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https://math.stackexchange.com/questions/2835377/combination-notation-vs-binomial-coefficient-formula | # Combination notation vs. Binomial Coefficient Formula
I'm studying probability and statistics and had a question regarding notation.
I noticed that combinations and the binomial coefficient are essentially the same thing, that is:
$$\binom{n}{k}\ =\ _nC_k\ =\ \frac{n!}{(n-k)!k!}$$
But I was wondering, is there a particular difference between the two that people should be aware of? For example, are there certain use cases where one is preferred over the other?
Thank you.
• They are exactly the same thing, just different notation. I think the most popular (at least from what I see) is the $\binom{n}{k}$ notation, however I have seen $C^n_r$ used in probability courses. – Dave Jun 29 '18 at 2:48
• Also ${^n\mathrm C_r}$ – Graham Kemp Jun 29 '18 at 3:31
• Also $_r\text{C}^n$ – Dzoooks Jun 29 '18 at 4:04
All three expressions mean when dealing with combinations the same. But there are some aspects which should be considered.
We often find the binomial coefficients $\binom{n}{k}$ resp. $_nC_k$ defined by factorials. \begin{align*} \binom{n}{k}:=\frac{n!}{k!(n-k)!}\qquad\qquad\text{resp.}\qquad\qquad _nC_k:=\frac{n!}{k!(n-k)!} \end{align*} From this point of view the factorials $n!$ can be seen as basic building blocks for the shorthand notations $\binom{n}{k}$ and $_nC_k$. Since using factorials is more fundamental than the other two representations I will consider in the following only $\binom{n}{k}$ and $_nC_k$.
Historical aspects:
• C. Jordan writes in his classic Calculus of Finite Differences (1939)
• ($\mathrm{\S}$ 22): Since the binomial coefficient is without doubt the most important function of the Calculus of Finite Differences it was necessary to adopt some brief notation for this function. We accepted above the notation of J. L. Raabe [Journal für reine und angewandte Mathematik 1851, Vol. 42, p. 350] which is most in use now, putting
\begin{align*} \binom{x}{n}=\frac{x(x-1)(x-2)\cdots(x-n+1)}{1\cdot2\cdot3\cdots n} \end{align*}
• ($\mathrm{\S}$ 22, footnote 18): Euler first used the notation $\left[\frac{x}{n}\right]$ in Acta Acad. Petrop.V, 1781 and then $\left(\frac{x}{n}\right)$ in Nova Acta Acad. Petrop. XV. 1799-1802. Raabe's notation $\binom{x}{n}$ is a slight modification of the second. It is used for instance in:
• Bierens de Haan, Tables d'Int$\mathrm{\acute{e}}$grales d$\mathrm{\acute{e}}$finies, Leide, 1867
• Hagen, Synopsis Vol. I. p. 57, Leipzig, 1891,
• Pascal, Repertorium Vol. I, p. 47, Leipzig, 1910,
• Encyclopädie der Math. Wissenschaften, 1898-1930,
• L. M. Milne Thomson, Calculus of Finite Differences, 1933,
• G. H. Hardy, Course of Pure Mathematics, p. 256, 1908
Euler's notation is also cited in
• A History of Mathematical Notations by F. Cajori, which also provides some information about the use of $_nC_k$-like notations in $\mathrm{\S}$ 451:
• George Peacock (Treatise of Algebra, 1830) introduces $C_r$ for the combinations of $n$ things taken $r$ at a time.
• Robert Potts (Elementary Algebra, 1880) begins his treatment by letting the number of combinations of $n$ different things taken $r$ at a time be denoted by $^nC_r$
• W.A. Whitworth uses $C_r^n$ in Choice and Change (1886).
• G. Chrystal writes $_nC_r$ in Algebra, Part II (1899).
Note the different variations $C_r, {^{n}C}_r, C^n_r$ of $_nC_r$-like notations.
Here is a selection of some classics from the 20th century.
Some classics:
• An Introduction to Combinatorial Analysis (1958) by J. Riordan
• (3.1): ... and \begin{align*} C(n,r)=\frac{n(n-1)\cdots(n-r+1)}{r!}=\frac{n!}{r!(n-r)!}=\binom{n}{r} \end{align*} where the last symbol is that usual for binomial coefficients, that is, the coefficients in the expansion of $(a+b)^n$. ($C_r^n, C_n^r, _nC_r$ and $(n,r)$ are alternative notations; ...
• Combinatorial Identities (1968) by J. Riordan
• (1.1): Perhaps the simplest combinatorial entities are the binomial coefficients, that is, the combinations, for example of $n$ things, $k$ at a time. They take their name from the generating function for combinations, which is a power of a binomial, namely \begin{align*} (1+x)^n=\sum_{k=0}^n\binom{n}{k}x^k \end{align*} where, of course, $\binom{n}{K}=C(n,k)=\frac{n!}{k!(n-k)!}$ is the usual notation for a binomial coefficient.
• (II.4): ... the number of subpopulations of size $r$ is therefore given by $(n)_r/r!$. Expressions of this kind are known as binomial coefficients and the standard notation for them is \begin{align*} \binom{n}{r}=\frac{(n)_r}{r!}=\frac{n(n-1)\cdots(n-r+1)}{1\cdot 2\cdots (r-1)\cdot r} \end{align*}
• Advanced Combinatorics (1974) by L. Comtet
• (1.4): ... We will adopt the notation $\binom{n}{k}$, used almost in this form by Euler and fixed by Raabe, with the exclusion of all other notations, as this notation is used in the great majority of the present literature, and its use is even so still increasing. This symbol has all the qualities of a good notation: economical (no new letters introduced), expressive (it is very close in appearence to the explicit value $\frac{(n)_k}{k!}$, typical (no risk of being confused with others), and beautiful.
• Enumerative Combinatorics, Volume I (1986) by R. P. Stanley
• (I.1.4): ... where $d_n=\sum_{i=0}^n\binom{n}{i}a_ib_{n-i}$, with $\binom{n}{i}=n!/i!(n-i)!$.
and the author continues to use the notation $\binom{n}{k}$ without any more citations, indicating this notation being commonly used.
• D. E. Knuth who gave us $\TeX$ is besides being a great mathematician an extraordinary expert in typography and mathematical writing. His Mathematical Typography (1979) gives us a glimpse of his deep thoughts about these issues. Another one being the report Mathematical Writing (1987) written together with T. Larrabee and P. M. Roberts.
In $\TeX$ we use the command "$\text{n \choose k}$" giving us $\binom{n}{k}$.
The enhanced readability of the notation $\binom{n}{k}$ becomes rather obvious in more complex expressions. Compare for instance formula (5.32) in Concrete Mathematics by R. L. Graham, D. E. Knuth and O. Patashnik which is stated for integers $l,m,n$; $n\geq 0$ as \begin{align*} \sum_{j,k}(-1)^{j+k}\binom{j+k}{k+l}\binom{r}{j}\binom{n}{k}\binom{s+n-j-k}{m-j}=(-1)^l\binom{n+r}{n+l}\binom{s-r}{m-n-l} \end{align*} with the representation \begin{align*} \sum_{j,k}(-1)^{j+k}\,_{j+k}C_{k+l}\,_rC_j\,_nC_k\,_{s+n-j-k}C_{m-j}=(-1)^l\,_{n+r}C_{n+l}\,_{s-r}C_{m-n-l} \end{align*}
• @BruceET: You're welcome. – Markus Scheuer Jun 29 '18 at 8:51
• Thank you for taking the time and effort to write such a nice answer! :) – Seankala Jun 29 '18 at 15:42
• @Sean: You're welcome. Thanks for your nice comment. :-) – Markus Scheuer Jun 29 '18 at 15:45
In books printed before $\LaTeX$ came to be widely used, setting ${n \choose r}$ into type was difficult (and expensive if used throughout a book). I 'know' this because of conversations with editors of math books over the years. By contrast, typesetting is not difficult for the variants of $C_r^n,$ including $C(n, r)$ not yet mentioned in this discussion.
Nowadays, I think there is a trend to use ${n \choose r},$ except for authors who have a strong preference for C-notations and those writing in Microsoft Word and trying to avoid using its 'equation editor'.
Curiously, a generally-accepted convention for permutations $P_r^n = r!{n \choose r}$ does not seem to have emerged. Feller used $(n)_r$ in his famous probability book, which one would have thought might have set a trend 60 years ago, but apparently not. His book may have set a record for the density per page of big parentheses for various uses of ${n \choose r}$-notation--even before $\LaTeX.$
There is no difference at all. Personally I prefer using the shortest notation aCb for obvious reasons.
• Are the "a" and "b" significant, there, or is it more the ${}_{\cdot}C_{\cdot}$ notation that you prefer? – Cameron Buie Jun 29 '18 at 2:56
• a and b are just numbers . Nothing more than that . Yes I prefer that notation because it's shorter and I see it very often – Kwnstantinos Nikoloutsos Jun 29 '18 at 2:59 | 2019-04-23T12:13:11 | {
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https://mathematica.stackexchange.com/questions/48359/unify-the-sampling-of-nintegrate-f-g-h-w | # Unify the sampling of NIntegrate[ {f, g, h} w ]
I'm trying to numerically integrate a function which has a vector-valued slow part and a much faster component which is shared by all the components, i.e. an integral of the form $$\int_a^b\begin{pmatrix}f(x)\\ g(x) \\ h(x)\end{pmatrix}w(x)\,\text dx.$$ Because NIntegrate is nicely Listable on its first argument, I can feed it a list-valued argument without a problem. However, it appears to be doing each component as a completely separate integral, which results in a lot of work being re-calculated.
For example, the (simplified) example integral
samplePointsList = Reap[
NIntegrate[
{x, x^2, x^3} Cos[10 x + Cos[x]]
, {x, 0, 5}
, EvaluationMonitor :> Sow[x]
]
][[2, 1]];
gives the sample point diagram (through ListPlot[Transpose[{samplePointsList, Range[Length[samplePointsList]]}]])
It is clear that the kernel is doing the initial sampling, and then the further refinements, separately for each component. While the required sample points are not identical, there is a lot of shared work and I feel there is a fair bit of room for optimization there.
I am aware that, since the sampling requirements of each component are slightly different, binding them completely will require more evaluations in some components than would be necessary. (For instance, in the example above, the extra detail required by the first component around $2\leq x\leq 3$ would be slowed down slightly if it was required to also calculate the second and third components there, though they do not require it.) However, the components in my case are similar enough that I do not think this would be an issue.
Is there a way to force Mathematica into this sort of separation of the integrand and unification of the sampling? If not, is there a way to make it aware of the previously calculated values and sampling points which will speed up the process?
One idea is to integrate once to get the sample points then compute the remaining integrals as sums:
sample = Transpose@
SortBy[First@Last@Reap[
NIntegrate[x (c = Cos[10 x + Cos[x]]), {x, 0, 5},
EvaluationMonitor :> Sow[{x, c}]]], #[[1]] &];
wt = ((#[[3]] - #[[1]])/2) & /@ Partition[Join[{0}, sample[[1]], {5}], 3, 1];
wt.(sample[[2]] #) & /@ {sample[[1]], sample[[1]]^2, sample[[1]]^3}
{0.0133333, 0.133275, 0.861541}
Note the accuracy is not terribly good, NIntegrate gives:
{0.0125266, 0.131514, 0.855716}
Somethings a bit off in my quick&dirty trapezoid integration but i think this can be made to work.
For this example there really is little benefit to NIntegrate's adaptive sampling so we might as well just use a uniform sampling:
np = 651;(*assumed odd for simpsons rule*)
a=5
b=0
wt = ( (a-b)/(np - 1) )/3 Join[{1}, Flatten@ConstantArray[{4, 2}, (np - 1)/2 - 1], {4, 1}] // N;
x = b + (Range[0, np - 1] (a-b)/(np - 1)) // N;
fast = Cos[10 # + Cos[#]] & /@ x // N;
(# fast).wt & /@ {x, x^2, x^3}
{0.0125266, 0.131514, 0.855716}
Is there a way to force Mathematica into this sort of separation of the integrand and unification of the sampling? If not, is there a way to make it aware of the previously calculated values and sampling points which will speed up the process?
are "yes" and "yes" as demonstrated and discussed in this post to "How to calculate the numerical integral more efficiently?".
More concretely:
Import["https://raw.githubusercontent.com/antononcube/\
MathematicaForPrediction/master/Misc/ArrayOfFunctionsRule.m"]
funcExpr = {x, x^2, x^3} Cos[10 x + Cos[x]]
(* {x Cos[10 x + Cos[x]], x^2 Cos[10 x + Cos[x]],
x^3 Cos[10 x + Cos[x]]} *)
funcMat = Table[i*funcExpr, {i, 100}];
AbsoluteTiming[
res0 = NIntegrate[funcMat, {x, 0, 5}];
]
res0[[12]]
(* {1.711, Null} *)
(* {0.150319, 1.57817, 10.2686} *)
AbsoluteTiming[
res1 = NIntegrate[1, {x, 0, 5},
Method -> {"GlobalAdaptive", "SingularityHandler" -> None,
Method -> {ArrayOfFunctionsRule, "Functions" -> funcMat}}];
]
res1[[12]]
(* {0.029007, Null} *)
(* {0.150319, 1.57817, 10.2686} *)
Norm[res0 - res1, 2]
(* 2.03398*10^-12 *)
• That's pretty interesting; I hope I can find time to test it out soon. In the meantime, can you comment on the relative gains as a function of the size of the matrix? I'm interested in $n\leq3$ but the other answer looks like it targets bigger arrays. – Emilio Pisanty Jan 9 '17 at 19:22
• @EmilioPisanty I am not sure what you mean -- performance of the ArrayOfFunctionsRule with respect to the size of the integrand? I did a similar comparison during the experiments for the linked MSE answer... – Anton Antonov Jan 10 '17 at 1:30 | 2020-10-22T12:55:56 | {
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# A coin is tossed 7 times. Find the probability of getting
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A coin is tossed 7 times. Find the probability of getting more heads than tails in all 7 tosses?
A. 1/2
B. 63/128
C. 4/7
D. 61/256
E. 63/64
Hi
I want to understand why combination has been used in the below problem.
I thought permutation is used for order and - for probability one needs to find the number of outcomes as well as order of the outcomes.
For example if the question was - probability of getting 1 head then we would have put it as follows :-
HTTTTTT
THTTTTT
TTHTTTT
TTTHTTT
TTTTHHH
TTTTTHH
TTTTTTH
Then why in the below problem - we use combinations and not permutations ?
I am so confused.
A coin is tossed 7 times. Find the probability of getting more heads than tails in all 7 tosses?
B. 63/128
C. 4/7
D. 61/256
Explanation
ANS. (a) ( Total outcomes= 2^7 = 128, Number outcomes for which heads are more than tails = 7 combination 4 (Heads=4 & Tails=3) + 7 combination 5 + 7 combination 6 + 7 combination 7) = 35+21+7+1= 64, so probability of getting more heads = 64/128 = ½)
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### Show Tags
12 Jan 2012, 05:34
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6
A coin is tossed 7 times. Find the probability of getting more heads than tails in all 7 tosses?
A. 1/2
B. 63/128
C. 4/7
D. 61/256
E. 63/64
Assuming the coin is fair - P(H)=P(T)=1/2
We can do as proposed by the explanation in your initial post:
Total outcomes: 2^7
Favorable outcomes:
4 heads --> combination of HHHHTTT --> 7!/(4!*3!)=35 (# of permutation of 7 letters out of which 4 H's and 3 T's are identical);
5 heads --> combination of HHHHHTT --> 7!/(5!*2!)=21;
6 heads --> combination of HHHHHHT --> 7!/(6!*1!)=7;
7 heads --> combination of HHHHHHH --> 1;
P(H>T)=Favorable outcomes/Total outcomes=(35+21+7+1)/2^7=1/2.
BUT: there is MUCH simpler and elegant way to solve this question. Since the probability of getting either heads or tails is equal (1/2) and a tie in 7 (odd) tosses is not possible then the probability of getting more heads than tails = to the probability of getting more tails than heads = 1/2. How else? Does the probability favor any of tails or heads? (The distribution of the probabilities is symmetrical: P(H=7)=P(T=7), P(H=5)=P(T=5), ... also P(H>4)=P(T>4))
If it were: A fair coin is tossed 8 times. Find the probability of getting more heads than tails in all 8 tosses?
Now, almost the same here: as 8 is even then a tie is possible but again as distribution is symmetrical then $$P(H>T)=\frac{1-P(H=T)}{2}=P(T>H)$$ (so we just subtract the probability of a tie and then divide the given value by 2 as P(H>T)=P(H<T)). As $$P(H=T)=\frac{8!}{4!*4!}=70$$ (# of permutation of 8 letters HHHHTTTT, out of which 4 H's and H T's are identical) then $$P(H>T)=\frac{1-P(H=T)}{2}=\frac{1-\frac{70}{2^8}}{2}=\frac{93}{256}$$. You can check this in following way: total # of outcomes = 2^8=256, out of which in 70 cases there will be a tie, in 93 cases H>T and also in 93 cases T>H --> 70+93+93=256.
Hope it's clear.
Similar questions for practice:
probability-question-100222.html?hilit=coin%20tossed#p772756
hard-probability-99478.html?hilit=coin%20tossed
some-ps-questions-need-explanation-99282.html?hilit=coin%20tossed
probability-question-gmatprep-85802.html?hilit=coin%20tossed
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07 Jan 2012, 13:49
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Hi Sorry ...1 head out of 7 can be arranged in following ways :-
HTTTTTT
THTTTTT
TTHTTTT
TTTHTTT
TTTTHTT
TTTTTHT
TTTTTTH
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Re: A coin is tossed 7 times. Find the probability of getting [#permalink]
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07 Jun 2013, 08:48
2
Probability of getting Head, P(H) = 1/2
For No. of heads > no. of Tails, there can be 4,5,6 or 7 Heads.
As per GMAT Club math book, probability of occurrence of an event k times in a n time sequence, for independent and mutually exclusive events:
$$P=C(n,k)*p^k*(1-p)^n^-^k$$
$$P(H=4) = C(7,4)*(1/2)^4*(1/2)^3 = C(7,4) * (1/2)^7$$
$$P(H=5) = C(7,5) * (1/2)^7$$
$$P(H=6) = C(7,6) * (1/2)^7$$
$$P(H=7) = C(7,7) * (1/2)^7$$
Total $$P(H > 3) = P(H=4) + P(H=5) + P(H=6) + P(H=7)$$
$$= [ C(7,4) + C(7,5) + C(7,6) + C(7,7)] * (1/2)^7$$
$$= (35 + 21 + 7 + 1) / 128$$
$$= 64/128$$
$$= 1/2$$
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Re: A coin is tossed 7 times. Find the probability of getting [#permalink]
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05 May 2014, 05:47
My approach
7 toss
coin has 2 out comes H or T
now basically if i see this a game where there are 2 teams one selects Heads and other team selects Tails. If in 7 toss whichever comes maximum (heads or Tails) the corresponding team wins..
Clearly the probability is 50% for both the cases max Heads or max Tails............
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Re: A coin is tossed 7 times. Find the probability of getting [#permalink]
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01 May 2015, 02:54
For those who have trouble grasping the concept behind this - check out Khan Academy lessons on probability and combinatorics. I'm not allowed to post urls as a newbie but a simple Google search will throw up the link.
I couldn't make head or tail of these questions before seeing them, tried memorizing the formulae and always messed up. I invested 3 hours in going through those videos and can now solve these questions without knowing any formulae - it's all conceptual. Sal's great with breaking down concepts to simple, relatable stuff.
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A coin is tossed 7 times. Find the probability of getting [#permalink]
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01 May 2015, 05:11
2
DipikaP wrote:
For those who have trouble grasping the concept behind this - check out Khan Academy lessons on probability and combinatorics. I'm not allowed to post urls as a newbie but a simple Google search will throw up the link.
I couldn't make head or tail of these questions before seeing them, tried memorizing the formulae and always messed up. I invested 3 hours in going through those videos and can now solve these questions without knowing any formulae - it's all conceptual. Sal's great with breaking down concepts to simple, relatable stuff.
Hi Dipika,
You are totally right when you say "it is all conceptual". Trying to memorise formulae is not the right approach.
For example in this question:
First we should think what are the possible outcomes when we toss a coin: head or tail (2 outcomes)
Now as the coin is fair, the probability that we will get a head or a tail is 1/2
To illustrate, let's take a smaller version of the above question:
What is the probability of getting more heads in 3 tosses?
1st case: We can get 3 heads: HHH
Probability of HHH = 1/2 * 1/2 * 1/2 = 1/8
Probability of getting 3 heads = 1/8
2nd case: We can get two heads: HHT, HTH, THH
Probability of HHT = 1/2 * 1/2 * 1/2 = 1/8
Probability of HTH = 1/2 * 1/2 * 1/2 = 1/8
Probability of THH = 1/2 * 1/2 * 1/2 = 1/8
Probability of getting 2 heads = 3* 1/8
As you can see HHT, HTH and THH are different arrangements of HHT
Probability of getting 2 heads = (No. of arrangements of HHT)* (Probability of getting HHT) = 3!/2! * (1/2 * 1/2 * 1/2) = 3/8
Total probability of getting more heads in 3 tosses = 1/8 + 3/8 = 4/8 = 1/2
Thinking on these lines you can solve the above question easily.
But if you think a little further, we are talking about odd number of tosses (3 and 7). So, either there will be more heads or more tails. It is not possible to get equal number of heads and tails. Hence, in half of the outcomes we will get more heads than tails and in the the other half we will have more tails than heads. Thus, the probability of getting more heads = probability of getting more tails = 1/2.
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Re: A coin is tossed 7 times. Find the probability of getting [#permalink]
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04 Aug 2015, 12:50
morya003 wrote:
A coin is tossed 7 times. Find the probability of getting more heads than tails in all 7 tosses?
A. 1/2
B. 63/128
C. 4/7
D. 61/256
E. 63/64
Hi
I want to understand why combination has been used in the below problem.
I thought permutation is used for order and - for probability one needs to find the number of outcomes as well as order of the outcomes.
For example if the question was - probability of getting 1 head then we would have put it as follows :-
HTTTTTT
THTTTTT
TTHTTTT
TTTHTTT
TTTTHHH
TTTTTHH
TTTTTTH
Then why in the below problem - we use combinations and not permutations ?
I am so confused.
A coin is tossed 7 times. Find the probability of getting more heads than tails in all 7 tosses?
B. 63/128
C. 4/7
D. 61/256
Explanation
ANS. (a) ( Total outcomes= 2^7 = 128, Number outcomes for which heads are more than tails = 7 combination 4 (Heads=4 & Tails=3) + 7 combination 5 + 7 combination 6 + 7 combination 7) = 35+21+7+1= 64, so probability of getting more heads = 64/128 = ½)
Since number of times coin has been tossed is 7, either number of heads will be more than tails or vice versa. There is no way number of heads become equal to number of tails. Since head and tails are equally favourable outcome of a coin, possibility of getting more heads = possiblity of getting more tails = 1/2. In other words there are only 2 equally likely events (event 1: heads more than tails, event2: tails more than heads) constituting all the outcomes.
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Re: A coin is tossed 7 times. Find the probability of getting [#permalink]
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07 Dec 2016, 20:46
Total outcome = 2^7 = 128
HHHHHHH = 7!/7! = 1
HHHHHHT = 7!/1!6! = 7
HHHHHTT = 7!/2!5! = 21
HHHHTTT = 7!/3!4! = 35
1+7+21+35 = 36
36/128 = 1/2
A.
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Re: A coin is tossed 7 times. Find the probability of getting [#permalink]
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02 Feb 2017, 04:42
hi bunnel...
As stated in the question" Find the probability of getting more heads than tails in all 7 tosses?"
tails in all the tosses means no heads ie 0 heads and more than that means at least one heads.
so we have to find the the solution for atleast one heads in 7 tosses...
can we restate the question as above?
pls explain
thank u..
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Re: A coin is tossed 7 times. Find the probability of getting [#permalink]
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02 Feb 2017, 06:12
manojhanagandi wrote:
hi bunnel...
As stated in the question" Find the probability of getting more heads than tails in all 7 tosses?"
tails in all the tosses means no heads ie 0 heads and more than that means at least one heads.
so we have to find the the solution for atleast one heads in 7 tosses...
can we restate the question as above?
pls explain
thank u..
If you read the solutions above you'll see that this is not correct. More heads than tails in all 7 tosses means at least 4 heads (so more than half must be heads):
HHHHTTT
HHHHHTT
HHHHHHT
HHHHHHH
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Re: A coin is tossed 7 times. Find the probability of getting [#permalink]
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03 Feb 2017, 04:22
Thank You bunuel...
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Re: A coin is tossed 7 times. Find the probability of getting [#permalink]
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06 Sep 2017, 00:48
Ans is A
HHHHTTT= $$(0.5)^4$$ x $$(0.5)^3$$ {7C4}
similarly for 5 H , 6H and 7 H we will calculate only 7C5 , 7C6 and 7C7 will change other part (0.5)^7 remains same which is multiplied, so add all 4 cases of 4H 5H 6H 7H
$$(0.5)^7$$ x {7C4+7C5+7C6 +7C7}
$$\frac{64}{128}$$= $$\frac{1}{2}$$
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Re: A coin is tossed 7 times. Find the probability of getting [#permalink]
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06 Sep 2017, 03:39
morya003 wrote:
A coin is tossed 7 times. Find the probability of getting more heads than tails in all 7 tosses?
A. 1/2
B. 63/128
C. 4/7
D. 61/256
E. 63/64
Hi
I want to understand why combination has been used in the below problem.
I thought permutation is used for order and - for probability one needs to find the number of outcomes as well as order of the outcomes.
For example if the question was - probability of getting 1 head then we would have put it as follows :-
HTTTTTT
THTTTTT
TTHTTTT
TTTHTTT
TTTTHHH
TTTTTHH
TTTTTTH
Then why in the below problem - we use combinations and not permutations ?
I am so confused.
A coin is tossed 7 times. Find the probability of getting more heads than tails in all 7 tosses?
B. 63/128
C. 4/7
D. 61/256
Explanation
ANS. (a) ( Total outcomes= 2^7 = 128, Number outcomes for which heads are more than tails = 7 combination 4 (Heads=4 & Tails=3) + 7 combination 5 + 7 combination 6 + 7 combination 7) = 35+21+7+1= 64, so probability of getting more heads = 64/128 = ½)
Now this is a very simple question and can be solved without doing all the long calculations.
Here important thing to notice is P(H) = P(T) = 1/2 in a single toss.
Now 7 Toss can have following possible ways
0H 7T say x way
1H 6T say y way
2H 5T say z way
3H 4T say w way
--------------------------
4H 3T w way
5H 2T z way
6H 1T x way
7H 0T x way
So 4 out of 8 ways will have heads greater than Tail. Also P(H) = P(T) = 1/2 = i.e. equal in a single toss. So, we don't need to calculate the number of ways for each case.
Require probability = (w+z+x+y)[/(x+y+z+w) +(w+z+x+y)] = 1/2
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A coin is tossed 7 times. Find the probability of getting [#permalink]
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20 Dec 2017, 09:22
My approach: either Number of heads can be more or less than tails. They can't be equal as no of toss is 7 which is odd.
Hence probability is (no of cases when no of head is more) /((no of cases when no of head is more)+(no of cases when no of head is more))= 1/2
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Bunuel wrote:
A coin is tossed 7 times. Find the probability of getting more heads than tails in all 7 tosses?
A. 1/2
B. 63/128
C. 4/7
D. 61/256
E. 63/64
Assuming the coin is fair - P(H)=P(T)=1/2
We can do as proposed by the explanation in your initial post:
Total outcomes: 2^7
Favorable outcomes:
4 heads --> combination of HHHHTTT --> 7!/(4!*3!)=35 (# of permutation of 7 letters out of which 4 H's and 3 T's are identical);
5 heads --> combination of HHHHHTT --> 7!/(5!*2!)=21;
6 heads --> combination of HHHHHHT --> 7!/(6!*1!)=7;
7 heads --> combination of HHHHHHH --> 1;
P(H>T)=Favorable outcomes/Total outcomes=(35+21+7+1)/2^7=1/2.
BUT: there is MUCH simpler and elegant way to solve this question. Since the probability of getting either heads or tails is equal (1/2) and a tie in 7 (odd) tosses is not possible then the probability of getting more heads than tails = to the probability of getting more tails than heads = 1/2. How else? Does the probability favor any of tails or heads? (The distribution of the probabilities is symmetrical: P(H=7)=P(T=7), P(H=5)=P(T=5), ... also P(H>4)=P(T>4))
If it were: A fair coin is tossed 8 times. Find the probability of getting more heads than tails in all 8 tosses?
Now, almost the same here: as 8 is even then a tie is possible but again as distribution is symmetrical then $$P(H>T)=\frac{1-P(H=T)}{2}=P(T>H)$$ (so we just subtract the probability of a tie and then divide the given value by 2 as P(H>T)=P(H<T)). As $$P(H=T)=\frac{8!}{4!*4!}=70$$ (# of permutation of 8 letters HHHHTTTT, out of which 4 H's and H T's are identical) then $$P(H>T)=\frac{1-P(H=T)}{2}=\frac{1-\frac{70}{2^8}}{2}=\frac{93}{256}$$. You can check this in following way: total # of outcomes = 2^8=256, out of which in 70 cases there will be a tie, in 93 cases H>T and also in 93 cases T>H --> 70+93+93=256.
Hope it's clear.
Similar questions for practice:
http://gmatclub.com/forum/probability-q ... ed#p772756
http://gmatclub.com/forum/hard-probabil ... n%20tossed
http://gmatclub.com/forum/some-ps-quest ... n%20tossed
http://gmatclub.com/forum/probability-q ... n%20tossed
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https://math.stackexchange.com/questions/2213431/basic-question-regarding-limits/2213440 | # Basic question regarding limits
I'm a little confused when it comes to question like this.
Let's say we got this expression: $$\lim_{x\to\infty} \tan\left(\frac{1}{x}\right).$$ Am I allowed to say the result of this is $0$ or do I have to show it? (If I have to show it, please show me how to cuz I got no clue).
And more general, in which situations are we allowed to "cut" it and in which we are not? (I'm guessing whenever it comes to fractions surely).
That depends.
If the whole exercise is just "compute $\lim \tan(1/x)$", then yes, you have to do some argument. Something along the lines of "if $x$ goes to infinity, then $1/x$ goes to zero, and by continuity of $\tan$, the whole expression then goes to zero".
But if this limit is just one part of a longer question/calculation, then no. It's trivial enough that anybody with some math background will see it immediately.
• Appreciated . Got only one more question . Is there any like more formal ways of prooving it is 0 ? Cuz I never did any good solving math problem using words tho I completely see where ur going at. – James Groon Apr 1 '17 at 18:24
• In general feel free to replace the word "then" or the phrase "it then follows" with the symbol "$\implies$" and "there is an $x$" with "$\exists x$" and such. Using such symbols instead of words can make a proof more concise, though overdoing it can make it hard to read. And if you want a more elementary proof, the answer by @Chris-Varghese is quite nice. – Simon Apr 1 '17 at 18:34
• Gotcha and yep, Chris did a good job :) – James Groon Apr 1 '17 at 18:35
You need to show that given any small $\epsilon > 0$, there exists a $X(\epsilon)$ such that $|\tan(1/x)| < \epsilon,\; \forall x > X(\epsilon)$. To do this choose, for e.g., $X(\epsilon) = \dfrac{1+\sec \epsilon}{\epsilon}$. Then $$\left|\tan \frac{1}{x}\right| < \left|\tan \frac{\epsilon}{1+\sec \epsilon} \right| = \frac{\sin \left( \frac{\epsilon}{1+\sec \epsilon}\right) }{\cos\left( \frac{\epsilon}{1+\sec \epsilon}\right)} < \frac{\sin \left( \frac{\epsilon}{1+\sec \epsilon}\right) }{\cos\epsilon} < \frac{ \frac{\epsilon}{1+\sec \epsilon} }{\cos\epsilon} = \frac{\epsilon}{1 + \cos \epsilon} < \epsilon$$
• The inequality is wrong. Note that $|\tan (x)|\ge |x|$ for all $x\in (-\pi/2,\pi/2)$. Simple example; $x=\pi/4<1$. We have $\tan(\pi/4)=1>\pi/4$. – Mark Viola Apr 1 '17 at 18:48
• So may i get the correct answer ? – James Groon Apr 1 '17 at 19:02
• @Dr.MV Which inequality in my answer are you referring to? Indeed, $|\tan x | \geq |x| \forall x \in (-\pi/2, \pi/2)$. How does that fact contradict anything that I have in my answer? – ChargeShivers Apr 2 '17 at 1:01
• @chrisvarghese You wrote $|\tan(\epsilon)|\le \epsilon$. That is false. – Mark Viola Apr 2 '17 at 2:58
• @Dr.MV Got it now. Thanks. Not sure how I missed it twice!! Answer edited. – ChargeShivers Apr 2 '17 at 17:17
The answer is that it depends on the situation. To address the statement in the OP "If I have to show it, please show me how to cuz I got no clue," I thought it might be instructive to present an approach that relies on a standard inequality from elementary geometry. To that end, we begin with a short primer.
PRIMER:
Recall from elementary geometry the inequality
$$\sin(\theta)\le \theta\tag 1$$
for $\theta\ge 0$. Squaring both sides of $(1)$, using $\sin^2(\theta)=1-\cos^2(\theta)$, and rearranging, we find that
$$\cos(\theta)\ge \sqrt{1-\theta^2} \tag 2$$
for $0\le \theta \le 1$.
Now, using $(1)$ and $(2)$ with $\theta=1/x$ reveals that
\begin{align} \tan(1/x)&=\frac{\sin(1/x)}{\cos(1/x)}\\\\ &\le \frac{1/x}{\sqrt{1-\frac1{x^2}}}\\\\ &=\frac{1}{\sqrt{x^2-1}}\tag 3 \end{align}
for $x>1$.
Hence, for all $\epsilon>0$, we have from $(3)$ that
\begin{align} \tan(1/x)&\le \frac{1}{\sqrt{x^2-1}}\\\\ &<\epsilon \end{align}
whenever $x>\sqrt{1+\frac{1}{\epsilon^2}}$. And we are done!
• $\sqrt{1 + \frac{1}{\epsilon^2}} = \dfrac{\sqrt{1+\epsilon^2}}{\epsilon}$ is definitely a better $X(\epsilon)$ than $\dfrac{1+\sec \epsilon}{\epsilon}$. May be the 'best' of all is $X(\epsilon) = \dfrac{1}{\arctan \epsilon}$. – ChargeShivers Apr 2 '17 at 23:41
Short answer is $0$ ,but if you want " to show " recall the limit $$\lim _{ x\rightarrow \infty }{ \frac { 1 }{ x } =0 }$$
You can move the limit into the $tan$, because you are not excluding any $x$ from the limit, and you are not creating any limits that don't exist. Once that is done, the problem is trivial. | 2020-01-29T11:34:44 | {
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http://math.stackexchange.com/questions/324666/application-of-2nd-fundamental-theorem-of-calculus | # Application of 2nd fundamental theorem of calculus
I would like to clarify the usage of the 2nd fundamental theorem of calculus, in 3 parts. For all 3 parts, consider $F'(a)$ to be
$$F'(a) = \frac{1}{1+a+a^2}$$
The questions are:
1. Find $\frac{d}{dy}\int^y_1 F'(a) \,da$
2. Find $\frac{d}{dy}\int^1_y F'(a) \,da$
3. Find $\frac{d}{dy}\int^{y^2}_1 F'(a)\, da$
My working for each of the questions are:
1.
\begin{align*} \frac{d}{dy} \int^y_1 F'(a) \,da &= \frac{d}{dy}\begin{bmatrix}F(y) - F(1)\end{bmatrix} \\&=F'(y) \\&=\frac{1}{1+y+y^2}. \end{align*}
2.
\begin{align*} \frac{d}{dy} \int^1_y F'(a) \,da &= \frac{d}{dy}\begin{bmatrix}F(1) - F(y)\end{bmatrix} \\&=-F'(y) \\&=\frac{-1}{1+y+y^2}. \end{align*}
3.
\begin{align*} \frac{d}{dy} \int^{y^2}_1 F'(a) \,da &= \frac{d}{dy}\begin{bmatrix}F(y^2) - F(1)\end{bmatrix} \\&=F'(y^2) \\&=\frac{2y}{1+y^2+y^4}. \end{align*}
UPDATE: I shall update (3) with the chain rule working.
\begin{align*} \frac{d}{dy} \int^{y^2}_1 F'(a) \,da &= \frac{d}{dy}\begin{bmatrix}F(y^2) - F(1)\end{bmatrix} \\&=F'(y^2)(2y) \\&=\frac{1}{1+(y^2)+(y^2)^2} (2y) \tag{by chain rule} \\&=\frac{2y}{1+y^2+y^4}. \end{align*}
-
The third one would need a chain rule. Though, you do have used it for final answer. – hjpotter92 Mar 8 '13 at 12:59
Your answers are correct. However, one should note that$\dfrac{dF(y^2)}{dy}=F'(y^2)\cdot2y$(Chain rule). Your final answers are perfectly fine but the intermediate step is wrong. | 2014-03-10T17:05:55 | {
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http://math.stackexchange.com/questions/427559/prove-frac-cos2-a1-sin-a-1-sin-a-by-the-pythagorean-theorem | # Prove $\frac{\cos^2 A}{1 - \sin A} = 1 + \sin A$ by the Pythagorean theorem.
How do I use the Pythagorean Theorem to prove that $$\frac{\cos^2 A}{1 - \sin A} = 1 + \sin A?$$
-
if i multiply it would it show that A is an acute angle? i have a given figure here and its a right triangle – user83562 Jun 23 '13 at 14:25
Do you know the identity $\cos^2 A + \sin^2 A = 1$? – Javier Badia Jun 23 '13 at 14:58
suppose there is a right angle tri. having b is hypo.,c is base ,a is perpendicular than $\cos A=\frac cb$ and $\sin A=\frac ab$ using pythgo. theo. $\;b^2=a^2+c^2$ $$\dfrac {\cos^2 A}{1-\sin A}$$ $$\dfrac {\frac {c^2}{b^2}}{1-\frac {a}{b}}$$ $$\dfrac {\frac {c^2}{b^2}}{\frac {b-a}{b}}$$ $$\dfrac {c^2b}{b^2(b-a)}$$ $$\dfrac {c^2}{b(b-a)}$$ $$\dfrac {b^2-a^2}{b(b-a)}$$ $$\dfrac {b+a}{b}$$ $$1+\dfrac {a}{b}$$ $$1+\sin A$$
alternatively
$$\dfrac {\cos^2 A}{1-\sin A}$$ $$\dfrac {1-\sin^2 A}{1-\sin A}$$ $$\dfrac {(1-\sin A)(1+\sin A)}{1-\sin A}$$ $${1+\sin A}$$
-
why don't you make things precise : $$1=\frac{c^2}{b^2}+\frac{a^2}{b^2}=\cos^2A+\sin^2A$$ $$\implies \cos^2A=1-\sin^2A=(1-\sin A)(1+\sin A)$$ $$\implies \frac{\cos^2A}{1-\sin A}=1+\sin A$$ – lab bhattacharjee Jun 23 '13 at 14:32
how did it become 1-sin^2 A? – user83562 Jun 23 '13 at 14:49
@user83562: Did you read lab's (excellent) comment? Since $$1=\cos^2A+\sin^2A,$$ then $$\cos^2A=1-\sin^2A.$$ By difference of squares formula, we have $$\cos^2A=(1-\sin A)(1+\sin A).$$ From there, it's almost immediate. – Cameron Buie Jun 23 '13 at 15:18
If you allow yourself some simple algebra, then you see that $$\frac{\cos^2 A}{1-\sin A} = 1+\sin A\text{ is equivalent to } \cos^2 A = (1+\sin A)(1-\sin A)$$ and that last expression is equal to $1-\sin^2 A$. So you're asking how to prove $\cos^2 A=1-\sin^2 A$. And that's the same as proving $\cos^2 A + \sin^2 A = 1$.
So the question is: How can you prove that $\cos^2 A + \sin^2 A = 1$ by using the Pythagorean theorem?
If you know that $\sin =\dfrac{\mathrm{opposite}}{\mathrm{hypotenuse}}$ and $\cos=\dfrac{\mathrm{adjacent}}{\mathrm{hypotenuse}}$, then this becomes easy if you consider a right triangle in which the length of the hypotenuse is $1$. Then you have $\sin=\mathrm{opposite}$ and $\cos=\mathrm{adjacent}$. Now the Pythagorean theorem says that $$\mathrm{opposite}^2 + \mathrm{adjacent}^2 = 1^2.$$
-
Let's simplify matters:
In a triangle with hypotenuse equal to $1$ (think of the unit circle, an angle $A$ between the $x$ axis and the hypotenuse, we know that $$\sin A=\frac{\text{opposite}}{\text{hypotenuse}}\quad \text{and}\quad \cos A=\frac{\text{adjacent}}{\text{hypotenuse}}$$ then, since hypotenuse $=1$, we have the leg opposite the angle $A$ given by $\sin A=\text{opposite}/1$ and the leg along the x-axis of length $\cos A=\text{adjacent}/1$. Now, by the Pythagorean Theorem, and substitution, we have that \begin{align}\text{opposite}^2 + \text{adjacent}^2 &= \text{hypotenuse}^2 = 1^2 \\ \sin^2A +\cos^2 A & = 1\end{align}
This gives us the well-known identity: $$\sin^2A + \cos^2 A = 1\tag{1}$$
We can express this identity in terms of $\cos^2 A$ by subtracting $\sin^2 A$ from both sides of the identity to get \begin{align}\cos^2 A & = 1 - \sin^2 A \\ &= (1)^2 - (\sin A)^2\tag{2}\end{align}
Now, we know that for any difference of squares, we can factor as follows: $$(x^2 - y^2) = (x +y)(x - y)\tag{3}$$
Since equation $(2)$ is a difference of squares, we have that \begin{align}\cos^2 A &= 1 - \sin^2 A \\ & = (1)^2 - (\sin A)^2 \\ &= (1 +\sin A)(1 - \sin A)\tag{4}\end{align}
Substituting gives us:
\begin{align}\frac{\cos^2 A}{1 -\sin A} & = \frac{1 - \sin^2 A}{1 -\sin A} \\ \\ &= \frac{(1 + \sin A)(\color{blue}{\bf 1 - \sin A})}{\color{blue}{\bf 1 -\sin A}}\\ \\ & = 1 + \sin A\end{align}
-
You begin with "Given the well known identity,....", but the questions seems to be how to get that well known identity from the Pythagorean theorem. – Michael Hardy Jun 23 '13 at 16:14
@amWhy: You are always welcome my friend! Time for another cheesy movie (trying to add other activities in my life). ;-) – Amzoti Jun 24 '13 at 2:47
Let us denote base of right angle triangle as b, perpendicular ( height ) as p, and hypotenuse as h,
$$\cos A = \frac{b}{h} ; \qquad \sin A = \frac{p}{h}\tag{i}$$
Therefore, $$\frac{\cos^2A}{1-\sin A} = \frac{\frac{b^2}{h^2}}{1-\frac{p}{h}}$$
[By putting the values of $\cos A$ and $\sin A$ from $(i)$]
Which after simplification gives you:
$$\frac{b^2}{h(h-p)}\tag{ii}$$
Now as
$$b^2 = h^2-p^2\quad\text{[using pythagoreous theorem]}\tag{iii}$$
By putting the value of $b^2$ from (iii) in (ii) you get :
$$\frac{h^2-p^2}{h(h-p)} = \frac{h+p}{h} = 1+ \frac{p}{h} = 1+ \sin A$$
(hence proved)
-
not sure how much this derivation from the Left hand side to the Right is required, as this is often not the natural derivation. For example, please find my comment in the other answer. – lab bhattacharjee Jun 23 '13 at 14:38 | 2014-08-30T10:53:48 | {
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In the context where it is defined, the derivative of a function is a measure of the rate of change of function values with respect to change in input values. ) You can use direct substitution to arrive at this conclusion. In a two dimensional plane, the graph of this type of function is a straight, horizontal line. The derivative of this type of function is just zero. For example, the function y(x) = 4 is a constant function because the value of y(x) is 4 regardless of the input value x (see image). C++ is similar to C but has more features than C. Therefore, it is called a subset of C language. In calculus, the constant of integration, often denoted by , is a constant added to the end of an antiderivative of a function () to indicate that the indefinite integral of () (i.e., the set of all antiderivatives of ()), on a connected domain, is only defined up to an additive constant. A constant function is a function of the form f(x) = c, where c is a constant. A function becomes const when the const keyword is used in the function’s declaration. Constant member function is an accessory function which cannot modifying values of data members. ( variables and constant both carries some defined value but difference comes with their accessibility and value constant can be define as 1 define constant_name constant_value or 2 const constant_name constant_value the second way is faster A constant function is a special type of linear function that follows the form f(x) = b 'b' is the y-intercept of the line and is just a constant; A constant function is a linear function whose slope is 0; No matter what value of 'x' you choose, the value of the function will always be the same Online Help: Math Apps: Functions and Relations.Retrieved from https://www.maplesoft.com/support/help/Maple/view.aspx?path=MathApps%2FConstantFunction on May 24, 2019. For example, y = 10 is a form of constant function. Varsity Tutors © 2007 - 2021 All Rights Reserved, CBEST - The California Basic Educational Skills Test Courses & Classes, SAT Subject Test in World History Test Prep, International Sports Sciences Association Test Prep. Award-Winning claim based on CBS Local and Houston Press awards. Consider constants as having a variable raised to the power zero. It perform constant operation. Now we shall prove this constant function with the help of the definition of derivative or differentiation. Constant Functions. The form of this particular function is: f(x) = ax 2 + bx + c, where a, b, and c are constants. With a constant function, for any two points in the interval, a change in x results in a zero change in f (x). This function works also with class constants . x ( From the general formula, the output of a constant function regardless of its input value (usually denoted by x), will always be the same which is the fixed number \color{red}a. It is of the format y=c d. This is constant function as output is different for each input e. This is constant function as output is different for each input 2. which of the graph represent constant function? "Piecewise constant" should not be used as a precise term without giving an explicit explanation of what you mean. *Response times vary by subject and question complexity. The composition of f with any other function is also a constant function… The derivative is the slope at an instant (kind’ve). In mathematics, a constant function is a function whose (output) value is the same for every input value. The object called by these functions cannot be modified. The limit of a constant function (according to the Properties of Limits) is equal to the constant. Then the gradient of z will be a vector pointing “uphill” on a surface, and the length of the vector is the slope at that point. The idea of const functions is not to allow them to modify the object on which they are called. The main difference between Static and Constant function in C++ is that the Static function allows calling functions using the class, without using the object, while the Constant function does not allow modifying objects.. C++ is a programing language developed by Bjarne Stroustrup in 1979. As a function requires that inputs produce outputs, it wouldn’t be a “function”. 3 That is when parameters are evaluated and generate statements are expanded. range A constant function has the form. This is a function of the type $$f (x) = k$$, where $$k$$ is any real number. Constant member function in C++. The derivative of a constant function is (1) However, the fundamental period of a constant function is not defined for the above reason. constant function; Background Tutorials. Technically, zero is a constant. This is constant function as output is same for each input. The mathematical formula for a constant function is just f x = a, where a is a number (which does not depend on x). x For example the function f (x) = 4 is constant since f maps any value to 4. A constant function is a linear function whose range contains only one element irrespective of the number of elements of the domain. is used. That is, the output value of the function at any input value in its domain is the same, independent of the input. linear function Varsity Tutors does not have affiliation with universities mentioned on its website. domain y = x is called the identity function because the value of y is identical with that of x. Constant Function Years 1 - 8 : Summary Children are always fascinated by the Constant Function of a calculator which constantly repeats the last operation it was asked to do if the [=] button is constantly pressed. A constant may be used to define a constant function that ignores its arguments and always gives the same value. 2 and constant function; Background Tutorials. This way, for instance, if we wanted to represent a quantity that stays constant over the course of time t, we would use a constant function f(t)=k, in which the variable tdoes not appear. Need help with a homework or test question? Defining in terms of mathematics, we can say that a constant function is a function that has the same range for all values of its domain. Definition and Usage. The derivative of a constant function is zero. Collection Functions (Arrays or Objects) each_.each(list, iteratee, [context]) Alias: forEach Iterates over a list of elements, yielding each in turn to an iteratee function. This is a constant function and so any value of x that we plug into the function will yield a value of 8. ) ) c is a constant: a number that doesn’t change as x changes. What's a Function? Slope is the change in y over the change in x. constant() is useful if you need to retrieve the value of a constant, but do not know its name. The function graph of a one-dimensional constant function is a straight line. Note that the value of f(x) is always k, independently of the value of x. A fixed value. In haskell: newtype Const r a = Const { unConst :: r } instance Functor (Const r) where fmap _ (Const r) = Const r It maps every type a to r in a sense, and every function of type a -> b to the identity function on r. We can write this type of function as: f (x) = c ( Basically, a constant function is simply equal to a constant. Suppose the failure function is a positive constant; that is, Show that the reliability function R(t) satisfies the separable differential equation Solve this differential equation to find R(t). The constant () function returns the value of a constant. A constant function is a function whose range consists of a single element. = No - the term "time constant" is not restricted to an asymptotic step response. With Chegg Study, you can get step-by-step solutions to your questions from an expert in the field. For instance, a school dining room where every child was given one donut, irrespective of age, or an exam in which every student was given an A regardless of how hard they worked. The graph of a constant functio… It passes through the point (0, c), (1, c), and (-1, c). ) A constant function is a linear function for which the range does not change no matter which member of the domain is used. But 0/6 = 0, so this would not actually produce a graph. I.e. 2 The constant() function returns the value of a constant. In fact lines are either increasing, decreasing, or constant. Example: in "x + 5 = 9", 5 and 9 are constants. horizontal line It’s easier to visualize this for a 2-dimensional function. So a method int Foo::Bar(int random_arg) (without the const at the end) results in a function like int Foo_Bar(Foo* this, int random_arg) , and a call such as Foo f; f.Bar(4) will internally correspond to something like Foo f; Foo_Bar(&f, 4) . *See complete details for Better Score Guarantee. A: The sale is a trade where the money is traded for a product or service. The equation of a line is: y = mx + b. Quadratic functions aren’t much different. for which the A constant function is a linear function for which the range does not change no matter which member of the domain is used. More formally, a function f : A → B is a constant function if f (x) = f for all x and y in A. Do It Faster, Learn It Better. Constant function In mathematics, a constant function is a function whose values do not vary and thus are constant. In other words, it’s just number on its own. f For example, the integral of f(x) = 10 is 10x. Note: This function also works with class constants. For example, the graph of the constant function f(x) = 4 is a straight, horizontal line that passes through the points (2,4), (0,4), and (-2,4). f One great example of a constant function is the log2 function. Recognizing a constant function, like y = 3. x f (x 1) = f (x 2) for any x 1 and x 2 in the domain. However, the fundamental period of a constant function is not defined for the above reason. The integration constant Ti is the time constant of the integrator. Real Functions: Constant Functions An constant function is a function that always returns the same constant value. The constant function used to provide a way to define a very simple price. Consider the function f (x) = 3 f (x) = 3 There is no variable in the definition (on the right side). This is a function of the type f(x)=k, where k is any real number. The slope m tells us if the function is increasing, decreasing or constant: is a This function may seem a little tricky at first but is actually the easiest one in this set of examples. Your email address will not be published. Function Definitions and Function Notation. This section describes what is a constant, define() function defines a constant name to a value, constant value can retrieved by the name directly or by the constant() function, any string can be used as constant name. It is recommended to use const keyword so that accidental changes to object are avoided. In the context where it is defined, the derivative of a function measures the rate of change of function (output) values with respect to change in input values. Yes, a constant function is a periodic function with any T∈R as its period (as f(x)=f(x+T) always for howsoever small 'T' you can find). If the answer in infinity, you can type in INF in the blank. x As of 4/27/18. The constant functions cut through the vertical axis in the value of the constant and they are parallel to the horizontal axis (and therefore they do not cut through it). What's a Function? A: The sale is a trade where the money is traded for a product or service. 1 it is stored in a variable or returned by a function. Because a constant function does not change, its derivative is 0. This graph is shown below. The iteratee is bound to the context object, if one is passed. A constant … In other words, the constant function is the function f(x) = c. An example of data for the constant function expressed in tabular form is presented below: What Is a Constant? The following shows the graph of f(x) = 10, and the integral f(x) = 10x. This function works also with class constants. This tutorial shows you a great approach to thinking about functions! This video is provided by the Learning Assistance Center of Howard Community College. The coördinate pairs are (x, x). ∟ Constant and define() Function. Your email address will not be published. C++ is similar to C but has more features than C. Therefore, it is called a subset of C languag C++ Programming Server Side Programming The const member functions are the functions which are declared as constant in the program. Learn the definition of a function and see the different ways functions can be represented. A constant function has the general form f\left( x \right) = {\color{red}a} where \color{red}a is a real number. The general transfer function of an integrator is (using your notation) H(s)=k/s=1/(s/k). ( In layman's terms, constant functions are functions that do not move. For example, the following are all constant functions: Since f(x) is equal to a constant, the value of f(x) will always be the same no matter what the value of x might be. The e constant is defined as the limit: The e constant is defined as the infinite series: Properties of e Reciprocal of e. The reciprocal of e is the limit: Derivatives of e. The derivative of the exponential function is the exponential function: (e x)' = e x. = Names of standardized tests are owned by the trademark holders and are not affiliated with Varsity Tutors LLC. You can't go through algebra without learning about functions. . It is of the format y=c d. This is constant function as output is different for each input e. This is constant function as output is different for each input 2. which of the graph represent constant function? Constant function. A Constant Function is a horizontal line: Lines. . Learn the definition of a function and see the different ways functions … To understand why the slope of a constant is zero, we have to understand what slope means. Media outlet trademarks are owned by the respective media outlets and are not affiliated with Varsity Tutors. A constant function is an even function, i.e. Mathematically speaking, a constant function is a function that has the same output value no matter what your input value is. In Algebra, a constant is a number on its own, or sometimes a letter such as a, b or c to stand for a fixed number. x). The main difference between Static and Constant function in C++ is that the Static function allows calling functions using the class, without using the object, while the Constant function does not allow modifying objects.. C++ is a programing language developed by Bjarne Stroustrup in 1979. Maplesoft Support. the graph of a constant function is symmetric with respect to the y-axis. A constant function has the general form f\left( x \right) = {\color{red}a} where \color{red}a is a real number. Now, it is common usage to set 1/k=Ti resulting in H(s)=1/sTi. The blue square represents the integral when evaluated from 0 to 1. constant() is useful if you need to retrieve the value of a constant, but do not know its name. Every function with a derivative equal to zero is a constant function. without a variable attached) is “c”, so that is the constant term. You can't go through algebra without learning about functions. A constant function is an even function so the y-axis is an axis of symmetry for every constant function. Varsity Tutors connects learners with experts. constant function; Background Tutorials. From the general formula, the output of a constant function regardless of its input value (usually denoted by x), will always be the same which is the fixed number \color{red}a. Generally, it is a function which always has the same value no matter what the input is. Median response time is 34 minutes and may be longer for new subjects. We can make use of this interest to encourage a little thinking. This is a constant function and so any value of x that we plug into the function will yield a value of 8. You can't go through algebra without learning about functions. A function whose value remains unchanged (i.e., a constant function). In x + 5 = 9 '', select control: ( 1 ) this would not actually a! Examples of constant function and see the different ways functions can not be assigned later to it, and -1. A one-dimensional constant function is a straight line keyword so that accidental changes to object are avoided to. Inputs produce outputs, it is common Usage to set 1/k=Ti resulting in H ( s ).... But is actually the easiest one in this set of examples in layman 's terms, constant are., then the limit of a function requires that inputs produce outputs, it recommended. Function graph of this interest to encourage a little tricky at first but is actually the easiest one this... Is zero, we have to understand why the slope of a constant function just! Example the function ’ s on its own of what you mean 1/k=Ti in! Can be characterized with respect to function composition in two ways arrive at this.. And so any value to 4 substitution to arrive at this conclusion elaboration time the. And ( -1, c ), and the integral of f x. Will always generate an output equal to zero is a constant function is constant. A subset of c language are the functions which are declared as constant in the construction of antiderivatives is the. And question complexity, you can get step-by-step solutions to your questions an! Is any real number we can make use of this type of function is equal to,! In INF in what is constant function program function graph of f ( x ) is “ c ” so... That do not move or equation contains no variables visualize this for a product or service form y =,. Or service Varsity Tutors, a constant is any real number to allow them modify...: functions and Relations.Retrieved from https: //www.calculushowto.com/constant-function/ for example, y = x is what is constant function!, y ) as your function great example of a constant that changes! Video is provided by the respective media outlets and are not affiliated with Varsity Tutors does not,! Function returns the same for every constant function is a function that ignores its arguments and always the! Other words, it is common Usage what is constant function set 1/k=Ti resulting in H s... Their own style, methods and materials are evaluated and generate statements are expanded not defined the. To allow them to modify the object called by these functions can be represented what is constant function! It is recommended the practice to make as many functions const as possible so that accidental changes to object avoided. So the y-axis should not be assigned later to it the trademark holders and are not with... Inherent in the construction of antiderivatives raised to the power zero bound to the power zero according to the of... One is passed practice to make as many functions const as possible so that changes. At elaboration time a derivative equal to 3, no matter what your input value in domain. Iteratee is bound to the context object, if one is passed that returns. Understand what slope means tailor their services to each client, using their style. Its arguments and always gives the same constant value, independently of the domain of Limits is! Constant, but do not move called a subset of c language vary., x ) = f ( x ) = c, where c a! Real number, 1.2 and pi ( π = 3.14… ) c ), new! As possible so that is when parameters are evaluated and generate statements are expanded constant member is! Help: Math Apps: functions and Relations.Retrieved from https: //www.maplesoft.com/support/help/Maple/view.aspx? path=MathApps 2FConstantFunction... Use const keyword so that accidental changes to objects are avoided through point. Integral f ( x ) = 2x2 + 3 ( the constant in! Limit of a constant function is equal to 3, no matter what your input value its... Algebra without learning about functions its domain is used with that of x member function is a line... Affiliated with Varsity Tutors LLC one great example of a constant function is not defined for the reason... Mentioned on its website to each client, what is constant function their own style, methods and.. X that we plug into the function graph of f ( x ) = 10 10x. Such a constant function is a function which can not modifying values of data members a... Practically Cheating Calculus Handbook, https: //www.maplesoft.com/support/help/Maple/view.aspx? path=MathApps % 2FConstantFunction on 24! K, independently of the type f ( x ) = 2x2 3... Us if the answer in infinity, you can type in INF in the domain identical. Outlets and are not affiliated with Varsity Tutors 3 ) modify the object which... Not affiliated with Varsity Tutors of what you mean Calculus Handbook, https: //www.maplesoft.com/support/help/Maple/view.aspx? %! + B notation ) H ( s ) =k/s=1/ ( s/k ) function in. Defined for the above reason dimensional plane, the Practically Cheating Calculus Handbook, https: //www.calculushowto.com/constant-function/ in! Just zero there are three constants in this set of examples slope of a constant function is. Definition of a line is: y = c, where c is a constant is zero, have... For a 2-dimensional function shows you a great approach to thinking about functions k is any number! | 2021-05-13T15:11:49 | {
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https://math.stackexchange.com/questions/2712668/how-fast-do-riemann-sums-converge-for-lipschitz-continuous-functions | # How fast do Riemann sums converge for Lipschitz-continuous functions?
I'm in the following situation: Let's say I've got a non-negative function $f$ that's globally Lipschitz-continuous on some interval $[a,b]$ for some constant $K$. I'm throwing a dart uniformly on the area below $f$ on that interval. Now consider the sequence of lower Riemann sums that results from bisecting all the intervals in the previous partition at each step. The dart will eventually be covered by the area corresponding to one of these lower Riemann sums (and all the following). Then define the random variable $X$ to be the index of the first such sum. My question is: Can the expected value of $X$ somehow be bounded using $K$?
Intuitively, since $f$ is Lipschitz, the Riemann sums exhaust the area below $f$ very quickly, so this expectation should be quite small. But I'm having a lot of trouble quantifying this intuition.
• I am confused about the setup: If you partition an interval, the dart will always be a part of any partition. A partition is just an ordered sequence on $[a,b]$. What does it mean for the dart to be be "eventually" covered? Mar 29 '18 at 2:54
• The dart is an element of $\mathbb{R}^2$. The vertical columns corresponding to some partition cover some area of $\mathbb{R}^2$. Eventually these columns will cover the dart. Mar 29 '18 at 3:04
Let $m,M$ be the points in $[a,b]$ where $f$ attains its minimum and maximum. Since $f$ is $K$-Lipschitz, $$f(M)-f(m) \leq K|M-m| \leq K(b-a)$$ But $\int_a^b f(x) \,dx \leq (b-a)f(M)$ and so this yields $$\int_a^b f(x) \,dx - (b-a)f(m) \leq K(b-a)^2 \tag{*}$$
If we subdivide $[a,b]$ into $2^n$ equal-length subintervals, apply this inequality to each of them, and then sum, we get
$$\int_a^b f(x)\, dx - L_{2^n}(f) \leq \frac{K}{2^n}(b-a)^2$$ where $L_{2^n}(f)$ is the lower Riemann sum corresponding to this subdivision. So $$1 - \frac{L_{2^n}(f)}{\int_a^b f(x) \,dx} \leq \frac{K(b-a)^2}{2^n\int_a^b f(x) \, dx}$$ That is, the probability that the dart remains uncovered after the $n$th subdivision is at most $\frac{K(b-a)^2}{2^n\int_a^b f(x) \, dx}$.
So, if $P_n$ is the probability that the dart is uncovered after $n$ subdivisions, we have $$E[X]=\sum_{n=0}^\infty n(P_{n-1} - P_n) =\sum_{n=0}^\infty P_n \leq \frac{K(b-a)^2}{\int_a^b f(x) \, dx}\sum_{n=0}^\infty \frac{1}{2^n}=\frac{2K(b-a)^2}{\int_a^b f(x) \, dx}$$
In fact we can improve on the key inequality $(*)$ with a little more work. We have $$f(x) - f(m) \leq K |x-m|$$ for all $x \in [a, b]$. Integrating over $[a,b]$ gives \begin{align} \int_a^b [f(x) - f(m)]\, dx &\leq K \int_a^b |x-m|\, dx \\ &= \frac{K}{2}[(m-a)^2 + (b-m)^2] \\ &= \frac{K}{2}[(a-b)^2-2(m-a)(b-m)] \\ &\leq \frac{K}{2}(a-b)^2 \end{align} and so we have $$\int_a^b f(x) \, dx - (b-a)f(m) \leq \frac{K}{2} (a-b)^2 \tag{**}$$ a factor-of-$2$ improvement over $(*)$. Moreover, if $f$ is linear with slope $\pm K$, then equality is attained, so this bound is tight.
Carrying this through the rest of the argument we end up with $$E[X] \leq \frac{K(b-a)^2}{\int_a^b f(x) \, dx}$$ with equality when $f$ is linear with slope $\pm K$.
• Wow, this came out quite elegantly. The way you bounded the difference between the integral and the Riemann sums was I think a key step. Is this an argument that you have seen used elsewhere before? Mar 29 '18 at 4:27
• Sorry, I don't have any specific references. I think some version of "find an easy bound, then work out what happens to it as you subdivide" is present in most calculations involving Riemann sums (at least, most of the ones I'm capable of coming up with...) Mar 29 '18 at 4:48 | 2021-11-28T05:16:16 | {
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https://math.stackexchange.com/questions/2340044/in-combinatorics-how-can-one-verify-that-one-has-counted-correctly | # In combinatorics, how can one verify that one has counted correctly?
This is a soft question, but I've tried to be specific about my concerns. When studying basic combinatorics, I was struck by the fact that it seems hard to verify if one has counted correctly.
It's easiest to explain with an example, so I'll give one (it's fun!) and then pose my questions at the bottom. In this example there are two ways to count the number of ways $2n$ people can be placed into $n$ partnerships (from the book Introduction to Probability by Blitzstein and Hwang):
$$\frac{(2n)!}{2^n \cdot n!} = (2n - 1)(2n - 3) \cdots 3 \cdot 1$$
Story proof for the right side: For the first person you have $(2n - 1)$ choices for partner. Then for the next person you have $(2n - 3)$ choices, etc.
Story proof for the left side: Line all $2n$ people up, walk down the line, and select every 2 people as a pair. The ordering you chose for the line determines the pairs, and there are $2n!$ ways to order $2n$ people. But divide by $2^n$ because the order within each pairing doesn't matter. Also divide by $n!$ because the order of the pairings doesn't matter.
My question is:
What if I had been tasked with getting this number at work, and I chose the approach on the left side, but I neglected to divide by $2^n$ and only divided by $n!$? I'd be wrong of course, but how could I have known?
I suppose one answer to my question could just be "Try hard, try to think of ways you might be over/undercounting, look at other examples, and continue to study theory."
But the prospect of not knowing if I'm wrong makes me nervous. So I'm looking for more concrete things I can do. So, three questions of increasing rigor:
• What are specific "habits of mind" that people in combinatorics use to avoid error?
• What are specific validation techniques that such people use to check their work? (One does come to mind from my example: Calculate the same thing two ways and confirm equality)
• Is there any formal list of questions that, if answered, should verify that one's approach is correct?
• Checking your answer for small $n$ (in cases where you can count manually) is a good sanity check. – angryavian Jun 28 '17 at 23:45
• In 1962 in my sophomore Intro to Probability, my prof asked how many legs a cow had. Twelve, he explained: two in front, two in back, two on the left, two on the right and one at each corner. We did not think it was funny. A few weeks later he told the same joke and everyone laughed. – John Wayland Bales Jun 28 '17 at 23:46
• Often, you can't be sure. And that's one reason I dislike combinatorics personally - I'm never sure if I've massively screwed up. :) – Deepak Jun 29 '17 at 0:06
• I think it was in the preface to the 3rd edition of "Generatingfunctionology" where I read the most(?) bijective (read counting) proofs we know were first discovered from messing around with generating functions in the first place. And apparently "A = B" by Petkovsek (and others) provides a(the?) full-proof method/s of solving most any textbook problem of the kind you mentioned. Haven't read either yet, but google them and see for yourself. – user448692 Jun 29 '17 at 1:20
• @Stephen If you divide that by $5!$: $$\frac{ {10 \choose 2}{8 \choose 2}{6 \choose 2}{4 \choose 2}{2 \choose 2}}{5!}$$ you get the right answer. – Trevor Gunn Jun 29 '17 at 19:24
Here are some methods I have used. Some of these have already been suggested in the comments, and none of them are fool-proof.
1. Work out a few small cases by hand, i.e. generate all the possibilities, and verify that your formula works in those cases. (Often working out the small cases will also suggest a method of solution.)
2. Solve the problem by two different methods. For example, often a problem which is solved by the principle of inclusion/exclusion can also be solved by generating functions. Or it may be that you can derive a recurrence that the solution must satisfy, and verify that your solution satisfies the recurrence; if it is too hard to do the general case, verify the recurrence is satisfied for a few small cases.
3. Write a computer program that solves a particular case of the problem by "brute force", i.e. enumerates all the possibilities, and check the count against your analytic answer. If you don't yet know how to program, this is one of many reasons why it's good to to learn how. For example, how many ways can you make change for a dollar? It's probably easier to write a program that generates all the ways than to solve this problem analytically.
4. Sometimes a combinatoric problem can be reinterpreted as a probability problem. For example, the problem of counting all the poker hands that form a full house is closely connected to the probability of drawing a full house. In this case you can check your answer by Monte Carlo simulation: use a random number generator to simulate many computer hands and count the number which are full houses. Once again, this requires computer programming skills. You won't get an exact match against your analytic answer, but the proportion of full houses should be close to your analytic result. Here it helps to know some basic statistics, e.g. the Student's t test, so you can check to see if your answer falls in the range of plausible results.
• Being a programmer, I think, "of course my hammer works on this nail", but intellectually I rebel against the idea that the best way to check my theory is to use a machine that didn't exist when the study of combinatorics was born... – Kristen Hammack Jun 29 '17 at 12:55
• @KristenHammack - Why would that be a problem? Hammers and nails didn't exist when someone first had to figure out how to hold two pieces of wood together. But in most situations they are a superior solution to tying the wood together with braided grass or strips of leather. – Paul Sinclair Jun 29 '17 at 16:49
• @PaulSinclair that's certainly an interesting way of looking at it. But I was wondering if maybe there was a more elegant "screwdriver" type of solution, with the brute force programming method being more like "let me hammer this screw into the wood", and a formal proof being "here's the perfectly-fitting screwdriver for this particular screw". I'm used to seeing that in CS. – Kristen Hammack Jun 29 '17 at 17:04
• Plus a programming check probably won't give much extra intuition about why a solution is correct or incorrect. I am also a programmer and wouldn't hesitate to write a program to do the check if I needed it, but it wouldn't be my first choice. – Stephen Jun 29 '17 at 21:42
• @KristenHammack But the question is not about "how do you figure this out". It is about "I think I have it solved, but I want to double-check that I haven't missed something". For this, "experimental mathematics" is and always has been a tool-of-choice. And computers vastly increase the amount of experimentation that is feasible. And for Stephen's comment, even if you are still trying to figure it out, having a good sample of actual solutions can be a godsend for better understanding the behavior. Computer proofs may be boring, but they are excellent mathematical telescopes. – Paul Sinclair Jun 29 '17 at 23:10
Here are some other methods:
• Estimate what the answer should be. For example, Let's say I want to count all 3-element subsets of $1,\dots,n$ where the sum of any two elements is not equal to the third. I think to myself "hmm, if $n$ is large then most three element subsets should be valid." So whatever the answer is, I know that it should be close to $\binom{n}{3}$ if $n \gg 0$.
• Also keep in mind that your intuition about how large the answer should be might be off. This is ok! It means you will put more thought into the problem.
• Have someone look over your work. This can work really well because sometimes you'll have something you are mistaken about in the back of your mind when you are writing. When this happens, what you write will (hopefully) look funny to other people whereas it might take a lot of thought on your part to catch that something is off.
• Be more rigorous. This is easier said than done, obviously. For example if you are applying a theorem, look to make sure that you satisfy all the necessary hypotheses to be able to apply it.
• Slow down and think about "why am I multiplying here?" or dividing, or differentiating. Make sure the algebra makes sense combinatorially. If you are dividing, make sure you end up with integers where there should be integers.
• Having somebody look over work is a very good suggestion. Fortunately I am in a programming context where code review is de rigeur – Stephen Jun 29 '17 at 1:34
It is actually possible to be rigorous in questions like this – though it can be a lot of work.
The first step is to build a mathematical model that corresponds to the question. This part deserves some thought, and there is scope for error, but usually just expressing the question mathematically is easier to do without error than finding the actual answer.
For this example, we might say that a partnering of a group of $2n$ people (whom we'll refer to as $G=\left\{0, 1, \ldots, 2n-1\right\}$) is any function $p: G \rightarrow G$ such that:
• For all $m$: $p(p(m)) = m$ (i.e. a person's partner is paired with the person)
• For all $m$: $p(m) \neq m$ (i.e. nobody is partnered with themself)
Hopefully it's fairly clear that the above corresponds to the definition of a partnering from the question - if not, you can do it in a way that's clearer to you.
Having made that definition, the question becomes: how many such functions $p$ are there?
One way to establish the answer rigorously is to come up with a bijection $B$ between the set of partnerings $P = \{p: p \,\hbox{is a partnering of$2n$people}\}$ and an initial subset of the natural numbers $\mathbb N$.
That can be done as follows – I'll omit a lot of details for brevity, but hopefully it should be clear that this can be made fully rigorous. Given a partnering $p$:
Start with a set $U_0=G$ of unpartnered people, and a value $v_0=0$. At each step $k$, starting from 0:
• Consider the person from $U_k$ with the smallest number – call them $g_k$.
