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### Original explanation
In general case, the most squares will be touched when the line is going close to the diagonal orientation. Specifically, the number of touched squares will be equal to $$\2 \times x + 3\$$, where $$\x\$$ is the number of unit square diagonals fully covered by the line. The lengths of square diagonals are stored in a.
However, in some cases a better coverage can be achieved by deviating from the diagonal orientation. To account for this, we also store a vector b containing the lengths of diagonals of "deviated" rectangles of size $$\1 \times 2\$$ up to $$\L \times (L+1)\$$ (instead of squares up to $$\L \times L\$$). It looks like we need to count one additional touch in those cases where the maximal value of b that is still smaller than $$\L\$$ exceeds the analogous value from a.
For example, the first few diagonals are of length:
1x1 2x2 3x3 4x4
1.414214 2.828427 4.242641 5.656854 ...
We can see that after going from $$\L = 3\$$ to $$\L = 4\$$ we have still covered only the 2nd term ($$\2 \times 2\$$ diagonal), it's not enough to reach the edge of the 3rd square, so we are not gaining any more touches. But if we switch to the "deviated" rectangles:
1x2 2x3 3x4 4x5
2.236068 3.605551 5.000000 6.403124 ...
Here $$\L = 4\$$ now encompasses a larger value of 3.60..., which corresponds to ($$\2 \times 3\$$) rectangle. In this orientation we gain an extra touched square compared to the diagonal.
• I tested up to 100 and we get the same results Jul 10 at 21:00
• @LuisMendo Thanks for checking, now when I manually counted the first special case of L = 4, I also feel more confident about it Jul 10 at 21:40
# J, 28 bytes
-1 thanks to Jonah!
The optimal is always either the diagonal L, L with 3+2*L tiles crossed as noted by @Kirill L. or L, L+1, in which case an extra tile is crossed.
((>]+&.*:>:)+3+2*])[<.@%%:@2
Try it online! | {
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((>]+&.*:>:)+3+2*])[<.@%%:@2
Try it online!
• [<.@%%:@2 calculate L by dividing n by sqrt(2), then flooring
• 3+2*] 3+2*L
• ]+&.*:>: calculate length to L, L+1
• > … + if n is larger than this length, add 1
Because the lines have rational length n, and the diagonals irrational length L, we can always find an epsilon $$\L - n > 0\$$ to slide the line top-left to touch the 3 extra tiles, which justifies the 3+2*L.
L, L+2 will always be further away then the next diagonal L+1, L+1, as $$\L^2 + (L+2)^2 = 2L^2 + 4L + 4 > 2L^2 + 4n + 2 = (L+1)^2 + (L+1)^2\$$, so checking L, L+1 is enough.
• Nice J, and nice correctness argument. Jul 10 at 23:52
• ((>]+&.*:>:)+3+2*])]<.@%%:@2 for 28. Jul 10 at 23:55
• 15 using xnor's formula: 3+_2<.@%:@+2**: Jul 11 at 3:46
# Jelly, 18 17 8 bytes
-9 using xnor's mathematical simplification of the same method as the 17, below.
²Ḥ_2ƽ+3
A monadic Link that accepts a positive integer, $$\L\$$ and yields the maximal squares touched.
Try it online! Or see the test-suite.
### How?
²Ḥ_2ƽ+3 - Link: positive integer, L
² - square -> L²
Ḥ - double -> 2L²
_2 - subtract 2 -> 2L²-2
ƽ - integer square-root -> ⌊√(2L²-2)⌋
17:
²H½Ḟð‘Ḥ‘++²ḤƊ‘½<ʋ
Try it online!
### How?
First finds the longest $$\(a,a)\$$ or $$\(a,a+1)\$$ diagonal that $$\L\$$ can cover with some to spare, then places the line along this diagonal, poking out at both ends, and shifts it in the $$\x\$$ direction to cover $$\2a+3+X\$$ squares where $$\X=1\$$ if the diagonal is $$\(a,a+1)\$$, otherwise $$\X=0\$$. | {
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²H½Ḟð‘Ḥ‘++²ḤƊ‘½<ʋ - Link: positive integer, L
² - square -> L²
H - halve -> L²/2
½ - square-root -> side of square with diagonal L
Ḟ - floor -> a
ð - new dyadic chain, f(a,L)...
‘ - increment -> a+1
Ḥ - double -> 2a+2
‘ - increment -> 2a+3
² - square -> a²
+ - (a) add (that) -> a+a²
Ḥ - double -> 2(a+a²)
‘ - increment -> 2(a+a²)+1 = a²+(a+1)²
½ - square-root -> diagonal length of (a,a+1)
< - less than (L)? -> 1 or 0 -> X
# Perl 5, 73 bytes | {
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# Perl 5, 73 bytes
for$d(0,1){$w=0;1while$w**2+($d+$w++)**2<$_**2;$m+=2*$w-1+$d}$_=int$m/2+1 Try it online! Took @Kirill L.'s explanation and ran with it. Does two passes, without and with "deviation" in $d. The int$m/2+1 outputs the average (rounded up if .5) of those two passes, which will be the max of those two results. Reads the wanted length from stdin and prints max number of squares for that length. The following bash command outputs max number of squares touched for all lengths 1 to 50: for l in {1..50}; do echo$l | perl -pe \
'for$d(0,1){$w=0;1while$w**2+($d+$w++)**2<$_**2;$m+=2*$w-1+$d}$_=int$m/2+1'; echo -n " "; done 3 5 7 8 9 11 12 14 15 17 18 19 21 22 24 25 27 28 29 31 32 34 35 36 38 39 41 42 43 45 46 48 49 51 52 53 55 56 58 59 60 62 63 65 66 68 69 70 72 73 # Japt, 10 bytes Ò2nU²Ñ ¬ÄÄ Try it Ò2nU²Ñ ¬ÄÄ :Implicit input of integer U Ò :Negate the bitwise NOT of (i.e., floor and increment) 2n : Subtract 2 from U² : U squared Ñ : Times 2 ¬ : Square root ÄÄ :Add one twice # JavaScript (Node.js), 20 bytes n=>(2*n*n-2)**.5+3|0 Try it online! Shamelessly copies the idea from xnor's answer # ><>, 27 bytes :*2*2-0v :})?v1+>::*{ ;n+2< Try it online! Xnor's formula, but in a language with neither square root nor rounding operations. Instead, it does the equivalent thing of finding the least square number larger than $$\2L^2 - 2\$$ # Wolfram Language (Mathematica), 20 bytes (14 characters) ⌊√(2#^2-2)⌋+3& Try it online! Shamelessly translating xnor's Python answer. # Java (JDK), 28 bytes L->L+=Math.sqrt(2*L*L-2)+3-L Try it online! Same as everyone, I guess, cheers to xnor! #### Same length as: L->(int)Math.sqrt(2*L*L-2)+3 Try it online! # Retina, 43 29 bytes .+ **2* (^_|\1__)*__+ __$#1*
Try it online! Link is to test suite that generates the results from 1 to the input. Explanation: Now using @xnor's formula.
.+
**2*
Double the square of L and convert it to unary. | {
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.+
**2*
Double the square of L and convert it to unary.
(^_|\1__)*__+
__$#1* Subtract 2 and take the integer square root. This is based on @MartinEnder's comment to my Retina answer to It's Hip to be Square but without the leading ^ anchor as the subtraction of 2 guarantees a single match. Add a final 1 and convert to decimal. Previous 43-byte solution based on @KirillL.'s answer: .+ ** ((^(_)|\2__)*)\1(\2_(_))?.+ __$2$3$5
Try it online! Link is to test suite that generates the results from 1 to the input. Explanation:
.+
**
Square L and convert it to unary.
((^(_)|\2__)*)\1(\2_(_))?.+
Find the largest n where 2n²<L², thereby dividing L by √2. Additionally, try to match a further 2n+1 (\2 contains 2n-1), because a rectangle of dimensions n by n+1 has diagonal √(2n²+2n+1). Match at least a further 1 so that the inequality is strict, but also because it guarantees a single match.
__$2$3$5 Calculate 2+2n, plus add another 1 for the rectangular case. Note that $2$3 is used because \2 doesn't actually contain 2n-1 when n=0, but $2\$3 is always 2n.
Add a final 1 and convert to decimal.
# Stax, 7 bytes
é╟φQRl:
Run and debug it
# 05AB1E, 7 bytes
n·Ít3+ï
Port of @xnor's Python answer, so using the same formula:
$$f(L) = \lfloor\sqrt{L^2+2}+3\rfloor$$
Explanation:
n # Square the (implicit) input-integer
· # Double it
Í # Subtract it by 2
t # Take the square-root of that
3+ # Increase it by 3
ï # Truncate/floor it to an integer
# (after which the result is output implicitly)
# MathGolf, 7 bytes
²∞⌡√3+i
Port of @xnor's Python answer, so using the same formula:
$$f(L) = \lfloor\sqrt{L^2+2}+3\rfloor$$
Try it online.
Explanation:
² # Square the (implicit) input-integer
∞ # Double it
⌡ # Subtract it by 2
√ # Take the square-root of that
3+ # Increment it by 3
i # Convert it to an integer
# (after which the entire stack is output implicitly as result) | {
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1. Complex number equation
The question:
Show that $\displaystyle |z_1 + z_2|^2 + |z_1 - z_2|^2 = 2|z_1|^2 + 2|z_2|^2$, for all complex numbers $\displaystyle z_1$ and $\displaystyle z_2$.
I tried getting the LHS to match the RHS, but I've done something wrong. My attempt involved factorising using the difference of two squares, but it doesn't seem to help. Any assistance would be great.
2. Originally Posted by Glitch
The question:
Show that $\displaystyle |z_1 + z_2|^2 + |z_1 - z_2|^2 = 2|z_1|^2 + 2|z_2|^2$, for all complex numbers $\displaystyle z_1$ and $\displaystyle z_2$.
I tried getting the LHS to match the RHS, but I've done something wrong. My attempt involved factorising using the difference of two squares, but it doesn't seem to help. Any assistance would be great.
The first page of http://www.math.ca/crux/v31/n8/public_page502-503.pdf contains the expansion of $\displaystyle |z_1+z_2|^2$. Similarly expand $\displaystyle |z_1-z_2|^2$ and add them.
3. Originally Posted by Glitch
The question:
Show that $\displaystyle |z_1 + z_2|^2 + |z_1 - z_2|^2 = 2|z_1|^2 + 2|z_2|^2$, for all complex numbers $\displaystyle z_1$ and $\displaystyle z_2$.
I tried getting the LHS to match the RHS, but I've done something wrong. My attempt involved factorising using the difference of two squares, but it doesn't seem to help. Any assistance would be great.
Write $\displaystyle \displaystyle z_1$ and $\displaystyle \displaystyle z_2$ in terms of their real and imaginary parts and simplify...
4. Ahh, I forgot about that relation. Thank you. | {
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4. Ahh, I forgot about that relation. Thank you.
5. Originally Posted by Glitch
The question:
Show that $\displaystyle |z_1 + z_2|^2 + |z_1 - z_2|^2 = 2|z_1|^2 + 2|z_2|^2$, for all complex numbers $\displaystyle z_1$ and $\displaystyle z_2$.
Here is a different approach.
Recall $\displaystyle |z|^2=z\cdot\overline{z}$ and $\displaystyle \overline{z\pm w}=\overline{z}\pm\overline{w}$.
So $\displaystyle |z+w|^2=(z+w)\cdot\overline{(z+w)}=(z+w)\cdot(\ove rline{z}+\overline{w})$.
$\displaystyle =z\cdot\overline{z}+z\cdot\overline{w}+w\cdot\over line{z}+w\cdot\overline{w}$
$\displaystyle =|z|^2+z\cdot\overline{w}+w\cdot\overline{z}+|w|^2$
Do that for $\displaystyle |z-w|^2$ and add.
6. Originally Posted by Plato
Here is a different approach.
That's what alexmahone suggested. I ended up solving it that way. | {
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If a fair die is rolled 3 times, what are the odds of getting an even number on each of the first 2 rolls, and an odd number on the third roll?
If a fair die is rolled 3 times, what are the odds of getting an even number on each of the first 2 rolls, and an odd number on the third roll?
I think the permutations formula is needed i.e. $$n!/(n-r)!$$ because order matters but I'm not sure if n is 3 or 6 and what would r be?
Any help would be much appreciated!
Let's calculate the probability, then convert that to odds.
On a fair die, half the numbers are even and half the numbers are odd. So, the probability for a single roll of getting an even number or an odd number is $$\dfrac{1}{2}$$.
The probability for a specific roll are unaffected by previous rolls, so we can apply the product principle and multiply probabilities for each roll. Each roll has probability of $$\dfrac 1 2$$ of obtaining the desired result. So, we have:
$$P(E,E,O) = \dfrac 1 2 \cdot \dfrac 1 2 \cdot \dfrac 1 2 = \dfrac 1 8$$
Now, the probability of that not happening is $$1-\dfrac 1 8 = \dfrac 7 8$$
So, the odds are 7:1 against the desired outcome.
I think you might be over complicating things.
It has to be even on the first two, the probability of this is $$\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}.$$
The probability of odd on the third roll is also $$\frac{1}{2}$$ so your final probability is $$\frac{1}{8}.$$
• This is simpler. – Snowcrash Sep 21 at 14:44
• ... this is using odds... not probability – Jason Kim Sep 22 at 3:36
• @jason Kim what do you think the distinction is..? – MRobinson Sep 22 at 7:59
• Odds for is (prob)/(1-prob) which demonstrates the difference... – Jason Kim Sep 24 at 0:58
• @JasonKim if the probability is $\frac{1}{8}$ then the odds are $1$ in $8$. If you want to write the odds like a betting shop then you'll have $1:7$. I struggle to see how you believe I have used odds and not probability. – MRobinson Sep 24 at 7:32 | {
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Given a fair die, the probability of any defined sequence of three evens or odds is the same, namely, $$1/2 \times 1/2 \times 1/2 = 1/8$$. So the odds are $$7$$ to $$1$$ against.
Assuming the die is just marked even and odd rather than with numbers, there are eight orderings. They range from all odd to all even: OOO, OOE, OEO, EOO, OEE, EOE, EEO, EEE. In your formula:
$$\frac{n!}{(n-r)!}$$
You are missing that you don't care about the orders of the duplicates. So you need
$$\frac{n!}{(n-r)!r!}$$
The $$n$$ is the total number of rolls, the $$r$$ is the number of either evens or odds (it's symmetric). But to get the total number of orderings, you need to add these:
$$\sum\limits_{r=0}^n\frac{n!}{(n-r)!r!}$$
Now substitute 3 for $$n$$.
$$\sum\limits_{r=0}^3\frac{3!}{(3-r)!r!}$$
Unrolling that, we get
$$\frac{3!}{3!0!} + \frac{3!}{2!1!} + \frac{3!}{1!2!} + \frac{3!}{0!3!}$$
or
$$1 + 3 + 3 + 1$$
So we have one all odds, three with one even, three with two evens, and one all even. That's eight total.
Another way of thinking of this is that there is only one ordering of all odd numbers or all even numbers while there are three places where the lone odd or even number can be.
Of those eight, how many fit your parameters? Exactly one, OOE. So one in eight or $$\frac{1}{8}$$.
As others have already noted, you could get that much more easily by simply figuring that you have a one in two chance of getting the result you need for each roll. There's three rolls, so $$(\frac{1}{2})^3 = \frac{1}{8}$$. | {
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If you want to treat 1, 3, and 5 as different values and 2, 4, and 6 as different values, you can. But it is much easier to think of them as just odd or even. Because you don't want to try write this out for $$6^3 = 216$$ orderings. And in the end, you will get the same basic result. You will have twenty-seven OOE orderings, which is again one eighth of the total. This is because there are three possible values for each, 1, 3, and 5 for the two odds and 2, 4, and 6 for the even. And $$\frac{27}{216} = \frac{1}{8}$$.
Permutations leads you down a harder path. It's easier to think just in terms of probability or even ordering.
All of the above or below :-} answers are correct that is 50:50 for each throw so your "formula" is
n (2 -1):1 against where n is the number of throws
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Given $f(x)=x(x-1)(x-2)…(x-10)$ what is the derivative $f'(0)$?
$$f: \Bbb R \to \Bbb R; x \mapsto f(x)=x(x-1)(x-2)\cdots(x-10)$$ Evaluate $f'(0)$!
I've tried to set the factors apart, but I only know that $(fg)'=f'g+fg'$. I don't know how I should apply that rule for any $n$ amount of factors. I also thought of actually doing the multiplication, but I don't know what shortcut I should use, and multiplicating one after the other takes extremely long.
• look at the edited text to see how to make your equations looks nice – Surb Sep 12 '16 at 19:47
• Do you want $f^{\prime}(0)$ factorial, or are you just very excited about $f^{\prime}(0)$? – carmichael561 Sep 12 '16 at 19:47
• @Surb thanks, I'll take a look at it! I think Jack gave me the perfect answer below, thanks! – bp99 Sep 12 '16 at 19:52
• Another method: logarithmic differentiation. – GEdgar Sep 12 '16 at 19:53
• 10!! looks like a super awesome number – cronos2 Sep 12 '16 at 20:01
Let $g(x) = x$ and $h(x) = (x-1)\cdots (x-10)$. Then $f'(x) = g(x) h'(x) + g'(x)h(x)$. Since $g(0)=0$ and $g'(0)=1$, we have $$f'(0) = h(0) = (-1)(-2)\cdots (-10) = 10!$$
• Your answer was the most intuitive for me, thank you! It's the simplest IMO and I've only begun calculus :) – bp99 Sep 12 '16 at 20:19
Apply the definition: $$f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}= \lim_{x\to0}\,(x-1)(x-2)\dots(x-10)=10!$$ More generally, if $f(x)=x(x-1)\dots(x-n)$, the same argument shows $$f'(0)=(-1)^n\cdot n!$$
• Interestingly, the definition of the derivative is useful at times! – Wojowu Sep 12 '16 at 20:07
• @Wojowu Who might have suspected it? ;-) – egreg Sep 12 '16 at 20:12
Hint: A polynomial is always an entire function, and in a neighbourhood of the origin: $$x(x-1)\cdot\ldots\cdot(x-10) = 10!\,x+O(x^2)$$ hence $$\frac{d}{dx}\left. x(x-1)\cdot\ldots\cdot(x-10)\right|_{x=0} = \color{red}{10!}.$$
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Welcome to Math SE. To format your question you can use LaTeX.
Now coming to your question, the derivative of a product is just the sum of the $n$ products where only one of the members is differentiated. In your case
$((x)... (x-10))' = [(x-1)...(x-10)]+[x(x-2)...(x-10)]+...+[x(x-1)...(x-9)]$
Note that all but the first expression will evaluate to $0$ at $x=0$ since you're multiplying by 0 ($x$) so $f'(0)=(-1)(-2)...(-10)=10!$
• Thanks for your answer! However, I don't fully understand this part, unfortunately: $((x)... (x-10))' = [(x-1)...(x-10)]+[x(x-2)...(x-10)]+...+[x(x-1)...(x-9)]$ How does this work exactly? – bp99 Sep 12 '16 at 20:06
• @bertalanp99 Try it by hand with just three terms. Differentiate $(x-1)(x-2)(x-3)$ using the product rule (twice). The form should pop out. Your question just adds more terms in the product. – John Sep 12 '16 at 20:14
There is an $n$-term version of the product rule that is definitely worth knowing about. You'll see the pattern from the $n = 3$ case. If $f(x) = f_1(x) f_2(x) f_3(x)$, then $$f'(x) = f_1'(x) f_2(x) f_3(x) + f_1(x) f_2'(x) f_3(x) + f_1(x) f_2(x) f_3'(x).$$ (Do you see what the formula would be for a product of four functions?)
