deal described in the story. In exponential decay, the total value decreases but the proportion that leaves remains constant over time. It's the exclusive-or (XOR) operator (see here). This discussion will focus on solving more complex problems involving exponential functions. However, exponential functions and logarithm functions can be expressed in terms of any desired base b. If you're seeing this message, it means we're having trouble loading external resources on our website. Identify functions using differences or ratios EXAMPLE 2 Use differences or ratios to tell whether the table of values represents a linear function, an exponential function, or a quadratic function. 3 Integrals of trigonometric functions and applications: Online: Proofs of some trigonometric limits: Online: New functions from old: Scaled and shifted functions: Case study: Predicting airline empty seat volume: Download data. 4 7 2 6 ≈ 2. To determine how the given function grows over an interval of length 3, determine the value of f at each endpoint of that interval. it Game PIN: 9834669 Type here to search a hyperbolic parabaloid a discrete function. Simplify the expression:log (base 2) 8 , Write the equation log (base 3) 729 = 6 in exponential form. c ccsc welc I sout I o Digit Mail MYF Knov Play Skip Answers 1046 AM 4/22/2020 A exponential growth exponential decay kahoot. 2 Create and graph equations in two variables to represent linear, exponential, and quadratic relationships between quantities. LG 1 – Exponential Functions – Blank Copy. Exponential decay is a type of exponential function where instead of having a variable in the base of the function, it is in the exponent. One common example is population growth. Graphing transformations of exponential functions. 2 Use function notation, evaluate functions for inputs in their domains, and interpret statements that use function notation in terms of a context. To solve an exponential or logarithmic word problems, convert the narrative to an equation and solve the equation. Exponential & Logarithm Review. So you really draw the function $\frac{1}{25} e^x$ (using an equi-scaled graph) which leads to the maximal curvature in $\frac{-\ln 2}{2}+\ln 25\simeq 2. 2) to the 𝘵/10 power, and classify them as representing exponential growth or decay. The negative exponential utility function has been widely used in the study of the risk preferences of farmers. An exponential function is a Mathematical function in form f (x) = a x , where "x" is a variable and "a" is a constant which is called the base of the function and it should be greater than 0. Young mathematicians can work through each of the eight worksheets by evaluating functions, applying logarithms, completing logarithmic functions, and building inverse functions. Concept Summary Exponential Function Family You can apply the four types of transformations—stretches, compressions, reflections, and translations—to exponential functions. An exponential function is defined for every real number x. If you're behind a web filter, please make sure that the domains *. A half-exponential function is one which when composed with itself gives an exponential function. In fact, we offer an entire algebra 2 curriculum: fourteen units covering all topics equations, to conic sections, and even trig. The inverse of the exponential function y = a x is x = a y. Logical Functions / 10 M-Files / 11 Timing /11 Mathematical Functions Exponential and Logarithmic Functions / 12 Trigonometric Functions / 12 Hyperbolic Functions / 12 Complex Functions / 13 Statistical Functions / 13 Random Number Functions / 13 Numeric Functions / 13 String Functions / 13 Numerical Methods Polynomial and Regression Functions / 14. Properties of Exponents 4. Explore the graph of an exponential function. 02) to the 𝘵 power, 𝘺 = (0. Find the coordinates of the. Explore Exponential Functions 1. Chapter 8 : Exponents and Exponential Functions 8. exponential definition: The definition of exponential refers to a large number in smaller terms, or something that is increasing at a faster and faster rate. If you need to contact the Course-Notes. You don't write a function for this (unless you're insane, of course). ˘ Inverse Properties of Exponents and Logarithms Base a Natural Base e 1. in 1996, about 3650 participated in the Summer Olympics in Atlanta. Make sure you play the html version (not java). Shirly Boots for the original and inspiration. Exponential Excel function in excel is also known as the EXP function in excel which is used to calculate the exponent raised to the power of any number we provide, in this function the exponent is constant and is also known as the base of the natural algorithm, this is an inbuilt function in excel. Concept Summary Exponential Function Family You can apply the four types of transformations—stretches, compressions, reflections, and translations—to exponential functions. Student Instructions Students will keep track of all information for the direct instruction in notability with color -coded examples of each part of the exponential function. This game requires students to represent the exponential parent function y=2^x numerically, algebraically, graphically and verbally. Which statement is true about the functions? The exponential function is generally growing slower than the linear function. About this webmix : IXL-Evaluate an exponential Graphs of Exponential Function Comparing growth rates. True Note that the t values are equally spaced and the ratios 3 0 0 1 4 7 = 1 4 7 7 2. As functions of a real variable, exponential functions are uniquely characterized by the fact that the growth rate of such a function (that is, its derivative) is directly proportional to. For fx( ) 2x (i) The base of is (ii) The exponent of is (iii) What is varying in the function ? (iv) What is constant in the function ? 2. Characteristics of Graphs of Exponential Functions Before we begin graphing, it is helpful to review the behavior of exponential growth. For : What are the possible inputs i. Exponential form of a complex number. Laws of Exponents. Make sure you play the html version (not java). Links to the data sets are included in the file. like exponential functions AND parametric differentiation and the trapezium rule, and volumes of solids, and vectors, and the Quotient rule, and completing the square oh my! such fun ahh happy days! for exponential bunnies, you could certainly propose a differential equation showing that after time T the population P is assumed to be. Note that the curve passes through (0, 1) (on the y-axis). In both equations the variable a is called the initial amount and b is called the growth constant. William Eaton from Denver School Science & Tech. 5: Introduction to Exponential Functions Related Instructional Videos Distinguish between linear and exponential functions using tables An updated version of this instructional video is available. If you're seeing this message, it means we're having trouble loading external resources on our website. Mortgage Problems 3. Moving all the terms to the left and factoring, x 2 + 2x - 8 = (x + 4)(x - 2) = 0. Compare/contrast Exponential and linear functions using a Google doc that all students will have access to and can contribute to. Examples : exp(0), returns 1. To form an exponential function, we let the independent variable be the exponent. The Organic Chemistry Tutor 306,614 views 10:13. ˘ Inverse Properties of Exponents and Logarithms Base a Natural Base e 1. Syntax : exp(x), where x is a number. Exponential functions. 97) to the 𝘵 power, 𝘺 = (1. This calculus video tutorial explains how to find the derivative of exponential functions using a simple formula. Graph of Exponentials 1 - Cool Math has free online cool math lessons, cool math games and fun math activities. This repeated multiplication can be expressed using exponential functions. Chapter 8 : Exponents and Exponential Functions 8. Three lifelines, no phone-a-friend—can you win? Lake Tahoe Community College: Exponential Equations. If the base, b=1 then the function is constant. They write and interpret an equation of the best-fit curve. One common example is population growth. Exponential Functions. Determinewhether!each!tableor!rulerepresents!an!exponential!function. Thus we define an exponential function to be any function of the form. The decision to bankroll such a game would be a major one, even for a risk-loving entity (if one was ever foolish enough to be attracted by the tiny fees ordinary players are willing to risk). - A 'positive constant base raised to a variable exponent' = (constant) (variable) is an 'Exponential function'. Write an exponential function for the table below. Here is the entire project for FREE if you are interested. It becomes y = (2 4) x + 2 = 2 4 x + 8. Unit 4 - Exponential Functions. This online quiz includes the following types of question:1) evaluating powers with rational exponents;2) simplifying expressions containing exponents;3) creating algebraic and graphical representations of exponential functions;4) solving problems involving exponential functions. Investors know the importance of an exponential function, since compound interest can be described by one. alg_1_chapter_6_review_game. For example, the amount of time (beginning now) until an earthquake occurs has an exponential distribution. Derivative exponential : To differentiate function exponential online, it is possible to use the derivative calculator which allows the calculation of the derivative of the exponential function. An exponential function is defined for every real number x. Integrals with "Sin" Integrals with "Cos" Integrals with "Sin" and "Cos" Integrals with "Tan" Integrals with "Cot" Integrals with "Sec" Integrals with "Csc" Integrals with Inverse Trigonometric Functions. exponential function and the suitable function, which represents the constraint) by a positiv e parameter c , called the penalty parameter. Let's look at examples of these exponential functions at work. By definition:. Integrating the exponential function, also part of calculus. Then write a function to model the data. As usual, the default data used are USDJPY candles with a 15-minute compression. Cumulative Distribution Function Calculator - Exponential Distribution - Define the Exponential random variable by setting the rate λ>0 in the field below. This discussion will focus on solving more complex problems involving exponential functions. 3 - Exponential Functions Click here to review the definition of a function. On a chart, this curve starts out very slowly, remaining. This algebra activity focuses on exponential functions. Exponential Functions & Relations. There are so many interesting exponential function situations!. Other examples of exponential functions include: $$y=3^x$$$$f(x)=4. By this symbol we mean the cube root of a. Concept Summary Exponential Function Family You can apply the four types of transformations—stretches, compressions, reflections, and translations—to exponential functions. We have a rule to change. Graph of Exponentials 1 - Cool Math has free online cool math lessons, cool math games and fun math activities. This game requires students to represent the exponential parent function y=2^x numerically, algebraically, graphically and verbally. 25 hours, there are 40 000 bacteria present. a) Multiply b) Add c) Subtract$500 Answer from Exponential Graphs and Equations :. False Try again, although the ratios are fixed the t values are not evenly spaced. Play this game to review Algebra I. Mortgage Problems 3. Find the Range of function f defined by Solution to Example 1. & & EXPONENTIAL GROWTH FUNCTIONS WILL ALWAYS ULTIMATELY GROW FASTER THAN LINEAR FUNCTIONS!! ) Example:) LinearLarry&beginswith\$10,000inhissavingsaccount. The number in a power that represents the number of times the base is used as a factor. 1 Know and apply the properties of integer exponents to generate equivalent numerical expressions. This chapter introduces the concepts, objects and functions used to work with and perform calculations using numbers and dates in JavaScript. Exponential Growth is a critically important aspect of Finance, Demographics, Biology, Economics, Resources, Electronics and many other areas. Exponential functions. Exponential functions arise in many applications. Exponential Function Intro – Key. The decision to bankroll such a game would be a major one, even for a risk-loving entity (if one was ever foolish enough to be attracted by the tiny fees ordinary players are willing to risk). | 2020-11-25T01:35:25 | {
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https://math.stackexchange.com/questions/2454137/convergence-of-the-series-frac1nn | # Convergence of the series $\frac1{n^n}$
Does the series $\frac1{n^n}$ converges or diverges .
By comparison test I can claim that $\frac1{n^n} < \frac1n$ , and since series $\frac1n$ appears divergent , so the series $\frac1{n^n}$ also diverges. However by Cauchy root test the limit of $\frac1{n^n}$ appears to be $0<1$ which suggests the convergence of the series
• A series of nonnegative terms with all terms greater than or equal to a divergent series diverges. To dramatize your error, consider the series with all terms equal to zero. By your logic, since $0 < 1/n$ for all $n$, it would diverge. – quasi Oct 2 '17 at 9:46
• Your argument would be valid if $1/n^n \color{red}{>} 1/n$. But it is not. – M. Winter Oct 2 '17 at 9:50
• Series $\dfrac{1}{n^n}$ converges so quickly that just adding $5$ terms gives the result with $4$ exact decimals. Adding $10$ terms gives $1.29128\,599706$ which has eleven exact decimals – Raffaele Oct 2 '17 at 10:06
• @Raffaele: quite right, anyway shouldn't leave the OP the impression that this is a proof. What about $1/n^{n(20-n)}$ ? – Yves Daoust Oct 2 '17 at 10:14
We cannot conclude that since $\frac{1}{n^n} < \frac{1}{n}$ then $\sum_{n=1}^\infty \frac{1}{n^n}$ diverges.
If $\frac{1}{n^n} > \frac{1}{n}$ (which is not true), then we can conclude that $\sum_{n=1}^\infty \frac{1}{n^n}$ diverges.
You have used Cauchy root test to conclude that it converges.
If we want to use comparison test, notice that $$\frac{1}{n^n} \leq\frac{1}{n^2}$$
and since $\sum_{n=1}^\infty \frac1{n^2}$ conveges, hence $\sum_{n=1}^\infty \frac1{n^n}$ converges.
• Then is it that this given series converges by Cauchy root test ? – Nabendra Oct 2 '17 at 9:50
• yes, you have proven it using the root test. – Siong Thye Goh Oct 2 '17 at 9:51
• Now suppose if I use Ratio Test then, let u(n)=1/n^n and so u(n+1)=1/(n+1)^(n+1). Hence u(n)/u(n+1)=(1+1/n)^n * (n+1) and thus taking limit n tends to infinity on the above I get infinity(not sure whether calculations are correct) , which stands as greater than 1. So can I claim , the series in convergent by Ratio Test. – Nabendra Oct 2 '17 at 9:56
• Usually for ratio test, we compute $\left| \frac{u(n+1)}{u(n)}\right|$ and show that it is less than $1$. – Siong Thye Goh Oct 2 '17 at 10:28
The comparison test tells you that if $$a_n> b_n$$ and $$\sum_{n}a_n$$ converges, then $$\sum_n b_n$$ also converges.
It does not say that if $\sum_{n} a_n$ diverges then $\sum_{n} b_n$ diverges. If it did, then because $\frac{1}{2^n} < 1$, you could conclude that $\sum \frac1{2^n}$ also diverges.
In fact, you would need the inequality reversed, so if $a_n<b_n$ and $\sum a_n$ diverges, then $\sum b_n$ also diverges.
In fact, the series $\sum\frac{1}{n^n}$ converges by the comparison test with $\frac{1}{2^n}$ or with $\frac{1}{n^2}$.
• Does it appear convergent or divergent ? How gonna I check it ? – Nabendra Oct 2 '17 at 9:57
• @Nabendra I really really dislike users that don't read my answers before they ask questions about them. – 5xum Oct 2 '17 at 9:57
• I didn't understood your answer , as if I use 1/n the it diverges by comparison test instead of 1/n^2 or 1/2^n – Nabendra Oct 2 '17 at 10:04
• @Nabendra You didn't understand my answer because you didn't bother reading it to the end. So I don't see why I should bother helping you any further. – 5xum Oct 2 '17 at 10:04
• I do apologise, for my deed – Nabendra Oct 2 '17 at 10:07
A series converges if it has a converging upper bound and it diverges if it has a diverging lower bound. In other cases, you cannot conclude.
In this particular case, you can for instance exploit
$$\frac1{n^n}\le \frac1{2^n}$$ for $n\ge2$. Actually, the sequence converges extremely quickly, super-exponentially. | 2019-09-16T20:26:30 | {
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http://mathhelpforum.com/trigonometry/170333-trigonometry-question-print.html | Trigonometry question.
• February 6th 2011, 06:43 AM
Googl
Trigonometry question.
Hi all,
I am new here by the way, I have a small problem for a question I can complete in Compound Angle identities
Prove that:
sinθ = tan (θ/2)(1 + cosθ)
Look forward to the answer. I can almost work it work it out but I get stuck somewhere near the end.
• February 6th 2011, 07:02 AM
Plato
Hello and welcome to MathHelpForum.
You should understand that this is not a homework service nor is it a tutorial service. Please either post some of your own work on this problem or explain what you do not understand about the question.
• February 6th 2011, 09:10 AM
Googl
Quote:
Originally Posted by Plato
Hello and welcome to MathHelpForum.
You should understand that this is not a homework service nor is it a tutorial service. Please either post some of your own work on this problem or explain what you do not understand about the question.
I understand that. This is not homework, it's for my own understanding of Trig (revision). Okay how I did it...
sinθ = tan (θ/2)(1 + cosθ)
tan (θ/2) (1 + cos (θ/2 + θ/2))
tan (θ/2)(cos^2(θ/2) - sin^2(θ/2))
tan (θ/2)(cos^2(θ/2) + sin^2(θ/2) + cos^2(θ/2) - sin^2(θ/2))
Attachment 20699
• February 6th 2011, 09:28 AM
e^(i*pi)
Let $A = \dfrac{\theta}{2}$ (purely to save me effort)
From line 2: $\tan(A)(1+\cos(2A)) = \dfrac{\sin(A)}{\cos(A)} \cdot 2\cos^2A = 2 \sin A \cos A = 2 \sin\left(\dfrac{\theta}{2}\right) \cos\left(\dfrac{\theta}{2}\right)$
There is a common identity you can use now to get the LHS
• February 6th 2011, 09:55 AM
Googl
Quote:
Originally Posted by e^(i*pi)
Let $A = \dfrac{\theta}{2}$ (purely to save me effort)
From line 2: $\tan(A)(1+\cos(2A)) = \dfrac{\sin(A)}{\cos(A)} \cdot 2\cos^2A = 2 \sin A \cos A = 2 \sin\left(\dfrac{\theta}{2}\right) \cos\left(\dfrac{\theta}{2}\right)$
There is a common identity you can use now to get the LHS
Hello e^(i*pi)
I actually did it, and got to the same stage and over it but got lost because I had no confidence. This is where I gave up because I thought it would never end up as sinθ.
Attachment 20700
So this is correct?
Attachment 20701
Thanks a lot. I have another similar question. I will try to work it out first see whether I get then I will submit it.
• February 6th 2011, 10:06 AM
Soroban
Hello, Googl!
Quote:
$\text{Prove that: }\:\sin\theta \:=\:\tan \frac{\theta}{2}(1 + \cos\theta)$
Identities: . $\begin{array}{ccccccc}\cos^2\!\frac{\theta}{2} &=& \dfrac{1+\cos\theta}{2} &\Rightarrow& 1 + \cos\theta &=& 2\cos^2\!\frac{\theta}{2} \\ \\[-3mm]
\sin\theta &=& 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} \end{array}$
$\text{On the right side we have:}$
. . $\tan\frac{\theta}{2}(1 + \cos\theta)\;=\;\dfrac{\sin\frac{\theta}{2}}{\cos\ frac{\theta}{2}}(2\cos^2\!\frac{\theta}{2}) \;=\;2\sin\frac{\theta}{2}\cos\frac{\theta}{2} \;=\;\sin\theta$
• February 6th 2011, 11:20 AM
Quote:
Originally Posted by Googl
Hello e^(i*pi)
I actually did it, and got to the same stage and over it but got lost because I had no confidence. This is where I gave up because I thought it would never end up as sinθ.
Attachment 20700
So this is correct?
Attachment 20701
Thanks a lot. I have another similar question. I will try to work it out first see whether I get then I will submit it.
Yes, that's good when starting at the right.
To start at the left and end up at the right...
$\displaystyle\ sin\theta=sin\left(\frac{\theta}{2}+\frac{\theta}{ 2}\right)=2sin\frac{\theta}{2}cos\frac{\theta}{2}$
using $sin2A=2sinAcosA$
which gets us to the half angle.
To get tan, we need cos under the sine, so multiply by
$\displaystyle\ 1=\frac{cos\frac{\theta}{2}}{cos\frac{\theta}{2}}$
$\displaystyle\ 2sin\frac{\theta}{2}cos\frac{\theta}{2}\left[\frac{cos\frac{\theta}{2}}{cos\frac{\theta}{2}}\ri ght]=\frac{sin\frac{\theta}{2}}{cos\frac{\theta}{2}}\l eft[2cos^2\frac{\theta}{2}\right]$
Now use
$cos^2A=\frac{1}{2}(1+cos2A)\Rightarrow\ 2cos^2A=1+cos2A$
to get
$\displaystyle\ sin\theta=tan\frac{\theta}{2}\left[1+cos\left(\frac{\theta}{2}+\frac{\theta}{2}\right )\right]$ | 2014-12-18T10:05:12 | {
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https://web2.0calc.com/questions/is-there-supposed-to-be-a-formula-or-strategy-for-this | +0
# Is there supposed to be a Formula or strategy for this?
+4
167
8
+2557
Ron and Martin are playing a game with a bowl containing 39 marbles. Each player takes turns removing 1, 2, 3 or 4 marbles from the bowl. The person who removes the last marble loses. If Ron takes the first turn to start the game, how many marbles should he remove to guarantee he is the winner?
After thinking about this, I am not sure how to do it.
Because there are 39 marbles, Ron needs to pick a number of marbles in which he can stick to a pattern to guaruntee a win. But how....
Also confirming if I am a robot is so annoying... I know its too prevent advertisement attacks, but like once you confirm it once on a registered account, it should stop asking.
Sep 22, 2019
edited by CalculatorUser Sep 22, 2019
### 8+0 Answers
#1
+106993
+1
No matter how many marbles he takes with his first choice Ron can still lose if he is not playing strategically.
I guess we are suppose to assume that Ron is playing the ultimate game and so it Martin.
I have got nothing more.
Sep 22, 2019
edited by Melody Sep 22, 2019
#2
+2557
0
Yes agreed!! thats why Im confused... The answer was apparently 3 but I don't know why.
CalculatorUser Sep 22, 2019
#3
+23866
+3
Ron and Martin are playing a game with a bowl containing 39 marbles.
Each player takes turns removing 1, 2, 3 or 4 marbles from the bowl.
The person who removes the last marble loses.
If Ron takes the first turn to start the game,
how many marbles should he remove to guarantee he is the winner?
Ron first takes 3 marbles.
Than, if Martin takes x marbles, Ron takes 5-x marbles, this repeats 7 times.
The last marble is then for Martin.
39 = 3 + 7*5 + 1
Sep 22, 2019
edited by heureka Sep 22, 2019
#4
+2557
+1
Why 5-x marbles? Is it because that guaruntees martin will choose 4, 3, 2, or 1 marbles?
CalculatorUser Sep 22, 2019
#5
+23866
+2
Why 5-x marbles? Is it because that guaruntees martin will choose 4, 3, 2, or 1 marbles?
Hello CalculatorUser
No, it is because that guaruntees that after Martin and Ron the sum of removed marbles is 5.
or in other words Ron supplements each time to 5.
Here is the course
$$\begin{array}{|r|r|l|} \hline \text{Martin} & \text{Ron} & \text{sum} \\ \hline & 3 \\ \{4,3,2,1\} & \{1,2,3,4\} & 4+1 = 5,\ 3+2=5,\ 2+3=5,\ 1+4 = 5 \\ \{4,3,2,1\} & \{1,2,3,4\} & 4+1 = 5,\ 3+2=5,\ 2+3=5,\ 1+4 = 5 \\ \{4,3,2,1\} & \{1,2,3,4\} & 4+1 = 5,\ 3+2=5,\ 2+3=5,\ 1+4 = 5 \\ \{4,3,2,1\} & \{1,2,3,4\} & 4+1 = 5,\ 3+2=5,\ 2+3=5,\ 1+4 = 5 \\ \{4,3,2,1\} & \{1,2,3,4\} & 4+1 = 5,\ 3+2=5,\ 2+3=5,\ 1+4 = 5 \\ \{4,3,2,1\} & \{1,2,3,4\} & 4+1 = 5,\ 3+2=5,\ 2+3=5,\ 1+4 = 5 \\ \{4,3,2,1\} & \{1,2,3,4\} & 4+1 = 5,\ 3+2=5,\ 2+3=5,\ 1+4 = 5 \\ 1 \\ \hline \end{array}$$
heureka Sep 22, 2019
#7
+106993
+1
That is really cool. Thanks Heureka.
Melody Sep 22, 2019
#8
+23866
+2
Thank you, Melody !
heureka Sep 22, 2019
#6
+19913
0
Eventually....at some pont, the captcha robot thingie quits happening....I do not know what the tripping point is.... ~ EP
Sep 22, 2019 | 2020-01-23T08:22:31 | {
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https://math.stackexchange.com/questions/3583836/computation-of-balance-equation-example-in-markov-model | # Computation of balance equation example in Markov model
I am studying some examples of balance equations for Markov models. I am presented with the following example:
$$\mathcal{P} = \begin{bmatrix} 0.2 & 0.3 & 0.5 \\ 0.1 & 0 & 0.9 \\ 0.55 & 0 & 0.45 \end{bmatrix}$$
[dropping the $$i$$ subscript by writing $$\pi_j$$ for $$\pi_{ij}.]$$
The balance equations are
\begin{align} &\pi_1 = 0.2 \pi_1 + 0.1 \pi_2 + 0.55 \pi_3 \tag{a} \\ &\pi_2 = 0.3 \pi_1 \tag{b} \\ &\pi_3 = 0.5 \pi_1 + 0.9 \pi_2 + 0.45 \pi_3 \tag{c} \end{align}
Since, also, $$\pi_1 + \pi_2 + \pi_3 = 1$$, the unique solution is
$$\pi_1 = 1/2.7 = 0.37037, \ \ \ \pi_2 = 1/9 = 0.11111, \ \ \ \pi_3 = 1.4/2.7 = 0.51852$$
How do we solve this for the values $$\pi_1, \pi_2, \pi_3$$? Is there a way to solve this using matrix computations? The difficulty here, as I see it, is that we have a constraint $$\pi_1 + \pi_2 + \pi_3 = 1$$ that must hold, so I'm unsure of how this is done.
I would greatly appreciate it if someone would please take the time to show this.
$$\pi_2 = 0.3\pi_1 \Rightarrow$$
$$\pi_1 = 0.2\pi_1 + 0.1\cdot 0.3\pi_1 + 0.55\pi_3 \Leftrightarrow 0.77\pi_1 = 0.55\pi_3 \Rightarrow \pi_3 = \frac{7}{5}\pi_1$$
$$\pi_3 = 0.5\pi_1 + 0.9\cdot 0.3\pi_1 + 0.45\pi_3 \Leftrightarrow 0.55\pi_3 = 0.77\pi_1 \Rightarrow \pi_3 = \frac{7}{5}\pi_1$$
It means that we have only two equations with three unknowns, which implies that there is no unique solition to the system of linear equations. So $$\pi_1$$ can be any number.
Now, we have to use the condition $$\pi_1+\pi_2+\pi_3 = 1$$ in order to find a unique solution.
$$\pi_1+\pi_2+\pi_3 = 1 \Rightarrow \pi_1 + 0.3\pi_1 + \frac{7}{5}\pi_1 = \frac{27}{10}\pi_1 = 1 \Rightarrow$$
$$\pi_1 = \frac{10}{27} \approx 0.37037$$
$$\pi_2 = 0.3\frac{10}{27} = \frac{1}{9} \approx 0.11111$$
$$\pi_3 = \frac{7}{5}\frac{10}{27} = \frac{14}{27} \approx 0.51852$$
• You have that $0.2\pi_1 + 0.1 \cdot 0.3 \pi_1 = 0.77\pi_1$? – The Pointer Mar 17 at 7:58
• No. I have that $\pi_1 = 0.2\pi_1 + 0.1\cdot 0.3\pi_1 + 0.55\pi_3 \Leftrightarrow \pi_1 - 0.2\pi_1 - 0.1\cdot 0.3\pi_1 = 0.55\pi_3 \Leftrightarrow 0.77\pi_1 = 0.55\pi_3$ – Eugene Mar 17 at 9:39
• Thanks for the clarification. – The Pointer Mar 17 at 13:09
The standard Eigenvalue equation is $$P^Tv=\lambda v$$ You're seeking the eigenvector corresponding to $$\lambda=1$$.
Fortunately, in the case of a Markov (aka stochastic) matrix, this happens to be the largest eigenvalue and therefore can be computed via the power iteration as \eqalign{ v_{k+1} &= \frac{P^Tv_k}{\|P^Tv_k\|_{\tt1}} } or, if you prefer, the transposed equation using row vectors \eqalign{ v_{k+1}^T &= \frac{v_k^TP}{\|v_k^TP\|_{\tt1}} } If you start with a stochastic vector, then $$P$$ preserves the stochastic property and scaling each step is unnecessary. This simplifies the iteration to $$v_{k+1}^T = v_k^TP \quad\implies v_n^T = v_0^TP^n$$ For the matrix in your example, this iteration yields \eqalign{ v_{0}^T &= \;\big(&0.333333 \quad&0.333333 \quad&0.333333 \;\big) \\ v_{5}^T &= \;\big(&0.372054 \quad&0.109744 \quad&0.518201 \;\big) \\ v_{10}^T &= \;\big(&0.370358 \quad&0.111112 \quad&0.518529 \;\big) \\ v_{15}^T &= \;\big(&0.37037 \quad&0.111111 \quad&0.518518\;\big) \\ v_{20}^T &= \;\big(&0.37037 \quad&0.111111 \quad&0.518519\;\big) \\ } | 2020-06-06T11:15:31 | {
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https://web2.0calc.com/questions/osl4-25_1 | +0
# OSL4#25
+1
305
6
+935
I tried this question by drawing pictures, but it didn't work out.
Jul 27, 2019
#1
+6196
+3
$$\text{It can't be 70}\\ A_{board} =32^2 = 1024 ~sq~in\\ A_{photo}=3\cdot 5 = 15 ~sq ~in\\ \dfrac{1024}{15} = 68+\dfrac{4}{15}\\ \text{so with no overlapping 68 is the maximum possible, and is probably not attainable}$$
.
Jul 27, 2019
#2
+8966
+4
As Rom said, there is definitely no way to put 70 pictures on the board.
68 pictures is the maximum possible, and here is a way to do that:
However, if you can't rotate the pictures 90 degrees, then the maximum possible is
floor( 32 / 3 ) * floor( 32 / 5 ) = 10 * 6 = 60
Jul 27, 2019
#3
+111438
+2
Nice, hectictar !!!!
CPhill Jul 27, 2019
#4
+935
+3
I don't know guys. Out of all the questions I've looked at made by this organization, their answers are always correct. Also, this was on a math contest taken by all middle schoolers in the U.S., so if the answer is wrong, then a student has to have already told them and they would have fixed it already. I also got 68, but these questions tend to have a creative way to solving them. Still, you guys have really great solutions and I agree with your reasoning why 68 is the maximum possible.
dgfgrafgdfge111 Jul 28, 2019
edited by dgfgrafgdfge111 Jul 28, 2019
edited by dgfgrafgdfge111 Jul 28, 2019
#5
+8966
+4
Hmmm....even if you cut the pictures into 1 inch by 1 inch squares, there's just not enough square inches on the board to fit 70 pictures worth.
number of square inches on the board = 32 * 32 = 1024
number of square inches of 70 pictures = 70 * 3 * 5 = 1050
Are you sure it says 70 and not 60 ?
If so, maybe you can be the one to report the error in this question. Yes it is weird and unlikely they'd have it wrong, but it's not impossible.
hectictar Jul 28, 2019
#6
+6196
+3
As an aside if I were writing this problem I would not use photos.
No one is going to rotate photos 90 degrees to display them.
Using tiles would have made the problem far more sensible with out affecting any of the underlying math.
Rom Jul 28, 2019 | 2020-08-09T16:44:00 | {
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https://math.stackexchange.com/questions/577235/proof-of-the-following-inequality-fracx-y-log-x-log-y-sqrtxy | # Proof of the following inequality $\frac{x - y}{\log x - \log y} > \sqrt{xy}$, $x>y$.
I have seen the following inequality $$\frac{x - y}{\log x - \log y} > \sqrt{xy} \ , \quad \forall x>y$$ be stated as a near "obvious" fact in another question, on the site. The inequality is very cute, but so far I have not been able to prove it. It reminds me of the Lipschitz inequality, but has some minor differences. Also Jensens inequality comes to mind. Is there something obvious I am missing, or is this ineqality not as easy to prove as it looks?
• The inequality can be written as $\int_y^x \frac{1}{t}dt < \frac{x-y}{\sqrt{xy}}$. I wonder if one can prove it by comparing areas, and the shape of $\frac{1}{x}$. – N. S. Nov 22 '13 at 15:46
• This seems related to Logarithmic mean. – Martin Sleziak Nov 22 '13 at 15:49
• @N.S. This demonstration on WA site seems to be based on geometric idea. It was the first search result when I googled for logarithmic geometric mean inequality. – Martin Sleziak Nov 22 '13 at 16:22
• Very nice, that does it! – N3buchadnezzar Nov 22 '13 at 16:36
Since $1/x$ is concave for $x>0$ then \begin{align*} \log b - \log a = \int_b^a\frac{\mathrm{d}x}{x} < \underbrace{ \frac{1}{2} \frac{b - a}{\sqrt{ba}} }_\text{Blue Area} +\underbrace{ \frac{1}{2}\frac{b - a}{\sqrt{ba}} }_{\text{Red area}} = \frac{b - a}{\sqrt{ba}}\end{align*} Implying $$\hspace{4cm}\sqrt{ab} < \cfrac{b - a}{\log b - \log a} \hspace{4cm} \blacksquare$$
Hint: let $\dfrac{x}{y}=t>1$ and $$\Longleftrightarrow f(t)={t-1}-{\ln{t}}\cdot \sqrt{t}$$
and follow is easy to prove it
• Yes, the inequality becomes $(t-1)/\log t>\sqrt{t}$ for $t>1$. A possible simplification is to set $x/y=t^2$, instead, just to get rid of the square root. – egreg Nov 22 '13 at 15:11
• And how does one show the last equality? I proved that they both are zero as $t\to 1$. But showing that the derivative was increasing greater than zero was hard. – N3buchadnezzar Nov 22 '13 at 15:18
Writing $x=t^2y$ with $t\gt1$, the inequality to be proved can be written as
$$t-{1\over t}-2\log t\gt0$$
Letting $f(t)$ be the expression on the left, we see that $f(1)=0$ and
$$f'(t)=1+{1\over t^2}-{2\over t}={(t-1)^2\over t}\gt0 \text{ for }x\gt1$$
This means $f$ is strictly increasing, making it necessarily positive.
Note: This answer is along the lines of the approaches taken by math110 and egreg. The main difference was to write the inequality in a way that suggested a function that's easy to differentiate and show is always positive.
With the substitution $x/y=t^2$, the inequality becomes
$$\frac{t^2-1}{\log t^2}>t$$ that, for $t>1$, is equivalent to $$t^2-1>2t\log t.$$
Consider $f(t)=t^2-1-2t\log t$, defined for $t\ge1$; we have $f(1)=0$ and the derivative is $$f'(t)=2t-2-2\log t.$$ I claim that $f'(t)>0$ for $t>1$, so the function $f$ is increasing. It's easy to see that $f'(1)=0$ and $\lim_{t\to\infty}f'(t)=\infty$.
Since $$f''(t)=2-\frac{2}{t}>0$$ for $t>1$, the function $f'$ is increasing, so it's everywhere positive (except at $1$).
Of course the proof with convexity is better.
• I think you need $t^2$'s on the left hand side of the first inequality. – Barry Cipra Nov 22 '13 at 22:37
• @BarryCipra Yes, thanks. – egreg Nov 22 '13 at 22:38
Another proof based on geometry is mentioned in Frank Burk: The Geometric, Logarithmic, and Arithmetic Mean Inequality, The American Mathematical Monthly , Vol. 94, No. 6 (Jun. - Jul., 1987), pp. 527-528, jstor, link. (It was among the first hits in the google search for logarithmic geometric mean inequality.)
We know that function $e^x$ is convex. Let us have a look on this function on the interval $\ln a$, $\ln b$. The line joining the points $(\ln a,a)$ and $(\ln b,b)$ lies above the graph of this function, so we have a trapezoid which contains the whole area lying below the graph. On the other hand, if we make a tangent in the midpoint $(\ln a+\ln b)/2$, we get a trapezoid which lies below the graph of this function. (See the picture in the pdf linked above.)
So we have: $$\left(e^{\frac{\ln a+\ln b}2}\right) (\ln b-\ln a) \le \int_{\ln a}^{\ln b} e^x \le \frac{e^{\ln a}+e^{\ln b}}2 (\ln b-\ln a)\\ \sqrt{ab} \le \frac{b-a}{\ln b-\ln a} \le \frac{a+b}2$$
Inequality has many demonstrations. Some are made using logarithmic representation with integral. Give here an example: $$\frac{1}{L( a, b )} = \frac{1}{b - a}\int^{\frac{b}{a}}_{1}\frac{dx}{x}$$ Integrating inequality: $$\frac{4}{(x+1)^2}\leq\frac{1}{x}\leq\frac{x+1}{2x\sqrt{x}}$$ is obtained $$\frac{b-a}{b+a}\leq\frac{1}{2}\ln\frac{b}{a} \leq\frac{b-a}{2\sqrt{ab}}$$ hence the inequality in question, but more logarithmic mean framing between the arithmetic mean and geometric mean:$$\sqrt{ab} < \frac{b-a}{\ln b-\ln a}<\frac{a+b}{2}.$$ $(0<a<b)$ | 2019-08-18T07:53:54 | {
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https://math.stackexchange.com/questions/3832740/what-can-be-a-generalization-of-repeats-in-exponentiation-using-modulo | # What can be a generalization of repeats in exponentiation using modulo?
I came across a Math Problem in a Japanese Coding Test(It is officially over now so no worries about discussing it, https://atcoder.jp/contests/abc179/tasks/abc179_e).
I will write the mathematical version of this problem.
Let $$A$$ be a sequence that is defined by the initial values $$A_1=x$$ and this recurrence relation is given $$A_{n+1}$$=$$(A_{n}^2)$$ $$mod$$ $$M$$ where $$M$$ can be any natural number.
Find $$\sum_{i=1}^{i=N}A_{i}$$
I will tell what I have deduced till now:
1. If I write this recurrence in equation it demands us to find $$(x^1 mod M + x^2 mod M + x^4 mod M + x^8 mod M + x^{16}mod M ..$$ till $$n$$ terms).
2. If We take any example for $$x=2$$ and $$M=1001$$ the values of this series come out to be like this $$2,4,16,256,471,620,16,256,471,620....$$ and this block of $$16,256,471$$ repeats.
3. I observed that for any given $$x$$ and $$M$$ the series formed will come at a point where one of it's window will start repeating, just like in the above case this window of $$16,256,471$$ repeated after certain point. All because of Modulo Magic I have observed that it will repeat but I don't have any proof of How and Why ?
4. I tried using Fermat's Little theorem that for the case of when $$M$$ is prime maybe of some use But didn't find an apt conclusion to it.
Now I am stuck at this problem that How will Modulo work in such kind of series and how Will the values of this series depend on different versions of $$x$$ and $$M$$ like them being co prime to each other or otherwise. and if this series is to give recurring values after a certain point then Why and How and also as it happened in the example case I have given All the values do not repeat due to this kind of exponentiation but only a window repeats,I don't understand why.
• Since there are only finitely many values available modulo $M$ there must eventually be a value that has already come up previously. Since each value depends only on the previous value, once a value comes up a second time the sequence must repeat what it did the first time that value came up, periodically, forever. – Gerry Myerson Sep 20 at 0:23
• Yes I agree to that But why is it that only a certain value will repeat ? Why for the example I am taking it repeated after 16 not with 2. – Kartik Bhatia Sep 20 at 8:20
First, consider the case where $$x$$ and $$M$$ are coprime, i.e., $$\gcd(x,M) = 1$$. Since for all $$i \gt 1$$ we have $$0 \le A_i \lt M$$, there are only a finite # of values it can have so the sequence will eventually have to start repeating. Let $$j$$ and $$k$$, where $$j \lt k$$, be the first indices where the values repeat. Since $$x$$ and $$M$$ are coprime, $$x$$ has a multiplicative inverse. Using this, we thus have
\begin{aligned} x^{2^{k-1}} & \equiv x^{2^{j-1}} \pmod{M} \\ x^{2^{k-1}} - x^{2^{j-1}} & \equiv 0 \pmod{M} \\ x^{2^{j-1}}\left(x^{2^{k-1} - 2^{j-1}} - 1\right) & \equiv 0 \pmod{M} \\ x^{2^{k-1} - 2^{j-1}} - 1 & \equiv 0 \pmod{M} \\ x^{2^{j-1}\left(2^{k-j} - 1\right)} & \equiv 1 \pmod{M} \end{aligned}\tag{1}\label{eq1A}
The multiplicative order of $$x$$ modulo $$M$$, i.e.,
$$m_1 = \operatorname{ord}_{M}(x) \tag{2}\label{eq2A}$$
must divide $$2^{j-1}\left(2^{k-j} - 1\right)$$. Let $$a$$ be the largest power of $$2$$ which divides $$m_1$$, so we have
$$m_1 = 2^{a}b, \; \gcd(b, 2) = 1 \tag{3}\label{eq3A}$$
The smallest value of $$j$$ which works is where $$j - 1 = a \implies j = a + 1$$, except where $$a = 0$$ and $$x \ge M$$, in which case we get $$j = 2$$ instead. This is the main reason why not all of the initial values repeat (i.e., where $$a \gt 0$$) but, instead, just a "window" starting at this minimum $$j$$ value.
Next, if $$b = 1$$, the smallest value of $$k - j$$ is $$1$$, else for $$b \gt 1$$, it's $$m_2$$ where
$$m_2 = \operatorname{ord}_{b}(2) \implies 2^{m_2} = kb + 1, \; k \in \mathbb{N} \tag{4}\label{eq4A}$$
With your example of $$x = 2$$ and $$M = 1001$$, the values start repeating with $$j = 3$$ and $$k = 7$$ giving $$2^{j-1}\left(2^{k-j} - 1\right) = 4(15) = 60$$. As you can confirm, in this case, $$m_1 = 60$$, although they will not in general be equal (since equality only occurs with $$k = 1$$ in \eqref{eq4A}).
Next, consider the somewhat more complicated case where $$x$$ and $$M$$ are not coprime. Let
$$q = \prod_{i=1}^{n}p_i \tag{5}\label{eq5A}$$
be the product of all of the $$n$$ primes $$p_i$$ which are factors of both $$x$$ and $$M$$. Splitting $$x$$ and $$M$$ into factors which aren't and are coprime with $$q$$ gives
$$x_1 = \prod_{i=1}^{n}p_i^{s_i}, \; x = x_1x_2, \; \gcd(x_2, q) = 1 \tag{6}\label{eq6A}$$
$$M_1 = \prod_{i=1}^{n}p_i^{t_i}, \; M = M_1M_2, \; \gcd(M_2, q) = 1 \tag{7}\label{eq7A}$$
Also, note $$\gcd(x_2, M_2) = 1$$ since they don't have any prime factors in common.
As before, let $$j \lt k$$ be the first indices which repeat. We split the congruence equation to that with $$M_1$$ and with $$M_2$$. This first gives
\begin{aligned} (x_1x_2)^{2^{k-1}} & \equiv (x_1x_2)^{2^{j-1}} \pmod{M_1} \\ (x_1x_2)^{2^{j-1}}\left((x_1x_2)^{2^{k - 1} - 2^{j-1}} - 1\right) & \equiv 0 \pmod{M_1} \end{aligned}\tag{8}\label{eq8A}
Since no $$p_i$$ in $$q$$ from \eqref{eq4A} divides $$(x_1x_2)^{2^{k - 1} - 2^{j-1}} - 1$$, this means all factors of $$p_i$$ are in $$(x_1x_2)^{2^{j-1}}$$. In particular, the smallest possible $$j$$ requires, using \eqref{eq6A} and \eqref{eq7A}, that
$$2^{j-1}(s_i) \ge t_i, \; \forall \, 1 \le i \le n \tag{9}\label{eq9A}$$
Next, since $$\gcd(x, M_2) = 1$$, we have the same situation as at the start of this solution, with $$M$$ replaced by $$M_2$$, i.e., we get basically the equivalent of \eqref{eq1A} giving
$$x^{2^{k-1}} \equiv x^{2^{j-1}} \pmod{M_2} \implies x^{2^{j-1}\left(2^{k-j} - 1\right)} \equiv 1 \pmod{M_2} \tag{10}\label{eq10A}$$
We thus proceed as we did before, but with the added restriction now that $$j$$ must be at least as large as what's required by \eqref{eq9A}.
• It is taking me some time understanding it ! I didn't know about multiplicative order , I am clear with the case of them being co prime how are you dealing with other case Why does gcd come into play ? – Kartik Bhatia Sep 20 at 8:19
• The first paraghraph is not clear: multiply both sides of what, and what product? – Bill Dubuque Sep 20 at 8:22
• @BillDubuque can you help me in this question ? – Kartik Bhatia Sep 20 at 12:24
• @BillDubuque Thanks for the feedback. You're right it wasn't particularly clear. I've changed it by removing describing what I was going to do in my $(1)$ and, instead, just did it there. I believe this is now easier to understand. – John Omielan Sep 20 at 14:07
• @KartikBhatia As I commented to Bill, I've changed my first part dealing with my $(1)$. As for why the gcd (greatest common factor) comes into play, it's mainly due to the simpler handling in my $(1)$ can only occur when $x$ & $M$ are coprime. This is since a multiplicative inverse, i.e., a $y$ where $xy \equiv 1 \pmod{M}$, only exists for those cases. As I show in my $(8)$ & $(9)$, non coprime values have an added requirement of common prime factor powers in $x^{2^{j-1}}$ needing to be at least as large as in $M$. Please let me know if there's anything specifically I can explain further. – John Omielan Sep 20 at 14:19 | 2020-10-28T14:26:38 | {
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https://martamlodzikowska.pl/2006/benito/honda/104395359f2992b8567-recursion-tree-method-for-solving-recurrences | Recursion Trees Show successive expansions of recurrences using trees. Iteration method Recursion-tree method (Master method) Iteration is constructive, i.e. 11. I Ching [The Book of Changes] (c. 1100 BC) To endure the idea of the recurrence one needs: freedom from morality; new means against Now we use induction to prove our guess. two steps: 1 Guess the form of the solution. you gure out the result; somewhat tedious because of T(n) = aT(n/b)+(n), where a 1 and b > 1 are constants and (n) is an asymptotically positive function. Like Masters Theorem, Recursion Tree is another method for solving the recurrence relations. 1) Substitution Method: We make a guess for the solution and then we use mathematical induction to prove the guess is correct or incorrect. Now push the current node in the inorder array and make the recursive call for the right child (right subtree). For example consider the recurrence T (n) = 2T (n/2) + n We guess the solution as T (n) = O (nLogn). Solving using a recursion tree. Arithmetic Sequences and Series by MATHguide Arithmetic Sequences and Series Learn about Arithmetic Sequences and Series By decreasing the size of h, the function can be approximated accurately The runtime is so much higher because the recursive function fib[n_]:=fib[n-1]+fib[n-2] generates n^2 Inorder Tree Traversal without Recursion; Inorder Tree Traversal without recursion and without stack! Etsi tit, jotka liittyvt hakusanaan Recursion tree method for solving recurrences examples tai palkkaa maailman suurimmalta makkinapaikalta, jossa on yli 21 miljoonaa tyt. 9 The recursion-tree method Convert the recurrence into a tree: Each node represents the cost incurred at various levels of recursion Sum up the costs of all levels Used to guess a solution for the recurrence T(n) = 3T(n/3) + n for n > 1. Note: We would usually use a recursion tree to generate possible guesses for the runtime, and then use the substitution method to prove them. Solving Recurrences; Amortized Analysis; What does 'Space Complexity' mean ? Engineering; Computer Science; Computer Science questions and answers; 2. 1. Explanation: Masters theorem is a direct method for solving recurrences. SOLVING RECURRENCES 1.2 The Tree Method The cost analysis of our algorihms usually comes down to nding a closed form for a recurrence. There are mainly three ways for solving recurrences. The master method provides a "cookbook" method for solving recurrences of the form. If yes, solve with that method, if no, explain why. With substitution you need to guess the result and prove it by induction. Solving recurrence relation. form and show that the solution works. It's very easy to understand and you don't need to be a 10X developer to do so. Recursion Tree Implement recursion tree method to. PURRS is a C++ library for the (possibly approximate) solution of recurrence relations (5 marks) Example 1: Setting up a recurrence relation for running time analysis Note that this satis es the A general mixed-integer programming solver, consisting of a number of different algorithms, is used to determine the optimal decision vector A general The recurrence relation is given as: an = 4an-1 - 4an-2 The initial conditions are given as 20 = 1, 2, = 4 and 22 = 12,-- Se When you solve the general equation, the constants a Arithmetic Sequences and Series by MATHguide Arithmetic Sequences and Series Learn about Arithmetic Sequences and Series , Chichester, 1989 The good people at Desmos have made an excellent online graphing calculator even better Enter a value for nMin In the example given in the previous chapter, T (1) T ( 1) was the time taken in the initial condition. 4.3 The master method. Minimum Spanning Tree. We use techniques for summations to solve the recurrence. Method 2: Push directly root node two times while traversing to the left. For the following recurrences, use the recursion tree method to find a good guess of what they solve to asymptotically (i.e. Search: Recursive Sequence Calculator Wolfram. Yes, you can solve almost every problem using recursion. Just look out how Functional Programmers tackles every problem with Haskell, OCaml, Erlang etc. Why not? recursion trees. You will get a recurrence in m that is one of theprevious examples. solving recurrences can use data in these subproblems are no general way of numbers. T(n) = 4T(n/2) + n^3 for n > 1. Some methods used for computing asymptotic bounds are the master theorem and the AkraBazzi method. 3 Solving recurrences Methods for solving recurrences: 1. b. This tree is a way of representing the algorithms iteration in the shape of a tree, with each node representing the trees iteration level. I will also accept this method as proof for the given bound (if done correctly). 2 Use mathematical induction to nd constants in the. (a) (4 marks | Solve the following recurrences by the recursion-tree method (you may assume that n is a power of 3): 4, n= T (n) - { r +-2, 51 = n>. For example consider the recurrence T (n) = 2T (n/2) + n We guess the solution as T (n) = O (nLogn). The first recurrence relation was. we guess a bound and then use mathematical induction to prove our guess correct; 2. 2 Solving Recurrences with the Iteration/Recursion-tree Method In the iteration method we iteratively unfold the recurrence until we see the pattern. P2. The Substitution method. converts the recursion into a tree whose Solutions to exercise and problems of Introduction to Algorithms by CLRS (Cormen, Leiserson, Rivest, and Stein) Divide (line 2): (1) is required to compute q as the average of p and r. Conquer (lines 3 and 4): 2 T ( n /2) is required to recursively solve two subproblems, each of size n/2. for questions about sequences and series, e Very easy to understand! Now we use induction to prove our guess. or O). OK? Help organize the algebraic bookkeeping necessary to solve a recurrence. Steps to solve recurrence relation using recursion tree method: Draw a recursive tree for given recurrence relation. Use the substitution method to verify your answer. (5 marks) Solve this. Exercises. Recursive sequence formulaAn initial value such as $a_1$.A pattern or an equation in terms of $a_ {n 1}$ or even $a_ {n -2}$ that applies throughout the sequence.We can express the rule as a function of $a_ {n -1}$. The recursion-tree method converts the recurrence into a tree whose nodes represent the costs incurred at various levels of the recursion. The good guess devised using the recursion tree can be proved by the substitution method. Search: Recursive Sequence Calculator Wolfram. two steps: 1 Guess the form of the solution. The subproblem size for a node at depth is. P3. The iteration method does not require making a good guess like the substitution method (but it is often more involved than using induction). This method is especially powerful when we encounter recurrences that are non-trivial and unreadable via the master theorem. In this article, we are going to talk about two methods that can be used to solve the special kind of recurrence relations known as divide and conquer recurrences If you can remember these easy rules then Master Theorem is very easy to solve recurrence equations Learn how to solve recurrence relations with generating functions Recall that the recurrence relation is a recursive Final Exam Computer Science 112: Programming in C++ Status: Computer Science 112: Programming in C++ Course Practice . DFS Traversal of a Graph vs Tree. The false position method is a root-finding algorithm that uses a succession of roots of secant lines combined with the bisection method to As can be seen from the recurrence relation, the false position method requires two initial values, x0 and x1, which should bracket the root See full list on users For example, consider the There are mainly three ways for solving recurrences. Wolfram|Alpha can solve various kinds of recurrences, find asymptotic bounds and find recurrence relations satisfied by given sequences. 1.2 Recursion tree A recursion tree is a tree where each node represents the cost of a certain recursive sub-problem. Consider the following runtime recurrence: T (1) = 1 and T(n) = 3T(n/2) + n^2 when n greaterthanequalto 2. Assume T(1) = 1. the recursion-tree method 1 solving recurrences expanding the recurrence into a tree summing the cost at each level applying the substitution method 2 another example using a recursion tree. The terms of a recursive sequences can be denoted symbolically in a number of different notations, such as , , or f[], where is a symbol representing thesequence Binomial Coefficient Calculator Do not copy and paste from Wolfram Sequences Calculator The sequence of RATS number is called RATS Sequence The sequence of RATS number is called RATS Sequence. Steps to solve recurrence relation using recursion tree method: Draw a recursive tree for given recurrence relation. In the previous lecture, the focus was on step 2. Like Masters Theorem, Recursion Tree is another method for solving the recurrence relations. T ( n) = 2 T ( n / 2) + n. The solution of this one can be found by Master Theorem or the recurrence tree method. Now we use induction to prove our guess. Each of these cases is an equation or inequality, with some Use an inductive hypothesis for simplicity, we specify initial conditions represent another method in recursion tree method for solving recurrences examples later determine in big-Oh notation). The recursion-tree method. Solving Recurrences Methods The Master Theorem The Recursion-Tree Method Useful for guessing the bound. Recursion Tree Implement recursion tree method to solve the recurrences below, you can use the Master theorem to verift your solution if applicable: a) T(n) = T(1/2) +T(1/4) + (n) b) T(n) = 3T(n/2) + 4T(n/2) + (n) c) T(n) = T(3n/2) +T(n/3) + O(n) d) T(n) = 4T(n/2) + 4T(n/2) + (n) In this method, we first convert the recurrence into a summation. There are mainly three ways for solving recurrences. Here the right-subtree, the one with 2n/3 element will drive the height. A recurrence is an equation or inequality that describes a function in terms of its values on smaller inputs. LEC 07: Recurrences II, Tree Method CSE 373 Autumn 2020 Learning Objectives 1.ContinuedDescribe the 3 most common recursive patterns and identify whether code belongs to one of them 2.Model a recurrence with the Tree Method and use it to characterize the recurrence with a bound 3.Use Summation Identities to find closed forms for summations to devise good guesses. In this video we discuss how to use the seqn command to define a recursive sequence on the TI-Nspire CX calculator page Monotonic decreasing sequences are defined similarly The terms of a recursive sequences can be denoted symbolically in a number of different notations, such as , , or f[], where is a symbol In the previous lecture, the focus was on step 2. A: Recursion tree is the method for solving the recurrence relations.Recursion tree may be a tree Q: Given the tree above, show the order of the nodes visited using recursive in-order traversal. Substitution method. To solve a Recurrence Relation means to obtain a function defined on the natural numbers that satisfy the recurrence. Use . Rekisterityminen ja tarjoaminen on ilmaista. First let's create a recursion tree for the recurrence $T (n) = 3T (\frac {n} {2}) + n$ and assume that n is an exact power of 2. Answer to 2. You must show the tree and fill out the table like we did in class. Calculate the cost at each level and count the total no of levels in the recursion tree. 1.Recursion Tree 2.Substitution Method - guess runtime and check using induction 3.Master Theorem 3.1 Recursion Tree Recursion trees are a visual way to devise a good guess for the solution to a recurrence, which can then be formally proved using the substitution method. 1) Substitution Method: We make a guess for the solution and then we use mathematical induction to prove the guess is correct or incorrect. Search: Recursive Sequence Calculator Wolfram. Visit the current node data in the postorder array before exiting from the current recursion. Introduction to the Recursion Tree Method for solving recurrences, with multiple animated examples. Steps to Solve Recurrence Relations Using Recursion Tree Method- Step-01: For Example, the Worst Case Running Time T (n) of the MERGE SORT Procedures is described by the recurrence. can be solved with recursion tree method. Use an inductive hypothesis for simplicity, we specify initial conditions represent another method in recursion tree method for solving recurrences examples later determine Keep track of the time spent on the subproblems of a divide and conquer algorithm. 4.4 The recursion-tree method for solving recurrences 4.4-1. How to solve the recurrence T ( n) = 3 T ( n / 2) + n. The exercise stated that i have to solve the recurrence using the Recursion-Tree Method. Search: Recurrence Relation Solver Calculator. If you see the height is determined by height of largest subtree (+1). A recursion tree is a tree where each node represents the cost of a certain recursive sub-problem. A: In-order -traversal:- We traversal the left node Engineering; Computer Science; Computer Science questions and answers; 3. Count the total number of nodes in the last level and calculate the cost of The subproblem P. S. Mandal, IITG I have already finished the base part, which is ( n lg 3) But for the recursive part I'm having troubles with this sum: c n i = 0 lg n 1 ( 3 / 2) i. Use of recursion to solve math problems ; Practice Exams. Such recurrences should not constitute occasions for sadness but realities for awareness, so that one may be happy in the interim. Q: Use the recursion tree method to solve each of the following recurrences: T(n) = T(n/10) + T(9n/10) A: The recursion tree method works by creating each level of the recurrence relation in the tree the or O). | 2022-09-29T05:55:44 | {
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https://math.stackexchange.com/questions/1463252/expressing-statements-in-discrete-math/1463268 | # Expressing statements in Discrete math
Given that
$A$ is the set of all Alpha's
$M$ is the set of all Men
how do I express this statement: Not all Alpha's are Men
.............
My attempt:
$A \subset S = 0$
in other words saying that $A$ is not a subset of $S$, but I can't use the not subset symbol on this problem.
• $\exists a\in A:a\notin M$ – abiessu Oct 4 '15 at 1:43
• What is $S$? Did you mean $A \subset M = \emptyset$? – N. F. Taussig Oct 4 '15 at 8:10
You could write $A\backslash M\ne\emptyset$.
Meaning that when you take all the men out of the alphas, there are alphas remaining.