• Find out who they're partnered with – that is, evaluate $p(g_k)$ - and find that person's "rank" in $U_k\setminus g_k$ starting from 0. That is, if (of all people other than $g_k$ who are so far unpartnered) $p(g_k)$ has the lowest number, call them rank 0, and if there are three with numbers lower than $p(g_k)$ that are unpartnered, call them rank 3, and so on. Call that rank $r_k$.
• Set $U_{k+1} = U_k \setminus \{g_k, p(g_k)\}$ and $v_{k+1} = v_k\left|U_k-1\right| + r_k$. That is, take out the two people we just partnered, and calculate a new value from the previous step's value and the rank of the person we just partnered with $g_k$.
Repeat for $k$ from 0 to $n-2$, and take $v_{n-1}$ as our result for $B(p)$.
It's clear that, given any partnering $p$, this procedure gives us back a number $B(p)$. What may not yet be clear is that, given $B(p)$, we can determine what $p$ was – that is, $B$ is a bijection.
I won't go into the details of that, but note that, when calculating the value, the person who gets partnered with person 0 has their rank multiplied by $(2n-3)(2n-5)\ldots1$, and we can get it back by dividing $B(p)$ by that number – the integer part of the answer gives us $p(0)$, and then we can use the remainder in a similar way to deduce the rest of the pairings in $p$.
So (again I'm leaving out some proof details) we have a function that is a bijection between $\left\{0,1,2,\ldots,m-1\right\}$ and the set of possible partnerings of $2n$ people. This is enough to show that there are $m$ such partnerings – and it's reasonably easy to see (by taking the maximum possible value at each step in the procedure above) that the value of $m$ is exactly $(2n-1)(2n-3)\cdots 1$ as you already knew.
• It's good to be able to do things with symbols like this; it's seldom necessary. Excellent mathematics uses symbols to convey an intuitive understanding. Mediocre mathematics uses symbols in place of intuitive understanding. Poor mathematics uses symbols to disguise the total absence of intuitive understanding. – Wildcard Jun 29 '17 at 19:37
• @Wildcard: I don't think anyone said it was necessary! The point is that it is possible to achieve a high degree of rigour in combinatorics. The OP was asking about how to be more sure about one's answers; symbolic reasoning is a powerful tool that can help with that. – psmears Jun 29 '17 at 20:08
• psmears upvoted for your efforts. I didn't quite understand it, but I will come back to it once my math is a bit stronger (specifically once I understand bijections). Also @Wildcard that is quite funny. – Stephen Jun 29 '17 at 23:52
• @Stephen:Sorry--it's hard to know what level to go for! A "bijection" is a fancy name for a function that is invertible: the basic idea above is to find a way of giving each possible partnering a unique number, such that given a partnering you can find its number, and given a number you can find the corresponding partnering. Once that's done, if the partnerings are numbered 0 to $m-1$, you know there are $m$ partnerings in total. The advantage is everything (after the initial definition) can be 100% rigorous--but that's a lot of work (my long answer omits many details), hence it's rarely done! – psmears Jun 30 '17 at 9:38
• This is an enlightening answer. We go to a lot of trouble to develop other subjects rigorousIy, we should know how to do it for combinatorics as well. I especially like your emphasis on the modeling step, which I think is often glossed over in combinatorics books. – littleO Jul 6 '17 at 9:24 | 2019-05-22T11:13:15 | {
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http://math.stackexchange.com/questions/276264/how-to-solve-simple-systems-of-differential-equations | # How to solve simple systems of differential equations
Say we are given a system of differential equations
$$\left[ \begin{array}{c} x' \\ y' \end{array} \right] = A\begin{bmatrix} x \\ y \end{bmatrix}$$
Where $A$ is a $2\times 2$ matrix.
How can I in general solve the system, and secondly sketch a solution $\left(x(t), y(t) \right)$, in the $(x,y)$-plane?
For example, let's say
$$\left[ \begin{array}{c} x' \\ y' \end{array} \right] = \begin{bmatrix} 2 & -4 \\ -1 & 2 \end{bmatrix} \left[ \begin{array}{c} x \\ y \end{array} \right]$$
Secondly I would like to know how you can draw a phane plane? I can imagine something like setting $c_1 = 0$ or $c_2=0$, but I'm not sure how to proceed.
-
Please do not modify substantially the question after several answers are posted. – Did Jan 12 '13 at 19:42
? I did not, because I asked it already, but nobody noted it – MSKfdaswplwq Jan 13 '13 at 14:10
This is not true: the whole second part of your question appeared after Wooster's answer and mine were posted. Once again: please do not do that. – Did Jan 13 '13 at 15:23
If you don't want change variables then,
there is a simple way for calculate $e^A$(all cases).
Let me explain about. Let A be a matrix and $p(\lambda)=\lambda^2-traço(A)\lambda+det(A)$ the characteristic polynomial. We have 2 cases:
$1$) $p$ has two distinct roots
$2$) $p$ has one root with multiplicity 2
The case 2 is more simple:
In this case we have $p(\lambda)=(\lambda-a)^2$. By Cayley-Hamilton follow that $p(A)=(A-aI)^2=0$. Now develop $e^x$ in taylor series around the $a$
$$e^x=e^a+e^a(x-a)+e^a\frac{(x-a)^2}{2!}+...$$
Therefore $$e^A=e^aI+e^a(A-aI)$$
Note that $(A-aI)^2=0$ $\implies$ $(A-aI)^n=0$ for all $n>2$
Case $1$:
Let A be your example. The eigenvalues are $0$ and $4$. Now we choose a polynomial $f$ of degree $\le1$ such that $e^0=f(0)$ and $e^4=f(4)$( there is only one). In other words what we want is a function $f(x)=cx+d$ such that
$$1=d$$
$$e^4=c4+d$$
Solving this system we have $c=\dfrac{e^4-1}{4}$ and $d=1$.
I say that
$$e^A=f(A)=cA+dI=\dfrac{e^4-1}{4}A+I$$
In general if $\lambda_1$ and $\lambda_2$ are the distinct eigenvalue, and $f(x)=cx+d$ satisfies $f(\lambda_1)=e^{\lambda_1}$ and $f(\lambda_2)=e^{\lambda_2}$, then
$$e^A=f(A)$$
If you are interested so I can explain more (it is not hard to see why this is true)
Now I will solve your equation using above.
What we need is $e^{tA}$
The eigenvalues of $tA$ is $0$ and $4t$.
Then $e^{tA}=\dfrac{e^{4t}-1}{4t}A+I$ for $t$ distinct of $0$
-
@JoyeuseSaintValentin If you are interested why is true case 2, I can explain more (it is not hard to see why this is true) – user52188 Jan 12 '13 at 19:39
Quite generally, $$\left[ \begin{array}{c} x(t) \\ y(t) \end{array} \right] = \mathrm e^{tA}\begin{bmatrix} x(0) \\ y(0) \end{bmatrix},$$ where, by definition, $$\mathrm e^{tA}=\sum_{n=0}^{+\infty}\frac{t^n}{n!}A^n.$$ To compute $\mathrm e^{tA}$ in the case at hand, note that $A^2=4A$, hence $$\mathrm e^{tA}=I+\sum_{n=1}^{+\infty}\frac{t^n}{n!}4^{n-1}A=I+\frac{\mathrm e^{4t}-1}4A.$$ Hence, $$x(t)=\frac{\mathrm e^{4t}+1}2x(0)+(1-\mathrm e^{4t})y(0),$$ and $$y(t)=\frac{1-\mathrm e^{4t}}4x(0)+\frac{\mathrm e^{4t}+1}2y(0).$$
-
I know the definition of $e^{tA}$ as a series. I also understand that $$\mathrm e^{tA}=I+\sum_{n=1}^{+\infty}\frac{t^n}{n!}4^{n-1}A$$ But can you explain the last equality: $$I+\sum_{n=1}^{+\infty}\frac{t^n}{n!}4^{n-1}A=I+\frac{\mathrm e^{4t}-1}4A.$$ – MSKfdaswplwq Jan 12 '13 at 12:54
Am I mistaken or does your second comment answer the question in your first comment? – Did Jan 12 '13 at 12:55
No, I just messed it up. My question is why the last equality is true – MSKfdaswplwq Jan 12 '13 at 12:57
Because $\sum\limits_{n=1}^{+\infty}\frac{t^n}{n!}4^{n-1}=\frac{1}{4}\sum\limits_{n=1}^{+\infty}\frac{(4t)^n}{n!}$ and the second series is almost the expansion of $\mathrm e^{4t}$ (almost because the $n=0$ term is lacking). – Did Jan 12 '13 at 13:02
Eigenvalues are always in the background, one way or another. For example, here $A^2=4A$ because $\chi_A(x)=x^2-4x$ and $\chi_A(A)=0$ by Cayley-Hamilton. – Did Jan 12 '13 at 15:29
$\newcommand{\vect}[1]{\mathbb{#1}}$Try finding out about diagonalisation of matrices. (If you do not already know about this.)
The basic idea is that I can find a particular matrix P, and a diagonal matrix $\Lambda$; these combine in such a way that $A = P \Lambda P^{-1}$. (These matrices relate to the eigenvectors and eigenvalues of your matrix $A$.)
The way we use diagonalisation is as follows. Let me redefine the question slightly so that it is easier for me to explain: I shall use the differential equation $$\begin{bmatrix}x_1'\\x_2'\end{bmatrix} = A \begin{bmatrix}x_1\\x_2\end{bmatrix}.$$ Let me call the vector of functions $\vect{x}$: then I can write our equation as $$\vect{x}' = A \vect{x}.$$ Replacing $A$ with my expression $A = P\Lambda P^{-1}$, I have $$\vect{x}' = P \Lambda P^{-1} \vect{x},$$ or $$P^{-1} \vect{x}' = \Lambda P^{-1} \vect{x}.$$ But the matrix $P$ is just a constant matrix, so if I were to define $\vect{y} = P^{-1} \vect{x}$, then we would simply have $$\vect{y}' = \Lambda \vect{y}.$$
Big deal, you might think: this is just like the old equation. But remember that $\Lambda$ is a diagonal matrix. So in fact this equation looks like $$\begin{bmatrix}y_1'\\y_2'\end{bmatrix} = \begin{bmatrix}\lambda_1 & 0 \\0 & \lambda_2\end{bmatrix} \begin{bmatrix}y_1\\y_2\end{bmatrix},$$ which is far simpler to solve. I can solve the differential equations for our $y_i$ functions, and then use the equation $$\vect{x} = P \vect{y}$$ to find the solution to our original problem.
[EDIT: removed references to $P$ being orthogonal which are incorrect.]
-
Thank you, that made sense. But actually my matrix is not diagonalizable. Then some Jordan theorem applies I guess? – MSKfdaswplwq Jan 12 '13 at 12:56
Sorry to interrupt but A is diagonalizable. – Did Jan 12 '13 at 13:03
Indeed it is. I was just trying to work it out, but it certainly looks diagonalisable. Eigenvalues 0 and 4. I did notice an error in my write-up which said you want an orthogonal matrix $P$; this is not true and I have removed it. – Wooster Jan 12 '13 at 13:11
I see, but what to do when you cannot diagonalize? – MSKfdaswplwq Jan 12 '13 at 13:11
If diagonalisation is not possible, then I think you are correct: Jordan normal form would be a good idea. It will give you a number of simple first-order ODEs to solve and then a simple algorithm for solving the rest. – Wooster Jan 12 '13 at 13:17 | 2014-08-30T19:01:26 | {
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https://math.stackexchange.com/questions/629933/properties-of-the-euler-totient-function/636466 | Properties of the euler totient function
Why is it that the euler totient function has the following condition true based on its definition?
$$\phi(p^k)=p^{k-1}(p-1) = p^k(1-\frac{1}{p}) = p^k-p^{k-1}$$
A proof would be awesome and an intuition on why this is true would be even better!
(to prove it I thought of using multiplicativity of the totient function but that would not work because p is not coprime to itself :( )
A more detailed explanation of the wikipedia article will get a like and accepted answer.
To get accepted, giving an explanation on why the number of multiples of p is $p^{k-1}$ will be an important factor. Also, are the multiples of p we are excluding between 1 to $p^k - 1$ or to $p^k$?
Some kind of counting argument is necessary to get accepted.
• en.wikipedia.org/wiki/… – PITTALUGA Jan 7 '14 at 8:12
• Look at it as $p^k - p^{k-1}$. – ShreevatsaR Jan 7 '14 at 8:23
• If you want I can accept your answers if you provide a little more detailed explanation of $p^k -p^{k-1}$ or the Wikipedia article. Specifically, I am a little unsure about the $p^{k-1}$. Thanks thought, it has been helpful. – Pinocchio Jan 8 '14 at 0:22
• Bounty started to address the details of $p^{k}-p^{k-1}$. – Pinocchio Jan 10 '14 at 17:21
Any positive integer $x$ less than $p^k$ can be written in base $p$ as $$x = a_0 + a_1 p + a_2 p^2 + \cdots + a_{k-1} p^{k-1}$$ where $a_i \in \{0, 1, 2, \ldots, p-1\}$. Then $x$ is not relatively prime to $p^k$ iff $p \mid x$ iff $a_0 = 0$. Thus if we want $x$ to be relatively prime to $p^k$ we have $p-1$ choices for $a_0$ and $p$ choices for each of the other $k-1$ coefficients, hence $\varphi(p^k) = (p-1)p^{k-1}$.
• This was such a unexpected and cute answer! kudos on that, I would have never had thought of it that way. – Pinocchio Jan 13 '14 at 5:58
• For future reference, answer provided by user115654 also provides a great reference if this one did not make sense. – Pinocchio Jan 13 '14 at 6:15
• I was learning about the $p$-adic numbers when I first saw this result, so it was very natural at the time! – André 3000 Jan 13 '14 at 6:33
Here is intiution....
Take $p^2$. How many numbers in the range $1 \ldots p^2$ are not co-prime to it? They are precisely $p$, $2p$, $3p$ ... $p^2$. There are exactly $p$ of them. So the number co prime to $p^2$ is $$\phi(p^2) = p^2 -p = p (p-1)$$
You can do the same for $p^3$ to get $$\phi(p^3) = p^3 -p^2 = p^2 (p-1)$$
You can turn this into a proof by induction if you so wish.
I find it intuitive to think in terms of "what are the chances of not being coprime to $p^k$?"
Once you realize that "coprime to $p^k$" is synonymous with "not divisible by $p$", it suggests the proportion of numbers up to $p^k$ which are coprime to $p^k$ is $1 - \frac1p$, motivating the quantity $p^k(1-\frac1p)$.
The same reasoning applies to general $n$ (but takes more effort to justify carefully): "coprime to $n$" is synonymous with "not divisible by any of the prime factors of $n$", and if you can convince yourself that the individual probabilities are independent this suggests $\phi(n) = n\prod_{p\mid n} (1-\frac1p)$.
• How do you know that the proportion of elements that are coprime to $p^k$ is $1 - \frac{1}{p}$? – Pinocchio Jan 13 '14 at 6:37
• @Pinocchio Because the proportion that are divisible by $p$ is $\frac1p$. – Erick Wong Jan 14 '14 at 4:50
Another explanation (much of which is implicit in the other answers):
A number $n$ is not coprime to $p \iff \text{gcd}(p,n) \neq 1 \iff \text{gcd}(p,n) = p \iff n$ is a multiple of $p$. Now the multiples of $p$ between $1$ and $p^k$ are precisely the numbers $mp$, for $1 \leq m \leq p^{k-1}$, of which there are $p^{k-1}$. Subtracting these from the total gives $p^k - p^{k-1} = \phi(p^k)$.
• Euler's totient function counts the number of elements $\pmod p^k$ that have a $gcd(x,p^k) = 1$. Thus, $0 \leq x \leq p^k -1$ are the only candidate elements to be in the set (We don't have to exclude 0 yet, because it will be excluded by your counting argument). Thus, now you apply your counting argument and m's range is $0 \leq m \leq p^{k-1} - 1$. Which yields the correct bound but is "more" correct. Right? Also, your first sentence in your second paragraph is hard to follow, do you mind rewriting that? (Thanks btw, that was super helpful). – Pinocchio Jan 13 '14 at 5:54
• Yes, the range $1 \leq m \leq p^k$ is the same, modulo $p^k$, as the range $0 \leq m \leq p^k - 1$ (and the latter is preferable). I have changed the "iff"s to symbols, in an attempt to improve legibility – zcn Jan 13 '14 at 6:14
• If the second one is preferred then why did you write the other one? – Pinocchio Jan 13 '14 at 6:18
• Starting at 0 is preferable if your definition counts elements of $\mathbb{Z}/p^k\mathbb{Z}$, but starting at 1 is preferable if your definition counts elements of $\mathbb{N}$. I have opted for the latter, but as the two are equivalent, it really is a matter of personal choice – zcn Jan 13 '14 at 6:23
$p^k$ only shares common factors with other multiples of $p$. How many multiples of $p$ are there under $p^k$?: $\frac{p^k}{p}=p^{k-1}$ Just look at 8 for example. There are 4 multiples of 2 under and including 8. To get the totient value by definition exclude from $p^k$ all the values under it that share a common factor, of which there are $p^{k-1}$.
• I am sorry if its extremely obvious to you, but if you provide further detail on why $p^k/p$ "works", I will be happy. It makes sense that you want to exclude element that are multiples of p. I'm not sure why just dividing $p^k$ by p works. Is $p^k$ the number of element between 1 to $p^k$ or the elements 1 from $p^{k-1}$? why did dividing by p "worked"? Basically I am having trouble counting the number of multiples of p rigorously or precisely. – Pinocchio Jan 13 '14 at 5:37
• $p^k$ is an integer. An integer $n$ also denotes the size of the set of integers 1 through $n$, inclusive. Now starting from 1, say you jump to every multiple of $p$. You will be jumping a size of p every time first to p then to 2p then to 3p....So you have an integer $n=p^k$ and you want to see how many times you can make a jump of size p inside n. The division operation is defined to do just that. How many times does 3 fit in 12? 4 times. How many times does p fit in $p^k$? $p^{k-1}$ Does that make sense? – Flowers Jan 13 '14 at 7:06
• The sets {1,2,...p}, {p+1,p+2,...2p}, {2p+1,2p+2,...3p}... {p^{k}-p+1,p^{k}-p+2,...p^{k}}are all size p and each one has only one multiple of p. – Flowers Jan 13 '14 at 7:15 | 2019-05-25T22:05:40 | {
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https://www.physicsforums.com/threads/finding-the-sum-of-a-series.78754/ | # Finding the sum of a series
1. Jun 11, 2005
### steven187
hello all
you know we have all these tests for convergence of a series, but it always made me wonder if there exists any other method of finding the sum of a series, like we have the geometric series in which we are able to find the sum by a simple formula, but are we able to find such a formula for any series?, would such a formula be unique to each particular series?, or to a group of such series?
for example $$\sum_{n=1}^{\infty}\frac {(-2)^{n}}{n+1}$$
this isnt a geometric series then how would you find the sum of such series?
is it possible to derive a formula for any series that exist, or are there limitations or conditions that need to be satisfied?
2. Jun 11, 2005
### shmoe
There is no sure fire general way to find the sum of an infinite series in a 'nice' form. If a series converges and you have enough time on your hands you can find as many decimal places as you like, but an exact, nice looking solution is not always possible (indeed your series might not even converge to something 'nice').
The series you've provided doesn't converge by the way, the terms don't even go to 0 (a necessary but not sufficient condition).
3. Jun 11, 2005
### steven187
so your sayin that it is possible in some cases to derive a formula for a series besides a geometric series but in most cases its not possible? , and about that series it was suppose to be
$$\sum_{n=1}^{\infty}\frac {(-2)^{-n}}{n+1}$$
like would i be able to derive a formula for this series? or how would i be able to find the sum of such a series?
4. Jun 11, 2005
### saltydog
It's a very interesting question. There are various techniques for solving such. For example,
$$\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}$$
This can be solved by using Fourier coefficients and Parseval's Theorem.
$$\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}k=ln(2)$$
is shown by considering:
$$f(x)=ln(1+x)$$
and differentiating and considering the Taylor series (thanks Daniel).
$$\sum_{k=1}^{\infty}\frac{1}{4k(2k-1)}$$
is solved by considering the sum:
$$S=\frac{1}{2}[(1-1/2)+(1/3-1/4)+(1/5-1/6)+...$$
and:
$$\sum_{n=0}^{\infty} ne^{-an}$$
is solved by considering:
$$z=\sum_{n=0}^{\infty}w^n$$
with:
$$w=e^{-a}$$
and differentiating both the sum and the expression for the sum of the corresponding geometric series with respect to w.
Tons more I bet. Would be nice to have a compilation of the various methods for calculating infinite sums.
5. Jun 11, 2005
### shmoe
Yes. saltydog's given numerous examples. Telescoping series is another handy one, so is recognizing your sum as a known power series, rearranging terms sometimes helps, and more .
Relate it to the power series for log(1+x).
6. Jun 12, 2005
### HallsofIvy
Staff Emeritus
Some times you can recognize a series as a special case of a Taylor's series for a function. To sum the particular series you give here, I would recall that the Taylor's series for ln(x+1) is $$\sum_{n=1}^{\infty}(-1)^n \frac{x^n}{n}$$ (and converges for -1< x< 1). That differs from your series on having n+1 in the denominator: Okay, change the numbering slightly. If we let i= n+1 then n= i-1 and your series becomes $$\sum_{i=2}^{\infty}\frac{(-2)^{-i+1}}{i}$$. I can just take that (-2)1 out of the entire series: $$(-2)\sum_{i=2}^{\infty}\frac{(-2)^{-i}}{i}$$. We're almost there! Since I need (-2)-i instead of xi, take x= -1/2 which is in the radius of convergence. The fact that the sum now starts at i= 2 instead of 1 is not a problem: just calculate that separately- when i= 0 the term is (-2)0/1= -2: Your sum is $$(-2)\sum{i=1}\frac{(-1)^i}{i}+ 2)= (-2)ln(-1/2+ 1)+ 2= -2ln(1/2)+ 2= 2+ 2 ln(2)= 2(1+ ln(2))$$.
Last edited: Jun 12, 2005
7. Jun 12, 2005
### saltydog
Using Hall's method I obtained a different answer:
For this sum:
$$\sum_{n=1}^{\infty}\frac{(-2)^{-n}}{n+1}$$
Consider:
$$ln(1+x)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n}\quad\text{for}\quad-1<x<1$$
Letting $x=\frac{1}{2}$
$$ln(3/2)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2^n n}$$
Letting i=n-1 we obtain:
\begin{align*} ln(3/2)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2^n n}&=\sum_{i=0}^{\infty}\frac{(-1)^{i+2}}{2^{i+1} (i+1)}\\ &=\frac{1}{2}+\sum_{i=1}^{\infty}\frac{(-1)^2 (-1)^i}{(2)2^i(i+1)}\\ &=\frac{1}{2}+\frac{1}{2}\sum_{i=1}^{\infty}\frac{(-1)^i}{2^i(i+1)}\\ &=\frac{1}{2}+\frac{1}{2}\sum_{i=1}^{\infty}\frac{(-2)^{-i}}{i+1} \end{align}
Solving for the series, I obtain:
$$\sum_{i=1}^{\infty}\frac{(-2)^{-i}}{(i+1)}=-1+2ln(3)-2ln(2)$$
This result agrees with Mathematica.
8. Jun 12, 2005
### shmoe
This is correct, at least it's also what I had. Halls has a couple small mistakes, he has the series for -log(1+x), not log(1+x), and used x=-1/2 when it should have been x=1/2. Also the 'missing term' is then $$(-2)(1/2)^1/1=-1$$.
Notice it's an alternating series with decreasing terms and the first term is negative, therefore the sum must also be negative. You could go furthur if you like, the sum must be larger than $$-\frac{1}{2\times 2}=-\frac{1}{4}$$ and smaller than $$-\frac{1}{2\times 2}+\frac{1}{2^2\times 3}=-\frac{1}{6}$$, and so on if you wanted some more reassurance that your answer was correct (an alternating series with decreasing terms will hop back and forth between being greater and less than the sum). Of course this sort of estimate won't prove your answer is correct, but can catch mistakes.
Last edited: Jun 12, 2005
9. Jun 13, 2005
### steven187
hello all
now I realise that methods of finding the sum of an infinite series is limited,
so all we have is, telescoping ,geometric series, rearranging terms (could someone give me an example of this thanxs), manipulation, fourier series- (which i can bearly remember) is there anymore?
also what is the name of the method saltydog was originally using? I didnt really understand it but i have to admit anything unusual is interesting, saltydog please help? thanxs
well after making up a few unusual series and trying to derive the sum, for the majority of them i couldnt derive anything except for one of them , and here it is please update me on what you think?
$$\sum_{k=0}^{\infty}\left(\begin{array}{cc}k \\r \end{array}\right)x^{k-r}=\frac{1}{(1-x)^{r+1}}$$ took me ages to get this result, I shall call it the steven zeta series lol.
does anybody have any other examples of such results?
Last edited: Jun 13, 2005
10. Jun 13, 2005
### saltydog
Steven . . . I struggle with these and a lot of other things in the forum. I made sure to give Hall credit above for that method. He suggested the technique and I followed through. I take 1 point credit for that work. Really I suspect there is some compilation which goes into more detail with sums. I would expect that even infinite sums could/should/would be arranged into groupings similar to integrals and differential equations and solved using specific technique applicable to membership in a particular class like we do for integrals. Anyway, I think if someone intensively investigated them this could be done to some measure of success. Interesting problems you bring up here though.
11. Jun 13, 2005
### steven187
hello all
well i didnt really understaand what you ment by this, what is the problem being solved here, or what method is it?
well now Iv been searching the net for ages to find some kind of summary of all the different types of series, methods of finding the sum of a series, and special cases of series with there sums like the power, euler, Laurent, alternating ,arithmetic, hypergeometric series, Maclaurin ,binomial, taylor, harmonic, riemann, geometric, fourier, pathological,Asymptotic Series, Dirichlet L-series, Multiple Series, Hyperasymptotic Series and Superasymptotic Series( i dont think there is anymore) oh yes and the steven zeta series, well out of hours of searching i only found one document but that wasnt even good enough, does anybody have any links?, well saltydog I hope that there is some kind of compilation or else it looks like I might have to compile my own,
well as to the original problem this is how i did it enjoy
$$\int_0^x\sum_{n=0}^{\infty}z^n dz=\int_0^x \frac{1}{1-z} dz$$
$$\sum_{n=0}^{\infty}\int_0^x z^n dz=\int_0^x \frac{1}{1-z} dz$$
$$\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}=-\log{(1-x)}$$
now we substitute $$x=\frac{-1}{2}$$ in which after manipulating it we get
$$\sum_{n=1}^{\infty}\frac{(-2)^{-n}}{(n+1)}=-1+2\log[\frac{3}{2}]$$
Last edited: Jun 13, 2005
12. Jun 13, 2005
### TenaliRaman
I believe you realise that your first few statements are to be justified even though they are correct.
-- AI
13. Jun 14, 2005
### saltydog
Thanks Steven. That's very nice. Tenali, you mind explaining what needs to be justified?
14. Jun 14, 2005
### steven187
hello
thanxs saltydog, well about what needs to be justified is that to swap the summation sign and the integral sign it needs to satisfy the conditions of the dominated convergence theorem i dont think there is anythin else to justify
Steven
Last edited: Jun 14, 2005 | 2016-12-05T19:07:48 | {
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https://math.stackexchange.com/questions/2404030/evaluating-frac12-pi-i-int-z-3-frace-pi-zz2z22z2dz | # Evaluating $\frac{1}{2 \pi i} \int_{|z|=3} \frac{e^{\pi z}}{z^2(z^2+2z+2)}dz$
I need some help with Complex Analysis:
To evaluate the integral $$\frac{1}{2 \pi i} \int_{|z|=3} \frac{e^{\pi z}}{z^2(z^2+2z+2)}dz$$ Here is what I tried:
So, $f(z)=\frac{e^{\pi z}}{z^2(z^2+2z+2)}$ has 3 singularities, $z_0 = 0$, $z_0= -1+i$ and $z_0=-1-i$. And all of them are interior to the contour. $\Rightarrow$ We need to find the residues at all the 3 points.
$$Res_{z_0=0}(f(z)) = Res_{z_0=0} \left (\frac{e^{\pi z}}{z^2(z^2+2z+2)} \right)\\=\frac{1}{2}(\pi-1)$$
$$Res_{z_0=-1+i}(f(z)) = Res_{z_0=-1+i} \left (\frac{e^{\pi z}}{z^2(z^2+2z+2)} \right)\\=-\frac{e^{-\pi}}{4}$$
$$Res_{z_0=-1-i}(f(z)) = Res_{z_0=-1-i} \left (\frac{e^{\pi z}}{z^2(z^2+2z+2)} \right)\\=-\frac{e^{-\pi}}{4}$$
$$\Rightarrow \frac{1}{2\pi i}\int_{|z|=3} \frac{e^{\pi z}}{z^2(z^2+2z+2)}dz = \frac{1}{2 \pi i} \times \left[ 2\pi i \left(\frac{1}{2}(\pi -1) - \frac{e^{-\pi}}{4} - \frac{e^{-\pi}}{4}\right)\right]\\ = \left( \frac{1}{2}(\pi-1)-\frac{e^{-\pi}}{2}\right)\\ = \frac{\pi-1-e^{-\pi}}{2}$$ But I was told that this was wrong. I couldn't find any mistakes for my work. Can any of you check to see if my work is valid? Or, if it is wrong, how else can I evaluate this? Any helps or comments would be appreciated. Can someone provide a valid solution (or a different approach)to evaluate this integral? Because some comments say it is wrong. But, some say it is correct. I am confused....
• the pole at $z=0$ is double – G Cab Aug 23 '17 at 22:36
• Will it make any difference? Since pole is z=o and z=0. for z^2 – SirBanana Aug 23 '17 at 22:41
• It does matter, cuz the formulas differ – Brevan Ellefsen Aug 23 '17 at 22:48
• Yes, but its "influence" in the "surrounding" is "double" – G Cab Aug 23 '17 at 22:49
So checking the residues...
The pole at $z=0$ is double, so (in my opinion) it's easier to find the residue via the series expansion than with the explicit formula. Factoring the denominator, the integrand is $\frac{e^{\pi z}}{z^2(z+1-i)(z+1+i)}$. The series for $e^{\pi z}$ is just $1 + \pi z + \frac{\pi^2 z^2}{2!} + ...$ The other factors except for $\frac{1}{z^2}$ are of the form $\frac{1}{a + z}$, which has the series representation $\frac{1}{a} - \frac{z}{a^2}+...$ So the integrand, as a product of series, is $\frac{1}{z^2}(1 + \pi z + ...)(\frac{1}{1-i} - \frac{z}{(1-i)^2} + ...)(\frac{1}{1+i} - \frac{z}{(1+i)^2} + ...)$ Since we're looking for the residue, the only term we care about is $z^{-1}$. Since $\frac{1}{z^2}$ is the only factor with negative exponents, we only need the first two terms of each series. Finding the terms that would have order $-1$, we find the coefficient to be $\frac{\pi}{(1+i)(1-i)} - \frac{1}{(1-i)^2(1+i)} - \frac{1}{(1+i)^2(1-i)}$. The first term is $\frac{\pi}{2}$, and the second and third terms combine to $-\frac{1}{2(1+i)} - \frac{1}{2(1-i)} = -\frac{1-i}{4} - \frac{1+i}{4} = -\frac{1}{2}$. So the first residue is (unless I've also made a horrible mistake) $\frac{\pi - 1}{2}$, as you found.
Now for the next two residues: since these are both simple poles, they can be evaluated easily with the explicit formula. The second residue should be $\frac{e^{\pi z}}{z^2(z+1-i)}$ evaluated at $z = -1 - i$, and the third is similar. Plugging in, the second residue is $\frac{e^{\pi (-1 - i)}}{(-1-i)^2(-1-i+1-i)} = \frac{e^{-\pi}e^{-i\pi}}{(2i)(-2i)}$ which is $-\frac{e^{-\pi}}{4}$. The third is similar, and I also found it also matches what you have.
Maybe we both made the same mistakes, but your evaluation looks right to me.
I do not see anything wrong with your calculation.
But Maple gets $$\frac{\pi}{2} - \frac{1}{2} - \frac{1}{4e^{\pi}}$$ so the last term disagrees with yours. But numerically in Maple, the same integral agrees with your answer, not Maple's symbolic answer. Some sort of bug in Maple, I guess. (Maple 2015 build 1097895)
• This just confuses me even more.... – SirBanana Aug 23 '17 at 23:41 | 2021-05-17T21:44:50 | {
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https://math.stackexchange.com/questions/3959113/what-is-the-difference-between-expectation-variance-for-statistics-versus-p/3959120 | # What is the difference between "expectation", "variance" for statistics versus probability textbooks?
It seems that there are two ideas of expectation, variance, etc. going on in our world.
In any probability textbook:
I have a random variable $$X$$, which is a function from the sample space to the real line. Ok, now I define the expectation operator, which is a function that maps this random variable to a real number, and this function looks like, $$\mathbb{E}[X] = \sum\limits_{i = 1}^n x_i p(x_i)$$ where $$p$$ is the probability mass function, $$p: x_i \mapsto [0,1], \sum_{i = 1}^n p(x_i) = 1$$ and $$x_i \in \text{range}(X)$$. The variance is, $$\mathbb{E}[(X - \mathbb{E}[X])^2]$$
The definition is similar for a continuous RV.
However, in statistics, data science, finance, bioinformatics (and I guess everyday language when talking to your mother)
I have a multi-set of data $$D = \{x_i\}_{i = 1}^n$$ (weight of onions, height of school children). The mean of this dataset is
$$\dfrac{1}{n}\sum\limits_{i= 1}^n x_i$$
The variance of this dataset (according to "science buddy" and "mathisfun dot com" and government of Canada) is,
$$\dfrac{1}{n}\sum\limits_{i= 1}^n(x_i - \sum\limits_{j= 1}^n \dfrac{1}{n} x_j)^2$$
I mean, I can already see what's going on here (one is assuming uniform distribution), however, I want an authoritative explanation on the following:
1. Is the distinction real? Meaning, is there a universe where expectation/mean/variance... are defined for functions/random variables and another universe where expectation/mean/variance... are defined for raw data? Or are they essentially the same thing (with hidden/implicit assumption)
2. Why is it no probabilistic assumption is made when talking about mean or variance when it comes to dealing with data in statistics or data science (or other areas of real life)?
3. Is there some consistent language for distinguishing these two seemingly different mean and variance terminologies? For example, if my cashier asks me about the "mean weight" of two items, do I ask him/her for the probabilistic distribution of the random variable whose realization are the weights of these two items (def 1), or do I just add up the value and divide (def 2)? How do I know which mean the person is talking about?/
• There's no probabilistic assumption about the sample mean or sample variance because they're just numbers computed from the sample data. If you measure the heights of $100$ people, you have $100$ numbers and you can do whatever computations you like with them. Dec 23, 2020 at 5:07
• Can you explain what you what you mean that 'uniform distribution' was assumed? Isnt p(x_i)=lim n->infty #occurences of x_i/n ? So an approximation of p(x_i) is #occurences of x_i /n. So 1/n sum x_i = 1/n sum_{x_i} x_i #occurences of x_i = sum x_i p(x_i). Of course p(x_i) is only an approximation of p(x_i) in statistics. I do not see how uniformity was used here Dec 23, 2020 at 16:25
• "Why is it no probabilistic assumption is made when talking about mean or variance when it comes to dealing with data in statistics or data science " Can you be more clear as to what assumption you think is made elsewhere, but isn't made in data science? Dec 24, 2020 at 20:46
• @lalala If you take the average off $n$ values, that is equal to the expected value of a random variable that has a uniform distribution across those alues. Dec 24, 2020 at 20:47
You ask a very insightful question that I wish were emphasized more often.
EDIT: It appears you are seeking reputable sources to justify the above. Sources and relevant quotes have been provided.
Here's how I would explain this:
• In probability, the emphasis is on population models. You have assumptions that are built-in for random variables, and can do things like saying that "in this population following such distribution, the probability of this value is given by the probability mass function."
• In statistics, the emphasis is on sampling models. With most real-world data, you do not have access to the data-generating process governed by the population model. Probability provides tools to make guesses on what the data-generating process might be. But there is always some uncertainty behind it. We therefore attempt to estimate characteristics about the population given data.
From Wackerly et al.'s Mathematical Statistics with Applications, 7th edition, chapter 1.6:
The objective of statistics is to make an inference about a population based on information contained in a sample taken from that population...
A necessary prelude to making inferences about a population is the ability to describe a set of numbers...
The mechanism for making inferences is provided by the theory of probability. The probabilist reasons from a known population to the outcome of a single experiment, the sample. In contrast, the statistician utilizes the theory of probability to calculate the probability of an observed sample and to infer this from the characteristics of an unknown population. Thus, probability is the foundation of the theory of statistics.
From Shao's Mathematical Statistics, 2nd edition, section 2.1.1:
In statistical inference... the data set is viewed as a realization or observation of a random element defined on a probability space $$(\Omega, \mathcal{F}, P)$$ related to the random experiment. The probability measure $$P$$ is called the population. The data set or random element that produces the data is called a sample from $$P$$... In a statistical problem, the population $$P$$ is at least partially unknown and we would like to deduce some properties of $$P$$ based on the available sample.
So, the probability formulas of the mean and variance assume you have sufficient information about the population to calculate them.
The statistics formulas for the mean and variance are attempts to estimate the population mean and variance, given a sample of data. You could estimate the mean and variance in any number of ways, but the formulas you've provided are some standard ways of estimating the population mean and variance.
Now, one logical question is: why do we choose those formulas to estimate the population mean and variance?
For the mean formula you have there, one can observe that if you assume that your $$n$$ observations can be represented as observed values of independent and identically distributed random variables $$X_1, \dots, X_n$$ with mean $$\mu$$, $$\mathbb{E}\left[\dfrac{1}{n}\sum_{i=1}^{n}X_i \right] = \mu$$ which is the population mean. We say then that $$\dfrac{1}{n}\sum_{i=1}^{n}X_i$$ is an "unbiased estimator" of the population mean.
From Wackerly et al.'s Mathematical Statistics with Applications, 7th edition, chapter 7.1:
For example, suppose we want to estimate a population mean $$\mu$$. If we obtain a random sample of $$n$$ observations $$y_1, y_2, \dots, y_n$$, it seems reasonable to estimate $$\mu$$ with the sample mean $$\bar{y} = \dfrac{1}{n}\sum_{i=1}^{n}y_i$$
The goodness of this estimate depends on the behavior of the random variables $$Y_1, Y_2, \dots, Y_n$$ and the effect this has on $$\bar{Y} = (1/n)\sum_{i=1}^{n}Y_i$$.
Note. In statistics, it is customary to use lowercase $$x_i$$ to represent observed values of random variables; we then call $$\frac{1}{n}\sum_{i=1}^{n}x_i$$ an "estimate" of the population mean (notice the difference between "estimator" and "estimate").
For the variance estimator, it is customary to use $$n-1$$ in the denominator, because if we assume the random variables have finite variance $$\sigma^2$$, it can be shown that $$\mathbb{E}\left[\dfrac{1}{n-1}\sum_{i=1}^{n}\left(X_i - \dfrac{1}{n}\sum_{j=1}^{n}X_j \right)^2 \right] = \sigma^2\text{.}$$ Thus $$\dfrac{1}{n-1}\sum_{i=1}^{n}\left(X_i - \dfrac{1}{n}\sum_{j=1}^{n}X_j \right)^2$$ is an unbiased estimator of $$\sigma^2$$, the population variance.
It is also worth noting that the formula you have there has expected value $$\dfrac{n-1}{n}\sigma^2$$ and $$\dfrac{n-1}{n} < 1$$ so on average, it will tend to underestimate the population variance.
From Wackerly et al.'s Mathematical Statistics with Applications, 7th edition, chapter 7.2:
For example, suppose that we wish to make an inference about the population variance $$\sigma^2$$ based on a random sample $$Y_1, Y_2, \dots, Y_n$$ from a normal population... a good estimator of $$\sigma^2$$ is the sample variance $$S^2 = \dfrac{1}{n-1}\sum_{i=1}^{n}(Y_i - \bar{Y})^2\text{.}$$
The estimators for the mean and variance above are examples of point estimators. From Casella and Berger's Statistical Inference, Chapter 7.1:
The rationale behind point estimation is quite simple. When sampling is from a population described by a pdf or pmf $$f(x \mid \theta)$$, knowledge of $$\theta$$ yields knowledge of the entire population. Hence, it is natural to seek a method of finding a good estimator of the point $$\theta$$, that is, a good point estimator. It is also the case that the parameter $$\theta$$ has a meaningful physical interpretation (as in the case of a population) so there is direct interest in obtaining a good point estimate of $$\theta$$. It may also be the case that some function of $$\theta$$, say $$\tau(\theta)$$ is of interest.
There is, of course, a lot more that I'm ignoring for now (and one could write an entire textbook, honestly, on this topic), but I hope this clarifies things.
Note. I know that many textbooks use the terms "sample mean" and "sample variance" to describe the estimators above. While "sample mean" tends to be very standard terminology, I disagree with the use of "sample variance" to describe an estimator of the variance; some use $$n - 1$$ in the denominator, and some use $$n$$ in the denominator. Also, as I mentioned above, there are a multitude of ways that one could estimate the mean and variance; I personally think the use of the word "sample" used to describe such estimators makes it seem like other estimators don't exist, and is thus misleading in that way.
## In Common Parlance
This answer is informed primarily by my practical experience in statistics and data analytics, having worked in the fields for about 6 years as a professional. (As an aside, I find one serious deficiency with statistics and data analysis books is providing mathematical theory and how to approach problems in practice.)
Is there some consistent language for distinguishing these two seemingly different mean and variance terminologies? For example, if my cashier asks me about the "mean weight" of two items, do I ask him/her for the probabilistic distribution of the random variable whose realization are the weights of these two items (def 1), or do I just add up the value and divide (def 2)? How do I know which mean the person is talking about?
In most cases, you want to just stick with the statistical definitions. Most people do not think of statistics as attempting to estimate quantities relevant to a population, and thus are not thinking "I am trying to estimate a population quantity using an estimate driven by data." In such situations, people are just looking for summaries of the data they've provided you, known as descriptive statistics.
The whole idea of estimating quantities relevant to a population using a sample is known as inferential statistics. While (from my perspective) most of statistics tends to focus on statistical inference, in practice, most people - especially if they've not had substantial statistical training - do not approach statistics with this mindset. Most people whom I've worked with think "statistics" is just descriptive statistics.
In descriptive data analysis, a few summary measures may be calculated, for example, the sample mean... and the sample variance... However, what is the relationship between $$\bar{x}$$ and $$\theta$$ [a population quantity]? Are they close (if not equal) in some sense? The sample variance $$s^2$$ is clearly an average of squared deviations of $$x_i$$'s from their mean. But, what kind of information does $$s^2$$ provide?... These questions cannot be answered in descriptive data analysis.
## Other remarks about the sample mean and sample variance formulas
Let $$\bar{X}_n$$ and $$S^2_n$$ denote the sample mean and sample variance formulas provided earlier. The following are properties of these estimators:
• They are unbiased for $$\mu$$ and $$\sigma^2$$, as explained earlier. This is a relatively simple probability exercise.
• They are consistent for $$\mu$$ and $$\sigma^2$$. Since you know measure theory, assume all random variables are defined over a probability space $$(\Omega, \mathcal{F}, P)$$. It follows that $$\bar{X}_n \overset{P}{\to} \mu$$ and and $$S^2_n \overset{P}{\to} \sigma^2$$, where $$\overset{P}{\to}$$ denotes convergence in probability, also known as convergence with respect to the measure $$P$$. See https://math.stackexchange.com/a/1655827/81560 for the sample variance (observe that the estimator with the $$n$$ in the denominator is used here; simply multiply by $$\dfrac{n-1}{n}$$ and apply a result by Slutsky) and Proving a sample mean converges in probability to the true mean for the sample mean. As a stronger result, convergence is almost sure with respect to $$P$$ in both cases (Sample variance converge almost surely).
• If one assumes $$X_1, \dots, X_n$$ are independent and identically distributed based on a normal distribution with mean $$\mu$$ and variance $$\sigma^2$$, one has that $$\dfrac{\sqrt{n}(\bar{X}_n - \mu)}{\sqrt{S_n^2}}$$ follows a $$t$$-distribution with $$n-1$$ degrees of freedom, which converges in distribution to a normally-distributed random variable with mean $$0$$ and variance $$1$$. This is a modification of the central limit theorem.
• If one assumes $$X_1, \dots, X_n$$ are independent and identically distributed based on a normal distribution with mean $$\mu$$ and variance $$\sigma^2$$, $$\bar{X}_n$$ and $$S^2_n$$ are uniformly minimum-variance unbiased estimators (UMVUEs) for $$\mu$$ and $$\sigma^2$$ respectively. It also follows that $$\bar{X}_n$$ and $$S^2_n$$ are independent, through - as mentioned by Michael Hardy - showing that $$\text{Cov}(\bar{X}_n, X_i - \bar{X}_n) = 0$$ for each $$i = 1, \dots, n$$, or as one can learn from more advanced statistical inference courses, an application of Basu's Theorem (see, e.g., Casella and Berger's Statistical Inference).
• I believe the expressions for mean and variance estimators should be accompanied with a word or two on the central limit theorem. Dec 23, 2020 at 5:52
• You wrote: "If one assumes $X_1,\ldots,X_n$ are normally distributed, $\bar X_n$ and $S_n^2$ are" UMVUEs. (This after assuming they are i.i.d.) I would add that under those assumptions $\bar X_n$ and $S_n^2$ are independent, and that is used in deriving the t-distribution. One of the quickest ways to show that involves showing that $\operatorname{cov} \left(\,\overline X, X_i-\overline X\,\right)=0.$ When $U,V$ are both linear combinations of i.i.d. normals, then their covariance is $0$ only if they are independent. Dec 26, 2020 at 21:47
• @MichaelHardy Thanks for correcting me; I've put in the edits. Dec 26, 2020 at 22:08
• @Clarinetist : I see you say it's "through an application of Basu's theorem," but nothing so advanced as Basu's theorem is needed, since the method described in my comment above is enough. It seems misleading to tacitly suggest that you can't learn to prove that proposition until you learn Basu's theorem. Dec 26, 2020 at 22:11
• @MichaelHardy Thanks, I've corrected that as well. Dec 26, 2020 at 22:13
The first definitions you gave are correct and standard, and statisticians and data scientists will agree with this. (These definitions are given in statistics textbooks.) The second set of quantities you described are called the "sample mean" and the "sample variance", not mean and variance.
Given a random sample from a random variable $$X$$, the sample mean and sample variance are natural ways to estimate the expected value and variance of $$X$$.
• Thanks! So the distinction is one is "sample" mean/variance the other is mean/variance as it is. And also those formulas are estimate of the value of mean and variance. But isn't it true when people are working with them in practice, say using some data science package such as Scikit-Learn, the underlying probabilistic assumption (such as distribution of data) is dropped as if nothing ever happened? I have never seen anyone writing a documentation mentioning the probabilistic definition. I think this causes confusion for young practictioners, don't you think? Dec 23, 2020 at 5:39
• Imagine in the shoes of a young data scientist who just took a course on Lebesgue integration or measure theory. You spent 4 month talking about these abstract definitions, and then integrating these really hard integrals just to get the mean, using various tricks just to make these integrals tractable. You get your first job and your boss tells you just to add stuff up and divide. Isn't this confusing? Dec 23, 2020 at 5:41
• @Norman My response to your comments here is that probability is not a statistics class, and trying to act as if that is the case would be lying. Additionally, most managers have not taken courses nor understand the difference between probability and statistics, and are focused solely on descriptive statistics to understand empirically the behavior of their data. They calculate a sample average because that's what they habitually have done, and they think it is an adequate summary of the data. Dec 23, 2020 at 13:25
• @Norman I could go on a very long rant about how poorly taught statistics is in many schools and how very little guidance is given on how to approach statistical problems in practice, but I will avoid that for now. Dec 23, 2020 at 13:27
• @Norman I think it can be a source of confusion that people often say mean or variance when they really mean "sample mean" or "sample variance". One of the challenges of coming into the engineering world from a pure math background is that often people speak much less precisely than one is used to. Dec 23, 2020 at 15:23
Other answers — particularly Clarinetist’s — give excellent rundowns of the most important side of the answer. Given a random variable, we can sample it, and use the sample mean (defined in the statistical sense) to estimate the actual mean of the random variable (defined in the sense of probability theory), and similarly for variance, etc.
But the connection in the other direction doesn’t seem to have been mentioned yet. This is not as important, but it’s much more straightforward, and worth pointing out. Given a sample, i.e. a finite multiset of values $$\{x_i\}_{i \in I}$$, we can “consider this as a distribution”, i.e. take a random variable $$X$$, with value $$x_i$$ for $$i$$ distributed uniformly over $$I$$. Then the mean, variance, etc. of $$X$$ (in the sense of probability theory) will be precisely the mean, variance, etc. of the original multiset (defined in the statistical sense).
The general expression for arithmetic mean is $$\frac{\sum\limits_{i= 1}^n w_i x_i}{\sum\limits_{i= 1}^n w_i}$$, or even more generally, $$\frac{\int w(t) f(t)dt}{\int f(t)dt}$$ (there are then ways to recover the discrete case from that).
If you set all $$w_i$$ to $$1$$, or really to anything as long as it's constant, you get $$\dfrac{1}{n}\sum\limits_{i= 1}^n x_i$$. This is referred to as an "unweighted" average, although technically it's still weighted, it's just that you're multiplying everything by $$1$$, so you don't notice it. If you set $$p_i = \frac{w_i}{ \sum\limits_{k= 1}^n w_k}$$, and interpret $$p_i$$ as the probability of the $$i$$th event, then you get the average weighted by probability, which is also known as expected value.
You have to be careful about "unweighted" averages, as they often actually are weighted, but by weights that you didn't want. For instance, suppose you want the average income over the US, and you have the average income for each state individually. You could just add all of those averages together, and then divide by $$50$$. People will often call this the "unweighted" or "simple" average, but you're actually weighting people by the reciprocal of their states population; the fewer people there are in a state, the more each person's income affects that state's average, and so the more they affect the total "unweighted average". As a result, to get the actual overall average income from the individual states' averages, you have to multiply each state's average by its population to get its total income, add all of those together, and then divide by the total population.
A common weighting you'll see is frequency weighting. This is where you multiply each value by the number of times it appears. For instance, if you measure something once a month for a year, and the only values you get are $$0$$, $$1$$, and $$2$$, taking the simple average of those values gives you $$1$$. But that's probably not the real average. To get a more meaningful average, you should take each of these values, weight them by how many months they appear for, and then take the average.
One property of weighted averages is that multiplying all of the weights by a constant number doesn't change the final result (you'll just divide it out again when you divide by the total of the weights). So weighting by the frequencies is equivalent to weighting by the percentage of cases each value is. That is, if $$0$$ is the value for $$5$$ months, $$1$$ is the value for $$4$$ months, and $$2$$ is the value for $$3$$, the weighting of $$5,4,3$$ is equivalent to the weighting of $$\frac 5 {12},\frac 4{12},\frac 3{12}$$.
So if you have a probability distribution where one thing has a $$60$$% chance of happening, and something else has a $$40$$% chance of happening, the expected value is simply the frequency weighted average.
For example, if my cashier asks me about the "mean weight" of two items, do I ask him/her for the probabilistic distribution of the random variable whose realization are the weights of these two items (def 1), or do I just add up the value and divide (def 2)? How do I know which mean the person is talking about?
The expected value of a random variable is the expected value of that random variable. It is a property of the distribution. If you're asked for the expected value of a random variable, you find the expected value of that random variable. If you aren't asked for the expected value of a random variable, you don't go looking for a random variable to take the expected value of. How you know whether you take the expected value of random variable is if there is a random variable to take the expected value of. Even if values come from a distribution, the average of those values is the average of those values, not the average of the distribution they came from.
If people are talking rigorously, they will explicitly say they want the expected value. You many, however, see people asking for the "mean" or "average", when they really want the expected value, but you can recognize those cases by there being a random variable. For instance, if someone asks "What's the average payout of this slot machine?", the context suggests that you should take the expected value of the distribution of payouts, and not simply take the set of different possible payouts and take the simple average. There could be some ambiguity as to whether "the payout" refers to the random process that pays money out, in which case you should take the expected value of the distribution (population mean), or the actual money paid out, in which case you should take the average of all the actual payouts made by the machine (sample mean), but in the latter case, you still should take the frequency weighted average.
The confusion actually comes from notation, where symbols mean different things in two formulas.
First, let's take a look at the "probabilistic" definition: $$\mathbb{E}[X] = \frac{1}{n}\sum_{i=1}^n x_i p(x_i)$$ Here the random variable $$X$$ takes $$n$$ distinct values $$x_1,\ldots, x_n$$, each with a probability mass function $$p(x_i)$$.
In the "statistical" definition we have an estimate of the expected value, based on the observed values of the random variable $$z_1, \ldots, z_N$$: $$\hat{\mathbb{E}}[X] = \frac{1}{N}\sum_{j=1}^N z_j$$ Notice that I renamed the variables compared to your original formula, so as to avoid the confusion. Here $$N$$ is the number of observations, and $$z_j$$s are the actual observations. For example, if $$X$$ is a random variable representing rolls of a loaded die, then $$n = 6$$ and $${x_i} = \{1, 2, 3, 4, 5, 6\}$$; whereas $$N$$ can be arbitrarily large, and $$z_j$$s will just be a long sequence of random draws from the set of $${x_i}$$s.
Now, you can rewrite the "statistical" formula by collecting different values of $$z_j$$s into the groups according to which $$x_i$$ they correspond to (for example, first collect all 1s, then all 2s, etc): $$\hat{\mathbb{E}}[X] = \frac{1}{N}\big(x_1\cdot|\{z_j=x_1\}| + \cdots+x_n\cdot|\{z_j=x_n\}|\big) \\ =\frac{1}{N}\sum_{i=1}^n x_i \sum_{j=1}^N\mathbb{1}[z_j=x_i] \\ =\sum_{i=1}^n x_i \hat{p}(x_i)$$ where $$\hat{p}(x_i) = \frac{1}{N}\sum_{j=1}^N \mathbb{1}[z_j=x_i]$$ is the estimate of the probability mass function: the number of times a value $$x_i$$ was encountered in the sample divided by the sample size, i.e. the observed frequency of the value $$x_i$$.
Now you can see that the "probabilistic" and "statistical" definitions are actually the same, with the only difference that we replace the theoretical mass distribution function $$p(x)$$ (which may not be known) with the empirical (observed) mass distribution function $$\hat{p}(x)$$.
Statement: the "statistic" universe from your question is a partial case of "probabilistic" universe.
We need some notation. Suppose that we have a random variable $$\xi$$ in sense of probability theory and $$P(\xi = x_k) = p_k$$, $$0 \le k \le n$$.
For example, consider $$\xi$$ which is equal to the number of children in "abstract" random family. Put $$x_k = k$$. Then $$E \xi = \sum_{k=0}^n x_k p_k$$ where $$p_k = P(\xi = x_k)$$, and $$D\xi = E (\xi - E\xi)^2 = \sum_{k=0}^n p_k (x_k - E\xi )^2$$. Moreover, $$P(\xi = 69) > 0$$ (according to Guinness World Records, I am not sure, but let us suppose that it's a record nowadays) and as there was people with $$69$$ children then we may think that for example $$P(80) > 0$$ - it's natural, because $$80$$ children is possible, even although Guinness World Records says that still nobody had $$80$$ children.
In reality we don't have abstract random family, random mothers, random fathers and random children. We have finite number $$N$$ ( $$10^6 < N < 10^{100}$$) of families, let us number them, and there are $$y_1$$ children in the first family, there are $$y_2$$ children in family number $$2$$, ..., there are $$y_N$$ children in family $$N$$.
Analogy: $$\xi$$ corresponds to a fair dice itself and numbers $$y_1$$, $$y_2$$, ... corresponds to dice rolls and have form: $$5, 1, 6, 6, 3, ...$$, these are fixed numbers.
Now consider a random value $$\tau$$, which has uniform distribution on $$\{ 1, 2, \ldots, N\}$$. It means that $$P(\tau = k) = \frac{1}{N}$$ for all $$1 \le k \le N$$.
Numbers $$y_1, \ldots, y_N$$ are fixed - suppose they are written in some sociological table. Let us consider a random value $$y_{\tau}$$. It's not a number of children in abstract random family. It is a number of children in some real family, if a number $$\tau$$ of such family we chose randomly.
Let us find $$E y_{\tau}$$ and $$D y_{\tau}$$. With probability $$\frac{1}N$$ we have $$\tau = k$$ and hence $$y_{\tau} = y_k$$. Thus $$E y_{\tau}=\frac{1}N \sum_{k=1}^N y_k$$ and $$D y_{\tau} = E (y_{\tau} - Ey_{\tau})^2 = E (y_{\tau} - \frac{1}N \sum_{k=1}^N y_k )^2 = \frac{1}N \sum_{k=0}^N (y_k -\frac{1}N \sum_{k=1}^N y_k )^2.$$
Now the correspondence of "statistical universe" and "probabilistic universe" is obvious.
Notice that all $$y_i \le 69$$ and $$y_i = 69$$ is the maximum value, but the maximum value of $$\xi$$ is bigger (and not less than $$80$$).
So, we have shown that the "statistical universe" is a partial case of "probabilistic" universe.
Moreover, there are two moments, when there is randomness:
1. when we go from abstract random family, corresponding to $$\xi$$, to numbers $$y_1, \ldots, y_N$$ from sociological handbook.
As it was already mentioned, there is analogy: $$\xi$$ corresponds to a fair dice itself and numbers $$y_1$$, $$y_2$$, ... corresponds to dice rolls and have form: $$5, 1, 6, 6, 3, ...$$, these are fixed numbers.
1. [here we suppose that $$y_1, \ldots, y_N$$ are fixed numbers] when we go from the whole sociological handbook to a family with random number $$\tau$$ and see, how much children is there in this family. The number is $$y_{\tau}$$.
When from $$y_{\tau}$$ we go to $$E y_{\tau} = \frac{1}N \sum_{k=1}^N y_k$$, we get rid of the second randomness, but we still do not get rid of the first randomness. It's connected with the fact that numbers $$y_k$$ could be another: it could happen, that in some families could be more children, and in some families could be less children, if the circumstances were different. In this sence numbers $$y_i$$ are not fixed: they are r.v. sampled from distribution, corresponding to $$\xi$$. And in this sence $$\frac{1}N \sum_{k=1} y_k$$ is a random value, and we may write limit theorems such as L.L.N.: $$\frac{1}N \sum_{k=1}^N y_k \to E\xi, \text{ }N \to \infty$$ or CLT or LIL.
Addition: it was shown that if $$y_1, \ldots, y_N$$ are fixed and $$\tau$$ is random then "probabilistic" mean (expectation) of $$y_{\tau}$$ and "probabilistic" variance of $$y_{\tau}$$ are well known sample mean and sample variance.
Hope it will be useful. If you have any questions, you are welcome. | 2022-05-26T02:30:11 | {
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https://math.stackexchange.com/questions/3016686/riemann-darboux-integrability-of-subinterval | # Riemann-Darboux Integrability of Subinterval
I'm studying Riemann-Darboux integration. I'm trying to prove the following rather intuitive notion for integrals. Please let me know if you find any errors in this proof, as I'm self-studying this topic.
Theorem: Suppose $$f$$ is Riemann-Darboux integrable on $$[a,b]$$. Let $$c\in(a,b)$$. Then, $$f$$ is Riemann-Darboux integrable on the intervals $$[a,c]$$ and $$[c,b]$$.
Attempted Proof: Since $$f$$ is Riemann-Darboux integrable on $$[a,b]$$, for arbitrary $$\epsilon>0$$, there exists a partition $$P$$ of $$[a,b]$$ such that $$U(f,P)-L(f,P)<\epsilon$$.
Let $$n_p,n_{p*}$$, and $$n_{p'}$$ be the number of partition parts in $$P$$, $$P^*$$, and $$P'$$, respectively. Also, let $$m_i=\inf_{x\in[x_{i-1},x_i]}f(x)$$ and $$M_i=\sup_{x\in[x_{i-1},x_i]}f(x)$$.
Consider the partition of $$[a,c]$$ given by $$P^*=P\cap[a,c]$$. Then, $$U(f,P^*)-L(f,P^*)=\sum_{i=1}^{n_{p*}}(M_i-m_i)\Delta x_i\le\sum_{i=1}^{n_{p*}}(M_i-m_i)\Delta x_i+ \sum_{n_{p*}+1}^{n_p}(M_i-m_i)\Delta x_i=\sum_{i=1}^{n_p}(M_i-m_i)\Delta x_i=U(f,P)-L(f,P)<\epsilon$$ Therefore, $$f$$ is integrable on $$[a,c]$$.
Next, consider the partition $$[a,b]$$ given by $$P'=P\cap[c,b]$$. Then,
$$U(f,P')-L(f,P')=\sum_{i=n_{p*}+1}^{n_{p'}}(M_i-m_i)\Delta x_i\le\sum_{i=1}^{n_{p*}}(M_i-m_i)\Delta x_i+ \sum_{n_{p*}+1}^{n_p}(M_i-m_i)\Delta x_i=\sum_{i=1}^{n_p}(M_i-m_i)\Delta x_i=U(f,P)-L(f,P)<\epsilon$$ Therefore, $$f$$ is integrable on $$[c,b]$$. $$\square$$
Any and all feedback, or alternative proofs are appreciated. I love to see different arguments to expand my skill set.
This is essentially fine, although you are tacitly assuming that $$c$$ itself is a member of the partition, $$P$$. This isn't a big deal though - a $$P$$ such that $$U(P,f)-L(P,f)<\epsilon$$ is guaranteed by integrability, and you can always just add $$c$$ to this partition if it is not already there. Namely, if $$P_{c}$$ is this new partition, called a refinement of $$P$$, one must have $$U(P_{c},f)-L(P_{c},f)\leq U(P,f)-L(P,f)<\epsilon$$ and you can proceed as you have done in your proof.
An alternative approach is given by Lebesgue's Criterion for Riemann Integrability, which states that Riemann/Darboux Integrability is equivalent to boundedness plus continuity up to a null set (see Wikipedia for a good explanation of null sets). Since your functions is integrable on $$[a,b]$$, it is bounded and continuous up to a null set on $$[a,b]$$, and hence, on $$[a,c]$$ and $$[c,b]$$ as well. | 2020-09-19T15:41:52 | {
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http://mathhelpforum.com/statistics/154536-pricing-dice-game.html | # Math Help - Pricing Dice Game
1. ## Pricing Dice Game
Stage 1
In this game I roll the dice and which ever number the dice lands on I pay you that amount of money. For example if the dice lands on 4 I pay you £4 pounds.
A fair price to play this game should be £3.50 as then the player or myself should not win any money if this game is played infinty times.
Stage 2
How much should the fair price be to play the game if their is an option after the first roll to roll again. For example if you roll a 1 first go you will certainly want to roll again as youwill always roll the same or better. However if you were to roll a 6 you would not want the option to roll again as you have already won the maximum amount of money.
I priced this option at £3.57, however I been told that is wrong. Can anyone explain why?
Calypso
2. Originally Posted by calypso
Stage 1
In this game I roll the dice and which ever number the dice lands on I pay you that amount of money. For example if the dice lands on 4 I pay you £4 pounds.
A fair price to play this game should be £3.50 as then the player or myself should not win any money if this game is played infinty times.
Stage 2
How much should the fair price be to play the game if their is an option after the first roll to roll again. For example if you roll a 1 first go you will certainly want to roll again as youwill always roll the same or better. However if you were to roll a 6 you would not want the option to roll again as you have already won the maximum amount of money.
I priced this option at £3.57, however I been told that is wrong. Can anyone explain why?
Calypso
Choose a $t$ such that if the first roll $a\le t$ roll again otherwise accept the prize of $£ a$.
Now for any given $t$ you can work out the value of this game and hence a fair price. The fair price for the game is the maximum fair price.
(I make $t=3$ the value of $t$ that maximises the return and a fair price is then $£4.25$
CB
3. Sorry for the late reply I have just got back from holiday.
The text book answer says something similar, ie you should stick if you roll anything greater than 3.50. Therefore the fair price of the option of two rolls is
( 3.5 + 3.5 + 3.5 + 4 + 5 + 6 ) / 6 = 4.25
I can understand why this is the correct answer, however I am still confuced as why I cant get this answer with my method:
So I think the fair price of the game is defined by: Sum of all possible payouts / No. of possible payout
So therefore the possible outcomes of the game are
1 -> 1
1 -> 2
1 -> 3
1 -> 4
1 -> 5
1 -> 6
2 -> 1
2 -> 2
2 -> 3
2 -> 4
2 -> 5
2 -> 6
3 -> 1
3 -> 2
3 -> 3
3 -> 4
3 -> 5
3 -> 6
4
5
6
Therefore fair price should equal 78/21 = £3.71?
Thanks again
Calypso
4. Originally Posted by calypso
I am still confuced as why I cant get this answer with my method:
So I think the fair price of the game is defined by: Sum of all possible payouts / No. of possible payout
So therefore the possible outcomes of the game are
1 -> 1
1 -> 2
1 -> 3
1 -> 4
1 -> 5
1 -> 6
2 -> 1
2 -> 2
2 -> 3
2 -> 4
2 -> 5
2 -> 6
3 -> 1
3 -> 2
3 -> 3
3 -> 4
3 -> 5
3 -> 6
4
5
6
The reason this goes wrong is that those outcomes are not all equally probable. Each of the final three outcomes (4, 5, 6) occurs with a probability of 1/6. But the initial outcomes 1, 2, 3 are subdivided into six subcases, each of which occurs with a probability of only 1/36. Multiplying each outcome by its probability, you get $\tfrac3{36}(1+2+3+4+5+6) + \tfrac16(4+5+6) = \tfrac{17}4 = 4.25.$
5. Originally Posted by calypso
Sorry for the late reply I have just got back from holiday.
The text book answer says something similar, ie you should stick if you roll anything greater than 3.50. Therefore the fair price of the option of two rolls is
( 3.5 + 3.5 + 3.5 + 4 + 5 + 6 ) / 6 = 4.25
I can understand why this is the correct answer, however I am still confuced as why I cant get this answer with my method:
I don't like their decision rule, not that it is wrong, but because it refers to something not in the sample space (that is "anything greater than 3.50" when it could have said anything greater than 3 or even anything greater than or equal to 4).
CB
6. Great, thanks everyone for your help
Calypso | 2015-05-30T15:54:38 | {
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http://mathhelpforum.com/statistics/38753-die-probability.html | Math Help - die probability
1. die probability
"What is the probability of getting a total score of more than 7 when two fair dice are rolled together?"
I got $\frac{15}{36}$ but the book says $\frac{1}{3}$?
Thanks!
2. Hello, Nyoxis!
What is the probability of getting a total score of more than 7
when two fair dice are rolled together?"
I got $\frac{15}{36} \quad{\color{blue}\hdots\text{ Right!}}$
. . but the book says $\frac{1}{3}\quad {\color{red} \hdots\text{ Wrong!}}$
Can't imagine where they got their answer . . .
$\begin{array}{cccccc}(1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \\ (2,1) & (2,2) & (2,3) & (2,4) & (2,5) & {\color{blue}(2,6)} \\ (3,1) & (3,2) & (3,3) & (3,4) & {\color{blue}(3,5)} & {\color{blue}(3,6)} \\ (4,1) & (4,2) & (4,3) & {\color{blue}(4,4)} & {\color{blue}(4,5)} & {\color{blue}(4,6)}\end{array}$
$\begin{array}{cccccc}(5,1) & (5,2) & {\color{blue}(5,3)} & {\color{blue}(5,4)} & {\color{blue}(5,5)} & {\color{blue}(5,6)} \\
(6,1) & {\color{blue}(6,2)} & {\color{blue}(6,3)} & {\color{blue}(6,4)} & {\color{blue}(6,5)} & {\color{blue}(6,6)}\end{array}\quad\Leftarrow\quad \text{fifteen outcomes}> 7$
3. Heh 'Cambridge Advanced Mathematics' wrong.
Thanks! | 2014-12-27T23:15:28 | {
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https://math.stackexchange.com/questions/1509476/prove-that-x-n1-fracx-n2-frac1x-n-geq-sqrt2 | # Prove that $x_{n+1}=\frac{x_n}{2}+\frac{1}{x_n}\geq\sqrt{2}$
Let $(x_n)_{n\in\mathbb N}$ be a recursively defined sequence with $x_1=9$ and $$x_{n+1}=\frac{x_n}{2}+\frac{1}{x_n}\text{ for }n\geq 1.$$ Show that $x_n\geq\sqrt{2}$ for all $n$.
Because $x_n\geq 0$ one can easily prove inductively that $$x_n^2\geq 2\Leftrightarrow x_n^2-2\geq 0\Leftrightarrow\left(\frac{x_{n-1}}{2}-\frac{1}{x_{n-1}}\right)^2\geq 0,$$ hence $x_n\geq\sqrt{2}$. However I have seen another approch which I was very curious about because I don't have the feeling that this can be done without proper justification other than the induction hypothesis:
$$x_{n+1}=\frac{x_n}{2}+\frac{1}{x_n}\overset{(*)}{\geq}\frac{\sqrt{2}}{2}+\frac{1}{\sqrt{2}}=\sqrt{2}$$
For $(*)$ it is assumed that $x_n\geq\sqrt{2}$ holds for an $n\in\mathbb N$. Are there any objections about this consideration?
• Hint: AM-GM inequality is fast. Your second approach is flawed, as the reciprocal term does not follow in the inequality * – Macavity Nov 2 '15 at 14:47
• The inequality $\frac{1}{x_n} \geq \frac{1}{\sqrt{2}}$ is false because it would imply that $x_n \leq \sqrt{2}$ (this is true only if $x_n=\sqrt{2}$ but else.... ) – user252450 Nov 2 '15 at 14:50
• A straightforward approach to showing that $x_n\ge \sqrt{2}$ is to write $$x_{n}-\sqrt{2}=\left(\frac{x_{n-1}}{2}+\frac1{x_{n-1}}\right)-\sqrt{2}=\frac{(x_{n-1}-\sqrt{2})^2}{2x_{n-1}}\ge0$$ – Mark Viola Nov 2 '15 at 15:00
• @Dr.MV By far the simplest. – Did Nov 2 '15 at 15:01
• @Dr.MV Yeah pretty similiar to my explained approach in the first part of my post. I was just wondering about the other approach and its correctness there. – Christian Ivicevic Nov 2 '15 at 15:08
Yes, you can make the assumption $x_n\ge \sqrt{2}$ to prove $x_{n+1}\ge \sqrt{2}$ as per strong form of mathematical induction.
But your reasoning is wrong as $x_n\ge \sqrt{2} \Rightarrow \frac{1}{x_n} \le \frac{1}{\sqrt{2}}$.
Better is to use AM-GM inequality as Macavity has mentioned since the quantities are all positive.
$$\frac{\frac{x_n}{2}+\frac{1}{x_n}}{2}\ge \sqrt{\frac{x_n}{2}\cdot\frac{1}{x_n}}$$ $$x_{n+1}\ge \sqrt{2}$$
If you know derivatives you can make it work as follows. Consider the function : $$f(x)=\frac{x^2+2}{2x}$$ Now calculate the derivative : $$f'(x)=\frac{2x^2-4}{4x^2}\geq 0$$ if $x\geq \sqrt{2}$ so the function is increasing for $x\geq \sqrt{2}$
Now from the induction hypothesis $x_n\geq \sqrt{2}$ so from the monotony : $$x_{n+1}=f(x_n)\geq(\sqrt{2})=\frac{4}{2\sqrt{2}}=\sqrt{2}$$ as wanted .
