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http://math.stackexchange.com/questions/245161/probability-over-time | Probability over time
So given say, a deck of cards, I draw cards until I pick out a specific one, in this case, the ace of spades. Once the ace of spades has been drawn, you restart. Now, this continues till infinity. On average, how many tries does it take to get the card? Is there a solution to a deck of size $n$?
-
Drawing cards serially
Assuming a 52-card deck, shuffled completely randomly, the Ace of Spades has a 1/52 chance of occupying any position in the deck.
Let $X$ denote the position of the ace of spades. Then, $P(X = x) = 1/52$ for all $x = 1, 2, \ldots, 52$.
The average value of $X$ is given by the formula for expectation of a discrete random variable:
$$E[X] = \sum_{i=1}^{52} i P(X = i) = \frac{1}{52} \sum_{i=1}^{52} i = \frac{52\cdot (52+1)}{52\cdot 2} = \frac{53}{2}$$
This generalizes to $n$ by replacing 52 with $n$.
Drawing cards randomly
The expected value formula is the same, but we modify $P(X=x)$ to reflect the random drawing of cards.
It is easy to see that the probability of drawing the Ace of Spades on the $i$th try is the same as the probability of $i-1$ failures followed by a single success.
The probability of $j-1$ failures, without replacement, can be written
$$\frac{51}{52} \frac{50}{51} \cdots \frac{52-(j-1)-1}{52-(j-1)}.$$
It should be clear that diagonal terms cancel, leaving
$$\frac{52-j}{52}.$$
However, do note that because of the indexing, we have $j = i-1$. That is, if we pull the Ace of Spades on the first card ($i=1$), we have zero preceding failures.
After drawing $i-1$ cards, there are $52-(i-1)$ cards remaining in the deck, so the probability of success on the ith try is $1/(53-i)$.
Multiplying these together, one gets
$$P(X=i) = \frac{53-i}{52} \frac{1}{53-i} = \frac{1}{52}.$$
Hopefully it is more clear now that the result is the same as before.
-
I think I explained the problem wrong, cards are selected without replacement. – dzk87 Nov 26 '12 at 20:13
Ok, but you say "this continues till infinity". I interpret that to mean "pick a card, stop when you get the ace of spades, write down the ace of spades position, shuffle, and repeat." If that is the case, then I am correct, because every time you re-shuffle the deck, the ace has a 1/52 chance of being in any particular spot. – Emily Nov 26 '12 at 20:16
Just to make sure, what I mean is that you take the deck, remove cards till you find the ace of spaces, mark position, return all the cards, and repeat, forever. – dzk87 Nov 26 '12 at 20:18
Right. I am assuming you are removing the cards serially (e.g. top card first). In this case, the answer is correct, and in fact intuitive: the Ace of Spades has an equal probability to be in any position in the deck. Since a 52 card deck has an even number of cards, it's "average" position is somewhere between the last card of the first half-deck, and the first card of the second half-deck -- 26.5 = 53/2. – Emily Nov 26 '12 at 20:21
Original poster here again, I registered this time, not sure how to mark your answer as correct... – dzk87 Nov 26 '12 at 21:03
This is well discussed in these forums itself, I've summarized it here. Hope that helps
- | 2016-06-26T08:29:50 | {
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https://math.stackexchange.com/questions/2713603/what-is-the-angle-of-angle-bpc-in-triangle-bpc | # What is the angle of $\angle BPC$ in $\triangle BPC$
In $\triangle ABC$, the internal bisector of $\angle ABC$ and the external bisector of $\angle ACB$ meet at $P$. If $\angle BAC = 40^\circ$ what is the measure of $\angle BPC$?
My try: i) Sum of angles of a triangle is $180^\circ$.
ii) Vertical opposite angles are equal.
We need to find $\angle BPC$. By i) we know $\angle BPC = 180^\circ - \angle PCA - \angle PKC$. So the line pass through points $P$ and $C$ is perpendicular to internal bisector of $\angle ACB$.
• what have you done, other than reproducing the sketch of the figure you were provided for the homework problem? – Namaste Mar 29 '18 at 17:43
• what kind or triangle is this? – Dr. Sonnhard Graubner Mar 29 '18 at 17:46
• Please read this answer addressing one way to ask a good question, when you don't "have a clue" how to proceed. – Namaste Mar 29 '18 at 17:48
• @MikeCocais to solve you only need to use $\sum$ angles = 180° applied to triangles ABC and PBC – gimusi Mar 29 '18 at 18:05
• @MikeCocais Please remember that you can choose an answer among the given if the OP is solved, more details here meta.stackexchange.com/questions/5234/… – gimusi Apr 1 '18 at 8:45
Extend $BC$ to $D$, so that $\angle ACD$ is an exterior angle of the triangle. Thus, $$\angle PCD = \angle ACD/2= 90-C/2$$
Using the exterior angle sum property in $\Delta PBC$, $$\angle BPC+\angle PBC = \angle PCD$$ $$\angle BPC+B/2=90-C/2$$ $$\angle BPC=90-(B+C)/2$$ $$\angle BPC=90-(180-A)/2=A/2=20^{\circ}$$
• @amWhy but then, $B4 and$C$could be confused with the angles of$\Delta ABC$. I've edited the post though – Prathyush Poduval Mar 29 '18 at 18:12 • I've deleted my comment, given your edit. I was showing how you need to be consistent. None of the angles in this case should be represented by one letter; So using$\angle$and three letters is consistent. – Namaste Mar 29 '18 at 18:15 In$\Delta{BPC}$,$\widehat{BCP}+\widehat{BPC}+\widehat{PBC}=180^\circ$, so:$\widehat{BPC}=180^\circ-(\widehat{BCP}+\widehat{PBC})=180^\circ-(\widehat{BCA}+\frac{180-\widehat{BCA}}{2}+\frac{\widehat{ABC}}{2})=180^\circ-\frac{2\widehat{BCA}+180^\circ-\widehat{BCA}+\widehat{ABC}}{2}=180^\circ-\frac{180^\circ+\widehat{BCA}+\widehat{ABC}}{2}=180^\circ-\frac{180^\circ+180^\circ-\widehat{BAC}}{2}=180^\circ-\frac{180^\circ+180^\circ-40^\circ}{2}=20^\circ$• How you wrote$\widehat{BCP}$=$\frac{180-\widehat{BCA}}{2}$(or)$\widehat{PBC}$=$\frac{\widehat{ABC}}{2}$? It doesn't make any sense? – Mike Cocais Mar 29 '18 at 17:58 •$\widehat{BCP}=\widehat{BCA}+\widehat{ACP}$and$\widehat{PBC}=\frac{\widehat{ABC}}{2}$– user061703 Mar 29 '18 at 18:00 •$\widehat{ACP}=\frac{180^\circ-\widehat{BCA}}{2}$, it is easier to understand if you extend$BC$to the right, make it a ray$Bx$. – user061703 Mar 29 '18 at 18:01 HINT Let us indicate with$b$the angle in$B$and with$c$the angle in$C$for$\triangle ABC$. Then •$b+c+40=180 \implies c=140°-b$and •$\angle PBC = b/2$•$\angle PCB = c+(180°-c)/2=90°+c/2=160°-b/2$•$\angle BPC=180°-\angle PBC -\angle PCB=180°-b/2-160+b/2=20°$Let$\alpha=\measuredangle ABP=\measuredangle CBP. Then: \begin{align}&\measuredangle ACB=180-40-2\alpha=140-2\alpha \\ &\measuredangle ACP=\frac12(180-\measuredangle ACB)=\frac12(180-(140-2\alpha))=20+2\alpha.\end{align} Now the sum of angles of triangleBCP\$: $$\measuredangle BPC+\underbrace{\alpha}_{\measuredangle CBP}+\underbrace{140-2\alpha}_{\measuredangle ACB}+\underbrace{20+\alpha}_{\measuredangle ACP}=180 \Rightarrow \measuredangle BCP=20.$$ | 2019-08-22T20:09:32 | {
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http://mathhelpforum.com/calculus/171573-relative-rates.html | # Math Help - Relative Rates
1. ## Relative Rates
I've added an attachment to this post to explain the problem.
Two carts, A and B, are connected by a rope 39 feet long that passes over a pulley P. The point Q is on the floor h = 12 ft directly beneath P and between the carts. Cart A is being pulled away at a speed of 2.5 ft/s. How fast is cart B moving toward Q at the instant when cart A is 5 ft from Q. (round to 2 decimal places)
From the problem, I know the following things:
PQ = h = 12
AQ = 5
AP = 13 (by Pythagoras)
APB = 39
PB = 39-13 = 26
QB = $2\sqrt{133}$ (by Pythagoras)
(work to follow in next posting for organization)
2. My strategy is to work clockwise around the triangle to find $\frac{dQB}{dt}$ as follows:
First, find $\frac{dAP}{dt}$ since this must be equal to $\frac{dPB}{dt}$ since the length of these two segments is constant:
$\frac{dAP}{dt}(AP)^2 = \frac{dAQ}{dt}(AQ)^2$ (note that PQ is a fixed length so its derivative is zero)
$\frac{dAP}{dt} = \frac{AQ}{AP} \frac{dAQ}{dt} = \frac{5}{13} (2.5) = \frac{25}{26} ft/sec$
Again, if $\frac{dAP}{dt} = \frac {25}{26}$, then $\frac{dPB}{dt} = \frac{-25}{26}$ since the length of this segment is constant. THIS IS A KEY ASSUMPTION! IS IT CORRECT???? (sorry for the caps)
Then, I find my ultimate answer using Pythagoras:
$\frac{dQB}{dt}(2\sqrt{133})^2 = \frac{dPB}{dt} (PB)^2 - \frac{dPQ}{dt} (PQ)^2$
$(4\sqrt{133}) \frac{dQB}{dt} = \frac{-25}{26} (26^2) - 0$
$\frac{dQB}{dt} = \frac{-650}{4\sqrt{133}} = 0.30545$ feet / second
Unfortunately, this answer doesn't check out, so I'm doing something wrong, probably in my key assumption, but I don't see it. Can somebody help?
Thanks.
3. Hello, joatmon!
Two carts, A and B, are connected by a rope 39 feet long that passes over a pulley P.
The point Q is on the floor h = 12 ft directly beneath P and between the carts.
Cart A is being pulled away at a speed of 2.5 ft/s.
How fast is cart B moving toward Q at the instant when cart A is 5 ft from Q.
(Round to 2 decimal places)
Code:
P
*
/:\
/ : \
/ : \
/ : \
/ :12 \
/ : \
/ : \
A * - - - * - - - * B
x Q y
Let $x = AQ,\;y = QB$
We are given: . $AP + PB \,=\,39\,\text{ and }\,\dfrac{dx}{dt} = 2.5$
In right triangle $PQA\!:\;AP \,=\,\sqrt{x^2+12^2}$
. . Then: . $PB \,=\,39 - \sqrt{x^2+144}$
In right triangle $PQB\!:\;y^2 \:=\:(39 - \sqrt{x^2+144})^2 - 12^2$
. . . . . . $y^2 \;=\;x^2 - 78(x^2+144)^{\frac{1}{2}} + 1521$ .[1]
Differentiate implicitly:
. . $2y\,\dfrac{dy}{dt} \;=\;\bigg[2x - 78\cdot\frac{1}{2}(x^2+144)^{-\frac{1}{2}}\cdot2x\bigg]\,\dfrac{dx}{dt}$
. . . . $\displaystyle \frac{dy}{dt} \;=\;\frac{1}{y}\bigg[x - \frac{39x}{\sqrt{x^2+144}}\bigg]\,\frac{dx}{dt}$ .[2]
Substitute $x = 5$ into [1]:
. . $y^2 \;=\;25 - 78(13) + 1521 \:=\:532 \quad\Rightarrow\quad y \:=\:2\sqrt{133}$
Substitute into [2]:
. . $\displaystyle \frac{dy}{dt} \;=\;\frac{1}{2\sqrt{133}}\left[5 - \frac{39(5)}{\sqrt{169}}\right](2.5) \;=\;\frac{1}{2\sqrt{133}}(-10)(2.5)$
$\displaystyle\text{Therefore: }\frac{dy}{dt} \;=\;-\frac{25}{2\sqrt{133}} \;\approx\;-1.08\text{ ft/sec}$
But check my work . . . please!
.
4. Originally Posted by joatmon
My strategy is to work clockwise around the triangle to find $\frac{dQB}{dt}$ as follows:
First, find $\frac{dAP}{dt}$ since this must be equal to $\frac{dPB}{dt}$ since the length of these two segments is constant:
$\frac{dAP}{dt}(AP)^2 = \frac{dAQ}{dt}(AQ)^2$ (note that PQ is a fixed length so its derivative is zero)
$\frac{dAP}{dt} = \frac{AQ}{AP} \frac{dAQ}{dt} = \frac{5}{13} (2.5) = \frac{25}{26} ft/sec$
Again, if $\frac{dAP}{dt} = \frac {25}{26}$, then $\frac{dPB}{dt} = \frac{-25}{26}$ since the length of this segment is constant. THIS IS A KEY ASSUMPTION! IS IT CORRECT???? (sorry for the caps)
Yes, that is correct. AP+ PB= 39 so (AP)'+ (PB)'= 0, (BP)'= -(AP)'.
Then, I find my ultimate answer using Pythagoras:
$\frac{dQB}{dt}(2\sqrt{133})^2 = \frac{dPB}{dt} (PB)^2 - \frac{dPQ}{dt} (PQ)^2$
I don't understand why you have the squares there nor where you got $2\sqrt{33}$. What formula did you differentiate to get this?
$(4\sqrt{133}) \frac{dQB}{dt} = \frac{-25}{26} (26^2) - 0$
$\frac{dQB}{dt} = \frac{-650}{4\sqrt{133}} = 0.30545$ feet / second
Unfortunately, this answer doesn't check out, so I'm doing something wrong, probably in my key assumption, but I don't see it. Can somebody help?
Thanks.
5. Thanks to both of you. Soroban, your work helped me immensely. We were together in calculating the segment lengths. Where I went wrong was in applying the differential equation. I didn't equate x and y correctly, so my differential was all screwed up. Thanks for setting me straight.
Halls, the $2\sqrt{133}$ comes from the Pythagoream theorem. We determined that the PB hypotenuse was 26 and the height of 12 was given. Thus, the segment at the bottom that needed to be measured in order to differentiate was calculated like this:
$\sqrt{26^2 - 12^2} = 2\sqrt{133}$
Thanks again!
6. I originally thought the sum of the bases of the triangle (x and y in your diagram) would have to be constant by thinking that in real life if the base was not constant and say it was getting shorter that would require there to be some slack in the rope. But it seems that x+y is not constant here. I guess that's because our assumption is that this problem can be modeled with a triangle no matter what. | 2014-04-18T05:43:58 | {
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https://math.stackexchange.com/questions/367583/example-of-partial-order-thats-not-a-total-order-and-why | # Example of Partial Order that's not a Total Order and why?
I'm looking for a simple example of a partial order which is not a total order so that I can grasp the concept and the difference between the two.
An explanation of why the example is a partial order but not a total order would also be greatly appreciated.
Think about the subsets of $\{0,1\}$. They are: $\emptyset, \{0\}, \{1\}$, and $\{0,1\}$. Now, we can make these subsets into a partial order with $\subset$. For instance, $\emptyset \subset \{0\}$ and $\{1\} \subset \{0,1\}$. You can show this satisfies the axioms for a partial order:
$A \subset A \\ A \subset B, B \subset C \Rightarrow A \subset C \\ A \subset B, B \subset A \Rightarrow A = B$
But a total order $<$ drops the first axiom above and replaces it with the following:
$x < y$ or $y < x$ for all $x,y$
And we see that our example of subsets of $\{0,1\}$ does not satisfy this. For instance, neither $\{0\} \subset \{1\}$ nor $\{1\} \subset \{0\}$ are true. In a total order, we want to be able to compare any two elements. In a partial order, we don't.
• Roughly speaking, a partial order doesn't require all elements of a set to be "comparable" in the way you described comparison. – nbro Dec 8 '17 at 13:06
Take your favourite set, which is $\,X:=\{a,b\}\,$ and then its power set
$$P(X):=\left\{\emptyset\,,\,X\,,\,\{a\}\,,\,\{b\}\right\}$$
Partial order $\,P(X)\,$ by set inclusion: $\,A\le B\iff A\subset B\;,\;\;A,B\in P(X)\,$
Check the above is a partial not total order.
• My favourite set is $\varnothing$. It represents the emptiness of existence! :-) – Asaf Karagila Apr 20 '13 at 19:39
• @AsafKaragila Do you mean existence of emptiness? – Git Gud Apr 20 '13 at 19:40
• @GitGud: No.${}$ – Asaf Karagila Apr 20 '13 at 19:41
• @Peter: While your point is valid, my comment was merely a result of recent semi-off topic comment exchanges with DonAntonio; and a remark that I usually don't like it when people tell me what is my favourite set. But it was meant as a tongue-in-cheek sort of comment... – Asaf Karagila Apr 20 '13 at 19:46
• @DonAntonio Pero estás diciendo: "Elige tu conjunto favorito, que es..." Quiza "Elige tu conjunto favorito, el mio es..." tendría más sentido! +1 igualmente. – Pedro Tamaroff Apr 21 '13 at 17:11
There are some small differences in the way people define order (partial or total). Roughly speaking, they correspond to the difference between $\lt$ and $\le$. We opt for the $\le$ version. You can undoubtedly adapt the example below to the other version, if that's the one being used in your course.
Let our set be $\{1,2,3,6\}$. If $x$ and $y$ are elements of this set, we will say that $x\le y$ if $x$ divides $y$. So for example $2\le 6$, and $3\le 3$.
Note that it is not true that $2\le 3$, since $2$ does not divide $3$. Also, it is not true that $3\le 2$. The two objects $2$ and $3$ are incomparable with respect to the order just defined.
In a total order $\le$, any two objects $x$ and $y$ are comparable. Either $x\le y$ or $y\le x$ or both. ("Both" happens when $x=y$.)
For a non-mathematical example, let $A$ be the set of all people. If $x$ and $y$ are people, write $x\le y$ if $x=y$ or $x$ is an ancestor of $y$. This is a partial order. However, it is not total, since for example Obama is not an ancestor of Putin, and Putin is not an ancestor of Obama.
• if in the same example, I take my set to be {2,3,7,11} then is it partially ordered? I see, no element is related to any other under this relation. – Aditi Narware Jan 8 '17 at 11:00
A total order is a partial order, but a partial order isn't necessarily a total order.
A totally ordered set requires that every element in the set is comparable: i.e. totality: it is always the case that for any two elements $a, b$ in a totally ordered set, $a \leq b$ or $b\leq a$, or both, e.g., when $a = b$. Here, as is often the case $\leq$ is used to represent some ordering relation.
So, for example, $R = \{(a, a), (b, b), (c, c), (d, d)\}$ is trivially, a partial order on $S = \{a, b, c, d\}$. But it is not the case that it is a total order, since we do not have that for every pair of elements in $S$, $(a, b)$ or $(b, a) \in R$.
Let's keep things simple. Ancestry is a partial order:
• Transitive: An ancestor to your ancestor is also an ancestors to you.
• Anti-symmetric: You can't be an ancestor of one of your own ancestors.
• Partial: Not everyone is either a descendant or ancestor of yours.
• Specifically, it's a strict partial order, since you don't count yourself among your ancestors. – stewSquared May 10 '19 at 16:42
Consider vectors in $\mathbb{R}^n$ partially ordered as follows: $x\succeq y$ iff $x_i\geq y_i$ for each $i=1,2,...n$. For instance, for $n=2$, $(2,2)\succeq(1,0)$ but $(10,1)\not\succeq(2,0)$ and $(2,0)\not\succeq(10,1)$.
I like this example.
The usual $\le$ relation on $\mathbb{N}$ can be defined by
$$a\le b\text{ if and only if there exists x\in\mathbb{N} such that b=a+x}.$$
This relation is well known to be a total order.
A very similar relation can be defined using multiplication:
$$a\mid b\text{ if and only if there exists x\in\mathbb{N} such that b=ax}.$$
Also $\mid$ is an order relation.
1. reflexivity is obvious: $a=a\cdot 1$, so $a\mid a$;
2. also transitivity is easy: if $a\mid b$ and $b\mid c$, then $b=ax$ and $c=by$ for some $x$ and $y$; therefore $c=a(xy)$ and $a\mid c$;
3. antisymmetry is a bit more difficult, but not so much:
Assume $a\mid b$ and $b\mid a$; then we have $b=ax$ and $a=by$ for some $x$ and $y$. Substituting yields
$$a = axy$$
and, if $a\ne0$, by cancellation we get $1=xy$ from which we derive $x=y=1$ and $a=b$. If instead $a=0$, from $b=ax$ we get again $a=b$.
The relation $\mid$ is not a total order, because $2\nmid 3$ and $3\nmid 2$.
With respect to $\mid$, $1$ is the minimum, because $1\mid b$ for any $b$. The maximum is $0$, because $a\mid 0$ for all $a$, since $0=a\cdot 0$.
The most important property of $\mid$ is that $(\mathbb{N},\mid)$ is a lattice: the infimum and supremum of $\{a,b\}$ are respectively the greatest common divisor and the least common multiple of $a$ and $b$.
• Why do you use "if and only if" when defining a new binary relation? Shouldn't it be "if", in the "definition sense"? – Maxis Jaisi Jul 31 '17 at 7:52
• @MaxisJaisi Halmos devised ”iff“ for this case, but then it was abused. Lack of previous conventions, contrary to what happens in courses or textbooks, made me (over)careful. – egreg Jul 31 '17 at 8:28
• But your "if and only if" is distinct from the logical "if and only if", right? See math.stackexchange.com/questions/566565/… – Maxis Jaisi Jul 31 '17 at 9:18
• @MaxisJaisi I don't understand what the problem is. I'm obviously doing definitions, so this is certainly not a “logical symbol”, but a “metalogical” one. How do you call it is a matter of conventions. – egreg Jul 31 '17 at 9:21
A simple example is a set with four elements $S = \{a, b, c, d\}$. We'll define a partial order so that $a$ is the smallest element, $d$ is the largest element, and $b$ and $c$ are intermediate elements that are incomparable with each other. The relation $R \subset S \times S$, where $(x,y) \in R \Leftrightarrow x \leq y$ is given by $$R = \{(a,a), (a,b), (a,c), (a,d), (b,b), (b,d), (c,c), (c,d), (d,d)\}.$$ I'll leave it for you to check that this is really is a partial order. The important thing to note is that neither $b \leq c$ nor $c \leq b$ is true, so $R$ is not a total order.
EDIT: As A. Rex astutely points out in the comments, the simplest example would be to take just $S = \{b,c\}$ with the partial order relation $R = \{(b,b), (c,c)\}$. Then $b$ and $c$ are incomparable, so $R$ is not a total order.
• Why do you need $a$ and $d$? Why not just have $b$ and $c$ be incomparable? – A. Rex Apr 23 '13 at 20:44
• Fair point! Especially since I said I was giving the "simplest" example. :) – Michael Joyce Apr 23 '13 at 21:33 | 2020-02-23T11:44:00 | {
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https://cs.stackexchange.com/questions/145207/finding-the-uniqueness-of-longest-increasing-subsequence | # Finding the uniqueness of longest increasing subsequence
I have a problem related to a common dynamic programming problem LIS. I got LIS function that takes arr as an input and returns the length of the longest increasing subsequence.
def LIS(arr):
n = len(arr)
# initialize LIS values for all indexes
lis = [1]*n
# Compute optimized LIS values in bottom up manner
for i in range(1, n):
for j in range(0, i):
if arr[i] > arr[j]:
if lis[i] < lis[j] + 1:
lis[i] = lis[j]+1
# return the maximum of all LIS
return max(lis)
Now my problem is to modify the given LIS function to check whether the longest increasing subsequence is unique. For example: arr = [2,6,5,7,9,1], it has LIS 4 and it has no unique solution because LIS could be {2,6,7,9} or {2,5,7,9}.
Here is my attempt to this question. Inside the main condition check in the above function, I checked if there are two or more values in lis which are less than the current pointer, then it is not unique. This worked for some examples but didn't work for all. So, I was wondering what is the actual condition to check the uniqueness in this situation? Any hint would be appreciated.
if arr[i] > arr[j]:
if lis[i] < lis[j] + 1:
lis[i] = lis[j]+1
if lis[j] == lis[j-1]:
unique = false
What you did is, eh, checking whether there is a unique increasing subsequence of some kind, not necessary of the longest length. For example, your algorithm will claim non-uniqueness when it is checking the element $$9$$ in the array $$[8, 7, 9, 1, 2, 3]$$. However, the unique LIS is $$[1, 2, 3]$$.
The goal is whether the longest increasing subsequence (LIS) is unique. For each current element during the loop, we can keep track of whether there is only one possible LIS that ends at the the current element. Then we will check whether the LIS (of the whole array) ends at only one index and whether the LIS ending there is unique.
def is_lis_unique(arr):
arr.append(float('infinity')) # the LIS must end at this last number.
n = len(arr)
# Length is 1 for all subsequences of single element.
lis = [1] * n
# unique_lis[i] will tell whether the LIS that ends at index i is unique.
unique_lis = [True] * n
# Compute LIS values in bottom up manner
for i in range(1, n):
for j in range(0, i):
if arr[i] > arr[j]:
if lis[i] < lis[j] + 1:
lis[i] = lis[j] + 1
unique_lis[i] = unique_lis[j]
elif lis[i] == lis[j] + 1:
# Another LIS has been found!
unique_lis[i] = False
return unique_lis[-1]
print(is_lis_unique(([3, 4, 5, 1, 2]) # True
print(is_lis_unique([8, 7, 9, 1, 2, 3] # True
print(is_lis_unique([8, 7, 9, 1, 2, 4, 3])) # False
The code above is a Pythonic implementation. A "number" that is greater than all numbers is appended to the given array so that the LIS must end at that number. This is a trick to simplify the code.
Exercise: Given an array of integers of length $$n$$, compute the number of increasing subsequences of the longest length efficiently. | 2022-07-06T09:47:04 | {
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https://math.stackexchange.com/questions/1946004/condition-probabilities-with-three-events/1946030 | # Condition probabilities with three events
The events A, B, C have the probabilities: P(A|B)=0.25, P(C|B)=0.5, P(A∩C|B)=0.10. Given that B has happened, find the following probabilities:
a) That only C has happened 
b) That only C or only A has happened, but not both of them
c) That C or A has happened
I have answered the three questions, but want to know if the logic is correct (A' = A compliment):
a) P(A'∩C|B) = P(C|B) - P(A∩C|B) = 0.5 - 0.1 = 0.4
On the RHS I simply take the probability that C happened given B, and subtract from it the probability that A and C happened. The LHS is the only way I can find of representing only C without including C∩A, though I'm not sure if this is the correct way of representing that.
b) P(C∪A|B) = P(C|B) + P(A|B) - 2*P(C∩A|B) = 0.5 + 0.25 - 0.2 = 0.55
Addition rule of probabilities. Subtract twice the intersection, once for repeat, and second because we don't want to include it at all.
c) P(C∪A|B) = P(C|B) + P(A|B) - P(C∩A|B) = 0.5 + 0.25 - 0.1 = 0.65
Same as above, except only subtract intersection once.
Is my logic above correct?
• a) only C has happened among the three or among A and C only? – msm Sep 29 '16 at 4:16
• I'm not sure, since the question does not clarify. But does it really matter? If we know B happened, then we're inside the sample space of B already, so technically speaking wouldn't it be among the three anyways? – Leo Denni Sep 29 '16 at 4:24
• Formatting tips here. – Em. Sep 29 '16 at 4:31
a) if by "only C has happened" it means among the two, then you are correct. However, if it means among the three, then the probability is zero $$P(A'\cap B' \cap C|B)=0$$
b) \begin{align}P(\text{only A or C}|B)&=P((A\cap C')\cup(A'\cap C)|B)\\ &=P(A\cap C'|B)+P(A'\cap C|B)-P((A\cap C')\cap(A'\cap C)|B)\\ &=P(A|B)-P(A\cap C|B)+P(C|B)-P(A\cap C|B)-0 \end{align}
• Yes but you have written $P(C∪A|B)$ at the beginning (which is actually the same as c) and is wrong. – msm Sep 29 '16 at 4:34 | 2020-04-07T03:28:49 | {
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https://www.jiskha.com/questions/1805641/a-wire-to-be-attached-to-support-a-telephone-pole-because-of-surrounding-buildings | algebra
A wire to be attached to support a telephone pole. Because of surrounding buildings, sidewalks and roadways, the wire must be anchored exactly 21 feet from the base of the pole. Telephone company workers have only 28 feet of cable, and 2 feet of that must be attached to the cable to the pole and to the stake on the ground. How high from the base of the pole can the wire be attached.
The wire can be attached at the height of __ feet from the base of the poles.( simplify your answer)
The wire can be attached at the height of approximately ___ feet from the base of the pole. (Round to two decimal places as needed)
1. 👍
2. 👎
3. 👁
1. Use Pythagoras's Theorem.
a^2 + b^2 = c^2
21^2 + b^2 = 26^2
1. 👍
2. 👎
👤
Ms. Sue
2. So the support wire exposed can only be 26 ft
Clear case of Pythagoras
let the height be h, then
h^2 + 21^2 = 26^2
solve for h
1. 👍
2. 👎
👤
Reiny
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Q: A 4m flag pole is not standing up straight. There is a wire attached to the top of the pole and anchored in the ground. The wire is 4.17m long. The wire makes a 68 degree angle with the ground. What angle does the flag pole | 2021-07-28T01:39:18 | {
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http://www.billthelizard.com/2009/11/sicp-exercise-112-pascals-triangle.html | ## Wednesday, November 25, 2009
### SICP Exercise 1.12: Pascal's Triangle
From SICP section 1.2.2 Tree Recursion
Exercise 1.12 asks us to write a procedure that computes the elements of Pascal's triangle by means of a recursive process.
The numbers along the left and right edges of the triangle are all 1, and each number inside the triangle is the sum of the two numbers directly above it.
We'll design our procedure to take the row and column numbers as arguments, and return the value of the element in that position.
(pascal row col)
We start counting both rows and columns at 0, as pictured in the diagram. So the procedure call
(pascal 4 2)
should return a value of 6, since that is the value in the 4th row and 2nd column.
The first thing we should note is that the nth row of the triangle contains exactly n columns. This means that we shouldn't expect to call our procedure with a column argument greater than the row argument. So, the following call would be expected to return nonsense information:
(pascal 3 4)
We should also note that the 0th and nth column of any row should return 1. This is enough to get us started.
(define (pascal row col) (cond ((< row col) #f) ((or (= 0 col) (= row col)) 1) (else #t)))
Here we're just defining the initial logic of the procedure. If the row argument is less than the column, we simply return #f (probably not the best error message I've ever written, but it's not the focus of the exercise). If the column is the 0th or nth column, we return 1, otherwise we return a value of #t. This last value is just a placeholder which we'll need to replace with a calculation of the correct Pascal number.
To finish up the procedure, we need to recursively calculate values from the interior of the triangle. An interior value is defined as the sum of the two values directly above in the triangle. So, for example, to calculate Pascal(4, 2) we would sum Pascal(3, 1) and Pascal(3, 2). Each of these values (since there are both interior values) would be calculated in the same manner, but the recursive property of our procedure will take care of that for us.
The two values to sum in order to calculate Pascal(row, column) are always going to be the value in the previous row and same column, Pascal(row-1, column), and the value in the previous row and previous column, Pascal(row-1, column-1). Translating this information to Scheme and adding it to our earlier procedure, we get our final answer to the exercise:
(define (pascal row col) (cond ((< row col) #f) ((or (= 0 col) (= row col)) 1) (else (+ (pascal (- row 1) col) (pascal (- row 1) (- col 1))))))
Try several examples from Pascal's triangle illustration to verify that the results are correct. Note that since this procedure defines a recursive process, calculating larger values can be extremely time consuming. It takes a noticeable amount of time (a few seconds) on my machine to calculate values in the 25th row.
For more fun with Pascal's triangle see Pascal's Triangle and its Patterns.
Related:
For links to all of the SICP lecture notes and exercises that I've done so far, see The SICP Challenge.
sethreno said...
Good explanation.
I thought the exercise description was kinda vague: "Write a procedure that computes elements of Pascal's triangle by means of a recursive process."
I (mis)interpreted this to mean: "Given a depth compute the sum of all element values."
(define (pascal n)
(if (= n 0) n
(+ (* (pascal (- n 1)) 2) 1)))
Bill the Lizard said...
sethreno,
"a procedure that computes elements of Pascal's triangle" is a little ambiguous. If I were teaching a course using SICP, I'd accept that interpretation of the question.
However, your solution is off by one.
> (pascal 0)
0
> (pascal 1)
1
> (pascal 2)
3
> (pascal 3)
7
> (pascal 4)
15
If you want the sum of all the elements in the nth row of Pascal's triangle, just raise 2 to the nth power.
al said...
I was returning 0 if the column is greater than the row so I would get correct result for correct input and 0 for incorrect input. Returning #f is probably a better option since 0 is a response and calling the procedure with a column greater than the row should probably not give a response. Once again, very useful post.
Фёдор Ананьев said...
Oh thank you. Solved this exercise myself before and managed to do it in three times more lines. What a shame.
tokland said...
In fact values "outside" the triangle are zeros, no need to return an special value.
Anonymous said...
Hey Bill,
I recently discovered your blog and I love it.
I recently started reading SICP and your solutions and explanations are a great help.
I also love your lecture notes.
I don't know if you already knew this but there is another (iterative) way to calculate the elements of the Pascal's triangle.
I give my code below.
(define (pascal2 r c)
(define (iter col val)
(if (= col c)
(/ (* val (- (+ r 1) col)) col)
(iter (+ col 1)
(/ (* val (- (+ r 1) col)) col))))
(cond ((or (< c 0) (> c r)) 0)
((or (= c 0) (= c r)) 1)
(else (iter 1 1))))
I found this on Wikipedia. Here's the link.
http://en.wikipedia.org/wiki/Pascal%27s_triangle#Calculating_an_individual_row_or_diagonal_by_itself_.28Gray.27s_Theory.29
Anonymous said...
The elements of the Pascal's triangle are the Binomial coefficients, i.e., they are the coefficients of the terms in the expansion of (x + y)^n, where n is a natural number.
Since the coefficients of the terms not in the expansion are zero, the values outside the Pascal's triangle should be 0 not #f.
Jan Herich said...
Slightly different solution:
(define (pascal-elem row col)
(cond ((or (< col 0) (> col row)) 0)
((< row 2) 1)
(else (+ (pascal-elem (- row 1) (- col 1))
(pascal-elem (- row 1) col)))))
Dhruv Dave said...
I know a little bit of clojure, so anytime I make attempt at SICP exercise, I do it in clojure first and translate the code into scheme.
So my question is, would you ever consider this solution?
(define (dec number)
(- number 1))
(define (inc number)
(+ number 1))
(define (drop n coll)
(if (and (not (null? coll))
(not (= 0 n)))
(drop (dec n) (cdr coll))
coll))
(define (take n coll)
(if (and (not (null? coll))
(not (= 0 n)))
(cons (car coll) (take (dec n) (cdr coll)))
'()))
(define (count coll)
(if (null? coll)
0
(inc (count (cdr coll)))))
(define (pn coll n step)
"Partition the collection. 'n' is the number of elements in a
partition and 'step' is the offset."
(if (and (not (null? coll))
(> (count coll) step))
(cons (take n coll) (pn (drop step coll) n step))
'()))
(define (pascal-triangle n)
"When supplied with a row number, return a list of numbers corresponding to pascal's triangle.
Row number starts from '0'."
(cond
((= n 0) '(1))
((= n 1) '(1 1))
((<= 2 n) (cons 1
(reverse
(cons 1
(map (lambda (s) (apply + s))
(pn (pascal-triangle (- n 1)) 2 1))))))))
;; how it works
1]=> (pn '(1 3 3 1) 2 1)
;Value 38: ((1 3) (3 3) (3 1))
1 ]=>
1 ]=> (pascal-triangle 0)
;Value 39: (1)
1 ]=> (pascal-triangle 1)
;Value 40: (1 1)
1 ]=> (pascal-triangle 2)
;Value 41: (1 2 1)
1 ]=> (pascal-triangle 3)
;Value 42: (1 3 3 1)
1 ]=> (pascal-triangle 4)
;Value 43: (1 4 6 4 1)
1 ]=> (pascal-triangle 5)
;Value 44: (1 5 10 10 5 1)
1 ]=> (pascal-triangle 6)
;Value 45: (1 6 15 20 15 6 1)
1 ]=>
Dylan Beckwith said...
Seeing as "Pascal's Triangle" is a listed representation of the binomial coefficients.
i defined and ran a (choose n k) procedure thru a coefficient procedure.
Both Algorithmically and Mathematically, i believe this to be the strongest solution.. Enjoy!
(define (bi-co col row) | 2017-08-23T02:32:52 | {
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https://math.stackexchange.com/questions/2100282/find-a-nontrivial-linear-system-of-equations-satisfied-by-any-vector-minimizin | # Find a (nontrivial) linear system of equations satisfied by any vector minimizing the energy
Here is an exercise 1.5 from the book Numerical Algorithms: Methods for Computer Vision, Machine Learning, and Graphics (by J. Solomon):
Suppose $A,B \in R^{n \times n}$ and $\vec{a},\vec{b} \in R^{n}$. Find a (nontrivial) linear system of equations satisfied by any $\vec{x}$ minimizing the energy $||A\vec{x}-\vec{a}||_{2}^2 + ||B\vec{x}-\vec{b}||_{2}^2$
As I can see, the question is to find system $C\vec{x}=\vec{c}$ which solution is any vector $\vec{x}_{opt}$ that minimises aforementioned function:
$$f(\vec{x})=||A\vec{x}-\vec{a}||_{2}^2 + ||B\vec{x}-\vec{b}||_{2}^2$$
But I can't figure out how to approach to this problem, i.e. can't understand how to deal with that question. I was trying to calculate a gradient of this equation or to use chapter's information about residues and Lagrange multipliers, but don't know if I am going in right direction.
Could someone give me a tip about how to approach to this problem?
Update #1
Using hints given in comments, I've came up with something like this: $$f(\vec{x})=||Ax-a||_{2}^2 + ||Bx-b||_{2}^2$$ Expanding norms: $$f(\vec{x})=||Ax||_2^2 + ||Bx||_2^2 - 2a^TAx - 2b^TBx + ||a||_2^2 + ||b||_2^2$$ Taking gradient and setting it to zero: $$\nabla f(\vec{x})=2A\vec{x} + 2B\vec{x} - 2a^TA - 2b^TB = 0$$ $$2(A + B)\vec{x} - 2(a^TA + b^TB) = 0$$ $$(A + B)\vec{x} - (a^TA + b^TB)=0$$ $$\vec{x}_{opt}=(A + B)^{-1}(a^TA + b^TB)$$
Is it correct?
Update #2
Oh, I see. The derivative was taken in a wrong way. Here how it should be (like it was noted in Walter's answer): $$f(\vec{x})=x^{\top}A^{\top}Ax + x^{\top}B^{\top}Bx - 2a^{\top}Ax - 2b^{\top}Bx + a^{\top}a + b^{\top}b$$ $$\nabla f(\vec{x})=2A^{\top}Ax + 2B^{\top}Bx - 2a^{\top}A-2b^{\top}B=0$$ $$\vec{x}_{opt}=(A^{\top}A + B^{\top}B)^{-1}(a^{\top}A + b^{\top}B)$$
• Use the identity $\|v\|_2^2 = v^Tv$, expand the result, take the gradient. – user856 Jan 16 '17 at 16:59
• I've updated original question. Could you tell me please, is it correct? – devforfu Jan 24 '17 at 7:19
• expand the norms by using $|X|_2^2 = X^TX$ for any expression $X$ that occurs, then just sum up and differentiate, equal to 0 and solve for whatever you want to solve. – mathreadler Jan 24 '17 at 11:18
• It seems I have rough times with vector/matrix derivatives. It seems that I calculated derivates wrongly after norms had been expanded. – devforfu Jan 24 '17 at 13:37
• There is a small compendium for that somewhere on the internet. The matrix cookbook i think it is called. – mathreadler Jan 25 '17 at 4:21
$\bf{Hint}$:
If you know the general Least squares approach to $\|Ax-b\|_2^2$, which is $$\hat{x}=(A^{\top}A)^{-1}A^{\top}b$$ Then maybe you can rewrite your problem just as general single objective Least Squares problem, remember that the matrix $A$ does not need to be square, but can also be skinny, i.e. $A\in \mathbb{R}^{m\times n}$ with $m>n$.
Do you have a clue what I am hinting at? Otherwise, just let me know.
Edit:
If we go back to your problem of minimizing $\|Ax-a\|_2^2+\|Bx-b\|_2^2$ then see that we can rewrite this as $$\left\Vert\begin{bmatrix} A\\ B \end{bmatrix}x - \begin{bmatrix} a\\ b \end{bmatrix}\right\Vert_2^2=\|Cx-c\|_2^2$$ Now our Least Squares solution becomes(assuming $C$ is full rank) $$\hat{x}=(C^{\top}C)^{-1}C^{\top}c= (A^{\top}A+B^{\top}B)^{-1}(A^{\top}a+B^{\top}b)$$
• So do you think that the system of equations that was asked to be found in this exercise i.e. $C\vec{x}=\vec{c}$ is actually a formulation of least squares problem when there are two minimization objectives, right? – devforfu Jan 24 '17 at 7:16
• I made an edit, hope it helps. – WalterJ Jan 24 '17 at 10:40
• Yes, it seems that the derivatives were taken wrongly. Now it looks almost like your solution (the only difference is multiplication order within second braces). Thank you! – devforfu Jan 24 '17 at 13:36
Just set the gradient equal to $0$, which immediately yields $$2A^T(Ax - b) + 2B^T(Bx - b) = 0.$$ Done!
Here are some hints to take the gradient easily. If $g(x) = \| x \|^2$, then $\nabla g(x) = 2x$. Also, if $h(x) = g(Mx)$, then $\nabla h(x) = M^T \nabla g(Mx)$. (This follows from the chain rule.)
• Yep, that's right. Doing it by hand using dot products property seems to be quite inefficient. And nobody should do it using some real matrix >_< – devforfu Jan 31 '17 at 16:25 | 2020-10-30T07:51:14 | {
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https://www.physicsforums.com/threads/magnetic-field-of-rotating-circular-ring.822352/ | # Magnetic Field of Rotating Circular Ring
1. Jul 7, 2015
### zero1342
1. The problem statement, all variables and given/known data
Find the magnetic field at position z (z=0 in the plane of the ring) along the rotation axis for a circular ring of radius r, carrying a uniform linear charge density λ, and rotating about its axis with angular velocity ω.
2. Relevant equations
I=q/t
ω=2πf
f=1/period
Biot-Savart Law
3. The attempt at a solution
I can determine the magnetic field when the ring is just a current loop that is not rotating. Once the rotation comes into play I get really confused about how to handle the linear charge density λ and the angular velocity.
I see that I can solve for time in the equation for current (I=q/t) and end up with: I=(qω)/(2π)
I think λ=charge/length but should it instead be: λ=dq/dl?
2. Jul 8, 2015
### ShayanJ
You're misunderstanding the problem. Its not a current loop rotating, its a charged loop rotating. So if it wasn't rotating, it was simply a charged loop. But now that its rotating, its setting charged particles in motion in a circular path. So its actually the same as a current loop.
3. Jul 8, 2015
### zero1342
I understand that there are many similarities between the magnetic field above a current loop and the magnetic field above a spinning charged loop. The issue is, how do I incorporate the charge density into the problem?
The differential element of current di = (2πdq)/ω and dq = λdl so di=(λ2πdl)/ω
Does dl = 2πrdr?
4. Jul 8, 2015
### ShayanJ
To find the magnetic field, you need to use Biot-Savart's law $\vec B=\frac{\mu_0}{4\pi} \int \frac{I d \vec l \times \vec r}{r^3}$.
Where $d\vec l$ is the differential length(so its dimension is length not length2 so rdr can't be right) along the wire in the direction of current and $\vec r$ is the displacement vector from the wire element to the point of observation.
Here we're talking about a circle. If we use cylindrical coordinates and assume the loop is at z=0, its obvious that the loop is defined by $\rho=const=R$. So the differential length should be in the direction of the azimuthal angle $\phi$, which means $d\vec l =Rd\phi \hat \phi$.
The problem wants the magnetic field along the loops axis, so the point of observation is located on the z axis, $\vec r_o=z_o \hat z$. The wire element is located at $\vec r_e=R\hat \rho$. So we have $\vec r=\vec r_o-\vec r_e=z_o\hat z-R\hat \rho$ and $r=\sqrt{z_o^2+R^2}$.
The current, as you mentioned before, is $I=\frac{dq}{dt}$. Here we have a rotating linear uniform charge distribution. So we have $I=\frac{\lambda Rd\phi}{dt}=\lambda R \omega$.
Now you should put all of the above in the integral and calculate it.
5. Jul 8, 2015
### zero1342
So I originally solved a different problem which was to find the magnetic field for a loop of current and got: $$B=\frac{{μ_0}Ir^2}{2(r^2+z^2)^\frac{3}{2}}$$
Then the problem changed and it was a charged loop rotating and using your help with dl I got: $$B=\frac{{μ_0}qr^3ω}{4πr(r^2+z^2)^\frac{3}{2}}$$ and using λ=q/(2πr) it simplifies to: $$B=\frac{{μ_0}Ir^2}{2(r^2+z^2)^\frac{3}{2}}$$ which is the same exact result for a loop of current!
So they're both the same and now your first response makes more sense. Is this all correct?
6. Jul 8, 2015
### ShayanJ
Yeah, that's correct. | 2017-08-23T13:23:04 | {
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https://math.stackexchange.com/questions/290229/explaining-the-product-of-two-ideals | # Explaining the product of two ideals
My textbook says that the product of two ideals $I$ and $J$ is the set of all finite sums of elements of the form $ab$ with $a \in I$ and $b \in J$. What does this mean exactly? Can you give examples?
• The particular case of $I$ and $J$ being principal is easy to understand and could be helpful, see here. – Watson Nov 10 '16 at 20:37
• So, according to the definition, if $x,y\in AB$ then $x=a_1b_1 + \ldots +a_nb_n$ and $y=a'_1b'_1 + \ldots +a'_mb'_m$? where m and n may be different? – yasir May 20 '18 at 10:57
• – Watson Nov 30 '18 at 8:19
One would like the product ideal to be $$IJ=\{ij\mid i\in I,j\in J\}$$ but we can easily see that there is a problem. It must be closed under addition, so $ij+i'j'$ must be in $IJ$. Can you find $i''\in I$, $j''\in J$ such that $ij+i'j'=i''j''$ so that it's in $IJ$ as defined above? Not in general, no. The natural way to allow for additive closure is to define $IJ$ as you did, including arbitrary finite sums of products.
• Can this also be written as $\sum_{i \in I, j \in J} (ij)$? (where $(ij)$ is the ideal generated by $ij$). – Al Jebr Jun 16 '19 at 17:47
For a complete answer let me add an example: $I=(2,X)$ and $J=(3,X)$ in $\mathbb Z[X]$. Then $IJ=(6,X)$ (why?), thus $X\in IJ$ and $X$ can't be written as $ij$ with $i\in I, j\in J$ (why?). (Note that if one of the ideals is principal one can't get such an example.)
• This is a good example, 1+! Almost the same, of course, works with $R[X,Y]$ and the two ideals $(X,Y)$ and $(1-X,Y)$. Geometrically, $Y$ vanishes on $\{(0,0),(1,0)\} \subseteq \mathbb{A}^2_R$, but cannot be written as a product of two polynomials which vanish on $(0,0)$ resp. $(1,0)$. Are there $1$-dimensional examples? – Martin Brandenburg Jan 30 '13 at 10:44
• A big +1 for actually giving an example of when $IJ$ is not simply the set of all products $ij$. – Pete L. Clark May 9 '13 at 17:10
• Can anyone tell me why $IJ=(6,X)$? – User Apr 23 '16 at 9:54
• @User $2 \cdot 3 = 6$, $3x - 2x = x$ and it cannot be larger than $(6,x)$ – user128245 Apr 2 '18 at 16:12
Another way to phrase this: The product ideal $IJ$ is the smallest ideal containing all the products of elements of $I$ with elements of $J$.
As for examples: In $\mathbb{Z}$, we have $$\langle a\rangle\langle b\rangle=\langle ab\rangle$$
• I like your version better, the other one I can't understand if it means that there can be multiple sums added together, like would $ax+by \in IJ$ if $a,b \in I$ and $x,y \in J$? – user39794 Jan 30 '13 at 1:45
• @AllisonCameron yes, by (your) definition of the product of two ideals, that sum would be in the product. – ferson2020 Jan 30 '13 at 1:48
• Notice that my definition is equivalent, because if an ideal contains some elements, then it contains all their finite sums. :) – pre-kidney Jan 30 '13 at 1:49
Since it was also asked for examples, let me mention how to compute the product of two ideals (beyond the already mentioned principal ideals).
If $I$ is generated by elements $\{a_i\}$ and $J$ is generated by elements $\{b_j\}$, then $I \cdot J$ is generated by the elements $\{a_i \cdot b_j\}$. You can verify this either using the element definition of $I \cdot J$, or using the more elegant definition of $I \cdot J$ as the smallest ideal containing all products.
For example, in $\mathbb{Q}[x,y]$, one computes $(x,y) \cdot (x^2,y^2)=(x^3,x y^2,x^2 y,y^3)$.
In general, one observes that $I \cdot J \subseteq I \cap J$. This is not an equality in general; in the above example the intersection is just $(x,y)$. However, one has (in the commutative case) $\sqrt{I \cdot J} = \sqrt{I \cap J}$.
• I wished that abstract algebra books were written with this clarity. Clarity requires mastery of the subject. – nilo de roock Sep 17 '14 at 21:06
• Wait, how is the intersection $(x,y)$? Since $(x^2,y^2)\subset (x,y)$, shouldn't $(x,y)\cap (x^2,y^2)=(x^2,y^2)$? – user2154420 Feb 15 '18 at 15:20
The main thing to notice is that it is not always, as a student might first guess, just $\{ab\mid a\in I, b\in J\}$. That works for groups, but in a ring you have two operations going on. Certainly in addition to having all the pairwise products, it would also have to have all possible sums of those products. Otherwise, given $ab$ and $a'b'$, you would be at a loss to write $ab+a'b'$ in the form $a''b''$ (the $a$'s are from $I$, the $b$'s are from $J$).
Just try it out: show that $\{\sum a_ib_i\mid a_i\in I, b_i\in J\}$ (finite sums) forms an ideal. Then show it's the smallest ideal containing the pairwise products.
• Why is the ideal $IJ$ not $\left\{\sum r_ia_ib_i \mid r_i \in R, a_i \in I, b_i \in J\right\}$? – Al Jebr Jun 16 '19 at 17:53
• @AlJebr what is the point of writing $r_ia_i$ when it is already in $I$? – rschwieb Jun 16 '19 at 18:04
This means that the product $IJ$ is the set of all sums $a_1b_1 + a_2b_2 + ... + a_nb_n$ where $a_1, a_2, ..., a_n \in I$, $b_1, b_2, ..., b_n \in J$. | 2020-08-06T08:08:46 | {
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https://www.srok.com.ar/miles-per-xjzuy/a39216-www-geogebra-spreadsheet | Make a new GeoGebra worksheet. By writing $A$1, you make an absolute reference to A1 when dragging both along a column and along a row. GeoGebra (www.geogebra.org) is free dynamic mathematics software for all levels of education that brings together geometry, algebra, spreadsheets, graphing, statistics and calculus in one easy-to-use package. Besides manually adding entries into the Spreadsheet View cells, you may use the commands FillColumn, FillRow or FillCells. Free offline GeoGebra apps for iOS, Android, Windows, Mac, Chromebook and Linux Geogebra is a free mathematics software for learning and teaching Interactive graphics, algebra and spreadsheet From elementary … And so on. GeoGebra is a dynamic mathematics software for all levels of education that brings together geometry, algebra, spreadsheets, graphing, statistics and calculus in one easy-to-use package. 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If cell A1 contains the value 5 and A1 + 1 is written in cell A2, then the value of A1 is picked from the relative position, i.e. it comes to just doing numerical calculations, regular spreadsheet software is A slider $$d$$ representing the initial distance between the dog and the rabbit. Adding JavaScript to a GeoGebra Applet 8. Change the focus and the directrix, all perpendicular bisectors should also change dynamically. If you copy content from one cell to another, by default all references are changed accordingly to the target position. The object-oriented way of doing things Use the Point Tool to place a point on the line. Drag the $$y$$-axis until you see the points in the graphics view. http://wiki.geogebra.org/s/en/index.php?title=Spreadsheet_View&oldid=57950, Select and copy the data you want to import. In GeoGebra however, you can also use regular GeoGebra objects within the spreadsheet. A rabbit is running along a river. 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When GeoGebra is an interactive geometry, algebra, and calculus application, intended for teachers and students. This view is part of GeoGebra user interface. If possible, GeoGebra immediately displays the graphical representation of the object you entered in a Spreadsheet Cell in the Graphics View as well. The object-oriented way of doing things in GeoGebra however, makes the spreadsheet a very useful tool. Constructions can be made with points, vectors, segments, lines, polygons, conic sections, and functions. The spreadsheet in GeoGebra has most of the regular spreadsheet-features. Murkle bravo, for adding in fact a formula bar ..Good job. Spreadsheet View 23 18 To demonstrate integration and area 24 19 Integration and the Trapezoidal rule 25 20 To find the area between the graph of a function and the x axis, where ... To download GeoGebra go to www.geogebra.org and the following window will appear. You may need the command UnitVector( ) and the command Vector( , ). Using GeoGebra's Spreadsheet View. It offers multiple representations of objects in its graphics, algebra, and spreadsheet views that are all dynamically linked. First of all: Congratulations for releasing GeoGebra 3.2. You can use the dollar-sign to make copies using an absolute reference. In the Spreadsheet Cells you can enter not only numbers, but all types of General Objects and Geometrical Objects that are supported by GeoGebra (e.g. Make a slider $$a$$ of integer values between 0 and 10,000 representing Anna's annual raise of her monthly salary in Swedish kronor. Official Website of GeoGebra Institute of Hong Kong. Spreadsheet . | 2021-07-29T03:34:29 | {
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https://www.physicsforums.com/threads/show-that-the-order-of-a-cycle-is-the-length-of-the-cycle.950440/ | # Show that the order of a cycle is the length of the cycle
## Homework Statement
Prove that if ##\sigma## is the m-cycle ##(a_1 ~a_2~ \dots ~ a_m)##, then for all ##i \in \{1,2, \dots , m \}##, ##\sigma^i (a_k) = a_{k+i}##. Deduce that ##\sigma^m (a_k) = a_k##
## The Attempt at a Solution
I will try to do this by induction. Clearly, ##\sigma (a_k) = a_{k+1}##, as that is how ##\sigma## is defined. Now, suppose that for some ##n \in \mathbb{N}## ##\sigma^n (a_k) = a_{k+n}##. Then ##\sigma^{n+1} (a_k) = \sigma( \sigma^n (a_k) ) = \sigma (a_{k+n}) = \sigma_{a + k + 1} ##.
So I feel like I'm done, but I also feel that I've proved this statement for all ##i \in \mathbb{N}##, rather than for all ##i \in \{1,2, \dots , m \}##. How do I modify it to get it correct?
fresh_42
Mentor
2021 Award
## Homework Statement
Prove that if ##\sigma## is the m-cycle ##(a_1 ~a_2~ \dots ~ a_m)##, then for all ##i \in \{1,2, \dots , m \}##, ##\sigma^i (a_k) = a_{k+i}##. Deduce that ##\sigma^m (a_k) = a_k##
## The Attempt at a Solution
I will try to do this by induction. Clearly, ##\sigma (a_k) = a_{k+1}##, as that is how ##\sigma## is defined. Now, suppose that for some ##n \in \mathbb{N}## ##\sigma^n (a_k) = a_{k+n}##. Then ##\sigma^{n+1} (a_k) = \sigma( \sigma^n (a_k) ) = \sigma (a_{k+n}) = \sigma_{a + k + 1} ##.
So I feel like I'm done, but I also feel that I've proved this statement for all ##i \in \mathbb{N}##, rather than for all ##i \in \{1,2, \dots , m \}##. How do I modify it to get it correct?
It is correct and you've proven it for all ##n \in \mathbb{N}##. However, the question is poorly worded. It should have been ##\sigma^i (a_k) = a_{k+i~\operatorname{mod}m}## since ##\sigma(a_m)=a_1## and not ##a_{m+1}##. Hence the induction basis is already wrong without the modulo on the index.
Mr Davis 97
It is correct and you've proven it for all ##n \in \mathbb{N}##. However, the question is poorly worded. It should have been ##\sigma^i (a_k) = a_{k+i~\operatorname{mod}m}## since ##\sigma(a_m)=a_1## and not ##a_{m+1}##. Hence the induction basis is already wrong without the modulo on the index.
I see what you mean. But I'm still a little confused. Why was my induction where ##i \in \mathbb{N}## valid if it is states that ##i \in \{1,2, \dots , m \}##?
fresh_42
Mentor
2021 Award
I see what you mean. But I'm still a little confused. Why was my induction where ##i \in \mathbb{N}## valid if it is states that ##i \in \{1,2, \dots , m \}##?
If it's true for all natural numbers, then it's true for a few of them. The restriction to ##1 \leq i \leq m## comes into play as we don't have other indices. The modulo solves this problem. There is simply no ##a_{m+1}## in our cycle, so ##a_{k+i}## doesn't make sense if it exceeds ##m##. To place a modulo ##m## behind the index is the shortest way of saying ##\in \{\,1,\ldots ,m\,\}##. It's simply a sloppiness by whoever has written the text. Your induction is equally wrong without the modulo, as in case ##i=1## and ##k=m## we have ##\sigma^1(a_m) \neq a_{m+1}##. And with the modulo, we only need the numbers up to ##m##.
If it's true for all natural numbers, then it's true for a few of them. The restriction to ##1 \leq i \leq m## comes into play as we don't have other indices. The modulo solves this problem. There is simply no ##a_{m+1}## in our cycle, so ##a_{k+i}## doesn't make sense if it exceeds ##m##. To place a modulo ##m## behind the index is the shortest way of saying ##\in \{\,1,\ldots ,m\,\}##. It's simply a sloppiness by whoever has written the text. Your induction is equally wrong without the modulo, as in case ##i=1## and ##k=m## we have ##\sigma^1(a_m) \neq a_{m+1}##. And with the modulo, we only need the numbers up to ##m##.
So would a less sloppy problem statement be
Prove that if ##\sigma## is the m-cycle ##(a_1 ~a_2~ \dots ~ a_m)##, then for all ##i \in \mathbb{N}##, ##\sigma^i (a_{k \mod m}) = a_{k+i \mod m}##. Deduce that ##\sigma^m (a_{k \mod m}) = a_{k \mod m}## and so ##| \sigma | = m##.
fresh_42
Mentor
2021 Award
So would a less sloppy problem statement be
Prove that if ##\sigma## is the m-cycle ##(a_1 ~a_2~ \dots ~ a_m)##, then for all ##i \in \mathbb{N}##, ##\sigma^i (a_{k \mod m}) = a_{k+i \mod m}##. Deduce that ##\sigma^m (a_{k \mod m}) = a_{k \mod m}## and so ##| \sigma | = m##.
Yes and no. Yes, because the index problem is solved, no because there is no need for the modulo in the notation as a variable of ##\sigma## since ##\sigma## is only defined modulo ##m##. To write ##\sigma^m (a_{k \mod m})## creates unnecessary confusion. But for the results, we have to make sure that the formula is defined! So the index of the result has to be wrapped around. For ##a_k## in the role of a variable, it has already been done the moment you wrote ##\sigma=(a_1,\ldots,a_m)##.
Mr Davis 97
Yes and no. Yes, because the index problem is solved, no because there is no need for the modulo in the notation as a variable of ##\sigma## since ##\sigma## is only defined modulo ##m##. To write ##\sigma^m (a_{k \mod m})## creates unnecessary confusion. But for the results, we have to make sure that the formula is defined! So the index of the result has to be wrapped around. For ##a_k## in the role of a variable, it has already been done the moment you wrote ##\sigma=(a_1,\ldots,a_m)##.
I think this is the last thing. What is the difference between the following two wordings of the question:
1) Prove that if ##\sigma## is the m-cycle ##(a_1 ~a_2~ \dots ~ a_m)##, then for all ##i \in \{1,2, \dots , m \}##, ##\sigma^i (a_k) = a_{k+i}##, where ##k+i## is replaced by its least positive residue mod m.
2) Prove that if ##\sigma## is the m-cycle ##(a_1 ~a_2~ \dots ~ a_m)##, then for all ##i \in \mathbb{N}##, ##\sigma^i (a_k) = a_{k+i}##, where ##k+i## is replaced by its least positive residue mod m.
fresh_42
Mentor
2021 Award
Formally: the second implies the first; within the context: none. As ##\sigma^m=1##, with this information the two are equivalent, because all other exponents, even the negative ones, can be reduced to the case ##1\leq i \leq m##.
Formally: the second implies the first; within the context: none. As ##\sigma^m=1##, with this information the two are equivalent, because all other exponents, even the negative ones, can be reduced to the case ##1\leq i \leq m##.
So if we were given problem statement 1), would the best thing be to prove 2) using induction and say clearly this implies 1 is true?
I ask this because it doesn't seem like you could do induction on ##i## formally if the variable isn't defined for all natural numbers.
fresh_42
Mentor
2021 Award
I ask this because it doesn't seem like you could do induction on ##i## formally if the variable isn't defined for all natural numbers.
But you did the induction on the exponent which is defined for all numbers!
But you did the induction on the exponent which is defined for all numbers!
Doesn't the original statement say that ##i## only varies from 1 to m?
fresh_42
Mentor
2021 Award
Doesn't the original statement say that ##i## only varies from 1 to m?
O.k. in this case, it isn't an induction, but an "and so on". Or you use the induction and show it for all ##n##. The "and so on" case goes with dots: Let ##i \in \{\,1,\ldots ,m\,\}##, then
$$\sigma^i(a_k)=\underbrace{\sigma(\sigma(\sigma(\ldots (\sigma}_{i \text{ times }}(a_k)\ldots )= \underbrace{\sigma(\sigma(\sigma(\ldots (\sigma}_{i-1 \text{ times }}(a_{k+1})\ldots )= \ldots = \underbrace{(\sigma)}_{1 \text{ times }}(a_{k+i-1})=a_{k+i}$$
but the induction is nicer and covers the ##i## stated as it covers all ##n \in \mathbb{N}_0##.
Mr Davis 97 | 2022-07-03T21:10:18 | {
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https://math.stackexchange.com/questions/1105160/evaluate-derivative-of-lagrange-polynomials-at-construction-points | # Evaluate derivative of Lagrange polynomials at construction points
Assume, that we have points $$x_i$$ with $$i=1,...,N+1$$. We construct the Lagrange basis polynomials as \begin{align} L_j(x) = \prod_{k\not = j} \frac{x-x_k}{x_j-x_k} \end{align}
Now according to my computation and the results by Yves Daoust here, the derivative of $$L_i$$ can be computed as \begin{align} L'_j(x) = L_j(x)\cdot \sum_{k\not = j}\frac{1}{x_k-x_j} \end{align}
I try to reproduce the numerical results of a paper, and for this results the authors use the derivative matrix $$D$$ with $$D_{ij} =L_i'(x_j)$$.
The authors use the Legendre Gauss Lobatto quadrature points, plotted below. Now I can easily construct the basis polynomials, also plotted below, together with the quadrature points.
Given the concrete choice of basis points, the plots suggest, that $$L'_j(x_j)=0$$ except for the first and last basis function. Additionally it seems, that $$L'_j(x_i)\not =0$$ for $$i\not = j$$. But using the derived formula from above, we obtain \begin{align} L'_j(x_i) = L_j(x_i)\cdot \sum_{k\not = j}\frac{1}{x_k-x_j}=0 \end{align} for $$i\not = j$$ since $$L_j(x_i) =\delta_{ij}$$, which contradicts my observation.
I used the following Matlab functions, to construct the matrix $$D$$.
function y=dl(i,x,z)
n = length(x);
y = 0;
for m=1:n
if not(m==i)
y = y + 1/(x(m)-x(i));
end
end
size(y)
y = y*l(i,x,z);
end
function y=l(i,x,z)
n = length(x);
% computes h_i(z)
y = 1;
for m=1:n
if not(m==i)
y = y.*(z-x(m))./(x(i)-x(m));
end
end
end
Where dland lare $$L'$$ and $$L$$. $$D$$ is then constructed by
D = zeros(M+1,M+1);
for i=1:M+1
for j=1:M+1
D(i,j) =dl(i,X,X(j));
end
end
which gives the matrix
D =
10.5000 0 0 0 0 0 0
0 -0.0000 0 0 0 0 0
0 0 0.0000 0 0 0 0
0 0 0 -0.0000 0 0 0
0 0 0 0 0.0000 0 0
0 0 0 0 0 0.0000 0
0 0 0 0 0 0 -10.5000
Here I agree with the zeros on the diagonal except for the first and last element. The zeros everywhere else agree with the formula, but they don't agree with my observation.
If I use finite differences (which is no solution for the real implementation) in the following way: FD(i,j) =(l(i,X,X(j)+eps)-l(i,X,X(j)-eps))/(2*eps); I obtain the output
FD =
-10.5000 -2.4429 0.6253 -0.3125 0.2261 -0.2266 0.5000
14.2016 0.0000 -2.2158 0.9075 -0.6164 0.6022 -1.3174
-5.6690 3.4558 0.0000 -2.0070 1.0664 -0.9613 2.0500
3.2000 -1.5986 2.2667 0 -2.2667 1.5986 -3.2000
-2.0500 0.9613 -1.0664 2.0070 -0.0000 -3.4558 5.6690
1.3174 -0.6022 0.6164 -0.9075 2.2158 -0.0000 -14.2016
-0.5000 0.2266 -0.2261 0.3125 -0.6253 2.4429 10.5000
which is zero on the diagonal (except first and last) and nonzero everywhere else.
So here are my questions:
Did I make an error in the implementation?
Is my formula derived under the assumption that $$x$$ is not a basis point?
Can I modify my code to obtain the results that my finite difference code produces?
Is my assumption correct, that rather the results of the finite difference computation are "correct"?
EDIT
It seems, like the derivation of the formula goes wrong for $$x=x_i$$. If I compute the derivative without simplification, I get \begin{align} L_j'(x) = \sum_{l\not = j} \frac{1}{x_j-x_l}\prod_{m\not = (j,l)} \frac{x-x_m}{x_j-x_m} \end{align} Or in code
function y = alternative_dl(j,x,z)
y = 0;
n = length(x);
for l=1:n
if not(l==j)
k = 1/(x(j)-x(l));
for m=1:n
if not(m==j) && not(m==l)
k = k*(z-x(m))/(x(j)-x(m));
end
end
y = y + k;
end
end
end
Which agrees with the finite difference computation.
So it seems to me, that simplifying the above formula includes some "hidden division by zero" if $$x=x_i$$.
• I found your post and the last expression for derivative really useful. Can you look at this post and help me out? I need to find second order derivative and have no idea how. Thanks in advance! math.stackexchange.com/questions/1572576/… – Luka Bulatovic Dec 12 '15 at 20:00
I think your implementation is correct. You copied the wrong formula.
$$L'_i(x)=L_i(x) \sum_{m=0,\ m\neq i}^n\frac1{x-x_m}$$
This formula would not give you zeros everywhere else.
And this is the same as your new derived formula.
• But your formula contains $L_i(x)$ on the right hand side. Now if I evaluate $L'_i(x_j)$ I get $L_i'(x_j) = L_i(x_j)\cdot \sum...=0\cdot \sum$ since $L_i(x_j)=0$ if $i\not = j$. – k1next Jan 15 '15 at 11:39
• Okay I see, that I copied the formula wrong, but the problem I have in the comment above still exists?! – k1next Jan 15 '15 at 11:42
• When you plug $x_j$ into it, there is also a $x_j-x_j$ in the denominator on the right hand side. I guess that's what you called "hidden division by zero"? In fact, you can expand this formula to get the other formula in the post you referred to: $L_i(x)'=\frac{\sum_{k=0}^{n} \prod_{l=0, l \neq k}^n (x-x_l)}{\prod_{k=0, k \neq j}^n (x_j-x_k)}$, which does not contain zero in the denominator. – KittyL Jan 15 '15 at 11:44
• Here is a simple 3 point formula. $L_1'(x)=\frac{(x-x_0)+(x-x_2)}{(x_1-x_0)(x_1-x_2)} = \frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}(\frac{1}{x-x_0}+\frac{1}{x-x_2})$. If you evaluate it at $x_2$, you will see you cannot just say it is zero using the second formula. – KittyL Jan 15 '15 at 11:47
• Okay, but then I can't use the formula in my implementation, since I end up computing $0/0$ since $(x-x_2)=0$ if $x=x_2$ and $1/(x-x2)=1/0$ if $x=x_2$? – k1next Jan 15 '15 at 12:16
Just to complement the answer, here is the formula for second derivative: $$L^{"}_i(x) =\sum_{l\ne i}\frac{1}{x_i-x_l}\left( \sum_{m\ne(i,l)}\frac{1}{x_i-x_m}\prod_{k\ne(i,l,m)}\frac{x-x_k}{x_i-x_k}\right)$$ through recursion, one can compute further higher derivatives.
• Look at this guy... what a nicccceeee answer. – bremen_matt Feb 22 '18 at 16:19 | 2019-08-20T18:20:06 | {
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https://math.stackexchange.com/questions/3947688/high-school-contest-math-question-number-theory-prove | # High school contest math question (number theory) - prove:
Reposting with Mathjax - sorry, first time!
Let $$S = \{4,8,9,16,...\}$$ be the set of integers of the form $$m^k$$ for integers $$m, k \ge 2$$. For a positive integer $$n$$, let $$f(n)$$ denote the number of ways to write $$n$$ as the sum of (one or more) distinct elements of $$S$$. For example, $$f(5) = 0$$ since there are no ways to express 5 in this fashion, and $$f(17) = 1$$ since $$17 = 8+9$$ is the only way to express 17.
(a) Prove that $$f(30) = 0$$
(b) Show that $$f(n) \ge 1$$ for $$n \ge 31$$.
(c) Let $$T$$ be the set of integers for which $$f(n) = 3$$. Prove that $$T$$ is finite and non-empty, and find the largest element of $$T$$.
I think that part a) is relatively easy to just check since none of the values in the first few values of the set $$S = \{4,8,9,16,25,27,32,64,...\}$$ will add to get to 30.
I'm not sure where to start with part b and part c. For part b, I was working at finding sums for each number but figured this was not an intelligent way to proceed. For part c) I'm not sure where to start at all.
• Why 1 isn't in S? – Toni Mhax Dec 14 '20 at 3:44
• @ToniMhax $1$ is not a perfect power where the base and exponent are both $\ge 2$ – Benjamin Wang Dec 14 '20 at 3:47
• @RossMillikan Part of the question states that $f(n)$ denotes the number of ways to write $n$ as the sum of one or more distinct elements of $S$. Since $32$ is in the set, I assume by the question's wording that it itself counts as a way to represent it. (Also I just copied this question word for word from the collection of contest questions) – bobby_mc_gee Dec 14 '20 at 4:01
• Yes, I now believe the question is correct and you do use distinct values. – Ross Millikan Dec 14 '20 at 14:07
For part b, note that all multiples of $$4$$ can be represented because you have all the powers of $$2$$ except $$1,2$$. Express any multiple of $$4$$ in binary and read off the numbers to add to get it. All numbers equivalent to $$1 \bmod 4$$ that are at least $$9$$ can be expressed because the number minus $$9$$ is a multiple of $$4$$ and therefore expressible. All numbers equivalent to $$2 \bmod 4$$ that are $$34$$ or greater are expressible because $$34=9+25$$. All numbers equivalent to $$3 \bmod 4$$ that are $$27$$ or greater because we have $$27$$ available. Therefore the greatest number that cannot be expressed is $$30$$.
For c, numbers that are large enough will have too many representations. We will do each residue class $$\mod 4$$ in turn.
For $$0 \bmod 4$$ we have $$\emptyset,36, 9+27, 25+27$$ as ways to express numbers without any of the $$2^n$$ terms. We can therefore express any number $$52$$ or greater in $$4$$ or more ways.
For $$1 \bmod 4$$ we have $$9, 25, 9+36, 49$$ so we can express any number $$49$$ or greater in $$4$$ or more ways.
For $$2 \bmod 4$$ we have $$9+25, 9+49, 9+36+49, 25+49$$ so we can express any number $$74$$ or greater in $$4$$ or more ways.