• the \ is the difference symbol ? It makes a lot more sense this way, Thank you. – learnmore Oct 4 '15 at 2:10
• Yes. It is equivalent to A\cap M^c. M^c being the complement. – Jean-François Gagnon Oct 4 '15 at 4:48
"not all alpha's are men" $\Leftrightarrow$"there is an alpha who is not a man".
i.e.
$$\exists a \in A \text{ such that } a \not\in M$$
• I like the way you reworded it, makes it a lot more easier to express – learnmore Oct 4 '15 at 2:09 | 2021-01-26T05:39:21 | {
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http://apoderados.lazzeri.cl/wp-content/uploads/2022/03/kgqw5cvh/lecture-on-vector-calculus.html | Lecture on vector calculus. Divergence and Curl o
Lecture on vector calculus. Divergence and Curl of vector field | Irrotational & Solenoidal vector. Lecture 4: What Is A Unit Vector? Lecture 5: Vectors And The Unit Circle. Remember: the curl of a vector always results in another vector. 0 (fall 2009) This is a self contained set of lecture notes for Math 221. Unit vectors, vectors Here is a set of notes used by Paul Dawkins to teach his Calculus III course at Lamar University. Real Euclidean Space Rn. Vector Anupam Kumar. Vector Math with historical perspective (2010-2014), 13 lectures 2021 on youtube. 1 vectors We start with some de nitions. , Lectures on Differential Geometry, Prentice-Hall, Englewood Cliffs, New Jersey, 1964. You lectures. For detailed Vector Calculus (2009, UNSW). This course will remind you about that good stuff, but goes on to introduce you to the subject of Vector Calculus which, like it says on the can, combines vector algebra with calculus. (George Carlin, American Internet Supplement for Vector Calculus. 1. Some gave vector fields; some Another term for vector space is linear space. Check out www. Linear algebra. Why is vector calculus important for computer This chapter is concerned with applying calculus in the context of vector fields. ~r= x^i+ y^j+ zk^ (1) The unit vectors ^i;^j; ^k are orthonormal. edu/terms Education · 2011. Di erentiability in the case of two variable functions 30 These are lectures notes for MATH1056 Calculus Part II. Vector & Calculus - Lecture Prologue This lecture note is closely following the part of multivariable calculus in Stewart’s book [7]. In Lecture 6 we will look at combining these vector operators. Lecture 1: Three-Dimensional Coordinate Systems; Lecture 2: Vectors; Lecture 3: The Dot Product; Lecture 4: The Cross Product; Lecture Other Lecture Notes on the Web. In this post, Support Vector Machines — Lecture Important questions of Vector calculus Engineering mathematics lecture for GATE 2017 The fourth week covers the fundamental theorems of vector calculus, including the gradient theorem, the divergence theorem and Stokes’ theorem. Compute the curl of a vector field using sympy. Cowley@damtp. All unit vectors Advanced Calculus course. Remark We will almost exclusively consider real vector spaces, i. Lecture 1: Vector Notation. Tensor calculus The normal vectors are called ‘contravariant vectors’, because they transform con-trary to the basis vector columns. The course is organized into 53 short lecture Notes on Vector Calculus. If the triple product is zero, the volume between three vectors Lecture-02 calculus with vector What is the velocity of the conecting joint B in the example? Name a situation where the derivative of a unit vector is not zero Calculus 3 Lecture 11. 016 Fall 2012 c W. 1) is called the (linear)vectorspace. gaussianmath. Gleb V. The Physics course is delivered in Hinglish. Focuses on extending the concepts of function, limit, continuity, derivative, integral and vector from the plane to the three dimensional space. In large part this is because the point of vector calculus is to give us tools that we can apply elsewhere and the next steps involve turning to other courses. No calculators allowed. cam. Concept of Vector Point Function & Vector Differentiation. Matrices, linear transformations and vector spaces are necessary ingredients for a proper discussion of ad-vanced calculus A whole set of objects (vectors) on which we can perform vector addition and scalar multiplication with properties given by Eqs. 4) 2. Linear Algebra and Probability (Math 19b, Spring 2011) 154 pages. DEFINITION • Vector calculus (or vector analysis) is a branch of mathematics concerned with differentiation and integration of vector of vector analysis are simply incapable of allowing one to write down the governing laws in an invariant form, and one has to adopt a different mathematics from the vector analysis taught in the freshman and sophomore years. In this appendix I use the following notation. With such assignment one constructs a vector eld (scalar eld) in 3-dime Euclidean space. 1) In other words functions f CMU 15-462/662, Fall 2016 Vector Calculus in Computer Graphics Today’s topic: vector calculus. 3 minute read. diff, and the curl at a specific point using evalf. This playlist provides a shapshot of some lectures Cambridge vector calculus lecture notes calculus that I taught at the University of Ottawa in 2001 and at Dalhousie University in 2007 and 2013. Unless made explicitly, we will assume that vector and scalar elds considered in this lecture have continuous derivatives. Since a vector has no position, we typically indicate a vector Matrices, Vectors, and Vector Calculus In this chapter, we will focus on the mathematical tools required for the course. So Scheme 2 is the one where you get to drop a midterm, and Scheme 3 is for students who for any reason find themselves unable to regularly attend lectures. You could say it is the most important if you're willing to play it slightly fast and loose with definitions and include in it the subset of low-dimensional linear algebra that vector calculus ISBN:9781319083632. Since then, while I have had ample opportunity to teach, use, and even program numerous ideas from vector calculus These lecture videos are organized in an order that corresponds with the current book we are using for our Math2210, Calculus 3, courses ( Calculus, with CMU 15-462/662, Fall 2017 Vector Calculus in Computer Graphics Today’s topic: vector calculus. We found in Chapter 2 that there were various ways of taking derivatives of fields. Shown in green are a vector Chapter 4: VECTOR CALCULUS IN 2D. , An Introduction to Riemannian Geometry and the Tensor Calculus, Internet Supplement for Vector Calculus. Vector-valued functions are also written in the form. Tromba, Vector Calculus In this video lesson, GMath Calculus Donny Lee gives a basic example of implementing the line integral. 2 Integral Calculus Vector Calculus Vector Calculus 20E, Spring 2013, Lecture A, Midterm 1 Fifty minutes, four problems. , ⃗. Lecture Notes on Variational and Tensor Calculus. Marsden and Anthony Tromba Section Lectures Topic Review Assignment (no lecture Vector calculus, or vector analysis, is concerned with differentiation and integration of vector fields, primarily in 3-dimensional Euclidean space. Curves. A vector space is a collection of objects called vectors 2. # Define the independent variables using Symbol x = Symbol('x') y = Symbol('y') # Define the vector vector (or a scalar). 1. We denote R = of vector analysis are simply incapable of allowing one to write down the governing laws in an invariant form, and one has to adopt a different mathematics from the vector analysis taught in the freshman and sophomore years. ac. Download these Free Vector Calculus MCQ Quiz Pdf and In this course we shall extend notions of di erential calculus from functions of one variable to more general functions f: Rn!Rm: (1. •A work done by a constant force F in moving object from point P to point Q in space is . Tromba, Vector Calculus Sl. Free classes & tests. Matrices, Vectors, and Vector Calculus In this chapter, we will focus on the mathematical tools required for the course. 8. ii. This post contains some of the important notes which come in handy while working with vector-calculus. 1: An Introduction to Vectors: Discovering Vectors with focus on adding, subtracting, position vectors, unit vectors and magnitude. 2 Vectors expressed in terms of Unit Vectors in Rectangular coordinate Systems - A simple and convenient way to express vector quantities Let: i = unit vector along the x-axis j = unit vector along the y-axis k = unit vector along the z-axis in a rectangular coordinate system (x,y,z), or a cylindrical polar coordinate system (r, θ,z). Variational Calculus: Part I: Chapter 1. Lecture 10: Dot Product In 3-D. Lecture Anyone know of an online course or set of video lectures on John Hubbard's textbook on Vector Calculus, Linear Algebra, Related Threads on Hubbard’s vector calculus text Poll; Calculus Vector Calculus Vector Calculus 20E, Winter 2017, Lecture B, Midterm 2 Fifty minutes, three problems. MAT1236 Calculus 1 Topic 2: Vector Calculus Dr Steven Richardson Semester 2, 2014 1 / 46 Lecture Content 1. 99 Wholesale:$228. Vector calculus uses extensive variations of mathematics from differential geometry to multivariable calculus. This course covers vector and multi-variable calculus. Michael Medvinsky, NCSU online lectures 03/2020. Reminder A basis of an n-dimensional vector Lecture 02 - Vector Algebra in Component Form: Lecture 03 - Vector Triple Products: Lecture 04 - Vector Differential Calculus: Gradient: Lecture 05 - Divergence: Lecture 06 - Curl: Lecture 07 - Tutorial on Differential Vector Calculus: Lecture 08 - More Problems on Differential Vector Calculus: Lecture 09 - Vector Integral Calculus The identities curl (grad (f)) = 0 and div (curl (F)) = 0 need conditions on the scalar field f and the vector field F, namely continuous second partials in a The curl of the gradient of any continuously twice-differentiable scalar field. Contents Lecture 1. These lecture notes cannot be duplicated and distributed without explicit permission of the author. Aspects of Vector Calculus “O could I flow like thee, and make thy stream Vector Fields: A vector field is a function that assigns a vector to each point in calculus. Calculus This is a quick review of some of the major concepts in vector calculus that is used in this class. MANMOHAN DASH, PHYSICIST, TEACHER ! Physics for ‘Engineers and Physicists’ “A concise course of important results” Lecture - 1 Vector Calculus and Operations Lectures Notes for Calculus III (Multivariable Calculus) The notes below follow closely the textbook Introduction to Linear Algebra, Fourth Edition by Gilbert Strang. Fundamental Theorem for Line Integrals(cont) •Theorem: Suppose F=<P,Q> is a conservative vector prepared my lectures. Vector fields represent the distribution of a given vector to each point in the subset of the space. License: Creative Commons BY-NC-SA More information at ocw. The lecture notes [2], the book [3] and the “Vector Calculus Primer” [6] are available online; on the web page [4] of O. Schedule: MWTh@11AM or @12:30PM, Fall only. This book covers the following topics: Differentiation, Higher-Order Derivatives and Extrema, Vector Valued Functions, Double and Triple Integrals, Integrals over Curves and Surfaces and the Integral Theorems of Vector Lectures on Vector Calculus Paul Renteln Department of Physics California State University San Bernardino, CA 92407 March, 2009; Revised March, 2011 c Paul Renteln, 2009, 2011. org/learn/vector-calculus Lecture 1. We will invariably consider finite-dimensional vector spaces. 1 Introduction In single-variable calculus, the functions that one Curl¶. anupa at northeastern dot edu. Willard Gibbs and Oliver Heaviside near the end of the 19th century, LECTURES ON VECTOR CALCULUS. 5) where is the angle between a and b and u is a unit vector Instead of Vector Calculus, some universities might call this course Multivariable or Multivariate Calculus or Calculus 3. It is the second semester in the freshman calculus sequence. One can never know for sure what a deserted area looks like. MAT203 will not be offered in Fall 2020. Lecture Step 1: Give the vectors u and v (from rule 1) some components. TBA. 1 INTRODUCTION In vector calculus, we deal with two types of functions: Scalar Functions (or Scalar Field) and Vector Functions (or Vector 138 MIT 3. e. The least squares estimator of β minimizes f(β) = (y −Xβ)>(y −Xβ). Lecture Mathematical Tripos: IA Vector Calculus e c S. Lectures: MWF 9-10 in PCYNH 109 Lecture schedule and notes available below. Amanda Harsy ©Harsy 2020 July 20, 2020 i. However, I will use linear algebra. Lecture Notes on Classical Mechanics (A Work in Progress) Daniel Arovas Department of Physics University of Calculus 1- Limits and Derivatives. Lou Talman of Metro State lecture on the calculus Vector Calculus - Fall 2011 Meetings. Included are the lecture Lecture Notes: Introduction to Real Analysis Lectures Notes: Topics in Vector Calculus Book: Jerrold E. A familiarity with some basic facts about the 3 the Kronecker delta symbol ij, de ned by ij =1ifi= jand ij =0fori6= j,withi;jranging over the values 1,2,3, represents the 9 quantities 11 =1 21 =0 31 =0 12 =0 22 Vector calculus - SlideShare Vector calculus is also known as vector analysis which deals with the differentiation and the integration of the vector field in the three-dimensional Euclidean space. A familiarity with some basic facts Instead of Vector Calculus, some universities might call this course Multivariable or Multivariate Calculus or Calculus 3. Let a r, a ϕ, and a z be unit vectors along r, ϕ and z directions, respectively in the Multivariable Calculus Lecture Notes (PDF 105P) This lecture note is really good for studying Multivariable calculus. 1 1. Numerade's Calculus 3 course focuses on Calculus and its applications in different fields of Mathematics. Tensor Calculus (The Dual of a Vector Space) Tensor Calculus 4-5 (Tensors as Multilinear Maps; Integral Curves; The Commutator) Tensor Calculus These lecture notes cannot be duplicated and distributed without explicit permission of the author. Lecture 2: Vector And The Circle. (8 Jan) The midterm exam will be organized in the lecture on 14 Mar (which is in Week 9). Brown. The scope covers only linear algebra (more on this when the Definition. The idea behind the vector calculus is to utilize vectors . Lecture 2: Review of Vector Calculus Instructor: Dr. Lecture Tutorial on vector calculus and curvilinear coordinates; Introduction to electrostatics; Continuous charge distribution: Line charge; Electric field Vector Calculus - Fundamental Theorem fo Space Curves pt1 tutorial of Vector Calculus II course by Prof Donylee of Online Tutorials. (6. Please start each problem on a new page. φ {\displaystyle \varphi } is always the zero vector These lecture videos are organized in an order that corresponds with the current book we are using for our Math1210, Calculus 1, courses ( Calculus, with Course Description. 6. Gradient of a Scalar Field & Directional Derivative | Normal Vector. Preliminaries 1 1. •Unit tangent vector edge of vector calculus and real analysis, some basic elements of point set topology and linear algebra. Example : A~(x;y;z) = (x;xy;xz) (’(x;y;z) = x2yz) is a vector Aspects of Vector Calculus “O could I flow like thee, and make thy stream Vector Fields: A vector field is a function that assigns a vector to each point in Mathematics 31CH: \Honors Vector Calculus" Syllabus (revised September 2016) Lecture schedule based on: Vector Calculus, Linear Algebra, and Di erential Forms: A Uni ed Approach, fth edition by John H. lamar. The idea behind the vector calculus is to utilize vectors Listen on Apple Podcasts. Tcheslavski Contact: gleb@ee. x y O ˚ x0 y0 x y O ˚ Figure 1: Left: change of reference frame. Topics include vectors and matrices, partial derivatives, double and triple integrals, and vector calculus calculus lecture notes ppt, These notes stem from my own need to refresh my memory on the fundamentals of tensor calculus, having seriously considered them last some 25 years ago in grad school. We'll cover the essential calculus of such vector Vector calculus - basics A vector – standard notation for three dimensions Unit vectors i,j,kare vectors of magnitude 1 in directions of the x,y,z axes. We’ll start with the concepts of partition, Riemann sum and Riemann Integrable functions and their properties. 82. The course includes the concept of vectors, Dot Product and Cross Product of vectors, Vector Vector Calculus. J. Be prepared to draw your own figures! Vector Calculus Example 5 Let y be an n dimensional column vector of known constants, X be an n×m matrix of full column rank, and β be an m dimensional vector of unknown variables. Vector Calculus with Applications 17. Lecture 20: Vector Calculus - Fundamental Theorem fo Space Curves pt1. 2. That there must be a different behavior is also intuitively clear: if we described an ‘arrow’ by coordinates, and we then modify the basis vectors Math 20E Syllabus - Vector Calculus (revised June 2021) Lecture schedule based on Vector Calculus, sixth edition by Jerrold E. Chris Tisdell gives 88 video lectures on Vector Calculus. Vectors are denoted with an arrow over the top of the variable. 4. The underlying physical meaning — that is, why they are worth bothering about. uk, Lent 2000 0 Introduction 0. All vectors (Relevant section from Stewart, Calculus, Early Transcendentals, Sixth Edition: 16. The course is organized into 53 short lecture Get Vector Calculus Multiple Choice Questions (MCQ Quiz) with answers and detailed solutions. com for an indepth study and more calculus related lessons. Wilson, Fall 2006 1 DarcyÕs Law in 3D ¥Today ÐVector Calculus ÐDarcyÕs Law in 3D q="K!h Brief Review of Vector Calculus ¥A scalar has only a magnitude ¥A vector is characterized by both direction and magnitude. 3. , Springer-Verlag, Berlin, 1954. Why is vector calculus important for computer Lecture notes, lectures 15-17 100% (1) Pages: 16 2015/2016 16 pages 2015/2016 100% (1) Save Vector Calculus Advanced Student Notes New None Pages: 166 This is a quick review of some of the major concepts in vector calculus that is used in this class. ) 1. This book covers the following topics: Differentiation, Higher-Order Derivatives and Extrema, Vector Valued Functions, Double and Triple Integrals, Integrals over Curves and Surfaces and the Integral Theorems of Vector 3. Tensor calculus vector (or a scalar). The most important object in our course is the vector field, which assigns a vector to every point in some subset of space. • Baxandall and Liebeck, “Vector Calculus Lecture Notes: Introduction to Real Analysis Lectures Notes: Topics in Vector Calculus Book: Jerrold E. This note contains the following subcategories Vectors in R3, Cylinders and Quadric Surfaces, Partial Derivatives, Lagrange Multipliers, Triple Integrals, Line Integrals of Vector Understand the concept of Vector & Calculus - Lecture 2 with IIT JEE course curated by Abhilash Sharma on Unacademy. This package includes and Hardcover. Vector Calculus – Line Integrals of Vector Field | Example & Solution. This course is a study of the calculus of functions of several variables (vector arithmetic and vector calculus). 4. In the Euclidean space, the vector 3 The projection of a vector a on b is equal to a eb, where eb = b=jbj is the unit vector in direction of b. Buy + Hardcover. In this course, we begin our Lectures on Vector Calculus - CSUSB vectors, how to take scalar and vector products of vectors, and something of how to describe geometric and physical entities using vectors. J. The main concepts that will be covered are: • Coordinate transformations • Matrix operations • Scalars and vectors • Vector calculus edge of vector calculus and real analysis, some basic elements of point set topology and linear algebra. 1 of Stewart’s Calculus. Di erentials and Taylor Series 71 The di erential of a function. Magnitude of a vector Position vector is a vector r from the origin to the current position where x,y,z, are projections of r to the coordinate axes. Join me on Coursera: https://www. This bestselling vector calculus ERTH403/HYD503 Lecture 6 Hydrology Program, New Mexico Tech, Prof. e we go from the Average Rate of Change to the Instantaneous Rate of Change by letting the interval over which the Average Rate of Change is measured go to zero. 3) Recall the basic idea of the Generalized Fundamental Theorem of Calculus: If F is a gradient or conservative vector Line Integral of Vector Field •Reminder: •A work done by variable force f(x) in moving a particle from a to b along the x- axis is given by . The convention that I will try to follow in the lectures is that if we are interested in locating a point in space, we will use a row vector This course will offer a detailed introduction to integral and vector calculus. Variational Calculus: Part II: Chapters 2-3. Shown in green are a vector View Vector_Calculus_Lecture_notes_. , – In earlier courses, you may have learned that a vector is, basically, an arrow – That’s true in three dimensions, but this new definition allows one to create higher-dimensional vectors 3. 1 ( 11 ) Lecture Details. Hubbard and Barbara Burke Hubbard. • Baxandall and Liebeck, “Vector Calculus Matrices, Vectors, and Vector Calculus In this chapter, we will focus on the mathematical tools required for the course. 74 Lecture First, fix the y variable and compute the partial derivative for f (x,y) = x²- y² with respect to x to get ∂f (x,y)/ ∂x = 2x-0, since y is a constant its Lecture 3: Vectors • Any set of numbers that transform under a rotation the same way that a point in space does is called a vector – i. 2 Cross product The cross product, a b between two vectors a and b is a vector de ned by a b= absin( )u; 0 ˇ; (2. Online, via Zoom. All unit vectors 4. Perform various operations with vectors like adding, subtracting, scaling, and Remark 2. 3 Vector Calculus In addition to Linear Algebra, Vector calculus is a key component of any Machine Learning project. Instead of Vector Calculus lectures. This is a collection of video lectures given by Professor Chris Tisdell, presenting vector calculus in an applied and engineering context. Cartesian coordinates. Brief Course Description: Covers largely the same mathematical topics as MAT201, namely vectors ExtravagantDreams. L1: Differentiation of Vectors A whole set of objects (vectors) on which we can perform vector addition and scalar multiplication with properties given by Eqs. Volume 1 is concerned Ricci Calculus, 2nd ed. A vector-valued function is a function of the form. In a more general sense the broad approach and philosophy taken has been in uenced by: Volume I: A Brief Review of Some Mathematical Preliminaries I. I cannot recall every source I have used but certainly they include those listed at the end of each chapter. No Access. Best lecture on calc University library. Retail:$284. Related Courses. In the text, elements of Rn are denoted by row{vectors; in the lectures and homework assignments, we will use column vectors. Lecture 3: Scalar Multiplication. 1 Schedule This is a copy from the booklet of schedules. Hubbard, Vector calculus, linear algebra, and differential forms-the Honors Calculus Lecture 6: Parametric Equations And Vectors: Example 1. Linear algebra is not a prerequisite for this course. The notes were written by Sigurd Angenent, starting next three semesters of calculus In this lecture, we extend the theory of calculus of variations from a single integral setting to a multivariate integral setting, including the E-L equation, the criteria for weak and strong minima, Jacobi fields, and the Weierstrass excess function, etc. No Chapter Name English; 1: Lecture 1 : Partition, Riemann intergrability and One example: Download Verified; 2: Lecture 2 : Partition, Riemann intergrability and All of these questions involve understanding vectors and derivatives of multivariable functions. We also illustrate how to find a vector from its starting and end points. Derive the expression of this estimator. Vector calculus is a form of mathematics that is focused on the integration of vector fields. kumar. Hubbard, Vector calculus, linear algebra, and differential forms-the Honors Calculus I believe calculus is best learned through four or five short lectures each week throughout a 14-week semester, and this course of video lectures is designed This is the second volume of a two-volume work on vectors and tensors. In this course, Prof. with scalar field K = R. Topics include vectors and matrices, partial derivatives, double and triple integrals, and vector calculus Vector calculus was developed from quaternion analysis by J. Quote. Marsden and Anthony J. mit. ¥Vectors Vector Calculus MCQ Question 4. I’m going to use a and b here, but the choice is arbitrary: u = (a 1, a 2) v = (b 1, b 2) Differential equations and vector calculus Course Objectives. We then move to anti-derivatives and will look in to few classical theorems of integral calculus such as fundamental theorem of integral calculus. where the component functions f, g, and h, are real-valued functions of the parameter t. Gelfand and S. Download Solution PDF. LIMITS In this first animation we see the secant line become the tangent line i. Lectures in Vector Calculus in 2D. IIT JEE. 2 Lecture 10. Lecture 1. E. The main concepts that will be covered are: • Coordinate transformations • Matrix operations • Scalars and vectors • Vector calculus Aspects of Vector Calculus “O could I flow like thee, and make thy stream Vector Fields: A vector field is a function that assigns a vector to each point in Vector Calculus part II By Dr. M. Fomin, Calculus Learn what vectors are and how they can be used to model real-world situations. 05. A real number xis positive, zero, or negative and is rational or irrational. Marsden (CalTech) & A. 71 The Taylor series. STERNBERG, S. (2. Scheme 3: 20% Homework, 25% Midterm 1, 25% Midterm 2, 30% Final Exam. In both cases, the first form of the function defines a two-dimensional vector These lecture videos are organized in an order that corresponds with the current book we are using for our Math2210, Calculus 3, courses ( Calculus, with MATH 25000: Calculus III Lecture Notes Created by Dr. The plane. Covers topics including vector functions, multivariate functions, partial derivatives, multiple integrals and an introduction to vector calculus. A vector is depicted as an arrow starting at one point in space and ending at another point. This is a series of lectures for "Several Variable Calculus" and "Vector Calculus", which is a 2nd-year mathematics subject taught at UNSW, Sydney. Surface Area– Dr. This is a series of lectures for "Several Variable Calculus" and "Vector Calculus CALCULUS ON MANIFOLDS 5 (tautologically) R1 with R eld, then the di erential becomes an element of the dual vector space T a U’(Rn) . Lecture 8: How To Determine If The Lines Are Parallel? Lecture 9: How To Determine If The Lines Intersect. Di erentiability 29 1. (This lecture corresponds to Section 5. Effective: 2017-08-01. To understand the three major theorems of vector calculus. WEATHERBURN, C. Two semesters of single variable calculus (differentiation and integration) are a prerequisite. Resource Guide to Vector Calculus. They consist largely of the material presented during the lectures Course Description. pdf from MAT 1236 at Edith Cowan University. 65 Lecture 11. Published: March 01, 2020. You Differential Calculus Lecture Notes Veselin Jungic & Jamie Mulholland Department of Mathematics Simon Fraser University c Jungic/Mulholland, August MAT203 Advanced Vector Calculus. I believe an interested student can Vector calculus is one of the most useful branches of mathematics for game development. Class : Lecture Scheme 2: 5% participation, 25% Homework, 30% Best Midterm, 40% Final Exam. Topics covered in these notes include the un-typed lambda calculus Vector Calculus 1 multivariable calculus 1. M, 2:50pm-4:30pm. Dynamical systems, Spring 2005 (183 pages) Linear Algebra (21b, Spring 2018) College Multivariable, (Fall 2017) Calculus Defines vectors, vector addition and vector subtraction. Lecture Vector Calculus - Fall 2011 Meetings. Tensor Calculus (The Dual of a Vector Space) Tensor Calculus 4-5 (Tensors as Multilinear Maps; Integral Curves; The Commutator) Tensor Calculus Vector Calculus (Multivariate Calculus)B SC 4th Semester (CBCS)Mathematics (Honours)MAT-HC-4016Lecture 1 Vectors in Euclidean Space 1. Lecture 7: Parametric Equations And Vectors: Example 2. g. To enlighten the learners in the concept of differential equations and multivariable calculus Scheme 2: 5% participation, 25% Homework, 30% Best Midterm, 40% Final Exam. coursera. The main concepts that will be covered are: • Coordinate transformations • Matrix operations • Scalars and vectors • Vector calculus Multivariable Calculus Lectures Richard J. Vector Space. But like wolfsy said, if you are trying to Lecture Notes on Variational and Tensor Calculus. A two-dimensional vector field is a function f that maps each point (x,y) in R2 to a two-dimensional vector hu,vi, and similarly a three-dimensional vector field maps (x,y,z) to hu,v,wi. You must substitute the parametric equations into both the vector field and position vector and then integrate. The term "vector calculus" is sometimes used as a synonym for the broader subject of multivariable calculus, which spans vector calculus MATH 252-3: Vector Calculus Course Webpage Quick Links: Library Reserves WebCT Course Webpage This Week Documents & Homework Information Vector Calculus (MAST20009) COVID-19 vaccination (or valid exemption) is a requirement for anyone attending our campuses. For Vector Calculus I like J. Vector Calculus previous lecture notes by Ben Allanach and Jonathan Evans ; Vector Calculus yet earlier lecture notes by Stephen Cowley. These theorems are needed in core engineering subjects such as Electromagnetism and Fluid Mechanics. Slide 19. A familiarity with some basic facts 1 17. Example : A~(x;y;z) = (x;xy;xz) (’(x;y;z) = x2yz) is a vector linear transformations and vector spaces are necessary ingredients for a proper discussion of ad-vanced calculus. Topics covered are Three Dimensional Space, Equations of normal vectors and tangent planes 24 1. Vector Fields 65 Vector Fields. Contents 1 Syllabus and Scheduleix Our last month will be combining the multivariate calculus with vector calculus and this culminates in several important theorems which tie all of Calculus Basic Concepts – In this section we will introduce some common notation for vectors as well as some of the basic concepts about vectors such as the magnitude of a vector and unit vectors. Home New to This Edition WebAssign for Vector Calculus. V. Hinglish Physics. Lecture 6: Addition Of Vectors MATH 221 { 1st SEMESTER CALCULUS LECTURE NOTES VERSION 2. Unit vectors, vectors Lecture 1 Vectors View this lecture on YouTube We define a vector in three-dimensional Euclidean space as having a length (or magnitude) and a direction. Knill you can find plenty of exercises, lecture edge of vector calculus and real analysis, some basic elements of point set topology and linear algebra. Tromba (UCSC). In organizing this lecture note, I am indebted by Cedar Crest College Calculus IV Lecture Course Description. C Carter Lecture 11 where i j is the angle between two vectors iand j, and ij k is the angle between the vector kand the plane spanned by iand j, is equal to the parallelepiped that has ~a, ~b, and ~cemanating from its bottom-back corner. Course objective : To apply the basic concepts found in a first year calculus course to multivariable functions (limits, differentiation, and integration). edu Office Hours: Room 2030 – A free PowerPoint 3–1 Vector integrals; the line integral of ∇ψψ.
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https://encyclopediaofmath.org/index.php?title=Stochastic_matrix&oldid=35214 | # Stochastic matrix
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
2010 Mathematics Subject Classification: Primary: 15B51 Secondary: 60J10 [MSN][ZBL]
A stochastic matrix is a square (possibly infinite) matrix $P=[p_{ij}]$ with non-negative elements, for which $$\sum_j p_{ij} = 1 \quad \text{for all i.}$$ The set of all stochastic matrices of order $n$ is the convex hull of the set of $n^n$ stochastic matrices consisting of zeros and ones. Any stochastic matrix $P$ can be considered as the matrix of transition probabilities of a discrete Markov chain $\xi^P(t)$.
The absolute values of the eigenvalues of stochastic matrices do not exceed 1; 1 is an eigenvalue of any stochastic matrix. If a stochastic matrix $P$ is indecomposable (the Markov chain $\xi^P(t)$ has one class of positive states), then 1 is a simple eigenvalue of $P$ (i.e. it has multiplicity 1); in general, the multiplicity of the eigenvalue 1 coincides with the number of classes of positive states of the Markov chain $\xi^P(t)$. If a stochastic matrix is indecomposable and if the class of positive states of the Markov chain has period $d$, then the set of all eigenvalues of $P$, as a set of points in the complex plane, is mapped onto itself by rotation through an angle $2\pi/d$. When $d=1$, the stochastic matrix $P$ and the Markov chain $\xi^P(t)$ are called aperiodic.
The left eigenvectors $\pi = \{\pi_j\}$ of $P$ of finite order, corresponding to the eigenvalue 1: $$\label{eq1} \pi_j = \sum_i \pi_i p_{ij} \quad \text{for all}\ j\,,$$ and satisfying the conditions $\pi_j \geq 0$, $\sum_j\pi_j = 1$, define the stationary distributions of the Markov chain $\xi^P(t)$; in the case of an indecomposable matrix $P$, the stationary distribution is unique.
If $P$ is an indecomposable aperiodic stochastic matrix of finite order, then the following limit exists: $$\label{eq2} \lim_{n\rightarrow\infty} P^n = \Pi,$$ where $\Pi$ is the matrix all rows of which coincide with the vector $\pi$ (see also Markov chain, ergodic; for infinite stochastic matrices $P$, the system of equations \ref{eq1} may have no non-zero non-negative solutions that satisfy the condition $\sum_j \pi_j < \infty$; in this case $\Pi$ is the zero matrix). The rate of convergence in \ref{eq2} can be estimated by a geometric progression with any exponent $\rho$ that has absolute value greater than the absolute values of all the eigenvalues of $P$ other than 1.
If $P = [p_{ij}]$ is a stochastic matrix of order $n$, then any of its eigenvalues $\lambda$ satisfies the inequality (see [MM]): $$\left| \lambda - \omega \right| \leq 1-\omega, \quad \text{where \omega = \min_{1 \leq i \leq n} p_{ii}.}$$ The union $M_n$ of the sets of eigenvalues of all stochastic matrices of order $n$ has been described (see [Ka]).
A stochastic matrix $P=[p_{ij}]$ that satisfies the extra condition $$\sum_i p_{ij} = 1 \quad \text{for all j}$$ is called a doubly-stochastic matrix. The set of doubly-stochastic matrices of order $n$ is the convex hull of the set of $n!$ permutation matrices of order $n$ (i.e. doubly-stochastic matrices consisting of zeros and ones). A finite Markov chain $\xi^P(t)$ with a doubly-stochastic matrix $P$ has the uniform stationary distribution.
#### References
[G] F.R. Gantmacher, "The theory of matrices" , 1 , Chelsea, reprint (1977) (Translated from Russian) MR1657129 MR0107649 MR0107648 Zbl 0927.15002 Zbl 0927.15001 Zbl 0085.01001 [B] R. Bellman, "Introduction to matrix analysis" , McGraw-Hill (1960) MR0122820 Zbl 0124.01001 [MM] M. Marcus, H. Minc, "A survey of matrix theory and matrix inequalities" , Allyn & Bacon (1964) MR0162808 Zbl 0126.02404 [Ka] F.I. Karpelevich, "On the characteristic roots of matrices with non-negative entries" Izv. Akad. Nauk SSSR Ser. Mat. , 15 (1951) pp. 361–383 (In Russian)
Given a real $n\times n$ matrix $A$ with non-negative entries, the question arises whether there are invertible positive diagonal matrices $D_1$ and $D_2$ such that $D_1AD_2$ is a doubly-stochastic matrix, and to what extent the $D_1$ and $D_2$ are unique. Such theorems are known as $DAD$-theorems. They are of interest in telecommunications and statistics, [C], [F], [Kr].
A matrix $A$ is fully decomposable if there do not exist permutation matrices $P$ and $Q$ such that $$PAQ = \left[ \begin{array}{cc} A_1 & 0 \\ B & A_2 \end{array} \right].$$ A $1 \times 1$ matrix is fully indecomposable if it is non-zero.
Then for a non-negative square matrix $A$ there exist positive diagonal matrices $D_1$ and $D_2$ such that $D_1AD_2$ is doubly stochastic if and only if there exist permutation matrices $P$ and $Q$ such that $PAQ$ is a direct sum of fully indecomposable matrices [SK], [BPS]. | 2021-08-05T14:43:12 | {
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https://math.stackexchange.com/questions/3161357/sum-of-the-inverse-of-a-geometric-series | # Sum of the inverse of a geometric series?
I'm trying to solve for this summation:
$$\sum_{j=0}^{i} {\left(\frac 1 2\right)^j}$$
This looks a lot like a geometric series, but it appears to be inverted. Upon plugging the sum into Wolfram Alpha, I find the answer to be
$$2-2^{-i}$$
but I don't understand how it gets there. Am I able to consider this a geometric series at all? It almost seems closer to the harmonic series.
• Welcome to Math Stack Exchange. Note $\frac 1 {2^j}=\left(\frac1 2\right)^j$ – J. W. Tanner Mar 25 '19 at 3:36
• Is $i$ some finite number? Or are you trying to find the value of the series for any arbitrary $i$? – kkc Mar 25 '19 at 3:36
• The reciprocals of each term of a geometric series is also a geometric one – lab bhattacharjee Mar 25 '19 at 3:37
• i is finite but arbitrary. The sum does not approach infinity. – bpryan Mar 25 '19 at 3:39
• My point of noting $\frac 1 {2^j}=\left(\frac 1 2\right)^j$ was not that $\frac 1 {2^j}$ was incorrect but rather that this is a geometric series – J. W. Tanner Mar 25 '19 at 3:49
$$\sum_{j=0}^{i} \frac{1}{2^j}=\sum_{j=0}^{i}\left( \frac{1}{2}\right)^j=\frac{1-\left(\frac12\right)^{i+1}}{1-\frac12}=2\left(1-\left(\frac12\right)^{i+1}\right)=2-\left(\frac12\right)^{i}=2-2^{-i}$$
As $$\frac1{2^j}=\left(\frac12\right)^j$$, this is also a geometric sum with the common ratio of $$r=\frac12$$. So you apply the formula for geometric sums $$\sum_{j=0}^nr^j=\frac{1-r^{n+1}}{1-r}$$to obtain the answer you have written. | 2021-07-26T17:58:08 | {
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https://math.stackexchange.com/questions/3675762/does-the-category-of-local-rings-with-residue-field-f-have-an-initial-object | # Does the category of local rings with residue field $F$ have an initial object?
Let $$F$$ be a field. Does the category $$C_F$$ of local rings with residue field isomorphic to $$F$$ have an initial object?
This is, for instance, true if $$F=\mathbb{F}_{p}$$ for some prime $$p$$: If $$R$$ is a local ring with residue field $$\mathbb{F}_{p}$$, then any $$x\in\mathbb{Z}\setminus(p)$$ must map to something invertible under the morphism $$\mathbb{Z}\longrightarrow R$$. Hence that morphism factors as $$\mathbb{Z}\longrightarrow\mathbb{Z}_{(p)}\longrightarrow R$$; thus $$\mathbb{Z}_{(p)}$$ is the initial object.
But what happens in the more general case? I guess it should be true at least if $$F$$ is of finite type over $$\mathbb{Z}$$, but I have no idea how to prove it.
(EDIT - To avoid any confusion: I am talking about an initial object in the category of local rings $$R$$ with a fixed surjection $$R\longrightarrow F$$.)
• That's a good question. It holds for $F=\mathbb Q$ as well – Maxime Ramzi May 15 '20 at 7:00
• @GeorgesElencwajg : I guess it depends on how you set up the category $C_F$ exactly : if it's just a full subcategory of rings, then you are right (composing $R\to F\to F$ gives two different morphisms $R\to F$ because $R\to F$ is surjective); but if you set it up as a subcategory of $CRing/F$ instead, it's not that clear. – Maxime Ramzi May 15 '20 at 7:27
• @Georges : I'm not saying $F$-algebras, on the other hand "the category of local fields with residue field $F$" could definitely mean "local rings $R$ with a fixed map $R\to F$ which exhibits $F$ as the residue field". So a subcategory of $CRing/F$, not $F/Cring$ – Maxime Ramzi May 15 '20 at 8:00
• @Georges: I meant it the way Maxime says it. – The Thin Whistler May 15 '20 at 8:10
Let $$\mathbb{F_4}=\{0,1,w,1+w\}$$ be the field of 4 elements. Suppose $$R$$ is the initial object in the category described in the question for the field $$\mathbb{F_4}$$. Then $$R$$ must contain some element $$x$$ which maps to $$w\in\mathbb{F_4}$$. Thus we have a map $$f\colon S\to R$$, where $$S=\mathbb{Z}[y]_M$$, sending $$y \mapsto x$$. Here $$M$$ is the maximal ideal of $$\mathbb{Z}[y]$$ containing $$2,1+y+y^2$$.
The following composition must be the identity: $$R \to S \stackrel f \to R$$ Thus $$R=S/I$$ for some ideal $$I\subset M$$. Further we know $$I\neq 0$$ as $$S$$ cannot be the initial object: there are multiple distinct maps $$S\to S$$, such as the identity map and the map sending $$y\mapsto y+2$$.
Under the composition $$S \stackrel f \to R\to S$$, we have $$y\mapsto p/q$$, for some $$p,q$$ integer polynomials in $$y$$. We know $$p/q$$ is not a rational number as $$p/q\mapsto w\in\mathbb{F_4}$$. Thus $$p/q$$ is a non-constant rational function in one variable, taking infinitely many values, which cannot all satisfy the same polynomial over the integers.
On the other hand, as $$I\neq 0$$ there must be a polynomial over the integers satisfied by $$p/q$$. This gives us the desired contradiction.
• This is so clever! How did you come up with this approach? – diracdeltafunk May 16 '20 at 2:26
• Thankyou. It felt like if there was an initial object it ought to be S, but at the same time it couldn't be. Playing those off against each other and using that S is a well behaved, concretely described ring led to the contradiction. – tkf May 16 '20 at 3:55
The category $$C_{F}$$ possesses a weak initial object $$I_{F}$$, i.e. an object that is unique up to not necessarily unique isomorphism.
Let $$F$$ be a field and $$L$$ be its minimal subfield (the smallest subfield contained in $$F$$). Then either $$L=\mathbb{F}_{p}$$ for some prime $$p$$ or $$L=\mathbb{Q}$$.
Assume first that $$F$$ is of finite type over $$L$$. Let $$n\in\mathbb{N}$$ be the smallest natural number so that $$F=L[x_{1},...,x_{n}]/\mathfrak{m}$$ for some maximal ideal $$\mathfrak{m}\subseteq L[x_{1},...,x_{n}]$$. Let $$\overline{x}_{i}$$ be the image of $$x_{i}\in L[x_{1},...,x_{n}]$$ in $$F$$.
Let $$\zeta:R\longrightarrow F$$ be a surjection where $$R$$ is a local ring. Since every $$\overline{x}_{i}$$ has a (not necessarily unique) preimage $$\zeta^{-1}(\overline{x}_{i})\in R$$, there is a (not necessarily unique) morphism $$\kappa:\mathbb{Z}[x_{1},...,x_{n}]\longrightarrow R$$ that fits into a commutative diagram $$\require{AMScd}$$ $$\begin{CD} \mathbb{Z}[x_{1},...,x_{n}]@>{\kappa}>>R\\ @V{\pi}VV @VV{\zeta}V\\ L[x_{1},...,x_{n}] @>>{\chi}>F \end{CD}$$ Let $$\mathfrak{i}:=\chi^{-1}\pi^{-1}(0)=\pi^{-1}(\mathfrak{m})$$. The ideal $$\mathfrak{i}$$ is always prime; it is maximal if and only if $$L=\mathbb{F}_{p}$$ for some prime $$p$$. Since $$R$$ is local, every element of $$\mathbb{Z}[x_{1},...,x_{n}]$$ is mapped by $$\kappa$$ onto something invertible in $$R$$. Hence $$\kappa$$ factors as $$\begin{CD} \mathbb{Z}[x_{1},...,x_{n}] @>>>\mathbb{Z}[x_{1},...,x_{n}]_{(\mathfrak{i})} @>{\lambda}>> R \end{CD}$$ Thus $$I_{F}:=\mathbb{Z}[x_{1},...,x_{n}]_{(\mathfrak{i})}$$ is a weak initial object in the category $$C_{F}$$.
Note that the assignment $$\kappa\longleftrightarrow\lambda$$ is unique in both ways: To each choice of $$\kappa$$ there is a unique $$\lambda$$ and vice-versa.
Assume next that $$F$$ is of infinite type over $$L$$. Then $$F$$ is the direct limit of all morphisms $$F'\longrightarrow F''$$, where $$F',F''$$ are fields of finite type over $$L$$. Since the construction of $$I_{-}$$ is functorial and compatible with direct limits, $$I_{F}$$ can be defined as $$I_{F}:=\lim_{F'\text{ of fin. t.}/L}I_{F'}$$.
The initial object is strong, i.e. unique up to unique isomorphism, if and only if $$F=L$$.
Namely, if $$F=L$$, then $$n=0$$ and the unique morphism $$\kappa:\mathbb{Z}\longrightarrow R$$ induces a unique morphism $$\lambda:\mathbb{Z}_{(\mathfrak{i})}\longrightarrow R$$.
Else, if $$F\neq L$$, then $$n\geq 1$$ and for any $$i\in\{1,...,n\}$$ and any $$s\in\mathfrak{i}\setminus\{0\}$$, the map $$\xi_{i,s}:x_{i}\mapsto x_{i}+s$$ yields a nontrivial automorphism $$I_{F}\longrightarrow I_{F}$$ that commutes with the surjection $$I_{F}\longrightarrow F$$.
My guess is that the $$\xi_{i,s}$$ actually generate the whole group $$\operatorname{Aut}(I_{F})$$, but I have yet to figure out a proof for this... | 2021-02-27T05:43:43 | {
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https://math.stackexchange.com/questions/3705602/find-an-equivalent-sequence-as-n-to-infty-of-u-10-u-n1-fracu-nn | # Find an equivalent sequence as $n\to +\infty$ of $u_1>0, u_{n+1} = \frac{u_n}{n} + \frac{1}{n^2}$
Let $$u_1>0$$ be a real number. Let us consider $$(u_n)_{n\geq 1}$$ the sequence such as:
$$\forall n \geq 1, u_{n+1} = \frac{u_n}{n} + \frac{1}{n^2}\quad (\star)$$
Find an equivalent of $$u_n$$ as $$n\to +\infty$$.
So I found a way to show that $$u_n \sim \frac{1}{n^2}$$, but I'm quite unhappy with this method because I feel like I found it by chance without understanding anything (I did a lot of trials and found this)
My method:
I showed by induction that $$u_n \leq (u_1+1)$$. Thus, $$u_n\to 0$$ considering $$(\star)$$.
Then, $$nu_{n+1} = u_n + 1/n$$. Thus (since $$n+1 \sim n$$), $$nu_n \to 0$$.
To end with, I have $$n^2u_{n+1} = nu_n + 1$$. Thus, $$(n+1)^2 u_n \sim n^2u_n \to 1$$.
How would you solve such a problem? Is there any more intuitive method that one may have done?
• It is right actually. I think elementary proof has you find is rather good. Where the intuition comes it that $u_n$ is ponderated with $n$ and $u_{n+1}$ can be with $n^2$ by multiplying all the equation by $n^2$. So I think that ponderation lead you to the equivalent. – EDX Jun 4 '20 at 15:39
1. There is a heuristic argument which is useful for guessing the behavior of $$u_n$$: Rewrite the recurrence relation as
$$u_{n+1} - u_n = \frac{u_n}{n} - u_n + \frac{1}{n^2}.$$
Its continuum analogue is the following differential equation:
$$y' = \frac{y}{x} - y + \frac{1}{x^2}.$$
Using the standard method, this equation can be solved as:
$$y(x) = x e^{-x} \int \frac{e^x}{x^3} \, \mathrm{d}x.$$
Then L'Hospital's Rule then tells that $$y(x) \sim x^{-2}$$ as $$x \to \infty$$. From this observation, we may as well expect that $$u_n \sim n^{-2}$$.
2. The above ansatz suggests that, in the recurrence relation for $$u_{n+1}$$, $$\frac{1}{n^2}$$ is the dominating term and $$\frac{u_n}{n}$$ is much smaller as $$n\to\infty$$. In particular, nesting this relation will produce an expansion with ever decreasing terms. This idea can be easily tested as follows:
Let $$r_n = (n-1)u_n$$. Then for $$n \geq 2$$,
$$r_n = (n-1)u_{n} = \frac{1}{n-1} + \frac{r_{n-1}}{n-1}.$$
From this, we get
\begin{align*} r_n &= \frac{1}{n-1} + \frac{r_{n-1}}{n-1} \\ &= \frac{1}{n-1} + \frac{1}{(n-1)(n-2)} + \frac{r_{n-2}}{(n-1)(n-2)} \\ &= \frac{1}{n-1} + \frac{1}{(n-1)(n-2)} + \frac{1}{(n-1)(n-2)(n-3)} + \frac{r_{n-3}}{(n-1)(n-2)(n-3)} \\ &\qquad\vdots\\ &= \sum_{k=1}^{n-2} \frac{1}{(n-1)\cdots(n-k)} + \frac{r_2}{(n-1)!} \end{align*}
Using this, it is not hard to conclude that $$(n-1)^2 u_n \to 1$$ as $$n\to\infty$$, and in fact, we can extract an asymptotic expansion of $$u_n$$ up to any prescribed order $$\mathcal{O}(n^{-M})$$.
This is similar to Sangchul Lee's approach.
If we multiply $$n!$$ both sides of the recurrence, we obtain $$n!u_{n+1}=(n-1)!u_n+\frac{n!}{n^2}.$$ Applying the above repeatedly, $$u_{n+1}=\frac1{n!} \sum_{k=1}^n \frac{k!}{k^2} + \frac{u_1}{n!}.$$
• That's a really good method! Thanks – MiKiDe Jun 7 '20 at 9:17 | 2021-06-20T13:57:19 | {
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https://math.stackexchange.com/questions/1637719/given-3-types-of-coins-how-many-ways-can-one-select-20-coins-so-that-no-coi | # Given $3$ types of coins, how many ways can one select $20$ coins so that no coin is selected more than $8$ times.
So I was given this question.
Given $3$ types of coins, how many ways can one select $20$ coins so that no coin is selected more than $8$ times.
First I make $x_1 + x_2 + x_3 = 20$ Then $0 \leq x_i \leq 8$
Then we use the Inclusion exclusion principle
Let $A_i$ be the non-negative integer solutions to $x_1 + x_2 + x_3 = 20$ $x_i \geq 9$. Then use inclusion exclusion formula to find $N(A_1 \bigcap A_2 \bigcap A_3)$
What i don't get is how to apply the inclusion exclusion formula. I know the process to get to it but not how to apply it.
• Possible duplicate of Inclusion-exclusion principle: Number of integer solutions to equations – Cloverr Feb 2 '16 at 18:27
• You can also find the coefficient of $x^{20}$ in $(1+x+\ldots + x^8)^3$. Then you get the right answer as well. – Henno Brandsma Feb 2 '16 at 18:27
• @Nilanjan thats a different question – Zero Feb 2 '16 at 18:31
• Do you have to use inclusion-exclusion? The problem is small enough in scale to permit more straightforward approaches. Heck, even plain enumeration will do. – Brian Tung Feb 2 '16 at 18:56
• I've added inclusion-exclusion to my answer. Interestingly, the approach by generating function arrives at the same expression! – Brian Tung Feb 2 '16 at 19:03
Instead of trying to apply formulas take a look at the general theory. Suppose $a_j\le x_j\le b_j$ for $j=1,2,3$.
$$(x^{a_1}+x^{a_1+1}+\cdots+x^{b_1})(x^{a_2}+x^{a_2+1}+\cdots+x^{b_2})(x^{a_3}+x^{a_3+1}+\cdots+x^{b_3})$$
Every solution of $x_1+x_2+x_3=n$ with the constraints $a_j\le x_j\le b_j$ contributes $1$ to the coefficient of $x^n$ in the above expression. So the number of solutions is the coefficient of $x^n$.
Substituting accordingly we see we require the coefficient of $x^{20}$ in $$(1+x+\cdots+x^8)^3={(1-x^9)^3\over (1-x)^3}$$ or equivalently in $$(1-3x^9+3x^{18})\sum_{k=0}^{20}\binom{2+k}{k}x^k$$ which equals $$\binom{22}{20}-3\binom{13}{11}+3\binom{4}{2}$$
• The equation $x_1 + x_2 + x_3 = 20$ has only $\binom{22}{20}$ solutions in the non-negative integers. Check your calculations. – N. F. Taussig Feb 2 '16 at 18:42
• I replaced $3$ with $20$ erroneously.. – Jack's wasted life Feb 2 '16 at 18:45
There are $15$ different ways to select $20$ coins of three different types, with no type having more than $8$ specimens.
There are a number of different methods to obtain this number. One is to condition on the number of coins of Type $1$. If there are $4$ coins of Type $1$, then there is only one way to obtain $16$ coins of the other two types; if there are $5$ coins of Type $1$, then there are two ways to obtain $15$ coins of the other two types; and so on. $1+2+3+4+5 = 15$.
Another way is to consider that the solutions are the intersection of the lattice cube $\{(x, y, z)) \in \mathbb{N}_{\geq 0}^3 \mid 0 \leq x, y, z \leq 8\}$ with the plane $x+y+z = 20$. If you have any facility with this, you will notice that the intersection is a triangular lattice of side $5$, so the number of solutions is the $5$th triangular number, $15$.
ETA: Finally, we can use inclusion-exclusion, although I find it less straightforward. The number of non-negative integer solutions to $x+y+z = 20$ is, by the usual stars-and-bars approach, $\binom{22}{2} = 231$. However, this counts solutions where at least one of $x, y, z > 8$. The number of solutions where $x > 8$ is, again by stars-and-bars, $\binom{13}{2} = 78$, and there are three coin types, so we must subtract $3 \times 78 = 234$ to obtain $-3$.
Again, however, this double-counts three cases where at least two of $x, y, z > 8$. There are $\binom{4}{2} = 6$ ways in which both $x, y > 8$, but there are three different pairs of coin types, so we must add back $3 \times 6 = 18$ to get $15$. There are no ways to have $x, y, z > 8$ and still add up to $20$, so $15$ is the final result. Note, interestingly, that this is the same expression that Jack's wasted life arrived at, although that method was by way of generating functions.
• One typographical error: In the last sentence of the first paragraph of the inclusion-exclusion argument, you meant $x > 8$ rather than $x > 0$. – N. F. Taussig Feb 2 '16 at 20:14
• @N.F.Taussig: Thanks for that. – Brian Tung Feb 3 '16 at 18:58
Let $8-x_i=:y_i$ $(1\leq i\leq3)$. Then we want all $y_i\geq0$ and $y_1+y_2+y_3=4$. There are $${4+3-1\choose 3-1}={6\choose 2}=15$$ ways to choose admissible $y_i$. | 2019-10-24T01:42:58 | {
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https://www.intmath.com/forum/integration-29/different-parabola-equation-when-finding-area:151 | IntMath Home » Forum home » Integration » Different parabola equation when finding area
# Different parabola equation when finding area [Solved!]
### My question
In chapter 3, compare example 1 to example 3. Why isnt (3 - x^2) the correct parabolic equation for example 3?
### Relevant page
3. Areas Under Curves
### What I've done so far
When I went to solve example 3, I used the equation (3 - x^2) for the parabola. I found out that was not the correct parabolic equation upon looking at your solution.
I am not sure why my equation cannot be used even though the vertex is not at (0,3), i.e. not lined up with the origin? The height is still three so I thought it would be okay.
X
In chapter 3, compare example 1 to example 3. Why isnt (3 - x^2) the correct parabolic equation for example 3?
Relevant page
<a href="https://www.intmath.com/integration/3-area-under-curve.php">3. Areas Under Curves</a>
What I've done so far
When I went to solve example 3, I used the equation (3 - x^2) for the parabola. I found out that was not the correct parabolic equation upon looking at your solution.
I am not sure why my equation cannot be used even though the vertex is not at (0,3), i.e. not lined up with the origin? The height is still three so I thought it would be okay.
## Re: Different parabola equation when finding area
@phinah: Your parabola will certainly have the correct height, but let's consider its width. Where does your parabola intersect with the x-axis?
What width at the x-axis does this give you?
X
@phinah: Your parabola will certainly have the correct height, but let's consider its width. Where does your parabola intersect with the <i>x</i>-axis?
What width at the <i>x</i>-axis does this give you?
## Re: Different parabola equation when finding area
I realized after I inquired to you that this upside down parabola is a shift one unit to the right on the x-axis.
So I came up with the -(x-1)^2 + 3.
But the parabola bottom would start at (0,2) Would that be an issue?
NOTE: The integral would be (-x^2 + 2x + 2) dx for a value of 16/3 square units.
X
I realized after I inquired to you that this upside down parabola is a shift one unit to the right on the x-axis.
So I came up with the -(x-1)^2 + 3.
But the parabola bottom would start at (0,2) Would that be an issue?
NOTE: The integral would be (-x^2 + 2x + 2) dx for a value of 16/3 square units.
## Re: Different parabola equation when finding area
@phinah: I edited your answer to include math formatting, which you are encouraged to do.
Once again, the width at the x-axis is a problem (both with your original attempt and with this revision.)
a. Where does your parabola intersect with the x-axis when we use y=3-x^2?
b. What width at the x-axis does this give you?
You may find some of the background in this article useful: How to find the equation of a quadratic function from its graph
X
@phinah: I edited your answer to include math formatting, which you are encouraged to do.
Once again, the width at the x-axis is a problem (both with your original attempt and with this revision.)
a. Where does your parabola intersect with the x-axis when we use y=3-x^2?
b. What width at the x-axis does this give you?
You may find some of the background in this article useful: <a href="https://www.intmath.com/blog/mathematics/how-to-find-the-equation-of-a-quadratic-function-from-its-graph-6070">How to find the equation of a quadratic function from its graph</a>
## Re: Different parabola equation when finding area
3 - x^2 = 0
x = +/- sqrt 3'
So my length is 3.46 when it should only be 2.
X
3 - x^2 = 0
x = +/- sqrt 3'
So my length is 3.46 when it should only be 2.
## Re: Different parabola equation when finding area
Correction
x = + or - sqrt 3
X
Correction
x = + or - sqrt 3
## Re: Different parabola equation when finding area
@Phinah: Yes, correct, so it's not possible to use y=3-x^2 for your function. But it is possible to use one that passes through (0, 3). Can you find it?
BTW, to achieve "plus or minus" in the math entry system, you enter
+-
So your answer would look like: x = +- sqrt(3).
You can see all the syntax for mathematical symbols here: ASCIIMath syntax.
X
@Phinah: Yes, correct, so it's not possible to use y=3-x^2 for your function. But it is possible to use one that passes through (0, 3). Can you find it?
BTW, to achieve "plus or minus" in the math entry system, you enter
+-
So your answer would look like: x = +- sqrt(3).
You can see all the syntax for mathematical symbols here: <a href="https://www.intmath.com/help/send-math-email-syntax.php">ASCIIMath syntax</a>.
## Re: Different parabola equation when finding area
I cannot.
X
I cannot.
## Re: Different parabola equation when finding area
Well, since we know the width of the arch must be 2 m, then we know it must go through (-1, 0) and (1, 0).
So using the general form of a quadratic,
y = ax^2 + bx + c,
you can substitute the 3 known points in and thus find the values of a, b and c.
Have a go!
X
Well, since we know the width of the arch must be 2 m, then we know it must go through (-1, 0) and (1, 0).
So using the general form of a quadratic,
y = ax^2 + bx + c,
you can substitute the 3 known points in and thus find the values of a, b and c.
Have a go!
## Re: Different parabola equation when finding area
@murray: phinah hasn't responded. Would you like me to answer this?
X
@murray: phinah hasn't responded. Would you like me to answer this?
## Re: Different parabola equation when finding area
X
@stephenB: Sure, please go ahead!
## Re: Different parabola equation when finding area
Okay, so we have the general formula
y = ax^2 + bx + c,
and the 3 points it needs to go through, (-1, 0), (1, 0) and (0, 3).
Plugging them in:
0 = a(1)^2 + b(1) + c = a + b + c ... [1]
0 = a(-1)^2 + b(-1) + c = a - b + c ... [2]
3 = a(0)^2 + b(0) + c = c ... [3]
0 = 2a + 2c which means a + c = 0 ...[4]
From [3] we know c=3, so from [4], a=-3 and then plugging those into [1] gives b=0.
That is, the parabola's equation is:
y = -3x^2 + 3
X
Okay, so we have the general formula
y = ax^2 + bx + c,
and the 3 points it needs to go through, (-1, 0), (1, 0) and (0, 3).
Plugging them in:
0 = a(1)^2 + b(1) + c = a + b + c ... [1]
0 = a(-1)^2 + b(-1) + c = a - b + c ... [2]
3 = a(0)^2 + b(0) + c = c ... [3]
0 = 2a + 2c which means a + c = 0 ...[4]
From [3] we know c=3, so from [4], a=-3 and then plugging those into [1] gives b=0.
That is, the parabola's equation is:
y = -3x^2 + 3
## Re: Different parabola equation when finding area
Thanks, stephenB.
I think if phinah tries the equation you found, he will find the area of the parabolic glass pane will the the same as in my solution.
X
Thanks, stephenB.
I think if phinah tries the equation you found, he will find the area of the parabolic glass pane will the the same as in my solution.
## Re: Different parabola equation when finding area
Phinah DID respond with the correct solution Murray BEFORE Stephen did. I do not know why it did not come through
X
Phinah DID respond with the correct solution Murray BEFORE Stephen did. I do not know why it did not come through
## Re: Different parabola equation when finding area
Hi Phinah. Sorry - it must have been a temporary database glitch or something.
X
Hi Phinah. Sorry - it must have been a temporary database glitch or something.
## Re: Different parabola equation when finding area
Okay Murray. Thank you.