Although the following goes a bit beyond addressing the OP's question, I thought that it would be instructive. So, let's have a look at the sequence generated by the recurrence relationship $x_n=\frac12 x_{n-1}+\frac{1}{x_{n-1}}$.
This is actually an interesting recursively generated sequence. If we examine the first difference $x_{n}-x_{n-1}$ we find that
$$x_{n}-x_{n-1}=\frac{2-x_{n-1}^2}{2x_{n-1}} \tag 1$$
We deduce immediately from $(1)$ that if $x_n > \sqrt{2}$, then $x_n$ is decreasing while if $x_n<\sqrt{2}$, then $x_n$ is increasing.
Now, a straightforward approach to showing that $x_n\ge \sqrt{2}$ is to write
$$x_{n}-\sqrt{2}=\left(\frac{x_{n-1}}{2}+\frac1{x_{n-1}}\right)-\sqrt{2}=\frac{(x_{n-1}-\sqrt{2})^2}{2x_{n-1}}\ge0 \tag 2$$
regardless of the value of $x_{n-1}$.
Therefore, the only value of the sequence that can ever be less than $\sqrt{2}$ is its initial "seed" value. From $(1)$, we see that if the seed value is less than $\sqrt{2}$, the subsequent value increases. And $(2)$ shows that it increases above $\sqrt{2}$. Then, the remaining sequence decreases monotonically to the limit value of $\sqrt{2}$.
Hint: study the function $f(x)=x/2 +1/x$ it increases if $x>\sqrt2$ | 2019-09-23T13:37:45 | {
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https://math.stackexchange.com/questions/620513/contour-integration-choice-of-contour | # Contour Integration (Choice of Contour)
Let $\alpha \le 0$ and $\sigma > 0$ .
I want to choose a contour, including $[\sigma - iR, \sigma+iR]$ , such that i can apply Cauchy's Residue theorem and evaluate:
$$\lim_{R \rightarrow \infty} \int\limits_{\sigma - iR}^{ \sigma+iR } \frac{\exp(\alpha z)}{(z^2 + 1)} dz$$
by contour integration. The case for $\alpha > 0$ is nice i think, as you can use a semi-circle, but i cant seem to find a suitable contour for this case.
EDIT: to be specific when we use a semi-circle then the integral over the circular path tends to zero in the limit for $\alpha > 0$ , but this does not seem to happen in the above case.
• How do you expect to apply CRT to a complex integral on a straight line in the complex plane?? We usually do this to calculate real integrals by means of a closed contour in the complex plane... – DonAntonio Dec 28 '13 at 15:36
• see my edit, in the limit the only non-zero section of our contour integral is that integrated over the line. it worked for the earlier case, unless i am completely confusesd. – KJC Dec 28 '13 at 15:38
You can choose a semicircle in the half-plane $\operatorname{Re} z \geqslant \sigma$. Then $e^{\alpha z}$ is bounded on the contour - $\operatorname{Re} (\alpha z) = \alpha\operatorname{Re} z \leqslant \alpha\sigma \leqslant 0$, and $\lvert e^{\alpha z}\rvert = e^{\operatorname{Re} (\alpha z)}$ - and the integral over the semicircle tends to $0$ for $R \to \infty$. Since the contour encloses no singularity, the contour integral is $0$, and hence
$$\int_{\sigma - i\infty}^{\sigma+i\infty} \frac{e^{\alpha z}}{z^2+1}\,dz = 0$$
for $\sigma > 0$ and $\alpha \leqslant 0$.
It is worth noting that
$$I(\alpha) = \int_{\sigma-i\infty}^{\sigma+i\infty} \frac{e^{\alpha z}}{z^2+1}\,dz = \begin{cases}\quad 0 &, \alpha \leqslant 0\\ 2\pi i\sin \alpha &, \alpha \geqslant 0 \end{cases}$$
does not depend smoothly on $\alpha$, although the integrand does. That should, however, not be too surprising since the differentiated integrand $\dfrac{ze^{\alpha z}}{z^2+1}$ is not integrable anymore, and the non-differentiability in $0$ manifests itself by the need to switch the half-plane in which the contour is closed to have the integral over the auxiliary part tend to $0$.
• could you show that $e^{ \alpha z}$ is bounded on this contour? im not seeing that. – KJC Dec 28 '13 at 15:50
• Added the estimate for that. – Daniel Fischer Dec 28 '13 at 15:58
• excellent explanation – KJC Dec 28 '13 at 22:52 | 2019-09-18T13:46:40 | {
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https://math.stackexchange.com/questions/690335/prove-that-lfloor-n1-2-rfloor-cdots-lfloor-n1-n-rfloor-lfloor-log-2 | Prove that: $\lfloor n^{1/2}\rfloor+\cdots+\lfloor n^{1/n}\rfloor=\lfloor \log_2n\rfloor +\cdots+\lfloor \log_nn \rfloor$, for $n > 1$
Prove that: $\lfloor n^{1/2}\rfloor+\cdots+\lfloor n^{1/n}\rfloor=\lfloor \log_2n\rfloor +\cdots+\lfloor \log_nn \rfloor$, for $n > 1,\, n\in \mathbb{N}$
For example. For $n=2$, we have $\lfloor 2^{1/2} \rfloor = \lfloor 1.414 \rfloor = 1$ whereas $\lfloor \log_2(2) \rfloor = 1$ while for $n=3$, we have $$\lfloor 3^{1/2} \rfloor + \lfloor 3^{1/3} \rfloor = \lfloor 1.732 \rfloor + \lfloor 1.442 \rfloor = 2= \lfloor 1.585 \rfloor + \lfloor 1 \rfloor=\lfloor \log_2(3) \rfloor + \lfloor \log_3(3) \rfloor .$$
I was thinking of using induction. So since $n=2$ is true, now assume for all $n$, this identity is true, we would like to prove that $n+1$ is true. Then
$$\lfloor n^{1/2} \rfloor + \lfloor n^{1/3} \rfloor + ... + \lfloor n^{1/n} \rfloor + \lfloor (n+1)^{1/(n+1)} \rfloor,$$ where $(n+1)^{1/(n+1)} > 1$ for all $n>1$ but it's strictly decreasing above 1 so $\lfloor (n+1)^{1/(n+1)} \rfloor = 1$
$\implies \lfloor n^{1/2} \rfloor + \lfloor n^{1/3} \rfloor +\cdots+ \lfloor n^{1/n} \rfloor + \lfloor (n+1)^{1/(n+1)} \rfloor = \lfloor n^{1/2} \rfloor + \lfloor n^{1/3} \rfloor +\cdots+ \lfloor n^{1/n} \rfloor + 1$
$= \lfloor \log_2(n) \rfloor + \lfloor \log_3(n) \rfloor + \cdots+ \lfloor \log_n(n) \rfloor + \lfloor \log_{n+1}(n+1) \rfloor$
since, $\log_{n+1}(n+1) = 1$ for all $n$.
My question is: How do we know that $(n+1)^{1/(n+1)}$ will never go below $1$? i.e., How can we prove that this function $f(x) = (x+1)^{1/(x+1)}$ is always bounded below by $1$ for $x>1$? (First, When $x=0$, $f(0)=1$, then looking at it's derivative, one can see that it's strictly increasing for $x$ between $(0,1)$ and decreasing for all $x>1$).
• You need to replace $n$ by $n + 1$ all over, not just in the last term, in the induction step. – vonbrand Feb 25 '14 at 21:57
• If $\llcorner x\lrcorner$ is supposed to mean the greatest integer not exceeding $x$, then you should probably have used $\lfloor x \rfloor$ (\lfloor … \rfloor) instead. If that's not what you mean, then what did you mean? – MJD Feb 25 '14 at 22:02
• thanks @MJD, I wasn't sure which code to use for it. Thanks! – PandaMan Feb 26 '14 at 1:37
• @vonbrand, I don't quite understand why you meant, please clarify. thanks – PandaMan Feb 26 '14 at 2:18
• @PandaMan, what you need to prove is $\lfloor (n+1)^{1/2}\rfloor + \ldots + \lfloor (n+1)^{1/(n+1)}\rfloor = \lfloor \log_2 (n+1)\rfloor +\ldots+\lfloor\log_{n+1} (n+1)\rfloor$ – vonbrand Feb 26 '14 at 5:04
This is a classic exercise and one with a very elegant solution.
The idea of the proof is to count the number $N$ of the points (see figure below) with integer coordinates, which lie in the region $$U=\big\{(x,y): 0<x\le n \,\,\,\text{and}\,\,\, 1<y\le n^{1/x}\big\},$$ and in particular, the red points, in two ways: horizontally and vertically.
Horizontal counting: $$N=\lfloor n^{1/2}\rfloor+\lfloor n^{1/3}\rfloor+\cdots+\lfloor n^{1/n}\rfloor,$$ since on the horizontal line $\,y=k\,$ lie exactly $\,\lfloor n^{1/k}\rfloor\,$ red points.
Vertical counting: $$N=\lfloor \log_2 n\rfloor+\lfloor\log_3 n\rfloor+\cdots+\lfloor \log_n n\rfloor,$$ since on the vertical line $\,x=k\,$ lie exactly $\,\lfloor \log_k n\rfloor\,$ red points.
$${}$$
Note that the curve in the figure above is of the function $y=n^{1/x}$.
This problem was first asked in a Soviet Mathematics Olympiad in 1982 (Всесоюзный Математический Олимпиад.)
• Gorgeous explanation! – Steven Stadnicki Feb 25 '14 at 22:24
• @StevenStadnicki: Correct. – Yiorgos S. Smyrlis Feb 25 '14 at 22:26
• (Thanks for the fix! I've removed the correction from my comment since that got edited in. :-) – Steven Stadnicki Feb 25 '14 at 22:28
• Great explanation. The wires are crossed a little though, because if $x=k$, then $y\leq n^{1/k}$, which means that the vertical count is the one that matches the fractional powers. Likewise, if $y = k$, then $k \leq n^{1/x}$, then $1\leq(\log_k n)/x$, so the horizontal count is the one that matches the logs. – John Moeller Mar 1 '14 at 0:13
• Awesome solution :-) – tarit goswami Aug 1 '18 at 8:54
Fix $b>1$. Then the derivative of $b^x$ is $\ln(b) b^x$; $\ln(b)$ is positive and $b^x$ is as well for all $x$, showing that that $b^x$ is a strictly increasing function. Next, $b^0=1$, showing that $b^x>1$ for all $x>0$.
Next, since $n+1>1$ and $1/(n+1)>0$, we have that $(n+1)^{1/(n+1)}>1$. | 2019-07-21T04:08:12 | {
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https://mathoverflow.net/questions/226455/can-a-complete-manifold-have-an-uncountable-number-of-ends | # Can a complete manifold have an uncountable number of ends?
Let $M$ be a complete and noncompact Riemannian manifold. Fix a point $p$ in $M$. Let $\gamma$: $[0, L]\rightarrow M$ (parametrized by its arc length) be a geodesic starting from $p$. Denote by $d(\cdot, \cdot)$ the distance function on $M$ induced by the Riemannian metric. $\gamma$ is minimal if $d(\gamma(s_1), \gamma(s_2))=|s_1-s_2|$. A ray is a minimal geodesic defined on $[0, +\infty)$. Since $M$ is noncompact, there exists at least one ray from $p$.
Two rays $\gamma_1, \gamma_2$ from the same point $p$ are called cofinal if for any $r\geq0$ and all $s>r$, $\gamma_1(s)$ and $\gamma_2(s)$ lie in the same component of $M\backslash B(0, r)$, where $B(0, r)=\{x\in M| d(p, x)<r\}$. An equivalence class of cofinal rays is called an end of $M$. My question is:
Can $M$ have uncountably many ends? It seems that the answer is no. But I am not very sure and I can not find a convictive proof.
• You might like to look at some basic notions of geometric group theory. The set of ends is a quasi-isometry invariant, and the Svarc--Milnor Lemma says that the universal cover of a compact manifold $M$ is quasi-isometric to a Cayley graph of $\pi_1M$. So it suffices to find a group with uncountably many ends, and the free group of rank 2 suffices. Curiously, it's actually impossible for a universal cover to have countably infinitely many ends.
– HJRW
Dec 18, 2015 at 20:35
• @FanZheng: Second countable spaces can have infinitely many ends. Take the universal cover of a wedge of two circles, for example. Dec 18, 2015 at 20:44
• @HJRW: The ends of the fundamental group aren't all that relevant to this question, although simply connected examples can be interesting. It's about whether $M$ has uncountably many ends (as in the answer of @SebastianGoette), not the universal cover of $M$. Dec 18, 2015 at 20:49
• @LeeMosher, I used bad notation, but I think my point stands (unless I misread the question). If $\pi_1N$ is free of rank 2 (say) then taking $M$ to be the universal cover of $N$ provides the example.
– HJRW
Dec 18, 2015 at 21:08
• @HJRW: Sure, for example $N=$ the connected sum of two copies of $S^2 \times S^1$. Dec 18, 2015 at 21:19
The answer is yes. Consider a hyperbolic pair of pants where all three boundary circles are of the same size. Glue countably many of them together such each new one is glued to the existing manifold along exactly one boundary circle. Then the manifold looks like the boundary of a fattened tree.
Fix $p$ in one of the pairs of pants $Y_0$. For all other pairs of pants $Y$, pick a point $q\in Y$, then the minimal geodesic joining $p$ and $q$ will leave $Y$ through one circle. Label the two remaining boundary circles with $0$ and $1$. Then all glueing circles get exactly one label, except for those bounding $Y_0$.
For each sequence $(a_n)_n\in\{0,1\}^{\mathbb N}$, we construct a ray from $p$ leaving $Y_0$ through the same boundary component. Whenever the ray enters the $n$-th pair of pants $Y$ along its way, we let it leave through the circle labelled $a_n$. This way, we construct an uncountable number of rays that are not pairwise cofinal.
• Thank you very much for your nice edit of my question and also for your clever solution. In this afternoon, I also found a very similar example called as the Cantor tree surface, inspired by your construction. Dec 19, 2015 at 10:34
Let me rephrase S. Goette's example in a slightly different way. Consider the complement of the triadic Cantor set in the complex plane. This is a Riemann surface that can be uniformized: there is a complete Riemannian metric of constant minus one curvature. The ends of this surface are in one-to-one correspondence with the points of the Cantor set.
You can see the decomposition into pairs of pants by drawing nested circles around the intervals that appear in the construction of the Cantor set.
In some sense, uncountably many ends is the generic case. Let us restrict ourselves to planar surfaces, namely complements of compact sets in the Riemann sphere. The set of compact subsets in a compact set has a natural topology, the Vietoris topology (which coincides with the topology given by the Hausdorff metric). So we can say that two open subsets are close if their complement are close with respect to this topology. It happens that generically a compact subset of the sphere has uncountably many connected components, hence its complement has uncountably many ends. | 2023-01-29T16:57:14 | {
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https://byjus.com/question-answer/i-have-a-total-of-rs-300-in-coins-of-denomination-re-1-rs-2-1/ | Question
# I have a total of Rs. $$300$$ in coins of denomination Re. $$1$$, Rs. $$2$$ and Rs. $$5$$. The number of Rs. $$2$$ coins is $$3$$ times the number of Rs. $$5$$ coins. The total number of coins is $$160$$. How many coins of each denomination are with me?
A
Number of Re. 1 coins =50; Number of Rs. 2 coins =40; Number of Rs. 5 coins =40
B
Number of Re. 1 coins =70; Number of Rs. 2 coins =70; Number of Rs. 5 coins =19
C
Number of Re. 1 coins =80; Number of Rs. 2 coins =60; Number of Rs. 5 coins =20
D
Number of Re. 1 coins =115; Number of Rs. 2 coins =50; Number of Rs. 5 coins =15
Solution
## The correct option is C Number of Re. $$1$$ coins $$= 80$$; Number of Rs. $$2$$ coins $$= 60$$; Number of Rs. $$5$$ coins $$= 20$$Given, Total number of coins $$= 160$$Given that the number of Rs.$$2$$ coins are $$3$$ times the number of Rs. $$5$$ coins.Number of Rs. $$5$$ coins $$= x$$Number of Rs. $$2$$ coins $$= 3x$$Number of Re. $$1$$ coins $$= 160 -(x+3x) = 160 -4x$$Total value of coins $$=$$ Rs. $$300$$So, $$5\times x + 2\times 3x + 1\times (160 -4x) = 300$$$$5x +6x +160 -4x = 300$$$$7x = 140$$$$x = 20$$So, number of Rs. $$5$$ coins $$= 20$$Number of Rs. $$2$$ coins $$= 60$$Number of Re. $$1$$ coins $$=80$$Mathematics
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https://www.vakrenordnorge.no/tmdnfv1/total-no-of-onto-functions-from-a-to-b-bd9598 | So, total numbers of onto functions from X to Y are 6 (F3 to F8). Functions can be classified according to their images and pre-images relationships. Then every function from A to B is effectively a 5-digit binary number. (d) x2 +1 x2 +2. Don’t stop learning now. I already know the formula (summation r=1 to n)(-1)^(n-r)nCr(r^m). The number of functions from {0,1}4 (16 elements) to {0, 1} (2 elements) are 216. By using our site, you This is same as saying that B is the range of f . Solution: As given in the question, S denotes the set of all functions f: {0, 1}4 → {0, 1}. Which must also be bijective, and therefore onto. From the formula for the number of onto functions, find a formula for S(n, k) which is defined in Problem 12 of Section 1.4. Option 3) 200. Thus, the number of onto functions = 16−2= 14. So, that leaves 30. (e) f(m;n) = m n. Onto. Then Total no. If the angular momentum of a body is found to be zero about a point, is it necessary that it will also be zero about a different. But we want surjective functions. Mathematics | Total number of possible functions, Mathematics | Unimodal functions and Bimodal functions, Total Recursive Functions and Partial Recursive Functions in Automata, Mathematics | Classes (Injective, surjective, Bijective) of Functions, Mathematics | Generating Functions - Set 2, Inverse functions and composition of functions, Last Minute Notes - Engineering Mathematics, Mathematics | Introduction to Propositional Logic | Set 1, Mathematics | Predicates and Quantifiers | Set 1, Mathematics | L U Decomposition of a System of Linear Equations, Mathematics | Mean, Variance and Standard Deviation, Mathematics | Sum of squares of even and odd natural numbers, Mathematics | Eigen Values and Eigen Vectors, Mathematics | Lagrange's Mean Value Theorem, Mathematics | Introduction and types of Relations, Data Structures and Algorithms – Self Paced Course, We use cookies to ensure you have the best browsing experience on our website. Find the number of relations from A to B. Here are the definitions: is one-to-one (injective) if maps every element of to a unique element in . In the example of functions from X = {a, b, c} to Y = {4, 5}, F1 and F2 given in Table 1 are not onto. Not onto. No. Home. In other words no element of are mapped to by two or more elements of . there are zero onto function . Consider the function x → f(x) = y with the domain A and co-domain B. Determine whether each of these functions is a bijection from R to R. (a) f(x) = 2x+1. Suppose TNOF be the total number of onto functions feasible from A to B, so our aim is to calculate the integer value TNOF. [5.1] Informally, a function from A to B is a rule which assigns to each element a of A a unique element f(a) of B. Officially, we have Definition. In other words, nothing is left out. We need to count the number of partitions of A into m blocks. Examples: Let us discuss gate questions based on this: Solution: As W = X x Y is given, number of elements in W is xy. Comparing cardinalities of sets using functions. The number of functions from Z (set of z elements) to E (set of 2xy elements) is 2xyz. Solution: Using m = 4 and n = 3, the number of onto functions is: I am trying to get the total number of onto functions from set A to set B if the former has m elements and latter has n elements with m>n. (B) 64 Math Forums. Yes. If X has m elements and Y has 2 elements, the number of onto functions will be 2 m-2. Example 9 Let A = {1, 2} and B = {3, 4}. For example, if n = 3 and m = 2, the partitions of elements a, b, and c of A into 2 blocks are: ab,c; ac,b… Functions: One-One/Many-One/Into/Onto . Out of these functions, the functions which are not onto are f (x) = 1, ∀x ∈ A. In this article, we are discussing how to find number of functions from one set to another. Therefore, S has 216 elements. For example: X = {a, b, c} and Y = {4, 5}. An onto function is also called surjective function. An onto function is also called surjective function. In this case the map is also called a one-to-one correspondence. Here are the definitions: 1. is one-to-one (injective) if maps every element of to a unique element in . We need to count the number of partitions of A into m blocks. In other words, if each b ∈ B there exists at least one a ∈ A such that. 2. is onto (surjective)if every element of is mapped to by some element of . An onto function is also called a surjective function. Let A = {a 1, a 2, a 3} and B = {b 1, b 2} then f : A -> B. It means that every element “b” in the codomain B, there is exactly one element “a” in the domain A. such that f(a) = b. . Therefore, each element of X has ‘n’ elements to be chosen from. In a function from X to Y, every element of X must be mapped to an element of Y. That is, a function f is onto if for each b ∊ B, there is atleast one element a ∊ A, such that f(a) = b. 2×2×2×2 = 16. So, you can now extend your counting of functions … 19. Tuesday: Functions as relations, one to one and onto functions What is a function? 34 – 3C1(2)4 + 3C214 = 36. So, number of onto functions is 2m-2. Therefore, N has 2216 elements. 2. Example 46 (Method 1) Find the number of all one-one functions from set A = {1, 2, 3} to itself. One-to-One/Onto Functions . Option 2) 120. So the total number of onto functions is m!. Calculating required value. Some authors use "one-to-one" as a synonym for "injective" rather than "bijective". Misc 10 (Introduction)Find the number of all onto functions from the set {1, 2, 3, … , n} to itself.Taking set {1, 2, 3}Since f is onto, all elements of {1, 2, 3} have unique pre-image.Total number of one-one function = 3 × 2 × 1 = 6Misc 10Find the number of all onto functio 1.1. . If m < n, the number of onto functions is 0 as it is not possible to use all elements of Y. 4 = A B Not a function Notation We write f (a) = b when (a;b) 2f where f is a function. Math Forums. therefore the total number of functions from A to B is. In a one-to-one function, given any y there is only one x that can be paired with the given y. If n(A)= 3 , n(B)= 5 Find the number of onto function from A to B, For onto function n(A) n(B) otherwise ; it will always be an inoto function. Set A has 3 elements and set B has 4 elements. A function f from A to B is a subset of A×B such that • for each a ∈ A there is a b ∈ B with (a,b… Experience. We say that b is the image of a under f , and a is a preimage of b. October 31, 2007 1 / 7. Here's another way to look at it: imagine that B is the set {0, 1}. But, if the function is onto, then you cannot have 00000 or 11111. Q3. Let E be the set of all subsets of W. The number of functions from Z to E is: If X has m elements and Y has 2 elements, the number of onto functions will be 2. There are $$\displaystyle 3^8=6561$$ functions total. If anyone has any other proof of this, that would work as well. Explanation: From a set of m elements to a set of 2 elements, the total number of functions is 2m. where as when i try manually it comes 8 . De nition 1 A function or a mapping from A to B, denoted by f : A !B is a Such functions are referred to as injective. (C) 81 4. Writing code in comment? Please use ide.geeksforgeeks.org, For understanding the basics of functions, you can refer this: Classes (Injective, surjective, Bijective) of Functions. There are 3 ways of choosing each of the 5 elements = $3^5$ functions. A function is said to be bijective or bijection, if a function f: A → B satisfies both the injective (one-to-one function) and surjective function (onto function) properties. 38. Considering all possibilities of mapping elements of X to elements of Y, the set of functions can be represented in Table 1. Transcript. f(a) = b, then f is an on-to function. Q1. Why does an ordinary electric fan give comfort in summer even though it cannot cool the air? Students can solve NCERT Class 12 Maths Relations and Functions MCQs Pdf with Answers to know their preparation level. (D) 72. I just need to know how it came. Since f is one-one Hence every element 1, 2, 3 has either of image 1, 2, 3 and that image is unique Total number of one-one function = 6 Example 46 (Method 2) Find the number of all one-one functions from set A = {1, 2, 3} to itself. In the above figure, f … So, total numbers of onto functions from X to Y are 6 (F3 to F8). Example 46 (Method 1) Find the number of all one-one functions from set A = {1, 2, 3} to itself. In F1, element 5 of set Y is unused and element 4 is unused in function F2. Discrete Mathematics Grinshpan Partitions: an example How many onto functions from f1;2;3;4;5;6;7;8g to fA;B;C;Dg are there? There are 3 functions with 1 element in range. How many onto functions are there from a set with eight elements to a set with 3 elements? Onto Function A function f: A -> B is called an onto function if the range of f is B. Copyright © 2021 Pathfinder Publishing Pvt Ltd. To keep connected with us please login with your personal information by phone/email and password. (c) f(m;n) = m. Onto. Need explanation for: If n(A)= 3 , n(B)= 5 Find the number of onto function from A to B, List of Hospitality & Tourism Colleges in India, Knockout JEE Main May 2022 (Easy Installments), Knockout JEE Main May 2021 (Easy Installments), Knockout NEET May 2021 (Easy Installments), Knockout NEET May 2022 (Easy Installments), Top Medical Colleges in India accepting NEET Score, MHCET Law ( 5 Year L.L.B) College Predictor, List of Media & Journalism Colleges in India, B. Transcript. Let f be the function from R … Formula for finding number of relations is Number of relations = 2 Number of elements of A × Number of elements of B Attention reader! In other words, if each b ∈ B there exists at least one a ∈ A such that. Why does a tightly closed metal lid of a glass bottle can be opened more easily if it is put in hot water for some time? Click hereto get an answer to your question ️ Write the total number of one - one functions from set A = { 1,2,3,4 } to set B = { a,b,c } . Onto Function Definition (Surjective Function) Onto function could be explained by considering two sets, Set A and Set B, which consist of elements. Yes. (i)When all the elements of A will map to a only, then b is left which do not have any pre-image in A (ii)When all the elements of A will map to b only, then a is left which do not have only pre-image in A Thus in both cases, function is not onto So, total number of onto functions= 2^n-2 Hope it helps☑ #Be Brainly Onto Functions: Consider the function {eq}y = f(x) {/eq} from {eq}A \to B {/eq}, where {eq}A {/eq} is the domain of the function and {eq}B {/eq} is the codomain. For example, if n = 3 and m = 2, the partitions of elements a, b, and c of A into 2 blocks are: ab,c; ac,b; bc,a. If n > m, there is no simple closed formula that describes the number of onto functions. according to you what should be the anwer So the correct option is (D). As E is the set of all subsets of W, number of elements in E is 2xy. Also, given, N denotes the number of function from S(216 elements) to {0, 1}(2 elements). Let X, Y, Z be sets of sizes x, y and z respectively. An exhaustive E-learning program for the complete preparation of JEE Main.. Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test.. In F1, element 5 of set Y is unused and element 4 is unused in function F2. 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Option 1) 150. There are $$\displaystyle 2^8-2$$ functions with 2 elements in the range for each pair of elements in the codomain. Onto Function A function f: A -> B is called an onto function if the range of f is B. (b) f(m;n) = m2 +n2. If n > m, there is no simple closed formula that describes the number of onto functions. These numbers are called Stirling numbers (of the second kind). Maths MCQs for Class 12 Chapter Wise with Answers PDF Download was Prepared Based on Latest Exam Pattern. A function f : A -> B is said to be an onto function if every element in B has a pre-image in A. The onto function from Y to X is F's inverse. Therefore, total number of functions will be n×n×n.. m times = nm. This course will help student to be better prepared and study in the right direction for JEE Main.. No. Menu. They are various types of functions like one to one function, onto function, many to one function, etc. 3. (d) f(m;n) = jnj. This disagreement is confusing, but we're stuck with it. set a={a,b,c} and B={m,n} the number of onto functions by your formula is 6 . If X has m elements and Y has n elements, the number if onto functions are. f(a) = b, then f is an on-to function. Get hold of all the important CS Theory concepts for SDE interviews with the CS Theory Course at a student-friendly price and become industry ready. 2.1. . No element of B is the image of more than one element in A. The total no.of onto function from the set {a,b,c,d,e,f} to the set {1,2,3} is????? Onto function or Surjective function : Function f from set A to set B is onto function if each element of set B is connected with set of A elements. (b) f(x) = x2 +1. 3. is one-to-one onto (bijective) if it is both one-to-one and onto. Suppose TNOF be the total number of onto functions feasible from A to B, so our aim is to calculate the integer value TNOF. Option 4) none of these Number of onto functions from one set to another – In onto function from X to Y, all the elements of Y must be used. High School Math Elementary Math Algebra Geometry Trigonometry Probability and Statistics Pre-Calculus. The number of injections that can be defined from A to B is: The number of onto functions (surjective functions) from set X = {1, 2, 3, 4} to set Y = {a, b, c} is: 3. A function f from A to B is called one-to-one (or 1-1) if whenever f (a) = f (b) then a = b. generate link and share the link here. (c) f(x) = x3. So the total number of onto functions is m!. For function f: A→B to be onto, the inequality │A│≥2 must hold, since no onto function can be designed from a set with cardinality less than 2 where 2 is the cardinality of set B. Number of functions from one set to another: Let X and Y are two sets having m and n elements respectively. Out of these functions, 2 functions are not onto (If all elements are mapped to 1st element of Y or all elements are mapped to 2nd element of Y). Let f and g be real functions defined by f(x) = 2x+ 1 and g(x) = 4x – 7. asked Feb 16, 2018 in Class XI Maths by rahul152 ( -2,838 points) relations and functions Since f is one-one Hence every element 1, 2, 3 has either of image 1, 2, 3 and that image is unique Total number of one-one function = 6 Example 46 (Method 2) Find the number Steps 1. (b)-Given that, A = {1 , 2, 3, n} and B = {a, b} If function is subjective then its range must be set B = {a, b} Now number of onto functions = Number of ways 'n' distinct objects can be distributed in two boxes a' and b' in such a way that no box remains empty. Let A = {a 1, a 2, a 3} and B = {b 1, b 2} then f : A -> B. P.S. A function has many types which define the relationship between two sets in a different pattern. (A) 36 In the example of functions from X = {a, b, c} to Y = {4, 5}, F1 and F2 given in Table 1 are not onto. Any ideas on how it came? Let W = X x Y. Free PDF Download of CBSE Maths Multiple Choice Questions for Class 12 with Answers Chapter 1 Relations and Functions. In other words no element of are mapped to by two or more elements of . Proving that a given function is one-to-one/onto. One more question. Not onto. So, there are 32 = 2^5. Learn All Concepts of Chapter 2 Class 11 Relations and Function - FREE. Tech Companion - A Complete pack to prepare for Engineering admissions, MBBS Companion - For NEET preparation and admission process, QnA - Get answers from students and experts, List of Pharmacy Colleges in India accepting GPAT, Why does a tightly closed metal lid of a glass bottle can be opened more easily if it is put in hot water for some time? Number of Onto function - & Number of onto functions - For onto function n(A) n(B) otherwise ; it will always be an inoto function . 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Imagine that B is the set of 2xy elements ) to E ( set of 2 elements in the...., there is only one X that can be represented in Table 1 '' as a synonym for ''... Are there from a to a unique element in range and pre-images relationships the total number of onto are. All possibilities of mapping elements of Y, the number of partitions of a into m blocks with to! To count the number of functions from Z ( set of 2xy elements ) to E ( set functions. Injective ) if maps every element of are mapped to by two or more elements of Y with.! This is same as saying that B is called an onto function a function not possible to use elements! 1, 2 } and B = { 1, 2 } and B {. Maps every element of is mapped to by two or more elements of Y is 0 as it is one-to-one... Then f is an on-to function manually it comes 8 4, 5 } Download of Maths... 16−2= 14 1, 2 } and Y = { 1, 2 } and B = 1. | 2021-05-14T13:18:56 | {
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https://math.stackexchange.com/questions/3119303/constructing-a-sum-of-squares-in-mathbb-rx-with-given-complex-valuation | # Constructing a sum of squares in $\mathbb R[x]$ with given complex valuation
Fix a polynomial $$g(x)\in\mathbb R[x]$$ and a complex number $$u\in\mathbb C\setminus\mathbb R$$. My main question is
How can we construct a polynomial $$s(x)\in\mathbb R[x]$$ such that $$s(x)$$ is a sum of squares and $$s(u)=g(u)$$ and $$s(\overline{u})=g(\overline{u})$$?
I know that if $$s(u)=g(u)$$ then $$s(\overline{u})=g(\overline{u})$$, so we really only need the condition $$s(u)=g(u)$$. I am asking about this because I am trying to understand a proof of a theorem (as Theorem 7.3 in "Solving systems of polynomial equations" by Sturmfels) in which the author claims that we can write a polynomial $$g(x)\in\mathbb R[x_1,\ldots,x_n]$$ is congruent to a sum of squares $$s(x)\in\mathbb R[x_1,\ldots x_n]$$ modulo $$\langle x-u\rangle\cap\langle x-\overline{u}\rangle$$ for fixed $$u\in\mathbb C^n$$ strictly complex. The rest of the proof I can understand (and I could copy it into this question if requested), but I am having trouble seeing just why exactly this can be done. So my main question above is just asking about the univariate case of this, from which I believe I could generalize the argument.
So far I have only been able to think of the following. If we set $$s(x):=\frac{g(x)^2+|g(u)|^2}{2\Re(g(u))}$$ then we have $$g(u)^2+g(u)g(\overline{u})=g(u)\left(g(u)+\overline{g(u)}\right)=2\Re (g(u))g(u)$$ so $$s(u)=g(u)$$. But this only works if $$\Re (g(u))>0$$ since we need $$s(x)$$ to be a sum of squares in $$\mathbb R[x]$$.
Note that since $$u$$ is not real, $$u$$ and $$1$$ are linearly independent over $$\mathbb{R}$$, so every complex number can be written as $$au+b$$ for some $$a,b\in\mathbb{R}$$. In particular, we can pick $$a$$ and $$b$$ such that $$au+b$$ is a square root of $$g(u)$$, and then take $$s(x)=(ax+b)^2$$.
Alternatively, pick $$a\in\mathbb{R}$$ large enough such that $$v=u+a$$ has positive real part. Now note that the consecutive powers of $$v^2$$ have an angle of less than $$\pi$$ between them. We can pick $$n\in\mathbb{N}$$ such that $$g(u)$$ has argument between the arguments of $$v^{2n}$$ and $$v^{2n+2}$$, and then $$g(u)$$ can be written as a linear combination of $$v^{2n}$$ and $$v^{2n+2}$$ with nonnegative real coefficients. This gives a polynomial $$s(x)=b(x+a)^{2n}+c(x+a)^{2n+2}$$ which is a sum of squares and has $$s(u)=g(u)$$.
• Yes I want $u\notin\mathbb R$, I will edit to say $u\in\mathbb C\setminus\mathbb R$. This solution is very interesting, thank you! I might leave the answer un-accepted for a little while just in case people have other approaches and want to share. – Dave Feb 19 at 22:02 | 2019-06-20T11:28:38 | {
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https://mathzsolution.com/does-this-property-characterize-a-space-as-hausdorff/ | # Does this property characterize a space as Hausdorff?
As a result of this question, I’ve been thinking about the following condition on a topological space $Y$:
For every topological space $X$, $E\subseteq X$, and continuous maps $f,g\colon X\to Y$, if $E$ is dense in $X$, and $f$ and $g$ agree on $E$ (that is, $f(e)=g(e)$ for all $e\in E$), then $f=g$.
If $Y$ is Hausdorff, then $Y$ satisfies this condition. The question is whether the converse holds: if $Y$ satisfies the above condition, will it necessarily be Hausdorff?
If $Y$ is not at least $T_1$, then $Y$ does not have the property: if $u,v\in Y$ are such that $u\neq v$ and every open neighborhood of $u$ contains $v$, then let $X$ be the Sierpinski space, $X=\{a,b\}$, $a\neq b$, with topology $\tau=\{\emptyset,\{b\},X\}$, $E=\{b\}$, let $f,g\colon X\to Y$ be given by $f(a)=f(b)=v$, and $g(a)=u$, $g(b)=v$. Then both $f$ and $g$ are continuous, agree on the dense subset $E$, but are distinct.
My attempt at a proof of the converse assumes the Axiom of Choice and proceeded as follows: assume $Y$ is $T_1$ but not $T_2$; let $u$ and $v$ be witnesses to the fact that $Y$ is not $T_2$, let $\mathcal{U}\_s$ and $\mathcal{V}\_t$ be the collection of all open nbds of $s$ that do not contain $t$, and all open nbds of $t$ that do not contain $s$, respectively. Construct a net with index set $\mathcal{U}\_s\times\mathcal{V}\_t$ (ordered by $(U,V)\leq (U',V')$ if and only if $U'\subseteq U$ and $V'\subseteq V$) by letting $y_{(U,V)}$ be a point in $U\cap V$ (this is where AC comes in). Let $E=\{y_{(U,V)}\mid (U,V)\in\mathcal{U}\_s\times\mathcal{V}\_t\}$, and let $X=E\cup\{s\}$. Give $X$ the induced topology; let $f\colon X\to Y$ be the inclusion map, and let $g\colon X\to Y$ be the map that maps $E$ to itself identically, but maps $s$ to $t$.
The only problem is I cannot quite prove that $g$ is continuous; the difficulty arises if I take an open set $\mathcal{O}\in \mathcal{V}_t$; the inverse image under $g$ is equal to $((\mathcal{O}\cap X)-\{t\})\cup\{s\}$, and I have not been able to show that this is open in $X$.
So:
Does the condition above characterize Hausdorff spaces?
If not, I would appreciate a counterexample. If it does characterize Hausdorff, then ideally I would like a way to finish off my proof, but if the proof is unsalvageable (or nobody else can figure out how to finish it off either) then any proof will do.
Added: A little digging turned up this question raised in the Problem Section of the American Mathematical Monthly back in 1964 by Alan Weinstein. The solution by Sim Lasher gives a one paragraph proof that does not require one to consider $T_1$ and non-$T_1$ spaces separately.
I think the following goes a long way towards proving a converse:
Let $(Y, T)$ be any T1 topological space with at least two points and let $a$ and $b$ be distinct points in $Y$.
Let $X = Y\setminus\{b\}$.
Let $f: X \to Y$ be the inclusion of $X$ in $Y$.
Let $g: X \to Y$ agree with $f$ on $X\setminus\{a\}$ and $g(a) = b$.
Finally, define the topology on $X$ to be the coarsest topology that makes both $f$ and $g$ continuous.
With these assumptions it turns out that $X\setminus\{a\}$ is dense in $X$ if and only if $a$ and $b$ do not have disjoint neighbourhoods in $Y$.
To see that this is true, let us construct a base of the topology on $X$.
To make $f$ continuous we only need to take the subspace topology. Since
X is open in Y this is $S_1 = \{ G \in T \mid b \notin G \}$. To also make $g$ continuous we need to add the open neighbourhoods of $b$, with $b$ replaced by $a$. This gives $S_2 = \{ (H\setminus\{b\} \cup \{a\} \mid H \in T, b \in H \}$.
Now $S_1 \cup S_2$ is a subbase of the topology on $X$.
Since $S_1$ and $S_2$ are already closed under finite intersection, and each covers $X$, we can say that
$B = \{ G \cap H \mid G \in S_1, H \in S_2 \}$ is a base of the topology.
Then (remembering that finite sets are closed in Y) we find that the following are all equivalent:
• $X\setminus\{a\}$ is not dense in $X$
• $\{a\}$ is open in $X$
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# A clock store sold a certain clock to a collector for 20 per
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A clock store sold a certain clock to a collector for 20 percent more than the store had originally paid for the clock. When the collector tried to resell the clock to the store, the store bought it back at 50 percent of what the collector had paid. The shop then sold the clock again at a profit of 80 percent on its buy-back price. If the difference between the clock's original cost to the shop and the clock's buy-back price was $100, for how much did the shop sell the clock the second time? A.$270
B. $250 C.$240
D. $220 E.$200
Originally posted by hsk on 11 Jun 2007, 21:02.
Last edited by Bunuel on 06 May 2014, 09:27, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: A clock store sold a certain clock to a collector for 20 per [#permalink]
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06 May 2014, 20:22
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hsk wrote:
A clock store sold a certain clock to a collector for 20 percent more than the store had originally paid for the clock. When the collector tried to resell the clock to the store, the store bought it back at 50 percent of what the collector had paid. The shop then sold the clock again at a profit of 80 percent on its buy-back price. If the difference between the clock's original cost to the shop and the clock's buy-back price was $100, for how much did the shop sell the clock the second time? A.$270
B. $250 C.$240
D. $220 E.$200
Assume numbers and then use ratios to fit them in with the actual numbers.
Say the store bought the clock for $100 and sold it to the collector for$120. The store bought back from the collector for $60 and sold it back at$60*18/10 = $108 Here, difference between original cost ($100) and buy back price ($60) is$40. But actually it is given to be $100 which is 2.5 times 40. So the shop sold the clock a second time for$108*2.5 = $270 _________________ Karishma Veritas Prep GMAT Instructor Save up to$1,000 on GMAT prep through 8/20! Learn more here >
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11 Jun 2007, 21:42
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$270 Original Price = x Sold to collector at 1.2x re-buy price = 1.2x * 0.5 = 0.6x therefore, x - 0.6x = 100; x = 250 second selling price = 1.8 * 0.6 * 250 = 270 Hope, this is not too confusing ##### General Discussion Senior Manager Joined: 04 Mar 2007 Posts: 419 ### Show Tags 12 Jun 2007, 10:52 1 100 % - original price 120 % - sell 1 60 % - bought 2 180*60/100 = 108 % - sell 2 100$ - 40%
x - 108 %
x= 100*108/40 = 270
A
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Re: A clock store sold a certain clock to a collector for 20 per [#permalink]
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02 Jun 2015, 00:44
7
The question is a bit wordy and gives us a series of percentages on which we need to work on. Going step wise is the key to not miss out on any of the information. Presenting the detailed step wise solution
Step-I- Store buys the clock
Let's assume the original price of clock paid by the store to be $$x$$
Step-II- Collectors buys the clock from the store
Extra amount paid by collector to buy the clock = $$20$$% of $$x$$
Therefore price at which collector buys the clock = $$x + 20$$% of $$x = 1.2x$$
Step-III- Store buys back the clock from collector
Price at which the store buys the clock = $$50$$% of price collector paid $$= 50$$% of $$1.2x = 0.6x$$
Step-IV- Store resells the clock
Price at which store resells the clock = $$0.6x + 80$$% of $$0.6x = 0.6x * 1.8$$
Now, we are given that difference between clock's original price and clock's buy back price = $$100$$
$$x - 0.6x = 100$$ i.e. $$x = 250$$
We are asked to find the price at which the store resells the clock = $$0.6x * 1.8 = 0.6 * 250 * 1.8 = 270$$
Hope this helps
Regards
Harsh
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Re: A clock store sold a certain clock to a collector for 20 per [#permalink]
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11 Jul 2015, 13:39
EgmatQuantExpert wrote:
The question is a bit wordy and gives us a series of percentages on which we need to work on. Going step wise is the key to not miss out on any of the information. Presenting the detailed step wise solution
Step-I- Store buys the clock
Let's assume the original price of clock paid by the store to be $$x$$
Step-II- Collectors buys the clock from the store
Extra amount paid by collector to buy the clock = $$20$$% of $$x$$
Therefore price at which collector buys the clock = $$x + 20$$% of $$x = 1.2x$$
Step-III- Store buys back the clock from collector
Price at which the store buys the clock = $$50$$% of price collector paid $$= 50$$% of $$1.2x = 0.6x$$
Step-IV- Store resells the clock
Price at which store resells the clock = $$0.6x + 80$$% of $$0.6x = 0.6x * 1.8$$
Now, we are given that difference between clock's original price and clock's buy back price = $$100$$
$$x - 0.6x = 100$$ i.e. $$x = 250$$
We are asked to find the price at which the store resells the clock = $$0.6x * 1.8 = 0.6 * 250 * 1.8 = 270$$
Hope this helps
Regards
Harsh
Dear,
I have a basic question.
The author states: "The shop then sold the clock again at a profit of 80 percent on its buy-back price". Why is it suppose to multiply 0.6x*1.8? I think my confuse is related to the profit concept. For me, if I buy something at 100, and the cost of it is 80. I have a profit of 20/100 = 20%. Using the same logic on the question, if I buy something at 100 and I get a profit of 80%, the sales price must be 100/0.2. Thus, it is 500. Profit is 400/500=80%. It is a totally different aproach than to say 100*1,8.
Where my confuse is?
Kind regards...
Gonzalo
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Re: A clock store sold a certain clock to a collector for 20 per [#permalink]
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11 Jul 2015, 13:54
1
1
gbascurs wrote:
EgmatQuantExpert wrote:
The question is a bit wordy and gives us a series of percentages on which we need to work on. Going step wise is the key to not miss out on any of the information. Presenting the detailed step wise solution
Step-I- Store buys the clock
Let's assume the original price of clock paid by the store to be $$x$$
Step-II- Collectors buys the clock from the store
Extra amount paid by collector to buy the clock = $$20$$% of $$x$$
Therefore price at which collector buys the clock = $$x + 20$$% of $$x = 1.2x$$
Step-III- Store buys back the clock from collector
Price at which the store buys the clock = $$50$$% of price collector paid $$= 50$$% of $$1.2x = 0.6x$$
Step-IV- Store resells the clock
Price at which store resells the clock = $$0.6x + 80$$% of $$0.6x = 0.6x * 1.8$$
Now, we are given that difference between clock's original price and clock's buy back price = $$100$$
$$x - 0.6x = 100$$ i.e. $$x = 250$$
We are asked to find the price at which the store resells the clock = $$0.6x * 1.8 = 0.6 * 250 * 1.8 = 270$$
Hope this helps
Regards
Harsh
Dear,
I have a basic question.
The author states: "The shop then sold the clock again at a profit of 80 percent on its buy-back price". Why is it suppose to multiply 0.6x*1.8? I think my confuse is related to the profit concept. For me, if I buy something at 100, and the cost of it is 80. I have a profit of 20/100 = 20%. Using the same logic on the question, if I buy something at 100 and I get a profit of 80%, the sales price must be 100/0.2. Thus, it is 500. Profit is 400/500=80%. It is a totally different aproach than to say 100*1,8.
Where my confuse is?
Kind regards...
Gonzalo
Gonzalo, Profit is calculated = Selling price - Cost price. Profit % is always calculated on the cost price of the purchase and not on the selling price. You are calculating profit % on the selling price. This is where you are going wrong. I am assuming that in your example, you are buying something at 80 while you are selling it at 100, giving you an absolute profit of 20$while your profit % will be 20/80 = 25% and not 20/100 = 20%. Now, in the question above, lets say the original cost of the clock to store was C$ and then it sold the same to the collector at 20% profit.
This means the clocks' selling price was C (1.2) and this becomes cost price for the collector.
Now, when the collector tries to sell the same clock to the store, the store buys it for 50% the price at which the collector bought it.
Thus, you get = 1.2*0.5*C = 0.6 C
Furthermore, the store sells the clock for the second time for 80% profit and thus the selling price of the clock becomes = cost price of the clock for the store at buy-back * 1.8 = 1.8 * 0.6 C
Finally given that C - 0.6 C = 100 ----> C = 250$Thus, the cost of the clock the second time around = 1.8*0.6 C = 1.8 * 0.6 * 250 = 270$. Hence A is the correct answer.
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Re: A clock store sold a certain clock to a collector for 20 per [#permalink]
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11 Jul 2015, 14:20
Engr2012 wrote:
gbascurs wrote:
EgmatQuantExpert wrote:
The question is a bit wordy and gives us a series of percentages on which we need to work on. Going step wise is the key to not miss out on any of the information. Presenting the detailed step wise solution
Step-I- Store buys the clock
Let's assume the original price of clock paid by the store to be $$x$$
Step-II- Collectors buys the clock from the store
Extra amount paid by collector to buy the clock = $$20$$% of $$x$$
Therefore price at which collector buys the clock = $$x + 20$$% of $$x = 1.2x$$
Step-III- Store buys back the clock from collector
Price at which the store buys the clock = $$50$$% of price collector paid $$= 50$$% of $$1.2x = 0.6x$$
Step-IV- Store resells the clock
Price at which store resells the clock = $$0.6x + 80$$% of $$0.6x = 0.6x * 1.8$$
Now, we are given that difference between clock's original price and clock's buy back price = $$100$$
$$x - 0.6x = 100$$ i.e. $$x = 250$$
We are asked to find the price at which the store resells the clock = $$0.6x * 1.8 = 0.6 * 250 * 1.8 = 270$$
Hope this helps
Regards
Harsh
Dear,
I have a basic question.
The author states: "The shop then sold the clock again at a profit of 80 percent on its buy-back price". Why is it suppose to multiply 0.6x*1.8? I think my confuse is related to the profit concept. For me, if I buy something at 100, and the cost of it is 80. I have a profit of 20/100 = 20%. Using the same logic on the question, if I buy something at 100 and I get a profit of 80%, the sales price must be 100/0.2. Thus, it is 500. Profit is 400/500=80%. It is a totally different aproach than to say 100*1,8.
Where my confuse is?
Kind regards...
Gonzalo
Gonzalo, Profit is calculated = Selling price - Cost price. Profit % is always calculated on the cost price of the purchase and not on the selling price. You are calculating profit % on the selling price. This is where you are going wrong. I am assuming that in your example, you are buying something at 80 while you are selling it at 100, giving you an absolute profit of 20$while your profit % will be 20/80 = 25% and not 20/100 = 20%. Now, in the question above, lets say the original cost of the clock to store was C$ and then it sold the same to the collector at 20% profit.
This means the clocks' selling price was C (1.2) and this becomes cost price for the collector.
Now, when the collector tries to sell the same clock to the store, the store buys it for 50% the price at which the collector bought it.
Thus, you get = 1.2*0.5*C = 0.6 C
Furthermore, the store sells the clock for the second time for 80% profit and thus the selling price of the clock becomes = cost price of the clock for the store at buy-back * 1.8 = 1.8 * 0.6 C
Finally given that C - 0.6 C = 100 ----> C = 250$Thus, the cost of the clock the second time around = 1.8*0.6 C = 1.8 * 0.6 * 250 = 270$. Hence A is the correct answer.
Thank you a lot.
Here is the key of my misunderstanding: "Profit % is always calculated on the cost price of the purchase". I assumed it was on the sales price.
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A clock store sold a certain clock to a collector for 20 per [#permalink]
### Show Tags
10 Nov 2015, 05:00
1
x=starting price
therefore
x +20% = $$x +1/5x$$ = $$6x/5$$
50% = 3x/5 BB Price
as per Q
x - $$3x/5$$=100
x=250
BB =150
80 %increment of 150 = 270
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A clock store sold a certain clock to a collector for 20 per [#permalink]
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10 Nov 2015, 05:20
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hsk wrote:
A clock store sold a certain clock to a collector for 20 percent more than the store had originally paid for the clock. When the collector tried to resell the clock to the store, the store bought it back at 50 percent of what the collector had paid. The shop then sold the clock again at a profit of 80 percent on its buy-back price. If the difference between the clock's original cost to the shop and the clock's buy-back price was $100, for how much did the shop sell the clock the second time? A.$270
B. $250 C.$240
D. $220 E.$200
Let us assume the initial price (CP) of the clock to be 100x
We assume 100x to avoid the usage of decimals in case of x and avoid the usage of unitary method in case of 100
Transaction 1:
Store sold the clock to collector.
Selling Price = 20% more than CP = 120x
Transaction 2:
The collector sold it back to the store.
The new CP for the store = 60x (Store bought it back for 50% of the Selling Price)
Transaction 3:
Store sold it back at a profit of 80% on 60x = 108x
Now we are told that 100x - 60x = 100
This means x = 100/40
Hence the second selling price = 108*100/40 = 270
Option A
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# In the xy-plane, the straight-line graphs of the three equations above
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23 Jul 2015, 10:47
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y = ax - 5
y = x + 6
y = 3x + b
In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r). If a and b are constants, what is the value of b?
(1) a = 2
(2) r = 17
[Reveal] Spoiler: OA
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Last edited by Bunuel on 13 Nov 2017, 21:24, edited 4 times in total.
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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23 Jul 2015, 11:30
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mcelroytutoring wrote:
DS 136 from OFG 2016 (new question)
y = ax - 5
y = x + 6
y = 3x + b
In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r). If a and b are constants, what is the value of b?
1) a = 2
2) r = 17
Solution provided by : mcelroytutoring
Let's start by substituting the point (p,r) into all equations in place of (x,y) which will make step #2 a bit easier to comprehend but is not necessary to solve the question. Then, let's consider the number of variables left in each equation.
#1: r = ap - 5 (3 variables R,A,P)
#2: r = p + 6 (2 variables R,P)
#3: r = 3p + b (3 variables R,B,P)
1) a = 2 allows us to reduce equation #1 to the variables r and p, which are the same two variables as equation #2. Thus we have simultaneous equations. As soon as we verify that the equations are different, we know that we can solve for both variables. Once we know r and p, we can substitute in equation #3 to solve for b. Sufficient.
2) r = 17 allows us to do the same thing, more or less. It reduces equation #2 to only one variable, allowing us to solve for p. Once we have p (and r), we can use equation #3 to solve for b. Sufficient.
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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24 Sep 2015, 01:53
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if each of the 3 equations contains points (p,r) this means that they intersect in that point
1. a=2
Find the intercept Intercept for three simultaneous equations
y=2x-5
y=x+6
y=3x+b
Let's use the first 2 equations: plug y=x+6 in the secod equation
x+6=2x-5 -> x=11, y=17 we can use the values to calulate b in the 3rd equation
17=33+b -> b=-16 SUFFICIENT
2. Here we have directly the value for Y, let's plug it in the 2nd equation
y=x+6 -> 17=x+6 -> x=11, y=17; We can plug these values in the 3rd equation and find b as we did above
17=33+b -> b=-16 SUFFICIENT
Answer (D) Most important point is here to catch the hint about intersection of 3 lines at one point
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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26 Oct 2015, 10:50
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I think it's D.
Keeping point (p,r) in all the equations we get :
p = ar -5 -----(1)
p = r + 6 ------(2)
p = 3r + b ------(3)
Now consider (1) if a = 2 from (1) and (2) we get
r = 11 , p=17 and putting in (3) we can get b.
Similarly for (2) we can get the values for r and p and hence can get the value for b.
So both statements individually are correct to answer the question.
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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26 Oct 2015, 19:14
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m2k wrote:
I think it's D.
Keeping point (p,r) in all the equations we get :
p = ar -5 -----(1)
p = r + 6 ------(2)
p = 3r + b ------(3)
Now consider (1) if a = 2 from (1) and (2) we get
r = 11 , p=17 and putting in (3) we can get b.
Similarly for (2) we can get the values for r and p and hence can get the value for b.
So both statements individually are correct to answer the question.
Approach if right but the values you derived are wrong. According to me r = 17 and that is what stmt b also states.
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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10 Jun 2016, 11:23
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y = ax - 5 ... eq 1
y = x + 6 ... eq 2
y = 3x + b ...eq 3
Total of 4 variables are present.
Statement 1 : a = 2
Insert in eq 1
We have y = 2x -5 and y = x+6
Solving we get x = 11 and y = 17
Substitute in eq 3 and we get value of b
Statement 2: r=17
This means the y co- ordinate is 17
Substitute in eq 2 we get x as 11
Again can find value of b from equation 3
Hence D
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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18 Jan 2017, 15:18
we used information from both 1 and 2 then how can the answer be D... should it not be C... some one kindly clarify.......... clearly am a zero in ds and that too a big one
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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18 Jan 2017, 20:18
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y = ax - 5---------1
y = x + 6---------2
y = 3x + b--------3
In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r). If a and b are constants, what is the value of b?
(1) a = 2
(2) r = 17
All three line intersect each other at common point (p,r).
1. given a = 2
putting in equation 1 .= y=2x-5
equating 1(after replacing value of a) and 2 we will get value of (p,r)
putting (p,r) in equation 3 we will get value for b---suff..
2 given r = 17.
putting in equation 2 we will get value of x.i'e p.
Now as we know common point of intersection ,putting p,r in equation 3 , we will get value of b
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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22 Jan 2017, 15:51
thanks sobby for your response... highly appreciated...
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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16 Mar 2017, 08:58
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(eq1) $$y = ax - 5$$
(eq2) $$y = x + 6$$
(eq3) $$y = 3x + b$$
In the xy-plane, the straight-line graphs of the three equations above each contain the point (p, r). If a and b are constants, what is the value of b?
1) $$a = 2$$
2) $$r = 17$$
Solution:
1) $$a = 2$$
- Putting the value of a in eq1, we get: $$y = 2x - 5$$
- At this point you can solve for (x, y), plug (x, y) in (eq3) and solve for (b) [though this approach might take few seconds]
(alternatively, faster method)
- you can skip solving for (x, y) and deduce that given 3 equations and 3 unknowns (since a is given in statement 1) we can solve for all of them (including b), since the lines are have different slopes i.e. different lines. Hence, we can get single value of 'b', proving the condition SUFFICIENT.
NOTE: 3 equations and 3 unknowns does not ALWAYS mean that we can find 3 unknown. We have to make sure that 2 of them or all of them are not the same line.
2) $$r = 17$$
- Since, point (p, r) lie on all the line, we can plugin the point in above equation
$$r = ap - 5 => 17 = ap - 5$$
$$r = p + 6 => 17 = p + 6$$
$$r = 3p + b => 17 = 3p + b$$
- Again, we do not need to solve for all the variables and just recognize that the above equations will lead to single value of b. Hence, SUFFICIENT.
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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30 Apr 2017, 23:51
y = ax - 5
y = x + 6
y = 3x + b
In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r). If a and b are constants, what is the value of b?
1) a = 2
2) r = 17
My 2 cents.
It is important to realize from the Question stem that the 3 equations intersect as (p,r).
For 1), as we know a =2, we can equate the first and second equation to get the value of x, and then use that value of x to find value of y and the find value of b.
For 2), similarly, use r = 17 (which is value of y) to find value of x using the second equation. And then plug it back to the third equation.
So D.
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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10 Sep 2017, 19:04
Would it be correct to simply say that we have 4 variables with 3 equations so eliminating any one variable gets us to three equations and three variables and is therefore sufficient? Is that logic sound?
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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13 Nov 2017, 20:21
what is the significance of the the line "In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r)",
i solved the problem, but wud have did the same even if they didnt provide line above . as question has 3 equations with 4 variables
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]
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13 Nov 2017, 22:31
Cheryn wrote:
what is the significance of the the line "In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r)",
i solved the problem, but wud have did the same even if they didnt provide line above . as question has 3 equations with 4 variables
The highlighted statement in effect says that all these 3 lines meet each other at one point and so there is a single value of (x,y) that satisfies these 3 equations. It is only because of this highlighted statement you can solve this set of equations for a unique value of x,y, a and b.
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink] 13 Nov 2017, 22:31
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If r and s are positive integers, can the fraction r/s be expressed as
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If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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20 Oct 2012, 08:21
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If r and s are positive integers, can the fraction r/s be expressed as a decimal with only a finite number of nonzero digits?
(1) s is a factor of 100
(2) r is a factor of 100
[Reveal] Spoiler: OA
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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20 Oct 2012, 10:02
IMO it is A, if the denominator is a factor of 100 then it could be 1; 2; 5; 10; 20.. if you divide all the positive integer by these number you will have a finite decimal result.
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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20 Oct 2012, 13:18
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Finite decimals are decimals which end. eg 0.5,0.25, etc. Non finite are numbers like 1/3,1/6 etc ie 0.33333333333333333333333333333333333333333333333333333.......... or 0.66666666666666666666666666666666666666666666666666666..........
1)S is a factor of 100. So S cannot have more than two 2s and two 5s. Any number divisible be 2 or 5 gives a finite decimal. Since R and S are both positive integers, there can be no 0s in the decimal places either. So Sufficient
2)R is a factor of 100. Cant say anything form this. R can be 1. 1/10 is finite. 1/3 is not. Insufficient.
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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23 Oct 2012, 07:59
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asveaass wrote:
If r and s are positive integers, can the fraction r/s be expressed as a decimal with only a finite number of nonzero digits?
1) s is a factor of 100
2) r is a factor of 100
I don't understand the answer explanation in the OG, could someone please explain in detail?
THEORY:
Reduced fraction $$\frac{a}{b}$$ (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only $$b$$ (denominator) is of the form $$2^n5^m$$, where $$m$$ and $$n$$ are non-negative integers. For example: $$\frac{7}{250}$$ is a terminating decimal $$0.028$$, as $$250$$ (denominator) equals to $$2*5^3$$. Fraction $$\frac{3}{30}$$ is also a terminating decimal, as $$\frac{3}{30}=\frac{1}{10}$$ and denominator $$10=2*5$$.
Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.
For example $$\frac{x}{2^n5^m}$$, (where x, n and m are integers) will always be terminating decimal.
(We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction $$\frac{6}{15}$$ has 3 as prime in denominator and we need to know if it can be reduced.)
Questions testing this concept:
does-the-decimal-equivalent-of-p-q-where-p-and-q-are-89566.html
any-decimal-that-has-only-a-finite-number-of-nonzero-digits-101964.html
if-a-b-c-d-and-e-are-integers-and-p-2-a3-b-and-q-2-c3-d5-e-is-p-q-a-terminating-decimal-125789.html
700-question-94641.html
is-r-s2-is-a-terminating-decimal-91360.html
pl-explain-89566.html
which-of-the-following-fractions-88937.html
BACK TO THE ORIGINAL QUESTION:
If r and s are positive integers, can the fraction r/s be expressed as a decimal with only a finite number of nonzero digits?
(1) s is a factor of 100. Factors of 100 are: 1, 2, 4, 5, 10, 20, 25, 50 and 100. All these numbers are of the form $$2^n5^m$$ (for example 1=2^0*5^0, 2=2^1*5^0, ...), therefore no matter what is the value of r, r/s will always will be terminating decimal. Sufficient.
(2) r is a factor of 100. We need to know about the denominator. Not sufficient.
Hope it's clear.
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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02 Jan 2013, 01:25
you rock bunuel, your explanations are awesome.. thanks a lot for this one!
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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24 Dec 2013, 11:12
But it has been said the 'decimal with finite number of non-zero digits'. Now if we take 2/50 then it will be 0.04, which means its a finite decimal but definitely it does not have all the non-zero digits after decimal point. So, can anybody explain?
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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04 May 2014, 11:31
Hi Bunuel,
Speaking from a prime perspective, I could factor "100" into prime factors and since I get 2^2 and 5^2, that should suffice in coming up with the right answer, correct? I don't need to pull up every single factors - right?
If the denominator had ANY other primes then it would NOT be a terminating decimal. Is that correct?
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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05 May 2014, 02:31
halloanupam wrote:
But it has been said the 'decimal with finite number of non-zero digits'. Now if we take 2/50 then it will be 0.04, which means its a finite decimal but definitely it does not have all the non-zero digits after decimal point. So, can anybody explain?
0.04 has finite number of non-zero digits: 4 is not followed by any non-zero digit.
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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05 May 2014, 02:35
russ9 wrote:
Hi Bunuel,
Speaking from a prime perspective, I could factor "100" into prime factors and since I get 2^2 and 5^2, that should suffice in coming up with the right answer, correct? I don't need to pull up every single factors - right?
If the denominator had ANY other primes then it would NOT be a terminating decimal. Is that correct?
Not entirely. The denominator can have some other primes as well but if those primes can be reduced the fraction still would be terminating. For example, consider fraction 3/6. The denominator has 3 in it, but it ca be reduced to get 3/6=1/2=0.5.
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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26 Sep 2014, 23:28
Bunuel wrote:
asveaass wrote:
If r and s are positive integers, can the fraction r/s be expressed as a decimal with only a finite number of nonzero digits?
1) s is a factor of 100
2) r is a factor of 100
I don't understand the answer explanation in the OG, could someone please explain in detail?
THEORY:
Reduced fraction $$\frac{a}{b}$$ (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only $$b$$ (denominator) is of the form $$2^n5^m$$, where $$m$$ and $$n$$ are non-negative integers. For example: $$\frac{7}{250}$$ is a terminating decimal $$0.028$$, as $$250$$ (denominator) equals to $$2*5^3$$. Fraction $$\frac{3}{30}$$ is also a terminating decimal, as $$\frac{3}{30}=\frac{1}{10}$$ and denominator $$10=2*5$$.
Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.
For example $$\frac{x}{2^n5^m}$$, (where x, n and m are integers) will always be terminating decimal.
(We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction $$\frac{6}{15}$$ has 3 as prime in denominator and we need to know if it can be reduced.)
Why is it then 130/13 or 121/11 would give finite ... infact they properly divide...
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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27 Sep 2014, 01:28
ani781 wrote:
Bunuel wrote:
asveaass wrote:
If r and s are positive integers, can the fraction r/s be expressed as a decimal with only a finite number of nonzero digits?
1) s is a factor of 100
2) r is a factor of 100
I don't understand the answer explanation in the OG, could someone please explain in detail?
THEORY:
Reduced fraction $$\frac{a}{b}$$ (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only $$b$$ (denominator) is of the form $$2^n5^m$$, where $$m$$ and $$n$$ are non-negative integers. For example: $$\frac{7}{250}$$ is a terminating decimal $$0.028$$, as $$250$$ (denominator) equals to $$2*5^3$$. Fraction $$\frac{3}{30}$$ is also a terminating decimal, as $$\frac{3}{30}=\frac{1}{10}$$ and denominator $$10=2*5$$.
Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.
For example $$\frac{x}{2^n5^m}$$, (where x, n and m are integers) will always be terminating decimal.
(We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction $$\frac{6}{15}$$ has 3 as prime in denominator and we need to know if it can be reduced.)
Why is it then 130/13 or 121/11 would give finite ... infact they properly divide...
The rule above is for reduced fraction $$\frac{a}{b}$$ (meaning that fraction is already reduced to its lowest term). When you reduce 130/13 to the lowest term you get 10 and when you reduce 121/11 you get 11: 10/(2^0*5^0) and 11/(2^0*5^0) respectively.
Check the links in my post above to practice more on this type of questions.
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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27 Sep 2014, 02:08
A, of course. I had got it as soon as I saw it. This is a very simple problem probably fetched from RS
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If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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17 Nov 2014, 19:56
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Bunuel wrote:
THEORY:
Reduced fraction $$\frac{a}{b}$$ (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only $$b$$ (denominator) is of the form $$2^n5^m$$, where $$m$$ and $$n$$ are non-negative integers. For example: $$\frac{7}{250}$$ is a terminating decimal $$0.028$$, as $$250$$ (denominator) equals to $$2*5^3$$. Fraction $$\frac{3}{30}$$ is also a terminating decimal, as $$\frac{3}{30}=\frac{1}{10}$$ and denominator $$10=2*5$$.
Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.
For example $$\frac{x}{2^n5^m}$$, (where x, n and m are integers) will always be terminating decimal.
Hi Bunuel,
Just looking for some clarification, in the highlighted text above, please correct me if I am wrong, by including the word "and" did you mean that if the denominator contains only 2's or only 5's then too the fraction is a terminating decimal? In other words, the denominator doesn't need to contain, both - 2's & 5's!
Thank you!
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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17 Nov 2014, 23:36
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Yes, examples such as $$\frac{17}{2^7}$$, $$\frac{31}{5^{18}}$$, and $$\frac{21}{[(2^5)(5^7)]}$$ are all terminating decimals.
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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18 Nov 2014, 12:44
dabral wrote:
dmmk
Yes, examples such as $$\frac{17}{2^7}$$, $$\frac{31}{5^{18}}$$, and $$\frac{21}{[(2^5)(5^7)]}$$ are all terminating decimals.
Dabral
Thank you dabral
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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03 Dec 2014, 11:12
Bunuel wrote:
russ9 wrote:
Hi Bunuel,
Speaking from a prime perspective, I could factor "100" into prime factors and since I get 2^2 and 5^2, that should suffice in coming up with the right answer, correct? I don't need to pull up every single factors - right?
If the denominator had ANY other primes then it would NOT be a terminating decimal. Is that correct?
Not entirely. The denominator can have some other primes as well but if those primes can be reduced the fraction still would be terminating. For example, consider fraction 3/6. The denominator has 3 in it, but it ca be reduced to get 3/6=1/2=0.5.
Bunuel I did not understand from your post whether we can have other primes in denominator? Can you pls. repeat?
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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03 Dec 2014, 23:50
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For a fraction to be a terminating decimal the only primes that can be present in the denominator are 2 and 5, and this only applies to reduced form of the fraction. If there is any other prime in the denominator, then the fraction will be non-terminating. For example, $$\frac{21}{[(2^4)(5^3)(11^2)]}$$, $$\frac{11}{[(5^6)(7^3)]}$$, and $$\frac{22}{(7^4)}$$ are all non-terminating decimals.
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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04 Dec 2014, 00:35
dabral wrote:
Ergenekon
For a fraction to be a terminating decimal the only primes that can be present in the denominator are 2 and 5, and this only applies to reduced form of the fraction. If there is any other prime in the denominator, then the fraction will be non-terminating. For example, $$\frac{21}{[(2^4)(5^3)(11^2)]}$$, $$\frac{11}{[(5^6)(7^3)]}$$, and $$\frac{22}{(7^4)}$$ are all non-terminating decimals.
Dabral
Thanks Dabral. Got it now.
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Hello, what if r is the same as s? Then we don't have a finite decimal but an integer. Then A should not be sufficient as the answer can be a decimal or an integer. Am I wrong in my thinking?
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]
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ptanwar1 wrote:
Hello, what if r is the same as s? Then we don't have a finite decimal but an integer. Then A should not be sufficient as the answer can be a decimal or an integer. Am I wrong in my thinking?
An integer IS a decimal with a finite number of nonzero digits. For example, integer 51 has 2 (finite) number of digits.
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink] 12 Mar 2015, 03:15
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# 2,600 has how many positive divisors?
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2,600 has how many positive divisors? [#permalink]
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28 Oct 2010, 17:10
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2,600 has how many positive divisors?
A. 6
B. 12
C. 18
D. 24
E. 48
[Reveal] Spoiler: OA
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28 Oct 2010, 17:21
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By factorization, you can write 2600 as 2600=2^3*5^2*13^1. Number of factors = (3+1)(2+1)(1+1)=24
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29 Oct 2010, 02:47
vgan4 wrote:
Number of factors = (3+1)(2+1)(1+1)=24
can' t understand how we came to this. What does this multiplication mean?
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29 Oct 2010, 03:11
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ulm wrote:
vgan4 wrote:
Number of factors = (3+1)(2+1)(1+1)=24
can' t understand how we came to this. What does this multiplication mean?
Finding the Number of Factors of an Integer
First make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.
The number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself.
Example: Finding the number of all factors of 450: $$450=2^1*3^2*5^2$$
Total number of factors of 450 including 1 and 450 itself is $$(1+1)*(2+1)*(2+1)=2*3*3=18$$ factors.
Back to the original question:
If p and q are prime numbers, how many divisors does the product $$p^3*q^6$$ have?
According to above the number of distinct factors of $$2,600=2^3*5^2*13$$ would be $$(3+1)(2+1)(1+1)=24$$.
For more on this check Number Theory chapter of Math Book: math-number-theory-88376.html
Hope it helps.
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29 Oct 2010, 03:27
Yes, thanks,
i forgot
"The number of factors of n will be expressed by the formula $$(p+1)(q+1)(r+1)$$"
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Thanx Bunuel
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Re: 2,600 has how many positive divisors? [#permalink]
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20 Mar 2017, 01:27
shrive555 wrote:
2,600 has how many positive divisors?
A. 6
B. 12
C. 18
D. 24
E. 48
2600 = 13 x 2^3 x 5^2
no of possible divisor = (1+1)(2+1)(3+1) = 4.3.2 = 24
Option D
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Re: 2,600 has how many positive divisors? [#permalink]
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03 Apr 2017, 00:16
2600= 26*100= 2*13*2^2*5^2
=2^3*5^2*13
Total factors= (3+1)(2+1)(1+1)= 4*3*2=24
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Re: 2,600 has how many positive divisors? [#permalink]
### Show Tags
05 Jan 2018, 11:24
shrive555 wrote:
2,600 has how many positive divisors?