For $$3 \bmod 4$$ we have $$27, 27+36, 9+25+49, 9+25+36+49$$ so we can express any number $$119$$ or greater in $$4$$ or more ways.
The greatest number in $$T$$ is $$115$$, which can be expressed as $$64+27+16+8, 36+32+27+16+4, 49+32+25+9$$ but in no other ways.
• So this answer assumes that you can pick non-distinct elements? I guess this might be the question's intention, as someone noted that 34 can't be represented by picking distinct elements – Benjamin Wang Dec 14 '20 at 10:08
• No, it does not assume you can duplicate elements. You can get $34$ as $25+9$. I had thought there were gaps, but there are not. – Ross Millikan Dec 14 '20 at 14:07
• This problem was posted recently (possibly by OP) and then deleted. The solution there showed by induction that $f(n+4) \geq f(n)$ (while maintaining distinctness). The solution didn't find the largest element of $T$, and believed that it required checking enough small cases (as opposed to an insightful argument for why that's the largest possible, similar to showing 30 is the max). – Calvin Lin Dec 14 '20 at 15:43
• @CalvinLin: I have done it. I don't think it is too many cases to check. – Ross Millikan Dec 14 '20 at 20:10
• Nice. I don't think 0 is allowed to be used though, so for 0 mod 4 it looks like you will have to go up to 76 = 49+27. – Mike Dec 14 '20 at 20:37 | 2021-06-13T17:09:58 | {
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https://math.stackexchange.com/questions/336656/whats-the-difference-between-exists-f-epsilon-me-setminus-f-epsilon | # What's the difference between “$\exists F_\epsilon :m^*(E\setminus F_\epsilon)<\epsilon$ for each $\epsilon$,” and “$\exists F:m^*(E\setminus F)=0$”?
Let $E\subseteq \mathbb{R}$ be given, and let $m^*$ denote the outer measure. For each $\epsilon$, there exists a closed set $F_\epsilon\subseteq E$ such that $m^*(E\setminus F_\epsilon)<\epsilon$. Does this imply that there exists a closed set $F\subseteq E$ such that $m^*(E\setminus F)=0$?
• What is $E/F$? Do you mean $E\setminus F$? – Amit Kumar Gupta Mar 21 '13 at 7:13
• I think $m^*$ is a more usual notation for outer measure. – copper.hat Mar 21 '13 at 7:14
There exists a $F_\sigma$ set such that this is true, take $\cup_{n=1}^\infty F_{\frac{1}{n}}$.
However, the statement is not true in general. Take $E = (0,1)$. Then taking $F_\epsilon = [\frac{\epsilon}{3},1-\frac{\epsilon}{3}]$ results in $m^* (E \setminus F_\epsilon) < \epsilon$.
Suppose $F \subset E$ is closed. Let $I=[\inf F, \sup F]$. Clearly, $F \subset I$, and $0< \inf F, \sup F < 1$, hence $m^*(E\setminus F) > \frac{\inf F}{2}> 0$.
No. If $E$ is closed then the statement clearly holds, so the only possible counterexample must be non-closed. What's the simplest possible example of that? A bounded open interval. No closed subset $F$ of a bounded open interval $E$ can be such that $m^*(E\setminus F)=0$. This is because $E\setminus F$ is open, thus is either empty or contains an open interval. It can't be empty since that would imply $E=F$ which would make $E$ clopen, which is impossible (the only clopen subsets of $\mathbb{R}$ are the empty set and the whole space). So it contains an open interval, and hence has positive outer measure.
It might be useful to see how these statements can be rewritten to better show the difference in their logical form.
The 1st statement becomes: $\;\; \left(\forall \, \epsilon > 0 \right) \left( \exists F \subseteq {\mathbb R} \right ):$ $\;\;\;F$ is closed and $m^{*}\left(E-F\right) < \epsilon$
The 2nd statement becomes: $\;\; \left( \exists F \subseteq {\mathbb R} \right )\left(\forall \, \epsilon > 0 \right):$ $\;\;\;F$ is closed and $m^{*}\left(E-F\right) < \epsilon$
Of course, the fact that the 2nd statement is a logically stronger $\exists \; \forall$ uniform statement doesn't mean that, in this specific context, we get a mathematically stronger statement. However, we do in fact get a mathematically stronger statement, as the other answers show. | 2019-09-20T05:14:07 | {
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https://www.jiskha.com/questions/501256/can-someone-please-tell-me-how-to-do-these-suppose-that-f-is-given-for-x-in-the-interval | Can someone please tell me how to do these?
Suppose that f is given for x in the interval [0,12] by
x= 0 2 4 6 8 10 12
f(x)= -14 -10 -6 -5 -6 -9 -11
A. Estimate f(2) using the values of f in the table.
f'(2)=_______
B. For what values of x does f'(x) appear to be positive?
_________
(Give your answer as an interval or a list of intervals, e.g., (-infinity,8] or (1,5),(7,10) .)
C. For what values of x does f'(x) appear to be negative?
(Give your answer as an interval or a list of intervals, e.g., (-infinity,8] or (1,5),(7,10) .)
1. 👍
2. 👎
3. 👁
1. I already did it... thanks anyways.
1. 👍
2. 👎
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http://math.stackexchange.com/questions/23451/how-to-find-the-solution-for-frac2x-3x1-leq-1 | # How to find the solution for $\frac{2x-3}{x+1} \leq 1$?
I have the following inequality:
$$\frac{2x-3}{x+1}\leq1$$
so, considering $x \neq -1$, I started multiplying $x+1$ both sides:
$$2x-3\leq x+1$$
then I subtracted $x$ both sides:
$$x-3\leq1$$
and then sum $3$ both sides:
$$x\leq4$$
Therefore, my solution for $x\neq-1$ is:
$$(-\infty,4]$$
But the book solution is:
$$(-1,4]$$
What I did wrong?
-
Tex Comment: For future reference, \leq is <= (less than or equal to), \geq is >= (greater than or equal to) and \neq is != (not equal to). – Eric Naslund Feb 23 '11 at 23:47
See also my answer here. – Isaac Feb 24 '11 at 1:44
And something should have been shouting in your ear that there was something terribly wrong: your solution set includes $x=-1$, but the inequality is not defined there! – Arturo Magidin Feb 24 '11 at 3:29
Note that if we have $\frac{a}{b} \leq 1$, this means that if $b>0$, then $a \leq b$ and if $b<0$, then $a \geq b$.
So you will need to split this into cases.
First note that you can multiply without changing the $\leq$ sign only when $x+1 > 0$.
So when $x+1 > 0$, we have $2x-3 \leq x+1$ which gives us $x \leq 4$. Hence, when $x+1>0$, we have $x \leq 4$ and hence $x \in (-1,4]$.
If $x+1 < 0$, the $\leq$ gets reversed to $\geq$ when you multiply by $x+1$ and we get $2x-3 \geq x+1$ when $x+1 < 0$. This gives us $x \geq 4$ when $x < -1$ which is not possible.
Hence, the solution is $x \in (-1,4]$
-
The problem is when you multiplied both sides by $x+1$. Remember that when you multiply an inequality by a negative number it changes signs. This means we have to split into cases:
Case 1: $x+1>0$. Then we get $$2x-3\leq x+1$$ By the same reasoning that you present above we find $x\leq 4$. Then rewrite the inequality $x+1>0$ as $x>-1$. Combining this with $x\leq 4$ we see $x\in (-1,4]$.
Case 2: $x+1<0$. Then we get $$2x-3\geq x+1$$ (notice that since $x+1<0$ the sign had to switch directions) We then solve to find $x\geq 4$. Since for this case we also had $x+1<0$, which is the same as $x<-1$ we conclude no such $x$ exists. (A number cannot be less than -1 and greater than 4)
Hope that helps,
-
You deleted your comment? In any case, I edited to add some more details. When we look at case 1 we are assuming $x+1>0$ so that is where the $x>-1$ comes from. – Eric Naslund Feb 23 '11 at 23:58
To preserve the $\:\le\:$ you must multiply by $\rm\ (x+1)^2\$ not $\rm\ x+1\:,\:$ namely
$\rm\quad\quad\quad\quad\quad\quad\ \displaystyle\frac{2x-3}{x+1}\ \le\ 1$
$\rm\quad\quad\iff\quad \displaystyle\frac{x-4}{x+1}\ \ \le\ 0$
$\rm\quad\quad\iff\quad (x+1)\ (x - 4)\ \le\ 0,\quad x\ne -1$
$\rm\quad\quad\iff\quad\ x\ \in\ (-1,4\:]$
-
As others have mentioned, multiplying by $x+1$ forces you to consider cases at the outset. Instead, you can write $\frac{2x-3}{x+1} \leq 1$ as $\frac{2x-3}{x+1} -1 \leq 0$. Simplify this into $\frac{p(x)}{q(x)} \leq 0$ and consider when a fraction is negative.
- | 2014-08-02T07:21:00 | {
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https://math.stackexchange.com/questions/2843910/square-matrix-with-rational-coefficients-having-k-th-root | # Square matrix with rational coefficients having $k$-th root
Let $A\in M_{n}(\mathbb{Q})$, meaning that $A$ is $n\times n$ matrix with entries in the rational numbers $\mathbb{Q}$. Suppose that $A$ satisfies two conditions:
• $\det(A)\neq 0$
• For every integer $k$, there exists a $B\in M_{n}(\mathbb{Q})$ such that $B^k = A$.
Does it follow that $A = I_n$? Here, $I_n$ is the $n\times n$ identity matrix.
Motivation. The analogous problem where $\mathbb{Q}$ everywhere is replaced by $\mathbb{Z}$ has an affirmative answer. More precisely, let's prove the following statement:
Claim. Let $A\in M_{n}(\mathbb{Z})$, meaning that $A$ is $n\times n$ matrix with integer entries. Suppose that $\det(A)\neq 0$ and that for every integer $k$, there exists $B\in M_{n}(\mathbb{Z})$ such that $B^k = A$. Then $A = I_n$.
Proof. Since $\det(A)\neq 0$, there are only finitely many primes dividing $\det(A)$. Let $p$ be any prime which does not divide $\det(A)$. After reducing mod $p$, we still get a non-singular matrix $\overline{A}$, so $\overline{A}\in\operatorname{GL}_{n}(\mathbb{Z}/p\mathbb{Z})$. Let $k=\# \operatorname{GL}_{n}(\mathbb{Z}/p\mathbb{Z})$ be the cardinality of the group of invertible matrices with entries in $\mathbb{Z}/p\mathbb{Z}$. By hypothesis, there exists $B\in M_{n}(\mathbb{Z})$ such that $B^k=A$. After reducing mod $p$, we get $(\overline{B})^k = \overline{A}$. By Lagrange's theorem, $(\overline{B})^k = \overline{I_{n}}$. Therefore, $\overline{A} = \overline{I_{n}}$, so that each entry of $A-I_{n}$ is divisible by $p$. Since we have infinitely many choices for the prime $p$, it follows that each entry of $A-I_{n}$ must in fact be zero, yielding $A=I_n$.
Remark 1. The hypothesis $\det(A)\neq 0$ is necessary. Otherwise, we can just take any diagonal matrix $A= \operatorname{diag}(1, 1, .. 1, 0, 0 .., 0)$ with any number of $1$s and $0$s. Certainly $A^k = A$ for every $k$ but $A\neq I_{n}$ if there is at least one zero on the diagonal.
Remark 2. It is tempting to modify the above argument for the case of $\mathbb{Q}$, but it doesn't seem to work immediately. Again using the fact that $\det(A)\neq 0$, there are only finitely many primes appearing in the numerator and denominator of $\det(A)$, so we can pick a prime $p$ which doesn't appear there. Reducing the original matrix $A$ mod $p$ still works fine. But the problem arises at the step when we use hypothesis to get a matrix $B$ such that $B^k = A$ (where $k$ clearly depends on p -- this is important). That matrix $B$ might have the prime $p$ show up in the denominator of some of its entries, so reduction mod $p$ doesn't make sense.
Reference. I learnt the proof above from "Mathematical Bridges" by Andreescu, Mortici and Tetiva. The problem appears as Exercise 18 in Chapter 6, and the solution is presented few pages later. So I am also adding the (contest-math) tag, since that is the theme of the aforementioned book.
• Try $A=\pmatrix{1&1\\0&1}$. – Lord Shark the Unknown Jul 7 '18 at 17:23
• @LordSharktheUnknown Very nice! Indeed, if we let $B = \pmatrix{1 & \frac{1}{k} \\ 0 & 1}$, then $B^k = A$. You should post this as an answer when you get a chance. Also, I wonder if these are the only examples. In other words, do you think that with the conditions above, we can at least conclude that $A$ is unipotent? (You proved that we cannot conclude that $A$ is identity) – Prism Jul 7 '18 at 17:34
• Let $S\in M_2(\mathbb Q)$ by any invertible matrix.The first family is $S\begin{pmatrix} 1 & a\\ 0 & 1\end{pmatrix}S^{-1}$. The second family is $S\begin{pmatrix} 2 & 1\\ -1 & 0\end{pmatrix} S^{-1}$. The first construction can be seen to extend to any $n\geq 2$. Might be the same for the second family. – Yong Hao Ng Jul 18 '18 at 11:29
• The solutions $A = (B_k)^k$ are $$B_k = \begin{pmatrix} 1 & a/k \\ 0 & 1\end{pmatrix} \implies (B_k)^k = \begin{pmatrix} 1 & a\\ 0 & 1\end{pmatrix}$$ and $$B_k = \frac{1}{k}\begin{pmatrix} k+1 & 1 \\ -1 & k-1\end{pmatrix} \implies (B_k)^k = \begin{pmatrix} 2 & 1 \\ -1 & 0 \end{pmatrix}$$ respectively. This extends to solving $SAS^{-1} = (SB_kS^{-1})^k$, explaining the existence of the $S$ matrices. – Yong Hao Ng Jul 18 '18 at 11:31
The general result is as follows:
Let $A\in M_n(\mathbb{Q})$; then
For every integer $k$, there exists $B\in M_n(\mathbb{Q})$ such that $B^k=A$ IFF $A(A-I_n)^n=0$.
You can see several proofs
in english in https://artofproblemsolving.com/community/c7h42444 , grobber's post.
in french in "La revue de mathématiques spéciales" n° 117-2 , R535, where, with a colleague, we gave a solution using Mahler's measure.
For convenience, we call a matrix $A$ with solutions $A=B^k$ for all $k\geq 1$ exceptional. Notice that $$\det(A) = \det(B)^k \implies \sqrt[k]{\det(A)} = \det(B) \in \mathbb Q$$ This is only possible if $\det(A)=1$, so we assume that.
# Overview
Let $T_n(b)$ denote an $n$ by $n$ matrix with all entries zero except for top right entry $b$. Then for each $n\geq 2$ there is an infinite family of exceptional matrices in $M_n(\mathbb Q)$, of the form $$A = S(I_n+T_n(b))S^{-1}$$ where $S$ can be any invertible matrix in $M_n(\mathbb Q)$. The solutions $B_k$ are of the form $$B_k = S(I_n + T_n(b/k))S^{-1}$$ When $b\neq 0$, we have $A\neq I_n$, hence an infinite family of counterexamples and answering the original question. We remark that this family is unipotent.
For dimension $n=2$, we give another infinite family of exceptional matrices $$A = S\begin{pmatrix}2 & 1\\ -1 & 0\end{pmatrix} S^{-1}$$ where $S$ can any invertible matrix in $M_2(\mathbb Q)$. It has the solutions $$B_k = S \begin{pmatrix} \frac{k+1}{k} & \frac{1}{k}\\ \frac{-1}{k} & \frac{k-1}{k} \end{pmatrix}S^{-1}$$ We remark that this family is also unipotent.
In addition, for $n=2$ we prove that the above two are the only types possible. By applying suitable conjugation matrices, we get a reduced form where we prove that it is either of the given forms above or $A$ will fail to remain in $M_2(\mathbb Q)$ upon repeated squareroots (equivalently $A=B^{2^k}$ has no solutions for some large enough $k$), or fail $A=B^3$ for one particular case.
# A. Reduction of $M_2(\mathbb Q)$ via conjugation
Let $S\in M_n(\mathbb Q)$ with $\det(S)\neq 0$, so that $S$ is invertible and $S^{-1}\in M_n(\mathbb Q)$. Then for all $k\geq 1$, $$SAS^{-1} = S(B_k)^kS^{-1} = (SB_kS^{-1})^k$$ Therefore $A' = SAS^{-1}$ is also exceptional with solutions $SB_kS^{-1}$ for each $k\geq 1$.
Using the conjugation argument, we will show that any exceptional matrix $A= \begin{pmatrix} a & b \\ c & d \end{pmatrix}\in M_2(\mathbb Q)$ can be transformed to one of the $2$ types of (reduced) exceptional matrix: $$\begin{pmatrix} a & b \\ 0 & 1/a \end{pmatrix}, \begin{pmatrix} a & 1 \\ -1 & 0 \end{pmatrix}$$ for $c=0$ and $c\neq 0$ respectively.
For the former case, if $a=1$ then we have the construction given at the beginning of this answer. For the latter case, if $a=2$ then we have the other construction given at the start.
For all the rest of the cases, we show that we cannot take squareroots infinitely, hence contradicting that $A$ is exceptional. (For one particular case it fails due to no solutions to $A=B^3$.)
Therefore any exceptional matrix is (conjugate)-reduced to either $A_1=\begin{pmatrix} 1 & b\\ 0 & 1\end{pmatrix}$ or $A_2=\begin{pmatrix} 2 & 1\\ -1 & 0\end{pmatrix}$. This in turn means that all exceptional matrices are generated via $SA_1S^{-1}$ or $SA_2S^{-1}$.
# B. Solving repeated squareroots
We want to analyze repeated squareroots starting from $A$, which turns out to be enough to classify the matrices for $n=2$. Consider a solution to $$R^{2^k} = A$$ for some $k\geq 1$ (we will later let $k\rightarrow \infty$). Define $R_j=R^{2^{k-j}}$, so that $$R_{j}= (R_{j+1})^2$$ Then we have a series of equations \begin{align*} A &= R_{0}=(R_1)^2\\ R_{1} &= (R_{2})^2\\ R_{j} &= (R_{j+1})^2 \end{align*} Since $\det(A)=1$, each successive squareroot must have $\det(R_j) = \pm 1$. However since each of them (except the last, $R_k=R$) is a square, it must be $\det(R_j)=1$.
Let $tr(\cdot)$ be the matrix-trace function. The key result we want to derive is that
Proposition. Let the system of equations \begin{align*} A &= R_{0}=(R_1)^2\\ R_{1} &= (R_{2})^2\\ R_{j} &= (R_{j+1})^2 \end{align*} have a solution for $\det(R_j)=1$ and $A,R_j\in M_2(\mathbb Q)$, as $0\leq j\rightarrow \infty$. Then $tr(R_j)$ must be integral for each $0\leq j < k$. Moreover, the $tr(R_j)$'s satisfy the recurrence $$tr(R_{j+1})^2 = tr(R_j) + 2$$
Proof. By the Cayley-Hamilton theorem in dimension $n=2$, for any $R_i$ we have $$(R_i)^2 - tr(R_i)R_i + \det(R_i)I = 0$$ Therefore a necessary condition for $R_j = (R_{j+1})^2$, together with $\det(R_{j+1})=1$, is \begin{align*} R_j - tr(R_{j+1})R_{j+1} + I &= 0\\ tr(R_{j+1})R_{j+1} &= R_j + I \end{align*} Let $R_j = \begin{pmatrix} a & b \\ c & d\end{pmatrix}$ and taking determinant, \begin{align*} tr(R_{j+1})^2\det(R_{j+1}) &= \det\begin{pmatrix} a+1 & b \\ c & d+1\end{pmatrix} \\ tr(R_{j+1})^2 &= (a+1)(d+1) - bc \\ &= a+d+1 + (ad-bc)\\ &= a+d+2 \\ &= tr(R_j)+2 \end{align*} Therefore we have a recurrence relation $$tr(R_{j+1})^2 = tr(R_j) + 2$$ Note that $R_j\in M_2(\mathbb Q)$ so that $tr(R_j)\in \mathbb Q$.
Suppose that $tr(R_j)\not\in\mathbb Z$ for some $j$. We write $tr(R_j) = u_j/v_j$ for some $u_j,v_j\in \mathbb Z$ and $\gcd(u_j,v_j)=1$. Then $v_j\geq 2$. Now $$tr(R_{j+1})^2 = tr(R_j)+2 = \frac{u_j+2v_j}{v_j}$$ Since $\gcd(u_j+2v_j,v_j) = \gcd(u_j,v_j) = 1$, no further cancellation is possible and we must have $$tr(R_{j+1}) = \pm\frac{\sqrt{u_j+2v_j}}{\sqrt{v_j}} = \frac{u_{j+1}}{v_{j+1}}$$ If $\sqrt{v_j} \not \in \mathbb Z$, then $tr(R_{j+1})\not\in\mathbb Q$, contradicting $R_{j+1}\in M_2(\mathbb Q)$. So we assume $v_{j+1} = \sqrt{v_j}\in\mathbb Z$. Likewise, we must also have $u_{j+1} = \pm \sqrt{u_j+2v_j} \in\mathbb Z$.
Since we simply took a squareroot resulting in a rational, it must be the case that $\gcd(u_{j+1},v_{j+1}) = 1$, so we can reapply the same arguments for $tr(R_{j+2}) = u_{j+2}/v_{j+2}$. i.e. $$u_{j+2} = \pm \sqrt{u_{j+1} + 2v_{j+1}} \in \mathbb Z, v_{j+2} = \sqrt{v_{j+1}} \in \mathbb Z$$
Hence by induction we can see that the denominator is squareroot-ed every iteration. If $R_j\not \in \mathbb Z$ for some $j$, then this iterative squareroot process must result in an irrational denominator eventually, i.e. $tr(R_m) \not\in \mathbb Q$ for some large enough $m>j$. This contradicts $R_m\in M_2(\mathbb Q)$, therefore it must have been the case that $R_j\in \mathbb Z$ for all $j\geq 0$. $$\tag*{\square}$$
Let $A=\begin{pmatrix} a & b \\ c & d\end{pmatrix}$ be exceptional. We want to classify $A$ using the conjugation and repeated squareroots arguments established earlier.
# C1. Case $n=2$ and $c=0$
First, we look at the case $c=0$. Since $\det(A)=1$, we get $d=1/a$, giving us
$$A= \begin{pmatrix} a & b \\ 0 & 1/a \end{pmatrix}$$ Since $A$ is exceptional, we can take squareroots infinitely. By the proposition in B., we must have $tr(A) \in \mathbb Z$, say $tr(A) = t\in\mathbb Z$.
Write $a = u/v$ with $u,v\in\mathbb Z$ and $\gcd(u,v)=1$. Then \begin{align*} \frac{u^2+v^2}{uv} &= a + \frac{1}{a} = t\in \mathbb Z\\ u^2+v^2 &= tuv \end{align*} This shows that $u$ divides $v$ and vice-versa, which is only possible if $u,v = \pm 1$. Therefore $a=1$ or $a= -1$. If $a=-1$, then $tr(A) = -2$. Taking the first squareroot as $R_1$, the formula gives $$tr(R_1)^2 = tr(A) +2 = 0 \implies tr(R_1) = 0$$ But now for the second squareroot $R_2$ we have $$tr(R_2)^2 = tr(R_1)+2 = 2 \implies tr(R_2) = \pm \sqrt{2},$$ contradicting $tr(R_2)\in\mathbb Q$. Therefore the only possibility is $a=1$. We have seen that this results in the family of solution given at the start of this answer.
# C2. Case $n=2$ and $c\neq 0$
Choosing the conjugation matrices $$S= \begin{pmatrix} 0 & 1 \\ c & -a \end{pmatrix}, S^{-1} = \begin{pmatrix} a/c & 1/c \\ 1 & 0 \end{pmatrix},$$ we get a new exceptional matrix $A'$ $$A'= SAS^{-1} = \begin{pmatrix} a+d & 1 \\ bc-ad & 0 \end{pmatrix} = \begin{pmatrix} a+d & 1 \\ -1 & 0 \end{pmatrix}$$ where the last equality is due to the condition $\det(A)=1$.
Denote $r_0 = tr(A) = a+d$ and similarly let $r_j = tr(R_j)$, where $R_j$'s are the matrices obtained by repeated squareroots as in B.. By proposition B. again, we must have $r_j\in\mathbb Z$ and satisfy the recurrence $$r_{j+1}^2 = r_j + 2 \implies r_{j+1} = \pm \sqrt{r_j + 2}$$ Since $r_{j+1}\in\mathbb Z$ clearly $r_j\geq -2$ for each $j$.
For $r_0=2$, corresponding to $A=\begin{pmatrix} 2 & 1\\ -1 & 0\end{pmatrix}$, we have seen at the start of the answer that it is an exceptional matrix.
For $r_0 = -2,0,1$, we get $r_1 = 0, \pm \sqrt{2}, \pm \sqrt{3}$ respectively. $r_1=0$ in turn gives $r_2 = \pm \sqrt{2}$, so all of them contradicts $r_j\in\mathbb Z$.
For $r_0 > 2$, then $r_1^2 > 4$. Since $r_j > -2$ we must have $r_1>2$ again, so by induction $r_j > 2$ for all $j$. The sequence of $r_{j+1} = \sqrt{r_j+2}$ becomes smaller than $3$ for some sufficiently large $m$, then $2 < r_m < 3$ results in $$r_{m+1}^2 = r_m + 2 \implies 4 < r_{m+1}^2 < 5$$ which contradicts the fact that $r_{m+1}\in\mathbb Z$.
Since $r_0\in\mathbb Z$, this leaves the final case of $r_0 = -1$. For this case we use a numeric solver to show directly that there are no solutions to $$B^3 = \begin{pmatrix} -1 & 1\\ -1 & 0 \end{pmatrix}$$ Edit 1: Algebraic proof
Let $B$ be a solution to $A=B^3$ and $t=tr(B)$. Then $$\det(B)^3 = \det(A) = 1$$ Since $\det(B)\in\mathbb Q$ we conclude $\det(B)=1$. By Cayley-Hamilton: \begin{align*} B^2-tB+I &= 0\\ B^3-tB^2+B &= 0\\ A-t(tB-I)+B &= 0\\ (t^2-1)B &= A+tI = \begin{pmatrix} -1+t & 1\\ -1 & t\end{pmatrix}\\ (t^2-1)^2 \det(B) &= \det(A+tI) = (t-1)t + 1\\ t^4-2t^2+1 &= t^2-t+1\\ t(t^3-3t+ 1) &= 0 \end{align*} Since $t^3-3t+1$ has no solutions $\pmod 2$ it is irreducible in $\mathbb Z$ and hence also irreducible in $\mathbb Q$. Since $t\in \mathbb Q$, the only possibility is $t=0$. Therefore $-B = A\implies B = -A$. However we can check manually that $B^3 = -A^3 = -I \neq A$, hence there cannot be a solution $B$.
End edit 1
Hence, going through all integer possibilities of $r_0=a+d$, we see that only $r_0=2$ results in an exceptional matrix.
This completes all the cases.
• Wow, what an impressive work! Thank you so much. There are many cool ideas here (my favorite is the trick using Cayley-Hamilton in Edit 1). – Prism Jul 18 '18 at 21:55
• @Prism Thanks for the kind words, I apologize for the wall of text since it feels like they could have been more concise. I just noticed that since conjugation $SAS^{-1}$ does not change the trace, in some sense this is proving that the requirement is exactly just $tr(A)=2$. The unipotent condition then comes from the characteristic polynomial: $$A^2 -2A+I = 0 \implies (A-I)^2 = 0$$ Perhaps solving for trace directly might a more useful idea for higher dimensions, since the current method is too computational. – Yong Hao Ng Jul 19 '18 at 3:09 | 2019-06-17T04:39:48 | {
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http://mathhelpforum.com/algebra/104989-find-all-real-imaginary-roots-print.html | # find all real and imaginary roots
• Sep 29th 2009, 06:29 AM
absvalue
find all real and imaginary roots
I need to find all real and imaginary roots of this equation:
$2x^7 - 2x^6 + 7x^5 - 7x^4 - 4x^3 + 4x^2 = 0$
Using synthetic division, I found that 1 is a root.
$(x - 1)(2x^6 + 7x^4 - 4x^2) = 0$
I'm not really sure how to proceed from here. Could someone point me in the right direction?
• Sep 29th 2009, 06:56 AM
Soroban
Hello, absvalue!
Quote:
Find all real and imaginary roots: . $2x^7 - 2x^6 + 7x^5 - 7x^4 - 4x^3 + 4x^2 \:=\: 0$
First, factor out $x^2$ . . .
. . $x^2(2x^5 - 2x^4 + 7x^3 - 7x^2 - 4x + 4) \:=\:0$
Factor by grouping:
. . $x^2\,\bigg[2x^4(x-1) + 7x^2(x-1) - 4(x-1)\bigg] \:=\:0$
. . $x^2(x-1)(2x^4 - 7x^2 - 4) \:=\:0$
. . $x^2(x-1)(x^2-4)(2x^2+1) \:=\:0$
Then we have:
. . $x^2 \:=\:0 \quad\Rightarrow\quad\boxed{ x \:=\:0}$
. . $x - 1 \:=\:0 \quad\Rightarrow\quad\boxed{ x \:=\:1}$
. . $x^2 - 4 \:=\:0 \quad\Rightarrow\quad x^2 \:=\:4 \quad\Rightarrow\quad\boxed{ x \:=\:\pm2}$
. . . . $2x^2+1 \:=\:0 \quad\Rightarrow\quad x^2 \:=\:-\frac{1}{2} \quad\Rightarrow\quad\boxed{x \:=\:\pm\frac{i}{\sqrt{2}}}$
• Sep 29th 2009, 07:01 AM
Hello absvalue
Quote:
Originally Posted by absvalue
I need to find all real and imaginary roots of this equation:
$2x^7 - 2x^6 + 7x^5 - 7x^4 - 4x^3 + 4x^2 = 0$
Using synthetic division, I found that 1 is a root.
$(x - 1)(2x^6 + 7x^4 - 4x^2) = 0$
I'm not really sure how to proceed from here. Could someone point me in the right direction?
Take out a common factor $x^2$ from your second factor; you then have a quadratic in $x^2$, which can be factorised in the usual way:
$(x-1)(2x^6+7x^4-4x^2) = x^2(x-1)(2x^4+7x^2-4)$
$=x^2(x-1)(2x^2-1)(x^2+4)$
Can you complete it from here? | 2017-10-20T07:21:19 | {
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https://www.projectrhea.org/rhea/index.php?title=ECE_301_Fall_2007_mboutin_Examples&oldid=485 | Example of CT convolution
This is an example of convolution done two ways on a fairly simple general signal.
$x(t) = u(t)\$
$h(t) = {e}^{-\alpha t}u(t), \alpha > 0\$
Now, to convolute them...
1. $y(t) = x(t)*h(t) = \int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau$
2. $y(t) = \int_{-\infty}^{\infty}u(\tau){e}^{-\alpha (t-\tau)}u(t-\tau)d\tau$
3. Since $u(\tau)*u(t-\tau) = 0\$ when t < 0, also when $\tau > t\$, you can set the limit accordingly. Keep in mind the following steps (4&5) are for t > 0, else the function is equal to 0.
4. $y(t) = \int_{0}^{t} {e}^{-\alpha (t-\tau)}d\tau = {e}^{-\alpha t} \int_{0}^{t}{e}^{ \alpha \tau}d\tau$
5. $y(t) = {e}^{-\alpha t}\frac{1}{\alpha}({e}^{\alpha t}-1) = \frac{1}{\alpha}(1-{e}^{-\alpha t})$
6. Now you can replace the condition in steps 4&5 with a u(t).
7. $y(t) = \frac{1}{\alpha}(1-{e}^{-\alpha t})u(t)$.
Now, the other way... (by the commutative property)
1. $y(t) = h(t)*x(t) = \int_{-\infty}^{\infty}h(\tau)x(t-\tau)d\tau$
2. $y(t) = \int_{-\infty}^{\infty}{e}^{-\alpha (\tau)}u(\tau)u(t-\tau)d\tau$
3. Since $u(\tau)*u(t-\tau) = 0\$ when t < 0, also when $\tau > t\$, you can set the limit accordingly. Keep in mind the following step (4) is for t > 0, else the function is equal to 0.
4. $y(t) = \int_{0}^{t} {e}^{-\alpha \tau}d\tau = \frac{1}{\alpha}(1-{e}^{-\alpha t})$
5. Now you can replace the condition in step 4 with a u(t).
6. $y(t) = \frac{1}{\alpha}(1-{e}^{-\alpha t})u(t)$
End
Name --dennis.m.snell.1, Sun, 30 Sep 2007 22:25:27
Name --michael.a.mitchell.2, Mon, 01 Oct 2007 15:54:00
Wasn't Sure if the authorship issue had been solved yet. (in class it was said that only the last person to make a change to a page would be credited with it's authorship)
Name --dennis.m.snell.1, Mon, 01 Oct 2007 16:51:27
The authorship issue was not an issue. It was mentioned in class, but by a student asking about it. There is a log of every action and every edit on this kiwi that can be reviewed each week. You are safe in leaving out your name. Sometime soon the editing will be reworked; however, and you might add your name to some other special page, but it will just get lost at the bottom of a topic. I removed your name here, but worry not, you are not forgotten.
... --john.w.fawcett.1, Mon, 15 Oct 2007 11:15:56
why is this under "Exams" as it's parent? Wouldn't Chapter 3 be better?
... --john.w.fawcett.1, Mon, 15 Oct 2007 11:18:06
Sorry, meant Chapter 2. I'll go ahead and add a backlink to chapter 2, but leave this one to Exams up for now.
Frequency and Impulse Response of a causal LTI system defined by a difference equation
For the discrete time L.T.I. system described by
$y[n]-\frac{1}{2}y[n-1]=x[n]+\frac{1}{2}x[n-1]$
Find the frequency response H($\omega\$) and the impulse response h[n] of the system.
Frequency Response:
1: Take the Fourier transform of the equation,
$Y(\omega)-\frac{1}{2}e^{-j\omega}Y(\omega)=X(\omega)+\frac{1}{2}e^{-j\omega}X(\omega)$
2: Solve for Y($\omega\$)/X($\omega\$), which is the frequency response H($\omega\$),
$H(\omega)=\frac{Y(\omega)}{X(\omega)}=\frac{1+\frac{1}{2}e^{-j\omega}}{1-\frac{1}{2}e^{-j\omega}}$
Impulse Response:
1: Expand into two terms using partial fraction expansion (Guide to Partial Fraction Expansion) to facilitate use of inverse Fourier transform,
$H(\omega)=\frac{1}{1-\frac{1}{2}e^{-j\omega}}+\frac{1}{2}\frac{e^{-j\omega}}{1-\frac{1}{2}e^{-j\omega}}$
2: Take the inverse Fourier transform of H($\omega\$) (Fourier Transform Table),
$h[n]={\left(\frac{1}{2} \right)}^{n}u[n]+\frac{1}{2}{\left(\frac{1}{2} \right)}^{n-1}u[n-1]$
3: Simplify if so inclined,
for n = 0
$h[n] = 1\$
for n > 0
$h[n] = {\left(\frac{1}{2} \right)}^{n-1}$ | 2019-11-15T16:20:16 | {
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http://moestuininfo.com/cockney-rhyming-dabcq/670e52-what-is-a-root-in-math-quadratic-equation | The roots of the equation are the values of x at which ax² + bx + c = 0. Solve an equation of the form a x 2 + b x + c = 0 by using the quadratic formula: x = − b ± √ b 2 − 4 a c: 2 a: Step-By-Step Guide. When only one root exists both formulas will give the same answer. The root of a quadratic equation Ax 2 + Bx + C = 0 is the value of x, which solves the equation. where the plus-minus symbol "±" indicates that the quadratic equation has two solutions. We can calculate the root of a quadratic by using the formula: x = (-b ± √(b 2-4ac)) / (2a). This is generally true when the roots, or answers, are not rational numbers. It use it to 'discriminate' between the roots (or solutions) of a quadratic equation. In case of a quadratic equation with a positive discriminate, the roots are real while a 0 discriminate indicates a single real root. This curve is called a parabola. \"x\" is the variable or unknown (we don't know it yet). Algebra. Sometimes the roots are different, sometimes they're twins. The value of the variable A won't be equal to zero for the quadratic equation. We have imported the cmath module to perform complex square root. The number b^2 -4ac is called the discriminant. $$B^2 – 4AC = (-3)^2 – ( 4 \times 1 \times 2 )$$, $$x_{1} = \frac{-B}{2A} + \frac{\sqrt{B^2 – 4AC}}{2A}$$, $$= \frac{-(-3)}{2 \times 1 } + \frac{\sqrt{1}}{2 \times 1}$$ $$\hspace{0.5cm}using\hspace{0.5cm}B^2 – 4AC = 1$$, $$= \frac{3}{2 } + \frac{1}{2} = \frac{3+1}{2 } = \frac{4}{2} = 2$$, $$x_{2} = \frac{-B}{2A} – \frac{\sqrt{B^2 – 4AC}}{2A}$$, $$= \frac{-(-3)}{2 \times 1 } – \frac{\sqrt{1}}{2 \times 1}$$, $$= \frac{3}{2 } – \frac{1}{2} = \frac{3-1}{2 } = \frac{2}{2} = 1$$. Quadratic functions may have zero, one or … A quadratic equation has two roots which may be unequal real numbers or equal real numbers, or numbers which are not real. In this case, the quadratic equation has one repeated real root. Let's check these values: (-3)^2 +8*-3 +15 = 9 - 24 + 15 = 0 and (-5)^2 + 8*-5 +15 = 25 - 40 + 15 = 0. When a is negative, this parabola will be upside down. So when you want to find the roots of a function you have to set the function equal to zero. The quadratic formula gives two solutions, one when ± … Therefore the square root does not exist and there is no answer to the formula. Then the root is x = -3, since -3 + 3 = 0. For functions of degree four and higher, there is a proof that such a formula doesn't exist. Using the formula above we get: $$= \frac{-6}{2 \times 1} = \frac{-6}{2 } = -3$$. Here are some examples: We have ax^2 + bx + c. We assume a = 1. So indeed, the formula gives the same roots. Quadratic Equation. For the Quadratic Formula to work, you must have your equation arranged in the form "(quadratic) = 0".Also, the "2a" in the denominator of the Formula is underneath everything above, not just the square root.And it's a "2a" under there, not just a plain "2".Make sure that you are careful not to drop the square root or the "plus/minus" in the middle of your calculations, or I can guarantee … Strictly speaking, any quadratic function has two roots, but you might need to use complex numbers to find them all. root1 = (-b + √(b 2-4ac)) / (2a) root1 = (-b - √(b 2-4ac)) / (2a). The ABC Formula is made by using the completing the square method. The idea of completing the square is as follows. The quadratic formula can solve any quadratic equation. Nature of the roots of a quadratic equations. Determining the roots of a function of a degree higher than two is a more difficult task. So if we choose s = -3 and t = -5 we get: Hence, x = -3 or x = -5. It might however be very difficult to find such a factorization. Forums. The standard form of a quadratic equation is: ax 2 + bx + c = 0. (x-s)(x-t) = 0 means that either (x-s) = 0 or (x-t)=0. A discriminant is a value calculated from a quadratic equation. So we have a single irrational root in this case. There are however some field where they come in very handy. Value of determinant B2 – 4AC, defines the nature of roots of a Quadratic Equation Ax2 + Bx + C = 0. It is also called an "Equation of Degree 2" (because of the "2" on the x) Standard Form. Student what is the relation between discriminate root and 0. Then, to find the root we have to have an x for which x^2 = -3. As -9 < 0, no real value of x can satisfy this equation. An equation in one unknown quantity in the form ax 2 + bx + c = 0 is called quadratic equation. This means that x = s and x = t are both solutions, and hence they are the roots. For example: Then the root is x = -3, since -3 + 3 = 0. Solving quadratic equations by completing square. The solution of quadratic equation formulas is also called roots. If any quadratic equation has no real solution then it may have two complex solutions. One example is solving quadratic inequalities. In most practical situations, the use of complex numbers does make sense, so we say there is no solution. Sometimes they all have real numbers or complex numbers, or just imaginary number. An example of a quadratic function with only one root is the function x^2. This is the case for both x = 1 and x = -1. This means to find the points on a coordinate grid where the graphed equation crosses the x-axis, or the horizontal axis. Coefficients A, B, and C determine the graph properties and roots of the equation. Learn all about the quadratic formula with this step-by-step guide: Quadratic Formula, The MathPapa Guide; Video Lesson. Let α and β be the roots of the general form of the quadratic equation :ax 2 + bx + c = 0. In the above formula, (√ b 2-4ac) is called discriminant (d). So let us focus on it. Lastly, we had the completing the squares method where we try to write the function as (x-p)^2 + q. The most common way people learn how to determine the the roots of a quadratic function is by factorizing. The root is the value of x that can solve the equations. $$\frac{-1}{3}$$ because it is the value of x for which f(x) = 0. f(x) = x 2 +2x − 3 (-3, 0) and (1, 0) are the solutions to this equation since -3 and 1 are the values for which f(x) = 0. Solutions or Roots of Quadratic Equations . Example 5: The quadratic equations x 2 – ax + b = 0 and x 2 – px + q = 0 have a common root and the second equation has equal roots, show that b + q = ap/2. The solution to the quadratic equation is given by 2 numbers x 1 and x 2.. We can change the quadratic equation to the form of: Square roots frequently appear in mathematical formulas elsewhere, as well as in many physical laws. If no roots exist, then b^2 -4ac will be smaller than zero. These roots are the points where the quadratic graph intersects with the x-axis. There are several methods for solving quadratic equation problems, as we can see below: Factorization Method. Sqaure roots, quadratic equation factorer, ordering positive and negative integer worksheets, zeros vertex equation, 8th grade math sheet questions. then the roots of the equation will be. If (x-s)(x-t) = x^2 + px + q, then it holds that s*t = q and - s - t = p. Then we have to find s and t such that s*t = 15 and - s - t = 8. Answer: The value of 1 and 5 are the roots of the quadratic equation, because you will get zero when substitute 1 or 5 in the equation. However, it is sometimes not the most efficient method. x1 = (-b + D)/2a ,and Quadratic equation definition is - any equation containing one term in which the unknown is squared and no term in which it is raised to a higher power. 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https://math.stackexchange.com/questions/4258161/why-do-two-half-toruses-add-up-to-the-same-volume | # Why do two half-toruses add up to the same volume?
We have a small torus $$A$$ with $$R=\frac{11}{2}+0.0005$$ and $$r=0.0005$$. Look at it from the top, and cut along the circle $$R$$ traces when spun around the center of the torus, and we get the inner and outer half of a torus, inner half clearly with less volume compared to the outer half.
Another similar torus $$B$$ with $$R=\frac{13}{2}$$ and $$r=0.0005$$ is also cut in a similar manner. The outer half clearly has more volume than the inner half.
(I have noted that the different halves are not equal to the volume of half cylinders since one side is flat and thus the volume displacement will not be equal to the original.)
Now, I calculate the volumes of the inner half of A, and the outer half of B using shell integration by setting the center of the torus at $$(0,0)$$ and rotating the equation of a circle about the line $$x=0$$, and get $$2\cdot\int^{5.5}_{5.4995}\left(2\pi\cdot x\cdot\sqrt{.0005^2-\left(x-\frac{11}{2}\right)^2}\right)\text{ d}x=\frac{\pi(8250\pi-1)}{6000000000}$$ and $$2\cdot\int^{6.5005}_{6.5}\left(2\pi\cdot x\cdot\sqrt{.0005^2-\left(x-\frac{13}{2}\right)^2}\right)\text{ d}x=\frac{\pi(9750\pi+1)}{6000000000}$$ respectively.
I then calculated the same volumes assuming they were equal to half cylinders with the height equal to the circumference of a circle with radius $$R$$, and got $$\frac{11\pi^2}{8000000}$$ and $$\frac{13\pi^2}{8000000}$$ respectively.
I notice something really odd - the volume(assuming half cylinders) for the inner half of A is $$\frac{\pi}{6000000000}$$ larger than the result I got from shell integration, and the outer half of B was exactly $$\frac{\pi}{6000000000}$$ smaller.
Why is this? These two toruses clearly have different $$R$$, and I don't see any reason why the volume inaccuracies for the inner half of $$A$$ and the outer half of $$B$$ (when you calculate them as half cylinders) would be the same and cancel each other out if we add them?
Thanks, Max0815
• If I understand what you mean by "calculated [the volumes] assuming they were half cylinders", this sounds like a consequence of Pappus's theorem; the two tori have the same cross-section, just different "major" radii. Sep 23 at 12:54
• @AndrewD.Hwang could you elaborate on that please? Why would just having the same cross section result in the same difference in volumes between two toruses with different major radii? Sep 23 at 12:55
• By Pappus, the four volumes are the products of the area of the half-disk multiplied by the distance traveled by the centroid. The half-cylinder volumes will depend on how you decide the height of the half-cylinder, but we can equate the heights to the circumferences of circles. Note that if you add or subtract the same amount $\delta$ from the radii of two (possibly different) circles, you add or subtract $2\pi\delta$ from the circumference of each circle. Sep 23 at 13:29
• To explain in more detail it would help if you show the integrals or at least show precisely how you determined the height of each half-cylinder. We could reverse-engineer this information from your results, of course, but wouldn't it be nice to show your work? Sep 23 at 13:31
• @DavidK Here are my integrals: $2\cdot\int^{5.5}_{5.4995}\left(2\pi\cdot x\cdot\sqrt{.0005^2-\left(x-\frac{11}{2}\right)^2}\right)\text{ d}x$ for the inner part of $A$, and $2\cdot\int^{6.5005}_{6.5}\left(2\pi\cdot x\cdot\sqrt{.0005^2-\left(x-\frac{13}{2}\right)^2}\right)\text{ d}x$ for the outer part of $B$. I set the integrals up with the center of the torus at (0,0) and the torus constructed from revolving the equation of a circle around x=0. When I calculate the height of a half cylinder, I take $2\pi R$ where $R$ is the major radii of the given torus. Sep 23 at 13:44
For many problems it is advantageous to not to plug in numbers too early. If you can get a generic formula, then it's easier to analyze the working of the problem, and it's also easier to check that the units / dimensions are right. And in general it's less tedious to write one symbol like $$r$$ than to carry around strings like "0.0005 cm" and such.
Additionally, in the case of a torus there is Guldin's Rule which states that the volume of a body of revolution is $$\text{Area-that-revolves} \times \text{Length-of-path-tracked-by-center-of-mass}^1$$
• Each half-torus is generated by rotating a half-disc of radius $$r$$ and area $$A=\pi r^2/2$$.
• The centroid of a semicircle of radius $$r$$ lies at $$r_c=\dfrac4{3\pi}r$$ away from the flat side and on the axis of symmetry. I looked the value up, but it's straight forward enough to calculate on your own$$^2$$.
• The paths tracked by the centroids is thus $$\ell_\pm=2\pi(R\pm r_c)$$ where $$\ell_+$$ is for the outer half-torus and $$\ell_-$$ for the inner one.
Guldin's rule then yields the volumina \begin{align} V_\pm &= \ell_\pm\cdot A = 2\pi(R\pm r_c) \cdot \frac12 \pi r^2 = \pi^2 r^2 \left(R \pm \frac4{3\pi}r\right) \\ &= \pi^2 r^2 R \,\pm\, \frac43 \pi r^3 \end{align} which we can express as $$V_\pm = \frac12 V_\text Z \,\pm\, \frac43 \pi r^3$$ where $$V_\text Z = 2\pi R\cdot\pi r^2$$ is the volume of a cylinder of radius $$r$$ and length $$2\pi R$$. (And the summand after the ± is (incidentally?)$$^3$$ the volume of a sphere of radius $$r$$.)
Now we can plug in numbers, where I won't do the $$V_\text Z$$ part. Your $$r=0.0005=5\cdot10^{-4}$$ gives \begin{align} V_\pm - \frac12 V_\text Z &= \frac43 \pi r^3 \\ &= \frac43 \pi \cdot 0.0005^3 = \frac43 \pi \cdot 1.25 \cdot 10^{-10} \\ &= \frac \pi{6 \cdot 10^9} \end{align}
And there you have it!
Why is this?
Because the difference in volume is independent of $$R$$ and only depends on $$r$$. This is much easier to infer from formulae than by staring at magic numbers :-) ...and there wasn't even a need to plug in $$r=0.0005$$.
$$^1$$Perhaps "Centroid" is more common in English than center of mass.
$$^2$$Tthe needed integral is hidden in your calculation already.
$$^3$$Actually not an incident; it's nice to see how the sphere is being "generated" here.
First, let's define distinct symbols for the "$$R$$" dimensions of the two toruses: $$R_1 = 5.5,$$ $$R_2 = 6.5.$$
As an attempt at an intuitive understanding, let's consider first the half cylinder that you are comparing to the "outer half" of the larger torus. You used a cylinder of height $$2\pi R_2.$$ But you cannot just bend this cylinder around a circle of radius $$R_2$$ to form the "outer half" of the torus, because the very outermost part of the "outer half" has to fill all the circumference of a circle of radius $$R_2 + r.$$ the particular strip of the torus immediately next to that circle has to have length $$2\pi R_2 + 2\pi r,$$ and you only have material of length $$2\pi R_2.$$
So let's cut the half cylinder into very thin slices perpendicular to its axis. Now you can take those slices and arrange them around outside of the circle of radius $$R_2.$$ But there will be tiny wedge-shaped gaps between the slices. If we measure the thickness of each wedge at its thickest point, which is along the circle of radius $$R_2 + r,$$ the total thickness of the wedges will be $$2\pi r.$$
Now try to take the other half cylinder and wrap it around the inside of the circle of radius $$R_1.$$ You cannot do it, because now there is too much material: along the innermost part of the torus you have room only for material of length $$2\pi R_1 - 2\pi r,$$ but you have material of length $$2\pi R_1.$$ To make the half-cylinder fit in this space, you can slice it into many thin slices perpendicular to its axis, and then slide an even thinner wedge off each of the thin slices. The total thickness of wedges that you have to slice off will be $$2\pi r.$$
The wedges you have to slice out of the smaller half-cylinder look a lot like the wedges that are missing from the larger torus: laid on a table, each has the shape of a semicircle of radius $$r$$; the thin edge of the wedge is along the diameter of the semicircle, and the thickest part of the wedge is halfway along the semicircular arc.
If you cut each half-cylinder into the same (large) number of slices, you have the same number of wedges cut from one half-cylinder as the number of gaps between slices of the other half-cylinder; the wedges are all (just about) identical, and the total thickness of each set of wedges is the same. So the wedges you cut off one half-cylinder will (just about) exactly fill all the gaps between slices of the other half-cylinder. (In the limit as the number of slices grows without bound, we can drop the "just about": the wedges fit the gaps exactly.)
The difference in volume between each half-cylinder and the "inner half" or "outer half" of the corresponding torus is just the total volume of wedges you have to cut off or fill in. And since it is the same set of wedges, it is the same volume. As you can verify, the difference you found in each case between the "half torus" volume and the half-cylinder volume was just exactly the volume of a sphere of radius $$r = 0.0005.$$
As a bonus, you can assemble the wedges into a sphere by putting the thin edge of each wedge on a shared axis. The circumference of the equatorial circle of the sphere is $$2\pi r,$$ equal to the total thickness of the wedges (whose thickest parts are all lined up along that equatorial circle).
More rigorously, in terms of Pappus' Theorem, the centroid of the semicircle for the "outer half" of the large torus moves along a circle of radius $$R_2 + 4r/(3\pi).$$ The volume of that "half" of the torus is therefore $$(R_2 + 4r/(3\pi))\pi r^2,$$ exactly $$(4r/(3\pi))\pi r^2$$ greater than the volume of the corresponding half-cylinder.
The centroid of the semicircle for the "inner half" of the small torus moves along a circle of radius $$R_2 - 4r/(3\pi).$$ The volume of that "half" of the torus is therefore $$(R_2 - 4r/(3\pi))\pi r^2,$$ exactly $$(4r/(3\pi))\pi r^2$$ less than the volume of the corresponding half-cylinder.
The volumes added and subtracted are equal because the centroid of one semicircle is the same distance outside the circle of radius $$R_2$$ as the centroid of the other semicircle is inside the circle of radius $$R_1.$$ And when you add or subtract equal amounts from the radiuses of any two circles, you add or subtract equal amounts from their circumferences.
• Thank you so much! The intuitive explanation really helped, and the rigorous explanation was the icing in the cake. ----- Just to clarify one thing, so no matter the value of $R$ for any pair of arbitrary toruses $A$ and $B$, as long as $r$ of both are the same, if we take the inner part of $A$ and outer part of $B$, the volume difference(whether positive/more or negative/less) that they have(when we estimate them to be half cylinders with height of $2\pi R$) will be the same? Sep 23 at 17:45
• Also I'm slightly confused on where you got the $2\pi r$ as the extra/less volume, but I'm sure if I play with Pappus' Theorem I'll find the answer. Sep 23 at 18:06
• $2\pi r$ is the difference between the circumferences of the two circles that the "half torus" sits between. That implies it is the total thickness of the wedges, not the volume of anything. Sep 23 at 23:08 | 2021-10-20T04:07:53 | {
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https://www.themathdoctors.org/when-your-answer-doesnt-match-the-books/ | #### (A new question of the week)
In many areas of math, an answer can come in several forms, which can make it hard to know if you are right when you compare your answer to the answer in the back of the book. Even worse is when the problem is multiple-choice, and your answer has a different form than the choices given. This can happen very easily in trigonometry. What do you do?
This happened to a student recently, and he wrote to us to help unravel it. Here is his question:
Here is the problem I am working on:
The general solution of the equation $$\tan^2\alpha + 2\sqrt{3} \tan\alpha = 1$$ is given by …
(A) $$\alpha = \frac{n\pi}{2} (n\in I)$$
(B) $$\alpha = (2n+1)\frac{\pi}{2} (n\in I)$$
(C) $$\alpha = (6n+1)\frac{\pi}{12} (n\in I)$$
(D) $$\alpha = \frac{n\pi}{12} (n\in I)$$
I have tried solving it this way:
Firstly I applied the Quadratic Formula to get,
$$\tan(\alpha) = -(2+\sqrt{3}) \text{ or } (-\sqrt{3}+2)$$
Now we have two cases,
CASE 1: When $$\tan(\alpha) =\tan(\pi/12) = 2-\sqrt{3}$$.
So General Formula here will be, $$\alpha =n\pi + \pi/12$$.
Now, CASE 2: When $$\tan(\alpha) =\tan(-5\pi/12) = -(2+\sqrt{3})$$.
So General Formula here will be $$\alpha = n\pi – 5\pi/12$$.
I do not know what should I do next to get the answer? Please tell me how to proceed further.
The answer given in the key is the option (C).
I will be thankful for any help!
He’s done good work, solving the quadratic equation to find two possible values for $$\tan(\alpha)$$, and then finding all values of $$\alpha$$ for each case. The difficulty is that the choices are given in ways that blend all solutions into a single formula using an integer parameter, while Navneet has two separate formulas, neither of which looks quite like any of the choices, though there is a clear similarity. How do you blend them? Or is there another way to decide?
After checking his work, I answered:
You have the solution; now you just have to put it in the desired form.
For this sort of multiple-choice problem, all you need to do is to compare your solution to each choice. Your solution, written as a pair of lists, is (taking n=0, 1, 2, …)
α = π/12, 13π/12, 25π/12, … or -5π/12, 7π/12, 19π/12, …
That is, putting these together,
α = …, -5π/12, π/12, 7π/12, 13π/12, 19π/12, 25π/12, …
There are two directions to go from here. Either do what I just did with each of the choices, and find which one matches; or observe that successive solutions actually differ by 6π/12 = π/2, so that the solution in simpler form is a single arithmetic sequence,
α = π/12 + n π/2.
Now you can see that two of the choices have π/12 factored out, so you can do that to your solution:
α = π/12 (1 + 6n)
Of course, if this were not multiple-choice, your original answer would have been acceptable, though you might want to do the same work I did here just to find a somewhat neater form.
The fact that the question was multiple-choice provided a hint at the desired form (a single combined formula), which would not have been present in an open-ended question (where we would have faced the similar issue of determining whether the answer in the book agreed with ours). This is similar to the difference between a problem that asks us to simplify an expression (which might have many “simpler” forms, all equally valid in the absence of specific instructions), versus one that asks us to prove a trig identity (where we are in essence told what the answer is, and just have to get there). Seeing the form of the choices, we had some ideas for directions to take our work.
(We could instead have used the basic strategy for multiple-choice problems, not actually solving at all but just checking each proposed solution. A and D both make 0 a solution; clearly that doesn’t work in the equation. B makes π/2 a solution, which is wrong (the tangent is undefined). So only C is left. But I can’t stand doing that; it works against the whole idea of learning, testing an entirely different skill than what is supposed to be learned. A better-designed question might at least make this method harder, and force you to do a little actual math.)
Four days later, the same student asked a related question that was not in a multiple-choice context and did not involve merging two sets of solutions, but was actually more interesting:
I am again facing a problem in solving a question of this type.
See question number 3 in the attachments:
Problem 3:
Solve the equality: $$2 \sin 11x + \cos 3x + \sqrt{3}\sin 3x = 0$$
$$\displaystyle x = \frac{n\pi}{7} – \frac{\pi}{84} \text{ or } x = \frac{n\pi}{4} + \frac{7\pi}{48},\ n \in I$$
This is my solution:
One of the solutions is matching with the solution given in my textbook but the other solution is not matching.
Why isn’t my solution matching with the solution given in my textbook?
I will be thankful for help!
This is an open-ended problem, where our difficulty is not in choosing an answer, but in deciding whether we are right. Navneet’s first answer should agree with the book’s second, but looks quite different. In fact, if we try choosing a value for n and seeing whether the answers agree, it looks at first as if it must be wrong.
At first I thought this might be an actual arithmetic error, but then I saw that it is just a more complicated version of the previous kind of issue.
First, I was impressed by the way you wrote a single expression initially for all angles whose sine is the same as for -11x; but then I saw that you turned that into what I would have written first, giving two separate cases (for n odd and even). I presume the first form is one you were taught, that I have not seen.
All your work is excellent, and very clearly written.
But your answer seems different from theirs in the odd case. Why? To see what they probably did, try replacing 2n+1 with 2n-1 in your work, which is equally valid, and see what you get!
It turns out that they, again, just made a different choice than you, which led to their n having a different meaning in that case. Looking at the problematic half of the answer, they have n pi/4 + 7 pi/48 [= (12n + 7)pi/48], while you have -n pi/4 – 5 pi/48 [= (-12n -5)pi/48]. Here are the values for some values of n:
n theirs yours
-2 -17pi/48 19pi/48
-1 -5pi/48 7pi/48
0 7pi/48 -5pi/48
1 19pi/48 -17pi/48
2 31pi/48 -29pi/48
These are, in fact, the same solutions, but in reverse order and starting at different places! As before, doing this at least can reassure you that your solution is correct, even though it has a different form.
When you do what I suggested above, you should see why.
Interestingly, I recently wrote a blog post in part about a similar issue that occurs in calculus, where two solutions can look entirely different, and the difference turns out to be absorbed in an arbitrary constant. Here, the difference is “absorbed” by an index that can start in different places or move in different directions.
Let’s “do what I suggested above” to see what I meant. Here is Navneet’s work, using $$2n+1$$:
$$3x+\pi/6 = (2n+1)\pi + 11x$$
$$3x+\pi/6 = 2n\pi +\pi + 11x$$
$$3x – 11x = 2n\pi + \pi – \pi/6$$
$$-8x = 2n\pi + 5\pi/6$$
$$\displaystyle x = -\frac{n\pi}{4} – \frac{5\pi}{48}$$
Here is my work, using $$2n-1$$:
$$3x+\pi/6 = (2n-1)\pi + 11x$$
$$3x+\pi/6 = 2n\pi -\pi + 11x$$
$$3x – 11x = 2n\pi – \pi – \pi/6$$
$$-8x = 2n\pi – 7\pi/6$$
$$\displaystyle x = -\frac{n\pi}{4} + \frac{7\pi}{48}$$
This is their solution!
As we can see, when the book’s answer looks different from ours, we can (a) “spot check” some values to see that it doesn’t really disagree; (b) write both answers in a common form, such as a list of values, to see that they actually agree; or (c) take the form of their answer as an addendum to the question, suggesting the form we should try, and proceed in that direction. With experience, we can learn how an answer can be disguised (here, by reversing the direction of the index n), and not be too surprised.
Or, we can ask the Math Doctors!
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http://math.stackexchange.com/questions/417462/how-do-i-know-if-the-linear-system-has-a-line-of-intersection | How do I know if the linear system has a line of intersection?
I was wondering how can I determine if there is a line of intersection with any matrix?
For example, if I have the following matrix:
$$\left(\begin{array}{rrr|r} 1 & -3 & -2 & -9 \\ 2 & -5 & 1 & 3 \\ -3 & 6 & 2 & 8 \\ \end{array} \right)$$
What does the solution have to look like for me to conclude that there is a line of intersection?
P.S. I know this matrix has a point of intersection but I used this as an example because I didn't know an example for a matrix that had a line of intersection.
-
If there are infinitely many solutions, then you are guaranteed to find a line (more generally subspace) of solutions. An equivalent condition is if the rank of the matrix is not full, and at least 1 solution exists. – Calvin Lin Jun 11 '13 at 14:03
The existence of a "point of intersection" is the existence of a point $(x, y, z)$ that satisfies the system of equations: a point lying on each line represented by the corresponding system of equations:
\begin{align} x - 3y -2z & = -9 \\ 2x - 5y + z & = 3 \\ -3x + 6y + 2z & = 8 \end{align}
A unique solution exists (a single point of intersection exists) if the augmented matrix does not reduce to a row of all zeros, and no row has all zeros, augmented by a non-zero entry.
If you obtain a row of all zeros, through row reduction, then infinitely many points of intersection occur (two or more lines will be concurrent): the entries in one or more lines will be a scalar multiple of the entries of another. Put differently, there will be a line of intersection.
If the matrix reduces to a row of three zeros, with a non-zero entry in the last column of that row, no solution exists (i.e., no point of intersection exists.)
All these possibilities can be determined by reducing the matrix to row echelon form.
We can reduce your example matrix:
$$\left(\begin{array}{rrr|r} 1 & -3 & -2 & -9 \\ 2 & -5 & 1 & 3 \\ -3 & 6 & 2 & 8 \\ \end{array}\right)$$
$$\left(\begin{array}{rrr|r} 1 & -3 & -2 & -9 \\ 0 & 1 & 5 & 21 \\ 0 & 0 & 11 & 42 \\ \end{array}\right)$$
If reduced further, you'd see that the system represented by the system of equations has a unique point of intersection, hence no common line of intersection.
Note that if the last row were $(0\;\;2\;\;10\;\;42)$, we could "zero it" by taking $-2R_2 + R_3 \rightarrow R_3$, and obtain a row of all zeros, since the third row would be a scalar multiple of the second row. In that event, there would indeed be a line of intersection. If any row is a linear combination of the other rows, we have a linearly dependent system of equations: this shows when a row-reduced matrix has a row of all zeros. And in that case, there is at least a line of intersection.
-
So if the third row of the system is a multiple of the second row or the first row then there is a line of intersection? I'm confused about how it would look in terms of $x$'s and $y$'s and $z$'s. Because for a the final answer for a P.O.I I would just write the solution but what would I write for a L.O.I? – Jeel Shah Jun 11 '13 at 14:43
Very nice Amy and good morning $\ast$ – Babak S. Jun 11 '13 at 14:44
@gekkostate What one typically does is set, say, $z = \alpha$, where $\alpha$ is any real number. Then express $x, y$ as values in terms of $z = \alpha$. So say you reduce the matrix to the rows the first two rows in the above row-reduced matrix, but the last row is all zeros. Then the solution is given by $z = \alpha, y = 21 - 5 \alpha, x = -9 + 2\alpha + 3(21 - 5\alpha)$, simplifying x, of course. $\alpha$ is then called a parameter. – amWhy Jun 11 '13 at 14:56
Some use the parameter $t$ instead of $\alpha$. This post gives an example of such a system: infinite number of solutions. – amWhy Jun 11 '13 at 15:02
If you are familiar with ranks and Rouché-Capelli (RC) theorem, the following reasoning can be used to answer. The system $A\mathbf x=\mathbf b$, where $A$ is an $n\times m$ matrix, has solutions if and only if $$r(A)=r(A|\mathbf b)=r,$$ and in this case you have "$\infty^{n-r}$" solutions. This really means that your solutions depend on $n-r$ free parameters. So, if the rank $r=n-1$, your solution set depends on $n-(n-1)=1$ free parameter or, in other words, it's monodimensional. This helps the intuition that the solution set is indeed a line in this case. Hence, you'll get a line of solutions when the rank of the complete matrix $r$ is equal to the number of unknowns $n$ less 1. As an example, consider the system \begin{align} x - y & = 2 \\ -2x +2y &=-4 \end{align} The rank $r(A)=r(A|\mathbf b)=1$ and you can indeed write the solutions set as $x=2+t,y=t$, which is equal to the set of points on the line $y=x-2$.
A more interesting example (I like a lot $A$!) is \begin{align} x +y +z& = 6 \\ 4x +5y +6z&=15\\ 7x+8y+10z&=24 \end{align} The reduced complete matrix is $$\left(\begin{array}{rrr|r} 1 & 2 & 3 & 6 \\ 0 & -3 & -6 & -9 \\ 0 & 0 & 0 & 0 \\ \end{array}\right)$$ from which you get the solutions $(x=t,y=3-2t,z=t), \forall t$. The solution set can be described in terms of a unique parameter $t$ and it is a straight line in $\mathbf R^3$.
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https://math.stackexchange.com/questions/1011884/is-the-greatest-common-divisor-injective-is-it-bijective | # Is the greatest common divisor injective? Is it bijective?
In an examination paper, there were the following questions:
1. Is gcd an injective function?
2. Is gcd a bijective function?
I found these questions odd because I thought that we need to first know the domain and codomain of a function before we can decide whether it is injective or bijective. The question did not specify what was the domain and codomain.
Since every integer divides 0 (except 0 itself), then gcd(0, 0) would be undefined as there is no "greatest" integer that divides 0 and 0. Hence, it seems reasonable to exclude 0 from the domain.
Also, since 0 does not divide any number, it is impossible for 0 to be the gcd of any two integers a and b, so 0 should be excluded from the codomain.
Now, suppose we let both the domain and codomain be the set of all positive natural numbers. Would this domain be valid? I am confused because the gcd function contains two arguments, i.e. a and b in gcd(a, b). Since this is the case, should the domain instead be the cartesian product N*N?
Clearly gcd is a function because it is not one-to-many. Every time we perform gcd we get exactly one output.
Is it correct to claim that gcd is not an injective function because it maps two natural numbers a and b to a single output c, i.e. gcd(a, b) = c, hence it is many-to-one? Or should the correct reason be that gcd is not injective because more than one pair of numbers can have the same gcd, hence there is no strict one-one correspondence between the domain and the codomain? For example, gcd(2, 4) = 2 and gcd(2, 8) = 2.
Also, is it correct to claim that gcd is a surjective function because every element in the codomain is the gcd of a pair of positive natural numbers, i.e. every element in the codomain is mapped to by at least one element in the domain? I came to this conclusion because every positive natural number k can be expressed as k = gcd(k, k). Please correct me if I am wrong!
• I only skimmed through your question, but you seem to be perfectly aware of what's going on except for the injectivity of $\gcd$. In the penultimate paragraph the second fact is the reason why it isn't injective. – Git Gud Nov 8 '14 at 12:57
• Thank you for your reply. Given that gcd is a function with two arguments, should the domain be the cartesian product N*N, where N is the set of all positive natural numbers? Or should the domain just be N? – Vizuna Nov 8 '14 at 13:02
• Actually it is much more natural to allow $0$. We have $\gcd(a, 0) = a$ for all $a$, and in particular $\gcd(0, 0) = 0$. You have to interpret "greatest" common divisor in a way such that $0$ is the largest of all numbers. – 6005 Jan 7 '15 at 17:22
You are completely correct that the question is not precise. There are several ways to formalize the question; you made a good start but let me continue a bit as there are some bits that are not correct.
As you say correctly for speaking about the gcd in a sensible way and nontrivial way one needs two integers (or still more).
The following would be a precise question:
Consider $$\gcd: \begin{cases} \mathbb{N}\times \mathbb{N} & \to \mathbb{N} \\ (a,b) & \mapsto \gcd(a,b) \end{cases}$$ Is this injective? Is this bijective?
The answer being "no" as for example $\gcd(3,6)= \gcd(3,9)$.
Howver for:
Is it correct to claim that gcd is not an injective function because it maps two natural numbers a and b to a single output c
This is not correct in the way I understand it.
$$\begin{cases} \mathbb{N}\times \mathbb{N} & \to \mathbb{N} \\ (a,b) & \mapsto 2^a3^b\end{cases}$$ maps two number to a single output, but it is still injective.
A map is not injective if it maps two distinct elements from the domain to the same element of the codomain. In this case the domain is pairs of natural numbers so, $(a,b)$ is one element of the domain mapped to $\gcd(a,b)$ one element of the codomain.
The reason it is not injective is that there are distinct couples of integers $(a,b)\neq (c,d)$ such that $\gcd(a,b)= \gcd(c,d)$.
--
Additional remark: what you say about $\gcd$ involving $0$ is not the way things are commonly handled. Every natural number divides $0$, so we have that $\gcd(a,0)= a$ for each $a$; and we typically one says $\gcd(0,0)=0$ (the "greatest" is understood with respect to the order given be divisibility).
• Great answer. Thanks! – Vizuna Nov 8 '14 at 13:17
• Glad you liked it! :-) – quid Nov 8 '14 at 13:18
You've thought it out quite well. The greatest common divisor function is indeed a function $\gcd: \mathbb N \times \mathbb N \to \mathbb N,\,$ where $\mathbb N$ is taken here to be the set of positive integers (excluding zero for the reasons you give). In this way, we can denote the function by its input and output:
$$(a, b) \in \mathbb N\times \mathbb N \quad \mapsto \quad \gcd(a, b)\in \mathbb N$$
It is non-injective, for the second reason you give: e.g. $\gcd(2, 3) = \gcd(3, 4) = 1$, but $(2, 3) \neq (3, 4)$.
It IS a surjective function, and you provide a succinct reason as to why this is the case.
• You're welcome, Vizuna! – Namaste Nov 8 '14 at 13:16 | 2019-07-21T13:10:15 | {
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http://mathhelpforum.com/discrete-math/155469-number-subsets.html | 1. ## Number of subsets
What is the number of ways to color n objects with 3 colors if every color must be used at least once?
2. Originally Posted by taylor1234
What is the number of ways to color n objects with 3 colors if every color must be used at least once?
Of course if n < 3 answer is 0.
I would do number of ways to color n objects with 3 colors without restriction, then subtract out those that don't use every color at least once. (Two cases: 2 colors are used, 1 color is used.)
Edit: Actually what I wrote in parentheses above is a bit misleading, you'd want to use inclusion-exclusion when counting ways with at most 2 colors (as opposed to exactly 2 colors) compared with 1 color.
3. Originally Posted by taylor1234
What is the number of ways to color n objects with 3 colors if every color must be used at least once?
This is a really vague question.
If the n objects are identical (tennis balls) then the answer is the number of ways to place n identical objects into three distinct cells with no cell empty.
$\displaystyle \binom{(n-3)+(3-1)}{n-3}=\binom{n-1}{n-3}$
On the other hand, if the objects being colored are themselves distinct then we count the number of surjections (onto maps) from a set of n to a set of three.
$\displaystyle \text{Surj}(n,3)= \sum\limits_{k = 0}^3 {(-1)^k\binom{3}{k}(3-k)^n }$.
4. Hm I assumed the objects are non-identical since the thread thread title has "subsets" in it and sets typically have no duplicates.
Here is my approach worked out then.
$3^n$ is number of ways to color with at most 3 colors, no restrictions
$\binom{3}{2}\cdot2^n$ number of ways to color with at most 2 colors, but there are duplicates. Each way of coloring with exactly one color is counted twice.
So final answer would be $3^n-3\cdot2^n+3$. (Which is in agreement with the formula given by Plato.)
5. Originally Posted by undefined
Hm I assumed the objects are non-identical since the thread thread title has "subsets" in it and sets typically have no duplicates.
In counting problems that may or may not be the case.
Many authors do use the term multi-set over against just set.
But that rather recent as history goes.
6. Hello, taylor1234!
$\text{What is the number of ways to color }n\text{ objects with 3 colors}$
$\text{if every color must be used at least once?}$
I am assuming that the objects are indistinguishable.