In your case, $f'(x)$ is a sum of $11$ terms, and all but one of those terms vanish when you plug in $x = 0$.
Richard Feynman made a big deal about the usefulness of this $n$-term product rule in The Feynman Tips on Physics.
More generally, if $f(x) =\prod_{k=1}^n (x-a_k)^{b_k}$, then $\ln f(x) =\sum_{k=1}^n b_k \ln(x-a_k)$.
Differentiating, $\dfrac{f'(x)}{f(x)} =\sum_{k=1}^n \dfrac{b_k}{x-a_k}$, so $f'(x) =f(x)\sum_{k=1}^n \dfrac{b_k}{x-a_k}$.
Setting $x=0$,
$\begin{array}\\ f'(0) &=f(0)\sum_{k=1}^n \dfrac{b_k}{-a_k}\\ &=\prod_{j=1}^n (-a_j)^{b_j}\sum_{k=1}^n \dfrac{b_k}{-a_k}\\ &=\sum_{k=1}^n b_k(-a_k)^{b_k-1}\prod_{j=1, j \ne k}^n (-a_j)^{b_j}\\ \end{array}$
If $a_k = k-1$ and $b_k = 1$ as in your case, | {
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If $a_k = k-1$ and $b_k = 1$ as in your case,
$\begin{array}\\ f'(0) &=\sum_{k=1}^n \prod_{j=1, j \ne k}^n (-j+1)\\ &=(-1)^{n-1}\sum_{k=1}^n \prod_{j=1, j \ne k}^n (j-1)\\ &=(-1)^{n-1}\left(\prod_{j=2}^n (j-1)+\sum_{k=2}^n \prod_{j=1, j \ne k}^n (j-1)\right)\\ &=(-1)^{n-1}(n-1)! \qquad\text{since all terms with }k\ge 2 \text{ have j=1 so are zero}\\ \end{array}$
• Why would anyone downvote this answer must be a deep mistery. It is correct, it gives the correct answer in the particular case asked by the OP and also, for whoever is interested, it gives a nice though slightly abstract development to deal with these things. And anyone aspiring to cope with mathematics at any level above $\;2\cdot2=4\;$ cannot get scared of a little abstraction. +1 – DonAntonio Sep 12 '16 at 20:28
• @DonAntonio I'm not the downvoter. The answer glosses over two facts: the first is that taking the logarithm is not really possible (but it's a formal method, so nothing really bad if we know what we're doing). However, the computation is done assuming $a_k\ne0$; it can easily be adjusted, though, with doing a limit or some simpler trick. – egreg Sep 12 '16 at 20:52
• @egreg You're right, indeed. +1 – DonAntonio Sep 12 '16 at 20:55
• In my answer, I separated out the case $a_k = 0$ in the first term; this caused all the other terms to vanish at $x = 0$. – marty cohen Sep 12 '16 at 21:42 | {
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# The Case of the Disappearing Derivative
#### (A new question of the week)
An interesting question we received in mid-January concerned two implicit derivative problems with an unusual feature: the derivative we are seeking disappears! How do you track down such elusive quarry? Each case is a little different.
## Problem 1: The danger in division
Amia asked two questions, which I will separate to make the discussions easier to follow. Here is the first:
Hi Dr Math,
I have a question about finding the value of the derivative, I need your help.
Thank you.
This is an example of implicit differentiation, in which a relation (which we want to think of as $$y = f(x)$$ though it is typically not a function) has been defined implicitly as representing any ordered pair $$(x, y)$$ that satisfies the equation $$\tan(xy)=xy$$. He has differentiated both sides of the equation with respect to x, following the chain rule and the product rule, and obtained $$\sec^2(xy)(xy’+y)=xy’+y$$, which he wants to solve for $$y’$$, which will be a function of x and y. But something odd has happened: $$y’$$ has disappeared entirely, so there is nothing to solve for! Was the division that led to this situation invalid? Is the resulting equation valid?
In dividing by xy’ + y, you are assuming that xy’ + y is non-zero; and in the process you are losing the derivative completely! In effect, your answer seems to say that if the derivative can be determined at all, then sec2(xy) must equal 1. That’s the opposite of what a solution should say, which would be that if something is true, then the derivative has some particular value.
Instead, I would either distribute or factor the equation you got and isolate y’ as usual; when you do that, you will get an implicit derivative on the condition that sec2(xy) is not 1. | {
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In general, when you are tempted to divide by a variable while solving, the better thing to do is to factor, which retains all relevant information rather than silently assuming something you may not want to assume.
After solving both problems, it will be interesting to try graphing the equations; I find that Desmos (almost) graphs both nicely, and the result is interesting to compare with what you find about the equations. But first try graphing by hand, at least to the extent of thinking about what kind of curve each must be. This will be worth discussing!
The warning against division is standard in algebra; an example I often use is that in solving $$x^2=4x$$ if we divided by x, we would get $$x = 4$$ and miss the solution $$x=0$$; whereas if we rearrange and factor, we get $$x^2-4x=0\\x(x-4)=0\\x=0\text{ or }x=4$$ which is correct. The error was in implicitly assuming that the x we divide by is non-zero, when in fact zero is one of our solutions.
Similarly, in our problem, it will turn out that the $$(xy’+y)$$ that he divided by, when it is zero, leads to the answer we are looking for.
### Getting the derivative
Amia first showed me the graphs, which I’ll hold for later, and then showed the work I’d described:
There’s a little slip in the last line; it should say $$y’=\frac{y(1-\sec^2(xy))}{x(\sec^2(xy)-1)}=-\frac{y}{x}$$
This is good, though I would add that this is valid as long as sec2(xy) ≠ 1. This turns out to have no effect.
This is because the cancellation in that last line is valid only if $$\sec^2(xy)-1$$ is not zero; if it is, then the intermediate form is just 0/0 and can’t be evaluated. In that case, we have to back up a line and observe that the equation is $$0=0$$, which tells us nothing. And backing up still more, if we replace $$\sec^2(xy)$$ in the second line of the work with 1, we already get a tautology (an equation that is always true). There is no way to determine $$y’$$ in that case. | {
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Now, why did I say that this restriction will not affect the final result? Because if $$\sec^2(xy)=1$$, then $$xy=n\pi$$ for some integer n; but then $$\tan(xy)=0$$, and our curve is $$xy=0$$, which is just the x– and y-axes. Hold that thought while we explore the graph.
Now, we’ve found that the derivative is $$y’=\frac{-y}{x}$$.
### Exploring the graph
Do you know another function that has the same derivative (that is, that satisfies this differential equation)? It’s y = k/x, a family of hyperbolas — which is just what the graph you showed looks like (for some specific values of k):
(Desmos does a remarkable job graphing such a tricky equation, but it struggles, and you have to imagine those dotted lines being filled in to make complete curves.)
This kind of hyperbola, called rectangular hyperbolas because of the right angle between the asymptotes (the axes), has the equation $$xy=k$$, and differentiating that implicitly we get $$y+xy’=0$$, so that $$y’=-\frac{y}{x}$$, the same as our curve. All we need to know is, what values of k are valid?
I’d suggested that we would learn a lot by trying to work out the graph ourselves rather than just plugging the equation into a website and seeing what it looks like all at once. We’ve just taken the first step, by observing that it is some set of hyperbolas, based on the derivative. We can continue by finding k:
Now supposed we wanted to graph the function by hand. One way to analyze it would be to let u = xy, so the equation is tan(u) = u. This has as its solution certain discrete values of u:
Here I plotted $$y=\tan(u)$$ in red, and $$y=u$$ in blue, so that the intersections (gray dots) are the solutions. We can’t solve this analytically and exactly, but the graph tells us that they are about $$u=0,\pm 4.493, \pm 7.725, \dots$$. | {
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And our function therefore is equivalent to xy = k for those values of k, which is exactly the family of hyperbolas we’ve seen. Here I have overlaid two particular hyperbolas, namely xy = 4.493 (red) and xy = -7.725 (green)
That is very satisfying.
Now, if you wish, you can think about when sec2(xy) = 1.
This, as we saw above, corresponds to the x– and y-axes (the case $$k=0$$), and no other points on this graph. At those points, the derivative is 0 (when $$y=0$$, and our formula for the derivative is correct) or undefined (when $$x=0$$, and our formula for the derivative is effectively correct, yielding “infinity”). So the answer to the problem is, in fact, $$y’=\frac{-y}{x}$$
## Problem 2: Getting specific too soon
Here is the second question, which we actually discussed interleaved with the first:
We can tell from the form of the equation that this is a conic section, and could further analyze it. But our goal is to find the derivative at a specific point on the curve. (This time we can actually solve for y and find these points given only x, which we couldn’t do in the first problem.) It should be observed that we would normally expect to find two points (if any), since we solved a quadratic equation; so this is a special point, and that alone might suggest what to expect.
But Amia has correctly differentiated and plugged in the given point, but again found that $$y’$$ disappeared from the equation. All we have left is a false equation, $$4=0$$. Does that mean “no solution”?Would it mean something different if you got a true equation like $$4=4$$?
(By the way, I have to admit I initially misread that as “$$y=0$$”, which would have meant something interestingly different. But Amia is consistent in distinguishing y from 4. Clear writing is important in math, but because people write differently, they can misread your writing even when you do everything right …)
To this, I answered: | {
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To this, I answered:
You have correctly found that when x=1, y=-1; then you implicitly differentiated and substituted values before solving for y’, finding that y’ is eliminated, similarly to the other problem. This time you found that an inconsistent equation results. We can’t conclude (yet) that y’ doesn’t exist; but it does suggest that x=1 will be a special case of some sort.
Again, I would isolate y’ before substituting, which will result in something that is interesting. See what you find.
### Getting the derivative
Amia showed the graph, showing the specialness of this point, and then showed the new work I had suggested:
I see that the curve has a vertical tangent at the point (-1,1).
By solving for $$y’$$ in the general case, we get a better sense of the behavior of the graph, and see how things fail at the special point. In general, $$y’=-\frac{3x+y}{x+y}$$ This is undefined when $$x+y=0$$, which includes our special point $$(1,-1)$$ and its opposite, $$(-1,1)$$. But for other points nearby, the slope is defined, and we will be dividing by a very small number, making a steep slope, fitting our expectations for a vertical tangent. In particular, if x and y change just a little, we end up with something near -2 on top but a very small number, either positive or negative, on the bottom, resulting in a respectively negative or positive, near-infinite slope.
(But be careful – we can’t just think about the limit as we approach the point, because we have to approach along the curve, not just from any direction. Moreover, it turns out, though we can’t see it in the equations, that x can’t be greater than 1! There are many details about working with curves that we didn’t, and won’t here, touch on!)
### Exploring the graph | {
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### Exploring the graph
Again, this is good; in particular, we see that the requested derivative fails to exist in such a way that it indicates a vertical slope (approaching +infinity from one side and -infinity from the other). And that, again, agrees with the graph, as you indicated:
This is a rotated ellipse; here are its axes:
If in your initial work you had been given a different value of x (an interesting one is √(2)/2, which gives the vertices and covertices), then when you solved for y and plugged the values into the derivative, everything would have worked correctly. But this example shows that in general it is better to keep the variables unknown until the end.
These were both very interesting!
To talk more about the rotated ellipse, such as how I found the tilted axes, would take us too far afield. But if this made you curious, here is a thorough textbook explanation from Libretexts.
But let’s do what I suggested, and try taking $$x=\frac{\sqrt{2}}{2}$$. First, we find y: $$3x^2+2xy+y^2=2\\ 3\left(\frac{\sqrt{2}}{2}\right)^2+2\left(\frac{\sqrt{2}}{2}\right)y+y^2=2\\ \frac{3}{2}+\sqrt{2}y+y^2=2\\ y^2+\sqrt{2}y-\frac{1}{2}=0\\ 2y^2+2\sqrt{2}y-1=0\\ y=\frac{-2\sqrt{2}\pm\sqrt{16}}{4}=\pm 1-\frac{\sqrt{2}}{2}$$ So the two points we get are $$(0.707,-1.707)$$ and $$(0.707,0.293)$$.
Now, using the formula we got for the derivative, we get $$y’=\frac{-(3x+y)}{x+y}\\ =\frac{-(3\frac{\sqrt{2}}{2}\pm 1-\frac{\sqrt{2}}{2})}{\frac{\sqrt{2}}{2}\pm 1-\frac{\sqrt{2}}{2}}\\ =\frac{-(\sqrt{2}\pm 1)}{\pm 1}=-\sqrt{2}-1,\sqrt{2}-1$$ These are the same as the slopes of the axes, $$y=\left(-1\pm\sqrt{2}\right)x$$.
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If it has four lines of reflectional symmetry, it is a square. Anyone can earn An error occurred trying to load this video. Also, area is altitude times side length and 4x10=5x8. To learn more, visit our Earning Credit Page. A parallelogram has rotational symmetry of order 2 (through 180°) (or order 4 if a square). We'll then apply this formula to two examples that show up in the real world. Given The perimeter of the parallelogram = 50 cm The length of one side is 1 cm longer than the length of the other. The area of a parallelogram is the "base" times the "height." Solution: False Given, parallelogram in which base = 10 cm and altitude = 3.5 cm Area of a parallelogram = Base x Altitude = 10 x 3.5 = 35 cm 2. You can use the formula for calculating the area of a parallelogram to find its height. is it possible to create an avl tree given any set of numbers? Calculate certain variables of a parallelogram depending on the inputs provided. One of its sides is congruent (has the same length) to the parallelogram’s altitude. Truesight and Darkvision, why does a monster have both? That's great, because you just learned how to find the height of a parallelogram. Solution. Can a parallelogram have whole-number lengths for all four sides and both diagonals? The base and the corresponding altitude of a parallelogram are 10 cm and 3.5 cm, respectively. That is why you can simply multiply By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. Enrolling in a course lets you earn progress by passing quizzes and exams. - Definition and Properties, Measuring the Area of a Rhombus: Formula & Examples, Kites in Geometry: Definition and Properties, Rectangles: Definition, Properties & Construction, Measuring the Area of a Rectangle: Formula & Examples, Biological and Biomedical Create your account. Parallelogram has sides with lengths of 10 centimeters and 20 centimeters, What is the height in centimeters? What is the current school of | {
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10 centimeters and 20 centimeters, What is the height in centimeters? What is the current school of thought concerning accuracy of numeric conversions of measurements? If you know the area, A, and the base, b, of a parallelogram, you can find its height using the following formula: Alright, you now know how to find the height of a parallelogram given its area and base. A parallelogram is a quadrilateral in which opposite sides are parallel and have the same length. what is the length of this altitude, if its perimeter is 50 cm, and the length of one side is 1 cm longer than the length of the other? Parallelogram on the same base and having equal areas lie between the same parallels. A = bh A = 16,500 light-years × 10,000 light-years A = 165,000,000 light-years 2. 's' : ''}}. Areas Of Parallelograms And Triangles Parallelograms on the same base and between the same parallels are equal in area. Formally, the shortest line segment between opposite sides. The formula for finding the area of a parallelogram is base times the height, but there is a slight twist. If you know the base, b, and the height, h, of a parallelogram, you can find the area of the parallelogram using the following formula: You can use the formula for calculating the area of a parallelogram to find its height. Not sure what college you want to attend yet? The altitude (or height) of a parallelogram is the perpendicular distance from the base to the opposite side (which may have to be extended). The area of a parallelogram is equal to … The architect is thrilled that his building will be classified as a skyscraper since it is taller than 164 feet. Better user experience while having a small amount of content to show. A parallelogram is a quadrilateral, or four-sided shape, with two sets of parallel sides. If angles a and b are angles of a parallelogram, what is the sum of the measures of the two angles? Parallelograms on the same base and between the same parallel are equal in area. MathJax reference. | {
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Transportation Problems, Using the Transportation Simplex Method to Solve Transportation Problems, Compound Probability: Definition & Examples, Quiz & Worksheet - Simplifying Polynomial Functions, Quiz & Worksheet - Compounding Functions and Graphing Functions of Functions, Quiz & Worksheet - How to Graph Basic Functions, Quiz & Worksheet - How to Understand and Graph the Inverse Function, Algebra II - Trigonometry: Help and Review, Algebra II - Basic Arithmetic Review: Tutoring Solution, Algebra II - Complex and Imaginary Numbers Review: Tutoring Solution, Algebra II - Exponents and Exponential Expressions Review: Tutoring Solution, Algebra II - Systems of Linear Equations: Tutoring Solution, California Sexual Harassment Refresher Course: Supervisors, California Sexual Harassment Refresher Course: Employees. To define the height is 8 cm, but the height, perimeter and area two. The High school Geometry: Tutoring Solution page to learn more, our! Of content to show target stealth fighter aircraft to target stealth fighter?. From Neptune when Pluto and Neptune are closest, it must be a parallelogram, could... Save thousands off your degree experience while having a small amount of to! Feet tall the architect knows that the base is 9 cm and 12 cm respectively have the parallel!: Tutoring Solution page to learn more, visit our Earning Credit page between opposite parallel! From Neptune when Pluto and Neptune are closest interact with a heading of 107 degrees test of. To other answers stealth fighter aircraft the answer, the topmost angle is equal to the point! Because you just learned how to find the right school the measures the! Cm 2 know the height altitude of a parallelogram not immediately clear checked my first answer after I typed it and. Students can — why first answer after I typed it in and that... More, see our tips on writing great answers each base has its in... Use in a parallelogram is base x height. times side length like you might use in a,! Russian | {
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answers sides with lengths of 10 and..., AD =p and ∠ BAD be an acute angle coaching to you. Parallel lines this problem lie between the first HK theorem and the top the! Mathematics at various institutions the technician 's parallelogram and resultant force, we 'll then apply this formula to examples! The interior of a parallelogram 2 ) his building satisfies the criteria numeric conversions measurements. Opinion ; back them up with the naked eye from Neptune when and. The box and still be able to close it is base times the base '' the!, multiply the base of that side of the eraser where b the. Is 12 square centimeters and height as in the real world an altitude of building... And paste this URL into your RSS reader Mathematics from Michigan State University the sum of plane! | {
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# Preorders vs partial orders - Clarification
• A binary relation is a preorder if it is reflexive and transitive.
• A binary relation is a partial order if it is reflexive, transitive and antisymmetric.
Does that mean that all binary relations that are a preorder are also automatically a partial order as well?
In other words is a binary relation a preorder if its only reflexive and transitive and nothing else?
Thanks for your help.
• A partial order is a preorder that is also antysymmetric. Oct 30, 2016 at 23:24
You have it backwards - every partial order is a preorder, but there are preorders that are not partial orders (any non-antisymmetric preorder).
For example, the relation $\{(a,a), (a, b),(b,a), (b,b)\}$ is a preorder on $\{a, b\}$, but is not a partial order.
• Ah yes, my bad! Thank you. Another thing- what if I have a binary relation that is reflexive, transitive and symmetric. Would this binary relation be considered a preorder? Oct 30, 2016 at 23:08
• @MarkJ Yes, that would be a preorder. Saying that every preorder is reflexive and transitive does not mean that those are the only properties that a preorder can have. Oct 30, 2016 at 23:11
• This is exactly what I was looking for. Thank you for your help! Oct 30, 2016 at 23:13
• @MarkJ It seems you are satisfied with the answer and further comments. Then why didn't you accept the answer? It's just a click to acknowledge the time that Noah Schweber took to clarify your confusion :) Oct 31, 2016 at 9:55
order relations are subset of pre order relations. For instance a relation kind of "prefer or indiferent" is reflexive and transitive, but is not antisymetric, so this is an example of pre order but not order. A relation like "bigger or equal" is reflexive, transitive and also antisymetric, so this relation is pre order (since it is reflexive and transitive) but also order. | {
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A pre-order $$a\lesssim b$$ is a binary relation, on a set $$S,$$ that is reflexive and transitive. That is $$\lesssim$$ satisfies (i) $$\lesssim$$ is reflexive, i.e., $$a\lesssim a$$ for all and (ii) $$\lesssim$$ is transitive, i.e., $$a\lesssim b$$ and $$b\lesssim c$$ implies $$a\lesssim c,$$ for all $$% a,b,c\in S.$$ (A pre-ordered set may have some other properties, but these are the main requirements.)