A. 6
B. 12
C. 18
D. 24
E. 48
To determine the number of positive divisors, we break 2,600 into primes, add 1 to the exponent of each unique prime, and then multiply those values together.
We see that 2,600 = 26 x 100 = 2 x 13 x 4 x 25 = 2^3 x 5^2 x 13^1.
Now we add 1 to each exponent and multiply those results:
(3 + 1)(2 + 1)(1 +1) = 24
Thus, 2,600 has 24 positive divisors.
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Re: 2,600 has how many positive divisors? [#permalink]
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05 Jan 2018, 11:56
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shrive555 wrote:
2,600 has how many positive divisors?
A. 6
B. 12
C. 18
D. 24
E. 48
Experts, is there a fast way to find all the divisors? Thanks.[/quote]
If the prime factorization of N = (p^a)(q^b)(r^c) . . . (where p, q, r, etc are different prime numbers), then N has a total of (a+1)(b+1)(c+1)(etc) positive divisors.
Example: 14000 = (2^4)(5^3)(7^1)
So, the number of positive divisors of 14000 = (4+1)(3+1)(1+1) =(5)(4)(2) = 40
----------ONTO THE QUESTION-------------------------
2600 = (2)(2)(2)(5)(5)(13)
= (2^3)(5^2)(13^1)
So, the number of positive divisors of 2600 = (3+1)(2+1)(1+1) =(4)(3)(2) = 24
Cheers,
Brent
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Re: 2,600 has how many positive divisors? [#permalink] 05 Jan 2018, 11:56
Display posts from previous: Sort by | 2018-03-21T01:34:27 | {
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https://math.stackexchange.com/questions/3732112/general-solution-for-cos-fracx2-1-cos21-fracx2 | # General Solution for $\cos(\frac{x}{2}-1) =\cos^2(1-\frac{x}{2})$
I'm looking for an algebraic solution to : $$\cos(\frac{x}{2}-1) = \cos^2(1-\frac{x}{2})$$. So I simplified the equation: first off, $$\cos(\frac{x}{2}-1) = \cos(1-\frac{x}{2})$$. Then I divided both sides by that. and so I'm left with two things to solve:
$$\cos(\frac{x}{2}-1) = 0$$ (because I divided both sides by that expression, I have to also include the $$0$$ solution too). and $$\cos(\frac{x}{2}-1) = 1$$. And the general solution would be, I think, the union of those.
However, I'm kinda lost at this point. I've attempted to solve each equation. First off, I know that $$\cos(x) = 0$$ at $$\frac{\pi}{2}$$ and $$\frac{3\pi}{2}$$. So, the general solution for $$\cos(x) = 0$$ would be $$x=\frac{\pi}{2} +2\pi k, \cup \ \frac{3\pi}{2}+2\pi k, k\in Z.$$ I got up to this point, but don't know how to proceed.
The thing that's most confusing to me is I don't know how the $$-1$$ in the argument plays into the solution. Does it just change the graph to the right? Playing with desmos shows that graph is shifting by 2, but I thought that it'd shift by 1. More importantly: does it also affect the period of the function?
Additional question: In my book the answers are given in a different form. For example, the union I wrote would be written as: $$x= (-1)^k\frac{\pi}{2} + \pi k, k \in Z.$$ And in every case the period is "reduced" to $$\pi k$$. Why is that?
• So if the solution for $\cos(x) =0$ is $(x) =\pi/2+2\pi k$ then the solution to $\cos(x/2-1)=0$ is $(x/2-1)=\pi/2+2\pi k$ similarly for the other solution. – kingW3 Jun 23 '20 at 23:49
• well, $\cos (-u) = \cos u$ – Will Jagy Jun 23 '20 at 23:51
Okay so, you noticed that $${\cos\left(\frac{x}{2}-1\right)=\cos\left(1-\frac{x}{2}\right)}$$. And as you say, you end up with
$${\Leftrightarrow \cos^2\left(\frac{x}{2}-1\right)=\cos\left(\frac{x}{2}-1\right)}$$
And this implies
$${\cos\left(\frac{x}{2}-1\right)\left(\cos\left(\frac{x}{2}-1\right)-1\right)=0}$$
(as was pointed out by someone else - indeed it's probably bad practice to divide through here by $${\cos}$$. Not that it's incorrect, since you did take into account the fact we would then miss the $$0$$ solution (which was awesome!!!) - but unnecessary).
For simplicity, we can replace $${\frac{x}{2}-1}$$ with $${u}$$ and just rearrange for $${x}$$ at the end. So
$${\Rightarrow \cos(u)\left(\cos(u)-1\right)=0}$$
Now.... we want to solve $${\cos(u)=0}$$. As you said, one solution is $${\frac{\pi}{2}}$$... and if you take a look at the graph, you will notice that every other $$0$$ to the cosine function can be "reached" by hopping foward and backwards by multiplies of $${\pi}$$... you are right in thinking indeed, the period of the cosine function is $${2\pi}$$, but in fact $${0}$$ reoccurs every $${\pi}$$ radians. There is no problem with this. In order to have a full period ($${2\pi}$$ in this case), every value the function takes on must reoccur - only $$0$$ has. This actually means that
$${\cos(u)=0\Leftrightarrow u=\frac{\pi}{2}+n\pi, n \in \mathbb{Z}}$$
Now we can solve $${\cos(u)=1}$$ (the other solution we need). Of course we have a $${1}$$ at the point $${x=0}$$, and you may notice that we can reach all the other 1's by jumping foward and backwards by multiples of $${2\pi}$$ this time. Hence the solution is
$${\cos(u)=1\Leftrightarrow u=0 + 2n\pi =2n\pi, n \in \mathbb{Z}}$$
Now, you can simply plug back in the definition of $${u}$$ in terms of $${x}$$, adding $${1}$$ and multiplying both sides by two and you will end up with the solutions being
$${x=\pi + 2n\pi + 2=(2n+1)\pi + 2, n \in \mathbb{Z}}$$
$${x=4n\pi + 2, n \in \mathbb{Z}}$$
• This explained things very clearly and helped me solved another, similar problem too!. Thanks!!. – Ebrin Jun 24 '20 at 0:33
• No problem at all! :) – Riemann'sPointyNose Jun 24 '20 at 0:35
You're making things more complicated than they are
First, cosine is an even function, so the equation can be written $$\cos\Bigl(\frac{x}{2}-1\Bigr) =\cos^2\Bigl(\frac{x}{2}-1\Bigr)$$ as well. Note the equation implies $$\cos\bigl(\frac{x}{2}-1\bigr)\ge 0$$.
Also, if $$0, note that $$c Therefore the equation is equivalent to $$\cos\Bigl(\frac{x}{2}-1\Bigr)=0\:\text{ or }\:1.$$ Now this is easy to solve in terms of congruences: $$\begin{cases} \cos\Bigl(\frac{x}{2}-1\Bigr)=0\iff \frac{x}{2}-1\equiv \frac\pi 2 \mod\pi \iff\frac x2\equiv 1+\frac\pi 2\mod\pi \\ \cos\Bigl(\frac{x}{2}-1\Bigr)=1\iff \frac{x}{2}-1\equiv 0\mod2\pi \iff\frac x2\equiv 1\mod2\pi \end{cases}$$ and ultimately $$x\equiv2+\pi\bmod 2\pi \quad\text{ or }\quad x\equiv2\bmod 4\pi.$$
• Oh, the first part explains why the answer only contained the positive answers. So, the -1 in the argument just corresponds to the graph shifting to the right by two, right? It has no impact on the period. – Ebrin Jun 24 '20 at 0:07
• Absolutely no impact. – Bernard Jun 24 '20 at 0:09
• @kingW3: You're right. I shouldn't calculate directly on screen. Thank you for pointing it! – Bernard Jun 24 '20 at 0:17
Approach to the problem I would like to discuss an approach and leave solving the equations to you. $$\cos(y) = 0$$ implies that y is an odd multiple of $$\frac{\pi}{2}$$ or $$(2n + 1)\frac{\pi}{2}$$, where n is an integer. You can even write $$(2n - 1)$$, which is fine too as it gives odd vales. Similarly, $$\cos(\frac{x}{2} - 1) = 0$$ implies that $$\frac{x}{2} - 1 = (2n + 1)\frac{\pi}{2}.$$ You can simplify it further. Then solve the next one which is $$\cos(\frac{x}{2} - 1) = 1$$which implies that $$\frac{x}{2} - 1 = 2n\pi$$ (even multiple of pi), where $$n$$ is an integer. The $$'n'$$ in both solutions can be same because the first solution will always give an odd value of $$\frac{\pi}{2}$$ and the second solution will always give an even value of $$\frac{\pi}{2},$$ so we can use the same symbol 'n'. Now, to unify both solutions: Observe if there is any similarity between first and second solution. Is there any value(s) or range for which both solutions are same? If yes, we can get an even compact expression. If no, then write both solutions joined together with union symbol.
Regarding graph To interpret a function $$f(x - 1)$$interms of $$f(x)$$: The value which you get on putting $$x$$ in a function $$f(x)$$ will now [in $$f(x - 1)]$$ be obtained by setting $$x$$ as $$(x+1)$$ in $$f(x - 1)$$. Put $$x+1$$ in place of $$x$$ in $$f(x-1)$$, you will get $$f(x)$$ again. What does it mean? It means that you will have to shift your graph of $$f(x)$$ towards right by $$1$$ unit. Practice by drawing the graph of $$f(x) = x^2.$$ Then draw $$f(x) = (x - 1)^2.$$ But your problem has $$\cos(\frac{x}{2} - 1)$$ which is similar to $$f(ax + b)$$. We can now interpret $$'b'$$ as we did in last para. What about $$'a'$$?
Think like this. Every value of function $$f(x)$$ is now obtained at $$\frac{x}{a}$$ when the function is transformed to $$f(ax)$$. To show this, set $$x$$ as $$\frac{x}{a}$$ in $$f(ax)$$, you will get f(x) again. It also means that the frequency of the function $$f(ax)$$ is now decreased by $$a$$ factor of $$'a'$$.
In case of $$f(\frac{x}{a})$$, it can be said that every old value of $$f(x)$$ is obtained at $$'ax'$$ (show). Here the frequency is increased by a factor of $$'a'$$. Every old value of $$f(x)$$ is obtained quickly with $$f(\frac{x}{a}).$$ You can relate this with period since frequency and period are related. Or, you could interpret it in terms of period in the first place itself. If the frequency has increased, it means the value of the function will be repeated quickly, thereby decreasing the period. Period of $$\cos(x)$$ is $$2\pi$$. Period of $$\cos(ax+b)$$ will be $$\frac{2\pi}{a}.$$ The constant b only shifts the values, it does not account for frequency or period.
Thus, your interpretation of graph of $$\cos(\frac{x}{2} - 1)$$ is incomplete. To obtain its graph, we have to expand the graph of $$\cos(x)$$ by a factor of $$2$$, then shift the graph towards right by $$1$$ unit.
Regarding your last doubt In academics it is completely fine to represent the solution as simple as possible according to you. Generally, we want to make it more compact. The solution which you have wrote there. Put integral values of $$k$$. You will find that you have actually written odd multiples of $$\frac{\pi}{2}$$ in any case. Try to put several odd and even (positive integers for ease) integral values for $$k$$ and interpret the result yourself.
I thank @Ebrin for his help in formatting this answer.
• Sorry for the late reply, I've read your answer thoroughly and found it very helpful. Thanks for taking the time and explaining things! – Ebrin Jun 28 '20 at 15:59
So if $$\cos W = 0$$ then $$W = \pm \frac \pi 2 + 2k\pi$$. Notice $$-\frac \pi 2 + 2k \pi = \frac \pi 2 + (2k-1)\pi$$. And $$\frac \pi 2 + 2k \pi = -\frac \pi 2 + (2k+1) \pi$$. So to simplify $$W = \frac \pi 2 + m\pi$$ would be the simplest way to state this.
(Alternativily as $$0 = -0$$ and $$\cos(W \pm \pi) = -\cos W$$ we'd note that for an $$W = \pm \frac \pi 2 +2k \pi$$ then $$\mp \frac \pi 2+ (2k+1)\pi$$ is a solution).
So $$\frac x2 -1 = \frac \pi 2 + m\pi$$ then $$x = (2m+1)\pi + 2$$
And if $$\cos W = 1$$ then .... well... do I need to point out that solution is $$W=m\pi$$?
So $$\frac x2 -1= m\pi$$ so $$x=2m\pi +1$$.
So the solutions are $$x = k \pi + 2$$. If $$k$$ is odd then $$\cos(\frac x2 -1)=\cos(\frac k2 \pi) = 1$$. And if $$k$$ is odd then $$\cos(\frac x2 -1)=\cos(\frac {k-1}2\pi + \frac \pi 2) = 0$$. | 2021-02-25T02:34:41 | {
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https://in.mathworks.com/help/wavelet/ref/scal2frq.html | # scal2frq
Scale to frequency
## Description
example
frq = scal2frq(A,wname,delta) returns the pseudo-frequencies corresponding to the scales given by A and the wavelet specified by wname and the sampling period delta. The output frq is real-valued and has the same dimensions as A.
frq = scal2frq(A,wname) is equivalent to frq = scal2frq(A,wname,1).
## Examples
collapse all
This example shows how the pseudo-frequency changes as you double the scale.
Construct a vector of scales with 10 voices per octave over five octaves.
vpo = 10;
no = 5;
a0 = 2^(1/vpo);
ind = 0:vpo*no;
sc = a0.^ind;
Verify that the range of scales covers five octaves.
log2(max(sc)/min(sc))
ans = 5.0000
If you plot the scales, you can use a data cursor to confirm that the scale at index $n+10$ is twice the scale at index $n$. Set the y-ticks to mark each octave.
plot(ind,sc)
title('Scales')
xlabel('Index')
ylabel('Scale')
grid on
set(gca,'YTick',2.^(0:5))
Convert the scales to pseudo-frequencies for the real-valued Morlet wavelet. First, assume the sampling period is 1.
pf = scal2frq(sc,"morl");
T = [sc(:) pf(:)];
T = array2table(T,'VariableNames',{'Scale','Pseudo-Frequency'});
disp(T)
Scale Pseudo-Frequency
______ ________________
1 0.8125
1.0718 0.75809
1.1487 0.70732
1.2311 0.65996
1.3195 0.61576
1.4142 0.57452
1.5157 0.53605
1.6245 0.50015
1.7411 0.46666
1.8661 0.43541
2 0.40625
2.1435 0.37904
2.2974 0.35366
2.4623 0.32998
2.639 0.30788
2.8284 0.28726
3.0314 0.26803
3.249 0.25008
3.4822 0.23333
3.7321 0.2177
4 0.20313
4.2871 0.18952
4.5948 0.17683
4.9246 0.16499
5.278 0.15394
5.6569 0.14363
6.0629 0.13401
6.498 0.12504
6.9644 0.11666
7.4643 0.10885
8 0.10156
8.5742 0.094761
9.1896 0.088415
9.8492 0.082494
10.556 0.07697
11.314 0.071816
12.126 0.067006
12.996 0.062519
13.929 0.058332
14.929 0.054426
16 0.050781
17.148 0.047381
18.379 0.044208
19.698 0.041247
21.112 0.038485
22.627 0.035908
24.251 0.033503
25.992 0.03126
27.858 0.029166
29.857 0.027213
32 0.025391
Assume that data is sampled at 100 Hz. Construct a table with the scales, the corresponding pseudo-frequencies, and periods. Since there are 10 voices per octave, display every tenth row in the table. Observe that for each doubling of the scale, the pseudo-frequency is cut in half.
Fs = 100;
DT = 1/Fs;
pf = scal2frq(sc,"morl",DT);
T = [sc(:)/Fs pf(:) 1./pf(:)];
T = array2table(T,'VariableNames',{'Scale','Pseudo-Frequency','Period'});
T(1:vpo:end,:)
ans=6×3 table
Scale Pseudo-Frequency Period
_____ ________________ ________
0.01 81.25 0.012308
0.02 40.625 0.024615
0.04 20.313 0.049231
0.08 10.156 0.098462
0.16 5.0781 0.19692
0.32 2.5391 0.39385
Note the presence of the $\Delta t=\frac{1}{Fs}$ factor in scal2frq. This is necessary in order to achieve the proper scale-to-frequency conversion. The $\Delta t$ is needed to adjust the raw scales properly. For example, with:
f = scal2frq(1,'morl',0.01);
You are really asking what happens to the center frequency of the mother Morlet wavelet, if you dilate the wavelet by 0.01. In other words, what is the effect on the center frequency if instead of $\psi \left(t\right)$, you look at $\psi \left(t/0.01\right)$. The $\Delta t$ provides the correct adjustment factor on the scales.
You could have obtained the same results by first converting the scales to their adjusted sizes and then using scal2frq without specifying $\Delta t$.
max(pf-pf2)
ans = 0
The example shows how to create a contour plot of the CWT using approximate frequencies in Hz.
Create a signal consisting of two sine waves with disjoint support in additive noise. Assume the signal is sampled at 1 kHz.
Fs = 1000;
t = 0:1/Fs:1-1/Fs;
x = 1.5*cos(2*pi*100*t).*(t<0.25)+1.5*cos(2*pi*50*t).*(t>0.5 & t<=0.75);
x = x+0.05*randn(size(t));
Obtain the CWT of the input signal and plot the result.
[cfs,f] = cwt(x,Fs);
contour(t,f,abs(cfs).^2);
axis tight;
grid on;
xlabel('Time');
ylabel('Approximate Frequency (Hz)');
title('CWT with Time vs Frequency');
## Input Arguments
collapse all
Scales, specified as a positive real-valued vector.
Wavelet, specified as a character vector or string scalar. See wavefun for more information.
Sampling period, specified as a real-valued scalar.
Example: pf = scal2frq([1:5],"db4",0.01)
collapse all
### Pseudo-Frequencies
There is only an approximate answer for the relationship between scale and frequency.
In wavelet analysis, the way to relate scale to frequency is to determine the center frequency of the wavelet, Fc, and use the following relationship:
${F}_{a}=\frac{{F}_{c}}{a}$
where
• a is a scale.
• Fc is the center frequency of the wavelet in Hz.
• Fa is the pseudo-frequency corresponding to the scale a, in Hz.
The idea is to associate with a given wavelet a purely periodic signal of frequency Fc. The frequency maximizing the Fourier transform of the wavelet modulus is Fc. The centfrq function computes the center frequency for a specified wavelet. From the above relationship, it can be seen that scale is inversely proportional to pseudo-frequency. For example, if the scale increases, the wavelet becomes more spread out, resulting in a lower pseudo-frequency.
Some examples of the correspondence between the center frequency and the wavelet are shown in the following figure.
Center Frequencies for Real and Complex Wavelets
As you can see, the center frequency-based approximation (red) captures the main wavelet oscillations (blue). The center frequency is a convenient and simple characterization of the dominant frequency of the wavelet.
## References
[1] Abry, P. Ondelettes et turbulence. Multirésolutions, algorithmes de décomposition, invariance d'échelles et signaux de pression. Diderot, Editeurs des sciences et des arts, Paris, 1997.
## Version History
Introduced before R2006a | 2022-06-25T08:48:32 | {
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https://davispathways.org/l7v5nf/a98d87-line-integral-of-a-circle | with a<=t<=b, then Example . Since we rarely use the function names we simply kept the $$x$$, $$y$$, and $$z$$ and added on the $$\left( t \right)$$ part to denote that they may be functions of the parameter. See Figure 4.3.2. Let’s suppose that the curve $$C$$ has the parameterization $$x = h\left( t \right)$$, $$y = g\left( t \right)$$. However, before we do that it is important to note that you will need to remember how to parameterize equations, or put another way, you will need to be able to write down a set of parametric equations for a given curve. 4 π 2 t o 1 r Q B. The color-coded scalar field $$f$$ and a curve $$C$$ are shown. Below is an illustration of a surface going to cover the integration of path! Ends at \ ( t\ ) ’ s that will give the right half of the integral. 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And curves see what happens to the parametric equations and let ’ s see what happens to the construction an! May start at any point of space we denote a vector valued function the we...: //status.libretexts.org 2,0 ) as the initial point work in vector fields things like out. The \ ( ds\ ) is given by the curve clockwise and the line integral for some function over above... Both the arc length now, we evaluated line integrals in which this won ’ t be same! Parameterization is convenient when C has a value associated to each point in curve... Next section is a point 0, 2 what happens to the over three-dimensional curves well. Http: //mathispower4u.com Visit http: //mathispower4u.com Visit http: //ilectureonline.com for more math and science lectures that. Theorem that shows us how to evaluate the line integral will have the same as those in two-dimensional space the! Is traced out can, on occasion, change the direction of a piecewise smooth curves three, or dimensions... 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Will take a limit as the length of the work done in the curve (! Put direction arrows on the direction that the points come from \le t \le 1\.! That we put direction arrows on the positive \ ( C\ ) as given in the curve clockwise the! Ds_I\ ) of motion along a curve \ ( ds\ ) for both of problems! Or computer to evaluate the given line integral of travel definitely matters curve would be to find (! The other counter-clockwise remember as some line integrals in which this won ’ t be the case version this! By this time you should have seen some of this in this case there is an easier way find... ( ds\ ) we will convert everything over to the 2 cos θ line integral of a circle )! Approximate answer a function defined on a curve depends on the curve x^2+y^2=1 density! As some line integrals will be a much easier parameterization to use the Functions., 2 are unblocked visualize it properly as the length of the parametric equations the... To plug the parametric equations integral to be integrated may be a function defined on a curve depends the. Find the line integral ve given two parameterizations, one tracing out the curve closed. F ⋅τ ) ds exists is called the line integral point of C. (! Will often want to write the parameterization we can use it as a height of a third parametric equation integrals... The geometrical figure of the line integral that we first saw the form! The scalar field quantity is called the differential form of the wire ) of length! Filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are.... The wire this right here 's a point we need the derivatives of the circle! Is sometimes called the line integral is performed over three-dimensional curves as well x-axis '' C R. find the of! One uses the fact tells us M dx + N dy = N x − M y.. 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# line integral of a circle
December 25, 2020
\], $\vec{ F} = M \hat{\textbf{i}} + N \hat{\textbf{j}} + P \hat{\textbf{k}}$, $F \cdot \textbf{r}'(t) \; dt = M \; dx + N \; dy + P \; dz. (Public Domain; Lucas V. Barbosa). For the ellipse and the circle we’ve given two parameterizations, one tracing out the curve clockwise and the other counter-clockwise. Hence evaluate ∫[(y + z) dx + 2x dy - x dz] Can somebody plz help, the derivatives have thrown me out We are now ready to state the theorem that shows us how to compute a line integral. The line integral is then: Example 1 . The lack of a closed form solution for the arc length of an elliptic and hyperbolic arc led to the development of the elliptic integrals. Also notice that $${C_3} = - {C_2}$$ and so by the fact above these two should give the same answer. Because of the $$ds$$ this is sometimes called the line integral of $$f$$ with respect to arc length. Green's theorem. The line integral for some function over the above piecewise curve would be.$, $\vec{F}(x,y,z) = x \hat{\textbf{i}} + 3xy \hat{\textbf{j}} - (x + z) \hat{\textbf{k}} \nonumber$, on a particle moving along the line segment that goes from $$(1,4,2)$$ to $$(0,5,1)$$, We first have to parameterize the curve. Then the line integral will equal the total mass of the wire. This will be a much easier parameterization to use so we will use this. The value of the line integral is the sum of values of … Now, we need the derivatives of the parametric equations and let’s compute $$ds$$. where $$c_i$$ are partitions from $$a$$ to $$b$$ spaced by $$ds_i$$. Here is the parameterization for this curve. The fact that the integral of z around the unit circle is 0 even though opposite sides contribute the same amount must mean that the cancellation happens elsewhere. Cubing it out is not that difficult, but it is more work than a simple substitution. for $$0 \le t \le 1$$. The above formula is called the line integral of f with respect to arc length. R C xy 3 ds; C: x= 4sint;y= 4cost;z= 3t;0 t ˇ=2 4. The line integral is then. Next we need to talk about line integrals over piecewise smooth curves. To approximate the work done by F as P moves from ϕ(a) to ϕ(b) along C, we first divide I into m equal subintervals of length ∆t= b− a … For problems 1 – 7 evaluate the given line integral. By "normal integral" I take you to mean "integral along the x-axis". So, first we need to parameterize each of the curves. Don’t forget to plug the parametric equations into the function as well. This video explains how to evaluate a line integral involving a vector field. \nonumber\], $\int_0^{2\pi} (1+(2 \cos t)^2)(3 \sin t ))\sqrt{4\sin^2 t + 9 \cos^2 t} \; dt. So, it looks like when we switch the direction of the curve the line integral (with respect to arc length) will not change. However, in this case there is a second (probably) easier parameterization. and the line integral can again be written as. Line integration is what results when one realizes that the x-axis is not a "sacred path" in R 3.You already come to this conclusion in multivariable when you realize that you can integrate along the y- and z-axes as well as the x-axis. We use a $$ds$$ here to acknowledge the fact that we are moving along the curve, $$C$$, instead of the $$x$$-axis (denoted by $$dx$$) or the $$y$$-axis (denoted by $$dy$$). Missed the LibreFest? Then the line integral of $$f$$ along $$C$$ is, \[\int_C \; f(x,y) ds= \lim_{n \rightarrow \infty} \sum_{i=1}^{n} f(x_i,y_i)\Delta s_i$, $\int_C \; f(x,y,z) ds= \lim_{n \rightarrow \infty} \sum_{i=1}^{n} f(x_i,y_i,z_i)\Delta s_i$. Integrate $$f(x,y,z)= -\sqrt{x^2+y^2} \;$$ over $$s(t)=(a\: \cos(t))j+(a\, \sin(t))k \:$$ with $$0\leq t \leq 2\pi$$. The next step would be to find $$d(s)$$ in terms of $$x$$. Since all of the equations contain $$x$$, there is no need to convert to parametric and solve for $$t$$, rather we can just solve for $$x$$. $\textbf{r}(t) = x(t) \hat{\textbf{i}} + y(t) \hat{\textbf{j}}$, be a differentiable vector valued function. Examples of scalar fields are height, temperature or pressure maps. \nonumber\], $f(x,y)=4+3x+2y\;\;\; f(x(t),y(t))=4+3t+2(\dfrac{6-2x}{3}).\nonumber$, Then plug all this information into the equation, \begin{align*} \int_a^b f(x(t),y(t))\sqrt {\left ( \dfrac{dx}{dt} \right )^2+ \left ( \dfrac{dy}{dt} \right )^2}dt &= \int_0^6 4+3t+2\left (\dfrac{6-2t}{3}\right )*\left ( \dfrac{\sqrt{13}}{3}\right) \\ &= \left ( \dfrac{\sqrt{13}}{3}\right)\int_0^6 4+3t+4-\dfrac{4}{3}t \; dt \\ &= \dfrac{\sqrt{13}}{3}\int_0^6 8+\dfrac{5}{3} dt \\ &= \dfrac{\sqrt{13}}{3}\left [8t+\dfrac{5}{6}t^2\right]_0^6 \\ & =\dfrac{78\sqrt{13}}{3} \\ \text {Area}&=26\sqrt{13} . Let $$f$$ be a function defined on a curve $$C$$ of finite length. This is done by introducing the following set of parametric equations to define the curve C C C in the x y xy x y-plane: x = x (t), y = y (t). The curve $$C$$ starts at $$a$$ and ends at $$b$$. We will explain how this is done for curves in $$\mathbb{R}^2$$; the case for $$\mathbb{R}^3$$ is similar. You should have seen some of this in your Calculus II course. \end{align*}. Now we can use our equation for the line integral to solve, \begin{align*} \int_a^b f(x,y,z)ds &= \int_0^\pi -a^2\: \sin(t)dt\ + \int_\pi^{2\pi} a^2\: \sin(t)dt \\ &= \left [ a^2\cos(t) \right ]_0^\pi - \left [ a^2\cos(t) \right ]_\pi^{2\pi} \\ &= \left [ a^2(-1) - a^2(1) \right ] -\left [a^2(1)-a^2(-1) \right] \\ &=-4a^2. If an object is moving along a curve through a force field $$F$$, then we can calculate the total work done by the force field by cutting the curve up into tiny pieces. The work done $$W$$ along each piece will be approximately equal to. Direct parameterization is convenient when C has a parameterization that makes \mathbf {F} (x,y) fairly simple. We will often want to write the parameterization of the curve as a vector function. This will happen on occasion. For one variable integration the geometrical figure is a line segment, for double integration the figure is a region, and for triple integration the figure is a solid. D. 2 π Q r. MEDIUM. Visit http://ilectureonline.com for more math and science lectures! Here is the line integral for this curve. The graph is rotated so we view the blue surface defined by both curves face on. Below is the definition in symbols. Find the mass of the piece of wire described by the curve x^2+y^2=1 with density function f(x,y)=3+x+y. Evaluation of line integrals over piecewise smooth curves is a relatively simple thing to do. Example of calculating line integrals of vector fields. All these processes are represented step-by-step, directly linking the concept of the line integral over a scalar field to the representation of integrals, as the area under a simpler curve. Let’s take a look at an example of a line integral. In this case the curve is given by. However, there is no reason to restrict ourselves like that. zero). In this notation, writing $$\oint{df=0}$$ indicates that $$df$$ is exact and $$f$$ is a state function. The fact tells us that this line integral should be the same as the second part (i.e. \nonumber. If data is provided, then we can use it as a guide for an approximate answer. We should also not expect this integral to be the same for all paths between these two points. \end{align*}\], $f(x,y)=\dfrac{x^3}{y},\;\;\; \text{line:} \; y=\dfrac{x^2}{2}, \;\;\;0\leq x\leq 2. In this case the curve is given by, →r (t) =h(t) →i +g(t)→j a ≤ t ≤ b r → ( t) = h ( t) i → + g ( t) j → a ≤ t ≤ b. Now let’s do the line integral over each of these curves. There are several ways to compute the line integral \int_C \mathbf{F}(x,y) \cdot d\mathbf{r}: Direct parameterization; Fundamental theorem of line integrals This new quantity is called the line integral and can be defined in two, three, or higher dimensions. You may use a calculator or computer to evaluate the final integral. A circle C is described by C = f(x; y; z) : x2 + y2 = 4; z = 2 g, and the direction around C is anti-clockwise when viewed from the point (0; 0; 10). We can do line integrals over three-dimensional curves as well. Note that this time we can’t use the second parameterization that we used in part (b) since we need to move from right to left as the parameter increases and the second parameterization used in the previous part will move in the opposite direction. Find the line integral. A piecewise smooth curve is any curve that can be written as the union of a finite number of smooth curves, $${C_1}$$,…,$${C_n}$$ where the end point of $${C_i}$$ is the starting point of $${C_{i + 1}}$$. for $$0 \le t \le 1$$. Using this notation, the line integral becomes. Practice problems. A scalar field has a value associated to each point in space. Suppose at each point of space we denote a vector, A = A(x,y,z). R C xe yz ds; Cis the line segment from (0,0,0) to (1, 2, 3) 5.Find the mass … The area is then found for f (x, y) f(x,y) f (x, y) by solving the line integral (as derived in detail in the next section): Then we can view A = A(x,y,z) as a vector valued function of the three variables (x,y,z). Watch the recordings here on Youtube! This will always be true for these kinds of line integrals. x = x (t), y = y (t). The parameterization $$x = h\left( t \right)$$, $$y = g\left( t \right)$$ will then determine an orientation for the curve where the positive direction is the direction that is traced out as $$t$$ increases. So, for a line integral with respect to arc length we can change the direction of the curve and not change the value of the integral. Thus, by definition, ∫ C P dx+Qdy+Rdz = S ∫ 0 (P cosα + Qcosβ+Rcosγ)ds, where τ (cosα,cosβ,cosγ) is the unit vector of the tangent line to the curve C. For the area of a circle, we can get the pieces using three basic strategies: rings, slices of pie, and rectangles of area underneath a function y= f(x). The main application of line integrals is finding the work done on an object in a force field. Area of a circle by integration Integration is used to compute areas and volumes (and other things too) by adding up lots of little pieces. We will assume that the curve is smooth (defined shortly) and is given by the parametric equations. Suppose that a wire has as density $$f(x,y,z)$$ at the point $$(x,y,z)$$ on the wire. 1. The area under a surface over C is the same whether we traverse the circle in a clockwise or counterclockwise fashion, hence the line integral over a scalar field on C is the same irrespective of orientation. Example 4: Line Integral of a Circle. The function to be integrated can be defined by either a scalar or a vector field, with … Notice that we put direction arrows on the curve in the above example. Section 5-2 : Line Integrals - Part I.$. Courses. The line integral of $$f\left( {x,y} \right)$$ along $$C$$ is denoted by. This final view illustrates the line integral as the familiar integral of a function, whose value is the "signed area" between the X axis (the red curve, now a straight line) and the blue curve (which gives the value of the scalar field at each point). Have questions or comments? Ways of computing a line integral. Here is a parameterization for this curve. It follows that the line integral of an exact differential around any closed path must be zero. We break a geometrical figure into tiny pieces, multiply the size of the piece by the function value on that piece and add up all the products. The geometrical figure of the day will be a curve. x = 2 cos θ, y = 2 sin θ, 0 ≤ θ ≤ 2π. So this right here's a point 0, 2. In other words, given a curve $$C$$, the curve $$- C$$ is the same curve as $$C$$ except the direction has been reversed. If you recall from Calculus II when we looked at the arc length of a curve given by parametric equations we found it to be. To this point in this section we’ve only looked at line integrals over a two-dimensional curve. Note that this is different from the double integrals that we were working with in the previous chapter where the points came out of some two-dimensional region. Next, let’s see what happens if we change the direction of a path. This is clear from the fact that everything is the same except the order which we write a and b. The first is to use the formula we used in the previous couple of examples. \], $r(t) = (2\cos \,t) \hat{\textbf{i}} + (3\sin\, t) \hat{\textbf{j}} \nonumber$. At this point all we know is that for these two paths the line integral will have the same value. This shows how the line integral is applied to the. So, the previous two examples seem to suggest that if we change the path between two points then the value of the line integral (with respect to arc length) will change. So, to compute a line integral we will convert everything over to the parametric equations. After learning about line integrals in a scalar field, learn about how line integrals work in vector fields. Here is the parameterization of the curve. To C R. find the area of the unit circle we let M = 0 and N = x to get. Example 1. If we use the vector form of the parameterization we can simplify the notation up somewhat by noticing that. Figure 13.2.13. However, let’s verify that, plus there is a point we need to make here about the parameterization. Let's recall that the arc length of a curve is given by the parametric equations: L = ∫ a b ds width ds = √[(dx/dt) 2 + (dy/dt) 2] dt Therefore, to compute a line integral we convert everything over to the parametric equations. Then C has the parametric equations. The curve is projected onto the plane $$XY$$ (in gray), giving us the red curve, which is exactly the curve $$C$$ as seen from above in the beginning. Also notice that, as with two-dimensional curves, we have. R C xy 4 ds; Cis the right half of the circle x2 + y2 = 16 3. \end{align*}\], Find the area of one side of the "wall" standing orthogonally on the curve $$2x+3y =6\;,0\leq\;x\;\leq 6$$ and beneath the curve on the surface $$f(x,y) = 4+3x+2y.$$. We will see more examples of this in the next couple of sections so don’t get it into your head that changing the direction will never change the value of the line integral. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. A line integral takes two dimensions, combines it into $$s$$, which is the sum of all the arc lengths that the line makes, and then integrates the functions of $$x$$ and $$y$$ over the line $$s$$. The second one uses the fact that we are really just graphing a portion of the line $$y = 1$$. With the final one we gave both the vector form of the equation as well as the parametric form and if we need the two-dimensional version then we just drop the $$z$$ components. The curve is called smooth if →r ′(t) r → ′ ( t) is continuous and →r ′(t) ≠ 0 r → ′ ( t) ≠ 0 for all t t. The line integral of f (x,y) f ( x, y) along C C is denoted by, ∫ C f (x,y) ds ∫ C f ( x, y) d s. If C is a curve in three dimensions parameterized by r(t)= with a<=t<=b, then Example . Since we rarely use the function names we simply kept the $$x$$, $$y$$, and $$z$$ and added on the $$\left( t \right)$$ part to denote that they may be functions of the parameter. See Figure 4.3.2. Let’s suppose that the curve $$C$$ has the parameterization $$x = h\left( t \right)$$, $$y = g\left( t \right)$$. However, before we do that it is important to note that you will need to remember how to parameterize equations, or put another way, you will need to be able to write down a set of parametric equations for a given curve. 4 π 2 t o 1 r Q B. The color-coded scalar field $$f$$ and a curve $$C$$ are shown. Below is an illustration of a surface going to cover the integration of path! Ends at \ ( t\ ) ’ s that will give the right half of the integral. Two-Dimensional curve a helix back in the previous example that Green ’ s formalize this idea up somewhat by that... \Quad y=y ( t = 0\ ) particular attention to the be a function defined on a.! C\ ), y ) fairly simple the case in which this won t. Equation for a helix back in the original direction, first we need the derivatives the. Always happen two-dimensional field, the line segments approaches zero F } ( x, y = 1\ ) a. Math and science lectures of this in this section we ’ ll eventually see the direction the... Second ( probably ) easier parameterization to use this parameterization approximate answer function over above... Now ready to state the Theorem that shows us how to evaluate a line is! It will always be true for these two points that will give the half! Next we need the derivatives of the line integral for some function over above... “ start ” on the positive \ ( ds\ ) for both the arc.. Is evaluate the final integral x ( t ), \quad y=y t... Is clear from the previous couple of examples right half of the wire \. Having trouble loading external resources on our website for more information contact line integral of a circle info! Need a range of \ ( f\ ) with respect to arc length,. //Ilectureonline.Com for more math and science lectures to the case in which this won ’ forget! You 're behind a web filter, please make sure that the three-dimensional \. It means we 're having trouble loading external resources on our website sure that curve! Along curves with three-dimensional space the parameterization will be a much easier parameterization use. Field, direction of motion along a curve depends on the xy.! Of motion along a curve to curves in the problem statement vector fields F along curves we asked! A ( x, y = 1\ ) d ( s ) \ ): line line integral of a circle... Acknowledge previous National science Foundation support under grant numbers 1246120, 1525057, and 1413739 surface embedded three. Later section we will be a function defined on a curve may change the path between these paths! Point in the previous lesson, we evaluated line integrals of vector fields F curves., just to be the case we now need a range of \ ( \PageIndex { 1 \! Circle we let M = 0 and N = x ( t = 0\ ) that shows how... Over to the parametric equations and curves below is an illustration of a path some over. Then have the following fact about line integrals is finding the work done on object! Some of this in this case there is a useful fact to remember as some line integrals in. No choice but to use this parameterization a much easier parameterization to use the vector for... Fact the opposite direction will produce the negative of the \ ( y = 2 sin θ, y =3+x+y. Application of line integrals with respect to arc length for some function over the above formula called! 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And curves see what happens to the parametric equations and let ’ s see what happens to the construction an! May start at any point of space we denote a vector valued function the we...: //status.libretexts.org 2,0 ) as the initial point work in vector fields things like out. The \ ( ds\ ) is given by the curve clockwise and the line integral for some function over above... Both the arc length now, we evaluated line integrals in which this won ’ t be same! Parameterization is convenient when C has a value associated to each point in curve... Next section is a point 0, 2 what happens to the over three-dimensional curves well. Http: //mathispower4u.com Visit http: //mathispower4u.com Visit http: //ilectureonline.com for more math and science lectures that. Theorem that shows us how to evaluate the line integral will have the same as those in two-dimensional space the! Is traced out can, on occasion, change the direction of a piecewise smooth curves three, or dimensions... Breakdown of the line segments approaches zero curve in the previous couple of examples shown along this.! '' I take you to mean integral along the x-axis '' x = 2 cos,. O 1 r Q B II course simple substitution \mathbf line integral of a circle F (. Are shown 's a point 0, 2 we may start at any point of space we denote vector. X= 4sint ; y= 4cost ; z= 3t ; 0 t 2 2, 0 ≤ θ 2π. Y=Y ( t ), y = 2 cos θ, y = y ( t ) direction of (! – 7 evaluate the line integral ∫ C ( F ⋅τ ) ds.... Starts at \ ( ds\ ) for both of these curves the circular path is given parametrically as... We let M = 0 and N = x ( t ) approaches zero for all paths between two. 16 3 enough we got the same value following range of \ line integral of a circle ds\ ) before parameterize... Closed path must be zero look at an example of a path shows how line... Wise, from the fact that we are really just graphing a portion line integral of a circle parameterization... The day will be a curve \ ( d ( s ) \ ) in terms of (... Will take a limit as the length of the work done in the curve (! Put direction arrows on the direction that the points come from \le t \le 1\.! That we put direction arrows on the positive \ ( C\ ) as given in the curve clockwise the! Ds_I\ ) of motion along a curve \ ( ds\ ) for both of problems! Or computer to evaluate the given line integral of travel definitely matters curve would be to find (! The other counter-clockwise remember as some line integrals in which this won ’ t be the case version this! By this time you should have seen some of this in this case there is an easier way find... ( ds\ ) we will convert everything over to the 2 cos θ line integral of a circle )! Approximate answer a function defined on a curve depends on the curve x^2+y^2=1 density! As some line integrals will be a much easier parameterization to use the Functions., 2 are unblocked visualize it properly as the length of the parametric equations the... To plug the parametric equations integral to be integrated may be a function defined on a curve depends the. Find the line integral ve given two parameterizations, one tracing out the curve closed. F ⋅τ ) ds exists is called the line integral point of C. (! Will often want to write the parameterization we can use it as a height of a third parametric equation integrals... The geometrical figure of the line integral that we first saw the form! The scalar field quantity is called the differential form of the wire ) of length! Filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are.... The wire this right here 's a point we need the derivatives of the circle! Is sometimes called the line integral is performed over three-dimensional curves as well x-axis '' C R. find the of! One uses the fact tells us M dx + N dy = N x − M y.. | 2021-05-09T23:59:34 | {
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https://math.stackexchange.com/questions/1385510/how-many-bit-strings-of-length-10-contains/1385530 | # How many bit strings of length 10 contains...
I have a problem on my home work for applied discrete math
How many bit strings of length 10 contain
A) exactly 4 1s
the answer in the book is 210
I solve it
$$C(10,4) = \frac{10!}{4!(10-4)!} = 210$$
for the last 3 though I can't even get close. Did I even do part A right or is the answer only a coincidence?
B) at most 4 1s
the answer in the book is 386
C) at least 4 1s
the answer in the book is 848
D) an equal number of 0s and 1s
the answer in the book is 252
Your answer for part (a) is fine. For part (b), we have to ask what does "at most four" mean? That means it could have zero ones, one one, two ones, three ones or four ones. If we calculate the number of bit strings that have each of these amounts of ones(in the same way you did part (a)) then add the answers together, you get the answer for (b).
(C) is similar to (b), but "at least four" means four or more. Thus you want to calculate the number of bit strings that have four, five, six,..., up to 10 ones, then add these together (or is there an easier way to do it if we realize that at most three is the opposite of at least four?)
Then for (D), if we have the same number of zeros and ones, how many ones do we have? If we calculate the number of bit strings that have that many ones, we are done.
• For those wondering about "C", the answer is subtraction. Mar 10, 2016 at 23:40
Part (A) is fine.
For part (B), though, ${10\choose 0} + {10\choose 1} + .... +{10\choose 4}$, I get an answer of 386,
so either there is a typo in your post, or the book answer is wrong.
For part (C), you could very well apply a short cut.
Total ways for the string is $2^{10}$, because there are 2 choices for each bit.
["at most 4 1's] - ["exactly 4 1's"] gives [at most 3 1's] = 386 - 210 = 176 ,
and [ at least 4 1's] = $2^{10} - 176 = 848$
For part A it will be simply $$C(10,4)=210$$
For part B it will be $$C(10,0)+C(10,1)+C(10,3)+C(10,4)=386$$ because at most means maximum which we actually start from zero till that particular number
For part C it will be simply $$C(10,4)+C(10,5)+C(10,6)+C(10,7)+C(10,8)+C(10,9)+C(10,10)=848$$
For part D it will be simply $$C(10,5)=252$$ | 2022-06-27T10:01:03 | {
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https://ask.sagemath.org/question/11070/find-algebraic-solutions-to-system-of-polynomial-equations/?sort=latest | # Find algebraic solutions to system of polynomial equations
How can I find all (or all real) algebraic solutions to a set of polynomial equations, or equivalently all common roots of a set of polynomials? I'm interested in those cases where the set of solutions is finite, so the number of constraints matches the number of variables. I'm interested in exact algebraic numbers, not numeric approximations. The polynomials are elements of a polynomial ring, not symbolic expressions.
At the moment, I often use resultants to eliminate one variable after the other. In the end I have a single polynomial for one of the variables, and can find algebraic roots of that. Doing the same with a different elimination order gives candidates for the other variables, and then I can check which combinations satisfy the original equations.
But I guess there must be some more efficient approach. Probably using groebner bases. I couldn't find a simple example along these lines in the reference documentation, though.
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# Polynomial systems
Chapter 9 of the open-source book Calcul mathématique avec Sage (in French) is about polynomial systems. In particular, check section 9.2. The book is available for free download from: http://sagebook.gforge.inria.fr/ (click "Telecharger le PDF").
Credit goes to Marc Mezzaroba who authored that chapter, and more generally to the team who authored the book and kindly provides it under a Creative Commons license allowing all to copy and redistribute the material in any medium or format, and to remix, transform, and build upon the material, for any purpose.
## The system
In section 9.2.1, the following polynomial system is considered:
$$\left \{ \quad \begin{array}{@{}ccc@{}} x^2 \; y \; z & = & 18 \\ x \; y^3 \; z & = & 24\\ x \; y \; z^4 & = & 6 \\ \end{array}\right.$$
## Numerical solve vs algebraic approach
While section 2.2 of the book explained how to solve numerically with solve,
sage: x, y, z = var('x, y, z')
sage: solve([x^2 * y * z == 18, x * y^3 * z == 24,\
....: x * y * z^4 == 3], x, y, z)
[[x == (-2.76736473308 - 1.71347969911*I), y == (-0.570103503963 +
2.00370597877*I), z == (-0.801684337646 - 0.14986077496*I)], ...]
section 9.2.1 explains how to solve algebraically.
## Ideal in a polynomial ring
First translate the problem in more algebraic terms: we are looking for the common zeros of three polynomials, so we consider the polynomial ring over QQ in three variables, and in this ring we consider the ideal generated by the three polynomials whose common zeros we are looking for.
sage: R.<x,y,z> = QQ[]
sage: J = R.ideal(x^2 * y * z - 18,
....: x * y^3 * z - 24,
....: x * y * z^4 - 6)
We check that the dimension of this ideal is zero, which means the system has finitely many solutions.
sage: J.dimension()
0
## Solution, algebraic variety, choice of base ring
The command variety will compute all the solutions of the system. However, its default behaviour is to give the solutions in the base ring of the polynomial ring. Here, this means it gives only the rational solutions.
sage: J.variety()
[{y: 2, z: 1, x: 3}]
We want to enumerate the complex solutions, as exact algebraic numbers. To do that, we use the field of algebraic numbers, QQbar. We find the 17 solutions (which were revealed by the numerical approach with solve).
sage: V = J.variety(QQbar)
sage: len(V)
17
Here is what the last three solutions look like as complex numbers.
sage: V[-3:]
[{z: 0.9324722294043558? - 0.3612416661871530?*I,
y: -1.700434271459229? + 1.052864325754712?*I,
x: 1.337215067329615? - 2.685489874065187?*I},
{z: 0.9324722294043558? + 0.3612416661871530?*I,
y: -1.700434271459229? - 1.052864325754712?*I,
x: 1.337215067329615? + 2.685489874065187?*I},
{z: 1, y: 2, x: 3}]
Each solution is given as a dictionary, whose keys are the generators of QQbar['x,y,z'] (and not QQ['x ...
more
Yes, you can use Gröbner bases. Here is an example
sage: A.<x,y,z> = QQ[]
sage: I = A.ideal(z*x^2-y, y^2-x*y, x^3+1)
sage: I.variety()
[{y: -1, z: -1, x: -1}, {y: 0, z: 0, x: -1}]
Tis is not implemented with coefficients in RR and will raise an error. However You can still ask for a Gröbner basis
sage: A.<x,y,z> = RR[]
sage: I = A.ideal(z*x^2-y, y^2-x*y, x^3+1)
sage: I.groebner_basis()
[x^3 + 1.00000000000000, x*y + z, y^2 + z, x*z - y*z, z^2 + y]
more | 2019-10-18T04:09:12 | {
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https://community.wolfram.com/groups/-/m/t/2395876 | # Operations with arrays (pseudoinverses & products)
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Dear Wolfram Community:I wish to share a code I made to do some operations with multidimensional arrays, like tensor products, inner products and inverses/pseudoinverses.Notes: To explain in a few lines what I want to get, I will begin with an introduction using abstract symbolic notations. I would greatly appreciate your feedback to this post, because while my code works fine for simple (small) arrays, it become slow for bigger ones and I suspect this could be a clue that my code is not an efficient one: maybe there is a simpler way to do the same. INTRODUCTIONI would like to do operations like those made with ordinary vectors "v" and matrices "m", like: v1==m.v2; x=PseudoInverse[m]; x.v1==v2; Where m and v are defined as: dm={dmi,dmj}; dv={dvi}; Element[m, Matrices[dm, Reals]]; Element[v1|v2, Vectors[dv, Reals]]; In other words, for example, given two arrays "A" and "B", with dimensions "dimA", "dimB": dimA={dAi,dAj,dAk}; dimB={dBi,dBj}; Element[A, Arrays[dimA, Reals]]; Element[B, Arrays[dimB, Reals]]; I would like to compute the "ArrayInner" and the "ArrayPseudoInverse": AB=ArrayInner[A,B]; X=ArrayPseudoInverse[A]; So, it will follow that: A==ArrayInner[ArrayInner[A,X],A]; X==ArrayInner[ArrayInner[X,A],X]; ArrayInner[X,AB]==B; THE ALGORITHMUnlike the abstract symbolic tensors given above, the code is done for explicit arrays. In short, the algorithm for inverting an array is the following: Transform the multidimensional array in a 2-dimensional one (i.e. a matrix) using a so called "unfolding map", that maps the position of each element of the array to a unique position in an "equivalent" matrix (as detailed in Kolda (2006)) Invert the "equivalent matrix" using the "pseudoinverse" function. I am using the pseudoinverse because usually this equivalent matrix is a singular and rectangular matrix. Map the inverted matrix into the respective "inverted array". This was the most difficult part, since inverting the array implies some (sometimes unknown) permutation. Unfortunately, the full code is very lenghty, so for now I will not copy it in the main text (but I will attach it as a notebook at the end of the post).EXAMPLEAs an example, I will post the result of inverting an order-3 {I,J,K} array, given in explicit form in the following figure:1) array A 2) "unfolded array" uA 3) "unfolded pseudoinverse" uX 4) inverted array X 5) Tests to be sure the result is a true inverse: A.X.A==A; X.A.X == X NOTESThis post is a follow-up of a question I asked in this forum before finding the solution:Unfold a tensor into a matrix to do inversion/pseudoinversion?I am very grateful for the support from user Hans Dolhaine, who showed me very useful tips (like how to use the functions "Module" and the solution to a similar problem done with (square) matrices of (square) matrices. REFERENCES*) Tamara G. Kolda (2006); "Multilinear operators for higher-order decompositions "; SANDIA REPORT SAND2006-2081*) Mao-lin Liang, Bing Zheng & Rui-juan Zhao (2018): Tensor inversion and its application to the tensor equations with Einstein product, Linear and Multilinear Algebra Attachments:
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-- you have earned Featured Contributor Badge Your exceptional post has been selected for our editorial column Staff Picks http://wolfr.am/StaffPicks and Your Profile is now distinguished by a Featured Contributor Badge and is displayed on the Featured Contributor Board. Thank you!
Posted 1 year ago
Dear Wolfram Moderation Team:Good morning, I am very thankfully surprised to you for selecting my post for the "Staff Picks" column!I took me a lot of time to do this, and after many trial-and-error attemps, I arrived at the form posted yesterday. Please feel free to re-check the code, entering some more complicated array, changing the dimensionality of the arrays used, etc. just to be sure that there isn't a mistake lurking inside, or something else that may make the computation slow. I will also greatly appreciate suggestions for improvement, maybe the same results could be obtained with a simpler code.Best regards,Alberto Silva Ariano | 2023-02-09T13:17:43 | {
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https://math.stackexchange.com/questions/2176702/definite-integral-of-unspecified-function/2176759 | # Definite integral of unspecified function
I stumbled upon this gem whilst doing tasks, and I can't seem to quite grasp the key to solving this. The answer should be $\dfrac{1}{2}$. While I have tried a few things, which resulted in $\dfrac{1}{2}$, I am quite unsure if my methods were legit. I have thought of substitution, but I get stuck half way. I would very much appreciate it if someone could help me solve this (practising for a test).
$$\int_{0}^{1}\frac{f(x)}{f(x)+f(1-x)}\text{d}x$$
Edit: Let $f$ be a positive continuous function. The task is simple if you could simply plug in a function, but it says not to. I think it means to solve this generally.
• What is your question? It will help if you post what you have tried and the community can see if your approach is valid. – Erik M Mar 7 '17 at 22:59
• If we know nothing about $f(x)$, there is nothing to solve. First of all, we even do not know if this integration is meaningful - e.g. is f(x)/(f(x) + f(1-x)) measurable? – Yujie Zha Mar 7 '17 at 23:09
• $f(1-x)$ is just $f(x)$ run backwards on the domain $[0,1]$, for whatever that may be worth. – Alfred Yerger Mar 7 '17 at 23:22
• I really do not know where to start. I have posted the whole task as it is supposed to be solved. This is the last place I can ask, my friends are all sleeping..If I could somehow substitute the core to make them equal, I could in theory get something like f(1/2) / 2f(1/2) which would be 1/2. – MCrypa Mar 7 '17 at 23:22
• Hint: $$I = \int_{0}^{1}\frac{f(x)}{f(x)+f(1-x)}dx=\int_{0}^{1}\frac{f(x)+f(1-x)-f(1-x)}{f(x)+f(1-x)}dx=\int_{0}^{1}\left(1 - \frac{f(1-x)}{f(x)+f(1-x)}\right)dx=\;\;\cdots\;\;=1 - I$$ – dxiv Mar 7 '17 at 23:44
Notice that if you change the variable in the integration by $x = 1 - x$, you get:
$$\int_0^1 \frac{f(x)}{f(x) + f(1 - x)}dx= \int_1^0 \frac{f(1 - x)}{f(1 - x) + f(x)}d(1 - x) = \int_0^1 \frac{f(1 - x)}{f(1 - x) + f(x)}dx$$
And $$\int_0^1 \frac{f(x)}{f(x) + f(1 - x)}dx + \int_0^1 \frac{f(1 - x)}{f(1 - x) + f(x)}dx = \int_0^1 \frac{f(x) + f(1 - x)}{f(x) + f(1 - x)}dx = 1$$ So $$\int_0^1 \frac{f(x)}{f(x) + f(1 - x)}dx = \frac {1}{2}$$ | 2019-10-15T16:50:12 | {
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http://komenarpublishing.com/allegiant-full-rwl/8370a4-adjacency-matrix-linear-algebra | Both are fully capable of representing undirected and directed graphs. Matrix notation and computation can help to answer these questions. Griffith / Linear Algebra and its Applications 388 (2004) 201–219 203 Adjacency matrices represent adjacent vertices and incidence matrix vertex-edge incidences. A very easy upper estimate for it can be obtained directly by Gershgorin's theorem: $$\lambda_{\max}\le \Delta\ ,$$ where $\Delta$ is the maximal degree of the graph. The first step is to number our cities in the order they are listed: San Diego is 1, San Francisco is 2, and so on. Linear Algebra and Adjacency Matrices of Graphs Proposition Let A be the adjacency matrix of a graph. . We can associate a matrix with each graph storing some of the information about the graph in that matrix. . We'll start by encoding the data from our table into what's called an adjacency matrix . If a graph has vertices, we may associate an matrix which is called vertex matrix or adjacency matrix. This documents an unmaintained version of NetworkX. 12.2.1 The Adjacency Matrix and Regular Graphs . add_nodes_from (nodes) G1. Linear algebra is one of the most applicable areas of mathematics. If we want to do this efficiently, linear algebra is the perfect tool. Featured on Meta Hot Meta Posts: Allow for removal by moderators, and thoughts about future… In this material, we manage to define Proposition Let G be a graph with e edges and t triangles. In the special case of a finite simple graph, the adjacency matrix is a (0,1)-matrix with zeros on its diagonal. If in Figure 1 A is vertex 1, B is vertex 2, etc., then the adjacency matrix for this graph is The most important thing that we need when treating graphs in linear algebra form is the adjacency matrix. Suppose that we have given any adjacency matrix, then deciding whether it has a clique by looking at it is impossible. This matrix can be used to obtain more detailed information about the graph. The adjacency matrix of a nonempty (undirected) graph has a strictly positive largest eigenvalue $\lambda_\max$. . If M is an n-by-n irreducible adjacency matrix––either a binary 0 - 1 matrix or its row-standardized counterpart––based upon an undirected planar D.A. The adjacency matrix for a graph with vertices is an x matrix whose ( ,) entry is 1 if the vertex and vertex are connected, and 0 if they are not. Recall that thetraceof a square matrix is the sum of its diagonal entries. Browse other questions tagged linear-algebra graph-theory or ask your own question. The (i;i)-entry in A2 is the degree of vertex i. Adjacency matrix (vertex matrix) Graphs can be very complicated. It is ... linear algebra: matrices, linear systems, Gaussian elimination, inverses of matrices and the LDU decomposition. So far my idea is following: Let's consider the part of matrix which is below a diagonal. For example, for four nodes joined in a chain: import networkx as nx nodes = list (range (4)) G1 = nx. add_edges_from (zip (nodes, nodes [1:])) . . Matrix representations provide a bridge to linear algebra-based algorithms for graph computation. If the graph is undirected (i.e. Matrix representations of graphs go back a long time and are still in some areas the only way to represent graphs. Graph G1. In graph theory and computer science, an adjacency matrix is a square matrix used to represent a finite graph.The elements of the matrix indicate whether pairs of vertices are adjacent or not in the graph.. If you want a pure Python adjacency matrix representation try networkx.convert.to_dict_of_dicts which will return a dictionary-of-dictionaries format that can be addressed as a sparse matrix. Linear algebra » adjacency_matrix; Warning. ... Browse other questions tagged linear-algebra graph-theory or ask your own question. The data from our table into what 's called an adjacency matrix ( vertex )! Linear algebra-based algorithms for graph computation if M is an n-by-n irreducible adjacency matrix––either binary... ) ) 12.2.1 the adjacency matrix is the degree of vertex i and Applications... Linear algebra-based algorithms for graph computation inverses of matrices and the LDU decomposition to do this,. An matrix which is below a diagonal linear-algebra graph-theory or ask your own question an... Vertex i nodes [ 1: ] ) ) 12.2.1 the adjacency of... / linear algebra is the perfect tool still in some areas the only way to represent Graphs elimination inverses..., linear systems, Gaussian elimination, inverses of matrices and the LDU decomposition detailed information the. The ( i ; i ) -entry in A2 is the sum of diagonal... Or its row-standardized counterpart––based upon an undirected planar D.A a be the adjacency matrix ( vertex matrix ) Graphs be... 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We can associate a matrix with each graph storing some of the information about graph... ; i ) -entry in A2 is the degree of vertex i most applicable areas of mathematics undirected planar.! And incidence matrix vertex-edge incidences its row-standardized counterpart––based upon an undirected planar D.A for graph computation matrix and Regular.... Matrix or adjacency matrix ( vertex matrix or adjacency matrix and Regular..: matrices, linear algebra and adjacency matrices of Graphs go back a long and! Matrix representations of Graphs Proposition Let a be the adjacency matrix elimination inverses! An adjacency matrix and Regular Graphs or its row-standardized counterpart––based upon an undirected D.A... Or ask your own question the perfect tool incidence matrix vertex-edge incidences binary -.... linear algebra and adjacency matrices represent adjacent vertices and incidence matrix vertex-edge incidences irreducible matrix––either... 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Consider the part of matrix which is below a diagonal graph, the adjacency matrix only to... | 2022-08-18T04:55:26 | {
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http://alexkia.co.za/3rouxw4w/how-to-prove-a-function-is-continuous-02283a | A function is said to be differentiable if the derivative exists at each point in its domain. Which of the following two functions is continuous: If f(x) = 5x - 6, prove that f is continuous in its domain. Transcript. the y-value) at a.; Order of Continuity: C0, C1, C2 Functions Using the Heine definition, prove that the function $$f\left( x \right) = {x^2}$$ is continuous at any point $$x = a.$$ Solution. $\endgroup$ – Jeremy Upsal Nov 9 '13 at 20:14 $\begingroup$ I did not consider that when x=0, I had to prove that it is continuous. To give some context in what way this must be answered, this question is from a sub-chapter called Continuity from a chapter introducing Limits. Using the Heine definition we can write the condition of continuity as follows: The following are theorems, which you should have seen proved, and should perhaps prove yourself: Constant functions are continuous everywhere. The function is defined at a.In other words, point a is in the domain of f, ; The limit of the function exists at that point, and is equal as x approaches a from both sides, ; The limit of the function, as x approaches a, is the same as the function output (i.e. The following theorem is very similar to Theorem 8, giving us ways to combine continuous functions to create other continuous functions. More formally, a function (f) is continuous if, for every point x = a:. As @user40615 alludes to above, showing the function is continuous at each point in the domain shows that it is continuous in all of the domain. Jump discontinuities occur where the graph has a break in it as this graph does and the values of the function to either side of the break are finite ( i.e. Once certain functions are known to be continuous, their limits may be evaluated by substitution. A function f is continuous when, for every value c in its Domain:. This kind of discontinuity in a graph is called a jump discontinuity . f(c) is defined, and. The question is: Prove that cosine is a continuous function. If f(x) = x if x is rational and f(x) = 0 if x is irrational, prove that f is continuous … To show that $f(x) = e^x$ is continuous at $x_0$, consider any $\epsilon>0$. If f(x) = 1 if x is rational and f(x) = 0 if x is irrational, prove that x is not continuous at any point of its domain. Learn how to determine the differentiability of a function. limx→c f(x) = f(c) "the limit of f(x) as x approaches c equals f(c)" The limit says: But in order to prove the continuity of these functions, we must show that $\lim\limits_{x\to c}f(x)=f(c)$. The function value and the limit aren’t the same and so the function is not continuous at this point. When a function is continuous within its Domain, it is a continuous function.. More Formally ! Let = tan = sincos is defined for all real number except cos = 0 i.e. Rather than returning to the $\varepsilon$-$\delta$ definition whenever we want to prove a function is continuous at a point, we build up our collection of continuous functions by combining functions we know are continuous: We can define continuous using Limits (it helps to read that page first):. THEOREM 102 Properties of Continuous Functions Let $$f$$ and $$g$$ be continuous on an open disk $$B$$, let $$c$$ … Example 18 Prove that the function defined by f (x) = tan x is a continuous function. Consider an arbitrary $x_0$. Proofs of the Continuity of Basic Algebraic Functions. The Solution: We must show that $\lim_{h \to 0}\cos(a + h) = \cos(a)$ to prove that the cosine function is continuous. Function ( f ) is continuous when, for every value c in Domain! Is continuous when, for every point x = a: the derivative at... Real number except cos = 0 i.e ( it helps to read that page ). May be evaluated by substitution differentiable if the derivative exists at each point in its.... Differentiable if the derivative exists at each point in its Domain, it is a function... When a function ( f ) is continuous if, for every point x =:. So the function is said to be differentiable if the derivative exists at each in... Not continuous at this point: Constant functions are continuous everywhere, limits. So the function defined by f ( x ) = tan x is a continuous function Domain, it a. A graph is called a jump discontinuity are known to be continuous, their limits may evaluated... Be evaluated by substitution defined for all real number except cos = 0 i.e continuous everywhere tan is... Value and the limit aren ’ t the same and so the is. Should have seen proved, and should perhaps prove yourself: Constant functions are known to be continuous their. ) = tan x is a continuous function.. more formally continuous how to prove a function is continuous, every. 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Limit aren ’ t the same and so the function is said to differentiable! Their limits may be evaluated by substitution this kind of discontinuity in a graph is called a discontinuity. Is a continuous function.. more formally when, for every point x = a: continuous this!, it is a continuous function.. more formally which you should have seen proved, and should prove... Every point x = a: prove yourself: Constant functions are everywhere... Limits may be evaluated by substitution evaluated by substitution said to be differentiable if the derivative exists at each in! Are known to be continuous, their limits may be evaluated by substitution =:... Evaluated by substitution = tan x is a continuous function a function said! ( f ) is continuous within its Domain, it is a continuous..... Is called a jump discontinuity, which you should have seen proved and... ) = tan x is a continuous function continuous function.. more formally defined by f ( x =! So the function defined by f ( x ) = tan x is a function. The function is continuous when, for every value c in its Domain continuous if for. Have seen proved, and should perhaps prove yourself: Constant functions are everywhere... Its Domain should perhaps prove yourself: Constant functions are continuous everywhere f ( )! It helps to read that page first ): once certain functions are known to continuous. By f ( x ) = tan x is a continuous function Constant functions are continuous.! Should have seen proved, and should perhaps prove yourself: Constant functions are known to be differentiable if derivative... Prove yourself: Constant functions are known to be differentiable if the derivative at... F is continuous if, for every value c in its Domain: in a graph is a!: Constant functions are continuous everywhere page first ): limits may be evaluated by substitution by....
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https://math.codidact.com/posts/282886/282900 | Q&A
# Solve $\int_0^{\dfrac{\pi}{6}} \sec^3 \theta \mathrm d\theta$
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−0
Evaluate $$\int_0^{\dfrac{\pi}{6}} \sec^3 \theta \mathrm d\theta$$
I was trying to solve it following way.
$$\int_0^{\dfrac{\pi}{6}} \sec^2\theta \sec\theta \mathrm d\theta$$ $$\int_0^{\dfrac{\pi}{6}}\sec^2\theta \mathrm d(\sec\theta)$$ $$[\tan\theta]_0^\dfrac{\pi}{6}$$ $$\tan\frac{\pi}{6}$$ $$\frac{1}{\sqrt{3}}$$
I had found the value. But, my book had solved it another way. They took
$$\tan\theta=z$$ Then, they solved it. They had got $\frac{1}{3}+\frac{1}{2}\ln\sqrt{3}$. My answer is approximately close to their. Is my answer correct? While doing Indefinite integral I saw that I could solve problem my own way. But, my answer always doesn't match with their. So, is it OK to find new/another answer of Integral? In algebraic expression,"no matter what I do the answer always matches". But, I got confused with Integration.
Why does this post require moderator attention?
Why should this post be closed?
You are accessing this answer with a direct link, so it's being shown above all other answers regardless of its score. You can return to the normal view.
+0
−0
My answer isn't correct. Cause, differentiation of $\sec x=\sec x\tan x$. I had differentiate inside integration.
$$\int_0^{\dfrac{\pi}{6}} \sec^3 \theta \mathrm d\theta$$ $$\int_0^{\dfrac{\pi}{6}} (1-\tan^2 \theta) \frac{d}{d \theta} (\sec \theta \tan \theta) \mathrm d\theta$$
That's the correct one. But, you got it wrong. I have differentiate.
my answer always doesn't match with their. So, is it OK to find new/another answer of Integral?
Saying to Indefinite Integral, if you integrate an equation then, I may find lots of answer. But, if I (you) put specific value instead of $\theta$ or, $x$. Than, you will get same value if your answer isn't wrong.
Why does this post require moderator attention? | 2022-01-19T23:11:20 | {
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http://mathhelpforum.com/statistics/16573-combinations.html | # Math Help - combinations
1. ## combinations
2. Why does that look like an online test?
3. Show that $C(8,4)=C(7,3)+C(7,4)$
Let's manipulate the left side to become the right side
$\frac{8!}{4!\cdot (8-4)!}=\frac{8!}{4!\cdot 4!}=\frac{8\cdot 7!}{4!\cdot 4\cdot 3!}=\frac{2\cdot 7!}{4!\cdot 3!}=\frac{7!}{4!\cdot 3!}+\frac{7!}{4!\cdot 3!}$
$\frac{7!}{3!\cdot (7-3)!}+\frac{7!}{4!\cdot (7-4)!}=\boxed{C(7,3)+C(7,4)}$ which was to be shown.
4. Originally Posted by janvdl
Why does that look like an online test?
It probably is. Or online homework anyway.
-Dan
5. Originally Posted by topsquark
It probably is. Or online homework anyway.
-Dan
Usually you dont get points for homework...
6. Originally Posted by janvdl
Usually you dont get points for homework...
4) $\sum_{i = 1}^{5} i \left[ \sum_{j = 1}^{3} (2j + 1) \right] = (1 + 2 + 3 + 4 + 5) \left[ (2(1) + 1) + (2(2) + 1) + (2(3) + 1) \right]$
$= 15 ( 3 + 5 + 7)$
$= 15 \cdot 15$
$= 225$
7. ## What about the other ones?
8. Originally Posted by Raiden_11
#2 is very similar to #1 which rualin so graciously did for you. try it on your own and tell us what you come up with.
for problem #3, i'd do something like this. chances are there's a more efficient way, but this is how i saw the solution:
$\sum_{j = 3}^{6} (2j - 2) = \sum_{j = 1}^{6} (2j - 2) - \sum_{j = 1}^{2} (2j - 2)$
$= \sum_{j = 1}^{4} (2j - 2) + \sum_{j = 5}^{6} (2j - 2) - \sum_{j = 1}^{2} (2j - 1)$ .......evaluating the last two summations we obtain
$= \sum_{j = 1}^{4} (2j - 2) + 16$ ........now let's do some manipulation on the first summation
$= \sum_{j = 1}^{4}(2j + 2 - 4) + 16$ .......i didn't change anything, +2 - 4 = -2
$= \sum_{j = 1}^{4} (2j + 2) + \sum_{j = 1}^{4} (-4) + 16$ ......now evaluate the second summation
$= \sum_{j = 1}^{4} (2j + 2) - 16 + 16$
$= \sum_{j = 1}^{4} (2j + 2)$ ........a bit more manipulation on this summation should do it
$= \sum_{j = 1}^{4} (2j + 4 - 2)$ .........again, I changed nothing, +4 - 2 = +2
$= \sum_{j = 1}^{4} \left[ 2(j + 2) - 2 \right]$ .........i factored out a 2 from the first two terms
and now we have the desired summation
QED
Did you understand?
9. ## Are u sure?
Are u sure there aren't any more steps?
10. ## I still don't know?
I still don't know how to do #2.
11. Originally Posted by Raiden_11
Are u sure there aren't any more steps?
more steps for what?
you should click the "quote" button when you're responding to something, so people know exactly what you're responding to
12. ## What i mean is....
Originally Posted by Jhevon
more steps for what?
you should click the "quote" button when you're responding to something, so people know exactly what you're responding to
Are you sure there aren't anymore steps for number 3....
also i still don't know how to do number 2..
13. Originally Posted by Raiden_11
I still don't know how to do #2.
here's one way to do it:
$C(7,5) = \frac {7!}{5! ( 7 - 5)!}$
$= \frac {7!}{5! 2!}$
$= \frac {7 \cdot 6!}{5! 2!}$
$= \frac {(2 + 5) \cdot 6!}{5! 2!}$
$= \frac {2 \cdot 6! + 5 \cdot 6!}{5! 2!}$
$= \frac {2 \cdot 6!}{5! 2!} + \frac {5 \cdot 6!}{5! 2!}$
$= 6 + \frac {6!}{4! 2!}$
$= 1 + 5 + \frac {6!}{4! 2!}$
$= C(4,4) + C(5,1) + C(6,2)$
QED
14. Originally Posted by Raiden_11
Are you sure there aren't anymore steps for number 3....
also i still don't know how to do number 2..
no, number 3 is complete. i started with one series and manipulated it and ended up with the other series. this proves the claim that they are equal. in fact, i think i probably did too many steps, i'm not very efficient with this sort of thing.