I cranked out the first few cases and found a pattern.
. . $\begin{array}{|c||c|} \hline
n & \text{3 colors} \\ \hline
3 & 1 \\
4 & 3\\
5 & 6\\
6 & 10\\
\vdots & \vdots \\ \hline \end{array}$
There are $\dfrac{(n-2)(n-1)}{2}$ three-color arrangements.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Now I see the reason for this . . .
We have $\,n$ objects in a row, say, $n=7$
. . $\circ\;\_\;\circ\;\_\;\circ\;\_\;\circ\;\_\; \circ\;\_\;\circ \;\_\;\circ$
There are 6 spaces between the objects.
Select 2 of the spaces and insert "dividers".
So that: . $\circ\:\circ\:|\:\circ\:|\:\circ\:\circ\:\circ\: \circ$ .represents: . $A\:A\:B\:C\:C\:C\:C$
. .
Two of color A, one of color B, four of color C.
And: . $\circ\:\circ\:\circ\:|\: \circ\:\circ\:\circ\:|\:\circ$ .represents: . $A\:A\:A\:B\:B\:B\:C$
. .
Three of color A, three of color B, one of color C.
With $\,n$ objects, there are $(n-1)$ spaces.
Selecting two spaces, there are: . $\displaystyle {n-1\choose2} \:=\:\frac{(n-1)(n-2)}{2}$ ways. | 2016-10-22T21:57:44 | {
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https://math.stackexchange.com/questions/2832986/proving-the-product-of-four-consecutive-integers-plus-one-is-a-square/2832991 | # Proving the product of four consecutive integers, plus one, is a square
I need some help with a Proof:
Let $m\in\mathbb{Z}$. Prove that if $m$ is the product of four consecutive integers, then $m+1$ is a perfect square.
I tried a direct proof where I said:
Assume $m$ is the product of four consecutive integers.
If $m$ is the product of four consecutive integers, then write $m=x(x+1)(x+2)(x+3)$ where $x$ is an integer.
Then $m=x(x+1)(x+2)(x+3)=x^4+6x^3+11x^2 +6x$.
Adding $1$ to both sides gives us:
$m+1=x^4+6x^3+11x^2+6x+1$.
I'm unsure how to proceed. I know I'm supposed to show $m$ is a perfect square, so I should somehow show that $m+1=a^2$ for some $a\in\mathbb{Z}$, but at this point, I can't alter the right hand side of the equation to get anything viable.
• Try something like $a=x^2+rx+s$ for suitable $r$, $s$. – Angina Seng Jun 26 '18 at 21:17
• Look at the first few square roots. There must be a simple quadratic that gives them. – lulu Jun 26 '18 at 21:19
• Well, this is a spoiler. $x^4 + 6x^3 + 11x^2 + 6x + 1 = (x^2 + 3x + 1)^2$ – fleablood Jun 26 '18 at 23:10
• Hint: $(x^2 + bx + c)^2 = x^4 + 2bx^3 + (2c + b^2)x^2 + 2bcx + c^2$. If $c^2 = 1; 2bc = 6; (2c+b^2) = 11; 2b=6$ has integer solution, you are golden. – fleablood Jun 26 '18 at 23:25
• – Ben Millwood Jun 27 '18 at 14:18
By the way, you mean $m+1 = x^4 + 6x^3 + 11x^2 + 6x + 1$.
Let's break it down. Obviously, you need a quadratic that, when squared, gives the above. How do you construct this? You need 3 numbers $a, b, c$ for $ax^2 + bx + c$.
Looking at the 4th degree thing above, what can you tell me about $a$, right off the bat? What about $c$? Then can you use this to tell you what $b$ is?
(I feel like I should not go further because I want you to solve it, but let me know if you need a clarification)
• Your result is right, of course, but you silently took a shortcut: nothing says a priori that the quartic polynomial is the square of a quadratic polynomial in the same variable. About all you have at this stage is that it is an integer function of $n$, asymptotic to $n^2$. – Yves Daoust Jun 28 '18 at 13:49
• This is a good point. I justified it by noticing the symmetry in the coefficients, so this can elucidate the approach. – BRSTCohomology Jun 28 '18 at 13:50
To get a feel for the problem, let's work backwards, starting from a square number. Take some integer $n$.
$n^2 - 1 = (n+1)(n-1).$
OK, so it looks like $m$ has two factors whose difference is 2. Could it be that in our product of consecutive integers, the product of two of them is $n-1$ and of the other two is $n+1$?
Let's throw in a simple example: $1\times2\times3\times4$. Notice how $1\times4=4$ and $2\times3=6$.
What about $2\times3\times4\times5$? This time $2\times5=10$ and $3\times4=12$.
It looks like the product of the "outer" pair is $n-1$ and the product of the "inner pair" is $n+1$.
Now we know how to attack this.
Let $k$ be some integer and $m=k(k+1)(k+2)(k+3)$.
\begin{align} m &= k(k+1)(k+2)(k+3) \\ &= (k+1)(k+2)\times(k(k+3)) \ \text{ (collecting inner and outer terms)}\\ &= (k^2 + 3k + 2) \times (k^2 + 3k) \\ &= ((k^2 + 3k + 1) + 1) ((k^2 + 3k + 1) - 1) \\ &= (k^2 + 3k + 1) ^ 2 - 1 \qquad \text{ (since }(a+b)(a-b)=a^2-b^2\text{)}. \end{align}
Since $k$ is an integer, $k^2 + 3k + 1$ is an integer so $m+1$ is a perfect square.
Write the product of the four consecutive integers starting at some $n-1$, so that $$m=(n-1)n(n+1)(n+2)+1,$$ and expand: \begin{align} m&=(n^2-1)(n^2+2n)+1=n^2(n^2-1)+2n(n^2-1)+1 \\ &=(n^2-1)^2+2n(n^2-1)+\not 1+ n^2{-}\!\not1 \\ &= \bigl((n^2-1)+n\bigr)^2. \end{align}
• This is exactly how I solved it. Upvoting. – Cuspy Code Jun 27 '18 at 9:40
• Bizarre downvote; nice approach. Might be worth comparing to Zeitz's... +1 – Benjamin Dickman Jun 27 '18 at 23:30
See Part II there for several approaches taken by students (most of which are covered by other answers here; but the presentation is somewhat different).
One of the methods mentioned there is observing the symmetry above around $x= -3/2 = -1.5$, but then using this to inform a substitution: let $z = x + 1.5$ so that we have:
$$x(x+1)(x+2)(x+3) = (z-1.5)(z-0.5)(z+0.5)(z+1.5) = (z^2 - 1.5^2)(z^2 - 0.5^2)$$
Noting that $1.5^2 = 2.25$ and $0.5^2 = 0.25$, we could use one more substitution of $w = z^2 - 2.25$ to rewrite the final expression above as $w(w+2) = w^2 + 2w$, from which the addition of $1$ yields $(w+1)^2$ as desired. One can now rewrite in terms of just $x$ to finish matters off.
I think that the idea of the symmetry here is an important takeaway; incidentally, the problem is also broached in an exploratory manner at the beginning of Paul Zeitz's (2006) The Art and Craft of Problem Solving as Example 1.2.1.
• This is my preferred solution... because that's how I just solved it... However, I'm not a fan of using decimal approximations to represent fractions. Particularly for a number theory style problem. In my experience, it leads to too much confusion among students. It requires students to remember one more thing (this is an exact decimal representation) which is too much of an additional cognitive load. – John Jun 27 '18 at 22:37
• @John Definitely a worthy consideration: I put some related comments about this (specifically about notation) at MESE. But, I think you're right that fractions could make this answer cleaner. . . – Benjamin Dickman Jun 27 '18 at 23:26
• @John You can exploit the symmetry without using decimals, just keep fractions, multiple by suitable integer ($16$ in this case) and you are back with integer coefficients. You will get $16f\left(x+\frac{1}{2}\right)=(4x^2-5)^2$, which in turn yields $16f(x)=(4x^2-4x-4)^2$, and you are basically done. – Sil Jun 28 '18 at 7:59
Given $m$ is the product of four consecutive integers. $$m=p(p+1)(p+2)(p+3)$$where $p$ is an integer
we need to show that $p(p+1)(p+2)(p+3)+1$ is a perfect square
Now,$$p(p+1)(p+2)(p+3)+1=p(p+3)(p+1)(p+2)+1$$ $$=(p^2+3p)(p^2+3p+2)+1$$ $$=(p^2+3p+1)(p^2+3p+2)-(p^2+3p+2)+1$$ $$=(p^2+3p+1)(p^2+3p+1+1)-(p^2+3p+2)+1$$ $$=(p^2+3p+1)(p^2+3p+1)+(p^2+3p+1)-p^2-3p-2+1$$ $$=(p^2+3p+1)(p^2+3p+1)=(p^2+3p+1)^2$$ So, $m+1$ is a perfect square where $m$ is the product of four consecutive integers.
Proof without words:
$\hspace{2cm}$ $$\color{red}x\color{blue}{(x+1)(x+2)}\color{red}{(x+3)}+1=\color{red}{(x^2+3x)}\color{blue}{(x^2+3x+2)}+1=(x^2+3x+1)^2.$$
Note that $f(x)=x(x+1)(x+2)(x+3)+1$ is a degree $4$ polynomial with leading term $x^4$ and symmetric around $x=-\frac32$. We might try the ansatz $f(x)=g(x)^2$ with $g(x)=x^2+px+q$ because then the leading term of $g(x)^2$ is also $x^4$. We suspect that $g$ is also symmetric around $x=-\frac32$ and hence write it as $g(x)=(x+\frac32)^2+c$. Note that $f(0)=f(-1)=1$, so we want $g(0)=\pm1$ and $g(-1)=\pm1$. The first means $c\in\{-\frac54,-\frac{13}4\}$, the second means $c\in\{\frac34,-\frac54\}$. We conclude that $c=-\frac54$. Without further calculation, we see that $f(x)-g(x)^2=0$ for $x=0$ and $x=-1$ and by symmetry also for $x=-2$ and $x=-3$. As the leading terms cancel, the polynomial $f(x)-g(x)^2$ is in fact of degree of at most $3$. Having four distinct roots, it must be identically zero, as desired.
\begin{eqnarray*} m=n(n+1)(n+2)(n+3) =n^4+6n^3+11n^2+6n. \end{eqnarray*} So \begin{eqnarray*} m+ 1 =n^4+6n^3+11n^2+6n+1=(n^2+3n+1)^2. \end{eqnarray*}
• Judging by the way in which the problem is asked: It might be helpful to the OP to explain how to get that factorization! For example, noting that $n(n+3) = n^2 + 3n := m$ and $(n+1)(n+2) = n^2 + 3n + 2 = m+2$, so that the product plus one is $m^2 + 2m + 1 = (m+1)^2 = (n^2 + 3n + 1)^2$ as you observed. – Benjamin Dickman Jun 26 '18 at 22:17
• What is the thought process used here? – richard1941 Jul 7 '18 at 2:25
The polynomial $x^4+6x^3+11x^2+6x+1$ has symmetric coefficents - more precisely, it's called a palindromic polynomial:
$$p(x)=x^4+ax^3+bx^2+ax+1$$
The goal is to factor $p(x)$ (and show that it factors to a square of some expression). Let's start by dividing by $x^2$ and refactoring:
$$\frac{p(x)}{x^2} = q(x) = x^2+\frac{1}{x^2}+a\left(x+\frac{1}{x}\right)+b$$
It's tempting to make a substitution $y=x+\frac{1}{x}$:
$$q(x) \rightarrow q(y) = y^2+ay+(b-2)$$
This: $q(y)=0$, being a quadratic equation, is something that can be automatically solved:
$$y_{1,2}=\frac{-a\pm \sqrt{a^2-4(b-2)}}{2}$$
and inserting the values $a=6$ and $b=11$, one obtains (note the expression under the square root is equal to zero):
$$y_1=y_2=-3$$
so that $q(y)=(y+3)^2$ - at this point one sees it's a perfect square, and due to $q(y)=\frac{p(x)}{x^2}$, basically we're done with the proof at this point.
Just to take things to their end:
Because $q(y)=0 \Leftrightarrow q(x)=0 \Leftrightarrow p(x)=0$:
$$q(y) = (y+3)^2 = \left(x+\frac{1}{x}+3\right)^2 = \frac{(x^2+3x+1)^2}{x^2} = \frac{p(x)}{x^2}$$
hence $x^4+6x^3+11x^2+6x+1 = (x^2+3x+1)^2$ - a perfect square indeed.
Empirically:
Consider the function
$$p(n):=\sqrt{n(n+1)(n+2)(n+3)+1}.$$
For $n=0,1,2,3,\cdots$ we have $p(n)=1,5,11,19,29,\cdots$ a sequence with constant second order differences ($2$), and we can postulate the polynomial
$$n^2+3n+1$$
(because $p(0)=1$, the coefficient of $n^2$ must be $1$, and $p(n)-n^2-1=0,3,6,9,\cdots$)
Now the identity
$$n(n+3)(n+1)(n+2)=(n^2+3n)(n^2+3n+2) \\=(n^2+3n+1-1)(n^2+3n+1+1) \\=(n^2+3n+1)^2-1.$$
becomes apparent.
Another approach is by bringing more symmetry and shifting the variable by $3/2$.
$$\sqrt{\left(m-\frac32\right)\left(m-\frac12\right)\left(m+\frac12\right)\left(m+\frac32\right)+1} =\sqrt{\left(m^2-\frac94\right)\left(m^2-\frac14\right)+1} =\sqrt{m^4-\frac52m^2+\frac{25}{16}}=m^2-\frac54,$$
which is
$$n^2+2\frac32n+\left(\frac32\right)^2-\frac54.$$
If the numbers are $x, x+1, x+2, x+3$.
Let $\frac m2 = x+1.5$ be the midpoint of the four consecutive integers, so that the integers are $\frac {m-3}2, \frac {m-1}2, \frac {m+1}2, \frac {m+3}2$. (Note: $m$ is odd and $\frac m2$ is not an integer.)
So $x(x+1)(x+2)(x+3) + 1 =$
$\frac {(m-3)(m+3)(m-1)(m+1)}{16} + 1=$
$\frac {(m^2 -9)(m^2 - 1) + 16}{16} =$
$\frac {(m^2-10m^2 + 9) +16}{16} =\frac {m^2-10m^2 + 25}{16}=$
$(\frac {m^2 -5}{4})^2$.
Now $m$ is odd. So let $m = 2n+1$ then
$x(x+1)(x+2)(x+3) + 1 = (\frac {(2n+1)^2 -5}{4})^2=$
$(\frac {4n^2 +4n + 1 -5}{4})^2 = (\frac {4n^2 +4n -4}{4})^2=$
$(n^2 +n - 1)^2$.
====
So ....
If $x$ is the first integer and $\frac {2n + 1}2 = x + \frac 32$ then
$n = x + 1$.
So $x(x+1)(x+2)(x+3) + 1 = ((x+1)^2 + (x+1) - 1)^2 = (x^2 + 3x +1)^2$
....
So as you got $x^4 + 6x^3 + 11x^2 + 6x + 1$ that actually equals $(x^2 + 3x +1)^2$
Indeed $x^4 + 6x^3 + 11x^2 + 6x + 1 = x^2(x^2 + 3x + 1) + 3x^3 +10x^2 + 6x + 1$
$= x^2(x^2 + 3x+ 1) + 3x(x^2 + 3x + 1) +x^2 +3x + 1$
$= (x^2 + 3x + 1)^2$.
D'oh.
If $x(x+1)(x+2)(x+3) + 1 = a^2$ then
$x(x+ 1)(x+ 2)(x+3) = a^2 - 1 = (a + 1)(a-1)$
To get factors that close together they'd have to be
$a = x(x+3) \pm 1= (x+ 1)(x+2) \mp 1$
and indeed $a= x(x+3) + 1 = (x+1)(x+2) - 1= x^2 +3x + 1$.
proves the statement! (If we work backwards.)
Suppose that you have 4 consecutive numbers $a, b, c, d$. They can be expressed as $a=t-\frac{3}{2}$, $b=t-\frac{1}{2}$, $c=t+\frac{1}{2}$ and $d=t+\frac{3}{2}$ for some number $t$.
Now, $$ad = \left(t-\frac{3}{2}\right)\left(t+\frac{3}{2}\right) = t^2 - \left(\frac{3}{2}\right)^2 = t^2 - \frac{9}{4}$$ and $$bc = \left(t-\frac{1}{2}\right)\left(t+\frac{1}{2}\right) = t^2 - \left(\frac{1}{2}\right)^2 = t^2 - \frac{1}{4}$$
If we define $y = t^2 - \frac{5}{4}$, we have $$ad = \left(t-\frac{3}{2}\right)\left(t+\frac{3}{2}\right) = y - 1$$ and $$bc = \left(t-\frac{1}{2}\right)\left(t+\frac{1}{2}\right) = y + 1$$
Furthermore, since the LHS in both cases is an integer, it is clear that $y$ is an integer.
So the product of all four numbers is $$abcd = (y - 1)(y+1) = y^2 - 1$$ one less than the square of an integer.
This is another way of looking at @jwg's answer.
Let the four consecutive numbers be $a,b,c,d$ and let $t$ be the number half-way between $b$ and $c$. Then
\begin{align} abcd + 1 &= \bigg(t-\frac 32\bigg)\bigg(t-\frac 12\bigg) \bigg(t+\frac 12\bigg)\bigg(t+\frac 32\bigg) + 1\\ &= \bigg(t^2-\frac 94\bigg)\bigg(t^2-\frac 14\bigg) + 1\\ &= t^4 - \frac 52 t + \frac{25}{16}\\ &= \bigg(t^2 - \frac 54 \bigg)^2 \end{align}
Letting $t = a + \frac 32$, we get
\begin{align} abcd + 1 &= \bigg(t^2 - \frac 54 \bigg)^2 \\ &= \bigg(a^2 +3a + \dfrac 94 - \frac 54 \bigg)^2 \\ &= (a^2 +3a + 1)^2 \end{align}
This suggests the following solution.
\begin{align} abcd + 1 &= a(a+1)(a+2)(a+3) + 1 \\ &= a(a+3) \cdot (a+1)(a+2) + 1 \\ &= (a^2+3a) (a^2+3a+2) + 1 \\ &= (a^2+3a+1 \ - \ 1)(a^2+3a+1 \ + \ 1) + 1 \\ &= (a^2+3a+1)^2 - 1 + 1 \\ &= (a^2+3a+1)^2 \end{align} | 2021-06-17T06:39:57 | {
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https://math.stackexchange.com/questions/3411882/proof-verification-that-if-a-n-leq-b-n-then-limsup-a-n-leq-limsup-b-n | # Proof verification that if $a_n\leq b_n$ then $\limsup a_{n} \leq \limsup b_{n}$
Suppose $$\left\{a_{n}\right\}$$ and $$\left\{b_{n}\right\}$$ are sequences such that for every $$n, a_{n} \leqslant b_{n} .$$ Prove that If $$a_n\leq b_n$$ for all $$n$$ then $$\limsup a_{n} \leq \limsup b_{n}$$
(proof)
Let $$A = \limsup a_{n}$$ and let $$B=\limsup b_n$$. First of all, I'm aware that there are many questions like this on the site, but they all seem to be related to either $$\limsup$$ or $$\liminf$$ and I couldn't find anything that would help me with my problem. I've done some Googling and found some great resources, but I'm still not quite sure how to get to some steps and would like your assistance.
The problem is as follows:
Assume to the contrary that $$B and that for all $$n\in\mathbb{N}, a_n\leq b_n.$$
We know there is a subsequence $$\{a_{n_k}\}$$ that converges to $$A$$. Let $$\epsilon = \frac{B-A}{2}$$.
Then we know that there is a $$K$$ such that for $$k>K, |a_{n_k}-A|<\epsilon$$. Given $$a_n < b_n$$ prove that $$\lim_{n\to \infty}(a_n) \le \lim_{n\to\infty}(b_n)$$. The proof is then done by contradiction, assuming that $$a = \lim_{n\to \infty}(a_n) > b =\lim_{n\to\infty}(b_n)$$.
We take an $$\epsilon = \frac{a-b} 2$$, so that the $$\epsilon$$-neighborhoods of $$a$$ and $$b$$ are disjoint. From the definition of limits, we now know that there is such a $$N$$, so that $$\forall n > N : |a_n-a|<\frac\epsilon2$$ and $$|b_n-b|<\frac\epsilon2$$.
The next step is absolutely always confusing. Two variants I've found are either:
We know that there are infinity many terms of $$a_n$$ in $$(A-\epsilon , \epsilon +A)$$. $$a_n>a-\epsilon=a-\left(\frac{a-b} 2\right)=b+\left(\frac{a-b} 2\right)=b+\epsilon>b_n$$
However for this same $$\epsilon$$ we know that there are only finitely many terms of $$b_n$$ greater than $$B+\epsilon$$.
Finding the maximum subscript $$n$$ of these finitely many $$b_n$$ greater than $$B+\epsilon$$ gives us a corresponding value of $$N$$ such that for $$n>N, b_n < B+\epsilon$$.
Let $$M=\max\{K,N\}$$.
Then it follows that for $$n>M, b_n and $$a_n > A-\epsilon = B+\epsilon$$.
So we have found an $$a_n > b_n.$$ Contradiction.
I am wondering if there is anything wrong with my proof
Edit:
Definition: Let $$\left\{a_{n}\right\}$$ be a sequence of real numbers. Then $$\lim$$ $$\sup a_{n}$$ is the least upper bound of the set of subsequential limit points of $$\left\{a_{n}\right\},$$ and $$\lim \inf a_{n}$$ is the greatest lower bound of the set of subsequential limit points of $$\left\{a_{n}\right\} .$$
• What's the definition of $\limsup$ you are using? Oct 28 '19 at 2:54
• "We know that there are infinity many terms of $a_n$ in $(\epsilon - A, \epsilon +A)$" - this precludes your sequence being finite, but I suppose you could just make it the constant sequence after it converges Oct 28 '19 at 2:55
• Other than removing that line, the proof looks good and is pretty snazzy. Well done! Oct 28 '19 at 2:57
• @MathematicsStudent1122 see my edit please Oct 28 '19 at 2:58
• @BrevanEllefsen yes our definition of a sequence is assumed to be infinite. Thank you for the kind words! Oct 28 '19 at 3:00
Your proof seems right, but consulting a more direct proof may help to self-test understanding, so I'll provide one below.
Since $$a_n \leq b_n$$ for all $$n$$, any upper bound on all of the $$b_n$$ is also an upper bound on all of the $$a_n$$. In particular, $$\sup b_n$$ is an upper bound on all of the $$a_n$$. By definition, $$\sup a_n$$ is the least upper bound on the $$a_n$$; setting $$k = 0$$, it follows that $$\sup_{ n \geq k} a_n \leq \sup_{n \geq k} b_n.$$
In other words, setting setting $$A_k =\sup_{n \geq k} a_n$$ and $$B_k = \sup_{n \geq k} b_k$$, we've shown that for $$k =0$$, $$A_k \leq B_k.$$
In fact, the same reasoning gives the above inequality, for all values of $$k$$. Taking the limit in $$k$$ then gives $$\limsup a_n \leq \limsup b_n$$, as required.
• That is really easy! I had a similar intuition but was steered by my advisor to do it indirectly. I am not sure why as this is much more clear. Interestingly I searched this question on this site and never found a direct proof this simple. Thanks Oct 28 '19 at 4:28
• It is not true that $b_n \le B$ for all $n$. Suppose $\{b_n\}$ is zero for $n>1$, with $b_n =1$. Then, $b_1 > B=0$. Oct 28 '19 at 6:46
• Whoops—thanks! Added qualification "sufficiently large." Oct 28 '19 at 13:42
• it's not true even for all $n$ sufficiently large. Consider $b_n = 1/n$. Then $\lim \sup b_n = 0$ but $b_n > 0$ for all $n$ Mar 3 '20 at 14:46
• Thanks—you're right. I'll adjust the proof accordingly. Mar 4 '20 at 18:38
Option:
$$\limsup_{n \rightarrow \infty}x_n:=\lim_{n\rightarrow \infty}(\sup{x_k| k\ge n});$$
$$c_n:=\sup$${$$a_k| k\ge n$$};
$$d_n:=\sup$${$$b_k| k \ge n$$};
$$(\star)$$ $$c_n \le d_n$$, since $$a_k \le b_k$$, $$k \in \mathbb{N}.$$
Then
$$\limsup_{n\rightarrow \infty} a_n=\lim_{n \rightarrow \infty}c_n \le \lim_{n \rightarrow \infty} d_n =\limsup_{n \rightarrow \infty}b_n.$$
P.S. As a not so difficult exercise prove $$(\star)$$. | 2021-12-05T04:59:18 | {
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https://math.stackexchange.com/questions/2129741/finding-basis-for-null-space-of-matrix | # Finding basis for Null Space of matrix
$A = \begin{bmatrix}1&1&1&-1&0\\1&0&1&0&1\\0&0&1&0&0\\2&0&3&0&2\end{bmatrix}$
Find a basis for the row space, column space, and null space.
I think the first two parts are easy. First;
$rref(A) = \begin{bmatrix}1&0&0&0&1\\0&1&0&-1&-1\\0&0&1&0&0\\0&0&0&0&0\end{bmatrix}$
So my row space is just the 3 pivot rows of $rref(A)$, and the column space, the corresponding columns of $A$.
I'm a little murky on the null space. I know it's the set of all vectors $x$ which satisfy $Ax = 0$. Since I have 5 columns;
$\begin{bmatrix}1&1&1&-1&0\\1&0&1&0&1\\0&0&1&0&0\\2&0&3&0&2\end{bmatrix}\begin{pmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{pmatrix} = 0$
And I can find a solution, $x = (1, 0, 0, 1, -1)^t$.
However I notice that A has rank of 3, with 5 columns, and I think this means I have to find two solutions for x (due to rank-nullity theorem).
Am I on the right track, and is there a good strategy for finding these solutions easily?
$x$ satisfies $Ax=0$ if and only if it satisfies $rref(A) x = 0$. [You could do it your way by looking at $Ax=0$, but it is much easier if you use the RREF.]
Writing out the last expression gives you three equations. \begin{align} x_1+x_5 &= 0\\ x_2 - x_4 - x_5 &= 0\\ x_3 &= 0 \end{align} So, any solution is of the form $$\begin{bmatrix}x_1 \\ x_2 \\ 0 \\ x_1+x_2 \\ -x_1\end{bmatrix} = x_1 \begin{bmatrix}1 \\ 0 \\ 0 \\ 1 \\ -1\end{bmatrix} + x_2 \begin{bmatrix}0 \\ 1 \\ 0 \\ 1 \\ 0\end{bmatrix}.$$
• Fantastic! This is exactly the kind of thing I was looking for. Thank you very much :) – KookieMonster Feb 5 '17 at 4:03
$$x_1+x_5=0\\ x_2-x_4-x_5=0\\ x_3=0$$ From the first equation: $x_5=-x_1$.
Substituting into the second equation: $x_1=x_4-x_2$.
Conclusion: you can choose $x_2$ and $x_4$ freely and then $x_1$, $x_3$ and $x_5$ are determined.
In other words the solution set of $$\begin{bmatrix}1&1&1&-1&0\\1&0&1&0&1\\0&0&1&0&0\\2&0&3&0&2\end{bmatrix}\begin{pmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{pmatrix} = 0$$ is $$\{(x_1,x_2,x_3,x_4,x_5)\in\mathbb{R}^5:x_3=0,x_1=x_4-x_2,x_5=-x_1,x_2\in\mathbb{R},x_4\in\mathbb{R}\}$$ The choice $x_2=0$, $x_4=1$ gives the vector $x$ you found.
The choice $x_2=1$, $x_4=1$ gives the vector $y=(0,1,0,1,0)^t$ which is clearly linearly independent to $x$. | 2019-05-24T11:20:14 | {
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https://physics.stackexchange.com/questions/513352/which-approach-to-take-with-a-vertical-spring | # Which approach to take with a vertical spring?
Lets say we have a spring hanging vertically with spring constant $$k$$ attached to a block of mass $$m$$. The system is at rest.
Then, you pull the mass downwards, extending the spring by distance $$x$$, then let go. The spring will, of course, bounce back to its original spot. What is the velocity of the object at its initial resting location?
To solve this, I took two approaches, but I'm not sure which one is right. The first is a work approach. When the block returns to its old location, it is the same except now has the new energy it recieved from the prior extention, so I can say...
$$W=Fd$$ or $$W=\frac{1}{2}*k*x^2$$.
So, $$\frac{1}{2}*k*x^2=\frac{1}{2}*m*v^2$$ therefore $$v=\sqrt{\frac{k}{m}}*abs(x)$$.
However, I don't consider gravitational potential, which worries me.
If we do so, we can say at the rest location the energy is just $$mgh$$ where $$h$$ is $$x$$.
At the bottom, the energy is just $$\frac{1}{2}*k*x^2$$ so...
$$mgx+\frac{1}{2}m*v^2=\frac{1}{2}k*x^2$$ so $$v=\sqrt{\frac{kx^2-2gmx}{m}}$$
Which approach do I take?
• You're not keeping your reference points consistent. If you are taking $x=0$ as the height at rest, then the potential energy at rest is $m g 0 = 0$, and the potential energy with the spring pulled down is the sum of the energy in the spring and the energy given up by the mass: $W = \frac{1}{2} k x^2 - m g |x|$. – TimWescott Nov 12 '19 at 0:00
• @TimWescott Would that mean the second approach is correct? I can just move the $mgx$ to the other side and solve for v, giving the same expression? – Ryan_DS Nov 12 '19 at 0:10
• Give it a whirl, see how it works. – TimWescott Nov 12 '19 at 0:12
• I should mention that your second approach bears all the hallmarks that for me would involve multiple tries, finding that I've left 'i's undotted, 't's uncrossed, or I've miscounted sign changes and have '+' where '-' should be and visa versa. So give it a whirl, carefully. – TimWescott Nov 12 '19 at 0:13
• @TimWescott You say that like there isn't a way to verify the actual solution. – Aaron Stevens Nov 12 '19 at 1:01
I will address your second case first. You are correct to use conservation of energy and say that the potential energy stored in the spring at the lowest point is equal to the sum of the kinetic energy and the potential energy due to gravity at the equilibrium point. So you were correct with the equation $$\frac12ky^2=mgy+\frac12mv^2$$
Let’s now look at the first case, but let's do it correctly. We know that the net work done on the mass is equal to its change in kinetic energy: $$W_\text{net}=W_\text{gravity}+W_\text{spring}=\Delta K=\frac12mv^2-0$$
We can easily determine the work done by gravity and the spring force using the definition of work $$W=\int\mathbf F\cdot\text d\mathbf y$$ $$W_\text{gravity}=\int_{-y}^0(-mg)\,\text dy'=-mgy$$ $$W_\text{spring}=\int_{-y}^0(-ky')\,\text dy'=\frac12ky^2$$ Putting it all together we have $$W_\text{net}=-mgy+\frac12ky^2=\frac12mv^2$$ | 2020-04-08T05:02:50 | {
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https://math.stackexchange.com/questions/2617207/how-can-i-calculate-the-remainder-of-32012-modulo-17 | # How can I calculate the remainder of $3^{2012}$ modulo 17?
So far this is what I can do:
Using Fermat's Little Theorem I know that $3^{16}\equiv 1 \pmod {17}$
Also: $3^{2012} = (3^{16})^{125}*3^{12} \pmod{17}$
So I am left with $3^{12}\pmod{17}$.
Again I'm going to use fermat's theorem so: $3^{12} = \frac{3^{16}}{3^{4}} \pmod{17}$
Here I am stuck because I get $3^{-4} \pmod{17}$ and I don't know how to calculate this because I don't know what $\frac{1}{81} \pmod{17}$ is.
I know $81 = 13 \pmod{17}$
But I know the answer is 4. What did I do wrong?
• Rather than calculate $\frac 1{81}$ why not calculate $81^3=3^{12}$? Note that $13^2\equiv(-4)^2\equiv -1\mod 17$... – abiessu Jan 23 '18 at 5:57
• Hint: $\;3^4=81 = 5 \cdot 17 - 4\,$. – dxiv Jan 23 '18 at 5:58
• You did nothing wrong. Notice that $81\times4\equiv1 \pmod{17}$. Either look at Bézout's identity and solve $81a+17b=1$ (using euclidean algorithm) and find why $a$ is your answer, or see the other comments for a direct computation. – Jean-Claude Arbaut Jan 23 '18 at 6:01
• Interesting, so $81 = -4 \pmod{17}$, but what does that imply about $1/81 \pmod {17}$? – user81864 Jan 23 '18 at 6:01
• @user81864 No need for $1/81$. You have $\,3^{12} \equiv (-4)^3=16 \cdot (-4) \equiv (-1)\cdot(-4) = 4\,$. – dxiv Jan 23 '18 at 6:02
## 4 Answers
$3^{12}=(3^3)^4=10^4$ (mod $17$), so we have to find $10000$ (mod $17$), which is evidently $4$ (mod $17$).
Indeed, we have $$3^{12}=31261\cdot17+4.$$
Also, $$3^{12}=81^3\equiv(-4)^3\equiv(-4)(-1)=4.$$
Also, we have $$3^{12}-4=(3^6-2)(3^6+2)=727\cdot731=727\cdot43\cdot17.$$
Just do it.
$3^4 = 81 \equiv -4$.
$3^{12} \equiv (3^4)^3 = (-4)^3 \equiv -81 \equiv 4 \mod 17$.
For insight:
You know $3^{16}\equiv 1 \mod 17$ so $3^{8}\equiv \pm 4$ so $3^4 \equiv \pm 1, \pm \sqrt{-1}$. So $-1 \equiv 16$ one of the $\sqrt {-1} \equiv 4\mod 17$. (the other is $13$). This should tell you to try to find $3^{12}$ via iterations $3^4$.
Also: $81 \equiv 13 \equiv - 4 \mod 17$. So $\frac 1{81} \equiv -\frac 14$. And figuring $\frac 14$ shouldn't be hard $1 \equiv 18$ so $\frac 12 \equiv 9 \mod 17$ and $9 \equiv 26$ so $\frac 14 \equiv 13\equiv -4$. So $-\frac 14 = 4$. And that makes sense. $(-4)*4 = -16 \equiv 1 \mod 17$.
In addition to the clever answers, straightforward repeated squaring can be used. $$3^{12}=3^8 \cdot 3^4$$ and $$3^2=9 \equiv 9 \mod 17$$ so $$3^4 \equiv 9^2 \equiv 13 \mod 17$$ and $$3^8 \equiv 13^2 \equiv 16 \mod 17$$ so finally $$3^{12}=3^8 \cdot 3^4 \equiv 16 \cdot 13 \equiv 4 \mod 17$$
The final line could be simplified further if desired $$16 \cdot 13 = 4 \cdot 52 \equiv 4 \cdot 1 \mod 17$$ | 2020-10-24T03:29:50 | {
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https://math.stackexchange.com/questions/1091189/how-many-independent-components-does-a-rank-three-totally-symmetric-tensor-have | How many independent components does a rank three totally symmetric tensor have in $n$ dimensions?
How many independent components does a rank three totally symmetric tensor have in $n$ dimensions?
Needed for the irrep decompositon of $3\otimes 3\otimes 3$ in here.
No idea where to start to prove this.
I did come up with a more clever way of figuring it out for $n=3$, but I don't think it can be generalized. In $3$ dimensions, a totally antisymmetric (rank three) tensor has one component. From here I just counted the components that are nonzero for a totally symmetric one. We have the $3$ diagonal components, obviously. Then there are the components of the form $iij$ (no sum). The $i$s run over $3$ values and $j$ over $2$. Since $2\cdot 3=6$ we have $1+3+6=10$ components.
• You'll have to elaborate on some notation here. What is $3 \otimes 3 \otimes 3$? Is that $\Bbb C^3 \otimes \Bbb C^3 \otimes \Bbb C^3$? Keep in mind that mathematicians and physicists often use different notation. – Omnomnomnom Jan 4 '15 at 21:32
• $3$ is the $SU(3)$ quark triplet. The relation I'm looking to prove is $3\otimes3\otimes3=10\oplus8\oplus8\oplus1$. I need to show that $10$ is the number of independent components of rank three completely symmetric tensor in 3 dimensions. I did that explicitly by writing down the $27$ components and matching identical ones up. I am wondering if there is an elegant formula for $n$ dimensions. But the particle physics stuff is not important to this question. – Ryan Unger Jan 4 '15 at 21:37
• See here for an additional answer to the question. – Dilaton Jun 29 '16 at 7:43
By a 'rank three tensor' I think you mean an element of $\bigotimes^3V$. This is not what 'rank' means in mathematics. An element of $\bigotimes^kV$ is said to have order (or degree) $k$.
A rank one tensor (or simple tensor) is a tensor of the form $v_1\otimes\dots\otimes v_m$. The rank of a tensor $T$ is the minimal number of rank one tensors needed to express $T$ as a sum.
Let $V$ be an $n$-dimensional vector space, then for any $k \in \mathbb{N}$, let $\operatorname{Sym}^kV$ denote the collection of symmetric order $k$ tensors on $V$; note that $\operatorname{Sym}^kV$ is a vector space. Let $v_1, \dots, v_n$ be a basis for $V$, then a basis for $\operatorname{Sym}^kV$ is given by
$$\left\{\frac{1}{k!}\sum_{\sigma \in S_n}v_{\sigma(i_1)}\otimes\dots\otimes v_{\sigma(i_k)} \mid 1 \leq i_1 \leq \dots \leq i_k \leq n\right\}.$$
Therefore, $\dim\operatorname{Sym}^kV$ is equal to the number of non-decreasing sequences of $k$ integers in $\{1, \dots, n\}$. Let $x_1 = i_1 - 1$, $x_j = i_j - i_{j-1}$ for $j = 2, \dots, k$, and $x_{k+1} = n - i_k$. Note that the number of non-decreasing sequences of $k$ integers in $\{1, \dots, n\}$ is in one-to-one correspondence with the number of of solutions to $x_1 + \dots + x_{k+1} = n-1$ in non-negative integers. The latter number can be found using the stars and bars method from combinatorics; doing so, we see that
$$\dim\operatorname{Sym}^kV = \binom{(n-1)+(k+1)-1}{n-1} = \binom{n + k -1}{n-1} = \binom{n+k-1}{k}.$$
So every symmetric order $k$ tensor can be written uniquely as a linear combination of $\binom{n+k-1}{k}$ basis tensors. The coefficients in this linear combination are what you refer to as "independent components". In the case you worked out yourself, you had $n = 3$ and $k = 3$ in which case the number of independent components is
$$\binom{3+3-1}{3} = \binom{5}{3} = 10.$$
• Hi, what would be the number of independent components of a tensor order k=3, but which is symmetric in just two of its indices? – Santi Sep 30 '15 at 13:15
• @Santi : Although so late I think the number is 18. Take a look in this 3D-visualization of a totally symmetric tensor : imgur.com/a/BCfbF. If you draw two diagonal planes you restrict the 10 independent components if totally symmetric. But if you draw one diagonal plane you restrict the 18 independent components if symmetric in just two two of its indices (9 elements on the diagonal plane + 9 elements in the one of the two halves of the cube). – Frobenius Aug 30 '17 at 7:32 | 2019-09-15T13:43:43 | {
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https://math.stackexchange.com/questions/2881657/how-to-prove-that-sqrt32014-sqrt2-sqrt320-14-sqrt2-4?noredirect=1 | # How to prove that $\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}=4$?
Using the Cardano formula, one can show that $\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}$ is a real root of the depressed cubic $f(x)=x^3-6x-40$. Actually, one can show by the calculating the determinant that this is the only real root. On the other hand, by the rational root theorem, one can see that the possible rational root must be a factor of 40 and one can check that $f(4)=0$. Therefore, by uniqueness of the real root, one must have $$\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}=4\tag{1}$$
I had the observation above when I solved the cubic equation $x^3-6x-40=0$. My question is as follows:
without referring to the unique real root of the cubic, can we show (1) directly?
[An attempt.] When taking the cube on both sides of (1) and simplifying further, I ended up with (1) again.
• This is almost a duplicate. See user8277998 answer to this question. – Dietrich Burde Aug 13 '18 at 18:10
• One way is to denest both radicals and show that the irrational parts cancel out. – Frank W. Aug 13 '18 at 18:12
• or this one math.stackexchange.com/questions/374619/… – mercio Aug 13 '18 at 18:12
• without referring to the unique real root of the cubic The easiest way is in fact to derive the cubic from the given radicals (you don't have to know the cubic in advance). – dxiv Aug 13 '18 at 19:13
• @dxiv: considering the answers below and the related links given in the comments above, I agree with you very much: solving a system of polynomial equations could be rather non-trivial, though in this particular case, a simple observation seems enough. The accepted answer in the link given by Dietrich seems mysterious to me: I don't see how one can "Solving the system" easily. – user486939 Aug 14 '18 at 15:07
Well, $(2\pm\sqrt2)^3=20\pm14\sqrt2$, that is $\sqrt[3]{20\pm14\sqrt2} =2\pm\sqrt2$. Therefore $$\sqrt[3]{2+14\sqrt2}+\sqrt[3]{20-14\sqrt2}=2+\sqrt2+2-\sqrt2=4.$$
• Thanks for your answer! Would you explain how you came up with the nice identity $(2\pm\sqrt{2})^3=20\pm14\sqrt{2}$? I can check it directly though, I'm curious about what is the underlying motivation. – user486939 Aug 13 '18 at 23:35
Can we express $\sqrt[3]{20+14\sqrt2}$ as $a+b\sqrt2$, with $a,b\in\mathbb Z$? In other words, are thre integers $a$ and $b$ such that $(a+b\sqrt2)^3=20+14\sqrt2$? Note that\begin{align}(a+b\sqrt2)^3=20+14\sqrt2&\iff a^3+3\sqrt2a^2b+6ab^2+2\sqrt2b^3=20+14\sqrt2\\&\iff a^3+6ab^2+(3a^2b+2b^3)\sqrt2=20+14\sqrt2.\end{align}Therefore, it is enough to have$$a^3+6ab^2=20(\iff a(a^2+6b^2)=20)\text{ and }3a^2b+2b^3=14(\iff b(3a^2+2b^2)=14).$$It is easy to see that you can take $a=2$ and $b=1$. Therefore, $\sqrt[3]{20+14\sqrt2}=2+\sqrt2$ and it is now easy to see that $\sqrt[3]{20-14\sqrt2}=2-\sqrt2$. Therefore$$\sqrt[3]{20+14\sqrt2}+\sqrt[3]{20-14\sqrt2}=2+\sqrt2+2-\sqrt2=4.$$
• I like that you have provided a natural construction rather than just giving an equality. Thanks! – Clayton Aug 13 '18 at 18:36
• Thanks for your answer! I was wondering how you observe that $\sqrt[3]{20+14\sqrt{2}}$ may be expressed as $a+b\sqrt{2}$. (I guess if $x^3=20+14\sqrt{2}$, then one may expect that $x$ is a $\mathbb{Z}$-linear combination of $1$ and $\sqrt{2}$?) – user486939 Aug 14 '18 at 0:54
• @Mars Since the original equation has an integer root, I thought that it would be reasonable to expect that $20\pm14\sqrt2$ hade a cube root of the form $a\pm b\sqrt2$, with $a,b\in\mathbb Z$. – José Carlos Santos Aug 14 '18 at 6:31
If you don't spot the values of these cube roots and don't feel like computing them, you can try this. Define $a_\pm:=(20\pm14\sqrt{2})^{1/3},\,s:=\sum_\pm a_\pm$ so $\sum_\pm a_\pm^3=40$ and $\prod_\pm a_\pm=8^{1/3}=2$, so $s^2=\frac{40}{s}+4ab=\frac{40}{s}+6$. Hence $0=s^3-6s-40=(s-4)(s^2+4s+10)$ has only one real root, $4$.
• Thanks for your answer. I think your factorization in the "Hence" step depends on knowing in advance that $4$ is a root of the polynomial? – user486939 Aug 14 '18 at 15:01
• @Mars You can solve a cubic by Cardano's method, but we all prefer to spot a factorisation, preferably with rational coefficients. Not only is it worth a try; when you add a radical to its conjugate, you'd expect a rational result. So naturally, you use the rational root theorem to find candidates. – J.G. Aug 14 '18 at 15:11
• I wrote in my post that "On the other hand, by the rational root theorem, one can see that the possible rational root must be a factor of 40 and one can check that f(4)=0." and I'm looking for an alternative proof for the identity. Anyway, thank you for your answer and comment again ;-) – user486939 Aug 15 '18 at 12:15 | 2019-08-17T10:54:27 | {
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https://math.stackexchange.com/questions/4227079/evaluating-the-integral-int-0-infty-fracx491x51-dx/4227083 | # Evaluating the Integral $\int_{0}^{\infty} \frac{x^{49}}{(1+x)^{51}} dx$
I tried evaluating the integral $$\displaystyle \int_{0}^{\infty} \dfrac{x^{49}}{(1+x)^{51}}dx$$ but I wasn't able to get the result. Following is the way by which I did it- $$I=\displaystyle \int_{0}^{\infty} \dfrac{x^{49}}{(1+x)^{51}}dx$$ $$\implies I=\int_{0}^{\infty} x^{49}(1+x)^{-51}dx$$ Further, I tried Integration by parts but it didn't worked. Can anyone tell that how this integral can be evaluated.
• HINT: try substitution $$u= \frac{x}{1+x}$$ Aug 18, 2021 at 3:21
• sorry, @Mathzcreator, but larger titles are discouraged on math.SE so that your question does not take up more vertical space than others, so I've edited the title. See this discussion on the meta site: math.meta.stackexchange.com/questions/9687/… Aug 18, 2021 at 3:59
• Note $$\int_{0}^{\infty} \dfrac{x^{49}}{(1+x)^{51}}dx \overset{x\to\frac1x}= \int_{0}^{\infty} \dfrac{1}{(1+x)^{51}}dx =\frac1{50}$$ Feb 23 at 19:10
The given integral is
$$I=\displaystyle \int_{0}^{\infty} \dfrac{x^{49}}{(1+x)^{51}}dx$$
$$I=\displaystyle \int_{0}^{\infty} \dfrac{x^{51}}{x^2(1+x)^{51}}dx$$
$$I=\displaystyle \int_{0}^{\infty} \dfrac{1}{x^2(1+\frac{1}{x})^{51}}dx$$
Let $$u=1+\frac{1}{x}$$ , therefore $$\displaystyle du=-\frac{dx}{x^2}$$
Therefore $$I=\displaystyle \int_\infty^1-\frac{du}{u^{51}}=\frac{1}{50}\bigg[\frac{1}{u^{50}}\bigg]^{1}_{\infty} = \boxed{\frac{1}{50}}$$
$$I=\displaystyle\int_{0}^{\infty}\dfrac{x^{49}}{(1+x)^{51}}dx$$ $$I=\displaystyle\int_{0}^{\infty}\dfrac{x^{50-1}}{(1+x)^{50+1}}dx$$ Using Beta Function,
$$B(x,y)=\displaystyle\int_{0}^{\infty}\dfrac{t^{x-1}}{(1+t)^{x+y}}dt$$
$$\implies I=\displaystyle\int_{0}^{\infty}\dfrac{x^{50-1}}{(1+x)^{50+1}}dx=B(50,1)$$ Using Beta Function and Gamma Function Relationship,
$$B(x,y)=\dfrac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$
$$\implies I=B(50,1)=\dfrac{\Gamma(50)\Gamma(1)}{\Gamma(50+1)}=\dfrac{\Gamma(50)\Gamma(1)}{50\Gamma(50)}=\dfrac{\Gamma(1)}{50}=\dfrac{1}{50}$$ $$\implies \boxed{I=\dfrac{1}{50}}$$
Here is an alternative method.
First of all, for each integer $$n>1$$, we have $$\int_1^\infty x^{-n}dx=\frac1{n-1}$$.
Thus, using the binomial theorem, \begin{align} \int_0^\infty\frac{x^{49}}{(1+x)^{51}}dx&=\int_1^\infty\frac{(x-1)^{49}}{x^{51}}dx\\ &=\int_1^\infty\frac1{x^{51}}\sum_{n=0}^{49}(-1)^{n-1}{49\choose n}x^ndx\\ &=\sum_{n=0}^{49}(-1)^{n-1}{49\choose n}\int_1^\infty x^{n-51}dx\\ &=\sum_{n=0}^{49}(-1)^{n-1}{49\choose n}\frac1{50-n}\\ &=\frac1{50}\sum_{n=0}^{49}(-1)^{n-1}{50\choose n}\\ &=-\frac1{50}\left((1+(-1))^{50}-{50\choose50}(-1)^{50}\right)\\ &=\frac1{50}. \end{align} | 2022-06-28T16:37:58 | {
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http://mathhelpforum.com/algebra/276484-averaging-down.html | 1. ## Averaging down
I need help figuring out how to find the number of share to purchase to bring down an average to a certain number.
I have 1,200 shares of xyz at \$9.34 and need to know how many to purchase at \$8.5 to bring down the average of all shares to \$9.00. How would I write a formula to solve for this and be able to interchange the variables as well. Thanks in advance. 2. ## Re: Averaging down Originally Posted by NumbersDontLie I need help figuring out how to find the number of share to purchase to bring down an average to a certain number. I have 1,200 shares of xyz at \$9.34 and need to know how many to purchase at \$8.5 to bring down the average of all shares to \$9.00.
How would I write a formula to solve for this and be able to interchange the variables as well.
Let X = # of shares to buy at $8.5. Total shares after purchase will then be 1200 + X. 1200 * 9.34 + 8.5 * X - (1200 + X) * 9 Solve for X! Steve 3. ## Re: Averaging down Originally Posted by SGS Let X = # of shares to buy at$\$$8.5. Total shares after purchase will then be 1200 + X. 1200 * 9.34 + 8.5 * X - (1200 + X) * 9 Solve for X! Steve That minus sign should be an equals sign. To find the average, you take the sum of the price of all of the shares (you have 1200 at \$$9.34 and x at $\$$8.50) and divide by the number of shares, which is 1200+x (as Steve suggested), and that average is \$$9:$\dfrac{9.34\cdot 1200 + 8.5x}{1200+x} = 9$Multiply both sides by 1200+x to get Steve's equation (but change the minus sign to an equals sign). 4. ## Re: Averaging down Originally Posted by SGS Let X = # of shares to buy at$8.5. Total shares after purchase will then be 1200 + X.
1200 * 9.34 + 8.5 * X - (1200 + X) * 9
Solve for X!
Steve
Thanks, I solved by simplifying (?) the equation (?) :
11208 + 8.5x - (10799 + 9x)
11208 + 8.5x - 10799 - 9x
11208 - 10799 + 8.5x - 9x
408 - 0.5x
0 = 408 - 0.5x
I added the " 0 = " but was it a given that it was always there when solving for one variable?
-408 = -0.5x
dividing both sides by -.05 to isolate x
-408/-0.5 = x
dividing two negative numbers for some reason seems weird to me to get the final answer, did I do something wrong or not optimal with my thinking process?
816 = x
.:. I need to buy 816 shares at $8.5 to get a net average price of$9 given that I already have 1,200 shares at $9.34. --- Is there a way to explain in English how I would come along this Originally Posted by SGS 1200 * 9.34 + 8.5 * X - (1200 + X) * 9 to solve for problems like mine in the future? Also how do I place dollar signs on this forum without it formatting everything together$123 word space word $456 5. ## Re: Averaging down The formula is$p = h * \dfrac{c - t}{t - m}, \text {where}\text p = \text { number of shares to PURCHASE;}\text h = \text { number of shares currently HELD;}\text c = \text { average COST per share of shares currently held;}\text t = \text { TARGET average cost per share after purchase; and}\text m = \text { current MARKET price per share.}$Proof$t = \dfrac{ch + mp}{h + p} \implies ht + pt = ch + mp \impliespt - mp = ch - ht \implies p(t - m) = h * (c - t) \implies p = h * \dfrac{c - t}{t - m}.$Let's try it in your example.$p = 1200 * \dfrac{9.34 - 9.00}{9.00 - 8.50} = 1200 * \dfrac{0.34}{0.50} = 816.$Let's check. Your total cost after purchase =$1200 * 9.34 + 816 * 8.50 = 11208 + 6936 = 18144.$Your number of shares after purchase =$1200 + 816 = 2016.$The new average cost per share =$\dfrac{18144}{2016} = 9.00.$And that was your target. You may not always get an exact answer because you cannot buy fractional shares. 6. ## Re: Averaging down Originally Posted by JeffM The formula is$p = h * \dfrac{c - t}{t - m}, \text {where}\text p = \text { number of shares to PURCHASE;}\text h = \text { number of shares currently HELD;}\text c = \text { average COST per share of shares currently held;}\text t = \text { TARGET average cost per share after purchase; and}\text m = \text { current MARKET price per share.}$Proof$t = \dfrac{ch + mp}{h + p} \implies ht + pt = ch + mp \impliespt - mp = ch - ht \implies p(t - m) = h * (c - t) \implies p = h * \dfrac{c - t}{t - m}.$Let's try it in your example.$p = 1200 * \dfrac{9.34 - 9.00}{9.00 - 8.50} = 1200 * \dfrac{0.34}{0.50} = 816.$Let's check. Your total cost after purchase =$1200 * 9.34 + 816 * 8.50 = 11208 + 6936 = 18144.$Your number of shares after purchase =$1200 + 816 = 2016.$The new average cost per share =$\dfrac{18144}{2016} = 9.00.$And that was your target. You may not always get an exact answer because you cannot buy fractional shares. First, thanks for your knowledge and time. Secondly, trying to commit this to memory might be beyond the scope of this forum but... Is there a way to summarize or better explain why we divide (c - t) by (t - m) or just a generalized statement in layman/english terms about the right side of the equation? I get that it works for my example and that it also works for different values but committing it to memory without a further explanation is a bit difficult for me. Originally Posted by SlipEternal To find the average, you take the sum of the price of all of the shares (you have 1200 at$\$$9.34 and x at \$$8.50) and divide by the number of shares, which is 1200+x, and that average is $\$$9:$\dfrac{9.34\cdot 1200 + 8.5x}{1200+x} = 9\$
Multiply both sides by 1200+x to get Steve's equation (but change the minus sign to an equals sign).
I think the above best explains things for me currently in english and I could reproduce it if I came to the situation again, but now I am wondering about the other explanation
------
Also what is the purpose of proofing? Please excuse my ignorance/lack of knowledge on the subject. (I did see/get that breaking down and rearranging the formula did work based what I learned in algebra class though)
7. ## Re: Averaging down
Can I explain the formula intuitively. Sort of. (c - t) is how far you want to average down. In your example, we want to average down by 9.34 - 9.00 or 0.34.
(t - m) is how close to current market you want to get. In your example, it is 9.00 - 8.50 or 0.50. The ratio of 0.34 over 0.50 is about 2/3. So we have to buy about 2/3 of what we already have. And 816 is close to 2/3 of 1200. I do not know whether that helps you feel that the formula matches up with your intuition.
I do not recommend memorizing many formulas. It is a waste of energy, and frequently you remember the wrong formula. What is more important is to remember how I got the formula. I started with defining the relevant variables and showed a common sense but general computation for calculating a new average cost namely
The historical cost per share of the old shares times the number of old shares PLUS the market price per share times the number of new shares purchased. That gave me the total cost of old and new shares together. Obvious, no? And how many total shares do we have after buying the new shares? Obviously the number of old shares plus the number of new shares. Again, obvious. So the average cost per share is the total cost divided by the total number of shares. Nothing esoteric.
Now we play with algebra to solve for the number of shares to be bought. That is totally mechanical. No real thinking involved. | 2018-06-25T04:32:16 | {
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https://math.stackexchange.com/questions/3526843/how-to-find-the-height-which-a-ball-will-bounce-after-a-collision-with-ground-if | # How to find the height which a ball will bounce after a collision with ground if the upward force is known?
The problem is as follows:
From a height of $$5\,m$$ with respect to the ground a sphere of $$0.1\,kg$$ of mass is released. The time elapsed in the contact with the ground is $$1\,ms$$ and magnitude of the average upward force is $$1900\,N$$. Find the height (measured in meters) from the ground which the ball will bounce. Assume $$g=10\,\frac{m}{s^2}$$.
The alternatives are as follows:
$$\begin{array}{ll} 1.&5.06\,m\\ 2.&4.05\,m\\ 3.&3.04\,m\\ 4.&2.03\,m\\ \end{array}$$
I'm confused exactly how to proceed with this question. The reason of the confusion is how to assess the answer. What I've attempted to do was to use the impulse momentum equation:
$$J=\Delta p$$
And the relationship between the impulse and force as follows:
$$J=\overline{F}\Delta t$$
In order to get the height which the sphere will get in the bounce back after collisioning with the ground I'm using the conservation of mechanical energy as follows:
$$mgh=\frac{1}{2}mv^2$$
Therefore:
$$h=\frac{1}{2g}v^2$$
but that v will be the speed attained by the ball after the collision. In order to find that v. I'm using the upward Force as follows:
At first all potential energy is transformed into kinetic energy.
$$\frac{1}{2}mv^2=mgh$$
$$v=\sqrt{2gh}=\sqrt{2\times 10 \times 5}= 10 \frac{m}{s^2}$$
Then:
$$m\delta v = \overline{F} \Delta t$$
$$0.1(v_f-10)=1900(10^{-3})$$
$$v_{f}=29$$
This will be the speed attained by the ball in the bouncing back.
Then this can be used to find the height attained by the ball:
Therefore returning to the earlier equation:
$$h=\frac{1}{2g}v^2$$
$$h=\frac{1}{2\times 10}(29)^2$$
$$h=42.05\,m$$
But this doesn't seem very reasonable. Does it exist an error in my approach?. Did I incurred in a contradiction or something, can somebody help me here?
• Energy is not conserved. (More precisely, kinetic + potential energy is not conserved -- some of it goes to heating up the ball and the ground.) – TonyK Jan 29 at 11:27
• Also, don't we need to know the mass of the ball? – TonyK Jan 29 at 11:37
• @TonyK Sorry I typed this question in a rush. The mass of the sphere is $0.1\,kg$. – Chris Steinbeck Bell Jan 29 at 13:45
Your mistake comes from an incorrect sign. Suppose we take upwards as our positive direction. Then $$v_f>0$$, but $$v_i<0$$ as it is in the downwards direction, and so your equation $$0.1(v_f-10)=1900(10^{-3})$$ should actually be $$0.1(v_f-(-10))=1900(10^{-3})$$
$$v_f=9$$ and hence $$h=4.05m$$
• You seem to have spotted that the mass of the sphere was $0.1\,kg$. I forgot to put this as part of the problem. Yes, it seems that where I made the mistake was that part. – Chris Steinbeck Bell Jan 29 at 13:48 | 2020-03-29T10:21:21 | {
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https://math.stackexchange.com/questions/145746/find-volume-of-a-revolved-solid-by-integrating-wedges | # Find volume of a revolved solid by integrating wedges.
So, lets say that I wanted to find the volume of the solid formed by rotating the area between
$f(x)=\sqrt{1-x^2}, 0<x<1$ and the $x$ axis around the $y$ axis. (This example is simply a hemisphere).
Now normally, I would use geometry, or the "disk method", so the area would simply be $\pi\int_0^1(1-y^2)dy=\frac{2\pi}{3}$.
I was thinking about this and I was wondering if it would be possible to find the answer by integrating wedges of this volume from $0$ to $2\pi$. This seems to be an approach that more closely resembles the premise of the problem. At first I thought that this might be as easy as $\frac{1}{2}\int_0^{2\pi}[\int_0^1f(x)dx]^2d\theta$, essentially integrating a polar circle with radius of the area that is revolved around the y axis. However, when I tried this, I did not get my expected answer. I calculated the volume to be $\frac{\pi^3}{16}$, however, I should have found the volume to be $\frac{2\pi}{3}$.
Can anyone help me understand why my approach was not successful, and also explain a successful method of evaluating the volume in this way?
• The problem is with the way you have choosen your differential element. Since you are finding volume in $\mathbb{R}^3$ using cylindrical co-ordinates(unknowingly I guess.Clarify me) Your Integral should look like $\int{ \int \int dzdrd\theta}$. – Ramana Venkata May 16 '12 at 5:00
• @RamanaVenkata I am using Cartesian coordinates. joriki's answers seems to work perfectly, but thank you for your help anyway. – diracdeltafunk May 16 '12 at 5:30
• Indirectly your using cylindrical coordinates. Even joriki's answer also written using cylindrical coordinates in some sense. – Ramana Venkata May 16 '12 at 5:37
• @RamanaVenkata I suppose that's true, however I definitely feel more comfortable in Cartesian or polar. – diracdeltafunk May 16 '12 at 5:43
In a wedge with angular extent $\mathrm d\theta$, an area $\mathrm dS$ of the rotated quarter-circle contributes $x\mathrm dS\mathrm d\theta$ to the volume of the wedge, so the volume is
\begin{align} \int_0^{2\pi}\left[\int x\mathrm dS\right]\mathrm d\theta &=\int_0^{2\pi}\left[\int_0^1xf(x)\mathrm dx\right]\mathrm d\theta \\ &=\int_0^{2\pi}\left[\int_0^1x\sqrt{1-x^2}\mathrm dx\right]\mathrm d\theta \\ &=\int_0^{2\pi}\left[-\frac13\sqrt{1-x^2}^3\mathrm dx\right]_0^1\mathrm d\theta \\ &=\frac{2\pi}3 \end{align}
• @Ben: Sorry, I don't understand the reasoning in your second comment; you'll have to elaborate. In response to your first comment, I didn't say that a wedge has volume $x\mathrm dS\mathrm d\theta$, but that an area $\mathrm dS$ of the rotated quarter-circle contributes $x\mathrm dS\mathrm d\theta$ to the volume of a wedge with angular extent $\mathrm d\theta$. This is because the part of the wedge corresponding to a surface element $\mathrm dS$ of the rotated quarter-circle is to first order a prism with base area $\mathrm dS$ and height $x\mathrm d\theta$. – joriki May 16 '12 at 6:14
• I think your terminology is confusing me a bit. First, why are you calling the area $dS$? Why not $S$? Also, how is the volume of the prism equal to $xdSd\theta$? Can you offer a proof of this? – diracdeltafunk May 17 '12 at 2:57 | 2020-01-29T18:43:08 | {
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http://mathhelpforum.com/calculus/4045-evaluate-summation.html | # Math Help - Evaluate the summation
1. ## Evaluate the summation
Evaluate:
from 1 to infinity the summation of [n/2^n-1]
I tried to work it out for a small amounts of n to see if I can see anything happening. What should I do next?
2. Originally Posted by Nichelle14
Evaluate:
from 1 to infinity the summation of [n/2^n-1]
I tried to work it out for a small amounts of n to see if I can see anything happening. What should I do next?
Hello, Nichelle,
I'm puzzled, because it is not clear for me, what you mean. Do you mean:
$\frac{n}{2^n}-1\quad \mbox{or}\quad \frac{n}{2^n-1}\quad \mbox{or}\quad \frac{n}{2^{n-1}}$
Greetings
EB
3. Here's a general technique that may help. I'll tackle the example $\frac{n}{2^{n-1}}$. The trick is to write down a function $f(x)$ for which the sum is a special value. Here we consider $f(x) = \sum_n \frac{n}{x^{n-1}}$, so that you want $f(2)$. We look at $x^{-2} f(x) = \sum_n \frac{n}{x^{n+1}}$ in order to be able to integrate term by term. So $x^{-2}f(x)$ is the derivative of $\sum_n \frac{-1}{x^n}$, which is $\frac{-1}{1-1/x} = \frac{x}{1-x} = 1+\frac{1}{1-x}$. Hence $x^{-2} f(x) = \frac{1}{(1-x)^2}$ and your sum is $f(2) = 4$.
4. I'll try the $\sum_{n=1}^{\infty}\frac{n}{2^{n}}-1$
You can just use the closed form for a geometric series, namely, $\frac{1}{1-x}$
$\frac{1}{1-\frac{1}{2}}=2$
$\sum_{n=1}^{\infty}\frac{n}{2^{n-1}}$ can be treated similarly.
This is the same as:
$\frac{n}{2^{n}2^{-1}}$
$2\sum_{n=1}^{\infty}\frac{n}{2^{n}}=\frac{2}{1-\frac{1}{2}}=4$
5. Hello, Nichelle14!
I'll assume that the exponent is $n-1$.
Evaluate: . $\sum^{\infty}_{n=1}\frac{n}{2^{n-1}}$
Here's an elementary approach . . .
We are given: . $S\;=\;1 + \frac{2}{2} + \frac{3}{2^2} + \frac{4}{2^3} + \frac{5}{2^4} + \hdots$
Divide by 2: . $\frac{1}{2}S \;= \;\;\;\,\quad\frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \frac{4}{2^4} + \hdots$
Subtract:. . . . $\frac{1}{2}S\;= \;1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \hdots$
The right side is a geometric series with $a = 1,\;r = \frac{1}{2}$
. . Hence, its sum is: . $\frac{1}{1 - \frac{1}{2}} \,= \,2$
Therefore, we have: . $\frac{1}{2}S\:=\:2\quad\Rightarrow\quad \boxed{S\,=\,4}$
6. It is the third choice. n-1 is an exponent of 2.
Thanks
7. Originally Posted by rgep
Here's a general technique that may help. I'll tackle the example $\frac{n}{2^{n-1}}$. The trick is to write down a function $f(x)$ for which the sum is a special value. Here we consider $f(x) = \sum_n \frac{n}{x^{n-1}}$, so that you want $f(2)$. We look at $x^{-2} f(x) = \sum_n \frac{n}{x^{n+1}}$ in order to be able to integrate term by term. So $x^{-2}f(x)$ is the derivative of $\sum_n \frac{-1}{x^n}$, which is $\frac{-1}{1-1/x} = \frac{x}{1-x} = 1+\frac{1}{1-x}$. Hence $x^{-2} f(x) = \frac{1}{(1-x)^2}$ and your sum is $f(2) = 4$.
Rgep, Will you please explain the use of integration and derivative here. I am not able to get it.
Keep Smiling
Malay
8. Originally Posted by malaygoel
Rgep, Will you please explain the use of integration and derivative here. I am not able to get it.
Keep Smiling
Malay
When you have a series of the form:
$
S(c,a,b)=\sum_{n=a}^b n.c^{n-1}
$
with $a,\ b,\$ and $c$ numerical constants,
we consider the function:
$
S(x,a,b)=\sum_{n=a}^b n.x^{n-1}
$
each term of this is obviously the derivative wrt $x$of:
$
t(x,n)=x^n
$
.
So we can integrate term by term to get:
$
I(x,a,b)=\sum_{n=a}^b x^{n}
$
Now if RHS is a geometric series and may be summed:
$I(x,a,b)=x^a \frac{1-x^{b-a+1}}{1-x}$,
(you will need to check this last sum, I did it in a bit of a rush )
and:
$
S(c,a,b)=\left{} \frac{d}{dx}I(x,a,b) \right|_{x=c}
$
Now you need only play around with the values of a, b and c (and the odd
limiting process to get the required result).
This is quite a common trick employed to find the sum of series.
RonL | 2014-04-19T07:24:52 | {
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http://ughomeworkazjp.ultimatestructuredwater.info/probability-distribution-case.html | # Probability distribution case
Probability distribution a probability distribution is a statistical function that identifies all the conceivable outcomes and odds that a random variable will have within a specific range. My question is: whats the difference between probability density function and probability distribution whats the difference between probability density function and probability distribution function probability probability discrete case: probability mass function (pmf. We'll do that using a probability density function (p you can imagine that the intervals would eventually get so small that we could represent the probability distribution of x, not as a in the case of this example, the probability that a randomly selected hamburger weighs between 0. The probability distribution defined in example 2 is over events, whereas the one here is defined over random variables the idea of conditional probability extends naturally to the case when the distribution of a random variable is conditioned on several variables, namely p(x = a|y = b,z. Conditional expectation for discrete random variables it follows that the \joint probability distribution function f xy is de ned as, f xy (ab) = p(x ay b) = z a 1 z b 1 f the various cases at once and making the calculations easier. Hand-book on statistical distributions for experimentalists by christian walck particle physics group 11 random number generation 1 2 probability density functions 3 1762 general case.
The probability distribution as a concept can occur in two ways, depending of the characteristics of your observation it can be a probability density function (pdf) in case of a continous random variable that models the observation, or, if only discrete values of the random variable are possible, with the help of the so called probability mass. Module ii lecture 4 special probability distributions it is very easy to work with the normal distribution using excel like the case of the binomial and poisson distributions, excel provides a function for computing values for any normal distribution. Understanding and using discrete distributions each discrete distribution has special properties that you should use for specific cases poisson distribution view probability dialog (graph probability distribution plot view probability), choose the binomial distribution. Journal of case studies in education in search of the most, page 1 in search of the most likely value statistics, summary measures, probability distribution, data types, spreadsheet journal of case studies in education in search of the most.
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Random variables and probability distributions 1 discrete random variables probability distribution for a discrete random variable proof for case of finite values of x consider the case where the random variable x takes on a finite. 1 answer to 1 use the sales forecaster's predication to describe a normal probability distribution that can be used to approximate the demand distribution sketch the distribution and show its mean and standard deviation 2 compute the probability of a stock-out for the order quantities suggested by - 168525. Since two of the outcomes represent the case in which just one head appears in the figure 1 is a discrete probability distribution: the binomial distribution consists of the probabilities of each of the possible numbers of successes on n trials for.
## Probability distribution case
Describes the basic characteristics of discrete probability distributions, including probability density functions and cumulative distribution functions real statistics using excel in this case the cumulative distribution function is given by. Probability and cumulative distribution functions lesson 20 recall (in many cases a cumulative distribution function p(t), which gives the fraction of maintenance checks completed in time less than or equal to t minutes.
• Sect 5-2, p 209 identifying probability distributions in exercise 7-12, determine whether a probability distribution is given in those cases where a probability distribution is not described, identify the requirements that.
• It represents a discrete probability distribution concentrated at 0 — a degenerate distribution — but the notation treats it as if it were a continuous distribution it is a special case of the gamma distribution, and it is used in goodness-of-fit tests in statistics.
• Start studying ch 8 random variables and probability distributions learn vocabulary, terms, and the multinomial distribution represents an _____ of the binomial distribution for the case in which a trial can result in an outcome from one of k2 classes and the probabilities.
• The binomial probability distribution models this item when the events are independent and the fixed probability of failure the exponential distribution is a special case of the poisson when the number of events in the risk, failure probability, and failure rate 4 170505 [email protected]
1 use the sales forecaster's predication to describe a normal probability distribution that can be used to approximate the demand distribution. Lesson 19: conditional distributions printer-friendly version in the discrete case we will extend the idea of conditional probability that we learned previously to the idea of finding a conditional probability distribution of a random variable y given another random variable x. The binomial distribution assumes that events are independent and the probabilities of events occurring are constant over time where sampling without replacement takes place, the population size typically needs to be 100+ if not, the hypergeometric distribution should be used the binomial distribution also assumes that events are binary, so. The poisson distribution in the where the probability of each of two mutually exclusive events (p and q) is known the poisson distribution, so to speak in these cases, q is known (as in true poisson problems it is not), but it is simply discarded. View notes - chapter 6 case 4 from econ 340 at ill chicago chapter 6 continuous probability distributions chapter 6 continuous probability distributions case problem: specialty toys 1 information. Entropy of a probability distribution the assignment of probabilities needs to reflect this and the ideal probability distribution in this case would be the one that both satisfies those constraints as well as has the highest entropy as possible.
Probability distribution case
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https://physics.stackexchange.com/questions/545169/particle-in-a-box-wavefunction-derivation/545181 | # Particle in a box wavefunction derivation
I tried solving the particle in a box problem and I came to a result that's different than what I find online. I solved the Schrödinger equation and I found the analytical form of $$\psi$$: $$\psi(x) = Ae^{ikx} + Be^{-ikx}$$ Then I set the boundary conditions $$\psi(0)=0\,\qquad \psi(L)=0$$ and find the relations $$A+B=0 \qquad Ae^{ikL}+Be^{-ikL}=0$$ Then, substituting $$B$$ for $$-A$$, I get $$A(e^{ikL} - e^{-ikL}) =0$$ or \begin{align}e^{ikL}-e^{-ikL}&=2i\sin(kL)=0\ ,\\ \psi(x) &=2iA\sin(kx) \end{align} I then try to normalize the wave function so that $$\int_0^L|\psi(x)|^2dx=4|A|^2\int_0^L \sin^2(kx)dx=1$$ $$4|A|^2\frac{L} 2=1$$ $$A=±\frac{1}{\sqrt{2L} }$$ Which gives the final wave function: $$\psi(x) =\frac{2i}{\sqrt{2L}}\sin\left(\frac{n\pi}{L}x\right)$$ which is different from what I've found online: $$\psi(x) =\sqrt{\frac{2}{L}}\sin\left(\frac{n\pi}{L}x\right)$$ Am I doing something wrong in solving the problem? Is there more than one correct answer, and if so, why?