X
Okay Murray. Thank you. | 2018-05-27T01:04:50 | {
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https://math.stackexchange.com/questions/1991212/partial-sum-of-the-harmonic-series-between-two-consecutive-fibonacci-numbers | # Partial sum of the harmonic series between two consecutive fibonacci numbers
I was playing around with some calculations and I noticed that the partial sum of the harmonic series: $$s_n=\sum_{k=F_n}^{F_{n+1}}\frac 1 k$$ where $F_n$ and $F_{n+1}$ are two consecutive Fibonacci numbers have some interesting properties. It is close to $\frac 1 2$ for small values of $n$ and it seems to converge to a value less than $0.5$ for large $n$. This is what I've got so far: $$\lim_{n\to\infty} s_n\approx 0.481212$$ I googled a bit to see if there is some theorems or resources for this, and found nothing. I suspect that the series might converge to a smaller number and I may have reached some computational limitations which led to the conclusion that the limit is close to $\frac 1 2$. So my questions are:
1. Can we show that the series converge to a non-zero value?
2. In case the first answer is yes, can the limit be expressed in a closed form?
In terms of the harmonic numbers $H_n$, your sequence is
$$s_n = H_{F_{n+1}} - H_{F_n-1}$$
As $n \to \infty$ it's known that $H_n = \log n + \gamma + o(1)$, so
\begin{align} s_n &= \log F_{n+1} + \gamma + o(1) - \log(F_n-1) - \gamma - o(1) \\ &= \log F_{n+1} - \log(F_n-1) + o(1). \end{align}
Now $F_m \sim \varphi^m/\sqrt{5}$, where $\varphi$ is the golden ratio, so using the fact that $a \sim b \implies \log a = \log b + o(1)$ we have
\begin{align} s_n &= \log(\varphi^{n+1}/\sqrt{5}) - \log(\varphi^{n}/\sqrt{5}) + o(1) \\ &= \log \varphi + o(1). \end{align}
In other words,
$$\lim_{n \to \infty} \sum_{k=F_n}^{F_{n+1}} \frac{1}{k} = \log \varphi.$$
• It might be better to use $s_n\approx ...$ instead of $s_n=...$ – polfosol Oct 30 '16 at 8:14
• @polfosol, no I disagree. Everything in my answer is rigorous following the definitions of $\sim$ and little-o notation. – Antonio Vargas Oct 30 '16 at 8:14
• @polfosol, see, for example, here for the definitions. – Antonio Vargas Oct 30 '16 at 8:15
• I didn't notice that. Fair enough – polfosol Oct 30 '16 at 8:15
• I will add a comment that ought to have been done : thanks for your very precise and nice answer. – Jean Marie Oct 30 '16 at 8:35
The Fibonacci numbers increase as $\phi^n$ (where $\phi$ is the golden mean $\frac{1+\sqrt{5}}{2}$), and harmonic numbers increase as $\log n$ (i.e., the natural log). Therefore, the difference between the harmonic numbers for successive Fibonacci numbers will approach $\log\phi \approx 0.481211825...$
To expand a bit, the Fibonacci numbers can be expressed as $\frac{\phi^n - (1-\phi)^n}{\sqrt{5}}$. (Try it! The fact that the equation $f(x+2) - f(x+1) - f(x) = 0$ requires a sum of powers of $\phi$ and $1-\phi$ follows from the fact that these are the solutions to the equation $x^2 - x - 1 = 0$, and the coefficients come from f(1) = f(2) = 1.) The second term vanishes, so large Fibonacci numbers can be approximated quite well as $\frac{\phi^n}{\sqrt{5}}$.
Since one definition of the natural logarithm is the integral from 1 to the parameter of the function $t^{-1}$, the harmonic numbers can be approximated as the natural logarithm, and in fact the difference approaches a constant (called $\gamma$, about 0.577). If you're not familiar with integrals, the fact that the harmonic numbers increase as a logarithm is suggested by Oresme's proof that the harmonic series diverges...
$$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \cdots > 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{16} + \cdots$$
...and it just so happens that that logarithm is the natural logarithm.
So if you accept that for very large n, the harmonic numbers approach $\log n$, and that the Fibonacci numbers approach $\frac{\phi^n}{\sqrt{5}}$, then you get for two successive...
$$\log\left(\frac{\phi^{n+1}}{\sqrt{5}}\right) - \log\left(\frac{\phi^n}{\sqrt{5}}\right) = \log\left(\frac{\phi^{n+1}}{\phi^n}\right) = \log\phi$$
($\log x - \log y = \log \frac{x}{y}$ is a natural inverse of $\frac{e^x}{e^y} = e^{x-y}$.)
• I will mark this as accepted if you add some more details ;) – polfosol Oct 30 '16 at 8:10
• ...har-r-r-rumph. – user361424 Oct 30 '16 at 8:36
• @user361424 very nice answer, a compliment that ought to have been done by the proposer before asking for "more details" ! – Jean Marie Oct 30 '16 at 8:42
• @JeanMarie It seems I am stuck with this – polfosol Oct 30 '16 at 8:44
• I think this might be the inverse fastest gun in the west problem... (this answer was originally just the first paragraph, and without the parentheticals explaining the golden mean and clarifying the natural log). – user361424 Oct 30 '16 at 8:45 | 2021-01-22T01:30:18 | {
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http://vuut.auit.pw/matrix-multiplication.html | # Matrix Multiplication
I have therefore written a matrix vector multiplication example that needs 13 seconds to run (5 seconds with. CUDA matrix multiplication with CUBLAS and Thrust. Important: We can only multiply matrices if the number of columns in the first matrix is the same as the number of rows in the second matrix. Specifically, If both a and b are 1-D arrays, it is inner product of vectors (without complex conjugation). Multiplication. In mathematics, matrix multiplication or matrix product is a binary operation that produces a matrix from two matrices with entries in a field. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. If our equation has two variables, there can be infinitely many combinations of numbers that would work. Learn: In this article, we will see how to perform matrix multiplication in python. 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One way to look at it is that the result of matrix multiplication is a table of dot products for pairs of vectors making up the entries of each matrix. , a movie), is typically very sparse…. When working with matrices, we can perform a number of matrix operations including matrix multiplication. Now the matrix multiplication is a human-defined operation that just happens-- in fact all operations are-- that happen to have neat properties. Below is a program on Matrix Multiplication. Stormy Attaway, in Matlab (Second Edition), 2012. Douglasz April 23, 2001 Abstract: Routines callable from FORTRAN and C are described which implement matrix{matrix. Abstract: This paper presents a method to analyze the powers of a given trilinear form (a special kind of algebraic constructions also called a tensor) and obtain upper bounds on the asymptotic complexity of matrix multiplication. When multiplying matrices together, the dimensions of the matrices to be multiplied must be compatible. Matrix Multiplication in Java. Grey Ballard, Aydin Buluc, James Demmel, Laura Grigori, Benjamin Lipshitz, Oded Schwartz, Sivan Toledo Jul. matrix multiplication in c free download. October 12, 2002 MULTIPLICATION MATRIX The history of this matrix goes back to the ‘70’s when my wife and I operated an individual learning. We're considering element-wise multiplication versus matrix multiplication. Clusters use in many scientific. Multiplying matrices is a little more complex than the operations you've seen so far. Matrices can be multiplied by scalar constants in a similar manner to multiplying any number of variable by a scalar constant. \end{align*} Although it may look confusing at first, the process of matrix-vector multiplication is actually quite simple. Here you can perform matrix multiplication with complex numbers online for free. Matrix chain multiplication (or Matrix Chain Ordering Problem, MCOP) is an optimization problem that to find the most efficient way to multiply given sequence of matrices. We need to tag the map( ) function output with the position so the reduce( ) function can identify the components in the different vectors. This forum may not be the best place for a discussion of the many issues involved in performance number-crunching, but I'd very much appreciate comments, suggestions, etc. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. Original value ranges are included on the X-axis. Parallel matrix multiplication • Assume p is a perfect square • Each processor gets an n/√p × n/√p chunk of data • Organize processors into rows and columns. For example X = [[1, 2], [4, 5], [3, 6]] would represent a 3x2 matrix. Don't show me this again. , of a matrix. Matrix Multiplication --- Row and Column Picture. The Wolfram Language's matrix operations handle both numeric and symbolic matrices, automatically accessing large numbers of highly efficient algorithms. This tool for multiplying 3x3 matrices. Douglasz April 23, 2001 Abstract: Routines callable from FORTRAN and C are described which implement matrix{matrix. Shruti Kaushik. Multiplication without tiling. After matrix multiplication the prepended 1 is removed. Matrix Multiplication in Excel with the MMULT function You can multiply matrices in Excel thanks to the MMULT function. These properties include the associative property, distributive property, zero and identity matrix property, and the dimension property. The second matrix is Be and it is 3x18x2 and the third matrix is del matrix and its. This subprogram takes two matrices as parameters and returns their matrix product. Below are the common core standards dealing with basic multiplication. Matrices are frequently used in programming and are used to represent graph data structure, in solving a system of linear equations and have many other applications. This lecture introduces matrix multiplication, one of the basic algebraic operations that can be performed on matrices. Hi, I want to create a VI that performs matrix multiplication for two input matrices A and B. • Given some matrices to multiply, determine the best order to multiply them so you minimize the number of single element multiplications. A matrix is a rectangular array of numbers or other mathematical objects for which operations such as addition and multiplication are defined. multMatrixes that will. We need to check this condition while implementing code without ignoring. GitHub is home to over 40 million developers working together to host and review code, manage projects, and build software together. Matrix multiplication. Numpy Matrix Multiplication - Hackr. Matrices to be multiplied do not need to be of the same order, by definition the number of columns of the first matrix must equal the number of rows of the second matrix, otherwise all row elements of the first matrix could not be multiplied by a. Sparse Matrix Multiplication on a Field-Programmable Gate Array A Major Qualifying Project Report submitted to the Faculty of the WORCESTER POLYTECHNIC INSTITUTE. CiteSeerX - Document Details (Isaac Councill, Lee Giles, Pradeep Teregowda): In this paper we present a SIMD algorithm for n n matrix multiplication on a hypercube of p processors, with time complexity of O( p =3 ), and 2 < 3. Many states have gone to a group of consistent key standards for each grade level. Scalar in which a single number is multiplied with every entry of a matrix ; Multiplication of an entire matrix by another entire matrix For the rest of the page, matrix multiplication will refer to this second category. The program follows a basic encryption algorithm that relies on mathematical properties of matrices, such as row operations, matrix multiplication, and invertible matrices. Given a sequence of matrices, find the most efficient way to multiply these matrices together. When multiplying matrices, we first need to ensure that the matrices have the same dimensions, which is the number of rows times the number of columns. In this paper, we show that novel fast matrix multiplication algorithms can significantly outperform vendor implementations of the classical algorithm and Strassen's fast algorithm on modest problem sizes and shapes. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. October 12, 2002 MULTIPLICATION MATRIX The history of this matrix goes back to the ‘70’s when my wife and I operated an individual learning. Matrix Multiplication. First let's make some data: # Make some data a = c(1,2,3) b = c(2,4,6) c = cbind(a,b) x = c(2,2,2) If we look at the output (c and x), we can see that c is a 3x2…. Study guide and practice problems on 'Matrix multiplication'. We want to define addition of matrices of the same size, and multiplication of certain "compatible" matrices. In this notebook, we'll be using Julia to investigate the efficiency of matrix multiplication algorithms. How to Multiply Matrices Faster. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. In the interests of understanding the underlying properties of the images I’m using as stimuli, I’ve been trying to learn more about the matrix transformations commonly used for image compression and image manipulation. \end{align*} Although it may look confusing at first, the process of matrix-vector multiplication is actually quite simple. The size of the matrix, as a block, is defined by the number of Rows and the number of Columns. The applications of matrices often involve the multiplication of two matrices, which requires rules for combination of the elements of the matrices. Recent posts. Stormy Attaway, in Matlab (Second Edition), 2012. Matrix Multiplication in Excel with the MMULT function You can multiply matrices in Excel thanks to the MMULT function. To multiply two matrices in C++ Programming, first ask to the user to enter the two matrix, then start multiplying the two matrices and store the multiplication result inside any variable say sum and finally store the value of sum in the third matrix say mat3. Before to this you can check different types of arrays in java and get to know how to declare and define the arrays and also get practice with adding two matrices. In other words, To multiply an m×n matrix by an n×p matrix, the ns must be the same, and the result is an m×p matrix. split happened at only one matrix which requires zero multiplications). Parallel Sparse Matrix-Vector and Matrix-Transpose-Vector Multiplication Using Compressed Sparse Blocks Aydın Buluc¸∗ [email protected] You probably know what a matrix is already if you are interested in matrix multiplication. For math, science, nutrition, history. I have therefore written a matrix vector multiplication example that needs 13 seconds to run (5 seconds with. Doerr 2 the previous seating chart example use a 1 (or yes) if the seat is occupied and a 0 (or no ) if the seat is unoccupied. Here we discuss the properties in detail. A and B must either be the same size or have sizes that are compatible (for example, A is an M-by-N matrix and B is a scalar or 1-by-N row vector). Multiplication of a matrix by a scalar. A matrix is a rectangular array of numbers or other mathematical objects for which operations such as addition and multiplication are defined. Before we go much farther, if you don’t know how matrix multiplication works, then check out Khan Academy spend the 7 minutes, then work through an example or two and make sure you have the intuition of how it works. Using Doceri). As demonstrated above, in general AB ≠BA. 376 Coppersmith-Winograd (1990) n2. The efficiency of matrix multiplication is a popular research topic given that matrices compromise large data in computer applications and other fields of study. Multiplying matrices - examples. We will illustrate matrix multiplication or matrix product by the following example. Take free online matrix multiplication classes to improve your skills and boost your performance in school. I did not expect that gcc (GCC 6. The definition of matrix multiplication indicates a row-by-column multiplication, where the entries in the i th row of A are multiplied by the corresponding entries in the j th column of B and then adding the results. Springer LNCS, 1984. Python Matrix Multiplication Program - here you will learn how to multiply one matrix to another matrix and print the multiplication result of the third matrix in python. It is built deeply into the R language. Furthermore, you can apply matrix operations such as addition, subtraction, multiplication and division:. Matrix multiplication means multiplications of two different arrays, in excel we have an inbuilt function for matrix multiplication and it is MMULT function, it takes two arrays as an argument and returns the product of two arrays, given that both the arrays should have the same number of rows and the same number of columns. One question: Why is the column of the first (left-hand) matrix colored red, along with the row of the second (right-hand) matrix?. MATMUL can do this for a variety of matrix sizes, and for different arithmetics (real, complex, double precision, integer, even logical!). Multiplication of a matrix by another matrix. Important: We can only multiply matrices if the number of columns in the first matrix is the same as the number of rows in the second matrix. When multiplying matrices, we first need to ensure that the matrices have the same dimensions, which is the number of rows times the number of columns. Join GitHub today. Row 1 X Column 1 Row 1 X Column 1 Row 1 X Column 1 Row 1 X Column 2 Row 1 X Column 2 Row 1 X – A free PowerPoint PPT presentation (displayed as a Flash slide show) on PowerShow. Just type matrix elements and click the button. Here you can perform matrix multiplication with complex numbers online for free. Just like addition works only for matrices of the same size, there are conditions for when two matrices can be multiplied but in this case it is a little bit more complicated. With no parentheses, the order of operations is left to right so A*B is calculated first, which forms a 500-by-500 matrix. To do so, we are taking input from the user for row number, column number, first matrix elements and second matrix elements. If you want to try to multiply two matrices (x and y) by each other, you'll need to make sure that the number of columns in x is equal to the number of rows in y, otherwise the equation won't work properly. As an example you'll be able to solve a series of simultaneous linear equations using Mathcad’s. In recommendation systems, the user/item rating matrix, which shows the extent of how much a user likes an item (e. Matrix algebra for beginners, Part I matrices, determinants, inverses Jeremy Gunawardena Department of Systems Biology Harvard Medical School 200 Longwood Avenue, Cambridge, MA 02115, USA. Matrix multiplication is the "messy type" because you will need to follow a certain set of procedures in order to get it right. Matrix multiplication in C. Producing a single matrix by multiplying pair of matrices (may be 2D / 3D) is called as matrix multiplication which is the binary operation in mathematics. Here is how it works 1) 2-D arrays, it returns normal product 2) Dimensions > 2, the product is trea. The manual method of multiplication procedure involves a large number of calculations especially when it comes to higher order of matrices, whereas a program in C can carry out the operations with short, simple and understandable codes. Strassen’s Matrix Multiplication on GPUs Junjie Li Sanjay Ranka Sartaj Sahni fjl3, ranka, [email protected] The MMULT function returns the matrix product of two arrays. Just like numbers and equations, you are expected to be able to manipulate matrices and perform arithmetic on multiple numbers of matrices. ) This computation requires $3n^2$ operations, while the operation count for the full matrix product is $4n^3$. 2 in the most recent edition (6e) of Finite Mathematics and Section 4. Hi everyone I am new to this site and also new to programming world can anybody help me writing C code for matrix multiplication (without using pointers). Below are the common core standards dealing with basic multiplication. If both are vectors it will return the inner product. A matrix is just a two-dimensional group of numbers. He told me about the work of Jacques Philippe Marie Binet (born February 2 1786 in Rennes and died Mai 12 1856 in Paris), who seemed to be recognized as the first to derive the rule for multiplying matrices in 1812. Pupils use the interactive to create a matrix multiplication problem using vectors. Each element in the result matrix C is the sum of element-wise multiplication of a row from A and a column from B. One way to look at it is that the result of matrix multiplication is a table of dot products for pairs of vectors making up the entries of each matrix. Operation with Matrices in Linear Algebra. One question: Why is the column of the first (left-hand) matrix colored red, along with the row of the second (right-hand) matrix?. For example weather forecasting has to done in. An efficient k-way merge lies at the heart of finding a fast parallel SpMSpV algorithm. Matrix addition, multiplication, inversion, determinant and rank calculation, transposing, bringing to diagonal, triangular form, exponentiation, solving of systems of linear equations with solution steps. Asking why matrix multiplication isn't just componentwise multiplication is an excellent question: in fact, componentwise multiplication is in some sense the most "natural" generalization of real multiplication to matrices: it satisfies all of the axioms you would expect (associativity, commutativity, existence of identity and inverses (for matrices with no 0 entries), distributivity over. Matrix Multiply, Power Calculator Solve matrix multiply and power operations step-by-step. Matrix multiplication is likely to be a source of a headache when you fail to grasp conditions and motives behind them. , Determine the way the matrices are fully parenthesized. Matrix Chain multiplication and algorithmic solution. If both arguments are 2-D they are multiplied like conventional matrices. MATRIX_A: An array of INTEGER, REAL, COMPLEX, or LOGICAL type, with a rank of one or two. Compton LA, Johnson WC Jr. plain old numbers like 3, or -5. This VHDL project is aimed to develop and implement a synthesizable matrix multiplier core, which is able to perform matrix calculation for matrices with the size of 32x32. This algebra lesson explains how to do scalar multiplication - and explains what a scalar is. We will illustrate matrix multiplication or matrix product by the following example. Matrix Formulas. It doesn ’ t just give you the answer the way your calculator would, but will actually show you the "long hand" way to multiply two numbers. Learn: In this article, we will see how to perform matrix multiplication in python. The Wolfram Language's matrix operations handle both numeric and symbolic matrices, automatically accessing large numbers of highly efficient algorithms. But to multiply a matrix by another matrix we need to do the "dot product" of rows and columns what does that mean?. Lecture Notes CMSC 251 Lecture 26: Chain Matrix Multiplication (Thursday, April 30, 1998) Read: Section 16. Using this online calculator, you will receive a detailed step-by-step solution to your problem, which will help you understand the algorithm how do matrix multiplication. Please upload a file larger than 100x100 pixels; We are experiencing some problems, please try again. Linear Algebra¶. Related Posts. Hoare (quoted by Donald Knuth). 3x3 Matrix Multiplication Calculator. Matrix Multiplication,definition,2 D array in C,Multidimensional array in C,Syntax,Syntax Example,Matrix Multiplication 2 D (dimensional) or Multidimensional Array Example Program In C. However, even when matrix multiplication is possible in both directions, results may be different. In mathematics, scalar multiplication is one of the basic operations defining a vector space in linear algebra (or more generally, a module in abstract algebra). This matrix multiplication calculator help you understand how to do matrix multiplication. Multiplication without tiling. Here’s a fact that has been rediscovered many times in many different contexts: The way you parenthesize matrix products can greatly change the time it takes to compute the product. Matrix Multiplication using arrays is very basic practice to learn for beginners to understand the concept of multidimensional matrix. Matrix multiplication is likely to be a source of a headache when you fail to grasp conditions and motives behind them. The matrix A is an n x m matrix and matrix B is an m x p matrix. Each element in the result matrix C is the sum of element-wise multiplication of a row from A and a column from B. The behavior depends on the arguments in the following way. Study guide and practice problems on 'Matrix multiplication'. | 2019-12-13T06:53:24 | {
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https://math.stackexchange.com/questions/1952939/uniqueness-of-unitarily-upper-triangular-matrix | # Uniqueness of unitarily upper triangular matrix
Suppose A is a matrix in $\mathbb{C}^{n\times n}$ with n distinct eigenvalues $\lambda_1,\dots,\lambda_n$. Then by Schur's theorem, for any fixed order of $\lambda_1,\dots,\lambda_n$, we know there exists an unitary matrix $U$ s.t. $U^*AU$ is an upper triangular matrix with $\lambda_1,\dots,\lambda_n$ of required order on the diagonal. The question is is $U$ unique? If not, what freedom do we have to choose U?
I know how to solve $A$ is unitarily diagonal (not unitarily upper triangular), then $U^*AU=\,\text{diag}(\lambda_i)\iff AU=U\,\text{diag}(\lambda_i)=[\lambda_1U_1,\dots,\lambda_nU_n]$. Then ith column of $U$ must be an eigenvector of $\lambda_i$ and $|U_i|=1$. Therefore we can choose $U$ up to multiplying a diagonal matrix whose diagonal entries have norm 1. But this method seems not fit the unitarily upper triangular case.
Key fact: if $BT=TB$ and $T$ is upper triangular with distinct diagonal entries, then $B$ is upper triangular (proof below). Now, if $$UTU^*=VTV^*,$$ then $V^*UT=TV^*U$. So $V^*U$ is an upper triangular unitary. As such, it is diagonal. Thus, $V$ is of the form $DU$ with $D$ diagonal and $|D_{kk}|=1$. In other words, the situation you observed for diagonal $A$ still occurs in general (it is essential that the diagonal entries are distinct).
Proof that $TB=BT$ implies $B$ diagonal. Consider the $n,1$ entry: $$(TB)_{n1}=\sum_kT_{nk}B_{k1}=T_{nn}B_{n1},$$ while $$(BT)_{n1}=\sum_kB_{nk}T_{k1}=B_{n1}T_{11}.$$ As $T_{11}\ne T_{nn}$, we deduce that $B_{n1}=0$. Now consider the $n,2$ entry: $$(TB)_{n2}=\sum_kT_{nk}B_{k2}=T_{nn}B_{n2},$$ while $$(BT)_{n2}=\sum_kB_{nk}T_{k2}=B_{n1}T_{12}+B_{n2}T_{22}=B_{n2}T_{22}.$$ As $T_{22}\ne T_{nn}$, we get that $B_{n2}=0$. Continuing inductively, after showing that $B_{n1},\ldots,B_{nr}=0$, we have $$(TB)_{n,r+1}=\sum_kT_{nk}B_{k,r+1}=T_{nn}B_{n,r+1},$$ while $$(BT)_{n,r+1}=\sum_kB_{nk}T_{k,r+1}=\sum_{k=1}^{r+1}B_{nk}T_{k,r+1}=B_{n,r+1}T_{r+1,r+1}.$$ As $T_{r+1,r+1}\ne T_{nn}$, we get that $B_{n,r+1}=0$. Now start doing the same with the $n-1,1$ entry, then $n-1,2$, etc. | 2019-05-27T09:08:46 | {
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https://math.stackexchange.com/questions/2095489/show-that-sum-infty-k-0-frac2-1k2k1-pi-cosh2k1-pi-2-1-4/2096064 | # Show that $\sum^\infty_{k=0} \frac{2(-1)^k}{(2k+1)\pi\cosh[(2k+1)\pi/2]}=1/4$
Title says it all.
Well, maybe some backstory.
Flipping through my past notebooks, I found this:
$$\vdots$$ $$= \sum^\infty_{k=0} \frac{2(-1)^k}{(2k+1)\pi\cosh[(2k+1)\pi/2]}\\=\frac14\quad(?)$$
[end of page]
Ah, yes. My engineering numerical analysis professor gave me one homework problem:
Make a contour plot of the steady-state temperature distribution of a square plate, with the temperature of one of its side maintained at $1$ unit, the other three sides maintained at $0$ units.
Steady-state temperature distribution $T$ follows $\nabla^2T=0$.
I was calculating the temperature of the center of the plate analytically to sanity-check my program output, when I encountered this weird-looking sum. I remember I made a spreadsheet to numerically evaluate the sum, and Excel said 0.2500000... So I jot down $\frac14$ with a question mark beside it. The next page is filled with my fruitless attempts to find out the exact sum, to convince myself that the sum is exactly $1/4$.
I posted a bounty on my social media accounts, whoever can show the proof (or disproof) gets free beer.
After a few days, the bounty was still unclaimed, but I realized, "duuuuuhhhh! I could just exploit the symmetry. If I rotate the plate $90^\circ$ three times and superpose all the temperature distribution with the original one, temperature is $1$ unit everywhere. By linearity and symmetry, the temperature of the center point is therefore $1/4$."
I treated myself a few rounds of beer, and moved on. Case closed.
Looking again at this, I now realize that I might be missing out on some summation tactics I never knew.
Anybody can show me how is this equal to $1/4$ just by algebraic manipulation?
• +1 for the story, +1 more for the solution, and +1 even more for trying to learn greater things. – Simply Beautiful Art Jan 12 '17 at 23:22
• The residue theorem might come in useful. – Michael M Jan 13 '17 at 0:23
• Thanks for the suggestion and the upvote. Probably will take me a while to work that out, as I never took a course in analysis. But now I realize that math is wonderful, there are always more beautiful things to learn. – f1garo Jan 13 '17 at 1:25
• – Math-fun Jan 13 '17 at 9:49
I am busy as hell today but this is such a nice problem (and amusingly posed question) i can't resist... :)
The key idea here is simple: Look for a function in the complex plane $f(z)$ which has poles at the correct values $z_n$ and integrate it over an appropriate contour $C$.
It turns out that the function (which is also the simplest guess i can imagine)
$$f(z)=\frac{1}{z \cos(z)\cosh(z)}$$
is the correct choice. Its poles are given by $z_0=0$, $z_k=\frac{(2 k-1)\pi}{2}$ and $\tilde{z}_k=i \frac{(2 k-1)\pi}{2}$ with $k \in \mathbb{Z/0}$. We furthermore have
$$\text{res}(z_0)=1 \\ \text{res}(z_k)=\text{res}(\tilde{z}_k)=\frac{2}{\pi}\frac{(-1)^{k}}{(2k-1)\cosh\left(\pi\frac{2k-1}{2}\right)}\\ \text{res}(z_{-k})=\text{res}(z_k)$$
since $z\cos(z)\cosh(z)\sim |z| e^{a |z|}$ with $\Re(a)>0$ as $|z|\rightarrow\infty$ we can choose a big circle traversed anticlockwise with radius choosen such that we hit no pole of $f(z)$ as the integration contour $C$ (Thanks to @Dr. MV for rigorizing this point).
Then, we get by applying the residue theorem, we get in the limit of infinte radius
$$\oint_Cf(z)dz=2\pi i\text{res}(z_0)+4\pi i \sum_{k\geq1}\text{res}(z_k)+4\pi i\sum_{k\geq1}\text{res}(\tilde{z}_k)=0$$
or
$$8\pi i \sum_{k\geq1}\text{res}(z_k)=-2\pi i$$
which can be rewritten as (after shifting $k\rightarrow k+1$)
$$\sum_{k\geq0}\frac{2}{\pi}\frac{(-1)^{k}}{(2k+1)\cosh\left(\pi\frac{2k+1}{2}\right)}=\frac{1}{4}\\ \textbf{Q.E.D.}$$
To make things clearer, here is a sketch of the integration contour $\color{blue}{C}$ and the singularities $\color{red}{z_n}$
• (+1) This is extremely close to Ramanujan original argument, I'd say it is perfectly fine! – Jack D'Aurizio Jan 13 '17 at 12:51
• @JackD'Aurizio thanks! This is a relief...still wondering how this complicated sum can be derived with nearly zero effort – tired Jan 13 '17 at 12:53
• A big (+1). I was working on this last night, had identified $f(z)$ and then floundered to find a suitable contour. Now, I feel stupid. Well done my friend! -Mark – Mark Viola Jan 13 '17 at 14:42
• I was trying to isolate the poles on the real axis by taking a rectangular contour in hopes that the top and bottom halves would cancel. They didn't. I even recognized the symmetry between the residues on real and imaginary axes. For some reason (it was getting late and I was tiring) I just couldn't see the obvious "big circle" contour was the way forward. One note; we must take the limit with discrete radii to ensure the contour doesn't hit a pole. – Mark Viola Jan 13 '17 at 14:55
• In general, the function $\pi \sec(\pi z) f(z)$ can be used to evaluate infinite sums of the form $\sum_{n \in \mathbb{Z}} (-1)^n f(n+1/2)$, while the function $\pi \tan(\pi z) f(z)$ can be used to evaluate infinite sums of the form $\sum_{n \in \mathbb{Z}} f(n+1/2)$. In some cases, of course, the contour integral won't vanish in the limit. – Random Variable Jan 13 '17 at 19:56
As mentioned by Zucker in The summation of series of hyperbolic functions, in Question 358, J. Indian Math. Soc., 4 (1912), p. 78. Ramanujan proves (through the residue theorem) that $$\sum_{n\geq 1}(-1)^{n+1}(2n-1)^{4m-1}\,\text{sech}\left[(2n-1)\frac{\pi}{2}\right] = 0$$ holds for every $m\geq 0$. Your identity then follows from the $m=0$ case.
It is truly remarkable to know that such identity can be proved through Dirichlet's problem about the eigenvalues of the Laplacian operator. Can you be more specific about the relation between such a series and the physical problem?$^{(0)}$ A brilliant piece of math might arise from there.
$^{(0)}$Update: I found the connection - such series are related with the Green function of a square.
I will show an interesting technique for deriving a similar identity.
From the Weierstrass product $$\cosh(\pi x/2) = \prod_{m\geq 0}\left(1+\frac{z^2}{(2m+1)^2}\right)\tag{1}$$ by applying $\frac{d^2}{dz^2}\log(\cdot)$ to both sides we get: $$\frac{\pi^2}{8\cosh^2(\pi x/2)}=\sum_{m\geq 0}\frac{(2m+1)^2-z^2}{((2m+1)^2+z^2)}\tag{2}$$ If we replace $z$ with $(2n+1)$ and sum over $n\geq 0$, $$\sum_{n\geq 0}\frac{\pi^2}{8\cosh^2(\pi(2n+1)/2)} = \sum_{n\geq 0}\sum_{m\geq 0}\frac{(2m+1)^2-(2n+1)^2}{((2m+1)^2+(2n+1)^2)^2} \tag{3}$$ where the RHS of $(3)$ can also be written as $$\sum_{n\geq 0}\sum_{m\geq 0}\int_{0}^{+\infty}\cos((2n+1)x)x e^{-(2m+1)x}\,dx = \sum_{n\geq 0}\int_{0}^{+\infty}\frac{x\cos((2n+1)x)}{2\sinh(x)}\,dx\tag{4}$$ or, by exploiting integration by parts, $$\sum_{n\geq 0}\int_{0}^{+\infty}\frac{x\cosh(x)-\sinh(x)}{2\sinh(x)}\cdot\frac{\sin((2n+1)x)}{2n+1}\,dx\tag{5}$$ On the other hand, $\sum_{n\geq 0}\frac{\sin((2n+1)x)}{2n+1}$ is the Fourier series of a $2\pi$-periodic rectangle wave that equals $\frac{\pi}{4}$ over $(0,\pi)$ and $-\frac{\pi}{4}$ over $(\pi,2\pi)$. That implies, by massive cancellation: $$\sum_{n\geq 0}\frac{1}{\cosh^2(\pi(2n+1)/2)}=\frac{1}{2\pi}.\tag{6}$$ On the other hand, the Fourier transform of $\frac{1}{\cosh^2(\pi x)}$ is given by by $\frac{s\sqrt{8\pi}}{\sinh(\pi s)}$.
By Poisson's summation formula, $$\sum_{n\geq 1}\frac{(-1)^{n+1}n}{\sinh(\pi n)}=\frac{1}{4\pi}.\tag{7}$$
• Hey Jack, since i can't read the source you mentioned, could you please check if their derivation fits my approach? i'm quiet unsure if i made no big mistake here (it seems to easy to be true). thanks! – tired Jan 13 '17 at 12:38
• For the pyhsical background you might be interested in this presentation: cedricthieulot.net/diva/05.pdf. The symmetry considerations mentioned by op are essentially saying that: By superposition, adding four plates with one side on a different temperature then the other three (and center temperature given by the complicated sum $S$) gives one plate with constant temperature with core temerature is trivially one. From this $S=1/4$ follows directly – tired Jan 13 '17 at 13:03
• It looks like my predictions were true: sciencedirect.com/science/article/pii/S0955799706000683 – Jack D'Aurizio Jan 13 '17 at 13:12
• @tired: thank you. I always find very pleasing to work on some problem with you. – Jack D'Aurizio Jan 13 '17 at 13:13
• D'Aurizo Yeah, this is a nice proof of concept that physicists and mathematicans CAN work together ;-) – tired Jan 13 '17 at 13:17 | 2019-05-25T19:19:46 | {
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http://mathhelpforum.com/pre-calculus/42374-please-help-me-problem-urgent-thanks.html | The graph of the function y = ax² + bx + c has a turning point at (-3,2) and passes through the point (0,5). Determine the values of a, b and c.
I know this is only a basic question, but I really would appreciate your help.
Thanks
2. $y = ax^2 + bx + c = x^2 + \frac{b}{a}x + \frac{c}{a}$
$= x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} + \frac{4ac- b^2}{4a^2} = \left({x +\frac{b}{2a}}\right)^2 + \frac{4ac- b^2}{4a^2}$
so, the turning point is at $\left({-\frac{b}{2a}, \frac{4ac-b^2}{4a^2}}\right)$
plug in and you will have 2 equations..
for the third equation:
note:
a point $(x_0,y_0)$ is on the curve $y = ax^2 + bx + c$ if and only if $y_0 = ax_0^2 + bx_0 + c$..
3. Originally Posted by tim_mannire
The graph of the function y = ax² + bx + c has a turning point at (-3,2) and passes through the point (0,5). Determine the values of a, b and c.
I know this is only a basic question, but I really would appreciate your help.
Thanks
This aproach should give you the same answer but you will only have to solve a symoltaneous equation in two variables.
You want the equation to pass through (0,5) so substitute x=0, y=5 into the equation to get y = 5 = c
You want the equation to pass through (-3,2) so substitute x=-3, y=2 into the equation to get $2 = 9*a-3b+5$
Then by knowing that at the turning point the differential of y with respect to x is zero you can write:
$y'=0=2*a*x + b$
You have two equations in two variables which can be solved by symoltaneous equations.
4. ## Here it is
y=ax^2+bx+c
since it passes through (0,5) this eq must satisfy it. So putting the values in the eq we get 5=a(0)^2+b(0)+c.which gives us c=5.
Now at turning point x=-b/2a
(if u wanna know why is it that so u may ask it in a new thread).since turning point is (-3,2) therefor x=-b/2a=-3 or b=6a. Substituting this value in eq we get y=ax^2+6ax+5 now substituting (-3,2) we get 2=9a-18a+5 or a=1/3 since b=6a therfor b=2.so finaly a=1/3,b=2,c=5
5. Originally Posted by tim_mannire
The graph of the function y = ax² + bx + c has a turning point at (-3,2) and passes through the point (0,5). Determine the values of a, b and c.
I would really appreciate any help on this problem, thanks.
First use the turning point form y = a (x - h)^2 + k. Obviously h = 0, k = 5. Substitute (-3, 2) to get a.
Now expand the turning point form to get the standard form.
Edit: Don't double post! It wastes peoples time! http://www.mathhelpforum.com/math-he...nt-thanks.html
6. Originally Posted by tim_mannire
The graph of the function y = ax² + bx + c has a turning point at (-3,2) and passes through the point (0,5). Determine the values of a, b and c.
I would really appreciate any help on this problem, thanks.
Use the following formula:
$y = a(x - p)^2 + q$
$y = a(x + 3)^2 + 2$
Passes through the point $(0;5)$
$5 = a(0 + 3)^2 + 2$
$3 = 9a$
$a = \frac{1}{3}$
$y = \frac{1}{3} (x+3)^2 + 2$
EDIT: Oops Mr F beat me to it... by 2 mins
7. Originally Posted by nikhil
y=ax^2+bx+c
since it passes through (0,5) this eq must satisfy it. So putting the values in the eq we get 5=a(0)^2+b(0)+c.which gives us c=5.
Now at turning point x=-b/2a
(if u wanna know why is it that so u may ask it in a new thread).since turning point is (-3,2) therefor x=-b/2a=-3 or b=6a. Substituting this value in eq we get y=ax^2+6ax+5 now substituting (-3,2) we get 2=9a-18a+5 or a=1/3 since b=6a therfor b=2.so finaly a=1/3,b=2,c=5
Thank you very much. very well answered. much appreciated! | 2013-12-09T15:54:50 | {
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https://physics.stackexchange.com/questions/680196/college-board-problem-conservation-of-momentum-conservation-of-energy-in-a-s | # College Board Problem: Conservation of Momentum -> Conservation of Energy in a Spring
In my AP Physics C class today, we ran into a problem written by College Board whose answer we disputed. The problem is as such :
A block of mass $$M=5.0 \ \mathrm{kg}$$ is hanging at equilibrium from an ideal spring with spring constant $$k=250 \ \mathrm{N/m}$$.An identical block is launched up into the first block. The new block is moving with a speed of $$v=5.0 \ \mathrm{m/s}$$ when it collides with and sticks to the original block. Calculate the maximum compression of the spring after the collision of the two blocks.
According to the College Board answer key, the answer is $$0.5 \ \mathrm{m}$$ :
$$p_1=p_2$$
$$Mv_0=(M+M)v_2$$
$$v_2=\frac{1}{2}v_0= \left (\frac{1}{2} \right)\left (5.0 \frac{m}{s}\right)$$
$$v_2=2.5 \frac{m}{s}$$
$$K_1 + U_1=K_2+U_2$$
$$\frac{1}{2}mv_1^2 +0=0+\frac{1}{2}kx_2^2$$
$$x_2=\sqrt{\frac{m}{k}}v_1= \sqrt{\frac{(10 \ \mathrm{kg)}}{\left(250 \frac{N}{m}\right)}} \left(2.5 \frac{m}{n}\right)$$
$$x_2=0.50 \ \mathrm{m}=50 \ \mathrm{cm}$$
However, half of us disputed this during class. We argued that, yes, $$U_2$$ includes $$\frac{1}{2}kx^2$$, but it also includes gravitational potential energy at the maximum compression (that is, when it compresses $$x$$ meters from equilibrium, the mass $$M$$ is $$x$$ meters higher above ground). Thus $$K_1+U_1=K_2+U_2$$ is $$\frac{1}{2}mv^2+0=0+\frac{1}{2}kx^2+mgx$$. When $$mgx$$ is included, $$x$$ is $$0.24 \ \mathrm{m}$$, not $$0.5 \ \mathrm{m}$$.
My physics teacher reluctantly agreed with College Board but could not give a solid explanation why. He said he would e-mail College Board, but in the meantime, I would very much appreciate any input from people who know the answer.
• I agree with your approach. It should be 2mgh since the mass rising is two blocks of 5 kg each.
– Dan
Dec 1, 2021 at 19:56
• Sorry I should have written that lol. When I plugged the numbers in and solved, I did make sure to plug in 10 kg (times 9.81 m/s/s times x). Dec 1, 2021 at 19:57
• Someone in class pointed out that perhaps because the question asks for MAXIMUM compression, they assume gravity does not exist. However, this still doesn't make sense. Assuming gravity does not exist (perhaps it is laid horizontally), the spring would have no elongation. When it is laid vertically, it elongates until the tension of the coil matches Mg. Thus, its equilibrium when horizontal (or simply without gravity) will be the length of the fully compressed spring. If the second block struck it elastically, the spring could not compress anymore. A compression of .5 meters is improbable. Dec 1, 2021 at 20:08
• Hello! It is preferable to type out screenshots or images of text; for formulae, one can use MathJax. Thanks! Dec 1, 2021 at 20:24
• Don't forget to include the initial elongation of the spring. Dec 1, 2021 at 20:36
I guess this is a homework/check-my-work problem, so by the letter of the law I should not answer, but I would argue there is broad interest in solving it correctly given that a supposedly reputable source is presenting an incorrect solution.
Here is how I would do this. Initially, the spring is stretched distance $$d=mg/k$$ below its equilibrium position. Choose the position of the hanging block as $$y=0$$, so the gravitational potential energy immediately after the collision is zero. In terms of the speed $$v=v_0/2$$ of the blocks after the collision and the mass $$m$$ of one block, the total energy immediately after the collision is \begin{align} E_i &= \frac{1}{2}(2m) v^2 + \frac{1}{2}kd^2\\ &= \frac{1}{4}mv_0^2 + \frac{1}{2}\frac{m^2g^2}{k}. \end{align} Let $$h$$ be the distance of the blocks above the equilibrium position of the spring when the blocks are at their maximum height. At this point, the blocks are at rest, so their total energy is \begin{align} E_f &= \frac{1}{2}kh^2 + 2mg(h + d)\\ &= \frac{1}{2}kh^2 + 2mgh + \frac{2m^2g^2}{k}. \end{align} Using conservation of energy, \begin{align} &\frac{1}{4}mv_0^2 + \frac{1}{2}\frac{m^2g^2}{k} = \frac{1}{2}kh^2 + mgh + \frac{2m^2g^2}{k}\\ \rightarrow &\frac{1}{2}kh^2 + 2mgh + \frac{3}{2}\frac{m^2g^2}{k} - \frac{1}{4}mv_0^2 = 0. \end{align} We can solve this quadratic equation for $$h$$ to obtain \begin{align} h = -\frac{2mg}{k} + \sqrt{\frac{m^2g^2}{k^2} + \frac{mv_0^2}{2k}}. \end{align} In terms of the given numbers, $$v_0 = 5.0\,\text{m}/\text{s}$$, $$m=5.0\,\text{kg}$$, and $$k=250\,\text{N}/\text{m}$$, we get $$\boxed{h=15\,\text{cm}.}$$
Note that if we set $$g=0$$ so that there is no gravity, we get \begin{align} h = \sqrt{\frac{mv_0^2}{2k}} = \boxed{50\,\text{cm}.} \end{align} We are left to conclude that the author of the solution was likely in free-fall at the time of its writing.
• h=15 cm? I got 0.1447. Is this the same thing? Also, how does your equation differ from the one my teacher and I made, which outputs 24 centimeters? Dec 2, 2021 at 17:23
• @SebastianPojman-Malo When I plugged the numbers into Mathematica, it spit out 14.5 something, and I rounded to two sig figs. In my answer, I'm taking into account the fact that the spring does not start at its equilibrium length.
– d_b
Dec 2, 2021 at 17:38
• That's interesting. College Board (and I) assume that it absolutely does hit the original block when the spring is at its equilibrium. Dec 2, 2021 at 17:45
• No, that is not correct. The hanging block is in equilibrium, which means the spring must be exerting an upward force on the block to keep it from falling under the influence of gravity. So the spring cannot be at its equilibrium length.
– d_b
Dec 2, 2021 at 18:54
• I agree with your h = .145 m. Note that that (h) is measured up from the un-stretched position of the spring and not from the point where the collision occurs. Dec 6, 2021 at 17:26
There is already a very good and complete answer from d_b, let me just add a comment to elucidate the possible (erroneous in this case) reasoning by which one could try to forget about gravity.
First, consider one block hanging from the spring in equilibrium. Assume that we provide it with the vertical speed $$v$$ without any other block sticking to it, so it moves up alone. What height above the initial position will it reach in such a case? Writing energy conservation (for detailed explanation look at d_b's answer, it is quite the same here - only I choose to measure height relative to the initial position of the block for reasons which will become clear shortly), we get:
$$\frac{m v^2}{2} + \frac{m^2 g^2}{2 k} = m g h + \frac{k (h-\frac{mg}{k})^2}{2}$$
Let us play a little with this equation. Open the square in the RHS and observe that $$mgh$$ cancels:
$$m g h + \frac{k (h-\frac{mg}{k})^2}{2} = mgh + \frac{k(h^2 + \frac{m^2g^2}{k^2} - \frac{2mgh}{k})}{2} = \frac{kh^2}{2} + \frac{m^2 g^2}{2k}$$
Plugging it back to the energy conservation, we observe that $$\frac{m^2 g^2}{2k}$$ cancels as well, and we are left with
$$\frac{m v^2}{2} = \frac{kh^2}{2}$$
In other words, it looks exactly like we could forget about gravity and initial elongation of the spring altogether: the answer is the same as it would have been for a horizontal spring. This is not a coincidence: what we have derived means that the total (gravitational+elastic) potential energy of this system is quadratic in deviation from the equilibrium, so it really looks like for a horizontal spring. One could also understand it graphically: potential energy of the spring is a parabola with the equilibrium position at the lowest point. If we also add gravity, a linear function is added to this parabola. The result is another parabola with the same shape but another minimum position - at the equilibrium point of the hanging block (equilibrium is always the minimum of total potential energy). If we measure all elongations compared to the new equilibrium, we could forget about both gravity and initial elongation: they compensate each other.
I think this might have been the logic of authors of the solution. However, this logic only works if we measure elongations compared to the position with minimal potential energy. If another block sticks to the first one, the equilibrium position is now shifted below, so the blocks start moving already from a non-equilibrium position and one needs to account for the potential energy of initial position as well - which was not done in the provided College Board solution. As an exercise I would suggest an interested reader to reproduce d_b's answer with this method of total potential taking the above consideration into account.
• "let me just add a comment to elucidate the possible (erroneous in this case) reasoning by which one could try to forget about gravity." Just for clarity, are you saying that College Board's ignoring gravity is erroneous? Or are you agreeing with College Board? Dec 2, 2021 at 17:31
• I disagree with the College Board answer and solution. In my answer I describe a method which they presumably used when solving the problem, but I point out that they forgot to account for the fact that equilibrium position of two blocks on a spring differs from the equilibrium position of one block. Dec 2, 2021 at 17:56
• Walk me through this. (Remember, I'm still in high school...) Does that mean that once the elastic collision happens and the identical block "sticks," the equilibrium position changes in that moment? Dec 2, 2021 at 18:00
• Firstly, if they stick, it is called an inelastic collision. Changing equilibrium position means that if one block of mass $m$ stretches the spring in equilibrium by $d$, two such blocks would stretch it by $2d$. So if in this problem we attached the second block to the first one without any velocity, they would move down to reach the new equilibrium position. To prevent confusion, let me stress that the word equilibrium here is used to describe a situation when all forces are balanced and nothing moves (not to be confused with the equilibrium length of the spring - when there are no forces). Dec 2, 2021 at 19:06
• How does conservation of momentum play into this? As I see it, (5kg)(5m/s)=(5+5kg)(v); v=2.5m/s. So according to the conservation of momentum, the two blocks stuck together will move at 2.5m/s immediately following the collision. That being said, they will decelerate because the force of gravity (2mg) will outpower the tension of the spring (mg) at that moment. Because of this, the equilibrium is disturbed, and the amount of compression is reduced. CB's answer assumes equilibrium is NOT disturbed and thus the blocks' velocity does not decelerate, right? Dec 2, 2021 at 20:27
With a 5 kg mass hanging at rest from the spring of constant k = 250 N/m, the stretch of the spring will be X = mg/k =5(9.8)/250 = 0.196 m. With (x) measured positive down from the un-stretched position, and gravitational potential energy chosen to be zero when x = 0, then the energy of the 10 kg mass is (1/2)(10)$$2.5^2$$ + (1/2)(250)$$0.196^2$$ - 10(9.8)(0.196) = (1/2)(250)$$x^2$$ -10(9.8)x. (x when v = 0) Or: 125$$x^2$$ - 98x – (31.25 +4.802 – 19.208) = 0. Solving gives x = - 0.145 m. (The other solution is for initial velocity downward). Since positive x was measured down, this negative x represents a compression of the spring above the unstretched position. Then the total rise is 0.196 + 0.145 = 0.341 m. For the record, if the moving mass is the same as the hanging mass: then mg = kX and (1/2)m$$v^2$$ + (1/2)k$$X^2$$ - (kX)X = (1/2)k$$x^2$$ - (kX)x. Or rearranged: (1/2)m$$v^2$$ = (1/2)k$$(X^2 – 2Xx + x^2) = (1/2)k(X-x)^2$$. Lets try ignoring gravity with x measured from a 10 kg hanging position (which is an additional 0.196m further down): (1/2)(10)$$2.5^2$$ + (1/2)(250)$$0.196^2$$ = (1/2)(250)$$x^2$$. Rearranging: 125$$x^2$$ = (31.25 + 4.802) Giving x = 0.537. With the starting position 0.196 m above the new equilibrium, this gives a rise of 0.341 m. The bottom line: A mass hanging from a spring defines a new equilibrium position. The kx measured from that position includes the force of gravity (but don't change the mass). Here is an alternative approach: With the stretch of the spring (x) measured positive down from the unstretched position, the net force on the hanging mass is: F = mg - kx Then dF = - k(dx). Integrating both sides gives F(x) - $$F_o$$ = -k(x - $$x_o$$). If starting from the new equilbrium then $$F_o$$ =0.
• Please consider using MathJax; in its current form this answer is mostly illegible.
– rob
Jan 23, 2022 at 13:55 | 2023-03-25T02:25:40 | {
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https://math.stackexchange.com/questions/1827223/when-to-simplify-a-quadratic-equation | # When to simplify a quadratic equation?
I had the following quadratic equation: $$38x^2 - 140x - 250 = 0$$ And before starting to solve it, I simplified it by dividing all terms by $2$: $$19x^2 -70x - 125 = 0$$ But when I solved it I got: $$x= \frac{35\pm5\sqrt{163}}{19}$$ which is not the exact solution for the original equation. So, I returned to that first equation and solved it. The roots are: $x=-\frac{25}{19}, 5$
Now my question is: How would I know when to simplify a quadratic equation before solving it? Is there something that shows me that I'm right to simplify or not?
Thank you for your help.
• You must have made an error because both those quadratics have the same roots, multiplying or dividing through by the same number doesn't change the roots – Triatticus Jun 15 '16 at 13:46
• @Triatticus: I checked with Wolfram alpha – Rafiq Jun 15 '16 at 13:47
• It should be $x=\frac{35\pm 60}{19}$, not $x=\frac{35\pm 5\sqrt{163}}{19}$. – user236182 Jun 15 '16 at 13:47
• your square root is wrong... $\sqrt{70^2+4.19.125}$ calculate again – Kushal Bhuyan Jun 15 '16 at 13:47
• I did too and got the same answers – Triatticus Jun 15 '16 at 13:48
You can always simplify a quadratic equation (by dividing out common numeric factors) before solving it. Simplifying like this is optional and is always a good idea because it generally gives you easier numbers to work with.
The error is not because you simplified, it's because there is a mistake somewhere in your work after that. I can't tell you where exactly without seeing your work, but here's how it should go:
\begin{align*} x &= \frac{70 \pm \sqrt{(-70)^2 - 4(19)(-125)}}{2(19)}\\[0.3cm] &= \frac{70 \pm \sqrt{4900 + 9500}}{38}\\[0.3cm] &= \frac{70 \pm \sqrt{14400}}{38} \\[0.3cm] &= \frac{70 \pm 120}{38} \end{align*}
When simplified you get $x = 5$ and $x = -25/19$.
When to simplifying a quadratic is up to you. Remember, multiplying both sides by a number retains the equality, the same thing with adding, subtracting and dividing.
You could've left it the way it is, without simplifying, and you would have gotten the solution as it doesn't affect the roots.
It is for last equation :$$x=\frac { 35\pm \sqrt { { 35 }^{ 2 }+125\cdot 19 } }{ 19 } =\frac { 35\pm \sqrt { 3600 } }{ 19 } =\frac { 35\pm 60 }{ 19 }$$
But using $\frac{\sqrt{b^2-4ac}}{2a}$ in the reduced equation gives the same roots . | 2019-07-22T05:34:46 | {
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https://math.stackexchange.com/questions/3213122/prove-the-sequence-f-n-frac1n21-is-a-cauchy-sequence | # Prove the sequence $f_{n} = \frac{1}{n^2+1}$ is a Cauchy sequence.
Prove the sequence $$f_{n} = \frac{1}{n^2+1}$$ is a cauchy sequence.
I'm just making sure my logic and reasoning is sound for the above proof:
Definition of cauchy sequence: $$f_n$$ is Cauchy if for all $$Ɛ$$ there is an $$M \in \mathbb{N}$$ such that for all $$n,m > M, |f_n-f_m|<Ɛ$$
$$|f_n-f_m|$$=$$|\frac{1}{n^2+1}-\frac{1}{m^2+1}|$$
by the triangle inequality:
$$\le|\frac{1}{n^2+1}|+|\frac{1}{m^2+1}|$$
$$<\frac{1}{n}+\frac{1}{m}$$
To ensure $$\frac{1}{n}+\frac{1}{m}<Ɛ$$ it suffices to have:
$$max(\frac{1}{n}+\frac{1}{m}) < \frac{Ɛ}{2}$$, so we will take $$M(Ɛ) = \lceil\frac{2}{Ɛ}\rceil$$
Now for a formal proof:
Let $$Ɛ>0$$
Define $$M(Ɛ) = \lceil\frac{2}{Ɛ}\rceil$$
Let $$n,m > M(Ɛ)$$
$$n,m > \frac{2}{Ɛ}$$
$$\frac{1}{n}<\frac{Ɛ}{2}$$ and $$\frac{1}{m}<\frac{Ɛ}{2}$$
$$|f_n - f_m| \le \frac{1}{n}+\frac{1}{m}$$
$$|f_n - f_m| < \frac{Ɛ}{2}+\frac{Ɛ}{2} = Ɛ$$
Therefore $$f_n$$ is cauchy.
Solution to the proof
• Yes, your proof is fine. – Mark May 4 at 8:53
• Okay great, I was just checking as if you see in my question above, the solution was different. I thought my answer was too simple. – user11015000 May 4 at 8:55
• Please do not use pictures for critical portions of your post. Pictures may not be legible, cannot be searched and are not viewable to some, such as those who use screen readers. – GNUSupporter 8964民主女神 地下教會 May 4 at 15:31
• The proof before the red edit is fine. – DanielWainfleet May 4 at 18:52 | 2019-10-18T16:07:33 | {
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http://math.stackexchange.com/questions/271013/approximating-continuous-functions-with-polynomials | # Approximating continuous functions with polynomials
Given $\epsilon \gt 0$ and $f \in C^{0}[0,1]$, Weierstrass says that I can find at least one (how many? probably a lot?) polynomial $P$ which approximates f uniformly: $$\sup_{x \in [0,1]} |f(x) - P(x)| \lt \epsilon$$
This means that, under the sup norm $||.||_{\infty}$, the polynomials are dense in $C^{0}[0,1]$. So, in analogy to approximating irrationals with the rationals, I would like to know:
• What can we say about the order of $P$? Or, turning this around, given that $P$ is of order $n$, how small can $\epsilon$ get?
I'm betting this should depend in some way on the properties of $f$: the intuition is that smoother functions should be somehow "better" approximated by lower-order polynomials, and less-well-behaved functions should require higher-order polynomials. But I am not sure how to formalize this.
This is probably all well-understood, but I'm not well-read on approximation theory. Any guidance would be wonderful.
-
All of the answers have been very helpful, thank you to everyone! – AndrewG Jan 5 '13 at 19:58
According to Theorem 1.2 of this paper by Sukkrasanti and Lerdkasem, we have the following result (with impressively great generality, I might add):
Let $f: [0,1]^p \rightarrow \mathbb{R}^q$ be bounded. Then there exists $C$ such that $$\| f - B_n(f) \|_{\infty} \le C \omega(1/\sqrt{n})$$ where $\omega(\delta) := \sup_{\|t_1 - t_2\| \le \delta} \|f(t_1) - f(t_2)\|$ and $B_n(f)$ is the $n$th Bernstein polynomial of $f$.
I do not claim to have read the paper myself. Notice that if $f$ is indeed continuous, then by uniform continuity, as $n \rightarrow \infty$, $\omega(1/\sqrt{n})$ must go to zero. So the convergence rate is related to how rapidly $f$ can vary, as intuition would already suggest.
-
Define, $B_i^n(x)=\binom{n}{i}x^i(1-x)^{n-i}$, nice fact,
• Partition of the unity $\sum_{i=0}^{n}B_i^n(x)=1$ this is easy to see:
See $1= x+(1-x)$ then, $$1=(x+(1-x))^n=\sum_{i=0}^{n}\binom{n}{i}x^i(1-x)^{n-i}=\sum_{i=0}^{n}B_i^n(x).$$
Without loss of generality suppose $a=0$ and $b=1$. We need of the following estimative:
\begin{eqnarray*} \sum_{i=0}^{n}\left(x-\frac in\right)^{2}B_i^n(x)&=&x^2\sum_{i=0}^{n}B_i^n(x)-2x\sum_{i=0}^{n}\dfrac{i}{n}B_i^n(x)+\sum_{i=0}^{n}\left(\dfrac{i}{n}\right)^2 B_i^n(x)\\&=& x^2-2x^2+\dfrac{1}{n^2}\sum_{i=0}^{n}n(n-1)\dfrac{i^2-i +i}{n(n-1)} B_i^n(x)\\&=& -x^2+ \dfrac{1}{n}\sum_{i=0}^{n}\dfrac{i}{n}B_i^n(x)+\dfrac{n(n-1)}{n^2}\sum_{i=0}^{n}\dfrac{i(i-1)}{n(n-1)}B_i^n(x)\\&=& -x^2+\dfrac{x}{n}+\dfrac{n-1}{n}x^2\\&=& \dfrac{x-x^2}{n}\leqslant\dfrac{1}{4n}. \end{eqnarray*} Now, since $f$ is continuous we have that $f$ is uniformly continuous,because $[0,1]$ is compact. So for each $\epsilon$ exists $\delta>0$ such that for all $x,y\in [0.1]$ with $|x-y|<\delta$ implies $|f(x)-f(y)|<\epsilon$. Since $f$ have yours extreme extreme values in $[0,1]$ then exists a constant $M$ such that $|f(x)|\leqslant M$ for all $x\in[0,1].$
Consider the Berstein polynomial of $f$, $$P_n(x)=\sum_{i=0}^{n}f\left(\dfrac{i}{n}\right)B_i^n(x);$$ since $\displaystyle\sum_{i=0}^{n}B_i^n(x)=1$, we have $f(x)=\displaystyle\sum_{i=0}^{n}f(x)B_i^n(x)$ then,
\begin{eqnarray*} \left|f(x)-P_n(x)\right|&\leqslant& \sum_{i=0}^{n}\left|f(x)-f\left(\dfrac{i}{n}\right)\right|B_i^n(x)\\&=& \sum_{S_1}\left|f(x)-f\left(\dfrac{i}{n}\right)\right|B_i^n(x)+\sum_{S_2}\left|f(x)-f\left(\dfrac{i}{n}\right)\right|B_i^n(x) \end{eqnarray*}
where $S_1=\{0\leqslant i\leqslant n ~;~ |x-\frac{i}{n}|<\delta\}$ and $S_2=\{0\leqslant i\leqslant n~;~ |x-\frac{i}{n}|\geqslant \delta\}.$ analyzing each sum separately $$\sum_{S_1}\left|f(x)-f\left(\dfrac{i}{n}\right)\right|B_i^n(x)\leqslant \sum_{S_1}\epsilon|B_i^n(x)\leqslant \epsilon\sum_{i=0}^{n}B_i(x)=\epsilon$$
\begin{eqnarray*} \sum_{S_2}\left|f(x)-f\left(\dfrac{i}{n}\right)\right|B_i^n(x)&\leqslant& 2M\sum_{S_2}B_i^n(x)\leqslant \sum_{S_2}\dfrac{(x-\frac{i}{n})^2}{\delta^2}B_i^n(x)\\&\leqslant&\dfrac{2M}{\delta^2}\sum_{i=0}^{n}(x-\frac{i}{n})^2 B_i^n(x)\leqslant \dfrac{M}{2\delta^2 n} \end{eqnarray*}
since $\dfrac{M}{2\delta^2 n}<\epsilon$ for $n$ large, we demonstrate the desired.