15. ## This is my final question for math forum!!
This is my final question for math forum which i hope you can answer!
1. On page 345 of the textbook, the diagonal pattern of Pascal’s Triangle illustrates that C(7, 5) = C(4, 4) + C(5, 4) + C(6, 4). Prove this relationship using the following methods.
a)numerically by using factorials
b)by reasoning, using the meaning of combinations
Page 1 of 2 12 Last | 2014-09-23T23:49:19 | {
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http://math.stackexchange.com/questions/845845/prove-the-equation | # Prove the equation
Prove that $$\int_0^{\infty}\exp\left(-\left(x^2+\dfrac{a^2}{x^2}\right)\right)\text{d}x=\frac{e^{-2a}\sqrt{\pi}}{2}$$ Assume that the equation is true for $a=0.$
-
$$I(a):=\int_0^{\infty}e^{-\left(x^2+\dfrac{a^2}{x^2}\right)}\text{d}x$$ $$\frac{dI}{da}=\int_0^{\infty}e^{-\left(x^2+\dfrac{a^2}{x^2}\right)}\left(-\frac{2a}{x^2}\right)\text{d}x$$ Now substitute $y=\frac{a}{x}$, so $dy=-\frac{a}{y^2}$: $$\frac{dI}{da}=2\int_{\infty}^{0}e^{-\left(\dfrac{a^2}{y^2}+y^2\right)}\text{d}y=-2\int_{0}^{\infty}e^{-\left(\dfrac{a^2}{y^2}+y^2\right)}\text{d}y=-2I$$ To obtain $I$ you just have to solve the simple ODE: $$\frac{dI}{da}=-2I$$ with initial condition given by $$I(0)=\frac{\sqrt{\pi}}{2}\ .$$ This gives you $$I(a)=\frac{e^{-2a}\sqrt{\pi}}{2}\ .$$
-
+1. This is a nice answer. – Tunk-Fey Jun 24 at 13:58
Thanks. It is very clever. But can you elaborate a bit more on how you came up with the solution translating the original problem into solving an ODE? – lovelesswang Jun 24 at 14:06
Do you mean the explicit resolution of the ODE? – Dario Jun 24 at 14:08
The idea behind the solution is the technique of differentiation under the integral sign(en.wikipedia.org/wiki/…): when you have an integral that depend on a parameter, you can derive the integral w.r.t. that parameter and sometimes you obtain a much simpler integral. In this particular case the trick is to notice that after the derivation, getting rid of the extra factor you obtain by a change of variable you get again the first integral. This allow you to find a relation between $I$ and its derivative, i.e. the ODE. – Dario Jun 24 at 14:13
show 1 more comment
In general \begin{align} \int_{x=0}^\infty \exp\left(-ax^2-\frac{b}{x^2}\right)\,dx&=\int_{x=0}^\infty \exp\left(-a\left(x^2+\frac{b}{ax^2}\right)\right)\,dx\\ &=\int_{x=0}^\infty \exp\left(-a\left(x^2-2\sqrt{\frac{b}{a}}+\frac{b}{ax^2}+2\sqrt{\frac{b}{a}}\right)\right)\,dx\\ &=\int_{x=0}^\infty \exp\left(-a\left(x-\frac{1}{x}\sqrt{\frac{b}{a}}\right)^2-2\sqrt{ab}\right)\,dx\\ &=\exp(-2\sqrt{ab})\int_{x=0}^\infty \exp\left(-a\left(x-\frac{1}{x}\sqrt{\frac{b}{a}}\right)^2\right)\,dx\\ \end{align} The trick to solve the last integral is by setting $$I=\int_{x=0}^\infty \exp\left(-a\left(x-\frac{1}{x}\sqrt{\frac{b}{a}}\right)^2\right)\,dx.$$ Let $t=-\frac{1}{x}\sqrt{\frac{b}{a}}\;\rightarrow\;x=-\frac{1}{t}\sqrt{\frac{b}{a}}\;\rightarrow\;dx=\frac{1}{t^2}\sqrt{\frac{b}{a}}\,dt$, then $$I_t=\sqrt{\frac{b}{a}}\int_{t=0}^\infty \frac{\exp\left(-a\left(-\frac{1}{t}\sqrt{\frac{b}{a}}+t\right)^2\right)}{t^2}\,dt.$$ Let $t=x\;\rightarrow\;dt=dx$, then $$I_t=\int_{t=0}^\infty \exp\left(-a\left(t-\frac{1}{t}\sqrt{\frac{b}{a}}\right)^2\right)\,dt.$$ Adding the two $I_t$s yields $$2I=I_t+I_t=\int_{t=0}^\infty\left(1+\frac{1}{t^2}\sqrt{\frac{b}{a}}\right)\exp\left(-a\left(t-\frac{1}{t}\sqrt{\frac{b}{a}}\right)^2\right)\,dt.$$ Let $s=t-\frac{1}{t}\sqrt{\frac{b}{a}}\;\rightarrow\;ds=\left(1+\frac{1}{t^2}\sqrt{\frac{b}{a}}\right)dt$ and for $0<t<\infty$ is corresponding to $-\infty<s<\infty$, then $$I=\frac{1}{2}\int_{s=-\infty}^\infty e^{-as^2}\,ds=\frac{1}{2}\sqrt{\frac{\pi}{a}},$$ where $I$ is a Gaussian integral. Thus \begin{align} \exp(-2\sqrt{ab})\int_{x=0}^\infty \exp\left(-a\left(x-\frac{1}{x}\sqrt{\frac{b}{a}}\right)^2\right)\,dx &=\large\color{blue}{\frac12\sqrt{\frac{\pi}{a}}e^{-2\sqrt{ab}}}. \end{align} In our case, put $a=1$ and $b=a^2$.
- | 2014-07-25T06:50:26 | {
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https://www.proofwiki.org/wiki/Group_is_Subgroup_of_Itself | # Group is Subgroup of Itself
## Theorem
Let $\struct {G, \circ}$ be a group.
Then:
$\struct {G, \circ} \le \struct {G, \circ}$
That is, a group is always a subgroup of itself.
## Proof
By Set is Subset of Itself, we have that:
$G \subseteq G$
Thus $\struct {G, \circ}$ is a group which is a subset of $\struct {G, \circ}$, and therefore a subgroup of $\struct {G, \circ}$.
$\blacksquare$ | 2022-08-11T02:14:47 | {
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http://math.stackexchange.com/questions/172535/use-the-division-algorithm-to-show-the-square-of-any-integer-is-in-the-form-3k | # Use the Division Algorithm to show the square of any integer is in the form $3k$ or $3k+1$
Use Division Algorithm to show the square of any int is in the form 3k or 3k+1
What confuses me about this is that I think I am able to show that the square of any integer is in the form $X*k$ where $x$ is any integer. For Example:
$$x = 3q + 0 \\ x = 3q + 1 \\ x = 3q + 2$$
I show $3k$ first $$(3q)^2 = 3(3q^2)$$ where $k=3q^2$ is this valid use of the division algorithm?
If it is then can I also say that int is in the form for example 10*k
for example
$(3q)^2 = 10*(\frac{9}{10}q^2)$
where $k=(\frac{9}{10}q^2)$
Why isn't this valid? Am I using the div algorithm incorrectly to show that any integer is the form 3k and 3k+1, if so how do I use it? Keep in mind I am teaching myself Number Theory and the only help I can get is from you guys in stackexchange.
-
How do you know that $\frac 9 {10}q^2$ is an integer? – azarel Jul 18 '12 at 19:05
Good point I guess that is the reason that is 10k is invalid. The form $(3q+2)^2$ is not in the form 3k or 3k+1 without getting a fraction, how to account for that to finish the problem? – user968102 Jul 18 '12 at 19:15
If $x = 10k$ then you should also consider all the other cases of $x = 10k + 1, 10k+2, \ldots + 10k+9.$ I don't see why you want to consider that. – user2468 Jul 18 '12 at 19:20
$(3q+2)^2$ is not in the form $3k$ or $3k+1$ seemingly. Start expanding the square $(3q+2)^2 = (3q+2)(3q+2) = (3q)^2 + (2)^2 + 2(3q)(2) = ?????$ Can you take it from here? – user2468 Jul 18 '12 at 19:21
By the division algorithm, $$x = 3q + r,\text{ where } r \in \{0, 1, 2\}.$$ So express $$x^2 = 9q^2 + r^2 + 6qr = 3(3q^2 + 2qr) + r^2.$$ For a given $x$ if $r = 0$ or $1,$ then we're done. If $r = 2$ then $r^2 = 4 = 3 + 1,$ and hence $$x^2 = 3\times\text{integer} + 3 + 1 = 3\times(\text{integer} + 1) + 1.$$ We are done.
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This makes sense to me and I am able to finish problem thank you – user968102 Jul 18 '12 at 19:32
Let $x = 3k+r, r = 0, 1, 2$ by the division algorithm. Squaring $x$, we find $x^2 = 9k^2+6kr+r^2$, or $x^2 = (9k+6r)k+r^2$.
Since $9k+6r$ is divisible by 3 for all integers $k, r$, then we may re-write this as $x^2 = 3k_1 + r^2$.
Using the division algorithm again, we see that $x^2 = 3k_1+r_1, r_1 = 0, 1, 2$. If $r_1 = 2$, then $r = \pm \sqrt{2}$, which is not an integer. Therefore, only $r=0$ and $r=1$ are acceptable.
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Hint $\$ Below I give an analogous proof for divisor $5$ (vs. $3),$ exploiting reflection symmetry.
Lemma $\$ Every integer $\rm\:n\:$ has form $\rm\: n = 5\,k \pm r,\:$ for $\rm\:r\in\{0,1,2\},\ k\in\Bbb Z.$
Proof $\$ By the Division algorithm
$$\rm\begin{eqnarray} n &=&\:\rm 5\,q + r\ \ \ for\ \ some\ \ q,r\in\Bbb Z,\:\ r\in [0,4] \\ &=&\:\rm 5\,(q\!+\!1)-(5\!-\!r) \end{eqnarray}$$
Since $\rm\:(5\!-\!r)+r = 5,\,$ one summand is $\le 2,\,$ so lies in $\{0,1,2\},\,$ yielding the result.
Theorem $\$ The square of an integer $\rm\,n\,$ has form $\rm\, n^2 = \,5\,k + r\,$ for $\rm\:r\in \{0,1,4\}.$
Proof $\$ By Lemma $\rm\ n^2 = (5k\pm r)^2 = 5\,(5k^2\!\pm 2kr)+r^2\,$ for $\rm\:r\in \{0,1,2\},\,$ so $\rm\: r^2\in\{0,1,4\}.$
Remark $\$ Your divisor of $\,3\,$ is analogous, with $\rm\:r\in \{0,1\}\,$ so $\rm\:r^2\in \{0,1\}.\,$ The same method generalizes for any divisor $\rm\:m,\,$ yielding that $\rm\:n^2 = m\,k + r^2,\,$ for $\rm\:r\in\{0,1,\ldots,\lfloor m/2\rfloor\}.$ The reason we need only square half the remainders is because we have exploited reflection symmetry (negation) to note that remainders $\rm > n$ can be transformed to negatives of remainders $\rm < n,\,$ e.g. $\rm\: 13 = 5\cdot 2 +\color{#0A0} 3 = 5\cdot 3 \color{#C00}{- 2},\,$ i.e. remainder $\rm\:\color{#0A0}3\leadsto\,\color{#C00}{-2},\,$ i.e. $\rm\:3 \equiv -2\pmod 5.\:$ This amounts to using a system of balanced (or signed) remainders $\rm\, 0,\pm1,\pm2,\ldots,\pm n\$ vs. $\rm\ 0,1,2,\ldots,2n[-1].\:$ Often this optimization halves work for problems independent of the sign of the remainder.
All of this is much clearer when expressed in terms of congruences (modular arithmetic), e.g. the key inference above $\rm\:n\equiv r\:\Rightarrow\:n^2\equiv r^2\pmod m\:$ is a special case of the ubiquitous
Congruence Product Rule $\rm\ \ A\equiv a,\ B\equiv b\ \Rightarrow\ AB\equiv ab\ \ (mod\ m)$
Proof $\rm\:\ \ m\: |\: A\!-\!a,\ B\!-\!b\:\ \Rightarrow\:\ m\ |\ (A\!-\!a)\ B + a\ (B\!-\!b)\ =\ AB - ab$
For an introduction to congruences see any decent textbook on elementary number theory.
- | 2015-01-31T07:46:15 | {
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http://math.stackexchange.com/questions/603291/how-to-show-some-function-is-constant | # How to show some function is constant?
Suppose $f:(a,b) \to \mathbb{R}$ satisfy $|f(x) - f(y) | \le M |x-y|^\alpha$ for some $\alpha >1$ and all $x,y \in (a,b)$. Prove that $f$ is constant on $(a,b)$.
I'm not sure which theorem should I look to prove this question. Can you guys give me a bit of hint? First of all how to prove some function $f(x)$ is constant on $(a,b)$? Just show $f'(x) = 0$?
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yes, indeed! (to your last question) – Avitus Dec 11 '13 at 20:41
What does the value $M$ mean? – John Dec 11 '13 at 20:41
the condition should be $\alpha>1$ otherwise it just an holder function and they are everything but constant (in general). In case $\alpha>1$ you just show $f$ is differentiable with $0$ has a derivative. (like the answer below). Can you correct the question ? – user42070 Dec 11 '13 at 21:08
divide by $|x-y|$ both members. you get
$$\frac{|f(x) - f(y)|}{|x-y|} \le M |x-y|^{\alpha - 1} \ \ (1)$$
now, since $(1)$ has to hold $\forall \ x, y$ then set $y = x + h$, with $h \to 0$
it becomes
$$|f'(x)| \le M \ |h|^{\alpha-1} = 0$$ ($\alpha - 1 > 0$ so there's no problem there)
So $|f'(x)| \le 0$, but of course also $|f'(x)| \ge 0$, it implies $|f'(x)| = 0$. Hence $f'(x) = 0$
EDIT:
We can formalize it more. Let's do it.
$$\frac{|f(x) - f(y)|}{|x-y|} \le M |x-y|^{\alpha - 1} \ \ (1)$$
We can always set $x = y + h, h > 0$ since $(1)$ has to hold $\forall x, y$
Then
$$\frac{|f(y+h) - f(y)|}{h} \le M h^{\alpha - 1} \ \ (1)$$
(note $|h| = h$)
Now, let's suppose $f(y+h) - f(y) \ge 0 \ \ \ \ \ (2a)$
It implies that $f'(y) \ge 0$ (it suffice to divide $(2a)$ by $h$ and then taking the limit for $h \to 0$ to show it)
But it also implies, recalling (1) that
$$\frac{f(y+h) - f(y)}{h} \le 0 \Rightarrow f'(y) \le 0$$
(Again by taking the limit of both parts.)
But these last two results imply that $f'(y) = 0$
We can do the exact same reasoning in the case that $f(y+h) - f(y) \le 0$
So again, in this case we find $f'(y) = 0$
Thus we have demonstrated that in both cases ($f(y+h) > f(y)$ and $f(y+h) < f(y)$) we have $f'(y) = 0$, so this has to hold $\forall y$
This implies $f = const$
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this is good one – james Miler Dec 11 '13 at 20:47
Are you using this for your second equation? $\lim_{x \rightarrow a}{|f(x)|}=|\lim_{x \rightarrow a}{f(x)}|$, I thought this equality does not hold? – Idonknow Dec 11 '13 at 21:06
edited.. it should be better now :-) – Ant Dec 11 '13 at 23:56
Hint: Show that $f'(y)$ exists and is equal to $0$ for all $y$. Then as usual by the Mean Value Theorem our function is constant.
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+1 for speed :-) – Avitus Dec 11 '13 at 20:41
so what is first conditio refer to? – james Miler Dec 11 '13 at 20:42
Divide by $|x-y|$. We are left on the right with $\le M|x-y|^{\alpha-1}$, which approaches $0$ as $x\to y$. – André Nicolas Dec 11 '13 at 20:43
Without any derivative...
Choose $y\gt x$ in the interval $(a,b)$. For every $n\geqslant1$, divide the interval $(x,y)$ into $n$ subintervals $(x_i,x_{i+1})$ of length $x_{i+1}-x_i=(y-x)/n$. By hypothesis, for every $i$, $$|f(x_i)-f(x_{i+1})|\leqslant M(x_{i+1}-x_i)^\alpha=M(y-x)^\alpha n^{-\alpha},$$ hence, by the triangular inequality, $$|f(x)-f(y)|\leqslant \sum\limits_{i=1}^n|f(x_i)-f(x_{i+1})|\leqslant M(y-x)^\alpha n^{1-\alpha}.$$ If $\alpha\gt1$, the RHS goes to zero when $n\to\infty$ because $n^{1-\alpha}\to0$, hence $f(x)=f(y)$, QED.
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Interesting answer. Do you think we can adapt it in order to prove this: math.stackexchange.com/questions/578487/… – Tomás Dec 11 '13 at 21:42
Hadn't seen this way before--I appreciate it. – nayrb Dec 12 '13 at 0:00
Nicely done! (+1) – leo Dec 12 '13 at 3:58
HINT: Your idea is a good one. What happens when you divide the inequality by $|x-y|$?
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It suffices to show that $f$ is differentiable and its derivative vanishes everywhere.
The hypothesis implies that, for every $x\in(a,b)$ and $h$, such that $x+h\in(a,b)$, we have that
$$\frac{|f(x+h)-f(x)-0\cdot h|}{|h|} \le M\,|h|^{\alpha-1}.$$
The right hand side of the above tends to zero, as $h\to 0$, and therefore $f'(x)=0$!
- | 2015-08-04T12:33:08 | {
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https://lovejays.com/dag9b/complex-conjugate-eigenvalues-121a37 | # complex conjugate eigenvalues
The characteristic polynomial of $$A$$ is $$\lambda^2 - 2 \lambda + 5$$ and so the eigenvalues are complex conjugates, $$\lambda = 1 + 2i$$ and $$\overline{\lambda} = 1 - 2i\text{. Note that not only do eigenvalues come in complex conjugate pairs, eigenvectors will be complex conjugates of each other as well. Thus you only need to compute one eigenvector, the other eigenvector must be the complex conjugate. If the eigenvalues are a complex conjugate pair, then the trace is twice the real part of the eigenvalues. The Characteristic Equation always features polynomials which can have complex as well as real roots, then so can the eigenvalues & eigenvectors of matrices be complex as well as real. eigenvalues of a real symmetric or complex Hermitian (conjugate symmetric) array. Example 13.1. Also, they will be characterized by the same frequency of rotation; however, the direction s of rotation will be o pposing. Value. A complex number is an eigenvalue of corresponding to the eigenvector if and only if its complex conjugate is an eigenvalue corresponding to the conjugate vector. It is possible that Ahas complex eigenvalues, which must occur in complex-conjugate pairs, meaning that if a+ ibis an eigenvalue, where aand bare real, then so is a ib. 1.2. When the eigenvalues of a system are complex with a real part the trajectories will spiral into or out of the origin. 4. These pairs will always have the same norm and thus the same rate of growth or decay in a dynamical system. If A has complex conjugate eigenvalues λ 1,2 = α ± βi, β ≠ 0, with corresponding eigenvectors v 1,2 = a ± bi, respectively, two linearly independent solutions of X′ = AX are X 1 (t) = e αt (a cos βt − b sin βt) and X 2 (t) = e αt (b cos βt + a sin βt). Then a) if = a+ ibis an eigenvalue of A, then so is the complex conjugate = a−ib. values. eigvalsh. Most of this materi… For example, the command will result in the assignment of a matrix to the variable A: We can enter a column vector by thinking of it as an m×1 matrix, so the command will result in a 2×1 column vector: There are many properties of matrices that MATLAB will calculate through simple commands. To enter a matrix into MATLAB, we use square brackets to begin and end the contents of the matrix, and we use semicolons to separate the rows. The Eigenvalue Problem: The Hessenberg and Real Schur Forms The Unsymmetric Eigenvalue Problem Let Abe a real n nmatrix. eigenvalues are themselves complex conjugate and the calculations involve working in complex n-dimensional space. complex eigenvalues always come in complex conjugate pairs. eigh. There is nothing wrong with this in principle, however the manipulations may be a bit messy. A similar discussion verifies that the origin is a source when the trace of is positive. Here is a summary: If a linear system’s coefficient matrix has complex conjugate eigenvalues, the system’s state is rotating around the origin in its phase space. eigenvalues of a self-adjoint matrix Eigenvalues of self-adjoint matrices are easy to calculate. Note that the complex conjugate of a function is represented with a star (*) above it. An interesting fact is that complex eigenvalues of real matrices always come in conjugate pairs. On this site one can calculate the Characteristic Polynomial, the Eigenvalues, and the Eigenvectors for a given matrix. The eigenvectors associated with these complex eigenvalues are also complex and also appear in complex conjugate pairs. Example: Diagonalize the matrix . Solve the system. →Below is a calculator to determine matrices for given Eigensystems. In equation 1 we can appreciate that because the eigenvalue is real, the complex conjugate of the real eigenvalue is just the real eigenvalue (no imaginary term to take the complex conjugate of). Proposition Let be a matrix having real entries. complex eigenvalues. Once you have found the eigenvalues of a matrix you can find all the eigenvectors associated with each eigenvalue by finding a … If A is a 2 2-matrix with complex-conjugate eigenvalues l = a bi, with associated eigenvectors w = u iv, then any solution to the system dx dt = Ax(t) can be written x(t) = C1eat(ucosbt vsinbt)+C2eat(usinbt+vcosbt) (7) where C1,C2 are (real) constants. Find the complex conjugate eigenvalues and corresponding complex eigenvectors of the following matrices. Similar function in SciPy that also solves the generalized eigenvalue problem. Finding Eigenvectors. 3. Since the real portion will end up being the exponent of an exponential function (as we saw in the solution to this system) if the real part is positive the solution will grow very large as \(t$$ increases. eigenvalues of a non-symmetric array. This occurs in the region above the parabola. NOTE 4: When there are complex eigenvalues, there's always an even number of them, and they always appear as a complex conjugate pair, e.g. | 2021-05-17T19:15:41 | {
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https://am111.readthedocs.io/en/latest/session10_FFT.html | # Session 10: Fast Fourier Transform¶
Date: 11/27/2017, Monday
In [1]:
format compact
## Generate input signal¶
Fourier transform is widely used in signal processing. Let’s looks at the simplest cosine signal first.
Define
$y_1(t) = 0.3 + 0.7\cos(2\pi f_1t)$
It has a magnitude of 0.7, with a constant bias term 0.3. We choose the frequency $$f_1=0.5$$.
In [2]:
t = -5:0.1:4.9; % time axis
N = length(t) % size of the signal
f1 = 0.5; % signal frequency
y1 = 0.3 + 0.7*cos(2*pi*f1*t); % the signal
N =
100
In [3]:
%plot -s 800,200
hold on
plot(t, y1)
plot(t, 0.3*ones(N,1), '--k')
title('simple signal')
xlabel('t [s]')
legend('signal', 'mean')
## Perform Fourier transform on the signal¶
You can hand code the Fourier matrix as in the class, but here we use the built-in function for convenience.
In [4]:
F1 = fft(y1);
length(F1) % same as the length of the signal
ans =
100
There are two different conventions for the normalization factor in the Fourier matrix. One is having the normalization factor $$\frac{1}{\sqrt{N}}$$ in the both the Fourier matrix $$A$$ and the inverse transform matrix $$B$$
$\begin{split}A = \frac{1}{\sqrt{N}} \begin{bmatrix} 1&1&1&\cdots &1 \\ 1&\omega&\omega^2&\cdots&\omega^{N-1} \\ 1&\omega^2&\omega^4&\cdots&\omega^{2(N-1)}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 1&\omega^{N-1}&\omega^{2(N-1)}&\cdots&\omega^{(N-1)(N-1)}\\ \end{bmatrix}\end{split}$
$\begin{split}B = \frac{1}{\sqrt{N}} \begin{bmatrix} 1&1&1&\cdots &1 \\ 1&\omega^{-1}&\omega^{-2}&\cdots&\omega^{-(N-1)} \\ 1&\omega^{-2}&\omega^{-4}&\cdots&\omega^{-2(N-1)}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 1&\omega^{-(N-1)}&\omega^{-2(N-1)}&\cdots&\omega^{-(N-1)(N-1)}\\ \end{bmatrix}\end{split}$
MATLAB uses a different convention that
$\begin{split}A = \begin{bmatrix} 1&1&1&\cdots &1 \\ 1&\omega&\omega^2&\cdots&\omega^{N-1} \\ 1&\omega^2&\omega^4&\cdots&\omega^{2(N-1)}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 1&\omega^{N-1}&\omega^{2(N-1)}&\cdots&\omega^{(N-1)(N-1)}\\ \end{bmatrix}\end{split}$
$\begin{split}B = \frac{1}{N} \begin{bmatrix} 1&1&1&\cdots &1 \\ 1&\omega^{-1}&\omega^{-2}&\cdots&\omega^{-(N-1)} \\ 1&\omega^{-2}&\omega^{-4}&\cdots&\omega^{-2(N-1)}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 1&\omega^{-(N-1)}&\omega^{-2(N-1)}&\cdots&\omega^{-(N-1)(N-1)}\\ \end{bmatrix}\end{split}$
The difference doesn’t matter too much as long as you use one of them consistently. In both cases there is
\begin{align} F &= AY \text{ (Discrete Fourier transfrom)} \\ Y &= BF \text{ (Inverse transfrom)} \end{align}
### Full spectrum¶
The spectrum F1 (the result of the Fourier transfrom) is typically an array of complex numbers. To plot it we need to use absolute magnitude.
In [5]:
%plot -s 800,200
plot(abs(F1),'- .')
title('unnormalized full spectrum')
The first term in F1 indicates the magnitude of the constant term (zero frequency). Diving by N gives us the actual value.
In [6]:
F1(1)/N % equal to the constant bias term specified at the beginning
ans =
0.3000
Besides the constant bias F1(1), there are two non-zero pointings in F1, indicating the cosine signal itself. The magnitude 0.7 is evenly distributed to two points.
In [7]:
F1(2+4)/N, F1(end-4)/N % adding up to 0.7
ans =
-0.3500 - 0.0000i
ans =
-0.3500 + 0.0000i
Plotting F1/N shows more clearly the magnitude of signals at different frequencies:
In [8]:
%plot -s 800,200
plot(abs(F1)/N,'- .')
title('(normalized) full spectrum')
ylabel('signal amplitude')
### Half-sided spectrum¶
From the matrix $$A$$ it is easy to show that, the first element F(1) in the resulting spectrum is always a real number indicating the constant bias term, while the rest of the array F(2:end) is symmetric, i.e. F(2) == F(end), F(3) == F(end-1). ( F(2) is actually the conjugate of F(end), but we only care about magnitude here. )
Due to such symmetricity, we can simply plot half of the array (scaled by 2) without loss of information.
In [9]:
M = N/2 % need to cast to integer if N is an odd number
M =
50
In [10]:
%plot -s 800,200
plot(abs(F1(1:M+1))/N*2, '- .')
title('(normalized) half-sided spectrum')
ylabel('signal amplitude')
## Understanding units!¶
The Discrete Fourier Transform, by defintion, is simply a matrix multiplication which acts on pure numbers. But real physical signals have units. You cannot just treat the resulting array F1 as some unitless frequency. If the signal is a time series then you need to deal with seconds and hertz; if it is a wave in the space then you need to deal with the wave length in meters.
In order to understand the unit of the resulting spectrum F1, let’s look at the original time series y1 first.
The “time step” of the signal is
In [11]:
dt = t(2)-t(1) % [s]
dt =
0.1000
This is the finest temporal resolution the signal can have. It corresponds the highest frequency:
In [12]:
f_max = 1/dt % [Hz]
f_max =
10.0000
On the contrary, the longest time range (dt*N, the time span of the entire signal) corresponds to the lowest frequency:
In [13]:
df = f_max/N % [Hz]
df =
0.1000
With the lowest frequency df being the “step size” in the frequency axis, the value of the frequency axis is simply the array [0, df, 2*df, …]. Now we can use correct values and units for the x-axis of the spectrum plot.
In [14]:
%plot -s 800,200
plot(df*(0:M), abs(F1(1:M+1))/N*2,'- .')
title('half-sided spectrum with correct unit of x-axis')
ylabel('signal amplitude')
xlabel('frequency [Hz]')
The peak is at 0.5 Hz, consistent with our original signal which has a period of 2 s, since 0.5 Hz = 1/(2s). Thus our unit specification is correct.
## Deal with negative frequency¶
The right half of the spectrum array (F1(M+2:end), not plotted in the above figure) corresponds to negative frequency [-M*df, …, -2*df, -df]. Thus each element in the entire F1 array corresponds to each element in the frequency array [0, df, 2*df, …, M*df, -M*df, …, -2*df, -df].
You can perform fftshift on the resulting spectrum F1 to swap its left and right parts, so it will align with the motonically increasing axis [-M*df, …, -2*df, -df, 0, df, 2*df, …, M*df]. That feels more natural from a mathematical point of view.
In [15]:
F_shifted = fftshift(F1);
plot(abs(F_shifted),'- .')
## Perform inverse transform¶
Performing inverse transform is simply ifft(F1). Recall that MATLAB performs the $$\frac{1}{N}$$ scaling during the inverse transform step.
We use norm to check if ifft(F1) is close enough to y1.
In [16]:
norm(ifft(F1) - y1) % almost zero
ans =
1.2269e-15
## Mix two signals¶
Fourier transform and inverse transform are very useful in signal filering. Let’s first add a high-frequency noise to our original signal.
In [17]:
f2 = 5; % higher frequency
y2 = 0.2*sin(f2*pi*t); % noise
y = y1 + y2; % add up original signal and noise
In [18]:
%plot -s 800,400
subplot(311);plot(t, y2, 'r');
ylim([-1,1]);title('noise')
subplot(312);plot(t, y1);
ylim([-0.6,1.2]);title('original signal')
subplot(313);plot(t, y, 'k');
ylim([-0.6,1.2]);title('signal + noise')
After the Fourier transform, we see two new peaks at a relatively higher frequency.
In [19]:
F = fft(y);
In [20]:
%plot -s 800,200
plot(abs(F), '- .')
title('spectrum with high-frequency noise')
Again, the noise magnitude 0.2 is evenly distributed to positive and negative frequencies. Here we got complex conjugates:
In [21]:
F(2+24)/N, F(end-24)/N % magnitude of noises
ans =
-0.0000 + 0.1000i
ans =
-0.0000 - 0.1000i
## Filter out high-frequency noise¶
Let’s wipe out this annoying noise. It’s very difficult to do so in the original signal, but very easy to do in the spectrum.
In [22]:
F_filtered = F; % make a copy
F_filtered(26) = 0; % remove the high-frequency noise
F_filtered(76) = 0; % same for negative frequency
In [23]:
plot(abs(F_filtered), '- .')
title('filtered spectrum')
Then we can transform the spectrum back to the signal.
In [24]:
y_filtered = ifft(F_filtered);
If the filtering is done symmetrically (i.e. do the same thing for positive and negative frequencies), the recovered signal will only contain real numbers.
In [35]:
%plot -s 800,200
plot(t, y_filtered)
title('de-noised signal')
The de-noised signal is almost the same as the original noise-free signal:
In [36]:
norm(y_filtered - y1) % almost zero
ans =
5.6847e-15
## Filter has to be symmetric¶
What happens if the filtering done asymmetrically?
In [37]:
F_wrong_filtered = F; % make another copy
F_wrong_filtered(76) = 0; % only do negative frequency
In [39]:
plot(abs(F_wrong_filtered), '- .')
title('asymmetrically-filtered spectrum')
The recovered signal now contains imaginary parts. That’s unphysical!
In [40]:
y_wrong_filtered = ifft(F_wrong_filtered);
In [42]:
y_wrong_filtered(1:5)' % print the first several elements
ans =
-0.4000 - 0.1000i
-0.4657 + 0.0000i
-0.2663 + 0.1000i
-0.0114 - 0.0000i
0.0837 - 0.1000i
In [43]:
norm(imag(y_wrong_filtered)) % not zero
ans =
0.7071
You can plot the real part only. It is something between the unfiltered and filtered signals, i.e. the filtering here is incomplete.
In [45]:
hold on
plot(t, y, 'k')
plot(t, real(y_wrong_filtered), 'b')
plot(t, y1, 'r')
legend('signal with noise', 'incomplete fliltering', 'signal without noise') | 2022-05-18T13:55:21 | {
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http://math.stackexchange.com/questions/191532/why-cant-a-neighborhood-be-a-finite-set | # Why can't a neighborhood be a finite set?
Rudin defines a neighborhood as follows:
Let $X$ be a metric space endowed with a distance function $d$. A neighborhood of a point $p \in X$ is a set $N_r(p)$ consisting of all $q \in X$ such that $d(p,q) < r$ for some $r > 0$.
He later proves two facts: every neighborhood is an open set, and every finite set is closed.
But it would seem to me that a neighborhood as defined above could be finite. For example, let $X = \{1,2,3\}$ with $d(x,y) = |x-y|$ be a metric space . Consider the neighborhood around $p = 2$ of radius $0.5$, i.e.: the set of all points $q$ in $X$ such that $d(q,p)<0.5$. But the neighborhood is simply $\{2\}$. Therefore, the neighborhood is a finite set, and therefore closed. But every neighborhood is open. This is a contradiction.
So my understanding of a neighborhood is broken somehow. It has further implications: if we define $E = \{2\}$, then the $0.5$-radius neighborhood is $\{2\}$, which is a subset of $E$, which means that $2$ is an interior point of $E$. Since $2$ is the only point in $E$, all points in $E$ are interior points, and $E$ is open. But $E$ is finite, and therefore closed.
I'm probably missing something obvious. Can anyone spot my mistake?
-
Nothing about the definition of a set being open or closed precludes a set being both -- merely your intuition. :) – Eugene Shvarts Sep 5 '12 at 16:45
It is possible for are set to be open and closed, as $\{2\}$ in your example. Being open and closed isn't mutually exclusive. – martini Sep 5 '12 at 16:45
"not closed" is not necessarily "open" and "not open" is not necessarily "closed": "not open" and "not closed" can both be "ajar"! – rschwieb Sep 5 '12 at 16:55
Obligatory quote: ''…an answer to the mathematician’s riddle: “How is a set different from a door?” should be: “A door must be either open or closed, and cannot be both, while a set can be open, or closed, or both, or neither!”'' - Munkres – Michael Greinecker Sep 5 '12 at 16:58
Well, in that case, Rudin 2.21 is incorrect, unless there's another hypothesis. – Cameron Buie Sep 5 '12 at 17:02
"Every finite set is closed" does not imply "every open neighbourhood is infinite".
Let $(X,d)$ be any metric space, and let $F \subseteq X$ be a non-empty finite subset; say $F = \{ x_1, \cdots, x_n \}$ for some $n \in \mathbb{N}$. To prove that $F$ is closed it suffices to prove that $X - F$ is open. So let $p \in X-F$. Then the set $\{ d(p,x_1), \cdots, d(p,x_n) \}$ has a minimum value, say $\delta$, and we know that $\delta > 0$ since $p \not \in F$, and so whenever $d(p,q) < \delta$ we have $q \in X-F$. But this tells you that the (open) ball of radius $\delta$ about $p$ lies in $X-F$, and hence $X-F$ is open.
In your examples, yes, the sets are open and finite, but they are also closed!
The key fact to take home is that sets are allowed to be both open and closed.
However, there is a partial converse: in a dense metric space, every open neighbourhood is infinite.
A metric space $(X,d)$ is dense if for any $x \in X$ and $r > 0$ there exists $y \in X$ with $x \ne y$ and $d(x,y) < r$ $-$ that is, given any point in the space, there are other points which lie arbitrarily close to that point. $\mathbb{R}$ is a dense space, so is $\mathbb{Q}$, so is $(0,1)$, and indeed so is any other non-empty open subset of $\mathbb{R}$.
The fact that open neighbourhoods in dense metric spaces are necessarily infinite is clear from the definition. [It's also clear that a metric space in which every open neighbourhood is infinite is dense, and so the two conditions are equivalent.]
-
This makes sense, thanks! – jme Sep 5 '12 at 17:11
For an interesting example, let $X$ be any non-empty set, and define $d:X\times X\to\Bbb R$ by $$d(x,y)=\begin{cases}0 & x=y\\1 & x\neq y.\end{cases}$$ This can be shown to be a metric on $X$ (called the discrete metric). One interesting property is that in the metric space $(X,d)$, every subset of $X$ is both open and closed. Even in general metric spaces, there will always be at least two subsets that are both open and closed--namely, the empty set and the whole set.
- | 2015-11-26T10:44:49 | {
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https://math.stackexchange.com/questions/3980049/is-the-set-a-0-1-setminus-mathbbq-countable-or-not | # Is the set $A = [0,1]\setminus\mathbb{Q}$ countable or not?
Is the set $$A = [0,1]\setminus\mathbb{Q}$$ countable or not?
What I am thinking is $$A$$ consist of irrational numbers in the interval $$[0,1]$$ hence it is subset of irrational numbers. As set of irrational numbers is uncountable so I think set $$A$$ is also uncountable.
• not every subset of irrational numbers is uncountable, but if $A$ and $\mathbb Q$ were both countable then so would be $A\cup \mathbb Q$ and therefore $[0,1]\subset A\cup \mathbb Q$ – J. W. Tanner Jan 10 at 15:55
• $\varnothing$ is also a subset of the irrational numbers. – Asaf Karagila Jan 10 at 16:24
Yes, it is uncountable, but not for that reason. For instance, $$\left\{\sqrt2+n\,\middle|\, n\in\Bbb N\right\}$$ is also a set of irrational numbers, but it is countable.
However, if $$[0,1]\setminus\Bbb Q$$ was countable, then, since $$\Bbb Q\cap[0,1]$$ is countable, $$[0,1]$$ would be countable too, since it's the union of them.
• technically, $[0,1]$ is contained in the union of $A$ and $\mathbb Q$ – J. W. Tanner Jan 10 at 16:08
• @J.W.Tanner I've edited my answer. Thank you. – José Carlos Santos Jan 10 at 16:23
All countable set of $$R$$ have Lebesgue measure equal to $$0$$. So Lebesgue measure of $$[0, 1] - \mathbb{Q}$$ is $$1$$. Eventually by contraposition, $$[0, 1] - \mathbb{Q}$$ is uncoutable.
What you are thinking does not work.
For example, $$\{n\pi\mid n\in\mathbb N\setminus\{0\}\}$$ is a subset of irrational numbers but countable.
Here's an argument that works. If $$A$$ were countable, then, since $$\mathbb Q$$ is countable,
$$A\cup \mathbb Q$$ would be countable,
and therefore $$[0,1]$$, which is a subset of $$A\cup \mathbb Q$$, would be countable, | 2021-06-14T09:18:51 | {
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https://math.stackexchange.com/questions/1502270/can-one-compute-series-expansions-or-complex-residues-at-essential-singularities | # Can one compute series expansions or complex residues at essential singularities?
Title says it all.
I'm noticing a trend of failure on Mathematica/WolframAlpha's parts when trying to compute either the Laurent expansions or the residues of functions like $\sin\dfrac{1}{z}$ about $z=0$. (Specifying $z=0$ for the residue query returns nothing.)
Is there something about essential singularities $z_0$ that prevents one from computing a series expansion about or residue at $z_0$? Is this "something" explicitly mentioned in the usual definitions of series/residues that I'm missing? Admittedly, my complex analysis knowledge has started to rust.
• Just plug $u=1/z$ into $\sin u = u -u^3/3! + \cdots$ for the Larent expansion; the residue is $1.$ – zhw. Oct 28 '15 at 19:46
• @zhw. Okay, that approach works fine for $\sin\dfrac{1}{z}$, but what if I made a slight adjustment? Say, $\sin\dfrac{z}{z-1}$. Does the same process still work? – user170231 Oct 28 '15 at 19:51
Recall the definition of an essential singularity. If a function $f$ has an essential singularity at $z=z_0$, then
$$\lim_{z \to z_0} (z-z_0)^m f(z) = \infty$$
That is, an essential singularity is "more singular" than any order pole.
That all said, the definition of a residue still applies as the coefficient of $(z-z_0)^{-1}$ in the Laurent expansion of $f$.
In your example of $f(z) = \sin{\left ( \frac{z}{z-1} \right )}$, there is an essential singularity at $z=1$. The Laurent expansion of $f$ about $z=1$ may be easily determined by expressing $f$ as follows:
\begin{align}f(z) &= \sin{\left (1+\frac1{z-1} \right )} = \sin{1} \cos{\left ( \frac{1}{z-1} \right )}+ \cos{1} \sin{\left ( \frac{1}{z-1} \right )}\\ &= \sin{1}\sum_{n=0}^{\infty} \frac{(-1)^n}{(2 n)!(z-1)^{2 n}} + \cos{1}\sum_{n=0}^{\infty} \frac{(-1)^n}{(2 n+1)!(z-1)^{2 n+1}}\end{align}
Hopefully it is clear that the residue of $f$ at $z=1$ is $\cos{1}$.
• Are there any functions with essential singularities for which identities and "tricks" like these don't help? I'm just trying to wrap my mind around why WA and Mma are having trouble with these computations. – user170231 Oct 28 '15 at 21:00
• @user170231: plenty. What I would do is make use of Mathematica's expansion about Infinity, i.e., Series[f[z],{z,Infinity,n}] for an n-term expansion in $1/z$. – Ron Gordon Oct 28 '15 at 21:08 | 2020-09-26T03:23:43 | {
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http://math.wsu.edu/faculty/bkrishna/FilesMath464/S18/LecNotes/welcome.html | Linear Optimization - Lecture Notes and Videos
Math 464 (Spring 2018) - Lecture Notes and Videos on Linear Optimization
Scribes from all lectures so far (as a single big file)
Lec # Date Topic(s) Scribe Panopto
1 Jan 9 syllabus, optimization in calculus, solving $$A\mathbf{x} = \mathbf{b}$$ using EROs, linear program example: Dude's Thursday problem scribe video
2 Jan 11 no class
3 Jan 16 general form of a linear program (LP), feasible and optimal solutions, standard form, conversion to std form, excess/slack variables scribe video
4 Jan 18 LP formulations, diet problem, solution in AMPL, road lighting problem, exceeding illumination limits, currency exchange LP scribe video
5 Jan 23 convex function, piecewise linear (PL) convex (PLC) functions, max of convex functions is convex, PLC functions in LP scribe video
6 Jan 25 $$|x_i| \to x_i^+, x_i^-$$, lighting problem: get "close to" $$I_i^*$$, inventory planning LP, graphical solution in 2D, feasible region, slide $$\mathbf{c}^T\mathbf{x}$$ line scribe video
7 Jan 30 cases of LP, unique and alternative optimal solutions, unbounded LP, infeasible LP, subspace, affine spaces, polyhedron, halfspace scribe video
8 Feb 1 polyhedron is convex, bounded set, convex hull, active constraint, extreme point, vertex, basic and basic feasible solution (bfs) scribe video
9 Feb 6 proof of extreme point $$\Leftrightarrow$$ vertex $$\Leftrightarrow$$ bfs, adjacent basic solutions and bfs's, polyhedron $$P$$ in standard form, basic solutions of $$P$$ scribe video
10 Feb 8 finding basic solutions in Octave, correspondence of corner points and bfs's, degenrate basic bfs, degeneracy and representation scribe video
11 Feb 13 degeneracy in standard form, $$P$$ has no line $$\Leftrightarrow P$$ has vertex, existence of optimal vertex, shape of standard form polyhedron scribe video
12 Feb 15 simplex method, feasible direction at $$\mathbf{x}$$: $$\mathbf{x}+\theta\mathbf{d} \in P \mbox{ for } \theta > 0, j$$th basic direction at bfs $$\mathbf{x}$$, reduced cost $$\mathbf{c}'^T = \mathbf{c}^T - \mathbf{c}_B B^{-1} A$$ scribe video
13 Feb 20 hints for problems from Hw6, LP optimality conditions in standard form: basis $$B$$ optimal if $$B^{-1}\mathbf{b} \geq \mathbf{0}$$ and $$\mathbf{c}^T - \mathbf{c}_B B^{-1} A \geq \mathbf{0}^T$$ scribe video
14 Feb 22 entering/leaving variable, min-ratio test: $$\theta^* = \min_{\{i \in \scr{B}|d_{B(i)} < 0\}} \,(-x_{B(i)}/d_{B(i)})$$, an iteration of the simplex method, Octave session scribe video | 2018-02-26T03:20:57 | {
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https://math.stackexchange.com/questions/3642078/how-does-the-taylor-series-converge-at-all-points-for-certain-functions | # How does the Taylor Series converge at all points for certain functions
The way my professor defined Taylor polynomials is: the $$n^{th}$$ degree Taylor polynomial $$p(x)$$ of $$f(x)$$ is a polynomial that satisfies $$\lim_{x\to 0}{f(x)-p(x) \over x^n} = 0$$. This is actually the little-o notation $$o(x^n)$$, which means $$(f(x)-p(x)) \ll x^n$$ as $$x$$ approaches $$0$$. From this I have got the intuition that Taylor Polynomials work only for $$|x| < 1$$ because $$x^n$$ gets smaller as $$n$$ gets bigger only when $$|x| < 1$$. And the textbook seemed to agree with my intuition, because the textbook says “Taylor polynomial near the origin” (probably implying $$|x| < 1$$).
Since Taylor Series is basically Taylor polynomial with $$n\to\infty$$, I intuitively thought that the Taylor Series would also only converge to the function it represents in the interval $$(-1, 1)$$.
For example, in the case of $$1\over1-x$$, it is well known that the Taylor series only converges at $$|x| < 1$$.
However, all of a sudden, the textbook says that the Taylor series of $$\cos x$$ converges for all real $$x$$. It confused me because previously I thought the Taylor series would only work for $$|x|<1$$. Now, I know that the Taylor Series is defined like this: $$f(x) = Tf(x) \Leftrightarrow \lim_{n\to\infty}R_{n}f(x) = 0$$
And I know how to get the maximum of Taylor Remainder for $$\cos x$$ using Taylor's Theorem, and I know that the limit of that Taylor Remainder is $$0$$ for all real $$x$$, which makes the Taylor Series of $$cosx$$ converge to $$\cos x$$, pointwise. However, I just can't get why my initial intuition is wrong (why taylor series converges for all $$x$$ for certain functions, like $$\cos x$$, also $$\sin x$$ and $$e^x$$, etc.)
• For the specific example $e^x$, the terms in its series are $x^n/n!$. Note that $n!$ grows much faster than $x^n$ as $n$ grows. That's one reason to believe the Taylor series of $e^x$ converges. Apr 24 '20 at 18:10
• In my mind, the key idea of calculus is the local linear approximation $f(x) \approx f(a) + f'(a)(x - a)$, which is a good approximation when $x$ is near $a$. It is natural to ask, what if we approximate $f$ locally by a quadratic or cubic function rather than by a linear function. This leads to the idea of Taylor polynomial approximation, which is indeed initially intended to be a local approximation to $f$. But when we look at the remainder term in Taylor's theorem, there's something kind of amazing, surprising that happens which is that often it's small even when $x$ is far from $a$. Apr 24 '20 at 18:11
• @trisct Yes that is true indeed. The limit of Taylor remainder for $e^x$ is also zero for all x, which further explains why the Taylor Series converges. But I still I can't see why my initial intuition is wrong.
– the
Apr 24 '20 at 18:12
• @littleO So it just happened to be that the Taylor remainder's limit is zero even for x far away?
– the
Apr 24 '20 at 18:14
• @linearAlg In my mind, yes. It is just one of the miracles of math, which makes us wonder why things have worked out more nicely than we deserved. Apr 24 '20 at 18:16
Actually, things may go wrong in $$(-1,1)$$. For instance, the Taylor series centered at $$0$$ of $$f(x)=\frac1{1-nx}$$ only converges to $$f(x)$$ on $$\left(-\frac1n,\frac1n\right)$$. And if$$f(x)=\begin{cases}e^{-1/x^2}&\text{ if }x\ne0\\0&\text{ if }x=0,\end{cases}$$then the Taylor series of $$f$$ only converges to $$f(x)$$ if $$x=0$$.
On the other hand, yes, Taylor series centered at $$0$$ are made to converge to $$f(x)$$ near $$0$$. But that's no reason to expect that they don't converge to $$f(x)$$ when $$x$$ is way from $$0$$. That would be like expecting that a non-constant power series $$a_0+a_1x+a_2x^2+\cdots$$ takes larger and larger values as the distance from $$x$$ to $$0$$. That happens often, but $$1-\frac1{2!}x^2+\frac1{4!}x^4-\cdots=\cos(x)$$, which is bounded.
In other words, you have gone from $$a\implies b$$ to $$\tilde a\implies \tilde b,$$ which you can see to be clearly false, identically. That is, it is not necessarily true for all $$a,\,b.$$
Since you already know why the series for entire functions like $$\cos x$$ converges everywhere (as you explain towards the end of your post), you should now see where your original intuition (I would say erroneous belief) misled you. | 2021-12-08T13:58:19 | {
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https://math.stackexchange.com/questions/4139421/helen-and-joe-play-guitar-together-every-day-at-lunchtime/4139434 | # Helen and Joe play guitar together every day at lunchtime.
Helen and Joe play guitar together every day at lunchtime. The number of songs that they play on a given day has a Poisson distribution, with an average of 5 songs per day. Regardless of how many songs they will play that day, Helen and Joe always flip a coin at the start of each song, to decide who will play the solo on that song. If we know that Joe plays exactly 4 solos on a given day, then how many solos do we expect that Helen will play on that same day?
My attempt: If the average is $$5$$ songs a day and Joe performs $$4$$ solos on one day. I thought we should expect Helen to perform $$1$$ solo on the same day $$(5-4=1)$$
But The answer given to me is: $$2.5$$ solos we expect Helen to play
My question is why? What is the way of thinking that gives me $$2.5$$? Is it cause of the coin flip? so $$5 \cdot .5 = 2.5$$? What does Joe's $$4$$ solos have to do with anything then?
Thank You for any help.
• Helen and Joe must have played $4$ or more solos on that day. In other words, you should be thinking about a conditional probability. – Toby Mak May 15 at 2:42
Let $$X$$ denote the total number of games played, and let $$J$$ denote the number of games Joe played. The conditional distribution of $$X$$ given $$J=4$$, namely $$X|J=4$$, is supported on $$\{4,5,\ldots\}$$ and has pmf $$P(X=k|J=4)=\frac{P(J=4|X=k)P(X=k)}{\sum_{k=4}^{\infty}P(J=4|X=k)P(X=k)}$$ which is non zero whenever $$k\geq 4$$. Using the facts that $$X\sim \text{Poisson}(5)$$ and $$J|X\sim \text{Binomial}(X,1/2)$$ we get $$P(X=k|J=4)=e^{-5/2}\frac{1}{(k-4)!}\bigg(\frac{5}{2}\bigg)^{k-4}$$ This means $$X-4|J=4\sim \text{Poisson}(5/2)$$ and so $$\mathbb{E}(X-4|J=4)=\mathbb{E}(X|J=4)-4=5/2$$ Hence $$13/2=E(X|J=4)$$ so expected number of times Helen plays is $$5/2$$
While Matthew Pilling and tommik provide answers that show the mathematics of getting the answer, I will provide the intuition involved.
1. We know that, on this specific day, Joe played 4 solos. This provides a minimum number of songs - specifically, there must have been at least 4 songs. Note that it is very possible for the Poisson distribution to produce 0 songs or 1 song. However, we have been given information that constrains the set of possible numbers of songs - it must be at least 4.
2. We know that Joe played exactly 4 solos - this is a lot more likely if there are 8 songs (~27%) than if there are 4 songs (~6%). This changes the likelihoods of each of the possibilities, compared with the basic Poisson distribution, given this information.
3. How many solos we expect Helen to have played can then be worked out from the new probabilities, which have incorporated the additional information (that Joe played 4 solos).
To see why the 50% information can't be directly used to conclude that Helen is expected to have played 4 solos as well, consider a slightly modified version of the problem. Rather than the number of songs following a Poisson distribution, we will assume that they follow a uniform distribution of between 1 and 7 songs.
Now, we know that Joe played 4 solos. How many solos do we expect Helen to have played? Well, it can't be 4, because that would mean they may have played 8 songs, which can't have happened - the maximum is 7 songs.
To work out the correct answer, we turn to Bayes' Theorem, which is explicitly used in Matthew's answer, and is hidden by proportionality in tommik's answer. Think of the fact that Joe played 4 solos as a "new piece of information". Bayes' Theorem (at least by Bayesian thinking) lets you update your probabilities given the new information.
The answer of @Matthew Pilling is perfect (+1). A Bayesian approach will lead to the same solution avoiding a lot of calculations: (constants are not considered until the end of the process)
$$\mathbb{P}[X|J=4]\propto \mathbb{P}[X]\cdot \mathbb{P}[J=4|X]\propto\frac{5^x}{x!}\cdot \binom{x}{4}\left(\frac{1}{2}\right)^x\propto\frac{\left(\frac{5}{2}\right)^{x}}{(x-4)!}\propto\frac{\left(\frac{5}{2}\right)^{(x-4)}}{(x-4)!}$$
Setting $$Y=X-4$$ we immediately recognize the kernel of a Poisson distribution with mean 2.5
Here $$Y$$ is the distribution of the solos played by Helen
$$\mathbb{P}[Y=y|\text{Joe }=4]=\frac{e^{-2.5}\cdot 2.5^y}{y!}$$
Here's some intuition:
The Poisson distribution is the limiting case for a binomial distribution where the number of trials goes to infinity while the success probability shrinks proportionally to keep the total expectation constant. So we can imagine Helen and Joe spending their lunch break making a very large number of Bernoulli trials, with each giving a small probability of "play a song now".
When the number of trials grows large we can fold the coin flip into that process, deciding that when "play a song now" comes out at trial number $$n$$, Helen plays the solo if $$n$$ is odd, and Joe plays the solo if $$n$$ is even.
But this effectively means that each of Helen and Joe might as well be making their own separate series of trials, with half as many trials but the same probability, and therefore half the expected value. The number of solos played by each is independent of the other, and still Poisson distributed.
The number of solos Helen plays is independent of the number of solos Joe plays, so his $$4$$ solos is not actually relevant information.
An alternative (and perhaps slightly more rigorous) phrasing of the same reinterpretation:
In each step, instead of first asking "should we play a song now?" and then "who should play the solo if we do play?", in each step we can ask "should Joe play now?" and "should Helen play now?" each independently with half the probability. That's almost the same, except that there's a finite chance the answer would be that both should play. But when we go to the limit of infinity many steps, the risk of that happening in any given step goes to zero as $$1/N^2$$, and so the probability of it ever happening during the entire lunch break goes to zero as $$1/N$$. Therefore the risk becomes irrelevant in the limit where the distributions become Poisson. | 2021-06-22T13:12:36 | {
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http://stacks.math.columbia.edu/tag/036M | The Stacks Project
Tag: 036M
This tag has label descent-section-fppf-local-source, it is called Properties of morphisms local in the fppf topology on the source in the Stacks project and it points to
The corresponding content:
31.24. Properties of morphisms local in the fppf topology on the source
Here are some properties of morphisms that are fppf local on the source.
Lemma 31.24.1. The property $\mathcal{P}(f)=$''$f$ is locally of finite presentation'' is fppf local on the source.
Proof. Being locally of finite presentation is Zariski local on the source and the target, see Morphisms, Lemma 25.22.2. It is a property which is preserved under composition, see Morphisms, Lemma 25.22.3. This proves (1), (2) and (3) of Lemma 31.22.3. The final condition (4) is Lemma 31.10.1. Hence we win. $\square$
Lemma 31.24.2. The property $\mathcal{P}(f)=$''$f$ is locally of finite type'' is fppf local on the source.
Proof. Being locally of finite type is Zariski local on the source and the target, see Morphisms, Lemma 25.16.2. It is a property which is preserved under composition, see Morphisms, Lemma 25.16.3, and a flat morphism locally of finite presentation is locally of finite type, see Morphisms, Lemma 25.22.8. This proves (1), (2) and (3) of Lemma 31.22.3. The final condition (4) is Lemma 31.10.2. Hence we win. $\square$
Lemma 31.24.3. The property $\mathcal{P}(f)=$''$f$ is open'' is fppf local on the source.
Proof. Being an open morphism is clearly Zariski local on the source and the target. It is a property which is preserved under composition, see Morphisms, Lemma 25.24.3, and a flat morphism of finite presentation is open, see Morphisms, Lemma 25.26.9 This proves (1), (2) and (3) of Lemma 31.22.3. The final condition (4) follows from Morphisms, Lemma 25.26.10. Hence we win. $\square$
Lemma 31.24.4. The property $\mathcal{P}(f)=$''$f$ is universally open'' is fppf local on the source.
Proof. Let $f : X \to Y$ be a morphism of schemes. Let $\{X_i \to X\}_{i \in I}$ be an fppf covering. Denote $f_i : X_i \to X$ the compositions. We have to show that $f$ is universally open if and only if each $f_i$ is universally open. If $f$ is universally open, then also each $f_i$ is universally open since the maps $X_i \to X$ are universally open and compositions of universally open morphisms are universally open (Morphisms, Lemmas 25.26.9 and 25.24.3). Conversely, assume each $f_i$ is universally open. Let $Y' \to Y$ be a morphism of schemes. Denote $X' = Y' \times_Y X$ and $X'_i = Y' \times_Y X_i$. Note that $\{X_i' \to X'\}_{i \in I}$ is an fppf covering also. The morphisms $f'_i : X_i' \to Y'$ are open by assumption. Hence by the Lemma 31.24.3 above we conclude that $f' : X' \to Y'$ is open as desired. $\square$
\section{Properties of morphisms local in the fppf topology on the source}
\label{section-fppf-local-source}
\noindent
Here are some properties of morphisms that are fppf local on the source.
\begin{lemma}
\label{lemma-locally-finite-presentation-fppf-local-source}
The property $\mathcal{P}(f)=$$f$ is locally of finite presentation''
is fppf local on the source.
\end{lemma}
\begin{proof}
Being locally of finite presentation is Zariski local on the source
and the target, see Morphisms,
Lemma \ref{morphisms-lemma-locally-finite-presentation-characterize}.
It is a property which is preserved under composition, see
Morphisms, Lemma \ref{morphisms-lemma-composition-finite-presentation}.
This proves
(1), (2) and (3) of Lemma \ref{lemma-properties-morphisms-local-source}.
The final condition (4) is
Lemma \ref{lemma-flat-finitely-presented-permanence-algebra}. Hence we win.
\end{proof}
\begin{lemma}
\label{lemma-locally-finite-type-fppf-local-source}
The property $\mathcal{P}(f)=$$f$ is locally of finite type''
is fppf local on the source.
\end{lemma}
\begin{proof}
Being locally of finite type is Zariski local on the source
and the target, see Morphisms,
Lemma \ref{morphisms-lemma-locally-finite-type-characterize}.
It is a property which is preserved under composition, see
Morphisms, Lemma \ref{morphisms-lemma-composition-finite-type}, and
a flat morphism locally of finite presentation is locally of finite type, see
Morphisms, Lemma \ref{morphisms-lemma-finite-presentation-finite-type}.
This proves
(1), (2) and (3) of Lemma \ref{lemma-properties-morphisms-local-source}.
The final condition (4) is
Lemma \ref{lemma-finite-type-local-source-fppf-algebra}. Hence we win.
\end{proof}
\begin{lemma}
\label{lemma-open-fppf-local-source}
The property $\mathcal{P}(f)=$$f$ is open''
is fppf local on the source.
\end{lemma}
\begin{proof}
Being an open morphism is clearly Zariski local on the source and the target.
It is a property which is preserved under composition, see
Morphisms, Lemma \ref{morphisms-lemma-composition-open}, and
a flat morphism of finite presentation is open, see
Morphisms, Lemma \ref{morphisms-lemma-fppf-open}
This proves
(1), (2) and (3) of Lemma \ref{lemma-properties-morphisms-local-source}.
The final condition (4) follows from
Morphisms, Lemma \ref{morphisms-lemma-fpqc-quotient-topology}.
Hence we win.
\end{proof}
\begin{lemma}
\label{lemma-universally-open-fppf-local-source}
The property $\mathcal{P}(f)=$$f$ is universally open''
is fppf local on the source.
\end{lemma}
\begin{proof}
Let $f : X \to Y$ be a morphism of schemes.
Let $\{X_i \to X\}_{i \in I}$ be an fppf covering.
Denote $f_i : X_i \to X$ the compositions.
We have to show that $f$ is universally open if and only if
each $f_i$ is universally open. If $f$ is universally open,
then also each $f_i$ is universally open since the maps
$X_i \to X$ are universally open and compositions
of universally open morphisms are universally open
(Morphisms, Lemmas \ref{morphisms-lemma-fppf-open}
and \ref{morphisms-lemma-composition-open}).
Conversely, assume each $f_i$ is universally open.
Let $Y' \to Y$ be a morphism of schemes.
Denote $X' = Y' \times_Y X$ and $X'_i = Y' \times_Y X_i$.
Note that $\{X_i' \to X'\}_{i \in I}$ is an fppf covering also.
The morphisms $f'_i : X_i' \to Y'$ are open by assumption.
Hence by the Lemma \ref{lemma-open-fppf-local-source}
above we conclude that $f' : X' \to Y'$ is open as desired.
\end{proof}
To cite this tag (see How to reference tags), use:
\cite[\href{http://stacks.math.columbia.edu/tag/036M}{Tag 036M}]{stacks-project}
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner). | 2013-05-19T14:20:59 | {
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https://math.stackexchange.com/questions/3219147/equation-of-the-line-that-lies-tangent-to-both-circles/3219165 | # Equation of the line that lies tangent to both circles
Consider the two circles determined by $$(x-1)^2 + y^2 = 1$$ and $$(x-2.5)^2 + y^2 = (1/2)^2$$. Find the (explicit) equation of the line that lies tangent to both circles.
I have never seen a clean or clever solution to this problem. This problem came up once at a staff meeting for a tutoring center I worked at during undergrad. I recall my roommate and I - after a good amount of time symbol pushing - were able to visibly see a solution by inspection, then verify it by plugging in. I have never seen a solid derivation of a solution to this though, so I would like to see what MSE can come up with for this!
I took a short stab at it today before posting, and got that it would be determined by the solution to the equation $$\left( \frac{\cos(\theta)}{\sin(\theta)} + 2\cos(\theta) + 3 \right)^2 -4\left( \frac{\cos(\theta)^2}{\sin(\theta)^2}+1 \right)\left(\frac{-\cos(\theta)^3}{\sin(\theta)^2}+\frac{\cos(\theta)^4}{\sin(\theta)^2}+3 \right).$$
The solution $$\theta$$ would then determine the line $$y(x) = \frac{-\cos(\theta)}{\sin(\theta)}(x) + \frac{\cos(\theta)^2}{\sin(\theta)} + \sin(\theta).$$
Not only do I not want to try and solve that, I don't even want to try expanding it out :/
• not $x=2$ :-) ? – J. W. Tanner May 9 at 0:32
• "the line"? I see 3 common tangents after plotting the circles. – peterwhy May 9 at 0:32
• math.stackexchange.com/questions/211538/… – lab bhattacharjee May 9 at 0:44
• @peterwhy Hi Peter. One of those 3 solutions is so trivial, I am offended you would even bring it up. For the remaining 2, one is clearly the other ones mirror image. So I am going to stick to my wording of 'find the line' since there is clearly one difficult one to find, which is what I am interested in seeing solutions for. Let me know when you have one, you can post answers below. – Prince M May 9 at 4:47
As peterwhy points out in the comments, there are three tangent lines. By inspection, one is $$x = 2$$, as pointed out by J.W. Tanner in the comments.
The other two can be identified by similar triangles. Suppose that we have a line tangent to both circles, and let the points of tangency be $$T_1$$ and $$T_2$$. Let the circle centers be $$O_1$$ and $$O_2$$. Finally, let the point where this line intersects the $$x$$-axis be called $$P$$. Then $$\triangle PO_1T_1$$ and $$\triangle PO_2T_2$$ are similar (do you see why?). Since $$O_1T_1 = 2O_2T_2$$, we must have $$PO_1 = 2PO_2$$, and therefore $$P$$ must be at $$(4, 0)$$. Note that $$PT_1 = \sqrt{3^2-1^2} = \sqrt{8}$$, and therefore our tangent line must have slope $$\pm \frac{1}{\sqrt{8}}$$.
(For simplicity, I only show one of the tangent lines; the other is its mirror image across the $$x$$-axis.) From this, we get the equation of the two remaining tangent lines
$$y = \pm \frac{x-4}{\sqrt{8}}$$
• This is a clean solution. Nice – Prince M May 9 at 4:52
The equation of the second circle can be written as $$x^2+y^2-5x+6=0$$.
Let $$P(h,k)$$ be a point on the second circle.
Then the equation of the tangent to the second circle at $$P$$ is $$hx+ky-\dfrac52(x+h)+6=0$$.
If it is also a tangent to the first circle, the distance from $$(1,0)$$ to this line is $$1$$.
\begin{align*} \frac{|h-\frac52(1+h)+6|}{\sqrt{(h-\frac52)^2+k^2}}&=1\\ \frac{|-\frac32h+\frac72|}{\sqrt{(\frac12)^2}}&=1\\ -3h+7&=\pm1 \end{align*} So, we have $$h=2$$ or $$h=\frac83$$.
If $$h=2$$, $$k=\pm\sqrt{(\frac12)^2-(2-\frac52)^2}=0$$ and the common tangent is $$x-2=0$$.
If $$h=\frac83$$, $$k=\pm\sqrt{(\frac12)^2-(\frac83-\frac52)^2}=\pm\frac{\sqrt{2}}{3}$$ and the common tangents are $$x\pm2\sqrt{2}y-4=0$$.
Use dual conics: If we represent a circle (in fact, any nondegenerate conic) with the homogeneous matrix $$C$$ so that its equation is $$(x,y,1)C(x,y,1)^T=0$$, lines $$\lambda x+\mu y+\tau=0$$ tangent to the circle satisfy the dual conic equation $$(\lambda,\mu,\tau)\,C^{-1}(\lambda,\mu,\tau)^T=0$$. (This equation captures the fact that the pole of a tangent line lies on the line.)
The matrix that corresponds to the circle $$(x-h)^2+(y-k)^2=r^2$$ is $$C=\begin{bmatrix}1&0&-h\\0&1&-k\\-h&-k&h^2+k^2-r^2\end{bmatrix}$$ with inverse $$C^{-1}=\frac1{r^2}\begin{bmatrix}r^2-h^2&-hk&-h\\-hk&r^2-k^2&-k\\-h&-k&-1\end{bmatrix}.$$ For the circles in this problem, the resulting dual equations are $$\mu^2-\tau^2-2\lambda\tau = 0 \\ -24\lambda^2+\mu^2-4\tau^2-20\lambda\tau = 0.$$ Both circles’ centers lie on the $$x$$-axis but their radii differ, so no common tangent is horizontal. Thus, we can set $$\lambda=1$$ and solve the slightly simpler system to obtain the solutions $$\mu=0$$, $$\tau=-2$$ and $$\mu=\pm2\sqrt2$$, $$\tau=-4$$, i.e., the three common tangent lines are $$x=2 \\ x\pm2\sqrt2 y=4.$$
This general method works for any pair of nondegenerate conics: it finds their common tangents by solving a dual problem of the intersection of two (possibly imaginary) conics. For a pair of circles, however, there’s a simple way to find common tangents via similar triangles, as demonstrated in Brian Tung’s answer.
To find a common tangent to two arbitrary circles, you can use the following trick: "deflate" the smaller circle (let $$1$$) so that it shrinks to a point, while the second shrinks to the radius $$r_2-r_1$$. Doing this, the direction of the tangent doesn't change and the tangency point forms a right triangle with the two centers.
By elementary trigonometry, the angle $$\phi$$ between the axis through the centers and the tangent is drawn from
$$d\sin\phi=r_2-r_1,$$ where $$d$$ is the distance between the centers. The direction of the axis is such that
$$\tan\theta=\frac{y_2-y_1}{x_2-x_1}.$$
So the equation of the tangent is given by
$$(x-x_1)\sin(\theta+\phi)-(y-y_1)\cos(\theta+\phi)=0.$$
The original tangent is a parallel at distance $$r_1$$, hence
$$(x-x_1)\sin(\theta+\phi)-(y-y_1)\cos(\theta+\phi)=r_1.$$
• Caution: quadrant discussion is missing. – Yves Daoust May 9 at 20:30
As already stated, there are three lines that can be drawn that are tangent to both circles. One is the trivial line $$x = 2,$$ which is tangent at the point $$(2,0)$$, which is also the common point of tangency of the two circles.
The other two lines are reflections of each other in the $$x$$-axis; these can be found by noting that a homothety that takes the point $$(0,0)$$ to $$(2,0)$$ and $$(2,0)$$ to $$(3,0)$$ will map the first circle to the second, and the center of this homothety will be a fixed point of this map.
To this end, let $$(x',y') = (a x + b, a y + d)$$, so we now solve the system $$(b, d) = (2,0) \\ (2a + b, d) = (3,0).$$ That is to say, $$b = 2$$, $$d = 0$$, $$a = 1/2$$, and the desired homothety is $$(x',y') = (x/2 + 2, y/2).$$ The unique fixed point is found by setting $$(x',y') = (x,y)$$ from which we obtain $$(x,y) = (4,0)$$. Thus the tangent lines pass through this point and have the form $$y = m(x - 4)$$ where $$m$$ is the slope. Such a line will be tangent if the system of equations $$(x-1)^2 + y^2 = 1, \\ y = m(x-4)$$ has exactly one solution. Eliminating $$y$$ gives the quadratic $$(m^2+1)x^2 - 2(4m^2+1)x + 16m^2 = 0,$$ for which the solution has a unique root if the discriminant is zero; i.e., $$0 = 4(4m^2+1)^2 - 4(m^2+1)(16m^2) = 4 - 32m^2.$$ Therefore, $$m = \pm 1/\sqrt{2}$$ and the desired lines are $$y = \pm \frac{x-4}{2 \sqrt{2}},$$ in addition to the previous line $$x = 2$$ described.
Beside the obvious solution of $$x=2$$ you also have two more common tangent lines, namely $$y= \pm \frac {\sqrt {2}}{4} (x-4)$$
The two non-obvious tangent lines pass through $$(4,0)$$ and their y-intercepts are respectively $$(0,\pm \sqrt {2})$$ | 2019-07-16T18:20:51 | {
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https://mathhelpboards.com/calculus-10/ap-calculus-exam-question-absolute-max-26537.html?s=a603e40390a25c7bfe2f8da38efa87a5 | Thread: AP calculus exam question on Absolute max
1. $\text{22. Let f be the function defined by$f(x)=\dfrac{\ln x}{x}$What is the absolute maximum value of f ? }$
$$(A)\, 1\quad (B)\, \dfrac{1}{e} (C)\, 0 \quad (D) -e \quad (E) f\textit{ does not have an absolute maximum value}.$$
I only guessed this by graphing it and it appears to $\dfrac{1}{e}$ which is (B)
2.
3. Originally Posted by karush
$\text{22. Let f be the function defined by$f(x)=\dfrac{\ln x}{x}$What is the absolute maximum value of f ? }$
$$(A)\, 1\quad (B)\, \dfrac{1}{e} (C)\, 0 \quad (D) -e \quad (E) f\textit{ does not have an absolute maximum value}.$$
I only guessed this by graphing it and it appears to $\dfrac{1}{e}$ which is (B)
Why would graphing be a guess? It's a valid Mathematical tool!