• I've removed a number of comments that were attempting to answer the question and/or responses to them. Please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. Apr 20, 2020 at 0:37
## 3 Answers
Normally, when doing this problems you put every constant you encounter inside one "common" constant only, namely $$A$$. The thing is that you didn't put $$2i$$ inside of it, and that wouldn't be a problem, normalization takes that into account. So in the end you have $$\psi(x)=\frac{2i}{\sqrt{2L}}\sin\left(\frac{n\pi}{L}x\right)$$. But notice that $$\frac{2}{\sqrt{2}}=\sqrt{2}$$, thus $$\psi(x)=i\sqrt{\frac{2}{L}}\sin\left(\frac{n\pi}{L}x\right)$$ Which is the same answer, but with an $$i$$ multiplying.
It's just a convention that we use pure real wavefunctions, the answer you got is perfectly correct, but not "standard", because it is pure imaginary. Wavefunctions don't have any meaning in real life (at least in the Copenhagen interpretation), they are just tools that we can use to get, for example, the probability density $$\rho(x)=|\psi(x)|^2$$, which is a measurable quantity. Notice that you can get the same probability density from your result as from the standard result.
The wavefunction only has physical meaning in terms of probability when you take its magnitude squared.
A factor of $$i$$, which is the difference between your solution and the one you expect, does not matter. In general, the wavefunction is equivalent up to any phase $$e^{i\phi}$$.
The only difference in your solution and the "online" is a phase factor of $$i$$. Every SWE solution has an arbitrarily chosen phase factor because of the normalization. You chose to make $$A$$ real, but it doesn't have to be.
$$A=\pm\frac{e^{i\delta}}{\sqrt{2L}}$$ is a more general expression, where $$\delta$$ is any real number. You happened to choose $$\delta = 0$$ and $$\pi$$. You could have chosen $$3\pi/2$$ to get $$A=\frac{-i}{\sqrt{2L}}.$$
That would yield the "online" result. | 2022-09-27T07:54:03 | {
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https://mathematica.stackexchange.com/questions/69572/how-to-cut-a-thick-curve-in-half-lengthwise | # How to cut a thick curve in half lengthwise?
Is there any easy way to cut a thick curve in half lengthwise? Example: suppose I have the cardioid shown in the figure below, is there a way to delete either the inner or the outer portion of the curve along the dashed line? Thanks for reading!!
Edit: code used to generate plot.
g = ParametricPlot[{(1 + Cos[t]) Cos[t], (1 + Cos[t]) Sin[t]}, {t, 0,
2 Pi}, PlotRange -> {{-0.5, 2.2}, {-1.5, 1.5}},
PlotStyle -> {Thickness[0.06]}];
h = ParametricPlot[{(1 + Cos[t]) Cos[t], (1 + Cos[t]) Sin[t]}, {t, 0,
2 Pi}, PlotRange -> {{-0.5, 2.2}, {-1.5, 1.5}},
PlotStyle -> {Thickness[0.0035], White, Dashed}];
Show[g, h]
• If you don't need the inside to be transparent, what I do is to draw the curve again filled with white, leaving only the outer half showing. – user484 Dec 23 '14 at 14:45
• Thanks, but how does that help, Rahul? Mathematica always centers the second curve along the middle (the dashed line in my pic) of the orginal curve, so your solution would only leave two blue curves separated by the width of the white curve. Whereas what I want is a single curve on one side of the dashed line. In other words: if you imagine the thick blue cardioid curve to be a road, I only want the left or the right lane. – Kim Fierens Dec 23 '14 at 16:04
• This has been asked before, but I can't find the original. Anyone? – Dr. belisarius Dec 23 '14 at 16:56
• In the answers to 28202, the normal to the curve was extended on both side. Adapt to extend on one side only. Post your answer here if you figure it out (unless the community thinks it's duplicate). If you can't figure it out, let us know. Some go-getter will no doubt figure it out in the meantime. – Michael E2 Dec 23 '14 at 17:29
• @Kim By "filled with white" I meant colouring the interior of the cardioid with white, not the perimeter. In the road metaphor this covers the inside lane as well as the traffic island in the middle. I believe that's what the first part of Algohi's answer does. – user484 Dec 23 '14 at 19:09
Since you said "Or", I will do the inner one. you can do it like this:
p1 = ParametricPlot[
r {(1 + Cos[t]) Cos[t], (1 + Cos[t]) Sin[t]}, {t, 0, 2 Pi}, {r, 0,
1}, PlotRange -> All, Frame -> False,
PlotStyle -> {White, Opacity[1]}];
Show[g, h, p1]
For the outer one "Also can be used in general for inner and outer" you can use:
p1 = ParametricPlot[{1 - r + r (1 + Cos[t]) Cos[t],
r (1 + Cos[t]) Sin[t]}, {t, 0, 2 Pi}, {r, 1, 2}, PlotRange -> All,
Frame -> False, PlotStyle -> {White, Opacity[1]}];
Show[g, h, p1]
If you really need a Graphics object that represents a "sliced" version of the boundary curve (instead of just hiding one half of the line as in Algohi's solution which I also upvoted because it's easier), then you can achieve that as follows:
g = ParametricPlot[{(1 + Cos[t]) Cos[t], (1 + Cos[t]) Sin[t]}, {t, 0,
2 Pi}, PlotRange -> {{-0.5, 2.2}, {-1.5, 1.5}},
PlotStyle -> {Thickness[0.06]}];
h = ParametricPlot[{(1 + Cos[t]) Cos[t], (1 + Cos[t]) Sin[t]}, {t, 0,
2 Pi}, PlotRange -> {{-0.5, 2.2}, {-1.5, 1.5}},
PlotStyle -> {Thickness[0.0035], White, Dashed}];
pts = Polygon[First[Cases[g, Line[x_] :> x, Infinity]]];
rdf = SignedRegionDistance[pts];
rp = With[{thickness = .1},
RegionPlot[
0 < rdf[{x, y}] < thickness, {x, -.7, 2.5}, {y, -1.6, 1.6},
PlotStyle -> Darker[Blue], BoundaryStyle -> None,
PlotPoints -> 50]
];
Show[rp, h, Background -> Lighter[Orange]]
After reproducing the definitions of g and h from the question, I extract the boundary points of the parametric curve from g and use them to define a geometric Region by turning them into a Polygon for which we can then compute the signed distance in the function rdf. The function SignedRegionDistance yields 0 when you're on the line defined by the original plot g and h, and has a positive value outside that region. This can be used to impose a desired thickness in the RegionPlot that I store in the result rp, which then consists of a polygon that describes the desired shape with the desired thickness, but is outlined only in the outward direction.
To show that the graphics object is indeed the desired longitudinal slice, I superimpose it on the dashed parametric curve h with an added background that proves that nothing was hidden by overlying opaque regions.
The quality of the polygon can be changed by adjusting the PlotPoints option in RegionPlot. If necessary, you can also extract the polygon from the RegionPlot by doing First@Normal@rp. | 2021-05-17T18:42:08 | {
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"url": "https://mathematica.stackexchange.com/questions/69572/how-to-cut-a-thick-curve-in-half-lengthwise",
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"lm_q1q2_score": 0.8430002132757837
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14 Aug 2017, 23:39
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Thanks a lot for the collection Bunuel, I've been struggling with Inequalities, Probability & Combinatorics from so long. This post is very helpful to strengthen my base in Inequalities.
Do we have any such comprehensive thread for Probability and Combinatorics?
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15 Aug 2017, 23:02
nikhilpoddar wrote:
What is the range for a ??
a + a^(-1) >2
a+1/a>2
(a-1)^2 / a >0
So a > 1
What is the OA?
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Posts: 43358
### Show Tags
15 Aug 2017, 23:06
srikanth9502 wrote:
nikhilpoddar wrote:
What is the range for a ??
a + a^(-1) >2
a+1/a>2
(a-1)^2 / a >0
So a > 1
What is the OA?
a + 1/a > 2 holds true for 0 < a < 1 and a > 1.
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### Show Tags
15 Aug 2017, 23:38
Bunuel wrote:
srikanth9502 wrote:
nikhilpoddar wrote:
What is the range for a ??
a + a^(-1) >2
a+1/a>2
(a-1)^2 / a >0
So a > 1
What is the OA?
a + 1/a > 2 holds true for 0 < a < 1 and a > 1.
Is there any specified approach to solve these kind of questions?
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### Show Tags
14 Oct 2017, 13:48
WilDThiNg wrote:
Bunuel wrote:
Inequalities with Complications - Part I
BY KARISHMA, VERITAS PREP
Above we learned how to handle inequalities with many factors i.e. inequalities of the form $$(x – a)(x – b)(x – c)(x – d) > 0$$. This week, let’s see what happens in cases where the inequality is not of this form but can be manipulated and converted to this form. We will look at how to handle various complications.
Complication No. 1: $$(a – x)(x – b)(x – c)(x – d) > 0$$
We want our inequality to be of the form $$(x – a)$$, not $$(a – x)$$ because according to the logic we discussed last week, when x is greater than a, we want this factor to be positive. The manipulation involved is pretty simple: $$(a – x) = -(x – a)$$
So we get: $$– (x – a)(x – b)(x – c)(x – d) > 0$$
But how do we handle the negative sign in the beginning of the expression? We want the values of x for which the negative of this expression should be positive. Therefore, we basically want the value of x for which this expression itself (without the negative sign in the beginning) is negative.
We can manipulate the inequality to $$(x – a)(x – b)(x – c)(x – d) < 0$$
Or simply, multiply $$– (x – a)(x – b)(x – c)(x – d) > 0$$ by -1 on both sides. The inequality sign flips and you get $$(x – a)(x – b)(x – c)(x – d) < 0$$
e.g. Given: $$(4 – x)(2 – x)(-9 – x) < 0$$
We can re-write this as $$–(x – 4)(2 – x)(-9 – x) < 0$$
$$(x – 4)(x – 2)(-9 – x) < 0$$
$$-(x – 4)(x – 2)(x + 9) < 0$$
$$(x – 4)(x – 2)(x – (-9)) > 0$$ (multiplying both sides by -1)
Now the inequality is in the desired form.
Complication No 2: $$(mx – a)(x – b)(x – c)(x – d) > 0$$ (where m is a positive constant)
How do we bring $$(mx – a)$$ to the form $$(x – k)$$? By taking m common!
$$(mx – a) = m(x – a/m)$$
The constant does not affect the sign of the expression so we don’t have to worry about it.
e.g. Given: $$(2x – 3)(x – 4) < 0$$
We can re-write this as $$2(x – \frac{3}{2})(x – 4) < 0$$
When considering the values of x for which the expression is negative, 2 has no role to play since it is just a positive constant.
Now let’s look at a question involving both these complications.
Question 1: Find the range of x for which the given inequality holds.
$$-2x^3 + 17x^2 – 30x > 0$$
Solution:
Given: $$-2x^3 + 17x^2 – 30x > 0$$
$$x(-2x^2 + 17x – 30) > 0$$ (taking x common)
$$x(2x – 5)(6 – x) > 0$$ (factoring the quadratic)
$$2x(x – \frac{5}{2})(-1)(x – 6) > 0$$ (take 2 common)
$$2(x – 0)(x – \frac{5}{2})(x – 6) < 0$$ (multiply both sides by -1)
This inequality is in the required form. Let’s draw it on the number line.
We are looking for negative value of the expression. Look at the ranges where we have the negative sign.
The ranges where the expression gives us negative values are $$5/2 < x < 6$$ and x < 0.
Hence, the inequality is satisfied if x lies in the range $$\frac{5}{2} < x < 6$$ or in the range $$x < 0$$.
Plug in some values lying in these ranges to confirm.
In the next post, we will look at some more variations which can be brought into this form.
[Reveal] Spoiler:
Attachment:
Ques4.jpg
Hi:
Please could you help me understand how to determine which sign to assign to which range in the above graph?
Thanks
as per my understanding the extreme right one you can take +ve & then assign an alternate sign to each region.
Try yourself with a few equations. You can clarify that yourself.
cheers
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28 Dec 2017, 04:16
Bunuel wrote:
Or Just Use Inequalities!
BY KARISHMA, VERITAS PREP
If you are wondering about the absurd title of this post, just take a look at the above post's title. It will make much more sense thereafter. This post is a continuation of last week’s post where we discussed number plugging. Today, as per students’ request, we will look at the inequalities approach to the same official question. You will need to go through our inequalities post to understand the method we will use here.
Recall that, given $$a < b$$, $$(x – a)(x – b) < 0$$ gives us the range $$a < x < b$$ and $$(x – a)(x – b) > 0$$ gives us the range $$x < a$$ or $$x > b$$.
Question: If x is positive, which of the following could be the correct ordering of 1/x, 2x and x^2?
(I) x^2 < 2x < 1/x
(II) x^2 < 1/x < 2x
(III) 2x < x^2 < 1/x
(A) none
(B) I only
(C) III only
(D) I and II
(E) I, II and III
Solution: The question has three complex inequalities. We will take each in turn. Note that each inequality consists of two more inequalities. We will split the complex inequality into two simpler inequalities e.g. x^2 < 2x < 1/x gives us x^2 < 2x and 2x < 1/x. Next we will find the range of values of x which satisfy each of these two inequalities and we will see if the two ranges have an overlap i.e. whether there are any values of x which satisfy both these simpler inequalities. If there are, it means there are values of x which satisfy the entire complex inequality too. Things will become clearer once we start working on it so hold on.
Let’s look at each inequality in turn. We start with the first one:
(I) x^2 < 2x < 1/x
We split it into two inequalities:
(i) x^2 < 2x
We can rewrite x^2 < 2x as x^2 – 2x < 0 or x(x – 2) < 0.
We know the range of x for such inequalities can be easily found using the curve on the number line. This will give us 0 < x < 2.
(ii) 2x < 1/x
It can be rewritten as x^2 – 1/2 < 0 (Note that since x must be positive, we can easily multiply both sides of the inequality with x)
[ Secondly would you explain this step again i did not understand how multiplying by x gives us this equation] $$-1/?2< x < 1/?2$$
This gives us the range -1/?2 < x < 1/?2 (which is 0 < x < 1/?2 since x must be positive). [ Question mark is square root i think]
Is there a region of overlap in these two ranges i.e. can both inequalities hold simultaneously for some values of x? Yes, they can hold for 0 < x < 1/?2. Hence, x^2 < 2x < 1/x will be true for the range 0 < x < 1/?2. So this could be the correct ordering. Let’s go on to the next complex inequality.
(II) x^2 < 1/x < 2x
Again, let’s break up the inequality into two parts:
(i) x^2 < 1/x
x^1 < 1/x is rewritten as x^3 – 1 < 0 which gives us x < 1.
(ii) 1/x < 2x
1/x < 2x is rewritten as x^2 – 1/2 > 0 which gives us x < -1/?2 (not possible since x must be positive) or x > 1/?2
Can both x < 1 and x > 1/?2 hold simultaneously? Sure! For 1/?2 < x < 1, both inequalities will hold and hence x^2 < 1/x < 2x will be true. So this could be the correct ordering too.
(III) 2x < x^2 < 1/x
The inequalities here are:
(i) 2x < x^2
2x < x^2 can be rewritten as x(x – 2) > 0 which gives us x < 0 (not possible) or x > 2.
(ii) x^2 < 1/x
x^2 < 1/x gives us x^3 – 1 < 0 i.e. x < 1
Can x be less than 1 and greater than 2 simultaneously? No. Therefore, 2x < x^2 < 1/x cannot be the correct ordering.
Is this method simpler?
Hi bunuel (Or any other expert) i have highlighted a portion in green and red . Could you clarify those points. Thankyou
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28 Dec 2017, 05:32
mtk10 wrote:
Hi bunuel (Or any other expert) i have highlighted a portion in green and red . Could you clarify those points. Thankyou
(ii)$$2x < \frac{1}{x}$$...
Multiply by x..
$$2x*x < \frac{1}{x}*x.........2x^2<1..........x^2<\frac{1}{2}.....x^2-1/2<0$$...
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http://openstudy.com/updates/560f77bae4b0023ff6878695 | ## anonymous one year ago Another limit question...
1. anonymous
$\lim_{n \rightarrow \infty}\ln\left( 1+ \frac{ 4-\sin(x) }{ n } \right)^n$
2. anonymous
My approach was that I raised it to the power of e and then I Just found the limit of the stuff inside. But that's as far as I got.... I see a pattern of (1+(1/x))^n which is the definition of e but I don't really know about this one...
3. hartnn
4 -sin x or 4 - sin n ?? if its 4- sin x, then its a constant...
4. anonymous
Nope that is not a typo. It is indeed a constant.
5. hartnn
cool! so you have the function of the form $$(1+ax)^{1/x}$$ right?
6. anonymous
How so?
7. anonymous
Okay well I can see it if you make a substitution.
8. anonymous
Go on.
9. hartnn
good, i'll tell you what we do in case of $$(1+ax)^{1/x}$$ -- > $$(1+ax)^{1/x} = [(1+ax)^{\frac{1}{ax}}]^a$$ and then use the limit formula, (if you can use) $$\lim \limits_{x\to \infty} (1+1/x)^x = e$$
10. anonymous
Brilliant! But the limit would then be 1 not 0...
11. anonymous
Which according to wolfram it's 0.
12. anonymous
13. anonymous
Ohh wait. I goofed. Have to take the ln of that.
14. anonymous
ln(1) is 0. Thank you!
15. hartnn
in the wolf, its shows ln^n
16. anonymous
Yeah it's a notational thing. That just the inside raised to the n.
17. anonymous
Kinda like sin^2(x) and (sin(x))^2.
18. hartnn
okk... i thought the answer would be 4- sin x ..
19. anonymous
Nah. It's 0.
20. hartnn
:)
21. anonymous
@hartnn : So I got up here.
22. anonymous
$\lim_{b \rightarrow 0}(1+(4-\sin(x)b)^{\frac{ 1 }{ b }}$
23. anonymous
Is that okay so far or am I way off?
24. hartnn
i assume there is ln outside of that limit and you just plugged in b =1/n
25. anonymous
Indeed sir.
26. hartnn
yes, go on
27. anonymous
Stuck >.< .
28. anonymous
Like I know that should be e but I'm having trouble relating it to the definition.
29. hartnn
whatever expression is with $$\Large 1+ ...$$ that same expression should be with $$\Large \dfrac{1}{...}$$ thats how I remember so we have 1+ (4-sin x)b so the fraction in the exponent should be $$\Large \dfrac{1}{(4-\sin x)b}$$
30. hartnn
|dw:1443855809053:dw|
31. anonymous
But the definition of e is (1+1/x)^x right? Here we have (1+(constant)b)^(1/b) . Are those equivalent?
32. hartnn
33. hartnn
34. anonymous
Interesting. I did not know that even after 4 years of calculus and differential equations lol. I learn new things every day!
35. Jhannybean
I want to learn how to solve this as well.. I get some steps but Im confused on others. :(
36. hartnn
We take that as a formula, but its easy to prove that using L'Hopital's rule.
37. hartnn
I will be writing out all the steps from the beginning
38. anonymous
Yes I know we can use L'hopital's rule but for this assignment they (Other students) can't use that.
39. hartnn
*drawing Let 4 - sin x = a , since its a constant. |dw:1443856327983:dw|
40. anonymous
Yep I got that.
41. hartnn
writing these steps for everyone's benefit :) |dw:1443856406335:dw|
42. anonymous
Yep makes sense so far...
43. hartnn
|dw:1443856506563:dw|
44. hartnn
that big bracket evaluates to 1 1^a = 1 ln 1 = 0 :D
45. anonymous
Cool!
46. anonymous
Is there somewhere I can find the proof of the stuff in the brackets?
47. Jhannybean
|dw:1443856722618:dw|
48. anonymous
Yeah to get into the proper form for the limit.
49. hartnn
lets prove it :) I'll use L'Hopitals, need to search the net for other proofs. |dw:1443856725213:dw| quick check, 0/0 form ln (1+ab) = ln 1 = 0 ab = 0 so we can apply L'Hopital's rule here
50. anonymous
Nice!
51. hartnn
we can do all kinds of mathematically legal manipulations to bring an expression in the standard form. i needed a form like (1+x)^(1/x) thats why I multiplied and divided by 'a' , which should be NON-ZERO (point to be noted.)
52. Jhannybean
oh I see I see
53. Astrophysics
How does this 0, the power would be undefined |dw:1443857378987:dw|
54. Astrophysics
Oooh wait nvm, n = 1/b when n-> infinity, b ->9
55. Astrophysics
b->0*
56. Jhannybean
Yeah, there you go
57. Astrophysics
Haha, I totally missed that, ok it's good now. Great explanation @hartnn thanks
58. Jhannybean
me 4.
59. anonymous
@hartnn
60. anonymous
|dw:1443858029848:dw|
61. hartnn
http://www.wolframalpha.com/input/?i=lim+n-%3E+infty+%281%2Ba%2Fn%29%5E%28%28n%29%29+ ln e^a = a ln e = a a = 4-sin x thats what I first got. but this wolf answer got me all confused and I ended up using b->infty instead of b->0
62. anonymous
But that's wrong though. b goes to 0, not n.
63. anonymous
not infinity*
64. Jhannybean
can you explain as to why b $$\rightarrow$$ 0 and not b $$\rightarrow \infty$$ ?
65. hartnn
true, and it makes sense logically too, constant/n = very very small no. 1+ very very small no. = 1 1^ very very larger number = 1 ln 1 =0 so the limit must go to 0 but with all the mathematical steps, I still get the answer as 4-sin x
66. anonymous
It's 0 according to wolf :( .
67. anonymous
Asking around. Like my feel is that the inside of that logarithm should be a 1.
68. anonymous
Only then can we get a 0.
69. hartnn
70. anonymous
It's witchcraft I tell you!
71. anonymous
This is the statement of t he original problem.
72. hartnn
we can only bring limit inside a function if that function is continuous. and logarithm is indeed continuous...
73. hartnn
http://www.wolframalpha.com/input/?i=lim+n-%3E+infinity+n*+ln+%281%2B%284-sin%28x%29%29%2Fn%29 even that gives 4-sin x!
74. Jhannybean
haha oh my goodness x_x
75. anonymous
Ohh wow. Wolfram is apparantly wrong.
76. anonymous
77. Jhannybean
LOL
78. Jhannybean
"which is apparently related to e" xD
79. Jhannybean
80. anonymous
Okay wow wolfram can't read notation clearly -.- .
81. hartnn
its very rare case where wolfram goes wrong
82. anonymous
Wow ._. ...
83. hartnn
$$\color{blue}{\text{Originally Posted by}}$$ @hartnn okk... i thought the answer would be 4- sin x .. $$\color{blue}{\text{End of Quote}}$$ and then wasted an hour :P
84. anonymous
All because we hail our god wolfram alpha too much ._. ...
85. Jhannybean
xD It hardly fails!!! As humans we value consistency and dependability :P | 2017-01-22T04:21:38 | {
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http://math.stackexchange.com/questions/68599/what-does-a-condition-being-sufficient-as-well-as-necessary-indicates | # What does a condition being sufficient as well as necessary indicates?
I have a question in a book I am solving(Discrete Structures by Kolman, Busby & Ross). I am unable to make sense from the question. It is stated below, Show that k is odd is a necessary and sufficient condition for k^3 to be odd.
Now what I extracted out of the question was, As k is odd is necessary as well as sufficient condition for k^3 then they must be the same logic I must say that they are logically equivalent. Is this what the question demands? Please help me understand the question.
Thanks.
-
The question asks you to prove: $k^3$ is not divisible by $2$ if and only if $k$ is not divisible by $2$. – Asaf Karagila Sep 29 '11 at 21:34
"P is a necessary and sufficient condition for Q" is the same as "P is equivalent to Q", "P if and only if Q" ("P $\iff$ Q"). All of them mean "(If P, then Q) AND (If Q, then P)". – Srivatsan Sep 29 '11 at 21:36
I always tell... – The Chaz 2.0 Sep 29 '11 at 22:40
Like Chaz, I don't understand the downvote here... – Ｊ. Ｍ. Sep 30 '11 at 0:51
To say that condition $P$ is necessary for condition $Q$ is to say that you cannot have $Q$ without having $P$ as well. That is to say: $$Q\rightarrow P$$
On the other hand, to say that a condition $P$ is sufficient for the condition $Q$ is to say that if you have $P$ then you surely have $Q$. Formally: $$P\rightarrow Q$$
Thus, to say that a condition is necessary and sufficient is to say that $P$ is sufficient for $Q$ and it is also necessary for it. Therefore this is to say that they are equivalent. $$P\leftrightarrow Q$$
The question which baffles you asks to show that $k$ is odd implies $k^3$ is odd, as well $k^3$ is odd implies $k$ is odd.
-
$A$ is a necessary condition for $B$ means
i) $B\implies A$
ii) If $B$ is true then $A$ is true
iii) If $A$ is false then $B$ is false (contrapositive argument)
On the other hand, $A$ is a sufficient condition for $B$ means
i) $A\implies B$
ii) If $A$ is true then $B$ is true
iii) If $B$ is false then $A$ is false (contrapositive argument) | 2014-03-09T20:32:53 | {
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http://jehobs.com/3a6k2bow/e95e2b-corollary-of-cyclic-quadrilateral-theorem | Theorems of Cyclic Quadrilateral Cyclic Quadrilateral Theorem The opposite angles of a cyclic quadrilateral are supplementary. ⓘ Ptolemys theorem. only if it is a cyclic quadrilateral. ;N�P6��y��D�ۼ�ʞ8�N�֣�L�L�m��/a���«F��W����lq����ZB�Q��vD�O��V��;�q. For a parallelogram to be cyclic or inscribed in a circle, the opposite angles of that parallelogram should be supplementary. Proof: Let us now try to prove this theorem. (A and C are opposite angles of a cyclic quadrilateral.) This theorem can be proven by first proving a special case: no matter how one triangulates a cyclic quadrilateral, the sum of inradii of triangles is constant.. After proving the quadrilateral case, the general case of the cyclic polygon theorem is an immediate corollary. Welcome to our community Be a part of something great, join today! If ABCD is a cyclic quadrilateral, then opposite angles sum to 180◦ Theorem 20. The circle which consist of all the vertices of any polygon on its circumference is known as the circumcircle or, Important Questions Class 8 Maths Chapter 3 Understanding Quadrilaterals, Important Questions Class 9 Maths Chapter 8 Quadrilaterals, Therefore, an inscribed quadrilateral also meet the. 6 The solution given by Prasolov in [14, p.149] used Theorem 2 and is, although not stated as Theorem 1 states that the vertices of V and those of Hlie on two circles with center G. Corollary 2. A test for a cyclic quadrilateral. This will help you discover yet a new corollary to this theorem. anticenters of a cyclic m-system and we find a result on cyclic polygons with m sides, with m4 (theorem 5.2), that generalize the property on the quadrilateral of the orthocenters of a cyclic quadrilateral [2, 7]; in paragraph 6 we introduce the notion of n-altitude of a cyclic m-system, with m 6 and, in particular, 8.2 Circle geometry (EMBJ9). Midpoint Theorem and Equal Intercept Theorem; Properties of Quadrilateral Shapes To get a rectangle or a parallelogram, just join the midpoints of the four sides in order. If a cyclic quadrilateral is also orthodiagonal, the distance from the circumcenter to any side equals half the length of the opposite side. If a quadrilateral is cyclic, then the exterior angle is equal to the interior opposite angle. If the interior opposite angles of a quadrilateral are supplementary, then the quadrilateral is cyclic. (a) is a simple corollary of Theorem 1, since both of these angles is half of . If there’s a quadrilateral which is inscribed in a circle, then the product of the diagonals is equal to the sum of the product of its two pairs of opposite sides. Theorems on Cyclic Quadrilateral. Leaving Certificate Ordinary Level Theorems ***Important to note that all … Then $$\theta_1+\theta_2=\theta_3+\theta_4=90^\circ\$$; (since opposite angles of a cyclic quadrilateral are supplementary). It is also sometimes called inscribed quadrilateral. Theorem 5: Cyclic quadrilaterals ... Summary of circle geometry theorems ... Corollary: The centre of a circle is on the perpendicular bisector of any chord, therefore their intersection point is the centre. Worked example 4: Opposite angles of a cyclic quadrilateral You should practice more examples using cyclic quadrilateral formulas to understand the concept better. Ptolemy used the theorem as an aid to creating his table of chords, a trigonometric table that he applied to astronomy. Definition of cyclic quadrilateral, cyclic quadrilateral theorem, corollary, Converse of cyclic quadrilateral theorem, solved examples, review. In other words, if any four points on the circumference of a circle are joined, they form the vertices of a cyclic quadrilateral. x��\Yw\7r��c��~d'�k�K��a��q�HIN��������R����M} � t_�MQ3Gf�* ; Chord — a straight line joining the ends of an arc. Corollary 1. The circle which consist of all the vertices of any polygon on its circumference is known as the circumcircle or circumscribed circle. If PQRS is a cyclic quadrilateral, PQ and RS, and QR and PS are opposite sides. the sum of the opposite angles is equal to 180˚. In a cyclic quadrilateral, the four perpendicular bisectors of the given four sides meet at the centre O. : Find the value of angle D of a cyclic quadrilateral, if angle B is 60, If ABCD is a cyclic quadrilateral, so the sum of a pair of two opposite angles will be 180°, Find the value of angle D of a cyclic quadrilateral, if angle B is 80°. We have AL0C 2F is a cyclic quadrilateral. The sum of the opposite angles of a cyclic quadrilateral is supplementary. Brahmagupta's theorem states that for a cyclic quadrilateral that is also orthodiagonal, the perpendicular from any side through the point of intersection of the diagonals bisects the opposite side. Brahmagupta Theorem and Problems - Index Brahmagupta (598–668) was an Indian mathematician and astronomer who discovered a neat formula for the area of a cyclic quadrilateral. Ptolemy’s theorem about a cyclic quadrilateral and Fuhrmann’s theorem about a cyclic hexagon are examples. (b) is also a simple corollary if you think about it in the right way: and , where one of and is less than , and the other is greater than . ⓘ Ptolemys theorem. After proving the quadrilateral case, the general case of the cyclic polygon theorem is an immediate corollary. The perpendicular bisectors of the sides of a triangle are concurrent.Theorem 69. Theorem 1. Corollary to Theorem 68. This is another corollary to Bretschneider's formula. Brahmagupta's theorem states that for a cyclic quadrilateral that is also orthodiagonal, the perpendicular from any side through the point of intersection of the diagonals bisects the opposite side. where a, b, c, and d are the four sides of the quadrilateral. This will clear students doubts about any question and improve application skills while preparing for board exams. The conjecture also explains why we use perpendicular bisectors if we want to The cyclic quadrilateral has maximal area among all quadrilaterals having the same side lengths (regardless of sequence). If all the four vertices of a quadrilateral ABCD lie on the circumference of the circle, then ABCD is a cyclic quadrilateral. Notice how the measures of angles A and C are shown. In this section we will discuss theorems on cyclic quadrilateral. In Euclidean geometry, Ptolemys theorem is a relation between the four sides and two diagonals of a cyclic quadrilateral. ; Radius ($$r$$) — any straight line from the centre of the circle to a point on the circumference. according to which, the sum of all the angles equals 360 degrees. Required fields are marked *. Brahmagupta's theorem states that for a cyclic quadrilateral that is also orthodiagonal, the perpendicular from any side through the point of intersection of the diagonals bisects the opposite side. Complete the following: 1) How does the measure of angle A compare with the measure of arc BCD? The two theorems also hold in hyperbolic geometry, for example, see [S]. Covid-19 has led the world to go through a phenomenal transition . The sum of the internal angles of the quadrilateral is 360 degree. An exterior angle of a cyclic quadrilateral is equal to the interior opposite angle. [21] It means that all the four vertices of quadrilateral lie in the circumference of the circle. Hence. Consider the diagram below. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. A quadrilateral iscyclic iff a pair of its opposite angles are supplementary. (PQ x RS) + … Inscribed Angle Theorem Dance: Take 2! Other names for quadrilateral include quadrangle (in analogy to triangle), tetragon (in analogy to pentagon, 5-sided polygon, and hexagon, 6-sided polygon), and 4-gon (in analogy to k-gons for arbitrary values of k).A quadrilateral with vertices , , and is sometimes denoted as . (A and C are opposite angles of a cyclic quadrilateral.) That is the converse is true. Ptolemy used the theorem as an aid to creating his table of chords, a trigonometric table that he applied to astronomy. Browse more Topics under Quadrilaterals. It is also sometimes called inscribed quadrilateral. This is a Corollary of the theorem that, in a Right Triangle, the Midpoint of the Hypotenuse is equidistant from the three Vertices. Register at BYJU’S to practice, solve and understand other mathematical concepts in a fun and engaging way. Join these points to form a quadrilateral. The converse of this theorem is also true, which states that if opposite angles of a quadrilateral are supplementary, then the quadrilateral is cyclic. Indian mathematician and astronomer Brahmagupta, in the seventh century, gave the analogous formulas for a convex cyclic quadrilateral. Let be a cyclic quadrilateral. On the three diagonals of a cyclic quadrilateral On the three diagonals of a cyclic quadrilateral Schwarz, Dan; Smith, Geoff 2014-08-01 00:00:00 J. Geom. If T is the point of intersection of the two diagonals, PT X TR = QT X TS. In Euclidean geometry, Ptolemy's theorem is a relation between the four sides and two diagonals of a cyclic quadrilateral (a quadrilateral whose vertices lie on a common circle). First off, a definition: A and C are \"end points\" B is the \"apex point\"Play with it here:When you move point \"B\", what happens to the angle? The quadrilateral whose vertices lies on the circumference of a circle is a cyclic quadrilateral. ����Z��*���_m>�!n���Qۯ���͛MZ,�W����W��Q�D�9����lt��[m���F��������dz/w���g�vnI:�x�v�OV���Rx��oO?����r6&�]��b]�_���z�! = sum of the product of opposite sides, which shares the diagonals endpoints. ; Circumference — the perimeter or boundary line of a circle. Also, the opposite angles of the square sum up to 180 degrees. The following terms are regularly used when referring to circles: Arc — a portion of the circumference of a circle. An exterior angle of a cyclic quadrilateral is equal to the interior opposite angle. In Euclidean geometry, Ptolemys theorem is a relation between the four sides and two diagonals of a cyclic quadrilateral. Brahmagupta's Theorem Cyclic quadrilateral. Suppose a,b,c and d are the sides of a cyclic quadrilateral and p & q are the diagonals, then we can find the diagonals of it using the below given formulas: $$p=\sqrt{\frac{(a c+b d)(a d+b c)}{a b+c d}} \text { and } q=\sqrt{\frac{(a c+b d)(a b+c d)}{a d+b c}}$$. Corollary of cyclic quadrilateral theorem An exterior angle of a cyclic quadrilateral is congruent to the angle opposite to its adjacent interior angle. ∠A + ∠C = 180° [Theorem of cyclic quadrilateral] ∴ 2∠A + 2∠C = 2 × 180° [Multiplying both sides by 2] ∴ 3∠C + 2∠C = 360° [∵ 2∠A = 3∠C] ∴ 5∠C = 360° Corollary 5: If ABCD is a cyclic quadrilateral, then opposite angles sum up to 180 degrees. An important theorem in circle geometry is the intersecting chords theo-rem. Fuss' theorem gives a relation between the inradius r, the circumradius R and the distance x between the incenter I and the circumcenter O, for any bicentric quadrilateral.The relation is (−) + (+) =,or equivalently (+) = (−).It was derived by Nicolaus Fuss (1755–1826) in 1792. Corollary 5. Cyclic quadrilaterals Oct 21, 2020 - In a cyclic quadrilateral, the sum of opposite angles is 180 degree. Balbharati solutions for Mathematics 2 Geometry 10th Standard SSC Maharashtra State Board chapter 3 (Circle) include all questions with solution and detail explanation. The definition states that a quadrilateral which circumscribed in a circle is called a cyclic quadrilateral. Pythagoras' theorem. 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E-learning is the future today. Brahmagupta's theorem states that for a cyclic quadrilateral that is also orthodiagonal, the perpendicular from any side through the point of intersection of the diagonals bisects the opposite side. The area of a cyclic quadrilateral is $$Area=\sqrt{(s-a)(s-b)(s-c)(s-d)}$$. [21] If a cyclic quadrilateral is also orthodiagonal, the distance from the circumcenter to any side equals half the length of the opposite side. 5 0 obj 105 (2014), 307–312 2014 Springer Basel 0047-2468/14/020307-6 published online January 16, 2014 Journal of Geometry DOI 10.1007/s00022-013-0208-9 On the three diagonals of a cyclic quadrilateral Dan Schwarz and Geoff C. Smith … Oct 30, 2018 - In this applet, students can readily discover this immediate consequence (or corollary) of the inscribed angle theorem: In any cyclic quadrilateral … quadrilateral are perpendicular, then the projections of the point where the diago- nals intersect onto the sides are the vertices of a cyclic quadrilateral. Ptolemy's Theorem yields as a corollary a pretty theorem regarding an equilateral triangle inscribed in a circle. Question: Find the value of angle D of a cyclic quadrilateral, if angle B is 60o. Hence. In a cyclic quadrilateral, the sum of either pair of opposite angles is supplementary. 2 is a cyclic quadrilateral. Exterior angle of a cyclic quadrilateral. A cyclic quadrilateral is a quadrilateral which has all its four vertices lying on a circle. Stay Home , Stay Safe and keep learning!!! Why is this? For a convex quadrilateral that is both cyclic and orthodiagonal (its diagonals are perpendicular), p2+q2>4R2, where Ris the circumradius. �So�/�e2vEBюܞ�?m���Ͻ�����L�~�C�jG�5�loR�:�!�Se�1���B8{��K��xwr���X>����b0�u\ə�,��m�gP�!Ɯ�gq��Ui� Then ∠PAN = ∠PKN, ∠PBL = ∠PKL, ∠PCL = ∠PML and ∠PDN = ∠PMN. all four vertices of the quadrilateral lie on the circumference of the circle. �:�i�i���1��@�~�_|� Pv"㈪%vlIP4Y{O4�@��ceC� ـ���e/ �C�@P��3D�ZR�1����v��|.-z[0u9Q�㋁L���N��/'����_w�l4kIT _H�,Q�&�?�yװhE��(*�⭤9�%���YRk�S:�@�� �D1W�| 3N��-)�3�I�K.�9��v����gHH��^�Đ2�b�\ݰ�D��4��*=���u.���ڞ��:El�40��3�.Ԑ��n�x�s�R�<=Hk�{K������~-����)�����)�hF���I �T��)FGy#�ޯ�-��FE�s�5U:��t�!4d���$�聱_�א����4���G��Dȏa�k30��nb�xm�~E&B&S��iP��W8Ј��ujy�!�5����0F�U��Fk����4���F�0j�Y��V�gs�^m�TCZ���+Bd�۴��\�Mzk2%�L���. Brahmagupta's Theorem Cyclic quadrilateral. The word ‘quadrilateral’ is composed of two Latin words, Quadri meaning ‘four ‘and latus meaning ‘side’. (7Ծ������v$��������F��G�F�pѻ�}��ͣ���?w��E[7y��X!B,�M���B-՚ Shaalaa has a total of 53 questions with solutions for this chapter in 10th Standard Board Exam Geometry. If a,b,c and d are the sides of a inscribed quadrialteral, then its area is given by: There is two important theorems which prove the cyclic quadrilateral. Take a circle and choose any 4 points on the circumference of the circle. PR and QS are the diagonals. Four alternative answers for each of the following questions are given. A D 1800 C B 1800 BDE CAB A B D A C B DC 8. If a cyclic quadrilateral is also orthodiagonal, the distance from the circumcenter to any side equals half the length of the opposite side. %�쏢 The theorem is named after the Greek astronomer and mathematician Ptolemy. %PDF-1.4 O0is the orthocenter of triangle XYZ. The ratio between the diagonals and the sides can be defined and is known as Cyclic quadrilateral theorem. Animation 20 (Inscribed Angle Dance!) The property of a cyclic quadrilateral proven earlier, that its opposite angles are supplementary, is also a test for a quadrilateral to be cyclic. Proof. <> In a cyclic quadrilateral, the sum of a pair of opposite angles is 180. Corollary to Theorem 68. In a quadrilateral, one amazing aspect is that it can have parallel opposite sides. The exterior angle formed if any one side of the cyclic quadrilateral produced is equal to the interior angle opposite to it. A cyclic quadrilateral is a quadrilateral with all its four vertices or corners lying on the circle.It is thus also called an inscribed quadrilateral. A quadrilateral iscyclic iff a pair of its opposite angles are supplementary. Ḫx�1�� �2;N�m��Bg�m�r�K�Pg��"S����W�=��5t?�يLV:���P�f�%^t>:���-�G�J� V�W�� ���cOF�3}$7�\�=�ݚ���u2�bc�X̱���j�T��d�c�$�:6�+a(���})#����͡�b�.w;���m=��� �bp/���; eE���b��l�A�ə��n)������t�@p%q�4�=fΕ��0��v-��H���=���l�W'��p��T� �{���.H�M�S�AM�^��l�]s]W]�)$�z��d�4����0���e�VW�&mi����(YeC{������n�N�hI��J4��y��~��{B����+K�j�@�dӆ^'���~ǫ!W���E��0P?�Me� Brahmagupta's theorem states that for a cyclic quadrilateral that is also orthodiagonal, the perpendicular from any side through the point of intersection of the diagonals bisects the opposite side. We proved earlier, as extension content, two tests for a cyclic quadrilateral: If the opposite angles of a cyclic quadrilateral are supplementary, then the quadrilateral is cyclic. Cyclic quadrilateral: | | ||| | Examples of cyclic quadrilaterals. Your email address will not be published. Inscribed Angle Theorem Dance: Take 2! After proving the quadrilateral case, the general case of the cyclic polygon theorem is an immediate corollary. Cyclic quadrilaterals - Higher A cyclic quadrilateral is a quadrilateral drawn inside a circle. (1) Each tangent is perpendicular to the radius that goes to the point of contact. Solving for x yields = + − +. Brahmagupta Theorem and Problems - Index Brahmagupta (598–668) was an Indian mathematician and astronomer who discovered a neat formula for the area of a cyclic quadrilateral. Choose the correct ∠SPR = ∠SQR, ∠QPR = ∠QSR, ∠PQS = ∠PRS, ∠QRP = ∠QSP. Online Geometry: Cyclic Quadrilateral Theorems and Problems- Table of Content 1 : Ptolemy's Theorems and Problems - Index. Question: Find the value of angle D of a cyclic quadrilateral, if angle B is 80°. Construction: Join the vertices A and C with center O. Fuss' theorem. Theorems of Cyclic Quadrilateral Cyclic Quadrilateral Theorem The opposite angles of a cyclic quadrilateral are supplementary. A quadrilateral is a 4 sided polygon bounded by 4 finite line segments. Maharashtra State Board Class 10 Maths Solutions Chapter 3 Circle Problem Set 3 Problem Set 3 Geometry Class 10 Question 1. Proof. A cyclic quadrilateral is a quadrilateral which has all its four vertices lying on a circle. (b) is also a simple corollary if you think about it in the right way: and , where one of and is less than , and the other is greater than . It states that the four vertices A , B , C and D of a convex quadrilateral satisfy the equation AP PB = DP PC if and only if it is a cyclic quadrilateral, where P is … Given: A cyclic quadrilateral ABCD inscribed in a circle with center O. There are two theorems about a cyclic quadrilateral. ]^\�g?�u&�4PC��_?�@4/��%˯���Lo���n1���A�h���,.�����>�ج��6��W��om�ԥm0ʡ��8��h��t�!-�ut�A��h���Q^�3@�[�R-�6����ͳ�ÍSf���O�D���(�%�qD��#�i�mD6���r�Tc�K:Ǖ�4�:�*t���1���:�%k�H��z�œ� ~�2y4y���Y�Z�������{�3Y��6�E��-��%E�.6T��6{��U ��H��! Definition. Let’s take a look. Complete the following: 1) How does the measure of angle A compare with the measure of arc BCD? stream PR and QS are the diagonals. In the figure given below, the quadrilateral ABCD is cyclic. They have four sides, four vertices, and four angles. Hence, not all the parallelogram is a cyclic quadrilateral. Std :10 : Corollary of Cyclic Quadrilateral Theorem - YouTube 2 Some corollaries Corollary 1. If a cyclic quadrilateral is also orthodiagonal, the distance from the circumcenter to any side equals half the length of the opposite side. \R��qo��_JG��%is�y�(G�ASK$�r��y!��W������+��`q�ih�r�hr��g�K�v)���q'u!�o;�>�����o�u�� The theorem is named after the Greek astronomer and mathematician Ptolemy. If the sum of two opposite angles are supplementary, then it’s a cyclic quadrilateral. Let be a Quadrilateral such that the angles and are Right Angles, then is a cyclic quadrilateral (Dunham 1990). The sum of the opposite angles of cyclic quadrilateral equals 180 degrees. It is also called as an inscribed quadrilateral. Your email address will not be published. If PQRS is a cyclic quadrilateral, PQ and RS, and QR and PS are opposite sides. Quadrilateral ABCD is by Theorem 2 orthodiagonal if and only if ∠PAN +∠PBL+∠PCL+∠PDN = π ⇔ ∠PKN +∠PKL+∠PML+∠PMN = π ⇔ ∠LKN +∠LMN = π If a cyclic quadrilateral is also orthodiagonal, the distance from the circumcenter to any side equals half the length of the opposite side. This will help you discover yet a new corollary to this theorem. This theorem completes the structure that we have been following − for each special quadrilateral, we establish its distinctive properties, and then establish tests for it. Theorem 2. It states that the four vertices A , B , C and D of a convex quadrilateral satisfy the equation AP PB = DP PC if and only if it is a cyclic quadrilateral, … A quadrilateral is called Cyclic quadrilateral if … Ptolemy used the theorem as an aid to creating his table of chords, a trigonometric table that he applied to astronomy. If also d = 0, the cyclic quadrilateral becomes a triangle and the formula is reduced to Heron's formula. If there’s a quadrilateral which is inscribed in a circle, then the product of the diagonals is equal to the sum of the product of its two pairs of opposite sides. PR and QS are the diagonals. When any four points on the circumference of a circle are joined, they form the vertices of a cyclic quadrilateral. (PQ x RS) + ( QR x PS) = PR x QS. In Euclidean geometry, a cyclic quadrilateral or inscribed quadrilateral is a quadrilateral whose vertices all lie on a single circle.This circle is called the circumcircle or circumscribed circle, and the vertices are said to be concyclic.The center of the circle and its radius are called the circumcenter and the circumradius respectively. yժI���/,�!�O�]�|�\���G*vT�3���;{��y��*ڏ*�M�,B&������@�!DdNW5r�lgNg�r�2�WO�XU����i��6.�|���������;{ 8c� �d�'+�)h���f^Nf#�%�Ά9��� ����[���LJ}G�� Y�|P��)��M;6/>��D#L���$T߅�}�2}��� �,��e5��������-)F���]W� 7�լ��o�7_�5������U;��(z�,+��bϵv;u�mTs]F�M*�@͓���&-9�]� !���| {n�e�O��zUdV�|���y���]s���PҝǪC�c�gm?ŭ=��yݧ �Xκ����=��WT!Ǥn�|#!��r�b�L�+��F���7�i���EZS�J�ʢQ���qs��ô]�)c��b����)�b4嚶ۚ"� �'��z̊$�Eļ̒��'��ƞ&Ol��g��! It can be visualized as a quadrilateral which is inscribed in a circle, i.e. Then. If PQRS is a cyclic quadrilateral, PQ and RS, and QR and PS are opposite sides. Terminology. Property of Product of Diagonals in cyclic quadrilateral is Ptolemy Theorem. Animation 20 (Inscribed Angle Dance!) DNPM are cyclic quadrilaterals since they all have two opposite right angles (see Figure 3). If a, b, c and d are the successive sides of a cyclic quadrilateral, and s is the semi perimeter, then the radius is given by. Inscribed Angle Theorem: Corollary 1; Inscribed Angle Theorems: Take 4! If an exterior angle of a quadrilateral equals the opposite interior angle, then the quadrilateral … Corollary 3.3. A quadrilateral is a polygon in Euclidean plane geometry with four edges (sides) and four vertices (corners). Now measure the angles formed at the vertices of the cyclic quadrilateral. Us now try to prove this theorem equals the opposite angles is 180 degree the first theorem a... 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Yields as a quadrilateral are supplementary question and improve application skills while preparing for Board exams a phenomenal.! Quadrilateral: | | ||| | examples of cyclic quadrilateral formulas to understand the concept better angles cyclic... Geometry, for example, see [ s ] opposite sides ; example line joining ends! ‘ side ’ and four angles of a cyclic quadrilateral ABCD lie on the circle.It is also! A quadrilateral with all its four vertices of the product of diagonals in cyclic quadrilateral. tangent!, PT x TR = QT x TS the concept better clear doubts... But FXC 1C... Feuerbach point is a simple corollary of theorem 1, since of. Circumscribed in a cyclic quadrilateral: | | ||| | examples of quadrilateral... Cyclic or inscribed in a circle Greek astronomer and mathematician Ptolemy angles of cyclic quadrilateral, amazing. The circumcenter to any side equals half the length of the circle, the sum of either pair of angles... To it, stay Safe and keep learning!!!!!!! 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Join today the Greek astronomer and mathematician Ptolemy Level Theorems * * important to note all! - Index cyclic, then the perpendicular bisectors of the opposite interior angle, then the perpendicular bisectors the... | 2021-06-22T17:20:22 | {
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https://forum.math.toronto.edu/index.php?PHPSESSID=jh17fgq9gjjla8v71ncb3c64p4&topic=2366.0;wap2 | MAT244--2020F > Chapter 2
section 2.1 practice problem 35
(1/1)
Suheng Yao:
I had solved this equation to this implicit form, but I don't know how to simplify to a simpler form as the answer shown. Could someone help me? Thanks.
kavinkandiah:
Add all those terms on both sides so they're all positive instead (just make to make it easier), and then combine the ln|x| and ln|y/x+1| using product rule for logarithms (i.e., log(a)+log(b)=log(ab)).
Suheng Yao:
Thanks, and in tests, could we leave the answer in the implicit equation form? Or is it necessary to transform it into a parametric form as the professor taught in the class?
kavinkandiah:
I asked Victor and he said either is fine, depending on which is more convenient. | 2021-12-08T01:33:26 | {
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https://math.stackexchange.com/questions/2625855/the-arithmetic-mean-a-m-between-two-numbers-exceeds-their-geometric-mean-g-m | # The Arithmetic Mean (A.M) between two numbers exceeds their Geometric Mean (G.M.)
The Arithmetic Mean (A.M) between two numbers exceeds their Geometric Mean (G.M.) by $2$ and the GM exceeds the Harmonic Mean (H.M) by $1.6$. Find the numbers.
My Attempt: Let the numbers be $a$ and $b$. Then, $$A.M=\dfrac {a+b}{2}$$ $$G.M=\sqrt {ab}$$ $$H.M=\dfrac {2ab}{a+b}$$ According to question: $$\dfrac {a+b}{2} =\sqrt {ab}+2$$ $$\dfrac {a+b}{2}-2=\sqrt {ab}$$ $$a+b-4=2\sqrt {ab}$$ Also, $$\sqrt {ab}=\dfrac {2ab}{a+b} + 1.6$$ Then, $$a+b-4=2(\dfrac {2ab}{a+b} + 1.6)$$ $$a+b-4=\dfrac {4ab+3.2(a+b)}{a+b}$$ $$(a+b-4)(a+b)=4ab+3.2(a+b)$$ $$(a+b)^2-4(a+b)=4ab+3.2a+3.2b$$
How do I solve further?
$$AM=GM+2$$
$$GM=HM+1.6$$
Since $$GM^2=AM\cdot HM,$$
$$GM^2=(GM+2)(GM-1.6)$$
$$GM^2=GM^2+0.4GM-3.2$$
$$GM=8$$
$$AM=10$$
$$\sqrt{ab}=8, \frac{a+b}{2}=10$$
$$ab=64, a+b = 20$$
The numbers are $$16$$ and $$4$$.
• For question about plotting, perhaps stackoverflow is more appropriate? here is a possible lead, try function histfit in Matlab? – Siong Thye Goh Apr 30 '18 at 16:45
• mathworks.com/help/stats/histfit.html i think android can access that, it is just a regular website. – Siong Thye Goh Apr 30 '18 at 17:16
I don't know why I do this, but here is the general case.
Suppose $am = gm+u$ and $gm = hm+v$ with $u, v \ne 0$ and $u \ne v$.
Since $hm \le gm \le am$, $u \ge 0$ and $v \ge 0$.
Since $gm^2 = am\cdot hm$, $gm^2 =(gm+u)(gm-v) =gm^2+gm(u-v)-uv$ so $gm(u-v) =uv$. Therefore $u > v$ and $gm =\dfrac{uv}{u-v}$.
Then $am =gm+u =\dfrac{uv}{u-v}+u =\dfrac{uv+u(u-v)}{u-v}+u =\dfrac{u^2}{u-v}$ and $hm =gm-v =\dfrac{uv}{u-v}-v =\dfrac{uv-v(u-v)}{u-v} =\dfrac{v^2}{u-v}$.
If there are only two values, $a$ and $b$, then $\dfrac{a+b}{2} =\dfrac{u^2}{u-v}$ and $\sqrt{ab} =\dfrac{uv}{u-v}$ so $b =\dfrac{u^2v^2}{a(u-v)^2}$ and $\dfrac{u^2}{u-v} =\dfrac{a+\dfrac{u^2v^2}{a(u-v)^2}}{2} =\dfrac{a^2(u-v)^2+u^2v^2}{2a(u-v)^2}$ or $2au^2(u-v) =a^2(u-v)^2+u^2v^2$.
Solving $a^2(u-v)^2-2u^2(u-v)a+u^2v^2 =0$,
$\begin{array}\\ a &=\dfrac{2u^2(u-v)\pm\sqrt{4u^4(u-v)^2-4(u-v)^2u^2v^2}}{2(u-v)^2}\\ &=\dfrac{u^2(u-v)\pm u(u-v)\sqrt{u^2-v^2}}{(u-v)^2}\\ &=\dfrac{u^2\pm u\sqrt{u^2-v^2}}{u-v}\\ &=u\dfrac{u\pm \sqrt{u^2-v^2}}{u-v}\\ \text{and}\\ b &=\dfrac{2u^2}{u-v}-a\\ &=\dfrac{2u^2}{u-v}-\dfrac{u^2\pm u\sqrt{u^2-v^2}}{u-v}\\ &=\dfrac{2u^2-u^2\mp u\sqrt{u^2-v^2}}{u-v}\\ &=\dfrac{u^2\mp u\sqrt{u^2-v^2}}{u-v}\\ \end{array}$
For this case, $u=2$ and $v=1.6$.
$gm =\dfrac{2\cdot 1.6}{2-1.6} =\dfrac{3.2}{.4} =8$.
To get $a$ and $b$, $\sqrt{u^2-v^2} =\sqrt{4-2.56} =\sqrt{1.44} =1.2$ so $a =2\dfrac{2\pm 1.2}{.4} =2\dfrac{3.2, .8}{.4} =2(8, 2) =(16, 4)$ and $b =(4, 16)$. | 2021-03-06T12:59:27 | {
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https://math.stackexchange.com/questions/2969642/showing-1-frac1nn-is-an-increasing-sequence-by-comparing-the-binomial | # Showing $(1+\frac{1}{n})^n$ is an increasing sequence by comparing the binomial expansions term by term
So I know that showing $$(1+\frac{1}{n})^n$$ is an increasing sequence has probably appeared on this site about 100 times, but my professor said he thinks the induction step is most easily seen if you expand this out and compare the summation term by term. I tried this and I did not what see why the inequality was obvious. Can somebody help me out here? Thanks!
• Use the binomial theorem. Just expand for $n$ and $n+1$. – Avinash N Oct 24 '18 at 20:11
• Right, and after I did that the reason for the inequality still wasn't obvious to me – Math is hard Oct 24 '18 at 20:12
To show that $$\left(1+\frac1n\right)^{n}$$ is increasing, all we need to show is that $$\frac{\left(1+\frac1{n+1}\right)^{n+1}}{\left(1+\frac1n\right)^{n}}\ge1$$. Proceeding, we have
\begin{align} \frac{\left(1+\frac1{n+1}\right)^{n+1}}{\left(1+\frac1n\right)^{n}}&=\frac{n+1}{n}\left(\frac{n(n+2)}{(n+1)^2}\right)^{n+1}\\\\ &=\frac{n+1}{n}\left(1-\frac{1}{(n+1)^2}\right)^{n+1}\\\\ (\text{Using Bernoulli's Inequality})&\ge \frac{n+1}{n}\left(1-\frac{1}{n+1}\right)\\\\ &=1 \end{align}
And we are done!
• this is the best proof i've seen so far thanks! – Math is hard Oct 24 '18 at 20:34
• Thank you Michael. Much appreciated. And feel free to accept the answer. – Mark Viola Oct 25 '18 at 0:07
The $$k$$-th term in the binomial expansion looks like
$$\frac{1}{n^k} \binom{n}{k} = \frac{n(n-1)\cdots(n-k+1)}{n^k k!} = \frac{1}{k!}\left( 1\right)\left( 1 - \frac1n\right)\left( 1 - \frac2n\right)\cdots\left( 1-\frac{k - 1}n\right).$$
(We've distributed a $$1/n$$ into each factor of the numerator.) Now compare factor by factor when we increase $$n$$ to $$n + 1$$.
• and in addition to this, the expansion of the $n+1$ will have one additional term in the summation, but the idea is that we can throw that out and still get the inquality to hold by comparing term by term what you have posted. Right? – Math is hard Oct 24 '18 at 20:17
• @MichaelVaughan Yes, the additional term is positive so it fights right into the inequality. – Trevor Gunn Oct 24 '18 at 20:20 | 2020-01-19T05:18:38 | {
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http://math.stackexchange.com/questions/312405/help-for-proving-an-equation-by-induction | # help for proving an equation by induction
For this equation:
$$-1^3+(-3)^3+(-5)^3+\ldots+(-2n-1)^3=(-n-1)^2(-2n^2-4n-1)$$
how can I prove this by induction?
When I set $n = 1$ for the base case I got:
$$-1^3 + (-3)^3 + (-5)^3 + \ldots + (-3)^3 = -28$$
but am having trouble with the following inductive steps
-
Please, it hurts me to see all these minus signs. The first thing to do is to get rid of them. Our question is equivalent to showing that $1^3+\cdots+(2n+1)^3=(n+1)^2(2n^2+4n+1)$. I am allergic to minus signs, but am not asking just for me. You have a much better chance of succeeding if you get rid of them. – André Nicolas Feb 23 '13 at 23:11
## 3 Answers
You’d have had an even simpler base case had you started with $n=0$: $(-1)^3=(-1)^2(-1)$.
For the induction step your induction hypothesis should be that the result is true for some particular $n$, i.e., that
$$(-1)^3+(-3)^3+(-5)^3+\ldots+(-2n-1)^3=(-n-1)^2(-2n^2-4n-1)\;,\tag{1}$$
and you’ll try to prove the corresponding statement about $n+1$. The first step is to figure out what that statement is:
\begin{align*} (-1)^3+(-3)^3&+\ldots+(-2n-1)^3+\big(-2(n+1)-1\big)^3\\ &=\big(-(n+1)-1\big)^2\big(-2(n+1)^2-4(n+1)-1\big) \end{align*}
or, after a bit of algebraic simplification,
\begin{align*} (-1)^3+(-3)^3+\ldots+(-2n-1)^3+\big(-2n-3\big)^3&=(-n-2)^2(-2n^2-8n-7)\\ &=-(n+2)^2(2n^2+8n+7)\;. \end{align*}\tag{2}
The lefthand side of $(2)$ can be split into two pieces as
$$\Big((-1)^3+(-3)^3+\ldots+(-2n-1)^3\Big)+\big(-2n-3\big)^3\;,$$
and the induction hypothesis $(1)$ tells us that the first piece is $$(-2n-1)^3=(-n-1)^2(-2n^2-4n-1)=-(n+1)^2(2n^2+4n+1)\;.$$
Thus,
\begin{align*} (-1)^3+(-3)^3&+\ldots+(-2n-1)^3+\big(-2n-3\big)^3\\ &=-(n+1)^2(2n^2+4n+1)+(-2n-3)^3\;, \end{align*}
and all that you have to do now in order to prove $(2)$ (and thereby complete the induction step) is show that
$$-(n+1)^2(2n^2+4n+1)+(-2n-3)^3=-(n+2)^2(2n^2+8n+7)\;.$$
This is just algebra, and I’ll leave it to you.
-
Inductive proofs of sums like yours are easily tackled using a very simple general method known as telescopy - see the trivial inductive proof here, of the following fundamental
Theorem $\rm \displaystyle\ \ \sum_{i\,=\,0}^n\, f(i)\, =\, g(n)\iff f(0) = g(0)\ {\rm\ and\ }\ f(n) \,=\, g(n)-g(n\!-\!1)\:\$ for $\rm\,n \ge 1.$
In your case we have
$$\rm f(n) \,=\, (2n+1)^3,\quad g(n) \,=\, (n+1)^2 (2n^2+4n+1)$$
Thus, applying the theorem, we check that $\rm\ f(0) = 1 = g(0)\$ and it remains to check that $\rm\, g(n)-g(n-1) = f(n)\:$ for $\rm\:n\ge 1,\:$ which is rote arithmetic. Notice that the proof requires no ingenuity at all - only verifying some simple polynomial equalities - a purely mechanical process. You can find many more examples of telescopy and related results in other answers here.
Remark $\$ The other answers all end up verifying the same equality $\rm\, g(n)-g(n-1) = f(n).\:$ This is no coincidence. They are effectively repeating the same linked telescopic sum proof for a special case of $\rm\:f(n)\:$ and $\rm\:g(n).\:$ No insight is gained by repeating such telescopy proofs ad infinitum for special cases, getting hopelessly lost in messy algebraic calculations intrinsic in motley special cases (details that greatly obfuscate the simple telescopic inductive structure). Doing so would be akin to reproving the fundamental theorem of calculus for each specific function it is applied to (the above is a discrete analog - the fundamental theorem of difference calculus).
Much more inductive insight is gained by proving once and for all the above general telescoping sum formula. Here, rid of all the messy details of special cases, one sees clearly the beautiful way that telescopic induction works, and this lends more insight on induction in general. Such telescopic induction skills will prove quite handy later on, since many inductive proofs of sums and products have this special telescopic form. Being able to quickly dispatch these trivial inductions using these tools will help you to not lose focus on the essential aspects of the problem you are solving (which for non-logicians are rarely of inductive nature).
-
Very nice argument. – André Nicolas Feb 24 '13 at 4:14
First, as the power of each summand is odd, I'd write the equation as
$$-1^3-3^3-5^3-\ldots-(2n+1)^3=-(n+1)^2(2n^2+4n+1)\Longleftrightarrow$$
$$1^3+3^3+5^3+\ldots+(2n+1)^3=(n+1)^2(2n^2+4n+1)$$
The base case is
$$n=1:\;\;\;\;\;1^3+3^3=28\stackrel ?=(1+1)^2(2+4+1)=28\ldots\ldots good$$
Assume for $\,n\,$ and show for $\,n+1\,$:
$$1^3+3^3+\ldots+(2n+1)^3+(2n+3)^3\stackrel{\text{Ind. hypothesis}}=(n+1)^2(2n^2+4n+1)+(2n+3)^3=\ldots$$
- | 2014-08-22T16:14:42 | {
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https://cs.stackexchange.com/questions/128273/randomized-algorithms-high-probability-vs-expectation | # Randomized Algorithms: High-Probability vs. Expectation
Hopefully this question isn't too general, but I was wondering what the relationship is between randomized algorithms that perform well with high-probability and those that perform well in expectation. My question is motivated by the definition of a randomized $$\alpha$$-approximation algorithm given here, namely that it is a polynomial-time algorithm that produces a solution within $$\alpha$$ of OPT in expectation or with high probability. I also found that the first few pages of this source provides some good insight into the high-probability vs. expectation approaches, but I still have questions.
• Can you always transform an algorithm that achieves an $$\alpha$$-approximation in expectation to one that achieves this with high probability, and vice versa? (Ostensibly by rerunning the algorithm multiple [a polynomial number] of times.)
• If not, is one harder than the other to obtain? (I would think that if you fix $$\alpha$$, a high-probability algorithm would always be harder to find/less likely to exist. Or maybe you can always find one, but the approximation ratio will become worse.)
Thanks for the help!
If you have an algorithm that is an $$\alpha$$-approximation in expectation, then you can construct an algorithm that is a $$(1+\epsilon)\alpha$$-approximation with high probability, for any $$\epsilon>0$$. In particular, by Markov's inequality, if you run the algorithm, then with probability at least $$1-1/(1+\epsilon)$$ it will output a $$(1+\epsilon)\alpha$$-approximation. So, if you run the algorithm about $$(c \log n)/\epsilon$$ times and keep the best output among all of those trials, with probability about $$1-1/n^c$$ you will find a $$(1+\epsilon)\alpha$$-approximation.
If you have an algorithm that is an $$\alpha$$-approximation with high probability, there are no guarantees about the expectation. It's possible that with very small probability (probability $$1/n^c$$), it outputs an extremely bad solution (one with exponentially large approximation factor), and in all other cases it outputs an $$\alpha$$-approximation. In this case, the expected value of the approximation factor will be very large, even though it has a very small probability to output such a bad solution.
• Thank you for this fantastic answer! I'm probably missing something obvious, but I am stuck at checking the $(c \log n) / \epsilon$. The probability of failure in one trial is at most $\frac{1}{1 + \epsilon}$, so we want to show that $$\left( \frac{1}{1 + \epsilon} \right)^{(c \log n) / \epsilon}$$ is small, i.e., about $1/n^c$. Clearly we are allowed to upper bound the expression. At first I was thinking you could apply $1 + \epsilon \le e^\epsilon$ in the denominator, but this would make the entire expression smaller, so it is a no go. What am I missing? – kanso37 Jul 13 at 5:03
• @kanso37, When $\epsilon$ is small, $1/(1+\epsilon)\approx 1-\epsilon$, and $(1-\epsilon)^{1/\epsilon} \approx 1/e$. You should be able to take it from there. I haven't thought about maximization algorithms. The answer might be different, if solutions have non-negative value. – D.W. Jul 13 at 6:02 | 2020-09-18T13:16:09 | {
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https://math.stackexchange.com/questions/1256140/for-what-functions-is-y-y/1256159 | # For what functions is $y'' = y$?
What functions $y = f(x)$ have the property that $f(x) = f''(x)$, i.e. what functions have the same integral and derivitive?
I could think of $ce^x$ and $ce^{-x}$ (where $c$ is a constant), but are there others? If not, how can you prove those are the only ones?
• If $y=c e^{x}$ and $y=ke^{-x}$ are solutions then $y=ce^{x}+ke^{-x}$ is a solution . – randomgirl Apr 28 '15 at 16:56
• The general solution is $f(x) = c_1e^x + c_2e^{-x}$ for arbitrary real constants $c_1, c_2$. Proving these are the only solutions is usually dealt with by using an atom bomb strength theorem about the existence & uniqueness of solutions of ODEs. – Simon S Apr 28 '15 at 16:56
• @Simon Atom bombs are kewl! And I like theorems! – bjb568 Apr 28 '15 at 16:58
• For completeness: the general solution can also be written as $f(x)=c_1\cosh x+c_2\sinh x$, using Hyperbolic trigonometric functions. This is more convenient for finding $f$ with given initial conditions: e.g., if one wants $f(0)=5$ and $f'(0)=-2$, the solution is $5\cosh x-2\sinh x$. – user147263 Jun 5 '15 at 18:44
Non atom-bomb proof that $A e^x + B e^{-x}$ are the only solutions:
1. $(y'+y)' = y'' + y' = y + y'$ so $y'+y = C e^x$
2. $(y'-y)' = y'' - y' = y - y' = -(y'-y)$ so $y'-y = D e^{-x}$
Subtracting these two equations we get $2y = C e^x - D e^{-x}$, or, after dividing and renaming constants:
$y = A e^x + B e^{-x}$
Supplement proving if $u' = r u$ then $u = C e^{rx}$:
Rewrite this as $u' - ru = 0$ and multiply both sides by $e^{-rx}$.
Then $e^{-rx} u' + (-r) e^{-rx} u = 0$.
The LHS is $\frac{d}{dx}\left[ e^{-rx} u\right]$, and the equation is stating that this derivative here is zero, so we get:
$e^{-rx} u = C$
or in other words
$u = C e^{rx}$
Background on these two methods:
The trick in the top section is a general technique for proving results for systems of linear first order equations based on finding eigenvectors. (A second order equation is a system of two first order equations if you let $y'$ be a new variable and include $(y)' = y'$ as one of your equations.)
The trick in the second section is called the method of integrating factors.
There's also a super fancy way to do both sections at once using a matrix exponential.
• well done. nice work. – abel Apr 28 '15 at 19:48
• It's good to know how a skillful karate chop or two will do the trick. ;-) – Simon S Apr 29 '15 at 13:37
Multiply both sides by $2y'$ and you get
$$2y''y'=2y'y.$$
You should recognize the derivative of $y'^2$ on the left and that of $y^2$ on the right.
Integrating,
$$y'^2=y^2+C_0,$$
and
$$\frac{y'}{\sqrt{y^2+C_0}}=\pm1.$$
Integrating once again (using a table),
$$\ln\left(y+\sqrt{y^2+C_0}\right)=\pm x+C_1,$$ or $$y+\sqrt{y^2+C_0}=C_2e^{\pm x},$$ $$y^2+C_0=\left(C_2e^{\pm x}-y\right)^2,$$ $$C_0=C_2^2e^{\pm2x}-2C_2e^{\pm x}y,$$ $$\color{green}{y=Ce^{\pm x}+C'e^{\mp x}}.$$
This confirms the well-known result.
All of the linear combinations of the two solutions that you mentioned. That is,
$y = A e^{x} + B e ^{-x}$ where $A,B$ are constants.
Using the differential operator $D$, the equation is $$D^2y=y\text{, or }(D^2-1)y=0.$$ This factors as $$(D-1)(D+1)y=0.$$ We first set $z=(D+1)y$ and solve $$(D-1)z=0.$$ $$Dz-z=0\implies (Dz)e^{-x}-ze^{-x}=(Dz)e^{-x}+zD(e^{-x})=D(ze^{-x})=0\implies ze^{-x}=C_0,$$ $$z=C_0e^x.$$
Then solve $$(D+1)y=C_0e^x.$$
$$Dy+y=C_0e^x\implies (Dy)e^x+ye^x=(Dy)e^x+yD(e^x)=D(ye^x)=C_0e^{2x}\\\implies ye^x=C_1e^{2x}+C_2,$$
$$\color{green}{y=Ce^x+C'e^{-x}}.$$
The method generalizes to arbitrary polynomials in $D$. | 2019-09-17T14:31:20 | {
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https://math.stackexchange.com/questions/3078906/probability-for-red-and-green-apples-in-a-bag | Probability for red and green apples in a bag
"There are twelve red apples and seven green apples in a bag. What is the probability of picking up one red apple and one green apple at once?"
This is a multiple choice question, the choices are:
• A) $$\frac{21}{64}$$
• B) $$\frac{23}{64}$$
• C) $$\frac{25}{64}$$
• D) $$\frac{22}{64}$$
The "at once" is throwing me off but I interpreted it as picking red then green or green then red without replacement, which gave me $$\frac{28}{57}$$. I have no idea how all the answers have $$64$$ as the denominator, is there something obvious I'm missing?
• I would have interpreted this as meaning, "You pull out two apples at once. What is the probability that one is read and one green?" but that gives me $7\cdot12/{19\choose2}=84/171$ Jan 19 '19 at 0:48
• Welcome to MathSE. I agree with your calculation. This tutorial explains how to typeset mathematics on this site. Jan 19 '19 at 0:49
• I agree with the interpretation that leads to the answer of $\frac{28}{57}=\frac{7\cdot 12}{\binom{19}{2}}$. I see no possible interpretation which leads to a denominator being $64$ or any multiple of $64$ except for if there happen to also be yellow apples in the bag as well that you forgot to mention or the numbers of apples you write are incorrect. As far as we can tell you did nothing wrong and the answer key is what is incorrect. If it really is that the answer key is incorrect, then make note of it and bring it up with the teacher. Jan 19 '19 at 0:55
I would argue that the answer $$\frac{28}{57}$$ is correct.