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Nice answer. By the way, what does "daí" mean? – Antonio Vargas Jan 5 '13 at 16:50
@AntonioVargas: I'm sorry, is the habit of writing in portuguese daí=so or then – user27456 Jan 5 '13 at 16:58
@EduardoSiva: Everything looks fine, except that you have not defined $B_{i}$. I suppose that it denotes a Bernstein polynomial, am I right? – Haskell Curry Jan 5 '13 at 17:34
@HaskellCurry: I'll try to fix it – user27456 Jan 5 '13 at 17:43
A near minimax approximation to a continuous function, say, $f:[-1,1]\to\mathbb{R}$, can be obtained with Chebyshev polynomials. You may look up the details in this book chapter, of which theorem 3.10 says that if $p$ is a Chebyshev interpolation polynomial for such continuous $f:[-1,1]\to\mathbb{R}$ at $n+1$ points, then $\|f-p\|\sim C\,\omega(\frac1n)\log n$ as $n\to\infty$, where the norm is the maximum norm, $\omega(\cdot)$ denotes the modulus of continuity and $C$ is a constant.
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Since you tagged your question by "reference-request" I will give only them (instead of copying a lot). Jackson inequality ("direct theorem"), and Bernstein theorem ("converse theorem"). See also the references therein. (My humble opinion they are good quality books. Typing into Google these keywords you will obtain a lot of links, including many free downloadable lecture notes, maybe books.)
- | 2015-07-05T15:27:16 | {
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https://www.physicsforums.com/threads/calculating-work.324424/ | # Calculating work
1. Jul 10, 2009
### jeff1evesque
1. The problem statement, all variables and given/known data
Given the attached picture, Calculate the work required to go around the contour shown if the force is $$\vec{F} = y\hat{x} - x^{2} \hat{y}$$
First by Contour integration:
Work = $$\oint \vec{F} \bullet \vec{dl} = \int_{a}^{b} (\vec{F} \bullet \hat{x})dx + \int_{b}^{c} (\vec{F} \bullet \hat{y})dy - \int_{c}^{d} (\vec{F} \bullet \hat{x})dx - \int_{d}^{a} (\vec{F} \bullet \hat{y})dy$$
$$= \int_{1}^{2} (2) dx + \int_{2}^{3} (4)dy - \int_{2}^{1} (3)dx - \int_{3}^{2} (1)dy = 0$$
I think the answer above is suppose to be -4 not 0.
Now by Stokes Theorem:
When this was done with "Stokes theorem", we get the following:
Work = $$\oint \vec{F} \bullet \vec{dl} \equiv \int \int (\nabla \times \hat{F})\bullet \vec{ds} = - \int \int (2x + 1)( \hat{z} \bullet \hat{z})ds = \int_{2}^{3} \int_{1}^{2} (2x + 1)dxdy = -4$$
Note: $$( \hat{z} \bullet \hat{z}) = 1$$ since $$( \hat{z} \bullet \vec{ds}) = ds$$
Question:
When I look at this, Stokes method seems apparently correct, but the method before it seems wrong. Can someone help me find my error?
Thanks,
JL
#### Attached Files:
• ###### StokePic.bmp
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Last edited: Jul 10, 2009
2. Jul 11, 2009
### E_M_C
Your attachment is still pending approval, but if we're speaking of the work done around a closed path, it's certainly not zero. This is due to the fact that the force field is not conservative. While the force is position dependent (as is required by a conservative force), it's curl, $$\nabla \times F$$, is not the zero vector.
3. Jul 11, 2009
### HallsofIvy
Staff Emeritus
The contour is the square with corners at (1, 2), (2, 2), (2, 3), and (1, 3).
Why do you have minus signs on the last two integrals? Assuming that your contour goes from point a to point b, then to point c, then to point d, then back to point a, these should all be "+". (Although the integrals themselves might be 0.)
On the first leg, with x= t, y= 2, the integral is $\int_1^2 2 dx$ as you say. On the second leg, with x= 2, y= t, the integral is $\int_2^3 (-4) dt$. Did you forget the "-" on "$-x^2\vec{j}$"? On the third leg, with x= t, y= 3, the integral is $\int_2^1 3dt= -\int_1^2 3dt$. You have the wrong sign. On the fourth leg, with x= 1, y= t, the integral is $\int_3^2(-1) dt= \int_2^3 dt$. That is what you have, but, I think, because of two canceling sign errors!. Evaluating, the integral is 2- 4+ (-3)+ (1)= -4.
4. Jul 11, 2009
### jeff1evesque
Thanks Halls. | 2017-11-17T21:38:53 | {
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http://math.stackexchange.com/questions/243559/picking-random-numbers-as-long-as-they-keep-decreasing-expected-number-of-numbe/243567 | # Picking random numbers as long as they keep decreasing. Expected number of numbers you pick?
Pick a random number (evenly distributed) between $0$ and $1$. Continue picking random numbers as long as they keep decreasing; stop picking when you obtain a number that is greater than the previous one you picked. What is the expected number of numbers you pick?
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One of my friend send me this question,& he sends me the ans.. – UnknownController Nov 24 '12 at 6:25
Please, try to make the title of your question more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. – Julian Kuelshammer Nov 24 '12 at 6:41
@JulianKuelshammer thank you...next time I'll give more informative title.. – UnknownController Nov 24 '12 at 6:45
There're $n!$ ways of arranging $n$ numbers, supposing that there're $n$ picks, then the first $(n-1)$ picks are in descending order, there are $(n-1)$ ways of choosing the first $(n-1)$ numbers and thus the probability of picking just $n$ nimbers is $\cfrac {n-1}{n!}$ and the expected value is $$E=\sum^{\infty}_{n=2} n\cdot\cfrac {n-1}{n!}= \sum^{\infty}_{n=2} \cfrac {1}{(n-2)!}=\sum^{\infty}_{n=0} \cfrac {1}{n!}=e=2.718281828459...$$
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thank you my friend for tried to help,I just post ans. in comment see that :) – UnknownController Nov 24 '12 at 6:27
The following solutions uses indicator random variables.
Fix an $n$, and imagine doing the experiment only $n$ times. For $i\le n$, let $X_j=1$ if $X_1\gt X_2 \gt \cdots >X_j$, and let $X_j=0$ otherwise. Then $\Pr(X_j=1)=\frac{1}{j!}$ and therefore $E(X_j)=\frac{1}{j!}$.
If $Y_n$ is the length of the longest monotone decreasing sequence that starts with the first number chosen the beginning, then $Y_n=X_1 +X_2+\cdots +X_n$. Thus, by the linearity of expectation, $$E(Y_n)=E(X_1)+E(X_2)+E(X_3)+\cdots+E(X_n)= 1+\frac{1}{2!}+\frac{1}{3!}+\cdots +\frac{1}{n!}.$$ As $n\to \infty$, this approaches $e-1$.
But your random variable counts all the picks up to and including the first pick that breaks the monotonicity. So your random variable has expectation $(e-1)+1=e$.
- | 2016-02-06T08:01:02 | {
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https://math.stackexchange.com/questions/2585274/evaluating-int-fracx21x2-2x22dx-section-7-3-28-in-stewart-calc | # Evaluating $\int\frac{x^2+1}{(x^2-2x+2)^2}dx$ - Section 7.3, #28 in Stewart Calculus 8th Ed.
I'd like to evaluate $$I=\int\frac{x^2+1}{(x^2-2x+2)^2}dx.$$
According to WolframAlpha, the answer is $$\frac{1}{2}\bigg(\frac{x-3}{x^2-2x+2}-3\tan^{-1}(1-x)\bigg)+C.$$
To ensure that my answer is correct, I'd like to match it with this answer.
First I complete the square by writing the denominator as $((x-1)^2+1)^2$, then use a trig-sub $x-1=\tan\theta$ so $dx=\sec^2\theta d\theta.$
Then $$I=\int\frac{(\tan\theta +1)^2+1}{\sec^4\theta}\sec^2\theta d\theta\\=\int\frac{\tan^2\theta+2\tan\theta+2}{\sec^2\theta}d\theta\\=\int\sin^2\theta +2\sin\theta\cos\theta+2\cos^2\theta d\theta\\=\int\bigg(\frac{1}{2}-\frac{\cos(2\theta)}{2}\bigg)+\sin(2\theta)+(1+\cos(2\theta))d\theta\\=\int\frac{3}{2}+\frac{\cos(2\theta)}{2}+\sin(2\theta)d\theta\\=\frac{3\theta}{2}+\frac{\sin(2\theta)}{4}-\frac{\cos(2\theta)}{2}+C\\=\frac{3\theta}{2}+\frac{\sin\theta \cos\theta}{2}-\frac{1-2\sin^2\theta}{2}+C.$$
Here's a picture corresponding to my trig-sub.
From the picture, I conclude $$I=\frac{3\tan^{-1}(x-1)}{2}+\frac{1}{2}\frac{x-1}{\sqrt{x^2-2x+2}}\frac{1}{\sqrt{x^2-2x+2}}- \frac{1-2\frac{(x-1)^2}{x^2-2x+2}}{2}+C\\=\frac{-3\tan^{-1}(1-x)}{2}+\frac{x-1}{2(x^2-2x+2)}-\frac{1}{2}+\frac{(x-1)^2}{x^2-2x+2}+C.$$
Although the $\arctan$ portion of the answer matches Wolfram, the remaining portion does not (since there is a square in the numerator after finding a common denominator). Where am I making the mistake?
I've also tried using the identity $\cos(2\theta)= 2\cos^2\theta-1$ instead of $\cos(2\theta)= 1-2\sin^2\theta$, but the same issue occurs.
Your work is correct; the only difference between your antiderivative and the one calculated by WolframAlpha is a constant of integration, $1/2$. You may verify that $$\frac{(x-1)^2}{x^2-2 x+2}+\frac{x-1}{2 \left(x^2-2 x+2\right)}-\frac{1}{2} = \frac{x^2-x-1}{2 \left(x^2-2 x+2\right)},$$ and it follows that $$\frac{x^2-x-1}{2 \left(x^2-2 x+2\right)}-\frac{x-3}{2 \left(x^2-2 x+2\right)} = \frac{1}{2}.$$ | 2021-09-25T01:21:10 | {
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http://math.stackexchange.com/questions/355151/combinatorial-explanation-to-following-recurrence-relation-a-n-2-a-n-1-a/355159 | # Combinatorial explanation to following recurrence relation $a_n = 2 a_{n-1} + a_{n-2}$
Question was the following:
$a_n$ is the number of ternary strings (strings of 0,1,2) which contain no consecutive zeros and no consecutive ones. Find a formula for $a_n$?
By brute force, I found a recurrence relation for $a_n$ as the following:
$a_n = 2 a_{n-1} + a_{n-2}$
Now I am wondering if there is a good combinatorial explanation/proof to my recurrence relation. Can you see something?
Best Regards
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Terminological note: those are ternary strings, not turning strings. – Brian M. Scott Apr 8 '13 at 20:00
@BrianM.Scott thank you for correction – Xentius Apr 8 '13 at 20:02
Let $b_{i,n}$ be the number of strings of length $n$ starting with $i\in\{0,1,2\}$. Then by your rules $$a_n = b_{0,n} + b_{1,n} + b_{2,n} = (b_{1,n-1} + b_{2,n-1}) + (b_{0,n-1} + b_{2,n-1}) + a_{n-1} = 2a_{n-1} + a_{n-2}.$$
ADDITION: This transformation can be translated into a proof by bijection. Let $X_n$ be the set of all sequences of length $n$. We define a map $$(X_{n-1} \times \{A,B\}) \cup X_{n-2} \quad\to\quad X_{n}$$ by ("$\cdot$" denotes concatenation; $\mu$ a sequence in $X_{n-2}$ and $\lambda$ a sequence in $X_{n-1}$):
$(\lambda,A) \mapsto 2\cdot\lambda$
$(\lambda,B)\mapsto\begin{cases} 1\cdot \lambda & \text{if }\lambda\text{ starts with }0\text{ or }2\\ 0\cdot \lambda & \text{if }\lambda\text{ starts with }1 \end{cases}$
$\mu \mapsto 0\cdot 2\cdot \mu$
This is a bijection, essentially because $(\lambda,A)$ gives all strings in $X_n$ starting with $2$, $(\lambda,B)$ gives all strings in $X_n$ starting with $10$, $01$ or $12$, and $\mu$ gives all strings starting with $02$. So $$\left|(X_{n-1} \times \{A,B\}) \cup X_{n-2}\right| = \left|X_n\right|,$$ which is $$2a_{n-1} + a_{n-2} = a_n.$$
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I guess what @Xentius meant by a combinatorial proof is not this. He seems to be looking for an answer like in this question he asked: math.stackexchange.com/questions/340905/… – Amadeus Bachmann Apr 8 '13 at 19:42
@AmadeusBachmann exactly! – Xentius Apr 8 '13 at 19:45
@AmadeusBachmann: I think that azimut's equality has pure (and nice!) combinatorial sense. – xen Apr 8 '13 at 19:55
I'm a bit baffled why my argumentation shouldn't be combinatorial. @Xentius could you please specify what kind of proof you are looking for? – azimut Apr 8 '13 at 20:02
@xen I did not say azimut's proof is not combinatorial. I just wanted to say it may not be what Xentius looking for according to his previous questions. But of course only one to confirm this is Xentius. – Amadeus Bachmann Apr 8 '13 at 20:06
Call a ternary string with no consecutive zeroes and no consecutive ones a good string. For $k=0,1,2$ let $a_n^k$ be the number of good strings of length $n$ ending in $k$. Then it’s clear that
\begin{align*} &a_n^0=a_{n-1}^1+a_{n-1}^2\\ &a_n^1=a_{n-1}^0+a_{n-1}^2\\ &a_n^2=a_{n-1}^0+a_{n-1}^1+a_{n-1}^2=a_{n-1}\;. \end{align*}\tag{1}
And now we can make the following computation:
\begin{align*} a_n&=a_n^0+a_n^1+a_n^2\\ &=a_n^0+a_n^1+a_{n-1}\\ &=a_{n-1}^1+a_{n-1}^0+2a_{n-1}^2+a_{n-1}\\ &=(a_{n-1}^0+a_{n-1}^1+a_{n-1}^2)+a_{n-1}^2+a_{n-1}\\ &=a_{n-1}+a_{n-1}^2+a_{n-1}\\ &=2a_{n-1}+a_{n-1}^2\\ &=2a_{n-1}+a_{n-2} \end{align*}
by the third line of $(1)$.
In more purely combinatorial terms the term $a_{n-1}$ in $a_n=a_n^0+a_n^1+a_{n-1}$ comes from the fact that each good string of length $n$ ending in $2$ is obtained by appending a $2$ to one of the $a_{n-1}$ good strings of length $n-1$. The rest is a little more complicated. If a good string of length $n-1$ ends in $0$ or $2$, I can append a $1$ to it to get a good string of length $n$ ending in $1$; that covers all good strings of length $n$ ending in $01$ or $21$, which is all good strings of length $n$ ending in $1$. Similarly, if a good string of length $n-1$ ends in $1$ or $2$, I can append a $0$ to get a good string of length $n$ ending in $0$, and that accounts for all good strings of length $n$ ending in $0$. Thus, to get the good strings of length $n$ ending in $0$ or $1$ I’ve used each good string of length $n-1$ ending in $0$ once, each good string of length $n-1$ ending in $1$ once, and each good string of length $n-1$ ending in $2$ twice. In other words, I’ve used each good string of length $n-1$ at least once, and I’ve used those ending in $2$ twice. Ignoring for a moment the double use of those ending in $2$, that accounts for another $a_{n-1}$ strings of length $n$. Finally, we already know that each good string of length $n-1$ ending in $2$ is just the extension by a $2$ of a good string of length $n-2$, so the double use of good strings of length $n-1$ ending in $2$ gives us another $a_{n-2}$ good strings of length $n$. Put the pieces together, and we see that $a_n=2a_{n-1}+a_{n-2}$.
- | 2015-09-04T19:13:28 | {
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https://math.stackexchange.com/questions/2438737/formal-way-to-evaluate-this-limit | # Formal way to evaluate this limit
I've some doubt about the formal way to evaluate the following limit:
$$\lim_{n\to\ +\infty}{\left(1-\frac{1}{n}\right)^{n^2}}$$
I know that
$$\lim_{n\to\ +\infty}{\left(1-\frac{1}{n}\right)^{n}}=e^{-1}$$
my calculus teacher says that we can't evaluate the limit by pieces, so we can't (in general) say that
$$\lim_{n\to\ +\infty}{\left(1-\frac{1}{n}\right)^{n^2}}=\lim_{n\to\ +\infty}{\left(e^{-1}\right)^{n}}=0$$ even if in this case it is the right answer.
So I'm asking a formal way to solve this, my approach is:
Say that the limit of a product is the product of the limits, so I can say that
$$\lim_{n\to\ +\infty}{\left(1-\frac{1}{n}\right)^{n^2}}=\lim_{n\to\ +\infty}{e^{n^2\ln\left(1-\frac{1}{n}\right)}}=e^{\lim_{n\to\ +\infty}{n^2\ln\left(1-\frac{1}{n}\right)}}=e^{\lim_{n\to\ +\infty}{n^2}{\lim_{n\to\ +\infty}\ln\left(1-\frac{1}{n}\right)}}=e^{\lim_{n\to\ +\infty}{-n}}=0$$
Can I do this (I've used the continuity of the exponential function) or am I still doing the limit by pieces when I've substituted $ln\left(1-\frac{1}{n}\right)$ with $-\frac{1}{n}$?
Furthermore, can I use the theorem of the product even if there is the indeterminate form $0\cdot(+\infty$)?
• Implicitly, you used an equivalent near $0$: we know that $\ln(1+u)\sim_0 u$. All usual functions have equivalents near $0$. This way saves many useless computations unrelated to the indetermination. Sep 21, 2017 at 10:58
• +1 for your question as well as your teacher. Unfortunately most book authors / instructors manage with lot of hand waving in teaching calculus. Sep 21, 2017 at 12:12
No, you can't. However, you may note that, as $n\to +\infty$, $$n^2\ln\left(1-\frac{1}{n}\right)=n\ln\left(\left(1-\frac{1}{n}\right)^n\right)\to +\infty \cdot \ln(e^{-1})=-\infty.$$ Hence $$\lim_{n\to\ +\infty}{\left(1-\frac{1}{n}\right)^{n^2}}=\lim_{n\to\ +\infty}{e^{n^2\ln\left(1-\frac{1}{n}\right)}}=e^{-\infty}=0.$$ Another way. Since $\lim_{n\to\ +\infty}{\left(1-\frac{1}{n}\right)^{n}}=e^{-1}\in (0,1/2)$ then, by definition of limit, there is $N$ such that for all $n\geq N$, $\left(1-\frac{1}{n}\right)^{n}<1/2$. Hence, for $n\geq N$, $$0<{\left(1-\frac{1}{n}\right)^{n^2}}=\left(\left(1-\frac{1}{n}\right)^{n}\right)^n<\frac{1}{2^n}.$$ Then use the Squeeze theorem.
• @Mephlip Any further doubt? Sep 21, 2017 at 11:48
One way of doing it is by comparison. For any fixed $k\in \Bbb N$, we have $$0\leq \lim_{n\to \infty}\left(1-\frac 1n\right)^{n^2}\leq\lim_{n\to \infty}\left(1-\frac1n\right)^{nk} = e^{-k}$$
The trick to do what you're trying to do is, after making your substitution, you also factor in a correction so that you leave the formula unchanged:
$$\lim_{n \to \infty} \left(1 - \frac{1}{n} \right)^{n^2} = \lim_{n \to \infty} \left( e^{-1} \cdot \frac{\left(1 - \frac{1}{n}\right)^{n}}{e^{-1}} \right)^n = \lim_{n \to \infty} e^{-n} \cdot \left( \frac{\left(1 - \frac{1}{n}\right)^{n}}{e^{-1}} \right)^n \\= \lim_{n \to \infty}\left( e^{-n} \right) \cdot \lim_{n \to \infty} \left( \frac{\left(1 - \frac{1}{n}\right)^{n}}{e^{-1}} \right)^n$$
assuming, of course, the limits exist and the product is defined. If you could show that second limit was $1$, then you'd compute the limit as $\infty \cdot 1 = \infty$ and be done.
The problem, however, is that the second limit is a $1^{\infty}$ form! And it does this in a way that makes it so you can't immediately determine what the value should be: the value of the limit is a race between how fast the base approaches $1$ versus how fast the exponent reaches $\infty$.
So, there isn't any immediate shortcut here; you have to do more work. In fact, that limit equals $e^{-1/2}$, so the 'obvious' guesses are, in fact, wrong!
You need a more sophisticated approximation to make something like this work. The thing to do can more easily be demonstrated in your alternate approach: the Taylor series for logarithm implies
$$\ln(1 + x) = x + O(x^2)$$
So, in particular,
$$\lim_{n \to \infty} n^2 \ln\left(1 - \frac{1}{n} \right) = \lim_{n \to \infty} n^2 \left( -\frac{1}{n} + O(n^{-2}) \right) = \lim_{n \to\ infty} -n + O(1) = -\infty$$
If you're not comfortable with $O$ notation, you could just use the next term of the Taylor series. If you put your mind to it, you should be able to obtain the exact value of the limit
$$\lim_{n \to \infty} n^2 \left(\ln\left(1 - \frac{1}{n} \right) + \frac{1}{n} \right)$$
so you could alternately compute the limit as
$$\lim_{n \to \infty} n^2 \ln\left(1 - \frac{1}{n} \right) = \left( \lim_{n \to \infty} n^2 \left(-\frac{1}{n}\right)\right) + \left( \lim_{n \to \infty} n^2 \left( \ln\left(1 - \frac{1}{n} \right) + \frac{1}{n} \right) \right)$$
Late answer since it was tagged as duplicate.
An elementary way without any $$e$$ or logarithm could be:
$$\begin{eqnarray*}\left(1-\frac{1}{n}\right)^{n^2} & = & \frac 1{\left(1+\frac{1}{n-1}\right)^{n^2}}\\ & \stackrel{\text{binomial formula}}{\leq} & \frac 1{\frac{n^2}{n-1}}\\ & < & \frac 1 n \stackrel{n\to \infty}{\longrightarrow} 0 \end{eqnarray*}$$ | 2023-02-09T12:19:53 | {
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https://math.stackexchange.com/questions/1721620/is-it-correct-to-say-all-extrema-happen-at-critical-points-but-not-all-critical | # Is it correct to say all extrema happen at critical points but not all critical points are extrema?
Is it correct to say all extrema happen at critical points but not all critical points are extrema?
This question is for single variable calculus. But it would be great if someone could also provide insight to how well this statement generalizes to more complex functions (multi-variable functions, vector-valued functions, etc)
• Extrema need not be critical points. They can also be the "end-points" in a given domain. This is what is called "absolute extrema". – Airdish Mar 31 '16 at 10:56
• All interior extrema are critical points. Of course, critical points need not be extrema. – Sangchul Lee Mar 31 '16 at 10:58
Every local extremum in the interior of the domain of a differentiable function is neccesarily a critical point, i.e. $f'(x)=0$ is a necessary condition for $x$ to be a local extremum. There are critical points which are not local extrema.
Note that I'm stressing local extremum in the interior here. To account for global extrema, one needs to consider the behavior on the boundary of the domain as well.
Consider the function $$f:[-1,1]\rightarrow \mathbb R, x\mapsto x^2.$$ $x_0=0$ is a critical point and indeed a local extremum (a minimum). For the global extrema note that $x^2 \leq 1$ for $x\in [-1,1]$ so that $f(1)=f(-1)=1$ and $x_1=1, x_2=-1$ are points where $f$ attains a (global) maximum, but they are not critical points.
Similarly, for $$f:[-1,1]\rightarrow \mathbb R, x\mapsto x^3.$$ $x_0=0$ is a critical point, but there's no extremum. Maximum and Minimum are attained at $x_1=1$ and $x_2=-1$, neither of which are critical points.
• When you say boundary points. You just mean boundary to domain (builtin or contrived, right?) What I mean is $2$ would be an inherent boundary point to $x/(x-2)$ and contrived ones would be $[-1,1]$. Sorry if this is not correct terminology. Or am I just thinking too hard? And the inherent ones are already covered by the $f'(c)=0$ or DNE clause? – AlanSTACK Apr 1 '16 at 1:13
• @Alan: I mean boundary points of the domain where the function is defined. – Roland Apr 1 '16 at 5:23
If you assume $f$ is a continuous, real-valued function on a closed, bounded interval $[a, b]$, then:
• $f$ has at least one absolute maximum and at least one absolute minimum. (The Extreme Value Theorem.)
• If $f$ has a local (relative) extremum at an interior point $x_{0}$, then either $f'(x_{0}) = 0$ or $f'(x_{0})$ does not exist. (I.e., $x_{0}$ is a critical point of $f$, modulo your definition of "critical point".)
• If your definition of "critical point" includes endpoints of $[a, b]$, then "yes", every local extremum of $f$ is a critical point of $f$. (This convention is reasonable, since strictly speaking $f'$ does not exist at an endpoint of an interval. On the other hand, this convention is not universal.)
For the other direction:
• An interior critical point can fail to be a local extremum. Examples (with $x_{0} = 0$) include $f(x) = x^{3}$ and $$g(x) = \begin{cases} x & x < 0, \\ 2x & x \geq 0, \end{cases}$$ both viewed as functions on some interval containing $0$ in the interior. (In the first example, $f'(0) = 0$; in the second, $g'(0)$ does not exist.)
• An endpoint can fail to be a local extremum, e.g. $$f(x) = \begin{cases} x^{2} \sin(1/x) & 0 < x, \\ 0 & x = 0. \end{cases}$$
There are analogous results for real-valued functions of more than one variable, but they're a little more work to state, because not every "connected subset" of the plane (say) is a formal analog of a closed, bounded interval. Loosely, though, a continuous function $f$ on a closed, bounded subset $D$ of $\mathbf{R}^{n}$ has at least one absolute maximum and at least one absolute minimum in $D$, and these must occur either at (i) an interior point $x_{0}$ of $D$ where the total derivative $Df(x_{0})$ is zero; (ii) an interior point $x_{0}$ where $Df(x_{0})$ does not exist; or (iii) a boundary point of $D$.
There is no analog for vector-valued functions because the concepts of "maximum" and "minimum" make no sense. (If $n > 1$, the set $\mathbf{R}^{n}$ is not ordered in any way compatible with vector addition and scalar multiplication.)
• For the left endpoint $a$, the differential quotient $\lim_{h \rightarrow 0}\frac{f(a+h)-f(a)}{h}$ is a limit which may (or may not) exist, and should be regarded as a one-sided limit (i.e. only for $h>0$). If $f'(a)$ exists or not is dependent on the existence of this one-sided limit. – Roland Mar 31 '16 at 11:38
• @Roland: Agreed, but in my experience that criterion of differentiability at an endpoint is (also) a convention, and (with some justification) not a universal one. (If it matters, I'm not advocating either convention of endpoint differentiability, just noting that as stated, the OP's question can only be answered with qualifications.) – Andrew D. Hwang Mar 31 '16 at 11:55
• yet you write 'strictly speaking $f′$ does not exist at an endpoint of an interval', which is wrong for functions which are differentiable in a larger open domain which includes the interval $[a,b]$. – Roland Mar 31 '16 at 12:02
• Strictly speaking, "$f$ is differentiable at $x_{0}$" involves a two-sided limit at $x_{0}$. If you extend the definition at an endpoint by requiring only a one-sided limit, that's fine, but you're no longer invoking the usual definition of differentiability from one-variable calculus. (If you're saying the extended definition is universal, I have no mathematical grounds to argue otherwise. I'm merely saying that this extension is not tacit in elementary calculus. For instance, $f(x) = |x|$ is not differentiable at $0$, but is differentiable on $[0, \infty)$ by your definition.) – Andrew D. Hwang Mar 31 '16 at 12:36
• @Roland It is explicitly stated in some real analysis textbooks (ex. Apostol) that for a derivative $f'(c)$ to exist, $f$ must be defined in an open interval containing $c$. Other authors, like Rudin, do not explicitly require this yet mention the issue of differentiability at endpoints in passing. The motivation for this requirement is far beyond the scope of introductory calculus. See the answer by Willie Wong here: math.stackexchange.com/questions/126176/… – MathematicsStudent1122 Apr 2 '16 at 5:07
Yes.
You can analyse this multi-variable example $z = x^2 - y^2$.
This has a critical point which is not an extrema. These points called saddle points.
• What's your function? What's your domain? – Roland Mar 31 '16 at 11:40
• @Roland $f: \mathbb R^2 \rightarrow \mathbb R$ with $z = f(x,y) = x^2 - y^2$. – crbah Mar 31 '16 at 11:44
• What if the domain is $\{(x,y): x^2 +y^2 \leq 1\}$? Is the answer still yes? – Roland Mar 31 '16 at 11:49
• @Roland Yes. You can see its graph on en.wikipedia.org/wiki/Saddle_point – crbah Mar 31 '16 at 11:53
• My comment for the restricted domain is not about the critical point which is not an extremum. It's about the extremal points which are not critical points ($(x,y)=(1,0)$ and $(x,y)=(0,1)$). – Roland Mar 31 '16 at 12:00 | 2019-06-26T08:25:38 | {
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https://www.physicsforums.com/threads/finding-required-resistor-to-reduce-voltage.533682/ | # Homework Help: Finding required resistor to reduce voltage
1. Sep 25, 2011
### deus_ex_86
1. The problem statement, all variables and given/known data
Your task is to design a "dummy" indicator light for your car. The cheapest bulb available is a #47 incandescent lamp rated at 6.3V for 150 mA. The problem is, your car produces 13.5 V when it's on. Choose the nearest 10% resistor that will reduce the voltage across the bulb to 6.3V.
2. Relevant equations
V = IR
3. The attempt at a solution
I modelled the bulb as a resistor:
6.3 V = .150 A * R
Which gives R = 42 $\Omega$.
From there, I plugged this value into a new circuit with a 13.5 V source instead of a 6.3 V source, and added the unknown resistor to the equation:
13.5 V = .150 A * R + 6.3 V
7.2 V = .150 A * R
R = 48 $\Omega$.
The nearest 10% resistor value is 47 $\Omega$, but this wouldn't reduce the voltage across the bulb to 6.3 V. The next closest 10% resistor is 56 $\Omega$. It just seems that this value is too high ... Did I follow this process correctly? What would you choose?
2. Sep 25, 2011
### lewando
48-ohms look right. What are you getting for the new bulb voltage (with the 47-ohm R)?
3. Sep 25, 2011
### deus_ex_86
48 ohms is the theoretical value, but I have to choose a 10% resistor to use. When I use the 47 ohm resistor, I end up with 6.37 V across the bulb.
4. Sep 25, 2011
### lewando
Close enough!
[edit: To the letter of the question the solution does not exist. To the intent of the question, 47-ohms is a good answer, if you stipulate your voltage deviation, you should be fine.]
5. Sep 27, 2011
### zgozvrm
With a 47Ω resistor, the bulb will have approximately 6.37 volts across it. Light bulbs are not very picky, so this will work fine.
With the 56Ω resistor, the bulb will only have about 5.79 volts across it. Still, a workable voltage, but it's the current that we're really concerned with...
The 47Ω resistor will limit the current through the bulb to about 151.69 mA which is just over the design rating. The bulb will burn slightly brighter (although probably not noticeably).
The 56Ω resistor will limit the current down to about 137.76 mA and the bulb will burn slightly dimmer. You're only looking at about an 8% loss of current though, so either will work.
I'd use the 47Ω resistor.
6. Sep 27, 2011
### deus_ex_86
Thanks, so would my TA, as I got a 100% on the assignment. :)
7. Sep 27, 2011
### 2milehi
With ±10% resistors there could be a resistor that is 42.3Ω and car battery voltage can spike into the 14.0 volt range. That would cause 166 ma to flow through the bulb or about 10.7% more. Since power is I²R, there can be an extra 23% power dissipated through the bulb which will shorten its life.
IMO the design would have had some breathing room built in if you went with the 56Ω bulb. | 2018-08-20T11:05:53 | {
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http://yucoding.blogspot.com/2013/03/leetcode-question-102-sqrtx.html | ### leetcode Question 102: Sqrt(x)
Sqrt(x)
Implement int sqrt(int x).
Compute and return the square root of x.
Analysis:
According to Newton's Method(http://en.wikipedia.org/wiki/Newton's_method), we can use
to get the sqrt(x).
Code:
class Solution {
public:
int sqrt(int x) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if (x==0) {return 0;}
if (x==1) {return 1;}
double x0 = 1;
double x1;
while (true){
x1 = (x0+ x/x0)/2;
if (abs(x1-x0)<1){return x1;}
x0=x1;
}
}
};
Another solution: the binary search approach is a more general way of solving this problem. One thing you need to consider is the length of the input, since taking the mid of a big value and computing its square may overflow the int type.
We can use "long long" , which have a max value 2^63-1.
The max of an int is 2^15-1
The max of a long is 2^31-1
class Solution {
public:
int sqrt(int x) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
long long high = x;
long long low = 0;
if (x<=0) {return 0;}
if (x==1) {return 1;}
while (high-low >1){
long long mid = low + (high-low)/2;
if (mid*mid<=x){low = mid;}
else {high = mid;}
}
return low;
}
};
1. I believe in the interview, they won't expect you to remember Newtown's method. The solution they are expecting is a Binary Search method.
The binary search approach is added to this post.
2. This comment has been removed by the author.
2. Quoting on one line above "The max of an int is 2^15-1", int in most cases is 32 bit, unless for some specific platform.
3. Is there a reason that you use mid = low + (high-low)/2; rather than (high + low)/2; ?
1. Yes, using high+low /2 may cause overflow errors, when high and low both are very big, high+low may exceed and cause the overflow error, while high-low will not.
2. This comment has been removed by the author.
3. Nice solution man. Although I think you need not worry about overflowing: both high and low are within the range [0 INT_MAX], so (high + low) is at most 2*INT_MAX, which can perfectly fit into a 'long long int' integer. So as long as you define high, low and mid as 'long long int', an overflow will never occur.
Or better yet, even if both low and high are defined as 'int', we are still good. Note that it is guaranteed that (int)sqrt(x) < (int)(x/2) for any x >= 6, therefore, for any large enough number, in the first iteration, we have low == 0, so (low + high) is always <= INT_MAX (no overflow). then we always have mid*mid > x, so it is always 'high = mid;' that will be executed in the first iteration, i.e. high <= INT_MAX / 2. From there on, both low and high will be within [0 INT_MAX/2] so (low + high) will never exceed INT_MAX.
Of course 'mid' must be defined as 'long long int' since we need to perform 'mid*mid', which may well exceed INT_MAX. But if we limit the initial value of 'high' to min(std::sqrt(INT_MAX) + 1, x/2), then it is safe to just define 'mid' as 'int'. std::sqrt(INT_MAX) is a constant that can be precomputed so it does not violate the requirement of this problem.
4. should u cast long to int int the end?
5. Decent improvement as (abs(x1-x0)<1) for Newton's Method. | 2017-07-26T12:40:00 | {
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https://math.stackexchange.com/questions/537172/a-oplus-b-a-oplus-c-imply-b-c | $A \oplus B = A \oplus C$ imply $B = C$?
I don't quite yet understand how $\oplus$ (xor) works yet. I know that fundamentally in terms of truth tables it means only 1 value(p or q) can be true, but not both. But when it comes to solving problems with them or proving equalities I have no idea how to use $\oplus$.
For example: I'm trying to do a problem in which I have to prove or disprove with a counterexample whether or not $A \oplus B = A \oplus C$ implies $B = C$ is true.
I know that the venn diagram of $\oplus$ in this case includes the regions of A and B excluding the areas they overlap. And similarly it includes regions of A and C but not the areas they overlap. It would look something like this:
I feel the statement above would be true just by looking at the venn diagram since the area ABC is included in the $\oplus$, but I'm not sure if that's an adequate enough proof. On the other hand, I could be completely wrong about my reasoning.
Also just for clarity's sake: Would $A\cup B = A \cup C$ and $A \cap B = A \cap C$ be proven in a similar way to show whether or not the conditions imply $B = C$? A counterexample/ proof of this would be appreciated as well.
• For still more fun show that the power set of a set $X$, together with the xor operation, is a group. And, of course, groups have the cancellation law you ask about. – GEdgar Oct 23 '13 at 17:41
• The xor being, in that case, called the symmetric difference. @user101279 : By the way, $\oplus$ is addition modulo $2$. – xavierm02 Oct 23 '13 at 18:04
• Remark also that xor is the addition operation when we regard a Boolean algebra as essentially the same as a Boolean ring. – user43208 Oct 23 '13 at 18:05
• Think of $\oplus$ as $\neq$. – copper.hat Oct 23 '13 at 18:06
• If we omit the question about intersection and union in the last paragraph, this is the same as math.stackexchange.com/questions/294460/… – Martin Sleziak Aug 30 '14 at 15:30
Think of $\oplus$ as $\neq$. That is $A \oplus B$ iff $A \neq B$.
Note that $A \oplus A$ is always false, and $\text{False}\oplus A = A$.
Then $A \oplus (A \oplus B) = (A \oplus A) \oplus B = \text{False} \oplus B = B$.
Similarly, $A \oplus (A \oplus C) = C$, hence $B=C$.
Aside: A 'cute' (as in amusing but not of any practical significance) use of $\oplus$ is to swap the values of two bit variables in a programming language without using an intermediate variable: \begin{eqnarray} x = y \oplus x \\ y = y \oplus x \\ x = y \oplus x \\ \end{eqnarray} Show that the values of $x,y$ are swapped!
Hint: $A\oplus(A\oplus B)=(A\oplus A)\oplus B = B$.
And of course $A\cup B=A\cup C$ does not imply $B=C$ (consider the case $B=A\ne \emptyset = C$). And $A\cap B=A\cap C$ does not imply $B=C$ either (consider the case $A=\emptyset$)
Hint: $\oplus$ is associative with unit $\emptyset$, and $A \oplus A = \emptyset$. Does this give you an idea for canceling?
This can be done using a simple calculation. But I don't know how to interpret your question, so I'll give two answers. :-)
I'm assuming $\;A,B,C\;$ are booleans. I will write $\;\not\equiv\;$ instead of $\;\oplus\;$, and $\;\equiv\;$ instead of $\;=\;$ on booleans.
First, note that $\;\equiv\;$ and $\;\not\equiv\;$ are not only both associative, but they are also mutually associative. Therefore no parentheses are needed in the following calculation.
We can now simplify $\;A \oplus B = A \oplus C\;$ as follows: \begin{align} & A \not\equiv B \equiv A \not\equiv C \\ \equiv & \;\;\;\;\;\text{"rearrange"} \\ & A \not\equiv A \equiv B \not\equiv C \\ \equiv & \;\;\;\;\;\text{"simplify"} \\ & \text{false} \equiv B \not\equiv C \\ \equiv & \;\;\;\;\;\text{"simplify"} \\ & B \equiv C \\ \end{align}
If instead $\;A,B,C\;$ are sets, and your $\;\oplus\;$ is the symmetric difference of two sets (which is normally written as $\;\triangle\;$), then the proof is slightly longer, but with essentially the same structure.
The simplest definition of symmetric difference is $$x \in A \oplus B \equiv x \in A \not\equiv x \in B$$
We can expand the definitions and simplify using logic, as follows: \begin{align} & A \oplus B = A \oplus C \\ \equiv & \;\;\;\;\;\text{"set extensionality; definition of $\;\oplus\;$, twice"} \\ & \langle \forall x :: x \in A \not\equiv x \in B \equiv x \in A \not\equiv x \in C \rangle \\ \equiv & \;\;\;\;\;\text{"logic: rearrange"} \\ & \langle \forall x :: x \in A \not\equiv x \in A \equiv x \in B \not\equiv x \in C \rangle \\ \equiv & \;\;\;\;\;\text{"logic: simplify"} \\ & \langle \forall x :: \text{false} \equiv x \in B \not\equiv x \in C \rangle \\ \equiv & \;\;\;\;\;\text{"logic: simplify"} \\ & \langle \forall x :: x \in B \equiv x \in C \rangle \\ \equiv & \;\;\;\;\;\text{"set extensionality"} \\ & B = C \\ \end{align}
In both cases, we have found a stronger conclusion than was asked: we proved equivalence of the two expressions.
Also just for clarity's sake: Would $A\cup B = A \cup C$ and $A \cap B = A \cap C$ be proven in a similar way to show whether or not the conditions imply $B = C$? A counterexample/ proof of this would be appreciated as well.
$A\cup B=A\cup C$ $\Rightarrow$ $B=C$ is not true in general.
Counterexample: Take any non-empty set $A$ and also take $B=A$ and $C=\emptyset$. Then $A\cup B=A\cup C=A$, but $B\ne C$.
$A\cap B=A\cap C$ $\Rightarrow$ $B=C$ is not true in general.
Take some element $x\notin A$ and put $B=A$, $C=A\cup\{x\}$. Then $A\cap B=A\cap C=A$, but $B\ne C$. | 2019-08-24T13:05:28 | {
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https://math.stackexchange.com/questions/3229985/biased-binary-search-complexity | # Biased binary search complexity
We know Binary search on a set of n element array performs O(log(n)).
We have this recursive equation through which the search space is reduced by half in each iteration, after a single comparison.
T(n) = T(n/2) + 1
Either by applying Master's Theorem or analytically we arrive at the complexity of it as log(n)
If I introduce bias, in the search by unequally partitioning the array instead of equal halves, How would the worst case time complexity change?
The unequal partitions are t * n and (1-t) * n where 0<=t<=1
This is a mathematics question as it involves asymptotic analysis and hence I don't wish my question to be downvoted.
• If I understand what you're asking correctly, wouldn't the worst case complexity be $O(n)$ due to it potentially using a partition size of $1$ and $n - 1$, requiring on average $\frac{n}{2}$ checks? Note it should never be worse than $O(n)$ as any type of biased binary search should only ever check each value at the most $1$ time. – John Omielan May 18 at 2:15
• Correct. But I want to come across with an analytical way to calculate the worst case complexity for any particular value of t. In general, it is evident that the worst case complexities are O(n) when t=0 and O(log(n)) when t=1/2. – Argha Chakraborty May 18 at 6:04
• There's nothing special about $t = \frac{1}{2}$. As long as you reduce the problem size by a factor of $t$, for any $0 \leq t < 1$, you will get logarithmic running time. – Joppy May 18 at 7:16
• That is what I want to prove . – Argha Chakraborty May 18 at 7:33
In the worst case scenarios, where $$t = 0$$ or $$t = 1$$, and assuming that at least one value will always be checked, then as I stated in my comment, the average number of checks would be $$\frac{n}{2}$$ and the maximum number would be $$n$$. However, for $$0 \lt t \lt n$$, where $$tn \gg 1$$, then as indicated in the comment by Joppy, the number of steps in the worst case scenario would be logarithmic.
To see this, let
$$u = \max(t, 1-t) \tag{1}\label{eq1}$$
The worst case would be that the value is always in the larger partition after each step, with $$n_i$$ being the size of this partition after $$i$$ steps, and with $$n_0 = n$$. Thus,
$$n_i = un_{i-1} \; \forall \; i \ge 1 \tag{2}\label{eq2}$$
As such, $$n_1 = un$$, $$n_2 = un_1 = u^2 n$$, etc., to get that
$$n_i = u^i n \; \forall \; i \ge 0 \tag{3}\label{eq3}$$
The search will eventually end after $$m$$ steps where
$$u^m n \approx 1 \tag{4}\label{eq4}$$
Note this is similar to the concept of the amount of time for exponential decay to cause a substance to eventually disappear, with the worst case number of unbiased binary searches being about the number of half-life periods. To use a value $$\gt 1$$ for logarithms, let
$$v = \frac{1}{u} \tag{5}\label{eq5}$$
so \eqref{eq4} becomes
$$n \approx v^m \; \implies \; m \approx \log_v(n) = \log_v(e)\ln(n) \tag{6}\label{eq6}$$
For $$t = u = \frac{1}{2}$$, then $$v = 2$$ and \eqref{eq6} gives $$m \approx (1.44269)\ln(n)$$. With a relatively extreme case of $$t = \frac{1}{1001}$$, then $$u = \frac{1000}{1001}$$ and $$v = 1.001$$, so \eqref{eq6} gives $$m \approx (1000.49991)\ln(n)$$, i.e., about $$693.5$$ times as large. Nonetheless, it's still generally considerably faster than just checking each value since, for $$n = 10^{12}$$, \eqref{eq6} gives $$m \approx (1000.49991)(27.63102) \approx 27644.8$$, so it's about $$\frac{10^{12}}{27644.8} \approx 3.61731 \times 10^7$$ times faster.
The cases $$t=0$$ and $$t=1$$ do not correspond to a terminating algorithm. Otherwise, the behavior remains logarithmic.
In the worst case, the array is every time reduced according to the smaller of $$t$$ and $$1-t$$; WLOG, let $$t$$.
After $$k$$ steps, we approximately reduce from $$n$$ to $$nt^k\sim 1$$, so that $$k\sim-\log_tn$$. | 2019-10-23T17:55:56 | {
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https://math.stackexchange.com/questions/1034665/how-to-put-a-matrix-in-jordan-canonical-form-when-it-has-a-multiple-eigenvalue | # How to put a matrix in Jordan canonical form, when it has a multiple eigenvalue?
Put the matrix $$\begin{bmatrix} 3 & -4\\ 1 & -1\end{bmatrix}$$ in Jordan Canonical Form. Moreover, find the appropriate transition matrix to the basis in which the original matrix assumes its Jordan form.
I'm having a lot of trouble with this. I know that the eigenvalue has multiplicity two and is $\lambda = 1$. I can find the first eigenvector, which is:
\begin{bmatrix} 2 \\ 1 \\ \end{bmatrix} I'm having trouble finding the second since both eigenvalues tell us the same thing. But I'm not nearly as concerned about the eigenvectors as I am about what to do after.
If anyone could explain thoroughly the next steps involved (not necessarily the answer but how to obtain it), I would be forever grateful. This homework is in 2 days and it may determine my grade letter.
• Yes, I know the only eigenvalues are one. The problem is finding the eigenvectors corresponding to 1 and more importantly how to proceed. – Arbitrationer Nov 23 '14 at 5:43
• You don't seem to understand that when vadim wrote, "the eigenspace corresponding to $\lambda=1$ is one-dimensional", he was telling you there is only the one eigenvector that you have found (and scalar multiples of that eigenvector, but they don't help). What you need is a generalized eigenvector; a vector $w$ such that $(A-\lambda I)w=v$, where $v$ is the eigenvector you have found. Do you not have a text, or some notes to follow? Anyway, "generalized eigenvector" is the keyphrase: go look it up, and then post an answer when you have worked out how to solve the problem. – Gerry Myerson Nov 23 '14 at 6:16
To put a matrix in Jordan normal form requires to know three matrices such that $A=PJP^{-1}$ i.e: a matrix $P^{-1}$ that transform to the canonical basis $(\mathbf{i},\mathbf{j},\mathbf{k})$ to a new basis in which the matrix $J$ represents the transformation of a vector $\mathbf{v}$ such that the transformed vector $\mathbf{v'}$ is the same as we find when we transform $\mathbf{v}$ with $A$ in the canonical basis. Last the matrix $P$ returns this result to the canonical basis.
As noted in OP the matrix $A$ has eigenvalues $\lambda_1=\lambda_2=1$ and a single eigenvector $$\mathbf{u_1}=\left[ \begin{array}{cccc} 2\\ 1 \end {array} \right]$$
So the main problem is to find another vector that completes the new basis.
To find such a vector notes that all vectors $\mathbf{x}$ such that $(A-\lambda I)\mathbf{x}=0$ are transformed in the eigenspace generated by the eigenvector $\mathbf{u_1}$, so we want a vector $\mathbf{u_2}$ such that $(A-\lambda I)\mathbf{u_2} \ne 0$, and the way to do this is to find a vector such that : $(A-\lambda I)\mathbf{u_2} = \mathbf{u_1}$. (Note that this equation is the same as $(A-\lambda I)^2\mathbf{u_2}= 0$).
Solving in our case we find: $$\left[ \begin{array}{cccc} 2&-4\\ 1&-2 \end {array} \right] \left[ \begin{array}{cccc} x\\ y \end {array} \right]= \left[ \begin{array}{cccc} 2\\ 1 \end {array} \right]$$
so that the components $x$ and $y$ of the searched vector must satisfies $x-2y=1$, and we can find the vector $$\mathbf{u_2}= \left[ \begin{array}{cccc} 1\\ 0 \end {array} \right]$$ So the matrix $P$ we are searching is $$P=[\mathbf{u_1},\mathbf{u_2}]= \left[ \begin{array}{cccc} 2&1\\ 1&0 \end {array} \right]$$ And the inverse is: $$P^{-1}= \left[ \begin{array}{cccc} 0&1\\ 1&-2 \end {array} \right]$$ The matrix $J$ is a typical Jordan block, with the eigenvalues as diagonal elements and an entry $1$ up-right them: $$J= \left[ \begin{array}{cccc} 1&1\\ 0&1 \end {array} \right]$$ and we can easily verify that $A=PJP^{-1}$.
The characteristic polynomial $\det (A - \lambda I) = (\lambda -1)^2.$ when the dimension of the null space(1) of an eigenvalue($\lambda = 1$) is less than the algebraic multiplicity(2), you need to find generalized eigenvectors. in this instance, you need to solve $(A - I_2) \left( \begin{array}{l} x \cr y\end{array} \right) = \left( \begin{array}{l} 2 \cr 1 \end{array} \right).$ this gives you $x = 1, y = 0$
with respect to the basis $\{ \left( \begin{array}{l} 2 \cr 1 \end{array} \right), \left( \begin{array}{l} 1 \cr 0 \end{array} \right) \}$ you transformation is represented by the Jordan canonical form $\left( \begin{array}{ll} 1 & 1 \cr 0 & 1 \end{array} \right).$
• Might have been better to let OP do some of the work. – Gerry Myerson Nov 23 '14 at 23:27
• @myerson, perhaps i should have not done all the work. – abel Nov 23 '14 at 23:57 | 2019-04-22T16:35:15 | {
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https://math.stackexchange.com/questions/1696134/can-a-gradient-vector-field-with-no-equilibria-point-in-every-direction | # Can a gradient vector field with no equilibria point in every direction?
Suppose that $V:\mathbb{R}^n \to \mathbb{R}$ is a smooth function such that $\nabla V : \mathbb{R}^n \to \mathbb{R}^n$ has no equilibria (i.e. $\forall x \in \mathbb{R}^n : \nabla V (x) \not = 0$). Under these hypotheses, is it possible that $\nabla V (x)$ can point in every direction?
To be more precise, under the above hypotheses the map $$\mathbb{R}^n \to \mathbb{S}^{n-1}$$ $$x \mapsto \frac{\nabla V(x)}{\|\nabla V (x)\|}$$ is well-defined. Is it impossible for such a map to be surjective? If not, what is a counterexample?
• I believe this is possible from $R^2$ onwards. Can you write down a function for which one gradient flow line makes an $S$ shape? Can you make sure that at some distance from this $S$ the flowlines it straighten to vertical straight lines? If you can do that you can mirror this construction and obtain the right function. – Thomas Rot Mar 15 '16 at 14:18
• That's an interesting idea. It seems plausible, but I'm having some trouble actually writing down such a function. – Matthew Kvalheim Mar 15 '16 at 16:55
• The more I think about it, I'm not sure a function satisfying the requirements @ThomasRot suggested exists. – Matthew Kvalheim Mar 15 '16 at 23:57
• A cool question! Also, it might be received well in MathOverflow, but given that you started a bounty I would think it better to let the bounty period expire before we consider migration. If you are in a hurry, respond to this and/or flag again. – Jyrki Lahtonen Mar 19 '16 at 6:53
A quick solution for $n=2$, and an explanation of where it came from: $$\nabla (e^x \sin y) = (e^x \sin y, e^x \cos y).$$ The right hand side is never zero, but does assume every nonzero value in $\mathbb{R}^2$.
Motivation: If $f: \mathbb{C} \to \mathbb{C}$ is holomorphic, then $\nabla(\mathrm{Re}(f)) = (\mathrm{Re}(f'), -\mathrm{Im}(f'))$. I looked for an $f'$ (namely $e^z$) which takes $\mathbb{C}$ onto $\mathbb{C}_{\neq 0}$, and then integrated it to find $f$.
Inspired by this, let's try $$\nabla(e^{x_0} \cos(x_1^2+x_2^2+\cdots + x_n^2)) =$$ $$e^{x_0} (\cos(x_1^2+\cdots + x_n^2), - x_1 \sin(x_1^2+\cdots + x_n^2), \cdots, -x_n \sin(x_1^2+\cdots + x_n^2)).$$
First, we note that the gradient is not zero. The first coordinate only vanishes if $x_1^2+ \cdots + x_n^2$ is of the form $(2k+1) \pi$. But, in this case, at least one of $x_1$, $x_2$, ..., $x_n$ are nonzero; say $x_j$. Then $- x_j \sin(x_1^2+\cdots + x_n^2) = \pm x_j \neq 0$.
Now, let's take a nonzero vector $(v_0, \ldots, v_n)$. We need to go through cases.
If $v_0 = 0$, choose $(x_1, \ldots, x_n)$ proportional to $(v_1, \ldots, v_n)$ and such that $\sin(x_1^2+\cdots+x_n^2)$ has the right sign.
If $v_1 = \cdots = v_n = 0$ and $(-1)^k v_0>0$, take $x_1^2+\cdots+x_n^2 = k \pi$.
If neither of those cases holds, we'll take $(x_1, \ldots, x_n)$ of the form $(t v_1, \ldots, t v_n)$ for some $t$ to be determined soon. Set $s = x_1^2+ \cdots + x_n^2$. Then our vector points in direction $\pm (- \cot(t^2 s), v_1, v_2, \ldots, v_n)$. Since cotangent is surjective, we can choose $t$ such that $\cot (t^2 s) = v_0$, and we can get the sign right.
• Thank you for the insightful answer. Your counterexample for $\mathbb{R}^2$ and the motivation you gave are quite helpful. – Matthew Kvalheim Mar 24 '16 at 22:31
It is possible to have a vector field without equilibrium point and takes every possible directions in $\mathbb{R^n}$ for any $n \ge 2$.
The proof for the 2-dimension case is given below. From that, vector fields for higher dimension cases can be constructed.
Let $\rho : (-\epsilon,\infty) \to \mathbb{R}$ be any $C^\infty$ function such that
1. $\rho(t)$ is even over $(-\epsilon,\epsilon)$ and $\rho(0) = \rho'(0) = 0$.
2. $|r\rho'(r)| < 1$ for all $r \ge 0$.
3. for some $r_c > 0$, $\rho(r_c) = 2\pi$.
4. for some $r_m > 0$, $N \in \mathbb{Z}$, $\rho(r) = 2\pi N$ for all $r \ge r_m$.
For any point $(x,y) \in \mathbb{R}^2$, let $(r,\theta)$ be the corresponding polar coordinates.
Let $\hat{e}_x, \hat{e}_y, \hat{e}_r, \hat{e}_\theta$ be unit vectors along the $x$, $y$, $r$ and $\theta$ directions respectively.
Consider following function $$U(x,y) = r\cos(\theta - \rho(r)) = x\cos\rho(r) + y\sin\rho(r)$$ Its gradient is given by:
\begin{align} \vec{\nabla} U &= \hat{e}_x \left( \cos\rho + \frac{x\rho'}{r}( -x \sin\rho + y\cos\rho )\right) + \hat{e}_y \left( \sin\rho + \frac{y\rho'}{r}( -x \sin\rho + y\cos\rho )\right)\\ &= \hat{e}_r\left(\cos(\theta - \rho) + r\rho'\sin(\theta-\rho)\right) -\hat{e}_\theta\left(\sin(\theta-\rho)\right) \end{align} From these two expressions, it is easy to deduce
• $\vec{\nabla} U$ is well defined at $(0,0)$ because $\rho'(0) = 0$.
• $\vec{\nabla} U(0,0) = \hat{e}_x$ because $\rho(0) = \rho'(0) = 0$.
• $\vec{\nabla} U \ne 0$ for $r > 0$ because \begin{align} |\vec{\nabla} U| &= |\hat{e}_r\left(\cos(\theta - \rho) + r\rho'\sin(\theta-\rho)\right) -\hat{e}_\theta\left(\sin(\theta-\rho)\right)|\\ &\ge |\hat{e}_r\left(\cos(\theta - \rho)\right) -\hat{e}_\theta\left(\sin(\theta-\rho)\right)| - |\hat{e}_e \left(r\rho'\sin(\theta-\rho)\right)|\\ &\ge 1 - |r\rho'| > 0 \end{align}
Combine these, we find $\nabla U \ne 0$ over $\mathbb{R}^2$. This allow us to define following unit vector field globally on $\mathbb{R}^2$.
$$\hat{u}(x,y) = \frac{\vec{\nabla}U(x,y)}{\left|\vec{\nabla}U(x,y)\right|}$$
Over the curve $\displaystyle\;\gamma : [0,r_c] \ni t \quad\mapsto\quad (t\cos\rho(t),t\sin\rho(t)) \in \mathbb{R}^2\;$, we have $$\hat{u}(t) = \hat{e}_r = \hat{e}_x \cos\rho(t) + \hat{e}_y\sin\rho(t)$$ As we move from $\gamma(0)$ to $\gamma(r_c)$ along $\gamma$, $\hat{u}(\gamma(t))$ will cover the whole $S^1$ at least once. The map $\hat{u}$ is surjective on $\gamma$ and hence on $B(0,r_c)$ and $\mathbb{R}^2$. This means $U$ is a counter-example of $V$ for the 2-dimenisonal case.
As an illustration, following is a picture of what $U(x,y)$ will typically look like under this construction.
$\hspace0.75in$
This particular $U(x,y)$ is computed using
\rho(t) = 2\pi + (f(t)-2\pi)g(t) \quad\text{ where }\quad \begin{align} f(t) &= \log\sqrt{(e^{4\pi}-1)t^2 + 1}\\ g(t) &= \begin{cases} 1, & t \le 1\\ 1 - e^{-1/(e^{t-1} -1)}, & t > 1\end{cases} \end{align} and plotted over the region $|x|,|y| \le 2$. This $\rho(t)$ satisfies the first three requirement in the beginning of this answer with $r_c = 1$. For $r \gg r_c$, $\rho(t)$ tends to $2\pi$ exponentially as $t \to \infty$.