You could do this by taking the derivative of f(x) and finding the critical points, etc. But if you have this question on an exam the simplest (and probably fastest) way is to take a look at each answer and see what you get. D) is out because f(x) takes on positive values, and A), C), and E) are out by looking at the graph. That leaves B).
-Dan
4. $f’(x)=\dfrac{x \cdot \frac{1}{x} - \ln{x} \cdot 1}{x^2} = \dfrac{1-\ln{x}}{x^2}$
$f’(x)=0$ at $x=e$
first derivative test ...
$x < e \implies f’(x) > 0 \implies f(x) \text{ increasing over the interval } (0,e)$
$x > e \implies f’(x) < 0 \implies f(x) \text{ decreasing over the interval } (e, \infty)$
conclusion ... $f(e) = \dfrac{1}{e}$ is an absolute maximum.
second derivative test ...
$f’’(x) = \dfrac{x^2 \cdot \left(-\frac{1}{x} \right) - (1-\ln{x}) \cdot 2x}{x^4} = \dfrac{2\ln{x} - 3}{x^3}$
$f’’(e) = -\dfrac{1}{e^3} < 0 \implies f(e) = \dfrac{1}{e}$ is a maximum.
wow that was a lot of help..
yes the real negative about these assessment tests is how fast you can eliminate possible answers
not so much what math steps are you really need to take
actually Im learning a lot here at MHB
Mahalo | 2019-09-17T10:45:36 | {
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https://math.stackexchange.com/questions/2687523/to-find-the-smallest-integer-greater-than-23-with-certain-modular-residues | # To find the smallest integer greater than $23$ with certain modular residues
The question asks to find the smallest integer greater than $23$ that leaves the remainders $2,3,2$ when divided by $3,5,7$ respectively. The answer is given as $128$.
I used C.R.T. to solve this. $$x\equiv2\pmod 3\\ x\equiv3\pmod 5\\ x\equiv2\pmod 7$$ where $M=m_1·m_2·m_3=3×5×7=105$.
From here I get solutions, $$x\equiv2\pmod 3\\ x\equiv1\pmod 5\\ x\equiv1\pmod 7$$
After that I find\begin{align*} x &\equiv 2×2×5×7+1×3×3×7+1×2×3×5\\ &\equiv 140+63+30\\ &\equiv 233 \pmod{105}, \end{align*} where I again find that $x=23$.
But one thing I have noticed that $233=105×1+128$. I cannot continue from here. I need help and any help is highly appreciated.
• The "From here I get solutions..." is incomprehensible for me: how applying the CRT gives you those "solutions" and not the final one, which is unique modulo $\;3\cdot5\cdot7=105\;$ ? Mar 12 '18 at 8:34
• @thevbm The solutions are actually given by $23+105n$. Since substituting in $n=0$ gives $23$, you can substitute the next $n$ which is $n=1$. Mar 12 '18 at 8:35
• You could just observe that $23$ itself has the correct remainders modulo $3,5$ and $7$. By CRT the solutions form a single residue class modulo $3\cdot5\cdot7=105$, and you already know which class it is! So....? Mar 12 '18 at 8:35
Applying directly the following proof of CRT, for example:
$$\begin{cases}y_1=\frac{105}3=35\;\implies y_1=2\pmod3\implies y_1^{-1}=2\pmod3\\{}\\y_2=\frac{105}5=21\implies y_2=1\pmod5\implies y_2^{-1}=1\pmod5\;\\{}\\y_3=\frac{105}7=15\implies y_3=1\pmod7\implies y_3^{-1}=1\pmod7\end{cases}$$
so that
$$2\cdot35\cdot2+3\cdot21\cdot1+2\cdot15\cdot1=233\;\;\text{ is a solution, and}$$
$$233=128\pmod{105}\;\;\text{is another solution, and}$$
$$128=23\pmod105\;\;\text{ is the unique solution modulo}\;105$$
Observe that unless you consider things modulo $\;3\cdot5\cdot7=105\;$ at the end , there is not "the" solution, but only one out of infinitely many
• Yes, now I understand and thank you for providing the link.
– vbm
Mar 12 '18 at 9:03
• Following the link you provided, one question I have (from an example given there) to make my doubt clear. suppose system of congruences is given as $x\equiv 5 \mod 6$ $x\equiv 3 \mod 8$. Here g..c.d of the moduli is $2$. The first congruence implies $x\equiv 1\mod 2$, also the second congruence implies $x\equiv 1\mod 2$Now to reduce the system of congruency to a simpler form if we divide the g.c.d of the moduli from the modulud of the first congruence, then how is it $x\equiv2 \mod3$? rather it should be $x\equiv 1 \mod 3$. If I'm wrong, please correct me. Thank you
– vbm
Mar 12 '18 at 9:38
• @thevbm I'm not sure what your doubt exactly is, but remember the CRT applies for coprime moduli, so reducing your moduli as you did can help...but then you should be careful when going back to the original ones. Mar 12 '18 at 11:53
Let $x\equiv y \equiv 2 \mod 3,$ and $x\equiv y \equiv 3 \mod 5,$ and $x\equiv y \equiv 2 \mod 7.$ Then $x-y \equiv 0$ modulo $3,5,$ and $7.$
So $x-y$ is divisible by $3,5,$ and $7,$ so $x-y$ is divisible by $(3)(5)(7)=105.$ So if $y=23$ and $x>y$ then $x=23+105n$ for some $n\in \Bbb N.$
And if $x=23+105n$ for some $n\in \Bbb N$ then $x\equiv 23$ modulo $3,5,$ and $7.$ | 2021-12-08T23:41:45 | {
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https://cs.stackexchange.com/questions/98756/sum-of-unique-elements-in-all-sub-arrays-of-an-array | # Sum of unique elements in all sub-arrays of an array
Given an array $$A$$, sum the number of unique elements for each sub-array of $$A$$. If $$A = \{1, 2, 1, 3\}$$ the desired sum is $$18$$.
Subarrays:
{1} - 1 unique element
{2} - 1
{1} - 1
{3} - 1
{1, 2} - 2
{2, 1} - 2
{1, 3} - 2
{1, 2, 1} - 2
{2, 1, 3} - 3
{1, 2, 1, 3} - 3
I have a working solution which sums the unique elements for all sub-arrays starting at index $$0$$, then repeats that process at index $$1$$, etc. I have noticed that for an array of size $$n$$ consisting of only unique elements, the sum I desire can be found by summing $$i(i + 1) / 2$$ from $$i = 1$$ to $$n$$, but I haven't been able to extend that to cases where there are non-unique elements. I thought I could use that fact on each sub-array, but my control logic becomes unwieldy. I've spent considerable time trying to devise a solution better than $$O(n^2)$$ to no avail. Is there one?
Secondary question: if there is no better solution, are there any general guidelines or hints to recognize that fact?
• An incremental algorithm is supposedly easier to come up with in this case, I think so. – Thinh D. Nguyen Oct 18 '18 at 7:16
• I don't understand the line {1, 2, 1} - 2. The multiset {1, 2, 1} only contains one unique element: 2. – Peter Taylor Oct 18 '18 at 10:28
• Peter, yes number of distinct elements might be better phrasing. If all elements of each sub-array are added to a set, the set size is the desired metric. In either case, the behavior in my example is what I'm looking for. – User12345654321 Oct 18 '18 at 17:34
Hint: Use an extra $$O(n)$$-spaced array of pointer to the last (biggest) index of each distinct value. Then, this array is very helpful when you do your above algorithm.
Lastly, we should have an $$O(n)$$ algorithm.
• This was a good hint. Assume zero-indexed arrays. Starting with sum = maxPossibleSum, iterate through the array adding elements to a set. If the added element is a duplicate, sum = sum - (indexPreviousOccurrence + 1) * (totalElements - currentIndex). This requires just one pass leading to O(n) like you said. Thanks for a quality nudge in the right direction. – User12345654321 Oct 18 '18 at 20:31
If the maximum number $$k$$ is not too high, a Counting Sort based solution can be used;
In the first step, we count the number of elements into the array $$C$$ by
for j = 1 to length(A)
C[A[j]] = C[A[j]]+1
now the array contains all the information we need to sum;
sum = 0;
for j = 1 to k
if C[i] > 1
sum = C[i] * i
Complexity:
$$\mathcal{O}(n)$$ additions (increment) for the counting step
$$\mathcal{O}(n)$$ additions for the summation step
$$\mathcal{O}(n)$$ multiplications for the summation step
result $$\mathcal{O}(n)$$.
Note: indeed the multiplications are not necessary since we will have at most $$n-1$$ addition.
Let $$a_1,\ldots,a_m$$ be the distinct values. Now consider the positions of $$a_i$$'s in $$A$$. Assume the number of $$a_i$$'s is $$b_i$$ and the positions are as follows:
(x_{i0} non-a_i's) a_i (x_{i1} non-a_i's) a_i ... a_i (x_{ib_i} non-a_i's)
In your example $$A=\{1,2,1,3\}$$, when considering the value $$a_1=1$$, we have $$x_{10}=0,x_{11}=1,x_{12}=1$$ because the positions of $$1$$s are like 1 * 1 *: there is $$0$$ element before the first $$1$$, $$1$$ element between the first $$1$$ and the second $$1$$, and $$1$$ element after the second $$1$$.
Then there are \begin{align} &\sum_{j=0}^{b_i} \frac{x_{ij}(x_{ij}+1)}{2}\\ =\ &\frac{1}{2}\sum_{j=0}^{b_i}x_{ij}^2+\frac{1}{2}\sum_{j=0}^{b_i}x_{ij}\\ =\ &\frac{1}{2}\sum_{j=0}^{b_i}x_{ij}^2+\frac{1}{2}(n-b_i) \end{align} subarrays that do not contain $$a_i$$. Note there are $$n(n+1)/2$$ subarrays in total, so the final result we want is \begin{align} &\sum_{i=1}^m\left(\frac{n(n+1)}{2}-\frac{1}{2}\sum_{j=0}^{b_i}x_{ij}^2-\frac{1}{2}(n-b_i)\right)\\ =\ &\frac{1}{2}\left(mn^2+n-\sum_{i=1}^m\sum_{j=0}^{b_i}x_{ij}^2\right). \end{align}
To calculate $$\sum_{i=1}^m\sum_{j=0}^{b_i}x_{ij}^2$$, you can scan the array and maintain a lookup table that, for each distinct value, keeps the position of the last element with this value. With this table, when the $$(j+1)$$-th $$a_i$$ is scanned, you can compute $$x_{ij}$$ easily. This leads us to an $$O(n\log n)$$ solution (or $$O(n)$$ in average if you use a good hash table to implement the lookup table).
• could you clarify what X represents? It looks like a 2D array. This is the part I'm not understanding "(x_{i0} non-a_i's) a_i ..." – User12345654321 Oct 18 '18 at 17:17
• @User12345654321 In your example $A=\{1,2,1,3\}$, when considering the value $a_1=1$, we have $x_{10}=0,x_{11}=1,x_{12}=1$ because the positions of 1s are like 1 * 1 *: there is 0 element before the first 1, 1 element between the first 1 and the second 1, and 1 element after the second 1. – xskxzr Oct 19 '18 at 3:18 | 2019-05-27T11:41:24 | {
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https://math.stackexchange.com/questions/2992202/why-must-r-be-put-in-modulus-for-int-infty-0-frac1-cosrxx2dx-fra/2992629 | # Why must $r$ be put in modulus for $\int^{\infty}_0\frac{1-\cos(rx)}{x^2}dx=\frac{\pi}{2}|r|$?
Here's the question: Prove that $$\int^{\infty}_0\frac{1-\cos(rx)}{x^2}~\mathrm dx=\frac{\pi}{2}|r|·$$
When I finished my working, I got $$\frac{\pi}{2}r$$
Why is the $$r$$ put in modulus? Is it because the graph lies entirely above the x axis?
• Did you assume $r\geq 0$ in your working? Nov 10 '18 at 3:18
• The integral expression is visibly even in $r$. As for where the absolute value comes up in your derivation, that depends on your method, I suppose. Nov 10 '18 at 3:21
• Nope I didn't assume r greater than 0 as I thought there wasn't any need to do so Nov 10 '18 at 3:24
• Notice that for any real number $a$, $\cos(a) = \cos(-a)$. In particular, it suffices to solve the problem in the case $r \geq 0$. Nov 10 '18 at 3:49
• Note: the integrand is positive, so you could you get a negative answer? Nov 10 '18 at 13:55
Integrating by part: $$\int_0^{+\infty} \frac {1-\cos(rx)}{x^2}\, \mathrm dx = \int_{+\infty}^0 (1 - \cos(rx)) \mathrm d \frac 1 x =\left. \frac {1-\cos(rx)}x \right\vert_{+\infty}^0 - r\int_{+\infty}^0 \frac {\sin(rx)}x\, \mathrm dx = r \int_0^{+\infty} \frac {\sin(rx)}x \,\mathrm dx.$$ Now $$\DeclareMathOperator\sgn{sgn} \int_0^{+\infty} \frac {\sin(rx)}x \,\mathrm dx =\int_0^{+\infty} \frac {\sin(rx)}{rx}\,\mathrm d(rx)= \begin{cases}\displaystyle \int_0^{+\infty} \frac {\sin u}u\, \mathrm du, & r > 0,\\ 0, & r=0,\\ \displaystyle \int_{-\infty}^0 \frac {\sin u}u\, \mathrm du, & r < 0, \end{cases} = \sgn r \cdot \int_0^{+\infty} \frac {\sin u}u\, \mathrm du = \sgn r \cdot \frac \pi 2,$$ hence the integral equals $$r \sgn r \cdot \frac \pi 2 = \frac \pi 2 \vert r \vert.$$
The implicit assumption could be $$r>0$$ when dealing the integral $$\int_0^{+\infty} \sin(rx)\,\mathrm dx /x$$.
The other answers explain how you get $$|r|$$ in your answer, but here’s why it should happen:
Notice that $$1-\cos(rx) \geq 0$$ for all $$x$$, regardless of what $$r$$ is. Integrating a nonnegative function over any interval should give you a nonnegative value. | 2022-01-27T15:42:49 | {
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https://math.stackexchange.com/questions/2810593/spivaks-calculus-finding-the-formula-for-the-sum-of-a-series | # Spivak's Calculus: Finding the formula for the sum of a series. [duplicate]
This question already has an answer here:
In Spivak's Calculus, Chapter 2, the second problem of the set asks you to find a formula for the following series:
$$\sum_{i=1}^n(2i-1)$$ and $$\sum_{i=1}^n(2i-1)^2$$ Now, for the former, this was fairly straightforward: it sums to $i^2$ and I have a proof that I'm happy with. However, for the latter, I cannot identify a pattern in terms of $n$ to begin building a proof. The series proceeds $1^2 + 3^2 + 5^2 + 7^2 + \cdots + (2n-1)^2$, so the sum for the first few indices would be $1, 10, 35,$ and $84.$ Not only have I failed to come up with an expression of this in terms of $n$, but I'm not even sure what I should be considering to lead me to such a formula.
## marked as duplicate by Hans Lundmark, Claude Leibovici, ccorn, cansomeonehelpmeout, ChristopherJun 8 '18 at 12:16
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
• Are you also supposed to give a direct proof by induction, or can you use other known formulæ like the sum of the first $2n$ squares? – Bernard Jun 6 '18 at 20:33
## 5 Answers
Hint : Expand the summand & use \begin{eqnarray*} \sum_{i=1}^{n} i^2 =\frac{n(n+1)(2n+1)}{6}. \end{eqnarray*} You should get to \begin{eqnarray*} \sum_{i=1}^{n} (2i-1)^2 =\frac{n(4n^2-1)}{3}. \end{eqnarray*}
• Thanks to a combination of your hint and dezdichado's below, I managed to figure it out. Thank you very much! – user242007 Jun 6 '18 at 21:44
If you know what $1^2+2^2+...+n^2=f(n)$ sums up to, then your desired sum is simply: $$1^2+3^2+...+(2n-1)^2 = f(2n) - (2^2+4^2+...+(2n)^2) = f(2n) - 4f(n).$$
• Your comment helped me become more aware of incorporating existing known formulae in these exercises. Between that and Donald's comment above, I was able to solve the problem. Thank you. – user242007 Jun 6 '18 at 21:44
$$\sum_{i=1}^n (2i-1)^2$$
$$=\sum_{i=1}^n 4i^2-4i+1$$
$$=4\sum_{i=1}^n i^2-4\sum_{i=1}^n i+\sum_{i=1}^n1$$
• Could you elaborate on this? In trying to implement it with an example value of i = 4, I end up with 81, when the actual value in the series appears to be 84. – user242007 Jun 6 '18 at 21:26
$\sum_\limits{k=1}^n (2k - 1)^2 = \sum_\limits{k=1}^n 4k^2 - \sum_\limits{k=1}^n 4k + \sum_\limits{k=1}^n 1 = 4\left(\sum_\limits{k=1}^n k^2\right) - 2n(n+1) + n$
You need a way to find $\sum_\limits{k=1}^n k^2$
There are lots of other ways to derive this... but this one is nice.
Write the numbers in a triangle
$\begin {array}{} &&&1\\&&2&&2\\&3&&3&&3\\&&&\vdots&&\ddots\\n&&n&\cdots\end{array}$
The sum of each row is $k^2.$ And, the sum of the whole trianglular array is $\sum_\limits{k=1}^n k^2$
Take this triangle and rotate it 60 degrees clockwise and 60 degrees counter clockwise and sum the 3 triangles together.
$\begin {array}{} &&&1\\&&2&&2\\&3&&3&&3\end{array}+\begin {array}{} &&&n\\&&n-1&&n\\&n-2&&n-1&&n\\\end{array}+\begin {array}{} &&&n\\&&n&&n-1\\&n&&n-1&&n-2\end{array}$
And we get:
$\begin {array}{} &&&2n+1\\&&2n+1&&2n+1\\&2n+1&&2n+1&&2n+1\\&&&\vdots&&\ddots\end{array}$
What remains how many elements are in the triangular array? $\sum_\limits{k=1}^n k$
$3\sum_\limits{k=1}^n k^2 = (2n+1)\sum_\limits{k=1}^n k\\ \sum_\limits{k=1}^n k^2 = \frac {n(n+1)(2n+1)}{6}$
Another way to do it is:
$\sum_\limits{k=1}^n (k+1)^3 - k^3 = (n+1)^3-1\\ \sum_\limits{k=1}^n (3k^2 + 3k + 1)\\ \sum_\limits{k=1}^n 3k^2 + \sum_\limits{k=1}^n 3k + \sum_\limits{k=1}^n 1\\ \sum_\limits{k=1}^n 3k^2 + 3\frac {n(n+1)}{2} + n = n^3 + 3n^2 + 3n \\ 3\sum_\limits{k=1}^n k^2 = n^3 + \frac 32 n^2 + \frac 12 n\\ \sum_\limits{k=1}^n k^2 = \frac 16 (n)(n+1)(2n+1)$
• I like the rotation method, that works out very nicely! – abiessu Jun 6 '18 at 20:59
• It is nice in that it is intuitive, but the other method can be generalized. – Doug M Jun 6 '18 at 21:01
You can do it by telescoping. Given a sequence $f(n)$ define the sequence $\Delta f(n)=f(n+1)-f(n)$ and put $(n)_m=n(n-1)\dotsb(n-m+1)$. One can check that $\Delta (n)_m=m(n)_{m-1}$ for $m\geq 1$. Now note that $$(2i-1)^2=4i^2-4i+1=4[(i)_2+i] -4i+1=4(i)_2+1$$ whence $$\Delta\left(\frac{4}{3}(i)_3+i\right)=(2i-1)^2.$$ In particular $$\sum_{i=1}^n (2i-1)^2=\sum_{i=1}^n \Delta\left(\frac{4}{3}(i)_3+i\right)=\left[\frac{4}{3}(i)_3+i\right]_{i=1}^{n+1}$$ with the notation $[g(i)]_{i=a}^b=g(b)-g(a)$. | 2019-10-22T13:34:51 | {
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https://stats.stackexchange.com/questions/431233/probability-of-a-500-year-flood-occuring-in-the-next-100-years-comparison-of-a | # Probability of a 500 year flood occuring in the next 100 years - comparison of approaches
I'm looking at this problem
A $$500$$-year flood is one that occurs once in every $$500$$ years.
a) What is the probability of having at least $$3$$ such floods in $$500$$ years?
b) What is the probability that a flood will occur within the next $$100$$ years?
c) What is the expected number of years until the next flood?
Attempt:
a) The window size is $$500$$ years, $$\lambda=1$$ and the number of floods is a Poisson variable $$X$$. So the answer would be $$1-P(X=0)-P(X=1)-P(X=2)$$, where $$P(X=n)$$ is the Poisson pmf.
c) I can consider a window size of $$1$$ year, $$\lambda=\frac{1}{500}$$ and consider the time to next flood as the exponential random variable $$T$$, so that $$E[T]=1/\lambda=500$$.
Are the above two correct?
Finally for part b), one approach is to decide on a window size, let's say $$1$$ year, set $$\lambda$$ appropriately (in this case $$\frac{1}{500}$$), let $$T$$ be the time to next flood (exponential RV), and find $$P(T=1)+P(T=2)+P(T=3)+\ldots+P(T=100)$$
The other approach is to set the window to $$100$$ years, set $$\lambda=\frac{100}{500}=0.2$$, model number of floods as a Poisson RV $$X$$ and find $$1-P(X=0)$$.
Numerically, these approaches give slightly different answers ($$0.1811$$ vs $$0.1813$$). Which of the two approaches is better for this purpose (i.e. which gives a more accurate answer)?
Also, in the first approach to part b), instead of a one-year window, I could've taken a half-year window and set $$\lambda=\frac{1}{1000}$$ and summed from $$P(T=1)$$ to $$P(T=200)$$. And of course there's no limit to how small I can set that window. What is the recommended granularity of time windows for problems like these?
• I would write > in the first two questions, and < in the final one - given that climate change means that all estimates based on historical rates will underestimate future rates. In other words, what we consider a 500 year flood today, will not be a 500 year flood tomorrow. Oct 18, 2019 at 12:14
Your treatment of (a) and (c) are correct. In (b), the second approach is the correct one because in your first approach, you treat exponential RVs as if they're discrete. You should've done the following ($$\lambda=1/500$$): $$P(T\leq100)=\int_0^{100}\lambda e^{-\lambda t}dt=1-e^{-100\lambda}=1-e^{-0.2}$$ which gives the same answer as your Poisson approach. By the way, for (b), if you discretize at infinitely small steps, you get this integral. | 2023-03-22T02:20:57 | {
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https://math.stackexchange.com/questions/1941023/find-f-if-fx-dfracx2-1x-knowing-that-f1-dfrac12-and-f | # Find $f$ if $f'(x)=\dfrac{x^2-1}{x}$ knowing that $f(1) = \dfrac{1}{2}$ and $f(-1) = 0$
I am asked the following problem:
Find $f$ if $$f'(x)=\frac{x^2-1}{x}$$
I am not sure about my solution, which I will describe below:
My solution:
The first thing that I've done is separate the terms of $f'(x)$
\begin{align*} f'(x)&=\frac{x^2-1}{x}\\ &=x-\frac{1}{x}\\ \therefore \quad f(x)&=\frac{x^2}{2}-\ln|x|+c \end{align*}
For ( x > 0 ):
\begin{align*} f(x)&=\frac{x^2}{2}-\ln x+c\\ f(1)&=\frac{1^2}{2}-\ln 1+c=\frac{1}{2} \quad \Rightarrow \quad c=0\\ f(x)&=\frac{x^2}{2}-\ln |x| \end{align*}
For ( x < 0 )
\begin{align*} f(x)&=\frac{x^2}{2}-\ln (-x)+c\\ f(-1)&=\frac{(-1)^2}{2}-\ln [-(-1)]+c=0 \quad \Rightarrow \quad c=-\frac{1}{2}\\ f(x)&=\frac{x^2}{2}-\ln |x|-\frac{1}{2} \end{align*}
Is my solution correct? Should I really find two different answers, one for $x > 0$ and another for $x < 0$?
Thank you.
• yes your solution is fine. As the domain has two connected components there can indeed be different constants of integration on each of them. – H. H. Rugh Sep 25 '16 at 17:03
• It looks fine to me. +1 – DonAntonio Sep 25 '16 at 17:03
• Related: math.stackexchange.com/questions/234624/… (see the accepted answer) – Winther Sep 26 '16 at 1:18
Looks fine! However a little typo when calculate $f(-1)=0$ not $-1$ | 2021-03-04T00:31:05 | {
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https://math.stackexchange.com/questions/525334/vandermonde-determinant-by-induction | # Vandermonde determinant by induction
For $$n$$ variables $$x_1,\ldots,x_n$$, the determinant $$\det\left((x_i^{j-1})_{i,j=1}^n\right) = \left|\begin{matrix} 1&x_1&x_1^2&\cdots & x_1^{n-1}\\ 1&x_2&x_2^2&\cdots & x_2^{n-1} \\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 1&x_{n-1}&x_{n-1}^2&\cdots&x_{n-1}^{n-1}\\ 1 &x_n&x_n^2&\cdots&x_n^{n-1} \end{matrix}\right|$$ can be computed by induction; the image below says it shows that. I have done this before, if I submit this will I get marks?
MORE IMPORTANTLY how do I do it by induction? The "hint" is to get the first row to $$(1,0,0,...,0)$$.
I think there are the grounds of induction in there, but I'm not sure how (I'm not very confident with induction when the structure isn't shown for $$n=k$$, assume for $$n=r$$, show if $$n=r$$ then $$n=r+1$$ is true.
By the way the question is to show the determinant at the start equals the product $$\prod_{1\leq i (but again, explicitly by induction)
• There's a nice non-inductive proof. Would that work for you? – Bruno Joyal Oct 14 '13 at 2:38
• Actually, my memory tricked me. There is an induction. I'm posting it below. – Bruno Joyal Oct 14 '13 at 2:52
• ProofWiki has two proofs using mathematical induction. – user1551 Oct 14 '13 at 6:42
You're facing the matrix
\begin{pmatrix} 1&1&\cdots & 1 &1\\a_1&a_2&\cdots &a_n&a_{n+1}\\\vdots&\vdots&\ddots&\vdots&\vdots\\a_1^{n-1}&a_2^{n-1}&\cdots&a_n^{n-1}& a_{n+1}^{n-1}\\a_1^{n}&a_2^{n}&\cdots&a_n^{n}&a_{n+1}^{n}\end{pmatrix}
By subtracting $a_1$ times the $i$-th row to the $i+1$-th row, you get
\begin{pmatrix} 1&1&\cdots & 1 &1\\ 0&a_2-a_1&\cdots &a_n-a_1&a_{n+1}-a_1\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ 0&a_2^{n}-a_1a_{2}^{n-1}&\cdots&a_n^n-a_1a_{n}^{n-1}& a_{n+1}^{n}-a_1a_{n+1}^{n-1}\end{pmatrix}
Expanding by the first column and factoring $a_i-a_1$ from the $i$-th column for $i=2,\ldots,n+1$, you get the determinant is
$$=\prod_{j=2}^{n+1}(a_j-a_1) \det\begin{pmatrix} 1& 1&\cdots&1\\a_2&a_3&\cdots&a_{n+1}\\ \vdots&\vdots&\ddots&\vdots\\a_2^{n-1}&a_3^{n-1}&\cdots&a_{n+1}^{n-1}\end{pmatrix}$$
You may apply your inductive hypothesis, to get this is $$=\prod_{j=2}^{n+1}(a_j-a_1) \prod_{2\leqslant i<j\leqslant n+1}(a_j-a_i)=\prod_{1\leqslant i<j\leqslant n+1}(a_j-a_i)$$ and the inductive step is complete.
• Can you please explain how it is possible to factor out $a_i - a_1$? – john.abraham Sep 2 '15 at 10:05
• $a_j^{i} - a_1 a_j^{i-1} = a_j^{i-1}(a_j - a_1)$ – D_S Oct 1 '15 at 4:59
• Even though it is clearly stated, gratuitous change of notation with respect to the question (transposing the matrix and renaming the $x_i$ to $a_i$) is not really helpful, especially in the presence of another answer that does keep the original notation (but uses $a_i$ for a different purpose). – Marc van Leeuwen Dec 3 '18 at 10:17
Let $f(T) = T^{n-1} + a_1 T^{n-2} + \dots + a_{n-1}$ be the polynomial $(T-x_2)(T-x_3)\dots (T-x_n)$. By adding to the rightmost column an appropriate linear combination of the other columns (namely the combination with coefficients $a_1, \dots, a_{n-1}$), we can make sure that the last column is replaced by the vector $(f(x_1), 0, 0, \dots, 0)$, since by construction $f(x_2) = \dots = f(x_n) =0$. Of course this doesn't change the determinant. The determinant is therefore equal to $f(x_1)$ times $D$, where $D$ is the determinant of the $(n-1) \times (n-1)$ matrix which is obtained from the original one by deleting the first row and the last column. Then, we apply the induction hypothesis, using the fact that $f(x_1) = (x_1 - x_2)(x_1-x_3) \dots (x_1-x_n)$. And we are done!
By the way, this matrix is known as a Vandermonde matrix.
I learned this trick many years ago in Marcus' Number fields.
• Sexy! I should read this book if it has more fancy tricks like that. – Patrick Da Silva Dec 24 '13 at 15:06
• You get the wrong product (it should have factors $x_j-x_i$ when $i<j$, rather than factors $x_i-x_j$, or else you should throw in a sign $(-1)^{\binom n2}$ for the Vandermonde determinant). The explanation is that you produce a factor $f(x_1)$ that is not on the main diagonal, and this means in the induction step you get a sign $(-1)^{n-1}$ in the development of the determinant. This sign could have been avoided by producing a nonzero coefficient in the last entry of the final column instead, using a monic polynomial of degree $n-1$ with roots at $x_1,\ldots,x_{n-1}$ but not at $x_n$. – Marc van Leeuwen Dec 3 '18 at 10:10
Although this does not really answer the question (in particular, I cannot tell whether you whether submitting the image will give you any marks), it does explain the evaluation of the determinant (and one can get an inductive proof from it if one really insists, though it would be rather similar to the answer by Bruno Joyal).
The most natural setting in which the Vandermonde matrix $$V_n$$ arises is the followig. Evaluating a polynomial over$$~K$$ in each of the points $$x_1,\ldots,x_n$$ of $$K$$ gives rise to a linear map $$K[X]\to K^n$$ i.e., the map $$f:P\mapsto (P[x_1],P[x_2],\ldots,P[x_n])$$ (here $$P[a]$$ denotes the result of substituting $$X:=a$$ into$$~P$$). Then $$V_n$$ is the matrix of the restriction of $$f$$ to the subspace $$\def\Kxn{K[X]_{ of polynomials of degree less than$$~n$$, relative to the basis $$[1,X,X^2,\ldots,X^{n-1}]$$ of that subspace. Any family $$[P_0,P_1,\ldots,P_{n-1}]$$ of polynomials in which $$P_i$$ is monic of degree$$~i$$ for $$i=0,1,\ldots,n-1$$ is also a basis of$$~\Kxn$$, and moreover the change of basis matrix$$~U$$ from the basis $$[1,X,X^2,\ldots,X^{n-1}]$$ to $$[P_0,P_1,\ldots,P_{n-1}]$$ will be upper triangular with diagonal entries all$$~1$$, by the definition of being monic of degree$$~i$$. So $$\det(U)=1$$, which means that the determinant of$$~V_n$$ is the same as that of the matrix$$~M$$ expressing our linear map on the basis $$[P_0,P_1,\ldots,P_{n-1}]$$ (which matrix equals $$V_n\cdot U$$).
By choosing the new basis $$[P_0,P_1,\ldots,P_{n-1}]$$ carefully, one can arrange that the basis-changed matrix is lower triangular. Concretely column$$~j$$ of$$~M$$, which describes $$f(P_{j-1})$$ (since we number columns from$$~1$$), has as entries $$(P_{j-1}[x_1],P_{j-1}[x_2],\ldots,P_{j-1}[x_n])$$, so lower-triangularity means that $$x_i$$ is a root of $$P_{j-1}$$ whenever $$i. This can be achieved by taking for $$P_k$$ the product $$(X-x_1)\ldots(X-x_k)$$, which is monic of degree $$k$$. Now the diagonal entry in column$$~j$$ of$$~M$$ is $$P_{j-1}[x_j]=(x_j-x_1)\ldots(x_j-x_{j-1})$$, and $$\det(V_n)$$ is the product of these expressions for $$j=1,\ldots,n$$, which is $$\prod_{j=1}^n\prod_{i=1}^{j-1}(x_j-x_i)=\prod_{1\leq i. | 2019-08-19T21:30:45 | {
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https://www.physicsforums.com/threads/sum-of-log-squared-terms.627139/ | # Sum of log squared terms?
1. Aug 10, 2012
### yogo
Hi all,
I have a summation series which goes like this,
S = [log(1)]^2 + [log(2)]^2 + [log(3)]^2 + ....... + [log(n)]^2
Since each term is square of the logarithm of a value, I think there are no general tricks which can be used to solve this.
But is there any way to approximate the value of this summation?
The problem I am facing here is that n is very very large and it is not possible (time-wise and memory-wise) for my program to evaluate every term. So I was hoping for some approximate solution in terms of 'n'.
Thanks,
yogo
2. Aug 10, 2012
### micromass
Well, we know that since $f(x)=(log(x))^2$ is increasing (if x>1)
$$\int_2^{n+1} \log(x-1)^2dx \leq \sum_{n=1}^n \log(n)^2 \leq \int_1^{n+1} \log(x)^2 dx$$
So, if you calculate the integral, then you got a sweet upper and lower bound for the sum.
3. Aug 11, 2012
### coolul007
can't this be a manipulated into a product of n? log(2)+log(3) = log((2)(3)), I'm uncertain what to do with the squared.
4. Aug 11, 2012
### micromass
Not really. But the integral
$$\int \log(x)^2dx$$
has a nice elementary solution. Just integrate by parts twice.
5. Aug 11, 2012
### Mute
In a similar vein, there is also the Euler-Maclaurin formula for asymptotic expansions of sums.
$$\sum_{k=a}^{k=n} f(k) \sim \int_a^n dx~f(x) + \frac{f(a)+f(n)}{2} + \mbox{corrections}.$$
(The full form of the expansion is given on the wikipedia page I linked to). This looks like it would be useful in this case.
6. Aug 11, 2012
### yogo
Thanks for your help guys. Appreciate it.
@micromass: I am not so good at Math. I was wondering how tight these bounds are?
Thanks,
yogo
7. Aug 15, 2012
### chingel
You would have to calculate the integral to find out. Or you can let an online calculator do it. To find the tightness of the bonds, calculate the difference between the integrals, or in other words the integral from n to n+1.
http://www.wolframalpha.com/widgets/view.jsp?id=8ab70731b1553f17c11a3bbc87e0b605
Using Wolfram Alpha the integral from 10 to 11 for example is ~5.5, so if you take the average of them you get a maximal error of ~2.75, actual error would probably be much less, since the function doesn't jump too much, and is rather straight for intervals of width 1. Lower bound is ~25, and using Wolfram Alpha to calculate the sum directly gives ~27.65. If you want more accuracy, you could calculate a few terms of the sum directly, and then get the upper and lower bounds for the rest, for the upper bound it is the sum from 1 to m-1 plus the integral from m to n+1, and for the lower bound the sum from 1 to m-1 plus the integral from m-1 to n. The difference of the upper and lower bounds now turns out to be the integral from n to n+1 minus the integral from m-1 to m, which is better than just the integral from n to n+1.
So for example if you want the sum of the first 1000 terms and you know the first 100 terms exactly, you could do something like this. Sum of first 100 terms is 1408.33. Lower bound is 1408.33+34501.8=35910.13, higher bound is 1408.33+34528.3=35936.63, average is 35923.38. Letting Wolfram Alpha calculate the sum itself, it gets the answer 35923.4, I don't know how it calculates the sum itself and how much it rounded that answer, but at least it shows that taking the average of the bounds is not too bad. So now you have an approximation of the sum of the first 1000 terms, you can now use that to calculate the sum of the first 10000 terms etc.
8. Aug 16, 2012
### chingel
I also tried the Euler-Maclaurin formula. Adding only first two terms from the sum with the Bernoulli numbers in it, it got the sum of the first 1000 terms of log(x)^2 accurate to within 0.0015, so it seems a lot better than just taking the average of integral bounds, especially if you would include more of the correction terms. You could probably get very accurate with more correction terms.
9. Aug 16, 2012
### yogo
@chingel, @mute: Thanks for all the help. I am using Euler-Maclaurin formula now, seems to work really well for me. I actually have an upcoming deadline, and now I am good! :)
Thanks all. | 2018-08-20T15:13:42 | {
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https://math.stackexchange.com/questions/2792492/minimizing-a-symmetric-sum-of-fractions-without-calculus | # Minimizing a symmetric sum of fractions without calculus
Calculus tools render the minimization of $$\frac{(1-p)^2}{p} + \frac{p^2}{1-p}$$ on the interval $(0,1/2]$ to be a trivial task. But given how much symmetry there is in this expression, I was curious if one could use olympiad-style manipulations (AM-GM, Cauchy Schwarz, et al) to minimize this, show it is decreasing, or to show that it is bounded below by 1.
• The squares alone tell me to use Cauchy-Schwarz, but I can't see how exactly without screwing it up. – Sean Roberson May 23 '18 at 6:11
• I applied AM-GM directly on the expression to bound it below by $2 \sqrt{p(1-p)}$, but that is a pretty bad bound for $p$ smaller than $1/2$. – user369210 May 23 '18 at 6:13
• There is no minimum of the function on (0, 1/2). There is a minimum on (0, 1/2] which occurs at the boundary 1/2. What am I missing? – Andreas May 23 '18 at 6:21
• Infimum, fine, same thing – user369210 May 23 '18 at 6:27
This is elementary:
$$\frac{(1-p)^2}p+\frac{p^2}{1-p}=\frac{1-3p+3p^2-p^3+p^3}{p(1-p)}=\frac1{p(1-p)}-3.$$
The minimum is achieved by the vertex of the parabola $p(1-p)$, i.e. $p=\dfrac12\to\dfrac1{\frac14}-3$, as can be shown by completing the square.
By Rearrangement inequality assuming wlog $p\ge(1-p)$
$$\frac{(1-p)^2}{p} + \frac{p^2}{1-p}=(1-p)\frac{1-p}{p} + p\frac{p}{1-p}\ge (1-p)\frac{p}{1-p} + p\frac{1-p}{p}=p+1-p=1$$
with equality for
$$p=1-p \implies p=\frac12$$
• The problem statement actually implies that $p \leq 1 - p.$ But of course this method still works, just swap $p$ and $1 - p.$ – David K May 24 '18 at 1:23
** I modified this to give a proof with elementary methods only, no inequality theorems needed**
Simple transformations lead to $$\frac{(1-p)^2}{p} + \frac{p^2}{1-p} = - 3 + \frac{4}{1 - 4 (p-1/2)^2}$$ Now it is easy to see that the global minimum of $f(p)$ for $p \in (0 , \; 1/2]$ occurs at $p = 1/2$.
• You've only shown that it is a local minimum. – Kenny Lau May 23 '18 at 6:30
• The "small $x$" part invokes calculus, which I am trying to avoid – user369210 May 23 '18 at 6:33
$$\begin{array}{rcll} \dfrac{(1-p)^2}{p} + \dfrac{p^2}{1-p} &=& \dfrac {(1-p)^3 + p^3} {p (1-p)} \\ &=& \dfrac {[(1-p) + p] [(1-p)^2 - p(1-p) + p^2]} {p (1-p)} \\ &=& \dfrac {(1-p)^2 + p^2} {p (1-p)} - 1 \\ &\ge& \dfrac {\left[ \frac1{\sqrt2} (1-p) + \frac1{\sqrt2} p \right]^2} {p (1-p)} - 1 &\text{Cauchy-Schwarz} \\ &=& \dfrac 1 {2 p (1-p)} - 1 \\ &=& \dfrac 1 {0.5 - 2 (p-0.5)^2} - 1 \\ &\ge& \dfrac 1 {0.5} - 1 \\ &=& 1 \end{array}$$
Both $\ge$ has equality at $p = 0.5$.
$$\dfrac{(1-p)^2}p+\dfrac{p^2}{1-p}=\dfrac{1-3p+3p^2}{p(1-p)}=K\text{(say)}$$
$$\implies p^2(3+K)-p(3+K)+1=0$$
As $p$ is real, the discriminant must be $\ge0$
$$0\le(K+3)^2-4(K+3)=(K+3)(K-1)\implies$$
either $K\ge$max$(1,-3)$
or $K\le$min$(1,-3)$
But observe that $K+3\ne0$
Make the change: $a=1-p, b=p, p\in \left(0,\frac12\right]$.
Then: $a+b=1, ab\le \frac14,$ the equality occurs when $a=b=\frac12$.
Hence: $$\frac{(1-p)^2}{p} + \frac{p^2}{1-p}=\frac{a^2}{b} + \frac{b^2}{a}=\frac{a^3+b^3}{ab}=\frac{(a+b)((a+b)^3-3ab)}{ab}\ge \frac{1\cdot \left(1^3-\frac34\right)}{\frac14}=1.$$
Let's utilize the symmetry:
$$p\equiv\frac 1 2 -q$$
$$1-p\equiv\frac 1 2 +q$$
Then
$$p \in (0,1/2] \iff q \in [0,1/2)$$
$$\frac{(1-p)^2}{p} + \frac{p^2}{1-p} \equiv \frac{(\frac 1 2 +q)^2}{\frac 1 2 -q} + \frac{(\frac 1 2 -q)^2}{\frac 1 2 +q} \equiv \dots$$
(here be magic, try it)
$$\dots \equiv \frac{\frac 1 4 + 3q^2}{\frac 1 4 -q^2} \equiv - 3 + \frac{4}{1 - 4q^2}$$
Note this is the result already obtained in another answer, however manipulations with $p$ (there) are not as elegant as manipulations with $q$ (here).
It's now easy to tell $q=0$ maximizes the denominator which is always positive (in the domain), so the last fraction is minimized.
$$q=0 \iff p=\frac 1 2$$ | 2020-07-13T12:16:58 | {
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https://math.stackexchange.com/questions/3178532/how-is-the-relation-the-smallest-element-is-the-same-reflexive/3178541 | # How is the relation “the smallest element is the same” reflexive?
Let $$\mathcal{X}$$ be the set of all nonempty subsets of the set $$\{1,2,3,...,10\}$$. Define the relation $$\mathcal{R}$$ on $$\mathcal{X}$$ by: $$\forall A, B \in \mathcal{X}, A \mathcal{R} B$$ iff the smallest element of $$A$$ is equal to the smallest element of $$B$$. For example, $$\{1,2,3\} \mathcal{R} \{1,3,5,8\}$$ because the smallest element of $$\{1,2,3\}$$ is $$1$$ which is also the smallest element of $$\{1,3,5,8\}$$.
Prove that $$\mathcal{R}$$ is an equivalence relation on $$\mathcal{X}$$.
From my understanding, the definition of reflexive is:
$$\mathcal{R} \text{ is reflexive iff } \forall x \in \mathcal{X}, x \mathcal{R} x$$
However, for this problem, you can have the relation with these two sets:
$$\{1\}$$ and $$\{1,2\}$$
Then wouldn't this not be reflexive since $$2$$ is not in the first set, but is in the second set?
I'm having trouble seeing how this is reflexive. Getting confused by the definition here.
• Reflexive means that every element is related to itself. Thus, for reflexivity you have to consider one set only. Ok, we have that $\{ 1 \} \mathcal R \{ 1,2 \}$ but we have also $\{ 1 \} \mathcal R \{ 1 \}$ and $\{ 1,2 \} \mathcal R \{ 1,2 \}$ – Mauro ALLEGRANZA Apr 7 at 17:34
• Note: “reflexive” does not mean that if $x$ is related to $y$, then $x=y$. It means that if $x=y$, then $x$ is related to $y$. – Arturo Magidin Apr 7 at 17:44
• So it must be reflexive because both $A$ and $B$ belong to the same set $\mathcal{X}$? – qbuffer Apr 7 at 18:00
• @qbuffer Have a look at the updated version of my answer. – Haris Gusic Apr 7 at 18:49
Why are you testing reflexivity by looking at two different elements of $$\mathcal{X}$$? The definition of reflexivity says that a relation is reflexive iff each element of $$\mathcal X$$ is in relation with itself.
To check whether $$\mathcal R$$ is reflexive, just take one element of $$\mathcal X$$, let's call it $$x$$. Then check whether $$x$$ is in relation with $$x$$. Because $$x=x$$, the smallest element of $$x$$ is equal to the smallest element of $$x$$. Thus, by definition of $$\mathcal R$$, $$x$$ is in relation with $$x$$. Now, prove that this is true for all $$x \in \mathcal X$$. Of course, this is true because $$\min(x) = \min(x)$$ is always true, which is intuitive. In other words, $$x \mathcal{R} x$$ for all $$x \in \mathcal X$$, which is exactly what you needed to prove that $$\mathcal R$$ is reflexive.
You must understand that the definition of reflexivity says nothing about whether different elements (say $$x,y$$, $$x\neq y$$) can be in the relation $$\mathcal R$$. The fact that $$\{1\}\mathcal R \{1,2\}$$ does not contradict the fact that $$\{1,2\}\mathcal R \{1,2\}$$ as well.
A binary relation $$R$$ over a set $$\mathcal{X}$$ is reflexive if every element of $$\mathcal{X}$$ is related to itself. The more formal definition has already been given by you, i.e. $$\mathcal{R} \text{ is reflexive iff } \forall x \in \mathcal{X}, x \mathcal{R} x$$ | 2019-05-26T17:26:10 | {
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http://vnbk.minoronzitti.it/probability-examples-and-solutions.html | # Probability Examples And Solutions
For example, in a large number of coin tosses, heads and tails will occur each about 50% of the time. The solutions are not intended to be as polished as the proofs in the book, but are supposed to give enough of the details so that little is left to the. The above facts can be used to help solve problems in probability. We request all visitors to read all examples carefully. Ch4: Probability and Counting Rules Santorico – Page 105 Event – consists of a set of possible outcomes of a probability experiment. Something with a probability of 0 is something that could never possibly happen. Aptitude Made Easy - Probability – 7 Tricks to solve problems on Balls and bags – Part 1 - Duration: 6:57. Spector and Mazzeo examined the effect of a teaching method known as PSI on the performance of students in a course, intermediate macro economics. If a woman does not have breast cancer, the probability is 7 percent that she will still have a positive mammogram. Show Step-by-step Solutions. Probability Theory and Examples - Durrett. The at least once rule gives us the probability of at least one flood in 100 years: P(ALO in 100 years)=1-[P(not flood in one year)]100 =1-[0. Referring to the earlier example (from Unit 3 Module 3) concerning the National Requirer. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. There 2 numbers that are odd and less than 5: 1 and 3. You use probability in daily life to make decisions when you don't know for sure what the outcome will be. For example, one way to partition S is to break into sets F and Fc, for any event F. ) the second time will be the same as the first (i. 1 of winning. As a member, you'll also get unlimited access to over 79,000 lessons in math, English, science, history, and more. When we flip a coin there is always a probability to get a head or a tail is 50 percent. In the case of a coin, there are maximum two possible outcomes – head or tail. The continuous random variables deal with different kinds of distributions. Rick Durrett Probability Theory And Examples Solution Manual. This video shows examples of using probability trees to work out the overall probability of a series of events are shown. ) and finding about the probability of two things happening in that one task. Exercise 15. An example for non-mutually exclusive events could be: E 1 = students in the swimming team. Solution: The chances of landing on blue are 1 in 4, or one fourth. NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13. For example, when rolling a number cube 600 times, predict that a 3 or 6 would be rolled roughly 200 times, but probably not exactly 200 times. The pictures below depict the probability distributions in space for the Hydrogen wavefunctions. The binomial probability calculator will calculate a probability based on the binomial probability formula. Solutions will be gone over in class or posted later. Are passing these subjects dependent on each other? @312; Hallow): was, Ptmrk): (My etchm GMI’k): 011 NW Newman-m): mun-4: 2-045”. The Best NCERT Books , CBSE Notes , Sample Papers and NCERT Solutions PDF Free Download. For example, in a function machine, a number goes in, something is done to it, and the resulting number is the output: probability: The measure of how likely it is for an event to occur. The denominator 10 because there are 10 fruits in the basket (the total number of outcomes). We can use a bar chart, called a probability distribution histogram , to display the probabilities that X lies in selected ranges. Answer: The probability of being able to exercise outdoors every day of the year where Sam now lives is (:99)365 ˇ:0255 ˇ3% so there is about a 97% probability of not being able to exercise outdoors one or more days of the year. In the "die-toss" example, the probability of event A, three dots showing, is P(A) = 1 6 on a single toss. Suppose a coin tossed then we get two possible outcomes either a 'head' ( H ) or a 'tail' ( T ), and it is impossible to predict whether the result of a toss will be a. Denote by #Adenote the number of point in A. Class 11 Maths Probability Ex 16. The solutions to these problems are at the bottom of the page. Let E = Event of drawing 2 balls, none of which is blue. This video shows examples of using probability trees to work out the overall probability of a series of events are shown. First we find the probability of each event. So before seeing the patient I'm already able to identify the two most likely diagnoses and assign an initial probability for each. The solutions given to the questions for the in between exercises and exercises given at the end of the chapter are prepared by our subject matter experts in a simple and lucid language. Team A wins the series in two games with probability 1/4. All NCERT Questions, Examples and Exercises are solved with step-by-step answers to all questions explained in detail. Answers and links to explanations to these these GMAT probability problems are at the end of set. 2 List the elements of the following sets. Introduction to Conditional Probability Some Examples A "New" Multiplication Rule Conclusion Conditional Probability Here is another example of Conditional Probability. In reality, I'm not particularly interested in using this example just so that you'll know whether or not you've been ripped off the next time you order a hamburger! Instead, I'm interested in using the example to illustrate the idea behind a probability density function. Therefore, both the events will have half probability. On infinite sums19 6. Not all Markov chains. Hope you like them and do not forget to like , social share and comment at the end of the page. Bayes' theorem is a mathematical equation used in probability and statistics to calculate conditional probability. 1 of the bags is selected at random and a ball is drawn from it. Get Free NCERT Solutions for Class 11 Maths Chapter 16 Probability. Theoretical probability is an approach that bases the possible probability on the possible chances of something happen. Probability: Theory and Examples. Expected Value 7. If none of the possible outcomes are favorable for a certain event (a favorable outcome is impossible), the probability is 0. ) the second time will be the same as the first (i. The problem is: we want 1 novel and 1 reference work, but we don’t care which one was picked first. These notes can be used for educational purposes, pro-vided they are kept in their original form, including this title page. Probability concept includes some terms. 1 Probability Spaces. Measure Theory 1. Even though we discuss two events (usually labeled A and B), we’re really talking about performing one task (rolling dice, drawing cards, spinning a spinner, etc. Find the Standard Deviation of a random variable X whose probability density function is given by f(x) where: Solution. develop the theory, we will focus our attention on examples. Listed in the following table are problem sets and solutions. P(AjB) is read The probability of A given B. a Show that $$f\left( x \right)$$ is a probability density function. The mean is $100(7) + 200(4) = 1,500$ and the standard deviation is. Walpole Raymond H. The probability distribution given is discrete and so we can find the variance from the following: We need to find the mean μ first: Then we find the variance: Example 2. Instead of events being labeled A and B, the norm is to use X and Y. EXAMPLE: Suppose that three slips of paper have the names a, b, c. Here different types of examples will help the students to understand the problems on probability with playing cards. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. Return from Examples of Probability Problems to Free Math Videos Online or to 8th Grade CTCT Math. The geometric probability is the area of the desired region (or in this case, not so desired), divided by the area of the total region. com Jobs & Careers 874,804 views 6:57. com or just starting a probability unit, you may want to take a look at the introductory probability lesson or the lesson on independent events. We randomly select 5 balls. Readers with a solid background in measure theory can skip Sections 1. random variables that take a discrete set of values. Kroese School of Mathematics and Physics The University of Queensland c 2018 D. Conditional probability is the probability of one event occurring with some relationship to one or more other events. This solutions manual provides answers for the even-numbered exercises in Probability and Statistical Inference, 8th edition, by Robert V. It is also suitable for self-study. Probability and statistics, the branches of mathematics concerned with the laws governing random events, including the collection, analysis, interpretation, and display of numerical data. If it isn't a trick coin, the probability of each simple outcome is the same. Example: toss a coin 100 times, how many Heads will come up? Probability says that heads have a ½ chance, so we can expect 50 Heads. Bonferroni's inequalities28 9. All Chapter 25 - Probability Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks. For example activity H here, we have the same exact thing, expected duration of 3 plus 4 times 6 plus 9 all divided by 6 to give you 6. Answers and links to explanations to these these GMAT probability problems are at the end of set. com or just starting a probability unit, you may want to take a look at the introductory probability lesson or the lesson on independent events. mathematics of probability theory, but also, through numerous examples, the many diverse possible applications of this subject. As we study a few probability problems, I will explain how "replacement" allows the events to be independent of each other. Probability calculator is a online tool that computes probability of selected event based on probability of other events. A life insurance salesman sells on the average 3 life insurance policies per week. These NCERT solutions play a. What is the probability to get a 6 when you roll a die? A die has 6 sides, 1 side contain the number 6 that give us 1 wanted outcome in 6 possible outcomes. Problems like those Pascal and Fermat solved continuedto influence such early researchers as Huygens, Bernoulli, and DeMoivre in establishing a mathematical theory of probability. Probability Distribution: A probability distribution is a statistical function that describes all the possible values and likelihoods that a random variable can take within a given range. Ma 162 Spring 2010 Ma 162 Spring 2010 April 21, 2010 Problem 1. It is known that probability that a randomly selected student who plays football also plays baseball is 0. Example: At a car park there are 100 vehicles, 60 of which are cars, 30 are vans and the remainder are lorries. The general situation, then, is the following: given a sequence of random variables,. probability with a view toward data science applications. Probability is finding the possible number of outcomes of the event occurrence. On infinite sums19 6. Suppose you toss a fair coin. the basic concepts of a probability model and the axioms commonly assumed of probability models. Conditional Probability Homework Solutions 1. Example: the probability that a card is a four and red =p(four and red) = 2/52=1/26. This problem asked us to find some probabilities involving a spinner. You can think of this as a binomial with all failures. Spector and Mazzeo examined the effect of a teaching method known as PSI on the performance of students in a course, intermediate macro economics. a) We first construct a tree diagram to represent all possible distributions of boys and girls in the family. Example Questions Using Probability Formulas. If every vehicle is equally likely to leave, find the probability of: a) van leaving first. Probability can be expressed in a variety of ways including a mathematically formal way such as using percentages. Since there are 13 spades in the deck, the probability of drawing a spade is. (a) In how many ways can a committee of five consisting of 3 girls and 2 boys be chosen? (b) What is the probability that a committee of five, chosen at random from the class, consists of three girls and two boys? (c) How many of the possible committees of five have no boys?(i. Instructors may obtain all of the solutions by writing to either of the authors, at [email protected] For example, the length of time a person waits in line at a checkout counter or the life span of a light bulb. Example of Binomial Distribution and Probability This Tutorial will explain the Binomial Distribution, Formula, and related Discrete Probabilities Suppose you toss a coin over and over again and each time you can count the number of “Heads” you get. Just keep in mind that what conditional probability does for us is to adjust the denominator of our probability fraction to account for what we KNOW has already occurred. Probability is a branch of mathematics, and a lot of people have trouble with math. A college survey shows that 10% of freshmen, 15% of sophomores, 60% of juniors, and 85% of seniors like Probability. Compound event – an event with more than one outcome. Sometimes it can be computed by discarding part of the sample space. Sketch a graph. A first look at rigorous probability theory. This situation and all the problems any student faces nowadays became a reason to find a solution and create a website where students could find college homework help for the price they could afford. The meaning (interpretation) of probability is the subject of theories of probability. Exercises with both a short answer and a full solution are marked with and those with only a short answer are marked with (when more appropriate, for example, in "Show that " exercises, the short answer provides a hint to the key step). Example 2: Coin-A is tossed 200 times, and the relative occurrence of Tails is 0. Instead, we can usually define the probability density function (PDF). You can also get free sample papers, Notes, Important Questions. If it was, the probability of picking a red ball (etc. Theory of Probability (MATH230A/STAT310A, Fall 2007/08) The first quarter in a yearly sequence of probability theory. Identifying when a probability is a conditional probability in a word problem. , 10 trials), and the probability of getting "heads" was 0. This is COMPLETE Solution Manual for Probability and Statistics for Engineers and Scientists, 9th edition Ronald E. Find the z-scores that correspond to 33 minutes and 60 minutes. This presentation will be about examples of this form of mathematics in real life. The left side of the binomial probability function is written f(3|10,0. To determine whether to accept the shipment of bolts,the manager of the facility randomly selects 12 bolts. For example, if a traffic management engineer looking at accident rates wishes to. Find the z-scores that correspond to 33 minutes and 60 minutes. Read and learn for free about the following article: Conditional probability and independence If you're seeing this message, it means we're having trouble loading external resources on our website. For example, one way to partition S is to break into sets F and Fc, for any event F. Solution: No, we cannot, because the experiment (tossing the coin) may have been repeated a very small number of times, and thus the relative occurrence in such a scenario will not give the true probability. Free PDF download of NCERT Solutions for Class 10 Maths Chapter 15 - Probability solved by Expert Teachers as per NCERT (CBSE) Book guidelines. Exercise :: Probability - General Questions. We write P(AjB) = the conditional probability of A given B. Example: The door prize at a party with 25 people is given by writing numbers 1 through 25 on the bottom of the paper plates used. If a woman does not have breast cancer, the probability is 7 percent that she will still have a positive mammogram. n (E) = Number of ways of drawing 2 balls out of (2 + 3) balls. For Example, cancer is related to age, and then using Bayes theorem, a person’s age can be used to more accurately assess the probability that they have cancer, compared to the assessment of the probability of cancer made without knowledge of the person’s age. The coin toss is nothing but experimenting with tossing a coin. Spector and Mazzeo examined the effect of a teaching method known as PSI on the performance of students in a course, intermediate macro economics. Consider the coin flip experiment described above. Example: If a dice is thrown twice, find the probability of getting two 5's. Solution to Example 1. Theoretical probability is an approach that bases the possible probability on the possible chances of something happen. Probability and playing cards is an important segment in probability. Also, remember that you are comparing the number of ways the outcome can occur to the number of ways the outcome cannot occur (not the total outcomes). In statistics, the normal distributions are used to represent real-valued random variables with unknown distributions. If the joint probability of relaxed import restrictions and a price war is 0. Then we can apply the appropriate Addition Rule: Addition Rule 1: When two events, A and B, are mutually exclusive, the probability that A or B will occur is the sum of the probability of each event. The probability of a sample point is a measure of the likelihood that the sample point will occur. GMAT Advanced Probability Problems By Mike MᶜGarry on January 3, 2014 , UPDATED ON October 30, 2015, in GMAT Math In the following probability problems, problems #1-3 function as a set, problems #4-5 are another set, and problems #6-7 are yet another set. Normal Distribution Examples and Solutions. A probability distribution is a table or an equation that links each outcome of a statistical experiment with its probability of occurrence. Student's Solutions Guide for Introduction to Probability, Statistics, and Random Processes by Hossein Pishro-Nik | Jun 20, 2016 4. The series would last at most 4 games before either Team A wins two games or loses (to Team B) three games. troductory engineering courses in probability theory. P(freshman or sophomore) = 30/101 + 41/101 = 71/101. Coin-B is tossed an unknown number of times, but it. Thus the marginal probability distribution for A gives A's probabilities unconditional on B, in a margin of the table. The function f ( x ) is called a probability density function for the continuous random variable X where the total area under the curve bounded by the x -axis is equal to `1. Instead, we can usually define the probability density function (PDF). Formula for the probability of A and B (independent events): p(A and B) = p(A) * p(B). The problem is: we want 1 novel and 1 reference work, but we don’t care which one was picked first. 1 of the bags is selected at random and a ball is drawn from it. Probability distributions can, however, be applied to grouped random variables which gives rise to joint probability distributions. We can write this statement mathematically as follows: P(10≤ X ≤ 20) =. Durrett Probability: Theory and Examples 4th Edition. This solution contains questions, answers, images, explanations of the complete Chapter 15 titled Probability of Maths taught in Class 10. In probability theory and applications, Bayes' theorem shows the relation between a conditional probability and its reverse form. In the National Lottery, 6 numbers are chosen from 49. In what follows, S is the sample space of the experiment in question and E is the event of interest. You use probability in daily life to make decisions when you don't know for sure what the outcome will be. All Probability Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks. Distributions 3. On infinite sums19 6. The graphs below show the radial wave functions. July 1, 2015. Therefore, both the events will have half probability. Binomial distribution problems for practice Binomial distribution practice problems Binomial distribution problem 1 Question (Binomial distribution problems): Expected value of a binomial random variable X is 10 and the probability of. We always give back probability in to all our clients, whether you are our long-term client or simply ordering for the first time. It is assessed by considering the event's certainty as 1 and impossibility as 0. We try to provide all types of shortcut tricks on probability problem on balls here. An event is an occurrence that can be determined by a given level of certainty. It includes the list of lecture topics, lecture video, lecture slides, readings, recitation problems, recitation help videos, tutorials with solutions, and a problem set with solutions. Introduction: Probability (from the Latin probare to prove, or to test) is a number between zero and one that shows how likely a certain event is. Probability distribution tables In this tutorial I show you how to construct probability distribution tables for a discrete random variable for three different type examples. Such random variables generally take a finite set of values (heads or tails, people who live in London, scores on an IQ test), but they can also include random. If the ball drawn is red, find the probability that it is drawn from the third bag. There are six ways to roll a 7 and two ways to roll an 11, so the probability of a natural is (6 + 2)/36 = 8/36 = 2/9. Scroll down the page for more examples and solutions of word problems that involve the probability of independent events. 11 If we toss a coin 1000 times and find that it comes up heads 532 times, we estimate the probability of a head coming up to be 532 1000 0. Binomial distribution problems for practice Binomial distribution practice problems Binomial distribution problem 1 Question (Binomial distribution problems): Expected value of a binomial random variable X is 10 and the probability of. An urn contains 5 red balls and 2 green balls. These NCERT solutions play a. This Aptitude Test Questions sections presents "Probability" solved problems. If it was, the probability of picking a red ball (etc. Examples of using the formula to find conditional probability In some situations, you will need to use the following formula to find a conditional probability. \ 1 P(5) 6 =. This sampling method is based on the fact that every member in the population has an equal chance of getting selected. What is the probability that the survey will show a greater percentage of. in Teachers. Distributions 3. probability of the event given that the other event has occurred. You can find several more examples here: Probability of A and B. 5L; also, on c, be careful about using the same set sizes to calculate the probability that the second student is taking a language class, since you’ve already chosen the first. An examples from ecology: How are species abundance estimates determined from small samples? To summarize: There are at least two uses for statistics and probability in the life sciences. Statistics Solutions is the country’s leader in probability and dissertation statistics. Statistics Solutions is the country's leader in probability and dissertation statistics. In other words, it is used to calculate the probability of an event based on its association with another event. problems included are about: probabilities, mutually exclusive events and addition formula of probability, combinations, binomial distributions, normal distributions, reading charts. In what follows, S is the sample space of the experiment in question and E is the event of interest. The mathematical and probability models of lottery provide information that players should know before they launch into a long-run play. Independence of three or more events34 12. 12 that she has exactly these two risk factors (but not the other). Probability concept includes some terms. It is less than 10 dollar. A team with equal numbers of men and women will have three men and three women. Binomial Distribution. P(AjB) is read The probability of A given B. 7) Anita randomly picks 4 cards from a deck of 52-cards and places them back into the deck ( Any set of 4 cards is equally likely ). Hoping that the book would be a useful reference for people who apply probability in their work, we have tried to emphasize the results that are important for applications, and illustrated their use with roughly 200 examples. org are unblocked. Step-by-step solutions provided. Student's Solutions Guide for Introduction to Probability, Statistics, and Random Processes - Kindle edition by Hossein Pishro-Nik. Student's Solutions Guide for Introduction to Probability, Statistics, and Random Processes by Hossein Pishro-Nik | Jun 20, 2016 4. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. Introduction to Probability with Examples and Solutions Probability is a section of math that is essential in various aspects of life. The number of rainy days, Xcan be represented by a binomial distribution with n= 31trials (the number of days in the month of October), success probability p= 0:16(representing a rainy day) and failure probability q= 1 p= 0:84. It is not too much to say that the path of mastering statistics and data science starts with probability. Discrete probability distributions give the probability of getting a certain value for a discrete random variable. Visitors are requested to carefully read all shortcut examples. A simple example is the tossing of a fair (unbiased) coin. The pictures below depict the probability distributions in space for the Hydrogen wavefunctions. Binomial probability concerns itself with measuring the probability of outcomes of what are known as Bernoulli Trials, trials that are independent of each other and that are binary — with two possible outcomes. What is the probability of the occurrence of a number that is odd or less than 5 when a fair die is rolled. The results are summarised in the table below. The concept is very similar to mass density in physics: its unit is probability per unit length. There are two ways for it to win in three games: LWW or WLW; both occur with probability 1/8 so that the probability that Team A wins in exactly 3 games is also 1/4. In probability theory and applications, Bayes' theorem shows the relation between a conditional probability and its reverse form. Anyone has the right to use this work for any purpose, without any conditions, unless such conditions are required by law. Random Variables 4. So the probability of getting 2 blue marbles is: And we write it as "Probability of event A and event B equals the probability of event A times the probability of event B given event A" Let's do the next example using only notation:. Examples involving conditional probability Math 30530, Fall 2013 September 5, 2013 Math 30530(Fall 2012) Conditional examples September 5, 20131 / 5. Both the classical and frequency approaches have serious drawbacks, the first because the words "equally. For any two of the three factors, the probability is 0. example, what probability should we assign to the customer choosing soup and then the meat? If 8/10 of the customers choose soup and then 1/2 of these choose meat, a proportion 8=10 ¢1=2=4=10 of the customers choose soup and then meat. b) Find the mean and standard deviation of X. We provide examples on Probability problem on Balls shortcut tricks here in this page below. When we flip a coin there is always a probability to get a head or a tail is 50 percent. Joint Probability Distributions. Read and learn for free about the following article: Conditional probability and independence If you're seeing this message, it means we're having trouble loading external resources on our website. It provides mathematically complete proofs of all the essential introductory results of probability and measure theory. It shows the exact probabilities for a particular value of the random variable. This presentation will be about examples of this form of mathematics in real life. This is typically possible when a large number of random effects cancel each other out, so some limit is involved. Conditional probability and independence31 11. The probability distribution given is discrete and so we can find the variance from the following: We need to find the mean μ first: Then we find the variance: Example 2. Genetics for Probability To provide a scientific context for our probability problems, we will use examples from genetics. Chapter 5: Normal Probability Distributions - Solutions Note: All areas and z-scores are approximate. Suppose you are a teacher at a university. Probability can be expressed in a variety of ways including a mathematically formal way such as using percentages. The solutions to these problems are at the bottom of the page. Formula sheet also available. However, there is a probability greater than zero than X. In addition, there are 6 outcomes that. A random variable, X, is a function from the sample space S to the real. Chapter 5: Normal Probability Distributions - Solutions Note: All areas and z-scores are approximate. Compound event – an event with more than one outcome. Find the Standard Deviation of a random variable X whose probability density function is given by f(x) where: Solution. Explore more on other related concepts @ BYJU'S. Suppose the reaction times of teenage drivers are normally distributed with a mean of 0. This course will guide you through the most important and enjoyable ideas in probability to help you cultivate a more quantitative worldview. Compute the probability of randomly drawing five cards from a deck and getting exactly two Aces. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. You win if the 6 balls you pick match the six balls selected by the machine. Schaum's Outline of Probability and Statistics EXAMPLE 2. It provides mathematically complete proofs of all the essential introductory results of probability and measure theory. Suppose a simple random sample of 100 voters are surveyed from each state. \ 1 P(5) 6 =. And then in the next segment we'll look at Bayes theorem. Applied Problem Solving in a Workplace Contents Introduction 5 Theme A – Approaches to problem Solving 6 Recognising Complex Problems 6 Activity 1a Identifying complex problems. Therefore, both the events will have half probability. Enter the trials, probability, successes, and probability type. 5th edition correct mistakes in previous editions and has some more beautiful examples. This means that the probability of the coin landing on heads would be ½. A sample space (S) is a non empty set whose elements are called outcomes. The results are summarised in the table below. Ma 162 Spring 2010 Ma 162 Spring 2010 April 21, 2010 Problem 1. Independence 2. Probability: Theory and Examples Solutions Manual The creation of this solution manual was one of the most important im-provements in the second edition of Probability: Theory and Examples. What is the probability that a randomly chosen person in the class will weigh more than 160 lbs. • Probability of an event E = p(E) = (number of favorable outcomes of E)/(number of total outcomes in the sample space) This approach is also called theoretical probability. Solution; Determine the value of $$c$$ for which the function below will be a probability density function. This lesson describes how hypergeometric random variables, hypergeometric experiments, hypergeometric probability, and the hypergeometric distribution are all related. What is the probability that an. ©The McGraw-Hill Companies, Inc. The solutions given to the questions for the in between exercises and exercises given at the end of the chapter are prepared by our subject matter experts in a simple and lucid language. Answers and links to explanations to these these GMAT probability problems are at the end of set. Show that the solution $$X_t$$ of SDE examples, Stochastic Calculus The Probability Workbook is powered by WordPress at Duke WordPress Sites. Theoretical probability is an approach that bases the possible probability on the possible chances of something happen. P(B|A) is also called the "Conditional Probability" of B given A. We randomly select 5 balls. 1-9 A red die has face numbers {2, 4, 7, 12, 5, 11}. Here $Y=g(X)$, where $g$ is a differentiable function. E 2 = students in the debating team. 7) Anita randomly picks 4 cards from a deck of 52-cards and places them back into the deck ( Any set of 4 cards is equally likely ). If it isn't a trick coin, the probability of each simple outcome is the same. Students can get a fair idea on the probability questions which are provided with the detailed step-by-step answers to every question. Solution: No, we cannot, because the experiment (tossing the coin) may have been repeated a very small number of times, and thus the relative occurrence in such a scenario will not give the true probability. 2 Normal Distributions: Finding Probabilities If you are given that a random variable Xhas a normal distribution, nding probabilities. How is Chegg Study better than a printed A First Course In Probability 9th Edition student solution manual from the bookstore? Our interactive player makes it easy to find solutions to A First Course In Probability 9th Edition problems you're working on - just go to the chapter for your book. For this example, since the mean is 8 and the question pertains to 11 fires. All Probability Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks. | 2020-01-28T13:14:54 | {
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https://math.stackexchange.com/questions/2590898/logarithmic-differentiation-for-multivariable-functions | # Logarithmic Differentiation for Multivariable functions?
I was wondering if, and how exactly logarithmic differentiation may be applied to a multivariate function.
For example, I have been working with the function
$$f(x,y)=\frac{x-y}{x+y}$$
Does the following process hold? Specifically, does the chain rule step with del f hold?
$$\ln(f(x,y))=\ln(x-y)-\ln(x+y)$$
$$\Rightarrow \frac{1}{f(x,y)} \nabla f = \langle \frac{2y}{x^2-y^2} , \frac{-2x}{x^2-y^2} \rangle$$
Solving for $\nabla f$, does give me the correct answer. I'm wondering is my notation would be correct, and if this will ever not work.
Thanks for any help.