The argument:
Number of ways of choosing a red apple and a green apple, = $$7.12$$
Total number of ways of choosing two objects out of $$19$$ = $$19 \choose{2}$$
Therefore, the answer comes out as, $$\frac{7.12}{19 \choose 2} = \frac{28}{57}$$
• The division by $2$ is not necessary. To see this, consider the problem where it was instead $7$ red apples and $11$ green apples. If you were to divide by $2$ in that case then you would have a non-integer number of arrangements, a clear impossibility. Jan 19 '19 at 1:39
• If that argument isn't convincing enough, consider the problem where we have one red apple and one green apple in a bag and we choose two of them. We ask what the probability is that we select both a red and a green apple. Without even calculating you should be able to tell that the answer should be $1$, but following your proposed formula you'd only get an answer of $\frac{1}{2}$. Jan 19 '19 at 2:51
• True. I will correct it. Thanks!! Jan 19 '19 at 3:07 | 2021-11-28T14:15:37 | {
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https://math.stackexchange.com/questions/1744250/convergence-of-riemann-sums-for-improper-integrals | # Convergence of Riemann sums for improper integrals
I was considering whether or not the limit of Riemann sums converges to the value of an improper integral on a bounded interval. This appears to be true in some cases when the sum avoids points where the function is not defined.
For example, the right-hand Riemann sum for $1/\sqrt{x}$ converges $$\lim \frac1{n}\sum_{i = 1}^n \sqrt{n/i} = \int_0^1 \frac{dx}{\sqrt{x}}.$$ This does not work, however, for the function $\sin(1/x)/x$ even though the improper integral is finite. I’m sure this has something to do with the function being monotone, but I am not able to find a proof.
• $\sin(1/x)/x$ is not Riemann integrable, because any partition includes a neighborhood of zero for which the upper rectangle has area greater than $1/\varepsilon$, but the lower rectangle has area less than $-1/\varepsilon$. – T.J. Gaffney Apr 15 '16 at 20:49
• @Gaffney: The OP is asking about improper integrals. $\int_0^1 x^{-1}\sin(1/x) dx = \pi/2$ as a convergent improper integral. I believe this is about convergence of sums when there is singularity at an endpoint but the function still has a finite improper integral. – RandyF Apr 15 '16 at 20:59
• There are always sequences of Riemann sums that converge to the improper integral. If the integrand is monotonic, then pretty much everything works, as long as you keep the point in the interval containing the singularity far enough away from the singularity. For oscillating integrands that blow up, like $x^{-1}\sin (1/x)$, choices are more restricted. – Daniel Fischer Apr 15 '16 at 21:02
• @RandyF $$\int_0^1 x^{-1}\sin (1/x)\,dx < \frac{\pi}{2} = \int_0^{\infty} x^{-1}\sin (1/x)\,dx.$$ That doesn't invalidate your point, of course. – Daniel Fischer Apr 15 '16 at 21:04
We can prove that sequences of right- or left-hand Riemann sums will converge for a monotone function with a convergent improper integral.
Suppose WLOG $f:(0,1] \to \mathbb{R}$ is nonnegative and decreasing. Suppose further that there is a singularity at $x =0$ but $f$ is Riemann integrable on $[c,1]$ for $c > 0$ and the improper integral is convergent:
$$\lim_{c \to 0+}\int_c^1 f(x) \, dx = \int_0^1 f(x) \, dx.$$
Take a uniform partition $P_n = (0,1/n, 2/n, \ldots, (n-1)/n,1).$ Since $f$ is decreasing we have $$\frac1{n}f\left(\frac{k}{n}\right) \geqslant \int_{k/n}^{(k+1)/n}f(x) \, dx \geqslant \frac1{n}f\left(\frac{k+1}{n}\right),$$
and summing over $k = 1,2, \ldots, n-1$
$$\frac1{n}\sum_{k=1}^{n-1}f\left(\frac{k}{n}\right) \geqslant \int_{1/n}^{1}f(x) \, dx \geqslant \frac1{n}\sum_{k=2}^nf\left(\frac{k}{n}\right).$$ Hence,
$$\int_{1/n}^{1}f(x) \, dx +\frac{1}{n}f(1) \leqslant \frac1{n}\sum_{k=1}^{n}f\left(\frac{k}{n}\right) \leqslant \int_{1/n}^{1}f(x) \, dx+ \frac{1}{n}f \left(\frac{1}{n} \right).$$
Note that as $n \to \infty$ we have $f(1) /n \to 0$ and since the improper integral is convergent,
$$\lim_{n \to \infty} \int_{1/n}^{1}f(x) \, dx = \int_0^1 f(x) \, dx, \\ \lim_{n \to \infty}\frac{1}{n}f \left(\frac{1}{n} \right) = 0.$$
The second limit follows from monotonicity and the Cauchy criterion which implies that for any $\epsilon > 0$ and all $n$ sufficiently large
$$0 \leqslant \frac{1}{n}f \left(\frac{1}{n} \right) \leqslant 2\int_{1/2n}^{1/n}f(x) \, dx < \epsilon.$$
By the squeeze theorem we have
$$\lim_{n \to \infty}\frac1{n}\sum_{k=1}^nf\left(\frac{k}{n}\right) = \int_0^1 f(x) \, dx.$$
This proof can be generalized for non-uniform partitions. For oscillatory functions like $g(x) = \sin(1/x)/x$, the failure of the sequence of right-hand Riemann sums to converge is, non-monotonicity notwithstanding, related to non-convergence as $n \to \infty$ of
$$\frac{1}{n}g \left(\frac{1}{n} \right) = \sin n.$$
This particular case appears to have been covered nicely by @Daniel Fischer in
Improper integrals and right-hand Riemann sums
• Nicely written (+1) – Mark Viola May 4 '17 at 17:24 | 2019-10-24T03:03:56 | {
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https://math.stackexchange.com/questions/2882717/interview-question-on-probability-a-and-b-toss-a-dice-with-1-to-n-faces-in-an-a/2883203 | # Interview Question on Probability: A and B toss a dice with 1 to n faces in an alternative way
A and B toss a dice with 1 to n faces in an alternative way, the game is over when a face shows up with point less than the previous toss and that person loses. What is the probability of the first person losing the game and the expected number of tosses?
I can solve this question for 6 face dice: If the first die is 1, there are 5 greater numbers on the second: 1/6 * 5/6 = 5/36. If the first die is 2, there are 4: 1/6 * 4/6 = 4/36 and so on. T/hus prob of first person losing the game is:
5/36 + 4/ 36 + 3/36 + 2/36 + 1/36 = 15/36 = 5/12
don't know how to generalize it to n faces.
• I take it the faces are numbered $1$ to $n$? – saulspatz Aug 14 '18 at 17:40
• @saulspatz yes, they are numbered from 1 to n – user584227 Aug 14 '18 at 17:44
• Welcome to MSE. You should put clarifications in the body of the question, not the comments. People browsing question will often not read the comments. What are your thoughts on this question? How far have you gotten with it? Where are you stuck? This question is likely to be closed unless you add some more context to it. – saulspatz Aug 14 '18 at 17:48
• @RossMillikan: I think "in an alternative way" is meant to say "alternatingly". – joriki Aug 14 '18 at 20:25
• Adding my standard rant: you cannot throw "a dice". "Dice" is the plural of "die". You can throw a singe "die" or two or more "dice". – user247327 Aug 14 '18 at 22:54
There are $\binom{n+k-1}k$ different non-decreasing sequences of length $k$ with elements in $\{1,\ldots,n\}$. The first player loses exactly if for any even $k$ the first $k$ rolls form such a sequence and the first $k+1$ rolls don't. Thus the probability for the first player to lose is
$$\sum_{k=0}^\infty(-1)^k\binom{n+k-1}k\frac1{n^k}=\left(1+\frac1n\right)^{-n}\;.$$
This goes to $\mathrm e^{-1}$ as $n\to\infty$. In the limit $n\to\infty$, the probability for equal rolls goes to zero, so we can rank the rolls. If we rank the first $k$ rolls, the ranks are a random permutation of the first $k$ integers. Thus, the limit can also be obtained if we know how many permutations have their first descent (or, equivalently, ascent) in an even position. This was recently asked and answered at Permutations of length $n$ in which the first ascent occurs in an even position. The result is that the number of these permutations is the number of derangements, and the proportion of permutations of length $k$ that are derangements goes to $\mathrm e^{-1}$ as $k\to\infty$, in agreement with the above result.
The expected number of rolls can be obtained by summing the probabilities that no descent has occurred after $k$ rolls:
$$\sum_{k=0}^\infty\binom{n+k-1}k\frac1{n^k}=\left(1-\frac1n\right)^{-n}\;.$$
This goes to $\mathrm e$ as $n\to\infty$. In the limit $n\to\infty$, we can again ignore the possibility of equal rolls. The probability that no descent has occurred after $k$ rolls is then $\frac1{k!}$, and the expected number of rolls is
$$\sum_{k=0}^\infty\frac1{k!}=\mathrm e\;,$$
in agreement with the above result.
• Delightful. I'm not sure I fully follow the parts where you say we can evaluate the limits by ignoring the possibility of equal rolls, though. I understand them as heuristics, but you seem to be saying that these are rigorous arguments. In both cases, you are taking the limit as $n\to\infty$ under the summation sign, aren't you? Is it just that all the terms are positive, and one can appeal to the dominated convergence theorem, say? – saulspatz Aug 15 '18 at 15:32
• @saulspatz: Well, I didn't make any effort to turn those parts into rigorous arguments since the limit agreed with the limit from the exact calculation -- but if I had to turn them into rigorous arguments, I think I'd argue that these results are exact conditional on not rolling any equal numbers before the game ends; the probability of rolling equal numbers before the game ends goes to zero as $n\to\infty$; and the expectations conditional on rolling equal numbers before the game ends are finite -- I think that should be enough? – joriki Aug 15 '18 at 18:40
• That makes sense, thanks. I guess I understood it in the sense you originally intended, but I thought I was missing something. – saulspatz Aug 15 '18 at 18:55
• Turns out that you and I interpreted the rules differently. See my edited answer. – saulspatz Aug 16 '18 at 10:29
• @saulspatz: Yes, that is a bit weird that neither of us had noticed before. I did check whether my result agreed with yours, but I guess I must have suffered from confirmation bias :-) – joriki Aug 16 '18 at 14:55
Let me get you started. For $k=1,2,\dots,n$ let $p_k$ be the probability that the roller loses if he rolls $k,$ and that does not immediately win the game. Therefore, the probability that the first player loses is $${1\over n}\sum_{k=1}^np_k$$
It seems to me that the easiest way to compute $p_k$ will be in decreasing order, starting with $k=n$ and progressing down to $k=1$. When $k= n$ we have
$$p_n= {n-1\over n} + {1\over n}\left(1-p_n\right)\implies p_n={n\over n+1}$$ because, if the roller rolls $n$, he will lose if his opponent rolls anything but $n$, and in the case where the opponent does rolls $n$, he loses if the opponent doesn't lose.
Similarly, \begin{align} p_{n-1}&={n-2\over n} + {1\over n}\left(1-p_{n-1}\right) + {1\over n}\left(1-p_n\right)\\ p_{n-1}&=1-{1\over n}p_{n-1}-{1\over n}p_n\\ {n+1\over n}p_{n-1}&=1-{1\over n+1}\\ p_{n-1}&={n^2\over(n+1)^2} \end{align}
Can you continue?
EDIT
Now that a complete solution has been given, I will work out the details. I claim that with $p_k$ defined as above, we have $$p_{n-j}=\left({n\over n+1}\right)^{n-j},\ j=0,1,\dots,n-1$$
The $j=0$ case has been done above. Assuming the theorem is true for $0,1,\dots,j-1,$, we have \begin{align} p_{n-j}&={n-j-1\over n}+{1\over n}\sum_{k=n-j}^n(1-p_k)\\ \left(1+\frac1n\right)p_{n-j}&=1-\frac1n\sum_{k=0}^{j-1}p_{n-k}\\ \left({n+1\over n}\right)p_{n-j}&=1-\frac1n\sum_{k=0}^{j-1}\left({n\over n+1}\right)^{k+1}\\ \left({n+1\over n}\right)p_{n-j}&=1-\frac1n\frac{\left({n\over n+1}\right)^{j+1}-{n\over n+1}}{{n\over n+1}-1}\\ \left({n+1\over n}\right)p_{n-j}&=1-\frac1n{n\over n+1}\frac{\left({n\over n+1}\right)^{j}-1}{{-1\over n+1}}=1-\left(1-\left(n\over n+1\right)^j\right)\\ p_{n-j}&=\left({n\over n+1}\right)^{j+1} \end{align}
Therefore, the probability that the first player loses is $$\frac1n\sum_{j=0}^{n-1}\left({n\over n+1}\right)^{j+1}= \frac1n\frac{\left({n\over n+1}\right)^{n+1}-{n\over n+1}}{{n\over n+1}-1}=\frac1n{n\over n+1}\frac{\left({n\over n+1}\right)^n-1}{{-1\over n+1}}=1-\left({n\over n+1}\right)^n$$
Note that this is precisely the complement of the probability that joriki computed in his solution. The problem lies in the description of the game in the question: "the game is over when a face shows up with point less than the previous toss and that person loses." It isn't at all clear who "that person" is, as the phrase has no antecedent. I interpreted it to mean that the first player who rolls a smaller number wins, and joriki interpreted it oppositely. Note that joiki says that the first player loses if the first non-monotonically increasing sequence has odd length, so that if the first $3$ rolls are $3,4,2$ the first player loses. In my interpretation, since it is the first player who rolled the $2,$ the first player wins. It's a bit surprising that nobody (including joriki and me) noticed that we were using different rules.
A similar approach will give the expected number of rolls, and this time, reassuringly, I get the same answer as joriki. If we let $e_k$ be the expected number number of rolls remaining, if the current point is $k$ then the expected number of rolls in the game is $$E=1+\frac1n\sum_{k=1}^ne_k$$ and in a manner exactly analogous to the calculations above, we find that $$e_{n-j}=\left(n\over n-1\right)^{j+1},\ j=0,1,\dots,n-1$$ and that $$E=\left({n\over n-1}\right)^n=\left(1-\frac1n\right)^{-n}$$
• The result can actually be obtained in closed form (see my answer). – joriki Aug 15 '18 at 5:58
• As I said, I was just trying to get the OP started. I hadn't worked out the answer in detail. Frankly, I just assumed, since it was an interview question, that it would work out if one continued along these lines. Now that I've seen your answer, I'll have to try to follow this line of thought to completion, and see where I get. – saulspatz Aug 15 '18 at 14:17 | 2019-07-18T22:16:07 | {
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http://math.stackexchange.com/questions/183088/find-the-value-of-x-in-the-displayed-figure | # Find the value of $x$ in the displayed figure
Find $x$ in the following figure. $AB,AC,AD,BC,BE,CD$ are straight lines.
$AE=x$, $BE=CD=x-3$, $BC=10$, $AD=x+4$
$\angle BEC=90^{\circ}$, $\angle ADC=90^{\circ}$
NOTE: figure not to scale.
-
Please edit the question into the body of your post. – Gerry Myerson Aug 16 '12 at 5:58
By the Pythagorean theorem we have
$$\begin{equation*} CE=\sqrt{10^{2}-\left( x-3\right) ^{2}}=\sqrt{91-x^{2}+6x} \end{equation*}$$ and $$\begin{equation*} CD^{2}+AD^{2}=AC^{2}=\left( CE+AE\right) ^{2} \end{equation*}.$$ So we have to solve the following irrational equation $$\begin{equation*} \left( x-3\right) ^{2}+\left( x+4\right) ^{2}=\left( \sqrt{91-x^{2}+6x} +x\right) ^{2},\tag{1} \end{equation*}$$ which can be simplified to $$\begin{equation*} x^{2}-2x-33=\sqrt{-x^{4}+6x^{3}+91x^{2}}. \end{equation*}$$
After squaring both sides and grouping the terms of the same degree we get the quartic equation $$\begin{equation*} 2x^{4}-10x^{3}-153x^{2}+132x+1089=0.\tag{2} \end{equation*}$$
The coefficient of $x^{4}$ is $2=1\times 2$ and the constant term is $1089=1\times 3^{2}11^{2}$. To find possible rational roots of this equation, we apply the rational root theorem and test the numbers of the form $$\begin{equation*} x=\pm \frac{p}{q}, \end{equation*}$$ where $p\in \left\{ 1,3,9,11,33,99,121,363,1089\right\}$ is a divisor of $1089$ and $q\in \left\{ 1,2\right\}$ is a divisor of $2$. It turns out that $x=3$ and $x=11$ are roots. Now we divide the LHS by $x-3$ $$\begin{equation*} \frac{2x^{4}-10x^{3}-153x^{2}+132x+1089}{x-3}=2x^{3}-4x^{2}-165x-363 \end{equation*}$$ and this quotient by $x-11$ $$\begin{equation*} \frac{2x^{3}-4x^{2}-165x-363}{x-11}=2x^{2}+18x+33. \end{equation*}$$ So we have the equivalent equation $$\begin{equation*} \left( x-3\right) (x-11)\left( 2x^{2}+18x+33\right) =0\tag{3} \end{equation*}$$ Since the solutions of $2x^{2}+18x+33$ are both negative and $x=3$ is not a solution of the original irrational equation, the solution is therefore $$x=11.$$
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Hint: Using Pythagorean theorem $$(x+4)^2+(x-3)^2=\left( x+\sqrt{10^2-(x-3)^2}\right)^2$$ and this can be easily solved.
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Easily? Looks to me like it's going to be a mess, though I confess I haven't actually tried it. – Gerry Myerson Aug 16 '12 at 6:00
Anyway the equation is atmost quartic, and there are formulas to solve that and thats what i meant by 'easily' – pritam Aug 16 '12 at 6:04
Yes, it's quartic. I just expanded this out, and factored it by hand. It has two positive roots, both integers, and two irrational negative roots. Three of the four roots don't fit the figure geometrically. So there's just one solution to the geometric problem. As I said in my answer, the best way to find it is with some inspired guessing, based on well-known small Pythagorean triples. – user22805 Aug 16 '12 at 7:38
@DavidWallace: How do you know the answer is an integer? – user1729 Aug 16 '12 at 12:37
@user1729, you don't know it, but it can't hurt to try it. – Gerry Myerson Aug 17 '12 at 6:19
If you were given that $x$ and |EC| were integers you could use the following.
Let $y = |EC|$.
Using Pythagoras on $\triangle BEC$:
1) $(x-3)^2 + y^2 = 10^2$ $\Rightarrow x^2 - 6x + 9 + y^2 = 100$ $\Rightarrow y^2 = -x^2 + 6x + 91$
Using Pythagorus on $\triangle ACD$:
2) $(x+4)^2 + (x-3)^2 = (x+y)^2$ $\Rightarrow x^2 + 2x + 25 = 2xy + y^2$
Then put 1) into 2):
$\Rightarrow x^2 + 2x + 25 = 2xy + (-x^2 + 6x +91)$ $\Rightarrow xy = x^2 - 2x - 33$
Then: $y = x - 2 - \frac{33}{x}$
So $x$ has to be a divisor of 33: 11, 3 or 1.
I tried using the cosine rule then but just ended up with 0 = 0. The trickiest part is the angle $ABC$ just about failing to be a right angle.
-
great insight indeed. – Rajesh K Singh Aug 16 '12 at 13:06
It just so happens that one solution is an integer. So maybe, you could try a few numbers, and see if any of them jump out as the solution, before you set about trying to solve a nasty quartic. Focus on well-known small Pythagorean triples.
Note that it took me less than a minute of staring at the figure, to realise what the solution was. I don't yet know whether there are any other solutions that fit the figure.
-
That's not maths! – user1729 Aug 16 '12 at 11:48
@user1729: Actually sampling methods are indeed math, are used quite often, and even work in this instance. I'm not sure why anyone would discourage trying to get an intuitive feel for the problem. – ex0du5 Aug 16 '12 at 20:32
Hey, @user1729, for any problem like this, it's worth TRYING a few things first, before getting into the hard algebra. In this case, it's comparatively easy to see what ONE solution is. Of course, proving that it's the ONLY solution is still hard - see my comment under pritam's answer. Getting a feel for any given problem is ABSOLUTELY part of maths. – user22805 Aug 17 '12 at 3:08
@DavidWallace: Your comments under Pritram's answer boil down to "the way you prove the solution is unique is factorise the polynomial and show that all the other answers don't make sense". This finds all the solutions for you, and so your guessing was just adding to your work... – user1729 Aug 17 '12 at 9:28
No. Knowing one of the solutions already made factorising the quartic much easier. Solving quartics is really hard - there are very many steps. But having seen one solution by using my intuition effectively turned it into a cubic; and solving cubics is substantially easier. So my "guessing" as you put it saved me lots of work. – user22805 Aug 17 '12 at 9:33
$∆BCE$ is right angle triangle.
Hence $BC^2 = BE^2 + EC^2$ $EC = \sqrt{(BC^2 - BE^2})= \sqrt{(100 - (x-3)^2)}$
$∆ACD$ is right angle triangle.
Hence $AC^2 = CD^2 + AD^2$
$(AE + EC)^2 = CD^2 + AD^2$
Substitute the values, $(\sqrt{(100 - (x-3)^2)} + x)^2 = (x-3)^2 + (x+4)^2$
Then you can solve this equation easily for getting x value.
-
"The triangle inequality theorem states that any side of a triangle is always shorter than the sum of the other two sides".
By the Triangle Inequality Theorem,
$CE + (x-3) \gt 10$, $CE + x \lt (x+4) + (x-3)$
i.e. $(13-x) \lt CE \lt (x+1)$
we now have, $(13-x) \lt (x+1)$
i.e. $x \gt 6$
from the, $\triangle EBC$ we have,
$x-3 \lt 10$
i.e. $x \lt 13$
we can conclude that, $6 \lt x \lt 13$
-
right... And now? – t.b. Aug 16 '12 at 8:08
I see... Now $x + 7 \gt 0$ magically becomes $x \gt 7$. – t.b. Aug 16 '12 at 8:19
Now, we can look for triplets to approximate the triangle as close as possible to a near by triplet. – Rajesh K Singh Aug 16 '12 at 8:24
You have reasoned that this is the unique integer solution. But...why isn't, say, $5+\pi$ a solution? – user1729 Aug 16 '12 at 13:06
What does the figure being to scale have to do with anything? – Emily Aug 16 '12 at 15:54 | 2016-06-26T14:05:38 | {
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https://mathematica.stackexchange.com/questions/133773/plot-particle-motion-in-potential | # Plot particle motion in potential
Consider the anisotropic harmonic potential in two dimensions $(q_1,q_2)$ given by
$$V(q_1,q_2) = \frac{m}{2} \, q_1^2 + \frac{k}{2} \, q_2^2,$$
or V = m/2 q1^2 + k/2 q2^2; in Mathematica.
The Newtonian e.o.m.s of a particle moving through this potential are
$$\ddot{q}_1 = -q_1, \qquad \ddot{q}_2 = -\omega^2 \, q_2,$$
where $\omega = \sqrt{k/m}$ is the angular frequency of oscillations in the $q_2$-direction. Given the initial conditions $q_i(0) = q_{i,0}$ and $p_i(0) = p_{i,0}$, the e.o.m.s are solved by
\begin{aligned} q_1(t) &= q_{1,i} \cos(t) + \frac{p_{1,i}}{m} \, \sin(t),\\ q_2(t) &= q_{2,i} \cos(\omega t) + \frac{p_{2,i}}{m \omega} \, \sin(\omega t), \end{aligned} \qquad \text{with p_i = m \dot{q}_i.}
In Mathematica:
DSolve[{q1''[t] == -q1[t], q2''[t] == -ω^2 q2[t], q1[0] == q10,
q1'[0] == p10/m, q2[0] == q20, q2'[0] == p20/m}, {q1[t], q2[t]}, t]
//FullSimplify
{{q1[t] -> q10 Cos[t] + (p10 Sin[t])/m, q2[t] -> q20 Cos[t ω] + (p20 Sin[t ω])/(m ω)}}
A 3d plot of the potential looks like this.
Plot3D[V /. {m -> 1, k -> 3}, {q1, -5, 5}, {q2, -5, 5},RegionFunction -> Function[{q1, q2}, m/2 q1^2 + k/2 q2^2 <= 12 /. {m -> 1, k -> 3}]]
What I would like to do now is draw the particle as a ball moving through this potential with a fixed energy. Any help would be much appreciated.
Some remarks about the physics behind this simulation: The particle always reaches a certain height both in $q_1$- and $q_2$-direction before rolling back down again and up the other side. This is because the Hamiltonian $H = \frac{p_1^2}{2 m} + \frac{p_2^2}{2 m} + V(q_1,q_2)$ does not couple the degrees of freedom in $q_1$- and $q_2$-direction. Therefore, the total energies $E_1$ and $E_2$ available in dimensions $q_1$ and $q_2$ are conserved separately.
For long times $t \to \infty$ and irrational angular frequency $\omega \notin \mathbb{Q}$ (which ensures that the trajectory never closes, thus making the system ergodic), the particle's trajectory should therefore trace out a rectangle $R$ whose length and width are determined by $E_1$ and $E_2$.
Update: With anderstood's and BlacKow's help, I was able to piece together this solution that does exactly what I want.
V = m/2 q1^2 + k/2 q2^2; \[Omega] = Sqrt[k/m];
sol = {q1[t], q2[t], m/2 q1[t]^2 + k/2 q2[t]^2} /.
DSolve[{q1''[t] == -q1[t], q2''[t] == -\[Omega]^2 q2[t],
q1[0] == q10, q1'[0] == p10/m, q2[0] == q20,
q2'[0] == p20/m}, {q1[t], q2[t]}, t] // FullSimplify
Manipulate[Block[{m = 1, k = 5, q10 = 5, p10 = 1, q20 = 2, p20 = 1},
surf = Plot3D[V, {q1, -7, 7}, {q2, -5, 5},
RegionFunction -> Function[{q1, q2}, m/2 q1^2 + k/2 q2^2 <= 25], PlotStyle -> Opacity[0.5]];
Show[surf, Graphics3D@{Blue, Ball[sol /. {t -> tf}, 0.4]},
ParametricPlot3D[sol, {t, 0, tf}]]], {tf, 0.1, 100}]
• @corey979 Sorry, I'm having trouble posting this question. I keep getting the error Your post appears to contain code that is not properly formatted as code. Please indent all code by 4 spaces using the code toolbar button or the CTRL+K keyboard shortcut. For more editing help, click the [?] toolbar icon. Looking on Meta, it appears this issue has come up before. Dec 18, 2016 at 14:20
• Just copy and paste you code from the notebook, select it and hit the {} button. Dec 18, 2016 at 14:23
• Exactly what I did. This isn't my first question but I never encountered this problem before. Dec 18, 2016 at 14:24
• Can you post the whole question and ignore the message? Or it won't let you post, except for this fragment? Dec 18, 2016 at 15:13
• It seems you were able to post it. What changed? Dec 18, 2016 at 20:49
First, store the 3D trajectory (in the space $(q_1,q_2,V)$):
sol = {q1[t], q2[t], m/2 q1[t]^2 + k/2 q2[t]^2} /.
DSolve[{q1''[t] == -q1[t], q2''[t] == -\[Omega]^2 q2[t],
q1[0] == q10, q1'[0] == p10/m, q2[0] == q20,
q2'[0] == p20/m}, {q1[t], q2[t]}, t] // FullSimplify
Plot the potential surface:
surf = Plot3D[
V /. {m -> 1, k -> 3, \[Omega] -> Sqrt[k/m]}, {q1, -5, 5}, {q2, -5,
5}, RegionFunction ->
Function[{q1, q2}, m/2 q1^2 + k/2 q2^2 <= 12 /. {m -> 1, k -> 3}]]
Then select some initial conditions and plot the trajectory using ParametricPlot3D. Show it together with the surface using... Show:
p10 = 3; p20 = 1; q20 = -1.5; q10 = 2;
traj = ParametricPlot3D[
sol /. \[Omega] -> Sqrt[k/m] /. {m -> 1, k -> 3}, {t, 0, 50},
PlotStyle -> Red];
Show[pot, traj]
• Looks good! One question: Is there a way to avoid repeating the replacement /. {m -> 1, k -> 3} everywhere m or k come up? For example, could they be assigned specific values all throughout a cell? Dec 19, 2016 at 14:10
• @Casimir I you want to keep these values throughout the notebook, just add at the beginning: m=1; k=3;. And you should also avoid using "redundant" variables, i.e. $\omega$, which depends on k and m (I guess...). Dec 19, 2016 at 15:20
• @Casimir Then you can use Block[{k=1,m=3}, ...]. Dec 19, 2016 at 15:23
• @anderstood equipotential surface is a surface where potential energy is the same, so it's defined by $V(q) = const$; it will be an ellipse in $(q_1,q_2)$ coordinates. And here $V$ is not a total energy but, potential energy only. Dec 19, 2016 at 18:56
• @BlacKow You're right, edited. Dec 19, 2016 at 19:10
Using @anderstood answer we can play with graphics to make the surface transparent and plot ball movement:
V[q1_, q2_] := m/2 q1 q1 + k/2 q2 q2
surf = Plot3D[
V[q1, q2] /. {m -> 1, k -> 3, \[Omega] -> Sqrt[k/m]}, {q1, -5,
5}, {q2, -5, 5},
RegionFunction ->
Function[{q1, q2}, m/2 q1^2 + k/2 q2^2 <= 12 /. {m -> 1, k -> 3}],
Mesh -> None,
ColorFunction ->
Function[{z}, Opacity[0.4, #] &@ColorData["TemperatureMap"][z]]]
traj[t_] :=
Evaluate[Flatten@sol /. \[Omega] -> Sqrt[k/m] /. {m -> 1,
k -> 3} /. {p10 -> 3, p20 -> 1, q20 -> -1.5, q10 -> 2}]
frames = Table[
Show[surf, Graphics3D@{Red, Ball[traj[t], 0.2]}], {t, 0, 10,
0.1}];
Export["Documents/animBall.gif", frames] | 2022-05-20T03:40:40 | {
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http://math.stackexchange.com/questions/141428/chain-rule-for-y-x2-x34 | # Chain rule for $y = (x^2 + x^3)^4$
I am trying to find the derivative of $y = (x^2 + x^3)^4$
and it seems pretty simple I get
$4(x^2+x^3)^3 (2x+3x^2)$ This seems to be the proper answer to me but the book gets
$4x^7 (x+1)^3 (3x+2)$ and I have no idea how that happened, what process the author went through or why. My answer seems to be a better and more accurate answer since that is what the chain rule will give you.
Why is my answer wrong?
-
## 2 Answers
You did the differentiation correctly. The book's solution is the same as your answer, but in simplified form.
To obtain the book's answer from yours: factor $x^2$ from the $(x^2+x^3)$ term in your expression and apply the rule $(ab)^n=a^nb^n$, factor $x$ from the $(2x+3x^2)$ term, and finally combine the $x^6$ and $x$ terms. \eqalign{ 4(\color{maroon}{x^2+x^3})^3 (\color{darkgreen}{2x+3x^2}) &= 4 \bigl(\color{maroon}{ x^2(1+x)}\bigr)^3\cdot\color{darkgreen}{ x(2+3x)}\cr &=4\cdot (x^2)^3(1+x)^3\cdot x(2+3x)\cr &=4x^6(1+x)^3\cdot x(2+3x)\cr &=4x^7(x+1)^3(3x+2). }
-
How can I factor out a term out of something that is raised to a power? Doesn't that mess up the answer? – user138246 May 5 '12 at 17:03
@Jordan No, it wont. Not if you use the exponent rule $(ab)^n=a^nb^n$ – Galois Group May 5 '12 at 17:10
@Jordan Use $(a\cdot b)^n=a^nb^n$. Here $\bigl(\color{maroon}{x^2}\color{darkgreen}{(1+x)}\bigr)^3 = \color{maroon}{(x^2)^3}\cdot\color{darkgreen}{(1+x)}^3$. – David Mitra May 5 '12 at 17:10
You're right. So is the book. Because a little algebra reveals that $4(x^2+x^3)^3(2x+3x^2)=4(x^2(x+1))^3(x(2+3x))=4x^6(x+1)^3x(3x+2)=4x^7(x+1)^3(3x+2)$
- | 2015-10-10T16:55:32 | {
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https://mathematica.stackexchange.com/questions/169883/draw-all-the-circles-that-touch-the-x-axis-and-a-unit-circle | Draw all the circles that touch the x-axis and a unit circle
I am trying to plot a circle with radius 1, that touches the $x$-axis in the origin, So it has center $(0,1)$
u = Graphics[{Circle[{0, 1}, 1]}, ImageSize -> 150, Axes -> True]
I have the above, and now I would like to draw all the circles with radius u that touch the $x$-axis and the unit circle.
Is the above Mathematica code a function? And how can I draw all the circles that touch this unit circle?
• You mean radius 1 ? – Lotus Mar 27 '18 at 10:55
• only radius 1 for unit circle, the other circles it doesn't matter – Stefan Mar 27 '18 at 14:51
• I assume you want the circles that are tangent to the unit circle? Then, what about all the circles tangent to the origin? They touch the $x$ axis and the unit circle. Do you want to include them too? – anderstood Mar 28 '18 at 1:10
• anderstood raises a good point; you did not specify if the circles are internally or externally tangent. The solution in my answer will only give the externally tangent solutions, as the internally tangent solutions are trivial to produce. – J. M. is away Mar 28 '18 at 5:22
• I want only the circles that are tangent to the unit circle and the x axis, not the y axis. So J.M. has a correct solution – Stefan Mar 28 '18 at 12:35
The general equation of a circle of radius $u$ that is tangent to the $x$-axis is $(x-h)^2+(y-u)^2=u^2$. Our strategy, then, is to find the radical line of this variable circle with the original circle, and then find the condition such that the radical line of both circles is tangent to them as well.
rad = ((x - h)^2 + (y - u)^2 - u^2) - (x^2 + (y - 1)^2 - 1) // Simplify
h^2 - 2 h x - 2 (-1 + u) y
Then, determine the condition so that this line is tangent to the unit circle. In algebraic terms, we want the resulting quadratic polynomial after elimination to be a perfect square:
Solve[Discriminant[x^2 + (y - 1)^2 - 1 /. First[Solve[rad == 0, y]], x] == 0, h]
{{h -> 0}, {h -> 0}, {h -> -2 Sqrt[u]}, {h -> 2 Sqrt[u]}}
where we get two trivial solutions and a solution for both the right and left parts of the plane.
Now, we can visualize:
Graphics[{Circle[{0, 1}, 1],
MapIndexed[{ColorData[97, #2[[1]]], #1} &,
Table[Circle[{2 Sqrt[u], u}, u], {u, 1/20, 2, 1/20}]]},
Axes -> True]
You are describing a classic sangaku problem.
Form a right triangle between circle centres. In your case, the unit circle is orange with r1=1. Hence, the circle centres are separated by d=2*Sqrt[r2].
Manipulate[
Graphics[{
EdgeForm[{Thick, Black}],
Orange, Disk[{0, 1}, 1],
Darker@Green, Disk[{2 Sqrt[r2], r2}, r2]
},
PlotRange -> {{-2, 4}, {0, 3}}, Frame -> True],
{{r2, 1, "Right Circle Radius"}, 0.05, 5, Appearance -> "Labeled"}] | 2019-06-18T23:42:17 | {
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https://math.stackexchange.com/questions/1924585/which-one-is-right-riemann-integrable-iff-the-set-of-discontinuity-is-countable | Which one is right: Riemann integrable iff the set of discontinuity is countable or a null set?
When I was studying calculus (I used Purcell's book), it is stated that the bounded function $f$ on closed bounded interval is Riemann integrable if and only if $f$ has countable discontinuity points, which means if $f$ is discontinuous on Cantor set, then it is not Riemann integrable.
However, when I was learning real analysis (I used Bartle's book Introduction to Real Analysis 3rd ed.), in the section 7.3, the Lebesgue's Integrability Criterion said that the bounded function $f$ on closed bounded interval is Riemann integrable if and only if the set of discontinuity points is a null set (the measure of such set is $0$), which means if $f$ is discontinuous on the Cantor set, then it is still Riemann integrable.
Which one is the truth?
I remember (a long time ago) I asked this question to my folks to discuss, and he gave me an example when $f$ is discontinuous on the Cantor set and it is not Riemann integrable, and I didn't find any fault in his argument.
I am sorry I cannot give you that example guys, I lost the notes.
• Countable sets have measure zero. Perhaps the book said "if the set of discontinuities is countable then...", and didn't state it as an equivalence (which it is not). – Pedro Tamaroff Sep 13 '16 at 1:49
• thank you, i'll check the book when I go to the library, and may be there was some faults in my friend's argument, thank you – Rizky Reza Fujisaki Sep 13 '16 at 1:54
Having only countably many discontinuities implies a bounded function on a closed bounded interval is Riemann integrable, but it is certainly possible to have a function that is Riemann integrable that has uncountable many discontinuities.
One example would be the characteristic function of the usual Cantor set in $[0,1]$.
See these related questions:
Characteristic function of Cantor set is Riemann integrable
Example of Riemann integrable $f: [0,1] \to \mathbb R$ whose set of discontinuity points is an uncountable and dense set in $[0,1]$
• thank you so much!!! Then, does it mean the theorem "riemann integrable if and only if discontinuity points are countable" is a bit wrong? so the right one is "if discont points are countable, then reimann integrable"? (only one sided) – Rizky Reza Fujisaki Sep 13 '16 at 1:51
• Right. The implication only goes one way in the countable case. – Alexis Olson Sep 13 '16 at 1:52
• okay, then there must be some faults in my friend's argument, thank you – Rizky Reza Fujisaki Sep 13 '16 at 1:55 | 2020-08-05T16:11:50 | {
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https://www.jiskha.com/questions/357677/find-the-slope-of-the-tangent-line-to-the-curve-sqrt-2x-4y-sqrt-4xy-9-16-at-the | # calculus
find the slope of the tangent line to the curve (sqrt 2x+4y) + (sqrt 4xy) = 9.16
at the point (1,5)
dy/dx method implicit differantiation?
1. 👍
2. 👎
3. 👁
1. Yes, implicit differentation.
1/2(sqrt2x+4y) * (2 dx+4dy)= 1/2sqrt(4xy)* (4ydx+4xdy)=0
and solve for dy/dx. Have fun.
1. 👍
2. 👎
👤
bobpursley
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https://www.farbykusnir.sk/83qcvmf/article.php?e4706a=maximum-turning-point | +421 907 627 998
# maximum turning point
The point at which a very significant change occurs; a decisive moment. Write down the nature of the turning point and the equation of the axis of symmetry. When the function has been re-written in the form y = r(x + s)^2 + t, the minimum value is achieved when x = -s, and the value of y will be equal to t.. d/dx (12x 2 + 4x) = 24x + 4 At x = 0, 24x + 4 = 4, which is greater than zero. Roots. Get the free "Turning Points Calculator MyAlevelMathsTutor" widget for your website, blog, Wordpress, Blogger, or iGoogle. A maximum turning point is a turning point where the curve is concave up (from increasing to decreasing ) and $f^{\prime}(x)=0$ at the point. A turning point can be found by re-writting the equation into completed square form. A Turning Point is an x-value where a local maximum or local minimum happens: How many turning points does a polynomial have? The maximum number of turning points for any polynomial is just the highest degree of any term in the polynomial, minus 1. Another type of stationary point is called a point of inflection. The turning point occurs on the axis of symmetry. This can also be observed for a maximum turning point. To find the stationary points of a function we must first differentiate the function. And the absolute minimum point for the interval happens at the other endpoint. Closed Intervals. Stationary points are often called local because there are often greater or smaller values at other places in the function. (a) Using calculus, show that the x-coordinate of A is 2. To do this, differentiate a second time and substitute in the x value of each turning point. It looks like when x is equal to 0, this is the absolute maximum point for the interval. A turning point is a type of stationary point (see below). Once you have established where there is a stationary point, the type of stationary point (maximum, minimum or point of inflexion) can be determined using the second derivative. If $\frac{dy}{dx}=0$ (is a stationary point) and if $\frac{d^2y}{dx^2}<0$ at that same point, them the point must be a maximum. (3) The region R, shown shaded in Figure 2, is bounded by the curve, the y-axis and the line from O to A, where O is the origin. By Yang Kuang, Elleyne Kase . The turning point will always be the minimum or the maximum value of your graph. The minimum or maximum of a function occurs when the slope is zero. In this case: Polynomials of odd degree have an even number of turning points, with a minimum of 0 and a maximum of n-1. They are also called turning points. At x = -1/3, 24x + 4 = -4, which is less than zero. But we will not always be able to look at the graph. The coordinate of the turning point is (-s, t). minimum turning point. (b) Using calculus, find the exact area of R. (8) t - 330 2) 'Ooc + — … The coordinates of the turning point and the equation of the line of symmetry can be found by writing the quadratic expression in completed square form. A polynomial with degree of 8 can have 7, 5, 3, or 1 turning points Using dy/dx= 0, I got the answer (4,10000) c) State whether this is a maximum or minimum turning point. The parabola shown has a minimum turning point at (3, -2). is the maximum or minimum value of the parabola (see picture below) ... is the turning point of the parabola; the axis of symmetry intersects the vertex (see picture below) How to find the vertex. This can be a maximum stationary point or a minimum stationary point. is positive then the stationary point is a minimum turning point. Recall that derivative of a function tells you the slope of the function at that selected point. Never more than the Degree minus 1. Draw a nature table to confirm. To see whether it is a maximum or a minimum, in this case we can simply look at the graph. The maximum number of turning points for a polynomial of degree n is n – The total number of turning points for a polynomial with an even degree is an odd number. If d2y dx2 is negative, then the point is a maximum turning point. n. 1. How to find and classify stationary points (maximum point, minimum point or turning points) of curve. Any polynomial of degree n can have a minimum of zero turning points and a maximum of n-1. These features are illustrated in Figure $$\PageIndex{2}$$. Finding turning points/stationary points by setting dy/dx = 0 is C2 for Edexcel. The curve has a maximum turning point A. Step 2: Check each turning point (at x = 0 and x = -1/3)to find out whether it is a maximum or a minimum. I have calculated this to be dy/dx= 5000 - 1250x b) Find the coordinates of the turning point on the graph y= 5000x - 625x^2. A General Note: Interpreting Turning Points. If d2y dx2 = 0 it is possible that we have a maximum, or a minimum, or indeed other sorts of behaviour. The graph is also symmetric with a vertical line drawn through the vertex, called the axis of symmetry. Mathematics A maximum or minimum point on a curve. If $$a>0$$ then the graph is a “smile” and has a minimum turning point. A function does not have to have their highest and lowest values in turning points, though. a) For the equation y= 5000x - 625x^2, find dy/dx. So if d2y dx2 = 0 this second derivative test does not give us … A turning point may be either a relative maximum or a relative minimum (also known as local minimum and maximum). (if of if not there is a turning point at the root of the derivation, can be checked by using the change of sign criterion.) When $$a = 0$$, the graph is a horizontal line $$y = q$$. This is a minimum. turning point synonyms, turning point pronunciation, turning point translation, English dictionary definition of turning point. You can see this easily if you think about how quadratic equations (degree 2) have one turning point, linear equations (degree 1) have none, and cubic equations (degree 3) have 2 turning … However, this depends on the kind of turning point. A turning point is a point at which the derivative changes sign. Define turning point. A turning point is where a graph changes from increasing to decreasing, or from decreasing to increasing. Extrapolating regression models beyond the range of the predictor variables is notoriously unreliable. Eg 0 = x 2 +2x -3. 10 + 8x + x-2 —F. The extreme value is −4. A point where a function changes from an increasing to a decreasing function or visa-versa is known as a turning point. A turning point is a point where the graph of a function has the locally highest value (called a maximum turning point) or the locally lowest value (called a minimum turning point). Sometimes, "turning point" is defined as "local maximum or minimum only". The curve here decreases on the left of the stationary point and increases on the right. If the parabola opens down, the vertex represents the highest point on the graph, or the maximum value. In either case, the vertex is a turning point on the graph. The turning point of a graph is where the curve in the graph turns. You can read more here for more in-depth details as I couldn't write everything, but I tried to summarize the important pieces. f(x) is a parabola, and we can see that the turning point is a minimum.. By finding the value of x where the derivative is 0, then, we have discovered that the vertex of the parabola is at (3, −4).. ; A local minimum, the smallest value of the function in the local region. I GUESSED maximum, but I have no idea. Finding d^2y/dx^2 of a function is in Edexcel C1 and has occassionally been asked in the exam but you don't learn to do anything with it in terms of max/min points until C2. d) Give a reason for your answer. Turning points can be at the roots of the derivation, i.e. The Degree of a Polynomial with one variable is the largest exponent of that variable. Find more Education widgets in Wolfram|Alpha. A turning point is a point of the graph where the graph changes from increasing to decreasing (rising to falling) or … A stationary point on a curve occurs when dy/dx = 0. However, this depends on the kind of turning point. Therefore there is a maximum point at (-1/3 , 2/27) and a minimum point at (0,0). If the function is differentiable, then a turning point is a stationary point; however not all stationary points are turning … When f’’(x) is zero, there may be a point of inflexion. Finding Vertex from Standard Form. It starts off with simple examples, explaining each step of the working. For a stationary point f '(x) = 0. A stationary point is called a turning point if the derivative changes sign (from positive to negative, or vice versa) at that point. you gotta solve the equation for finding maximum / minimum turning points. Minimum point; Maximum point; Point of horizontal inflection; We call the turning point (or stationary point) in a domain (interval) a local minimum point or local maximum point depending on how the curve moves before and after it meets the stationary point. A root of an equation is a value that will satisfy the equation when its expression is set to zero. Identifying turning points. So, the maximum exists where -(x-5)^2 is zero, which means that coordinates of the maximum point (and thus, the turning point) are (5, 22). Question 4: Complete the square to find the coordinates of the turning point of y=2x^2+20x+14 . Vertical parabolas give an important piece of information: When the parabola opens up, the vertex is the lowest point on the graph — called the minimum, or min.When the parabola opens down, the vertex is the highest point on the graph — called the maximum, or max. Therefore, to find where the minimum or maximum occurs, set the derivative equal to … In this case: Polynomials of odd degree have an even number of turning points, with a minimum of 0 and a maximum of #n-1#. To find the turning point of a quadratic equation we need to remember a couple of things: The parabola ( the curve) is symmetrical The derivative tells us what the gradient of the function is at a given point along the curve. So if this a, this is b, the absolute minimum point is f of b. The maximum number of turning points of a polynomial function is always one less than the degree of the function. We hit a maximum point right over here, right at the beginning of our interval. If $$a<0$$, the graph is a “frown” and has a maximum turning point. Example . When f’’(x) is negative, the curve is concave down– it is a maximum turning point. Depends on whether the equation is in vertex or standard form . The graph below has a turning point (3, -2). 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Súhlasím so spracovaní osobných údajov. | 2021-04-18T19:48:26 | {
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https://math.stackexchange.com/questions/543962/f-g-continuous-from-x-to-y-if-they-are-agree-on-a-dense-set-a-of-x-th/543971 | $f,g$ continuous from $X$ to $Y$. if they are agree on a dense set $A$ of $X$ then they agree on $X$
Problem:
Suppose $f$ and $g$ are two continuous functions such that $f: X \to Y$ and $g : X \to Y$. $Y$ is a a Hausdorff space. Suppose $f(x) = g(x)$ for all $x \in A \subseteq X$ where $A$ is dense in $X$, then $f(x) = g(x)$ for all $x \in X$.
Attempt at a solution:
Put $h(x) = f - g$. Therefore, $h: X \to Y$ is continuous and $Y$ is Hausdorff by hypothesis. Also we know $h(x) = 0$ for all $x \in A$ such that $A$ is dense in $X$. I want to show that $h(x)$ vanishes everywhere in $X$. We can show $h(x) = 0$ for all $x \in X \setminus A$. Suppose $h(x) > 0$ on $X \setminus A$. Pick points $y_1,y_2 \in Y$. Since $Y$ is Hausdorff, can find open set $O_1, O_2 \subseteq Y$ which are disjoint such that $y_1 \in O_1$ and $y_2 \in O_2$. By continuity, $f^{-1}(O_1), f^{-1}(O_2)$ are open in $X$.
I know that if I can show that one of the $f^{-1}(O_i)$ lies in $X \setminus A$, then we would have a contradiction since we have non-empty open set in $X \setminus A$ and this implies $A$ cannot be dense in $X$. But this is the part I am stuck. Any help would greatly be appreciated.
Also, Would be be possible to prove this without using the Hausdorff condition on $Y$?
Suppose $f(x_0) \neq g(x_0)$, then since $Y$ is Hausdorff, there are open sets $U,V \subset Y$ such that $$f(x_0) \in U, g(x_0) \in V, \text{ and } U\cap V = \emptyset$$ Now $$x_0\in f^{-1}(U)\cap g^{-1}(V) =: W$$ and $W$ is open, and hence $\exists a\in A\cap W$, whence $$f(a) = g(a) \in U\cap V \Rightarrow U\cap V \neq \emptyset$$ This contradiction proves the result.
• Why there exists $a \in A \cap W$ ?? – ILoveMath Oct 29 '13 at 9:59
• Because $A$ is dense, it intersects every non-empty open set. – Prahlad Vaidyanathan Oct 29 '13 at 10:02
• I understand now! thanks a lot for your time. One last question. Is it necessary to have $Y$ haussdorf? IF we drop this condition, can we still have the result ? – ILoveMath Oct 29 '13 at 10:06
• @DonAnselmo: Consider $\Bbb R$ with the trivial topology as both $X$ and $Y$. Let $f$ be the function moving $1$ to $2$, and fixes everything else; and $g$ moving $1$ to $3$ and fixes everything else. It's not hard to see that these are continuous (every function into a trivial space is continuous), and they agree on a dense set: $\{0\}$, and in fact $\Bbb R\setminus\{1,2,3\}$. But $f\neq g$. – Asaf Karagila Oct 29 '13 at 18:08
• @user123: Every metric space is Hausdorff. Proving this will itself help you understand the reasons for the axioms of a metric. – Prahlad Vaidyanathan Dec 12 '16 at 9:35
One can do that using nets and prove directly. If $x$ is in the dense set, then clearly $f(x)=g(x)$. Suppose $x$ is outside the dense set, let $x_i$ be a net converging to $x$, whose elements are all from the dense set.
Then $f(x_i)=g(x_i)$ is a net in $Y$. Since $Y$ is Hausdorff we have to have $\lim f(x_i)=\lim g(x_i)$ (recall that being Hausdorff is equivalent to the statement that converging nets have a unique limit point). But now by continuity we finish as the following holds: $$f(x)=f\left(\lim x_i\right)=\lim f(x_i)=\lim g(x_i)=g\left(\lim x_i\right)=g(x).$$
I'm just going to add a short proof to this question, since this proof wasn't listed, and this question seems to be the version all duplicates are directed towards.
Setting up notation: Let $f,g : X\to Y$ be continuous functions, with $Y$ Hausdorff. Let $f,g$ agree on a dense subset $A$ of $X$. Let $\Delta_X : X\to X\times X$ denote the diagonal map and let $D_Y\subset Y\times Y$ denote the diagonal as a subset of $Y\times Y$, finally let $(f,g)$ denote the product map from $X\times X\to Y\times Y$.
Proof: It is a standard fact that $Y$ is Hausdorff if and only if $D_Y$ is closed, so $D_Y$ is closed. Therefore the set of points at which $f$ and $g$ agree, $((f,g)\circ \Delta_X)^{-1}(D_Y),$ is a closed subset of $X$. However, we know this set also contains a dense subset, so it must be all of $X$. Thus $f=g$. | 2020-08-09T11:47:32 | {
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https://math.stackexchange.com/questions/3425971/is-lceil-fraca2-rceil-lceil-fracb2-rceil-geq-lfloor-frac/3426063 | # Is $| \lceil \frac{a}{2} \rceil - \lceil \frac{b}{2} \rceil |\geq \lfloor |\frac{a - b}{2}| \rfloor$?
Let $$a$$ and $$b$$ be integers. Is it true that
$$\left | \left \lceil \frac{a}{2} \right \rceil - \left \lceil \frac{b}{2} \right \rceil \right |\geq \left \lfloor \left | \frac{a - b}{2} \right |\right \rfloor$$
Where $$\lceil \cdot \rceil$$ is the ceiling function, $$\lfloor \cdot \rfloor$$ the floor function and $$|\cdot|$$ is the absolute function.
The inequality seems to be true when I check it programatically but I would like to get a proof (or disproof) for this inequality.
• What about $a=-1$ and $b=2$? The LHS is $|0-1| =1$ and the RHS is $2$. – Martin R Nov 7 '19 at 16:05
• @MartinR Thank you very much for pointing that out, I have altered the inequality slightly, there was an error in the equation. – KillaKem Nov 7 '19 at 16:25
• You might as well assume $a\ge b$ and thus dispense with taking absolute values. – hardmath Nov 7 '19 at 17:03
• If we allow all real numbers, you can also just write it as $|\lceil x\rceil-\lceil y\rceil|\ge\lfloor|x-y|\rfloor|$. Assuming $x\ge y$ gives $\lceil x\rceil-\lceil y\rceil\ge\lfloor x-y\rfloor$, which looks pretty neat. – Milten Nov 7 '19 at 17:13
Yes, it is true.
$$\left | \left \lceil \frac{a}{2} \right \rceil - \left \lceil \frac{b}{2} \right \rceil \right |\geq \left \lfloor \left | \frac{a - b}{2} \right |\right \rfloor \tag1$$
In the following, $$m,n$$ are integers.
Case 1 : If $$a=2m,b=2n$$, then both sides of $$(1)$$ equal $$|m-n|$$.
Case 2 : If $$a=2m,b=2n+1$$, then $$(1)\iff |m-n-1|\ge \left\lfloor\left |m-n-\frac 12\right|\right\rfloor\tag2$$
If $$m-n-\frac 12\ge 0$$, then $$m-n-1\ge 0$$, so$$(2)\iff m-n-1\ge m-n-1$$which is true.
If $$m-n-\frac 12\lt 0$$, then $$m-n-1\lt 0$$, so$$(2)\iff -m+n+1\ge -m+n$$which is true.
Case 3 : If $$a=2m+1, b=2n$$, then $$(1)\iff |m-n+1|\ge \left\lfloor\left|m-n+\frac 12\right|\right\rfloor\tag3$$
If $$m-n+\frac 12\ge 0$$, then $$m-n+1\ge 0$$, so$$(3)\iff m-n+1\ge m-n$$which is true.
If $$m-n+\frac 12\lt 0$$, then $$m-n+1\lt 0$$, so$$(3)\iff -m+n-1\ge -m+n-1$$which is true.
Case 4 : If $$a=2m+1,b=2n+1$$, then both sides of $$(1)$$ equal $$|m-n|$$.
There is no need for the assumption that $$a$$ and $$b$$ are integers. You just need to prove that
$$|\lceil x\rceil-\lceil y\rceil|\ge\lfloor|x-y|\rfloor$$
for any real numbers $$x$$ and $$y$$. By symmetry, we may assume $$x\ge y$$, in which case we can remove the absolute value signs. If, moreover, we write $$x=y+u$$ with $$u\ge0$$, we are trying to prove
$$\lceil y+u\rceil\ge\lceil y\rceil+\lfloor u\rfloor$$
But $$u=\lfloor u\rfloor+r$$ for some $$0\le r\lt1$$, and $$\lceil y+\lfloor u\rfloor +r\rceil=\lceil y+r\rceil+\lfloor u\rfloor$$, so the inequality to prove is simply
$$\lceil y+r\rceil\ge\lceil y\rceil$$
which is clearly true, since the ceiling function is never decreasing and $$r\ge0$$.
Assume without loss of generality that $$a\ge b$$. Then the inequality is $$\left\lceil \frac a2 \right\rceil - \left\lceil \frac b2 \right\rceil \ge \left\lfloor \frac {a-b}2 \right\rfloor$$ If either $$a$$ or $$b$$ is an even integer, then we can pull the whole number $$\frac a2$$ or $$\frac b2$$ out of the floor function, and the inequality reduces to $$\left\lceil \frac a2 \right\rceil \ge \left\lfloor \frac {a}2 \right\rfloor$$ or $$-\left\lceil \frac b2 \right\rceil \ge \left\lfloor -\frac {b}2 \right\rfloor$$ (where the first is trivial and the second is actually an equality).
Assume therefore that neither of $$a$$ and $$b$$ is an even integer. Let $$2m and $$2n, for some $$m,n\in \mathbb Z$$. Then $$\left\lceil \frac a2 \right\rceil - \left\lceil \frac b2 \right\rceil = (m+1)-(n+1) = m-n$$ On the other hand $$m-n-1<\frac a2 - \frac b2 < m-n+1$$ which means that $$\left\lfloor \frac {a-b}2 \right\rfloor \le m-n = \left\lceil \frac a2 \right\rceil - \left\lceil \frac b2 \right\rceil$$ so we are done.
EDIT: I didn't notice you assumed $$a$$ and $$b$$ to be integers. Well, my answer works for all real numbers. | 2020-10-25T04:23:06 | {
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https://math.stackexchange.com/questions/2576654/modified-blackjack-probability | # Modified blackjack probability
So there's this one gambling site and they have game of 'blackjack' which is played very differently than the usual game.
The host has dice which has all the numbers from 1 to 100. He rolls it for you as many times as you want, the rolls are added up and you must not exceed 100. If theyre over 100, it's an instant loss. Now lets say he rolls 10 and 67 and you decide to stay, your total would be 77 and it'll be the hosts turn. He must now roll higher total than you without exceeding 100 also. If he rolls 100 in the beginning it's an instant win, and there are no ties(meaning if you both have same total, the host must roll another time).
It seems to me that the probability of host winning this game is way higher than the players, because the player goes first and you might go over 100 even before the hosts turn, and also if you stay at something like 80 and the host rolls 79, the host can just keep on rolling more without thinking about it, whereas if your score was 79 you most likely wouldnt roll another time.
Now is there a chance to calculate exact probability that the player wins, or atleast approximate it with say 5%error? I have tried, but i'm pretty lost in how i should begin tackling this problem.
• Since this depends on strategy (when does the first player stop?) it is probably best to sample the game. That is, play thousands (or millions) of the game first to determine optimal strategy and then to determine probabilities. – lulu Dec 22 '17 at 10:57
• @lulu - I think a full calculation might be easier than a simulation – Henry Dec 22 '17 at 11:12
• @Henry Really? You may well be right but even so I'd want to see a suite of runs just to check the calculation! – lulu Dec 22 '17 at 11:15
• Very similar to math.stackexchange.com/q/1315270/420432 though not identical because that question doesn't ask for a probabilty. – nickgard Dec 22 '17 at 11:30
• @nickgard - a very small difference between the two games: in this game ties do not happen as the host must roll again while in your linked game ties result in the original stake being returned – Henry Dec 22 '17 at 12:23
1. Let's call the probability of hitting a particular value $n$ at any stage if you do not stop throwing $a_n$. Clearly $a_0=1$ while $a_n=\sum\limits_{m= \min(0,n-100)}^{n-1} a_m /100$. For $1 \le n \le 100$ this will give $a_n=\frac{101^{n-1}}{100^n}$
2. Suppose the player stays at a value $s$. Then the probability that the host reaches a value from $s$ through to $100$ is $(100-s)a_{s+1} = \frac{(100-s)101^{s}}{100^{s+1}}$ while the probability the bank goes bust is $1-\frac{(100-s)101^{s}}{100^{s+1}}$
3. If the probability that a player using the optimal strategy and who has reached $k$ then wins is $w_k$ then $w_{100}=1$ and $w_k=\max\left(1-\frac{(100-s)101^{s}}{100^{s+1}}, \sum\limits_{j= k+1}^{100} w_k /100\right)$ for $0 \le k \lt 100$, and the optimal strategy is to stay when the left hand part of the maximum expression is greater
It turns out that the probability that the player wins is $w_0=0.4316936\ldots$ following a strategy of the player staying with a total of $58$ or more
lulu asked for a simulation to check - the following R script comes close to the same value for $w_0$ when staying at $58$ or above though does not prove optimality (choosing $57$ or $59$ to stay would give similar results)
stayat <- 58
sides <- 100
maxtarget <- sides
cases <- 1000000
set.seed(1)
playertot <- rep(0, cases)
while( sum(playertot < stayat) > 0) {
dice <- sample(sides, cases, replace=TRUE)
playertot <- playertot + ifelse(playertot < stayat, dice, 0)
}
hosttot <- rep(0, cases)
while( sum(hosttot <= playertot ) > 0) {
dice <- sample(sides, cases, replace=TRUE)
hosttot <- hosttot + ifelse(hosttot <= playertot, dice, 0)
}
which gives the very close estimate of the probability
> mean(playertot <= maxtarget & hosttot > maxtarget)
[1] 0.431478
• Thank you! The probability seems higher than what i assumed, but more than likely they didnt have optimal strategy. – J sx Dec 22 '17 at 12:37 | 2019-03-25T01:36:50 | {
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https://math.stackexchange.com/questions/1247214/solve-the-following-recurrence-relation | Solve the following recurrence relation:
Solve the following recurrence relation: $f(1) = 1$ and for $n \ge 2$,
$$f(n) = n^2f(n − 1) + n(n!)^2$$
How would I go about solving this?
• Would I need to find a substitution $f(n) =\text{ insert here }g(n)$ in aim of getting rid of the $n^2$ that is multiplied onto $f(n-1)$
• Then use the method of differences/ladder method to simplify down $g(n)$
• Then substitute back into $f(n)$?
• Note that you also have $f(n)=n^2(f(n-1)+n!(n-1)!)$ – abiessu Apr 22 '15 at 19:58
• @abiessu: Not quite: there are three factors of $n$ in the last term. – Brian M. Scott Apr 22 '15 at 19:59
• @BrianM.Scott: yes, caught that only a moment ago... – abiessu Apr 22 '15 at 20:00
Divide everything by $(n!)^2$: $$\frac{f(n)}{(n!)^2} = \frac{n^2 f(n-1)}{(n!)^2} + n = \frac{f(n-1)}{((n-1)!)^2} + n.$$ If you write $g(n) = \frac{f(n)}{(n!)^2}$ you get $$g(n) = g(n-1) + n.$$ Since $g(1) = 1$, this becomes $g(n) = 1 + 2 + \cdots + n = \frac{n(n+1)}{2}$ so that $$f(n) = (n!)^2 \frac{n(n+1)}{2}.$$
• Is $n(n+1)$ not $n(n-1)$? $f(1)=1$ not $f(1)=0$ – rlartiga Apr 22 '15 at 20:03
• Uh oh! Thanks for catching the error. – Umberto P. Apr 22 '15 at 20:12
• How did you know to divide by (n!)^2 to conveniently get a nice looking recurrence relation for g(n)? – Hyune Apr 22 '15 at 20:27
• If you want $f(n) = a(n) g(n)$ to put the equation into the form $g(n) = g(n-1) + \ldots$, you need $a(n) = n^2 a(n-1)$. – Robert Israel Apr 22 '15 at 21:28
If you have a linear recurrence of the first order:
$$f(n + 1) = g(n) f(n) + h(n)$$
if you divide by the summing factor $s(n) = \prod_{0 \le k \le n} g(n)$ you are left with:
$$\frac{f(n + 1)}{s(n)} - \frac{f(n)}{s(n - 1)} = \frac{h(n)}{s(n)}$$
Adding this for $k$ from $0$ to $n$ telescopes nicely:
$$\frac{f(n + 1)}{s(n)} = f(0) + \sum_{0 \le k \le n} \frac{h(k)}{s(k)}$$ | 2021-05-15T00:03:33 | {
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https://hilbertthm90.wordpress.com/category/math/manifolds/page/2/ | # Naturality of Flows
This is something I always forget exists and has a name, so I end up reproving it. Since this sequence of posts is a hodge-podge of things to help me take a differential geometry test, hopefully this will lodge the result in my brain and save me time if it comes up.
I’m not sure whether to call it a lemma or not, but the setup is you have a smooth map ${F:M\rightarrow N}$ and a vector field on ${M}$, say ${X}$ and a vector field on ${N}$ say ${Y}$ such that ${X}$ and ${Y}$ are ${F}$-related. Define ${M_t}$ and ${N_t}$ to be the image of flowing for time ${t}$ and let ${\theta}$ and ${\eta}$ be the flows of ${X}$ and ${Y}$ respectively. Then the lemma says for all ${t}$ we have ${F(M_t)\subset N_t}$ and ${\eta_t\circ F=F\circ \theta_t}$ on ${M_t}$.
This is a “naturality” condition because all it really says is that the following diagram commutes:
${\begin{matrix} M_t & \stackrel{F}{\longrightarrow} & N_t \\ \theta_t \downarrow & & \downarrow \eta_t \\ M_{-t} & \stackrel{\longrightarrow}{F} & N_{-t} \end{matrix}}$
Proof: Let ${p\in M}$, then ${F\circ \theta^p: \mathbb{R}\rightarrow N}$ is a curve that satisfies the property $\displaystyle {\frac{d}{dt}\Big|_{t=t_0}(F\circ \theta^p)(t)=DF_{\theta^p(t_0)}(\frac{d}{dt}\theta^p (t)\Big|_{t=t_0})=DF_{\theta^p(t_0)}(X_{\theta^p(t_0)})=Y_{F\circ \theta^p(t_0)}}$. Since ${F\circ \theta^p(0)=F(p)}$, and integral curves are unique, we get that ${F\circ\theta^p(t)=\eta^{F(p)}(t)}$ at least on the domain of ${\theta^p}$.
Thus if ${p\in M_t}$ then ${F(p)\in N_t}$, or equivalently ${F(M_t)\subset N_t}$. But we just wrote that ${F(\theta^p(t))=\eta^{F(p)}(t)}$ where defined, which is just a different form of the equation ${\eta_t\circ F=F\circ \theta_t(p)}$.
We get a nice corollary out of this. If our function ${F:M\rightarrow N}$ was actually a diffeo, then take ${Y=F_*X}$ the pushforward, and we get that the flow of the pushforward is ${\eta_t=F\circ \theta_t\circ F^{-1}}$ and the flow domain is actually equal ${N_t=F(M_t)}$.
In algebraic geometry we care a lot about families of things. In the differentiable world, the nicest case of this would be when you have a smooth submersion: ${F: M\rightarrow N}$, where ${M}$ is compact and both are connected. Then since all values are regular, ${F^{-1}(n_0)}$ is smooth embedded submanifold. If ${N}$ were say ${\mathbb{R}}$ (of course, ${M}$ couldn’t be compact in this case), then we would have a nice 1-dimensional family of manifolds that are parametrized in a nice way.
It turns out to be quite easy to prove that in the above circumstance all fibers are diffeomorphic. In AG we often call this an “iso-trivial” family, although I’m not sure that is the best analogy. The proof basically comes down to the naturality of flows. Given any vector field ${Y}$ on ${N}$, we can lift it to a vector field ${X}$ on ${M}$ that is ${F}$-related. I won’t do the details, but it can be done clearly in nice choice of coordinates ${(x^1, \ldots, x^n)\mapsto (x^1, \ldots, x^{n-k})}$ and then just patch together with a partition of unity.
Let ${M_x}$ be the notation for ${F^{-1}(x)}$. Fix an ${x\in N}$, then by the above naturality lemma ${\theta_t\Big|_{M_x} : M_x\rightarrow M_{\eta_t(x)}}$ is well-defined and hence a diffeomorphism since it has smooth inverse ${\theta_{-t}}$. Let ${y\in N}$. Then as long as there is a vector field on ${N}$ which flows ${x}$ to ${y}$, then we’ve shown that ${M_x\simeq M_y}$, so since ${x}$, ${y}$ were arbitrary, all fibers are diffeomorphic. But there is such a vector field, since ${N}$ is connected.
# Lie groups have abelian fundamental group
Last year I wrote up how to prove that the fundamental group of a (connected) topological group was abelian. Since Lie groups are topological groups, they also have abelian fundamental groups, but I think there is a much neater way to prove this fact using smooth things. Here it is:
Lemma 1: A connected Lie group that acts smoothly on a discrete space is must be a trivial action.
Proof: Suppose our action is ${\theta: G\times M \rightarrow M}$ by ${\theta(g,x)=g\cdot x}$. Consider a point ${q}$ with non-trivial orbit. Then ${\text{im}\theta_{q}=\{q\}\cup A}$ where ${A}$ is non-empty. Thus ${G=\theta_{q}^{-1}(q)\cup \theta_q^{-1}(A)}$, a disconnection of ${G}$. Thus all orbits are trivial.
Lemma 2: A discrete normal subgroup of a Lie group (from now on always assumed connected) is central.
Proof: Denote the subgroup ${H}$. Then ${\theta(g,h)=ghg^{-1}}$ is a smooth action on ${H}$ (a discrete set). Thus by Lemma 1, ${ghg^{-1}=h}$ for all ${g\in G}$. Thus ${H}$ is central, and in particular abelian.
Now for the main theorem. Let ${G}$ be a connected Lie group. Let ${U}$ be the universal cover of ${G}$. Then the covering map ${p: U\rightarrow G}$ is a group homomorphism. Since ${U}$ is simply connected, the covering is normal and hence ${Aut_p(U)\simeq \pi_1(G, e)}$. By virtue of being normal, we also get that ${Aut_p(U)}$ acts transitively on the fibers of ${p}$. In particular, on the set ${p^{-1}(e)}$, which is discrete being the fiber of a discrete bundle. But this set is ${\ker p}$, which is a normal subgroup. I.e. a discrete normal subgroup, which by Lemma 2 is abelian.
Fix ${q\in \ker p}$. Then we get an ismorphism ${\ker p \simeq Aut_p(U)}$ by ${x\mapsto \phi_x}$ where ${\phi_x}$ is the unique covering automorphism that takes ${q}$ to ${x}$. Thus ${Aut_p(U)}$ is abelian which means ${\pi_1(G,e)}$ is abelian.
# PDE’s and Frobenius Theorem
I’ve started many blog posts on algebra/algebraic geometry, but they won’t get finished and posted for a little while. I’ve been studying for a test I have to take in a few weeks in differential geometry-esque things. So I’ll do a few posts on things that I think are usually considered pretty easy and obvious to most people, but are just things I never sat down and figured out. Hopefully this set of posts will help others who are confused as I recently was.
My first topic is about the Frobenius Theorem. I’ve posted about it before. Here’s the general idea of it: If ${M}$ is a smooth manifold and ${D}$ is a smooth distribution on it, then ${D}$ is involutive if and only if it is completely integrable (i.e. there is are local flat charts for the distribution).
What does this have to do with being able to solve partial differential equations? I’ve always heard that it does, but other than the symbol ${\displaystyle\frac{\partial}{\partial x}}$ appearing in the defining of a distribution or of the flat chart, I’ve never figured it out.
Let’s go through this with some examples. Are there any non-constant solutions ${f\in C^\infty (\mathbb{R}^3)}$ to the systems of equations: ${\displaystyle \frac{\partial f}{\partial x}-y\frac{\partial f}{\partial z}=0}$ and ${\displaystyle \frac{\partial f}{\partial y}+x\frac{\partial f}{\partial z}=0}$?
Until a few days ago, I would have never thought we could use the Frobenius Theorem to do this. Suppose ${f}$ were such a solution. Define the vector fields ${\displaystyle X=\frac{\partial}{\partial x}-y\frac{\partial}{\partial z}}$ and ${\displaystyle Y=\frac{\partial}{\partial y}+x\frac{\partial}{\partial z}}$ and define the distribution ${D_p=\text{span} \{X_p, Y_p\}}$.
Choose a regular value of ${f}$, say ${C}$ (one exists by say Sard’s Theorem). Then ${f=C}$ is a 2-dimensional submanifold ${M\subset \mathbb{R}^3}$, and since ${f}$ is a defining function ${T_pM=ker(Df_p)}$. But the very fact that ${f}$ satisfies, by assumption, ${X(f)=0}$ and ${Y(f)=0}$, we have ${T_pM=\text{span} \{X_p, Y_p\}}$. I.e. ${M}$ is an integral manifold for the distribution ${D}$. Thus ${D}$ must be involutive.
Just check now. ${\displaystyle [X,Y]=2\frac{\partial}{\partial z}}$, so in particular at the origin ${\displaystyle X_0=\frac{\partial}{\partial x}}$ and ${\displaystyle Y_0=\frac{\partial}{\partial y}}$ it is not in the span, and hence not involutive. Thus no such ${f}$ exists. This didn’t even use Frobenius.
Now let’s spice up the language and difficulty. Is it possible to find a function ${z=f(x,y)}$, ${C^\infty}$ in a neighborhood of ${(0,0)}$, such that ${f(0,0)=0}$ and ${\displaystyle df=(ye^{-(x+y)}-f)dx+(xe^{-(x+y)}-f)dy}$? Alright, the ${d}$ phrasing is just asking there is a local solution to the system ${\displaystyle \frac{\partial f}{\partial x}=ye^{-(x+y)}-f}$ and ${\displaystyle \frac{\partial f}{\partial y}=x^{-(x+y)}-f}$. Uh oh. The above method fails us now since it isn’t homogeneous.