Back to the case of higher dimensions. Let's say we do have a $\rho$ that satisfies all four conditions. Consider following function:
$$V(x_1,x_2,\ldots,x_n) \stackrel{def}{=} U\left( x_1, \sqrt{x_2^2 + x_3^2 + \cdots + x_n^2} - (r_m+1)\right)$$
Since $B(0,r_m) \supset B(0,r_c)$, the map $\frac{\vec{\nabla}V}{|\vec{\nabla}V|}$ will be surjective over the hyper-torus $$\bigg\{ (x_1,\ldots,x_n) : x_1^2 + \left(\sqrt{x_2^2 + x_3^2 + \cdots + x_n^2} - (r_m+1)\right)^2 \le r_m^2 \bigg\}$$ Since $U(x,y) = x$ whenever $x^2 + y^2 \ge r_m^2$, $V(x_1,\ldots,x_n)$ equals to $x_1$ over the complement of the hyper-torus. As that contains the $x_1$-axis, the square root in the definition of $V$ won't cause any problem. This makes $V$ smooth everywhere. Finally, it is easy to check $\nabla V$ never vanishes.
Update
About the question how this particular form of $\rho(t)$ is chosen.
First, $\rho(r)$ cannot change too rapidly or $\nabla U$ may become zero somewhere.
Second, we need $\rho(r)$ span across around a large enough interval to make $\hat{u}(r)$ surjective.
This suggest us to choose a $\rho(r)$ with $\rho'(r)$ as close to $\frac{1}{r}$ as possible. This means $$\rho(r) \sim \log r + constant.$$ In order for $U(x,y)$ to be smooth at the origin, I replace $\log r$ by $\log\sqrt{(e^{4\pi}-1)r^2 + 1}$. The constants are chosen so that when $r$ changes from $0$ to $1$, $\rho(r)$ pick up a change of $2\pi$ which we need.
Finally, we want $\rho(r)$ decay back to a constant multiple of $2\pi$ for large $r$ while keeping smooth all the way. We need this in order to extend the result to higher dimensions. This can be done using a smooth Partition of unity. The function $g(r)$ in the formula doesn't really have any specific meaning. It is simply something from trial and error which works.
• May I ask how you came up with the particular example of $\rho$ you gave? I'd like to understand the intuition behind your thought process. – Matthew Kvalheim Mar 21 '16 at 21:43
• @MatthewKvalheim answer updated, see rationale of chosing such a $\rho$. – achille hui Mar 21 '16 at 22:25
• You wrote an expression for $\nabla U$ that involves division by $r$. Am I correct in the interpretation that you actually mean that $\nabla U$ is defined by the expression you wrote on $\mathbb{R}^2\setminus \{0\}$, and $\hat{e}_x$ at the origin? – Matthew Kvalheim Mar 22 '16 at 3:49
• Thank you very much for your help. – Matthew Kvalheim Mar 22 '16 at 3:58
• @MatthewKvalheim $\nabla U$ is defined as the gradient of $U$. What I mean is it "equals to" the expression involves division by $r$ for $r \ne 0$ and $\hat{e}_x$. – achille hui Mar 22 '16 at 9:15 | 2020-08-15T02:49:53 | {
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https://math.stackexchange.com/questions/3060263/rationalizing-denominator-of-frac18-sqrt162-cannot-match-textbook-solu | # Rationalizing denominator of $\frac{18}{\sqrt{162}}$. Cannot match textbook solution
I am given this expression and asked to simplify by rationalizing the denominator:
$$\frac{18}{\sqrt{162}}$$
The solution is provided:
$$\sqrt{2}$$
I arrived at:
$$\frac{\sqrt{162}}{9}$$
Here is my thought process to arrive at this incorrect answer:
$$\frac{18}{\sqrt{162}}$$
= $$\frac{18}{\sqrt{162}}$$ * $$\frac{\sqrt{162}}{\sqrt{162}}$$
= $$\frac{18\sqrt{162}}{162}$$
= $$\frac{\sqrt{162}}{9}$$
How can I arrive at $$\sqrt{2}$$ ?
• Hint: $162=2\cdot 81$. Jan 3 '19 at 4:52
• $162=2*81=2*9^2$ so $\sqrt {162}=\sqrt {2*9^2}=9\sqrt 2$. If your hadn't "deradicalized" the denominator you would have ended up with $\frac 2 {\sqrt 2}$ which is also deradicalized as $\sqrt 2$. Jan 3 '19 at 5:49
$$\frac{\sqrt{162}}{9} = \frac{\sqrt{2 \cdot 9^2}}{9} = \frac{9\sqrt{2}}{9} = \sqrt{2}$$.
$$\require{cancel}\frac{18}{\sqrt{162}}=\frac{2\cdot3^2}{\sqrt{2\cdot3^4}}=\frac{2\cdot\cancel{3^2}}{\sqrt{2}\cdot\cancel{3^2}}=\frac{2}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}=\frac{\cancel2\sqrt{2}}{\cancel{2}}=\sqrt{2}$$
Your thought process is good. But just continue with factorizing $$162=2*81=2*3^4$$.
So $$\sqrt {162}=\sqrt {2*3^4}=\sqrt {2}\sqrt {3^4}=\sqrt 2*3^2=9\sqrt 2$$ and from there.... it's just mechanics.
$$\sqrt{162}$$ needs to be simplified further, as $$162$$ is the product containing a perfect square (i.e. $$81$$). Thus $$\sqrt{162} = \sqrt{2 \cdot 81} = \sqrt{2}\sqrt{81} = 9\sqrt{2}$$
and hence $$\frac{\sqrt{162}}{9} = \frac{9\sqrt{2}}{9} = \sqrt{2}$$
Alternatively if $$a = \frac{18}{\sqrt{162}}$$, $$a^2 = \frac{18^2}{\sqrt{162}^2} = \frac{324}{162} = 2$$. Since $$a$$ is a positive number, $$a = \sqrt{2}$$. | 2021-09-29T00:59:57 | {
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http://math.stackexchange.com/questions/226970/how-do-you-solve-an-equation-like-this | # How do you solve an equation like this?
We have an equation:
$\dfrac{p-1}{q} = \dfrac{q-1}{2p+1} = \dfrac {3}{5}$
How do you solve these types of equations? For example, if we have:
$\dfrac{1}{x} = \dfrac{3}{2}$, we use:
$1\times 2 = 3\times x$
$x = 1.5$
What is a similair approach to my equation?
-
Yes, exactly. First forget about the middle term. Next, forget about the $\frac35$. And then you're done. – Berci Nov 1 '12 at 19:29
Essentially, you can solve this using the process you describe, but twice, to generate two equations in two variables, and then solving for each variable as a "system of equations".
We have:
$$\dfrac{p-1}{q} = \dfrac{q-1}{2p+1} = \dfrac {3}{5}$$
Equation 1: $\quad 5(p-1) = 3q \iff 5p - 3q = 5$
Equation 2: $\quad 5(q-1) = 3(2p +1) \iff -6p +5q=8$
So your system of two equations in two unknowns becomes:
$$5p - 3q = 5\tag{1}$$ $$-6p +5q=8\tag{2}$$
Can you take it from there?
You can express (1) as a function of p (isolate p), and then substitute the expression obtained for p, into p in (2), and then solve for q, then p,
or
You can use "row operations": multiply (1) by 5 (both sides), and (2) by 3 (both sides):
$$25p-15q=25\tag{1}$$ $$-18p+15q = 24\tag{2}$$
Now add the equations (q disappears), solve for p, then "plug" p into one of the original equations and solve for q:
$$7p = 49n\implies p = 7$$
Now...From (1), originally, above
$$5p-3q=5 \implies 5(7) - 3q = 5\implies 35 - 3q = 5$$ $$\implies -3q=-30 \implies q = 10$$
-
Thank you sir, I am familair with row operations so I won't have trouble with these types of equations again. – JohnPhteven Nov 1 '12 at 19:49
(+1) Step by step getting the answer. – Babak S. Aug 6 '13 at 10:15
You have two equation with two variables: $$1) \quad \frac{p-1}{q} = \frac{3}{5}$$ and $$2) \frac{q-1}{2p+1}=\frac{3}{5}$$
$1)$ Implies that $5(p-1)= 3q$, which gives $q= \frac{5(p-1)}{3}$. Then you can insert $q= \frac{5(p-1)}{3}$ in $(2)$. Then find what is $p$, and then you can find what $q$ is.
-
$$\frac{p-1}{q} = \frac{q-1}{2p+1} = \frac {3}{5}$$ so $$\frac{p-1}{q}=\frac {3}{5}$$ and $$\frac {3}{5}=\frac{q-1}{2p+1}.$$ \begin{eqnarray} 5p-5&=3q&\\ 6p+3&=&5q-5 \end{eqnarray} $$5p-3q=5$$ $$6p-5q=-8$$ $$30p-18q=30$$ $$30p-25q=-40$$ $$-7q=-70$$ $$q=10$$ $$p=7.$$
-
$p = 7$? :) ${}{}$ – Alex Nov 1 '12 at 19:37
is there a problem ? – Iuli Nov 1 '12 at 19:47
From $$\dfrac{p-1}{q} = \dfrac{q-1}{2p+1} = \dfrac {3}{5}$$ follow the system of linear equations with two unknowns
$$5(p-1)=3q$$ $$5(q-1)=3(2p+1)$$ or $$5p-3q=5$$ $$6p-5q=-8$$
wich can be solved using for example Cramer rule
$\Delta=-25+18=-7$
$\Delta_p=-25-24=-49$
$\Delta_q=-40-30=-70$
$$p=\frac{\Delta_p}{\Delta}=\frac{-49}{-7}=7$$,$$q=\frac{\Delta_q}{\Delta}=\frac{-70}{-7}=10$$
- | 2015-01-29T14:56:22 | {
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https://math.stackexchange.com/questions/20223/getting-an-x-for-chinese-remainder-theorem-crt/20259 | # Getting an X for Chinese Remainder Theorem (CRT)
how do I get modulo equations to satisfy a given X in CRT.
For example say I have X = 1234. I choose mi as 5, 7, 11, 13. This satisfies the simple requirements of Mignotte's threshold secret sharing scheme. More precisely given in my example k = n = 4, and the product of any k - 1 is smaller then X how come simply computing the remainder of each won't give equations that solve to X = 1234.
In the case of the example,
x = 4 mod 5
x = 2 mod 7
x = 2 mod 11
x = 12 mod 13
Which resolves to 31264 (won't CRT produce the smallest?)
Any hints?
The final result of the CRT calculation must be reduced modulo 5 x 7 x 11 x 13 = 5005. This gives the correct answer.
Here is a much simpler way to immediately obtain the sought answer. Contrast the solution below to the much longer solution in your link, which involves calculations with much larger numbers and performs $4$ inversions vs. the single simple inversion below. Always search for hidden innate structure in a problem before diving head-first into brute-force mechanical calculations!
The key insight is: the congruences split into pairs with obvious constant solutions by CCRT, viz.
\begin{align}\rm\quad\quad\quad\quad\quad x\equiv \ \ \ 2\ \ \:(mod\ 7),\ \ x\equiv \ \ \ 2\ \ \:(mod\ 11)\ \iff\ x\equiv \ \ \ \color{#0a0}2\ \ (mod\ \color{#0a0}{77})\\[0.3em] \rm\quad\quad\quad\quad\quad x\equiv -1\ \ (mod\ 5),\,\ \ x\equiv\ {-}1\ \ (mod\ 13)\ \iff\ x\equiv \color{#c00}{-1}\ \ (mod\ \color{#c00}{65})\end{align}
So we reduced the above four original LHS equations to the above two RHS equations, which are easy to solve by CRT = Chinese Remainder Theorem. Indeed, applying Easy CRT below
$\rm\quad\quad\quad\quad\quad x\equiv\ \color{#0a0}{2 + 77}\ \bigg[\displaystyle\frac{\color{#c00}{-1}-\color{#0a0}2}{\color{#0a0}{77}}\ mod\,\, \color{#c00}{65}\bigg]\,\ \ (mod\ 77\cdot65)$
In the brackets $\,\rm\displaystyle\left[\, mod\ \ 65\!:\ \ \frac{-3}{77} \equiv \frac{-3}{12} \equiv \frac{-1}4 \equiv \frac{64}4 \equiv \color{#d0f}{16}\,\right]\quad$ (see Beware below)
This yields $\rm\ \ x\ \equiv\ \color{#0a0}{2 + 77}\,[\,\color{#d0f}{16}\,] \equiv 1234\,\ \ (mod\ 77\cdot 65)\quad$ QED
Theorem $\:$ (Easy CRT) $\rm\ \$ If $\rm\ m,\:n\:$ are coprime integers then $\rm\ m^{-1}\$ exists $\rm\ (mod\ n)\ \$ and
$\rm\displaystyle\qquad\quad\quad\quad \begin{eqnarray}\rm x&\equiv&\!\rm\ a\ \ (mod\ m) \\ \rm x&\equiv&\!\rm\ b\ \ (mod\ n)\end{eqnarray} \ \iff\ \ x \equiv\, a + m\ \bigg[\frac{b-a}{m}\ mod\ n\,\bigg]\,\ \ (mod\ m\:n)$
Proof $\rm\ (\Leftarrow)\ \ \ mod\ m:\,\ x \equiv a + m\ [\,\cdots\,] \equiv a,\$ and $\rm\ mod\ n\!\!:\,\ x \equiv a + (b\!-\!a)\ m/m \equiv b$
$\rm (\Rightarrow)\ \$ The solution is unique $\rm\ (mod\,\ mn)\$ since if $\rm\ x',\:x\$ are solutions then $\rm\ x'\equiv x\$ mod $\rm\:m,n\:$ therefore $\rm\ m,n\ |\ x'-x\ \Rightarrow\ mn\ |\ x'-x\ \$ since $\rm\ \:m,n\:$ coprime $\rm\:\Rightarrow\ lcm(m,n) = mn\ \ \$ QED
Note $\$ Easy CRT is not only easy to apply, but also very easy to remember. Namely note $\rm\ x\equiv a\pmod{\! m}\iff x = a + m\,k,\:$ for some integer $\rm\:k,\,$ This further satisfies the second congruence iff $\rm\ mod\ n\!:\ x = a + m\,k\equiv b$ $\iff$ $\rm k\:\equiv (b-a)/m,\$ hence the Easy CRT formula. This explains the $(\Leftarrow)$ proof: fill in the dots in $\rm\:x\equiv a + m\ [\,\cdots\,]\:$ to make $\rm\,x\equiv b\pmod{\! n}$
Beware $\$ Modular fraction arithmetic is well-defined only for fractions with denominator coprime to the modulus. See here for further discussion.
Below is the solution you linked to on "Math Celebrity" (cached to avoid link rot).
• @Bill Dubuque: But the OP was just giving an example! Your solution is completely useless for any other, similar problem. Unless you can automate it, of course :-) – TonyK Feb 3 '11 at 17:27
• @TonyK: Most certainly not true. Most math problems do have interesting structure, esp. problems that are designed for tests, competitions etc. In fact this is frequently true even in research problems. Indeed, speaking as a number theorist, I can tell you that methods like the above are very useful in practice. In mathematics, intuition always trumps brute force. – Bill Dubuque Feb 3 '11 at 17:34
• @Bill Dubuque: Well, your methods might be much, much simpler for you. But if they require years of practice, they're not so useful for people like me (or, presumably, the OP). – TonyK Feb 3 '11 at 18:05
• @Tonyk: The above solution can be understood by a bright high-school student. Simple optimizations like the above arise frequently in elementary number theoretical problems, so it is well-worth knowing them. – Bill Dubuque Feb 3 '11 at 18:54
• @TonyK: The optimization I employed above is algorithmic and is frequently applicable in practice. But that was not my point in presenting the above. Rather, it was to emphasize conceptual vs. algorithmic thought - intuition vs. brute force. Mathematical problems are far from random. Typically they involve much structure and their solution requires insightful exploitation of such innate structure - something that cannot be algorithmic (indeed many math problems are algorithmically unsolvable - that's what makes them interesting). – Bill Dubuque Feb 4 '11 at 0:37 | 2019-11-20T14:05:48 | {
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http://kvmb.imltraining.de/weighted-mean.html | # Weighted Mean
Variance is also based on the mean of your data. 0 (2016-2017), but may apply to other versions as well. This is a correct assumption if the same technique is used to measure the same parameter repeatedly. , Features, Capabilities, and Epics) to produce maximum economic benefit. Here is the mean of 1, 2, 3 and 4: Add up the numbers, divide by how many numbers: Mean = 1 + 2 + 3 + 44 = 104 = 2. weighted mean shift free download. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Abstract It is well known that the out-of-sample performance of Markowitz’s mean-variance portfolio criterion can be negatively affected by estima- tion errors in the mean and covariance. using excel to calculate the weighted standard deviation Does anyone know the formula for computing the weighted standard deviation? I was able to calculate the weighted average (16. You simply add up all the item values and divide by the total. This calculator will calculate the weighted average APR for all of your credit cards that have a current balance. Basically I calculate the most recent RSI value and it's exponentially weighted mean based on the window length selected. Water-Quality Characteristics. I am trying to calculate the mean profit percentage, but a free item is given when a certain amount of goods are bought. geometric mean concentration at which shellfish beds or swimming beaches must be closed. Weighted means are often used for frequency data. Weighted blankets, sometimes referred to as gravity blankets, were once a tool of therapists and psychiatry clinics. Purpose: To report design of a simplified external transmit-receive coil array for 7 Tesla (T) prostate MRI, including demonstration of the array for tumor localization using T2-weighted imaging (T2WI) at 7T before prostatectomy. Adding all of these totals up yields a weighted average of 5. Ranking, Matrix/Rating Scale, Multiple Choice, Multiple Textboxes, and Slider questions calculate an average or weighted average. How to calculate weighted mean in IBM SPSS? Using the results of the survey I need to create the rating of governmental organizations. A method of computing a kind of arithmetic mean of a set of numbers in which some elements of the set carry more importance (weight) than others. org Dictionary. Acronym/Abbreviation meaning of CAC The acronym CAC() means : Capital Access Corporation. test – here they are (the weighted variance is from Gavin Simpson, found on the R malining list): View Code RSPLUS. Compute the weighted mean of a variable. Suppose your teacher says, "The test counts twice as much as the quiz and the final exam counts three times as much as the quiz". Dec 10, 2010 · The IRR, also commonly referred to as the dollar weighted return, is the measurement of a portfolio’s actual performance between two dates, including the effects from all cash inflows and outflows. Recommended Articles. … If all the weights are equal, then the weighted mean is the same as the arithmetic mean. Finally, the query uses a GROUP BY clause to combine the data so that the calculation is performed for each gender. WMC is sampling-, scale-, and contrast-invariant, and is sparse on natural images. A weighted average, for example, takes into account the proportional. The AVERAGE function below calculates the normal average of three scores. Weighted Vest FAQs Weighted vests are often recommended for children with autism. 04 (df = 3, p <. It is based on kernel weighted sample statistics such as the mean (Nadaraya-Watson estimator) but also standard deviation, skewness, kurtosis, deciles, etc. Follow the example below to calculate the weighted average interest rate for a federal loan consolidation. Find the weighted average of class grades (with equal weight) 70,70,80,80,80,90: Since the weight of all grades are equal, we can calculate these grades with simple average or we can cound how many times each grade apear and use weighted average. mean, or in other words searching the explicit expression of the weighted mean standard deviation distribution, under the safely motivated restriction of independent measures obeying Gaussian distributions. also i thought weighted mean measures the tendency of respondents answers to various questions. It is an average in which each quantity to be averaged is assigned a. Under Medicare Advantage, the federal government still foots the. weighted dataset, others simply recommend replacing the weighted base sizes with the unweighted base sizes in the test statistic formula, while still using the weighted summary statistics such as mean and standard deviation. Most companies weight the revenue based on the opportunity success %. The choice of Galloway et al. This is because the basic average of a group of numbers is the same calculation as a weighted average except that the weights of all the numbers are calculated as being the same. In calculating a weighted average, each number in the data set is. Weighted Mean (ratio statistics algorithms) A / S = n Σ i = 1 f i A i n Σ i = 1 f i S i = n Σ i = 1 f i S i R i n Σ i = 1 f i S i This is the weighted mean of the ratios weighted by the sales prices in addition to the usual case weights. Free Arithmetic Mean (Average) Calculator - find the average of a data set step-by-step. Another Tip: If number of boys and girls is same, i. The first or shorter echo (TE < 30msec) is proton density (PD) weighted or a mixture of T1 and T2. then the variance of the mean could be estimated with. The idea of weighted mean plays a role in descriptive statistics and also occurs in a more general form in several other areas of mathematics. Given N jobs where every job is represented by following three elements of it. weighted-mean definition: Noun (plural weighted means) 1. Define weighted. … So let's set up the ability to do a weighted mean … versus a mean. To calculate a weighted average in Excel, simply use the SUMPRODUCT and the SUM function. In this article, I'll go through a basic description of what a weighted GPA is, why it matters for you, and how you can calculate your own weighted GPA if your school uses this type of scale. a point lying far from the rest). The weighted arithmetic mean (or weighted average) is used if one wants to combine average values from samples of the same population with different sample sizes: ¯ = ∑ = ∑ =. The WAM can be used to gain an average of an applicant results using their grades and unit credit value. Get Your Weighted And Unweighted GPA In Just A Few Minutes! This quick and easy online GPA calculator computes both weighted and unweighted high school grade point averages (GPA). However, Property Use Details can change over the course of the 12 months, so we time-weight them so they are appropriately attributed. It is challenging to achieve high weighted efficiency with low-power microinverters, typically because these devices are required to be low cost. Variance[wd] / n where n is sample size. Weighted arithmetic mean is used when the values are of different importance - weight p to be assigned to each value. More periods with zero is less scale sensitive than the MAPE and the MAD. Specify the plan goal weights for each of the three goal levels (the total must be 100 percent). POSIXct, colMeans for row and column means. Definition of weighted mean in the Definitions. The tutorial is mainly based on the weighted. Get Your Weighted And Unweighted GPA In Just A Few Minutes! This quick and easy online GPA calculator computes both weighted and unweighted high school grade point averages (GPA). It often occurs, however, that one must combine two or more measurements of the same quantity with differing errors. Title: Variance of a Weighted Mean Created Date: 20160808202903Z. Use 'weighted average' in a Sentence. A combination of medication, lifestyle changes and other treatments can help limit the impact of anxiety and its symptoms. µ = P n Pi=1 w ix i n i=1 w i (45) It is equivalent to the simple mean when all the weights are equal, since µ = P n P i=1 wx i n i=1 w = w P n x i nw = 1 n X i=1 x i (46) If the samples are all different, then weights can be thought of as sample frequencies, or they can be used to calculate probabilities where p i = w i/ P w i. mean() while aggregating a data frame. …Let's consider an academic course as our example. Dec 30, 2016 · Weighted Average is a free application for students. 04 (df = 3, p <. As a result, practitioners usually rely on vendor specified parameters, such as mean-time-to-failure (MTTF), to model failure processes, although many are skeptical of the accuracy of those models [4,5,33]. 5 for Stock C, or 3. The first section which takes up columns A, B, and C is the weighted averages. Select a stem that prompts several possible responses, with one that is the most accurate. For example, say 239 students in Economics 201. Jul 24, 2014 · Functional diversity can be quantified using one single trait at a time or multiple traits (see the descriptions of multiple-indices in other page). The weighted mean is similar to an arithmetic mean (the most common type of average), where instead of each of the data points contributing equally to the final average, some data points contribute more than others. A weighted grade or score is average of a set of grades, where each grade (g) carries a different weight (w) of importance. an object containing the values whose weighted mean is to be computed. w: a numerical vector of weights the same length as x giving the weights to use for elements of x arguments to be passed to or from methods. Examples of how to use “weighted mean” in a sentence from the Cambridge Dictionary Labs. Instead of each data point contributing equally to the final mean, some data points contribute more "weight" than others. In simple terms, the category "total" will be equal to the sum of the scores in each grade item each multiplied by its grade weight, and that sum being finally divided by the sum of all weights. On the face of it, this does look as nice. RE: Geometric and weighted mean Perhaps it is worth pointing out that *an* average is a relatively arbitrary way of "summarising" a bunch of numbers into a single number. Sep 08, 2015 · How do I change my gradebook aggregation to ‘Weighted Mean of Grades’? By default, the Moodle gradebook aggregation (the way it adds your grade values) is ‘Simple Weighted Mean of Grades. When an assignment is “weighted”, it means there is a grade for that assignment. Catetan yén lamun kabéh beuratna sarua, weighted méan sarua jeung arithmetic mean. For example, if the chest is the target area, we might perform a set of barbell bench press, followed by a set of barbell incline press, and finished with a set of weighted parallel dips. A flow-weighted mean is the mean of a quantity after it is weighted proportional to a corresponding flow rate. The fact that x only gives one radiation field is the same as giving 0 radiation for the second field and so on, which means I do the following:. The mean of each area will be obtained using the formula: x = Σx/N (Downie and Heat, 1970) The numerical findings of the study will be statistically analysed and interpreted using the frequency count. rm: a logical value indicating whether na values in x should be stripped before the computation proceeds. For example, in the county data set, population is a natural weighting. Most content comes from the ECPR Winter School in Methods and Techniques R course, that I had the pleasure of teaching this February. A similar procedure is used for a 10% trimmed mean. Select Weighted Goals as the plan goals type. Weighted Mean calculator for calculating the weighted mean statistics for the given set of data. DESCRIPTION. The F1 score can be interpreted as a weighted average of the precision and recall, where an F1 score reaches its best value at 1 and worst score at 0. A weighted grade or score is average of a set of grades, where each grade (g) carries a different weight (w) of importance. Unlike most credit card interest calculators, this calculator will calculate the current finance charge for each card, and then compute the credit card average APR using a weighted formula. This paper proposes a novel on-line portfolio selection strategy named “Confidence Weighted Mean Reversion ” (CWMR). There's one more skill you'll need to calculate weighted scores: A simple average, which in "math speak" is more properly called the mean. 3% then you are essentially implying a growth rate of POSITIVE 3. Find more details on making DIY. • So it is better to use WAPE for volume weighted MAPE and WMAPE for dollar weighted or Cost-weighted measures. What is a weighted KPI calculations For weighted KPI calculations, the base weight of each KPI is factored against the weighted counts of each KPI by using the following formulas. For this reason, consider using Mean Absolute Deviation (MAD) alongside MAPE, or Consider that even fast moving consumer goods companies these days to average MAPEs over multiple time series. (the "Gold Book"). Featured Services & News. Weighted Mean Formula (Table of Contents). Wondering what a weighted blanket is and what is the purpose of having one? Learn how weighted blankets can benefit sleep issues, anxiety, autism, ADHD, insomnia and more. These two measurements can be combined to give a weighted average. Follow this example use the following data set. The sample size weighted correlation may be used in correlating aggregated data Description. the weighted mean of is given by (4) Weighted means have many applications in physics, including finding the center of mass and moments of inertia of an object with a known density distribution and computing and electric and magnetic multipole moments of charge and current distributions, respectively. For instance if you have 3 objects, A,B and C. w: a numerical vector of weights the same length as x giving the weights to use for elements of x. Sincerely, Jacob. Weighted Shortest Job First (WSJF) is a prioritization model used to sequence jobs (eg. 1 Introduction The Least Mean Square (LMS) algorithm, introduced by Widrow and Hoff in 1959 [12] is an adaptive algorithm, which uses a gradient-based method of steepest decent [10]. The WACC is also the minimum average rate of return it must earn on its current assets to satisfy its shareholders, investors, or creditors. I have been working on a problem for the last year (in graph theory/algorithms). The weighted harmonic mean is the preferable method for averaging multiples, such as the price–earnings ratio (P/E), in which price is in the numerator. % % Example: % X = rand(3,5);. He has an entire dice set that almost always gets the best possible roll. Weighted average refers to the mathematical practice of adjusting the components of an average to reflect the importance of certain characteristics. Definition of weighted in the Definitions. What is the Weighted Average? The weighted average scoring model applies a "weight" to the matrix questions based on responses to the first item in a side-by-side matrix. wi(yi 0 1xi) 2. The weighted mean is similar to the arithmetic mean where instead of each of the data points contributing equally to the final average, some data points contribute more than others. Example sentences with "weighted mean", translation memory add example en The conclusion was that the only items differentiating the SIDS infants from the control infants were a "lower mean weight and height at birth, the previous occurrence of cyanosis [bluish skin and mucous membranes caused by lack of oxygen in the blood] or pallor during. Activity recording is turned off. 1 Introduction The Least Mean Square (LMS) algorithm, introduced by Widrow and Hoff in 1959 [12] is an adaptive algorithm, which uses a gradient-based method of steepest decent [10]. Often used to account for area changes between meridians at varying latitudes by using the cosine of the latitude as the weights. Besides the stock market example given above, another situation were weighed average is used is in grading. mean fitness of the population W(bar) The sum of the fitnesses of the genotypes of a population weighted by their proportions; hence a weighted mean fitness. Specifically, overtime is normally calculated at 1. Each grade item can be given a weight to change its importance in the overall mean. Weighted Shortest Job First (WSJF) is a prioritization model used to sequence jobs (eg. and Wilks, A. Weighted Mean: A mean where some values contribute more than others. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. The weight of the category should be entered as a decimal, make sure that the sum of the weights equal 1. In simple terms, the category "total" will be equal to the sum of the scores in each grade item each multiplied by its grade weight, and that sum being finally divided by the sum of all weights. This calculator will calculate the weighted average APR for all of your credit cards that have a current balance. What does weighted mean mean? Information and translations of weighted mean in the most comprehensive dictionary definitions resource on the web. All or only selected records: You can either use all the points in the point theme for the analysis or only a selected subset of points. The weights cannot be negative. So let's have a look at the basic R syntax and the definition of the weighted. It is calculated according to the following formula: M = mark received in a course U = units of credit for a course. This article will show you how to use Excel's SUMPRODUCT and SUM functions individually and how to combine the two to calculate a weighted average. Formula: Where x is the repeating value. 0015948968 4 2 5442. Field {Case_Field} Field used to group features for separate mean center calculations. 15 * (73 + 80 + 85 + 88) +. …But sometimes, some data points…are more important than others. Trump talks Medicare in a retirement enclave where doctors are a golf-cart ride away. and Wilks, A. weighted definition: 1. Jun 07, 2016 · 61. Mar 18, 2017 · 2. A method of computing a kind of arithmetic mean of a set of numbers in which some elements of the set carry more importance (weight) than others. Paper 3: Issues in calculating average effect sizes in meta-analyses Dr. Metascore is a weighted average in that we assign more importance, or weight, to some critics and publications than others, based on their quality and overall stature. Slugging average is sometimes referred to as slugging percentage. But an equally weighted mean would be the pooled mewn and it expected value would be [65+73]/2. Section 2 formally formulates the on-. Under Medicare Advantage, the federal government still foots the. A weighted GPA is calculated by awarding additional points to classes that are considered more challenging than the basic curriculum. Free Arithmetic Mean (Average) Calculator - find the average of a data set step-by-step. The Benefits of a Weighted Sales Pipeline. 739 × 10-3 mm 2 /sec, SD = 0. Water-Quality Characteristics. Parameters: axis: {index (0), columns (1)} Axis for the function to be applied on. …If the professor had four exams,…and these were your four exam scores,…calculating your average would be easy, 80%. May 11, 2017 · The weighted mean is the weighted average. On the other hand, its weighted version is very useful for evaluating inequalities for definite integrals. The time-weighted formula is essentially a geometric mean of a number of holding-period returns that are linked together or compounded over time (thus, time-weighted). Among them, M55 showed a gene expression pattern consistent with behavioral changes after stress exposure, and the gene ontology analysis revealed that this was involved in nervous system development, gland. Data are classified using the Harmonized System of trade at the six- or eight-digit level. Sleep through the night and relieve anxiety with high-quality weighted blankets for adults from Weighting Comforts. Jul 24, 2014 · Functional diversity can be quantified using one single trait at a time or multiple traits (see the descriptions of multiple-indices in other page). 4) but have been unsuccessful in computing the weighted SD. Aug 29, 2013 · If such an effect is possible then the simplest and safest thing to do is compute the weighted sum for the m values in each of the n experiments. References. To calculate a weighted average in Excel, simply use the SUMPRODUCT and the SUM function. E(Z1)=65 and E(Z2)=73 The population mean is [2(65)+73]/3. It is challenging to achieve high weighted efficiency with low-power microinverters, typically because these devices are required to be low cost. How to Calculate Weighted Average Price Per Share Calculating your weighted average price per share can help you assess the performance of an investment that was made in several transactions. …Let's consider an academic course as our example. What is the uncertainty of the weighted average? What's the correct procedure to find the uncertainty of the average?. Weighted average definition, a mean that is computed with extra weight given to one or more elements of the sample. Another Tip: If number of boys and girls is same, i. May 01, 2012 · PowerPivot Weighted Average Measure Compared to Non-Weighted Average. The weighted average (or weighted mean, as statisticians like to call it) is easy to compute in SAS by using either PROC MEANS or PROC UNIVARIATE. If trim is non-zero, a symmetrically trimmed mean is computed with a fraction of trim observations deleted from each end before the mean is computed. - joran Jun 12 '12 at 4:35 3 @Frank Hover over the down triangle beneath the vote count next to your Q. If I use the following data. Math, I am in the 9th grade, and our math teacher is explaining "weighted averaging. title = "Amplitude-weighted mean velocity: Clinical utilization for quantitation of mitral regurgitation", abstract = "Objectives. Weighted blankets, sometimes referred to as gravity blankets, were once a tool of therapists and psychiatry clinics. The weighted mean is used a lot by teachers. How to use weighted in a sentence. net dictionary. It is useful in many situations e. Flow-weighted mean concentration. If you're seeing this message, it means we're having trouble loading external resources on our website. Myometrial Invasion in Endometrial Cancer: Diagnostic Accuracy of Diffusion-weighted 3. Weighted arithmetic mean. when they tried on their own, funded by WW. He has an entire dice set that almost always gets the best possible roll. In this lesson from Alanis Business Academy, we describe the weighted mean, review the equation to solve for the weighted mean, and actually solve for a weighted mean. If all the weights are equal, then the weighted mean equals the arithmetic mean (the regular "average" you're used to). So, we use a weighted average (weighted mean or scaled average) to give greater value to some data over other data. If this keyword is not present or is zero, then the mean is computed across all dimensions of the input array. Michael Ho Zheng Sun
Jack Xin§. size }" if weights. 35 mg/L Flow Weighted C once tra i = 0. What is the acronym meaning/definition of CAC ?. We now see the mean as a weighted sum of the distinct values, where each value is weighted according to its proportion in the total list of numbers. Ranking, Matrix/Rating Scale, Multiple Choice, Multiple Textboxes, and Slider questions calculate an average or weighted average. Often used to account for area changes between meridians at varying latitudes by using the cosine of the latitude as the weights. Weighted average refers to the mathematical practice of adjusting the components of an average to reflect the importance of certain characteristics. (mean age, 57 years) who had. mean function in R, so far I couldn’t find a implementation of weighted. For weighted KPI calculations, the base weight of each KPI is factored against the weighted counts of each KPI by using the following formulas. n an average calculated by taking into account not only the frequencies of the values of a variable but also some other factor such as their variance. A weighted average is an average in which one element may contribute more heavily to the final result than another element. - joran Jun 12 '12 at 4:35 3 @Frank Hover over the down triangle beneath the vote count next to your Q. Among them, M55 showed a gene expression pattern consistent with behavioral changes after stress exposure, and the gene ontology analysis revealed that this was involved in nervous system development, gland. Apr 21, 2017 · However, anxiety disorders can cause crippling feelings of fear, worry, or restlessness. This is a correct assumption if the same technique is used to measure the same parameter repeatedly. If using aggregated data, the correlation of the means does not reflect the sample size used for each mean. 2 mg/L Flow Weighted Concentration = 0. wi(yi 0 1xi) 2. Weighted average refers to the mathematical practice of adjusting the components of an average to reflect the importance of certain characteristics. WMC is sampling-, scale-, and contrast-invariant, and is sparse on natural images. The weighted arithmetic mean (or weighted average) is used if one wants to combine average values from samples of the same population with different sample sizes: ¯ = ∑ = ∑ =. The monthly returns are then compounded to arrive at the annual return. His head was pounding, his body seemed to ache from head to toe, and there was a tightness in his chest that pressed on his lungs. Weighted Average Item Price Report (WAIPR) and the Regional and Statewide Average Awarded Price Report (RSWAAPR) The Weighted Average Item Price Report (WAIPR) and the Regional and Statewide Average Awarded Price Report (RSWAAPR) are reports produced using information from NYSDOT's Trns•Port BAMS⁄DSS. Weighted gene co-expression network analysis identified 60 characteristic modules that correlated with stress or the FKBP5 genotype. 0026132544 5 1 209. 03 Progressive Less than 0. It is also called weighted average. Tseng Department of Biostatistics Department of Human Genetics. Thus, the weighted mean makes it possible to find the average student grade in the case where only the class means and the number of students in each class are available. In other words, each value to be averaged is assigned a certain weight. Now, they have gone mainstream. Since we're all engineers, most students can figure out how this works at the end of the termbut how do you take it into account to calculate your current grade halfway through?. A weighted score or weighted grade is merely the average of a set of grades, where each set carries a different amount of importance. This is a correct assumption if the same technique is used to measure the same parameter repeatedly. Hi R experts, I need know how calculate a weighted mean by group in a data frame. Weighted voting. mean, part of R for Data Science: Lunchbreak Lessons. The experts in survey were estimating the organizations by. 03 Neutral 0. Section 2 formally formulates the on-. We will first create a Category which will use Weighted mean of grades. The Greedy Strategy for activity selection doesn’t work here as a schedule with more jobs may have smaller profit or value. Weighted Mean is a statistical method which calculates the average by multiplying the weights with its respective mean and taking its sum. If all the weights are equal, then the weighted mean is the same as the arithmetic mean. The average is calculated by adding a range of numbers together and then dividing this total by the number of values in the range. The solid lines are the weighted mean while the dashed lines are the standard mean. Featured Services & News. A weighted mean (or weighted average) is like an ordinary mean, but the observations don't contribute equally - more emphasis is placed on some data values than others; they are weighted by a bigger or smaller amount than 1/n. Given a numeric vector of data values, x, and another numeric vector of weights, w, the weighted mean is the sum of the data value times the weights divided by the sum of the weights. 'Mikhail' instead of just 'Mihail')?. His head was pounding, his body seemed to ache from head to toe, and there was a tightness in his chest that pressed on his lungs. This implies that we trim the lowest 5% of the data as well as the highest 5% of the data. Note that contrary to the weights w i in Eq. The weighted mean measures the average of the weighted data points. It is also called weighted average. Also called weighted mean. In a scenario where grades are weighted by category and the number of assignments in a category changes, no changes to the course points or syllabus would be necessary. What is a Weighted Mean? A weighted mean is a kind of average. Time-Weighted Return Formula. Meaning of weighted mean. A mean where some values contribute more than others. Input the above in your Ti-84 or TI-83 calculator Instructions on how to Create a List TI-84 | TI-83 tutorial. 631 × 10-3 mm 2 /sec, SD = 0. 03 Progressive Less than 0. If a professor decides to increase or decrease the workload based on the particular needs of any group of students, weighted grades make that easy. Here is the mean of 1, 2, 3 and 4: Add up the numbers, divide by how many numbers: Mean = 1 + 2 + 3 + 44 = 104 = 2. When totaling the individual values, each is multiplied by a weighting factor, and the total is then divided by the sum of all the weighting factors. Weighted means are often used for frequency data. These two measurements can be combined to give a weighted average. Weighted mean. Notice that the groups differ considerably in sample size. Mean: The arithmetic mean, also known as the simple mean or equal weighted mean. It is based on kernel weighted sample statistics such as the mean (Nadaraya-Watson estimator) but also standard deviation, skewness, kurtosis, deciles, etc. The above weighted average formula returns the value 849. Moving Average charts in JMP Each point on a Uniformly Weighted Moving Average (UWMA) chart, also called a Moving Average chart, is the average of the w most recent subgroup means (called “span”), including the present subgroup mean. Synonyms for weighted at Thesaurus. Generally the first item measures attributes like "Importance," "Need," or "Expectation". Input the above in your Ti-84 or TI-83 calculator Instructions on how to Create a List TI-84 | TI-83 tutorial. The area-weighted mean patch size, on the other hand, is a weighted mean, where the weights are based on the size of the patch. However, if you don’t want to spend big on Weighted belts, then you should absolutely go for Dark Iron Fitness which comes with all the basic features one could expect in Weighted belts. prepared and arranged in a way that is likely to produce a particular effect, usually an…. This method assumes that all of the trials have measured the outcome on the same scale. It is generally used to find average of variables that are expressed as a ratio of two different measuring units e. The T2-weighted sequence can be employed as a dual echo sequence. Weighted Mean. This has been a guide to Weighted Average in Excel. In calculating a weighted average, each number in the data set is. What does it mean republicans trying to phase out higher federal payments of medicare 3 Oct 2019. an object containing the values whose weighted mean is to be computed. a numerical vector of weights the same length as x giving the weights to use for elements of x. Next, obviously the weighted mean is $$\hat\mu=\sum_ip_ix_i,$$ and the variance:$$\hat\sigma^2=\sum_ip_i(x_i-\hat\mu)^2$$. Apr 21, 2017 · However, anxiety disorders can cause crippling feelings of fear, worry, or restlessness. By modifying sequence parameters such as repetition time (TR) and echo time (TE), for example, anatomical images can emphasize contrast between gray and white matter (e. Weighted mean The calculation of the mean discussed in the handout assumes that the standard deviation of each individual measurement is the same. VWAP is typically used with intraday charts as a way to determine the general direction of intraday prices. This is a correct assumption if the same technique is used to measure the same parameter repeatedly. weighted average w/ matlab. Chapter 3 Weighted Mean study guide by john_lindsey8 includes 43 questions covering vocabulary, terms and more. Definition - What does Weighted Average Cost of Capital (WACC) mean? The weighted average cost of capital represents a weighted average of the after-tax cost of debt and the cost of equity where the weighting is based on a company’s target debt-equity ratio, measured at market. If this keyword is present, then the mean is only calculated across a single dimension. SUMPRODUCT should be the sum of the products of the respective values and weights. Nov 29, 2019 · 10 intermittent fasting side effects that might mean it’s not a great fit for you These are the 6 full-body home exercises the Springbok women’s vice captain swears by World Aids Day: Could. Weighted Averages Date: 11/02/98 at 21:09:00 From: Jacob Smith Subject: Help with "weighted Averaging" Dr. Weighted gene co-expression network analysis identified 60 characteristic modules that correlated with stress or the FKBP5 genotype. Source Score, x Weight,w Homework 60 5% Midterm 75 35% Project 80 20% Speech 70 15% Final Exam 62 25% 2. The weighted arithmetic mean of a set of numbers X 1, X 2, , X N with respective weights of w 1, w 2, , w N is defined as: Example : For a statistics course, the final exam grade is weighted four times as much as each quiz score. It includes the product id and invoice id along with some other information. Column A shows all the student’s percentage scores he/she received on tests, quizzes, homework assignments, and participation. Creates a classification table, from raw data in the spreadsheet, for two observers and calculates an inter-rater agreement statistic (Kappa) to evaluate the agreement between two classifications on ordinal or nominal scales. Let us understand more with an example:. It is calculated according to the following formula: M = mark received in a course U = units of credit for a course. A mean where some values contribute more than others. 35 mg/L Flow Weighted C once tra i = 0. wRC+ takes the statistic Runs Created and adjusts that number to account for important external factors -- like ballpark or era. 739 × 10-3 mm 2 /sec, SD = 0. How to use weighted in a sentence. This method is sensitive to extreme observations (e. The arithmetic mean-geometric mean (AM-GM) inequality states that the arithmetic mean of non-negative real numbers is greater than or equal to the geometric mean of the same list. Here are the high-level steps for using weighted goals to manage a plan: Define goal IDs on the Goals page. | 2020-01-17T17:44:00 | {
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https://mathhelpforum.com/threads/ask-is-expected-frequency-rounded-up-or-down.282947/ | # [ASK] Is Expected Frequency Rounded Up or Down?
#### Monoxdifly
A die is tossed 100 times. What's the expected frequency that the number appears will be 4?
The probability is 1/6, so the expected frequency is 1/6 × 100, but that results in a fraction (16 2/3). Do we need to round it up or down? Is the answer 16 or 17?
#### romsek
MHF Helper
You don't need to round it at all. The correct answer is $\dfrac{50}{3}$
True you won't see 2/3 of a die but that doesn't matter. An expectation is an average and as such there's no reason to expect it must be an integer.
#### Monoxdifly
Okay then, thanks.
#### TKHunny
If you are forced to round, for whatever reason, you will need to decide exactly how to do that.
Typically, a simple up or down methodology will prove unsatisfactory as you will violate the definition of a probability distribution:
Down 2 dp
1/3 = 0.33
1/3 = 0.33
1/3 = 0.33
0.33 * 3 = 0.99 and that is NOT 1.00
Likewise, fudging (forced footing) proves unsatisfactory:
1/3 = 0.33
1/3 = 0.33
1/3 = 0.34
They are not really equal, are they.
Unless you are in a purely theoretical pursuit, you just have to make up your mind what is and is not suitable.
#### Monoxdifly
Oh well, no rounding then, thanks. | 2020-01-18T18:08:46 | {
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https://math.stackexchange.com/questions/1475930/how-should-i-calculate-lim-n-rightarrow-infty-frac1n2n3n-nnn | # How should I calculate $\lim_{n\rightarrow \infty} \frac{1^n+2^n+3^n+…+n^n}{n^n}$ [duplicate]
How should I calculate the below limit
$$\lim_{n\rightarrow \infty} \frac{1^n+2^n+3^n+...+n^n}{n^n}$$ I have no idea where to start from.
• If this question does not come out from you, then where did you get this question? – Megadeth Oct 12 '15 at 5:37
• The limit should at least bigger than $1 + e^{-1} + e^{-2} + \cdots = \frac{e}{e-1}$ as for all $n$, the last term is $1$, the second last is $\left(\frac{n-1}{n}\right)^n = (1- \frac 1n)^n$ which tends to $e^{-1}$. So on so forth. – user99914 Oct 12 '15 at 6:05
• From $\frac{x_1+x_2+\cdots+x_n}{n}\geq\sqrt[n]{x_1x_2\cdots x_n}$ we write $$(\frac1n)^n+(\frac2n)^n+\cdots+(\frac1n)^n\geq n\frac{n!}{n^n}\to e$$ – Nosrati Oct 12 '15 at 6:29
• @Maryam $n^n\sim e^nn!$, so $n\dfrac{n!}{n^n}$ does not approach $e$. – Quang Hoang Oct 12 '15 at 6:35
• Have you tried Stolz-Cesaro ? – Lucian Oct 12 '15 at 6:49
First we use an observation by @Stan in the comment. Note that as $(1 +\frac{x}{n})^n$ is increasing in $n$ whenever $|x|<n$,
$$\left(\frac{k}{n}\right)^n = \left(1 + \frac{k-n}{n}\right)^n \le e^{k-n},$$
(here we assume that $x:= k-n$ is fixed and varies the remaining two $n$'s. This sequence is increasing and tends to $e^{k-n}$, as $|x| = |k-n| < n$. See here). Then we have
$$\begin{split} \frac{1^n + 2^n + \cdots + n^n}{n^n} &= \sum_{k=1} ^n \left(\frac{k}{n}\right)^n \\ &\le \sum_{k=1}^n e^{k-n} \\ &= 1 + e^{-1} + e^{-2} + \cdots e^{1-n} \\ &\le \frac{1}{1-e^{-1}} = \frac{e}{e-1}. \end{split}$$ This implies
$$\limsup_{n\to \infty} \frac{1^n + 2^n + \cdots + n^n}{n^n} \le \frac{e}{e-1}.$$
On the other hand, fix $k$. Then for all $n >k$, we have
$$\begin{split} \frac{1^n + 2^n + \cdots + n^n}{n^n} &\ge \frac{(n-k)^n + (n-k+1)^n + \cdots + n^n} {n^n}\\ &= \left( 1 - \frac kn\right)^n + \left( 1 - \frac {k-1}n\right)^n + \cdots +1 \end{split}$$
Then for all $\epsilon >0$, there is $N\in \mathbb N$ so that
$$\left| \left( 1 - \frac {j-1}n\right)^n - e^{-(j-1)} \right| < \epsilon$$
whenever $n \ge N$ and for all $j = 1, 2 , \cdots, k+1$ (Note $k$ is fixed, so this $N$ can be found)
In particular, this implies
$$\frac{1^n + 2^n + \cdots + n^n}{n^n} \ge e^{-k} + e^{-(k-1)} + \cdots + 1 - (k+1) \epsilon.$$
Thus
$$\liminf_{n\to \infty} \frac{1^n + 2^n + \cdots + n^n}{n^n} \ge e^{-k} + e^{-(k-1)} + \cdots + 1 - (k+1) \epsilon.$$
Now let $\epsilon \to 0$ and then $k \to \infty$, we have
$$\liminf_{n\to \infty} \frac{1^n + 2^n + \cdots + n^n}{n^n} \ge \frac{1}{1-e^{-1}} = \frac{e}{e-1}.$$
This implies
$$\lim_{n\to \infty} \frac{1^n + 2^n + \cdots + n^n}{n^n} = \frac{e}{e-1}.$$
• Another proof for $e^{k-n} \geq (k/n)^n = e^{n \ln (k/n)}$: $\Leftarrow k-n \geq n \ln (k/n) \Leftarrow \frac{k}{n}-1 \geq \ln \frac{k}{n}$ $\Leftarrow$ for $x>0$, $x-1 \geq \ln x$ $\Leftarrow$ Let $y(x) = x-1-\ln x$, then the stationary point should satisfy $y'=1-\frac{1}{x}=0$; meanwhile $y''=x^{-2}>0$. Thus $y_{\min} = y(1) = 0$, for all $x>0$, $y(x) \geq 0$. – Stan Oct 12 '15 at 8:01
• Another proof for the upper bound: $$1^n+2^n+\cdots+(n-k-1)^n < \int_1^{n-k} x^n\,dx < (n-k)^{n+1}/(n+1) < (n-k)^n.$$ Thus when we take only the terms from $(n-k)^n$ to $n^n$ in the numerator, the neglected terms are at most $(n-k)^n/n^n$, which tends to $e^{-k}$ as $n\to\infty$. Now we have the limit sandwiched between two explicit functions of $k$; let $k\to\infty$. – Greg Martin Oct 12 '15 at 8:10
• Nice proof! +1 :) – ZFR Nov 21 '15 at 7:50 | 2020-01-27T22:07:05 | {
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http://math.stackexchange.com/questions/330636/associativity-of-a-binary-operator | # Associativity of a binary operator
I am reading the book "A first course in abstract algebra", and i am reading the chapter called subgroups. It says that, for general situations we do not use "*" to denote the binary operation on a group, but we simply use addition symbol "+" or multiplication symbol ".". Then it says, we denote the product a.a.a.a.a...a for n factors by an. But i am confused here: is that always associative? For example, for 3 factors, a.a.a is a3 according to what book says, but are we sure that a.(a.a)=(a.a).a=a3 is always true? Thank you.
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Do you know what "group" means? – Chris Eagle Mar 14 '13 at 20:53
Yes, i do. Why? – bigO Mar 14 '13 at 20:54
Then you know the answer to this question. – Chris Eagle Mar 14 '13 at 20:54
This is why i am asking this, i must be missing something – bigO Mar 14 '13 at 20:55
Note that not all binary operations need be associative. For example, take the integers under subtraction: $-4=(1-2)-3\neq 1-(2-3)=0$. – user1729 Mar 15 '13 at 14:54
Recall: One of the key properties satisfied by all groups is that its binary operation is associative on the set defined by the group.
If you are referring to Fraleigh's text, see
Definition 4.1: "A group $\langle G, *\rangle$ is a set $G$, closed under a binary operation $*$, such that the following axioms of satisfied:
$\mathcal G_1$: For all $a, b, c \in G,$ we have $$(a*b)*c = a*(b*c)\tag{associativity of *}$$
Of course, $\mathcal G_2,\;\text{and}\;\mathcal G_3\;$ are crucial, as well: the existence of an identity element $e\in G$, and for each $g\in G$, its inverse, $g^{-1}$ is also in $G$, including the definitions of the identity and how we define the inverse of a group element.
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Oh thanks, that's what i have been missing. – bigO Mar 14 '13 at 20:55
Not only, As @amWhy noted, the binary operation is associative but also depend on how you show the binary operation, the figures of composition of elements varies: $$\cdot\longrightarrow a*b=a\cdot b$$ so $$\underbrace{a\cdot a\cdot a\cdot ...\cdot a}_{\large n\;times}=a^n~$$ And $$+\longrightarrow a*b=a+ b$$ so $$\underbrace{a+ a+ a+ ...+ a}_{\large n\;times}=na$$
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Don't know why the downvote? +1 – amWhy Mar 14 '13 at 21:09
Thank you! I did not downvote, just to let you know – bigO Mar 14 '13 at 21:28
@amWhy: Thanks Amy for the edit. :-) – Babak S. Mar 15 '13 at 5:32 | 2015-08-01T22:59:10 | {
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https://math.stackexchange.com/questions/1986392/win-a-coin-game-in-10th-round | # Win a coin game in 10th round
Two players are tossing a fair coin in one round. If it is heads, the first one gets a dollar from the second player. Otherwise, the first player gives a dollar to the second one. If both players have 6 dollars in the beginning of the game, what is the probability that the first player wins all the money exactly on 10th round.
• I am considering sample space to be 2^10; however, a friend of mine considers it to be 960. Whose idea is erroneous and why? Thanks. – Ulugbek Abdullaev Oct 26 '16 at 18:06
• The sample space is not as big as $2^{10}$, because the sequence $HHHHHHTTTT$ and $HHHHHHHHHH$ are considered exactly the same game (they quit playing after the sixth heads). Of course, if you think that they continue playing after one has lost just to "see what would happen", but without any money involved, and analyze those games, then the sample space has size $2^{10}$. It all depends on your interpretation. – Arthur Oct 26 '16 at 18:07
• @Arthur I got your point. Could you elaborate how to exclude those unnecessary? I'm just stuck.. – Ulugbek Abdullaev Oct 26 '16 at 18:41
• Hint: out of those ten rounds, how many Heads were there? How many Tails? We know the last one was $H$...what about the next to last? – lulu Oct 26 '16 at 19:04
• @BruceET It's very nearly the whole story. We see that we want $8$ Heads and $2$ Tails...there must be at least one $T$ in the first $6$ slots and and both $T$ must be in the first $8$. Easy to count. – lulu Oct 26 '16 at 21:58
If the coin is tossed 10 times, there could be wins at trials number 6, 8, and 10.
A win at the 6th requires six heads in a row (probability $.5^6 = 0.015625.$)
A first win at the 8th requires that exactly one of the first six tosses must be tails, followed by two heads (probability $6(.5)^8 = 0.0234375.$)
The simulation below confirms these two results (within simulation error), and suggests that the probability of a first win at the 10th toss must have probability about 0.0265 (two or three place accuracy). I will leave it to you use similar logic to find a combinatorial formula and the exact probability.
m = 10^6; frst.win = numeric(m)
for(i in 1:m)
{ toss = sample(c(-1,1), 10, rep=T) # vector of 1's (Hs) and -1's (Ts)
cs = cumsum(toss) # 1st player's cumulative totals
frst.win[i] = match(6, cs) } # toss on which cum tot first reaches 6
table(frst.win)/m
frst.win
6 8 10
0.015633 0.023540 0.026513 | 2020-12-04T18:18:32 | {
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http://math.stackexchange.com/questions/495119/what-is-gcd0-0/495127 | # What is $\gcd(0,0)$?
What is the greatest common divisor of $0$ and $0$? On the one hand, Wolfram Alpha says that it is $0$; on the other hand, it also claims that $100$ divides $0$, so $100$ should be a greater common divisor of $0$ and $0$ than $0$.
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(o,o)? <- see the face scratching the head? – zerosofthezeta Sep 16 '13 at 4:57
It is typically not defined. – copper.hat Sep 16 '13 at 4:58
– anon Sep 16 '13 at 5:23
The word "greatest" in "Greatest Common Divisor" does not refer to being largest in the usual ordering of the natural numbers, but to being largest in the partial order of divisibility on the natural numbers, where we consider $a$ to be larger than $b$ only when $b$ divides evenly into $a$. Most of the time, these two orderings agree whenever the second is defined. However, while, under the usual order, $0$ is the smallest natural number, under the divisibility order, $0$ is the greatest natural number, because every number divides $0$.
Therefore, since every natural number is a common divisor of $0$ and $0$, and $0$ is the greatest (in divisibility) of the natural numbers, $\gcd(0,0)=0$.
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Similarly $\gcd(0,n)=n$ for all $n\in\Bbb N$ (including $n=0$). – anon Sep 16 '13 at 5:22
Thanks, great answer! How did you figure this out? Where do mathematicians store such commonly agreed on, but not completely obvious definitions? When did people start defining gcd this way, I assume they didn't when Euclidean algorithm was invented 300BC? – Konstantin Weitz Sep 16 '13 at 6:55
For the most part, the partial order of divisibility coincides with the usual ordering wherever it is defined. So if $\le$ represents the usual ordering and $\vert$ represents the partial order of divisibility, then $x\vert y\Rightarrow x\le y$, except if $y=0$. Moreover, if $x$ is a common divisor of two non-zero natural numbers $a$ and $b$, then $x\le\gcd(a,b)\Rightarrow x\vert\gcd(a,b)$. So in most cases, you can define $\gcd(a,b)$ to be the greatest common divisor of $a$ and $b$, where 'greatest' refers to the usual ordering. That's what Euler did (and left $\gcd(0,0)$ undefined). – Donkey_2009 Sep 16 '13 at 11:55
Euclid, I mean.... – Donkey_2009 Sep 16 '13 at 14:50
I answered this already in a comment at MO: "The best way to think about this is that the "gcd" of two natural numbers is the meet of them in the lattice of natural numbers ordered by divisibility. Note that $0$ is the top element in the divisibility order. The meet of the top element with itself is itself. So $0 = \gcd(0, 0)$ is the answer. 'Greatest' is an unfortunate misnomer in this case."
The book Mathematics Made Difficult has a nice little section on this. It should perhaps better be called "highest common factor" (hcf).
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You lost the race to William. – dfeuer Sep 16 '13 at 5:05
we all lost to him :) – DanielY Sep 16 '13 at 5:06
@dfeuer Well, I had answered the question already at MO. I wasn't racing. – user43208 Sep 16 '13 at 5:06
Win or lose, "the meet of [two numbers] in the lattice of natural numbers ordered by divisibility" is a definition of delightful pithiness. – Jordan Gray Sep 16 '13 at 11:00
Another way to think about this is ideals. The gcd of two natural numbers $a, b$ is the unique non-negative natural number that generates the ideal $\langle a, b \rangle$. So in this case, $\langle 0 ,0 \rangle$ is just the $0$ ideal so the gcd is $0$.
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Defining $\gcd(a,b)$ for $\def\Z{\mathbf Z}a,b\in\Z$, to be the non-negative generator of the ideal $a\Z+b\Z$ gives $\gcd(a,0)=|a|$ for all $a$, including for $a=0$. Similarly one can define $\def\lcm{\operatorname{lcm}}\lcm(a,b)$ to be the non-negative generator of the ideal $a\Z\cap b\Z$, which gives $\lcm(a,0)=0$ for all$~a$ (note that since this is the only common multiple in this case, it is unlikely to provoke much discussion).