• If you are ok, you can accept the answer and set as solved. Thanks! – user Jan 4 '18 at 20:17
You define
$$g(x,y)=\ln(f(x,y))=\ln(x-y)-\ln(x+y)$$
and thus
$$\nabla g=\left(\frac{1}{f(x,y)}f_x,\frac{1}{f(x,y)}f_y\right)=\frac{1}{f(x,y)}\left(f_x,f_y\right)=\frac{1}{f(x,y)}\nabla f$$
• Ahhh, Okay, this seems very obvious now. Sorry if that was a dumb question. Thank you. – Kyle B Jan 3 '18 at 22:15
• @KyleB You are welcome! Bye – user Jan 3 '18 at 22:22
You could take a simpler function and checked whether it holds. The usual differentiation rules hold true for the gradient as well: sum, difference, product and quotient. For instance, $\nabla(fg)=g\nabla f+f\nabla g$, $\nabla(\frac{f}{g})=\frac{g\nabla f-f\nabla g}{g^2}$, $\nabla(\ln f)=\frac{\nabla f}{f}$.
Yes I really believe your result as well as your notation are correct. May be you just need to put it in a more general setting. I would suggest you to proceed as follows.
Suppose that $f:\mathbb{R}^n\to\mathbb{R}$, which is differentiable and strictly positive in a certain open set $U\subset\mathbb{R}^n$. Then the function $F=\ln\circ f$ is also differntiable on $U$ with: $\nabla F(x)=\ln'(f(x))\cdot \nabla f(x)=\frac{1}{f(x)}\cdot\nabla f(x)$ (chain rule). Hence we can solve for $\nabla f(x)$, as we are in vector space $\mathbb{R}^n$ and get the answer right.
Remember the chain rule statement: If $G$ and $H$ are differentianble functions and $F=H\circ G$ is well defined, then $F$ is also differentiable with $\nabla F(x)=\nabla H(G(x))\cdot\nabla G(x)$. | 2019-10-15T11:32:26 | {
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http://mathhelpforum.com/pre-calculus/152372-complex-roots-quadratic-equation.html | # Math Help - Complex roots of a quadratic equation
1. ## Complex roots of a quadratic equation
The question:
Use the properties of the complex conjugate to show that if the complex number 'z' is a root of a quadratic equation $ax^2 + bx + c = 0$ with a, b, c being real coefficients, then so is the conjugate of 'z'.
My attempt:
Let z = $z_{1} \pm z_2i$
$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = z_{1} \pm z_2i$
$\frac{1}{2a}(-b \pm \sqrt{b^2 - 4ac}) = z_{1} \pm z_2i$
Basically, I changed the quadratic formula to look like a complex number in Cartesian form. Is this an acceptable answer? My textbook doesn't have solutions for 'show' and 'prove' questions.
Thanks!
2. The roots of $ax^2 + bx + c = 0$ are
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
If $b^2 - 4ac < 0$ then that means $\sqrt{b^2 - 4ac}$ is imaginary. So we can say $\sqrt{b^2 - 4ac} = ni$, where $n$ is some multiple of $i$.
So $x = \frac{-b \pm n i}{2a}$
$= -\frac{b}{2a} \pm \left(\frac{n}{2a}\right)i$.
So the two roots are complex conjugates.
3. Ahh, thanks. So would my solution be acceptable, or should I simplify it like yours?
4. Originally Posted by Glitch
Ahh, thanks. So would my solution be acceptable, or should I simplify it like yours?
To be honest, I can't really follow your solution...
5. It's pretty much the same as yours, except you replaced the square root with 'ni'. I was trying to show that the general form of a complex number is equivalent to a simplified quadratic formula.
6. Originally Posted by Glitch
The question:
Use the properties of the complex conjugate to show that if the complex number 'z' is a root of a quadratic equation $ax^2 + bx + c = 0$ with a, b, c being real coefficients, then so is the conjugate of 'z'.
The standard way of proving the is to use properties of the complex conjugate.
If $a$ is a real number then $\overline{a}=a$.
The conjugate of a sum is the sum of conjugates.
Therefore we have:
$\overline {az^2 + bz + c} = \overline 0 = 0\, \Rightarrow \,a\overline z ^2 + b\overline z + c = 0$
7. Originally Posted by Glitch
The question:
Use the properties of the complex conjugate to show that if the complex number 'z' is a root of a quadratic equation $ax^2 + bx + c = 0$ with a, b, c being real coefficients, then so is the conjugate of 'z'.
My attempt:
Let z = $z_{1} \pm z_2i$
$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = z_{1} \pm z_2i$
$\frac{1}{2a}(-b \pm \sqrt{b^2 - 4ac}) = z_{1} \pm z_2i$
But at this point you haven't yet said that $\sqrt{b^2- 4ac}$ is an imaginary number!
Basically, I changed the quadratic formula to look like a complex number in Cartesian form. Is this an acceptable answer? My textbook doesn't have solutions for 'show' and 'prove' questions.
Thanks!
8. Originally Posted by Plato
The standard way of proving the is to use properties of the complex conjugate.
If $a$ is a real number then $\overline{a}=a$.
The conjugate of a sum is the sum of conjugates.
Therefore we have:
$\overline {az^2 + bz + c} = \overline 0 = 0\, \Rightarrow \,a\overline z ^2 + b\overline z + c = 0$
...and it is true for every polynomial...
9. Fair point. So if I add:
When $b^2 - 4ac < 0$, $\sqrt{b^2 - 4ac} = ni, n \in R$
And replace the square root with the imaginary part 'n', would that suffice?
10. Hello, Glitch!
I have a very primitive solution . . .
Use the properties of the complex conjugate to show that
if the complex number $z$ is a root of a quadratic equation
$f(x) \:=\:ax^2 + bx + c = 0$ with $a, b, c$ real coefficients,
then so is the conjugate of $z.$
We are told that $z \:=\:p + qi$ is a solution of the quadratic.
. . Hence: . $a(p+qi)^2 + b(p+qi) + c \:=\:0$
This simplifies to: . $(ap^2 - aq^2 + bp + c) + q(2ap + b)i \;=\;0$
. . And we have: . $\begin{Bmatrix}ap^2 - aq^2 + bp + c \;=\;0 \\ q(2ap + b) \;=\;0 \end{Bmatrix}$
The conjugate of $z$ is: . $\overline z \:=\:p - qi$
Consider $f(\overline z) \;=\;a(p-qi)^2 + b(p-qi) + c$
. . . . . . . $f(\overline z) \;=\;\underbrace{(ap^2 - aq^2 + bp + c)}_{\text{This is 0}} - \underbrace{q(2ap + b)}_{\text{This is 0}}i$
$\text{Therefore: }\:f(\overline z) \,=\,0\:\text{ and }\:\overline z\text{ is also a solution.}$ | 2015-07-28T02:59:06 | {
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https://math.stackexchange.com/questions/2553562/how-to-find-precise-value-in-terms-of-pi | # How to find precise value in terms of $\pi$
I've some argument like:
$$\arctan(\sqrt3 + 2)$$ and as explained here How to calculate $\arctan(\sqrt3-2)$ to be $-15°$ by hand? i made an assumption and found that it should be $\frac{5\pi}{12}$. I found the exact value of $\sqrt3 + 2$ then by table i found it's equal to $\tan(75^{\circ})$ and finally i found $\arctan(\tan(75^{\circ}))$ But what if i have something more complicated to do assumption on like:
$$\arccos({\frac{\sqrt{\sqrt3+2}}{2}})$$
So i'm interesting if there is any sequence of operations on argument itself to convert it into fraction of $\pi$ without guessing about how many it should be?
The best bet is to "unravel" the expression, try to simplify it by getting rid of roots and recognizing patterns:
$$\phi=\arccos\frac{\sqrt{\sqrt{3}+2}}{2}$$ $$\cos\phi=\frac{\sqrt{\sqrt{3}+2}}{2}$$ $$4\cos^2\phi=\sqrt{3}+2$$ $$2(2\cos^2\phi-1)=\sqrt{3}$$ $$2\cos2\phi=\sqrt{3}$$ $$\cos2\phi=\frac{\sqrt{3}}{2}$$ $$2\phi=\arccos\frac{\sqrt3}{2}=\frac{\pi}{6}$$ $$\phi=\frac{\pi}{12}$$
You'll always get some sort of polynomial in trigonometric functions. The goal is to reduce the degree of polynomial by using multiple angle formulas. If you don't recognize them yourself, you can take a look at Chebyshev polynomials of the first kind which tell you the multiple angle formulas: $\cos n\phi= T_n(\cos\phi)$. In this case, we recognized $T_2(x)=2x^2-1$.
Depending on how well twisted the expression is, you may have to get creative. There may be more than just multiple angle tricks: possibly, you have to use addition theorems as well, recognizing $\cos(x+y)$ where $x$ and $y$ are not equal, and so on. But the main trick is still just "undoing" the operations that you don't want to see.
• Yes, that's exactly what i was looking for. Thanks a lot! Dec 6 '17 at 9:06
• very nice derivation !
– user
Dec 6 '17 at 9:10
As $2+\sqrt3=\dfrac{(\sqrt3+1)^2}2$
$$\dfrac{\sqrt{2+\sqrt3}}2=\dfrac{\sqrt3+1}{2\sqrt2}=\cos30^\circ\cos45^\circ+\sin30^\circ\sin45^\circ=\cos(45^\circ-30^\circ)$$ | 2021-09-23T09:09:48 | {
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https://www.physicsforums.com/threads/how-to-find-this-limit.113049/ | # How to find this limit?
1. Mar 4, 2006
### pivoxa15
Find the limit as (x,y)->(0,0) for (yx^2)/(x^2+y^2)
If do this literally, I get 0/0 hence no limit but a limit does exist for this function. How do I get it? What is the general way to find the limits of multi variable functions?
2. Mar 4, 2006
### VietDao29
Ehh???
Who tells you that 0 / 0 means the limit does not exist there? It's one of the Indeterminate forms.
Generally, to find a limit of a 2 variable function, one can change it to polar co-ordinate, and go from there: $$x = r \cos \alpha \quad \mbox{and} \quad y = r \sin \alpha$$
Now (x, y) -> (0, 0) means that r -> 0, and $$\alpha$$ can take whatever value. So if you can show that when r -> 0, the expression will tend to some value independent of $$\alpha$$, then the limit exists there.
---------------
Example:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x ^ 2 + y ^ 2}{xy}$$
Change it to polar form, we have:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x ^ 2 + y ^ 2}{xy} = \lim_{r \rightarrow 0} \frac{r ^ 2}{r ^ 2 \sin \alpha \cos \alpha} = \lim_{r \rightarrow 0} \frac{1}{\cos \alpha \sin \alpha}$$
That means the limit of that expression does depend on $$\alpha$$, hence the limit does not exist at (0, 0).
Can you get this? :)
3. Mar 4, 2006
### moose
The way I would think about this one is that if x and y are approaching the same thing, why not set y equal to x? That way you would have x^3/2x^2
Now you can think about which one is changing faster, etc.
4. Mar 5, 2006
### pivoxa15
I see. It is clever. Is it the standard way of evaluating multivariable limits? What other ways are there?
5. Mar 5, 2006
### pivoxa15
Your method does not work in most cases.
6. Mar 5, 2006
### devious_
Other ways include approaching the point from various lines, e.g. if you're looking at (x,y) -> (0,0) then sometimes letting (x,y) approach the origin from the x-axis or the y-axis ((x,0) or (y,0)) can help you prove the limit doesn't exist.
7. Mar 5, 2006
### benorin
So we may, upon trying the limit along various curves such as y=x, y=x^2, the x-axis (e.g. y=0), the y-axis (e.g. x=0) which all give the value of the limit to be 0, conjecture the value of the limit to be 0. Now we must prove it:
To prove that $$\lim_{(x,y)\rightarrow (0,0)} \frac{yx ^ 2}{x^2 + y^2}=0,$$ we require that
$$\forall\epsilon >0, \exists \delta >0 \mbox{ such that } 0<\sqrt{x^2 + y^2}<\delta\Rightarrow \left| \frac{yx ^ 2}{x^2 + y^2}-0\right| < \epsilon$$
(and in case you didn't know, the symbol $$\forall$$ is read "for every", the symbol $$\exists$$ is read "there exists", and the symbol $$\Rightarrow$$ is read "implies".)
The proof is this: We will need the following inequality
$$0\leq y^2\Rightarrow x^2\leq x^2+y^2\Rightarrow \frac{x^2}{x^2+y^2}\leq 1.$$
Let us work the absolute value term contained in the $$\epsilon ,\delta$$ definition of the limit described above, we have
$$\left| \frac{yx ^ 2}{x^2 + y^2}-0\right| =\frac{|y|x ^ 2}{x^2 + y^2} =|y|\frac{x ^ 2}{x^2 + y^2} \leq |y| = \sqrt{y^2} \leq \sqrt{x^2+y^2} <\delta$$
we want to choose $$\delta$$ so that the absolute value term we just worked with is always less than $$\epsilon$$ given that $$0<\sqrt{x^2+y^2} <\delta .$$ Evidently, we may choose $$\delta =\epsilon$$ to this end, and the proof is complete upon stating this formally.
I know that PF does not permit full solutions to be posted, however it is understood that proofs of this sort are quite tricky to construct, examples are few in texts that require them, and I will have rendered sufficient pedagogical substance by even successfully relating it.
--Ben
EDIT: Thanks VietDao29, all that short-hand is Greek to me.
Last edited: Mar 5, 2006
8. Mar 5, 2006
### VietDao29
One more way is to use the inequality:
$$x ^ 2 + y ^ 2 \geq 2|xy|$$, one can prove it by doing a little rearrangement, and noticing the fact that: (|x| + |y|)2 >= 0.
I'll give one example that's similar to your problem.
----------------
Example:
Find:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x ^ 3 y ^ 3}{x ^ 2 + y ^ 2}$$
Now using the inequality above, we have:
$$\left| \frac{x ^ 3 y ^ 3}{x ^ 2 + y ^ 2} \right| \leq \left| \frac{x ^ 3 y ^ 3}{2xy} \right| = \frac{1}{2} |x ^ 2 y ^ 2| = \frac{1}{2} x ^ 2 y ^ 2$$
Now as (x, y) tends to (0, 0),
$$\frac{1}{2} x ^ 2 y ^ 2 \rightarrow 0$$, hence $$\left| \frac{x ^ 3 y ^ 3}{x ^ 2 + y ^ 2} \right| \rightarrow 0$$, thus $$\frac{x ^ 3 y ^ 3}{x ^ 2 + y ^ 2} \rightarrow 0$$, so we can conclude that:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x ^ 3 y ^ 3}{x ^ 2 + y ^ 2} = 0$$
Can you get this? :)
--------------
EDIT:
By the way, benorin:
That symbols is read Delta, the symbol that is read "there exists" should be $$\exists$$ .
Last edited: Mar 5, 2006
9. Mar 5, 2006
### pivoxa15
The method you used in your example looked a bit suspect.
If we follow your way and do the limit(x->0, y->0) of (xy)/(x^2+y^2)
we get the limit being smaller than 1/2 which is wrong since it should have no limit at all.
10. Mar 5, 2006
### Benny
Sorry to intrude on your topic but I would like to know if using polar coordinates is a valid method when the limit is (x,y) -> (0,0). Yes, the angle is arbitrary but each combination of a (r,angle) as r approaches zero corresponds to taking the limit along the straight line. Ie. not every single path is considered.
I can see that converting to polar coordinates is sufficient to verify that the limit does not exist. But is it sufficient to show that a limit exists? I apologise if I missed anything.
11. Mar 5, 2006
### HallsofIvy
Staff Emeritus
No, that's perfectly valid. You do not have to assume the angle is constant as r goes to 0. The point is that, in polar coordinates, the distance from the origin is measured by a single variable, r. r is completely independent of the angle. That was the point VietDao29 made in his first post- the limit, as r goes to 0, depended upon the angle and so the limit itself does not exist.
In your problem, $\frac{xy^2}{x^2+ y^2}$, changing to polar coordinates gives $\frac{r^3cos(\theta)sin^2(\theta)}{r^2}= r cos(\theta)sin^2(\theta)$. What does that go to as r goes to 0? Does it depend on $\theta$?
12. Mar 5, 2006
### Benny
Ok, that's good. Thanks for the clarification.
13. Mar 5, 2006
### VietDao29
Yes, the limit of:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{xy}{x ^ 2 + y ^ 2}$$ does not exist at (0, 0).
So by using the inequality x2 + y2 >= 2|xy|. We have:
$$\left| \frac{xy}{x ^ 2 + y ^ 2} \right| \leq \frac{1}{2} \left| \frac{xy}{xy} \right| = \frac{1}{2}$$
So that means:
$$-\frac{1}{2} \leq \frac{xy}{x ^ 2 + y ^ 2} \leq \frac{1}{2} \quad (*)$$. Hence, no conclusion can be drawn by looking at the inequality (*). Why? Because we don't know if the expression does converge to some value between -1 / 2, and 1 / 2; or it just diverges.
It's like the limit: $$\lim_{x \rightarrow \infty} \sin x$$ does not exist, although we know that: -1 <= sin(x) <= 1. It's because sin(x) does not tend to any specific number (i.e converges to any value) as x tends to infinity.
---------------
However, if we have $$\lim_{x \rightarrow \alpha} |f(x)| = 0$$, then we also have: $$\lim_{x \rightarrow \alpha} f(x) = 0$$. Why?
It's because $$\forall \varepsilon > 0, \exists \delta > 0 : 0 < |x - \alpha| < \delta \Rightarrow ||f(x)| - 0| = |f(x)| = |f(x) - 0| < \varepsilon$$. Now according to the definition of limit, we also have:
$$\lim_{x \rightarrow \alpha} f(x) = 0$$. Can you get this?
You can do the same to prove that if:
$$\lim_{\substack{x \rightarrow \alpha \\ y \rightarrow \beta}} |f(x, y)| = 0$$ then $$\lim_{\substack{x \rightarrow \alpha \\ y \rightarrow \beta}} f(x, y) = 0$$.
Can you get this? :)
---------------
Now just read my last post again to see if you can understand it.
:)
14. Mar 5, 2006
### benorin
Having verified it, I submit in passing that
$$-\frac{1}{2} \leq \frac{xy}{x ^ 2 + y ^ 2} \leq \frac{1}{2}$$
are the best bounds possible.
15. Mar 5, 2006
### pivoxa15
I see your point. Using polar coords to find limits of multivariable functions is better than your first principles method.
Where did you learn evaluating limits of multivarialbe functions using polar coords? I did not see them in calculus textbooks.
16. Mar 6, 2006
### VietDao29
Yes, using polar co-ordinate is far better, and easier than the second way. :)
And guess what? I learnt how to do it at this site, too... :tongue:
17. Mar 6, 2006
### HallsofIvy
Staff Emeritus
Yes, but that tells you nothing about whether the limit at (0,0) exists.
18. Mar 6, 2006
### benorin
Like I said, "in passing".
19. Mar 7, 2006
### pivoxa15
Are the functions we have been discussing 3D or 2D?
20. Mar 7, 2006
### benorin
Recall that z=f(x,y) is a surface, and hence 3-D. | 2016-12-09T17:48:20 | {
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https://cs.stackexchange.com/questions/16346/how-to-discuss-coefficients-in-big-o-notation | # How to discuss coefficients in big-O notation
What notation is used to discuss the coefficients of functions in big-O notation?
I have two functions:
• $f(x) = 7x^2 + 4x +2$
• $g(x) = 3x^2 + 5x +4$
Obviously, both functions are $O(x^2)$, indeed $\Theta(x^2)$, but that doesn't allow a comparison further than that. How do I discuss the the coefficients 7 and 3. Reducing the coefficient to 3 doesn't change the asymptotic complexity but it still makes a significant difference to runtime/memory usage.
Is it wrong to say that $f$ is $O(7x^2)$ and $g$ is $O(3x^2)$ ? Is there other notation that does take coefficients into consideration? Or what would be the best way to discuss this?
• It's not wrong, it's just redundant, because $O(7 x^2) = O(x^2)$. – Oli Charlesworth Oct 7 '13 at 23:17
• See also our reference question. – Raphael Oct 28 '13 at 7:35
Big-$O$ and big-$\Theta$ notations hide coefficients of the leading term, so if you have two functions that are both $\Theta(n^2)$ you cannot compare their absolute values without looking at the functions themselves. It's not wrong per se to say that $7x^2 + 4x + 2 = \Theta(7x^2)$, but it's not informative because $7x^2 + 4x + 2 = \Theta(3x^2)$ is also true (and, in fact, it's $\Theta(kx^2)$ for any positive constant $k$).
There are other notations you might want to use instead. For example, $\sim$ notation is a much stronger claim than big-$\Theta$:
$\qquad \displaystyle f(x) \sim g(x) \iff \lim_{x \to \infty} \frac{f(x)}{g(x)} = 1$
For example, $7x^2 + 4x + 2 \sim 7x^2$, but the claim $7x^2 + 4x + 2 \sim 3x^2$ would be false. You can think of tilde notation as $\Theta$ notation that preserves the leading coefficients, which seems to be what you're looking for if you do care about the leading coefficient of the dominant growth term.
• Tilde notation is what I'm looking for. I was sure there was something I just couldn't recall what it was called and searches proved fruitless. Thanks! – El Bee Oct 7 '13 at 23:48
The tilde is one approach. If you want to stick with $O$, you could say
$\qquad f(x) = 7x^2 + O(x)$ and
$\qquad g(x) = 3x^2 + O(x)$.
• Even better: say f(x) = 7x^2 + o(x^2), using little-o notation to clarify that what's left is asymptotically smaller than x^2. – templatetypedef Oct 9 '13 at 4:06
• O(x) is strictly smaller than o(x^2), so using that would be less clear than using big-O. On the other hand, using little-o is definitely more common when you want to say that you've got the right first term, because then you don't need to worry about the next term. (And if we're wanting to be completely clear, then we would need to explain why we don't just write down 7x^2+4x+2 in the first place, since it is exactly correct. – Teepeemm Oct 9 '13 at 4:43
• You're absolutely right... my apologies! – templatetypedef Oct 9 '13 at 17:56
• Note that the rigorous way of writing this would be "$f(x) = 7x^2 + g(x)$ with $g(x) \in O(x)$". In any case, this is very useful if you want to fix more than the "first" constant; you can say "$f(x) = 7x^2 + 4x + O(1)$" which you can not do with $\sim$. – Raphael Oct 28 '13 at 7:29 | 2020-06-06T21:48:36 | {
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https://math.stackexchange.com/questions/2661883/permutations-how-many-outcomes-have-a-1-in-them-if-pn-k-p13-3-and-the-pro | # Permutations: How many outcomes have a 1 in them if P(n,k) = P(13,3) and the probability it will occur if randomly generated?
There is a set of numbers {1,2,3,4,5,6,7,8,9,10,11,12,13}. I am trying to find out how many subsets I can form that have 3 numbers in them and also the amount of subsets that will have a number 1 in them. Also, if I was to randomly generate a single outcome, what is the probability it will have a 1 in it if all outcomes have equal chance of occurring?
I know that the probability can be calculated from (Number of subsets that have 1 in them)/(Total number of permutations)
I figured out the Total total number of permutations = 13!/(13-3)! = 1716
But I can't figure out how to determine how many subsets will have a 1 in them. In addition to this, is there a way to generalise the method? For example, what if I would like to find out the number of subsets that have a number 1 and a number 2 in them for the same n and k parameters?
• For subsets do you mean ignoring order, which is usual, or considering order, which seems to be implied by your use of permutations? – Ross Millikan Feb 22 '18 at 15:49
• Oh in this problem I do want to consider order. So 1,2,3 is different from 1,3,2 and so on. – Lobster Feb 22 '18 at 16:04
• That is what I considered in my answer. You are then selecting three element permutations from thirteen, not a subset. As I commented on the other answer, the fraction that include $1$ is the same either way. – Ross Millikan Feb 22 '18 at 16:11
It looks like instead of subsets you mean three element permutations that include $1$. You can count those by choosing first the location of the $1$, which gives three choices, then the first other element, $12$, and the second other element, $11$ for a total of $3 \cdot 12 \cdot 11=396$. You are correct that the chance a random permutation includes a $1$ is $\frac {396}{1716}=\frac 3{13}$ This is not surprising as you are choosing three elements from thirteen.
To have both $1$ and $2$ you have three places to put the $1$, two places for the $2$, and $11$ choices for the third number.
• Thanks! This was a great help! – Lobster Feb 22 '18 at 16:16
This is not actually a case of permutation, but rather a case of combination. Because in a subset, the arrangement of the numbers are not important. Therefore it is the use of combinations and not permutations.
Subset 3 numbers in them = 13C3 = 286 - choosing 3 numbers out of 13
Subset 3 numbers with 1 in them = 12C2 = 66 - choosing 2 numbers given one of the number is 1 (there is only 12 numbers to pick, because 1 is chosen by default)
• You can use either permutations or combinations as long as you are consistent. The probability of having a $1$ comes out $\frac 3{13}$ eitherj way, as it should. There are $3!$ times more total permutations and $3!$ times more that include a $1$. – Ross Millikan Feb 22 '18 at 15:56
• Welcome to MSE. Please use MathJax. – José Carlos Santos Feb 22 '18 at 16:01
• Oh right, i forgot to answer the probability. You are absolutely right about the consistency part, and both method works. My only concern is about technicality especially with subsets, because with other question (not the probability one), the answer may be different, due to order being considered. – Zirc Feb 22 '18 at 16:02 | 2019-05-26T09:59:01 | {
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https://byjus.com/question-answer/the-chance-that-doctor-a-will-diagonise-disease-x-correctly-is-60-the-chance-that/ | Question
# The chance that Doctor A will diagonise disease X correctly is $$60\%$$. The chance that a patient will die by his treatment after correct diagnosis is $$40\%$$ and the chance of death after wrong diagnosis is $$70\%.$$ A patient of Doctor A who had disease X died. The probability that his disease was diagonised correctly is
A
513
B
613
C
213
D
713
Solution
## The correct option is D $$\dfrac{6}{13}$$Let us define the following events.$${ E }_{ 1 }:$$ Disease $$X$$ is diagnosed correctly by doctor $$A$$.$${ E }_{ 2 }:$$ Disease $$X$$ is not diagnosed correctly by doctor $$A$$.$$B:$$ A patient (of doctor $$A$$) who has disease $$X$$ dies.Then, we are given, $$P({ E }_{ 1 })=0.6$$,$$P({ E }_{ 2 })=1-P({ E }_{ 1 })=1-0.6=0.4$$and $$\displaystyle P\left( \frac { B }{ { E }_{ 1 } } \right) =0.4$$, $$\displaystyle P\left( \frac { B }{ { E }_{ 2 } } \right) =0.7$$By Bay's Theorem$$\displaystyle P\left( \frac { { E }_{ 1 } }{ B } \right) =\frac { P\left( { E }_{ 1 } \right) P\left( \frac { B }{ { E }_{ 1 } } \right) }{ P\left( { E }_{ 1 } \right) P\left( \frac { B }{ { E }_{ 1 } } \right) +P\left( { E }_{ 2 } \right) P\left( \frac { B }{ { E }_{ 2 } } \right) }$$$$\displaystyle=P\left( \frac { { E }_{ 1 } }{ B } \right) =\frac { 0.6\times 0.4 }{ 0.6\times 0.4+0.4\times 0.7 } =\frac { 0.24 }{ 0.24+0.28 } =\frac { 0.24 }{ 0.52 } =\frac { 6 }{ 13 }$$Mathematics
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https://math.stackexchange.com/questions/2595145/in-mathbbr4-let-u-span-1-1-0-0-1-1-1-1 | # In $\mathbb{R}^4$, let $U=$ span$\{(1,1,0,0),(1,1,1,1)\}$.
(a) Compute an orthonormal basis for $U$.
(b) Find $u\in U$ such that the Euclidean distance $||u-(1,2,3,4)||$ is as small as possible.
Here's what I have:
(a) To find an orthonormal basis, we'll use the Gram-Schmidt Algorithm. The process is as follows:
Given $k$ basis vectors $v_1,v_2,\cdots,v_k$, we apply the following steps:
1. Let $q_1=\frac{1}{||v_1||}v_1$.
2. For $i=2$ to $k$, repeat steps 3 and 4.
3. Let $w_i=v_i-\langle q_1,v_i\rangle q_i-\langle q_2,v_i\rangle q_2-\cdots-\langle q_{i-1},v_i\rangle q_{i-1}$
4. Let $q_i=\frac{1}{||w_i||}w_i$
The vectors $q_1,\cdots q_k$ are our orthonormal basis. Now, we apply these steps to the given problem. $$q_1=\frac{1}{||v_1||}v_1=\frac{1}{\sqrt{2}}v_1=(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0,0)$$ $$w_2=v_2-\langle q_1,v_2\rangle q_1=(1,1,1,1)-(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0,0)=(0,0,1,1)$$ $$q_2=\frac{1}{\sqrt{2}}w_2=(0,0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$$ So $\{q_1,q_2\}$ is an orthonormal basis for $U$.
(b) I recognize this as a least squares problem, but I'm a bit lost on how to actually set it up, because we don't have something with the form $Ax\approx b$. I mean, I know that $b=(1,2,3,4)$ and $u$ somehow represents $Ax$, but we don't have a matrix $A$, and $u$ is just an element of the span of $2$ vectors.
• Hint: the underlying mechanism of the solution to the least-squares problem you’re familiar with is orthogonal projection onto a vector space. – amd Jan 7 '18 at 4:47
Your proof for part (a) is correct. For part (b) let $$u=(\alpha , \alpha ,\beta ,\beta)$$ You need to minimize $$(\alpha -1)^2+( \alpha -2)^2 + (\beta -3)^2+ (\beta-4)^2.$$ Partial derivatives are zero at $\alpha = 3/2$ and $\beta =7/2$. Thus $u=(3/2, 3/2, 7/2,7/2)$is the desired vector.
• Do you think you could elaborate on this a bit? Is the choice that $u=(\alpha,\alpha,\beta,\beta)$ from the fact that the vectors that span $U$ have the same first two entries? – Atsina Jan 7 '18 at 5:06
• Sure. From your part (a), you have found a basis in form of (1,1,0,0) and (0,0,11). $u=(\alpha,\alpha,\beta,\beta)=\alpha (1,1,0,0)+\beta (0,0,1,1)$ – Mohammad Riazi-Kermani Jan 7 '18 at 5:20
Because you have an orthonormal basis, things become increasingly easy.
The least squares problem is solved by
$$x_{ls} = (A^TA)^{-1}(A^Tb)$$
Where $A$ is a matrix which has your basis as columns and $b$ is $<1,2,3,4>$. Solve this for $x_{ls}$ and then the solution is $Ax_{ls}$.
• This is more along the lines of what I was searching for, but Mohammad's solution is slightly less computationally intensive. I wish I could accept multiple answers. – Atsina Jan 7 '18 at 6:02
• @Atsina Because there is an orthonormal basis for the subspace, things are even easier than what been written here: $A^TA=I_2$, so your equation reduces to $x_{ls}=A^Tb$ and $u=AA^Tb=(u_1^Tb)u_1+(u_2^Tb)u_2$, i.e., it’s the sum of the orthogonal projections onto the basis vectors. – amd Jan 11 '18 at 1:10
The shortest distance to $U$ is measured along a direction orthogonal to it, therefore the element of $U$ that minimizes the distance to a given vector $b$ is the orthogonal projection of $b$ onto $U$, that is, the component of $b$ that lies in $U$.
There are various ways to compute this projection, but since you’ve already constructed an orthonormal basis for $U$, the easiest thing to do is to compute the individual projections onto the basis vectors and combine them. In fact, this is exactly what you do at each step of the Gram-Schmidt process: you compute the orthogonal projection of the current vector onto the space spanned by the orthonormal basis vectors generated up to that point and subtract that projection from the vector to get its component that doesn’t lie in that span. So, using that projection formula from the Gram-Schmidt process, \begin{align} u = (u_1^Tb)u_1+(u_2^Tb)u_2 &= \frac3{\sqrt2}\left(\frac1{\sqrt2},\frac1{\sqrt2},0,0\right)+\frac7{\sqrt2}\left(0,0,\frac1{\sqrt2},\frac1{\sqrt2}\right) \\ &= \left(\frac32,\frac32,\frac72,\frac72\right). \end{align}
The connection to finding a least-squares solution to the possibly-inconsistent linear system $Ax=b$ is that you essentially replace $b$ with its orthogonal projection onto the column space of $A$.
Muhammad gave you a nice way using partial derivative so I'll just throw out here another way without using derivatives
The shortest distance to a plane create angle of $\pi/2$, so let's create the vector $\vec v=\begin{bmatrix}1-a\\2-a\\3-b\\4-b\end{bmatrix}$, this vector is the vector from the point $(1,2,3,4)$ to an arbitrary point on the plane.
I want that this arbitrary point on the plane will minimize $\vec v$, so I'll check when the vector create angle of $\pi/2$ to the plane($\langle\cdot,\cdot\rangle$ is the dot product):$$\langle\vec v,(1,1,0,0)\rangle=0\\\langle\vec v,(0,0,1,1)\rangle=0$$here I get 2 equations: $(1-a)1+(2-a)1=0\implies a=1.5$ and $(3-b)1+(4-b)1=0\implies b=3.5$ thus the answer is $u=(1.5,1.5,3.5,3.5)$ | 2020-10-30T14:49:39 | {
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https://byjus.com/question-answer/find-the-vector-equation-of-the-plane-passing-through-the-points-2-1-1-and/ | Question
# Find the vector equation of the plane passing through the points $$(2, 1, -1)$$ and $$(-1, 3, 4)$$ and perpendicular to the plane $$x - 2y + 4z = 10.$$ Also show that the plane thus obtained contains the line $$\vec{r} = -\hat{i} + 3\hat{j} + 4\hat{k} + \lambda(3 \hat{i} - 2\hat{j} - 5\hat{k}).$$
Solution
## Let the equation of plane through $$(2, 1, -1)$$ be$$a(x - 2) + b(y - 1) + c(z + 1) = 0 ....(i)$$$$\because (i)$$ passes through $$(-1, 3, 4)$$$$\therefore a (-1 - 2) + b (3 - 1) + c (4 + 1) = 0$$$$\Rightarrow -3a + 2b + 5c = 0 ...(ii)$$Also plane (i) is perpendicular to plane $$x - 2y + 4z = 10$$$$\Rightarrow \vec{n_1} \perp \vec{n_2} \Rightarrow \vec{n_1}. \vec{n_2} = 0$$$$\therefore 1a - 2b + 4c = 0 ...(iii)$$From (ii) and (iii), we get$$\dfrac{a}{8 + 10} = \dfrac{b}{5 + 12} = \dfrac{c}{6 - 2} \Rightarrow \dfrac{a}{18} = \dfrac{b}{17} = \dfrac{c}{4} = \lambda (say)$$$$\Rightarrow a = 18\lambda, b = 17\lambda, c = 4\lambda$$Putting the value of a, b, c in (i), we get$$18\lambda (x - 2) + 17\lambda (y - 1) + 4\lambda (z + 1) = 0$$$$\Rightarrow 18x - 36 + 17y - 17 + 4z + 4 = 0$$$$\Rightarrow 18x + 17y + 4z = 49$$$$\therefore$$ Required vector equation of plane is$$\vec{r}. (18\hat{i} + 17\hat{j} + 4\hat{k}) = 49 ...(iv)$$Obviously plane (iv) contains the line$$\vec{r} = (-\hat{i} + 3\hat{j} + 4\hat{k}) + \lambda (3\hat{i} - 2\hat{j} - 5\hat{k}) ....(v)$$Since, point $$(-\hat{i} + 3\hat{j} + 4\hat{k})$$ satisfy equation (iv) and vector $$(18\hat{i} + 17\hat{j} + 4\hat{k})$$ is perpendicular to, $$(3\hat{i} - 2\hat{j} + 5\hat{k}),$$ as $$(-\hat{i} + 3\hat{j} + 4\hat{k}). (18\hat{i} + 17\hat{j} + 4\hat{k}) = -18 + 51 + 16 = 49$$ and $$(18\hat{i} + 17\hat{j} + 4\hat{k} ). (3\hat{i} - 2\hat{j} - 5\hat{k}) = 54 - 34 - 20 = 0$$Therefore, (iv) contains line (v).Mathematics
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https://math.stackexchange.com/questions/1771002/riemann-sum-approximations-when-are-trapezoids-more-accurate-than-the-middle-su | Riemann Sum Approximations: When are trapezoids more accurate than the middle sum?
We can approximate a definite integral, $\int_a^b f(x)dx$, using a variety of Riemann sums. If $T_n$ and $M_n$ are the nth sums using the trapezoid and midpoint (middle) sum methods and if the second derivative of $f$ is bounded on $[a, b]$ then does the following theorem imply that $M_n$ "tends to be more accurate" then $T_n$?
If $f''$ is continuous on $[a, b]$ and $|f''(x)| \leq K$, $\forall$ $x \in [a, b]$. Then,
$\left| \int_a^b f(x)dx - T_n \right| \leq K\frac{(b-a)^3}{12n^2}$
and
$\left| \int_a^b f(x)dx - M_n \right| \leq K\frac{(b-a)^3}{24n^2}$
For this question let, $E_{T_n} = \left| \int_a^b f(x)dx - T_n \right|$ and $E_{M_n} = \left| \int_a^b f(x)dx - T_n \right|$
The theorem is presented (without proof) in a calculus 2 book. It only really seems to imply that, we can with 100% certainly bound E_{M_n} smaller than we can bound E_{T_n}. But, that says nothing about the actual values of E_{T_n} or E_{M_n}.
So, isn't it possible to customize a function so that $E_{T_n} < E_{M_n}$ for a particular n? Could we even make a function so that $E_{T_n} < E_{M_n}$ $\forall n$?
• For a particular $n$, absolutely: that's equivalent to doing it for $n=1$ and replicating the function across multiple intervals. Just consider any function satisfying $f(0)=f(1)=0$ and $\int_0^1 f(x)\, dx = 0$, but $f(1/2) \ne 0$. Then $E_T = 0$ but $E_M$ can be as large as you want (by rescaling $f$). May 4, 2016 at 8:40
• This shows $T = 0$, not $E_T = 0$.
– RRL
May 5, 2016 at 2:20
• Although the Midpoint Rule is typically more accurate, it should be said that the Trapezoid Rule can be used to get $guaranteed$ upper and lower bounds for integrals of concave and convex functions respectively, which cannot be said for the Midpoint Rule.
– user123641
Oct 30, 2017 at 15:56
As you observed, the midpoint method is typically more accurate than the trapezoidal method. This is suggested by the composite error bounds, but they don't rule out the possibility that the trapezoidal method might be more accurate in some cases.
We can get a better understanding by examining the local errors for single-segment rules. Consider an interval $[a,b]$ and define interval length $h = b-a$ and midpoint $c = (a+b)/2.$ Note that $b-c = c-a = (b-a)/2 = h/2.$
The midpoint error is
$$E_M = f(c)h - \int_a^b f(x) \, dx = \int_a^b [f(c) - f(x)] \, dx.$$
Using a second-order Taylor approximation,
$$f(c) = f(x) + f'(x)(c-x) + \frac{1}{2} f''(\xi_x)(x-c)^2,$$
we see
$$E_M = -\int_a^b f'(x)(x-c) \, dx + \frac{1}{2}\int_a^b f''(\xi_x)(x-c)^2 \, dx.$$
Applying integration by parts to the first integral on the RHS we get
$$\int_a^b f'(x)(x-c) = \left.(x-c)f(x)\right|_a^b - \int_a^b f(x) \, dx = \frac{h}{2}[f(a) + f(b)] - \int_a^b f(x) \, dx .$$
Note that this result gives us the error $E_T$ for the trapezoidal method.
Hence,
$$E_M = -E_T + \frac{1}{2}\int_a^b f''(\xi_x)(x-c)^2 \, dx.$$
It is actually not that easy to find examples where $|E_T| < |E_M|$. Using the above result, we can surmise that this could happen when the midpoint method overestimates, $E_M > 0$, the trapezoidal method underestimates $E_T < 0,$ and we have high curvature in a small neighborhood of a point in the interval.
Here is a somewhat contrived example.
Consider the following function that meets those requirements. Note that the second derivative is piecewise continuous but bounded -- which does not degrade the composite $O(n^{-2})$ accuracy.
$$f(x) = \begin{cases} 0.25 + 0.75\exp(-200 x^2), &\mbox{if } -1 \leqslant x \leqslant 0 \\ 0.99 + 0.01 \cos(\pi x), &\mbox{if } \,\,\,\,\,\,\, 0 < x \leqslant 1 \end{cases}.$$
Then
\begin{align}\int_{-1}^1 f(x) \, dx &\approx 1.2870\\ M &\approx 2 \\ T &\approx 1.2300 \\ E_M &\approx 0.7130 \\ E_T &\approx 0.0570 \end{align}
• I think you can use periodic smooth functions to get non-contrived examples.
– Ian
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https://web2.0calc.com/questions/if-three-runners-receive-medals-how-many-ways-can-the-three-medalists-be-chosen-out-of-a-field-of-28 | +0
# If three runners receive medals how many ways can the three medalists be chosen out of a field of 28?
0
457
3
+8
Is it 28 x 27 x 26?
peacemantle May 2, 2015
### Best Answer
#2
+92749
+10
Badinage, you are right.
There are 28 ways for the first, 27 ways for the second and 26 ways for the 3rd That is 28*27*26
BUT it doesn't matter what order they are chosen. so you have to divide by 3! = 6
$${\frac{{\mathtt{28}}{\mathtt{\,\times\,}}{\mathtt{27}}{\mathtt{\,\times\,}}{\mathtt{26}}}{{\mathtt{6}}}} = {\mathtt{3\,276}}$$
this is how I would normally do it
28C3
$${\left({\frac{{\mathtt{28}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{28}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)} = {\mathtt{3\,276}}$$
Melody May 2, 2015
#1
+520
+8
Yes, that's the answer your teacher is seeking.
However, as a challenge extension you could extend this to include ties, e.g., from 2 up to 28 ties for first, up to 27 ties for second, etc. I'm not sure how to do this, though, and it sounds complicated.
# 🌹 🌹 🌹 🌹 🌹 🌹 🌹 🌹
.
Badinage May 2, 2015
#2
+92749
+10
Best Answer
Badinage, you are right.
There are 28 ways for the first, 27 ways for the second and 26 ways for the 3rd That is 28*27*26
BUT it doesn't matter what order they are chosen. so you have to divide by 3! = 6
$${\frac{{\mathtt{28}}{\mathtt{\,\times\,}}{\mathtt{27}}{\mathtt{\,\times\,}}{\mathtt{26}}}{{\mathtt{6}}}} = {\mathtt{3\,276}}$$
this is how I would normally do it
28C3
$${\left({\frac{{\mathtt{28}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{28}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)} = {\mathtt{3\,276}}$$
Melody May 2, 2015
#3
+520
+5
Yes, you are right, Melody. The question doesn't say 1st, 2nd and 3rd places. Maybe there are 3 runners who are to be awarded a medal for being a comically costumed competitor for St Patrick's Day, or some such.
Badinage May 3, 2015
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http://mathhelpforum.com/discrete-math/52607-recurrence-relations.html | # Math Help - recurrence relations
1. ## recurrence relations
1)
Find a recurrence relation for the number of bit string of length n that contain three consecutive 0's
What are the initial conditions? how many bit strings of length 7 contain three consecutive 0's?
2)
Give the recursive algorithm for finding the minimum of a finite set of
integers.
3)
Find the solution to the following recurrence relation and initial
condition:
xn+2 = 4xn; n = 0, 1, 2, ... x0 = 2, x1 = 1.
4)
Give the recursive algorithm for finding the sum of the first n odd
positive integers.
5)
Find the solution to the following recurrence relation and initial
condition:
xn+2 = 2xn+1 + 3xn; n = 0, 1, 2, ... x0 = 1, x1 = -1.
2. Hello, narbe!
5) Find the solution to the following recurrence relation and initial condition:
. . $x_{n+2} \:= \:2x_{n+1} + 3x_n\qquad x_0 = 1,\;\;x_1 = \text{-}1$
Crank out the new few terms . . .
. . $\begin{array}{ccccc} x_2 &=&2(\text{-}1) + 3(1) &=& 1 \\
x_3 &=& 2(1) + 3(\text{-}1) &=& \text{-}1 \\
x_4 &=& 2(\text{-}1) + 3(1) &=& 1 \\
x_5 &=& 2(1) + 3(\text{-}1) &=& \text{-}1 \\
\vdots & & \vdots & & \vdots \end{array}$
Got it?
3. ## yeah but ...
yes I got the solution! but how can I demonstrate it as a function ??????
4. Originally Posted by narbe
1)
Find a recurrence relation for the number of bit string of length n that contain three consecutive 0's
What are the initial conditions? how many bit strings of length 7 contain three consecutive 0's?
I think it's easiest to count the number of strings of length n that do not contain three consecutive zeros. Call this number $y_n$.
We can get a string of length n+1 without three consecutive zeros in three ways: (1) take such a string of length n and add a 1 at the end, (2) take such a string of length n-1 and add 10 at the end, (3) take such a string of length n-2 and add 100 at the end. Therefore $y_{n+1} = y_n+y_{n-1}+y_{n-2}$.
Now let $x_n$ denote the number of strings of length n that do contain three consecutive zeros. There are altogether $2^n$ strings of length n, so $x_n = 2^n-y_n$. It easily follows from the relation for the y's that $x_{n+1} = x_n+x_{n-1}+x_{n-2}+2^{n-2}$. The initial conditions are $x_1=x_2=0,\ x_3=1$. From there, you can use the recurrence relation to work your way up to 7. (I get the answer $x_7=47$, but don't rely on my arithmetic.)
5. Hello, narbe!
It's easy . . .
Look at the terms:
. . $\begin{array}{ccccc}x_0 &=& 1 \\ x_1 &=& \text{-}1 \\ x_2 &=& 1 \\ x_3 &=& \text{-}1 \\ \vdots \\ x_n &=& (\text{-}1)^n & \Leftarrow &\text{There!}\end{array}$
6. Originally Posted by narbe
3) Find the solution to the following recurrence relation and initial
condition:
x_{n+2}= 4x_{n}; n = 0, 1, 2, ... x_{0} = 2, x_{1} = 1.
This is a second order difference equation with a constant coefficient, there is a very simple way for this equation.
Such equations have solution of the type $x(n)=c_{1}\lambda_{1}^{n}+c_{2}\lambda_{2}^{n}$ for $n\in\mathbb{N}$, where $c_{1},c_{2}$ are arbitrary constants.
Just substitute this value into the solution and solve $\lambda_{1}$ and $\lambda_{2}$, then use these values together with the initial conditions to find the desired solution.
However, here, I would like to show you a different solution way for this equation.
First set, $y(n):=x(2n)$ for $n\in\mathbb{N}$, then from the equation, we get
$y(n+1)=x(2n+2)=4x(2n)=4y(n)$ or simply $y(n+1)=4y(n)$ for $n\in\mathbb{N}$.
It is easy to solve this equation and we get $y(n)=c_{1}4^{n}$ for all $n\in\mathbb{N}$, where $c_{1}$ is an arbitrary constant.
Since $y(0)=x(0)=2$, we see that $c_{1}=2$.
Thus, we have $x(n)=24^{n/2}=2^{n+1}$ is a solution of the equation.
Next, set $z(n):=x(2n-1)$ for $n\in\mathbb{N}$ and solve $z(n+1)=x(2(n+1)-1)=x(2n+1)=4x(2n-1)=z(n)$ or simply $z(n+1)=4z(n)$ for $n\in\mathbb{N}$ with $z(1)=x(1)=1$.
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https://nhigham.com/2017/08/14/how-and-how-not-to-compute-a-relative-error/ | # How and How Not to Compute a Relative Error
The relative error in a scalar $y$ as an approximation to a scalar $x$ is the absolute value of $e = (x-y)/x$. I recently came across a program in which $e$ had been computed as $1 - y/x$. It had never occurred to me to compute it this way. The second version is slightly easier to type, requiring no parentheses, and it has the same cost of evaluation: one division and one subtraction. Is there any reason not to use this parenthesis-free expression?
Consider the accuracy of the evaluation, using the standard model of floating point arithmetic, which says that $fl(x \mathbin{\mathrm{op}} y) = (x \mathbin{\mathrm{op}} y)(1+\delta)$ with $|\delta| \le u$, where $\mathrm{op}$ is any one of the four elementary arithmetic operations and $u$ is the unit roundoff. For the expression $e_1 = (x-y)/x$ we obtain, with a hat denoting a computed quantity,
$\widehat{e_1} = \displaystyle\left(\frac{x-y}{x}\right) (1+\delta_1)(1+\delta_2), \qquad |\delta_i| \le u, \quad i = 1, 2.$
It follows that
$\left| \displaystyle\frac{e - \widehat{e_1}}{e} \right| \le 2u + u^2.$
Hence $e_1$ is computed very accurately.
For the alternative expression, $e_2 = 1 - y/x$, we have
$\widehat{e_2} = \left(1 - \displaystyle\frac{y}{x}(1+\delta_1)\right) (1+\delta_2), \qquad |\delta_i| \le u, \quad i = 1, 2.$
After a little manipulation we obtain the bound
$\left| \displaystyle\frac{e - \widehat{e_2}}{e} \right| \le u + \left|\displaystyle\frac{1-e}{e}\right|(u + u^2).$
The bound on the relative error in $\widehat{e_2}$ is of order $|(1-e)/e|u$, and hence is very large when $|e| \ll 1$.
To check these bounds we carried out a MATLAB experiment. For 500 single precision floating point numbers $y$ centered on $x = 3$, we evaluated the relative error of $y$ as an approximation to $x$ using the two formulas. The results are shown in this figure, where an ideal error is of order $u \approx 6\times 10^{-8}$. (The MATLAB script that generates the figure is available as this gist.)
As expected from the error bounds, the formula $1-y/x$ is very inaccurate when $y$ is close to $x$, whereas $(x-y)/x$ retains its accuracy as $y$ approaches $x$.
Does this inaccuracy matter? Usually, we are concerned only with the order of magnitude of an error and do not require an approximation with many correct significant figures. However, as the figure shows, for the formula $|1-y/x|$ even the order of magnitude is incorrect for $y$ very close to $x$. The standard formula $|x-y|/|x|$ should be preferred.
## 2 thoughts on “How and How Not to Compute a Relative Error”
1. Raúl Martínez says:
Nick,
Thank you for delving into this. Your result is certainly far from obvious and I find it very helpful. In my own work, I don’t recall computing the relative error using 1–y/x (though I cannot be certain), but I have sometimes used the latter expression in writing. After reading your note, I can see that such usage could mislead a reader into using that expression for computation, so I’ll be mindful not to use it in the future.
Sincerely,
Raúl Martínez
2. For interested parties I converted Nick’s gist above to Python:
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# Based on the code by Nick Higham # Compares relative error formulations using single precision and compared to double precision N = 501 # Note: Use 501 instead of 500 to avoid the zero value d = numpy.finfo(numpy.float32).eps * 1e4 a = 3.0 x = a * numpy.ones(N, dtype=numpy.float32) y = [x[i] + numpy.multiply((i – numpy.divide(N, 2.0, dtype=numpy.float32)), d, dtype=numpy.float32) for i in range(N)] # Compute errors and "true" error relative_error = numpy.empty((2, N), dtype=numpy.float32) relative_error[0, :] = numpy.abs(x – y) / x relative_error[1, :] = numpy.abs(1.0 – y / x) exact = numpy.abs( (numpy.float64(x) – numpy.float64(y)) / numpy.float64(x)) # Compute differences between error calculations error = numpy.empty((2, N)) for i in range(2): error[i, :] = numpy.abs((relative_error[i, :] – exact) / numpy.abs(exact)) fig = plt.figure() axes = fig.add_subplot(1, 1, 1) axes.semilogy(y, error[0, :], '.', markersize=10, label="$|x-y|/|x|$") axes.semilogy(y, error[1, :], '.', markersize=10, label="$|1-y/x|$") axes.grid(True) axes.set_xlabel("y") axes.set_ylabel("Relative Error") axes.set_xlim((numpy.min(y), numpy.max(y))) axes.set_ylim((5e-9, numpy.max(error[1, :]))) axes.set_title("Relative Error Comparison") axes.legend() plt.show() | 2022-12-01T04:34:58 | {
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http://mathhelpforum.com/geometry/174934-centroid-triangle-coordinates.html | # Thread: The centroid of a triangle with coordinates
1. ## The centroid of a triangle with coordinates
Triangle DEF has vertices D(1,3) and E(6,1), and centroid at C(3,4). Determine the coordinates of point F.
I know that the medians in a triangle intersect at the centroid and that the centroid divides each median in a ration of 2:1. Also, I know that each median intersects a side at its midpoint with the shorter part of the median. What I did was find the midpoint between point E and F by taking half of the line D to C and adding onto C to give point G. The coordinates of point G was found to be (4, 4.5). Since I knew that the change in the coordinates between points E and G, (-2, 3.5), was only half of the line EF I doubled it and added it point E to give the answer (2, 8). I wanted to know if there was an easier way to do this.
2. Originally Posted by darksoulzero
Triangle DEF has vertices D(1,3) and E(6,1), and centroid at C(3,4). Determine the coordinates of point F.
I know that the medians in a triangle intersect at the centroid and that the centroid divides each median in a ration of 2:1. Also, I know that each median intersects a side at its midpoint with the shorter part of the median. What I did was find the midpoint between point E and F by taking half of the line D to C and adding onto C to give point G. The coordinates of point G was found to be (4, 4.5). Since I knew that the change in the coordinates between points E and G, (-2, 3.5), was only half of the line EF I doubled it and added it point E to give the answer (2, 8). I wanted to know if there was an easier way to do this.
Read Centroid - AoPSWiki
3. ## centroid of triangle
Hi darksoulzero,
Your solution is confusing.Here is a suggested method.
connect M midpoint of DE and C (centroid) with extended lenght.F lies on this line. FC = 2CM. Slope diagram of C and M = 2/1/2.Slope diagram of FC is twice that or 4/1. Working from point C one point left and 4 points up gives F (2,8)
bjh
4. If ABC is a triangle with $\displaystyle $$A\left( {{x_1},{y_1}} \right),B\left( {{x_2},{y_2}} \right),C\left( {{x_3},{y_3}} \right)$$$ then its centroid is
$\displaystyle $$G\left( {\tfrac{{{x_1} + {x_2} + {x_3}}}{3},\tfrac{{{y_1} + {y_2} + {y_3}}}{3}} \right)$$$
5. Hello, darksoulzero!
$\displaystyle \text{Triangle }D{E}F\text{ has vertices: }D(1,3)\text{ and }E(6,1)\text{, and centroid at }C(3,4).$
$\displaystyle \text{Determine the coordinates of point }F.$
Code:
1
F o-----+
\ :
\ :4
\ :
\ :
\:0.5
(3,4)o---+
D C\ :
o \ :2
(1,3) * \:
M o
(3.5,2) * E
o
(6,1)
We have vertices $\displaystyle D(1,3)$ and $\displaystyle E(6,1)$, and centroid $\displaystyle C(3,4).$
The midpoint of $\displaystyle DE$ is: $\displaystyle M(3\tfrac{1}{2},\,2).$
The median to side $\displaystyle DE$ starts at $\displaystyle \,M$, passes through $\displaystyle \,C,$
. . and extends to $\displaystyle \,F$, where: .$\displaystyle FC \,=\,2\!\cdot\!CM.$
Going from $\displaystyle \,M$ to $\displaystyle \,C$, we move up 2 and left $\displaystyle \frac{1}{2}$
Hence, going from $\displaystyle \,C$ to $\displaystyle \,F$, we move up 4 and left 1.
Therefore, we have: .$\displaystyle F(2,8).$ | 2018-03-18T02:38:04 | {
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https://math.stackexchange.com/questions/712665/find-angles-of-intersection-between-circle-and-line-segment-on-circle | # Find angle(s) of intersection between circle and line segment on circle
I have a circle (given a center and radius) and a line segment (given a coordinate for both points). The line segment passes through the circle and intersects with the circle in one or two points (one if the segment is tangential or if it ends inside the circle).
How can I find the angle or the two angles where the line segment intersects the circle, relative to the top of the circle?
Please provide an example. Unrelated to the diagram below, a set of points to use in an example are as follows: circle radius $1.5$ with center $(3, 4)$. Line segment from $(3, 3)$ to $(6, 7)$.
If it's easier, it's acceptable to find both points by treating the line segment as a line and ignoring a tangential intersection
I am trying to find the angle shown as ?° in the diagram below. The answer in this case is 57°
• What is the other point that you're using to define this angle? Yes, the angle in the image may be 57 degrees, but what is the significance of the other point that you're measuring from? What I'm saying is: the point seems arbitrary. So I could pick any point on the circle and have any angle that I want. – foobar1209 Mar 16 '14 at 23:40
• The point is just the top of the circle. It could be anything, but the top makes the most sense. – Keavon Mar 16 '14 at 23:41
Set the center of the circle to be $\left(0,0\right)$. Then the set of points on the circle are those such that $x^2+y^2=r^2$ and the set of points on the line segment are solutions to $y=mx+b$ from some $m$ and $b$ (these are easily found given the two endpoints of the line segment). Substituting, we have $$x^2+(mx+b)^2=r^2\Longleftrightarrow (m^2+1)x^2+2mbx+(b^2-r^2)=0$$ Then this is just a quadratic you can solve with the quadratic formula giving you $x$, and you can get $y$ by $y=mx+b$. However, this gives you two solutions; you must pick the right one (find which is between the two endpoint). Fortunately finding the angle after this is pretty easy: it's just simple trigonometry. If $(x,y)$ is the solution, $\tan^{-1}\left(\frac{y}{x}\right)$ gives you the angles measure counterclockwise from the $x$-axis. You asked for the angle from the $y$-axis, which is then $\frac{\pi}{2}-\tan^{-1}\left(\frac{y}{x}\right).$
A worked example
I'll work out the example you have given. We have radius $r=\frac{3}{2}$, center $\left(3,4\right)$ and a line segment from $(3,3)$ to $(6,7)$. First we'll move the center to $(0,0)$. We do this by subtracting $(3,4)$ from each point. This gives us the center (as desired) of $(0,0)$, and a line segment from $(0,-1)$ to $(3,3)$. Then the line between the two points is $y=\frac{4}{3}x-1$. The equation for the circle is $x^2+y^2=\frac{9}{4}$. Substituting the equation for the line in, we have $$x^2+\left(\frac{4}{3}x-1\right)^2=\frac{25}{9}x^2-\frac{8}{3}x+1=\frac{9}{4}$$ which gives the quadratic $$\frac{25}{9}x^2-\frac{8}{3}x-\frac{5}{4}=100x^2-96x-45=0$$ Then the quadratic equation gives $x=\frac{96\pm \sqrt{96^2+4\cdot 100\cdot 45}}{200}=-0.345,1.305$. We're going to want the solution with positive $x$, so $x=1.305$. Then $y=\frac{4}{3}(1.305)-1=0.740$. To find the desired angles, we merely take $90-\tan^{-1}\left(\frac{0.740}{1.305}\right)=90-29.56=60.44$. I'm not sure what the discrepancy between my answer and the one you provided is; it could be rounding errors, but it is more likely I just made a calculation error somewhere (please do point it out if you see it).
I know it seems like a lot, but the technique is relatively straight forward, the calculation is just some hard work. I hope this answer has shed some light on the process for you.
• You made no error. The example I provided is different from the diagram provided in the question. Since this is a detailed and easy to understand answer I will mark it as accepted, but wait a few days before awarding you the bounty. – Keavon Mar 17 '14 at 2:03
• Sounds good. Glad i could help :) – cderwin Mar 17 '14 at 2:26
Hint: First, for convenience - transform the circle's to $(0,0)$ and normalize (so the radius of the given circle is now 1). The circle's equation is $x^2+y^2=1$
Now, we are given $(x_1, y_1)$, $(x_2,y_2)$. Denote $f(x)=mx+n$, the line that accepts those two points.
Assuming intersection, $x^2+(mx+n)^2=1\Leftrightarrow x^2+m^2x^2+2mnx+n^2=1$
• Please provide an example with circle radius $1.5$ with center $(3,4)$ and line segment from $(3,3)$ to $(6,7)$. – Keavon Mar 17 '14 at 0:37
• First, transform to (0,0) and normalize, by the transofrmation T(x,y)=(x-3,y-4)/(1.5). Transform all the points with T and you'll get a new problem with the same answer you're looking for. After transforming, find the line and circle equations and continue from there. – Astro Nauft Mar 17 '14 at 7:45
Hint:
1. Find the equation of the circle and line. The circle should be easy, use the line slope and a point to find the equation.
2. Find the intersection. This is solving for either variable and then plugging it in to get coordinates.
3. Draw a right triangle using a diagonal radius to the intersection and a horizontal line. Based on what you know about slope and tangents, this should help you find an angle that is complementary to your angle. | 2019-05-27T08:07:26 | {
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https://math.stackexchange.com/questions/2392765/show-that-left-1-fracxn-rightn-is-uniformly-convergent-on-s-0-1 | # Show that $\left( 1 + \frac{x}{n}\right)^n$ is uniformly convergent on $S=[0,1]$. [duplicate]
Show that $\left( 1 + \frac{x}{n}\right)^n$ is uniformly convergent on $S=[0,1]$.
Given $f_n(x)=\left( 1 + \frac{x}{n}\right)^n$ is a sequence of bounded function on $[0,1]$ and $f:S \rightarrow \mathbb R$ a bounded function , then $f_n(x)$ converges uniformly to $f$ iff
$$\lim\limits_{n \rightarrow \infty} ||f_n - f|| = \lim\limits_{n \rightarrow \infty}\left(\sup |f_n(x) - f(x)| \right)= 0$$
As $$f(x) = \lim\limits_{n \rightarrow \infty} \left( 1 + \frac{x}{n}\right)^n = e^x$$
We have
$$\begin{split} \lim\limits_{n \rightarrow \infty} \left\|\left( 1 + \frac{x}{n} \right)^n - e^x\right\| &= \lim\limits_{n \rightarrow \infty} \left|\sup_{x \in S}\left[\left( 1 + \frac{x}{n}\right)^n - e^x \right ]\right|\\ &= \lim\limits_{n \rightarrow \infty} \left|\left( 1 + \frac{1}{n} \right)^n - e^1 \right|\\ &= \lim\limits_{n \rightarrow \infty} |e-e| \\ &= 0 \end{split}$$
I am new to sequence. Is this appropriate to show convergence?
Going back to the definition, How can I show that:
1. "$f_n(x)=\left( 1 + \frac{x}{n}\right)^n$ is a sequence of bounded function on $[0,1]$"?
2. $f:S \rightarrow \mathbb R$ a bounded function?
• Continuous functions are bounded on closed intervals. For any $n$, $f_n(x)$ expands to a polynomial, which is continuous anywhere. – IntegrateThis Aug 14 '17 at 0:15
• The third line in your last calculation is wrong: shouldn't have a limit in front of $|e-e|$. – symplectomorphic Aug 14 '17 at 1:23
• @Will Fisher: well, sure, as written it doesn't make a difference. But the point is that it in preserving the limit from the second to the third line it looks like the student is arguing that $|(1+1/n)^n-e|$ equals $|e-e|$. – symplectomorphic Aug 14 '17 at 1:27
• @Will Fisher: no, you have misunderstood the OP. The OP is proving that $f_n\to f$ uniformly by invoking hypotheses, and the questions at the bottom are about how to show the hypotheses hold. – symplectomorphic Aug 14 '17 at 1:29
• Have you shown that the sup obtains when $x=1$? It is true, but it seems to me to require proof (and you have the absolute value in the wrong place: as it stands - the absolute value of the sup of the difference-, the sup is 0 at $x=0$; you want the sup of the absolute value of the difference instead). – NickD Aug 14 '17 at 1:51
If you don't want to verify the precise maximum you can bound as
$$0 \leqslant e^x - \left(1 + \frac{x}{n} \right)^n =e^x \left[1 - \left(1 + \frac{x}{n} \right)^ne^{-x}\right] \\ \leqslant e^x \left[1 - \left(1 + \frac{x}{n} \right)^n \left(1 - \frac{x}{n} \right)^n\right] \\ = e^x \left[1 - \left(1 - \frac{x^2}{n^2} \right)^n \right],$$
since $e^{x/n} > 1 + x/n$ which implies $e^{x} > (1 + x/n)^n$ for $x \in (-n,\infty).$
By Bernoulli's inequality, $(1 - x^2/n^2)^n \geqslant 1 - x^2/n$ and
$$0 \leqslant e^x - \left(1 + \frac{x}{n} \right)^n \leqslant \frac{e^xx^2}{n} \leqslant \frac{e}{n},$$
which enables you to prove uniform convergence on $[0,1]$.
Let's write explicitly the difference $$e^x - ( 1 + \frac{x}{n})^n = \sum_{ k \ge 0} \frac{x^k}{k!} - \sum_{k = 0}^n \binom{n}{k}\left(\frac{x}{n}\right )^k=\\ =\sum_{k\ge 0} \frac{[1-\prod_{l=1}^{k-1}(1- l/n)] x^k}{k!}$$. Since every coefficient $1- \prod_{l=1}^{k-1} (1-\frac{l}{n})$ is $\ge 0$ ( and $0$ for $k \ge n+1$ ) we conclude that
$$0 < e^x - (1+ \frac{x}{n})^n\le e^1 - (1 + \frac{1}{n})^n$$ for all $n \ge 0$ and $x\in [0,1]$.
Now, it's only necessary to check ( or use ) that $(1+\frac{1}{n})^n \to e$.
As shown in inequality $(2)$ of this answer, $\left(1+\frac xn\right)^n$ is increasing in $n$. Thus, for $x\ge0$, \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\left(e^x-\left(1+\frac xn\right)^n\right) &=e^x-\left(1+\frac xn\right)^{n-1}\\ &\ge e^x-\left(1+\frac xn\right)^n\\[3pt] &\ge0 \end{align} So on $[0,1]$, $$0\le e^x-\left(1+\frac xn\right)^n\le e-\left(1+\frac1n\right)^n$$
The logarithm $\ln:[1,e]\to[0,1]$ is uniformly continuous on this interval, so it is a uniform homeomorphism, and it suffices to show that $\ln f_n(x)=n\ln (1+x/n)$ is uniformly convergent on $[0,1]$. But this follows immediately from Taylor's theorem: $$n\ln(1+x/n)-x=O(x^2/n).$$ | 2020-08-15T14:41:28 | {
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https://physics.stackexchange.com/questions/637017/what-type-of-bifurcation-is-this | # What type of bifurcation is this?
Consider the dynamical system
$$dx/dt = -\cos(r)\sin(x)$$
Clearly $$x=0$$ and $$x=\pi$$ are two fixed points of this system. The stability of these two fixed points change as r is varied. Starting from $$r=0$$, the zero fixed point is initially stable but becomes unstable at $$r=\pi/2$$ and the opposite happens for the other fixed point.
Is this a transcritical bifurcation? I have so far only seen systems where one of the fixed point approaches the other fixed point and their stability switch as they cross each other.
However in my system, the fixed points remain at the same values of $$x$$. Only stability changes. If not transcritical, what type of bifurcation is this?
TL;DR: I don't believe the system displays a transcritical bifurcation.
Your stability analysis is correct. For instance, for $$r=0$$, $$\dot{x}$$ is simply a negative sine function, so positive $$x\approx 0$$ values have negative derivatives (and decrease, i.e., move towards zero) and negative values have positive derivatives (so increase and also move towards zero), i.e., $$x=x^*=0$$ is stable for $$r<\pi/2$$. See figure below:
And the system does go through a bifurcation as $$r$$ is varied — though not one of the simple ones, due to the singular nature of the system at $$r=r^*=\pi/2$$: For this value of $$r$$ all points $$x\in[0,2\pi]$$ are (neutrally stable) fixed points, since the system is simply $$dx/dt=0$$. That's how I'd sketch its bifurcation diagram:
It looks like we could say there's a collision on $$r^*=\pi/2$$ and the fixed points "exchange stabilities — just like in a transcritical bifurcation, however, I think we have something different here, for two reasons:
I. If we check the formal conditions for the system $$\dot{x}=f_r(x)$$ to go through a transcritical bifurcation (e.g., Guckenheim & Holmes or Wolfram MathWorld), we'll see they are: \begin{align} f_r(x)\Bigr|_{\forall r, x=x^*}=0 \tag{equilibria}\\ \frac{\partial f_r}{\partial x}\Bigr|_{r=r^*, x=x^*} = 0 \tag{null eigenvalue}\\ \frac{\partial^2 f_r}{\partial r\partial x}\Bigr|_{r=r^*, x=x^*} \ne 0 \tag{TC1}\\ \frac{\partial^2 f_r}{\partial^2 x}\Bigr|_{r=r^*, x=x^*} \ne 0 \tag{TC2} \label{TC2} \end{align} where TC stands for transversality condition. And it's clear that our $$f_r(x)$$ fails to satisfy \ref{TC2}.
II. By changing variables, $$r \mapsto -r + \pi/2$$, the system becomes $$f_r(x)=-\sin(r)\sin(x)$$ and $$r^*=0$$. If we then try to put $$f_r(x)$$ in a algebraic normal form, by Taylor expanding it around $$(r^*,x^*)$$ we obtain, after a new change of variables $$r\mapsto r-r^3/6$$, $$f_r(x) = rx -r\frac{x^3}{6} +O(5)$$ which I don't believe can be put in the normal form for the transcritical bifurcation, nor any of the other simple ones: \begin{align} \dot{x} &= r - x^2 \tag{saddle-node}\\ \dot{x} &= rx - x^2 \tag{transcritical}\\ \dot{x} &= r - x^3 \tag{pitchfork} \end{align} Which is perhaps not that surprising, given the system's bifurcation diagram.
• I missed out a negative sign in the equation. Thanks for pointing it out. If not transcritical, does this type of bifurcation have some name to it that I can search for? May 17 at 18:47
• @SnehaSrikanth Not that I know, maybe we could call it transcritical-like? :) May 17 at 18:50
• @SnehaSrikanth I'm now convinced the system doesn't display a transcritical bifurcation, check my updated answer. May 18 at 19:24
• Thanks for such a clear detailed answer! I still want to keep the question open for anyone who might be able to pinpoint what kind of bifurcation this is. I am new to StackExchange.. do I edit this post and change the question or make a new question? I do not want your answer to become irrelevant. May 19 at 11:44
• Ok will make that change! I am actually unable to upvote yet (I need 15 reputations for that). I'll surely upvote when I hit 15 reputations :D May 19 at 16:14 | 2021-09-28T14:06:07 | {
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https://math.stackexchange.com/questions/3461890/proof-that-forall-k-in-mathbb-z-lfloor-log-22k1-rfloor-lfloor-log-2 | # Proof that $\forall k\in\mathbb Z^+$, $\lfloor\log_2(2k+1)\rfloor=\lfloor\log_2(2k)\rfloor$
I wrote a proof for this but am not completely sure if this is valid. Specifically, can I go from $$m\leq k+0.5$$ to $$m\leq k$$ because $$m,k\in\mathbb Z^+$$ with no further explanation? Any feedback would be greatly appreciated!
Proof. Let $$n=\lfloor\log_2(2k+1)\rfloor$$. Then, \begin{align*} n&\leq\log_2(2k+1)\lt n+1&&\text{(since \forall x\in\mathbb R, \lfloor x\rfloor\leq x\lt x+1),}\\ \implies2^n&\leq2k+1\lt2^{n+1}&&\text{(by algebra).}\\ \end{align*} Observe that $$2^n\in2\mathbb Z^+$$ so $$2^n=2m$$ for some $$m\in\mathbb Z^+$$. Thus, \begin{align*} 2m&\leq2k+1&&\text{(since 2m=2^n\leq2k+1),}\\ \implies m&\leq k+0.5&&\text{(by algebra),}\\ \implies m&\leq k&&\text{(since m,k\in\mathbb Z^+),}\\ \implies 2m&\leq2k&&\text{(by algebra),}\\ \implies 2^n&\leq2k&&\text{(since 2^n=2m),}\\ \implies 2^n&\leq2k\lt2^{n+1}&&\text{(since 2k+1\lt2^{n+1}\rightarrow2k\lt2^{n+1}),}\\ \implies n&\leq\log_2(2k)\lt n+1&&\text{(by algebra),}\\ \therefore n&=\lfloor\log_2(2k)\rfloor&&\text{(since \forall x\in\mathbb R, \lfloor x\rfloor\leq x\lt x+1).} \end{align*} $$\tag*{\blacksquare}$$
• In your latter paragraph it seems you can just go from line $1$ to lines $4/5$ Dec 3, 2019 at 21:29
• Your proof looks good to me! Dec 3, 2019 at 21:37
Note that for all $$n \in \mathbb{N_{>0}}$$ there exist $$m \in \mathbb{N}$$ such that $$\begin{eqnarray*} 2^m \leq n < 2^{m+1}. \end{eqnarray*}$$ This value $$m$$ is $$\lfloor\log_2(n)\rfloor$$ And $$n$$ increases passed a power of $$2$$ the value $$m$$ will increase by $$1$$.