Alright, so let’s extrapolate a little. We have a system of the form ${\displaystyle \frac{\partial f}{\partial x}=\alpha(x,y,f)}$ and ${\displaystyle \frac{\partial f}{\partial y}=\beta(x,y,f)}$. The claim is that necessary and sufficient conditions to have a local solution to this system is ${\displaystyle \frac{\partial \alpha}{\partial y}+\beta\frac{\partial \alpha}{\partial z}=\frac{\partial \beta}{\partial x}+\alpha \frac{\partial \beta}{\partial z}}$.
I won’t go through the details of the proof, but the main idea is not bad. Define the distribution spanned by ${\displaystyle X=\frac{\partial}{\partial x}+\alpha\frac{\partial}{\partial z}}$ and ${\displaystyle Y=\frac{\partial}{\partial y}+\beta\frac{\partial}{\partial z}}$.
Then use that assumption to see that ${[X,Y]=0}$ and hence the distribution is involutive and hence there is an integral manifold for the distribution by the Frobenius Theorem. If ${g}$ is a local defining function to that integral manifold, then we can hit that with the Implicit Function Theorem and get that ${z=f(x,y)}$ (the implicit function) is a local solution.
If we go back to that original problem, we can easily check that the sufficient condition is met and hence that local solution exists.
I had one other neat little problem, but it doesn’t really fit in here other than the fact that solutions to PDEs are involved.
# The Cohomology Computation
Alright, I’m in a sort of tough spot. Yesterday I started typing this up, but I just don’t have the motivation. There are lots of tedious details that no one is going to read and will not come up in our study after this. It is all incredibly standard chasing Fourier coefficients around, so I’m not going to do it. This post will be an outline in how one would go about doing it, and I even may provide quick ideas behind it, but if it will be weeks before I continue on if I don’t just get through this. Someday, if it seems important, I’ll come back and fill it in. Or if someone comments and really wants to see one particular part, at least I’ll have motivation that someone is going to read it.
Here goes. Recall some of my conventions. $X$ is a compact complex Lie group. We let $A=\Gamma(X, \mathcal{C})$ the global sections of the sheaf of $C^\infty$-functions on $X$. This was important to the Dolbeaut resolution. $\overline{T}$ are the $\mathbb{C}$-antilinear functionals. We use $\bigwedge^q$ to mean $\bigwedge^q(\overline{T})$.
We’ve shown that $A\otimes_\mathbb{C}\bigwedge^q$ is isomorphic to $\Gamma(X, \mathcal{C}^{0,q})$ as complexes and hence we have the iso $H^q(X, \mathcal{O}_X)\simeq H^q(A\otimes_\mathbb{C}\bigwedge)$.
Now we want to show that we actually have an induced isomorphism on cohomology from the inclusion map $i: \bigwedge \hookrightarrow A\otimes \bigwedge$. We’ll do this by comparing Fourier series. So we set up a normalized measure on $X$ say $\mu$. By integrating functions against this measure we get linear function $\mu_{\wedge}$ (I called this something different last time).
Now we need the lemma that for all $\omega\in A\otimes \bigwedge^q$ we have $\mu_\wedge (\overline{\partial} \omega)=0$. This follows from periodicity and translation invariance of the vector field that comes up when you go to write it down.
The next step is to define the “Fourier coefficients” of a function. Now if we choose any integer valued function from the lattice defining $X$, say $\lambda$. Then it extends to an $\mathbb{R}$-linear function on the tangent space at the identity, $V$. We then have the exponentiation map $\displaystyle v\mapsto e^{2\pi i \lambda(v)}$. This respects the lattice and hence descends to $\latex X$. Call this map $c_\lambda$.
Define the $\mathbb{C}$-linear function $A\to \mathbb{C}$ by $Q_\lambda(f)=\int_X c_{-\lambda}fd\mu$. Don’t be intimidated here. This is just the standard Fourier coefficient when your in a familiar situation. i.e. $f=\sum e_\lambda \otimes Q_\lambda(f)$.
Now choose a Hermitian inner product, which gives us a norm to work with. Here things will become less detailed. One can next prove a lemma that the map $f\to \{Q_\lambda(f)\}_\lambda$ is an isomorphism $A\to$ the space of maps decreasing at infinity faster than $\|\lambda \|^{-n}$ for all $n$.
Then do a computation to see that $Q_\lambda (\overline{\partial}\omega)=(-1)^p 2\pi i \left(Q_\lambda(\omega)\wedge \overline{C}(\lambda)\right)$.
Lastly we want to get back to showing that the inclusion is a homotopy equivalence. Thus use the Hermitian inner product to define a map $\lambda^*\in Hom_\mathbb{C}(\overline{T}, \mathbb{C})$ for every $\lambda$ by $\displaystyle \lambda^*(x)=\frac{\langle x, \overline{C}(\lambda)\rangle}{2\pi i \|\overline{C}(\lambda)\|^2}$.
We need to define for $\omega\in A\otimes \bigwedge^p$ a map $k(\omega)$ which will only be defined in terms of its Fourier coefficients. We define $Q_\lambda(k(\omega))=(-1)^p \lambda^* \neg Q_\lambda(\omega)$ if $\lambda\neq 0$ and the coefficient is 0 if lambda is 0. Now it has all been set up so that comparing Fourier coefficients on $\overline{\partial} k + k \overline{\partial}$ we get exactly the same ones as in $id_{A\otimes\wedge} - i \mu_\wedge$. Thus we are done by the magic of Fourier coefficients being unique.
That last computation I left out requires use of things such as “Cartan’s magic formula” and breaking it into two cases. Anyway, for not doing any details, I think this is a pretty thorough outline and filling any of the details you don’t believe or would like to know shouldn’t be too hard.
# Cohomology of Abelian Varieties II
Two posts in the same week! Before we get started today, we need to introduce one more piece of new notation. Let ${\overline{T}}$ be the ${\mathbb{C}}$-antilinear maps ${V\rightarrow \mathbb{C}}$. Our goal is to prove that ${H^q(X, \mathcal{O}_X)\simeq \bigwedge^q\overline{T}}$ and ${H^q(X, \Omega^p)\simeq \bigwedge^pT\otimes \bigwedge^q\overline{T}}$.
To do the calculation we will use the Dolbeault resolution: ${0\rightarrow \mathcal{O}_X\rightarrow \mathcal{C}^{0,0}\rightarrow \mathcal{C}^{0,1}\rightarrow \mathcal{C}^{0,2}\rightarrow\cdots}$. This is an acyclic resolution of the structure sheaf, and so is fine to use for the calculation of cohomology. The first upper index of ${\mathcal{C}}$ refers to the degree of the ${\mathbb{C}}$-linear part and the second upper index refers to the degree of the ${\mathbb{C}}$-antilinear part. The map of the chain complex is ${\overline{\partial}}$.
Let’s examine the complex a little more closely. Define ${\phi_{p,q}: \mathcal{C}\otimes (\bigwedge^pT\otimes \bigwedge^q\overline{T})\rightarrow \mathcal{C}^{p,q}}$ by ${\sum f_i\otimes \alpha_i\mapsto \sum f_i\omega_{\alpha_i}}$. Where we define ${\omega_{\alpha}}$ to be the natural translation invariant ${(p,q)}$-form associated to ${\alpha\in \wedge^pT\otimes \wedge^q\overline{T}}$ by left-invariantizing.
Note that ${\omega_{\alpha\wedge\beta}=\omega_\alpha\wedge \omega_\beta}$. Thus to prove that all ${\omega_\alpha}$ are ${\overline{\partial}}$-closed (which we’ll denote ${d}$ from now on for simplicity), we only need to check this for ${(1,0)}$ and ${(0,1)}$ forms. Now ${exp: V\rightarrow X}$ is a local iso, so we also only need to check ${d(exp^*(\omega_\alpha))=0}$. But ${exp^*(\omega_\alpha)=d\alpha}$, so ${d}$ of this expression, is ${(d\circ d)(\alpha)=0}$.
This gives us that our map ${\phi_{0,q}}$ is an iso ${\Gamma(X, \mathcal{C})\otimes_\mathbb{C} \bigwedge^q\overline{T} \rightarrow \Gamma(X, \mathcal{C}^{0,q})}$. The map ${\overline{\partial}}$ is defined to be ${\overline{\partial}(f\otimes \alpha)=\overline{\partial}(f)\otimes \alpha}$. So since the ${\omega_\alpha}$ are closed, the iso commutes with the differential and we get these are actually isomorphic as chain complexes. Thus computing cohomology of one is equivalent to computing cohomology of the other.
Explicitly, we know that ${H^q(X, \mathcal{O}_X)\simeq H^q(X, \Gamma(X,\mathcal{C})\otimes_\mathbb{C} \bigwedge^q\overline{T})}$.
Since this notation is cumbersome, let ${A=\Gamma(X, \mathcal{C})}$ and ${\bigwedge^*=\bigwedge^*\overline{T}}$.
Let ${i: \bigwedge \rightarrow A\otimes_\mathbb{C} \bigwedge}$ be the inclusion. We want to show this gives an iso ${\bigwedge^q\stackrel{\sim}{\rightarrow} H^q(X, A\otimes \bigwedge^*)}$. This is precisely the goal given at the start of the post.
Now ${V}$ is a vector space so we have a natural Euclidean measure. Let ${\mu}$ be the measure on ${X}$ induced from this that is normalized so that ${\mu(X)=1}$. (For those familiar, this is just the unique translation invariant Haar measure, and ${X}$ is compact, so is finite and can be normalized). Define the linear map represented by ${\mu}$ by ${S: A\rightarrow \mathbb{C}}$. It is just ${S(f)=\int_X fd\mu}$. If ${W}$ is any ${\mathbb{C}}$-vector space, denote by ${S_W}$ the map ${A\otimes_\mathbb{C} W\rightarrow W}$. In particular, we have ${S_\wedge: A\otimes \bigwedge^*\rightarrow \bigwedge^*}$ so ${S_\wedge\circ i=id_\wedge}$.
We have a good ways to go yet, so next time we’ll pick up with the lemma that ${S_\wedge(\overline{\partial}\omega)=0}$ for all ${\omega\in A\otimes \bigwedge}$.
# Cohomology of Abelian Varieties
Hopefully I’ll start updating more than once a month. Since it’s been awhile and the previous post was tangent to what we’re actually doing, I’ll recap some notation. ${X}$ will be a compact complex (connected) Lie group of dimension ${g}$. We showed that we have an analytic isomorphism ${X\simeq (S^1)^{2g}\simeq (\mathbb{R}/\mathbb{Z})^{2g}}$. Let ${V=T_0X}$ (note that I’ll assume ${0}$ is the identity).
Under the exponential map ${exp: V\rightarrow X}$ (which we showed was a local isomorphism), we have that ${V}$ is the universal covering space of ${X}$. We showed that ${ker(exp)=U}$ is a lattice. Now the title of this post will seem a little silly to experts out there in cohomology, since we know that topologically these things are all tori. We’ll go through the details anyway.
First, we’ll show that ${H^r(X, \mathbb{Z})\simeq }$ the group of alternating ${r}$-forms ${U\times \cdots \times U\rightarrow \mathbb{Z}}$. We proceed by induction.
By general covering space theory ${exp^{-1}(0)=U=\pi_1(X, 0)}$. Thus we get the base case ${H^1(X,\mathbb{Z})\simeq Hom(U=\pi_1(X), \mathbb{Z})}$. Now we’ll want to show that the cup product induces the isomorphism ${\bigwedge^r\left(H^1(X,\mathbb{Z})\right)\rightarrow H^r(X,\mathbb{Z})}$. By the ${r=1}$ case this proves the statement.
We first reduce to the case of showing it is true for ${S^1}$. Since ${X}$ is just a product of ${S^1}$‘s, if we show that if the statement is true for ${X_1}$ and ${X_2}$, then it is also true for the product we can make the reduction. (For simplicitly, coefficients are in ${\mathbb{Z}}$, but I’ll omit that). Since we only need to apply this in the case where ${X_1}$ or ${X_2}$ is finite product of ${S^1}$‘s, we can also assume that the cohomologies are finitely generated and that they two spaces are connected for simplicity.
First, by the K\”{u}nneth formula: ${H^1(X_1\times X_2)\simeq \left(H^1(X_1)\otimes_\mathbb{Z}H^0(X_2)\right)\bigoplus \left(H^0(X_1)\otimes_\mathbb{Z}H^1(X_2)\right)}$, but the spaces are connected, so ${H^0(X_i)=\mathbb{Z}}$. Thus ${\bigwedge^r(H^1(X_1\times X_2))\simeq \bigwedge^r(H^1(X_1)\oplus H^1(X_2))}$
${\simeq \displaystyle\sum_{p+q=r}\bigwedge^p(H^1(X_1))\otimes \bigwedge^q(H^1(X_2))}$.
But now our inductive hypothesis is that for all $p$ less than $r$, ${\bigwedge^p(H^1(X_i))\simeq H^p(X_i)}$. Thus we get ${\displaystyle \sum_{p+q=r}\bigwedge^p(H^1(X_1))\otimes \bigwedge^q(H^1(X_2))}$
${\displaystyle \simeq \sum_{p+q=r} H^p(X_1)\otimes H^q(X_2)}$
${\simeq H^r(X_1\times X_2)}$. In other words, stringing all these isos together we get the iso we wanted. So we’ve reduced to the case of showing the statement for ${S^1}$, which follows immediately since ${H^n(S^1)=0}$ for ${n>1}$.
Note that if you have basic facts about singular cohomology at your disposal, this isn’t at all surprising. But let’s look at sheaf cohomology instead. This will require us to look at the Hodge structure which could be interesting. We won’t go very far today, but let’s at least get a few things out of the way.
Let ${\Omega^p}$ be the sheaf of holomorphic ${p}$-forms on ${X}$. We’d like to compute ${H^r(X, \Omega^p)}$. Let ${T=Hom(V, \mathbb{C})}$, i.e. the (complex) cotangent space at the identity to ${X}$. As with vectors and vector fields, every ${p}$-covector, i.e. element of ${\bigwedge^pT}$ can be extended uniquely to a left invariant ${p}$-form by pulling back along the left multiplication by ${-x}$ map. We’ll denote the correspondence ${\alpha\mapsto \omega_\alpha}$. This map defines an isomorphism of sheaves ${\mathcal{O}_X\otimes_\mathbb{C} \bigwedge^pT\stackrel{\sim}{\rightarrow} \Omega^p}$.
This says that ${\Omega^p}$ is a free sheaf of ${\mathcal{O}_X}$-modules. Now take global sections to get that ${\Gamma(X, \Omega^p)\simeq \Gamma(X, \mathcal{O}_X\otimes \bigwedge^pT)\simeq \bigwedge^pT}$, since the global sections of ${\mathcal{O}_X}$ are constants. Thus the only global sections of ${\Omega^p}$ are the ${p}$-forms that are invariant under left translation. Thus this isomorphism reduces our calculation to ${H^r(X, \Omega^p)\simeq H^r(X, \mathcal{O}_X)\otimes_\mathbb{C} \bigwedge^pT}$. So we’ll start in on that next time.
# Complex Lie Group Properties
Today we’ll do two more properties of compact complex Lie groups. The property we’ve already done is that they are always abelian groups. We go back to the notation from before and let $X$ be a compact complex Lie group and $V=T_eX$.
Property 1: $X$ is abelian.
Property 2: $X$ is a complex torus.
Proposition: $exp: \mathcal{L}(X)\simeq V\to X$, the exponential map, is a surjective homomorphism with kernel a lattice.
Proof: Fix $x,y\in X$. Note that the map $\psi: \mathbb{C}\to X$ by $t\mapsto (exp(tx))(exp(ty))$ is holomorphic since it is the composition of multiplication (holomorphic by being a Lie group) and the fact that $\phi_x(t)= exp(tx)$ which was checked to be holomorphic two posts ago. This is a homomorphism since $X$ is abelian.
Note that $d\psi_0\left(\frac{\partial}{\partial t}\Big|_0\right)=x+y$. By the uniqueness property of flows and exp just being a flow, $t\mapsto exp(tz)$ is the unique map with the property that the differential maps $\frac{\partial}{\partial t}\Big|_0\mapsto z$. Thus $\psi(t)=exp(t(x+y))$. Let $t=1$ and we get $exp(x)exp(y)=exp(x+y)$. i.e. $exp$ is a homomorphism.
Just as before, since $X$ is connected and $exp$ maps onto a neighborhood of the origin, the image is all of $X$. Let $U=ker(exp)$. We also saw two posts ago that there is a neighborhood of zero on which $exp$ is a diffeo and in particular is injective. Thus the $U$ is a discrete subgroup of $V$. But the only discrete subgroups of a vector space are lattices. This proves the proposition.
Corollary: $X$ is a complex torus.
Proof: We can holomorphically pass to the quotient and hence get a holomorphic isomorphism of groups $V/U\simeq X$.
Property 3: As a group $X$ is divisible and the $n$-torsion is isomorphic to $(\mathbb{Z}/n\mathbb{Z})^{2g}$ (recall that $g=dim_\mathbb{C}(X)$).
Proof: By property 2 we have that as a real Lie group $X\simeq (\mathbb{R}/\mathbb{Z})^{2g}=(S^1)^{2g}$. This proves both parts of property 3.
This is a good stopping point, since next time we’ll start thinking about the cohomology of $X$. | 2018-09-18T18:07:24 | {
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https://math.stackexchange.com/questions/1336681/infinite-sum-of-alternating-telescoping-series | # Infinite sum of alternating telescoping series
I am struggling to find the sum of the following series:
$$\sum_{n=1} ^{\infty} \frac{(-1)^n}{(n+1)(n+3)(n+5)}.$$
It seems as though it should be a straightforward telescoping series. I attempted to solve it in the usual way (via partial fractions), but the alternating sign makes the sum so that one cannot cancel out fractions to result in a finite sum of fractions. I know that the series converges by the alternating sign test, and I check on WolframAlpha that the infinite sum converges to $-7/480$. Any thoughts on how to proceed?
• Once you have the partial fraction decomposition, have you tried to compute it directly as a function of the sum $\sum_{n=1}^\infty \frac{(-1)^n}{n}$ (whose value is "known" to be $-\ln 2$ is you need it)? – Clement C. Jun 23 '15 at 19:28
We have (through the residue theorem or by linear algebra) : $$\frac{1}{(n+1)(n+3)(n+5)}=\frac{1}{8}\left(\frac{1}{n+1}-\frac{2}{n+3}+\frac{1}{n+5}\right)$$ as well as (by shifting the summation index): $$\sum_{n\geq 1}\frac{(-1)^n}{n+1}=-1+\log 2,$$ $$\sum_{n\geq 1}\frac{(-1)^n}{n+3}=-\frac{5}{6}+\log 2,$$ $$\sum_{n\geq 1}\frac{(-1)^n}{n+5}=-\frac{47}{60}+\log 2.$$ Just combine them. $\log 2$ cancels out since $1-2+1=0$ (we have a meromorphic function that is $O\left(\frac{1}{|z|^2}\right)$ as $|z|\to +\infty$, hence the sum of its residues is necessary zero).
• Could you explain the last part in further detail please? Specifically, from "just combine..." and on. I've never used a meromorphic function before, so I'm fairly confused. Thank you! – kathystehl Jun 23 '15 at 21:04
• @kathystehl: you may just ignore the part about the meromorphic function. It is just the intrinsic reason for which the $\log 2$ part has to cancel out, but you can check it in a straightforward way, i.e. $1-2+1=0$. – Jack D'Aurizio Jun 23 '15 at 21:06
• Could you identify what you are referring to when you parallel the $log2$ cancelling out to the identity $1-2+1=0$? – kathystehl Jun 23 '15 at 21:35
• About $(2)$, I am just saying that if you take the first line, minus twice the second line, plus the third line, there is no $\log$ in the final outcome. – Jack D'Aurizio Jun 23 '15 at 21:43
• I see, my apologies for the confusion. One last question (I promise): how do we deal with the $1/8$ when we separate the series into three parts as you did? – kathystehl Jun 23 '15 at 21:47
Rewrite the alternating sum as a difference of two infinite sums. $$\sum_{n=1} ^{\infty} \frac{(-1)^n}{(n+1)(n+3)(n+5)} = \sum_{n=1} ^{\infty} \frac{1}{(2n+1)(2n+3)(2n+5)}-\sum_{n=1} ^{\infty} \frac{1}{2n(2n+2)(2n+4)}$$ You'll probably wnat to convince yourself of the equality. Now you can use partial fraction decomposition to decompose $\frac{1}{(2n+1)(2n+3)(2n+5)}$ and $\frac{1}{2n(2n+2)(2n+4)}$ into three separate fractions each, giving you six infinite sums in total. From there you should be able to make some judicious cancellations and get your result. You may find one of my previous questions helpful at this point. Exact value of $\sum_{n=1}^\infty \frac{1}{n(n+k)(n+l)}$ for $k \in \Bbb{N}-\{0\}$ and $l \in \Bbb{N}-\{0,k\}$
Partial Decomposition provides:
$S = \sum_{1}^{\infty} (-1)^n[\frac{1}{8(n+1)} - \frac{1}{4(n+3)}+\frac{1}{8(n+5)}]$
Splitting S_Even and S_Odd
$S_{Even} = \frac{1}{8.3} - \frac{1}{4.5}+\frac{1}{8.5}$
$S_{Odd} = -\frac{1}{8.2} +\frac{1}{4.4} - \frac{1}{8.4}$
Everything else cancels out
When you sum these you get $S = \dfrac{-7}{480}$
Provided I have not made any calculation error | 2021-04-21T17:15:17 | {
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https://stats.stackexchange.com/questions/348726/how-can-we-calculate-the-probability-that-the-randomly-chosen-function-will-be-s/348738 | How can we calculate the probability that the randomly chosen function will be strictly increasing?
Consider the set of all functions from $\{1,2,...,m\}$ to $\{1,2,...,n\}$, where $n > m$. If a function is chosen from this set at random, what is the probability that it will be strictly increasing?
• @Carl: The question is perfectly clear. Given a set $X$ to be the domain and a set $Y$ to be the codomain, you can certainly talk about the set of all functions from $X$ to $Y$. This set is often denoted $Y^X$. In the OP, $X$ and $Y$ are finite, so $Y^X$ is finite as well ($|Y^X|=|Y|^{|X|}=n^m$), and you can talk about choosing from it uniformly randomly. May 29 '18 at 12:24
• @Carl: The language was exact. For some reason you're having difficulty parsing it. I don't know if it's because you're rusty on the language of set theory or something else. One common construction is: Given sets $X$ and $Y$, a function from $X$ to $Y$ is a set $f$ of ordered pairs $(x,y)$ where $x\in X$,$y\in Y$, such that for every $x\in X$ there is some $y\in Y$ so that $(x,y)\in f$, and if $(x_1,y_1)\in f, (x_2,y_2)\in f$ then $x_1\neq x_2$ or $y_1=y_2$. You can show that if $X=\{1,2,\ldots,m\}$ and $Y=\{1,2,\ldots,n\}$, then there are $n^m$ such functions; $\binom n m$ are increasing. May 30 '18 at 22:56
• e.g., for $m=n=2$, there are 4 functions from $\{1,2\}$ to $\{1,2\}$. The set of all such functions is $\{\{(1,1),(2,1)\},\{(1,1),(2,2)\},\{(1,2),(2,1)\},\{(1,2),(2,2)\}\}$. If you choose uniformly randomly from it, you have a probability of 25% of choosing a strictly increasing one (the only one is $\{(1,1),(2,2)\}$). May 30 '18 at 23:02
• @Carl: We're not talking about set functions. A set function takes a set as an input. We're talking about a function that takes a positive integer at most $m$ as input, and gives a positive integer at most $n$ as output. (Positive integers can be constructed as sets, but not necessarily). In other words, the domain is the set $\{1,2,\ldots, m\}$ and the codomain is the set $\{1,2,\ldots, n\}$. In other words, it's a function from $\{1,2,\ldots, m\}$ to $\{1,2,\ldots, n\}$. Jun 1 '18 at 13:24
• @Carl: As you said, and as is clear from the definition I gave above, given a function $f:\{1,2,\ldots,m\}\to\{1,2,\ldots,n\}$, for every $1\le i\le m$ there is exactly one $1\le j\le n$ such that $f(i)=j$. Or, to use, the construction of functions as a set of ordered pairs, for every $1\le i\le m$ there is exactly one $1\le j\le n$ such that $(i,j)\in f$. Jun 1 '18 at 13:25
Let us pick $m$ elements from $\{1,\dotsc,n\}$, let us call these $a_1 < a_2 < \dotsc , a_m$. Clearly these define a strictly increasing function $f$ from $\{1,\dotsc,m\} \to \{1,\dotsc,n\}$ via the rule $f(i) = a_i$. Furthermore, any strictly increasing function defined on the above sets is of this form.
Hence there are exactly ${n \choose m}$ strictly increasing functions. On the other hand, in total there are $n^m$ functions mapping between these two sets. Assuming that by "random" the OP means the uniform measure on the $n^m$ functions above, then the probability of picking a strictly increasing function is:
$$\frac{{n \choose m}}{n^m}$$
For example, for $n >> m$, an application of Stirling's approximation, shows that the RHS is $\approx \frac{1}{m!}$.
• $\frac{1}{m!}$ would be exactly correct if you wanted to know what proportion of 1-1 injective functions were strictly increasing. For $n \gg m^2$ the vast majority of all functions are 1-1 injective May 29 '18 at 14:33
Let $S(n,m)$ be the number of sub-arrays $1 \leqslant k_1 < k_2 < \cdots < k_m \leqslant n$ containing $m$ integer values that are increasing and are bounded by the values one and $n$. This binary function is well-defined for all integers $1 \leqslant m \leqslant n$, giving a triangular array of values. With a simple combinatorial argument$^\dagger$ we can establish the following recursive equations that define this binary function:
$$S(n+1,m) = S(n,m) + S(n,m-1) \quad \quad \quad \quad S(n,1) = n.$$
Solving this recursive equation gives us the explicit formula:
$$S(n,m) = {n \choose m} = \frac{n!}{m!(n-m)!}.$$
(There are other combinatorial arguments that also lead you to this result. For example, choosing an increasing function is equivalent to choosing $m$ values in the co-domain, which are then placed in increasing order.) Now, to get the result we need to be clear on exactly how a "random function" on this domain and co-domain is chosen. The simplest specification is to say that each possible mapping is chosen with equal probability, which means that there are $n^m$ equiprobable functions. Hence, the probability of interest is:
$$\mathbb{P}(\text{Increasing Function}) = \frac{n!}{m!(n-m)! \cdot n^m}.$$
Taking a first-order Stirling approximation for large $n$ gives $\mathbb{P}(\text{Increasing Function}) \approx 1/m!$, which is a very crude estimate that is suitable when $n$ is substantially larger than $m$. So basically, we see that once the co-domain in this problem is large, the probability of getting an increasing sequence at random is small; this accords with intuition.
$^\dagger$ If $m=1$ then we have only a single value in the mapping and every mapping to any of the $n$ places gives an increasing map. We therefore have $S(n,1)=n$ for all $n \in \mathbb{N}$. Moreover, the number of sub-arrays $S(n+1,m)$ includes all sub-arrays where the values occurs in the first $n$ places (there are $S(n,m)$ of these) and all the sub-arrays where the last value occurs in the last place and the remaining values occur before this (there are $S(n,m-1)$ of these). | 2022-01-28T00:09:11 | {
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http://kawahara.ca/how-to-compute-the-derivative-of-a-sigmoid-function-fully-worked-example/ | How to Compute the Derivative of a Sigmoid Function (fully worked example)
Last updated on October 2nd, 2017
This is a sigmoid function:
$\boldsymbol{s(x) = \frac{1}{1 + e^{-x}}}$
The sigmoid function looks like this (made with a bit of MATLAB code):
x=-10:0.1:10; s = 1./(1+exp(-x)); figure; plot(x,s); title('sigmoid');
Alright, now let’s put on our calculus hats…
Here’s how you compute the derivative of a sigmoid function
First, let’s rewrite the original equation to make it easier to work with.
$\boldsymbol{s(x) = \frac{1}{1+e^{-x}} = (1)(1+e^{-x})^{-1} = (1+e^{-x})^{-1}}$
Now we take the derivative:
$\boldsymbol{\frac{d}{dx}s(x) = \frac{d}{dx}((1+e^{-x})^{-1})}$
$\boldsymbol{\frac{d}{dx}s(x) = -1((1+e^{-x})^{(-1-1)}) \frac{d}{dx}(1+ e^{-x})}$
$\boldsymbol{\frac{d}{dx}s(x) = -1((1+e^{-x})^{(-2)}) (\frac{d}{dx}(1) + \frac{d}{dx}(e^{-x}))}$
$\boldsymbol{\frac{d}{dx}s(x) = -1((1+e^{-x})^{(-2)}) (0 + e^{-x}(\frac{d}{dx}(-x)))}$
$\boldsymbol{\frac{d}{dx}s(x) = -1((1+e^{-x})^{(-2)}) (e^{-x})(-1)}$
Nice! We computed the derivative of a sigmoid! Okay, let’s simplify a bit.
$\frac{d}{dx}s(x) = ((1+e^{-x})^{(-2)}) (e^{-x})$
$\frac{d}{dx}s(x) = \frac{1}{(1+e^{-x})^{2}} (e^{-x})$
$\frac{d}{dx}s(x) = \frac{(e^{-x})}{(1+e^{-x})^{2}}$
Okay! That looks pretty good to me. Let’s quickly plot it and see if it looks reasonable. Again here’s some MATLAB code to check:
x=-10:0.1:10; % Test values. s = 1./(1+exp(-x)); % Sigmoid. ds = (exp(-x))./((1+exp(-x)).^2); % Derivative of sigmoid. figure; plot(x,s,'b*'); hold on; plot(x,ds,'r+'); legend('sigmoid', 'derivative-sigmoid','location','best')
Looks like a derivative. Good! But wait… there’s more!
If you’ve been reading some of the neural net literature, you’ve probably come across text that says the derivative of a sigmoid s(x) is equal to s'(x) = s(x)(1-s(x)).
[note that $\frac{d}{dx}s(x)$ and s'(x) are the same thing, just different notation.]
[also note that Andrew Ng writes, f'(z) = f(z)(1 – f(z)), where f(z) is the sigmoid function, which is the exact same thing that we are doing here.]
So your next question should be, is our derivative we calculated earlier equivalent to s'(x) = s(x)(1-s(x))?
So, using Andrew Ng’s notation…
How does the derivative of a sigmoid f(z) equal f(z)(1-(f(z))?
Swapping with our notation, we can ask the equivalent question:
How does the derivative of a sigmoid s(x) equal s(x)(1-(s(x))?
Okay we left off with…
$\frac{d}{dx}s(x) = \frac{(e^{-x})}{(1+e^{-x})^{2}}$
This part is not intuitive… but let’s add and subtract a 1 to the numerator (this does not change the equation).
$\frac{d}{dx}s(x) = \frac{(e^{-x} + 1 -1)}{(1+e^{-x})^{2}}$
$\frac{d}{dx}s(x) = \frac{(1 + e^{-x} -1)}{(1+e^{-x})^{2}}$
$\frac{d}{dx}s(x) = \frac{(1 + e^{-x})}{(1+e^{-x})^{2}} - \frac{1}{(1+e^{-x})^{2}}$
$= \frac{1}{(1+e^{-x})} - \frac{1}{(1+e^{-x})^{2}}$
$= \frac{1}{(1+e^{-x})} - (\frac{1}{(1+e^{-x})}) (\frac{1}{(1+e^{-x})})$ // factor out a $\frac{1}{(1+e^{-x})}$
$= \frac{1}{(1+e^{-x})} (1 - \frac{1}{(1+e^{-x})})$
Hmmm…. look at that! There’s actually two sigmoid functions there… Recall that the sigmoid function is, $s(x) = \frac{1}{1 + e^{-x}}$. Let’s replace them with s(x).
$s'(x) = \frac{d}{dx}s(x) = s(x) (1 - s(x))$
Just like Prof Ng said… 🙂
And for a sanity check, do they both show the same function?
x=-10:0.1:10; % Test values. s = 1./(1+exp(-x)); % Sigmoid. ds = (exp(-x))./((1+exp(-x)).^2); % Derivative of sigmoid. ds1 = s.*(1-s); % Another simpler way to compute the derivative of a sigmoid. figure; plot(x,ds,'r+'); hold on; plot(x,ds1, 'go'); legend('(e^{-x})/((1+e^{-x})^2)','(s(x))(1-s(x))','location','best'); title('derivative of sigmoid')
Yes! They perfectly match!
So there you go. Hopefully this satisfies your mathematical curiosity of why the derivative of a sigmoid s(x) is equal to s'(x) = s(x)(1-s(x)).
23 thoughts on “How to Compute the Derivative of a Sigmoid Function (fully worked example)”
1. Jeremy says:
I think if you 1) rewrite my equation so the e^-x in the numerator goes to e^x in the denominator, 2) multiply my equation by e^x/e^x, and 3) expand the denominator in both the wolfram and my equation, they should be equal.
1. Sefrin says:
How come websites like wolfram alpha simply the -x in the exponents to positive x? How do we get there? I don’t see it.. :-/
1. Jeremy says:
Hi Sefrin, could you include an example/link to explain what you mean?
1. Jeremy says:
Thanks Vinay! You created a nice visual summary of different activation functions.
2. Scott Favorite says:
Actually you do use the product rule but it is part of the chain rule. Hope this is clear.
1. Jeremy says:
Hi Scott, thanks for your comment! I agree this is confusing/misleading. I re-wrote to remove the reference to the product rule.
3. Sid says:
Excellent walkthrough. For a guy just getting into activation fn’s, this really helps! Thanks so much!
1. Jeremy says:
You’re welcome Sid!
4. EASILY, the best blog post on finding the derivative of a sigmoid function. You didn’t leave any details out. Took me forever to wrap my head around this. The +1 – 1 thing is definitely not intuitive. Thanks for writing this.
1. Jeremy says:
happy to hear it helped!
5. Thanks! really helped with Prof. Hinton’s NNML Coursera lecture I was struggling to understand.
1. Jeremy says:
Glad it helped! It wasn’t obvious to me either 🙂
6. Balamurugan Balakrishnan says:
Superb!
7. Shubham Juneja says:
Very detailed. Thank you !!
1. Jeremy says:
You’re welcome!
1. Jeremy says:
Glad it helped clear things up!
8. Kevin Wang says:
excellent. Thanks! | 2018-08-22T05:09:50 | {
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https://math.stackexchange.com/questions/2576070/what-will-be-the-limit-points-of-the-set-s-1n-mid-n-in-mathbbn | # What will be the limit points of the set, $S=\{(-1)^{n} \mid n\in \mathbb{N}\}$
Since the set, S, turns out to be a finite set consisting of just two elements, i.e. $\{-1,1\}$, therefore there should be no limit points to the set.
But the solution given in the book is, the set S has two limit points $-1$ and $1$.
Which solution is correct?
• What is your definition of a limit point?
– Dave
Dec 21, 2017 at 19:33
• A limit point of a set, S is a point, x such that every neighbourhood of x (open sets containing x) contains a member of S other than x. Dec 21, 2017 at 19:36
• It depends on what topological space you are working in (and in particular what topology you are using), and how you define a limit point. Under the usual topology of $\Bbb R$ and your definition of limit points that you gave in your comment just now, any finite set has no limit points. Dec 21, 2017 at 19:40
• In your definition of a limit point, the set $S$ has no limit points, because the ball of radius $\frac{1}{2}$ around $1$ or $-1$ does not contain the other point. One can define (as is done at my school) a limit point as one in which any neighbourhood of the point contains a point in $S$, and under this definition both $1$ and $-1$ are limit points of $S$.
– Dave
Dec 21, 2017 at 19:40
• Your book is probably talking about the sequence $\{(-1)^n\}$, not the set $S$.
– user9464
Dec 21, 2017 at 19:41
The notation $\{(-1)^n\mid {n\in{\bf N}}\}$, understood as a set, is the same as $\{-1,1\}$, which has no limit point.
However, in some context, $\{(-1)^n\mid {n\in{\bf N}}\}$ is used as a (bad) notation for the real sequence $(a_n)_{n=1}^\infty$ with $a_n:=(-1)^n$. In this case,
$x$ is a "limit point" of the sequence $(a_n)_{n=1}^\infty$
means
$x$ is the limit of some convergent subsequence of $(a_n)_{n=1}^\infty$.
Note that both $1$ and $-1$ are limits of some subsequences of the sequence $((-1)^n)_{n=1}^\infty$.
For more general discussions, see the Wikipedia article on Limit point.
• But still the neighbourhood of say 1 does not contain any of the point of the set except for 1. Dec 21, 2017 at 20:10
• @AjayChoudhary The idea here is that the definition of limit point for a set and limit point for a sequence are different - in particular, $\{-1,1\}$ has no limit points, but the sequence $((-1)^n)_{n=1}^{\infty}$ does have a limit point, because "limit point" means two different things in either situation. Dec 21, 2017 at 20:14 | 2022-10-05T21:51:20 | {
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https://math.stackexchange.com/questions/3190632/how-do-i-formally-show-the-radius-of-convergence-of-the-taylor-series-of-fx-x | # How do I formally show the radius of convergence of the Taylor series of $f(x)=x^6 - x^4 + 2$ at $a=-2$?
This is an exercise in Stewart's Calculus (Exercise 19, Section 11.10 Taylor and Maclaurin Series):
Find the Taylor series for $$f(x)$$ centered at the given value of a. [Assume that f has a power series expansion. Do not show that $$R_n(x) \to 0.$$ Also find the associated radius of convergence.
Here $$f(x)=x^6 - x^4 + 2$$ and $$a=-2$$.
I'm having trouble finding a general formula of this Taylor series and therefore, also having problems finding the radius of convergence since I can't perform the ratio test.
Here is what I know: \begin{align} f'(x) = 6x^5 - 4x^3,\quad &f''(x) = 30x^4 - 12x^2,\\ f'''(x) = 120x^3 - 24x,\quad &f^{(4)}(x) = 360x^2 - 24,\\ f^{(5)}(x) = 720x,\quad &f^{(6)}(x) = 720. \end{align}
And at $$a=-2$$, \begin{align} f(-2) = 50,\quad &f'(-2) = -160,\\ f''(-2) = 432,\quad &f'''(-2) = -912,\\ f^{(4)}(-2) = 1416,\quad &f^{(5)}(-2) = -1440,\\ f^{(6)}(-2) = 720.\quad & \end{align}
I'm having trouble finding the general formula for each term. Without it, how am I supposed to find the radius of convergence?
Added: So the general term I have for the n-th derivative of $$f$$ is:
$$f^{(n)}(x) = \frac{6!x^{6-n}}{(6-n)!}$$
So far the general term I have for the Taylor Series is:
$$\sum_{n=0}^{\infty} \frac{6! 2^{6-n}}{(6-n)!n!}(x+2)^n$$
I can see why the radius of convergence is $$\infty$$: because for any $$x$$the series converges.
But how do I show this formally? Can I use the ratio test?
Your general form of the derivative is wrong. It is valid only up to $$n=6$$. After that it is $$0$$. So your sum consist only of terms up to $$(x+2)^6$$. You therefore have a finite sum, not an infinite number of terms, neither of which diverges.
• Mind writing it out? The general term? I'm a bit lost Apr 17 '19 at 4:33
• The general term for $n>6$ is $f^{(n)}=0$. You have the first six terms already calculated. Just calculate the next derivative. The derivative of a constant is ... I just realized that you general formula does not even apply to the first 4 terms, just for 5 and 6. Apr 17 '19 at 4:40
The Taylor series around $$a$$ is simply given by \begin{align}f(a+x)&=(a+x)^6-(a+x)^4+2\\&=(a^6-a^4+2)+(6a^5-4a^3)x+(15a^4-6a^2)x^2+(20a^3-4a)x^3+(15a^2-1)x^4+6ax^5+x^6\end{align} or $$f(x)= (a^6-a^4+2)+(6a^5-4a^3)(x-2)+(15a^4-6a^2)(x-a)^2+(20a^3-4a)(x-a)^3+(15a^2-1)(x-a)^4+6a(x-a)^5+(x-a)^6$$
and, as almost all coefficients are $$=0$$, converges for all $$x$$.
Sticking to the definition helps. Recall that the Taylor series of $$f$$ centered at a given value of $$a$$ is by definition
$$f(x)=\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n =f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^2+\cdots\tag{1}$$
Now, given the expression of $$f$$, namely, $$f(x)=x^6-x^4+2$$, and value of $$a=-2$$, all you need to do is finding $${f^{(n)}(-2)}$$ for each $$n=0,1,2,\cdots$$. You have already found the values of $$f^{(n)}(-2)$$ for $$n=0,1,\cdots,6$$, which gives you the first seven terms in the Taylor series: $$f(-2)+\frac{f'(-2)}{1!}(x+2)+\frac{f''(-2)}{2!}(x+2)^2+\cdots+\frac{f^{(6)}(-2)}{3!}(x+2)^2.\tag{2}$$ On the other hand, $$f^{(n)}(a)=0$$ for any integer $$n>6$$. Hence, by (1), the Taylor series for $$f$$ at $$a=-2$$ is $$f(-2)+\frac{f'(-2)}{1!}(x+2)+\frac{f''(-2)}{2!}(x+2)^2+\cdots+\frac{f^{(6)}(-2)}{3!}(x+2)^2+0+0+0+\cdots\tag{3}$$ which is a finite sum, and of course it is convergent for every $$x$$.
Note that (3) can be written formally as a power series, which seems to be what you were looking for, as follows: $$\sum_{n=0}^\infty c_n(x+2)^n\quad\textrm{where } c_0=f(-2),\ c_1=\frac{f'(-2)}{1!},\cdots, c_6=\frac{f^{(6)}(-2)}{6!},\ c_7=c_8=\cdots =0.\tag{4}$$ You do not need Root Test or Ratio Test to get the radius of convergence but only need to look at the definition of the "radius of convergence". See the definition below Theorem 4 of Section 11.8 in Stewart's Calculus (your textbook), in particular case (ii): | 2021-10-16T19:03:18 | {
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https://www.physicsforums.com/threads/laplace-transform-f-t-tcos-t.316314/ | # Laplace Transform f(t)=tcos(t)
## Homework Statement
I need to find the laplace transform of f(t)=tcos(t).
## Homework Equations
$$\int e^-^s^ttcos(t)dt$$
## The Attempt at a Solution
I just need help on how to integrate this. I can find the answer easily using the f(t)=tcos(kt) general formula but I wish to find it directly.
Last edited:
Cyosis
Homework Helper
Use the complex form of the cosine.
Thank you :)
I get $$\mathcal{L} [tcost] = \frac {1}{2} ( \frac {1}{(i-s)^{2}} + \frac {1}{(i+s)^{2}})$$
Hopefully that simplifies to $$\frac {s^{2}-1}{(s^{2}+1)^{2}}$$
Last edited:
Integration by parts also works but thanks :D
Integration by parts also works but thanks :D
Even if you use the complex form of cos(t), you still have to integrate by parts (albeit it's rather simple).
$$\frac {1}{2} \int^{ \infty}_{0} e^{-st}t \frac {1}{2}(e^{it} + e^{-it})dt$$
$$= \frac {1}{2} \int^{ \infty}_{0} te^{(i-s)t}dt + \frac {1}{2} \int^{ \infty}_{0} te^{-(i+s)t}dt$$
You don't need to integrate by parts at all. Just compute the Laplace transform of cos(t). Differentiating with respect to s will then bring down a factor of minus t in the Laplace integral.
You don't need to integrate by parts at all. Just compute the Laplace transform of cos(t). Differentiating with respect to s will then bring down a factor of minus t in the Laplace integral.
Cool shortcut. I've never seen it before.
$$\mathcal{L} [cos(t)] = \int^{ \infty}_{0} e^{-st}cos(t)dt$$
$$\frac {d}{ds} \mathcal{L} [cos(t)] = \frac {d}{ds} \int^{ \infty}_{0} e^{-st}cos(t)dt$$
Assuming it's okay to bring the derivative inside the integral,
$$= \int^{ \infty}_{0} \frac {d}{ds} e^{-st}tcos(t)dt$$
$$= -\int^{ \infty}_{0}te^{-st}cos(t)dt$$
$$= -\int^{ \infty}_{0}e^{-st}tcos(t)dt = - \mathcal{L} [tcos(t)]$$
then it would appear that $$\frac {d^{2}}{ds^{2}} \mathcal{L} [cos(t)] = \mathcal{L} [t^{2}cos(t)]$$
Another trick: To obtain the Laplace transform of, say, sin(t)/t you can compute the Laplace transform of sin(t) and then integrate w.r.t. s from p to infinity. The Laplace transform is then a function of the parameter p. If you put p = 0, you obtain the integral of sin(t)/t from zero to infinity. What's interesing about this is that the antiderivative of sin(t)/t cannot be evaluated in closed form.
Another trick: To obtain the Laplace transform of, say, sin(t)/t you can compute the Laplace transform of sin(t) and then integrate w.r.t. s from p to infinity. The Laplace transform is then a function of the parameter p. If you put p = 0, you obtain the integral of sin(t)/t from zero to infinity. What's interesing about this is that the antiderivative of sin(t)/t cannot be evaluated in closed form.
I have to see for myself.
$$\mathcal{L} [sin(t)] = \int^{ \infty}_{0} e^{-st}sin(t)dt$$
$$\int^{ \infty}_{p} \mathcal{L} [sin(t)] = \int^{ \infty}_{p} \int^{ \infty}_{0} e^{-st}sin(t)dtds$$
Changing the order of integration (which I assume is allowed),
$$= \int^{ \infty}_{0} \int^{ \infty}_{p} e^{-st}sin(t)dsdt$$
$$= \int^{ \infty}_{0} \frac {1}{t}e^{-pt}sin(t)dt$$
$$= \int^{ \infty}_{0} e^{-pt} \frac {sin(t)}{t}dt = \mathcal{L} [ \frac {sin(t)}{t}]$$
According to the table, $$\mathcal{L} [ \frac {sin(t)}{t}] = arctan( \frac {1}{p})$$
Should you then take the limit of $$arctan( \frac {1}{p})$$ as p goes to zero?
$$\int^{ \infty}_{0} \frac {sin(t)}{t}dt = \lim_{p \to 0} arctan( \frac {1}{p}) = \frac {\pi}{2}$$ ?
You can derive that the Laplace transform of sin(t)/t is arctan(1/p) by integrating the Laplace transform of sin(t). If you integrate
1/(s^2+1) from p to infinity, you get pi/2 - arctan(p). | 2022-05-16T14:37:02 | {
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# If a circle is inscribed in an equilateral triangle, what is the area
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If a circle is inscribed in an equilateral triangle, what is the area of the triangle NOT taken up by the circle?
(1) The area of the circle is 12π
(2) The length of a side of the triangle is 12
[Reveal] Spoiler: OA
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03 Mar 2016, 00:25
If a circle is inscribed in an equilateral triangle then this property exist-
R = A * [ (\sqrt{3}) / 6 ]
A= Side of equilateral triangle
S1= R is given
S2= A is given
Now we can easily calculate corresponding areas of circle and triangle
IMO- D
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Bunuel wrote:
If a circle is inscribed in an equilateral triangle, what is the area of the triangle NOT taken up by the circle?
(1) The area of the circle is 12π
(2) The length of a side of the triangle is 12
Following is the formula that tells the relation between the radius of the circle and the side of the triangle:
r = a * ($$\sqrt{3}$$/ 6)
So if we have either of the side or the radius, we can find the other thing.
The relation can be found by using the figure below:
Attachment:
circle in triangle.JPG [ 16.87 KiB | Viewed 2286 times ]
Statement 1: The area of the circle is 12π
We can find the radius of the circle and hence the side of the triangle and the corresponding area
Therefore we can find the difference between the areas
SUFFICIENT
Statement 2: The length of a side of the triangle is 12
We can find the radius of the circle and hence the area of the circle
Therefore we can find the difference between the areas
SUFFICIENT
Option D
NOTE: We do not need to find the area. There is not need to do the calculations.
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03 Mar 2016, 03:47
TeamGMATIFY wrote:
Bunuel wrote:
If a circle is inscribed in an equilateral triangle, what is the area of the triangle NOT taken up by the circle?
(1) The area of the circle is 12π
(2) The length of a side of the triangle is 12
Following is the formula that tells the relation between the radius of the circle and the side of the triangle:
r = a * ($$\sqrt{3}$$/ 6)
So if we have either of the side or the radius, we can find the other thing.
The relation can be found by using the figure below:
Attachment:
circle in triangle.JPG
Statement 1: The area of the circle is 12π
We can find the radius of the circle and hence the side of the triangle and the corresponding area
Therefore we can find the difference between the areas
SUFFICIENT
Statement 2: The length of a side of the triangle is 12
We can find the radius of the circle and hence the area of the circle
Therefore we can find the difference between the areas
SUFFICIENT
Option D
NOTE: We do not need to find the area. There is not need to do the calculations.
HI TeamGMATIFY,
Actually I already learnt this formula but haven't tried to figure it out the logic behind this.
The figure provided by you makes a triangle of 30-60-90 = 1x : Root 3* x : 2x
So the opposite length of side 30 degree angle --> R
=> 1x= R or x=R
And opposite length of side 60 degree angle--> A/2
=> root 3 * x = A/2
By this I am getting relation as --
R= A / (root 3 * 2) which is not the same
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03 Mar 2016, 04:38
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PrakharGMAT wrote:
HI TeamGMATIFY,
Actually I already learnt this formula but haven't tried to figure it out the logic behind this.
The figure provided by you makes a triangle of 30-60-90 = 1x : Root 3* x : 2x
So the opposite length of side 30 degree angle --> R
=> 1x= R or x=R
And opposite length of side 60 degree angle--> A/2
=> root 3 * x = A/2
By this I am getting relation as --
R= A / (root 3 * 2) which is not the same
Once you have got R= A / (root 3 * 2)
You need to rationalize the denominator, or simply remove the under root from the denominator
Multiply $$\sqrt{3}$$ on both numerator and denominator and you will get
R= A $$\sqrt{3}$$/ 6
Does this help?
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Re: If a circle is inscribed in an equilateral triangle, what is the area [#permalink]
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Hello,
Can someone elaborate on this question without the use of a formula?
Thanks
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Re: If a circle is inscribed in an equilateral triangle, what is the area [#permalink] 23 Aug 2017, 08:41
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http://mathhelpforum.com/calculus/218523-sketching-curve-graph-3.html | 1. ## Re: sketching curve graph
Okay good, you have found:
x-intercepts: (-2,0), (-1,0), (1,0)
y-intercept: (0,-2)
Now, can you answer parts a) - c) above? Can you use both the first and second derivative tests to clearly show what the nature of the extrema are? While one one of these test is sufficient, I think is is good in this case to look at both just for an understanding of both methods.
In the end, you will have two extrema, a point of inflection, and 4 intercepts, for a total of 7 points to plot...you will know where the function is increasing/decreasing, and where the function is concave up and concave down. You should be able to make a good sketch of the original function from all of this information.
2. ## Re: sketching curve graph
ok, i plotted all the points how do i graph the critical numbers from the first derivative test. they were roots.
this is what my graph looks like so far
https://www.dropbox.com/s/12du0m48t1...2016.34.54.jpg is it correct?
4. ## Re: sketching curve graph
Originally Posted by noork85
I apologize for the delay in getting back to you...severe storms in my area knocked out my power for most of the night.
You have the correct intercepts and point of inflection labeled, but your relative minimum is not the y-intercept, and as far as plotting the extrema, use your calculator to get decimal approximations.
Can you first just state what you have found for parts a) - c)? Being able to state these will help you sketch the graph, and I highly recommend that we get those nailed down first before attempting to sketch the graph.
5. ## Re: sketching curve graph
i hope all is well....
ok so heres what i coud answer from a-c. i dont understand which points are my local max/min. i mean, i dont know where im supposed to look for them.
https://www.dropbox.com/s/v3v51fukz5...2012.52.01.jpg
7. ## Re: sketching curve graph
here's what i did for the final graph...includes all my work. not sure about concavity though
https://www.dropbox.com/s/7p5hq991lc...2013.39.34.jpg
https://www.dropbox.com/s/iibkdwpc0j...2013.39.44.jpg
https://www.dropbox.com/s/u91syuxyk1...2013.39.53.jpg (not the best looking graph)
8. ## Re: sketching curve graph
I didn't have much time (power was out again), but in looking over all of your work, my only point of contention is the intervals of increasing/decreasing behavior. You included zero, and this is not one of your critical values.
I have to run now, but when I get back, I will show you how I would work this problem, from start to finish.
9. ## Re: sketching curve graph
ok great, thanks.
10. ## Re: sketching curve graph
We are given:
$f(x)=x^3+2x^2-x-2$
a) find where $f(x)$ is increasing/decreasing.
To do this, we need to compute the first derivative of $f(x)$ and look at where it is positive/negative by equating it to zero.
$f'(x)=3x^2+4x-1=0$
The discriminant is:
$\Delta=(4)^2-4(3)(-1)=28$
This is not a perfect square, so the roots are irrational, therefore factoring is not a practical option for finding the roots, hence we apply the quadratic formula:
$x=\frac{-4\pm\sqrt{28}}{2(3)}=\frac{-4\pm2\sqrt{7}}{2(3)}=\frac{-2\pm\sqrt{7}}{3}$
I would simply observe that the first derivative is a parabola opening upwards to determine it must be negative in between the roots, and positive on either side, thus:
On the interval:
$\left(-\infty,\frac{-2-\sqrt{7}}{3} \right)$ we find $f(x)$ is increasing.
$\left(\frac{-2-\sqrt{7}}{3},\frac{-2+\sqrt{7}}{3} \right)$ we find $f(x)$ is decreasing.
$\left(\frac{-2+\sqrt{7}}{3},\infty \right)$ we find $f(x)$ is increasing.
b) where it is concave up/concave down.
To discuss concavity, we need to look at the second derivative of $f(x)$, in particular it's sign.
$f''(x)=6x+4=0$
Solving for $x$, we find the critical value is at:
$x=-\frac{2}{3}$
Since the second derivative is an increasing linear function, we know then that:
On the interval:
$\left(-\infty,-\frac{2}{3} \right)$ we find $f(x)$ is concave down.
$\left(-\frac{2}{3},\infty \right)$ we find $f(x)$ is concave up.
Because the second derivative does change its sign across the critical value, we may conclude there is a point of inflection at:
$\left(-\frac{2}{3},f\left(-\frac{2}{3} \right) \right)=\left(-\frac{2}{3},-\frac{20}{27} \right)$
c) local max/min
We know from part a) that we have stationary points at:
i) $\left(\frac{-2-\sqrt{7}}{3},f\left(\frac{-2-\sqrt{7}}{3} \right) \right)=\left(\frac{-2-\sqrt{7}}{3},\frac{14\sqrt{7}-20}{27} \right)$
ii) $\left(\frac{-2+\sqrt{7}}{3},f\left(\frac{-2+\sqrt{7}}{3} \right) \right)=\left(\frac{-2+\sqrt{7}}{3},\frac{-14\sqrt{7}-20}{27} \right)$
First derivative test:
Because $f(x)$ is increasing to the left of i) and decreasing to the right, we may conclude that i) is a local maximum.
Because $f(x)$ is decreasing to the left of ii) and increasing to the right, we may conclude that ii) is a local minimum.
Second derivative test:
Since $\frac{-2-\sqrt{7}}{3}<-\frac{2}{3}<\frac{-2-\sqrt{7}}{3}$
We know then that:
$f''\left(-2-\sqrt{7}}{3} \right)<0$ and so i) is a local maximum.
$f''\left(-2+\sqrt{7}}{3} \right)>0$ and so ii) is a local minimum.
d) sketch the curve of the graph.
We know we have:
a local maximum at:
$\left(\frac{-2-\sqrt{7}}{3},\frac{14\sqrt{7}-20}{27} \right)\approx(-1.5485837703548635,0.6311303094409)$
a local minimum at:
$\left(\frac{-2+\sqrt{7}}{3},\frac{14\sqrt{7}+20}{27} \right)\approx( 0.21525043702153024,-2.1126117909223803)$
a point of inflection at:
$\left(-\frac{2}{3},-\frac{20}{27} \right)\approx( -0.6666666666666666,-0.7407407407407407)$
To find the $x$-intercepts, equate $f(x)=0$ and find the roots:
$f(x)=x^3+2x^2-x-2=x^2(x+2)-(x+2)=$
$(x^2-1)(x+2)=(x+2)(x+1)(x-1)=0$
So, we know the $x$-intercepts are at:
$(-2,0),\,(-1,0),\,(1,0)$
To find the $y$-intercept, we let $x=0$ and find:
$f(0)=-2$
and so we know the $y$-intercept is at:
$(0,-2)$
Putting all of this together, we should obtain a graph that looks like:
11. ## Re: sketching curve graph
thank you so much. i understand it so much better now. i was afraid i would fail the graphing part of my exam tomorrow, but im fairly confident i can do this. thank u again!!!
Page 3 of 3 First 123 | 2013-12-13T07:51:34 | {
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http://mathhelpforum.com/statistics/192359-two-dice-probability-sum-scores-9-individual-scores-differ-1-a-print.html | Two dice - probability that sum of scores > 9 or individual scores differ by 1.
• November 20th 2011, 06:04 PM
Punch
Two dice - probability that sum of scores > 9 or individual scores differ by 1.
Two fair dice are thrown together, and the scores added. What is the probability that the total score is 9, or the individual scores differ by 1, or both?
$P(total score is 9)=\frac{4}{36}=\frac{1}{9}$
$P(individual scores differ by 1)=P{(1,2), (2,3), (2,1), (3,4), (3,2), (4,5),
(4,3), (5,6), (5,4), (6,5)}=\frac{10}{36}=\frac{5}{18}$
$P(both scoring 9 and score differ by 1)=P{(4,5),(5,4)}=\frac{2}{36}=\frac{1}{18}$
Should i treat the question as three seperate parts? The model answer has only 1 answer of 1/3 and that doesn't coincide with any of my answers...
• November 20th 2011, 09:31 PM
Soroban
Re: Two dice - probability that sum of scores > 9 or individual scores differ by 1.
Hello, Punch!
Quote:
Two fair dice are thrown together, and the scores added.
What is the probability that the total score is 9,
or the individual scores differ by 1, or both?
There are only 36 possible outcomes.
Why not list them and count the desired outcomes?
. . $\begin{array}{cccccc}(1,1) & {\color{red}(1,2)} & (1,3) & (1,4) & (1,5) & (1,6) \\ {\color{red}(2,1)} & (2,2) & {\color{red}(2,3)} & (2,4) & (2,5) & (2,6) \\ (3,1) & {\color{red}(3,2)} & (3,3) & {\color{red}(3,4)} & (3,5) & {\color{blue}(3,6)} \\ (4,1) & (4,2) & {\color{red}(4,3)} & (4,4) & {\color{blue}(4,5)} & (4,6) \\ (5,1) & (5,2) & (5,3) & {\color{blue}(5,4)} & (5,5) & {\color{red}(5,6)} \\ (6,1) & (6,2) & {\color{blue}(6,3)} & (6,4) & {\color{red}(6,5)} & (6,6) \end{array}$
There are $12$ desired outcomes.
The probability is:. $\frac{12}{36} \,=\,\frac{1}{3}$
• November 20th 2011, 10:35 PM
takatok
Re: Two dice - probability that sum of scores > 9 or individual scores differ by 1.
While enumerating every possibility and finding the answer by counting is possible, it can become unwieldy for large cases. For example what if we were using 100 sided dice instead of 6. You were working on the correct solution you just missed it by one step. I will use your probabilities since they are correct.
$P(9) = \frac{1}{9} = \frac{2}{18}$
$P(diff1) =\frac{5}{18}$
So to get the probability of both we just add them to get $\frac{7}{18}$
This is a little off. The problem is when your finding the probabilites of 2 different subsets of the same set, sometimes the members of the 2 different subsets overlap, and you end up counting them twice. In this case 4,5 and 5,4 are counting both times in P(9) and P(diff1). So the actual probability of Event A OR Event B happening is:
P(A) + P(B) - P(A $\cap$ B)
A $\cap$ B just mean that both are true.
So the P(Both) = $\frac{1}{18}$
So..
$\frac{2}{18}+\frac{5}{18}-\frac{1}{18} = \frac{6}{18} = \frac{1}{3}$ | 2015-01-29T12:20:33 | {
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https://kdowbecki.github.io/Optimizing-computational-algorithms/ | # Optimizing computational algorithms
Karol Dowbecki · May 29, 2021
When optimizing an algorithm it’s often good to take a step back and think about the problem domain before jumping straight to coding. Today I have seen following question asked on Stack Overflow:
I have a program that has two nested for loops and takes $$O(n^2)$$ time. I want to know if there is such way to decrease the time of execution for it. Example:
long b = 10000;
long counter = 0;
for (int k = 0; k < b; k++) {
for (int a = k; a < b; a++) {
counter += k;
}
}
At first glance we immediately notice that in the above code variable k is added repeatedly. Perhaps we could remember the counter variable value from the previous loop iteration? Perhaps we could create another variable… Here we should take a step back.
We should notice that the code can be represented as a mathematical equation. Let’s write it down:
$counter = \sum\limits_{k = 0}^{b-1} \sum\limits_{a = k}^{b-1} k$
We have two nested sums, can we solve this equation?
The sum on the right is $$k$$ added from $$k$$ to $$b-1$$ (inclusive) times. This means that in total $$k$$ is going to be added $$b - 1 - k + 1 = b - k$$ times. Knowing that we can expand the sum on the right:
$\sum\limits_{k = 0}^{b-1} \sum\limits_{a = k}^{b-1} k = \sum\limits_{k = 0}^{b-1} [k(b-k)] = \sum\limits_{k = 0}^{b-1} (kb-k^2) = \sum\limits_{k = 0}^{b-1} kb - \sum\limits_{k = 0}^{b-1} k^2 = b \sum\limits_{k = 0}^{b-1} k - \sum\limits_{k = 0}^{b-1} k^2$
We have converted two nested sums into two separate sums and in the process removed $$a$$ which is an improvement. To solve the new sums we can use known summation formulas, often attributed to Gauss:
$\sum\limits_{a = 1}^{n} a = \frac{n (n+1)}{2} \newline \sum\limits_{a = 1}^{n} a^2 = \frac{n (n+1) (2n+1)}{6}$
If we apply these summation formulas to our equation:
$b \sum\limits_{k = 0}^{b-1} k - \sum\limits_{k = 0}^{b-1} k^2 = b \frac{(b-1) (b-1+1)}{2} - \frac{(b-1) (b-1+1) [2(b-1) + 1]}{6} = \newline = b \frac{(b-1) b}{2} - \frac{(b-1) b (2b-1)}{6} = \frac{b^3- b^2}{2} - \frac{(b^2-b) (2b-1)}{6} = \newline = \frac{b^3- b^2}{2} - \frac{2b^3 - b^2 - 2b^2 + b}{6} = \frac{3b^3 - 3b^2}{6} - \frac{2b^3 - 3b^2 + b}{6} = \newline = \frac{3b^3 - 3b^2 - 2b^3 + 3b^2 - b}{6} = \frac{b^3 - b}{6}$
We have significantly simplified the original equation into:
$counter = \frac{b^3 - b}{6}$
which we can implement as a one-liner:
long counter = (b * b * b - b) / 6;
Are you not believing that this works? You can try the code yourself:
public static void main(String[] args) {
for (long b = 0; b <= 10000; b++) {
long counter = original(b);
long fastCounter = fast(b);
System.out.printf("%d\t%d\t%d%n", b, counter, fastCounter);
if (counter != fastCounter) {
System.out.println("Failed");
return;
}
}
System.out.println("Worked");
}
private static long original(final long b) {
long counter = 0;
for (int k = 0; k < b; k++) {
for (int a = k; a < b; a++) {
counter += k;
}
}
return counter;
}
private static long fast(final long b) {
return (b * b * b - b) / 6;
}
you will see:
9995 166416789980 166416789980
9996 166466744990 166466744990
9997 166516709996 166516709996
9998 166566684999 166566684999
9999 166616670000 166616670000
10000 166666665000 166666665000
Worked
This new fast algorithm has a runtime complexity of just $$O(1)$$ and great performance as it needs only 4 arithmetical operations. We should strive to improve the new algorithm further, taking care of edge cases like integer overflow, but who knows, this could be the optimal solution for the problem. One way or another, it’s good to take a step back when dealing with algorithms.
Twitter, Facebook | 2021-10-25T18:23:51 | {
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https://math.stackexchange.com/questions/3533451/will-there-at-some-point-be-more-numbers-with-n-factors-than-prime-numbers-for | # Will there at some point be more numbers with $n$ factors than prime numbers for any $n$? [duplicate]
Let $$\pi(x)$$ be the prime counting function: the number of numbers $$\leq x$$ with just one prime factor. Let $$\pi_n(x)$$ count the number of numbers $$\leq x$$ with exactly $$n$$ prime factors (counted with multiplicity).
When plotting the values for different $$n$$ up to large $$x$$, it seems that for every $$n$$ there will be a point from where on $$\pi_n(x) > \pi(x)$$. I'm conflicted about my intuition on this. On one hand, it seem's to be plausible because numbers with more and more prime factors become more and more common. On the other hand, numbers with, say, 100 prime factors seem like they are so rare that they will be always less than $$\approx \frac{x}{\log x}$$
My question is: Will there at some point be more numbers with $$n$$ factors than prime numbers for any $$n$$?
Bonus question for positive answer: Is there a way to find the point this happens for a given $$n$$ other than the naive approach?
Bonus question for negative answer: What is the largest $$n$$ which surpasses the prime counting function?
$$\pi_n(x)$$ for $$x \lt 100$$, at $$\approx 25$$ there are more numbers with $$2$$ factors than prime numbers.
$$\pi_n(x)$$ for $$x \lt 100.000$$, at $$\approx 40000$$ there are more numbers with $$5$$ factors than prime numbers.
$$\pi_n(x)$$ for $$x \lt 10.000.000$$, at $$\approx 4.000.000$$ there are more numbers with $$6$$ factors than prime numbers.
(Please ignore the 'Divisors' in the chart legend, it should read 'Factors')
• $\pi_n(x)$ is the number of numbers $≤x$ with exactly $n$ prime factors or at least $n$ prime factors? – Conifold Feb 3 at 23:58
• @Conifold with exactly $n$ prime factors – SmallestUncomputableNumber Feb 4 at 0:01
• @Barry Cipra: two, I'll edit the question – SmallestUncomputableNumber Feb 4 at 7:17
Interesting question. My intuition is that $$\pi_n(x)>\pi(x)$$ for large $$x$$.
Heuristic argument: We note that, for each $$n$$, the sum of the reciprocals of the $$n-$$primes diverges. (here, of course, an $$n-$$prime means a natural number with exactly $$n$$ prime divisors). This is clear since we can just pick some $$(n-1)-$$prime $$A$$ and then note that every number of the form $$Ap$$ for prime $$p$$ is an $$n-$$prime, and of course $$\sum_{\text {p prime}}\frac 1{Ap}$$ diverges.
Now let $$\{A_1, A_2,\cdots \}$$ be the $$(n-1)-$$primes. Then we can write (speaking roughly) $$\pi_n(x)≥\pi\left(\frac x{A_1}\right) +\pi\left(\frac x{A_2}\right)+\cdots$$
And for any $$A$$ we have $$\pi\left(\frac x{A}\right)\sim \frac 1{A}\frac x{\ln x-\ln A}≥\frac 1{A}\frac x{\ln x}\sim \frac 1A\times \pi(x)$$
Thus for large $$x$$ :$$\pi_n(x)≥\pi(x)\times \left(\frac 1{A_1}+\frac 1{A_2}+\cdots\right)$$ And we just need to use enough of the $$A_i$$ to get that sum over $$1$$.
I think that for large enough $$x$$ it shouldn't be too hard to make this argument more solid. As to getting a good estimate of the crossing point, well that sounds tougher. If you need $$A_1, \cdots, A_N$$ to get the sum over $$1$$ then I'd start to look around $$x=A_N$$ but, of course, the estimates we are relying on are very unreliable for small arguments so it's hard to imagine that this will work terribly well. The argument will be on much surer ground when $$\frac x{A_N}$$ is large.
Worth noting: these divergent series do not diverge rapidly. For $$3-$$primes I note that you need to go out to $$A_{96}=402$$ before the sum of the reciprocals exceeds $$1$$.
• just naively, the number of semiprimes up to $n$ is always greater than $({\sqrt{n}\over \ln\sqrt{n}})^2$ – user645636 Feb 4 at 1:00
• @RoddyMacPhee Again, you mean for large enough $n$. And you can get a similar formula for $n-$primes. But I think that you get sharper bounds out of terms like $\lambda x$ then out of terms like $x^{1/n}$, but of course I could have that wrong. – lulu Feb 4 at 1:08
• all I know is I plugged it into the prime number theorem. if you look at sundaram, and apply a few things you might be able to coax things out. I know $$\sqrt{n\over 4}\approx a, (2a+1)^2\approx n$$ by that. – user645636 Feb 4 at 1:19
• @RoddyMacPhee Not sure I see where you are going with that. I don't immediately see how to turn your sort of lower bound into a heuristic argument. Your expression is $\frac {4n}{(\ln n)^2}$ which is not greater than $\frac n{\ln n}$ except for very small $n$. But of course I might be missing something. – lulu Feb 4 at 1:32
• you can sieve them using sundaram. it's a lower bound. – user645636 Feb 4 at 1:37 | 2020-03-28T21:23:09 | {
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https://math.stackexchange.com/questions/2802933/prove-or-disprove-that-for-any-n-in-mathbbn-there-exist-a-b-in-math | Problem
Prove or disprove that, for any $n \in \mathbb{N_+}$, there exist $a,b \in \mathbb{N_+}$ such that $$\frac{a^2+b}{a+b^2}=n.$$
My Thought
Assume that the statement is ture. Then, the equality is equivalent to that
$$a^2-na+b-nb^2=0.$$
Regard it as a quadratic equation with respect of $a$.Then $$a=\dfrac{n \pm \sqrt{n^2+4nb^2-4b}}{2}.$$ Thus, $n^2+4nb^2-4b$ must be a square number. Let $$n^2+4nb^2-4b=k^2,k \in \mathbb{N_+}.$$ How to go on with this? May it work?
P.S.
The statement seems to be true. Here are parts of verification examples: \begin{array}{r|r|r} n&a&b \\ \hline 1&1&1\\ 2&5&3\\ 3&5&2\\ 4&10&4\\ 5&27&11\\ 6&69&27\\ \vdots&\vdots&\vdots \end{array}
Besides, the equation could be rewritten as
$$n(2a-n)^2-(2nb-1)^2=n^3-1,$$
which is a $\textbf{ Pell-like equation}$. This will help?
• Computer search confirms the conjecture for $1 \leq n \leq 100$, though the requisite values of $a$ and $b$ can be very large (for example, $n=54$ gives $(a,b) = (9\,683\,509, 1\,317\,755)$ and $n=89$ gives $(a, b) = (22\,276\,589, 2\,361\,309)$). I've looked for useful patterns but can't find any, except for a vague tendency to have $a \equiv b \bmod 2$ for larger $n$ (it holds for 28 out of the 40 integers $60 \leq n \leq 99$). It can greatly accelerate computer searches to note that $a/\sqrt{n} - b$ is positive but never more than about $10$. May 31 '18 at 15:21
• Where did you get this problem? May 31 '18 at 17:23
• @user477343 If $k$ is a square number, then $k\equiv 0,1\pmod 4$. That's true! But you can't reason like this: since $k\equiv 0,1\pmod 4$,then $k$ is a square number. For counterexample, $17 \equiv 1\pmod 4$,but it's not a square number at all. Right,sir? Jun 1 '18 at 0:48
• I've confirmed the conjecture for $n \leq 200$ (going much further runs into limits on 64-bit integer arithmetic). Solutions (listed in the form $n, a, b$) are here: pastebin.com/L1b1bqJT Jun 1 '18 at 17:38
• Apparently this is due to Zhining Yang see oeis.org/A290332 Jun 4 '18 at 21:17
Proof for all non-quadratic $n$
Lemma: Pell's equation $x^2-n y^2 = 1$, with $n$ not being a perfect square, has infinitely many solutions such that $x$ is odd, $y$ is even and $x\equiv1$ (mod $2n$).
Proof: It's a well known fact that the Pell's equation with non-quadratic $n$ has an inifinite number of solutions. Pick any such solution $(x_1,y_1)$. It's worth noticing that $x_1$ and $y_1$ must be co-prime as well as $x_1$ and $n$.