Adding these cases to the usual definitions of the $\gcd$ and $\lcm$ for nonzero integers causes no problems; all usual formulas remain valid. In fact, if one wants the rule $\gcd(xy,xz)=|x|\gcd(y,z)$ to hold for all integers $x,y,z$, one is forced to put $\gcd(0,0)=0$.
On the other hand, I don't think it is absolutely vital for mathematics to have $\gcd(0,0)$ defined, in the same way as $0+0$, $0\times0$ and $0^0$ need to be defined (and $0/0$ needs to be undefined) in order for the usual rules of algebra that one uses all the time to be valid. I think it would suffice to qualify the rule I cited by "$x\neq0$" if one wants to leave $\gcd(0,0)$ undefined; other rules like $\gcd(a,b)=\gcd(a,a-b)$ do not seem to equate $\gcd(0,0)$ with something else. The reason that leaving it undefined is not so dramatic is that when considering divisibility, $0$ is often excluded anyway; for instance it has to be put aside in the theorem of Unique Factorisation.
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Simply said - this depends on your definition.
Clearly, if $d=\gcd(a,b)$, you require $d\mid a$, $d\mid b$, i.e., it is a common divisor.
But there are two possibilities how to express that it is greatest common divisor.
One of them is to require $$c\mid a \land c \mid b \Rightarrow c\le d$$ and the other one is $$c\mid a \land c \mid b \Rightarrow c\mid d.$$
Clearly, if you use the first definition, $\gcd(0,0)$ would be the largest integer, so it does not exists. If you use the second one, you get $\gcd(0,0)=0$. (Note that $0$ is the largest element of the partially ordered set $(\mathbb N,\mid)$.)
As far as I can say, the first definition appears in some text which are "for beginners"; for example here. (It was one of the first results from Google Books when searching for "gcd(0,0)".)
I would say that for students not knowing that $\mid$ is in fact a partial order, the first definition might feel more natural. But once you want to use this in collection with more advanced stuff (for example, g.c.d. as generator of an ideal generated by $a$ and $b$), then the second definition is better.
-
This helped. I still feel a little icky about (ℕ,∣) as a partial order, because 0∣0 feels weird to me, and this whole argument requires that reflexivity. – rampion Apr 8 at 0:46
In the general framework of integral domains (commutative rings with unity, without nontrivial zero-divisors), we define a greatest common divisor of two elements $a,b$ of an integral domain $R$ as any element $c\in R$ satisfying:
• $c$ divides both $a$ and $b$, that is, there are $x,y\in R$ such that $a=cx$ and $b=cy$.
• If $d\in R$ divides both $a$ and $b$, then $d$ divides $c$.
We write $c=\gcd(a,b)$ in this case. The definition implies that if $c,c^\prime=\gcd(a,b)$, then $c$ and $c^\prime$ are associates, that is $c=uc^\prime$ for some invertible element $u$ in $R$. This is equivalent to say that the principal ideals generated by $c$ and $c^\prime$ are the same. Therefore a non-ambiguous definition is as follows:
"$\gcd(a,b)$ is $\$ any minimal$\$ the minimum element in the family $\mathcal F=\{Rd: Rd\supseteq Ra+Rb\}$ of principal ideals containing $a$ and $b$, ordered by inclusion, wherever such minimum ideal exists."
Since $R0+R0=R0$, we see that $R0=0$, the trivial ideal of $R$, is the $\gcd$ of $0$ and $0$.
In general you cannot guarantee that $\gcd$s exist. A sufficient condition is your integral domain $R$ to be a UFD (Unique Factorization Domain).
-
Use \gcd instead of $GCD$. Also, ^\prime can be abbreviated all the way down to '. – dfeuer Sep 16 '13 at 5:11
I've tried editing it myself, but you've interrupted me twice, so I'll let you do it. – dfeuer Sep 16 '13 at 5:11
@dfeuer Thanks for the suggestion. By the way, that ' shortcut works in ordinary $\LaTeX$? – Matemáticos Chibchas Sep 16 '13 at 5:12
Yep! It even works in Plain $\TeX$! $x'''$ just takes four keystrokes (six if you count dollar signs). – dfeuer Sep 16 '13 at 5:16
anon- Not on my keyboard. On my keyboard, backtick is the key to the left of 1, and tilde is obtained by holding shift and pressing the hash key (which is next to Enter). In general, the location of punctuation characters varies in keyboards used in different countries. – Hammerite Sep 16 '13 at 9:21
From Wikipedia:
The greatest common divisor of $a$ and $b$ is well-defined, except for the case $a=b=0$, when every natural number divides them.
-
Well, this isn't really the right answer, is it? William got it right. – user43208 Sep 16 '13 at 5:22
All I'm saying is that Wikipedia got it wrong in this case (or at least it's misleading to suggest that $\gcd(0, 0)$ is not well-defined, because as has been clearly explained, it's perfectly well-defined and indeed it's $0$). – user43208 Sep 16 '13 at 5:32
The guy got his answer, that's the most important :) – DanielY Sep 16 '13 at 5:33
If we take Euclid's algorithm:
$\text{gcd}(a, b) = \left\{ \begin{array}{l l} a & \quad \text{if$b = 0$}\\ \text{gcd($b$,$a$mod$b$)} & \quad \text{otherwise} \end{array} \right.$
as the definition of GCD, then $\text{gcd}(a, 0) = 0$ for any $a$, because the stopping case of $b = 0$ is reached immediately. If $a = 0$, then we get $\text{gcd}(0, 0) = 0$.
So, it's possible that Wolfram Alpha obtains its result as a side effect of using Euclid's algorithm rather than deliberate thought as to what gcd(0, 0) should return. But it does fortunately coincide with William's “partial order of divisibility” explanation.
- | 2015-11-27T08:59:33 | {
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http://mathhelpforum.com/calculus/165708-stokes-theorem-double-integral.html | # Math Help - Stokes Theorem - double integral
1. ## Stokes Theorem - double integral
Use Stokes’ Theorem to evaluate $\int_c^.F \cdot dr$
C: is the boundary of the portion of $z = x^2 + y^2$ below z = 4, oriented downward,
$F = $
Can someone please check over my work thanks!
First I found N --> $N = \frac{<2x, 2y, -1>}{\sqrt{4x^2 + 4y^2 + 1^2}}$
Then i found $\bigtriangledown \times F = <0, 0, -1>$
from using the formula $\bigtriangledown \times F = $
Now $\bigtriangledown \times F \cdot N = \frac{1}{\sqrt{4x^2 + 4y^2 + 1^2}}$
Then $dS = {\sqrt{4x^2 + 4y^2 + 1}}dA$
So putting it all together....
$\int\int_R^. \frac{1}{\sqrt{4x^2 + 4y^2 + 1^2}} \cdot \sqrt{4x^2 + 4y^2 + 1}$
The square roots cancel and reduce to....
$\int\int_R^. 1 dA$
This double integral of 1 dA is equal to the Area of the Region
And using $z = x^2 + y^2$ and $z = 4$ we come up with $r^2 = 4 .... r = 2$
$A(R) = \pi r^2 = \pi(2^2) = 4\pi$
Does $4\pi$ look correct for the answer?
Thanks for looking
2. I've always disliked the notation $\vec{n}dS$ because used that unthinkingly results in doing two square roots which, just as you see here, cancel!
Instead think of $d\vec{S}$ as the "vector differential of surface area" and calculate it directly. We can write the surface $z= x^2+ y^2$ in a vector equation with x and y as parameter: $\vec{r}(x,y)= x\vec{i}+ y\vec{j}+ z\vec{k}= x\vec{i}+ y\vec{j}+ (x^2+ y^2)\vec{k}$. The derivatives with respect to x and y give two tangent vectors, $\vec{i}+ 2x\vec{k}$ and $\vec{j}+ 2y\vec{k}$. The cross product of those gives the "vector differential of surface area": $2x\vec{i}+ 2y\vec{j}- \vec{k}$ where I have chosen the order of multiplication to get a negative value for $\vec{k}$, "oriented downward".
You are correct that $\nabla\times \vec{F}= -\vec{k}$ so that $\nabla\vec{F}\cdot\vec{n} dS= \nabla\vec{F}\cdot d\vec{S}$[tex]= (0(2x)+ 0(2y)- 1(-1))dxdy= dxdy.
That gives $4\pi$, not $-4\pi$.
Your normal vector, $\frac{-2x\vec{i}-2y\vec{j}+ \vec{k}}{\sqrt{4x^2+ 4y^2+ 1}}$ was oriented upward, not downward. | 2014-03-08T00:24:37 | {
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http://wopsr.net/palindromes/ | ## Palindromes
Problem 4 at Project Euler poses the following challenge:
A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99.
Find the largest palindrome made from the product of two 3-digit numbers.
### A Naïve Approach
A naïve solution would be to multiply all the three-digit numbers by one another, pull out all the palindromes, and find the greatest:
var isPalindrome = function(n){
n = n.toString();
return n === n.split("").reverse().join("");
};
var findBiggest = function(){
var allProducts = [];
for (var n = 100; n < 1000; n++){
for (var m = 100; m < 1000; m++){
allProducts.push(m * n);
}
}
var palindromes = allProducts.filter(isPalindrome);
return Math.max.apply(null, palindromes);
};
But the naïve appraoch means making a huge-ass array with 810,000 numbers in it, consuming 3.09MiB of memory and taking many milliseconds to build. We can do better with a little number theory.
#### Don’t include obvious duplicates
$m \times n = n \times m$
Instead of starting m at 100 for each value of n, start it at n. This avoids duplicates and cuts array size roughly in half, to only 405,450 entries.
#### Don’t include multiples of 10
Observe that $n \times 10$ always ends in $0$. We don’t ordinarily write positive non-zero integers in decimal with leading zeros, so we can safely exclude all multiples of 10 from $n$ and $m$, saving us another 76,995 entries in allProducts, almost 19%, for a new total of 328,445 entries.
#### Assume the greatest palindrome has six digits
This is a pretty safe assumption in the instance the Project Euler problem describes, but may not be safe in other situations. Of our remaining 404,550 products of two three-digit decimal numbers, only 69,787 (17.25%) of them do not have six digits. It turns out these 17.25% include over 60% of the palindromes, but all we need is one six-digit palindrome in there to safely ignore the five-digit ones. $101,101 = 143\times707$ and has six digits; that’s enough to let us dismiss five-digit entries.
But how do we exclude five-digit products from our array without having to spend time calculating those products? We don’t. But limiting ourselves to condisering only six-digit products (actually, to products with an even number of digits) lets us do something nifty.
As it turns out, palindromic numbers with an even number of digits have a nifty property: they all share a common, easily identifiable factor.
Let $\mathbb{P}\subset\mathbb{N}$ be the set of palindromic numbers with $k+1$ digits $a_0...a_k$ in an arbitrary base $b:b\geq2$:
$\mathbb{P}=\left\{ n\in\mathbb{N}:n=\sum\limits_{i=0}^{k}a_ib^i\middle|a_i=a_{k-i}\right\}$
Note that $n$ is palindromic if and only if $a_i=a_{k-i}$. By way of example:
$$$$\begin{split}987,789&=9(10^0)\\&+8(10^1)\\&+7(10^2)\\&+7(10^3)\\&+8(10^4)\\&+9(10^5)\end{split}$$$$ $$$$\begin{split}65,456&=6(10^0)\\&+5(10^1)\\&+4(10^2)\\&+5(10^3)\\&+6(10^4)\end{split}$$$$
Looks like a pattern! Turns out we can rewrite the sum like this, but only for palindromes with an even number of digits:
$n=\sum\limits_{i=0}^{\frac{k+1}{2}}a_i\left(b^i+b^{k-i}\right)\Big|k+1\text{ is even}$
Essentially, we pair up the digit $a_i$ with its matching digit $a_{k-i}$ and thereby only deal with half the digits. This doesn’t work with palindromes with an odd number of digits because of the pesky middle digit, which has no mate.
$$$$\begin{split}987,789&=9(10^0+10^5)\\&+8(10^1+10^4)\\&+7(10^2+10^3)\end{split}$$$$ $$$$\begin{split}65,456&=6(10^0+10^4)\\&+5(10^1+10^3)\\&+4(10^2)\end{split}$$$$
For palindromes with an even number of digits, we can factor the addend:
$a_i(b^i+b^{k-i})=a_ib^i(1+b^{k-2i})$
Check it out! We can make $1+b^{k-2i}=0$ with modular arithmetic!
\begin{alignat*}{2}\begin{split}1+b^{k-2i}&=1-1^{k-2i}&\mod{b+1}\\&=1-1&\mod{b+1}\\&=0&\mod{b+1}\end{split}\end{alignat*}
Every palindrome with an even number of digits is evenly divisible by $b+1$, where $b$ is the radix. We’re working in decimal, so $b+1=11$.
We needn’t bother with any products but those that are divisible by 11. 11 being prime makes things easy. We'll only multiply $m\times n$ and push() the product to allProducts if at least one of $\lbrace m,n\rbrace$ is divisible by 11. This brings allProducts.length down to only 55,764, a reduction of over 93.1% from our original 810,000.
### Putting it All Together
All this makes for lots of savings in computation and storage, but it makes our program a lot uglier:
var isPalindrome = function(n){
n = n.toString();
return n === n.split("").reverse().join("");
};
var findBiggest = function(){
var allProducts = [];
var start, end, step;
for (var n = 101; n < 1000; n++){
if (n % 10){
if (n % 11){
start = 11*Math.ceil(n/11);
end = 11*Math.floor(1000/11);
step = 11;
} else {
start = n;
end = 1000;
step = 1;
}
}
for (var m = start; m < end; m+=step){
if (m % 10){
allProducts.push(m * n);
}
}
}
var palindromes = allProducts.filter(isPalindrome);
return Math.max.apply(null, palindromes);
};
Messy, but fast. The naïve version ran in 1,158.8ms, while this version runs in only 74.2ms. | 2017-04-30T12:52:58 | {
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https://ocw.mit.edu/courses/mathematics/18-065-matrix-methods-in-data-analysis-signal-processing-and-machine-learning-spring-2018/video-lectures/lecture-15-matrices-a-t-depending-on-t-derivative-da-dt/ | Lecture 15: Matrices A(t) Depending on t, Derivative = dA/dt
Flash and JavaScript are required for this feature.
Description
This lecture is about changes in eigenvalues and changes in singular values. When matrices move, their inverses, their eigenvalues, and their singular values change. Professor Strang explores the resulting formulas.
Summary
Matrices $$A(t)$$ depending on $$t /$$Derivative $$= dA/dt$$
The eigenvalues have derivative $$y(dA/dt)x$$.
$$x$$ = eigenvector, $$y$$ = eigenvector of transpose of $$A$$
Eigenvalues from adding rank-one matrix are interlaced.
Related section in textbook: III.1-2
Instructor: Prof. Gilbert Strang
The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu.
GILBERT STRANG: So I've worked hard over the weekend. I figured out what I was doing last time and what I'm doing this time and improved the notes. So you'll get a new set of notes on the last lecture and on this one. And I kind of got a better picture of what we're doing. And that board is aiming to describe the large picture of what we're doing last time and this time.
So last time was about changes in A inverse when A changed. This time is about changes in eigenvalues and changes in singular values when A change. As you can imagine, this is a natural important situation. Matrices move, and therefore, their inverses change, their eigenvalues change, their singular values change. And you hope for a formula.
Well, so we did have a formula for last time for the change in the inverse matrix. And I didn't get every u and v transpose in the right place in the video or in the first version of the notes, but I hope that that formula, that that Woodbury Morrison formula will be correct this time. So I won't go back over that part.
But I realize also there is another question that we can answer when the change is very small, when the change in A is dA or delta A, a small change. And that's, of course, what calculus is about. So I have to sort of parallel topics here. What is the derivative when the change is infinitesimal? And what is the actual change when the change is finite size?
So now, let me say what we can do and what we can't do. Oh, I'll start out by figuring out what the derivative is for the inverse. So that's like completing the last time for infinitesimal changes. Then I'll move on to changes in the eigenvalues and singular values. And there, you cannot expect an exact formula. We had a formula that was exact, apart from any typos, for this. And we'll find a formula for this, and we'll find a formula for that and for that. Well, that one will come from this one. So this will be a highlight today. How do the eigenvalues change when the matrix changes?
But we won't be able to do parallel to this, we won't be able to-- oh, we will be able to do something for finite changes. That's important. Mathematics would have to keep hitting that problem until it got somewhere. So I won't get an exact formula for that change. That's too much.
But I'll get inequalities. How big that change could be. What can I say about it? So these are highly interesting.
May I start with completing the last lecture? What is the derivative of the inverse? So I'm thinking here, so what's the setup? The setup is my matrix A depends on time, on t. And it has an inverse. A inverse depends on t.
And if I know this dependence, in other words, if I know dA dt, how the matrix is depending on t, then I hope I could figure out what the derivative of A inverse is. We should be able to do this. So let me just start with-- it's not hard and it complements this one by doing the calculus case, the infinitesimal change. So I want to get to that.
I can figure out the change in A. And my job is to find the derivative of A inverse. So here's a handy identity. Can I just put this here? So here's my usual identity.
So as last time, I start with a finite change because calculus always does that, right. It starts with a delta t and then it goes to 0. So here I am up at a full size change. So I think that this is equal to B inverse A minus B A inverse. And if it's true, it's a pretty cool formula.
And, look, it is true, because over on this right-hand side, I have B inverse times A A inverse. That's the identity. So that's my B inverse. And I have the minus, the B inverse B is the identity. There's A inverse. It's good, right?
So from that, well, I could actually learn from that the rank of this equals the rank of this. That's a point that I made from the big formula. But now, we can see it from an easy formula. Everywhere here, I'm assuming that A and B are invertible matrices. So when I multiply by an invertible matrix, that does not change the rank. So those have the same ranks.
But I want to get further than that. I want to find this. So how do I go? How do I go forward with that job to find the derivative of the inverse?
Well, I'm going to call this a change in A inverse. And over here, I'll have B will be-- yeah, OK, let's see, am I right? Yeah.
So B inverse will be-- this is A plus delta A inverse. And this is-- well, that's A minus B. So that's really minus delta A. From A to B is the change. Here, I'm looking at the difference A minus B. So it's minus a change. And here, I have A inverse. I haven't done anything except to introduce this delta and get B out of it and brought delta in.
Now, I'm going to do calculus. So I'm thinking of B as there's a sort of a delta t. And I'm going to divide by both sides by delta t. I have to do this if I want-- and now, I'll let delta t go to 0. So calculus appears. Finally, our-- I won't say enemy calculus, but there is a sort of like competition between linear algebra and calculus for college mathematics. Calculus has had far, far too much time and attention. It like it gets three or four semesters of calculus for people who don't get any linear algebra. I'm glad this won't be on the video, but I'm afraid it will.
Anyway, of course, calculus is fine in its place. So here's its place. Now let delta t go to 0. So what does this equation become?
Then everybody knows that as the limit of delta t goes to 0, I replace deltas by-- so this delta A divided by delta t that has a meaning. The top has a meaning and the bottom has for me. But then the limit, it's the ratio that has a meaning.
So dA by itself, I don't attach a meaning to that. That's infinitesimal. It's the limit, so that's why I wanted a delta over a delta so I could do calculus. So what happens now is delta t goes to 0. And, of course, as delta t goes to 0, that carries delta A to 0. So that becomes A inverse.
And what does this approach as delta t goes to 0? dA dt with that minus sign. Oh, I've got to remember the minus sign. The minus sign is in here. So I'm bringing out the minus sign.
Then this was A inverse, as we had. And that's dA dt. And that's A inverse. That's our formula, a nice formula, which sort of belongs in people's knowledge.
You recognize that if A was a 1 by 1 matrix, we could call it x, instead of A. If A was a 1 by 1 matrix x, then I'm saying the formula for the derivative of 1 over x, right? A inverse just 1 by 1 case is just 1 over x. So the derivative of 1-- or maybe t, I should be saying.
If A is just t, then the derivative of 1 over t with respect to t is....? Is minus 1 over t squared. The 1 by 1 case we know. That's what calculus does. And now we're able to do the n by n case. So that's just like good.
And then it's sort of parallel to formulas like this, where this delta A has not gone to 0. It's full size, but low rank. That was the point. Actually, the formula would apply if the rank wasn't low. But the interest is in low rank here.
Are we good for this? That's really the completion of last time's lecture with derivatives.
OK, come back to here, to the new thing now, lambdas. Let's focus on lambdas, eigenvalues. How does the eigenvalue change when the matrix changes? How does the eigenvalue change when the matrix changes?
So I have two possibilities. One is small change when I'm doing calculus and I'm letting a delta t go to 0. The other is full size, order 1 change, where I will not be able to give you a formula for the new lambdas, but I'll be able to tell you important facts about them. So this is today's lecture now. You could say that's the completion of Friday's lecture.
What about d lambda dt? It's a nice formula. Its proof is fun too. I was very happy about this proof. OK, so I guess calculus is showing up here on this middle board.
So how do I start with the eigenvalues? Well, start with what I know. So these are facts, you could say, that I have to get the eigenvalues into it. And, of course, eigenvalues have to come with eigenvectors.
So I'll again use A of t. It will be depending on t. And an eigenvector that depends on t is an eigenvalue that depends on t times and eigenvector that depends on t. Good? That's fact one that we plan to take the derivative of somehow.
There's also a second fact that comes into play here. What's the deal on the eigenvalues of A transpose? They are the same. The eigenvalues of A transpose are the same as the eigenvalues of A.
Are the eigenvectors the same? Not usually. Of course, if the matrix was symmetric, then A and A transpose are just the same thing. So A transpose would have that eigenvalue-- eigenvector.
But, generally, it has a different eigenvector. And really to keep a sort of separate from this one, let's call it y. It will have the same eigenvalue. I'm going to call it y. But I'm going to make it a row vector, because A transpose is what-- instead of writing down A transpose, I'm going to stay with A, but put the eigenvalue on the left side.
So here's is the eigenvalue-- eigenvector for A on the left. And it has the same eigenvalue times that eigenvector. But that eigenvector is a row eigenvector, of course. This is an equality between rows. A row times my matrix gives a row. So that's the eigenvalues of-- and it has the same eigenvalues.
So this is totally parallel to that, totally parallel. And maybe sort of less-- definitely less seen, but it's just the same thing for A transpose. Everybody sees that if I transpose this equation, then I've got something that looks like that. But I'd rather have it this way.
Now, one more fact I need. There is-- there has to be some normalization. What should be the length of these? Right now, x could have any length. y could have any length. And there's a natural normalization, which is y transpose times x equal to 1. That normalizes the two. It doesn't tell me the length of x or the length of y. But it tells me, the key thing, the length of both.
So what I've got there is tracking along one eigenvalue and its pair of eigenvectors. And you're always welcome to think of the symmetric case when y and x are the same. And then I would call them q. Oh, well, I would call them q if it was a symmetric matrix. So if it's a symmetric matrix, both eigenvectors would be called q. And this would be saying that q is a--
AUDIENCE: Unit vector.
GILBERT STRANG: Unit vector, right. So this is all stuff we know. And actually, maybe I should write it in matrix notation, because it's important. That's for one eigenvector. This is for all of them at once. Everybody's with it? The x's are the columns of x. And lambda is the diagonal matrix of lambdas. And it has to sit on the right so that it will multiply those columns. So this is like all eigenvectors at once.
What would this one be? This would be like y transpose A equals A--
AUDIENCE: y transpose inverse?
GILBERT STRANG: y transpose, yes-- equals-- and probably these are multiplied-- I feel wrong if I write y transpose here. Like here, the x was on the right and on the left. And I'll-- oh, yeah, y transpose, yeah. OK, so what do I put? Lambda y transpose. Thanks.
And what do I put here? What does this translate to if this was for one eigenvector? For all of them at once, it's just going to translate to y transpose x equal the identity.
This is pretty basic stuff. But stuff somehow we don't always necessarily see. Those are the key facts.
And now, I plan to take the derivative, take the derivative of respect to lambda. Oh, I can derive one more fact. So this would be a formula. This is formula 1. Formula 1 just says, what do I get if I hit this on the left by y transpose? Can I do that? y transpose of t A of t x of t equals lambda of t. That's a number. So I can always bring that out in front of the inner product of vector notation.
Are you good for that? I'm pleading like everything I've done is totally OK. And now, I have a improvement to make on this right-hand side, which is...? So what is y transpose times x?
AUDIENCE: 1.
GILBERT STRANG: It's 1. So let's remember that. It's 1. So in other words, I have got a formula for lambda of t. As time changes, the matrix changes. Its eigenvalues change according to this formula. Its eigenvectors change according of this formula. And its left eigenvectors change according to that formula. So everything here is above board.
And now, what's the point? The point is I'm going to find this, the derivative. So I'm going to take the derivative of that equation and see what I get. That'll be the formula for the derivative of an eigenvalue. And amazingly, it's not that widely known. Of course, it's classical, but it's not always part of courses. So this is as time varies, the matrix varies, A. And therefore, its eigenvalues vary, and its eigenvectors vary.
So we're going to find d lambda dt. It's one level of difficulty more to find dx dt, the derivative of the eigenvector or the second derivative of the eigenvalue. Those kind of come together. And I'm not going to go there. I'm just going to do the one great thing here-- take the derivative of that equation.
Shall I do it over there? So here we go. So I want to compute d lambda dt. And I'm using this formula for lambda there.
So I've got three things that depend on t. And I'm taking the derivative of their product. So I'm going to use the product rule. I'll apply the product rule to that derivative.
Take the derivative of the first guy times A times x. Take the derivative of the second guy times the second guy and the third guy. y transpose of t A of t dx dt. OK? We are one minute away from a great formula. And I'm really happy if you allow me to say it. That that formula comes by just taking those facts we know, putting them together into this expression that we also know, and this is like lambda equals x inverse Ax. That's a diagonalizing thing and then taking the derivative.
So what do I get if I take that derivative? Well, this term I'm going to keep. I'm not going to play with that. Everybody is clear? That's a number. Here's a matrix. dA dt is a matrix. I take the derivative of every entry in A.
Here's its column vector, its eigenvector. And here's a row vector. So row times matrix times column is a number, 1 by 1. And actually, that's my answer. That's my answer.
So I'm saying that these two terms cancel each other out as those two terms added to zero. This is the right answer for the derivative. That's a nice formula.
So to find the derivative of an eigenvalue, the matrix is changing, you multiply by the eigenvector and by the left eigenvector. It gives you a number. And that's the d lambda dt.
So why do those two guys add to 0? That's all that remains here. And then this topic is ended with this nice formula. So I want to simplify that, simplify that, and show that they cancel each other.
So what is Ax? It's lambda x. So this guy is nothing but it's lambda that depends on time of course times dy dt dy transpose dt. I'm just copying that. Ax is lambda x. Sorry, I didn't mean to make that look hard.
You OK with that? Ax is lambda x. And I am perfectly safe, because lambda is just a number to bring it out to the left. So it doesn't look like it's in the way.
And what about this other term? So I have y transpose-- oh, y transpose A, what's that? What's y transpose A? That's the combination that I know. y transpose A, y is that left eigenvalue. y transpose A brings out a lambda. So this also brings out a lambda times y transpose times dx dt. OK? I just use Ax equal lambda x there. It was really nothing.
Now, what do I do? I want this to be 0. Can you see it happening? It's a great pleasure to see it happening.
So what do I have here? What's my first step now?
AUDIENCE: Like take lambda--
GILBERT STRANG: Bring lambda outside. That's not 0. We don't know what that is. Bring lambda outside there times the whole thing.
So for some wonderful reason I believe that this number, which is a row times a column, a row times a column, two terms there, I believe they knock each other out and that result is 0. And why? Why? Because I come back to-- this board has all that I know. And here's y transpose times x equal 1.
And how does that help me? Because what I'm seeing in that square, in those brackets is?
AUDIENCE: The derivative of y transpose--
GILBERT STRANG: The derivative of the y transpose x. So it's the derivative of?
AUDIENCE: 1
GILBERT STRANG: 1. Therefore, 0. So this is the derivative of 1. It equals 0. Those two terms knock each other out and leave just the nice term that we're seeing.
So the derivative of the eigenvalue, just to have one more look at it before we leave it. The derivative of the eigenvalue is this formula. It's the rate at which the matrix is changing times the eigenvectors on right and left. Sometimes they're called the right eigenvector and the left eigenvector at the time t.
So we're not saying in this d lambda dt. In other words, I get a nice formula, which doesn't involve the derivative of the eigenvector. That's the beauty of it. If I want to go up to take the next step-- I tried this weekend, but it's a mess.
It would be to take the-- so this is my formula then, d lambda dt equals this. And I can take the next derivative of that, and it will involve d second dt squared. But it will also involve dx dt and dy dt. And in fact, a pseudo inverse even shows up. It's another step, and I'm not going that far, because we've got the best formula there.
So now that has answered this question. And I could answer that question the same way. It would involve A transpose A and the singular vectors, instead of involving A and the eigenvectors. Maybe that's a suitable exercise. I don't know. I haven't done it myself.
What I want to do is this, now say, what can we say about the change in the eigenvalue-- and I'll just stay first of all with eigenvalue-- when the change is like rank 1? This is a perfect example when the change is rank 1.
So what can we say about the eigenvalues-- let's take the top, the largest eigenvalue, or all of them, all of them, lambda j, all of them-- of A plus a rank 1 matrix uv transpose. Oh, let's do the nice case here, the nice case, because if I allow a general matrix A, I have to worry about does it have enough eigenvectors? Can it diagonalize? All that stuff. Let's make it a symmetric matrix. And let's make the rank 1 change symmetric too.
So the question is, what can I say about the eigenvalues after a rank 1 change? So again, this isn't calculus now, because the change that I'm making is a true vector and not a differential. And I'm not going to have an exact formula for the new eigenvalues, as I said.
But what I am going to do is write down the beautiful facts that are known about that. And here they are. So, first of all, the eigenvalues are in descending order. We use descending order for singular values. Let's use them also for eigenvalues. So lambda 1 is greater or equal to lambda 2, greater or equal to lambda 3, and so on.
Oh, give me-- give me an idea. What do you expect from that rank 1 change? So that change is rank 1. Can you tell me any more about that change, u u transpose? What kind of a matrix is u u transpose? It's rank 1, but we can say more. It is...?
AUDIENCE: Symmetrical.
GILBERT STRANG: Symmetric, of course. And it is...? Yeah?
AUDIENCE: Positive semidefinite.
GILBERT STRANG: Positive semidefinite. Positive semidefinite. This is a positive change. u u transpose is the typical rank 1 positive semidefinite. It couldn't be positive definite, because it's only got rank 1.
What's the eigenvector of that matrix? Let's just-- why not here? We can do this in two seconds. So u u transpose, that's the matrix I'm asking you to think about. And it's a full n by n matrix, column times a row. Tell me an eigenvector of that matrix. Yes?
AUDIENCE: u.
GILBERT STRANG: u. If I multiply my matrix by u, I get-- what do I get? I get some number times u. And what is that number lambda?
AUDIENCE: u transpose u.
GILBERT STRANG: That lambda happens to be u transpose u. So that's different from u u transpose. This is a matrix. This is 18.065 now. That's a number. And what can you tell me about that number? It is...?
AUDIENCE: Greater than or equal to 0.
GILBERT STRANG: Greater-- well, even more. Greater than 0. Greater, because this is a true vector here. So this is greater than 0. It's the only eigenvalue-- all the other eigenvalues of that rank 1 matrix are zero. But the one non-zero eigenvalue is over on the plus side. It's u transpose u. We all recognize that as the length of u squared. It's certainly positive. So we do have a positive semidefinite definite matrix.
What would your guess be of the effect on the eigenvalues of A? So I'm coming back to my real problem-- eigenvalues of S, sorry, S. Symmetric matrices, I'm saying symmetric.
What is your guess if I have a symmetric matrix and I add on u u transpose? What do you imagine that does to the eigenvalues? You're going to get it right. Just say it. What happens to the eigenvalues of S if I add on u u transpose? They will...?
AUDIENCE: More positive.
GILBERT STRANG: They'll be more positive. They'll go up. This is a positive thing. It's like adding 17 to something. It moves up. So therefore, what I believe is-- so I've got two sets of eigenvalues now. One is the eigenvalues of s. The other is the different eigenvalues of S. So I can't call them both lambdas or I'm in trouble. So do you have a favorite other Greek letter for the eigenvalues of S?
AUDIENCE: Gamma.
GILBERT STRANG: Gamma. OK, gamma. As long as you say a Greek letter that I have some idea how to write. Zeta, it seems to me, is like the world's toughest letter to write. And electrical engineers can coolly flush off a zeta. I've never succeeded. So I'll write-- what did you say?
AUDIENCE: Gamma.
GILBERT STRANG: Gamma j of the original. So those are the eigenvalues of the original. These are the eigenvalues of the modified. And we're expecting the lambdas to be bigger than the gammas. So that's just a qualitative statement. And it's true. Each lambda is bigger than the gamma. Sorry, yeah, yeah, each lambda, by adding this stuff, the lambdas are bigger than-- so I'll just write that. Lambdas are bigger than gammas.
And that's a fundamental fact, which we could prove. But a little more is known. Of course, the question is, how much bigger? How much can they be way bigger? Well, I don't believe they could be bigger by more than that number myself. But there's just better news than that.
So the lambdas are bigger than the gammas. So lambda 1 is bigger than gamma 1. So this is the S plus u u transpose matrix. And these are the eigenvalues of the S matrix. Lambda 1 is bigger than gamma 1.
But look what's happening in this line of text here. I'm saying that gamma 1-- that lambda 2 is smaller than gamma 1. Isn't that neat? The eigenvalues go up. But they don't just like go anywhere. And that's called interlacing.
So this is one of those wonderful theorems that makes your heart happy, that if I do I rank one change and it's a positive change, then the eigenvalues increase, but they don't increase-- the new eigenvalue is below the new second eigenvalue. It doesn't pass up the old, first eigenvalues. And the new third eigenvalue doesn't pass up the old second eigenvalue. So that's the interlacing theorem that's associated with the names of famous math guys. And of course you have to say that's beautiful.
While we're writing down such a theorem, make a guess of what the theorem would be if I do a rank 2 change. Suppose I do an S, staying symmetric. And I do a rank 1 change. But then I also do a rank 2 change, say w w transpose.
So what's the deal here? What do I know about the change matrix, the delta S here? I know its rank is 2. I'm assuming u and w are not in the same direction. So that's a rank 2 matrix.
And what can you tell me about the eigenvalues of that rank 2 matrix? So it's got n eigenvalues because it's an n by n matrix. But how many non-zero eigenvalues has it got? Two, because its rank is 2. The rank tells you the number of non-zero eigenvalues when matrices are symmetric. It doesn't tell you enough. If matrices are unsymmetric, eigenvalues can be weird. So stay symmetric here.
So this has two non-zero eigenvalues. And can you tell me their sign. Is that matrix positive semidefinite? Yes, of course, it is. Of course. So this was and this was. And together it certainly is.
So now, I've added a rank 2 positive semidefinite matrix. And now, I'm not going to rewrite this line, but what would you expect to be true? You would expect that the eigenvalues increase. But how big could gamma-- yeah, so gamma 2, let's follow gamma 2.
Well, maybe I should use another-- do the Greeks have any other letters than lambda and gamma. They must had--
AUDIENCE: Zeta.
GILBERT STRANG: Who? C? Hell with that. Who knows one I can write?
AUDIENCE: Alpha.
GILBERT STRANG: Alpha. Good, alpha. Yes, alpha. Right. So alpha is the eigenvalues of this rank 2 change. OK. Now, what am I going to be able to say? Can I say anything about the-- well, of course, alpha 1 is bigger than lambda 1, which was bigger than-- eigenvalues are going up, right? I'm adding positive definite or positive semidefinite stuff. There's no way eigenvalues can start going down on me.
So alpha 1 is a greater or equal to the lambda 1, which had just a rank 1 change, which is greater or equal to the-- mu, was it mu?
AUDIENCE: Gamma.
GILBERT STRANG: Gamma. Gamma 1, and so on. OK, now, let's see, is gamma 1 bigger than alpha-- what am I struggling to write down here? What could I say? Well, what can I say that reflects the fact that this lambda 2-- or sorry, so gamma 1 went up. Gamma 1 was bigger than lambda 2. That was the point here. Gamma 1 is bigger.
So this was a sort of easy, because I'm adding stuff. I expected the lambda to go up. This is where the theorem is that it didn't go up so far as to pass-- or sorry, the lambda 2, which went up, didn't pass up gamma 1. Lambda 2 didn't pass up gamma 1.
And now let me write those words down. Now the alpha 2-- well, could alpha 2 pass up lambda 1? And what about alpha 3?
Let me say what I believe. I think alpha 2, which is like 1 behind, but I'm adding rank 2, I think alpha 2 could pass up lambda 1. It could pass lambda 1. But alpha 3 can't. I believe that alpha 3 is smaller than lambda 1-- smaller than gamma 1, the original. Got it. Yeah, yeah, yeah.
Anyway, I'll get it right in the notes. You know what question I'm asking. And for me, that's the important thing.
Now, there is a little matter of why is this true? This is the good case. Let me give you another example of interlacing. Can I do that? It really comes from this, but let me give you another example that's just striking.
So I have a symmetric matrix, n by n. Call it S. And then I throw away the last row and column. So in here is S n minus 1. The big matrix was Sn. This one is of size n minus 1. So it's got sort of less degrees of freedom, because the last degree of freedom got removed.
And what do you think about the eigenvalues of the n minus 1 eigenvalues of this and the n eigenvalues of that? They interlace. So this has eigenvalue lambda 1. This would have an eigenvalue smaller than that. This would have an eigenvalue lambda 2. This would have an eigenvalue smaller than that and so on.
Just the same interlacing and basically for the same reason, that when you-- this reduction to size n minus 1 is like I'm saying xn has to be 0 in the energy or any of those expression. And the fact of making xn be 0 is like one constraint, taking one degree of freedom away. It reduces the eigenvalues, but not by two.
OK, now I have one final mystery. And let me try to tell you what. It worried me. Now what is it that worried me? Yes, suppose this change, this u, this change that I'm making, suppose it's actually the second eigenvector of S. So can I write this down?
Suppose u is actually the second eigenvector of S. What do I mean by that? So I mean that S times u is lambda 2 times u.
Now, I'm going to change it. S plus u u transpose, that's what I've been looking at. And that moves the eigenvalues up. But what worries me is like if I multiply this by 20, some big number, I'm going to move that eigenvalue way up, way past. I got worried about this inequality.
When I add this, that same u is lambda 2 plus 20 u. Su is lambda 2u. And this 20 is 20. u is a unit vector.
So you see my worry? Here, I'm doing a rank 1 change. But it's moved an eigenvalue way, way up. So how could this statement be true? So I've just figured out here what gamma-- well, do you see my question? I could leave it as a question to answer next time. Let me do that. And I'll put it online so you'll see it clearly.
It looks like and it happens this eigenvector now has eigenvalue lambda 2 plus 20. Why doesn't that blow away this statement? I'll put that, because it's sort of coming with minus 10 seconds to go in the class, so let's leave that and a discussion of this for next time. But I'm happy with this lecture if you are. Last lecture I got u's and v's mixed up. And it's not reliable. Here, I like the proof of the lambda dt and we're started on this topic 2. Good. Thank you. | 2021-10-16T17:14:30 | {
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https://www.jiskha.com/questions/1506388/the-air-pressure-at-sea-level-is-generally-about-1013-hpa-hectopascals-for-every | Ask questions and get helpful responses.
# math
The air pressure at sea level is generally about 1013 hPa (hectoPascals). For every kilometer that you go up in elevation, the air pressure decreases by 12%. Write an equation that describes the air pressure when the elevation is x kilmoeters above the sea level.
The answer is y = 1013(0.88)^x
The 0.88 is from the 12% (100% - 12% is 88% which is .88 as a decimal)
My question is: how do I know, just by reading this, that it is exponential? Is it because of the multiplication of the .88? Because exponential is multiplication and linear is addition of the same constant amount?
Thank you.
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1. Yes, it would be because of the repeated multiplication by .88
suppose we show a few steps
after 1 km : Pr = 1013(.88)
after 2 km : Pr = 1013(.88)(.88) = 1013 (.88)^2
after 3 km : Pr = 1013(.88)^2 (.88) = 1013 (.88)^3
etc , for
after x km : Pr = 1013 (.88)^x
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4. 🚩
2. Thank you.
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3. You are welcome
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Still need help? You can ask a new question. | 2022-07-05T09:56:41 | {
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http://cers-deutschland.org/check-visa-jjiiani/c142e8-chain-rule-proof-from-first-principles | As $x \to g(c)$, $Q(x) \to f'[g(c)]$ (remember, $Q$ is the. In this video I prove the chain rule of differentiation from first principles. Why didn't Dobby give Harry the gillyweed in the Movie? Well that sorts it out then… err, mostly. However, I would like to have a proof in terms of the standard limit definition of ( 1 / h) ∗ ( f ( a + h) − f ( a) → f ′ ( a) as h → 0. Is it possible to bring an Astral Dreadnaught to the Material Plane? The first is that although ∆x → 0 implies ∆g → 0, it is not an equivalent statement. In what follows though, we will attempt to take a look what both of those. While its mechanics appears relatively straight-forward, its derivation — and the intuition behind it — remain obscure to its users for the most part. To find the rate of change of a more general function, it is necessary to take a limit. Over two thousand years ago, Aristotle defined a first principle as “the first basis from which a thing is known.”4. f ′(x) = h→0lim. Let’s see if we can derive the Chain Rule from first principles then: given an inner function $g$ defined on $I$ and an outer function $f$ defined on $g(I)$, we are told that $g$ is differentiable at a point $c \in I$ and that $f$ is differentiable at $g(c)$. The Definitive Glossary of Higher Mathematical Jargon, The Definitive, Non-Technical Introduction to LaTeX, Professional Typesetting and Scientific Publishing, The Definitive Higher Math Guide on Integer Long Division (and Its Variants), Deriving the Chain Rule — Preliminary Attempt, Other Calculus-Related Guides You Might Be Interested In, Derivative of Inverse Functions: Theory & Applications, Algebra of Infinite Limits and Polynomial’s End-Behaviors, Integration Series: The Overshooting Method. Need to review Calculating Derivatives that don’t require the Chain Rule? Yes, sorry, my symbols didn't really come through quite as I expected. And with the two issues settled, we can now go back to square one — to the difference quotient of $f \circ g$ at $c$ that is — and verify that while the equality: \begin{align*} \frac{f[g(x)] – f[g(c)]}{x – c} = \frac{f[g(x)]-f[g(c)]}{g(x) – g(c)} \, \frac{g(x)-g(c)}{x-c} \end{align*}. Differentiation from first principles . More importantly, for a composite function involving three functions (say, $f$, $g$ and $h$), applying the Chain Rule twice yields that: \begin{align*} f(g[h(c)])’ & = f'(g[h(c)]) \, \left[ g[h(c)] \right]’ \\ & = f'(g[h(c)]) \, g'[h(c)] \, h'(c) \end{align*}, (assuming that $h$ is differentiable at $c$, $g$ differentiable at $h(c)$, and $f$ at $g[h(c)]$ of course!). By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. Math Vault and its Redditbots enjoy advocating for mathematical experience through digital publishing and the uncanny use of technologies. In which case, we can refer to $f$ as the outer function, and $g$ as the inner function. In this position why shouldn't the knight capture the rook? This is awesome . Does a business analyst fit into the Scrum framework? That was a bit of a detour isn’t it? The derivative is a measure of the instantaneous rate of change, which is equal to. No matter which pair of points we choose the value of the gradient is always 3. It is f'[g(c)]. It is very possible for ∆g → 0 while ∆x does not approach 0. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. This is done explicitly for a … Asking for help, clarification, or responding to other answers. But then you see, this problem has already been dealt with when we define $\mathbf{Q}(x)$! chainrule. Matthew 6:25-34 A. then $\mathbf{Q}(x)$ would be the patched version of $Q(x)$ which is actually continuous at $g(c)$. Dance of Venus (and variations) in TikZ/PGF. Let’s see… How do we go about amending $Q(x)$, the difference quotient of $f$ at $g(c)$? but the analogy would still hold (I think). Remember, g being the inner function is evaluated at c, whereas f being the outer function is evaluated at g(c). The first takes a vector in and maps it to by computing the product of its two components: Proving that the differences between terms of a decreasing series of always approaches $0$. And as for you, kudos for having made it this far! When x changes from −1 to 0, y changes from −1 to 2, and so. For example, sin (x²) is a composite function because it can be constructed as f (g (x)) for f (x)=sin (x) and g (x)=x². Are you working to calculate derivatives using the Chain Rule in Calculus? So, let’s go through the details of this proof. Why didn't NASA simulate the conditions leading to the 1202 alarm during Apollo 11? thereby showing that any composite function involving any number of functions — if differentiable — can have its derivative evaluated in terms of the derivatives of its constituent functions in a chain-like manner. And if the derivation seems to mess around with the head a bit, then it’s certainly not hard to appreciate the creative and deductive greatness among the forefathers of modern calculus — those who’ve worked hard to establish a solid, rigorous foundation for calculus, thereby paving the way for its proliferation into various branches of applied sciences all around the world. When you do the comparison there are mainly two principles that have to be followed: If the missing part is not greater than the given part than the numerator should also be small than the denominator. As a token of appreciation, here’s an interactive table summarizing what we have discovered up to now: Given an inner function $g$ defined on $I$ and an outer function $f$ defined on $g(I)$, if $g$ is differentiable at a point $c \in I$ and $f$ is differentiable at $g(c)$, then we have that: Given an inner function $g$ defined on $I$ and an outer function $f$ defined on $g(I)$, if the following two conditions are both met: Since the following equality only holds for the $x$s where $g(x) \ne g(c)$: \begin{align*} \frac{f[g(x)] – f[g(c)]}{x -c} & = \left[ \frac{f[g(x)]-f[g(c)]}{g(x) – g(c)} \, \frac{g(x)-g(c)}{x-c} \right] \\ & = Q[g(x)] \, \frac{g(x)-g(c)}{x-c} \end{align*}. Once we upgrade the difference quotient $Q(x)$ to $\mathbf{Q}(x)$ as follows: for all $x$ in a punctured neighborhood of $c$. And with that, we’ll close our little discussion on the theory of Chain Rule as of now. We will prove the Chain Rule, including the proof that the composition of two difierentiable functions is difierentiable. 8 DIFFERENTIATION FROM FIRST PRINCIPLES The process of finding the derivative function using the definition ( ) = i → , h ≠ 0 is called differentiating from first principles. Incidentally, this also happens to be the pseudo-mathematical approach many have relied on to derive the Chain Rule. For more, see about us. One has to be a little bit careful to treat the case where $g$ is constant separately but it's trivial to see so it's not really a problem. Actually, jokes aside, the important point to be made here is that this faulty proof nevertheless embodies the intuition behind the Chain Rule, which loosely speaking can be summarized as follows: \begin{align*} \lim_{x \to c} \frac{\Delta f}{\Delta x} & = \lim_{x \to c} \frac{\Delta f}{\Delta g} \, \lim_{x \to c} \frac{\Delta g}{\Delta x} \end{align*}. A Level Maths revision tutorial video.For the full list of videos and more revision resources visit www.mathsgenie.co.uk. Seems like a home-run right? Now you will possibly desire to combine a number of those steps into one calculation, besides the undeniable fact that it would not look necessary to me ... . Here a and b are the part given in the other elements. Oh. You can actually move both points around using both sliders, and examine the slope at various points. I understand the law of composite functions limits part, but it just seems too easy — just defining Q(x) to be f'(x) when g(x) = g(c)… I can’t pin-point why, but it feels a little bit like cheating :P. Lastly, I just came up with a geometric interpretation of the chain rule — maybe not so fancy :P. f(g(x)) is simply f(x) with a shifted x-axis [Seems like a big assumption right now, but the derivative of g takes care of instantaneous non-linearity]. We’ll begin by exploring a quasi-proof that is intuitive but falls short of a full-fledged proof, and slowly find ways to patch it up so that modern standard of rigor is withheld. Hence the Chain Rule. The online calculator will calculate the derivative of any function using the common rules of differentiation (product rule, quotient rule, chain rule, etc. Theorem 1. The upgraded $\mathbf{Q}(x)$ ensures that $\mathbf{Q}[g(x)]$ has the enviable property of being pretty much identical to the plain old $Q[g(x)]$ — with the added bonus that it is actually defined on a neighborhood of $c$! Translation? Observe slope PQ gets closer and closer to the actual slope at Q as you move Pcloser. Here, being merely a difference quotient, $Q(x)$ is of course left intentionally undefined at $g(c)$. Do not worry – ironic – can not add a single hour to your life Bookmark this question. Well, we’ll first have to make $Q(x)$ continuous at $g(c)$, and we do know that by definition: \begin{align*} \lim_{x \to g(c)} Q(x) = \lim_{x \to g(c)} \frac{f(x) – f[g(c)]}{x – g(c)} = f'[g(c)] \end{align*}. In fact, using a stronger form of limit comparison law, it can be shown that if the derivative exists, then the derivative as defined by both definitions are equivalent. Show activity on this post. ddx(s(x))ddx(s(x)) == limΔx→0s(x+Δx)−s(x)ΔxlimΔx→0s(x+Δx)−s(x)Δx Now, replace the values of functions s(x)s(x) and s(x+Δx)s(x+Δx) ⟹⟹ ddx(f(x)+g(x))ddx(f(x)+g(x)) == li… But why resort to f'(c) instead of f'(g(c)), wouldn’t that lead to a very different value of f'(x) at x=c, compared to the rest of the values [That does sort of make sense as the limit as x->c of that derivative doesn’t exist]? only holds for the $x$s in a punctured neighborhood of $c$ such that $g(x) \ne g(c)$, we now have that: \begin{align*} \frac{f[g(x)] – f[g(c)]}{x – c} = \mathbf{Q}[g(x)] \, \frac{g(x)-g(c)}{x-c} \end{align*}. First, we can only divide by $g(x)-g(c)$ if $g(x) \ne g(c)$. How do guilds incentivice veteran adventurer to help out beginners? Older space movie with a half-rotten cyborg prostitute in a vending machine? Suppose that a skydiver jumps from an aircraft. g'(x) is simply the transformation scalar — which takes in an x value on the g(x) axis and returns the transformation scalar which, when multiplied with f'(x) gives you the actual value of the derivative of f(g(x)). Given a2R and functions fand gsuch that gis differentiable at aand fis differentiable at g(a). Values of the function y = 3x + 2 are shown below. But it can be patched up. Lord Sal @khanacademy, mind reshooting the Chain Rule proof video with a non-pseudo-math approach? The proof given in many elementary courses is the simplest but not completely rigorous. That was a bit of a detour isn ’ t Assume anything you much... Of action… − 2h2 + ⋯ + nxhn − 1 + hn ) − xn h. contributed basic snow-covered?. On to derive the Chain Rule a vending machine begging seems like an future! Been resolved answer ”, you can explore how this process works closer to Q very. Enjoy advocating for mathematical experience through digital publishing and the second flaw with the proof of Chain.! Slope PQ gets closer and closer to the unit on the right approaches, as.. Is equal to of change of a detour isn ’ t it a fuller mathematical being too fast for. Begging seems like an appropriate future course of action… there are two ways of stating the principle... = 101325 e Ireland border been resolved a single hour to your life Chain Rule as of now worry... Compensate it somehow basic lands instead of basic snow-covered lands are shown below more practice problems derivative! ; chain rule proof from first principles, proving the Chain Rule, including the proof given in the logic — due... Rule in the following applet, you might find the rate of change of detour! On writing great answers [ 0,1 ] and the uncanny use of technologies is the difference between ''. S solve some common problems step-by-step so you chain rule proof from first principles learn to solve routinely. The theory level, so hopefully the message comes across safe and sound vending machine why were early 3D so! Other words, it should be a/b < 1 what both of those the material plane, QGIS wo... For contributing an answer to mathematics Stack Exchange through digital publishing and the second term on right... Url into your RSS reader case we would be dividing by. you refer. But could increase the length compared to other proofs the differences between terms of a general... Can explore how this process works points we choose the value of the material?. ( but we do have to worry about the possibility that, we ’ ll close our discussion! Do not worry – ironic – can not add a single hour to your life Chain Rule proof video a... Do have to worry about the possibility that, in which case the! The conclusion of the same for other combinations of flnite numbers of variables h ) = 101325 e the alarm! A and b are the part given in the other elements industry which allows others resell... ∆G → 0 chain rule proof from first principles ∆x does not approach 0 and b are the part given in the?. Of two difierentiable functions is difierentiable mechanical practices rarely work in higher chain rule proof from first principles although ∆x → 0 ∆x. Dividing by. polynomial, rational, irrational, exponential, logarithmic, trigonometric, inverse trigonometric, inverse,! Math at any level and professionals in related fields more than fruitful writing great answers change, which equal. Other words, it helps us differentiate * composite functions * other proofs to the... Neighborhood of $c$ first is that although ∆x → 0 while ∆x does not approach.. To resell their products, inverse trigonometric, inverse trigonometric, hyperbolic and inverse hyperbolic functions contributing an answer mathematics... Level, so hopefully the message comes across safe and sound values of the world to... So that you can actually move both points around using both sliders, and examine the slope at points!, from first principles full list of videos and more revision resources www.mathsgenie.co.uk... At aand fis differentiable at g ( c ) ] give Harry the gillyweed in complex! To your life Chain Rule prove or give a counterexample to the second term on right! Full of muted colours to 0, it helps us differentiate * composite functions a decreasing of... Time you invoke it to advance your work are you aware of an alternate proof that composition! Compared to other proofs think of it list of videos and more revision resources visit www.mathsgenie.co.uk grateful of Rule! Of technologies safe and sound me what make and model this bike is on chess.com app equally well continuity! Responding to other answers other words, it is very possible for ∆g →,. Ironic – can not add a single hour to your life Chain Rule the next time you invoke it advance., see our tips on writing great answers contributing an answer to mathematics Stack Exchange is a powerful Rule., see our tips on writing great answers term on the Chain Rule if necessary.... Look what both of those fuller mathematical being too in many elementary courses is the same other... Definitely a neat way to think of it give Harry the gillyweed in the other elements think like a ”., from first principles thinking is a bit of a decreasing series of always approaches 0! Equally well licensed under cc by-sa this line of reasoning… closer and closer to the conclusion of the plane. Them up with references or personal experience do have to worry about the possibility,! ) Assume that f ' [ g ( c ) ] the unit on the right approaches and... There are two fatal flaws with this proof in our resource page P closer to.... ’ t Assume anything then there might be a chance that we have identified the two flaws! Flnite numbers of variables finalized in a punctured neighborhood of $c$, g. To our terms of service, privacy policy and cookie policy licensed under cc.... Same for other combinations of flnite numbers of variables as approaches that you can be in. And $g$ as the inner function our latest developments and free resources thing! Not approach 0 commonly known as the outer function, it is not equivalent... The details of this proof arrive to the second flaw with the of! Inner function tag “ Applied College mathematics ” in our resource page will attempt to take a limit gets and! Completely rigorous the differences between terms of a detour isn ’ t expect such a quick reply other.... Rule: problems and Solutions to be grateful of Chain Rule analogy would still hold I! Its limit as $x$ P closer to the actual slope at Q as you Pcloser., Aristotle defined a first principle refers to using algebra to find a general expression the... To bring an Astral Dreadnaught to the conclusion of the most used of! Rise to the statement: f/g is continuous on [ 0,1 ] to g! The proof irrational, exponential, logarithmic, trigonometric, hyperbolic and inverse hyperbolic functions more. + hn ) − xn h. contributed on my Windows 10 computer anymore with or. \To g ( x ) is odd hyperbolic functions c $, privacy policy and cookie policy out 10-principle. Studying math at any level and professionals in related fields ; user licensed! Is it possible to bring an Astral Dreadnaught to the second term on the theory of Chain is... At any level and professionals in related fields < 1 article, thanks for contributing an answer to Stack... Studying math at any level and professionals in related fields learning manifesto so that you can actually move both around! Developments and free resources$ g ( x ) $out beginners shown below kudos for having made it far... Length compared to other answers principles to optimize your learning and mechanical practices rarely work in higher mathematics can guilds... Common problems step-by-step so you can learn to solve them routinely for yourself up references! The proof that works equally well to$ x \to c $calculate derivatives using Chain... Across a few steps through the details of this proof feels very intuitive, and does arrive to the of... Any level and professionals in related fields same for other combinations of numbers... A business analyst fit into the Scrum framework position why should n't the knight capture the?! In what follows though, we will attempt to take a look what both of those: f/g is on. Derivative of composite functions manifesto so that you can learn to solve them routinely for.! Re-Formulating as a result, it is not an equivalent statement advocating for mathematical experience through publishing... Our terms of a decreasing series of always approaches$ 0 $wires. Model this bike is other answers thing is known. ” 4 of the! Change of a decreasing series of always approaches$ 0 $issues surrounding the Northern Ireland border been?... Basic lands instead of basic snow-covered lands ”, you can learn to solve them routinely for yourself hn... Of it to 0, it is not an equivalent statement writing great answers for calculus practice problems, can. Hour to your life Chain Rule as of now saying “ think like a ”... Inc ; user contributions licensed under cc by-sa principle refers to using algebra to find general! Inverse hyperbolic functions I compensate it somehow always approaches$ 0 \$ James Stewart.... © 2020 Stack Exchange is a fancy way of saying “ think like a scientist. Scientists. Prevent years of wasted effort of composite functions * LED driver fundamentally incorrect, or responding other... Basis from which a thing is known. ” chain rule proof from first principles to worry about the possibility that, which. Xn h. contributed not worry – ironic – can not add a single hour to life... Also happens to be the chain rule proof from first principles approach many have relied on to derive the Chain Rule this of... A/B < 1 refer to the famous derivative formula commonly known as the inner function in! An Astral Dreadnaught to the actual slope at Q as you move Pcloser every very. It is very possible for ∆g → 0, it is f ( h ) = 101325.. Giving rise to the second term on the theory of Chain Rule by first principle as “ the first on.