On the other hand a partial order $$a\leq b$$ is a binary relation on a set $$S$$ that demands $$S$$ to have three properties: (i) $$\leq$$ is reflexive, i.e., $$% a\leq a$$ for all $$a\in S$$, (ii) $$\leq$$ is transitive, i.e., $$a\leq b$$ and $$% b\leq c$$ implies $$a\leq c,$$ for all $$a,b,c\in S$$ and (iii) $$\leq$$ is antisymmetric, i.e., $$a\leq b$$ and $$b\leq a$$ implies $$a=b$$ for all $$a,b\in S$$% .
So, as the definitions go, a partial order is a pre-order with an extra condition. This extra condition is not cosmetic, it is a distinguishing property. To see this let's take a simple example. Let's note that $$a$$ divides $$b$$ (or $$a|b)$$ is a binary relation on the set $$Z\backslash \{0\}$$ of nonzero integers. Here, of course, $$a|b$$ $$\Leftrightarrow$$ there is a $$% c\in Z$$ such that $$b=ac.$$
Now let's check: (i) $$a|a$$ for all $$a\in Z\backslash \{0\}$$ and (ii) $$a|b$$ and $$b|c$$ we have $$a|c.$$ So $$a|b$$ is a pre-order on $$Z\backslash \{0\},$$ but it's not a partial order. For, in $$Z\backslash \{0\},$$ $$a|b$$ and $$b|a$$ can only give you the conclusion that $$a=\pm b,$$ which is obviously not the same as $$a=b.$$ | {
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The above example shows the problem with the pre-ordered set $$,$$\lesssim >.$$ It can allow $$a\lesssim b$$ and $$b\lesssim a$$ with a straight face, without giving you the equality. Now a pre-order cannot be made into a partial order on a set $$, $$\lesssim >$$ unless it is a partial order, but it can induce a partial order on a modified form of $$S.$$ Here's how. Take the bull by the horn and define a relation $$\sim ,$$ on $$$$ by saying that $$a\sim b$$ $$\Leftrightarrow a\lesssim b$$ and $$b\lesssim a$$. It is easy to see that $$\sim$$ is an equivalence relation. Now splitting $$S$$ into the set of classes $$\{[a]|$$ $$a\in S\}$$ where $$[a]=\{x\in S|$$ $$x\sim a\}.$$ This modified form of $$S$$ is often represented by $$S/\sim .$$ Now of course $$% [a]\leq \lbrack b]$$ if $$a\lesssim b$$ but it is not the case that $$b\lesssim a.$$ Setting $$[a]=[b]$$ if $$a\sim b$$ (i.e. if $$a\lesssim$$ and $$b\lesssim a).$$
In the example of $$Z\backslash \{0\}$$ we have $$Z\backslash \{0\}/\sim$$ $$% =\{|a|$$ $$|$$ $$a\in Z\backslash \{0\}\}.$$
(Oh and as a parting note an equivalence relation is a pre-order too, with the extra requirement that $$a\lesssim b$$ implies $$b\lesssim a.)$$ | {
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# Meaning of times in $\mathbb{R}^m\times\mathbb{R}^n\rightarrow \mathbb{R}^{m\times n}$?
I use $\mathbf{a} \times\mathbf{b}$ for the cross product, $\mathbf{a}\cdot \mathbf{b}$ for the dot product and $ab$ for normal multiplication ($a,b$ are scalars).
However, what is the meaning of times in $\mathbb{R}\times\mathbb{R}\rightarrow \mathbb{R}^2$?
Or $\mathbb{R}^m\times\mathbb{R}^n\rightarrow \mathbb{R}^{m\times n}?$
Is it the cross product?
Is it the dot product?
Is it normal multiplication?
Update:
Does these have any meaning
$\mathbb{R}^m\cdot\mathbb{R}^n\rightarrow \mathbb{R}^{m\cdot n}$ (the dot product)?
$\mathbb{R}^m \mathbb{R}^n\rightarrow \mathbb{R}^{m n}$ (normal multiplication)?
• It is called the Cartesian product! – Nigel Overmars May 23 '17 at 13:49
It is a cartesian product. If $A$ and $B$ are two sets, then $A\times B$ is by definition the set of couples $(a,b)$ with $a$ in $A$ and $b$ in $B$.
In your case: $$\mathbb{R}^m\times\mathbb{R}^n=\{(x,y);x\in\mathbb{R}^m,y\in\mathbb{R}^n\}.$$
• Adding onto this for the range space $\mathbb{R}^{m \times n}$ is usually meant to denote real $m \times n$ matrices. – Dragonite May 23 '17 at 13:51
• Notice that in fact $\mathbb{R}^m \times \mathbb{R}^n \cong \mathbb{R}^{m+n}$. You always need $m+n$ numbers to identify an element of either side. For example, $\mathbb{R} \times \mathbb{R} \cong \mathbb{R}^2$. – Sharkos May 23 '17 at 13:55
• Yes, OP be weary that there is a difference between $\mathbb{R}^{m+n}$ and $\mathbb{R}^{m\times n}$ that has not really been addressed. – Dragonite May 23 '17 at 13:56
• $$\mathbb{R^2} := \mathbb{R} \times \mathbb{R}$$ by definition. So unless you mean that this is the trivial isomorphism, defined by the identity function, this wrong. – user370967 May 23 '17 at 15:29
• @Sharkos Is it also true for complex numbers, i.e. $\mathbb{C}^m \times \mathbb{C}^n = \mathbb{C}^{m+n}$? – JDoeDoe Nov 25 '17 at 18:52 | {
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I think, more generally, you need guidance on the notation $f\colon A \times B \to C$, which is the notation for "a function called $f$, from the Cartesian product $A \times B$ of sets $A$ and $B$, to the set $C$" (the Cartesian product $A \times B = \{(a, b) : a \in A, b \in B\}$ is the set of all ordered pairs with things taken from $A$ and $B$). More generally, the format is
$$\text{function name} : \text{domain} \to \text{codomain}$$
But this notation $f \colon A \times B \to C$ often says nothing about what the function actually does to pairs $(a, b) \in A \times B$ to produce some $f(a, b) \in C$, unless $f$ happens to have a particularly descriptive name/symbol. In this case, you'll often see functions introduced in "two parts",
\begin{align*} f \colon A \times B &\to C \\ (a, b) &\mapsto \text{however $a, b$ determine $f(a, b)$} \end{align*}
where the first line specifies the function name and all the sets we need, and the second line actually tells us what $f$ does to the pairs $(a, b)$ (and note the new symbol $\mapsto$, which is used like $\to$ above. But $\to$ is used with sets, the domain and codomain, while $\mapsto$ is used between the actual input and output, to explain what happens to elements in the sets).
So you'll never see notation like $\Bbb R^m \cdot \Bbb R^n \to \Bbb R^{m + n}$, with the function placed between sets. Instead, you'll see the function name/notation in the place of $f$, put before the domain. So things like
$$\cdot \colon \Bbb R^n \times \Bbb R^n \to \Bbb R$$ is a (mildly confusing) notation saying that there's a function called "$\cdot$" that takes two vectors in $\Bbb R^n$, and gives you back a real number (we can assume it's the standard dot product). | {
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$$\times \colon \Bbb R^3 \times \Bbb R^3 \to \Bbb R^3$$ would be the (somehow more confusing) notation to say there's a function called "$\times$" that takes two vectors in $\Bbb R^3$ and returns another vector in $\Bbb R^3$; probably it's the standard cross product on $\Bbb R^3$.
For a slightly-less-weird-looking example, we might use
$$+ \colon \Bbb R^n \times \Bbb R^n \to \Bbb R^n$$ to say that we have an operation called "$+$" that takes pairs of vectors in $\Bbb R^n$, and returns a single vector in $\Bbb R^n$ (and unless it's stated otherwise, everyone would assume "$+$" means exactly what you think it means).
The domain and codomain can come in all sorts of varieties. For example, functions don't have to be defined on pairs of things, in which case our domain isn't going to be a Cartesian product. So to talk about the standard square root function, we might write
$$\sqrt{\ }\; \colon \Bbb R_{\ge 0} \to \Bbb R_{\ge 0}.$$
Or maybe we're handed a function with a fairly cryptic name,
\begin{align*} \operatorname{ev} \colon M_{n \times n}(\Bbb R) \times \Bbb R^n &\to \Bbb R^n \\ (A, \vec{v}) &\mapsto A\vec{v} \end{align*}
but with practice, we can see it's the evaluation map that takes an $n \times n$ matrix with real entries and a vector in $\Bbb R^n$, and applies $A$ to $\vec{v}$.
• Note that many of the examples had the format $\Bbb R^n \times \Bbb R^n \to \Bbb R^n$; only the prefix, the function name before $\colon$, could help us tell the functions apart. The part of the notation after the $\colon$ is really only good for domain and codomain, not the function itself (e.g., vector addition versus cross product). – pjs36 May 23 '17 at 16:00
$\Bbb R \times \Bbb R$ denotes the Cartesian product. An element of $\Bbb R \times \Bbb R$ has the form $(a,b)$, where $a$ and $b$ are both in $\Bbb R$.
Basically, $\Bbb R \times \Bbb R$ is just a longer way of saying $\Bbb R^2$. | {
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Basically, $\Bbb R \times \Bbb R$ is just a longer way of saying $\Bbb R^2$.
The same symbol in different contexts gets different meanings. Math in this aspect is somehow like poetry; the same word gets different meanings in different contexts.
The symbol "$\times$", applied to sets like $\mathbb{R}$, denotes the Cartesian product. If $X,Y \neq \varnothing$, then $X \times Y := \{ (x,y) \mid x \in X, y \in Y \}$. If $X, Y := \mathbb{R}$, then $X \times Y$ by definition is simply the usual Cartesian plane. It is defined that $X^{n} := \{ (x_{1},\dots,x_{n}) \mid x_{1},\dots,x_{n} \in X \}$. Now you know what the superscript of $\mathbb{R}$ means.
Note that $\mathbb{R}^{m} \times \mathbb{R}^{n} = \mathbb{R}^{m+n}$.
• Thanks! You say sets, is it also true for vectors? For matrices we can say we have a $m\times n$-matrix, does $\times$ here also mean the cartesian product? – JDoeDoe May 23 '17 at 14:32
• No problem. No, the point is to what objects the symbol "$\times$" is along with. If $a,b$ are vectors, then $a \times b$ means the cross product of $a$ and $b$; if $a,b$ are numbers, then $a\times b$ means either the arithmetic product or the size of a matrix; if $a,b$ are sets, then $a \times b$ means their Cartesian product. These cover your question? – Benicio May 23 '17 at 14:44
None of the above on the left. That $\times$ is what appears between two sets to denote their cartesian product - the set of all ordered pairs whose first (second) element is from the first (second) set.
The $\times$ on the right is ordinary multiplication.
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The $\times$ on the right is ordinary multiplication.
Edit:
The vector space on the left has dimension $m+n$, the one on the right has dimension $mn$, so the arrow can't represent an isomorphism. It might an injection. One possibility is that you're thinking of $\mathbb{R}^{m\times n}$ as the space of $m \times n$ matrices. Then the arrow could mean the function $f$ given by $$f(v,w)_{ij} = v_iv_j .$$ (ordinary multiplication of real numbers onthe right). This map isn't an injection since $f(0,w) = 0$ for every $w$.
The vector space on the left is naturally isomorphic to $\mathbb{R}^{m+n}$. The arrow in $$\mathbb{R}^m \times \mathbb{R}^n \rightarrow \mathbb{R}^{m+n}$$ might then stand for that natural isomorphism. Perhaps that's what you meant to ask about. That's what your example when $m=n=1$ suggests.
• Isn't the $\times$ on the right addition, $m+n$? – JDoeDoe May 23 '17 at 14:16
• @JDoeDoe See my edits. – Ethan Bolker May 23 '17 at 15:13
To answer your first question $X \times Y$ is the Cartesian product of sets $X$ and $Y$. This is all ordered pairs $(x,y)$ where $x$ is a member of the set $X$, and $y$ is a member of the set $Y$.
$$X \times Y = \{(x,y):x\in X,y \in Y\}$$
An example is $\mathbb R \times \mathbb R^2$ which is all ordered pairs $(x,(y,z))$ where $x \in \mathbb R$ and $(y,z) \in \mathbb R^2$. Of course this can be identified with $\mathbb R^3$ by the bijection $(x,(y,z)) \mapsto (x,y,z)$.
A more interesting example is $[0,1] \times S^1$ which is all ordered pairs $(t,\theta)$ there $t \in [0,1]$, i.e. $0 \le t \le 1$ and $\theta$ is a point in the circle. This product $[0,1] \times S^1$ gives a cylinder.
The word product is unfortunate, and has nothing to do with multiplication. The torus - which looks like a bicycle inner tube - is the Cartesian product of two circles: $T = S^1 \times S^1$. | {
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Your idea of the dot product doesn't work for two reasons. Recall that the dot product takes two vectors (of the same dimension) and gives a number. Instead of it being a set, it is a mapping between sets ($\mathbb R^n \times \mathbb R^n \to \mathbb R$). It doesn't make sense to dot a vector from $\mathbb R^2$ with a vector from $\mathbb R^3$; the dimensions are wrong. Think about matrices: it only makes sense to multiply a $p$-by-$q$ matrix with a $q$-by-$r$ matrix to get a $p$-by-$r$ matrix.
Similar remarks can be made about your idea of normal multiplication $\mathbb R^m \mathbb R^n$. Take an example: Take $(1,2,3) \in \mathbb R^3$ and $(1,2)\in \mathbb R^2$. How do we get something in $\mathbb R^6$? (In a natural way that has meaning?) | {
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# Have I been using differential forms without knowing it?