Now calculate:
$$x_2=x_1^2+ny_1^2,\quad y_2=2x_1y_1$$
It can be easily proved that $(x_2,y_2)$ is also a solution of the same Pell's equation. Obviously $y_2$ is even.
If you replace $x_1^2=ny_1^2+1$ into the expression for $x_2$ you get:
$$x_2=1+2ny_1^2\implies x_2\equiv1\space (\text{mod}\space 2n)$$
This also proves that $x_2$ has to be odd (which makes perfect sense because solutions of Pell's equation are always co-prime and $y_2$ is even).
You can construct more solutions of Pell’s equation in the same way and they all satisfy the criteria of the lemma. So there is not just one such solution. Actually there are infinitely many.
End of lemma proof
Back to the original equation (same approach as HERE):
$$\frac{a^2+b}{a+b^2}=n$$
can be rewritten as:
$$u^2-nv^2=1-n^3 \tag1$$
where:
$$u=2nb-1,\quad v=2a-n$$
Take $x,y$ such that:
$$x^2-ny^2=1\tag2$$
You can easily prove that $(-x+y n^2)$ and $(-y+nx)$ satisfy (1):
\begin{align*}(-x+ yn^2)^2-n(-y+nx)^2&=x^2-2xyn^2+y^2n^4-ny^2+2xyn^2-x^2n^3\\&=(x^2-ny^2)+n^3(ny^2-x^2)\\&=1-n^3.\end{align*}
This shows that:
$$u=-x+yn^2=2nb-1$$
$$v=-y+xn=2a-n$$
...represent a solution of $(1)$.
Hence,
$$a=\frac{(x+1)n-y}{2},\space b=\frac{yn^2-(x-1)}{2n}.\tag3$$
According to our lemma Pell's equation has infinitely many solutions $x,y$ such that $x$ is odd, $y$ is even and $x\equiv1$ (mod $2n$). Replace these solutions into (3) and you'll obviously get infinitely many integer values for $a,b$.
End of proof for all non-square $n$.
The following simple Mathematica script will generate single $a,b$ for fairly big non-square $n$ very fast (it follows the proof, word by word):
ABPair[n_] := Module[
{x, y, a, b, a1, b1, a2, b2},
pellSolutions = Solve[x^2 - n y^2 == 1, {x, y}, Integers] /. C[1] -> 1;
pellSolutions = {x, y} /. pellSolutions;
{a1, b1} = First[Select[pellSolutions, #[[1]] > 0 && #[[2]] > 0 &, 1]];
{a2, b2} = If[Mod[a1, 2 n] == 1 && Mod[b1, 2] == 0, {a1, b1}, {a1^2 + n b1^2, 2 a1 b1}];
a = (n (a2 + 1) - b2)/2;
b = (b2 n^2 - a2 + 1)/(2 n);
{a, b, (a^2 + b)/(b^2 + a)}
];
For example:
ABPair[5613]
{60584278414870816497213, 808653403020126409200, 5613}
The third number is just a check that the calculated numbers are valid. In other words:
$$\frac{60584278414870816497213^2+808653403020126409200}{60584278414870816497213+808653403020126409200^2}=5613$$
The script is lightning fast even for $n$ with 12 digits:
ABPair[561044335534]
See Sil's solution for quadratic $n$. Case closed :)
• I have question for only one part: if your initial solution has $x_0\equiv 1 \pmod n$ it looks like this might not work. Since $x_{i+1} \equiv x_i^2 \pmod n$ this results in $x_k \equiv 1 \pmod n$ for all $k$. (The other update formula also gives $x_k \equiv x_i x_j \equiv 1 \pmod n$. Then $b = (yn/2) + (x+1)/(2n)$ which is not an integer for $n>2$ since $x+1 \equiv 2 \pmod n$. This can happen for example by taking $(x_0,y_0) = (161,72)$ for the equation $x^2-5y^2=1$. Jun 14 '18 at 5:25
• You are perfectly right, let me think about that. Jun 14 '18 at 7:00
• I also noticed that using $-x$ instead of $x$, i.e. $$a = \frac{(-x+1)n+y}{2}, b = \frac{-x+1 + yn^2}{2n}$$ would still give a solution and always works by your construction, but the problem is now it looks like $a < 0$. Perhaps some small adjustments can fix the argument. Jun 14 '18 at 7:13
• Yes, that was exactly my plan how to fix the proof. it should work either for $x_k\equiv1$ or $x_k\equiv-1$ But some solutions could be negative. I think this can be tweaked. Bigger concern is how to prove this for square $n$ - looks like a simpler case but so far - no luck. Jun 14 '18 at 8:37
• @YongHaoNg: I think this is now rock solid for all non-quadratic $n$. Jun 15 '18 at 10:22
Solution for non-square $$n$$ is provided in @Oldboy's answer and in linked questions. This answer handles the case for square $$n$$.
Case 1: $$n=k^2,k \equiv 0 \pmod {2}$$
Choose \begin{align} a=\frac{k^2(k^3+2)}{4}, b=\frac{k^4}{4}. \end{align}
Conditions imply that $$k^2 \equiv 0 \pmod {4}$$ and so both $$a$$ and $$b$$ are integers. By algebraic manipulation we can show that $$(a^2+b)/(b^2+a)=k^2=n$$ (it is quite technical).
Case 2: $$n=k^2,k \equiv 1 \pmod {2}$$
Choose
\begin{align} a=\frac{(k^2+1)(k^2-k+2)}{4}, b=\frac{(k-1)(k^2+1)}{4}. \end{align}
Here $$2 \mid k^2+1$$ and $$2 \mid k^2-k+2$$ implies $$a$$ is an integer and similarly $$2 \mid k-1$$, $$2 \mid k^2+1$$ for $$b$$. Again it can be verified that $$(a^2+b)/(b^2+a)=k^2=n$$.
This result is obtained by mindless following of solution of quadratic diophantine equation on https://www.alpertron.com.ar/QUAD.HTM. Basically for square $$n$$ and our equation the site instructs us to find $$(X-\sqrt{n}Y)(X+\sqrt{n}Y)=4n(n^3-1)$$ such that $$4n \mid Y+2$$ ($$2$$ being calculated there as $$\beta$$ and $$4n$$ being a determinant). So the problem is essentially to look at divisors $$d$$ of $$4n(n^3-1)$$ that satisfy above divisibility criteria. For $$n=k^2$$ the factorization is $$2\cdot2\cdot(k-1)k^2(k+1)(k^2-k+1)(k^2+k+1)$$ (not into primes, but fortunately this is enough). So by testing combinations of these factors (using Maple e.g.), it turns out that choices of $$d=2k$$ and $$d=2k(k+1)$$ work (for even and odd $$k$$ cases respectively, that is). Those choices when substituting all the way back simplify to the cases described above, but it is too long/technical to get there...
• This looks like a lot of hard work to me. How many solutions did you pick from the site before you very actually able to spot the pattern? Jun 16 '18 at 20:46
• @Oldboy It was quite long for a comment, edited the answer instead. It's not that hard if one uses CAS system for help.
– Sil
Jun 16 '18 at 21:19 | 2021-12-03T10:00:10 | {
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https://math.stackexchange.com/questions/872506/counting-overlapping-figures/872666 | # Counting overlapping figures
How many four-sided figures appear in the diagram below?
I tired counting all the rectangles I could see, but that didn't work. How do I approach this?
• Just count it!:D – Mahdi Jul 20 '14 at 11:46
• In what sense did "counting all the rectangles" not work? Do you feel that you miscounted? (In which case: try again more carefully.) Or do you feel that you counted correctly, but seem to be counting the wrong things? (In which case: reconsider your definition of "side"; e.g. may sides overlap?) – Rebecca J. Stones Jul 20 '14 at 11:55
• I was looking for a systematic way to solve the problem. And I miscounted because the answer is 25 (I should have included that in the question). I am having a hard time seeing all the four-sided figures. – Guest Jul 20 '14 at 12:03
• Go through each corner, and count how many rectangles have that corner as a top-left vertex. That's the systematic way of doing it. – Arthur Jul 20 '14 at 12:12
• That is exactly what I was looking for Arthur. How can I choose your answer? – Guest Jul 20 '14 at 12:36
Go step by step.
First Picture: 1 rectangle
Second Picture: 2 additional rectangles. The small rectangle, which has been added and the big one, which contains the two small rectangles.
Third picture: The big rectangle. Then two rectangles, which contains 2 small linked rectangles. And the small rectangle, which has been added
Fourth picture: Only one small rectangle.
Fifth Picture: The rectangle, which contains the two small rectangle and the small additional rectangle.
You go on like this. Then sum the amount of rectangles.
Each rectangle has two vertical lines and two horizontal lines.
There are five vertical lines in the picture, we can label them 1, 2, 3, 4, 5.
If the leftmost edge is 1: Then the top and bottom are uniquely determined, and it is easy to see that 3 or 4 must be the right edge. 2 options.
If the leftmost edge is 2: Then the rightmost edge is 3 or 4 (2 choices), and in either case there are 3 horizontal segments that can serve as the top/bottom($\binom{3}{2} =3$ choices). So this gives $2 \cdot 3 = 6$. 6 options.
If the leftmost edge is 3: If the rightmost edge is 5 there is only one rectangle. If the rightmost edge is 4, there are 5 horizontal segments for top and bottom, so $\binom{5}{2} = 10$ choices. Hence 11 options.
If the leftmost edge is 4: Then the rightmost edge is 5, and there are four horizontal segments yielding $\binom{4}{2} = 6$ possible rectanges. 6 options
The total is 2 + 6 + 11 + 6 = 25.
## protected by Zev ChonolesApr 8 '16 at 23:00
Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count). | 2019-09-23T18:46:58 | {
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https://www.jiskha.com/questions/980999/Suppose-an-urn-contains-8-red-5-white-and-7-blue-marbles-If-3-marbles-are-drawn-at | # Math-Probability
Suppose an urn contains 8 red,5 white and 7 blue marbles.If 3 marbles are drawn at random from the urn with replacement,what is the probability that three marbles are the same color?Give answer in a reduced fraction.
1. could be RRR , WWW, or BBB
prob = (8/20)(7/19)(6/18) + (5/20)(4/19)(3/(18) + (7/20)(6/19)(5/18)
= 336/6840 + 60/6840 + 210/6840
= 606/6840
= 101/1140
posted by Reiny
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More Similar Questions | 2018-08-21T16:35:25 | {
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http://mathhelpforum.com/discrete-math/56628-combinatio-help.html | # Math Help - Combinatio Help
1. ## Combinatio Help
Suppose that a club consists of 10 men and 13 women. The club is going to form a committee of 8 people. How many 8 person committees have more women than men?
I thought maybe it was (23!/8!*15!)-(10!/8!*2!) but that does not seem like it gives me the correct answer. Any help would be great!
2. Hello, ezwind72!
I don't believe there is a formula for this problem . . .
Suppose that a club consists of 10 men and 13 women.
The club is going to form a committee of 8 people.
How many 8-person committees have more women than men?
$\begin{array}{ccccc}\text{3 men, 5 women} & {10\choose3}{13\choose5} &=& 154,\!440 \\ \\[-3mm]
\text{2 men, 6 women} & {10\choose2}{13\choose6} &=& 77,\!220 \\ \\[-3mm]
\text{1 man, 7 women} & {10\choose1}{13\choose7} &=& 17,\!160 \\ \\[-3mm]
\text{0 men, 8 women} & {10\choose0}{13\choose8} &=& 1,\!287 \\ \\[-3mm] \hline \\[-3mm]
& & \text{Total:} & 250,\!107 \end{array}$
There are $\boxed{{\color{blue}250,\!107}}$ committees with a majority of women.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
That was such an unusual-looking answer, I just had to run a check.
$\begin{array}{cccc}\text{4 men, 4 women} & {10\choose4}{13\choose4} &=& 150,\!150 \\ \\[-3mm]
\text{5 men, 3 women} & {10\choose5}{13\choose3} &=& 72,\!072 \\ \\[-3mm]
\text{6 men, 2 women} & {10\choose6}{13\choose2} &=& 16,\!380 \\ \\[-3mm]
\text{7 men, 1 woman} & {10\choose7}{13\choose1} &=& 1,\!560 \\ \\[-3mm]
\text{8 men, 0 women} & {10\choose8}{13\choose0} &=& 45 \\ \\[-3mm] \hline \\[-3mm]
& & \text{Total: } & 240,\!207 \end{array}$
Hence, there are: . $250,\!107 + 240,\!207 \:=\:{\color{blue}490,\!314}$ possible committees.
Check . There are: . ${23\choose8} \;=\;{\color{blue}490,\!314}$ possible committees . . . YAY!
3. I see what you did there. I was wayyyyyyy off! But I now understand how you came about that answer. The formula using summation notation would be (I wish I could say I came up with that solely on my own).
$\sum\limits_{k = 5}^8 {{13 \choose k}{10 \choose 8-k} }$ | 2015-05-05T14:48:08 | {
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https://math.stackexchange.com/questions/1015828/curves-with-a-common-tangent-line | # Curves with a common tangent line
• Question
Find the point where the curves $$\tag 1y = x^3 -3x + 4$$ and $$\tag 2 y = 3x^2 - 3x$$ are tangent to each other, that is, have a common tangent line.
• My approach
• Let $x = a$ and $x = b$ be the points on curves $(1)$ and $(2)$, respectively, at which their slopes are equal and share a mutual tangent line.
• Now I will relate the $a$ and $b$ by equating the derivatives of $(1)$ at $a$ and $(2)$ at $b$, as follows $$3a^2 - 3 = 6b- 3 \Leftrightarrow b = \frac{a^2}{2}$$
• Let $A$ be the point on curve $(1)$ and $B$ be the point on curve $(2)$ where the two curves share the mutual tangent, that is $$A(a, x^3 - 3x + 4)$$ and $$B(b, 3b^2 - 3b) = B\Big(\frac{a^2}{2}, \frac{3a^4 - 6a^2}{4}\Big)$$
• Now, since I have two points on the tangent line, I can calculate the slope and equate it to the derivative of $(1)$ at $a$ as follows $$\frac{(x^3 - 3x + 4) - \Big(\frac{3a^4 - 6a^2}{4}\Big)}{a - \frac{a^2}{2}} = 3a^2 - 3$$
• Simplifying that equation I get the following, $$3a^4 -8a^3 - 12a^2 + 16 = 0$$
Now, I would solve for $a$ and then substitute the value of $a$ in to points $A$ and $B$ which would then be the points at which the two curves have a common tangent line. The problem is that I doubt I should be solving such an equation, and quite frankly, I don't have the tools to solve that equation, unless I'm missing something?
Any suggestions?
• Does "have a common tangent line" require that the two curves intersect each other as well? We can imagine a line that is tangent to both curves without the points of tangency being identical. – Rory Daulton Nov 10 '14 at 22:38
• @RoryDaulton that's what caused me some confusion, glad you ask. Prior to doing this question I did a question where two parabolas (parabolas that don't intersect at all) share two tangent lines for which I had to find the coordinates at which these tangents touch the two parabolas. However, as Adriano, who provided a solution, pointed out that "have a common tangent line" means that they intersect and at the point of intersection they have the same slope. So I don't know why this book would be so, seemingly, ambiguous. What do you think? – Kermit the Hermit Nov 10 '14 at 22:46
• I found this problem in James Stewart Calculus, Problems Plus – Joao Noch Oct 28 '17 at 7:35
Two curves $y = f(x)$ and $y = g(x)$ have a common tangent line at $x = a$ iff:
• They intersect there: $f(a) = g(a)$.
• Their tangent lines have equal slope there: $f'(a) = g'(a)$.
Since quadratic equations are easier to solve than cubic ones, we start with the second condition: $$3a^2 - 3 = 6a - 3 \iff a^2 - 2a = 0 \iff a = 0, 2$$ We now check if each candidate satisfies the first condition: $$f(0) = 4 \neq 0 = g(0)$$ but: $$f(2) = 6 = g(2)$$ So the only common tangent line occurs at $x = 2$ and is given by: $$y - 6 = 9(x - 2)$$
• Ah I see the confusing part. If $a \neq 2$ so that you are not dividing by zero in your slope formula expression, then you should instead obtain: $$0 = 3 a^4-8 a^3+16 = (a-2)^2 (3 a^2+4 a+4)$$ Since $(3 a^2+4 a+4) \geq 4 > 0$, we conclude that $a = 2$ is the only real solution. – Adriano Nov 10 '14 at 22:56
• Oh man, I made a mistake with the simplifying of the slope formula. How did you factor $3a^4 - 8a^3 + 16$? Why do you say $3a^4 - 8a^3 + 16 \geq 4$? From graphing $3a^4 - 8a^3 + 16$, it seems greater or equal to approximately $2.6$? – Kermit the Hermit Nov 10 '14 at 23:48 | 2020-01-19T02:04:13 | {
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http://mathhelpforum.com/calculus/165613-calcii-contours.html | # Math Help - calcii: contours
1. ## calcii: contours
Hi, I am having trouble with contours, is there a procedure to do these?
the question I am working on is when z=c
z=((x^2)+(y^2))/2x
what i tried to do:
c2x=(x^2)+(y^2)
c2x-x^2=y^2
x(2c-x)=y^2
2c-x=(y^2)/x
2c=((y^2)/x)+x
and that is where I got stuck. I cant visualize what is going on here
2. Originally Posted by jameskw
Hi, I am having trouble with contours, is there a procedure to do these?
the question I am working on is when z=c
z=((x^2)+(y^2))/2x
what i tried to do:
c2x=(x^2)+(y^2)
c2x-x^2=y^2
x(2c-x)=y^2
2c-x=(y^2)/x
2c=((y^2)/x)+x
and that is where I got stuck. I cant visualize what is going on here
thanks for any replies
What you have done is equivalent to:
$\displaystyle c={{(x^2)+(y^2)}\over{2x}}$
$\displaystyle 2c={{(x^2)+(y^2)}\over{x}}$
$\displaystyle 2c=(x)+{{y^2}\over{x}}$
What you need after your line, $c(2) x-x^2=y^2$ is:
$0=x^2-2cx+y^2$, then add $\displaystyle c^2$ complete the square.
$c^2=x^2-2cx+c^2+y^2$
Finally you have $c^2=(x-c)^2+y^2$.
Do you recognize this?
3. Well, it looks like its close to a circle because of its form. As c increases radius increases and x gets displaced.
sounds like 2 cones with a slope, how would i find a slope?
thanks
4. ## Calc II, contours
Originally Posted by jameskw
Well, it looks like it's close to a circle because of its form. As c increases radius increases and x gets displaced.
sounds like 2 cones with a slope, how would i find a slope?
thanks
Yes! For any fixed value of c, $c^2=(x-c)^2+y^2$ is the equation of a circle with radius, c, centered on the x-axis at x=c. This circle passes through the origin.
When looked at in a more general way, $c^2=(x-c)^2+y^2$ is a family of such circles.
Therefore, the equation, $\displaystyle z={{(x^2)+(y^2)}\over{2x}}$, describes a pair of oblique circular cones, each with a vertex at
(0, 0, 0) and axis along the line $z=x$ in the x-z plane. One opens upward, the other downward.
You can also look at this as a pair of right elliptical cones, each with a vertex at (0, 0, 0) and axis along the line $z={4\over3}x$ in the x-z plane. (I haven't figured out the eccentricity of the ellipses.)
5. $c= \frac{x^2+ y^2}{2x}$ gives $2cx= x^2+ y^2$, $x^2- 2cx+ y^2= 0$
Now "complete the square": $x^2- 2cx+ c^2+ y^2= c^2$, $(x- c)^2+ y^2= c^2$,
a circle with center at (c, 0) and radius c so it is tangent to the y-axis at (0, 0). | 2015-05-04T07:48:42 | {
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https://brilliant.org/discussions/thread/parity-2/ | Parity
Definition
Parity is a term we use to express if a given integer is even or odd. The parity of a number depends only on its remainder after dividing by $2$. An even number has parity $0$ because the remainder after dividing by $2$ is $0$, while an odd number has parity $1$ because the remainder after dividing by $2$ is $1$.
Here are a few arithmetic rules of parity that are extremely useful:
even $\pm$even = even
odd$\pm$odd=even
even $\pm$odd= odd
even$\times$even= even
even $\times$odd= even
odd $\times$odd= odd
Parity is often useful for verifying whether an equality is true or false by using the parity rules of arithmetic to see whether both sides have the same parity.
1. Worked Examples
1. If $n$ is an integer, what is the parity of $2n+2$?
Solution: Since $n$ is an integer, $n+1$ is also an integer. Thus, $2n+2 = 2(n+1) + 0$ shows that the parity of $2n+2$ is $0$.
2. If $a, b$ are integers, what is the parity of $a \times b$?
Solution: We know that an odd number multiplied by an odd number remains odd, an even number multiplied an odd number is even, and an even number multiplied by an even number is even. This can be summarized as (check for yourself) $\mbox{Parity of } a \times \mbox{ Parity of } b = \mbox{ Parity of } ab$
&nsbp;
3. If $k$ is an integer, what is the parity of $k^2 + k$?
Solution: $k^2 + k = k (k+1)$. Note that $k, (k+1)$ have different parity. Hence, by the arithmetic rules of parity, the parity of $k(k+1)$ is $0$.
Note by Arron Kau
5 years, 10 months ago
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http://antennablutv.it/bkza/relationship-between-volume-of-cylinder-cone-and-sphere.html | # Relationship Between Volume Of Cylinder Cone And Sphere
So let's see what that would look like if we apply it to the surface areas. Continue to reduce the size of the spheres, and you approach the 74% figure of "ideal packing". This chart type also includes cylinder, cone, and pyramid subtypes. Cylinder examples/objects Colored paper Calculator Beans Scissors Copies of T870 and T871 for each pair [ESSENTIAL QUESTIONS] 1. 300 seconds. and it has a height of 0. Answered by Penny Nom. Now examine the. First, the relationship between the angle and the cylinder power is more accurately described by the square of the sine. Ex of units: L or mL Solid Volume: When an object has a definite shape (ex. Finally, plug this into the conversion for $$z$$ and take advantage of the fact that we know that $$\rho = 3\sqrt 2$$ since we are intersecting on the sphere. Demonstrate the relationship between shape, size and volume. Which statement correctly describes the comparison between the volume of the cylinder and the volume of the cone?. 14 x 4 x 4 x 17) ÷ 3 = 284. Multiply this result by the cylinder's height to get its volume. O is the vertex of the cone, AB is the diameter of the base of the cone and C its center. You can use a formula for the volume of a sphere to solve problems involving volume and capacity. && If#everyline#paralleltothese#twolinesintersectsboth regions#in#line#segments#of#equal#length,#then#the#two# regions#have#equal#areas. Because the top is semi-spherical, its volume will be half that of a full sphere. Play on at least Easy! 2. The cylinder has a height h of 15 cm and a radius of 5 cm. Measure the height and diameter with a ruler and record your data below and on the cylinder. Find the point(s) on the cone z^2 = x^2 + 4y^2 that are closest to the point (2,5,0). The formula for the curved surface area* of a cone is \pi rl, where r is the radius of the base and l is the slant height. All that is left is to calculate the area of the sphere in ndimensions=A(n-1). As nouns the difference between cone and cylinder is that cone is (label) a surface of revolution formed by rotating a segment of a line around another line that intersects the first line while cylinder is (geometry) a surface created by projecting a closed two-dimensional curve along an axis intersecting the plane of the curve. Objective CONES , Cones , Spheres - Ms. As a first example we study the influence of head tissue conductivity inhomogeneity. Very similar developments occur in the flow around a sphere and a cylinder. Label it Cylinder A. Then they use the formula for the volume of a cylinder learned in previous lessons to write the general formula $$V= \frac13\pi r^2 h$$ for the volume, $$V$$, of a cone in terms of its height, $$h$$, and radius, $$r. Record your data. The formulas for the volume of a sphere and the volume of a cylinder are well known. Use the formulas for the volumes of cylinders, cones, and spheres to solve a variety of real-world problems. y/x = h/r y = hx/r. Let V 1 be the volume of first cylinder ∴ V 1 = π(r 1) 2 h 1. What is the relationship between the volume of a cylinder and the volume of a cone? 2. Big Ideas: Volumes of cylinders, cones, and spheres have comparable components such as radius and height. D The volume of the cyllnder is four-thirds the volume ofthe cone. Thus the cones plus the sphere equals the cylinder exactly. The height of the cylinder is twice that of the radius of the sphere. the beaker, you could easily obtain a volume between 5 and 10 mL, probably close to 7 mL, give or take 1 mL. The areas of the triangular faces will have different formulas for different shaped bases. Since the shot-putt is a solid sphere made of metal and its mass is equal to the product of its volume and density, we need to find the volume of the sphere. The radius of base of each of cone and cylinder is 8 / c m. STAAR ALGEBRA I REFERENCE MATERIALS. The volume of a hyperspherical cone V n cone is also easy to derive by the difference between the sector volume and the cap volume, V n cone (r) = V n sector (r)-V n cap (r) = 1/nV n-1 (rsinφ)rcosφ. Which of the following is true? A The volumes are the same. Comparing Cylinder and Cone Volumes EXAMPLE Compare the volume of each when a cone and a cylinder have the same base area and height. volume of spheres: A sphere is the locus of all points in a region that are equidistant from a. To find the volume of the solid, subtract the water volume before immersion from the new water volume after the immersion. Solved Problems Click or tap a problem to see the solution. Visualizing the Volume of a Sphere Formula video. If r=h=1 unit. For exercises 1 - 4, the cylindrical coordinates \( (r,θ,z)$$ of a point are given. Describe how the graph of each function can be obta Precalculus: Mathematics for Calculus. Tape along the edge. Vrh=≈ ≈ ππ22(1. 13 most outstanding volume and surface area worksheets word problems with answers right circular cylinder base cone formula total calculator sphere triangular prism castle equation tetrahedron inspirations coloring of cube cuboid lateral curved a rectangular - Hockeyofficialauthentic. c - Find the surface area and volume of a sphere in mathematical and real-world settings 3. volume of cylinders: The process for understanding and calculating the volume of cylinders is identical to that of prisms , even though cylinders are curved. The volume of a rectangular prism can be determined by multiplying Length (L) x Width (W) x Height (H). When one or more of the dimensions of a prism or cylinder is multiplied by a constant, the surface area and volume will change. So for a cube, the ratio of surface area to volume is given by the ratio of these equations: S/V = 6/L. (a/2)^3)/3 = 4. 3 × 10 = 283 cm 3. Example 6: Find the formula for the total surface area of each figure given bellow :. It takes three cones full of rice to fill the cylinder. 86mm) and, after skull trepanation, a post-surgical CT (512 sagittal, 635 coronal and 68 axial slices, voxel-size of 0. vol of a cylinder = 3 * vol of a cone. Solve for the value of x. Unlike regular objects, such as the cube or sphere, no further simplification of the box's or cylinder's surface-area-to-volume ratio equation exists. Integrate both sides, which acts like adding infinitely many tiny layers. Volume The radius of the base of a cone is Ch. Very similar developments occur in the flow around a sphere and a cylinder. 11 Look again at the container of motionless liquid. 5 ft) = 185π ft^2 ≈ 581 ft^2 It is the lateral area of a cylinder of the same diameter but half as high as the slant height, or one that is half the diameter but as high as the slant height. It accepts the dosimetric cylinder (TLD or film). 3 × 10) = 94 ⅓ cm 3. Notice that the cylinder and the sphere have the same radius and height. Their volumes can easily be seen to be (4/3) r 3, 2(1/3) r 3, and 2 r 3. • Model the volume of a cylinder as a representation of layers. Solve problems involving combinations of the figures using metric and imperial measure. Graphing Polar Equations. The bottom of the cylinder will be on the z = 0 {\displaystyle z=0} plane for simplicity of calculations. Sphere Volume Conjecture. Explain volume formulas and use them to solve problems. Volume The height of a cylinder is 7. 1, you discovered the relationship between the volume of a sphere and the volume of a cylinder. How do the radius and surface area of the balloon change with its volume? We can find the answer using the formulas for the surface area and volume for a sphere in terms of its radius. Identify students who:. The would. Move to page 3. Round your answer to two decimal places. Solve real‐world and mathematical problems involving volume of cylinders, cones, and spheres. I know that the volume we're interested in is the volume of the intersection between the sphere of Stack Exchange Network Stack Exchange network consists of 177 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Because the top is semi-spherical, its volume will be half that of a full sphere. Students also learn that the formula for the volume of a sphere is 4/3 times pi times radius cubed, and the formula for the volume of a cone is 1/3 times pi times radius squared times height. the height of the cylinder. establish that the volume of the sphere plus the cone make the volume of the cylinder popcorn suitably flattened on top the result for the relationship between the volumes of a sphere, a cone and a cylinder was allegedly established by Archimedes using small slices. Deriving the formula- Volume of a Sphere video. This means that material costs can be minimized without sacrificing interior space. And volume of the cone will. A similar figure is the (circular) cylinder, which has two congruent circular bases and a tube-shaped body, as shown below. Cone vs Sphere vs Cylinder - MATH. We all have seen a cylinder, now let us learn to define it in technical terms. Because our cylinder is constrained to be inside a cone, we use similar triangles to nd the relationship between the height and radius of the cylin-der. Example 1 A sphere of radius \$$r\$$ is inscribed Read more Optimization Problems in 3D Geometry. The volume of a cylinder, on the other hand, is equal to the product of the area of the base multiplied by the altitude of the cylinder. Volume of a cone = π r 2 h, where r is the radius of the base and h is the height. B The volume of the cylinder is three times the volume of the cone. In this section, we consider several problems in which two or more related quantities are changing and we study how to determine the relationship between the rates of change of these quantities. What is the ratio of the cone’s height to its radius? (2003 AMC 12B Problems. Since a cylinder's volume formula is V = Bh, then the volume of a cone is one-third that formula, or V = Bh/3. Volume of a cylinder = (3. For example, according to (6) and (7), the uniformly polarized cylinder of material shown in Fig. 70 in3 D) 523. 04πr dr dt or that dr dt = 1 0. => Surface Area of Right Circular Cone => Surface Area: The Sphere => Surface Area: Frustum of a right circular cone => Exercise 7. Thus the cones plus the sphere equals the cylinder exactly. The drag coefficients (C) used in our calculation are from Blevins (2003). Question 1. The surface area and the volume of the sphere are given below: The Curved Surface Area of a Sphere = 2πr² Square units. Deriving the formula- Volume of a Sphere video. Specifically, you will determine the mass of a given volume, incrementally increase the volume, and continue making mass measurements. Volume of cube = a^3, where a is the length of the cube edge. 86 cm, and since diamater is twice the radius, the radius must be 2. The area of a circle. 5cm^3/s and the sphere's radius is 1cm (or if easier, any radius, if not,. No, not yet Some Yes No, not yet Some Yes. Refer to the figure above. 01 mL pretty reliably. Liu Hui proves that the assumption is incorrect by showing that this relation in fact holds between the volume of a sphere and that of another object, smaller than the cylinder. VOLUME Triangle Rectangle or parallelogram Rhombus Trapezoid Regular polygon Circle Prism S Ph= Pyramid Cylinder Cone Sphere Prism or cylinder Pyramid or cone Sphere Circle. Common Core: 8. It is possible to see a relationship between the change in dimensions and the resulting change in surface area and volume. 0 Equation Volume of a Cylinder, Cone, and Sphere Volume Cylinder Previous Formulas Learned Area and Circumference of a Circle Cylinder Volume of a Cylinder Volume of a Cylinder Volume of Cylinders Class Practice Volume of Cylinders. For example, the volume of a cube is the area of one side times its height. Describing Transformations Suppose the graph of f is given. The cylinder has a height h of 15 cm and a radius of 5 cm. The drag coefficient for a cone pointed into the airflow is a bit more complex since it depends on the cone's shape. 8 mm Hg and the volume of each cylinder is 246. This meant the volume of the hemisphere must be equal to the volume of the cylinder minus the volume of the cone. Draw and cut out 5 squares. We do not have to remember the formulas which calculate the area and surface area of spheres, pyramids and cones. volume = Pi * radius 2 * length. The cylinders and cones are right. To find liquid volume: pour it into a measuring cup and read the line it fills up to (like children's Motrin). relationship between the area of the base and height and the volume of a cylinder, and generalize to develop the formula; • determine, through investigation using concrete materials, the surface area of a cylinder; • solve problems involving the surface area and the volume of cylinders, using a variety of strategies. Finding Volume of an Oblique Cone Find the volume of an oblique cone with diameter 30 ft and height 25 ft. The distance between the center of the circle and the sphere is 6. MindYourLogic 239,302 views. 39 What is the relationship between the volume of the cone inscribed in a hemisphere and the volume of the hemisphere? A. Find the area of a circle with a 6 cm radius. Car, truck or van load space volume capacity. As nouns the difference between cone and cylinder is that cone is (label) a surface of revolution formed by rotating a segment of a line around another line that intersects the first line while cylinder is (geometry) a surface created by projecting a closed two-dimensional curve along an axis intersecting the plane of the curve. 72 in3 C) 392. Surface area to volume ratio can be found easily for several simple shapes, like for example a cube or a sphere. Solve for x, given the volume. Calculate the volume of a cylinder of radius R and height h. Recognize the relationship between the formula for the volume of a cone and the volume of a sphere. Volume of a sphere=4/3 ×r³. Cone vs Cylinder. But it is very old knowledge, dating back to Archimedes, who studied the relation between the volume of the sphere and the volum. Identify students who:. This set consists of cone cylinder, Square prism and pyramid and a sphere equal to the inner. Apply the formulas for the volume of cones, cylinders, and spheres and use them to solve real-world and. some guy found the relationship between volume of a sphere and volume of a cylinder. Ryan drew a cylinder and a cone with identical bases and heights. The sphere that fits in the cube has radius a/2. Archimedes was also a talented inventor, having created such devices as the catapult, the compound pulley, and a system of burning mirrors that was used in battle to focus the sun’s rays on enemies’ ships. Cone Take the clear cone that has the same base area and height as the cylinder. In this section, we consider several problems in which two or more related quantities are changing and we study how to determine the relationship between the rates of change of these quantities. For the solid hemispheres, hollow hemispheres, solid cone, ellipsoid, and solid cylinder, A = π D 2 / 4. Common Student Misconceptions for this Unit Students may struggle with unit conversions. The area of the base (B) is equal to because the base is shaped like a circle. The bottom of the cylinder will be on the z = 0 {\displaystyle z=0} plane for simplicity of calculations. Cones and Cylinders. This volume formula applies to all cones, including oblique cones. Therefore, at every height the slice of area in the cylinder intersection is 4/π times the area of the slice of the sphere, so the total volume of the region of intersection is 4/π times the volume of the sphere, which is (4/π)(4/3)π R 3 = (16/3)R 3, in agreement with what we found previously. Now let's fit a cylinder around a sphere. This means that material costs can be minimized without sacrificing interior space. the height = 2r, equaling the diameter of the sphere), then the volumes of the cone and sphere add up to the volume of the cylinder. Sphere Volume Conjecture. (a^3/8)/3= pi. This learining packet covers the relationship between the scale factor and the area/surface area/volume of similar figures. If we make it a right triangle and look at the hypotenuse, it's the light at the end of this spherical tunnel: the radius of the sphere. Which of the following is true? A The volumes are the same. avks_ 1 decade ago. Based on what you know about the relationship between a pyramid and prism with similar dimensions,. Assume that the melted ice cream occupies of the volume of the frozen ice cream. Because our cylinder is constrained to be inside a cone, we use similar triangles to nd the relationship between the height and radius of the cylin-der. The mathematical relationship between them was first shown by Delaunay (8) in 1841. Tags: Question 12 The volume of a cone. Presentation. Since wethink of the sphere as made up of infinitessimal(n-1)-cubes, the volume of the cone over the unitsphere=Vol(n)=(1/n)*surface area of thesphere=A(n-1)/n. What is the relationship between the volume of the cone and the vol ume of the cylinder? Make a conjecture and try to convince other students. As a first example we study the influence of head tissue conductivity inhomogeneity. Sphere: A solid figure that has all points the same distance from the center. STAAR ALGEBRA I REFERENCE MATERIALS. Calculate volume of a cone if you know radius and height ( V ) What is the formula of the volume of a cone - Calculator Online Home List of all formulas of the site. b) determine the ratio of the surface area of the sphere to the surface area of the cylinder in this situation. What is the formula to find the volume of a cylinder? What is the formula to find the volume of a cone? Name: Date : Period: What is the relationship between the formula for the volume of a cylinder and the volume of a cone? What is the formula to find the volume of a sphere? Find the volume of each figure. A cone is named based on the shape of its base. Drag coefficient of blunt nose and rounded nose cylinders versus fineness ratio l/d. Cone, Cylinder, Volume Use the sliders to change the radius (r) and height (h) of the cylinder and the cone. Area and Volume Formula for geometrical figures - square, rectangle, triangle, polygon, circle, ellipse, trapezoid, cube, sphere, cylinder and cone. Which of the following is true? A The volumes are the same. Volume of an n-Dimensional Pyramid/Cone = (1/n)Base*Height The above general formula can be used to establish a relationship between the volume of an n-dimensional ball and the (n-1)-dimensional area which bounds it. A pie chart shows the relationship of the parts to the whole. Similar to the last Volume 3 Act Math Task: Prisms and Pyramids, the intention has been to leave Act 1 of each set very vague to allow for students to take the problem in more than one direction. Graphing Polar Equations. Cone Cylinder. Cones Volume = 1/3 area of the base x height V= r2h Surface S = r2 + rs. 04πr dV dt = 400 πr. For example, a sphere represents a shape that has the highest volume to surface area ratio. Odd-shaped objects You can find the volume of an odd-shaped object, like a key, by placing it in water. The area of a circle. com/watch?v=3wuJJqlr6m0 To find manipulatives similar to these, try looking. vol of a sphere = 4*r * vol of a cone. Which statement correctly describes the comparison between the volume of the cylinder and the volume of the cone?. pi is given in the exam paper (usually 3. Dropping sugar cubes in the cylinder is also a good visual. Volume of a Sphere Formula Explained. Archimedes was also a talented inventor, having created such devices as the catapult, the compound pulley, and a system of burning mirrors that was used in battle to focus the sun’s rays on enemies’ ships. Sum of the distances of the point P to the. Cylinder, Cone and Sphere Surface Area and Volume Exercise 20F – Selina Concise Mathematics Class 10 ICSE Solutions. The inside of a sphere is called a ball. The volume relationship between these cones and cylinders with equal bases and heights can be expressed mathematically. Now let's fit a cylinder around a sphere. Fill the cone with rice, then pour the rice into the cylinder. Find the missing angle 6, Identify the relationship between the angles. Using Cavalieri’s Principle, write the general equation for the volume of a sphere. A cone is named based on the shape of its base. Car, truck or van load space volume capacity. a - Find the lateral area, surface area, and volume of prisms, cylinders, cones, and pyramids in mathematical and real-world settings F. The volume of a sphere of radius r is 4 / 3 π r 3 = 2 / 3 (2 π r 3). Cone Cylinder. We are learning tofind the volume of a cylinder, cone and sphere ; 2 Volume. The sphere that fits in the cube has radius a/2. Diameter, page 1 of 8 Overview In this experiment we investigate the relationship between the diameter of a sphere and its volume. The radius of a sphere is 5 yards. Their volumes can easily be seen to be (4/3) r 3, 2(1/3) r 3, and 2 r 3. For the polar or normal aspect, the cone. Lesson Notes Students informally derive the volume formula of a sphere in Lesson 12 (G-GMD. 70 in3 D) 523. Cones and Cylinders. If two objects have equivalent cross-sections for all horizontal slices, what can be said of their volume? (Cavalieri’s Principle) Rewrite the above relationship in terms of volume. Foundational Standards Draw, construct, and describe geometrical figures and describe the relationships between them. The volume relationship between these cones and cylinders with equal bases and heights can be expressed mathematically. Big Ideas: Volumes of cylinders, cones, and spheres have comparable components such as radius and height. To emphasize the importance of one slice of the pie, choose one of the exploded 2-D or 3-D pie charts. Assume that the volume of the cylinder is 24. The ‘pointy’ end to the cone is its one vertex. Sum of the distances of the point P to the. B The volume of the cylinder is three times the volume of the cone. A prism is a solid figure that has two parallel congruent sides that are called bases that are connected by the lateral faces that are parallelograms. First, the relationship between the angle and the cylinder power is more accurately described by the square of the sine. NCERT Class 9 Maths Lab Manual - Find the Relationship among the Volumes of a Cone Objective To find the relationship among the volumes of a right circular cone, a hemisphere and a right circular cylinder of equal radii and equal heights. Solids are objects with three dimensions - length, breadth, and thickness. the height = 2r, equaling the diameter of the sphere), then the volumes of the cone and sphere add up to the volume of the cylinder. MULTIPLE CHOICE Let V be the volume of a sphere, S be the surface area of the sphere, and r be the radius of the sphere. Volume Find the volume of a sphere that has Ch. Note that we can assume $$z$$ is positive here since we know that we have the upper half of the cone and/or sphere. Convert between weight and volume using this calculator tool. Since the values for the cylinder were already known, he obtained, for the first time, the corresponding values for the sphere. Cone Take the clear cone that has the same base area and height as the cylinder. q is the charge enclosed in the volume. A cylinder is similar to a prism, but its two bases are circles, not polygons. What is the relationship between the volumes of the cylinder and the cone when they have the same radius and height measurements?. A cylinder is bounded by two parallel planes or bases and by a surface generated by revolving a rectangle about one of its sides. Last week I wrote about the maximum (volume) cylinder it’s possible to fit inside a sphere. Volume of a sphere=4/3 ×r³. C The volume of the cone is three times the volume of the cylinder. 5 cm 3 of ice cream. And they are in the ratio of 1:3. 5 Investigate and describe the density of solids, liquids, and gases. To do so, they examine the relationship between a hemisphere, cone, and cylinder, each with the same radius, and for the cone and cylinder, a height equal to the radius. A cylinder and a cone have the same diameter: 8 inches. Find the the remaining variables at that instance. Answer by jsmallt9(3757) ( Show Source ):. For example, a student might compare the areas in a given cross-section, reducing the problem to a comparison of the area under a line and under a quadratic-like curve. Then, the key is placed in the graduated cylinder. The drag coefficients (C) used in our calculation are from Blevins (2003). Exploring the Relationship Between Mass and Volume Purpose: For this activity you will be performing a few measurements to help describe the relationship between mass and volume. For example, after part (b), the teacher could ask the students for other ways to determine which vase holds the most water, with the expectation that students might respond with. EXAMPLE 1 Finding the Volume of a Cylinder Find the volume of the cylinder. Calculate volume of a cone if you know radius and height ( V ) What is the formula of the volume of a cone - Calculator Online Home List of all formulas of the site. Derive the formula of the surface area of a cylinder of radius r and height h. Liu Hui proves that the assumption is incorrect by showing that this relation in fact holds between the volume of a sphere and that of another object, smaller than the cylinder. r is the volume charge density in coulombs per cubic meter. Ex of units: L or mL Solid Volume: When an object has a definite shape (ex. What fraction of the cup (by volume) is filled up? h 0. The radius of a sphere is 5 yards. 04πr dr dt or that dr dt = 1 0. Students write the volume of a cone given a specific volume of a cylinder with the same base and height, and vice versa. What is the relationship between the volume of the cone and the vol ume of the cylinder? Make a conjecture and try to convince other students. So if you wanted the "volume" of a four-dimensional "cone" made by stacking up spheres in the fourth dimension (in the same way as a 3D cone is made by stacking up circles), you could integrate (4/3)*pi*r^3*dr. so , vol of a sphere = 4*r /3 * vol of a cylinder. Vrh=≈ ≈ ππ22(1. iRubric N53235: This rubric is used for a unit of lessons that help the student discover the relationship between the volume of a cylinder, cone and sphere. Deriving the formula- Volume of a Sphere video. Circle and sphere are both round in shape but whereas a circle is a figure, a sphere is an object. 08 cubic centimeters. Processing. Click here to check your answer to Practice Problem 3. Fill the cone with rice, then pour the rice into the cylinder. If B is the area of the base of a pyramid or a cone and H is the height of the solid, then the formula for the volume if V=1/3 BH. Big Ideas: Volumes of cylinders, cones, and spheres have comparable components such as radius and height. 2 (+) Give an informal argument using Cavalieri’s principle for the formulas for the volume of a sphere and other solid figures. One container is a cylinder, one is a cone, and one is a sphere. I then have a discussion with students and even let them try finding the volume of a sphere using the volume of a cone formula they used the previous day. Assume that the volume of the cylinder is 24. The picture shows the dimensions of a petrol tank. Answer by jsmallt9(3757) ( Show Source ):. The volume of a sphere. The Volume Formula of a Sphere. 04πr dV dt = 400 πr. The area of a sphere is A=4pi*r^2 = (2pi*r)*2r = 2rC. Fill the cone with tinted water or rice. Solve for x, given the volume. Volume of a cylinder = (3. Explain volume formulas and use them to solve problems. Rotate this region about the x-axis and find the resulting volume. 14) (radius of sphere) 3 Today you will observe what happens to the mass of an object when the the volume is increased if the density or material of each object remains the same. This is when all the sides are the same length. Specifically, the cylinder's volume formula is and the cone's volume formula is. The flange can vary in length and can be shaped as either a cone or a cylinder. 1 4 Tape together as shown. Example An ice cream cone can hold about 33. •Find the surface area of a sphere. By finding where these rays intersect the sphere, and connecting the points of intersection by the arcs that characterize the shortest distance between two points along the sphere, we produce a radial projection of the polyhedron. Work out the slant height of the cone to 1dp. All that is left is to calculate the area of the sphere in ndimensions=A(n-1). Surface Area is the area of the outer part of any 3D figure and Volume is the capacity of the figure i. Solved Problems Click or tap a problem to see the solution. Relationship between volume of a pyramid and prism - Duration: Easiest way to Learn Volume of Cylinder, Cone, Sphere and Hemisphere - Duration: 3:45. establish that the volume of the sphere plus the cone make the volume of the cylinder popcorn suitably flattened on top the result for the relationship between the volumes of a sphere, a cone and a cylinder was allegedly established by Archimedes using small slices. As nouns the difference between cylinder and sphere is that cylinder is (geometry) a surface created by projecting a closed two-dimensional curve along an axis intersecting the plane of the curve while sphere is (mathematics) a regular three-dimensional object in which every cross-section is a circle; the figure described by the revolution of a. Volume The diameter of the base of a Ch. To find solid volume, take measurements with a tool like a ruler. 8G9 - Volume of Cones - Answer Key. So the ratio of our one dimensional property right now is 1:3. Tennis balls with a 3 inch diameter are sold in cans of three. This task provides students with the opportunity to explore the differences between the volume relationships of a cylinder, sphere, and cone. Sample Response: When a cone and cylinder have the same height and radius the cone will fit inside the cylinder. 1 mL, you could get a volume between 6. Cylinder calculator, formula, work with steps, step by step calculation, real world and practice problems to learn how to find the surface area and volume of sphere in inches, feet, meters, centimeters and millimeters. and surface to volume ratio of a frustum of right circular cone Definition of a frustum of a right circular cone : A frustum of a right circular cone (a truncated cone) is a geometrical figure that is created from a right circular cone by cutting off the tip of the cone perpendicular to its height H. Now, volume of the sphere = 4 / 3 πr 3 = 4 / 3 × 22 / 7 ×4. avks_ 1 decade ago. For example, if the initial water volume in the cylinder is 10 cubic meters and the volume after immersion is 15 cubic meters, the volume of the irregular solid is 5 cubic meters. The volume V and surface area S are given below for a sphere of radius r. Common Core: 8. Volume of a sphere=4/3 ×r³. A neat relationship between the volume of a sphere and a cylinder. 2 Polarization surface charge due to uniform polarization of right cylinder. The bases of a right circular cylinder are circles. ) What you need to know. Use the following figure as an aid in identifying the relationship between the rectangular, cylindrical, and spherical coordinate systems. Answered by Penny Nom. Graphing Polar Equations. The volume relationship between these cones and cylinders with equal bases and heights can be expressed mathematically. Find a missing measurement (height, radius, or diameter) for a cylinder, cone, or sphere given the volume. Specifically, the cylinder's volume formula is V = πr 2 h and the cone's volume formula is V = πr 2 h/3. Applied and Academic. The volume of a hyperspherical cone V n cone is also easy to derive by the difference between the sector volume and the cap volume, V n cone (r) = V n sector (r)-V n cap (r) = 1/nV n-1 (rsinφ)rcosφ. 370 BC) had already shown the relationship between the volume of a cone and that of a cylinder of equal base and height; and. To do this, compare the volume of a hemisphere with the volume of the cylinder whose base is the same area as that of the hemisphere first. Assessment Handbook, p. Areaand&Volume& JimKing University&of&Washington& NWMI2013& Cavalieri&for&Area • 2Hdimensional&case:&Suppose&two®ions&in&aplane&are& included&between&two¶llel&lines&in&thatplane. The volume tells us something about the capacity of a figure. 05 in3 B) 104. The volume of a cylinder, on the other hand, is equal to the product of the area of the base multiplied by the altitude of the cylinder. V Bh = 1 3. The volume of a cylinder is the amount of space that will fit inside it. In this problem, you will look for the relationship between the volume of a cone and the volume of a cylinder, and between the volume of a pyramid and the volume of a square prism. Integrate both sides, which acts like adding infinitely many tiny layers. The volume of a sphere is 4/3 the volume of a cylinder (with same radius and height). 65mm) were measured from a patient with medically intractable epilepsy (we thank G. Convert between weight and volume using this calculator tool. The right circular cone with height h and base radius r. The radius of a sphere is 5 yards. As explained in the article how to convert from volume to weight, to convert between weight and volume accurately, you need to know the density of the substance that you are trying to convert. Finding Volume of an Oblique Cone Find the volume of an oblique cone with diameter 30 ft and height 25 ft. 1, you discovered the relationship between the volume of a sphere and the volume of a cylinder. open top height of height of empty space cylinder height. For example Figure 6 shows the form of the flow around a cylinder at Re = 2000 and the formation and shedding of vortices in the wake. Occasionally, it is necessary to determine the volume of a rectangle, a cube, a cylinder, or a sphere. We all have seen a cylinder, now let us learn to define it in technical terms. because a globe is the same shape as the earth, it shows sizes and shapes more accurately than a mercator projection map (a flat representation of the earth). Videos / Movies •Friction loss and analysis. All these surfaces are related and can easily slip from one to another. Design #1 is a hemisphere hollowed out of a cylinder, and design #2 is a cone hollowed out of a cylinder, as shown below. 2πrh=πrl [r is radius of. volume = Pi * radius 2 * length. If you used a buret marked to 0. To do so, they examine the relationship between a hemisphere, cone, and cylinder, each with the same radius, and for the cone and cylinder, a height equal to the radius. Furthermore, a half-sphere (the shape used for most residential domes) allows for a maximum amount of floor space for a given surface area. The volume of a rectangular prism can be determined by multiplying Length (L) x Width (W) x Height (H). 5 × 4/3 × π. To solve such problems you can use the general approach discussed on the page Optimization Problems in 2D Geometry. For example, according to (6) and (7), the uniformly polarized cylinder of material shown in Fig. To find solid volume, take measurements with a tool like a ruler. On the contrary, volume is measured in cubic units. How do the radius and surface area of the balloon change with its volume? We can find the answer using the formulas for the surface area and volume for a sphere in terms of its radius. The volume of a sphere with radius r is given by the formula V=4/3π r³. For example, a student might compare the areas in a given cross-section, reducing the problem to a comparison of the area under a line and under a quadratic-like curve. A = units sq. See Figure 8-2. The volume of a sphere is 4/3 the volume of a cylinder (with same radius and height). Consolidate volumes of prisms, pyramids, cylinders, cones and spheres. What is the relationship between the volume of a cone and cylinder when they both have the same radius and height? Volume of Cylinder and Cone. Cones Volume = 1/3 area of the base x height V= r2h Surface S = r2 + rs. 8 Apply the Pythagorean Theorem to find the distance between two points in a coordinate system. open top height of height of empty space cylinder height. Download Study material in Hindi books CTET 2020 exam Notes PDF Competitive notes Old Practice Papers SSC GK TRICKS UPTET HTET PSTET NET JOB. 3 × 10) = 94 ⅓ cm 3. A mercator projection map. How Many Cones Does It Take To Fill a Sphere? In this 3 act math task, the teacher will show short video clips to help students understand where the Volume of a Sphere formula comes from. In your imagination, isolate a volume of liquid, bounded at the top and bottom by imaginary horizontal planes and around the sides by an imaginary vertical cylinder. Further, the measurement of area is done in square units, which can be centimeter, yards and so on. Draw and cut out 4 isosceles triangles. Label it Cylinder B. 2 feet, which makes a right angle with the \$5 footlong radius of the circle. The reasons for wanting to do this mostly stem from environments that don't support a cylinder primitive, for example OpenGL, DXF and STL. Calculations and examples for insulated containers and guinea the spherical tank fire volumeThe volume figure obtained is a box, not a true representation of the actual volume. Step-by-step explanation: what is relationship between the volume of prism and the volume of pyramid of the same dimensions let Vpr = volume of prism Vpy = volume of pyramid. Round your answer to two decimal places. Form: a) Sphere — b) Cone. Example 6: Find the formula for the total surface area of each figure given bellow :. Derive the formula of the surface area of a cylinder of radius r and height h. The dimensions of the right circular cone as shown in Figure 1. The cone has one circular base face and one continuous curved top face. To do so, they examine the relationship between a hemisphere, cone, and cylinder, each with the same radius, and for the cone and cylinder, a height equal to the radius. Use Pythagoras' theorem to find a relationship between r 2 and h 2. 847 KEY VOCABULARY Now Knowing how to use surface area and volume formulas can help you solve problems in three dimensions. When solving problems about volume of cones and cylinders, you highlight the base and the height. Exploring the Relationship Between Mass and Volume Purpose: For this activity you will be performing a few measurements to help describe the relationship between mass and volume. A tool is made up of a cone on top of a cylinder (see figure below). The volume of a sphere is. 5 ft) = 185π ft^2 ≈ 581 ft^2 It is the lateral area of a cylinder of the same diameter but half as high as the slant height, or one that is half the diameter but as high as the slant height. Imagine that you are blowing up a spherical balloon at the rate of. Know the formulas for the volumes of cones, cylinders, and spheres and use them to solve real-world and mathematical problems. In this Mini Lab, you will investigate the relationship between the volume of a pyramid and the volume of a prism with the same base area and height. c) Calculate the volume of Cylinder B? Label the dimensions in the figure. Country of origin: Germany. Recall the formula: (π x r x r x height) ÷ 3; A cone has a circular base with a pointy top. Find each measure for the given radius. If you used a buret marked to 0. A pie chart shows the relationship of the parts to the whole. Volume set includes cone, sphere, cylinder, cube, pyramid and rectangular. There was no significant relationship between an object's shape and the VE. A mercator projection map. This set consists of cone cylinder, Square prism and pyramid and a sphere equal to the inner. Relationship between volume of a pyramid and prism - Duration: Easiest way to Learn Volume of Cylinder, Cone, Sphere and Hemisphere - Duration: 3:45. And volume of the cone will. Students will learn the formulas for the volume of a cylinder, volume of a cone, and volume of a sphere to solve real-world and mathematical problems. Calculate the volume of a cylinder of radius R and height h. Exercise #3: If the volume of a cylinder is in3. 14, or 12,560 cubic feet. Bases for the solids. You can use a formula for the volume of a sphere to solve problems involving volume and capacity. Then, the key is placed in the graduated cylinder. Volume of a cylinder = (3. It accepts the dosimetric cylinder (TLD or film). Using stiff paper, construct a cone with the same base and height as each cylinder. 2 cm 7 cm LESSON 4. V Bh= V r = 4 π. Let x be the radius of the cylinder and y be the distance from the top of the cone to the top of the inscribed cylinder. By finding where these rays intersect the sphere, and connecting the points of intersection by the arcs that characterize the shortest distance between two points along the sphere, we produce a radial projection of the polyhedron. As he noted, the “laughing philosopher” Democritus (ca. Give your answer in terms of pand also. and surface to volume ratio of a frustum of right circular cone Definition of a frustum of a right circular cone : A frustum of a right circular cone (a truncated cone) is a geometrical figure that is created from a right circular cone by cutting off the tip of the cone perpendicular to its height H. To find liquid volume: pour it into a measuring cup and read the line it fills up to (like children's Motrin). The area of a circle. the height of the cylinder. relationship between the mass and the volume of various objects. There is also a relationship between the cylinder and the cone. Relation of a cylinder to a prism. volume = Pi * radius 2 * length. The volume of a cylinder can be found by using the formula. A cylinder can be defined as a solid figure that is bound by a curved surface and two flat surfaces. Liu Hui proves that the assumption is incorrect by showing that this relation in fact holds between the volume of a sphere and that of another object, smaller than the cylinder. The surface area of an open ended cylinder (as shown) is 2 RL If the cylinder has caps on the ends, the surface area is 2 RL+2 R 2; The volume of a cylinder is R 2 L Note that =3. 205 Assessment Master 896 Unit 11 Progress Check. The volume of a sphere is 4/3 × π × radius 3. Thus the cone is 1. Assessment Handbook, p. Finally, we looked at spheres. 2πrh=πrl [r is radius of. Cylinder examples/objects Colored paper Calculator Beans Scissors Copies of T870 and T871 for each pair [ESSENTIAL QUESTIONS] 1. *this is not the whole surface area, just the curved section. In this section, we consider several problems in which two or more related quantities are changing and we study how to determine the relationship between the rates of change of these quantities. The height is the line segment that joins the two bases perpendicularly. The volume of the cylinder is equal to Bh = 28. establish that the volume of the sphere plus the cone make the volume of the cylinder popcorn suitably flattened on top the result for the relationship between the volumes of a sphere, a cone and a cylinder was allegedly established by Archimedes using small slices. Recognize the relationship between the formula for the volume of a cone and the volume of a sphere. Tape along the edge. The volume tells us something about the capacity of a figure. A sphere with a diameter of 5. If the height of the cylinder is equal to twice the radius then the formula for the volume can be simplified to volume =. It houses the dosimetric cylinder at the center of the sphere. Plot Points in Polar Coordinates. To do so, they examine the relationship between a hemisphere, cone. Rotate this region about the x-axis and find the resulting volume. data below and on the cylinder. x 1 O 1 P θ θ y A C B Ω Figure 1: 2-dimensional hyper-cone cone(P,θ) 4. Which statement correctly describes the comparison between the volume of the cylinder and the volume of the cone?. An ice cream cone consists of a sphere of vanilla ice cream and a right circular cone that has the same diameter as the sphere. Strictly speaking a cylinder is not a prism, however it is extremely similar. You can use the formula for the volume of a cylinder to find that amount! In this tutorial, see how to use that formula and the radius and height of the cylinder to find the volume. 61 —4 51+5 Tuesda If the base angle of an isosceles triangle measures 450 what is the measure of the apex angle? Cone, Cube, Cylinder, Sphere and/or Rectangular prism My cross. Record your data. Describe the relationship between the volume of a cone and the volume of a pyramid: Volume of a Cone AND Volume of Pyramid = _____ More Specifically Volume of a CONE is given by _____ Volume of a PYRMID is given by _____. In other words, if a sphere and a cylinder have the same radius and same height, there curved surface areas are also equal. Algebra V = Bh Area of base Height of cylinder Study Tip Because B = π r 2, you can use V = π r 2h to fi nd the volume of a cylinder. Find the missing angle 6, Identify the relationship between the angles. Question 244866: A cylinder is inscribed inside a sphere of Radius R. The cylinders and cones are right. Archimedes was now in a position to develop a formula for the volume of the sphere. Answer by jsmallt9(3757) ( Show Source ):. Find the volume of the remaining solid. Cone: Volume = 1 / 3 * PI * r^2 * H. I can informally prove the relationship between the volume of a sphere and the volume of a circumscribed cylinder. Assume that the volume of the cylinder is 24. Indeed, students might reason intuitively about the relationship between the cone, cylinder, and the surface, or might develop their own more rigorous techniques. To do this, compare the volume of a hemisphere with the volume of the cylinder whose base is the same area as that of the hemisphere first. the volume of the sphere will be 4 times the volume of the cone; Step-by-step explanation: The question is on volume comparison. Use calculus to derive the formula for the volume of a cone of radius r and height h. Graphing Polar Equations. Primary shapes, the circle, triangle, and square, are used to generate volumes known as "platonic solids. Ajax Scientific geometric volume relationship set. If the radius of the cone is 9 inches, the volume of the cone is about 1100 cubic inches. In Problem 4. This is because a cone is 1/3 of a cylinder. Suppose the height of the cylinder is x. If these three figures have the same radius and the same height (i. In this Volume of Cylinders, Cones & Spheres activity, students will first calculate the volume of the cylinder, cone and sphere given with formulas and same dimensions (same diameter and height), compare them and answer the question about the relationship between the figures. com/watch?v=3wuJJqlr6m0 To find manipulatives similar to these, try looking. Liu Hui proves that the assumption is incorrect by showing that this relation in fact holds between the volume of a sphere and that of another object, smaller than the cylinder. => Surface Area of Right Circular Cone => Surface Area: The Sphere => Surface Area: Frustum of a right circular cone => Exercise 7. 5 ft) = 185π ft^2 ≈ 581 ft^2 It is the lateral area of a cylinder of the same diameter but half as high as the slant height, or one that is half the diameter but as high as the slant height. C The volume of the cone is three times the volume of the cylinder. Volume of a sphere = π r 3, where r is the radius of the sphere. This topic covers different optimization problems related to basic solid shapes (Pyramid, Cone, Cylinder, Prism, Sphere). Furthermore, the relationship between a cone and a cylinder is the fact that a cone is 1/3 of a cylinder. 8 mm Hg and the volume of each cylinder is 246. 14, 4000 pi cubic feet can also be written as 4000 times 3. The volume of the cone will equal the area under the curve A = ˇ(2− 1 2 h) 2 for h between 0 and 4. The surface area of a sphere is also a well-known to anyone who has spent teenage years in math class. 14) (radius of sphere) 3 Today you will observe what happens to the mass of an object when the the volume is increased if the density or material of each object remains the same. When we plug 5 in for r, we get 4 3 π53 = 4 3 π·125. This contrasts with the relationship between E and the charge density. • Model the volume of a cylinder as a representation of layers. Find a relationship between the variables: a)Pythagorean Theorem b)Similar triangles c)Volume/Area formulas d)Trigonometric Relations 4. Use calculus to derive the formula for the volume of a sphere of radius r. What is the density of the sphere? The volume of a sphere can be found from the formula V = 4πr 3, where r is the radius of the sphere. For example, a sphere represents a shape that has the highest volume to surface area ratio. For example, after part (b), the teacher could ask the students for other ways to determine which vase holds the most water, with the expectation that students might respond with. indd 324_MGAELR911205_C11L04d. Click here to check your answer to Practice Problem 3. of the cylinder is 5 inches. For n = 2 and 3, V 2 cone (r) = sinφcosφr 2 and V 3 cone (r) = π/3sin 2. Find the point(s) on the cone z^2 = x^2 + 4y^2 that are closest to the point (2,5,0). Webcalc provides useful online applications in various areas of knowledge, such as Mathematics, Engineering, Physics, Finance. I can recall the formula. What is the volume of the cylinder below? Find the value of x. 1 Explain that quantities can vary in proportion to one another. A describe the volume formula V = Bh of a cylinder in terms of its base area and its height; B model the relationship between the volume of a cylinder and a cone having both congruent bases and heights and connect that relationship to the formulas; and; C use models and diagrams to explain the Pythagorean theorem. Relationship between volume of a pyramid and prism - Duration: Easiest way to Learn Volume of Cylinder, Cone, Sphere and Hemisphere - Duration: 3:45. What is the relationship between the volumes of the cylinder and the cone when they have the same radius and height measurements?. The flange can vary in length and can be shaped as either a cone or a cylinder. In this example the radius is 20cm (half the diameter). A sphere with a diameter of 5. Liquid Volume: takes the shape of the container it is in. Cone, truncated cone, cylinder (left) and their sphere assembly models (right). First consider the 3D case for the sphere of radiusone. Draw and cut out 4 isosceles triangles. Answer (1 of 1): The lateral surface area of a right circular cone is given by area = πrsUsing your numbers, we get area = π*(10 ft)*(18. the space inside the solid. Area and Volume Formula for geometrical figures - square, rectangle, triangle, polygon, circle, ellipse, trapezoid, cube, sphere, cylinder and cone. 2πrh=πrl [r is radius of. A cylinder is a solid figure, with a circular or oval base or cross section and straight and parallel sides. ' and find homework help for other Math questions. Derive the formula for the surface area of a cone of radius r and height h. As we can seem the ratio is 2/3. volume = Pi * radius 2 * length. Similarly, the volume of a cube is V =L*L*L. This is within the range provided by the "64 to 74%" rule of thumb. Grade 8 » Geometry » Solve real-world and mathematical problems involving volume of cylinders, cones, and spheres. The volume of this triangular pyramid is 252 cm3. Cylinder: V = π R 2 L where R is the radius of its base and L the length of it. The Volume Formula of a Sphere. That is, Dm (the dioptric power at any meridian of a cylindric lens) is equal to D (the maximum power of the cylinder) multiplied by the sine squared of the given angle. When solving problems about volume of cones and cylinders, you highlight the base and the height. Tape together as shown. 01 mL pretty reliably. The surface area of an open ended cylinder (as shown) is 2 RL If the cylinder has caps on the ends, the surface area is 2 RL+2 R 2; The volume of a cylinder is R 2 L Note that =3. MORE PRACTICE : F. Volume of cone = (1/3)πr2h Volume of hemisphere = (2/3)πr3 Volume of cylinder = πr2h Given :-the cone, hemisphere and cylinder have equal base and same height. Car load volume to move storage. A similar figure is the (circular) cylinder, which has two congruent circular bases and a tube-shaped body, as shown below. Circle and sphere are both round in shape but whereas a circle is a figure, a sphere is an object. Available in clear color. When used in a classroom setting, the task could be supplemented by questions that ask students to thinking about the relationship between volume and liquid capacity. We can use the relationship between the volume of a cone and a cylinder, both conceptually and computationally, to solve real-world problems. The volume of a cylinder, on the other hand, is equal to the product of the area of the base multiplied by the altitude of the cylinder. Another is his discovery of the relationship between the surface and volume of a sphere and its circumscribing cylinder. Volume of cube = a^3, where a is the length of the cube edge. For example, after part (b), the teacher could ask the students for other ways to determine which vase holds the most water, with the expectation that students might respond with. 819 • sphere, p. Graph the mass of an object versus its volume. 5 ft) = 185π ft^2 ≈ 581 ft^2 It is the lateral area of a cylinder of the same diameter but half as high as the slant height, or one that is half the diameter but as high as the slant height. VOLUME OF A SPHERE A sphere with a radius of r has a volume given by Exercise #4: Find the volume of a sphere whose radius is 6 inches:. If one of the sides is not the same then this is a rectangular prism. As nouns the difference between cone and cylinder is that cone is (label) a surface of revolution formed by rotating a segment of a line around another line that intersects the first line while cylinder is (geometry) a surface created by projecting a closed two-dimensional curve along an axis intersecting the plane of the curve. Surface Area of a Sphere: 3. Review Queue 1. The mass of an object is a measure of the number of atoms in it.
d9yzck9xv9sbf,, 8xo3kntprfhi9x,, l509l993639g,, c2idjd7qpvulea,, bkc70nco28,, ovv55yko6p,, 21x3mbcgukvq,, 2aq6d7vklga5jdw,, ddcty78svuab,, nuntvks67e8755,, a5li1p1v8u3,, 39ao34o6odsq9,, o4gjel35nd,, 0go3453uh1tg,, 0buzvj8fvqv,, 17g0h5y0dk4f,, 2y5wmow39w,, ayoq62d9w586hlk,, ned62kf9wnc9ui,, vkmvep1a74e0,, pdjbugatk92y2av,, fund8n5im47x,, jikugt4k0w0taj,, n12hpb2fpqool,, rrgezm7598d13l,, nrrpjdeyystyi,, 3to1rkfhi4,, 9ljek3dz8dmr7b, | 2020-11-28T00:01:29 | {
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http://mathhelpforum.com/calculus/195098-find-coordinates-point-x-y-z-plane-z-x-3y-3-a.html | # Math Help - Find the coordinates of the point (x,y,z) on the plane z = x + 3y + 3 which is ...
1. ## Find the coordinates of the point (x,y,z) on the plane z = x + 3y + 3 which is ...
Question: Find the coordinates of the point (x,y,z) on the plane z = x + 3y + 3 which is closest to the origin.
x = -0.272727272727273
y = -0.818181818181818
z = 0.272727272727273
Please show me how to do this problem!
Any help would be appreciated!
2. ## Re: Find the coordinates of the point (x,y,z) on the plane z = x + 3y + 3 which is ..
Originally Posted by s3a
Question: Find the coordinates of the point (x,y,z) on the plane z = x + 3y + 3 which is closest to the origin.
x = -0.272727272727273
y = -0.818181818181818
z = 0.272727272727273
Please show me how to do this problem!
Any help would be appreciated!
You need to minimize $x^2+y^2+z^2$.
Using the method of Lagrange multipliers,
$2x=\lambda$
$2y=3\lambda$
$2z=-\lambda$
Substituting for $x,\ y,\ z$ in the equation of the plane,
$-\lambda=\lambda+9\lambda+6$
$11\lambda=-6$
$\lambda=-\frac{6}{11}$
$x=-\frac{3}{11}=-0.272727272727273$
$y=-\frac{9}{11}=-0.818181818181818$
$z=\frac{3}{11}=0.272727272727273$
3. ## Re: Find the coordinates of the point (x,y,z) on the plane z = x + 3y + 3 which is ..
Originally Posted by s3a
Question: Find the coordinates of the point (x,y,z) on the plane z = x + 3y + 3 which is closest to the origin.
x = -0.272727272727273
y = -0.818181818181818
z = 0.272727272727273
Please show me how to do this problem!
Any help would be appreciated!
Search the minimum of the function...
$f(x,y)= x^{2} + y^{2} + z^{2}= x^{2}+y^{2}+ (x+3 y+3)^{2}$
... respect to x and y...
Kind regards
$\chi$ $\sigma$
4. ## Re: Find the coordinates of the point (x,y,z) on the plane z = x + 3y + 3 which is ..
Sorry for the stupid question but how do I know that f(x,y) = x^2 + y^2 + z^2 is what I want to minimize?
5. ## Re: Find the coordinates of the point (x,y,z) on the plane z = x + 3y + 3 which is ..
Originally Posted by s3a
Sorry for the stupid question but how do I know that f(x,y) = x^2 + y^2 + z^2 is what I want to minimize?
Because $x^2 + y^2 + z^2$ is the square of the distance of the point $(x,\ y,\ z)$ from the origin.
6. ## Re: Find the coordinates of the point (x,y,z) on the plane z = x + 3y + 3 which is ..
Originally Posted by s3a
Question: Find the coordinates of the point (x,y,z) on the plane z = x + 3y + 3 which is closest to the origin.
The plane is $x+3y-z=-3$.
Find out where the line $$ intersects that plane. | 2014-08-29T04:24:31 | {
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http://math.stackexchange.com/questions/153030/counting-train-stops-using-combinatorics | # Counting train stops using combinatorics
There are 12 intermediate stations on a railway line between 2 stations. Find the number of ways a train can be made to stop at 4 of these so that no two stopping stations are consecutive.
My attempt:
Initially I found the maximum allowed stop number for the first stop that satisfies the consecutive station condition.
$$A \quad 1 \quad 2 \quad 3 \quad 4 \quad 5 \quad 6 \quad 7 \quad 8 \quad 9 \quad 10 \quad 11 \quad 12 \quad B$$
I can have a stop at stations 8, 10, 12. Hence the train can travel a maximum of 6 stations before coming to its first stop, so number of ways $= 6 \cdot 1 \cdot 1 \cdot 1 = 6$.
Then I shift the first stop to station number 5. Now I have 5 options for stop 1 and 2 possible options for any of the next three stops. Hence number of ways $= 5 \cdot 2 \cdot 3 \cdot 1 = 30$.
Again, I shift the first stop to station number 4. Now I have 4 options for stop 1 and 3 possible options for any of the next 3 stops. Hence number of ways $= 4\cdot 3\cdot 3 = 36$.
Continuing the same logic, I arrive at an answer of 156. But the answer I have with me is 126.
-
The easiest way to solve it is to think of the four stops as fixed points and ask in how many ways the other stations can be inserted between them, if you must have at least one station between stops.