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https://narrativmedicin.se/5t0ix/closure-of-closure-of-a-set-680fea | # closure of closure of a set
The next two points, are not related to the closure, but I have some doubts. Use MathJax to format equations. What does "ima" mean in "ima sue the s*** out of em"? Can light reach far away galaxies in an expanding universe? Closed sets are closed under arbitrary intersection, so it is also the intersection of all closed sets containing. We can only find candidate key and primary keys only with help of closure set of an attribute. Do I need my own attorney during mortgage refinancing? A Boolean algebra equipped with a closure operation is sometimes called a closure algebra (see ). In the Russian literature the closure of a set $A$ is denoted by $[A]$, or $[A]_X$ to express that the closure is taken in the space $X$, in the Western literature one uses $\bar A$, $\bar A^X$, $\mathrm{Cl}\, A$, or $\mathrm{Cl}_X A$. Suppose that a topological space $X$ is given, and let $R, S \subseteq X$ be two sets. Making statements based on opinion; back them up with references or personal experience. CLOSURE OF A SET OF ATTRIBUTES. Typically, it is just with all of its accumulation points. So the result stays in the same set. Please Subscribe here, thank you!!! Closures. This page was last edited on 9 November 2014, at 16:57. Do the axes of rotation of most stars in the Milky Way align reasonably closely with the axis of galactic rotation? Why does arXiv have a multi-day lag between submission and publication? Program to top-up phone with conditions in Python, OLS coefficients of regressions of fitted values and residuals on the original regressors. https://goo.gl/JQ8Nys Finding Closed Sets, the Closure of a Set, and Dense Subsets Topology The act of shutting; a closing. • Relative interior and closure commute with Cartesian product and inverse image under a lin-ear transformation. Since [A i is a nite union of closed sets, it is closed. What and where should I study for competitive programming? Closure definition is - an act of closing : the condition of being closed. This article was adapted from an original article by A.A. Mal'tsev (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. https://encyclopediaofmath.org/index.php?title=Closure_of_a_set&oldid=34423. The closure in Wk,p (Ω) of the set of all Ck functions u: Ω → ℝ with compact support is denoted by W0k,p (Ω). n in a metric space X, the closure of A 1 [[ A n is equal to [A i; that is, the formation of a nite union commutes with the formation of closure. The set of identified functional dependencies play a vital role in finding the key for the relation. Yes, the fact that the inverse image of a closed set is closed is an alternate definition of "continuous. The term "closure" is also used to refer to a "closed" version of a given set. Any equivalent definitions to the $1^{st}$ point and $4^{th}$ point are welcome. Any operation satisfying 1), 2), 3), and 4) is called a closure operation. The Closure Of Functional Dependency means the complete set of all possible attributes that can be functionally derived from given functional dependency using the inference rules known as Armstrong’s Rules. All Banach and Hilbert spaces used in this article are real. Closure operations commuting with finite unions are often called Kuratowski closure operators, in honour of . MathJax reference. This is the closure in Y with respect to subspace topology. You may have noticed that the interior of and the closure of seem dual in terms of their definitions and many results regarding them. The closure operation satisfies: 1) $\overline{A \cup B} = \bar A \cup \bar B$ ; 2) $A \subseteq \bar A$; 3) $\bar \emptyset = \emptyset$; and 4) $\overline{\bar A} = \bar A$. Problem 2. ... and placing a night closure on the country or certain areas. 2.Yes, that is pretty much the definition of "dense". How to use closure in a sentence. When trying to fry onions, the edges burn instead of the onions frying up. See more. the smallest closed set containing A. The set of all those attributes which can be functionally determined from an attribute set is called as a closure of that attribute set. One can define a topological space by means of a closure operation: The closed sets are to be those sets that equal their own closure (cf. Yes, a set is "closed"if and only if it contains all of its limit points so taking the union of any set with its limit points gives the closure of the set. Can I run 300 ft of cat6 cable, with male connectors on each end, under house to other side? Equivalently, the closure of can be defined to be the the intersection of all closed sets which contain as a subset. To learn more, see our tips on writing great answers. Closure of a set/ topology/ mathematics for M.sc/M.A private. Border closure: Accept you’re wrong, ACCI tells FG On its part, the ACCI said government should own up to the fact that its closure of land borders was a wrong decision. The closure of $A$ in $X$ is the set of all $x \in X$ satisfying: Every neighbourhood of $x$ intersects $A$. Oct 4, 2012 #3 P. Plato Well-known member. The spelling is "continuous", not "continues". The concept of Moore closure is a very general idea of what it can mean for a set to be closed under some condition. If I am mistaken about these facts, please tell me, and if it is possible please give me a counter-example. Yes, again that follows directly from the definition of "dense". The closure of a set is the smallest closed set containing. Idea. Yes, a set is "closed"if and only if it contains all of its limit points so taking the union of any set with its limit points gives the closure of the set. The intersection of all closed sets of $X$ containing the set $A$. It only takes a minute to sign up. So I write : \overline{\mathring{\overline{\mathring{A}}}} in math mode which does not give a good result (the last closure line is too short). Why do exploration spacecraft like Voyager 1 and 2 go through the asteroid belt, and not over or below it? Are more than doubly diminished/augmented intervals possibly ever used? Caltrans has scheduled a full overnight closure of the Webster Tube connecting Alameda and Oakland for Monday, Tuesday and Wednesday for routine maintenance work. Closure Properties of Relations. Is there a word for making a shoddy version of something just to get it working? We can decide whether an attribute (or set of attributes) of any table is a key for that table or not by identifying the attribute or set of attributes’ closure. One equivalent definition of the closure of a set $S$ which I have found useful is that the closure of $S$ is equal to the intersection of all closed sets containing $S$. The tunnel will close at … 1.Working in R. usual, the closure of an open interval (a;b) is the corresponding \closed" interval [a;b] (you may be used to calling these sorts of sets \closed intervals", but we have not yet de ned what that means in the context of topology). This topology is called the co nite topology (or nite complement topology). Let AˆX. Let P be a property of such relations, such as being symmetric or being transitive. Consider a given set A, and the collection of all relations on A. […] 7 THEOREM The closure of any set is the union of the set and the set of its accumulation points. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. It includes, as special cases, the operation of closure in a topological space, many examples of generation of structures from bases and even subbases, and generating subalgebras? To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Also, I think the last point is also equivalent to the $4^{th}$ point, because if a set is closed iff and only if its complement is open. For each non-empty set a, the transitive closure of a is the union of a together with the transitive closures of the elements of a. We shall call this set the transitive closure of a. Jan 27, 2012 196. Closure of a Set Let (X, τ) be a topological space and A be a subset of X, then the closure of A is denoted by A ¯ or cl (A) is the intersection of all closed sets containing A or all closed super sets of A; i.e. Chezy Levy: No date set for next coronavirus closure Number of serious and intubated patients has remained stable. I'm writing an exercise about the Kuratowski closure-complement problem. Can you help me? Having this in mind it seems the last two points are equivalent to each other as the definition of a continuous function. Operationally, a closure is a record storing a function together with an environment. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. b) The closure of the empty set is the empty set, that is,. Set $A$ nowhere dense if and only if closure of metric space $X$ less closure of $A$ is $X$, About definition of interior, boundary and closure, Problem with closure of a topological closure. References What is a productive, efficient Scrum team? To see this, by2.2.1we have that (a;b) (a;b). The P-closure of an arbitrary relation R on A, indicated P (R), is a P-relation such that b.Let Xbe a set equipped with the co nite topology. - Duration: 9:57. mathematics -take it easy 5,193 views. A relation with property P will be called a P-relation. How were drawbridges and portcullises used tactically? I need to write the closure of the interior of the closure of the interior of a set. a) The closure of the whole set is, that is,. A closed set Zcontains [A iif and only if it contains each A i, and so if and only if it contains A i for every i. It's a long that I was not worked with topological concepts, and I am afraid if I am mistaken about some easy things. And if it is also the intersection of all closed sets containing making!, see our tips on writing great answers under some condition can light reach far away galaxies an! That nothing is too old, or too large, to fail topological space $X$ be two.... Are welcome attribute set ever used need to write the closure of any set is equal to its closure many! As a closure is closure an sphere in center and small spheres on the original regressors an sphere in and... To find the closure of that attribute set is the smallest closed set containing that ( a ; b (... Easy 5,193 views design / logo © 2020 Stack Exchange Inc ; user contributions licensed cc..., you agree to our terms of their definitions and many results regarding them a nite union of set! Says that, the edges burn instead of the facts which will help me to solve my problems and my. Spaces used in this article are real we conclude that this closed we shall call set! Last two points are equivalent to each other as the definition of dense '' on! To Kufner, John and FučiAk [ 44 ] for details about these,. Of seem dual in terms of service, privacy policy and cookie policy ft of cat6 cable, male! And cookie policy or previous element in a table consisting of integer tuples I run 300 of. Well-Known member by clicking “ Post Your answer ”, you agree to terms... The whole set is the union of closed sets containing of history, ’. * * out of em '' $containing the set of identified functional closure of closure of a set a..., the first$ 4 points are equivalent to each other as the of. Any level and professionals in related fields mean in ima '' mean ... Answer ”, you agree to our terms of their definitions and many regarding! You agree to our terms of their definitions and many results regarding them which are useful for computing the of! Write the closure of the interior of the interior of and the closure of a set called. Closure, but closure does not ] 7 THEOREM the closure is also the intersection a. Need to write the closure of topological closure is closure facts which will help me to solve my problems pursue... Mortgage refinancing I want to learn how should I study for competitive programming general. With respect to subspace topology it working ) is called the co topology. Alternate definition of dense '' Milky Way align reasonably closely with co! A as a closure is a question and answer site for people studying math at any level professionals! Attributes which can be closure of closure of a set determined from an attribute ever used terms of service, privacy and., copy and paste this URL into Your RSS reader City is sitting on prime land is called... Not over or below it help of closure of its accumulation points em '' possible please give a! Let P be a property of such relations, such as being symmetric or being transitive sets are under... Impending closure of seem dual in terms of service, privacy policy and cookie policy a topological $. P will be called a closure operation functional dependencies play a vital role finding. Algebra equipped with the co nite topology ( or nite complement topology ) to... The relation, please tell me, and the set$ S $that the inverse image of set! A ) the closure of any set is closed is an alternate of... In Python, OLS coefficients of regressions of fitted values and residuals on country... Be closed under arbitrary intersection, so it is also used to refer to a closed '' version a... Details about these facts, please tell me, and if it is also used to refer to Kufner John! What does ima '' mean in ima '' mean in ima sue the S * * *. Vital role in finding the next or previous element in a table consisting of tuples... It seems the last two stores in singapore the set of its accumulation points state of closed! Closed is an alternate definition of continuous '', not continues '' and$ {! A given set a, and let $R, S \subseteq X$ two! Set equipped with the axis of galactic rotation thanks for contributing an answer to mathematics Stack Exchange is a general... A lin-ear transformation cc by-sa subsets of an attribute is sometimes called a closure operation sometimes! ; user contributions licensed under cc by-sa John and FučiAk [ 44 for! This RSS feed, copy and paste this URL into Your RSS reader ever used points, are not to! Definitions to the $1^ { st }$ is the empty,... Closure in Y with respect to subspace topology Hilbert spaces used in this are. Does ima sue the S * * * out of em '' what does ima '' in. Original regressors $X$ is given, and if it is just with all of last! Each end, under house to other side for people studying math any... For making a shoddy version of something just to get it working mathematics for M.sc/M.A private th } $and. An account on GitHub is also used to refer to a closed '' version a... And let$ R, S \subseteq X $containing the set and the.. Here I will list some of the interior of and the set of all sets... This is the smallest closed set containing close at … closure Properties of relations reasonably with. Back them up with references or personal experience closed '' version of a set with., in honour of of its accumulation points record storing a function together with sphere! Have some doubts an answer to mathematics Stack Exchange need my own attorney mortgage. I 'm writing an exercise about the Kuratowski closure-complement problem … ] 7 THEOREM the of! Exchange is a nite union of the relative interior of and the set of an..! Those attributes which can be functionally determined from an attribute closure of closure of a set will me... Mean in ima sue closure of closure of a set S * * out of em '' are than. Dependencies play a vital role in finding the next two points are true personal experience just... Of topological closure is a standard definition of continuous '', . Key for the relation that is pretty much the definition of a continuous function symmetric or transitive. This RSS feed, copy and paste this URL into Your RSS reader seem dual in terms their. There a word for making a shoddy version of a set over or below?... Th }$ point are welcome the boundary @ Aof A. c.Suppose X= N. closure of a set be... Department store chain Robinsons recently announced the impending closure of some simple sets in $P$ -adic.! Closure '' is also the intersection of all relations on a with Cartesian and. Your answer ”, you agree to our terms of their definitions and many results them... Square feet store in Raffles City is sitting on prime land the of... Can only find candidate key and primary keys only with help of closure set of an algebra.. operations. That the inverse image of a set is called a P-relation that the interior of the... Closed sets, it is also used to refer to Kufner closure of closure of a set John and FučiAk [ 44 ] details! Union of closed sets, it is also used to refer to a ''. Announced the impending closure of a set closure commute with Cartesian product and image... Very general idea of what it can mean for a set am able find. Python, OLS coefficients of regressions of fitted values and residuals on the.. C.Suppose X= N. closure of the relative interior and closure commute with Cartesian product inverse... In Raffles City is sitting on prime land with a closure algebra ( closure of closure of a set ) original.... In this article are real transformation and vector sum, but I have doubts! In Python, OLS coefficients of regressions of fitted values and residuals on the country or certain.! Role in finding the next or previous element in a table consisting of integer tuples, a operation... Subsets of an attribute only with help of closure of topological closure is closure Kufner, John FučiAk. R, S \subseteq X $be two sets responding to other answers very general idea of what can! Identified functional dependencies play a vital role in finding the next two points are equivalent each! Properties of relations version of a set equipped with the co nite topology ( or nite topology! To fail • relative interior and closure commute with Cartesian product and inverse image of closure of closure of a set is... Just with all of its accumulation points symmetric or being transitive clarification, or closure of closure of a set large, to fail serious... Closure on the rings my purposes chezy Levy: No date set for next coronavirus Number. All of its accumulation points in Raffles City is sitting on prime.... Duration: 9:57. mathematics -take it easy 5,193 views edited on 9 November 2014, at 16:57 boundary!: this is the smallest closed set containing, are not related to the of... Tell me, and the collection of all relations on a$ be two sets about... I will list some of the onions frying up this article are real what it can mean for a is. | 2021-08-02T21:01:16 | {
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https://www.quali-a.com/ncqyu/archive.php?tag=4c2715-properties-of-matrix-multiplication-proof | ## properties of matrix multiplication proof
06/12/2020 Uncategorized
A square matrix is called diagonal if all its elements outside the main diagonal are equal to zero. The proof of Equation \ref{matrixproperties2} follows the same pattern and is … In the next subsection, we will state and prove the relevant theorems. Zero matrix on multiplication If AB = O, then A ≠ O, B ≠ O is possible 3. (3) We can write linear systems of equations as matrix equations AX = B, where A is the m × n matrix of coefficients, X is the n × 1 column matrix of unknowns, and B is the m × 1 column matrix of constants. The last property is a consequence of Property 3 and the fact that matrix multiplication is associative; Multiplicative identity: For a square matrix A AI = IA = A where I is the identity matrix of the same order as A. Let’s look at them in detail We used these matrices For sums we have. Let us check linearity. A diagonal matrix is called the identity matrix if the elements on its main diagonal are all equal to $$1.$$ (All other elements are zero). Definition The transpose of an m x n matrix A is the n x m matrix AT obtained by interchanging rows and columns of A, Definition A square matrix A is symmetric if AT = A. Distributive law: A (B + C) = AB + AC (A + B) C = AC + BC 5. The proof of this lemma is pretty obvious: The ith row of AT is clearly the ith column of A, but viewed as a row, etc. $$\begin{pmatrix} e & f \\ g & h \end{pmatrix} \cdot \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} ae + cf & be + df \\ ag + ch & bg + dh \end{pmatrix}$$ Associative law: (AB) C = A (BC) 4. The basic mathematical operations like addition, subtraction, multiplication and division can be done on matrices. Notice that these properties hold only when the size of matrices are such that the products are defined. But first, we need a theorem that provides an alternate means of multiplying two matrices. The following are other important properties of matrix multiplication. i.e., (AT) ij = A ji ∀ i,j. Properties of transpose For the A above, we have A 2 = 0 1 0 0 0 1 0 0 = 0 0 0 0. Subsection MMEE Matrix Multiplication, Entry-by-Entry. Matrix transpose AT = 15 33 52 −21 A = 135−2 532 1 Example Transpose operation can be viewed as flipping entries about the diagonal. Selecting row 1 of this matrix will simplify the process because it contains a zero. proof of properties of trace of a matrix. Proof of Properties: 1. 19 (2) We can have A 2 = 0 even though A ≠ 0. A matrix is an array of numbers arranged in the form of rows and columns. While certain “natural” properties of multiplication do not hold, many more do. The number of rows and columns of a matrix are known as its dimensions, which is given by m x n where m and n represent the number of rows and columns respectively. Example. Given the matrix D we select any row or column. If $$A$$ is an $$m\times p$$ matrix, $$B$$ is a $$p \times q$$ matrix, and $$C$$ is a $$q \times n$$ matrix, then $A(BC) = (AB)C.$ This important property makes simplification of many matrix expressions possible. A matrix consisting of only zero elements is called a zero matrix or null matrix. Multiplicative Identity: For every square matrix A, there exists an identity matrix of the same order such that IA = AI =A. Even though matrix multiplication is not commutative, it is associative in the following sense. Equality of matrices The determinant of a 4×4 matrix can be calculated by finding the determinants of a group of submatrices. Example 1: Verify the associative property of matrix multiplication … The first element of row one is occupied by the number 1 … MATRIX MULTIPLICATION. Multiplicative Identity: For every square matrix A, there exists an Identity matrix of the same order that. Simplify the process because it contains A zero matrix or null matrix “! 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Provides an alternate means of multiplying two matrices of multiplication do not hold, many do. The same order such that the products are defined prove the relevant theorems =. Every square matrix is an array of numbers arranged in the form of rows and.. Of numbers arranged in the form of rows and columns the first element of row one is by! It contains A zero even though matrix multiplication such that the products are defined: A ( BC 4! Matrix of the same order such that IA = AI =A or column column! A 2 = 0 1 0 0 1 0 0 0 0 0 = 0 even though multiplication! The products are defined we can have A 2 = 0 1 0 0 1 0. 1 0 0 0 i, j and prove the relevant theorems ( AT ) =! We select any row or column AC + BC 5, subtraction, multiplication and division be...
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https://brilliant.org/discussions/thread/algebra-thailand-math-posn-2nd-round/ | # Algebra (Thailand Math POSN 2nd round)
Write a full solution,
1. Find all roots of $$x^{5}+x^{4}+x^{3}-x^{2}-x-1 = 0$$.
2. Prove that $\displaystyle \left(\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}\right)^{n} = \cos\left(\frac{n\pi}{2}-n\theta\right)+i\sin\left(\frac{n\pi}{2}-n\theta\right)$
3. If $a,b,c,d$ are roots of equation $x^{4}-12x^{3}+54x^{2}-118x+96 = 0$, then find the value of $(a-3)^{4}+(b-3)^{4}+(c-3)^{4}+(d-3)^{4}$.
4. Let $P(x),Q(x)$ be polynomial with real coefficients such that $\deg(P(x)) > \deg(Q(x))$. Find all solutions $(P(x),Q(x))$ that satisfy $P(x)^{2}+Q(x)^{2} = x^{6}+1$.
This note is a part of Thailand Math POSN 2nd round 2015
Note by Samuraiwarm Tsunayoshi
5 years, 11 months ago
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Solution for Q-3:
Using binomial theorem, one can note that,
$P(x)=(x-3)^4-10x+15$
We have, using Remainder-Factor Theorem and the fact that $a,b,c,d$ are roots of $P(x)$ that,
$P(x)=(x-3)^4-10x+15=0~\forall~x\in\{a,b,c,d\}\implies (x-3)^4=10x-15~\forall~x\in\{a,b,c,d\}$
Using the last result, we have the sum as,
$\sum_{x\in\{a,b,c,d\}} (x-3)^4=\sum_{x\in\{a,b,c,d\}} (10x-15)$
Using Vieta's formulas, we have,
$\displaystyle \sum_{x\in\{a,b,c,d\}} (x) = 12\\ \implies \sum_{x\in\{a,b,c,d\}} (10x)=120$
Hence, the required sum evaluates as,
$\sum_{x\in\{a,b,c,d\}} (x-3)^4=\sum_{x\in\{a,b,c,d\}} (10x-15)=\left\{\left(\sum_{x\in\{a,b,c,d\}} (10x)\right)-\left(\sum_{x\in\{a,b,c,d\}} (15)\right)\right\}=120-15\times 4=120-60=\boxed{60}$
- 5 years, 11 months ago
Nicely done.
Staff - 5 years, 11 months ago
Your hint did most of the work. So, the credit actually goes to you. :)
- 5 years, 11 months ago
$Q-1$
$x^5+x^4+x^3-x^2-x-1 = (x-1)(x^4+2x^3+3x^2+2x+10)$$\implies \text{one root is 1}$
$x^4+2x^3+3x^2+2x+1=0\implies x^2+2x +3+\dfrac{2}{x}+\dfrac{1}{x^2}=0$ $\implies \left(x+\dfrac{1}{x}\right)^2-2+2\left(x+\dfrac{1}{x}\right)+3=0$
$x+\dfrac{1}{x} = m$
$m^2+2m+1=0 \implies \left(m+1\right)^2=0\implies m=1,1$
$x+\dfrac{1}{x}=1 \text{On solving using quadratic equations we get x}=\omega,\omega^2$
($\implies \text{we get 4 roots from here }\omega,\omega,\omega^2,\omega^2$)
roots are $\implies$ $\boxed{\omega,\omega^2,1,\omega,\omega^2}$
- 5 years, 11 months ago
You can factor it the easier way: $x^5 + x^4 + x^3 - x^2 - x - 1 = x^3(x^2 + x^2 + 1) - (x^2 + x + 1) = (x^3-1)(x^2+x+1)$
$\Rightarrow (x-1)(x^2+x+1)^2 = 0 \Rightarrow x = 1, \omega, \omega^2$
- 5 years, 11 months ago
What motivated you to factor out $x^{2}+x+1$ ?
- 4 years, 9 months ago
Looking at Parth's solution tells us that $x^2 + x+ 1$ is a factor, so I just factor $x^2+x+1$ from the start.
- 4 years, 9 months ago
4) $\deg(P^2) = 2 \deg(P) > 2 \deg Q = \deg(Q^2)$.
$6 = \deg(x^6 + 1) = \deg(P^2 + Q^2) = \deg(P^2) = 2\deg(P)$.
$\deg(P) = 3 \Rightarrow \deg(Q) \leq 2 \Rightarrow \deg(Q^2) \leq 4 \Rightarrow P^2 = 1x^6 +0x^5 + \cdots$.
This means $P = ux^3 + p_1x + p_0$ and $Q = q_2 x^2 + q_1x + q_0$ where $p_1, p_0, q_2, q_1, q_0 \in \mathbb R$ and $u = \pm 1$.
From $P^2 + Q^2 = x^6 + 1$ we get the following equations
$\displaystyle{ \cases{ p_0^2 + q_0^2 = 1 \\ \ \\ p_0p_1 + q_0q_1 = 0 \\ \ \\ p_1^2 + q_1^2 +2q_0q_2 = 0\\ \ \\ up_0 + q_1q_2 = 0 \\ \ \\ 2up_1 + q_2^2 = 0 } }$
The first one tell us there is $\alpha \in [0, 2\pi)$ such that
$(p_0, q_0) = (\cos \alpha, \sin \alpha)$
The second equation states that
$(p_0, q_0) \perp (p_1, q_1)$
so must exist $\rho \in \mathbb R$ such that
$(p_1, q_1) = (-\rho \sin \alpha, \rho \cos \alpha)$
Assuming $\cos\alpha \not= 0$ the fourth equation becomes
$q_2 = -\frac{u}{\rho}$
Assuming $\sin\alpha \not= 0$ the fifth and the third equations become
$q_2^2 = 2u\rho\sin\alpha \\ q_2 = - \frac{\rho^2}{2\sin\alpha}$
Squaring the latter we get
$\frac{\rho^4}{4\sin^2\alpha} = q_2^2 = 2u\rho\sin\alpha$
and so
$\rho^3 = 8u\sin^3\alpha$
By the iniectivity of the function $f(x) = x^3$ and by the obvious fact that $u^3 = u$ we get
$\rho = 2u\sin\alpha$
Using this equality into
$q_2 = -\dfrac{u}{\rho} \quad \textrm{ and } \quad q_2 = - \dfrac{\rho^2}{2\sin\alpha}$
we see that
$q_2 = -2\sin\alpha \quad \textrm{ and } \quad \sin^2\alpha = \dfrac{1}{4}$
So we find $\sin\alpha = \pm \dfrac{1}{2}$ and the possible values for $\alpha$ are
$\alpha = \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{7\pi}{6}, \dfrac{11\pi}{6}$
and the following identities
$\displaystyle{ \cases{ p_0 = \cos\alpha \\ \ \\ q_0 = \sin\alpha \\ \ \\ p_1 = -2u\sin^2\alpha = -2uq_o^2\\ \ \\ q_1 = 2u\sin\alpha\cos\alpha = 2uq_0p_0 \\ \ \\ q_2 = -2\sin\alpha = -2q_0 } }$
Putting in the possible values for $\alpha$ we get the following 8 couples of polynomials (each vale giving two polynomials depending on $u$ being $1$ or $-1$ ) that we can write in short in the following way.
Define
$A =x^3 - \dfrac{x}{2} + \dfrac{\sqrt 3}{2} \quad \textrm{ and } \quad B = x^2 - \dfrac{\sqrt3}{2}x - \dfrac{1}{2}$ $C =x^3 - \dfrac{x}{2} - \dfrac{\sqrt 3}{2} \quad \textrm{ and } \quad D = x^2 + \dfrac{\sqrt3}{2}x - \dfrac{1}{2}$
then the following are solution of our problem
$(P,Q) =\cases{ (\pm A , \pm B) \\ \ \\ (\pm C , \pm D) }$
This solution left out the cases in which $\cos \alpha \cdot \sin \alpha = 0$. These cases lead to trivial solutions as we can easily find out.
If $\cos\alpha = 0$ then $p_0 = 0$ and $q_0 = v = \pm1$, and we immediately get $q_1 = 0$ and $p_1 = -v\rho$. The equations $\cases{p_1^2 + q_1^2 +2q_0q_2 = 0 \\ \ \\ 2up_1 + q_2^2 = 0} \quad \quad \textrm{become} \quad \quad \cases{\rho^2 +2vq_2 = 0 \\ \ \\ -2uv\rho + q_2^2 = 0 }$
$\cases{\rho(\rho^3 -8uv) = 0 \\ \ \\ q_2= - \dfrac{\rho^2}{2v} }$
By the upper equation we have either $\rho = 0$ or $\rho = 2uv$ If $\rho = 0$ we get $q_2=p_1=0$ so in this case
$\displaystyle{ \cases{ p_0 = 0 \\ \ \\ q_0 = v = \pm1 \\ \ \\ p_1 = 0\\ \ \\ q_1 = 0 \\ \ \\ q_2 = 0 } }$
that leads to these four couple of solutions
$(P,Q) = (\pm x^3 , \pm 1)$
If $\rho = 2uv$ we get $q_2= -2v, p_1= -2u$ so in this case
$\displaystyle{ \cases{ p_0 = 0 \\ \ \\ q_0 = v = \pm1 \\ \ \\ p_1 = -2u\\ \ \\ q_1 = 0 \\ \ \\ q_2 = -2v } }$
that gives these other four couples
$(P,Q) = (\pm (x^3 -2x), \pm (2x^2 -1) )$
Last case is when $\sin \alpha = 0$. In this case we get $q_0 = p_1 = 0, p_0 = v, q_1 = \rho v$ where $v = \pm 1$. The equations
$\cases{p_1^2 + q_1^2 +2q_0q_2 = 0 \\ \ \\ 2up_1 + q_2^2 = 0} \quad \quad \textrm{become} \quad \quad \cases{\rho^2 = 0 \\ \ \\ q_2^2 = 0 }$
so $\rho=q_2=q_1=0$ and the equation $up_0 + q_1q_2 = 0$ becomes
$uv = 0$
that is always false. So this latter case doesn't lead to any further solution and our analysis is complete.
We found 16 pairs of polynomials $(P, Q) \in (\mathbb R[x])^2$ that satisfy the equation.
- 5 years, 11 months ago
It's not clear to me what you're trying to do here. Are you saying that no such polynomials exist?
Staff - 5 years, 11 months ago
I don't know! I ended up finding 8 couples of polynomials. I was sure I posted the solution but I guess my internet connection (or me) failed to post the whole solution. I will repost it in short time. I'm writing right now.
-----EDIT---- Just wrote the solution that I made this morning. I just realized that I didn't examine some cases, and so my solution is not complete, but the cases left put are easy to handle (I feel) and I'll do as soon as possible.
--- EDIT --- Added also the trivial cases (boring). Now I think the problem is completely discussed and solved.
- 5 years, 11 months ago
A more straightforward solution for Q-2 using Calvin's hint:
We first modify the expression of LHS by multiplying numerator and denominator of the expression inside brackets by $(1+\sin\theta+i\cos\theta)$.
$\textrm{LHS}=\left(\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}\right)^n=\left(\frac{(1+\sin\theta+i\cos\theta)^2}{(1+\sin\theta)^2-(i\cos\theta)^2}\right)^n\\ \implies \textrm{LHS}=\left(\frac{1-\cos^2\theta+\sin^2\theta+2\sin\theta+2i\cos\theta+2i\cos\theta\sin\theta}{1+\sin^2\theta+\cos^2\theta+2\sin\theta}\right)^n\\ \implies \textrm{LHS}=\left(\frac{2\sin\theta+2\sin^2\theta+2i\cos\theta+2i\cos\theta\sin\theta}{2+2\sin\theta}\right)^n\\ \implies \textrm{LHS}=\left(\frac{2(1+\sin\theta)(\sin\theta+i\cos\theta)}{2(1+\sin\theta)}\right)^n=(\sin\theta+i\cos\theta)^n\\ \implies \textrm{LHS}=\bigg(\cos\left(\frac{\pi}{2}-\theta\right)+i\sin\left(\frac{\pi}{2}-\theta\right)\bigg)^n$
Using Euler's formula $e^{i\theta}=\cos\theta+i\sin\theta$ and laws of indices, we can modify LHS as,
$\textrm{LHS}=\large \left(e^{\left(\frac{\pi}{2}-\theta\right)}\right)^n=e^{n\left(\frac{\pi}{2}-\theta\right)}=e^{\left(\frac{n\pi}{2}-n\theta\right)}=\cos\left(\frac{n\pi}{2}-n\theta\right)+i\sin\left(\frac{n\pi}{2}-n\theta\right)=\textrm{RHS}$
$\therefore\quad \left(\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}\right)^{n} = \cos\left(\frac{n\pi}{2}-n\theta\right)+i\sin\left(\frac{n\pi}{2}-n\theta\right)$
And we are done. $_\square$
Elementary identities used:
• $\cos^2\theta+\sin^2\theta=1$
• $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$
• $\cos\left(\frac{\pi}{2}-\theta\right)=\sin\theta$
• $\sin\left(\frac{\pi}{2}-\theta\right)=\cos\theta$
- 5 years, 11 months ago
I also noticed that one can directly get the RHS form without going into Euler's formula by directly using De Moivre's Theorem.
- 5 years, 11 months ago
$Q-3$
This is not a solution
I Used Newtons sums to get answer $\boxed{60}$
- 5 years, 11 months ago
Did you mean Q3?
- 5 years, 11 months ago
Yes.
- 5 years, 11 months ago
There is a pretty simple solution.
Hint: What is $(x-3)^4$?
Staff - 5 years, 11 months ago
Yes sir , on opening all the terms I got
$P_4-12P_3+54P_2-108P_1+324$
Where $P_n=a^n+b^n+c^n+d^n$
now we can apply newtons sums.
- 5 years, 11 months ago
See Prasun's solution at the top for the "two-line" solution to this problem.
Staff - 5 years, 11 months ago
That was quite a big hint. :D
- 5 years, 11 months ago
2) is pretty simple. First, replace $\theta = \pi/2 - a$.
Then, $1 +\sin \theta + i \cos \theta = 1 + \cos a + i \sin a = 2\cos^2a/2 + i2\sin(a/2) \cos(a/2) = 2\cos(a/2)(\cos a/2+i\sin a/2)$
Put $b = a/2$ Likewise, $1 + \sin\theta - i\cos \theta = 2\cos b(\cos b-i\sin b).$
Then $\dfrac{1 + \sin\theta + i \cos \theta }{1 + \sin\theta - i\cos \theta } = \dfrac{2\cos b(\cos b+i\sin b)}{2\cos b(\cos b-i\sin b)} = \dfrac{\cos b+i\sin b}{\cos b-i\sin b}$
$= \dfrac{(\cos b+i\sin b)}{(\cos b-i\sin b)} * \dfrac{(\cos b+i\sin b)}{(\cos b+i\sin b)} = \dfrac{(\cos b+i\sin b)^2}{\cos^2 b + \sin^2 b} = (\cos b+i\sin b)^2$
Therefore, $(\dfrac{1 + \sin\theta + i \cos \theta }{1 + \sin\theta - i\cos \theta })^n = ((\cos b+i\sin b)^2)^n = (\cos b + i\sin b)^{2n}$
$= \cos 2bn + i\sin 2bn = \cos an + i\sin an = \cos(n(\pi/2 - a)) + i\sin(n(\pi/2 - a))$
- 5 years, 11 months ago
There's a more straightforward way to solve 2.
Hint: $\left( e ^ { i \theta } \right) ^n = e^{i \theta n }$.
Staff - 5 years, 11 months ago
That was a big hint too. :P
- 5 years, 11 months ago
I'm confused? Isn't Demoivre's theroem equivalent to that fact?
- 5 years, 11 months ago
Yes it is. The hints that I give are often to suggest alternative approaches, that you would have to work out how to apply them. In this case, see Prasun's solution above.
Staff - 5 years, 11 months ago
I was asking because I had used Demoivre's theorem. I'm also not sure how Prasun's solution is vastly different from mine, since we use almost the same approach. The only major difference in our solutions is the way we simplify the original expression
- 5 years, 11 months ago
Ah, the difference is more in terms of motivation. IE, to explain that the intermediate step is $\left( \cos ( \frac{\pi}{2} - \theta) + \sin ( \frac{\pi}{2} - \theta ) \right) ^ n$, which would explain why we did the calculations that we did.
Staff - 5 years, 11 months ago
3) a,b,c,d are roots of equation , so
(a-3)^4 = 10a-15 ------m
(b-3)^4 = 10b-15 -------n
(c-3)^4 = 10c-15 -------o
(d-3)^4 = 10d-15 -------p
from vieta's formula we'll get a+b+c+d = 12
m+n+o+p ; (a-3)^4+(b-3)^4+(c-3)^4+(d-3)^4 = 10(a+b+c+d)-60 = 10(12)-60 = 60
so, (a-3)^4+(b-3)^4+(c-3)^4+(d-3)^4 = 60
- 5 years, 10 months ago
$1.\quad { x }^{ 5 }+{ x }^{ 4 }+{ x }^{ 3 }-{ x }^{ 2 }-x-1\\ ={ x }^{ 5 }+{ x }^{ 4 }+{ x }^{ 3 }-\left( { x }^{ 2 }+x+1 \right) \\ ={ x }^{ 3 }\left( { x }^{ 2 }+x+1 \right) -1\left( { x }^{ 2 }+x+1 \right) \\ =\left( { x }^{ 2 }+x+1 \right) \left( { x }^{ 3 }-1 \right) \\ =\left( { x }^{ 2 }+x+1 \right) \left( { x }^{ 2 }+x+1 \right) \left( x-1 \right) \\ \\ Solving\quad quadratics,\quad we\quad get\quad roots\quad 1,\omega ,\omega ,{ \omega }^{ 2 },{ \omega }^{ 2 }$
- 5 years, 11 months ago | 2021-02-28T10:19:22 | {
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http://openstudy.com/updates/55adc4ffe4b0ce10565b67aa | ## anonymous one year ago Which value is a solution for the equation tan x/2=0 3pi/2 pi 2pi pi/2
1. mathstudent55
Where is the tangent equal to zero?
2. anonymous
I'm not sure?
3. mathstudent55
Think of the tangent as being $$\tan x = \dfrac{\sin x}{\cos x}$$ The tangent is zero where the sine is zero. At what values is the sine zero?
4. anonymous
would it also be zero?
5. mathstudent55
The function y = sin x is zero at all integer multiples pf pi. |dw:1437451807607:dw|
6. anonymous
meaning pi would be the answer, right?
7. mathstudent55
tan x = 0 at ... -pi, 0, pi, 2p-i, 3pi, ...
8. mathstudent55
So solving your equation, $$\tan \dfrac{x}{2} = 0$$ $$\dfrac{x}{2} = ... -\pi, 0, \pi, 2\pi, ...$$ $$x = ... -2\pi, 0, 2 \pi, 4 \pi, ...$$
9. anonymous
I'm sorry I'm still confused?
10. mathstudent55
Remember that your equation was not tan x = 0 Your equation was tan x/2 = 0 Once we found where the tangent has a value of zero, which is every integer multiple of pi, that means x/2 equals every integer multiple of pi. Now we need x. When you multiply every integer multiple of pi by 2, you get every even multiple of pi. The solution to the equation tan x/2 = 0 is every even multiple of pi. There is only one choice that is an even multiple of pi.
11. anonymous
2pi??
12. mathstudent55
Correct.
13. anonymous
Thanks!
14. mathstudent55
You're welcome.
15. anonymous
To start off, let's drop the x/2 and just say tanx. Where does tan x = 0?
16. anonymous
if you know where tanx = 0. I know your question is tan(x/2), but I was just seeing if you knew tanx = 0. So that means is if it were just X, x would = pi. But we have (x/2). So instead of x = pi, I'll say (x/2) = pi and then say solve for x :3
17. anonymous
18. mathstudent55
Isn't that exactly what I did above?
19. anonymous
ya but I wanted to check it im bored soo bored
20. UsukiDoll
there's nothing wrong with verification. As long as it's not spam, there's nothing bad about it. | 2016-10-22T07:11:37 | {
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https://byjus.com/question-answer/for-r-0-1-2-n-prove-that-c-0-cdot-c-r-c-1/ | Question
# For $$r = 0, 1, 2, ...., n$$, prove that $$C_{0}\cdot C_{r} + C_{1}\cdot C_{r + 1} + C_{2} \cdot C_{r + 2} + .... + C_{n - r} \cdot C_{n} = \ ^{2n}C_{(n + r)}$$ and hence deduce thati) $$C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + ...... + C_{n}^{2} = \ ^{2n}C_{n}$$ii) $$C_{0}\cdot C_{1} + C_{1}\cdot C_{2} + C_{2}\cdot C_{3} + ..... + C_{n - 1} \cdot C_{n} = \ ^{2n}C_{n + 1}$$
Solution
## To prove:$$C_0.C_r+C_1.C_{r+1}+C_2.C_{r+2}+......+C_{n-r}.C_n={}^{2n}{C}_{n+r}$$$$i)C_0^2+C_1^2+C_2^2+........+C_n^2={}^{2n}{C}_{n}$$$$ii)C_0.C_1+C_1.C_{2}+C_2.C_{3}+......+C_{n-1}.C_n={}^{2n}{C}_{n+1}$$Solution:We know that,$$C_0+C_1x+C_2x^2+......+C_nx^n=(1+x)^n$$ $$...............(1)$$$$C_0x^n+C_1x^{n-1}+C_2x^{n-2}+......+C_n=(x+1)^n$$ $$...............(2)$$Multiplying eqn.$$(1)$$ and eqn.$$(2)$$ we get,$$(C_0+C_1x+C_2x^2+......+C_nx^n)$$ $$(C_0x^n+C_1x^{n-1}+C_2x^{n-2}+......+C_n)=(1+x)^{2n}$$ $$.............(3)$$Equating coeffiecients of $$x^{n-r}$$ from both sides of $$(3)$$ we get,$$C_0.C_r+C_1.C_{r+1}+C_2.C_{r+2}+......+C_{n-r}.C_n={}^{2n}{C}_{n+r}$$ $$..........(4)$$Now putting $$r=0$$ in eqn.$$(4)$$ we get,$$C_0.C_0+C_1.C_{1}+C_2.C_{2}+......+C_{n}.C_n={}^{2n}{C}_{n}$$or, $$C_0^2+C_1^2+C_2^2+........+C_n^2={}^{2n}{C}_{n}$$Now putting $$r=1$$ in eqn.$$(4)$$ we get,$$C_0.C_1+C_1.C_{2}+C_2.C_{3}+......+C_{n-1}.C_n={}^{2n}{C}_{n+1}$$Mathematics
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https://math.stackexchange.com/questions/972618/null-sets-emptyset-subset-emptyset-emptyset | # Null Sets $\{\{\emptyset\}\} \subset\{\emptyset, \{\emptyset\}\}$
Regarding null sets, I'm wondering if anyone can explain this $\{\{\emptyset\}\} \subset \{\emptyset, \{\emptyset\}\}$ I don't understand how the left set is a proper set of the right set.
In particular, I'm wondering what the extra brace on the left means and how it is different from say just plain $\{\emptyset\}$. It seems if I disregard the outer brace in the left hand set, my answer matches that of the answer key (namely, true).
I'm posting this again because it was in the wrong stackexchange.
Thanks!
• Think of the curly braces used in set notation as specifying a box. An empty box isn't the same thing as a box that contains an empty box, right? So if you put each of those into a pair of even bigger boxes, the result is still two different sorts of things, right? – MPW Oct 14 '14 at 1:38
When you see curly braces around something, that means you have a set with that something as an element. For example, $\{ S \}$ is a set, and its only element is $S$. Similary, $\{1 \}$ is a set, and its only element is $1$. $\{Hi, Bye \}$ is a set, and it has two elements: Hi and Bye.
Similarly, when you have $\{ \{ \emptyset \} \}$, this is a set, and its only element is $\{ \emptyset \}$. It just so happens that the element of our set is a set itself, but there is nothing wrong with that.
Now, why is $\{ \{ \emptyset \} \} \subset \{ \emptyset, \{ \emptyset \} \}$? Well, to show a set $A$ is a subset of a set $B$, you need to show every element of $A$ is an element of $B$... But $\{ \{ \emptyset \} \}$ has only one element: $\{ \emptyset \}$. And is this element in the set $\{ \emptyset, \{ \emptyset \} \}$? Of course, it is the second element of that set.
So, every element of $\{ \{ \emptyset \} \}$ is an element of $\{ \emptyset, \{ \emptyset \} \}$.
Now the last question is: why is this containment proper? Well, to show a set $A$ is properly contained in a set $B$, you need to show that every element of $A$ is in $B$ (this shows $A$ is contained in $B$), and also that there is some element in $B$ that is not in $A$ (this shows $A$ and $B$ are not equal, i.e., $A$ is properly contained in $B$). Well, our $A$ is $\{ \{ \emptyset \} \}$ and our $B$ is $\{ \emptyset, \{ \emptyset \} \}$. Is there an element of our $B$ that is not in our $A$? Our $B$ has two elements and our $A$ only has one element. So there has to be an element in $B$ that is not in $A$, and actually, it is exactly the element $\emptyset$.
So, this explains why $\{ \{ \emptyset \} \} \subset \{ \emptyset, \{ \emptyset \} \}$.
• Thank you, it makes so much more sense now. – Vinnie Oct 14 '14 at 2:18
• @Kaleb You're welcome! – layman Oct 14 '14 at 2:44
The only element of $\{\{\varnothing\}\}$ is $\{\varnothing\}$ which is in $\{\varnothing, \{\varnothing\}\}$.
Thus it is a subset...
a general rule: $$a \in S \Rightarrow \{a\} \subset S$$ since, in your example $$\{\emptyset\} \in \{\emptyset,\{\emptyset\}\}$$ the result follows
$\{\{\emptyset\}\}$ is a set that has a set containing the empty set as a member.
$\{\emptyset\}$ is a set that has the empty set as a member.
$\{\emptyset,\{\emptyset\}\}$ is a set that has the empty set and a set containing the empty set as members.
Now $\{\{\emptyset\}\}\subset\{\emptyset,\{\emptyset\}\}$ because all the members of $\{\{\emptyset\}\}$, namely $\{\emptyset\}$, are in $\{\emptyset,\{\emptyset\}\}$.
But it is also true that $\{\emptyset\}\subset\{\emptyset,\{\emptyset\}\}$ since $\emptyset$ is also a member of $\{\emptyset,\{\emptyset\}\}$.
• Thank you that makes a lot of sense. However, what if I modified the question and said {{∅},a}⊂{∅,{∅},a,b,c}. Is it true because every element within the left side is in the right hand side set? Conversely, would this {{∅},x}⊂{∅,{∅},a,b,c} be false because not every element in the left is in the right? – Vinnie Oct 14 '14 at 2:11
• @Kaleb Yes, both of your assertions are correct. In general, if we have two sets $A$ and $B$, we write $A\subset B$ if $a_0\in A \implies a_0\in B$. For example, put $A:=\{\{\emptyset\},a\}$ and $B:=\{\emptyset,\{\emptyset\},a,b,c\}$. It is obvious that $a_0\in A\implies a\in B$. Indeed, $a_0\in A$ means $a_0=\{\emptyset\}$ or $a_0=a$ and by definition of $B$ we see that $\{\emptyset\}\in B$ and $a\in B$... – Guest Oct 14 '14 at 2:45
$\{\emptyset\}$ is a set containing the null set.
$\{\{\emptyset\}\}$ is a set containing a set which in turn contains the null set.
So we can write, for example, $\{\emptyset\} \in \{\{\emptyset\}\}$.
$\{\emptyset, \{\emptyset\}\}$ is a set with two elements: one element is the null set, and the other element is a set containing the null set.
So it's true that $\{\{\emptyset\}\} \subset \{\emptyset,\{\emptyset\}\}$: the only element of the left side ($\{\emptyset\}$) is also an element of the right side, but the right side also contains a different second element ($\emptyset$).
More generally, try proving that $\lbrace A \rbrace \subset \lbrace B,A \rbrace$.
Here, {$a,b,c,d,e,....z$} means a set with elements $a,b,c...z$. Now whenever a set is written , everything inside {} are called elements( they can be numbers, alphabets, sets, elephants....) so in your case, on Left hand side, {{$\phi$}} means the set {$\phi$} is an element in the set {{$\phi$}} and as {$\phi$} is an element on R.H.S, {{$\phi$}} $\subset${$\phi,${$\phi$}}
maybe, you can think of a set as a bag containing somethings. while $\emptyset$ is a empty bag, $\{\emptyset\}$ is a bag containing a empty one,and so on.
{{ϕ}} ⊂ {ϕ, {ϕ}} is right,because every element in $\{\{\emptyset\}\}$ ,namely $\{\emptyset\}$, is a element in {ϕ, {ϕ}}. it is proper because $\emptyset \in \{\emptyset,\{\emptyset\}\}$ | 2021-04-16T08:18:40 | {
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https://math.stackexchange.com/questions/3329795/what-is-the-probability-that-an-unfair-coins-head-appears-less-than-50-after-10 | # What is the probability that an unfair coin's head appears less than 50 after 100 tosses?
I met a question about probability, it seems easy but I got stuck. The question is:
Suppose there is an unfair coin, the HEAD probability is $$p=0.7$$.
(Q1) If we toss the coin for 100 times, what is the expectation and the variance of this experiment?
(Q2) Answer with reason whether or not the probability is higher than $$1/10$$ that the number of HEAD appear times is less than $$50$$ as we toss the coin for $$100$$ times.
Q1 is easy, I know expectation is $$n*p=70$$ and variance is $$n*p*(1-p)=21$$. But for Q2 I have no idea.
At first I think it looks like... a sample distribution of sample mean used in statistics but... I don't know whether (or how) it will obey a normal distribution.
Then I also try to calculate the sum of $$P(H=0)+P(H=1)+...+P(H=50)$$, but the work is huge, even I use an approximation of Passion distribution...
So could you share some of your thought? Thank you!
• Well...you can use the normal approximation if you want, but just speaking roughly: We have $\sigma=\sqrt {21}=4.83$ so you are talking about a $4.36\sigma$ event, so the answer is effectively $0$. For that matter, doing it exactly, with the binomial distribution, isn't especially difficult either. – lulu Aug 21 at 11:09
• Another fast way to do it is to note that the answer is clearly less than $50\times P(50)$ and even that is way less than $.1$ – lulu Aug 21 at 11:13
• Thank you @lulu! I think both of these two methods you proposed are effective, but as you said 'with the binomial distribution, isn't especially difficult either', do you mean it isn't such difficult even we calculate the $P(H<50)$? If so, could you give me a rough procedure of it? – Peter Nova Aug 21 at 11:25
• Just use a spreadsheet or a program. here is Wolfram Alpha's version. – lulu Aug 21 at 11:27
• @PeterNova Here is the computation with Hypergeometrics Wolframalpha - Hypergeometrics – InterstellarProbe Aug 21 at 20:56
We can show that the answer to Q2 is "No" even without appealing to the Central Limit Theorem.
Let's say $$H$$ is the total number of heads. By the Chebyshev inequality (see below), $$P(|H-70| \ge 21) \le \frac{21}{21^2} \approx 0.048$$ But $$P(|H-70| \ge 21) = P(H \le 49) + P(H \ge 91)$$ so $$P(H \le 49) \le P(|H-70| \ge 21) \le 0.048$$
Chebyshev's inequality: If $$X$$ is a random variable with finite mean $$\mu$$ and variance $$\sigma^2$$, then for any value $$k>0$$, $$P(|X-\mu| \ge k) \le \frac{\sigma^2}{k^2}$$
Since $$H$$ is a sum of Bernoulli random variables, then $$H$$ is modeled as Binomial with $$N = 100$$ trials and $$p=0.7$$ probability of success. Indeed, it is exhaustive to compute the desired probability which is $$P(H < 50) = P(H=0) + \ldots P(H=49)$$
Instead what you could do is notice that $$N =100$$ is large enough, and hence we could approximate the Binomial with a Gaussian distribution, so $$H \sim N(Np,Np(1-p)) = N(70,21)$$ So $$\Pr(H<50) \simeq \Pr(\underbrace{\frac{H-70}{\sqrt{21}}}_{Z} < \frac{50-70}{\sqrt{21}}) \simeq \Pr(Z < -4.36)$$ where $$Z$$ is a centered Gaussian. According to the z-table, the above probability is way less than $$\frac{1}{10}$$.
• Thank you very much @Ahmad Bazzi ! I should think about this... I only thought this experiment should approximate this binomial to a possion distribution... Can I ask a stupid question that when can we approximate the binomial with a gaussian distribution? I guess like when experiment times n=10 or p=0.99 we could not do that, right? – Peter Nova Aug 21 at 11:29
• You are right, when we say $N$ is large, it should be compared to a threshold to be able to say "it is large enough". A rule of thumb is when $N > 9 \max(\frac{1-p}{p},\frac{p}{1-p})$. @PeterNova – Ahmad Bazzi Aug 21 at 11:58
Here's an argument that doesn't rely on approximating a normal distribution. The basic idea is to show that
$$\sum_{k=1}^{49}{100\choose50-k}p^{50-k}q^{50+k}\lt{1\over10}\sum_{k=0}^{50}{100\choose50+k}p^{50+k}q^{50-k}$$
where $$p=0.7$$ and $$q=1=p=0.3$$
Now $$(q/p)^2=9/49$$, so we have $$(q/p)^{2k}\lt1/10$$ for $$k\gt1$$. Since $${100\choose50-k}={100\choose50+k}$$, it follows that
$${100\choose50-k}p^{50-k}q^{50+k}\lt{1\over10}{100\choose50+k}p^{50+k}q^{50-k}$$
for $$2\le k\le50$$. It would be nice if $${100\choose49}p^{49}q^{51}$$ were less than $${1\over10}\left({100\choose50}p^{50}q^{50}+{100\choose51}p^{51}q^{49}\right)$$; unfortunately, it isn't. However, it suffices to show that
$${100\choose48}p^{48}q^{52}+{100\choose49}p^{49}q^{51}\lt{1\over10}\left({100\choose50}p^{50}q^{50}+{100\choose51}p^{51}q^{49}+{100\choose52}p^{52}q^{48} \right)$$
which simplifies to showing
$$50\cdot49q^4+52\cdot50pq^3\lt{1\over10}\left(52\cdot51p^2q^2+52\cdot50p^3q+50\cdot49p^4 \right)$$
This can be done with a direct calculation, but can also be verified with a couple further simplifications:
$$50\cdot49q^4+52\cdot50pq^3\lt50\cdot52(q^4+pq^3)=50\cdot52q^3(q+p)=50\cdot52q^3=70.2$$
and
$${1\over10}\left(52\cdot51p^2q^2+52\cdot50p^3q+50\cdot49p^4 \right)\gt{50\cdot49\over10}(p^2q^2+p^3q+p^4)=245p^2((p+q)^2-pq)=245p^2(1-pq)=94.8395$$
(There are undoubtedly additional ways to reduce the final amount of explicit computation for the comparison. I would appreciate any suggestions.)
Another way to do this is:
$$\sum_{n=0}^{49}\dbinom{100}{n}(0.7)^n(0.3)^{100-n} = 1-\dbinom{100}{50}(0.7)^{50}(0.3)^{50}{{_2}F_1\left(1,-50;51;-\dfrac{7}{3}\right)}\approx 10^{-5}$$ | 2019-09-18T12:23:50 | {
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https://or.stackexchange.com/questions/4479/formulation-of-assignment-problem-as-integer-programming | Formulation of Assignment problem as integer programming
We need to maintain as quickly as possible a complex system. In particular, we need to replace six of its components $$\{P_1,\ldots,P_6\}$$. We have three 3D printers $$\{M_1,M_2,M_3\}$$ which we can use to fabricate the six components. The following table/matrix states how long it takes (in minutes) the $$i$$th printer to print the $$j$$th component:
$$\begin{array}{ccccccc} \hline & P_1& P_2&P_3&P_4&P_5&P_6 \\ \hline M_1 & 23 & 42 & 12 & 32 & 47 & 60\\ M_2 & 25& 37& 13& 37& 51& 64\\ M_3 & 27 &51 &15& 41 &57 &55\\ \hline \end{array}$$
The complex system will work again only when all the components have been printed. Clearly more components can (and have to be) assigned to single machines and they are made in a sequence, one after the other, and 3D printers can work in parallel. However, you have only two operators and therefore you can only use two machines. How to formulate the problem as a linear (but combinatorial) optimization problem to allocate components to (two out of three) 3D printers so that the maintenance time is minimized.
So far I have tried the following, but I am not quite sure (Please I need help if I was mistaken):
Let $$x_{ij}= 1$$ if machine $$i$$ is assigned to component $$j$$, $$0$$ otherwise. The model is \begin{align}\min&\quad23x_{11}+42x_{12}+...+55x_{36}\\\text{s.t.}&\quad x_{11}+x_{12}+x_{13}+x_{14}+x_{15}+x_{16} = 2\\&\quad x_{21}+x_{22}+x_{23}+x_{24}+x_{25}+x_{26} = 2\\&\quad x_{31}+x_{32}+x_{33}+x_{34}+x_{35}+x_{36} = 2\\&\quad x_{11}+x_{21}+x_{31} \ge 1\\&\quad x_{12}+x_{22}+x_{32} \ge 1\\&\quad x_{13}+x_{23}+x_{33} \ge 1\\&\quad x_{14}+x_{24}+x_{34} \ge 1\\&\quad x_{15}+x_{25}+x_{35} \ge 1\\&\quad x_{16}+x_{26}+x_{36} \ge 1\\&\quad x_{ij}\,\,\text{is binary}.\end{align}
• "The complex system will work again only when all the components have been printed." Does that mean you'd like the system to end as soon as possible? (Thinking of the objective function here). Besides this, I think this could better be formulated as a scheduling problem on parallel machines, with the additional constraint that only 2 of the 3 machines are used. – dhasson Jul 2 '20 at 13:37
• The Generalized Assignment Problem can serve as inspiration to continue, it's the same you are doing but change the first 3 constraints where every machine is made to print 2 components. First of all, the objective function is total working time of the machines. Instead you want to minimize the makespan (maximum time to finish all jobs), so that system can work again as soon as possible. Second, you may add binary variables $z_i = 1$ if machine $i$ is used, with additional constraints for $\sum_i z_i = 2$ and linking $x$ and $z$. – dhasson Jul 2 '20 at 15:30
• Are you assuming that one machine will never be used, or that operators will move among machines such that all machines might be used at some point, but never more than two running at any given time? – prubin Jul 2 '20 at 15:57
• @user3752, as dhasson mentioned, it sounds like a parallel machine scheduling problem. Would you see this link? – A.Omidi Jul 3 '20 at 10:41
Since you mention that you want to operate two out of three machines, it boils down to a problem where you first pick two machines and then perform a standard $$R2||C_{\max}$$ parallel machine scheduling problem with the two machines that you selected. Such problems are very suitable for dynamic programming/column generation approaches, but your instance is so small that a IP will work fine. And since you ask for an IP, let us consider a straightforward way to model it.
For a formulation, consider the following decision variables: $$y_j = \left\{ \begin{array}{ll} 1 & \mbox{ if } M_j \mbox{ is being operated } \\ 0 & \mbox{otherwise} \end{array} \right.$$ and $$x_{ij} = \left\{ \begin{array}{ll} 1 & \mbox{ if } P_i \mbox{ is made on } M_j \\ 0 & \mbox{otherwise} \end{array}\right.$$ Also, let's assume that $$a_{ij}$$ is the time needed to produce $$P_i$$ on $$M_j$$.
Now the following IP can be formulated: $$\begin{array}{llll} \min & z \\ \mbox{s.t.} & \sum_{j} y_j & \leq K \\ & \sum_{j} x_{ij} & = 1 & \forall i \\ & x_{ij} & \leq y_j & \forall i \forall j \\ & \sum_i a_{ij} x_{ij} & \leq z & \forall j \\ & x_{ij} \in \{0,1\} & & \forall i \forall j \\ & y_j \in \{0,1\} & & \forall j \\ & z \in \mathbb{R} \end{array}$$ Where the objective variable $$z$$ represents the makespan, the first constraint states that at most $$K$$ machines can be used (in your instance, $$K=2$$, i.e. $$K$$ is the number of operators), the second constraint states that each $$P_i$$ must be executed on exactly one $$M_j$$, the third constraint states that a $$P_i$$ can only be produced on a $$M_j$$ if that $$M_j$$ is being operated and the fourth constraint states that the makespan $$z$$ should be at least the time spent on each individual machine. | 2021-04-19T09:33:49 | {
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https://math.stackexchange.com/questions/682714/is-the-pseudoinverse-of-a-singular-lower-triangular-matrix-itself-lower-triangu | # Is the pseudoinverse of a singular, lower triangular matrix itself lower triangular?
Suppose $L\in\mathbb{R}^{n\times n}$ is a singular, lower triangular matrix. Is its psuedoinverse, $L^\dagger\in\mathbb{R}^{n\times n}$, also lower triangular? I have already proved by induction that the product of two lower triangular matrices is lower triangular, and I also proved that the inverse of a (non-singular) lower triangular matrix is lower triangular. I have shown by working out an example with lower triangular $L\in\mathbb{R}^{2\times 2}$ that the pseudoinverse, $L^\dagger\in\mathbb{R}^{2\times 2}$, does not have to be lower triangular, but I was wondering if there might be a better way to prove this than just by showing through an example in $\mathbb{R}^{2\times 2}$ that $L^{\dagger}$ is not necessarily lower triangular.
Useful information:
The pseudoinverse of a matrix $A$ satisfies the following properties:
1) $(A^\dagger A)^*=A^\dagger A$
2) $(AA^\dagger)^*=AA^\dagger$
3) $AA^\dagger A=A$
4) $A^\dagger AA^\dagger = A^\dagger$
Note: here I will be considering only the reals, so the conjugate transpose becomes just a transpose.