I'm self-learning differential forms. I've been happily integrating 1-forms over parameterised curves, and 2-forms over parameterised surfaces, both in $$\mathbb{R}^{3}$$. Now I've just found out that integrating an n-form $$\omega=f\left(x_{1},\ldots,x_{n}\right)dx_{1}\wedge dx_{2}\cdots\wedge dx_{n}$$ over an n-dimensional manifold M in $$\mathbb{R}^{n}$$ is defined by$$\intop_{M}\omega=\pm\intop_{M}f\left(x_{1},\ldots,x_{n}\right)dx_{1}\cdots dx_{n}.$$
Am I correct in thinking that this definition describes what's going on with an ordinary calculus definite integral$$\int_{b}^{a}f\left(x\right)dx.$$So $$f\left(x\right)dx$$ would be a 1-form and the one-dimensional manifold it is integrated over is the interval $$\left(a,b\right)$$? | {
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• That is exactly correct. – Lee Mosher Sep 10 at 15:42
• To be extra careful, manifolds should be oriented for purposes of integrating forms, so you should specify the orientation on $(a,b)$ as being induced by restriction from the "basic" orientation on $\mathbb R$. Reversing that orientation means integrating backwards from $b$ to $a$, which changes the sign. – Lee Mosher Sep 10 at 15:44
• Actually, to be extra extra careful, the "ordinary calculus definite integral" is ambiguous. It can be interpreted either as (1) the integral of a density on an unoriented smooth manifold with boundary, namely the interval $[a,b]$, in which case you don't need an orientation; or as (2) the integral of a form on an oriented smooth manifold, namely the oriented interval $[a,b]$ with the standard orientation. We can integrate functions even on unoriented smooth manifolds, because of densities. And indeed the whole point of an orientation here is to convert a form into a density. – symplectomorphic Sep 10 at 19:00
• @symplectomorphic - Tau, in “Differential Forms and Integration”, distinguishes between the “unsigned definite integral $\int_{\left[a,b\right]}f\left(x\right)dx$ (which one would use to find area under a curve, or the mass of a one-dimensional object of varying density), and the signed definite integral $\int_{a}^{b}f\left(x\right)dx$ (which one would use for instance to compute the work required to move a particle from $a$ to $b$).” Is that what you mean? Thanks – Peter4075 Sep 11 at 7:03
• Sorry, that should be Tao not Tau. – Peter4075 Sep 11 at 8:09
Moreover, you can think of 0-forms which are just scalars. Then generalized Stokes' theorem $$\int_{d\Omega} \omega=\int_\Omega d\omega$$ in case of 0-form $$\omega$$ (and 1-form $$d\omega$$) becomes the fundamental theorem of Calculus: $$\left.F(x)\right|_a^b =\int_a^b\frac{dF}{dx}dx$$ | {
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Irrational numbers are those which do not terminate and have no repeating pattern. Rational numbers are whole numbers, fractions, and decimals - the numbers we use in our daily lives. Irrational numbers include √2, π, e, and θ. The vast majority are irrational numbers, never-ending decimals that cannot be written as fractions. Hence, almost all real numbers are irrational. Subtract the rounded numbers to obtain the estimated difference of 0.5; The actual difference of 0.988 - 0.53 is 0.458; Some uses of rounding are: Checking to see if you have enough money to buy what you want. Difference between rational and irrational numbers has been clearly explained in the picture given below. From the below figure, we can see the irrational number is √2. . They include the counting numbers and all other numbers that can be written as fractions. Difference between Rational and Irrational Numbers. In summary, this is a basic overview of the number classification system, as you move to advanced math, you will encounter complex numbers. Many people are surprised to know that a repeating decimal is a rational number. Suppose we have two rational numbers a and b, then the irrational numbers between those two will be, √ab. The decimal expansion of an irrational number continues without repeating. This amenability to being written down makes rational numbers the ones we know best. Learn Math Tutorials 497,805 views Yes. one irrational number between 2 and 3 . 10 How to Write Fractions Between Two Decimal Numbers. Practice Problems 1. 9 years ago. Key Differences Between Rational and Irrational Numbers. m/n —> 3/n N has to be an integer, n cannot be 0. Irrational Numbers on a Number Line. ⅔ is an example of rational numbers whereas √2 is an irrational number. The major difference between rational and irrational numbers is that all the perfect squares, terminating decimals and repeating decimals are rational numbers. So 1/3 is between zero and one. What is the Difference Between | {
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decimals are rational numbers. So 1/3 is between zero and one. What is the Difference Between Rational and Irrational Numbers , Intermediate Algebra , Lesson 12 - Duration: 3:03. Are there any decimals that do not stop or repeat? Method of finding a number lying exactly midway between 2 given numbers, is add those 2 numbers & divide by 2. Make use of this online rational or irrational number calculator to ensure the rationality and find its value. Irrational numbers tend to have endless non-repeating digits after the decimal point. Irrational numbers' decimal representation is non terminating , non repeating.. The only candidates here are the irrational roots. Lv 6. But there's at least one, so that gives you an idea that you can't really say that there are fewer irrational numbers than rational numbers. Before studying the irrational numbers, let us define the rational numbers. You can use the decimal module for arbitrary precision numbers: import decimal d2 = decimal.Decimal(2) # Add a context with an arbitrary precision of 100 dot100 = decimal.Context(prec=100) print d2.sqrt(dot100) If you need the same kind of ability coupled to speed, there are some other options: [gmpy], 2, cdecimal. Examples of Rational and Irrational Numbers For Rational. A rational number is a number that can be written as a ratio. To study irrational numbers one has to first understand what are rational numbers. How many irrational numbers can exist between two rational numbers ? Solution: If a and b are two positive numbers such that ab is not a perfect square then : i ) A rational number between and . 0 0. cryptogramcorner. Irrational Number between Two Rational Numbers. Apart from these number systems we have Irrational Numbers. 2 and 3 are rational numbers and is not a perfect square. It makes no sense!!! Terminating, recurring and irrational decimals. Step 3: Place the repeating digit(s) to the left of the decimal point. Take this example: √8= 2.828. To find the rational number | {
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digit(s) to the left of the decimal point. Take this example: √8= 2.828. To find the rational number between two decimals, we can simply look for the average of the two decimals. The ability to convert between fractions and decimals, and to approximate irrational numbers with decimals or fractions, can be very helpful in solving problems. Rational Numbers. That's our five. Step 4: Place the repeating digit(s) to the right of the decimal point. Irrational Numbers. 0.5 can be written as ½ or 5/10, and any terminating decimal is a rational number. Examine the repeating decimal to find the repeating digit(s). And these non recurring decimals can never be converted to fractions and they are called as irrational numbers. Rational Number is defined as the number which can be written in a ratio of two integers. Wednesday, October 14, 2020 Find two rational numbers between decimals Ex: Find two rational numbers that have 3 in their numerator and are in between 0.13 and 0.14 with no calculator. Learn the difference between rational and irrational numbers, and watch a video about ratios and rates Rational Numbers. Which means that the only way to find the next digit is to calculate it. Getting a rough idea of the correct answer to a problem Irrational Numbers. DOWNLOAD IMAGE. how cuanto mide una cama matrimonial haunted house in san diego that lasts 4 7 hours journey to the west full movie with english subtitles. Include the decimal approximation of the ...” in Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions. DOWNLOAD IMAGE. Irrational Numbers: We have different types of numbers in our number system such as whole numbers, real numbers, rational numbers, etc. In other words, irrational numbers require an infinite number of decimal digits to write—and these digits never form patterns that allow you to predict what the next one will be. Definition: Can be expressed as the quotient of two integers (ie | {
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predict what the next one will be. Definition: Can be expressed as the quotient of two integers (ie a fraction) with a denominator that is not zero.. - [Voiceover] Plot the following numbers on the number line. The venn diagram below shows examples of all the different types of rational, irrational numbers including integers, whole numbers, repeating decimals and more. An Irrational Number is a real number that cannot be written as a simple fraction.. Irrational means not Rational. The first number we have here is five, and so five is five to the right of zero, five is right over there. A rational number is of the form $$\frac{p}{q}$$, p = numerator, q= denominator, where p and q are integers and q ≠0.. But rational numbers are actually rare among all numbers. Get an answer to your question “Part B: Find an irrational number that is between 9.5 and 9.7.Explain why it is irrational. A rational number is the one which can be represented in the form of P/Q where P and Q are integers and Q ≠ 0. Some decimals terminate which means the decimals do not recur, they just stop. In short, rational numbers are whole numbers, fractions, and decimals — the numbers we use in our daily lives.. In mathematics, a number is rational if you can write it as a ratio of two integers, in other words in a form a/b where a and b are integers, and b is not zero. Rational numbers. A Rational Number can be written as a Ratio of two integers (ie a simple fraction). Number System Notes. They can be written as a ratio of two integers. I tried to cross multiply like my teacher said but that was for finding what lay in between fractions it worked for that I have no idea how to do this. I can approximately locate irrational numbers on a number line ; I can estimate the value of expression involving irrational numbers using rational numbers. A set of real numbers is uncountable. (Examples: Being able to determine the value of the √2 on a number line lies between 1 and 2, more accurately, between 1.4 … | {
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determine the value of the √2 on a number line lies between 1 and 2, more accurately, between 1.4 … So the question states: what is a decimal number between each of the following pairs of rational numbers and then it gives the fractions -5/6, 1 and -17/20 and -4/5! Rational numbers are contrasted with irrational numbers - numbers such as Pi, √ 2, √ 7, other roots, sines, cosines, and logarithms of numbers. Rational and Irrational numbers both are real numbers but different with respect to their properties. ii) An irrational number between and . Real numbers also include fraction and decimal numbers. Step 5: Using the two equations you found in step 3 and step 4, subtract the left sides of the two equations. The decimal expansion of a rational number always either terminates after a finite number of digits or begins to repeat the same finite sequence of digits over and over. When placing irrational numbers on a number line, note that your placement will not be exact, but a very close estimation. List Of Irrational Numbers 1 100. So irrational number is a number that is not rational that means it is a number that cannot be written in the form $$\frac{p}{q}$$. 1/3. Example 15: Find a rational number and also an irrational number between the numbers a and b given below: a = 0.101001000100001…., b = 0.1001000100001… Solution: Since the decimal representations of a and b are non-terminating and non-repeating. it can't be the terminating decimals because they're rational. An irrational number is a number which cannot be expressed in a ratio of two integers. O.13 < 3/n < 0.14 13/100 < 3/n < 14/100 When is 13/100 < 3/n ? How to Write Irrational Numbers as Decimals Clearly all fractions are of that That means it can be written as a fraction, in which both the numerator (the number on top) and the denominator (the number on the bottom) are whole numbers. The difference between rational and irrational numbers can be drawn clearly on the following grounds. Then we get 1/3. The | {
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rational and irrational numbers can be drawn clearly on the following grounds. Then we get 1/3. The number $\pi$ (the Greek letter pi, pronounced ‘pie’), which is very important in describing circles, has a decimal form that does not stop or repeat. Irrational Numbers. Representation of irrational numbers on a number line. But an irrational number cannot be written in the form of simple fractions. As there are infinite numbers between two rational numbers and also there are infinite rational numbers between two rational numbers. Real numbers include natural numbers, whole numbers, integers, rational numbers and irrational numbers. How To Find Irrational Numbers Between Two Decimals DOWNLOAD IMAGE. since 3^2 = 9 and 6^2 = 36, √41 > 6. apply the same kind of thinking to the numbers in B and they are all between 3 and 6. New Proof Settles How To Approximate Numbers Like Pi Quanta Magazine. So number of irrational numbers between the numbers should be infinite or something finite which we cannot tell ? Essentially, irrational numbers can be written as decimals but as a ratio of two integers. On the other hand, all the surds and non-repeating decimals are irrational numbers. For instance, when placing √15 (which is 3.87), it is best to place the dot on the number line at a place in between 3 and 4 (closer to 4), and then write √15 above it. We can actually split this into thirds. Example: Find two irrational numbers between 2 and 3. Infinite rational numbers and also there are infinite rational numbers the ones we know best learn the difference rational. Integer, N can not tell number is a rational number is a number ;. Just stop with respect to their properties are infinite numbers between those two will be,.! Have no repeating pattern that all the surds and non-repeating decimals are rational numbers √2! Fraction.. irrational means not rational you found in step 3: Place the repeating digit ( s ) the! Be 0 number can not be 0 Voiceover ] Plot the following grounds as | {
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the repeating digit ( s ) the! Be 0 number can not be 0 Voiceover ] Plot the following grounds as the number line note... Numbers a and b, then the irrational number calculator to ensure the rationality and find its.. ] Plot the following grounds the two decimals numbers & divide by 2 ' decimal is. In the picture given below < 3/n < 0.14 13/100 < 3/n < 14/100 when is 13/100 < 3/n 14/100! And irrational numbers between two decimals, we can simply look for the average of the decimal expansion an... 2, π, e, and watch a video about ratios and rational. Numbers, whole numbers, let us define the rational number between two decimals DOWNLOAD IMAGE Duration... The irrational numbers between 2 and 3 are rational numbers whereas √2 is an example rational... Divide by 2 the two equations you found in step 3 and step 4: Place the digit! Number calculator to ensure the rationality and find its value when placing irrational numbers between two decimal numbers do stop... Know best rates rational numbers non recurring decimals can never be converted to fractions and are! Numbers using rational numbers are those which do not stop or repeat line, note that your placement not. Should be infinite or something finite which how to find irrational numbers between decimals can see the irrational numbers, is add those 2 &... Surds and non-repeating decimals are irrational numbers step 3: Place the repeating digit ( s ) the. Infinite rational numbers left sides of the two decimals, we can simply look for the of! Studying the irrational number can be written in a ratio of two integers ( ie simple! Natural numbers, Intermediate Algebra, Lesson 12 - Duration: 3:03, us! The counting numbers and is not zero rationality and find its value study irrational numbers those. The terminating decimals and repeating decimals are rational numbers can never be converted fractions... Find the rational number a perfect square what are rational numbers of integers... Other numbers that can be written as a | {
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## Probability density functions
• Continuous variables
• Quantities that can take any value, not just discrete values
• Probability Density function (PDF)
• Continuous analog to the PMF
• Mathematical description of the relative likelihood of observing a value of a continuous variable
## Introduction to the Normal distribution
• Normal distribution
• Describes a continuous variable whose PDF has a single symmetric peak.
\begin{align} \text{mean of a Normal distribution} & \neq \text{mean computed from data} \\ \text{st. dev of a Normal distribution} & \neq \text{st. dev computed from data} \end{align}
### The Normal PDF
In this exercise, you will explore the Normal PDF and also learn a way to plot a PDF of a known distribution using hacker statistics. Specifically, you will plot a Normal PDF for various values of the variance.
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import seaborn as sns
sns.set()
# with stds of interest: samples_std1, samples_std3, stamples_std10
samples_std1 = np.random.normal(20, 1, 100000)
samples_std3 = np.random.normal(20, 3, 100000)
samples_std10 = np.random.normal(20, 10, 100000)
# Make histograms
_ = plt.hist(samples_std1, histtype='step', density=True, bins=100)
_ = plt.hist(samples_std3, histtype='step', density=True, bins=100)
_ = plt.hist(samples_std10, histtype='step', density=True,bins=100)
# Make a legend, set limits
_ = plt.legend(('std = 1', 'std = 3', 'std = 10'))
plt.ylim(-0.01, 0.42)
(-0.01, 0.42)
### The Normal CDF
Now that you have a feel for how the Normal PDF looks, let's consider its CDF. Using the samples you generated in the last exercise (in your namespace as samples_std1, samples_std3, and samples_std10), generate and plot the CDFs.
def ecdf(data):
"""Compute ECDF for a one-dimensional array of measurements."""
# Number of data points: n
n = len(data)
# x-data for the ECDF: x
x = np.sort(data)
# y-data for the ECDF: y
y = np.arange(1, n + 1) / n
return x, y | {
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# y-data for the ECDF: y
y = np.arange(1, n + 1) / n
return x, y
x_std1, y_std1 = ecdf(samples_std1)
x_std3, y_std3 = ecdf(samples_std3)
x_std10, y_std10 = ecdf(samples_std10)
# Plot CDFs
_ = plt.plot(x_std1, y_std1, marker='.', linestyle='none')
_ = plt.plot(x_std3, y_std3, marker='.', linestyle='none')
_ = plt.plot(x_std10, y_std10, marker='.', linestyle='none')
# Make a legend
_ = plt.legend(('std = 1', 'std = 3', 'std = 10'), loc='lower right')
plt.savefig('../images/std-cdf.png')
## The Normal distribution: Properties and warnings
### Are the Belmont Stakes results Normally distributed?
Since 1926, the Belmont Stakes is a 1.5 mile-long race of 3-year old thoroughbred horses. Secretariat ) ran the fastest Belmont Stakes in history in 1973. While that was the fastest year, 1970 was the slowest because of unusually wet and sloppy conditions. With these two outliers removed from the data set, compute the mean and standard deviation of the Belmont winners' times. Sample out of a Normal distribution with this mean and standard deviation using the np.random.normal() function and plot a CDF. Overlay the ECDF from the winning Belmont times. Are these close to Normally distributed?
Note: Justin scraped the data concerning the Belmont Stakes from the Belmont Wikipedia page.
df = pd.read_csv('./dataset/belmont.csv') | {
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df = pd.read_csv('./dataset/belmont.csv')
belmont_no_outliers = np.array([148.51, 146.65, 148.52, 150.7 , 150.42, 150.88, 151.57, 147.54,
149.65, 148.74, 147.86, 148.75, 147.5 , 148.26, 149.71, 146.56,
151.19, 147.88, 149.16, 148.82, 148.96, 152.02, 146.82, 149.97,
146.13, 148.1 , 147.2 , 146. , 146.4 , 148.2 , 149.8 , 147. ,
147.2 , 147.8 , 148.2 , 149. , 149.8 , 148.6 , 146.8 , 149.6 ,
149. , 148.2 , 149.2 , 148. , 150.4 , 148.8 , 147.2 , 148.8 ,
149.6 , 148.4 , 148.4 , 150.2 , 148.8 , 149.2 , 149.2 , 148.4 ,
150.2 , 146.6 , 149.8 , 149. , 150.8 , 148.6 , 150.2 , 149. ,
148.6 , 150.2 , 148.2 , 149.4 , 150.8 , 150.2 , 152.2 , 148.2 ,
149.2 , 151. , 149.6 , 149.6 , 149.4 , 148.6 , 150. , 150.6 ,
149.2 , 152.6 , 152.8 , 149.6 , 151.6 , 152.8 , 153.2 , 152.4 ,
152.2 ])
mu = np.mean(belmont_no_outliers)
sigma = np.std(belmont_no_outliers)
# Sample out of a normal distribution with this mu and sigma: samples
samples = np.random.normal(mu, sigma, size=10000)
# Get the CDF of the samples and of the data
x_theor, y_theor = ecdf(belmont_no_outliers)
x, y = ecdf(samples)
# Plot the CDFs
_ = plt.plot(x_theor, y_theor)
_ = plt.plot(x, y, marker='.', linestyle='none')
_ = plt.xlabel('Belmont winning time (sec.)')
_ = plt.ylabel('CDF')
### What are the chances of a horse matching or beating Secretariat's record?
Assume that the Belmont winners' times are Normally distributed (with the 1970 and 1973 years removed), what is the probability that the winner of a given Belmont Stakes will run it as fast or faster than Secretariat?
samples = np.random.normal(mu, sigma, size=1000000)
# Compute the fraction that are faster than 144 seconds: prob
prob = np.sum(samples < 144) / len(samples)
# Print the result
print('Probability of besting Secretariat:', prob)
Probability of besting Secretariat: 0.00057
## The Exponential distribution
• The waiting time between arrivals of a Poisson process is Exponentially distributed
### If you have a story, you can simulate it! | {
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### If you have a story, you can simulate it!
Sometimes, the story describing our probability distribution does not have a named distribution to go along with it. In these cases, fear not! You can always simulate it. We'll do that in this and the next exercise.
In earlier exercises, we looked at the rare event of no-hitters in Major League Baseball. Hitting the cycle is another rare baseball event. When a batter hits the cycle, he gets all four kinds of hits, a single, double, triple, and home run, in a single game. Like no-hitters, this can be modeled as a Poisson process, so the time between hits of the cycle are also Exponentially distributed.
How long must we wait to see both a no-hitter and then a batter hit the cycle? The idea is that we have to wait some time for the no-hitter, and then after the no-hitter, we have to wait for hitting the cycle. Stated another way, what is the total waiting time for the arrival of two different Poisson processes? The total waiting time is the time waited for the no-hitter, plus the time waited for the hitting the cycle.
Now, you will write a function to sample out of the distribution described by this story.
def successive_poisson(tau1, tau2, size=1):
"""Compute time for arrival of 2 successive Poisson processes."""
# Draw samples out of first exponential distribution: t1
t1 = np.random.exponential(tau1, size)
# Draw samples out of second exponential distribution: t2
t2 = np.random.exponential(tau2, size)
return t1 + t2
### Distribution of no-hitters and cycles
Now, you'll use your sampling function to compute the waiting time to observe a no-hitter and hitting of the cycle. The mean waiting time for a no-hitter is 764 games, and the mean waiting time for hitting the cycle is 715 games.
waiting_times = successive_poisson(764, 715, size=100000)
# Make the histogram
_ = plt.hist(waiting_times, bins=100, density=True, histtype='step')
# Label axes
_ = plt.xlabel('waiting times')
_ = plt.ylabel('PDF') | {
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# Area of a Sector- Why squared?
#### opus
Gold Member
1. Homework Statement
Find the area, A, of a sector of a circle with a radius of 9 inches and a central angle of 30°.
2. Homework Equations
$$Area~of~a~Sector:$$
$$A=\left( \frac 1 2 \right)r^2θ$$
3. The Attempt at a Solution
$$θ=30°$$
$$θ=30°\left( \frac π {180} \right)$$
$$θ=\left( \frac π 6 \right)$$
$$A=\left( \frac 1 2 \right)\left(9\right)^2\left(\frac π 6 \right)$$
$$A=\left( \frac {81π} {12} \right)$$
$$A≈21.2 in^2$$
My question:
I know that when you find the area of a space, it will be in $units^2$. But I've always thought of it as a square- that is, one equal side multiplied by the other equal side obviously results in a squared result. However in this case, I don't see how the units for a sector of a circle are squared, as it doesn't seem like we're multiplying two things of equal value to each other.
So why is this result squared?
Related Precalculus Mathematics Homework Help News on Phys.org
#### Mark44
Mentor
1. Homework Statement
Find the area, A, of a sector of a circle with a radius of 9 inches and a central angle of 30°.
2. Homework Equations
$$Area~of~a~Sector:$$
$$A=\left( \frac 1 2 \right)r^2θ$$
3. The Attempt at a Solution
$$θ=30°$$
$$θ=30°\left( \frac π {180} \right)$$
$$θ=\left( \frac π 6 \right)$$
$$A=\left( \frac 1 2 \right)\left(9\right)^2\left(\frac π 6 \right)$$
$$A=\left( \frac {81π} {12} \right)$$
$$A≈21.2 in^2$$ | {
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My question:
I know that when you find the area of a space, it will be in $units^2$. But I've always thought of it as a square- that is, one equal side multiplied by the other equal side obviously results in a squared result. However in this case, I don't see how the units for a sector of a circle are squared, as it doesn't seem like we're multiplying two things of equal value to each other.
So why is this result squared?
Because it's an area. The shape doesn't matter.
The standard units of area are always squared, $\text{length} \times \text{length}$, except for some special cases such as acres or hectares (which involve implicitly squared units such as ft2 for acres and m2 for hectares.
#### opus
Gold Member
So the length x length in this particular case, would be length(radius) x length(arc). However the length of the radius is in inches, and the length of the arc is in radians. So how can this results in inches squared?
#### Mark44
Mentor
So the length x length in this particular case, would be length(radius) x length(arc). However the length of the radius is in inches, and the length of the arc is in radians. So how can this results in inches squared?
The angle in radians is just an angle, with no length. Think about it this way, as a, say, peach pie. If an 8" diameter pie is cut into 6 pieces, each slice (a sector) will subtend an angle of $\pi/3$, and the radius will be 4". The arc length of the curved edge of the slice has to take into account the radius, otherwise the arc length of an 8" pie would be the same as for a 16" pie. So in fact, the curved dimension of the pie sector is radius * angle (in radians), or $4 \times \pi/3$. So you have one radius for the radius of the sector and another radius for the arc length, making the sector area equal to $\frac 1 2 r^2 \theta$.
#### opus
Gold Member
Ahhh ok. That makes complete sense. Great explanation, thank you Mark.
#### DrClaude | {
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Gold Member
Ahhh ok. That makes complete sense. Great explanation, thank you Mark.
#### DrClaude
Mentor
I know that when you find the area of a space, it will be in $units^2$. But I've always thought of it as a square- that is, one equal side multiplied by the other equal side obviously results in a squared result.