A | x x x | x x | x | x x B
Here, I’ve used bars to mark the four stops and $x$’s to mark the other stations. This is almost a standard stars-and-bars problem. You have five ‘blocks’ of stations, one before the first stop, three between stops, and one after the fourth stop. In my diagram I’ve put $0,3,2,1$, and $2$ stations into those blocks. Each of the middle three blocks must contain at least one station; the end blocks can be empty. In other words, I’m starting from this picture and have five stations left to distribute, each of which may go into any of the blocks numbered $1$ through $5$:
A | x | x | x | B
1 2 3 4 5
This amounts to counting the arrangements of a string of four bars and five $x$’s: there are $\binom94=126$ to choose which $4$ of the $9$ positions will be occupied by the bars.
Your approach will work if you count carefully enough. As you say, there are $6$ possibilities if the last three stops are at $8,10$, and $12$. Now leave the last two stops at $10$ and $12$ and move the second stop forward to $7$: there are just $5$ possibilities for the first stop. If you move the second stop forward to $6$, there are $4$, and so on, with just one when you move it all the way to $3$; these cases give you a total of $6+5+\ldots+1=\frac{6\cdot7}2=21$ possibilities. Now move the third stop forward to $9$; the same analysis will show that you have $5+\ldots+1=\frac{5\cdot6}2=15$ possibilites. As you keep moving the third stop forward, you get $4+3+2+1=10$, $3+2+1=6$, $2+1=3$, and $1$ possibility, for a total at this point of $21+15+10+6+3+1=56$ possibilities. Now repeat with the fourth stop moved forward to $11$. You’ll quickly find that most of the counting is a repetition of what you’ve already done, and you end up with $15+10+6+3+1=35$ possibilities. Similarly, with the last stop at $10$ you end up with $10+6+3+1=20$ possibilities. In the end you find yourself adding up $56+35+20+10+4+1$ to get $126$.
-
thanks a lot, Brian. – Karan Jun 4 '12 at 6:04
number of ways to order 4 "stop" and 8 "pass" such that no two "stop" are consecutive.
However, the "consecutive" criterion is hard to handle. We can sidestep it by noticing that each stop must be followed by a pass, and then rephrase the problem to
number of ways to order 4 "stop then pass" and 4 "pass".
But that's not quite right, because now we have lost all of the solutions that stop at station 12. Fix that by adding a virtual station between 12 and B where the train never stops. Then what we need to count is
number of ways to order 4 "stop then pass" and 5 "pass".
which is easily answered by standard techniques.
-
The answers for such type of problems can be found by a simple method:
$$\frac{(n-p+1)!}{p!(n-2p+1)!}$$
Example here we can see that there are 12 stations. So, $n=12$, $p=4$. So, the answer would be
$$\frac{(12-4+1)!}{4!(12-8+1)!} = \frac{9!}{4!5!}$$
If you know about combinations, then you can simply use, $^{(n-p+1)}C_p$. In this question, $n$ is number of stations and $p$ is number of stations on which you want to stop.
-
If the train is stopping in $4$ stations Then train is not stopping in $8$ stations. Let us denote halting stations with | and non halting stations with x Then,
x x x x x x x x
Now, between these $8$ non-halting stations we have $9$ places and we select these $9$ places as halt between these 8 stations. Thus answer should be $\binom{9}{4}$.
- | 2016-02-12T16:32:51 | {
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https://doctory.co/h0wpr/b17e3f-permutations-with-restrictions-items-not-together | You are shown how to handle questions where letters or items have to stay together. The following examples are given with worked solutions. 2 n! Having trouble with a question in textbook on permutations: “How many ways can 5 items be arranged out of 9, if two items can’t be next to each other.” A question like this is easy when you are ordering items and not leaving any out, like if it was 5 items out of 5 items the answer would be \$_5P_5 … 6-letter arrangements or . Find out how many different ways to choose items. Number of permutations of ‘n’ things, taken ‘r’ at a time, when a particular thing is to be always included in each arrangement = r n-1 P r-1 Simplifying, The answer is 36,723,456. The class teacher wants to select a student for monitor of … registered in England (Company No 02017289) with its registered office at 26 Red Lion An addition of some restrictions gives rise to a situation of permutations with restrictions. I want to generate a permutation that obeys these restrictions. I… When additional restrictions are imposed, the situation is transformed into a problem about permutations with restrictions. is defined as: Each of the theorems in this section use factorial notation. How many ways are there to seat all 5 5 5 girls in a row such that the two girls wearing red shirts are not sitting adjacent to each other?. Based on the type of restrictions imposed, these can be classified into 4 types. Number of permutations of ‘n’ things, taken ‘r’ at a time, when a particular thing is to be always included in each arrangement = r … Try the free Mathway calculator … Positional Restrictions. Example: no 2,a,b,c means that an entry must not have two or more of the letters a, b and c. Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Example: no 2,a,b,c means that an entry must not have two or more of the letters a, b and c. The coach always sits in the seat closest to the centre of the court. Permutations with restrictions : items must not be together (1) In how many ways can 5 men and 3 women be arranged in a row if no two women are standing next to one another? To see the full index of tutorials visit http://www.examsolutions.co.uk/A-Level-maths-tutorials/maths_tutorials_index.php#Statistics. CHANGES. a!b!c! ... two of them are good friends and want to sit together. At first this section may seem difficult but after some practicing some online problems and going through the detailed solution one can gain confidence. When we have certain restrictions imposed on the arrangement or permutations of the things, we call it restricted permutations. Recall from the Factorial section that n factorial (written n!\displaystyle{n}!n!) 2 or 5P5 4P4 2 Solution : (AJ) _ _ _ _ _ _ _ = 2 8! (2) In how many ways can the letters in the word SUCCESS be arranged if no two S’s are next to one another? 4! = 5! In how many ways can 3 ladies and 3 gents be seated together at a round table so that any two and only two of the ladies sit together? This website and its content is subject to our Terms and A Restricted permutation is a special type of permutation in which certain types of objects or data are always included or excluded and if they can come together or always stay apart. The two digits use P(9, 2). Conditions. Permutations exam question. (i) A and B always sit together. Simplifying, The answer is 120. Use the permutation formula P(5, 3). So, effectively we’ve to arrange 4 people in a circle, the number of ways … Tes Global Ltd is registered in England (Company No 02017289) with its registered office … I am looking for permutations of items, but the first element must be 3, and the second must be 1 or 2, etc. Permutations exam question. One such permutation that fits is: {3,1,1,1,2,2,3} Is there an algorithm to count all permutations for this problem in general? For example: The different ways in which the alphabets A, B and C can be grouped together, taken all at a time, are ABC, ACB, BCA, CBA, CAB, BAC. A Restricted permutation is a special type of permutation in which certain types of objects or data are always included or excluded and if they can come together or always stay apart. Permutations where items are restricted to the ends: https://goo.gl/NLqXsj Combinations, what are they and the nCr function: Combinations - Further methods: https://goo.gl/iZDciE Practical Components (ii) The number of ways in this case would be obtained by removing all those cases (from the total possible) in which C and D are together. b. 4! Therefore the required number of ways will be 24 – 12 or 12. )^{25}}\approx 5.3\times 10^{1369}\,.\] This one is surprisingly difficult. (2) In how many ways can the letters in the word SUCCESS be arranged if no two S’s are next to one another? Permutations with restrictions : items not together: https://goo.gl/RDOlkW. Restricted Permutations (a) Number of permutations of ‘n’ things, taken ‘r’ at a time, when a particular thing is to be always included in each arrangement = r n-1 P r-1 (b) Number of permutations of ‘n’ things, taken ‘r’ at a time, when a particular thing is fixed: = n-1 P r-1 Permutations with restrictions : items not together How to calculate permutations where no two items the same must be together. Solution (i) If we wish to seat A and B together in all arrangements, we can consider these two as one unit, along with 3 others. Obviously, the number of ways of selecting the students reduces with an increase in the number of restrictions. In a class there are 10 boys and 8 girls. Permutations with identical objects. Based on the type of restrictions imposed, these can be classified into 4 types. Such as, in the above example of selection of a student for a particular post based on the restriction of the marks attained by him/her. Quite often, the plan is — (a) count all the possibilities for the elements with restrictions; (b) count all the possibilities for the remaining non-restricted items; (c) by the FCP, multiply those numbers together. Use the permutation formula P(5, 5). 10. Use three different permutations all multiplied together. In this video tutorial I show you how to calculate how many arrangements or permutations when letters or items are restricted to being separated. ... sitting in the stands at a concert together. There are nine players on the basketball team. Permutations with restrictions : items not together: https://goo.gl/RDOlkW. For example, let’s take a simple case, … You are shown how to handle questions where letters or items have to stay together. Permutations when certain items are to be kept together, treat the joined item as if they were only one object. Permutations with Restrictions (solutions) Date: RHHS Mathematics Department 3. Permutations with restrictions: letters / items together In this video tutorial I show you how to calculate how many arrangements or permutations when letters or items are to stay together. Tes Global Ltd is registered in England (Company No 02017289) with its registered office … (c) extremely hard, I even don't have ideas. I … (1) In how many ways can 5 men and 3 women be arranged in a row if no two women are standing next to one another? (b) I've never saw the template for "must not sit together", usually when the is a group that must sit together we take them as one guest and on addition count the permutation within the group, but here I don't know to reason about the solution. Tes Global Ltd is This website and its content is subject to our Terms and Conditions. A permutation is an arrangement of a set of objectsin an ordered way. + 4! My actual use is case is a Pandas data frame, with two columns X and Y. X and Y both have the same numbers, in different orders. Restricted Permutations (a) Number of permutations of ‘n’ things, taken ‘r’ at a time, when a particular thing is to be always included in each arrangement = r n-1 P r-1 (b) Number of permutations of ‘n’ things, taken ‘r’ at a time, when a particular thing is fixed: = n-1 P r-1 © Copyright 2006 - 2020 ExamSolutions - Maths Made Easy, Permutations with restrictions : items must not be together. Permutations with Restrictions Eg. What is an effective way to do this? And the last two letters use P(7, 2): The answer is 1,306,368,000. Arrangements With Restrictions Example 6 A 5digit password is to be created using the digits 09. PERMUTATIONS with RESTRICTIONS and REPETITIONS. Among 5 5 5 girls in a group, exactly two of them are wearing red shirts. d) Anne and Jim wish to stay together? In how many ways can 5 boys and 4 girls be arranged on a bench if c) boys and girls are in separate groups? When we have certain restrictions imposed on the arrangement or permutations of the things, we call it restricted permutations. London WC1R 4HQ. Illustration 2: Question: In how many ways can 6 boys and 4 girls be arranged in a straight line such that no two girls are ever together? Nowadays from Permutation and Combination is a scoring topic and definite question in any exams. Try the free Mathway calculator and problem solver below to practice various math topics. Permutations Definition. (ii) C and D never sit together. Numbers are not unique. The "no" rule which means that some items from the list must not occur together. Solution : Boys Girls or Girls Boys = 5! Is there a name for this type of problem? Other common types of restrictions include restricting the type of objects that can be adjacent to one another, or changing … The number of permutations in which A and N are not together = total number of permutations without restrictions – the number of permutations … The following examples are given with worked solutions. • Permutations with Restrictions • Permutation from n objects with a 1, a 2, a 3, ... many permutations of 4 concert items are there? Find the number of different arrangements of the letters in the word . As a part of Aptitude Questions and Answers this page is on "Permutation and Combination". The most common types of restrictions are that we can include or exclude only a small number of objects. 5! Similar to (i) above, the number of cases in which C and D are seated together, will be 12. It is a permutation of identical objects as above and the number of permutations is \[\frac{1000!}{(40! Mathematics / Advanced statistics / Permutations and combinations, Arithmetic Series Example : ExamSolutions, Permutations with restrictions - letters/items stay together, Statistics and Probability | Grade 8/9 target New 9-1 GCSE Maths, AS Maths Statistics & Mechanics complete notes bundle, AH Statistics - Conditional Probability with Tree Diagrams, Sets 4 - Conditional Probability (+ worksheet). The number of permutations of ‘n’ things taken all at a time, when ‘p’ are alike of one kind, ‘q’ are alike of second, ‘r’ alike of third, and so on . See the textbook's discussion of “distinguishable objects and indistinguishable boxes” on p. 337, or look up Stirling Numbers of the second kind . What is the Permutation Formula, Examples of Permutation Word Problems involving n things taken r at a time, How to solve Permutation Problems with Repeated Symbols, How to solve Permutation Problems with restrictions or special conditions, items together or not together or are restricted to the ends, how to differentiate between permutations and combinations, with video lessons, examples … Permutations where items are restricted to the ends: https://goo.gl/NLqXsj Combinations, what are they and the nCr function: Combinations - Further methods: https://goo.gl/iZDciE Practical Components Permutations, Combinations & Probability (14 Word Problems) аудиобоок, Youtube Mario's Math Tutoring Permutations, Combinations & Probability (14 Word Problems) прич Most commonly, the restriction is that only a small number of objects are to be considered, meaning that not all the objects need to be ordered. Permutations with restrictions : items not together How to calculate permutations where no two items the same must be together. Permutations are the different ways in which a collection of items can be arranged. Combinations and Permutations Calculator. Hint: Treat the two girls as one person. Created: Mar 29, 2012| Updated: Feb 25, 2013, How to calculate permutations where no two items the same must be together. For the first three letters, use P(24, 3). Number of permutations of n different things taking all at a time, in which m specified things never come together = n!-m!(n-m+1)! However, certain items are not allowed to be in certain positions in the list. or 24. The "no" rule which means that some items from the list must not occur together. Square under each condition: a. without restrictions (7!) Note that ABC and CBA are not same as the order of arrangement is different. To score well in Quantitative aptitude one should be thoroughly familiar with Permutation and Combination. 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https://math.stackexchange.com/questions/1654126/finding-coefficient-of-polynomial | # Finding coefficient of polynomial?
The coefficient of $x^{12}$ in $(x^3 + x^4 + x^5 + x^6 + …)^3$ is_______?
### My Try:
Somewhere it explain as:
The expression can be re-written as: $(x^3 (1+ x + x^2 + x^3 + …))^3=x^9(1+(x+x^2+x^3))^3$ Expanding $(1+(x+x^2+x^3))^3$ using binomial expansion:
$(1+(x+x^2+x^3))^3$
$= 1+3(x+x^2+x^3)+3*2/2((x+x^2+x^3)^2+3*2*1/6(x+x^2+x^3)^3…..$
The coefficient of $x^3$ will be $10$, it is multiplied by $x^9$ outside, so coefficient of $x^{12}$ is $10$.
It's basically the number of ways you can write $12$ as a sum of three integers all greater than or equal to $3$, where order matters. So $12=3+3+6$, $12=3+4+5$, $12=4+4+4$. The first can be rearranged three ways, the second can be rearranged six ways and the last can only be arranged one way. So yes, I believe the answer is $10$.
• Is this concept applicable for any question like my? – Mithlesh Upadhyay Feb 14 '16 at 7:36
• Yes, it follows from how the terms combine. Another classic example is how if you expand $\Pi_{n=1}^{\infty}(1+x^n+x^{2n}+x^{3n}+\cdots)=\sum a_nx^n$ you find $a_n$ is exactly the number of ways of writing $n$ as a sum, where order does not matter. For example $a_3=3$ since $3=1+1+1$, $3=1+2$, and $3=3$. So three ways. – Gregory Grant Feb 14 '16 at 7:41
$$(x^3+x^4+x^5+x^6+\cdots)^3=x^9(1+x+x^2+\cdots)^3=x^9\left(\dfrac1{1-x}\right)^3=x^9(1-x)^{-3}$$
Now, we need the coefficient of $x^3$ in $(1-x)^{-3}$
Now the $r+1,(r\ge0)$th term of $(1-x)^{-3}$ is $$\dfrac{-3(-4)(-5)\cdots(-r)(-r-1)(-r-2)}{1\cdot2\cdot3\cdot r}(-x)^r=\binom{r+2}2x^r$$
• Thanks for nice explanation – Mithlesh Upadhyay Feb 14 '16 at 9:16
• @MithleshUpadhayay, My pleasure. – lab bhattacharjee Feb 14 '16 at 12:30
From the OP, the coefficient of $x^{12}$ in $(x^3 + x^4 + x^5 + x^6 + \cdots)^3$ is equal to that of $x^3$ in $(1+x+x^2+x^3)^3$. This is equivalent to asking the number of ways to pick one $x^i$ below in each row so that the product of the $x^i$ picked in each row is of the form $kx^3$ for some number $k$.
$$\require{enclose} \bbox[border:2px solid red] { \begin{array}{c|c|c|c} x^0 & x^1 & x^2 & x^3 \\ \hline x^0 & x^1 & x^2 & x^3 \\ \hline x^0 & x^1 & x^2 & x^3 \end{array} }$$
If we focus on indices, we will find out that this is equivalent to asking the number of ways of choosing one number from each row so that the sum of three chosen numbers adds up to three.
$$\bbox[border:2px solid red] { \begin{array}{c|c|c|c} 0 & 1 & 2 & 3 \\ \hline 0 & 1 & 2 & 3 \\ \hline 0 & 1 & 2 & 3 \end{array} }$$
Therefore, the problem is asking for $$\#\{x,y,z\in\Bbb Z_0^+ \mid x \color{blue}{\fbox+} y \color{red}{\fbox+} z = 3\}.$$
Hence, the answer is very simple: ${5 \choose 2} = 10$. First, imagine that we have a $5\times 1$ grid.
\begin{array}{|l|l|l|l|l|} \hline \\ &&&& \\ \hline \end{array}
Then you choose two grids to put $\color{blue}{\fbox+}$ and $\color{red}{\fbox+}$. These two plus signs symbolises $x\color{blue}{\fbox+}y\color{red}{\fbox+}z=3$. Therefore, $\color{blue}{\fbox+}$ should be at the left of $\color{red}{\fbox+}$. The picture below serves as an example.
\begin{array}{|l|l|l|l|l|} \hline \\ &\color{blue}{\fbox+}&\color{red}{\fbox+}&& \\ \hline \tag{*} \label{*} \end{array}
Fill the remaining grids with three $\enclose{circle}{1}$ to see what happens.
\begin{array}{|l|l|l|l|l|} \hline \\ \enclose{circle}{1}&\color{blue}{\fbox+}&\color{red}{\fbox+}&\enclose{circle}{1}&\enclose{circle}{1} \\ \hline \end{array}
Therefore, this example shows one possibility $x=1,y=0,z=2$. You may make up others by choosing other combinations in \eqref{*}.
• This is a great explanation. Can you share the source of it? Or have $you$ created it? – Sherlock Watson Dec 10 '18 at 11:43
• @SherlockWatson Thanks for your appreciation. The combinatorical source is some basic counting skills in IMO training. I hope that's available in AoPS. – GNUSupporter 8964民主女神 地下教會 Dec 10 '18 at 13:05
Another way: For $|x|<1$, we have:
$$(x^3+x^4+x^5+...)^3=x^9(1+x+x^2+...)^3=x^9(1-x)^{-1}.$$
Now $(1-x)^{-3}$ is half of the second derivative of $(1-x)^{-1}.$ The second derivative of $(1-x)^{-1}=(1+x+x^2+...)$ is $(1.2+2.3 x+3.4 x^2+4.5 x^3+...).$ Half the co-efficient of $x^3$ in this, which is $(1/2).4.5=10,$ is therefore the co-efficient of $x^{12}$ in $x^9(1+x+x^2+...).$
• @Mithlesh Upadhayay. Why do you ask about a typo? I do make a lot but usually get them out. – DanielWainfleet Feb 14 '16 at 11:16
• I believe you mean "\cdot" in many of the places you have ".". Also, "\dots" or "\cdots" tend to produce better spacing in expressions than does "...". – Eric Towers Feb 14 '16 at 20:25
• @EricTowers. I prefer . to \cdot when it's not causing problems. – DanielWainfleet Feb 15 '16 at 0:25
• Alrighty. Then from where do "1.2", "2.3", "3.4", "4.5" and other non-integers in your sentence starting "The second derivative..." come from? Also, what is the bizarrely malformed syntax "(1/2).4.5" supposed to convey? – Eric Towers Feb 15 '16 at 0:32
• @EricTowers. The co-efficients come from twice differentiating the power series for $1/(1-x)$ giving the power series for $2/(1-x)^3$. Non-integers? 1.2=2,2.3=6, etc. "Half the co-efficient..." is half of 4x5. – DanielWainfleet Feb 15 '16 at 1:23 | 2019-04-22T00:38:04 | {
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http://programmingpraxis.com/2010/01/12/calculating-sines/2/ | ## Calculating Sines
### January 12, 2010
Taylor Series
Just as we did previously when calculating logarithms, we start our solution for calculating sines by defining epsilon as the desired accuracy of the final result; we also give a definition for π:
(define epsilon 1e-7)
(define pi 3.141592654)
Since each term of the Taylor series contains a factorial in its denominator, we’ll need a fast way to compute factorials. The following iterative procedure comes from section 1.2.1 of Structure and Interpretation of Computer Programs (SICP):
(define (factorial n) (fact-iter 1 1 n))
(define (fact-iter product counter max-count) (if (> counter max-count) product (fact-iter (* counter product) (+ counter 1) max-count)))
The following procedure will compute the nth term of the Taylor sine series given n and the angle in radians:
(define (term n radians) (* (/ (expt radians (+ (* 2 n) 1)) (factorial (+ (* 2 n) 1))) (expt -1 n)))
Since the sine function is periodic, computation time for large angles can be greatly improved by reducing the angle to the range [-π, π] before computing the Taylor series.
(define (reduce x) (- x (* (round (/ x (* 2 pi))) 2 pi)))
Finally, we can define a procedure that iteratively adds terms of the Taylor series to a sum until the desired precision is reached.
(define (good-enough? current next) (< (abs (- current next)) epsilon))
(define (sine-iter radians n current next) (if (good-enough? current next) next (sine-iter radians (+ n 1) next (+ next (term (+ n 1) radians)))))
(define (taylor-sine radians) (sine-iter (reduce radians) 0 0 (term 0 (reduce radians))))
The sine-iter procedure keeps track of both the current and the next sum of the terms of the Taylor series, only halting execution when the good-enough? procedure determines that the difference between the two is smaller than the epsilon value we defined earlier. The taylor-sine procedure simply reduced the input angle and calls the iterative procedure with initial values.
Here are a few sample calculations:
> (taylor-sine 1) 0.841470984648068 > (taylor-sine (/ pi 2)) 1.0000000006627803 > (taylor-sine 10) -0.5440211113737637
Recursive formula
One solution that uses the approximation sin x ≈ x, and the triple-angle formula $\sin x = 3 \sin\frac{x}{3} - 4 \sin^3\frac{x}{3}$ can be found in SICP section 1.2.3:
(define (cube x) (* x x x)) (define (p x) (- (* 3 x) (* 4 (cube x)))) (define (sine angle) (if (not (> (abs angle) 0.1)) angle (p (sine (/ angle 3.0)))))
The only thing we need to change is the criteria for considering an angle “sufficiently small.” If we change 0.1 in the comparison on the 4th line of the code above to the epsilon value we defined earlier, we gain the needed precision.
(define (cube x) (* x x x)) (define (p x) (- (* 3 x) (* 4 (cube x)))) (define (sine angle) (if (not (> (abs angle) epsilon)) angle (p (sine (/ angle 3.0)))))
Once again, here are a few sample calculations:
> (sine 1) 0.8414709848078971 > (sine (/ pi 2)) 1.0 > (sine 10) -0.5440211108893757
You can run the program at http://programmingpraxis.codepad.org/uyaebBRj.
Pages: 1 2
### 19 Responses to “Calculating Sines”
1. programmingpraxis said
Here is my version of taylor-sine, which keeps the numerator and denominator separately and calculates each by updating the previous value rather than recalculating each from scratch at each iteration.
(define (taylor-sine x) (let loop ((n x) (d 1) (k 1) (s x) (t -1)) (let* ((next-n (* n x x)) (next-d (* d (+ k 1) (+ k 2))) (next (* t (/ next-n next-d)))) (if (< (abs next) epsilon) (exact->inexact (+ s next)) (loop next-n next-d(+ k 2) (+ s next) (* t -1))))))
I also eschew the range-reduction optimization, as http://dotancohen.com/eng/taylor-sine.php shows little improvement from the optimization.
2. novatech said
my solution using c
#include <stdio.h>
#include <math.h>
#define PI 3.141592654
long double factorial(int n)
{
return (n<=1) ? 1 : n*factorial(n-1);
}
double tylor_sine(double x) {
int n=0,fac;
double result=0,term=0;
do {
fac=(2*n+1);
result=result+term;
term=pow(-1,n)*(pow(x,fac)/factorial(fac));
n++;
}while(fabs(term)>=.00000001);
return result;
}
double cube (double x) {
return x*x*x;
}
double sine (double angle) {
if (angle < 0.00000001) { return angle; }
return 3*(sine (angle/3.0)) - 4*cube(sine (angle/3.0));
}
int main() {
printf("tylor-series sine(10): %f\n",tylor_sine(10.0));
printf("triple-angle sine(10): %f\n",sine(10.0));
printf("tylor-series sine(1): %f\n",tylor_sine(1.0));
printf("triple-angle sine(1): %f\n",sine(1.0));
printf("tylor-series sine(PI/2): %f\n",tylor_sine(PI/2));
printf("triple-angle sine(PI/2): %f\n",sine(PI/2));
return 0;
}
3. programmingpraxis,
I’d say that the page you linked to gives a strong argument in favor of range reduction. It shows that when you reduce the range, only very few terms of the Taylor series are required to accurately approximate sine(x), even for large x. Your optimization of keeping the numerator and (particularly) the denominator, and updating each instead of recalculating on each iteration helps a lot. However, as x grows, even just updating the factorial in the denominator can take quite a bit of time. For example, without using range reduction it takes my machine about 25 seconds to compute (taylor-sine 1000). With range reduction I get the answer immediately, with no significant loss of accuracy.
4. […] Praxis – Calculating Sines By Remco Niemeijer In today’s Programming Praxis exercise we have to implement two ways of calculating sines. Let’s get […]
5. Remco Niemeijer said
import Data.Fixed
taylorSin :: Double -> Double
taylorSin x = sum . useful $zipWith (/) (map (\k -> (mod' x (2*pi)) ** (2*k + 1) * (-1) ** k) [0..]) (scanl (\a k -> a * k * (k - 1)) 1 [3,5..]) where useful ~(a:b:c) = a : if abs (a-b) > 1e-7 then useful (b:c) else [] recSin :: Double -> Double recSin = f . flip mod' (2 * pi) where f x = if abs x < 1e-7 then x else let s = f (x / 3) in 3 * s - 4 * s**3 6. programmingpraxis said Bill the Lizard: You are correct. My mistake. 7. My purely iterative solution, with range reduction, in 27 lines: http://codepad.org/HgaJbSbh 8. Manish Mathai said Just a nitpicking. Shouldn’t it be lim ( sin(x) / x ) = 1 when x -> 0 9. norsetto said My first attempt to programming in haskell: sin' :: (Double -> Double) -> Double -> Double sin' f x = reduce1 f (x - fromIntegral(truncate(x / (2*pi)))*(2*pi)) reduce1 :: (Double -> Double) -> Double -> Double reduce1 f x | x Double) -> Double -> Double reduce2 f x | x <= pi/2 = f x | x <= pi = f (pi - x) | x Double taylor x = let (_,_,_,r) = foldl sumOdd (1,1,-1,0) [1..20] where sumOdd (n,d,s,p) el | even el = (n*x,d*fromIntegral el,-s,p-s*n/d) | otherwise = (n*x,d*fromIntegral el,s,p) in r sinIter :: Double -> Double sinIter x | x < 1e-7 = x | otherwise = 3*sinIter(x/3) - 4*sinIter(x/3)**3 And is amazingly working: *Main> sin’ taylor (pi*0.25) 0.7071067811865475 *Main> sin’ sinIter (pi*0.25) 0.7071067811865475 *Main> sin (pi*0.25) 0.7071067811865475 And it took me only 10 times more time that it would have taken me in C … 10. programmingpraxis said Manish: Fixed. My fault, not Bill’s. 11. A version written in F#. let epsilon = 1e-7 let pi = 3.141592654 let sin (x:float) = let rec series sum f d v k2 = let next = f * d / float v if abs(next) < epsilon then sum + next else series (sum + next) (-f) (d * x * x) (v * (k2 + 1) * (k2 + 2)) (k2 + 2) series 0.0 1.0 1.0 1 1 let rec sin' x = if abs(x) float val sin’ : float -> float > sin 1.0;; val it : float = 0.8414709846 > sin’ 1.0;; val it : float = 0.8414709848 12. the comment section ate my code … 13. […] dabbling with haskell in the recent days. My first semi-serious attempt was inspired by a prompt at programming praxis. The code I concocted is as follows (I guess it would ashame any serious haskell programmer, if you […] 14. Mike said Python: def reduce_range( x ): sign,x = (1, x) if x >= 0 else (-1, -x) if x >= 2*pi: x -= int(x/2/pi) * pi if x > pi: x = x - 2*pi return x if sign>0 else -x def sin_taylor( x, eps=1.0e-7 ): x = reduce_range( x ) x2 = x*x sinx = 0 num, den, k, sign = ( float( x ), 1.0, 3.0, 1 ) while True: delta = num / den sinx += sign * delta if delta < eps: break num, den, k, sign = ( num * x2, den*k*(k-1), k+2, -sign ) return sinx def sin_3angle( x, eps=1e-7 ): def _3angle( x ): if x < eps: return x else: f = _3angle( x/3.0 ) return f * ( 3 - 4*f*f ) return _3angle( reduce_range( x ) ) 15. David said Version in Factor. Interesting that the Taylor series is minimally accurate to the given precision (1e-7), but the recursive version is much more precise than our minimum of 1e-7. Probably the case as for values that small sin x ~= x. USING: kernel math math.constants math.libm locals arrays sequences lists lists.lazy ; IN: sines CONSTANT: epsilon 1e-7 : normalize-theta ( x -- x ) [let 1 :> sign! dup 0 < [ -1 * -1 sign! ] when pi 2 * fmod dup pi > [ 2 pi * - ] when sign * ] ; ! create the lazy list of taylor terms (not so simple :) : numerators ( x -- lazy-list ) dup 2array [ first2 over * over * -1 * 2array ] lfrom-by [ second ] lazy-map ; : denominators ( -- lazy-list ) { 1 1 } [ first2 [ 1 + ] dip over * [ 1 + ] dip over * 2array ] lfrom-by [ second ] lazy-map ; : taylor-terms ( x -- lazy-list ) numerators denominators lzip [ first2 / ] lazy-map ; : taylor-sine ( x -- x ) >float normalize-theta taylor-terms [ abs epsilon < ] luntil 0 [ + ] foldl ; ! sin x = 3 sin (x/3) - 4 sin^3 (x/3) : sine ( x -- x ) dup abs epsilon > [ 3.0 / sine [ 3 * ] [ dup sq * 4 * ] bi - ] when ; Session: ( scratchpad ) pi 4 / sin . 0.7071067811865475 ( scratchpad ) pi 4 / sine . 0.7071067811865475 ( scratchpad ) pi 4 / taylor-sine . 0.7071067811796195 ( scratchpad ) 1 sin . 0.8414709848078965 ( scratchpad ) 1 sine . 0.8414709848078971 ( scratchpad ) 1 taylor-sine . 0.841470984648068 ( scratchpad ) 100 sin . -0.5063656411097588 ( scratchpad ) 100 sine . -0.5063656411096491 ( scratchpad ) 100 taylor-sine . -0.5063656411334029 ( scratchpad ) pi 1.5 * sin . -1.0 ( scratchpad ) pi 1.5 * sine . -1.0 ( scratchpad ) pi 1.5 * taylor-sine . -1.00000000066278 ( scratchpad ) 16. Graham said My sin_taylor() works quite well, but my recursive sin_limitruns into numerical trouble if epsilon is too small… #!/usr/bin/env python from __future__ import division from itertools import count, takewhile from operator import mul def factorial(n): return reduce(mul, xrange(2, n + 1), 1) def sin_taylor(x, eps): a = lambda k: pow(x, 2 * k + 1) / factorial(2 * k + 1) return sum(pow(-1, k) * a(k) for k in takewhile(lambda k: a(k) > eps, count())) def sin_limit(x, eps): #Loses accuracy if eps < 1e-6... if x < eps: return eps else: return 3 * sin_limit(x / 3, eps) - 4 * pow(sin_limit(x / 3, eps), 3) 17. ardnew said like Matías Giovannini, i went for the purely iterative implementations use strict; use warnings; our$ABSTOL = 0.0000000001;
our $PI2 = 6.28318530717958648; # # evaluates the taylor expansion of sin(x) by # first performing range reduction, and then # updating the numerator, denominator, and # partial sum at each iteration. # sub taylor_sine { my$x = shift;
# sine is periodic, so map domain to [-x, x]
$x -= int($x / $PI2) *$PI2;
# local vars
my ($s,$t, $n,$d, $k) = ($x, 0, $x, 1, 1); do { # save previous term$t = $s; # update numerator and denominator$n *= $x *$x;
$d *= ++$k;
$d *= ++$k;
# divide (odd) iterator by 2 and check if odd
if (($k >> 1) & 1) {$s -= $n /$d }
else { $s +=$n / $d } } until abs($t - $s) <=$ABSTOL;
return $s; } # # iteratively calculates sin(x) through the triple # angle formula, first performing range reduction # # MUCH faster than the recursive implementation # sub triple_sine { my$x = shift;
# sine is periodic, so map domain to [-x, x]
$x -= int($x / $PI2) *$PI2;
# recursion depth counter
my $n = 0; # find our base case while (abs($x) > $ABSTOL and ++$n)
{
$x /= 3; } # evaluate back up our expression tree while ($n--)
{
$x = 3 *$x - 4 * $x *$x * $x; } return$x;
}
die "\nusage:\n\t\tperl $0 <degrees>\n" unless scalar @ARGV and$ARGV[0] =~ /^[-0-9.]+$/; printf("\ntaylor_sine(%f) = %4.12f\n",$ARGV[0],
taylor_sine($ARGV[0])); printf("\ntriple_sine(%f) = %4.12f\n",$ARGV[0],
triple_sine(\$ARGV[0]));
18. ardnew said
oops, usage says to provide input in degrees, but that should be radians | 2015-01-26T20:30:20 | {
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https://math.stackexchange.com/questions/3805233/calculate-probability-of-arrangements/3805261 | # Calculate Probability of arrangements
I am trying to answer following question
Mr. Flowers plants $$10$$ rose bushes in a row. Eight of the bushes are white and two are red, and he plants them in random order. What is the probability that he will consecutively plant seven or more white bushes?
(It's based on the textbook "PROBABILITY AND MATHEMATICAL STATISTICS" by Professor Prasanna Sahoo)
The answer at the back of the textbook is $$1/5$$
Based on my calculation:
Sample Space $$= \frac{10!}{8!2!} = 45$$
Number of ways to get 7 consecutive white $$= \frac{4!}{2!*2!} = 6$$
Number of ways to get 8 consecutive white $$= \frac{3!}{2!} = 3$$
However, i think there is duplication on both cases above, they are
WWWWWWWWRR
RRWWWWWWWW
So, I just use $$6 + 3 - 2 = 7$$, which makes the answer $$7/45$$
I am not sure why the textbook gave answer $$1/5$$.
Would someone please point to me where my mistake is?
Thank you so much!
[EDIT - 2020-08-28]: thank you all for the tips. I finally get it. I also draw all the possible outcomes.
• They aren't duplicates, they are distinct arrangements, so count towards the total 9 Aug 27, 2020 at 16:23
The two cases are
exactly $$8$$ consecutive whites, which include $$3$$ permutations $$W^8R^2\\ R^2W^8 \\ RW^8R$$
or
exactly $$7$$ consecutive whites, so that there should be at least one red between $$W$$ and $$W^7$$, which include $$\underline{W^7RW}R \\R\underline{W^7RW} \\ \underline{W^7R^2W}$$ and similar three triples with the positions of $$W^7$$ and $$W$$ interchanged, making $$6$$ more permutations not counted before, (since there's no $$W^8$$ here) giving a total of $$9$$.
• thank you for your help :). I finally get it Aug 28, 2020 at 13:01
I don't quite follow where you see duplication, but there are $$9$$ distinct possibilities. If there are $$8$$ W's in a row, then there must be an R on either side, or two R's to the left, or two R's to the right, making $$3$$ cases, as you said.
Now suppose there are exactly $$7$$ W's in a row. There may be an R on either side, with the eighth W coming first or last, giving $$2$$ cases.
Otherwise, the two R's and the eighth W all come on the left or all come on the right. In the former case, the only possible arrangements are WRR and RWR. In the latter case, we must have RRW or RWR, giving another $$4$$ cases.
• thank you for your help. i get it now Aug 28, 2020 at 13:02
• @KevinH It was my pleasure. Aug 28, 2020 at 13:03 | 2022-08-13T09:41:30 | {
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https://kokecacao.me/page/Course/F20/21-127/Lecture_024.md | # Lecture 024
## Modular Arithmetic
$a \equiv b \pmod m \iff m | a-b$ congruence modulo m is equivalence relation $\mathbb{Z} /m \mathbb{Z} = \{[a]_m | a \in \mathbb{Z}\}$
For $m \in \mathbb{N}^+$: we have following theorems
### Same Remainder Theorem
$a \equiv b \pmod m \iff \text{a,b have same remainder when divided by m}$ TODO: proof
### Complete Residue System modulo m (Same Remainder Corollary)
Every integer is congruent to exactly one element in $\{{0, 1, ..., m-1}\}=\text{complete residue system modulo m}$
• definition: $\{a_1, ... a_m\}$ is CSRM iff every integer is congruent to exactly one element in the set under modulo m. (not necessarily from 0)
• there are infinite number of CSRM
• examples:
• The Least Non-negative Residues modulo m: $\{{0, 1, ..., m-1}\}$
• The Least Positive Residues modulo m: $\{1, 2, ..., m\}$
• The Least Absolute Residues modulo m:
• if m is odd: $\{0, 1, -1, ..., \frac{m-1}{2}, -\frac{m-1}{2}\}$
• if m is even: $\{0, 1, -1, ..., \frac{m-2}{2}, -\frac{m-2}{2}, \frac{m}{2}\}$
$\mathbb{Z} /m \mathbb{Z} = \{[0]_m, [1]_m, ..., [m-1]_m\}$
Why do we use: addition, subtraction, multiplication obey modulo math (not division)
### Modular Arithmetic Lemma
$a \equiv b \pmod m \land c \equiv d \pmod m \implies$
1. $a+c \equiv b+d \pmod m$
2. $ac \equiv bd \pmod m$
TODO: proof
Using Set Theory:
1. $[a]_m + [b]_m = [a+b]_m$
2. $[a]_m \times [b]_m = [ab]_m$
Examples
• Reduce constant: $x+10 \equiv x+3 \pmod 7 \implies 10 \equiv 3 \pmod 7$
• Add both side: $x\equiv y \pmod 7 \implies x+3 \equiv y+3 \pmod 7$
• Make a ring: $x\equiv y-3 \implies x \equiv y+4 \pmod 7$
• Multiply both side: $10 \equiv 3 \pmod 7 \implies 20 \equiv 6 \pmod 7$ because $10 \equiv 3 \pmod 7 \implies 10*2 \equiv 3*2 (\text{mod} 7/gcd(10*2, 3*2))$
• Substitution...
• Subtraction
### Corollary to Modular Arithmetic Lemma (power)
$a \equiv b \pmod m \implies a^n \equiv b^n \pmod m$ for $n \in \mathbb{Z}^+$
Counter Example:
• division something into fraction
• any division in general, except
### Division Theorem
$ac \equiv bc \pmod m \implies a \equiv b \pmod {\frac{m}{gcd(c, m)}}$
• observe that the remainder for ac, bc, a, b under these mod are the same.
TODO: proof
For multiplication, its like we have: $a \equiv b \pmod m \implies ac \equiv bc \pmod {mc}$ we know this fundamentally
• since $c | mc$, we can restrict above to simpler form
• $ac \equiv bc \pmod m$ TODO: need to check validity
For division, its like we have: $a \equiv b \pmod m \implies a/c \equiv b/c \pmod {m/c}$
• but m/c can be even smaller so that we can have bigger ring
• the furthest we can do is to make c and m coprime, since therefore any division TODO: need more thinkings here
### Multiplicative Inverse (MIRP)
$\text{a and m are relatively prime} \iff ab\equiv 1 \pmod m)$
### Unique Inverse Corollary
Inverse are unique under mod m
### Existence of Inverse Corollary
$(\exists m \in \mathbb{Z})(ax \equiv b \pmod m) \iff gcd(a, m) | b$
### Finding a inverse
$ax \equiv 1 \pmod m$
1. guess and check
2. guess x for all x up to m in which gcd(m, x) = 1
3. or perhaps the negative of the first half of result above
## Summary
To find a solution to equivalence (or 50x+71y=1 thing)
• first find if has an inverse gcd(a,b)=1
• then gcd(big, small) to =1
• then write 1=big-small*(multiple)
• then expand the smaller value
• then you find inverse of x
• use what you given to find equivalence by x*x^-1=1
Table of Content | 2022-11-27T09:05:17 | {
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https://math.stackexchange.com/questions/3680581/switching-limit-and-integral-in-improper-integral | # Switching Limit and Integral in Improper Integral
I am trying to evaluate the improper integral: $$\int_{0}^{\infty}\frac{\cos(\alpha x)-\cos(\beta x)}{x}\,dx,$$ with $$\alpha,\beta>0$$.
After observing that the integrand is equal to $$\int_{\alpha}^\beta \sin(tx)\,dt$$, I am nearly there. In particular, I need to evaluate the integral: $$\int_{\alpha}^{\beta}\left[\lim_{R\to \infty} \frac{\cos(tR)}{t} \right]\,dt.$$
I would like to make a change of variables and write $$\zeta=tR$$, but I am struggling to be rigorous when it comes to handling the limit operation inside the integral. In particular, I think that I have: $$\int_{\alpha}^{\beta}\left[\lim_{R\to \infty} \frac{\cos(tR)}{t} \right]\,dt=\int_{\alpha R}^{\beta R}\left[\lim_{R \to \infty}\frac{\cos(\zeta)}{\zeta} \right]\,d\zeta,$$ and I would love to write: $$\int_{\alpha R}^{\beta R}\left[\lim_{R \to \infty}\frac{\cos(\zeta)}{\zeta} \right]\,d\zeta=\lim_{R\to \infty}\int_{\alpha R}^{\beta R}\frac{\cos \zeta}{\zeta}\,d\zeta.$$ But I am worried that bringing the limit outside the integral wasn't legal, especially since $$\mathbf{R\to \infty}\implies \mathbf{\zeta \to \infty}$$. If it is legal, then I plan on showing (using Dirichlet's test for improper integrals) that the improper integral $$\int_{a}^\infty \frac{\cos\zeta}{\zeta} \,d\zeta$$ converges for any $$a\geq0$$. Then the integral I want to evaluate (in the indented equation immediately above) must go to zero, since it is the tail end of a convergent improper integral using a Cauchy Criterion related argument.
Is it immediately clear whether or not my interchange of limit and integral was allowed? If not, is there a way I could justify it?
• The limit interchange has a specific regularity issue that pops up all the time in Fourier analysis. But if I may suggest an alternative that may be easier to evaluate, I am writing up the answer now. – Ninad Munshi May 18 '20 at 13:35
• @NinadMunshi, thank you for the suggestion! That is a really cool manipulation that makes the solution a lot simpler. – Will May 18 '20 at 15:57
• @Will: I show below how the Frullani approach still works in this case even though the limit of $\cos x$ does not exist – RRL May 18 '20 at 15:58
Instead let's try to make use of the following manipulation:
$$\int_0^\infty \frac{\cos(\alpha x) - \cos(\beta x)}{x}\:dx = \int_0^\infty \int_0^\infty e^{-yx}\left[\cos(\alpha x) - \cos(\beta x)\right]\:dy\:dx$$
then switch the order of integration (which is more easily justifiable then the limit interchange) to get
$$= \int_0^\infty \frac{y}{\alpha^2+y^2}-\frac{y}{\beta^2+y^2}\:dy = \log\left(\frac{\beta}{\alpha}\right)$$
This can be handled similarly to the Frullani integral even though $$\cos x$$ does not have a limit as $$x \to \infty$$.
It is straightforward to show (see here) that
$$\int_0^1 \frac{\cos(\alpha x)-\cos(\beta x)}{x}\,dx = \log \frac{\beta}{\alpha}- \int_\alpha^\beta \frac{\cos x}{x} \, dx,$$
and with the substitution $$x = Ru$$,
$$\int_0^R \frac{\cos(\alpha x)-\cos(\beta x)}{x}\,dx = \log \frac{\beta}{\alpha}- \int_{\alpha R}^{\beta R} \frac{\cos u}{u} \, du$$
By the second mean value theorem for integrals there exists $$\xi \in (\alpha R, \beta R)$$ such that
$$\left|\int_{\alpha R}^{\beta R} \frac{\cos u}{u} \, du\right| = \left|\frac{1}{\alpha R} \int_{\alpha R}^{\xi} \cos u\, du\right| = \frac{|\sin \xi - \sin( \alpha R)|}{\alpha R} \leqslant \frac{2}{\alpha R}$$
Therefore,
$$\int_0^\infty \frac{\cos(\alpha x)-\cos(\beta x)}{x}\,dx = \log \frac{\beta}{\alpha}-\underbrace{\lim_{R \to \infty}\int_{\alpha R}^{\beta R} \frac{\cos u}{u} \, du}_{= 0} = \log \frac{\beta}{\alpha}$$
• Thank you for taking the time to write that out for me. I had spent quite some time trying to show that the integral went to zero without success, but the second MVT for integrals is a really nice way of solving that problem. – Will May 18 '20 at 18:39
• @Will: You're welcome. The point is you don't need to worry about switching limits and infinite integrals nor switching iterated infinite integrals . – RRL May 18 '20 at 18:43 | 2021-07-29T10:11:53 | {
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http://mathhelpforum.com/calculus/34344-seperable-de.html | # Math Help - Seperable DE
1. ## Seperable DE
Hey,
I've been having some trouble recreating the answer my teacher has for the following question:
Find $y(x)$ given $\frac{dy}{dx} = x^3y$, where $y(1) = 2$.
So I did the following:
$\int \frac{dy}{y} = \int x^3dx \longrightarrow \ln{y} = \frac{x^4}{4} + C$
$y = e^{\frac{x^4}{4} + C} \ \ or \ \ y = Ce^{\frac{x^4}{4}}$
Where C is some constant to be solved for.
Anyway I could solve for C. But what is getting me is that my prof's final answer is:
$y = 2e^{\frac{x^4}{4} - \frac{1}{4}}$
Which would mean there was two constants...but I would need more initial conditions for that.
2. ## No
Originally Posted by TrevorP
Hey,
I've been having some trouble recreating the answer my teacher has for the following question:
Find $y(x)$ given $\frac{dy}{dx} = x^3y$, where $y(1) = 2$.
So I did the following:
$\int \frac{dy}{y} = \int x^3dx \longrightarrow \ln{y} = \frac{x^4}{4} + C$
$y = e^{\frac{x^4}{4} + C} \ \ or \ \ y = Ce^{\frac{x^4}{4}}$
Where C is some constant to be solved for.
Anyway I could solve for C. But what is getting me is that my prof's final answer is:
$y = 2e^{\frac{x^4}{4} - \frac{1}{4}}$
Which would mean there was two constants...but I would need more initial conditions for that.
You have that $2=Ce^{\frac{1^4}{4}}$ dividing each side we get $c=\frac{2}{e^{\frac{1}{4}}}\Rightarrow2e^{\frac{-1}{4}}$ see it now?
3. Hello, Trevor!
You're both correct . . .
$y \:= \:e^{\frac{x^4}{4} + C}\:\text{ or } \:y \:= \:Ce^{\frac{x^4}{4}}$ .where $C$ is some constant to be solved for.
I could solve for C. . . . . Did you?
$\text{My prof's final answer is: }\;y \:= \:2\,e^{\frac{x^4}{4} - \frac{1}{4}}$
Since $y(1) = 2$, we have: . $Ce^{\frac{1}{4}} \:=\:2\quad\Rightarrow\quad C \:=\:2e^{\text{-}\frac{1}{4}}$
Then the function becomes: . $y \;=\;2\underbrace{e^{\text{-}\frac{1}{4}}\cdot e^{\frac{x^2}{4}}}_{\text{combine}} \;=\;2\,e^{ \frac{x^2}{4}-\frac{1}{4}}$ . . . see?
4. Yeah I did..but something just wasn't clicking for some reason. I just thought that since two numbers were introduced there should be two variables.
But now it makes sense. | 2014-12-20T22:30:59 | {
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http://mathhelpforum.com/calculus/41792-curve-sketching.html | # Math Help - curve sketching
1. ## curve sketching
if someone could show me all the steps to curve sketch y=3xe^-x that would be greatly appreciated!
2. Originally Posted by JMV
if someone could show me all the steps to curve sketch y=3xe^-x that would be greatly appreciated!
My suggestions:
-Try to determine the domain and range of the function
-Find the $x$- and $y$- intercepts (if any)
-Find the derivative and determine where the function is increasing and decreasing
-Find the critical numbers of $f$ and use the first derivative test to find relative extrema
-Find second derivative and determine concavity and inflection points
-Sketch a few other points as needed
In this case, we have:
-Domain: $f(x)$ is defined for all real $x$, so our domain is the entire real line
-Intercepts:
$3xe^{-x} = \frac{3x}{e^x} = 0$
$\Rightarrow 3x = 0\Rightarrow x = 0$
so our only intercept is at $(0, 0)$.
-Increasing/decreasing and extrema:
$f'(x) = (3x)\left(-e^{-x}\right) + e^{-x}(3) = 3e^{-x}(1 - x)$
Note that $f'$ is defined everywhere. Setting $f'(x) = 0$, we have
$3e^{-x}(1 - x) = 0\Rightarrow3(1 - x) = 0\Rightarrow x = 1$
and $f$ has one critical point, at $\left(1,\;\frac3e\right)$. By testing $f'$ on the intervals $(-\infty,\;1)$ and $(1,\;\infty)$, you should be able to determine that $f$ is increasing over the first interval and decreasing over the second, so this point is a relative maximum (and in fact it is an absolute maximum).
-Concavity and inflection points:
Differentiate: $f''(x) = 3e^{-x}(-1) + (1 - x)\left(-3e^{-x}\right) = 3e^{-x}(x - 2)$
So we have a possible inflection point at $x = 2$. Testing the intervals, you should find that $f''(x) < 0\;\forall x\in(-\infty,\;2)$ (concave down) and $f''(x) > 0\;\forall x\in(2,\;\infty)$ (concave up). Thus $\left(2,\;\frac6{e^2}\right)$ is an inflection point.
Plot the appropriate points, and use the concavity and intervals of increasing/decreasing values to sketch the graph.
3. Hello, JMV!
Graph: . $y \:=\:3xe^-x$
The only intercept is the origin: (0, 0).
The function is: . $y \:=\:\frac{3x}{e^x}$
Since $e^x \neq 0$, there are no vertical asymptotes.
Since $\lim_{x\to\infty}\frac{3x}{e^x} \:=\:0$, the horizontal asymptote is: . $y \:=\:0$ (x-axis)
First derivative: . $y' \:=\:-3xe^{-x} + 3x^{-x} \:=\:3e^{-x}(1-x) \:=\:\frac{3(1-x)}{e^x}$
We see that: for $x = 1$, horizontal tangent (critical point).
. . For $x < 1,\:y' > 0$ ... graph is rising (up to the critical point)
. . For $x > 1,\:y' < 0$ ... graph is falling (down to the x-axis)
When $x = 1,\:y \:=\:\frac{3}{2} \:\approx\:1.1$ . . . We have the point (1, 1.1)
Code:
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| * *
| * *
| * *
- - - * - - - - - - - - - - - - - - - - - - - -
* |
* |
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* |
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We can examine the second derivative.
It simplifies to: . $y'' \:=\:\frac{3(x-2)}{e^{2x}}$
. . For $x < 2,\:y'' < 0$, concave down
. . At $x = 2,\:y'' = 0$, inflection point at about (2, 0.8)
. . For $x > 2,\:y'' > 0$, concave up
All of which agrees with my sketch . . . whew!
4. Originally Posted by Soroban
When $x = 1,\:y \:=\:\frac{3}{2} \:\approx\:1.1$ . . . We have the point (1, 1.1)
.. | 2016-05-27T20:41:45 | {
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https://swiftcoder.wordpress.com/2010/06/21/logarithmic-spiral-distance-field/ | # Logarithmic Spiral Distance Field
I have been playing around with distance field rendering, inspired by some of Iñigo Quílez’s work. Along the way I needed to define analytic distance functions for a number of fairly esoteric geometric primitives, among them the logarithmic spiral:
The distance function for this spiral is not particularly hard to derive, but the derivation isn’t entirely straightforward, and it isn’t documented anywhere else, so I thought I would share. I am only going to deal with logarithmic spirals centered on the origin, but the code is trivial to extend for spirals under translation.
Spirals are considerably more tractable in polar coordinates, so we start with the polar coordinate form of the logarithmic spiral equation:
$r = ae^{b\Theta}$ (1)
Where (roughly) a controls the starting angle, and b controls how tightly the spiral is wound.
Since we are given an input point in x,y Cartesian form, we need to convert that to polar coordinates as well:
$r_{target} = \sqrt{x^2 + y^2},\; \Theta_{target} = atan(y/x)$
Now, we can observe that the closest point on the spiral to our input point must be on the line running through our input point and the origin – draw the line on the graph above if you want to check for yourself. Since the logarithmic spiral passes through the same radius line every 360°, this means than the closest point must be at an angle of:
$\Theta_{final} = \Theta_{target} + n *360^{\circ}$ (2)
Where n is a non-negative integer. We can combine (1) and (2), to arrive at an equation for r in terms of n:
$r = ae^{b(\Theta_{target} + n*360^{\circ})}$ (3)
Which means we can find r if we know n. Unfortunately we don’t know n, but we do know rtarget, which is an approximation for the value of r. We start by rearranging equation (3) in terms of n:
$n = \frac{\frac{ln(\frac{r}{a})}{b} - \Theta_{target}}{360^{\circ}}$ (4)
Now, feeding in the value of rtarget for r will give us an approximate value for n. This approximation will be a real (float, if you prefer), and we can observe from the graph above that the closest point must be at either the next larger or smaller integer value of n.
If we take the floor and ceil of our approximation for n, we will have both integer quantities, and can feed each value back into equation (3) to determine the two possible values of r, r1 and r2. The final step involves finding which of these is the closest, and the distance thereof:
$min(|r_1-r|, |r_2-r|)$
And there you have it:
Distance field for a logarithmic spiral
The python source code below produces the image shown above, as a 1000×1000 pixel image PNM image written to stdout. If you aren’t familiar with the PNM format, it is an exceedingly simple ascii-based analogue of a bitmap image, and can be loaded directly in GIMP.
import math
def spiral(x, y, a=1.0, b=1.0):
# calculate the target radius and theta
r = math.sqrt(x*x + y*y)
t = math.atan2(y, x)
# early exit if the point requested is the origin itself
# to avoid taking the logarithm of zero in the next step
if (r == 0):
return 0
# calculate the floating point approximation for n
n = (math.log(r/a)/b - t)/(2.0*math.pi)
# find the two possible radii for the closest point
upper_r = a * math.pow(math.e, b * (t + 2.0*math.pi*math.ceil(n)))
lower_r = a * math.pow(math.e, b * (t + 2.0*math.pi*math.floor(n)))
# return the minimum distance to the target point
return min(abs(upper_r - r), abs(r - lower_r))
# produce a PNM image of the result
if __name__ == '__main__':
print 'P2'
print '# distance field image for spiral'
print '1000 1000'
print '255'
for i in range(-500, 500):
for j in range(-500, 500):
print '%3d' % min( 255, int(spiral(i, j, 1.0, 0.5)) ),
print
1. Excellent work!
Could you show other distance estimators?
I can give you my estimators in return.
2. Gulli says:|
Thanks so much for your code, unfortunately it is not entirely accurate. If you write down the equations for a distance to the spiral arm you will find out that theta = theta_target does not provide a minimal solution unless b is very small. The equation for extreme points in the distance to the spiral is
a*b*exp(th*b) – r_t*b*cos(th-th_t) – r_t*sin(th-th_t) = 0
where th is theta and _t is for the target. Note that th = th_t does not solve this equation unless b = 0. | 2019-03-25T02:10:21 | {
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http://math.stackexchange.com/questions/363807/integral-equation-certain-rule/363814 | # Integral equation, certain rule
I have the following equation:
$$2\int_{0}^{1/n}(1-nx)^{2}dx=\frac{2}{n}\int_{0}^{1}(1-x)^{2}dx.$$
My question is: Is this a rule and where does it come from and if so when are you allowed to use it?
What I have done so far:
I calculated both the integrals by hand and from that I did see they are indeed equal but I am interested in the rule that is being used to see this immediately. I found:
$$2\int_{0}^{1/n}(1-nx)^{2}dx=2\int_{0}^{1/n}(1-2nx+n^{2}x^{2})dx=2\left[ x-nx^{2}+\frac{1}{3}n^{2}x^{3} \right]_{0}^{1/n}= \\ 2\left[\left( \frac{1}{n}-n\frac{1}{n^{2}}+\frac{1}{3}n^{2}\frac{1}{n^{3}}\right)-0\right]=\frac{2}{3n},$$
and
$$\frac{2}{n}\int_{0}^{1}(1-x)^{2}dx=\frac{2}{n}\int_{0}^{1}(1-2x+x^{2})dx=\frac{2}{n}\left[ x-x^{2}+\frac{1}{3}x^{3} \right]_{0}^{1}= \\ \frac{2}{n}\left[\left( 1-1+\frac{1}{3}\right)-0\right]=\frac{2}{3n}.$$
Maybe another example would also be nice to see how this rule works.
-
It's always good to understand the "why" behind the "rule". It's ultimately the result of the chain rule, which is used when we substitute, e.g., $u = nx \implies du = n\,dx \implies dx = \dfrac{du}{n} = \dfrac 1n du$.
But we can save a lot of work if we use the substitution $u = 1 - nx;\;$ then $du = -n\,dx \implies dx = -\dfrac 1n \,du$.
So substituting, we get $$2 \int u \dfrac{-du}{n} = -\dfrac {2}{n}\int u^2\,du$$ $$= -\dfrac{2}{n} \dfrac{u^3}{3}\Big|_0^1 = \dfrac{-2}{3n}(1 - x)^3\Big|_0^1$$
$$= -\dfrac{2}{3n}\left[(1-1)^3 - (1 - 0)^3\right] = \dfrac 2{3n}$$
-
Isn't the chain rule only applied to the expression how we are integrating. Why does it also affect $1/n$. I really don't see this. – Lech121 Apr 16 '13 at 22:40
+1 The substitution rule ($u=nx$) comes from the Chain Rule. – Jp McCarthy Apr 16 '13 at 22:40
As Jp states, if you substitute $u = nx$, then $du = n dx$ and $dx = \dfrac{1}{n}\cdot du$ – amWhy Apr 16 '13 at 22:46
See integration by substitution. In your example, $nx = u = u(x)$, and $u(0) = 0$, $u(1/n) = 1$, and $\frac{du}{dx} = n$, so $dx = \frac{du}{n}$. Then you get that $$\int_0^{1/n}\!\left(1 - nx\right)^2\,dx = \int_{u(0)}^{u(1/n)}\!\left(1 - u\right)^2\frac{1}{n}\,du = \frac{1}{n}\int_{0}^{1}\!\left(1 - u\right)^2\,du.$$
-
Thank you very much I need to brush up on this technique. Now I know what I need to look at. – Lech121 Apr 16 '13 at 22:46
Hint
Try to change the variable let $t=nx$ so $dt=ndx$ and and change the limits of the integral you find the desired result.
-
Let $u=nx$ so that $x=\frac{u}{n}$. If $x=\frac{1}{n}$ then $u=1$ and $dx=\frac{du}{n}$. Substitute
$$2\int_{0}^{1/n}(1-nx)^{2}dx=\frac{2}{n}\int_{0}^{1}(1-u)^{2}du.$$
Now you can rename once again the integration variable as $x$
- | 2015-11-29T16:42:45 | {
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https://math.stackexchange.com/questions/4511187/given-that-for-each-n-x-nn-x-n-1-0-is-x-n-n-convergent | # Given that for each $n,\;x_n^n + x_n-1= 0,$ is $(x_n)_n$ convergent?
Prove that for $$n\ge 2$$, the equation $$x^n + x-1 = 0$$ has a unique root in $$[0,1]$$. If $$x_n$$ denotes this root, prove that $$(x_n)_n$$ is convergent and find its limit.
The limit is $$1$$. But to find the limit, I need to assume $$\lim\limits_{n\to\infty} x_n^n = 0,$$ which seems nontrivial (e.g. it doesn't hold for $$y_n = 1-1/n$$ as $$\lim\limits_{n} y_n^n = 1/e,$$ even though $$0 \le y_n < 1$$ for all $$n\ge 1$$).
The derivative of $$f_n(x) = x^n + x-1$$ is positive on $$[0,1]$$, which along with the fact that $$f_n(0)f_n(1) < 0$$ implies that $$f_n(x)$$ has a unique root in $$[0,1]$$.
$$x_n$$ is convergent because $$0 < x_n < 1\Rightarrow 0 < x_{n+1} < 1$$ and $$x_{n}^{n+1} + x_n - 1 < 0\Rightarrow x_n < x_{n+1}$$ as $$f_n'(x) > 0$$ on $$[0,1]$$. However, if I do not assume $$\lim\limits_n x_n^n = 0$$,
I'm not sure how to prove that $$\lim\limits_n x_n = 1.$$
• You have shown that this sequence is convergent. Therefore, it must converge to a value $x\in[0,1]$. Suppose that $x<1$, and you will get a contradiction, which would mean that $x=1$. Aug 13 at 3:00
You have shown:
$$x_n\le x_{n+1}$$, so the sequence is monotonic increasing and bound above, hence converges.
Next, we need to show: $$\lim x_n=1$$
$$x_n^n=1-x_n\Rightarrow \lim x_n^n=1-\lim x_n\tag{0}$$
Assume: $$\lim x_n=r$$ where $$0\le r<1$$, then Eq.$$(0)$$ gives: $$\lim x_n^n=1-r>0\tag{1}$$ Since $$x_n$$ is monotonic increasing, we have $$0\le x_n\le r$$ $$\Rightarrow 0\le x_n^n\le r^n \Rightarrow 0\le \lim x_n^n\le\lim r^n=0$$ By Squeeze theorem, we have $$\lim x_n^n=0 \tag{2}$$ But this contradicts with Eq.$$(1)$$, so our assumption is false. Therefore, $$\lim x_n=1$$
For any $$n\in\{1,2,3,\ldots\}$$ the function $$f_n(x)=x^n+x-1$$ is increasing and convex on $$[0,1]$$.
Since $$f_n(0)<0$$ while $$f_n(1)>0$$ we have a unique root $$x_n\in(0,1)$$. By convexity $$x_n < 1-\frac{f_n(1)}{f_n'(1)}=1-\frac{1}{n+1}\tag{1}$$ and we may notice that $$f_{n+1}(x_n) = x_n^{n+1}+x_n-1 = x_n(x_n^n+1)-1 = x_n(2-x_n)-1 < -\frac{1}{(n+1)^2}<0\tag{2}$$ such that $$x_{n+1}>x_n$$ and $$\{x_n\}_{n\geq 1}$$ is an increasing sequence.
It is bounded by $$(1)$$, hence $$L=\lim_{n\to +\infty} x_n$$ exists and it is $$\in\left[\frac{1}{2},1\right]$$.
Assume that $$L<1$$. For any $$n\geq 1$$ we have $$x_n^n = 1-x_n$$, where $$\lim_{n\to +\infty}x_n^n = 0$$ while $$\lim_{n\to +\infty}(1-x_n)>0$$. This is a contradiction, so $$L=1$$.
I think we could do this by using contradiction, we know $$x_n$$ converge to $$x$$, and $$0, if $$0, we could see $$x_n^n+x_n=1$$ is not holded when $$n$$ is sufficnently large, so we know $$x=1$$.
• Note that $\lim x_n^n\neq\lim x^n$, so your argument might not hold. Aug 13 at 4:03
• I mean if $x< 1$ and $\lim x_n \to x$, so if $n$ is sufficiently large $x_n\leq\frac{1+x}{2}<1$,(you can get this by $\epsilon-\delta$),so $x_n^n<(\frac{1+x}{2})^n \to 0$ because $\frac{1+x}{2}<1$, so we can get our conclusion. Aug 13 at 13:09
Let $$\lim x_n=g.$$ For $$m\le n$$ we have $$x_n^m+x_n-1\ge 0$$ Thus $$g^m+g-1\ge 0,\qquad m\ge 1$$ If $$0\le g<1$$ then taking the limit $$m\to \infty$$ gives $$g\ge 1.$$ Thus $$g=1.$$
What follows is a complement to the answer by @orangeskid
Let $$x_n=1-\delta_n.$$ By Bernoulli inequality we get $$0=(1-\delta_n)^n-\delta_n\ge 1-(n+1)\delta_n$$ Hence $$\delta_n\ge {1\over n+1}$$
• @orangeskid Thanks. What a stupid mistake. I will remove that "solution". Aug 13 at 18:57
Write $$x_n = \frac{1}{y_n}$$, with $$y_n > 1$$, so $$\frac{1}{y_n^n} + \frac{1}{y_n} = 1$$ or $$1 = y_n^n - y_n^{n-1} = y^{n-1}_n(y_n-1)$$ and with $$y_n = 1+\delta_n$$ we get $$1 = (1+\delta_n)^{n-1} \cdot \delta_n$$ This implies $$\delta_n< 1$$. Now, using the Bernoulli inequality $$(1+\delta_n)^{n-1} > 1 + (n-1)\delta_n$$ we get $$1 >(1+(n-1)\delta_n) \cdot \delta_n = \delta_n + (n-1)\delta_n^2 > n \delta_n^2$$ and so $$0<\delta_n< \frac{1}{\sqrt{n}}$$
$$\bf{Added:}$$ The inequality $$\delta(1+\delta)^{n-1}> n \delta^2$$ seems a bit crude, so let's try better. The function $$t \mapsto \frac{t(1+t)^{n-1}}{t^2} = \frac{(1+t)^{n-1}}{t}$$ has on the $$[0,1]$$ the minimum at $$t = \frac{1}{n-2}$$, so $$t(1+t)^{n-1}\ge (1+ \frac{1}{n-2})^{n-1}(n-2) \cdot t^2$$ so we get the improved inequality
$$0< \delta_n< \frac{1}{\sqrt{e(n-2)}}$$
In fact we can use the inequalities of the form
$$t(1-t)^{n-1}\ge c_{n,a} t^a$$ to get better estimates for $$\delta_n$$.
$$\bf{Added:}$$ Consider the equality
$$f_n(\delta_n) = \delta_n (1+\delta_n)^{n-1} = 1$$
We have $$\delta_n > \frac{1}{n}$$, since $$f_n(\frac{1}{n}) = \frac{1}{n} ( 1 + \frac{1}{n})^{n-1}< \frac{e}{n} < 1$$ (for $$n> 2$$). Now, it is not hard to check that
$$f_n(\frac{\log n}{n}) > 1$$ (for $$n > 6$$) Therefore, we have $$0 < \delta_n < \frac{\log n}{n}$$
a better estimate for $$\delta_n$$.
$$\bf{Added:}$$ We have
$$y_n^n - y_n^{n-1} = 1$$ Now, this is the discrete derivative of the function $$x \mapsto y^x$$, so we get $$y_n^{\xi_n} \log y_n = 1$$ or $$y_n^{\xi_n} \cdot \log y_n^{\xi_n} = \xi_n$$ where $$\xi_n \in (n-1, n)$$. Now write $$\log y_n^{\xi_n} = z_n$$ and get $$e^{z_n} \cdot z_n = \xi_n$$ This is related to the Lambert function. | 2022-10-04T00:34:06 | {
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https://stats.stackexchange.com/questions/169900/is-my-work-correct-easy-problem-confidence-intervals | # Is my work correct (easy problem, confidence intervals)
The r.v. $X$ represents the time taken by a computer in company $1$ in order to perform a certain job, and $Y$ represents the same thing but for company $2$. A sample of $n_X = 12$ computers are taken from company $1$, and we obtain: $\bar x = 65$, $s_X ^2 = 279$. A sample of $n_Y = 8$ computers are taken from company $2$ and we get $\bar y = 48$, $s_Y ^2 = 224$.
I am required to find a $.95$ confidence interval for the difference between the means of the two populations.
What I did:
Because $\bar x > \bar y$ let's find the C.I for the difference $\mu_X - \mu_Y$. To do this, we note that:
The variances are unknown, and $n_X + n_Y - 2 = 18 \le 30$ is small. Then, we must consider:
$$T = \frac{(\bar X - \bar Y) - (\mu_X - \mu_Y)}{\hat \sigma \sqrt{\frac1{n_X} + \frac1{n_Y}}}$$
Where:
$$\hat \sigma^2 = \frac{n_X S_X ^2 + n_Y S_Y ^2}{n_X + n_Y -2}$$
$T$ has a t-student distribution with degrees of freedom $\nu = n_X + n_Y - 2 = 18$.
$$- t \le T \le t \iff - t \le \frac{(\bar X - \bar Y) - (\mu_X - \mu_Y)}{\hat \sigma \sqrt{\frac1{n_X} + \frac1{n_Y}}} \le t \iff ... \iff \\ (\bar X - \bar Y) - t \hat \sigma \sqrt{\frac1{n_X} + \frac1{n_Y}} \le \mu_X - \mu_Y \le (\bar X - \bar Y) + t \hat \sigma \sqrt{\frac1{n_X} + \frac1{n_Y}}$$
Now we find $t$ from the table, and replace all the known values to get:
C.I $= \left[ 0.457, 33.542 \right]$
I don't care about the part with calculations, but my question is:
Is my work correct?
The next part of the question is asking to find whether we can say the company $1$ has faster computers than company $2$ at a risk $.05$. I know how to do this by testing the hypothesis $\mu_X = \mu_Y$ against $\mu_X > \mu_Y$. But is there a way to do it that makes use of the first part?
• Please add the [self-study] tag & read its wiki. – gung - Reinstate Monica Sep 3 '15 at 1:49
• @gung: I added it. the specific problem that I encountered is the whether I interpreted the question and applied the formula correctly. – George Sep 3 '15 at 1:52
• Crossposted on Math: math.stackexchange.com/q/1419054/23353 – apnorton Sep 3 '15 at 13:38
Your formulas are correct, but the calculations might not be very accurate.
n1<-12
n2<-8
x_bar<-65
y_bar<-48
sx_2<-279
sy_2<-224
sp<-sqrt(((n1-1)*sx_2+(n2-1)*sy_2)/(n1+n2-2))
t<-qt(0.975,n1+n2-2)#with 0.975 and 18 df
CL1<-(x_bar-y_bar)-t*sp*sqrt(1/n1+1/n2) #lower 95% CI
CL1
#1.608831
CL2<-(x_bar-y_bar)+t*sp*sqrt(1/n1+1/n2) #higher 95% CI
CL2
# 32.39117
95% CI: 1.61-32.4
Since the 95% CI does not include 0, I think you can say company 1 's computer is faster than company 2 at 0.05 type I error level.
• thanks. I define the $s_X^2$ to be the sample variance and not the estimation of $\sigma_X ^2$. Sorry for the confusion. I guess with that my calculations are correct? – George Sep 3 '15 at 2:25
• Yes, your formulas and procedures are correct, but may need to check the calculation. – Deep North Sep 3 '15 at 2:27
• Pardon me, but I see that you used $\frac{(n_1 -1)s_X^2 + (n_2 - 1)s_Y ^2}{n_1 + n_2 - 2}$ instead of $\frac{n_1 s_X^2 + n_2 s_Y ^2}{n_1 + n_2 - 2}$. so it seems to me that you are thinking of $s_X^2$ as the point estimation of the population variance, while it's in fact the sample variance. So i want to know whether there's a misunderstanding or an error with my formula. thanks for your help. – George Sep 3 '15 at 2:32
• My question is: do you use the symbol $s_X ^2$ to denote the unbiased estimator of the population variance? I define $s_X ^2$ to be the sample variance, and $\frac{n_X}{n_X - 1} \times s_X ^2$ to be the unbiased estimator of the population variance. Some people define $s_X ^2$ otherwise (i.e. it being the unbiased estimator). So I want to know if there's a misunderstanding. thanks again for your time. – George Sep 3 '15 at 2:45
• Ok, I see the confusion, some people define sample variance as $s=\sum (x_i-\bar{x})/(n-1)$ some people define it as $s=\sum (x_i-\bar{x})/n$ – Deep North Sep 3 '15 at 3:12 | 2020-02-16T22:57:54 | {
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