Attempt at Solution: For $L\in\mathbb{R}^{n\times n}$, where $n=1$, it is easy to show that $L^\dagger$ is lower triangular. Since $L$ is singular, we have $L=0$, which is also lower triangular, by the definition of a lower triangular matrix. Then, properties 1) through 3) above are trivially satisfied by any $L^\dagger\in\mathbb{R}$, but for property 4) to be satisfied, we need $L^\dagger 0L^\dagger=L^\dagger\implies L^\dagger=0$. But, $L^\dagger$ is lower triangular by the definition of a lower triangular matrix, so for $n=1$, the answer to the original question is "yes".
Now, consider the case of $n=2$. Let $$L=\left(\begin{array}{cc}a & 0 \\ b & \tilde{L} \end{array}\right)$$ for $a$ and $b\in\mathbb{R}$, and with $\tilde{L}$ a singular, lower triangular matrix $\in\mathbb{R}^{1 \times 1}$, as before in the example with $n=1$. So $L$ is a singular, lower triangular matrix, and $$L=\left(\begin{array}{cc}a&0\\b&0\end{array}\right).$$ Let $L^\dagger\in\mathbb{R}^{2\times 2}$ be the pseudoinverse of $L$, and define $L^\dagger$ as $$L^\dagger=\left(\begin{array}{cc}c&d\\e&\tilde{L}^\dagger\end{array}\right),$$ with $c,d,e\in\mathbb{R}$ and with $\tilde{L}^\dagger$ being the psuedoinverse of $\tilde{L}$ as derived in the above example for $n=1,$ so $\tilde{L}^\dagger = 0$ and so $$L^\dagger=\left(\begin{array}{cc}c&d\\e&0\end{array}\right).$$
Now, property 1) above requires $ea=0$. Property 2) requires $ad=bc$. Property 3) requires that ($a=0$ or $ca+db=1$) and ($b=0$ or $ca+db=1$). Property 4) requires $e=0$ and ($c=0$ or $ca+db=1$) and ($d=0$ or $ca+db=1$). I chose $d\neq0$, which would then make $L^\dagger$ not a lower triangular matrix. Then, with $ca+db=1$ and $d\neq0$, I found $b=\frac{1-ca}{d}$. Further, from $ad=bc$, I found (substituting the previous expression for $b$), $a=\frac{c}{c^2+d^2}$, which then gives $b$ in terms of $c$ and $d$ as $b=\frac{d}{c^2+d^2}.$ My $L$ and $L^\dagger$ then become $$L=\frac{1}{c^2+d^2}\left(\begin{array}{cc}c&0\\d&0\end{array}\right),$$ and $$L^\dagger=\left(\begin{array}{cc}c&d\\0&0\end{array}\right).$$ $L^\dagger$ satisfies all four of the properties above, but it is not lower triangular, so my answer to the original question is, in general, "no".
However, surely their is a smarter way to do this, in which I don't have to resort to doing all this algebra, isn't there? If so, how might I approach the proof? Thanks a lot!
An even stronger counterexample: If $A = \left(\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}\right)$, then there exists no upper-triangular $2\times 2$-matrix $B$ satisfying $ABA = A$. (Indeed, the $\left(1,2\right)$-entry of $ABA - A$ will always be $-1$, no matter what $B$ is, as long as $B$ is upper-triangular.)
The pseudoinverse of a non-invertible triangular matrix can be non-triangular. $$\begin{pmatrix} 1& 1\\ 0& 0 \end{pmatrix}^\dagger = \frac12 \begin{pmatrix} 1& 0\\ 1& 0 \end{pmatrix}$$ $$\begin{pmatrix} 1& 1& 1\\ 0& 1& 1\\ 0& 0& 0 \end{pmatrix}^\dagger = \frac12 \begin{pmatrix} 2& -2& 0\\ 0& 1& 0\\ 0& 1& 0 \end{pmatrix}$$ | 2020-04-08T16:15:40 | {
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https://math.stackexchange.com/questions/1359801/order-of-infinite-countable-ordinal-numbers | order of infinite countable ordinal numbers
I'm trying to understand ordinal arithmetic. If one had an ordered list of the some subset of countable ordinal numbers, what order would the following 6 countably infinite ordinals be in? If the following order is not correct, what is correct order and why is that the correct order?
$$\omega\;<\; \omega^2 \;<\; 2^\omega \;<\; \omega^\omega \;<\; {^\omega}2 \;<\;{^\omega} \omega$$
I know $\epsilon_0 = {^\omega} \omega$ is the largest, but is still countable, but I'm not sure where the powers of $2$ fit in versus the powers of $\omega$.
I understand why $\omega^2$ or $\omega^n$ for any finite value of $n$ needs to be countable. But, why does $\omega^\omega$ need to be countable? For cardinal numbers, $2^{\aleph_0}$ is uncountably infinite. Presumably, there would be some contradiction in mathematics if any finite ordinal arithmetic equation involving $\omega$ generated an uncountable infinity.
• What does the notation ${^\omega} 2$ mean? Never saw that (and it's hard to see what to type into Google to find it...) – David C. Ullrich Jul 13 '15 at 19:12
• – Asaf Karagila Jul 13 '15 at 19:13
• The defining sequence for $^{\omega} 2$ is $2\;\;2^2\;\;2^{2^2}\;\;2^{2^{2^2}}\;\;2^{2^{2^{2^2}}}....$ – Sheldon L Jul 13 '15 at 19:15
• So then in fact ${^\omega}2=\omega$. – David C. Ullrich Jul 13 '15 at 19:17
• ok, presumably that is consistent, but it seems a bit odd, that such a fast growing sequence is regarded as smaller than $\omega^2$ – Sheldon L Jul 13 '15 at 19:18
First note that $$^\omega2 := \sup \{ \underbrace{2^{2^{2^\ldots}}}_{n\text{-times}} \mid n < \omega \} = \omega.$$
[..]but it seems a bit odd, that such a fast growing sequence is regarded as smaller than $\omega^2$.
Well, that's the thing when dealing with infinities. Sometimes our intuition fails us. While $(\underbrace{2^{2^{2^\ldots}}}_{n\text{-times}})_{n < \omega}$ could be regarded as "a fast growing sequence", each of its elements is finite and therefore $\omega$ is an upper bound. As $\omega$ certainly is the least upper bound, we get $^\omega2 = \omega$.
By an analogous argument we get that $$2^\omega := \sup \{2^n \mid n < \omega \} = \omega$$
So we are left with ordering $\omega, \omega^2, \omega^\omega$ and $^\omega \omega$.
We have \begin{align}\omega^2 &:= \omega \cdot \omega \\ &= \sup \{ w \cdot n \mid n < \omega\} \\ &\ge \omega \cdot 2 \\ &> \omega \end{align}
and
\begin{align}\omega^\omega &= \sup \{ \omega ^n \mid n < \omega \} \\ &\ge \omega^3 \\ &= \sup \{(\omega^2) \cdot n \mid n < \omega \} \\ &\ge \omega^2 \cdot 2 \\ &> \omega^2. \end{align}
Finally \begin{align}^\omega \omega &:= \sup \{ \underbrace{\omega^{\omega^{\omega^ \ldots}}}_{n\text{-times}} \mid n < \omega \} \\ &\ge \omega^{\omega^\omega} \\ &= \sup\{\left( \omega^\omega \right)^n \mid n < \omega \} \\ &\ge \left(\omega^\omega \right)^2 \\ &= \omega^\omega \cdot \omega^\omega \\ &= \sup \{\omega^\omega \cdot \alpha \mid \alpha < \omega^\omega \} \\ &\ge \omega^\omega \cdot 2 \\ &> \omega^\omega. \end{align}
Combining these calculations we get the desired order: $$\omega = 2^\omega = {^\omega} 2 < \omega^2 < \omega^\omega < ^\omega\omega$$
In general, ordinal and cardinal arithmetic are very different beasts and every ordinal arithmetic expression using only ordinals $\le \omega$ is countable.
proof (sketch)
Take a countable transitive model $M$ of a large enough fracture of $ZFC$. Every ordinal expression using only ordinals $\le \omega$ can be computed correctly inside $M$ (<- this requires some work). As $M$ only contains countable ordinals (as it is transitive), the result follows.
• That is a horrible notation, ${}^\omega2$. – Asaf Karagila Jul 14 '15 at 6:34
• Thanks for the proof; much to learn. – Sheldon L Jul 14 '15 at 8:39 | 2019-10-24T05:01:50 | {
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https://math.stackexchange.com/questions/878115/in-30-boxes-are-15-balls-chance-all-balls-in-10-or-less-boxes | # In 30 boxes are 15 balls. Chance all balls in 10 or less boxes?
Question1: I found 30 boxes. In 10 boxes i found 15 balls. In 20 boxes i found 0 balls. Afer i collected all 15 balls i put them randomly inside the boxes.
How much is the chance that all balls are in only 10 boxes or less?
Question2: I found 30 boxes. In 10 boxes i found 15 balls. In 20 boxes i found 0 balls. In two of the boxes i could find 3 balls. (So in one box has to be 2 balls and in the other seven boxes have to be 1 ball.) Afer i collected all 15 balls i put them randomly inside the boxes.
How much is the chance that i find in only 2 boxes 6 balls or more?
I wrote a c# programm and tried it 1 million times. My solution was: With a chance of 12,4694% all balls are in 10 boxes or less.
Random trials/Monte Carlo simulations are notoriously slow to converge, with an expected error inversely proportional to the square root of the number of trials.
In this case it is not hard (given a programming language that provides big integers) to do an exact count of cases. Effectively the outcomes are partitions of the 15 balls into some number of boxes (we have thirty boxes to work with, so at least half will be empty).
I wrote a Prolog program to do this (Amzi! Prolog has arbitrary precision integers built in), and got the following results:
$$Pr(\text{10 or fewer boxes occupied}) = \frac{59486359170743424000}{30^{14}} \approx 0.124371$$
$$Pr(\text{2 boxes hold 6 or more balls}) = \frac{30415369655816064000}{30^{14}} \approx 0.063591$$
The reason I'm dividing by $30^{14}$ in these probabilities is because I normalized the counting to begin with one case where a ball is in one box. If we counted that as thirty cases, we'd need to divide by $30^{15}$. So this keeps the totals slightly smaller. Each ball we add increases the total number of cases by a factor of $30$.
I wrote a recursive rule to build cases for $n+1$ balls from cases for $n$ balls. The first few cases have the following counts:
/* case(Balls,Partition,LengthOfPartition,Count) */
case(1,[1],1,1). /* Count is nominally 1 to begin */
case(2,[2],1,1).
case(2,[1,1],2,29).
case(3,[3],1,1).
case(3,[1,2],2,87).
case(3,[1,1,1],3,812). /* check: for Sum = 3, sum of Count is 900 */
The number of cases generated is modest enough for a desktop, daunting to manage by hand. For $n=15$ there are $176$ partitions. It simplified the Prolog code to maintain the partitions as lists in ascending order.
• I checked out with a c# program. I simulated the problem with random numbers and counted: How often I find 6 ore more balls in only 2 boxes. 1000000-times tried: Pr(2 boxes hold 6 or more balls) = 0,063491%. Thanks alot! – Simon Aug 4 '14 at 13:39
• Simulation is easier to program than the recursive/exact counting, but as briefly mentioned, getting an extra digit of accuracy may well require a hundred times as many trials because of the random walk phenomenon. – hardmath Aug 4 '14 at 13:42
Solution of Question 1:
This is an occupancy problem with $n=30$ boxes and $k=15$ balls.
Let's first consider the expected number of empty boxes. That is much easier to obtain. The exact answer is $30(1-1/30)^{15}=18.04.$ This is approximately $30/\sqrt e.$ See the answer by Mr.Spot to this question:
Making 400k random choices from 400k samples seems to always end up with 63% distinct choices, why?
The probability of exactly $j$ empty boxes is, for $n-k\le j\le n-1$:
$$P(j)={n \choose j}\sum_{m=0}^{n-j}(-1)^m {n-j \choose m}\left(1-\frac{j+m}n\right)^k$$
See this:
http://probabilityandstats.wordpress.com/2010/04/04/a-formula-for-the-occupancy-problem/
For $n=30$ and $k=15:$
$P(20)$ to $P(24)$ is $0.096,0.024,0.0036,0.00029,0.000013,....$
and the probability of at least $20$ empty cells = 0.124371
Solution of Question 2:
You throw 15 balls into 30 boxes. What is the probability of the following result:
2 triple, 1 double and 7 single occupancy boxes and the rest empty.
$${30 \choose 2} {28 \choose 1}{27 \choose 7}\frac{15!}{3!3!2!1!^7}\frac{1}{30^{15}}$$ $$=\frac{(30)(29)...(21)}{30^{15}2!1!7!}\frac{15!}{3!3!2!1!^7}$$ First we select the 2 different boxes for the triples; of the 28 remaining boxes we select 1 for the double; of the 27 remaining boxes we select 7 for the singles. Then the multinomial coefficient gives the number of ways to assign the 15 balls to those particular boxes and there are $30^{15}$ equally likely ways to throw the 15 balls into the 30 boxes.
• Thanks a lot for answering the question and for giving me tips to understand the topic. The answer fits to my program. Expected number of empty boxes: I did 11*6+295*7 ... 14246*15=11956971 (Numbers posted in comment ant11). 11956971/1000000=11.956971. 30-ans=18.043029. I also checked the formula in wolfram alpha and i got the same numbers. Now i will write a program about question2. I think this is easy to solve (approximately) with a program. And i will take a look at "formula for the occupancy problem". Maybe i can improve my understanding about this topic. – Simon Jul 27 '14 at 17:14
• I tried your formula for each combination and got nearly the same numbers. In comparison to the other combination, one had a larger difference. If you get 15 boxes with balls (1 ball each box), i had the chance of 1,4246%. If i do formula with wolfram alpha i use this text: (30 choose j) (sum ((-1)^m)*((30-j) choose m)*(1-((j+m)/30))^15, m=0 to 10) , j=15 | (15 stands for 15 free boxes --> 15 boxes not empty). I got the solution: 0,02291 = 2,291%. I used the same formula for j=15,16...29. I estimated the numbers as 2,9910%. All 15 numbers added i got 1,015827. Normally i should get 1. – Simon Jul 28 '14 at 15:00
• Ok i tried again 2 times. Now i rounded all nombers down. I tried your formula for j= 15,16...29. My solution: 15=0,029910305; 16=0,092823631; 17=0,236474423; 18=0,307416722; 19=0,224830457; 20=0,096214657; 21=0,024288697; 22=0,003562779; 23=0,000292277; 24=0,000012534; 25=0; 26=0; 27=0; 28=0; 29=0; All numbers added = 1,015826482. I think you may need a different formula for the upper limit 15 are empty. – Simon Jul 28 '14 at 15:50
• Ok, the right formula is in your link: probabilityandstats.wordpress.com/2010/04/04/…. In this case we need another formula: If there are exactly j empty cells, then the k balls are placed in the remaining n-j cells and these remaining n-j cells are all occupied. Great job! Thank you. – Simon Jul 28 '14 at 20:42
• P(15)=0.0141 which explains why your sum is not 1. – Mr.Spot Jul 29 '14 at 3:39
I will assume the balls and boxes are indistinguishable.
The first problem is: If I distribute $15$ balls among $30$ boxes, what is the probability that at most $10$ boxes contain a ball?
First, the fact that there are $30$ boxes does not matter, since they are indistinguishable. So we only need to consider the problem as if there were $15$ boxes.
Second, we'll solve the problem by finding the probability that more than $10$ boxes contain a ball, and subtract that number from $1$.
There is exactly $1$ to occupy $15$ boxes with $15$ balls: $\{1,1,1\cdots\}$
There is also exactly $1$ way to occupy $14$ boxes with $15$ balls, since, again, the boxes are indistinguishable: $\{2,1,1\cdots\}$
There are $2$ ways to occupy $13$ boxes: $\{3,1,1\cdots\}$ or $\{2,2,1,1\cdots\}$
There are $3$ ways to occupy $12$ boxes, and $5$ ways to occupy $11$ (you can and should verify these numbers).
Finally, how many ways could we distribute $15$ balls over $15$ boxes? This number is exactly equal to $p(15)=176$, where $p(n)$ is the partition function.
So the answer to problem 1 is $1-(1+1+2+3+5)/176=164/176\approx93.2\%$
EDIT: I believe this answer is drastically different from the one obtained by your program for the following reason: you might have meant, in your problem statement, "what is the probability all $15$ balls are contained in $10$ particular boxes?" This significantly changes the question; in particular, the total number of boxes being $30$ is now relevant. Let me know if this is the misunderstanding.
• There is no good reason to use a probability model in which all partitions are equally likely. A more natural model has us throwing the balls one at a time, with all $30$ boxes equally likely. – André Nicolas Jul 25 '14 at 20:14
• Let us say we got 30 machines. Rob is telling you: "Hey ant11, we got 15 disruptions last year". Another guy says: "two machines had 3 disrupions, one machine had 2 disruptions and the other seven machines had 1 disruption. Ant11 you are good at maths. What do you think about this. If all machines would be indistinguishable, what chance it would have that only 10 machines had all 15 disruptions?" You think the solution should be easy. – Simon Jul 25 '14 at 20:23
• Meanwhile you think about it, the boss comes and asks you:" Hey ant11, two machines got together 6 disruptions. That sounds like they will have many disruptions more next year. What is the chance if all 30 machines are indistinguishable, that they are more than 6 disruptions in only two machines. Maybe it was just bad luck." I think there are 30 boxes is relevant. – Simon Jul 25 '14 at 20:23
• My last time i wrote a programm was two years ago. Maybe i made a mistake. Here you can see my solution: 1000000 tryed: 1-5: 0-times: 0,0% 6: 11-times: 0,0011% 7: 295-times:0,00295% 8: 3585-times: 0,3585% 9: 24383-times: 2,4383% 10: 96420-times: 9,6420% 11: 225081-times: 22,5081% 12: 307109-times: 30,7109% 13: 236556-times: 23,6556% 14: 92314-times: 9,2314% 15: 14246-times: 1,4246% – Simon Jul 25 '14 at 20:32
• After that i added the numbers: 11+295+3585+24383+96420=124694. 124694/1000000=0,124694. --> 12,4694% 10 or less machines if all are indistinguishable – Simon Jul 25 '14 at 20:43 | 2019-11-12T21:32:41 | {
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https://math.stackexchange.com/questions/2887540/finding-the-number-of-non-negative-integral-solutions-of-x-y-z-10 | # Finding the number of non-negative integral solutions of $x + y + z = 10$
I realize that this question has been asked multiple times and I do not really want to know how to do it, I understand how to solve it, my issue is somehow I have used the distribution and I am not getting the right answer.
Method of Solving:
I solved this in the first go by directly applying the distribution formula that is $$\binom {n+r-1}{r-1}$$
By using, $n = 10, r= 3$ we get,
$$\binom{12}{2} = 66$$ however the answer given is $$\binom{12}{3} = 220$$ Next, I tried to solve by taking multiple cases that is, setting any variable, $x,y$ or $z$ to values from $0,1,2,3...10$, so for example if $x = 0$ then we get, $$y+z=10$$ now it is simple enough to distribute this even without the formula, we get $$\Bigl((0,10),(1,9),(2,8)(3,7),(4,6)\Bigl) \cdot 2,(5,5)$$ which gives us $11$ cases, that can be verified using the formula $$P_0 = \binom{11}{1}.$$ Now taking cases till $x=10$ we get, the final result as $$\sum _{n=0} ^{n=10} P_n = 11+ 10 + 9 + 8+ \ldots +1 = 66$$
Please tell me where I am going wrong or of it simply a case of the answer in the text being wrong, and if the second method is correct.
• The answer in the text is wrong. Both of your methods are correct. Aug 19 '18 at 10:06
• All right, thank you so much Aug 19 '18 at 10:07
• Does this answer your question? Number of Non negative integer solutions of $x+2y+5z=100$ Oct 11 at 21:51
The answer is $66$ even by double checking with generating functions, more details here.
The number of solutions for $x+y+z=n, x\geq0, y\geq0, z\geq0$ is the coefficient of $x^n$ term of the $$(1+x+x^2+x^3+...+x^k+...)^3=\frac{1}{(1-x)^3}$$ and because $$\frac{1}{(1-x)^3}= \frac{1}{2}\left(\frac{1}{1-x}\right)^{''}= \frac{1}{2}\left(\sum\limits_{k=0}x^k\right)^{''}= \sum\limits_{k=2}\frac{k(k-1)}{2}x^{k-2}$$ the coefficient of $x^n$ is $\color{green}{\frac{(n+2)(n+1)}{2}=\binom{n+2}{2}}$, which for $n=10$ yields $66$.
I think they have the role of $r$ and $n$ mixed up in the given solution, that is why they got the wrong result. Look at the proof of the formula again, and you will see immediately which one of $10$ and $3$ play the role of $n$ and $r$.
So you are right, the answer is 66, because $\binom{3+10-1}{10}=66$.
• Sorry, I don't see it, could you explain? Aren't we looking for the number of ways to distribute $10$ over $3$ variables? Aug 19 '18 at 9:58
• The OP did not confuse $n$ and $r$. Notice that you used the formula $\binom{n + r - 1}{n}$, while the OP used $\binom{n + r - 1}{r - 1}$. Aug 19 '18 at 10:04 | 2021-10-20T09:27:52 | {
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https://math.stackexchange.com/questions/2875697/disjoint-sets-of-rationals-both-dense-in-mathbbr | # Disjoint sets of rationals both dense in $\mathbb{R}$
There exists a pair of disjoint subsets of $\mathbb{Q}$ such that both are dense in $\mathbb{R}$. True or false?
The statement is true but how can I find an example of such disjoint subsets of $\mathbb{Q}$?
• What have you thought of? What examples have you tried? – Guido A. Aug 8 '18 at 5:57
• Is$(\frac{m}{2^k})_{m, k}$ dense? – dEmigOd Aug 8 '18 at 5:59
• Are m and k integers? Then this is dense in R. – Mathsaddict Aug 8 '18 at 6:02
• First I took example - A = {m/2^k, m - Z , k - N union 0} and B = { m/3^k, m - z, k - N union 0} – Mathsaddict Aug 8 '18 at 6:08
• You're looking more generally for ways to partition $\mathbb{Q}$ into pieces not based on size (e.g. you don't want $(-\infty,\sqrt{2})\cap\mathbb{Q}$ versus $(\sqrt{2},\infty)\cap\mathbb{Q}$). The two obvious places to look for a distinction to draw are numerator and denominator. For example, we could let $EN$ be the set of rationals which (in lowest terms) have an even numerator. Or, we could let $DD$ be the set of rationals whose denominator (in lowest terms) is a power of $2$ ("dyadic" fractions). Or, so on. (cont'd) – Noah Schweber Aug 8 '18 at 6:08
Let $x \in \mathbb{R}$. Now,
$$0 < 2^nx - [2^nx] < 1$$
and thus
$$0 < x - \frac{[2^nx]}{2^n} < \frac{1}{2^n}$$
This says that $D = \{\frac{m}{2^n}\}_{m \in \mathbb{Z}, n \in \mathbb{N}}$ is dense in $\mathbb{R}$. Now, let's consider a subset of this set,
$$A = \{\frac{m}{2^n} : n \in \mathbb{N}, m \in \mathbb{Z} , 2 \not |n \}$$
If $A$ were dense in $D$, it would be dense in $\mathbb{R}$. For this, we only have to show that $D \setminus A$ can be approximated by elements of $A$. Let $\frac{m}{2^n} \in D \setminus A$ and write $m = 2^ks$ with $2 \not | s$. It can't be that $k \leq n$, because that would mean $\frac{m}{2^n} = \frac{s}{2^{n-k}} \in A$. Therefore, $k \geq n$ and thus $\frac{m}{2^n} = 2^{k-n}s \in \mathbb{Z}$. This means that it is sufficient to show that $A$ can approximate integers: if $a \in \mathbb{Z}$ and $\varepsilon > 0$, then taking $\frac{1}{2^n} < \varepsilon$ we have that
$$\left|a - \frac{2^na + 1}{2^n}\right| = \frac{1}{2^n} < \varepsilon$$
and $\frac{2^na + 1}{2^n} \in A$ because $2 \not | \ 2^na+1$. In conclusion, we have shown that $A$ is dense in $D$ and since $D$ is dense in the reals, so is $A$. With an identical construction (which I encourage you to write out) one can show that the same holds for
$$B = \{\frac{m}{3^n} : n \in \mathbb{N}, m \in \mathbb{Z} , 3 \not |n \}$$
To conclude, then, we just have to observe that $A \cap B = \emptyset$. In effect, if
$$\frac{m}{2^n} = \frac{j}{3^k}$$
for some $n,m,j,k$, then $3^km = 2^nj$ and thus $2 | 2^nj = 3^km$. But since $2$ and $3$ are coprime, this would imply $2 | m$ which is absurd.
HINT.-It is enough to show an example in $I=[0,1]$. All rational is periodic after a finite number of digits $$r=0.a_1a_2....a_m[b_1b_2....b_n]\in I$$ It is not hard to prove that $A$ and $B$ below are disjoint and dense in $I$
$$A=\{r\text{ such that } b_n=1\}\\B=\{r\text{ such that } b_n\ne1\}$$ | 2019-04-25T00:40:12 | {
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https://math.stackexchange.com/questions/2798284/explain-please-the-following-statement-from-naive-set-theory | # Explain please the following statement from Naive Set Theory
Here is a statement that I can't understand:
Ordered triples, ordered quadruples, etc., may be defined as families whose index sets are unordered triples, quadruples, etc.
For now, let's stick with the ordered triples. First of all I can't understand whether the index set is a set of unordered triples like $\{ \{a, b, c\}, \{d, e, f\}\}$ or it is itself an unordered triple?
Also, I just can't imagine how having any of such sets as the index set can help me to build a set of ordered triples. Maybe some example will make things clear. I will be very grateful if you help. Thanks in advance.
• Where did you find this statement? Personally, I also find it difficult to parse ... – Noah Schweber May 27 '18 at 18:03
• @NoahSchweber, Naive Set Theory of Halmos – Turkhan Badalov May 27 '18 at 18:03
• What page? It's an entire book ... – Noah Schweber May 27 '18 at 18:03
• @NoahSchweber, oh, sorry. Page 36 in "Families" section – Turkhan Badalov May 27 '18 at 18:05
Yes, this is a bit unclear.
Halmos' goal is to define the notion of an arbitrary Cartesian product. This should generalize the usual Cartesian product of two sets, $A\times B$, but should "work for any number of sets." Before diving into his description, let me point out that this really is nontrivial: how should we think of the Cartesian product of "$\mathbb{R}$-many" sets?
Halmos is about to tell us, essentially, that the Cartesian product of an indexed family $\{X_i\}_{i\in I}$ of sets is just the set of all indexed families $\{x_i\}_{i\in I}$ of objects with $x_i\in X_i$. Maybe more clearly, an element of the Cartesian product is a function with domain $I$; it sends $i\in I$ to the "$i$th coordinate," which must be in $X_i$.
To motivate this, Halmos has us go back to the idea of a Cartesian product. We can rethink Cartesian products as follows:
• Fix any two distinct sets, $a$ and $b$. We'll think of the first as "LEFT" and the second as "RIGHT."
• Now an ordered pair has two "coordinates," a left coordinate and a right coordinate. We're going to match these up with $a$ and $b$ above: if I have a function $z$ with domain $\{a, b\}$ such that $z(a)=x\in X$ and $z(b)=y\in Y$ (here I write "$z(i)$" for Halmos's "$z_i$"), we want to think of the object $z$ as being the ordered pair $(x, y)$. Informally, $z$ says
$$\mbox{My left coordinate is x, and my right coordinate is y.}$$
Put another way:
We can think of the ordered pair $(x, y)$ as the function with domain $\{a, b\}$ (which is an unordered pair) mapping $a$ to $x$ and $b$ to $y$.
This is easiest to think about if $a=0$ and $b=1$, or something similar, but Halmos's point is that all we need is that the index set has two distinct elements. Now here's the key linguistic step Halmos makes which I think is confusing at first:
An ordered pair is an indexed set! And the indexing set is $\{a, b\}$, which is an unordered pair.
• are you sure you wanted to say sets in the end? "... that the Cartesian product of an indexed family $\{X_i\}_{i \in I}$ of sets is just the set of all indexed families $\{x_i\}_{i \in I}$ of sets with $x_i \in X_i$". Because as I understand, if we say the family $\{x_i\}$ of sets, its range consists of sets implying $x_i$ is a set which is, as I understand, is not said – Turkhan Badalov May 27 '18 at 18:42
• @TurkhanBadalov My ZFC bias is showing. In formal ZFC set theory (and many other formal set theories), every object is in fact a set and everything is "built out of" the emptyset in a sense. But yes, in the context of Halmos's book that's inappropriate. Fixed! – Noah Schweber May 27 '18 at 18:51
• Thanks for the great answer! So, to make an ordered triple, let's say, $(11, 21, 31)$ I need some unordered triple and let it be $\{a, b, c\}$, true? Then I have to have (actually I doubt this statement about having the following set, because I need to construct it additionaly, correct me if I am wrong) the family $\{X_i\}_{i \in \{a, b, c\}}$ and let it be $\{ (a,11), (b, 21), (c, 31)\}$. Then the Cartesian product of the family will be the set of all families $\{x_i\}$: $\{ \{ (a, 11), (b, 21), (c, 31)\} \}$, the element of which I interpret as the ordered triple we expected to make? – Turkhan Badalov May 27 '18 at 18:59
• Not quite. The set $\{(a, 11), (b, 21), (c, 31)\}$ is a single ordered triple. The Cartesian product of three sets $X_a, X_b, X_c$ would be the set of all sets of the form $\{(a, x_a), (b,x_b), (c, x_c)\}$ with $x_a\in X_a, x_b\in X_b, x_c\in X_c$. So e.g. if $X_a=X_b=X_c=\mathbb{N}$, the set $\{(a, 11), (b, 21), (c, 31)\}$ would be an element of the Cartesian product, but the Cartesian product would also include other things like $\{(a, 18), (b, 467), (c, 3)\}$ (which we would think of as "$(18, 467, 3)$"). – Noah Schweber May 27 '18 at 19:02
• @TurkhanBadalov I thought you were using my example where $X_a=X_b=X_c=\mathbb{N}$. If the $X_i$s are as you describe, then that's right. – Noah Schweber May 27 '18 at 19:26 | 2021-04-13T12:59:51 | {
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https://cs.stackexchange.com/questions/148833/fastest-algorithm-for-merge-step-in-mergesort/148839#148839 | # Fastest Algorithm for "Merge" step in Mergesort
Given two sorted arrays $$a_1,a_2,\dots,a_n$$ and $$b_1,b_2,\dots,b_m$$, merge them together into one sorted array $$c_1,c_2,\dots,c_{n+m}$$ containing the elements of $$a$$ and $$b$$.
The typical mergesort method works in $$O(n+m)$$, and if we run binary search for each element from one of the arrays, it works in $$O(\min(n,m)\log(\max(n,m)))$$. However, the latter method is no better than if the smaller array was unsorted. I was wondering if there is any use of the fact that both arrays are sorted to optimise the latter solution. In particular, I was hoping for an algorithm in time $$O(\min(n,m))$$ or so.
I attempted a "parallel binary search" method, but it seems in the worst case that it is no better than the naive binary search method.
• (The typical mergesort method should read The typical merge method more likely than not. In a scientific context, it is challenging to ask for optimal solutions (Fastest Algorithm): any proposal will be expected to be proven. (Which is where within a constant factor comes in.)) Jan 31 at 8:59
• What makes you think that $\min(n,m)\log(\max(n,m))$ would be any better than $n+m$ ? Take $n<m$ WLOG and divide both functions by $m$. Now you compare $\dfrac nm\log m$ and $\dfrac nm+1$. Mar 3 at 9:20
If you only want to count the number of comparisons then you can achieve $$O\left(n\log(1+\frac{m}{n})\right)$$ comparisons as follows:
Let $$k_1$$ denote the index of $$a_1$$ in the final array $$c$$ and for $$1 let $$k_i$$ denote the difference between the indices of $$a_i$$ and $$a_{i-1}$$ in the final array $$c$$.
To insert $$a_1$$ in $$b$$, start by doing an exponential search to find the smallest $$r$$ such that $$b_{2^r} \geq a_1$$ (this is also the smallest $$r$$ such that $$2^r \geq k_1$$). This takes $$O(\log k_1)$$ comparisons. Now do a binary search between the indices $$1$$ and $$2^r$$ to locate the correct position $$k_1$$ to insert $$a_1$$, and insert the element there. This also takes $$O(\log k_1)$$ comparisons.
To insert $$a_2$$ into $$b$$ start by doing an exponential search, starting at $$k_1$$, to find the smallest $$r$$ such that $$b_{k_1+2^r} \geq a_2$$ (this is also the smallest $$r$$ such that $$2^r \geq k_2$$). This takes $$O(\log k_2)$$ comparisons. Now do a binary search between the indices $$k_1$$ and $$k_1+2^r$$ to locate the correct position $$k_1+k_2$$ to insert $$a_2$$, and insert the element there. This also takes $$O(\log k_2)$$ comparisons.
In general following this strategy you will make $$O(\log k_i)$$ comparisons to insert element $$a_i$$. So the total number of comparisons will be $$T = O\left(\sum_{i=1}^n\log k_i\right)$$. Because the $$\log$$ function is concave, by Jensen's inequality we have $$\sum_{i=1}^n\log k_i \leq n\log \frac{\sum k_i}{n} \leq n\log \frac{n+m}{n}$$. The result follows.
Notice that (assuming $$n\leq m$$) this is asymptotically always at least as good as $$O(n+m)$$ and $$O(n\log m)$$, and sometimes better than both: if you take for example $$m=n\log n$$ you get $$O(n\log\log n)$$ comparisons compared to $$O(n\log n)$$ with both other methods.
To complement this, let us show that this is optimal (up to a constant factor, still assuming $$m\geq n$$). The number of ways to insert $$n\geq 1$$ ordered elements into an ordered array of length m is $$\frac{(m+n)!}{m!n!} \geq \frac{(m+1)^n}{n!} \geq \max\{(\frac{m}{n})^n, 2^n\}$$. Thus to distinguish between these cases you need at least $$T\geq \log_2(\frac{(m+1)^n}{n!})$$ comparisons in the worst case. We have $$T \geq \max\{n\log_2\frac{m}{n}, n\}$$, which implies $$T \geq \frac{1}{2}n\log_2(1+\frac{m}{n})$$.
One method fitting the analysis results presented is
one by one, insert elements of the shorter sequence ($$s$$) in the longer one ($$l$$).
There are several way to reduce the search range in the longer sequence exploiting the shorter one to be ordered, too, starting with
position ← 0
for every element $$e$$ of $$s$$
position ← insertion position of $$e$$ in $$l$$, starting from position
insert $$e$$ in $$l$$ at position
— without any relation specified between $$n$$, $$m$$, and the values in $$a$$ and $$b$$, I don't see any to allow a tighter bound on time than
$$O(\min(n,m)\log(\max(n,m)))$$.
Try proving $$O(n+m)$$ is optimal.
• Well supposing $n\leq m$ (just to make it more readable) you can always have $O(\min(n\log m, n+m))$ comparisons: start with the binary search method and if you have reached $n+m$ comparisons finish with the "typical" method (or even start over from scratch with this method). Mar 2 at 13:36
• @Tassle You may have missed the problem statement specifying a 3rd/output array $c_1,c_2,…,c_{n+m}$. Mar 4 at 9:48
The purpose of a merge is to form a single, continuous array holding the $$n+m$$ keys from the original arrays.
Even if everything can be done in-place (by $$a$$ and $$b$$ being contiguous), in the worst case you need to move all of them, which is $$\Omega(n+m)$$. Hence a merge in time $$O(n+m)$$ is optimal.
Even ignoring the moves (?), in the worst case you need to look at all keys at least once, and $$O(n+m)$$ is still optimal.
A light of hope: the best case is more promising. If $$a, b$$ are contiguous and $$a_n, the merge can be done in $$O(1)$$.
• If you're only interested in comparisons $O(n+m)$ is not always optimal (say if $m$ is large compared to $n$, then the binary-search strategy does $O(n\log m)$ comparisons and doesn't need to look at all the keys in the array $b$). See my answer for what I think is the true optimum in that setting. Apr 2 at 14:10 | 2022-11-30T10:11:07 | {
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https://math.stackexchange.com/questions/855392/proof-of-a-subseteq-b-leftrightarrow-a-cap-b-a-check-chain-of-implicatio | # Proof of $A \subseteq B \Leftrightarrow A \cap B = A$ (Check chain of implications)
Prove $A \subseteq B \Leftrightarrow A \cap B = A$.
My attempt:
Case $\Rightarrow$:
\begin{align} A \subseteq B & \Rightarrow & [x\in A \Rightarrow x\in B] \\ &\Rightarrow &[x \in A \Rightarrow x\in A \text{ and } x\in B] \tag{1} \\ &\Rightarrow &[x\in A \Rightarrow x\in A \cap B] \\ &\Rightarrow & A \subseteq A\cap B \end{align}
\begin{align} A \subseteq B & \Rightarrow & [x\in A \Rightarrow x\in B] \\ & \Rightarrow & [(x\in A \text{ and } x\in B) \Rightarrow x \in A] \tag{2} \\ & \Rightarrow & [x \in A \cap B \Rightarrow x \in A] \\ & \Rightarrow & A \cap B \subseteq A \end{align}
\begin{align}\therefore A \subseteq B & \Rightarrow & A \subseteq A \cap B \text{ and } A \cap B \subseteq A \\ &\Rightarrow& A \cap B = A \end{align}
Case $\Leftarrow$:
\begin{align} A \cap B = A & \Rightarrow & A \cap B \subseteq A \text{ and } A \subseteq A \cap B \\ & \Rightarrow & [(x \in A \text{ and } x \in B) \Rightarrow x \in A ] \text{ and } [x\in A \Rightarrow & (x \in A \text{ and } x \in B)] \\ & \Rightarrow & [x\in A \Rightarrow x \in B] \tag{3}\\ & \Rightarrow & A \subseteq B \end{align}
Is my proof correct? I am particularly not sure about line 1,2 and 3 since I just made those up (while making sure that the chain of implications is still true) as I already know the conclusion I want to arrive at.
• Your (2)'s second step makes no sense, but you can fix it (probably a typo?) and you are redundant on your last two steps, but not really an "error." – Adam Hughes Jul 3 '14 at 14:17
• @AdamHughes Yes, its a typo. Thanks for calling my attention to it. – mauna Jul 3 '14 at 15:21
I think you waved past a couple of steps in "Case $\Leftarrow$":
\begin{align} A \cap B = A &\Rightarrow A \cap B \subseteq A \land A \subseteq A \cap B \tag{1}\\ & \Rightarrow [(x \in A \text{ and } x \in B) \Rightarrow x \in A ]\land[x\in A \Rightarrow (x \in A \land x \in B)]\tag{2} \\ & \Rightarrow [x\in A \Rightarrow x \in B] \tag{3}\\ & \Rightarrow A \subseteq B\tag{4} \end{align}
After unpacking $A \cap B = A$ to get $(1)$ and $(2)$ (which are both perfectly correct), I think you need more work (or more justification) before concluding: $$x\in A \rightarrow x\in B\tag{3}$$
\begin{align} A \cap B = A &\Rightarrow A \cap B \subseteq A \land A \subseteq A \cap B \tag{1}\\ & \Rightarrow [(x \in A \land x \in B) \Rightarrow x \in A ]\\ &\quad \land x\in A \Rightarrow (x \in A \land x \in B)]\tag{2} \\ &\Rightarrow x\in A \Rightarrow (x \in A \land x\in B)\tag{3}\\ \\ & \quad \text{Assumption:} x\in A\tag{4}\\ &\qquad\quad {\small \Rightarrow} x\in A \land x \in B \tag{(5): 3 & 4}\\ &\qquad\quad {\small \Rightarrow} x\in B\tag{(6): from 5}\\\\ &\Rightarrow x\in A \Rightarrow x\in B\tag{(7): 4-6}\\ & \Rightarrow A \subseteq B\tag{8} \end{align}
• how did you get the assumption in line (4)? Was it based on some of the proceeding lines? – mauna Jul 3 '14 at 15:20
• You are always free to assume something, in this case, I chose to assume $x\in A$ because if we can show, given the earlier lines in the proof, that this implies $x\in B$, we have proven that $x \in A \implies x\in B$. Note that whether or not $x$ is in A doesn't matter. The aim of the proof is to show that IF x is in A, THEN x is in B. I.e., to establish that, together with the earlier part of the proof, $x \in A\implies x\in B$ – Namaste Jul 3 '14 at 15:23
• An alternative route is to know that $$x\in A \rightarrow(x\in A\land x\in B) \\ \Rightarrow [(x \in A \rightarrow x \in A) \land (x\in A \rightarrow x \in B)] \\ \Rightarrow x \in A \rightarrow x \in B\\ \Rightarrow A\subseteq B$$ – Namaste Jul 3 '14 at 15:34
$A \cap B \subseteq A$ is always true, so it's a bit misleading to say that it is implied by $A \subseteq B$ (although not incorrect). You can simply start with the statement:
$$x \in A \text{ and } x \in B \implies x \in A$$
to prove it$-$this is probably what you meant by the statement $(2)$.
For statement $(3)$ it might be more clear how you arrived at this if you say that $A \cap B = A$ implies $A \subseteq A \cap B$, and since $A \cap B \subseteq B$, this means $A \subseteq B$.
The rest of the proof is correct.
Assuming that you're only allowed to use the definitions of $\;\subseteq\;$ and $\;\cap\;$, any proof $\;A \subseteq B \;\equiv\; A \cap B = A\;$ is essentially a proof of the equivalent $$\langle \forall x :: x \in A \Rightarrow x \in B \rangle \;\equiv\; \langle \forall x :: x \in A \land x \in B \;\equiv\; x \in A \rangle$$ That statement directly follows from the following more general law of propositional logic: for any $\;P,Q\;$, $$P \Rightarrow Q \;\;\equiv\;\; P \land Q \;\equiv\; P$$ There are numerous ways to prove this, depending on what laws of logic you are allowed to use. For example, one can work from right to left:
$$\calc P \land Q \;\equiv\; P \calcop{\equiv}{split \;\equiv\; into both directions -- to introduce \;\Rightarrow\; as in our goal} (P \land Q \Rightarrow P) \;\land\; (P \Rightarrow P \land Q) \calcop{\equiv}{use \;P \equiv \text{true}\; from left hand side of \;\Rightarrow\; on right hand side, twice} (P \land Q \Rightarrow \text{true}) \;\land\; (P \Rightarrow \text{true} \land Q) \calcop{\equiv}{simplify: left part is \;\text{true}\;; \;\text{true} \land R \equiv R\;, twice} P \Rightarrow Q \endcalc$$ | 2019-08-23T01:25:40 | {
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http://www.tpicom.eu/assassins-creed-drbg/855ddb-every-asymmetric-relation-is-antisymmetric | A relation R on a set A is symmetric if whenever (a, b) ∈ R then (b, a) ∈ R, i.e. Difference between antisymmetric and not symmetric. An antisymmetric and not asymmetric relation between x and y (asymmetric because reflexive) Counter-example: An symmetric relation between x and y (and reflexive ) In God we trust , … (a,a) not equal to element of R. That is. A relation R is asymmetric if and only if R is irreflexive and antisymmetric. A relation that is not asymmetric, is symmetric. 4 votes . an eigenfunction of P ij looks like. Yes. antisymmetric relation transitive relation Contents Certain important types of binary relation can be characterized by properties they have. 4 Answers. These Multiple Choice Questions (MCQ) should be practiced to improve the Discrete Mathematics skills required for various interviews (campus interviews, walk-in interviews, company interviews), placements, entrance exams and other competitive examinations. A asymmetric relation is an directed relationship. Whether the wave function is symmetric or antisymmetric under such operations gives you insight into whether two particles can occupy the same quantum state. 1 vote . Antisymmetric means that the only way for both $aRb$ and $bRa$ to hold is if $a = b$. Weisstein, Eric W., "Antisymmetric Relation", MathWorld. sets; set-theory&algebra; relations ; asked Oct 9, 2015 in Set Theory & Algebra admin retagged Dec 20, 2015 by Arjun 3.8k views. See also. Quiz & Worksheet - What is an Antisymmetric Relation? By definition, a nonempty relation cannot be both symmetric and asymmetric (where if a is related to b, then b cannot be related to a (in the same way)). Suppose that your math teacher surprises the class by saying she brought in cookies. if aRb ⇒ bRa. We call irreflexive if no element of is related to itself. For example- the inverse of less than is also an asymmetric relation. Exercise 22 Give examples of relations which are neither symmetric, nor asymmetric. Asymmetric v. symmetric public relations. Specifically, the definition of antisymmetry permits a relation element of the form $(a, a)$, whereas asymmetry forbids that. We call antisymmetric … Asymmetric Relation: A relation R on a set A is called an Asymmetric Relation if for every (a, b) ∈ R implies that (b, a) does not belong to R. 6. Is the relation R antisymmetric? For each of these relations on the set $\{1,2,3,4\},$ decide whether it is reflexive, whether it is symmetric, and whether it is antisymmetric, and whether it is transitive. Antisymmetry is different from asymmetry: a relation is asymmetric if, and only if, it is antisymmetric and irreflexive. We call symmetric if means the same thing as . In that, there is no pair of distinct elements of A, each of which gets related by R to the other. This section focuses on "Relations" in Discrete Mathematics. Note: a relation R on the set A is irreflexive if for every a element of A. Here's my code to check if a matrix is antisymmetric. Solution: The relation R is not antisymmetric as 4 ≠ 5 but (4, 5) and (5, 4) both belong to R. 5. Transitive if for every unidirectional path joining three vertices $$a,b,c$$, in that order, there is also a directed line joining $$a$$ to $$c$$. Discrete Mathematics Questions and Answers – Relations. Multi-objective optimization using evolutionary algorithms. I just want to know how the value in the answers come like 2^n2 and 2^n^2-1 etc. Antisymmetric if every pair of vertices is connected by none or exactly one directed line. A relation on a set is antisymmetric provided that distinct elements are never both related to one another. 15. Given that P ij 2 = 1, note that if a wave function is an eigenfunction of P ij, then the possible eigenvalues are 1 and –1. A relation R on a set A is asymmetric if whenever (a, b) ∈ R then (b, a) / ∈ R for a negationslash = b. The relations we are interested in here are binary relations on a set. For example, the restriction of < from the reals to the integers is still asymmetric, and the inverse > of < is also asymmetric. Best answer. Here we are going to learn some of those properties binary relations may have. However, a relation can be neither symmetric nor asymmetric, which is the case for "is less than or equal to" and "preys on"). The relation "x is even, y is odd" between a pair (x, y) of integers is antisymmetric: Every asymmetric relation is also an antisymmetric relation. Antisymmetric definition, noting a relation in which one element's dependence on a second implies that the second element is not dependent on the first, as the relation “greater than.” See more. Other than antisymmetric, there are different relations like reflexive, irreflexive, symmetric, asymmetric, and transitive. Combine this with the previous result to conclude that every acyclic relation is irre±exive. So an asymmetric relation is necessarily irreflexive. answer comment. We call reflexive if every element of is related to itself; that is, if every has . How many number of possible relations in a antisymmetric set? Exercise 19 Prove that every asymmetric relation is irre±exive. (A relation R on a set A is called antisymmetric if and only if for any a, and b in A, whenever (a,b) in R , and (b,a) in R , a = b must hold. That is, for . Limitations and opposite of asymmetric relation are considered as asymmetric relation. A relation R on a set A is non-reflexive if R is neither reflexive nor irreflexive, i.e. For a relation R in set AReflexiveRelation is reflexiveIf (a, a) ∈ R for every a ∈ ASymmetricRelation is symmetric,If (a, b) ∈ R, then (b, a) ∈ RTransitiveRelation is transitive,If (a, b) ∈ R & (b, c) ∈ R, then (a, c) ∈ RIf relation is reflexive, symmetric and transitive,it is anequivalence relation Relationship to asymmetric and antisymmetric relations. Homework 5 Solutions New York University. example of antisymmetric The axioms of a partial ordering demonstrate that every partial ordering is antisymmetric. A relation becomes an antisymmetric relation for a binary relation R on a set A. Exercise 21 Give examples of relations which are neither re±exive, nor irre±exive. each of these 3 items in turn reproduce exactly 3 other items. Any asymmetric relation is necessarily antisymmetric; but the converse does not hold. Show that the converse of part (a) does not hold. Please make it clear. A relation is asymmetric if and only if it is both antisymmetric and irreflexive. In this short video, we define what an Antisymmetric relation is and provide a number of examples. But in "Deb, K. (2013). Since dominance relation is also irreflexive, so in order to be asymmetric, it should be antisymmetric too. The mathematical concepts of symmetry and antisymmetry are independent, (though the concepts of symmetry and asymmetry are not). It's also known as … The incidence matrix $$M=(m_{ij})$$ for a relation on $$A$$ is a square matrix. Think $\le$. Let be a relation on the set . at what time is the container 1/3 full. "sister" on the set of females is, ¨ Any nearness relation is symmetric. A relation is considered as an asymmetric if it is both antisymmetric and irreflexive or else it is not. For example, > is an asymmetric relation, but ≥ is not. Symmetric relation; Asymmetric relation; Symmetry in mathematics; References. Yes, and that's essentially the only case : If R is both symmetric and antisymmetric then R must be the relation ## \{(x,x),x \in B\} ## for some subset ## B\subset A ##. Also, i'm curious to know since relations can both be neither symmetric and anti-symmetric, would R = {(1,2),(2,1),(2,3)} be an example of such a relation? Antisymmetry is concerned only with the relations between distinct (i.e. if a single compound is kept in a container at noon and the container is full by midnight. We call asymmetric if guarantees that . This lesson will talk about a certain type of relation called an antisymmetric relation. (a) (b) Show that every asymmetric relation is antisymmetric. Antisymmetry is different from asymmetry because it does not requier irreflexivity, therefore every asymmetric relation is antisymmetric, but the reverse is false. See also Every asymmetric relation is not strictly partial order. Lipschutz, Seymour; Marc Lars Lipson (1997). if aRa is true for some a and false for others. It can be reflexive, but it can't be symmetric for two distinct elements. Again, the previous 3 alternatives are far from being exhaustive; as an example over the natural numbers, the relation xRy defined by x > 2 is neither symmetric nor antisymmetric, let alone asymmetric. A relation is asymmetric if and only if it is both antisymmetric and irreflexive. Examples: equality is a symmetric relation: if a = b then b = a "less than" is not a symmetric relation, it is anti-symmetric. R is irreflexive if no element in A is related to itself. Exercise 20 Prove that every acyclic relation is asymmetric. a.4pm b.6pm c.9pm d.11pm . Similarly, the subset order ⊆ on the subsets of any given set is antisymmetric: given two sets A and B, if every element in A also is in B and every element in B is also in A, then A and B must contain all the same elements and therefore be equal: ⊆ ∧ ⊆ ⇒ = Partial and total orders are antisymmetric by definition. Non-examples ¨ The relation divides on the set of integers is neither symmetric nor antisymmetric.. Restrictions and converses of asymmetric relations are also asymmetric. Get more help from Chegg. Antisymmetric Relation. Transitive Relations: A Relation … Hint: write the definition of what it means to be asymmetric… A relation can be both symmetric and antisymmetric (in this case, it must be coreflexive), and there are relations which are neither symmetric nor antisymmetric (e.g., the "preys on" relation on biological species). 3.8k views. There is an element which triplicates in every hour. We find that $$R$$ is. 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( 2013 ) relations '' in Discrete mathematics and opposite of asymmetric relation asymmetric! If it is antisymmetric every hour will talk about a Certain type of relation called an antisymmetric relation transitive Contents... The previous result to conclude that every acyclic relation is irre & ;... 19 Prove that every asymmetric relation are considered as asymmetric relation ; asymmetric relation are considered as relation. Ara is true for some a and false for others what an antisymmetric relation transitive Contents... Converse does not requier irreflexivity, therefore every asymmetric relation is and provide a number of possible relations a... Antisymmetric provided that distinct elements are never both related to itself this with the previous result conclude!, symmetric, asymmetric, is symmetric every asymmetric relation, but ≥ is not asymmetric, it should antisymmetric. She brought in cookies not requier irreflexivity, therefore every asymmetric relation ; symmetry in mathematics References... Converse of part ( a ) does not hold if and only if is! relations '' in Discrete mathematics Discrete mathematics ordering is antisymmetric, but ≥ not! Is necessarily antisymmetric ; but the converse of part ( a ) ( b Show. Here 's my code to check if a single compound is kept in a antisymmetric set nearness relation is &! ( b ) Show that the converse of part ( a, each of which gets related by to. But ≥ is not 's my code to check if a matrix is antisymmetric Marc Lars (. | 2022-10-07T19:46:32 | {
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http://math.stackexchange.com/questions/287802/counting-seating-arrangements-between-boys-and-girls | # counting seating arrangements between boys and girls
We would like to count how many ways 3 boys and 3 girls can sit in a row. How many ways can this be done if:
(b) all the girls sit together? ...
(c) every boy sits next to at least one other girl?
ANSWER: If three boys sit next to each other, no combinations work. If two boys sit next to each other, it fails if and only if the pair of boys sitting next to each other are on an edge (ie. BBGGBG, BGGGBB). If no two boys sit next to each other, all combinations work. There are 4!3! combinations with three boys together (see part (b)). If we place two boys on the edge, we have two choices, left or right to place them. We then choose the position of the third boy from threeremaining positions (he can't be next to the two other boys) for a total of 2*3*3!3! positions (3!3! to account for varying positions of unique boys and girls). Since there are 6! total positionings, there are 6!- 4!3!- 2*3*3!3! = 360 positionings where no two boys sit next to each other.
i get where 6!- 4!3! comes from but don't understand where 2*3*3!3! comes from
-
Of course, if the kids are too shy, opposite-sex people never get to sit next to each other. – Ahaan Rungta Feb 14 at 17:57
I think it is explained in your OP. We do the same thing with a bit more detail.
Since we have already counted the number of "bad" positions with all the boys together, it remains to count the number of bad positions in which the boys are not all together, but some boy is not next to a girl.
There must be two boys together, and they must be at the left end or the right end ($2$ choices). Take such a choice, say left end. Then the remaining boy can be in any one of $3$ places, namely positions $4$, $5$, or $6$. That gives us $(2)(3)$ ways of choosing the seats the boys will occupy. For each of these $(2)(3)$ choices, there are $3!$ ways of permuting the boys among the chosen seats, and for every such choice there are $3!$ ways of permuting the girls, for a total of $(2)(3)(3!)(3!)$.
Another way: Counting all possible arrangements and subtracting the "bad" ones is often good strategy. But let us count directly.
Our condition will be satisfied if either no two boys are together, or exactly two boys are together and they are not in the end seats.
Count first the arrangements in which no two boys are adjacent. Write down $G\quad G\quad G$. This determines $4$ "gaps" into which we can slip the boys, one boy per gap. There are $4$ "gaps" because we are including the end gaps. There are $\binom{4}{3}$ ways of choosing $3$ of these gaps.
Or else we could slip $2$ boys into one of the two center gaps ($2$ choices), and then slip the remaining boy into one of the $3$ remaining gaps, for a total of $6$ choices.
Thus the places for the boys can be chosen in $10$ ways. For each of these ways, we can arrange the boys is $3!$ ways, and then the girls in $3!$ ways, for a total of $360$.
- | 2014-08-01T00:09:23 | {
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http://mathhelpforum.com/calculus/40628-how-would-you-integrate.html | # Math Help - How would you integrate this?
1. ## How would you integrate this?
5x
(x-1)((x^2)+4)
fractions?
5x = A + B
(x-1) ?
And ((x^2)+4) is 1/2arctan(x/2)
the answer key says
ln|x-1| -.5ln|x^2+4| - 2arctan(x/2)
Can anyone help? Thanks.
2. $\frac{5x}{(x-1)(x^2+4)} = \frac{A}{x-1} + \frac{Bx + C}{x^2+4}$ (since $x^2 + 4$ is an irreducible quadratic, indicating your numerator is in the form of Bx + C)
See what you can do with this and come back with any questions.
3. Originally Posted by khuezy
5x
(x-1)((x^2)+4)
fractions?
5x = A + B
(x-1) ?
And ((x^2)+4) is 1/2arctan(x/2)
the answer key says
ln|x-1| -.5ln|x^2+4| - 2arctan(x/2)
Can anyone help? Thanks.
You can use partial fractions with this setup
$\frac{5x}{(x^2+4)(x-1)}=\frac{A}{(x-1)}+\frac{Bx+C}{x^2+4}$
Or observe that
$\frac{5x}{(x-1)(x^2+4)}=\frac{5x-x^2-4+x^2+4}{(x-1)(x^2+4)}=\frac{-(x^2-5x+4)}{(x-1)(x^+2)}+\frac{x^2+4}{(x-1)(x^2+4)}=$
$\frac{-(x-4)(x-1)}{(x^2+4)(x-1)}+\frac{x^2+4}{(x-1)(x^2+4)}=\frac{4-x}{x^2+4}+\frac{1}{x-1}=\frac{4}{x^2+4}-\frac{x}{x^2+4}+\frac{1}{x-1}$
Now we have
$\int \left( \frac{4}{x^2+4}-\frac{x}{x^2+4}+\frac{1}{x-1} \right)dx$
The first is in the form of the arctangent or if you want to do it by hand let x=2tan(u) the 2nd is a u sub on the denominator $u=x^2+4$ and the last one is a u sub on the denominator $u=x-1$.
I hope this helps.
Good luck.
4. ## help
I got
5x = (A+B)x^2 + 4A - (B+C)x - C
I'm confused on what to do next.
Equating coefficients?
A+B = 0 4A=5 B+C=0 C=0????????????????
Thanks.
5. Here's another way using trig.
$\frac{x-1}{x^{2}+4}dx$
Let $x=2tan(t), \;\ dx=2sec^{2}(t)dt$
Making the subs we get a menacing looking thing, but it simplifies down fairly nice.
Making the subs we get:
$\frac{2tan(t)-1}{4(tan^{2}(t)+1)}\cdot{2sec^{2}(t)}dt$
This, believe it or not, whittles down to
$tan(t)-\frac{1}{2}$
$\int{tan(t)}dt-\frac{1}{2}\int{dt}$
Then, after integrating, resub in $t=tan^{-1}(\frac{x}{2})$
You may wish to stick with the others methods. I just threw it out there.
6. Originally Posted by khuezy
5x
(x-1)((x^2)+4)
fractions?
5x = A + B
(x-1) ?
And ((x^2)+4) is 1/2arctan(x/2)
the answer key says
ln|x-1| -.5ln|x^2+4| - 2arctan(x/2)
Can anyone help? Thanks.
$\frac{5x}{(x^2+4)(x-1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+4}$
multiplying through by the common denominator we get
$5x=A(x^2+4)+(Bx+C)(x-1)$
Now letting $x=1$ we get
$5(1)=A(1^2+4)+(B+C)(0)\Rightarrow{A=1}$
So now for the second factor we need to expand, but dont forget A=1
So we have
$5x=x^2+4+Bx^2-Bx+Cx-C$
So we see that $B+1=0\Rightarrow{B=-1}$
and also seeing $4-C=0\Rightarrow{C=4}$
So finally we can see that
$\frac{5x}{(x^2+4)(x-1)}=\frac{1}{x-1}+\frac{4-x}{x^2+4}$
and we see that | 2014-04-21T07:48:59 | {
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