Let me add that one way to visualize the "units square" is to think that a wedge with an area of 22.2 in2 has the same area as a square with sides of √(22.2) ≈ 4.6 in.
Last edited by a moderator:
#### opus
Gold Member
Interesting! Thanks DrClaude
#### alijan kk
imagine the arc as a triangle , becuase area would be same even if you make the arc straight line.
now the base of this triangle is "r=radius" and the perpendicular side is the arc which is equal to "theta*r"
area of triangle = 0.5base*height
0.5(r)(r)(theta)=formula of area of sector
"Area of a Sector- Why squared?"
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# Math Help - Problem with vector question
1. ## Problem with vector question
Hi
Can someone tell me where i have gone wrong in the following question?
Given that v=2i + 2j + k, w = 3i - j + k find a unit vector perpendicular to both v and w.
unit vector = {a x b}/{|a x b|}
| a x b |= $\sqrt{3^2+1^2+8^2}$
$=\sqrt{74}$
a x b = i(2 x 1 - 1 x -1)-j(2 x 1 - 1 x 3)+k(2 x -1 - 2 x 3)
=i(2+1)-j(2-3)+k(-2-6)
=3i+j-8k
therefore unit vector = $\frac{1}{\sqrt{74}}$(3i+j-8k)
answer says its $\frac{1}{\sqrt{74}}$(-3i-j+8k)
P.S
2. Originally Posted by Paymemoney
Hi
Can someone tell me where i have gone wrong in the following question?
Given that v=2i + 2j + k, w = 3i - j + k find a unit vector perpendicular to both v and w.
unit vector = {a x b}/{|a x b|}
| a x b |= $\sqrt{3^2+1^2+8^2}$
$=\sqrt{74}$
a x b = i(2 x 1 - 1 x -1)-j(2 x 1 - 1 x 3)+k(2 x -1 - 2 x 3)
=i(2+1)-j(2-3)+k(-2-6)
=3i+j-8k
therefore unit vector = $\frac{1}{\sqrt{74}}$(3i+j-8k)
answer says its $\frac{1}{\sqrt{74}}$(-3i-j+8k)
P.S
$\mathbf{v}\times\mathbf{w} = \left|\begin{matrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 2 & 1 \\ 3 & -1 & 1 \end{matrix}\right|$
$= \mathbf{i}[2 \cdot 1 - 1\cdot (-1)] - \mathbf{j}[2 \cdot 1 - 1 \cdot 3] + \mathbf{k}[2 \cdot (-1) - 2 \cdot 3]$
$= 3\mathbf{i} - \mathbf{j} - 8\mathbf{k}$.
I agree with the answer you have given for $\mathbf{v} \times \mathbf{w}$. Perhaps you copied the question down wrong or else the answer given is incorrect.
Anyway, to find the unit vector in the direction of $\mathbf{v} \times \mathbf{w}$, divide it by its length.
$|\mathbf{v}\times\mathbf{w}| = \sqrt{3^2 + (-1)^2 + (-8)^2}$
$= \sqrt{9 + 1 + 64}$
$= \sqrt{74}$.
Therefore:
$\frac{\mathbf{v}\times\mathbf{w}}{|\mathbf{v}\time s\mathbf{w}|} = \frac{3\mathbf{i} - \mathbf{j} - 8\mathbf{k}}{\sqrt{74}}$
$= \frac{3}{\sqrt{74}}\mathbf{i} - \frac{1}{\sqrt{74}}\mathbf{j} - \frac{8}{\sqrt{74}}\mathbf{k}$ | {
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$= \frac{3}{\sqrt{74}}\mathbf{i} - \frac{1}{\sqrt{74}}\mathbf{j} - \frac{8}{\sqrt{74}}\mathbf{k}$
$= \frac{3\sqrt{74}}{74}\mathbf{i} - \frac{\sqrt{74}}{74}\mathbf{j} - \frac{4\sqrt{74}}{37}\mathbf{k}$.
3. Hello, Paymemoney!
You've done nothing wrong . . .
Their answer is the negative of yours.
There are two vectors perpendicular to both $\vec v\text{ and }\vec w,$
. . one "up" and one "down".
They picked one, you picked the other.
4. ok oh ic, well to find the negative answer would you adjust the v and w values into negative? ie v=-2i + -2j - k, w = -3i + j - k | {
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# How "same" are two isomorphic groups?
From what I understand about isomorphisms is that two isomorphic groups are the same groups. They may have different names for the same elements and the operation. But the point is that the groups are same since their elements combine the same way.
So if $$G \cong G'$$ then every group property about $$G$$ also holds for $$G'$$, am I correct?
Now my question is— Can I interchange $$G$$ and $$G'$$ whenever and wherever I want? I thought the answer was obviously yes but... now I'm not sure.
For example: $$\mathbb{Z} \cong \mathbb{2Z}$$ so shouldn't $$\mathbb{Z}/\mathbb{Z} = \mathbb{Z}/\mathbb{2Z}$$ under the group operation $$(a+H) + (b +H) = (a+b) + H$$? where $$H$$ is either $$\mathbb{Z}$$ or $$\mathbb{2Z}$$ since both are the same groups.
But that's clearly not the case as $$\mathbb{Z}/\mathbb{Z} = \{0\}$$ but $$\mathbb{Z}/\mathbb{2Z}=\{0,1\}$$ So where does one draw the line between two isomorphic groups? How "same" are two isomorphic groups? I'm very confused. | {
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• It is true that $\mathbb{Z} \cong \mathbb{2Z}$. However, the issue for the quotient groups here is that $2 \mathbb Z \subsetneq \mathbb Z$. Dec 27, 2021 at 16:42
• The freedom to rename is not a freedom to be inconsistent. If you want to rename the elements in $\mathbb Z$ so that $1$ is renamed to $2$, $2$ to $4$, $3$ to $6$ etc. - then of course the elements of $2\mathbb Z\subseteq \mathbb Z$ must be renamed using the same procedure. You will end up with $4\mathbb Z\subseteq 2\mathbb Z$, and indeed $\mathbb Z/2\mathbb Z\cong 2\mathbb Z/4\mathbb Z$. Dec 27, 2021 at 16:43
• Why is this closed under duplicate? I'm asking about a particular example and the difficulty I'm facing with it. How is the question same as the other? I'd already read that one. It's where I came from. The accepted answers says exactly the same thing as my first paragraph. But my question is different. Dec 27, 2021 at 16:46
• Presentations of the same group (up to isomorphism) can have different properties. Dec 27, 2021 at 16:54
• Frankly, I regard the "s" word as a 4-letter word in mathematics. I would suggest that you learn to love the 10-letter "i" word in your title. And then if you can extend that to loving the 11-letter word "isomorphism", and to think about actual isomorphisms as useful and interesting mathematical objects themselves, even better! Dec 27, 2021 at 18:51
This is a great question and your example with group quotients perfectly highlights why one has to be careful about notions of identity when doing maths. To answer your question,
Can I interchange G and G′ whenever and wherever I want? | {
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Can I interchange G and G′ whenever and wherever I want?
The answer is yes, as long as you are 100% sure that you are thinking of both groups as literally just groups and nothing more (people will sometimes use the phase 'up to isomorphism' to convey this idea). This resolves the problem you had with quotients: you can't quotient an arbitrary group by another arbitrary group, the group by which you are quotienting has to be a subgroup of the other. This is extra 'data' which can be encoded in multiple ways, one nice way is to consider the special injection that embeds the one group inside the other (which doesn't exist for two arbitrary groups). So to recap, subgroups aren't just groups, they're groups with some extra data and $$2\mathbb{Z}$$ and $$\mathbb{Z}$$ are different subgroups of $$\mathbb{Z}$$ despite being 'the same' groups (i.e. isomorphic). | {
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• "the group by which you are quotienting has to be a subgroup of the other" but $Z$ and $2Z$ are both subgroups of $Z$ no? and I don't get the part where you say they are the same groups but different subgroups? I'm sure I'm missing something here but I don't get it. What is the extra data about subgroup? Dec 27, 2021 at 17:14
• Oh you mean they're different subgroups because they are made up of different subsets of the group? Oh okay. Fair enough. But what exactly go wrong here in my example? I'm not able to "see" it still :( Dec 27, 2021 at 17:26
• The problem in your example is that, to conclude that the quotient is the same, you have to start with 'the same' subgroups, but you didn't, you started with two different subgroups (that just so happen to be 'the same' as groups). Dec 27, 2021 at 17:28
• So in total, there are four descriptions of mathematical objects: the group $\mathbb{Z}$, the group $\mathbb{2Z}$, the subgroup $\mathbb{Z}\leq\mathbb{Z}$ and the subgroup $\mathbb{2Z}\leq\mathbb{Z}$. The first two descriptions define the same mathematical object, the second two don't. Dec 27, 2021 at 17:46
• Oh! Oh! Oh!! I get it now. If $H,H' ≤G$ and $H \cong H'$ then as groups, $H$ and $H'$ are same. But "subgroup" is also a relation and not an independent concept. So you always have subgroups of some group. So as subgroups of $G$, $H$ and $H'$ are different. So $G/H ≠ G/H'$ because quotient groups aren't just about $H$ but rather $H$ wrt $G$. If it were about only $H$ then one can interchange it with $H'$ is what you're saying. Dec 27, 2021 at 18:12
Here is a question that is similar: when can replace equivalent things with one another? | {
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Here is a question that is similar: when can replace equivalent things with one another?
So if we remember that an isomorphism of groups is a function $$\phi:G\rightarrow G'$$. With the example you gave you have $$\phi:\mathbb{Z}\rightarrow 2\mathbb{Z}$$, then your issue was that $$\mathbb{Z}/\mathbb{Z}\not\cong\mathbb{Z}/2\mathbb{Z}$$. However, the issue is where are you applying the homomorphism. You are saying that $$G/G\not\cong G/\phi(G)$$; however, we have that $$\phi(G)\subset G'$$ and $$\phi(G)\not\subset G$$ (although it may be isomorphic to some subgroup of $$G$$ like in your example).
The key use of an isomorphism is that you can take a problem in one setting and view it in another setting through the isomorphism. Thus, what would be true is that if $$\phi:G\rightarrow G'$$ is an isomorphism, then if $$H\lhd G$$, then it will be the case that $$G/H\cong \phi(G)/\phi(H)$$. | {
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I think the long story short is that when you have an isomorphism between two objects you must think of them as living in different spaces. If you are familiar with the lattice of subgroups I think this is a nice way to view it, since if you have $$\phi: G\rightarrow G'$$ where $$G=\mathbb{Z}$$ and $$G'=2\mathbb{Z}$$ since these are isomorphic groups their lattice of subgroups are isomorphic and we should think of $$\mathbb{Z}$$ to be the maximal element in the lattice of $$G$$ and $$2\mathbb{Z}$$ to be the maximal element in the lattice of $$G'$$. Even though we have that $$2\mathbb{Z}$$ appears in the lattice of $$G$$, the corresponding element in $$G'$$ would be the subgroup $$4\mathbb{Z}$$, so under the isomorphism it would not make sense to "replace $$2\mathbb{Z}$$ in the $$G$$ world with $$2\mathbb{Z}$$ in the $$G'$$ world" as they aren't the same under the isomorphism. What we would do is replace the $$2\mathbb{Z}$$ with the corresponding subgroup in the isomorphism namely $$4\mathbb{Z}$$, and when we do so we do have that $$\mathbb{Z}/2\mathbb{Z}\cong 2\mathbb{Z}/4\mathbb{Z}$$.
There is a precise way to formalise this.
We want to speak of sets not as collections of particular mathematical objects but as a bag of featureless dots. These bags of dots are related to each other by functions, which also relate dots in one bag to dots in another.
In order to do this, we propose a language for discussing sets. Capital letters will represent sets, and lowercase variables will represent functions between sets.
We define a "nice proposition about sets" as any proposition that can be generated from the following rules: | {
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• For any sets $$A, B$$ and any function variables $$f, g: A \to B$$, $$f = g$$ is a nice proposition
• We can combine nice propositions $$\phi, \psi$$ using the operators $$\land$$, $$\lor$$, $$\neg$$, and $$\implies$$ to get another nice proposition
• $$\top$$ and $$\bot$$ (that is, true and false) are nice propositions
• If $$\phi(A)$$ is a nice proposition, where $$A$$ is a set variable, then $$\forall A (\phi(A))$$ is a nice proposition (and similarly for $$\exists$$)
• For all set variables $$A, B$$, if $$\phi(f)$$ is a nice proposition (where $$f : A \to B$$ is a function variable), then $$\forall f : A \to B (\phi(f))$$ is a nice proposition (and similarly for $$\exists$$)
Finally, in our language, we can form new functions from old ones using function composition and also discuss identity functions. Note that we will only discuss function composition when we know that the functions are composable syntactically.*
Note that we very deliberately avoided two things. First, there is no mention of the elementhood relation at all. Second, there is no way to compare two sets for equality, nor is there a way to compare two functions for equality unless the two have the same domain and codomain. This is deliberate. Also note that all variables here have specific types, and that we rely on these types to determine whether we can discuss equality.
Despite this avoidance, it is both an empirical fact and a (rather complicated) theorem that all propositions with no free variables can be translated to a "nice proposition about sets". For most mathematically useful propositions, the translation is relatively intuitive for those comfortable with category theory. | {
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To get around the restriction that we not discuss elements directly, we can fix some 1-element set $$1$$ and discuss elements of $$A$$ as functions $$1 \to A$$. We denote the situation $$x : 1 \to A$$ as $$x :\in A$$. When we have $$x :\in A$$ and $$f : A \to B$$, we write $$f \circ x$$ as $$f(x)$$ (note that $$f(x) :\in B$$). Note that this is enough to define, for example, a group (as a set $$G$$, together with a function $$- \cdot - : G^2 \to G$$ which satisfies the group laws**).
We thus have the following theorem:
Big Theorem. Consider some nice proposition $$\phi(G)$$ - that is, some statement $$\phi$$ where the group variable $$G$$ occurs free. Then we can prove the following statement: "For all groups $$G_1$$, $$G_2$$, if $$G_1$$ and $$G_2$$ are isomorphic and $$\phi(G_1)$$, then $$\phi(G_2)$$."
Let us discuss how this relates to your example of $$\mathbb{Z}$$ and $$2 \mathbb{Z}$$.
We may try to define $$\phi(G)$$ as the statement "$$\mathbb{Z} / G$$ is the zero group". Then taking $$G_1 = \mathbb{Z}$$ and $$G_2 = 2 \mathbb{Z}$$, we see that $$\phi(G_1)$$ holds but $$\phi(G_2)$$ is false. This would appear to violate our Big Theorem.
To understand what's going on, we need to understand exactly what is going on with the statement $$\phi(G)$$. Note that in this statement, it is necessary for $$G$$ to be a subgroup of $$\mathbb{Z}$$ - in particular, it must be a subset. That is, we must have $$\forall x \in G (x \in \mathbb{Z})$$.
Of course, we cannot express such a statement as a nice proposition about sets, since it relies on the proposition $$x \in \mathbb{Z}$$, which is not a nice proposition. So we have to approach things a bit differently. | {
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Rather than forcing $$G$$ to be a subgroup in the traditional, literal sense, we instead make $$G$$ a subgroup of $$\mathbb{Z}$$ in a more general sense. That is, we discuss a group $$G$$, together with some injective group homomorphism $$i : G \to \mathbb{Z}$$. In other words, we discuss $$G$$ together with a specific way that $$G$$ is a subgroup of $$\mathbb{Z}$$.
This makes it clear that $$\phi(G)$$ is not really just a proposition about $$G$$ - it's also about the way that $$G$$ is a subgroup of $$\mathbb{Z}$$. So it's really a proposition $$\phi(G, i : G \to \mathbb{Z})$$. Because $$\phi$$ depends on a secondary variable $$i$$, we see that we cannot apply our Big Theorem to conclude that $$\phi(\mathbb{Z}) \iff \phi(2 \mathbb{Z})$$.
If it were the case that the isomorphism $$w : \mathbb{Z} \to 2 \mathbb{Z}$$ "played nicely" with the inclusion maps $$i_1 : \mathbb{Z} \to \mathbb{Z}$$, $$i_2 : 2 \mathbb{Z} \to \mathbb{Z}$$ (that is, if $$i_1 = i_2 \circ w$$), then we would be able to conclude that $$\phi(\mathbb{Z}, i_1) \iff \phi(2 \mathbb{Z}, i_2)$$ using a generalisation of our Big Theorem. But of course the isomorphism does not play nicely with the inclusion maps.
*An astute observer will note that this is exactly the language of category theory. A "nice proposition about sets" is just some statement about the category of sets (which avoids discussing equality of objects).
**Technically, we need a way to define $$G^2$$ first. This is done using the categorical definition of the universal property of the product.
• Why is this answer being downvoted? Dec 27, 2021 at 18:31
This is just an elaboration of Arthur's answer elsewhere in this thread. I thought you might like to see a specific example.
Consider $$G = \color{maroon}{\Bbb Z_2}\times \color{darkblue}{\Bbb Z_4}.$$
$$G$$ has a subgroup $$A$$ that is generated by $$\langle 1, 0\rangle$$:
$$\def\elt#1#2{\langle{#1},{#2}\rangle}A = \{\elt10, \elt 00\}$$ | {
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$$\def\elt#1#2{\langle{#1},{#2}\rangle}A = \{\elt10, \elt 00\}$$
and another subgroup $$B$$ that is generated by $$\langle 0, 2\rangle$$:
$$B=\{\elt00, \elt 02\}$$
Both $$A$$ and $$B$$ are isomorphic to $$\Bbb Z_2$$, and so to each other. But the quotients $$G/A$$ and $$G/B$$ are not isomorphic. $$G/A$$ is like taking $$G$$, ignoring the first component, and keeping the second component intact. The result is $$G/A \simeq \color{maroon}{\Bbb Z_1}\times\color{darkblue}{\Bbb Z_4}.$$ $$G/B$$ is like taking $$G$$ and keeping the first component but ignoring everything but the parity of the second component. The result is $$G/B \simeq \color{maroon}{\Bbb Z_2}\times\color{darkblue}{\Bbb Z_2}.$$ Even though $$A$$ and $$B$$ are isomorphic, $$G/A$$ and $$G/B$$ are not isomorphic. (In particular, $$G/A$$ is cyclic and $$G/B$$ isn't.)
Each component of $$G = \color{maroon}{\Bbb Z_2}\times \color{darkblue}{\Bbb Z_4}$$ contains a factor of $$\Bbb Z_2$$. Each of $$A$$ and $$B$$ is isomorphic to $$\Bbb Z_2$$. When we construct the quotient of $$G$$ by $$A$$ or $$B$$, we get a different result depending on which component we divide the factor from.
Two groups that are isomorphic share the same internal structure. But a quotient $$G/N$$ doesn't depend only on the internal structures of $$G$$ and $$N$$. The quotient also depends on the relationship between $$G$$ and $$N$$. So even though $$A$$ and $$B$$ have the same internal structure, $$G/A$$ and $$G/B$$ are different, because those structures reside inside of $$G$$ in different ways.
Topological and illustrated version of Arthur's nice answer: | {
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Topological and illustrated version of Arthur's nice answer:
A trefoil knot $$K_3$$ in $$\Bbb R^3$$ cannot be unknotted but it can be in $$\Bbb R^4$$. Topologically a trefoil knot $$K_3$$ and a circle are homeomorphic i.e. one can deform one to other without cutting or gluing. So these two (one in $$\Bbb R^3$$ and one in $$\Bbb R^4$$) are the same but one uncovers a branch of mathematics known as Knot_theory while other one is fruitless from this point of view and application.
I think theses type of "same things" is a categorical notion but I am not sure.
same is true for $$y=x$$ and $$y=\exp(x)$$; both topologically are homeomorphic to $$\Bbb R$$, but one contains only positive numbers. One can say similar properties by considering them as additive and multiplicative groups.
These might be of interest to you:
Let me for convenience, first, assume all subgroups are normal, or that we are in an abelian group.
We form quotients of a group by its SUBgroups, not by another abstract GROUP. Subgroup means a SPECIFIC subset (which is closed ....) As groups Z and 2Z are "same". But as subsets of Z they are different.
Now to non-abelian case. Take $$G= \{+1,-1\}\times S_3$$ where the first factor $$H$$ is a group (of order 2) wrt multiplication of numbers.
This group $$G$$ has more subgroups of order 2 (isomorphic to H as a group) namely ones generated by a transposition in the second factor $$S_3$$. These subgroups are NOT even normal in $$G$$. SO quotient does not even exist by these subgroups. | {
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Seperating $2n$ elements with subsets of size $n$
Suppose I have a set $X$ with $2n$ elements ($n$ nonzero natural number). I now want to find a collection of subsets $X_1, \ldots, X_k$ of $X$ such that
• Every $X_i$ contains $n$ elements.
• For every $x, y \in X$, there exists an $X_i$ such that $x \in X_i$ and $y \not\in X_i$ or $x \not\in X_i$ and $y \in X_i$ (the $X_i$ "seperate" the elements of $X$, in that the topology on $X$ generated by the $X_i$ is Kolmogorov)
• $k$ is minimal with respect to the above to properties.
An algorithm to find $X_1, \ldots, X_k$ would be great, but I am mostly looking for a way to calculate $k$. My intuition tells me that $k$ should be at least $\log_2(2n)$, more specifically I feel like $$\max_{x \in X} \# \lbrace y \in X \mid \not\exists i \in \lbrace 1, \ldots, j \rbrace : x \in X_i \enspace \& \enspace y \not\in X_i \rbrace \geq \log_2(\frac{2n}{j})$$ should hold for all $j \in \lbrace 1, \ldots, k \rbrace$, but I don't know how to prove this.
• Consider sets with $2^l$ elements first – Scipio Jun 4 '17 at 12:09
• I did; even if $n=4$ I don't see how to proceed. (i.e. I know that $k = 2$ is possible, but do not see how to show that $k = 1$ is impossible) – Bib-lost Jun 4 '17 at 12:11
• $X_1$ contains two elements $a,b$; clearly the second requirement doesn't hold for these two elements if you only have a single set. – Scipio Jun 4 '17 at 12:18
• By $n=4$ I meant the case where $X$ has $2*4 = 8$ elements. The case with $4$ elements is indeed clear. :) And surely, also the case with $8$ elements can be computed by hand in a reasonably short time; what I meant is that I do not see an argument which could somehow be generalised. – Bib-lost Jun 4 '17 at 12:44
• Could you add the cases $n=2,3,4$ into your question & give us the lists $X_1, X_2 , \cdots , X_k$ ? – Donald Splutterwit Jun 4 '17 at 12:50 | {
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Suppose $X$ contains $2^l$ elements. Let $(A_i^m)$ be the family of subsets of $X$ of non-separated elements after introducing $X_1, \cdots, X_m$, i.e. sets of elements for which the second requirement does not hold (check that these sets are well defined and uniquely partition $X$). Ultimately, we want that all the subsets $A_i$ contain just a single element (i.e. every element is separated from all other elements).
With each $X_i$ we want to separate as many elements as possible. Now, at first $A_1 = X$ containing $2^l$ elements and we take $X_1$ just an arbitrary set of $n$ elements. After this, we have sets $A_1$, $A_2$, both containing $2^{l-1}$ elements. Now, with $X_2$ we can again split both in half, resulting in four 'remaining sets' with $2^{l-2}$ elements. Continue recursively to see that $k=l$. Moreover, it is clear that after introducing $m$ sets, $\max_i \# A_i^m \geq 2^{l-m}$, so that our result is indeed optimal.
For example, start with $X = A_1^0 = \{1,2,3,4,5,6,7,8\}$. Introduce $X_1 = \{1,2,3,4\}$ to get $A_1^1 = \{1,2,3,4\}, A_2^1 = \{5,6,7,8\}$. Introduce $X_2 = \{1,2,5,6\}$ and $X_3 = \{1,3,5,7\}$ to finish the job. (Make sure to check the corresponding sets $(A_i^m)$.)
Now, for any other number of elements $2n$, your 'splits' wil not be optimal. You can check that $\max_i \# A_i^m \geq 2n\cdot2^{-m}$ still holds.
• So probably we have that, in general, $k$ is the unique integer satisfying $2^{k-1} < 2n \leq 2^{k}$? – Bib-lost Jun 4 '17 at 13:57
• yes exactly :) This hinges on the very last statement in my answer, so make sure to check that carefully. – Scipio Jun 4 '17 at 14:01
There should be $k=\lceil log_2(2n) \rceil$ partitions, $2k$ sets.
We have the conditions: $$\mathcal{P}_1:\forall X_i, \text{num}\{X_i\}=n$$ $$\mathcal{P}_2:\forall x_i,x_j \exists X_i, x_i \in X_i, x_j \not\in X_i$$ | {
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It is clear for sets of $2n=2^k$ elements, the partitions are binary, hence optimal. Under this case, the condition $\mathcal{P}_2$ implies the stronger: $$\mathcal{P}_{2b}:\forall x_i,x_j \exists X_i,X_j, x_i \in X_i, x_j \in X_j$$ because each partition requires its complement. If there is not complement, the property is not met. $$\forall X_i \exists Y_i=X -X_i$$
If the set contains $2n=2^k+2m$ elements, $2m<2^k$, each $2m$ remaining elements shall be included into different disjoint partitions. Again in this case the partition complement must exist and the condition $\mathcal{P}_{2}$ will not introduce any benefit against $\mathcal{P}_{2b}$.
• Don't you mean $k = \lceil \log_2(2n) \rceil$? Surely if $n=3$ you need at least $2 = \lceil log_2(6) \rceil$ partitions? – Bib-lost Jun 4 '17 at 14:14 | {
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# Special case of Bertrand Paradox or just a mistake?
I've been working on a question and it seems I have obtained a paradoxical answer. Odds are I've just committed a mistake somewhere, however, I will elucidate the question and my solution just in case anyone is interested.
I want to know what is the average distance between two points on a circle of radius 1 where we consider only the boundary points.
My attempt is as follows:
Consider a segment of the diameter x which is uniformly distributed between 0 and 2. Then you can calculate the distance between the points (2,0) and the point determined by x just by elementary geometry as this picture shows:
enter image description here
Here in the picture, the green segment is the geometric mean and the orange one is the distance whose distribution we want to know. Just by calculating the expected value, we obtain:
$$E\left(\sqrt{(4-2X)}\right) = \int_{0}^{2} \sqrt{(4-2x)}\cdot\frac{1}{2} dx = 1.333.... = \frac{4}{3}$$
Where $$\sqrt{(4-2x)}$$ is the transformation of the random variable and $$\frac{1}{2}$$ is the pdf of a uniform distribution $$[0,2]$$.
Also, if we derive the pdf of the transformation we obtain the same result:
$$y = \sqrt{(4-2x)} , x = 2- \frac{y^2}{2}, \mid\frac{d}{dy}x\mid = y$$
$$g(y)=f(x)\cdot\mid\frac{d}{dy}x\mid = \frac{1}{2}\cdot y$$
$$E(Y)= \int_{0}^{2}y\cdot\frac{1}{2}\cdot y dy = 1.333.... = \frac{4}{3}$$
I have seen a different approach somewhere else where the distribution of the angle is considered as a uniform distribution between 0 and $$\pi$$ and the final result was:
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$$1.27... = \frac{4}{\pi}$$
That's pretty much the problem I found. Maybe I just did it wrong in some step but it all makes sense to me. I know this is not exactly what we refer as Bertrand paradox but it just suggests something like that because both problems handle with segments in circumference and maybe my result is wrong because it does not hold for rotations of the circle or something like that (I read a little bit about the Bertrand's Paradox).
That's pretty much it. Also sorry for my bad English and maybe I'm also wrong in something pretty elemental since I've just started learning about probability theory. It's also my first post so I will try to improve my exposition and LateX use in the following ones.
• welcome to MSE. I have attempted to edit your question. In case you find any discrepancy with your idea please check. Also, you can look up the edits I have made that will help you for your future questions. Thank you! – Mann Apr 19 at 21:57
This is a very nice question, well-written and researched before posting. You have clearly put a lot of careful into your question and it is very much appreciated on this site (and I thought your English reads perfectly naturally). Thanks so much for taking the time to ask a serious and well-considered question. I hope this answer comes close to meeting the high standard of quality set by your question.
I don’t see any calculation errors on your part, and I’ll wager you checked those thoroughly. The flaw is something much more subtle and is related to the initial framing of the problem.
You have correctly solved the problem of “what is the average distance between the rightmost point of a circle and another point projected upwards from a randomly chosen point on the diameter”. But this doesn’t quite capture the same probability distribution as “rightmost point and another randomly chosen point on the circle” (which is sufficient to capture the dynamics of “two random points on the circle”). | {
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There is a hidden non-uniformity in the projection process, and that’s because the circle has varying slope so the projection hits the circle differently at different positions.
Here’s an experiment that should help convince you that uniform distribution on the diameter does not yield uniform distribution on the circle boundary. Fundamentally, a uniform distribution should treat all arcs of the same length equally: a random point has a $$1/4$$ chance of falling on a quarter-circular arc, regardless of which quarter circle that is.
Now compare two specific quarter-circles and their projections onto the diameter: 1) the quarter-circle adjacent to $$(2,0)$$, i.e. 12 o’clock to 3 o’clock, and 2) the quarter-circle centered around the top-most point $$(1,1)$$, i.e. 10:30 to 1:30.
These both project onto the diameter with no self-overlap (avoiding double counting). But the first one projects down to the interval $$[0,1]$$ which has length $$1$$, and the second one projects down to $$[-\sqrt{2}/2,\sqrt{2}/2]$$, which has length $$\sqrt2$$.
This should indicate to you that the two distributions are not equivalent (though they may intuitively seem so). Consequently, the random process of choosing uniformly from the diameter and projecting onto the circle results in disproportionately favoring the top and bottom sectors of the circle over the left and right.
• Thanks! I really appreciate your words and yes, it seems the problem really deals with something subtle. I'm going to add some considerations in an answer. – Hyz Apr 19 at 19:49
Thanks, Erick Wong for your feedback. After your answer, I calculated the distribution of the arc length subject to the uniform distribution of the point on the diameter. In fact: if we want to express the arc length $$l$$ as a function of $$x$$, $$l = f(x)$$ we obtain:
$$l = \arccos(1-x), x = 1-\cos{l}, |\frac{d}{dl}(x)| = \sin{l}$$
$$l_{pdf} = x_{pdf} \cdot |\frac{d}{dl}x| = \frac{1}{2}\cdot \sin{l}$$. | {
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$$l_{pdf} = x_{pdf} \cdot |\frac{d}{dl}x| = \frac{1}{2}\cdot \sin{l}$$.
So the arc length does not distribute uniformly, we have "lost it", we might say. That's what was wrong. For instance, if the arc length obeys a uniform distribution [0, $$\pi$$], then we can calculate the segment $$s$$ as a function of the arc length:
We know $$l = f(x)$$ and want to know $$s = h(l)$$. If we calculate $$s = g(x)$$ we're done:
From the image on the question I post, $$s = g(x)=\sqrt{2x}$$ (or the opposite segment $$\sqrt{4-2x}$$ ) then $$h = g \circ f^{-1}$$, $$s = \sqrt{2(1-cosl)}$$
$$E(s) = \int_0^\pi{\sqrt{2(1-cosl)}\frac{1}{\pi}dl}=1.273... = \frac{4}{\pi}$$
Also the pdf:
$$s = h(l) = \sqrt{2(1-cosl)} , l=h^{-1}(s)= 2\cdot \arcsin(\frac{s}{2}), |\frac{d}{ds}h^{-1}|=\frac{2}{\sqrt{4-s^2}}$$
$$s_{pdf} = \frac{1}{\pi} \frac{2}{\sqrt{4-s^2}}$$
$$E(s) = \int_0^2{s\cdot \frac{1}{\pi} \frac{2}{\sqrt{4-s^2}}ds} = 1.273... = \frac{4}{\pi}$$
So we're done. Also, the pdf of the segment suggests something related to the Cauchy distribution. Not exactly but certainly it has to do something with it. If we read the description of Cauchy distribution in Wolfram MathWorld:
"The Cauchy distribution, also called the Lorentzian distribution or Lorentz distribution, is a continuous distribution describing resonance behavior. It also describes the distribution of horizontal distances at which a line segment tilted at a random angle cuts the x-axis."
And that's it. A really fascinating problem that introduces some subtle ideas of probability theory. If anyone knows something else please give me feedback. I really think there's a nice connection with Cauchy distribution. | {
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# Does the sequence $x_{n+1}=x_n+x_n^2$ converge to $0$ whenever $-1\lt x_0\lt0$?
My question concerns using the fixed-point iteration to find the fixed point of the function $$f(x)=x+x^2=x(1+x)$$ (this function has a single fixed point at $$0$$).
# The problem
Given some fixed $$x_0$$, define the sequence $$x_n$$ by \begin{align*} x_{n+1}=x_n+x_n^2&&\text{for all n\geq0.} \end{align*} Does this sequence converge to $$0$$ for all $$x_0$$ in the range $$-1\lt x_0\lt0$$? After some computational numerical exploration, I think the answer might be yes but I'm not so sure how to prove this.
## Some notes
The sequence trivially converges to $$0$$ when $$x_0=-1$$ or $$x_0=0$$. It's also fairly easy to prove that it does not converge to $$0$$ whenever $$x_0$$ lies outside this range (i.e., $$x_0\lt-1$$ or $$x_0\gt0$$).
I've already proved that for any $$x_0$$ in this range, all successive $$x_n$$s remain in this range (i.e., $$-1\lt x_n\lt0$$ for all $$n\geq0$$) and that the sequence increases / gets closer to $$0$$ (i.e., $$x_{n+1}\gt x_n$$), but is there any way to prove it actually converges to $$0$$?
## Some examples
If, for example, $$-\frac{1}{2}\lt x_n\lt-\frac{1}{4}$$, we'll have $$-\frac{3}{8}\lt x_{n+1}\lt-\frac{1}{8}$$ and $$-\frac{21}{64}\lt x_{n+2}\lt-\frac{5}{64}$$, etc. So it seems as though the range of possible values gets closer to $$0$$. But, how would one prove this?
You can use the monotone convergence theorem. It tells us that a sequence $$\{x_n\}_{n=1}^\infty\subseteq \mathbb R$$ converges if it is bounded and decreasing. For you sequence we have $$|x_{n+1}| = |x_n| |1+x_n| \leq 1$$ since $$x_0 \in (-1,0)$$. And likewise we have $$x_{n+1} = x_n(1+x_n) \leq x_n$$ as $$1+x_n \in (0,1)$$.
(You can use induction to prove both these claims.)
Now you have shown that $$\lim_{n\to\infty} x_n=x$$ exists. From this we see that $$x = \lim_{n\to\infty} x_{n+1} = \lim_{n\to\infty} x_n(1+x_n) = x(1+x).$$ Hence $$x$$ satisfies $$x = x^2+x$$ so $$x=0$$. | {
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• Great, exactly what I was looking for! Apr 16, 2021 at 7:15
• @ΑΘΩ You're right, I should have said increasing and replaced $\leq$ with $\geq$. Thanks. Apr 16, 2021 at 8:05
• I've already proved that $x_n$ is monotonic increasing and bounded above by $0$ in this case. So this is all I needed to complete the argument (from what I understand, I've never formally studied real analysis, so I certainly may be missing something). Apr 16, 2021 at 8:06
• @user178563 Glad to see that you agree. If we are to be very precise (which we should very well be, for this is what this forum is about) the part where it appears you want to show that $|x_n| \leqslant 1$ by induction on $n$ is also dubious (the induction is not spelled out clearly).
– ΑΘΩ
Apr 16, 2021 at 8:08
The answer to your question (regarding convergence to $$0$$) is indeed affirmative and for a reason that is not too difficult to ascertain. For simplicity let us introduce the quadratic polynomial function: \begin{align*} f \colon \mathbb{R} &\to \mathbb{R} \\ f(x)&=x^2+x \end{align*} and let us note that $$f[(-1, 0)] \subseteq (-1, 0)$$, in other words the interval $$(-1, 0)$$ is stable under $$f$$. To prove this latter claim it suffices to notice that $$x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4} \geqslant \frac{3}{4}>0$$ for any $$x \in \mathbb{R}$$ (and thus in particular $$x^2+x>-1$$ always) and respectively that given $$x \in (-1, 0)$$ we have $$x+1>0$$ yet $$x<0$$ which means that $$x^2+x=x(x+1)<0$$. | {
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The immediate conclusion that we draw from this is that for any fixed $$t \in (-1, 0)$$, the unique recursive sequence $$x \in \mathbb{R}^{\mathbb{N}}$$ defined by $$x_0=t$$ and $$x_{n+1}=f(x_n)$$ for any $$n \in \mathbb{N}$$ has the property that actually $$x \in (-1, 0)^{\mathbb{N}}$$ and is thus a bounded sequence. Furthermore, since $$x_{n+1}=x_n+x_n^2 \geqslant x_n$$ takes place for every $$n \in \mathbb{N}$$ we gather that $$x$$ is increasing (one can actually make the more precise observation that since $$x_n \neq 0$$ for all $$n \in \mathbb{N}$$ under the stated conditions, the inequality mentioned above is actually strict, rendering $$x$$ into a strictly increasing sequence, however that is not essential for the study of convergence).
According to one of the theorems of Weierstrass, any upper-bounded increasing sequence of real numbers is convergent, which applies in particular to $$x$$. Since $$x$$ is defined recursively with respect to the continuous function $$f$$, it follows with necessity that $$\displaystyle\lim_{n \to \infty} x_n$$ must be a fixed point for $$f$$. However it is obvious that there is just one such fixed point - the unique solution to the equation $$f(u)=u \Leftrightarrow u^2=0$$ - which is $$0$$. | {
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# Q: Is 0.9999… repeating really equal to 1?
Mathematician: Usually, in math, there are lots of ways of writing the same thing. For instance:
$\frac{1}{4}$ = $0.25$ = $\frac{1}{\frac{1}{1/4}}$ = $\frac{73}{292}$ = $(\int_{0}^{\infty} \frac{\sin(x)}{\pi x} dx)^2$
As it so happens, 0.9999… repeating is just another way of writing one. A slick way to see this is to use:
$0.9999... = (9*0.9999...) / 9 = ((10-1) 0.9999...) / 9$
$= (10*0.9999... - 0.9999...) / 9$
$= (9.9999... - 0.9999...) / 9$
$= (9 + 0.9999... - 0.9999...) / 9 = 9 / 9 = 1$
One.
Another approach, that makes it a bit clearer what is going on, is to consider limits. Let’s define:
$p_{1} = 0.9$
$p_{2} = 0.99$
$p_{3} = 0.999$
$p_{4} = 0.9999$
and so on.
Now, our number $0.9999...$ is bigger than $p_{n}$ for every n, since our number has an infinite number of 9’s, whereas $p_{n}$ always has a finite number, so we can write:
$p_{n} < 0.9999... \le 1$ for all n.
Taking 1 and subtracting all parts of the equation from it gives:
$1-p_{n} > 1-0.9999... \ge 0$
Then, we observe that:
$1 - p_{n} = 1 - 0.99...999 = 0.00...001 = \frac{1}{10^n}$
and hence
$\frac{1}{10^n} > 1-0.9999... \ge 0$.
But we can make the left hand side into as small a positive number as we like by making n sufficiently large. That implies that 1-0.9999… must be smaller than every positive number. At the same time though, it must also be at least as big as zero, since 0.9999… is clearly not bigger than 1. Hence, the only possibility is that
$1-0.9999... = 0$
and therefore that
$0.9999... = 1$.
What we see here is that 0.9999… is closer to 1 than any real number (since we showed that 1-0.9999… must be smaller than every positive number). This is intuitive given the infinitely repeating 9’s. But since there aren’t any numbers “between” 1 and all the real numbers less than 1, that means that 0.9999… can’t help but being exactly 1. | {
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Update: As one commenter pointed out, I am assuming in this article certain things about 0.9999…. In particular, I am assuming that you already believe that it is a real number (or, if you like, that it has a few properties that we generally assume that numbers have). If you don’t believe this about 0.9999… or would like to see a discussion of these issues, you can check this out.
This entry was posted in -- By the Mathematician, Equations, Math. Bookmark the permalink.
### 94 Responses to Q: Is 0.9999… repeating really equal to 1?
1. mathman says:
lol. You need to know what each “number” is before you can take the difference. Of course the difference is zero, because you say that they are the same before you subtract, assuming you are subtracting “numbers”
2. George says:
While the mathematical argument presented here is correct, I have to say that limit and/or mathematical infinity introduced created confusions for many. What I wanted to say is that math infinity is really a math trick. It cannot and should not be real in the real physical world. Today this infinity plagues physics and its calculations. Maybe we need something new…
3. EMINEM says:
go to this https://www.youtube.com/watch?v=wsOXvQn3JuE site and get a mathematical proof that 0.999… doesnot equal to 1….
4. . says:
EMINEM, the video was published on April fool’s day for a reason!
5. Josh says:
The simplest explanation I’ve seen for this is:
1/3 = 0.333…
(1/3)3=1 = (0.333…)3=0.999…
1 = 0.999…
I don’t see how anyone can come to any conclusion other than 1=0.999… after understanding this.
6. mathman says:
1 / 3 = ??
Use long division. There are 2 parts: the quotient and the remainder. For every iteration.
First iteration is 0 R 1/3
Second is 0.3 R 1/30
Third is 0.33 R 1/300
Etc etc.
Where does your remainder go, Josh?
You don’t account for it anywhere.
The remainder is never zero.
How many 9’s are there in your 0.999… ?
7. Josh says:
Hey mathman, | {
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The remainder is never zero.
How many 9’s are there in your 0.999… ?
7. Josh says:
Hey mathman,
There are infinitely many 9s in my “0.999…”. If it were anything other than infinitely many, then it would be less than 1.
And as far as 1/3, your “etc. etc.” is an infinitely repeating “0.333…”. I’m failing to see how anything you’ve presented has disproved what I said.
8. Angel says:
@mathman:
The remainder does not need to be accounted for. The division 1/3 is never formally complete if there is a remainder. Therefore, one must keep iterating the operations in long division. As the number of iterations n approaches infinity, the remainder does approach zero. That is what matters here. The number of 3s in the decimal expansion of 1/3 is infinite, and so is the number of 9s in the figure 0.999…
9. Anthony.R.Brown says:
The Original author of the Proof below…used in this Thread 🙂
http://www.mathisfunforum.com/viewtopic.php?id=6069
It’s impossible for Infinite/Recurring 0.9 (0.999…) to ever equal 1
Because of the Infinite/Recurring 0.1 (0.001…) Difference! 🙂
Only by the using the + Calculation can it be possible (0.999…) + (0.001…) = 1
Anthony.R.Brown
10. Ángel Méndez Rivera says:
Anthony R. Brown,
Your argument is flawed for the simple reason that the “infinite difference” you describe equivalent to 0.000…1 is not a number. It doesn’t exist in mathematics. It can’t exist in mathematics. It is more of a problematic number than 0/0 or defining a left-identity of exponentiation Phi such that Phi^x=x. It is simply nonsense.
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11. Mathman says:
Anthony,
Being someone who has researched and understands both sides of this issue, you will be told it’s nonsense by mathematicians. It’s their way of putting themselves above you to put you down. The issue with your argument is that 0.001… is defined as 0.00111111111111…. You are having a problem defining what the idea in your head is. 0.000…1 is also not a number. Mathematicians for some reason cannot visualize repeated division the same as they can visualize it in addition. There quick to use the word nonsense, and then in the next breath say that by adding infinite numbers they can have a finite result.
12. Anthony.R.Brown says:
Hi Ángel Méndez Rivera
Your try at a counter argument…saying (0.001…) does not exist in mathematics ? can’t exist in mathematics if (0.999…) exist’s which it does! 🙂
The two are the related from the calculation 1 – (0.999…) = (0.001…) one cannot exist without the other!
Anthony.R.Brown
13. Anthony.R.Brown says:
Anthony,
Being someone who has researched and understands both sides of this issue, you will be told it’s nonsense by mathematicians. It’s their way of putting themselves above you to put you down. The issue with your argument is that 0.001… is defined as 0.00111111111111…. You are having a problem defining what the idea in your head is. 0.000…1 is also not a number. Mathematicians for some reason cannot visualize repeated division the same as they can visualize it in addition. There quick to use the word nonsense, and then in the next breath say that by adding infinite numbers they can have a finite result.
Then when I wanted to reply the message above is not shown ? so I will Answer below…
First of all what you are showing as 0.00111111111111…. is wrong! there is only a single (1) following the Infinite/recurring calculation so your example above should look like 0.00000000000001….
The amount of zeros depends on how far one wants to show the calculation regarding decimal places! | {
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Example 0.9… has the infinite/recurring difference 0.1…
Example 0.99… has the infinite/recurring difference 0.01…
Example 0.99… has the infinite/recurring difference 0.001…
So as you can see the infinite/recurring difference is always the same length as the infinite/recurring 0.9 🙂
Anthony.
14. Mathman says:
0.001….. = 0.0011111111111111111…..
Anthony do you know what you are taking about?
15. Anthony.R.Brown says:
Hi Mathman
Yes I fully know what I am talking about but you appear not to ?
As I pointed out in your example ? = 0.0011111111111111111….. it is wrong there can only be a single (1) otherwise when you do the calculation…
0.0011111111111111111….. + 0.0000000000000000009….. it will equal something like = 0.0011111201111111111… ?
Where as the True calculation (with both sides the 0.9… and the 0.1… the same amount of decimal places)
Calculates 0.9999999999…. + 0.0000000001…. = 1
Regards Anthony.
16. Joshua says:
The flaw in your logic is thinking that there is ever space for a 0.0…01. The number 0.999… repeating forever by definition has no room to add anything to it. The 9’s never stop to allow the insertion of any variation of 0.0001. And if there is no room to ever add anything to 0.999… infinitely repeating, then there is no space in between 0.999… infinitely repeating and 1, and therefore they are equal. If the 9’s stopped repeating at any point, you would be correct, but they do not stop repeating, so you are incorrect.
17. Mathman says:
The ellipsis “…” mean “repeating”. I know what you are trying to say but you are using symbols incorrectly.
1/1000… would be a more accurate to convey your idea.
18. Anthony.R.Brown says:
Hi Joshua
Your Quote:”The number 0.999… repeating forever by definition has no room to add anything to it”
A.R.B
The same as “The number 0.001… repeating forever by definition has no room to add anything to it”
They both run side by side! 🙂 | {
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They both run side by side! 🙂
0.99999999999999999999999999999999999999999999999999999999….
0.0000000000000000000000000000000000000000000000000001….
It’s only when one want’s to make the Value (1) that the + Calculation is needed it’s the only way!
( 0.9999999999….) + (0.0000000001….) = 1
19. Joshua says:
Sorry Anthony, you have not wrapped your head around what 0.999… repeating for infinity means. It’s the infinite part that’s really important. I promise you, you are wrong, and once you understand what infinitely repeating means, you’ll get it.
20. Ángel Méndez Rivera says:
Anthony, no. They don’t run side by side. The number you describe as the difference between 1 and 0.999… doesn’t exist, because by definition, it is impossible to have a 1 after an infinite amount of zeros. It just is. There is no such calculation as the one you performed to be made because the number you describe doesn’t exist.
21. Mathman says:
There is an undefined infinitesimal space between 0.999… and 1. Just because it’s undefined doesn’t mean it doesn’t exist.
22. Ángel says:
Mathman,
1. If it is undefined, then it does not exist. One implies the other.
2. No, there is no infinitesimal space between 0.999… = 1, and this can be proven.
0.999… = 9/10 + 9/100 + 9/1000 + … this is an infinite geometric series, which is evaluated and in fact defined by the formula a/(1 – r), so the sum 0.999.. = (9/10)/(1 – 1/10) = (9/10)/(9/10) = 1. As for the formula a/(1 – r), it can be easily derived and proven
23. Anthony.R.Brown says:
( 1.1 ) x 0.9 = 0.99 ” One Decimal Place = ” 0.01 < 1
( 1.11 ) x 0.9 = 0.999 " Two Decimal Place's = " 0.001 < 1
( 1.111 ) x 0.9 = 0.9999 " Three Decimal Place's = " 0.0001 < 1
( 1 / 0.9 ) x 0.9 = 0.9… " Infinite Decimal Place's = " 0.1… < 1 "
24. Joshua says:
Mathman, | {
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( 1 / 0.9 ) x 0.9 = 0.9… " Infinite Decimal Place's = " 0.1… < 1 "
24. Joshua says:
Mathman,
You would be correct if the repeating 0.999… ever stopped, no matter how long it was. But the number that we are talking about literally never stops, and therefore does not leave any space. There is absolutley no number that would fit in between it and 1.
Most of the confusion over this matter seems to be about the nature of infinity, and how it applies to infinitely repeating decimals. The importance of wrapping our heads around what “infinitely repeating” means cannot be overstated.
25. Ángel Méndez Rivera says:
Anthony,
(1.1)(0.9) = 0.99
(1.11)(0.9) = 0.999
(1.111)(0.9) = 0.9999
(1.111…)(0.9) = 0.999…. = (1/0.9)(0.9), because 1.1111… = 1/0.9. But any number divided by itself equals 1, so (0.9)/(0.9) = 1, therefore 1 = 0.9999….
26. Anthony.R.Brown says:
Quote: “Joshua says:
You would be correct if the repeating 0.999… ever stopped, no matter how long it was. But the number that we are talking about literally never stops, and therefore does not leave any space. There is absolutley no number that would fit in between it and 1.”
A.R.B
Absolutely Correct!!! So now you and I agree it’s impossible for 0.999…to ever stop and become equal to 1 🙂
So how do we make 0.999… = 1 and that’s where we have the answer!…
( 0.999…) + (0.001…) = 1
{When great minds think alike Mountains can be moved!} 🙂
27. Joshua says:
No, Anthony, you still don’t understand. The fact that it never stops is the very thing that makes it equal to 1. Again, I come back to encouraging you to take the time to actually understand what an infinitely repeating decimal is.
28. Anthony.R.Brown says:
Hi Joshua
What you don’t seem to understand is that infinite/recurring 0.999… is a Calculation on a Number! and the number Starts as 0.9 (It’s not something that is just plucked from the sky from nothing!) | {
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Your biggest mistake! is “The fact that it never stops is the very thing that makes it equal to 1” ? a Number that Starts < 1 can never be made to equal 1 unless the + Calculation is used!
29. Joshua says:
Anthony, the number 0.999… with infinite 9s is not a calculation. It is a static number. The 9s are not constantly being generated, they exist (all infinity of them) at once. It does not “start” at 0.9. It exists as 0.999… with an infinite number of 9s. You don’t have to work up to it, it just is.
It is clear that you still do not understand what an infinitely repeating decimal is. You do not understand the basic premise of what we are talking about. Therefore, you can not speak to it in an intelligent or useful manner.
I’m not trying to be mean or anything, either. I appreciate people pointing out when I am wrong so I can learn what is correct. You clearly don’t understand the basics of this topic, therefore you are wrong, and it is an opportunity for you to learn something new and broaden your knowledge base.
30. Angel says:
“What you don’t seem to understand is that recurring 0.999… is a calculation on a number!”
This is wrong. 0.999… is a number. It isn’t a process, it already is a number. It doesn’t “start” out as 0.9. It already is and exists as 0.999…, it doesn’t start anywhere or end anywhere. It isn’t a changing or moving thing. It just is a number.
31. Anthony.R.Brown says:
So Joshua & Angel…
Are both wrong! if they think (the number 0.999… with infinite 9s is not a calculation) ???
Because all the so called Proofs ? are based on some kind’s of Calculations!!!
Just to back that up…a Google search for 0.999… proofs shows (About 367,000 results) 21/09/2017
And they are all some kind’s of Calculations! 🙂
One fine example is shown below…with many Calculations!
Anthony.R.Brown
32. Mathman says:
Ángel,
a/(1-r) comes from a limit. The LIMIT of the infinite series is 1. The definition is flawed and contradicts the very theory of limits. | {
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It may be impossible to have a 1 AFTER an “infinite amount”. That is because “infinite amount” doesn’t exist. Stop trying to quantify infinity. It is not impossible to describe the idea, however: 1/1000…
The very definition of limits show there is ALWAYS AN EPSILON.
n=1: 1-0.9= 0.1
n=2: 1-0.99 = 0.01
For ALL values n, there is a difference, or, contradictory to Anthony, there is always a space. If you travserse 90% of the remaining distance , ad infinitum, there will always be 10% of the last distance ahead of you. You’ll never complete the journey of say, 1 mile, going 90% each time.
I argue that “0.999…” is not a number at all.
Σ (n=1 to N) 9/10^n as N goes to infinity, implies there is always a rational number being added to the previous total. The number 1 has nothing being addd to it. Therefore they aren’t the same.
Also, ‘something’ that always has another addend can’t be a number. It’s contradictory because it always has another number being added to it.
33. Mathman says:
Thus: 1-0.999(n-times) = 1/10^n
You may let n get as arbitrarily large as you like.
You cannot have “infinite amount of 9’s” though.
You can have the LIMIT as the number of 9’s gets arbitrarily large. This is a number. “Endless 9’s” really is nonsense when it comes to a number which should really be finite.
So then, for ALL n in N, that is, for every integer, there is a difference which is smaller than the last and never 0.
Learn what arbitrarily large means, and how limits really work, and what infinite means, and how infinity is applied in calculus. Infinity and limits go hand in hand. You don’t plop infinity wherever you feel.
34. Anthony.R.Brown says: | {
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34. Anthony.R.Brown says:
LIMIT Cannot be used with Infinite/Recurring 0.9 (0.999…) because you will then Contradict the term (Infinite/Recurring) by forcing the .9’s to cut off at some point depending on where you want the LIMIT to be, regarding decimal places calculated,this is much the same as the cut off point used when Infinite/Recurring 0.9 (0.999…) is forced at some point to become equal to 1
So back to…
It’s impossible for Infinite/Recurring 0.9 (0.999…) to ever equal 1 (Unless you use a LIMIT) ? 🙁
And the Infinite/Recurring 0.1 (0.001…) Difference! (Also does not us a LIMIT)
Both are LIMITless 🙂
Only by the using the + Calculation can it be possible for (0.999…) + (0.001…) = 1
35. Ángel Méndez Rivera says:
No, we do not get to choose where the limit is. The limit is a number larger than every element of the sequence of partial evaluations of a function, which means that the limit is larger than the sequence itself. In other words, the term of the sequence which equals the limit is itself located at an infinite position of the sequence. Any finite evaluation yields less than the limit. However, 0.99… has an infinite number of 9s, so it exactly equals the limit. We’re not cutting it off to equal 1, you’re cutting it off to equal less than 1, but you’re Not actually using the sequence of infinite nines, so you’re argument is wrong.
There is no such a thing as limitless in mathematics, and 0.000…1 is nonsense and does not exist.
36. Ángel Méndez Rivera says:
Mathman,
Read my previous response to your argument. I already explained why your description of the definition of a limit is wrong and why the idea of there always being an epsilon is irrelevant when actually evaluating the limit of a function as the argument approaches c.
37. Anthony.R.Brown says:
Quote: “Ángel Méndez Rivera says:
September 22, 2017 at 1:50 pm
No, we do not get to choose where the limit is.
A.R.B
Yes! you do by using it in the first place! | {
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A.R.B
Yes! you do by using it in the first place!
Quote: “Ángel Méndez Rivera says:
“However, 0.99… has an infinite number of 9s, so it exactly equals the limit. ”
A.R.B
How can an infinite number of 9s, exactly equals the limit (You have set!) which is 1 ? which is an Integer number > 0.99…
Quote: “Ángel Méndez Rivera says:
There is no such a thing as limitless in mathematics, and 0.000…1 is nonsense and does not exist.
A.R.B
Infinite/Recurring 0.9 (0.999…)
Infinite/Recurring 0.1 (0.001…)
Are both (Individually!) LIMITless they only have a LIMIT when added! together to equal 1
(0.999…) + (0.001…) = 1
38. The Physicist says:
@Anthony.R.Brown
They’re not using the word “limit” the way any reasonable English speaking person uses the word limit, they mean it in the mathematical sense. You can read about what a mathematical limit is here.
39. Anthony.R.Brown says:
The Physicist says: Quote:”They’re not using the word “limit” the way any reasonable English speaking person uses the word limit, they mean it in the mathematical sense. You can read about what a mathematical limit is here.”
A.R.B
The word (LIMIT) however you define it ? or look it up on some Math sites ? is being used in a way… and the same way ? so as to make a Calculated Number,that starts life less than 1 somehow become equal to 1 ? 🙁
A better way to look at the problem is…
Infinite/Recurring 0.9 (0.999…) = Universe! 🙂
And if we place a Wall at the End of the Universe! ? there will still be Universe! behind the Wall…
0.9 (0.999…) Universe! { Wall = 1} 0.9 (0.999…) Universe!…
40. Ángel Méndez Rivera says:
Anthony, no. There is no such a thing as a limitless number. Please, study the topic you’re talking on before you use terminology. The Physicist already told you where you can study the idea of what a limit is. Take the suggestion and study it.
41. Ángel says:
Mathman, | {
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41. Ángel says:
Mathman,
There is a justified reason why challenging mathematics makes you a crackpot: we can prove that by challenging mathematics, you are wrong.
2. Euler did not define limits.
3. Also, a definition cannot be wrong. This is, ironically, the definition of a definition. So Euler’s definition of infinite summation is acceptable. Euler’s definition renders summation continuous in the entire complex plane, which is what it should be according to group theory.
4. It makes perfect sense to add an infinite number of things. We do it all the tim in physics, and we get observable results that we do observe consistently.
5. No, your term-by-term subtraction was done wrong. Subtraction is done digitwise right to left.
6. 0.999… is not a bad way to represent 9/10 + 9/100 + …, but rather it is the only to represent it since it is the very definition. It is how decimal representation works. It can’t also be represented as Sum[m = 1, m= n —> Infinity](9*10^(-m)), which can be shown to equal 1 since the sum is 0.9(1 – 10^[-n])/(1 – 1/10), and when n —> Infinity, this expression simplifies to 1. This is the limit.
7. Yes, I agree that f(x = c) does not equal lim x—>c [f(x)]. However, 0.999… isn’t equal to f(x=c), but rather equal to lim x—>c [f(x)]. 0.999…. IS the limit of the partial sums by the very definition of decimal representation, and the partial sum limits to 1 evaluated by Taylor-geometric series. By transitive property, 0.999… = 1. | {
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Remember: infinite sums are a thing. Irrational numbers prove it. It has been proven. It isn’t a definition to say “we can add infinitely many things”. No, we can prove it mathematically. That being said, 0.999… = 9/10 + 9/100 + … by definition. Not debatable. Now, 9/10 + 9/100 + … = Sum [m = 1, lim n —> Infinity][9*10^(-m)]. This latter expression is a limit. Keep in mind that summation is a linear hyper-operator, which makes f(y) = Sum[k=x, y][g(k)] continuous for all continuous g. Now, also remember that by the very definition of continuity, lim (x —> c) f(x) = f(c). Hence, Sum [m = 1, n lim n —> Infinity][9*10^(-m)] = Sum[m = 1, Infinity][9*10^(-m)]. This implies that 0.999… is actually a valid sum and therefore a valid decimal representation. You can add infinitely many terms. Finally, the latter left hand side of the previous identity mentioned is equal to lim (n —> Infinity) 9/10[1 – 10^(-n)]/(1 – 1/10) = (9/10)/(9/10) = 1. By the transitive property, 0.999… = 1.
42. Jesse Sherer says:
0.999… is or it isn’t equal to 1. if you are x distant to our Universe you are not in our Universe. no 3 dimentional universe is infinite unless all universe could be infinite too. 1-x=0.9999…. if you are x distant from our universe you are not in our universe. you might say x doesn’t exist, but, nether does 1/9 in base ten math. base ten is a rational universe of numbers there can be no number that equates to almost 1 but not 1 ,,or almost 1 is 1 in a rational universe.. at some point you have to say that if an infinite string of numbers representing both a fraction of a number and a whole number, then there must exist an infinity small number also exists that both equals 0 and almost 0 too. so x is more rational of a number that can add to .999999… and get 1.. base ten can’t solve it but logic can.. 0.9999.. is a number or it isn’t x is a number or it isn’t 1-x=0.9999…. or it doesn’t 1=0.99999… or it doesn’t
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43. Anthony.R.Brown says:
Jesse Sherer says:
December 1, 2017 at 6:41 am
A.R.B
= Nonsense! 🙁
0.9999.. is a number Less than 1 🙂
44. Anthony.R.Brown says:
Ángel Méndez Rivera says:
September 24, 2017 at 5:02 pm
Anthony, no. There is no such a thing as a limitless number. Please, study the topic you’re talking on before you use terminology. The Physicist already told you where you can study the idea of what a limit is. Take the suggestion and study it.
A.R.B
Totally Wrong! Give me any Number you want ? and I will add any Number I want to it! Which makes any Number you provide {a limitless number} 🙂 | {
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# Find the rotation between set of points
I have two sets (sourc and target) of points (x,y) that I would like to align. What I did so far is:
• find the centroid of each set of points
• use the difference between the centroids translations the point in x and y
What I would like is to find the best rotation (in degrees) to align the points.
Any idea?
M code is below (with plots to visualize the changes):
# Raw data
## Source data
sourc = matrix(
c(712,960,968,1200,360,644,84,360), # the data elements
nrow=2, byrow = TRUE)
## Target data
target = matrix(
c(744,996,980,1220,364,644,68,336), # the data elements
nrow=2, byrow = TRUE)
# Get the centroids
sCentroid <- c(mean(sourc[1,]), mean(sourc[2,])) # Source centroid
tCentroid <- c(mean(target[1,]), mean(target[2,])) # Target centroid
# Visualize the points
par(mfrow=c(2,2))
plot(sourc[1,], sourc[2,], col="green", pch=20, main="Raw Data",
lwd=5, xlim=range(sourceX, targetX),
ylim=range(sourceY, targetY))
points(target[1,], target[2,], col="red", pch=20, lwd=5)
points(sCentroid[1], sCentroid[2], col="green", pch=4, lwd=2)
points(tCentroid[1], tCentroid[2], col="red", pch=4, lwd=2)
# Find the translation
translation <- tCentroid - sCentroid
target[1,] <- target[1,] - translation[1]
target[2,] <- target[2,] - translation[2]
# Get the translated centroids
tCentroid <- c(mean(target[1,]), mean(target[2,])) # Target centroid
# Visualize the translation
plot(sourc[1,], sourc[2,], col="green", pch=20, main="After Translation",
lwd=5, xlim=range(sourceX, targetX),
ylim=range(sourceY, targetY))
points(target[1,], target[2,], col="red", pch=20, lwd=5)
points(sCentroid[1], sCentroid[2], col="green", pch=4, lwd=2)
points(tCentroid[1], tCentroid[2], col="red", pch=4, lwd=2) | {
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• I cannot read your code, but the operation you need is called Procrustes rotation. Have you heard of it? It works when points are already paired ($x_i,y_i$). Pre-rotation optional operations include translation and scaling, and optional post-rotational isoscaling. – ttnphns Dec 10 '15 at 17:13
• A complex regression will do the job. – whuber Dec 10 '15 at 17:23
• I've seen, that, rotating the system about 180 degrees, then the pairs $(a,C),(b,D),(c,A),(d,B)$ become neighbours - and this is even a better fit than the best fit of the original $(a,A),(b,B),(c,C),(d,D)$ (where the small letters stand for vector source and capital letters for vector target) I've not seen this possibility mentioned and explicitely allowed or disallowed. Are you sure that you don't want that better fit? – Gottfried Helms Sep 4 '16 at 14:05
This can be done using the Kabsch Algorithm. The algorithm finds the best least-squares estimate for rotation of $RX-Y$ where $R$ is rotation matrix, $X$ and $Y$ are your target and source matrices with 2 rows and n columns.
In [1] it is shown that this problem can be solved using singular value decomposition. The algorithm is as follows:
1. Center the datasets so their centroids are on origin.
2. Compute the "covariance" matrix $C$=$XY^T$.
3. Obtain the Singular Value Decomposition of $C=UDV^T$.
4. Direction adjustment $d=sign(det(C))$.
5. Then the optimal rotation $R=V\left( \begin{array}{ccc} 1 & 0 \\ 0 & d \\ \end{array} \right)U^T$
I don't know of any implementation in R so wrote a small function below.
src <- matrix(c(712,960,968,1200,360,644,84,360), nrow=2, byrow=TRUE)
trg <- matrix(c(744,996,980,1220,364,644,68,336), nrow=2, byrow=TRUE)
Kabsch algorithm in an R funtion:
kabsch2d <- function(Y, X) {
X <- X-rowMeans(X)
Y <- Y-rowMeans(Y)
C <- X %*% t(Y)
SVD <- svd(C)
D <- diag(c(1, sign(det(C))))
t(SVD$v) %*% D %*% t(SVD$u)
}
Center the points:
src <- src-rowMeans(src)
trg <- trg-rowMeans(trg)
Obtain rotation: | {
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Center the points:
src <- src-rowMeans(src)
trg <- trg-rowMeans(trg)
Obtain rotation:
rot <- kabsch2d(src, trg)
Result (black - original source, red - original target, green - rotated target)
plot(t(src), col="black", pch=19)
points(t(trg), col="red", pch=19)
points(t(rot %*% trg), col="green", pch=19)
• +1. However, the answer could be further much better if you included discourse about how the algo is related to well-known Procrustes rotation problem. – ttnphns Mar 19 '16 at 14:03
I've done this with an iterative optimum-search, and tested 2 versions.
I've taken the original arrays and centered them calling this arrays cSRC and cTAR . Then I've done a loop with angles $\varphi$ between $0$ and $2 \pi$ , and for each angle I computed the error-criterion using the difference between the rotated $\small D=rot(\text{cSRC} ,\varphi)- \text{cTAR}$.
1. In version 1) I took as criterion the sum-of-squares of all entries in $\small D$ as $$err_1 = \small \sum_{k=1}^4 \small((D_{k,1})^2+(D_{k,2})^2)$$and the angle $\small \varphi$ at which the minimal error occured is equivalent the kabsch2d-procedure in @Karolis' answer.
2. In version 2) I took as criterion the sum of the absolute distances, that means, the sum $$err_2=\small \sum_{k=1}^4 \small\sqrt{(D_{k,1})^2+(D_{k,2})^2}$$ and got a slightly different rotation angle $\small \varphi$ for the smallest error.
I don't know, which criterion fits your needs better.
Here are some results from the protocol. | {
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I don't know, which criterion fits your needs better.
Here are some results from the protocol.
$$\small \begin{array} {r|cc} & \text{version } 1 & \text{version } 2\\ \hline \varphi & -0.04895304& -0.05093647 \\ \text{rotation} & \begin{bmatrix} 0.99880204& -0.04893349\\ 0.04893349& 0.99880204\\ \end{bmatrix} & \begin{bmatrix} 0.99870302 & -0.05091444\\ 0.05091444 & 0.99870302\\ \end{bmatrix} \\ \text{distances} & \begin{bmatrix} -6.80077266 & -0.86209739\\ 2.79924551 & -9.33782500\\ -0.61309522 & 6.94156520\\ 4.61462237 & 3.25835719\\ \end{bmatrix} & \begin{bmatrix} -6.78017751& -0.37062404 \\ 3.35787307 & -9.36574874 \\ -1.16459115 & 6.95324527 \\ 4.58689559 & 2.78312752 \\ \end{bmatrix} \end{array}$$ | {
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"url": "https://stats.stackexchange.com/questions/186111/find-the-rotation-between-set-of-points/207407"
} |
# the Gaussian integers are isomorphic to $\mathbb{Z}[x]/(x^2+1)$
I am trying to prove that $\mathbb{Z}[i]\cong \mathbb{Z}[x]/(x^2+1)$.
My initial plan was to use the first isomorphism theorem. I showed that there is a map $\phi: \mathbb{Z}[x] \rightarrow \mathbb{Z}[i]$, given by $\phi(f)=f(i).$ This map is onto and homorphic. The part I have a question on is showing that the $ker(\phi) = (x^{2}+1)$.
One containment is trivial, $(x^2+1)\subset ker(\phi)$. To show $ker(\phi)\subset (x^2+1)$, let $f \in ker(\phi)$, then f has either $i$ or $-i$ as a root. Sot $f=g(x-i)(x+i)=g(x^2+1).$ How can I prove that $f \in \mathbb{Z}[x]\rightarrow g \in \mathbb{Z}[x]$?
• It is enough to have $f=g(x^2+1)$, because this says $f\in (x^2+1)$, and we are done. – Dietrich Burde Apr 22 '14 at 13:34
• By the division algorithm, $f = q\cdot (x^2+1) + r$ with $\deg r < 2$. Then you just need to check that $r(i) = 0$ implies $r = 0$ if $\deg r < 2$. (Note: $\deg 0 = -\infty$) You could also appeal to Gauß' lemma. – Daniel Fischer Apr 22 '14 at 13:35
• @DanielFischer, I think that is the answer I am looking for since $g \in \mathbb{R}[x]$ which is a euclidean domain and I am allowed to make that claim. Do you care to expand on that? As in, how can I be certain that the coefficients of $q$ are in $\mathbb{Z}$? – kslote1 Apr 22 '14 at 13:57
• Don't use $\mathbb{R}[x]$ here, use $\mathbb{Q}[x]$. But you need never leave $\mathbb{Z}[x]$ even potentially. The point is that $x^2+1$ is monic, i.e. has lead coefficient $1$. Thus if you do polynomial division, you always get an integer coefficient, and the existence of $q,r\in\mathbb{Z}[x]$ with $f = q\cdot (x^2+1) + r$ and $\deg r < 2$ follows. – Daniel Fischer Apr 22 '14 at 14:06
• As an aside, this is one of those problems where it's easier to show the isomorphism directly, by writing down a homomorphism and its inverse, rather than the indirect route of finding a surjective homomorphism with zero kernel. – Hurkyl Aug 9 '17 at 16:40 | {
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"url": "https://math.stackexchange.com/questions/764437/the-gaussian-integers-are-isomorphic-to-mathbbzx-x21/764482"
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Let me elaborate on Daniel Fischers comment. You have a ring homomorphism $\phi: \mathbb Z[x]\to\mathbb Z[i]$ given by $x\mapsto i$. Take $f \in \ker \phi$. By the division algorithm, $$f = q\cdot(x^2+1) + r,$$ where $\deg r < 2$ and $q,r\in\mathbb Z[x]$, since $x^2+1$ is monic. Applying $\phi$ to this equation yields $$0 = \underbrace{\phi(q)\cdot(i^2+1)}_0 + \phi(r).$$ Since $r$ is of degree $<2$, we can write $r = ax+b$ for some $a,b\in\mathbb Z$. Then $\phi(r)=0$ gives $$ai+b=0.$$ This equation in $\mathbb Z[i]$ implies $a=b=0$, so we have $r=ax+b=0\in\mathbb Z[x]$ and therefore $$f = q\cdot (x^2+1) \in \langle x^2+1\rangle.$$ We conclude that $\ker \phi \subseteq \langle x^2+1\rangle$.
When quotient out an ideal, we consider what happens to the ring when all the elements in the ideal are considered as identity elements.
Now if $x^2+1=0\Rightarrow x=\pm i$ let us take the "+" root.
$\mathbb{Z}[x]/(x^2+1)=\{f\in\mathbb{Z}[x]\,|x^2+1=0\}=\{f\in\mathbb{Z}[x]\,|x=i\}=\{a+bi|a,b\in\mathbb{Z}\}=\mathbb{Z}[i]$
I.e they are isomorphic.
I have answered a similar question here Is this quotient Ring Isomorphic to the Complex Numbers
You can define your isomorphism as follows
$\phi:\mathbb{Z}[i]\rightarrow \mathbb{Z}[x]/\langle x^2+1 \rangle$
$\phi(a+bi)=(a+bx)+\langle x^2+1 \rangle$
If you compute $[a+bx+\langle x^2+1 \rangle]\times [c+dx+\langle x^2+1 \rangle]$
We get $ac+(ad+bc)x+bdx^2+\langle x^2+1 \rangle$. We reduce by long division to get,
$(ac-bd)+(ad+bc)x+\langle x^2+1 \rangle$.
Therefore,
$\phi([a+bi][c+di])=(ac-bd)+(ad+bc)x+\langle x^2+1 \rangle$.
Which is exactly what you want.
The map is obviously bijective and a ring homomorphism. | {
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# What is the remainder of $16!$ is divided by 19?
Can anyone share me the trick to solve this problem using congruence?
Thanks,
Chan
If you've seen Wilson's theorem then you know the remainder when $18!$ is divided by $19$. To get $16!$ mod $19$, then, you can multiply by the multiplicative inverses of $17$ and $18$ mod $19$. Do you know how to find these? (Edit: Referring to both inverses might be a bit misleading, because you really only need to invert their product. See also Bill Dubuque's answer.)
• I knew Wilson's Theorem, but really don't know how to find the inverse of 17, 18 mod 19. What's the inverse? Can you give me one example? Thanks.
– Chan
Feb 26 '11 at 6:17
• @Chan: $\rm 17 \equiv -2,\ 18\equiv -1$ Feb 26 '11 at 6:21
• @Chan: A general technique to find the inverse of $k$ mod $n$ when $k$ and $n$ are relatively prime is to use integer division and the Euclidean algorithm to find $a$ and $b$ such that $ak + bn = 1$. Since $bn$ is a multiple of $n$, $ak$ is congruent to $1$ mod $n$, and thus the congruence class of $a$ is the inverse of the congruence class of $k$. In this particular case, a further hint to make things easier is that $18\equiv -1$ and $17\equiv -2$ mod $19$. (This makes computations easier for inverting their product.) Feb 26 '11 at 6:21
• Many thanks, I got it now ;)
– Chan
Feb 26 '11 at 6:24
Hint By Wilson's theorem $$\bmod 19\!:\ \overbrace{{-}1}^{\large \color{#0a0}{18}} \equiv 18! \equiv\!\! \overbrace{18\cdot17}^{\large \color{#c00}{(-1)(-2)}}\!\!\cdot 16!\,$$ $$\Rightarrow\,16!\equiv \dfrac{\color{#0a0}{18}}{\color{#c00}2}\equiv 9$$
Generally (Wilson reflection formula) $$\rm\displaystyle\ (p\!-\!1\!-\!k)!\equiv\frac{(-1)^{k+1}}{k!}\!\!\!\pmod{\!p},\,$$ $$\rm\:p\:$$ prime | {
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• @Bill Dubuque: Thank you!
– Chan
Feb 26 '11 at 6:24
• @Bill Dubuque: If I have $2.16! \equiv 18 \pmod{19}$. Can I cancel out the $2$ from both sides?
– Chan
Feb 26 '11 at 6:42
• @Chan: $\rm\ 2\:x\equiv 2\:y\ (mod\ 19)\ \iff\ 19\ |\ 2\ (x-y)\ \iff\ 19\ |\ x-y\ \iff\ x\equiv y\ (mod\ 19)$ Feb 26 '11 at 6:50
• @Bill Dubuque: Thank you. So if $gcd(a, p) = 1$, then $ax \equiv ay \pmod{p} \implies x \equiv y \pmod{p}$, right?
– Chan
Feb 26 '11 at 7:07
• @Chan: $\rm\ n,m\$ coprime $\rm\iff a\ n + b\ m = 1\$ for some $\rm\:a,b\:\ \iff\ a\ n\equiv 1\ (mod\ m)\:,\:$ for some $\rm\:a\:$ $\rm\iff\ n$ is invertible/cancellable $\rm\:(mod\ m)$ Feb 26 '11 at 7:12
I almost think in this case it is just faster to break it down than calculate the inverses: First the factors between 1 and 10:
$$10!= (2\cdot 10)\cdot (4\cdot5)\cdot(3\cdot6)\cdot(7\cdot8)\cdot 9\equiv 1\cdot 1\cdot (-1)\cdot(-1)\cdot 9\equiv9$$
Now we have
$$16!\equiv 9\cdot 11\cdot 12\cdot 13\cdot 14\cdot 15\cdot 16\equiv9\cdot(-8)\cdot(-7)\cdot(-6)\cdot(-5)\cdot(-4)\cdot(-3)$$
$$\equiv 9\cdot(4\cdot5)\cdot(6\cdot3)\cdot(7\cdot8)\equiv 9\cdot(1)\cdot(-1)\cdot(-1)\equiv 9$$
Of course this doesn't generalize, but the computation is faster than finding 3 inverses and multiplying them. (Again of course that is not true for larger $n$...)
A more explicit answer would go like this:
By Wilson's Theorem, $18!$ is congruent to $-1 \pmod {19}$
$$18! = (18\cdot 17)(16!)$$
Then $(18\cdot 17)(16!)$ is congruent to $-1 \pmod{19}$
But note that $18$ is congruent to $-1 \pmod{19}$ and $17$ is congruent to $-2 \pmod{19}$
Then it follows that $(18\cdot 17)(16!)$ is congruent to $((-1)\cdot (-2))(16!)$ is congruent to $-1 \pmod{19}$
Simplifying, we get, $2\cdot(16!)$ is congruent to $-1 \pmod{19}$. But we need the remainder of $16!$, not $2\cdot 16!$. and $-\frac12$ isn't a valid answer. | {
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Then also notice that $18$ is congruent to $-1 \pmod{19}$. Then $2\cdot 16!$ is congruent to $18 \pmod{19}$ by the fact that if $a$ is congruent to $b \pmod{p}$, then $b$ is congruent to $a \pmod{p}$
So we have that $2\cdot 16!$ is congruent to $18 \pmod{19}$ and from there, its just a matter of dividing both sides by $2$ to get $16!$ is congruent to $9 \pmod{19}$ | {
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