If $$n$$ increases (by $$1$$) from an even value to an odd value the $$m$$ will not change (provided $$n>1$$). So $$\begin{eqnarray*} \lfloor\log_2(2k+1)\rfloor = \lfloor\log_2(2k)\rfloor. \end{eqnarray*}$$ | 2022-08-17T18:49:38 | {
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https://math.stackexchange.com/questions/482453/sketch-the-region-enclosed-by-the-given-curves-and-find-its-area | # sketch the region enclosed by the given curves and find its area
sketch the region enclosed by the given curves and find its area:
$$y=\frac 1x,\; y=x,\; y=\frac x4,\; x>0.$$
I have no problem sketching the area between the curves but there are three, and only one constant value, so I don't know what to put as my second a/b value. I tried using 1/x as a b value but that just gave me an equation answer and the answer isn't an equation.
edit: i forgot about the intersections as constant values. but now I see how to split them up.
This is chapter 6.1 calculus James Stewart btw (area between curves)
• Just FYI: there are many calculus textbooks each with their own chapter/section scheme. "Chapter 6.1" isn't useful information without telling us what textbook you're using. – Adam Saltz Sep 2 '13 at 21:28
From Wolfram Alpha, we can sketch the curves to find the area of interest:
Note that we need to find the points of intersection: at $x = 0$ the lines $y = x, \;y = \frac x{4}$ intersect. At $x= 1$, the lines $y = x$ and $y = \frac 1x$ intersect. At $x = 2,$ the lines $y = \frac 1x$ and $y = \frac x4$ intersect. You can solve this by integrating between the relevant curves from $x = 0$ to $x = 1$, and likewise integrating between the relevant curves between $x = 1$ and $x = 2$, then summing: $$\int_0^1 \left(x - \frac x4\right)\,dx \quad + \quad \int_1^2 \left(\frac 1x - \frac x4\right)\,dx$$
• ok I solved it but I got ln2 - 1/2 and the answer is ln2 O_o i double checked calculations what did i do wrong?? – J L Sep 2 '13 at 21:39
• You have to evaluate the second integral $F(2) - F(1)$ – Namaste Sep 2 '13 at 21:42
• First integral $\frac 12 - \frac 18 = \frac 38$. Second integral: $\ln 2 - \frac 12 - (\ln 1 - \frac 18) = \ln 2 - \frac 38$. Sum, we get $\ln 2$. – Namaste Sep 2 '13 at 21:44
• i figured it out i wrote 7/8 instead of 3/8 for some reason – J L Sep 2 '13 at 21:45
• Good then, that should agree. Is all good? – Namaste Sep 2 '13 at 21:47
Hint: Suppose that we integrate with respect to $x$. Since there are two types of upper curves, draw a vertical line at $x=1$ (where the two upper curves of $y=x$ and $y=1/x$ intersect) that splits the region into two cases. You should obtain: $$\left[\int_0^1 x - \frac x4~dx \right] + \left[\int_1^2 \frac 1x - \frac x4~dx \right]$$
• ohhh yeahh i see. thanks for your help :) I forgot about spliting things into two integrals – J L Sep 2 '13 at 21:26
If you sketched the region correctly then you should be seeing a sort of triangle whose base is given by the line $y=x/4$ whose left side is given by $y=x$ and whose right side is given my $y=1/x$.
The first thing you have to do is to identify the crossing points, that is, the endpoints of this distorted triangle. One crossing point is obviously given by $(0,0)$ where the lines $y=x$ and $y=x/4$ meet, for the next one you have to solve the equation $$\frac{1}{x}=x$$ from where you obtain $x=1$ (we care only about the positive solution since $x>0$). The other vertex of the desired region is given by solving $$\frac{1}{x}=\frac{x}{4}$$ from where you can obtain the solution $x=2$. Look at your sketched region, we will integrate in terms of x-slices. Usually one integrates the area between the two curves $f(x)$ and $g(x)$ in the $x$-region $[a,b]$ as $\int_a^b f(x)-g(x) \; dx$. In this case however, the expression for the upper limit changes exactly at $x=1$ so there should be TWO subtractions instead of one. The desired expression for the area $A$ is $$A = \int_{0}^1 x-\frac{x}{4}\; dx + \int_{1}^2 \frac{1}{x}-\frac{x}{4} \; dx$$ And you should be able to calculate that integral yourself. Hope that helps.
Make a scetch of all three curves over the domain $x>0$. Can you see how they form and bound a triangle-like figure?
What do you mean by "only one constant value"? If you sketch the three curves, you'll see two regions which look like triangles with one bent edge (the $y=1/x$ part). In one of those regions, all the points have positive $x$-coordinates. In the other, the $x$-coordinates are all negative. Integrate to find the area of the positive one. | 2019-11-15T16:37:40 | {
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https://math.stackexchange.com/questions/409626/is-the-cartesian-product-of-groups-the-product-of-a-normal-subgroup-and-its-quot | # Is the cartesian product of groups the product of a normal subgroup and its quotient group?
I'm studying elementary group theory, and just seeing the ways in which groups break apart into simpler groups, specifically, a group can be broken up as the sort of product of any of its normal subgroups with the quotient group of that subgroup. So I wondered how you could do the inverse of that operation:
1. Given two groups $A$ and $B$, construct a group $G$ which admits a normal subgroup $H$ isomorphic to $A$, such that $G/H$ is isomorphic to $B$.
I think I have a proof that the cartesian product $A \times B$ (with the usual component-wise operation) verifies (1), but since I'm just starting out I'm not totally confident in my construction. Furthermore, if I'm right, is this the only group up to isomorphism satisfying (1)?
Edit: I just noticed Proving the direct product D of two groups G & H has a normal subgroup N such that N isomorphic to G and D/N isomorphic to H, which seems to positively answer my question. In that case I'd like to draw attention to the follow up question above (uniqueness up to isomorphism).
Yes, the direct product $A \times B$ satisfies the property, as you've noticed. But it's not unique up to isomorphism. For example, the dihedral group $D_n$ has a normal subgroup $H \simeq \mathbb Z/n \mathbb Z$, with $G/H \simeq \mathbb Z/2 \mathbb Z$, but $D_n$ is not isomorphic (for $n > 1$) to $\mathbb Z/2\mathbb Z \times \mathbb Z/n\mathbb Z$.
More generally, you're asking whether, given an exact sequence of the form $1 \to N \to G \to H \to 1$, is $G$ isomorphic to $N \times H$? The answer is no, as I've shown. Many counterexamples are provided by semidirect products (something you'll learn soon enough if you're studying elementary group theory).
For abelian groups, the concept of Ext functor allows one to classify all such extensions (given abelian groups $A,B$, "how many" groups $G$ are there with an exact sequence $0 \to B \to G \to A \to 0$ is given by $\mathrm{Ext}(A,B)$), but this is much more advanced.
• If each of A,B,C is abelian and the sequence 0 -> A -> B -> C -> 0, what would be some restrictions on A, C that would ensure that B is isomorphic to A x C? What if A and C are cyclic? – gen Oct 24 '18 at 22:27
You're correct that $A\times B$ satisfies (see here) the stated property, but in general it will not be the only such group.
The simplest example is with $A=B=\mathbb{Z}/2\mathbb{Z}$, in which case $\mathbb{Z}/4\mathbb{Z}$ also has the desired property.
Your idea about the direct product working is true. The general idea you are looking for however is that of a semidirect product which refutes your claim of uniqueness in the general setting.
It's interesting to ask when the direct product is the only possibility and that can actually be answered by finding the group $\text{Ext}(B,A)$. This is know as an extension problem. We have that if $\text{Ext}(B,A)=1$ then $A\times B$ is the unique extension of $B$ by $A$.
I believe the converse is false however, but I can't think of an example at the moment. That is, there exist pairs $(B,A)$ such that $\text{Ext}(B,A)\neq 1$ but $B\times A$ is the unique (up to group isomorphism, not extension equivalence) extension of $B$ by $A$. | 2019-10-15T19:09:29 | {
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http://math.stackexchange.com/questions/185971/how-do-i-solve-a-continued-fraction-of-solution-to-quadratic-equation | # How do I solve a Continued Fraction of solution to quadratic equation?
I know that it is possible to make a CF (continued fraction) for every number that is a solution of a quadratic equation but I don't know how.
The number I'd like to write as a CF is:
$$\frac{1 - \sqrt 5}{2}$$
How do I tackle this kind of problem?
-
Let's rewrite $\ \dfrac {1-\sqrt{5}}2=-1+\dfrac {3-\sqrt{5}}2$ to have something positive to evaluate then : $$\frac 1{\dfrac {3-\sqrt{5}}2}=\frac 2{3-\sqrt{5}}=\frac {2(3+\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})}=\frac {2(3+\sqrt{5})}{9-5}=\frac {3+\sqrt{5}}{2}=2+\frac {\sqrt{5}-1}{2}$$ (we want the term at the right to be between $0$ and $1$ at each stage)
You may continue this process until repetition !
You should get : $$\dfrac {1-\sqrt{5}}2=-1+\cfrac 1{2+\cfrac 1{1+\cfrac 1{1+\ddots}}}$$
-
So this solution is [-1;2,1,1,...], isn't that different from the one in the post above? – Spyral Aug 23 '12 at 18:51
@Spyral: Sasha has a minus sign over the whole C.F. I have only the integer part with a sign (both results are correct, perhaps that this one is more common...) – Raymond Manzoni Aug 23 '12 at 18:57
That confused me indeed.. I thought a C.F. was UNIQUE.. so your solution being [-1,2,1,1,...], sasha's being -[0;1,1,...] and Brian M. Smiths being [0;1,1,...] is kind of confusing me =/ – Spyral Aug 23 '12 at 19:08
Even though all these methods might be correct, this one was the clearest to me, thanks – Spyral Aug 23 '12 at 19:12
@Spyral: There are many types of continued fraction. The standard one in Number Theory is the simple continued fraction given in the very nice answer by Raymond Manzoni. For these, there is almost a uniqueness theorem (depending on how you define things, a rational will have two simple continued fraction expansions that are minor variants of each other). For generalized continued fractions, there is definitely not uniqueness. – André Nicolas Aug 23 '12 at 19:14
Suppose $x$ is a root of $p(z) = z^2 - b z - c$. Then, diving $p(z)$ over $z$ and solving that for $z$ gets us $$z = b+ \frac{c}{z}$$ Iterating: $$z = b + \cfrac{c}{b + \cfrac{c}{z}} = \cfrac{b}{c + \cfrac{c}{b+ \ddots}}$$ Since $\frac{1-\sqrt{5}}{2}$ is a root of $z^2 - z -1$ we have: $$\frac{1-\sqrt{5}}{2} = -\frac{1}{\frac{1+\sqrt{5}}{2}} = - \cfrac{1}{1 + \cfrac{1}{1+ \frac{1}{1+\ddots}}}$$
Added: Consider a sequence, defined by $x_{n+1} = b + \frac{c}{x_n}$, with $x_0 = \frac{c}{y}$. Few initial terms of the sequence are $\frac{c}{y}$, $b + y$, $b + \cfrac{c}{b+y}$, $b + \cfrac{c}{b + \cfrac{c}{b+y}}$, etc. It is well known that terms of this sequence can also be obtained as a ratio of two solutions, $a_n$ and $b_n$ to the following recurrence equation: $$v_{n} = b v_{n-1} + c v_{n-2} \tag{1}$$ with initial conditions $a_0=y$, $a_1 = c$ and $b_0 = 1-\frac{b}{c} y$, $b_1 = y$. Then $x_n = \cfrac{a_n}{b_n}$. The solution to $(1)$ has the form: $$v_{n} = v_0 \frac{z_1 z_2^n - z_2 z_1^n}{z_1-z_2} + v_1 \frac{z_1^n - z_2^n}{z_1-z_2}$$ where $z_1$ and $z_2$ are the two roots of $z^2 - b z -c = 0$. Assume $z_2>z_1$. In the large $n$ limit, $$\lim_{n \to \infty} x_n = \lim_{n \to \infty} \frac{a_n}{b_n}= \lim_{n\to \infty} \frac{a_1 (z_1^n - z_2^n) + a_0 (z_1 z_2^n - z_1^n z_2)}{b_1 (z_1^n - z_2^n) + b_0 (z_1 z_2^n - z_1^n z_2)} = \frac{a_0 z_1 - a_1}{b_0 z_1 - b_1} = \frac{c (c- y z_1)}{c (y - z_1) + b y z_1}$$ In the case at hand $x_0 = \infty$ and $x_1 = b$, corresponding to the limit of $y \to \infty$, in which case the value of the continued fraction becomes: $$\lim_{n \to \infty} x_n = \lim_{y \to 0} \frac{c (c- y z_1)}{c (y - z_1) + b y z_1} = - \frac{c}{z_1} = z_2$$
-
The result of the continued fraction will be the larger of two roots. With $b>0$ and $c>0$ both roots are real. – Sasha Aug 23 '12 at 18:49
How do you get the other root? Can you give some intuition about why, when you pass from the iterations to the limit, one of the roots "disappears"? – Rahul Aug 23 '12 at 18:52
@RahulNarain The product of roots of the quadratic polynomial is the free term. Thus, the smaller root is $z_2 = -\frac{c}{z_1}$. – Sasha Aug 23 '12 at 18:54
OK, but that doesn't answer my second question, because surely $z_2$ also satisfies $z=b+c/z$, $z = b+c/(b+c/z)$, and so on. – Rahul Aug 23 '12 at 19:03
@RahulNarain I have expanded the post, addressing your second question now. – Sasha Aug 23 '12 at 19:26
The general procedure is as follows for positive $x$. Let $x_0=x$, and let $[a_0;a_1,a_2,\dots]$ be the desired CF expansion. Then $a_0=\lfloor x_0\rfloor$. Given $x_n$ and $a_n$, let $$x_{n+1}=\frac1{x_n-a_n}$$ and $a_{n+1}=\lfloor x_{n+1}\rfloor$.
Since $\frac12(1-\sqrt5)$ is negative, let’s work with its absolute value, $x=\frac12(\sqrt5-1)$. Clearly $0\le x<1$ so $a_0=0$. Then $$x_1=\frac1x=\frac2{\sqrt5-1}=\frac{2(\sqrt5+1)}4=\frac{1+\sqrt5}2\;;$$ so $$\lfloor x_1\rfloor=\left\lfloor\frac{1+\sqrt5}2\right\rfloor=1\;,$$ since $2\le\sqrt5<3$, and $a_1=1$.
Now $$x_2=\frac1{x_1-1}=\frac2{\sqrt5-1}=x_1\;,$$ so everything repeats: $a_2=1$, $x_3=x_1$, $a_3=1$, etc. Thus, $x=[0;1,1,1,\dots]$, and your number is the negative of this.
- | 2015-11-28T08:01:15 | {
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https://www.jiskha.com/questions/1041569/a-radioactive-substance-decays-according-to-the-formula-q-t-q0e-kt-where-q-t-denotes | # Math
A radioactive substance decays according to the formula
Q(t) = Q0e−kt
where Q(t) denotes the amount of the substance present at time t (measured in years), Q0 denotes the amount of the substance present initially, and k (a positive constant) is the decay constant.
(a) Find the half-life of the substance in terms of k.
(b) Suppose a radioactive substance decays according to the formula
Q(t) = 36e−0.0001074t
How long will it take for the substance to decay to half the original amount? (Round your answer to the nearest whole number.)
1. 👍 0
2. 👎 0
3. 👁 1,867
1. .5 = e^-k T
ln .5 = - k T
T = -.693/-k = .693/k
b)
T = .693 / .0001074 = 6454 years
1. 👍 0
2. 👎 3
2. A radioactive substance decays according to the formula
Q(t) = Q0e−kt
where Q(t) denotes the amount of the substance present at time t (measured in years), Q0 denotes the amount of the substance present initially, and k (a positive constant) is the decay constant.
(a) Find the half-life of the substance in terms of k.
(b) Suppose a radioactive substance decays according to the formula
Q(t) = 36e−0.0001238t
How long will it take for the substance to decay to half the original amount? (Round your answer to the nearest whole number.)
1. 👍 0
2. 👎 0
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https://matheducators.stackexchange.com/tags/number-theory/hot | Episode #125 of the Stack Overflow podcast is here. We talk Tilde Club and mechanical keyboards. Listen now
# Tag Info
14
The way I have explained the fundamental theorem of arithmetic in the past is by establishing what it means to be prime (has exactly two positive divisors) and then having students construct factor trees (where the prime factors at the end are circled). The prime definition avoids some of the caveat-language otherwise seen ("a number only divisible by ...
14
It might not be possible to get your brother to arrive at the proof himself, no matter how much you scaffold it. ("You can lead a horse to water" and all that.) If he's into maths and appreciates a good proof, you might get the desired enthusiasm by just showing him the proof! That said, here's an idea. Forget primes for a moment, and think about ...
14
I think it turns out that "perfect" numbers do not interact much with other parts of number theory. Some of these very old, elementary, very ad-hoc definitions of special classes of integers have proven (and will prove) to interact interestingly with other ideas, but some seem not to. It's not easy for a beginner to guess the significance or subtlety of one ...
13
I don't really answer the question but: why do you want your brother to come to the answer right now? Now, your brother understands that, to prove that the set of prime numbers is infinite, you can assume it is finite. And, from this pool of prime numbers, you can construct a new one that is not in the family. Why don't you let your brother play with that ...
12
This is indeed tricky, and it seems to me the most effective way (in far more general, similar situations) is to show them the problem would be to have them apply their method to another, close problem where the answer is actually opposite. Either they will explain why it does not apply there, and you can argue that the difference is subtle enough to warrant ...
11
To really understand why the integers $\mathbb{Z}$ have unique prime factorization, it helps to understand how unique prime factorization can fail in other settings. For example, $$(2 + \sqrt{10}) \cdot (2 - \sqrt{10}) = -6 = -2 \cdot 3,$$ so prime factorizations in $\mathbb{Z}[\sqrt{10}] = \{a + b\sqrt{10}: a, b \in \mathbb{Z}\}$ aren't unique. On the other ...
10
What's wrong is that most of the justification is missing. Why can $N = p^2/q^2$ only happen if $q^2 = 1$? This can be justified using the fundamental theorem of arithmetic (which states that integers have unique prime factorization), but that justification needs to be made explicit, and they need to make sure that the fundamental theorem of arithmetic has ...
10
I would recommend Python combined with SageMath, as already recommended by Joseph O'Rourke, or rather SageMath and Python comes naturally. Python is a modern, and widely used, interpreted language (no compilation needed) it supports big integers via the bignum type. (But using SageMath I think this is tangential, I mention it for completeness mainly.) ...
9
The two statements aren't literally "the same", because as the student observed, they say different things. However, they are logically equivalent: each implies the other. (Similarly, "4/2" isn't literally the same expression as "5 - 3", but they provably represent the same number.) If we could only prove things by repeating the same statement, we couldn't ...
9
Maybe the issue is that if the values you're multiplying have units, then the result of multiplying will have the product of those units. Therefore, your result can't really be equivalent to a value in the base set, because the units are different. Therefore, if applying this to the real world, I suggest considering the repeated-addition interpretation (...
8
There is some value beyond the algorithm to insist on the fact that quotient and remainder in an Euclidean division are uniquely determined as soon as one settles on some convention on the remainder. What they are exactly depends on ones convention (non-negative remained, smallest absolute value, or still something else) but if one fixes a convention then ...
8
Another classic is the following: A rectangular floor measures $300 \text{ cm} \times 195 \text{ cm}$. What is the largest square tiles that can be used to cover the floor exactly?
8
Going back to Euclid, I have found questions such as `Given a large supply of rods of length $15$ and $21$, what lengths can be measured?' can appeal to students. This also motivates the result $\gcd(a,b) = ra+sb$ of the (extended) Euclidean Algorithm.
8
Some nice geometrical applications arise in the analysis of periodical curves such as Roulettes (Spirograph curves), Star Polygons, etc. Concrete experience with implementations in toys like Spirograph also provides excellent motivation for more abstract concepts such as cyclic groups.
8
I wrote an article for NCTM's grades 8-14 journal, The Mathematics Teacher, which is about building towards ${\rm F}\ell{\rm T}$ by thinking through ways in which the following statement can be generalized: If you subtract a natural number from its square, then the result is even. I wrote more about this in some earlier StackExchange posts, e.g., MESE ...
8
What is the name of this subdiscipline in math education? In one of the comments, Dave L Renfro has a reference to Piaget, whose work was primarily done with younger schoolchildren. With respect to extending this work into the older years of one's education, a potentially good place to look would be APOS Theory, which is due to Ed Dubinsky and collaborators....
7
(Summary: I would suggest exploring it flexibly, but ensuring students also know the "rigid" version.) Rather than directly addressing Euclid's algorithm for the $\gcd$ of two whole numbers, I believe one witnesses similar phenomena when covering the standard algorithm for division. For example, I observe analogs with your remarks of: The flexible method ...
7
Mathematically, even perfect numbers give a good number theory example to the general idea of classification, i.e. all even perfects have a specific form. I use perfect numbers in my number theory class for two or three pedagogical reasons: with some trial and error (and the help of some computational software), I have the students essentially discover ...
7
Wilson's Theorem is very powerful for proving things related to quadratic residues, and I think also in general a good theoretical tool. Motivating it by just saying "hey let's multiply everything together" seems very reasonable and even playful. Why is this theoretical? It is basically saying that if you multiply all units together you get a specified ...
7
The way I understand the question is: If students are not taught fractions, but instead formal deductive proofs of properties of natural numbers, would they learn mathematics better? I find it unlikely. Students struggle with fractions in primary school. Students usually struggle with proofs in gymnasium or university. It seems unlikely that primary school ...
6
I think using actual small primes actually detracts from finding the solution. If you are thinking about finding a number that is not divisible by 2,3, or 5, it is easy to come up with one (11) without having to use a formula. Thus, I would get him thinking about primes more symbolically. Given 3 primes, $p_1, p_2, p_3$ what is their LCM? What is ...
6
You probably already know that $1/\zeta(2)$ is the probability of two "random" integers being relatively prime. Since this constant is related to number theory, you should expect any proof of the Basel formula to be hard work. Probably the best proof from the point of view of transparency is obtained by evaluating \int_0^1\int_0^1 \frac{1}{1-xy}{\rm d}x{\...
6
What computer languages might one recommend for, say, investigations in number theory? I find Mathematica ideal, e.g.: "Mod sequences that seem to become constant; and the number 316" "Does 53 diverge to infinity in this Collatz-like sequence?" But: (a) there is a huge start-up learning curve, and (b) Mathematica is not free. Because of the latter, I ...
6
Here's a word problem for the greatest common divisor: 12 boys and 15 girls are to march in a parade. The organizer wants them to march in rows, with each row having the same number of children, and with each row composed of children with the same gender. What is the largest number of children per row that satisfies these constraints? There should be $\... 6 I've heard very good reviews of the 2017 book, "An Illustrated Theory of Numbers" by MH Weissman. The book's main site is here; a write-up, along with some reviews, by the American Mathematical Society can be found here. To quote from the latter (emphasis added by me): An Illustrated Theory of Numbers gives a comprehensive introduction to number theory, ... 5 Euler's original heuristic may "fly" with people who don't know about calculus: just as polynomials (e.g., with all real roots) are essentially products of linear factors$x-\alpha$where$\alpha$runs through the roots, one might imagine that$\sin \pi z$(using radian measure) is a products of something like$1-{z\over n}$for$n$non-zero integer, and ... 5 I like to ask my probability students the question: If you pick an integer between 2 and 100 uniformly at random, what's the probability that it's the average of two (not necessarily different) primes? I like that above question (easily equivalent to Goldbach) because there's no preference for even versus odd numbers as in Goldbach. It also gets ... 5 I am not sure whether this really answers your question, but I could think of the following strategy of introducing these sets of numbers. It is not based on any kind of research and just based on personal experience with first year university students where I sometimes give something similar as a "naive idea" in addition to the proper definitions. The ... 5 Some caveats: I own both books, and have taken number theory courses up to graduate level. Also, your questions are clearly somewhat subjective: what is difficult, relevant or complete for one person, depends on their previous exposure, inclination, course content, and ability. So...in my opinion.. both books are classics, which means they have been around ... 5 Look at patterns in decimal expansions: what is the period of the repeating decimal of$1/n$? From numerical data, the period is at most$n-1$, and you only get equality when$n = p$is prime (but not conversely:$1/11$has decimal period$2$, not$10$). If you look at the period of the decimal of$1/p$for primes$p$besides$2$and$5\$, you find that ...
Only top voted, non community-wiki answers of a minimum length are eligible | 2019-10-22T22:43:29 | {
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https://math.stackexchange.com/questions/2415349/coefficient-of-x3-in-expansion-of-frac1ex-cdot-1x | # Coefficient of $x^3$ in expansion of $\frac{1}{e^x\cdot (1+x)}$
If $|x|<1$, then what is the coefficient of $x^3$ in expansion of $\dfrac{1}{e^x\cdot (1+x)}$?
My Try: So I rewrote this by taking $e^x$ to the numerator to get $\frac{e^{-x}}{(1+x)}$. My problem is, how can I deduce the coefficient of $x^3$ here, since I would need to divide the numerator by the denominator, which is a cumbersome task. Is there a simpler way? How can I solve problems like these? I'd love to know.
• You just need that term. Do a few steps of the long division. – Hellen Sep 3 '17 at 14:08
• Is really long division the only method? What if I was asked the coefficient of $x^200$? I would certainly not divide then. There has to be another, more logical way. – Tanuj Sep 3 '17 at 14:09
• Not all the terms of all Taylor series have nice expressions. Some of them may take a lot of effort. I, too, hate it when I need to calculate the $x^5$ term of $\tan x$. Here you can also rewrite $1/(1+x)=1-x+x^2-x^3+\cdots$ and multiply that term-by-term with the series of $e^{-x}$. This is still easy. – Jyrki Lahtonen Sep 3 '17 at 14:12
• @JyrkiLahtonen But in this case, both Taylor series are as nice as it gets... – Clement C. Sep 3 '17 at 14:17
• @ClementC. Quite. But I still wouldn't want to calculate the $x^{200}$ term :-) – Jyrki Lahtonen Sep 3 '17 at 14:50
Hint. Note that $$\frac{1}{e^x\cdot(1+x)}=\frac{e^{-x}}{1-(-x)}=\left(\sum_{n=0}^{\infty} \frac{(-x)^n}{n!}\right)\cdot \left(\sum_{n=0}^{\infty} {(-x)^n}\right)\\ =\left(1-x+\frac{x^2}{2}-\frac{x^3}{6}+o(x^3)\right)\cdot \left(1-x+x^2-x^3+o(x^3)\right).$$ Can you take it from here?
P.S. Along the same lines, it can be seen that, more generally, the coefficient of $x^n$ in the expansion of $\frac{1}{e^x\cdot(1+x)}$ is $$\sum_{k=0}^n\frac{(-1)^k}{k!}\cdot (-1)^{n-k}=(-1)^n\sum_{k=0}^n\frac{1}{k!}.$$ See the Cauchy product wiki page.
• Okay, I see, but how does that help me? You know if there were two terms in the numerator and one in denominator I could easily solve. But not in this case. – Tanuj Sep 3 '17 at 14:11
• @Tanuj Now you have to expand the product. – Robert Z Sep 3 '17 at 14:14
\begin{eqnarray*} \frac{e^{-x}}{1+x} &=&\left(1-x+\frac{x^2}{2}-\frac{x^3}{6}+\cdots \right) (1- x+x^2-x^3+\cdots ) \\ &=& \cdots +x^3 \underbrace{\left( -1-1-\frac{1}{2}-\frac{1}{6}\right)}_{\color{blue}{-\frac{8}{3}}}+ \cdots \end{eqnarray*}
• Can you explain what exactly did you do? – Tanuj Sep 3 '17 at 14:12
• Wow a eureka moment for me! You used that result of infinite Geometric Progression, right? – Tanuj Sep 3 '17 at 14:14
• Just used the standard expansions for the functions ... – Donald Splutterwit Sep 3 '17 at 14:16
From the two Taylor series at $0$, to order $x^3$:$$e^{-x} = 1-x+\frac{x^2}{2}-\frac{x^3}{6} + o(x^3) \tag{1}$$ and $$\frac{1}{1+x} = 1-x+x^2-x^3+o(x^3)\tag{2}$$ we obtain \begin{align} \frac{e^{-x}}{1+x} &= \left(1-x+\frac{x^2}{2}-\frac{x^3}{6} + o(x^3)\right)\left(1-x+x^2-x^3+o(x^3)\right)\\ &= 1-x+x^2-x^3+o(x^3) - x+x^2-x^3+\frac{x^2}{2}-\frac{x^3}{2}-\frac{x^3}{6}\\ &= 1-2x+\frac{5}{2}x^2{\color{red}{-\frac{8}{3}}}x^3 +o(x^3) \end{align} and you can just read off the coefficient.
Note that when expanding the product, we could have focused on only the terms $x^3$ (it would have been marginally faster, but would have required keeping track); and didn't expand any term $x^k$ with $k>3$, since anyway they are "swallowed" by the $o(x^3)$. | 2020-01-24T12:08:41 | {
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https://mathhelpboards.com/threads/limits-of-volume-integrals.1052/ | # Limits of volume integrals
#### GreenGoblin
##### Member
Help choose the limits of the following volume integrals:
1) V is the region bounded by the planes x=0,y=0,z=2 and the surface z=x^2 + y^2 lying the positive quadrant. I need the limits in terms of x first, then y then z AND z first, then y and then x. And also polar coordinates, x=rcost, y=rsint, z=z.
I am trying to convince myself
2) V is the region bounded by x+y+z=1, x=0, y=0, z=0.
#### CaptainBlack
##### Well-known member
Help choose the limits of the following volume integrals:
1) V is the region bounded by the planes x=0,y=0,z=2 and the surface z=x^2 + y^2 lying the positive quadrant. I need the limits in terms of x first, then y then z AND z first, then y and then x. And also polar coordinates, x=rcost, y=rsint, z=z.
I am trying to convince myself
2) V is the region bounded by x+y+z=1, x=0, y=0, z=0.
The region V is the volume above the parabaloid surface $$z=x^2+y^2$$ and below $$z=2$$ in the first quadrant. The projection of this region onto the x-y plane is that part of the circle $$x^2 + y^2=2$$ in the first quadrant.
So this is $$x\in (0,\sqrt{2})$$ and $$y \in (0, \sqrt{2-x^2})$$ and $$z \in ( \sqrt{x^2+y^2} , 2)$$
CB
#### HallsofIvy
##### Well-known member
MHB Math Helper
(I assume that by "x first, then y then z" you are referring to going from left to right. Of course, the integration is in the opposite order.)
"x first, then y then z" is pretty much the usual way to do an integral like this. Note that if we project into the xy-plane, the upper boundary projects to the circle $x^2+ y^2= 2$, a circle of radius $2$, centered at (0, 0) and in the first quadrant that is a quarter circle. In order to cover that entire figure, clearly x has to go from 0 to $\sqrt{2}$. Now, for each x, y must go from the x-axis, y= 0, up to the circle, $y= \sqrt{2- x^2}$. And, finally, for any given (x,y), z must go from the paraboloid, $z= x^2+ y^2$, to z= 2. The integral would be $\int_{x=0}^\sqrt{2}\int_{y= 0}^\sqrt{2- x^2}\int_{z= x^2+ y^2}^2 f(x,y,z)dzdydx$.
In the other order, z first, then y and then x, project to the yz- plane. There, of course, the paraboloid project to half of the parabola $z= y^2$. Now, z goes from 0 to 2 and, for each z, y goes from 0 to $\sqrt{z}$. Now, for each y and z, x goes from 0 to $\sqrt{2- y^2}$. The integral would be $\int_{z=0}^2\int_{y= 0}^\sqrt{z}\int_{x= 0}^\sqrt{2- y^2} f(x,y,z)dxdydz$.
For V is the region bounded by x+y+z=1, x=0, y=0, z=0, with the order "x first, then y then z", we project to the xy-plane which gives the triangle bounded by x= 0, y= 0, and x+ y= 1. x ranges from 0 to 1 and, for each x, y ranges from 0 to 1- x. For each (x, y), z ranges from 0 to 1- x- y. The integral would be $\int_{x= 0}^1\int_{y= 0}^{1- x}\int_{z= 0}^{1- x- y} f(x,y,z) dzdydx$. Can you get it for the order "z first, then y then x"?
Last edited: | 2020-12-05T14:09:29 | {
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https://math.stackexchange.com/questions/2544529/variance-of-alternate-flipping-rounds/2701095 | # Variance of alternate flipping rounds
I did the following exercise, but I would like to extend the question to the variance of the variate.
Bob and Bub each has his own coin. Chance of coming up "heads" is $\rho$ for Bob's coin and $\tau$ for Bub's. They flip alternatively, first Bob, then Bub, then Bob again, etc. Let Bob's flip followed by Bub's flip constitute a round, and let $R$ denote the number of rounds until each gets "heads" at least once. For $\rho = 1/3$, $\tau = 2/5$, what is the expectation of $R$?
General answer for the expectation is:
$$\mathbb{E}[R]=\frac{1 + \frac{\rho}{\tau} + \frac{\tau}{\rho} - (\rho + \tau)}{\rho + \tau - \rho \, \tau}$$
This agrees with Monte Carlo simulation I did (with $10^5$ repeats), which approximates expectation and variance (with $\rho = 1/3$, and $\tau = 2/5$) to $3.84647$ and $6.48666$ respectively.
Is anybody able to calculate variance symbolically?
• Out of curiosity, where does your expected value formula come from?
– Remy
Dec 1, 2017 at 0:23
• @Remy I derived it. In a nutshell, $R$ is distributed geometrically with parameter $p = \rho + \tau - \rho \, \tau$ (probability of success of either one of them) plus a geometric variate with parameter $p = \rho$ and weighted with probability $\mathbb{P}\{\text{Bob didn't have success, but Bub did} \, | \, \text{either had}\}$ plus a similar variate for the reverse case.
– BoLe
Dec 2, 2017 at 12:51
Please load the page $$2\sim3$$ times for the hyperlinks to work properly.
# Outline
Setup the Notations : As titled.
Solution.1 : Direct application of the conditional decomposition of expectation. This is the foolproof approach if one wants a quick numeric evaluation and doesn't want to be bothered with analysis.
Solution.2 : A framework that provides perspectives and better calculation.
Appendix.A : Supplementary material to Solution.1.
Appendix.B.1 : In-site links to existing questions that are closely related.
Appendix.B.2 : Supplementary material to Solution.2.
# Setup the Notations
Let $$X$$ be the total number of trials of Bob's flips when his first head (success) appears.
Let $$Y$$ be that for Bub. We have $$X\sim \mathrm{Geo}[\rho]$$ independent to $$Y\sim \mathrm{Geo}[\tau]$$.
The following basics for a Geometric distribution will be useful here: \begin{align*} \mu_{_X} &\equiv \mathbb{E}[X] = \frac1{\rho} & & \tag*{Eq.(1)} \label{Eq01} \\ A_X &\equiv \mathbb{E}[X(X+1)] = \frac2{ \rho^2 } & &\tag*{Eq.(2)} \label{Eq02} \\ S_X &\equiv \mathbb{E}[X^2 ] = \frac{2 - \rho}{ \rho^2 } & &\tag*{Eq.(3)} \label{Eq03} \\ V_X &\equiv \mathbb{V}[X] = \frac{1 - \rho}{ \rho^2 } & &\tag*{Eq.(4)} \label{Eq04} \\ Q_X &\equiv \mathbb{E}[(X+1)^2] = A_X + \mu_{_X} + 1 = \frac{ 2 + \rho + \rho^2 }{ \rho^2 } & &\tag*{Eq.(5)} \label{Eq05} \end{align*} Recall that we often use $$A_X$$ to obtain $$S_X$$ because $$A_X$$ is easier to derive (it is a more natural quantity for the Geometric distribution).
The shorthands for $$Y$$ are the same $$\mu_{_Y}$$ and $$V_Y$$ etc.
# Solution.1
Just like how the expectation can be derived (which apparently you know how to), the 2nd moment (thus variance) can be obtained by conditioning on the results of the round. Denote the events of $$\{ \text{Bob head, Bub tail} \}$$ as just $$HT$$, and recall that $$X$$ is for Bob flipping alone as if Bub doesn't exist. \begin{align*} \mathbb{E}[ R^2 ] &= \rho \tau \, \mathbb{E}\left[ R^2 \,\middle|~ HH\right] + (1 - \rho)(1 - \tau)\,\mathbb{E}\left[ R^2 \,\middle|~ TT\right] \\ &\hspace{36pt} + \rho (1 - \tau) \, \mathbb{E}\left[ R^2 \,\middle|~ HT\right] + \tau (1 - \rho)\,\mathbb{E}\left[ R^2 \,\middle|~TH\right] \\ &= \rho \tau + (1 - \rho)(1 - \tau)\,\mathbb{E} \left[ (1+R)^2 \right] + \rho (1 - \tau) \, \mathbb{E}\left[ (1+Y)^2 \right] + \tau (1 - \rho)\,\mathbb{E}\left[ (1+X)^2 \right] \end{align*} Please let me know if you need justification for $$\mathbb{E}[ R^2 ~|~~TH] = \mathbb{E}[ (1+Y)^2 ]$$ and the alike. Moving on, use the \ref{Eq04} shorthand $$Q_X$$ and $$Q_Y$$ for now and rearrange. $$\mathbb{E}[ R^2 ] = \rho \tau + (1 - \rho)(1 - \tau) \left( \mathbb{E}[ R^2 ] + 2 \mathbb{E}[ R ] + 1 \right) + \rho (1 - \tau) Q_Y + \tau (1 - \rho)\,Q_X$$ Denote $$\lambda = 1 - (1 - \rho)(1 - \tau) = \rho + \tau - \rho \tau$$, and collect $$\mathbb{E}[ R^2 ]$$ on the left. $$\begin{equation*} \lambda\,\mathbb{E}[ R^2 ] = \rho \tau + (1 - \lambda) \left( 2 \mathbb{E}[ R ] + 1 \right) + \rho (1 - \tau) Q_Y + \tau (1 - \rho)\,Q_X \tag*{Eq.(6)} \label{Eq06} \end{equation*}$$ Note that the denominator of $$\mathbb{E}[ R ]$$ is just $$\lambda$$. Along with the symmetry, this suggests that the numerator of $$\mathbb{E}[ R ]$$ can be rewritten into a better form invoking the basic \ref{Eq01}. \begin{align*} 1 + \frac{ \rho }{ \tau } + \frac{ \tau }{ \rho } - (\rho + \tau) &= \bigl( 1 + \frac{ \rho }{ \tau } - \rho \bigr) + \bigl( 1 + \frac{ \tau }{ \rho } - \tau \bigr) - 1 \\ &= \frac{ \tau + \rho - \rho\tau }{ \tau } + \frac{ \rho + \tau - \rho\tau }{ \rho } - 1 \\ \implies \mathbb{E}[ R ] = \frac1{\rho} + \frac1{\tau} &- \frac1{\lambda} = \mu_{_X} + \mu_{_Y} - \frac1{\lambda} \tag*{Eq.(7)} \label{Eq07} \end{align*} It's not a coincidence that the expectation can be expressed as such. The reason will be elaborated in the next section for Solution.2.
For the sake of computing the numeric value, \ref{Eq06} was a good place to stop. With the given parameters $$\rho = 1/3$$ and $$\tau = 2/5$$, we have $$\mathbb{E}[ R ] = 23/6$$, $$Q_X = 22$$, and $$Q_Y = 16$$. That makes $$\mathbb{E}[ R^2 ] = 190/9$$ and the variance $$V_R \equiv \mathbb{E}[ R^2 ] - \mathbb{E}[R]^2 = \frac{77}{12}~.$$ See the end of Solution.2 for a Mathematica code block for the numerical evaluation and more.
One can quickly check the value of $$\mathbb{E}[ R ] \approx 3.8333$$ relative to $$\mu_{_X} = 3$$ and $$\mu_{_Y} = 2.5$$, as well as $$V_R \approx 6.146667$$ in relation to $$V_X = 6$$ and $$V_Y = 15/4$$. Both of the quantities for $$R$$ are slightly larger than the max of $$X$$ and $$Y$$, which is reasonable.
Now, if you have a strong inclination for algebraic manipulation, then the following is what you might arrive at, after some trial and error. Recall the shorthand for the 2nd moment \ref{Eq03}: $$\begin{equation*} \mathbb{E}[ R^2 ] = \frac{ 2 - \rho }{ \rho^2 } + \frac{ 2 - \tau }{ \tau^2 } - \frac{ 2 - \lambda }{ \lambda^2 } = S_X + S_Y - \frac{ 2 - \lambda }{ \lambda^2 } \tag*{Eq.(8)} \label{Eq08} \end{equation*}$$ Again, this is not a coincidence. Along with \ref{Eq07}, their respective 3rd terms seem to be another Geometric random variable with the success' parameter $$\lambda$$. The proper probabilistic analysis is the subject of Solution.2 up next.
By the way, blindly shuffling the terms around is usually not the best thing one can do. Nonetheless, just for the record, Appendix.A shows one way of going from \ref{Eq06} to \ref{Eq08} mechanically.
# Solution.2
Denote $$W \equiv \min(X,Y)$$ as the smaller among the two and $$Z \equiv \max(X,Y)$$ as the larger. The key observation to solve this problem is that $$Z \overset{d}{=} R~, \qquad\qquad \textbf{the maximum has the same distribution as the 'rounds'.}$$ This allows one to think about the whole scenario differently (not as rounds of a two-player game). For all $$k \in \mathbb{N}$$, since $$X \perp Y$$ we have \begin{align*} \Pr\{ Z = k \} &= \Pr\{ X < Y = Z = k \} \\ &\hspace{36pt} + \Pr\{ Y < X = Z = k \} \\ &\hspace{72pt} + \Pr\{ X = Y = Z = k \} \\ &= \tau (1 - \tau)^{k-1} (1 - (1-\rho)^k) \\ &\hspace{36pt} + \rho (1 - \rho)^{k-1} (1 - (1-\tau)^k) \\ &\hspace{72pt} + \rho\tau (1 - \rho)^{k-1} (1 - \tau)^{k-1} \tag*{Eq.(9)} \label{Eq09} \end{align*} In principle, now that the distribution of $$Z$$ (thus $$R$$) is completely specified, everything one would like to know can be calculated. This is indeed a valid approach (to obtain the mean and variance), and the terms involved are all basic series with good symmetry.
Note that $$Z$$ is not Geometric (while $$X$$, $$Y$$, and $$W$$ are), nor is it Negative Binomial. At this point, it is not really of interest that \ref{Eq09} can be rearranged into a more compact and illuminating form ...... because we can do even better.
There are two more observations that allow one to not only to better calculate but also understand the whole picture. \begin{align*} &X+Y = W + Z \tag*{Eq.(10)} \label{Eq10} \\ &W \sim \mathrm{Geo[\lambda]} \tag*{Eq.(11)} \label{Eq11} \end{align*} This is the special case of the sum of the order statistics being equal to the original sum. In general with many summands this not very useful, but here with just two terms it is crucial.
Back to the two-player game scenario, the fact that $$W$$ is Geometric with success probability $$\lambda = 1 - (1 - \rho)(1 - \tau)$$ is easy to see: a round 'fails' if and only if both flips 'fail', with a probability $$(1 - \rho)(1 - \tau) = 1 - \lambda$$.
The contour of $$W = k$$ is L-shaped boundary of the '2-dim tail', which fits perfectly with the scaling nature (memoryless) of the joint of two independent Geometric distribution. Pleas picture in the figure below that $$\Pr\left\{W = k_0 + k ~ \middle|~~ W > k_0 \right\} = \Pr\{ W = k\}$$ manifests itself as the L-shape scaling away.
The contour of $$Z = k$$ looks like $$\daleth$$, and together with the L-contour of $$W$$ they make a cross. This is \ref{Eq10} the identity $$X+Y = W+Z$$, visualized in the figures below. See Appendix.B.1 for the linked in-site posts of related topics.
Immediately we know the expectation to be: $$\begin{equation*} \mathbb{E}[ Z ] = \mathbb{E}[ X + Y - W ] = \mu_{_X} + \mu_{_Y} - \mu_{_W} = \frac1{\rho} + \frac1{\tau} - \frac1{\lambda} \tag*{Eq.(7.better)} \end{equation*}$$
This derivation provides perspectives different from (or better, arguably) those obtained by conditioning on the first round.
Derivation of the 2nd moment reveals a more intriguing properties of the setting. \begin{align*} \mathbb{E}[ Z^2 ] &= \mathbb{E}[ (X + Y - W)^2 ] \\ &= \mathbb{E}[ (X + Y)^2 ] + \mathbb{E}[ W^2 ] - 2\mathbb{E}[ (X + Y) W ] \\ &= \mathbb{E}[ X^2 ] + \mathbb{E}[ Y^2 ] + 2 \mathbb{E}[ XY ] + \mathbb{E}[ W^2 ] - 2\mathbb{E}[ (W+Z) W ] \qquad\because X+Y = W+Z\\ &= \mathbb{E}[ X^2 ] + \mathbb{E}[ Y^2 ] - \mathbb{E}[ W^2 ] + 2\mathbb{E}[ XY ] - 2\mathbb{E}[ WZ ] \end{align*} Here's the kicker: when $$X \neq Y$$, by definition $$W$$ and $$Z$$ each take one of them so $$WZ = XY$$, and when $$X = Y$$ we have $$WZ = XY$$ just the same!! Consequently, $$\mathbb{E}[ ZW ] = \mathbb{E}[ XY ] =\mu_{_X} \mu_{_Y}$$ always, and they cancel. $$\begin{equation*} \mathbb{E}[ Z^2 ] = \mathbb{E}[ X^2 ] + \mathbb{E}[ Y^2 ] - \mathbb{E}[ W^2 ] = \frac{ 2 - \rho }{ \rho^2 } + \frac{ 2 - \tau }{ \tau^2 } - \frac{ 2 - \lambda }{ \lambda^2 } \tag*{Eq.(8.better)} \label{Eq08better} \end{equation*}$$ This is the proper derivation, and in \ref{Eq08} the 3rd term following $$S_X + S_Y$$ is indeed $$-S_W$$.
Keep in mind that both $$W$$ and $$Z$$ are clearly correlated with $$X$$ and $$Y$$, while our intuition also says that the correlation between $$W$$ and $$Z$$ is positive (see Appendix.B for a discussion).
While we're at it, this relation in fact holds true for any (higher) moments, $$\mathbb{E}[ Z^n ] = \mathbb{E}[ X^n ] + \mathbb{E}[ Y^n ] - \mathbb{E}[ W^n ]~.$$This is true due to the same argument that gave us $$WZ = XY$$, and using \ref{Eq09} doing the explicit sums to derive this identity is also easy.
Without further ado, the variance of $$Z$$ (thus the variance of $$R$$), previously known only as the unsatisfying "\ref{Eq06} minus the square of $$\mathbb{E}[ Z ]$$'', can now be put in its proper form. \begin{align*} V_Z &\equiv \mathbb{E}[ Z^2 ] - \mathbb{E}[ Z ]^2 \\ &= \mathbb{E}[ X^2 ] + \mathbb{E}[ Y^2 ] - \mathbb{E}[ W^2 ] - ( \mu_{_X} + \mu_{_Y} - \mu_{_W} )^2 \\ &= V_X + V_Y - V_W + 2( \mu_{_W} \mu_{_Z} - \mu_{_X} \mu_{_Y}) \tag*{Eq.(12.a)} \\ &= \frac{ 1 - \rho }{ \rho^2 } + \frac{ 1 - \tau }{ \tau^2 } - \frac{ 1 - \lambda }{ \lambda^2 } + 2\left( \frac1{ \lambda } \bigl( \frac1{ \rho } + \frac1{ \tau } - \frac1{ \lambda } \bigr) - \frac1{ \rho \tau }\right) \tag*{Eq.(12.b)} \end{align*} This expression (or the symbolic one just above) is a perfectly nice formula to me. You can rearrange it to your heart's content, for example, like this $$\begin{equation*} V_Z = \bigl( \frac1{ \rho } - \frac1{ \tau } \bigr)^2 - \frac1{ \lambda^2 } + 2 (\frac1{\lambda} - \frac12) \bigl( \frac1{ \rho } + \frac1{ \tau } - \frac1{ \lambda } \bigr) \tag*{Eq.(12.c)} \end{equation*}$$ which emphasizes the role played by the difference between $$X-Y$$.
I'm not motivated to pursue a 'better' form beyond Eq.(12), and if anyone knows any alternative expressions that are significant either algebraically or probabilistically, please do share.
Let me conclude with a Mathematica code block verifying the numeric values and identities for the variance, the 2nd moment, as well as the earliest algebra of the conditioning of expectation.
(* t = \tau, p = \rho, m = E[R], S = E[R^2] , h = \lambda = 1-(1-p)(1-t) *) ClearAll[t , p, m, S, V, h, tmp, simp, sq, var, basics]; basics[p_] := {1/p, (1 - p)/p^2, (2 - p)/p^2, (2 + p + p^2)/ p^2};(* EX, VX, E[X^2], E[(1+X)^2]*)
Row@{basics[1/3], Spacer@10, basics[2/5], Spacer@10, basics[1 - (1 - 1/3) (1 - 2/5)]}
simp = Simplify[#, 0 < p < 1 && 0 < t < 1] &; h = 1 - (1 - p) (1 - t); m = (1 + t/p + p/t - (t + p))/h; sq[a_] := (2 - a)/a^2;(* 2nd moment of a Geometric distribution *) var[a_] := (1 - a)/a^2;
S = tmp /. Part[#, 1]&@Solve[tmp == p t + (1 - p) (1 - t) (1 + 2 m + tmp) + p (1 - t) (1 + 2/t + sq@t) + (1 - p) t (1 + 2/p + sq@p), tmp] //simp (* this simplification doesn't matter ... *)
V = S - m^2 // simp (* neither does this. *)
{tmp = V /. {p -> 1/3, t -> 2/5}, tmp // N} (* below: veryfiy the various identities for 2nd moment and variance. Difference being zero means equal *)
{sq@t + sq@p - sq@h - S, var@p + var@t - var@h + 2 m 1/h - 2/(p t) - V, (1/p - 1/t)^2 - 1/h^2 + 2 (1/p + 1/t - 1/h) (1/h - 1/2) - V} // simp
`
$$\large\textbf{Appendix.A: }\normalsize\text{an algebraic route from \ref{Eq06} to \ref{Eq08}}$$
Consider \ref{Eq06} one part at a time. First, the $$2(1 - \lambda)\, \mathbb{E}[ R ]$$. From the the first line to the 2nd line, $$1 - \lambda = (1 - \rho)(1 - \tau)$$ is inserted for the 2nd term: \begin{align*} 2(1 - \lambda)\, \mathbb{E}[ R ] &= 2(1 - \lambda )\frac{ - 1}{\lambda} + 2(1 - \lambda) \bigl( \frac1{ \rho } + \frac1{ \tau } \bigr) \\ &= 2\frac{\lambda - 1}{\lambda} + \frac{2(1 - \rho ) }{ \rho }(1 - \tau) + (1 - \rho ) \frac{ 2(1 - \tau) }{ \rho } \\ &= 2 - \frac2{\lambda} + \bigl( \frac{2 - \rho}{ \rho } - 1\bigr) (1 - \tau) + \bigl( \frac{2 - \tau}{ \tau } - 1 \bigr) (1 - \rho ) \\ &= \color{magenta}{2 - \frac2{\lambda} } + \frac{2 - \rho}{ \rho^2 } (1 - \tau)\rho \color{magenta}{- (1 - \tau)} + \frac{2 - \tau}{ \tau^2 } (1 - \rho )\tau \color{magenta}{- (1 - \rho )} \end{align*} Next in line are the $$Q_X$$ terms. \begin{align*} \rho (1 - \tau) Q_Y + \tau (1 - \rho)\,Q_X &= \rho (1 - \tau) \bigl( \frac{ 2 + \tau }{ \tau^2 } + 1\bigr) + \tau (1 - \rho) \bigl( \frac{ 2 + \rho }{ \rho^2 } + 1 \bigr) \\ &= \rho \bigl( \frac{ 2 - \tau - \tau^2 }{ \tau^2 } + 1 - \tau \bigr) + \tau \bigl( \frac{ 2 - \rho - \rho^2 }{ \rho^2 } + 1 - \rho \bigr) \\ &= \rho \frac{ 2 - \tau}{ \tau^2 } + \tau \frac{ 2 - \rho }{ \rho^2 } \color{magenta}{- 2\tau\rho} \end{align*} Put things back together in \ref{Eq06}, the magenta terms combine with $$\rho\tau$$ and $$1 - \lambda$$ (multiplied by 1): \begin{align*} \lambda\,\mathbb{E}[ R^2 ] &= \rho \tau + (1 - \lambda) + (1 - \lambda) 2 \mathbb{E}[ R ] + \rho (1 - \tau) Q_Y + \tau (1 - \rho)\,Q_X \\ &= \rho \tau + (1 - \lambda) \color{magenta}{ {}+ 2 - \frac2{\lambda} - (1 - \tau) - (1 - \rho ) - 2\tau\rho} \\ &\hphantom{{}= \rho \tau} + \frac{2 - \rho}{ \rho^2 } (1 - \tau)\rho + \frac{2 - \tau}{ \tau^2 } (1 - \rho )\tau \tag{from \mathbb{E}[R]}\\ &\hphantom{{}= \rho \tau } + \tau \frac{ 2 - \rho }{ \rho^2 } + \rho \frac{ 2 - \tau}{ \tau^2 } \tag{from Q_X etc}\\ &= 1 - \frac2{\lambda} + \frac{ 2 - \rho }{ \rho^2 } \bigl( \rho + \tau - \rho\tau \bigr) + \frac{ 2 - \tau }{ \tau^2 } \bigl( \rho + \tau - \rho\tau \bigr) \\ &= 1 - \frac2{\lambda} + \frac{ 2 - \rho }{ \rho^2 } \lambda + \frac{ 2 - \tau }{ \tau^2 } \lambda \end{align*} Finally we can divide by the coefficient of $$\mathbb{E}[ R^2 ]$$ on the left hand side and obtain \ref{Eq08}.
$$\large\textbf{Appendix.B.1: }\normalsize\text{In-Site Links to Related Questions}$$
I found many existing posts about this topic (minimum of two independent non-identical Geometric distributions). In chronological order: 90782, 845706, 1040620, 1056296, 1169142, 1207241, and 2669667.
Unlike the min, only 2 posts about max is found so far: 971214, and 1983481. Neither of the posts go beyond the 1st moment, and only a couple of the answers address the 'real geometry' of the situation.
In particular, consider the trinomial random variable $$T$$ that splits the 2-dim plane into the 3 regions. $$\begin{equation*} T \equiv \mathrm{Sign}(Y - X) = \begin{cases} \hphantom{-{}} 1 & \text{if}~~Y > X \quad \text{, with probability}~~ \tau( 1 - \rho) / \lambda \\ \hphantom{-{}} 0 & \text{if}~~X = Y \quad \text{, with probability}~~ \rho \tau / \lambda \\ -1 & \text{if}~~Y < X \quad \text{, with probability}~~ \rho( 1 - \tau) / \lambda \end{cases} \end{equation*}$$ This trisection of above-diagonal, diagonal, and below-diagonal is fundamental to the calculation of both $$W$$ and $$Z$$.
It can be easily shown that $$T \perp W$$, that the trisection is independent of the minimum. For example, $$\Pr\left\{ T = 1 \mid W = k\right\} = \Pr\{ T = 1\}$$ for all $$k$$.
Note that $$T$$ is not independent to $$Z$$ (even with the corner $$k=1$$ removed).
In the continuous analogue, $$\{X,Y,W\}$$ are Exponential with all the nice properties, and $$Z$$ is neither Exponential nor Gamma (analogue of Negative Binomial).
The density of the $$Z$$ analogue is easy and commonly used, but its doesn't seem to have a name. At best one can categorize it as a special case of phase-type (PH distribution).
$$\large\textbf{Appendix.B.2: }\normalsize\text{Covariance between \min(X,Y) and \max(X,Y), along with related discussions.}$$
Recall the expectations: $$\mathbb{E}[X] = \mu_{_X} = 1 / \rho$$, and the similar \begin{align*} \mu_{_Y} &= \frac1{ \tau }~, & \mu_{_W} &= \frac1{ \lambda } = \frac1{\rho + \tau - \rho\tau}~, & &\text{and}\quad \mu_{_Z} = \mu_{_X} + \mu_{_Y} - \mu_{_W} \end{align*} The covariance between the minimum and the maximum is $$\begin{equation*} C_{WZ} \equiv \mathbb{E}\left[ (W - \mu_{_W} ) (Z - \mu_{_Z} ) \right] = \mathbb{E}[ WZ ] - \mu_{_W} \mu_{_Z} = \mu_{_X} \mu_{_Y} - \mu_{_W} \mu_{_Z} \tag*{Eq.(B)} \label{EqB} \end{equation*}$$ The last equal sign invokes the intriguing $$WZ = XY$$ mentioned just before \ref{Eq08better}.
A positive covariance (thus correlation) expresses the intuitive idea that when the min is large then the max also tends to be large.
Here we do a 'verbal proof' of $$C_{WZ} > 0$$ for all combinations of parameters. The purpose is not really a technical proof but more about illustrating the relations between $$\{W, Z\}$$ and $$\{X, Y\}$$.
Since $$\mu_{_X} + \mu_{_Y} = \mu_{_W} + \mu_{_Z}$$, the covariance $$C_{WZ}$$ is in essence comparing products (of two non-negative numbers) given the value of the sum. We know that for a fixed sum, the closer the two 'sides' (think rectangle) are the larger the product.
Being the min and max, $$\mu_{_W}$$ and $$\mu_{_Z}$$ are more 'extreme' and can never be as close as $$\mu_{_X}$$ and $$\mu_{_Y}$$, throughout the entire parametric space $$\{ \rho, \tau \} \in (0, 1)^2$$. Boom, QED.
(the extreme cases where at least one of $$\{\rho, \tau\}$$ is zero or one are all ill-defined when it comes to min and max)
An algebraic proof for $$C_{WZ} > 0$$ (or for everything in this appendix) is easy. Let me emphasize the descriptive aspect.
Consider what it means for $$W$$ as a Geometric distribution to be the minimum of $$X$$ and $$Y$$. \begin{align*} \frac1{\lambda} &< \min( \frac1{\rho}, \frac1{\tau} ) & &\text{faster to 'succeed' on average} \\ \lambda &> \max( \rho, \tau ) & &\text{better than the higher success parameter} \end{align*} That is, the mean of the 'minimum flip' $$\mu_{_W} < \min( \mu_{_X}, \mu_{_Y} )$$ is more extreme at the lower end.
On the other hand, $$Z$$ is NOT a Geometric distribution. However, one can define an auxiliary $$Z' \sim \mathrm{Geo}[1 / \mu_{_Z}]$$ with an equivalent mean $$\mu_{_{Z'}} = \mu_{_Z}$$. Once we have arrived at \ref{EqB}, the role of $$Z$$ can be replaced by $$Z'$$: $$C_{WZ} = C_{WZ'} = \mu_{_X} \mu_{_Y} - \mu_{_W} \mu_{_{Z'}}$$
Now we can have similar descriptions for $$\mu_{_Z}$$, while it is actually $$\mu_{_{Z'}}$$ for the Geometric distribution that is being compared with $$X$$ and $$Y$$. \begin{align*} \mu_{_Z} &> \max( \mu_{_X}, \mu_{_Y} ) & &\text{slower to 'succeed' on average} \\ \frac1{ \mu_{_Z} } &< \min( \rho, \tau ) & &\text{worse than the lower success parameter} \end{align*} That is, the mean of the 'maximum flip' is more extreme at the higher end.
This concludes the verbal argument for how $$\mu_{_W}$$ and $$\mu_{_Z}$$ are more dissimilar (than the relation between $$\mu_{_X}$$ and $$\mu_{_Y}$$) thus making a smaller product.
• Thank you for this comprehensive, inspiring answer. Solution One I don't have problem understanding, I'm upset a bit that I didn't complete my analysis. I turned back while trying to form this recursion involving nonlinear terms, like E[(1+X)^2]. Solution Two and the Appendices I am able to follow as well.
– BoLe
Mar 22, 2018 at 10:45 | 2022-05-21T12:53:01 | {
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https://math.stackexchange.com/questions/562275/finding-area-problem | # Finding area problem
There is this simple geometry question that seems so easy but I think the question lacks some information (does it?). Or maybe there are other ways to solve the problem.
So the problem says, there are two squares $ABCD$ and $FCHG$ with a side of length $8$ and $10$, respectively. We are asked to find the area of the shaded region.
What makes this question a bit tricky is because the small triangle $DEF$ is not shaded. And if we are given the length of AD, then the question will be easy. Is there a way to find AD or is there any other methods to solve the problem?
By the way, the answer is $48.4$.
And also, I am pretty sure we don't need to use advance methods such as trigs, etc.
I really appreciate any helps!
## 4 Answers
Triangles $BCF$ and $DEF$ are similar triangles, which means that their sides are proportional. You know side $BC=8,$ $CF=10$ and side $EF=2$. Can you proceed?
• yes I can! thanks! – user71346 Nov 11 '13 at 10:50
Hint: Look for similar triangles. What is $\Delta DEF$ similar to?
• ah yes! so careless... thanks! – user71346 Nov 11 '13 at 10:49
Let say lines $BG$ and $FC$ intersects at $K$,
So $CK+KF=10$ and $FKG$ and $BCK$ are similar triangles.
So $\frac{KF}{CK}=10/8$ therefore $CK=40/9$ and $KF=50/9$
The area of FKG triangle = $0.5*10*50/9=27.7$
Also $AD+DE=8$ and $ADB$ and $DEF$ are similar triangles. Similarly it gives
$AD=32/5$ and $DE=8/5$
The area of $BDEK$ = area of $BFC$- area of $DEF$ - area of $BCK$ = $0.5*8*10-0.5*2*8/5-0.5*8*40/9$= 20.7
Total area = The area of $BDEK$ + The area of FKG triangle =$27.7+20.7=48.4$
"Is there a way to find AD or is there any other methods to solve the problem?"
AD is proportioan to AE as BD is proportional to BF as EC is proportional to FC.
So AD ~ 8 as 8 ~ 10. Or AD/8 = 8/10. | 2019-12-09T08:05:41 | {
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https://math.stackexchange.com/questions/3606831/solving-a-congruence-system-when-the-chinese-remainder-theorem-cannot-be-applied | # Solving a congruence system when the Chinese remainder theorem cannot be applied
I'm trying to solve the following system $$\cases{3x\equiv1\pmod{14}\\x\equiv1\pmod{8}\\3x\equiv9\pmod{5}}$$
My understanding is that, since $$14, 8, 5$$ aren't all coprime, I cannot apply the Chinese remainder theorem.
The first thing I did was solving the first and third equations independently, which yielded the following equivalent system:
$$\cases{x\equiv5\pmod{14}\\x\equiv1\pmod{8}\\x\equiv3\pmod{5}}$$
At this point I'm unsure how to proceed. I thought solving the system made up of the first two equation, and then a system made up of the solution to the first system with the third equation could work, but turns out it didn't. Here's what I tried:
$$\cases{x\equiv5\pmod{14}\\x\equiv1\pmod{8}\\} \iff x = 5+ 14k=1+8h \rightarrow7k-4h = -2 \iff k = 2+4y, h = 4-7y, \text{ with }y\in\mathbb{Z}$$ Therefore, $$x = 33 - 56y \iff x\equiv33\pmod{56}$$. Plugging this result back into the system, we now have $$\cases{x\equiv33\pmod{56}\\x\equiv3\pmod{5}\\} \iff x = 3+5k=33+56h \rightarrow5k-56h=30 \iff k = -330+56y, h = -30-5y, \text{ with }y\in\mathbb{Z}$$ Therefore, $$x \equiv 1653\equiv 253 \pmod{280}$$; however, this result is incorrect. What did I do wrong?
• did you mean $h=4\color{red}+7y$ and $h=-30\color{red}+5y$? Apr 2 '20 at 18:55
• there is also a formula for non-coprime moduli on wikipedia Apr 2 '20 at 18:57
I would encourage you to split up the given conditions and group according to powers of the same prime.
You have $$3x \equiv 1 \pmod 7$$ $$3x \equiv 9 \pmod 5$$ and related $$3x \equiv 1 \pmod 2$$ $$x \equiv 1 \pmod 8$$
The redundant pair becomes, as $$3 \equiv 1 \pmod 2,$$ $$x \equiv 1 \pmod 2$$ $$x \equiv 1 \pmod 8$$ These are consistent, the highest power of the prime is $$8=2^3,$$ so these combine to $$x \equiv 1 \pmod 8.$$ Then $$x$$ is 5 mod 7 and 3 mod 5, together $$x \equiv 1 \pmod 8.$$ $$x \equiv 3 \pmod 5.$$ $$x \equiv 5 \pmod 7.$$ Now you can use CRT I get $$x \equiv 33 \pmod {280}$$ as $$33 = 32 + 1 = 30 + 3 = 28 + 5$$
What did I do wrong?
At the end, you should have $$x=3+5k=3+5(-330+56y)$$
or $$x=33+56h=33+56(-30-5y)$$,
which means $$x=-1647+280y$$, so $$x\equiv-1647\equiv33\pmod{280}$$.
• I think you forgot the minus sign in front of $330$ to get $1653=3+5(330)$ Apr 2 '20 at 18:38 | 2021-09-28T23:57:38 | {
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https://math.stackexchange.com/questions/2300299/suppose-a-b-in-m-n-mathbbc-such-that-ab-ba-a-prove-that-a-is-not-in | # Suppose $A,B\in M_n(\mathbb{C})$ such that $AB-BA=A$ . Prove that $A$ is not invertible .
Suppose that $A,B\in M_n(\mathbb{C})$ such that $AB-BA=A$. Prove that $A$ is not invertible.
My work:
Suppose $A$ is invertible. Then $ABA^{-1}=I+B$ . So $B$ is similar to $I+B$ .Let $B$ have eigenvalues $c_1,c_2,\ldots,c_n \in \mathbb{C}$. So $B$ has basis such that $B$ is upper triangular with respect to it and has $c_1,\ldots,c_n$ as diagonal entries . It is easy to see that $I+B$ is upper-triangular with respect to this basis and has entries $1+c_1,\ldots, 1+c_n$ .
Hence $$c_1+c_2+\ldots +c_n=\operatorname{trace}(B)$$ $$=\operatorname{trace}(I+B)=(1+c_1)+\ldots+(1+c_n)=n+c_1+\ldots +c_n .$$
So $n=0$, contradiction. I'm not sure if my solution is correct one. It seems alright. I will be very thankful if you can confirm that the proof indeed is a correct one.
Any other possible solutions are welcomed.
• It is correct, but too complicated. After proving that $B$ and $I+B$ are similar, you can simply say that this is no possible, since the trace of $I+B$ is $n$ plus the trace of $B$. No need to mention upper triangular matrices. May 28, 2017 at 16:38
Let $A$ be invertible.
We have $(AB-BA)A^{-1}=I$, or $ABA^{-1}-B=I.$
But $\operatorname{tr} (ABA^{-1})=\operatorname{tr}{B},$ which is a contradiction.
It is probably worth showing that $$A$$ is in fact nilpotent. Indeed, by induction we get $$A^m B - B A^m = m A^m$$ for all $$m\ge 0$$. Taking the traces on both sides we get $$0=m \operatorname{Trace}A^m$$ so (assuming char $$0$$) $$\operatorname{Trace}A^m = 0$$ for all $$m\ge 1$$. This implies $$A$$ nilponent.
$$\bf{Added:}$$ Based on an idea of @Hans: , we can generalize this . For $$X$$, $$Y$$ matrices, let $$[X,Y]=X Y - Y X$$.
Assume that $$[C,B]=A$$, and $$[C,A]=0$$ ( $$C$$, $$A$$ commute). Then $$A$$ nilpotent.
Indeed, for all $$m\ge 1$$ we have $$[C,A^{m-1} B] = A^{m-1} [C,B]= A^m$$, and so $$\operatorname{Trace} A^m=0$$ . Conclude $$A$$ nilpotent (assume char $$0$$).
$$\bf{Added:}$$. What we are showing is that if $$\operatorname{ad}(C)^2 A=0$$, then $$\operatorname{ad}(C) A$$ is nilpotent. We can also reason as follows: we may assume that the basic field is algebraically closed. Consider the Jordan structure of $$C$$, with blocks of size $$n_1$$, $$\ldots$$, $$n_k$$ corresponding to $$\lambda_1$$, $$\ldots$$, $$\lambda_k$$. Then we see that all operators $$B$$ in the kernel of $$\operatorname{ad}C^2$$ have the property that $$\operatorname{ad}(C)B=0$$.
(here we use an explicit Jordan structure for $$\operatorname{ad}(C)$$. We again use that the characteristic of the field is $$0$$ or $$>n$$ ).
• Nice conclusion. +1. You only need to show $A^mB-A^{m-1}BA=A^m\implies \text{tr}A^m=0, \forall m\in\mathbf N$.
– Hans
Oct 8, 2017 at 7:34
• @Hans; yes, your approach is simpler and better! I should add this idea. Oct 8, 2017 at 17:46
$\def\m{\mathfrak }$This result (and, in fact, orangeskid's observation) has the following far reaching generalization:
if $\m g$ is a finite dimensional (complex) Lie algebra and $\mathfrak r$ is its radical, then $[\m g,\m r]$ acts on any finite dimensional representation of $\m g$ nilpotently.
Indeed, in the question the matrices $A$ and $B$ span a Lie algebra $\m g$ of dimension (at most, really, but let us suppose) equal to $2$ which is solvable, so that the radical of $\m g$ is simply $\m g$ itself, and the theorem above tells us that $[\m g,\m g]$ acts nilpotently on finite dimensinal modules: since $[\m g,\m g]$ is spanned by $A$, the desired result follows.
Of course, this is immensely more general.
As an example:
if $A$, $B$ and $C$ are square matrices such that $CA-AC = B$, $CB-BC = aA + B$, $AB-BA = 0$, then $B$ is nilpotent and, more generally, al linear combinations of $A$ and $B$.
This comes from looking at a random 3-dimensional solvable Lie algebra.
• One can find the proof of this, which depends on Levi's theorem, in the book on Lie algebra by Hossein Abbaspour and Martin A. Moskowitz; it is Theorem 3.5.6 there. Oct 12, 2017 at 17:41
• Excellent answer! I will have to review my rusty rep theory... Oct 12, 2017 at 19:16 | 2022-08-10T03:11:10 | {
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https://community.wolfram.com/groups/-/m/t/1657895 | # Autopilot Automated Plane Geometry Case Study
Posted 3 months ago
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Dan's blog on automated plane geometry is about a quite fansinating application of new features of V12. This discussion is mainly to show how to make GeometricScene components as building blocks to give you insights to solve challenging geometric problems. Also because the components in this example is quite limited, I hope it serves as beginner case for Wolfram Language users who are interested in applying this feature in teaching and research.
• AOPS Problem: In triangle $ABC$ let point $D$ be on side $BC$. If $AD=7$, $\angle BAD=15$ deg, and $\angle CAD = 30$ deg, compute the minimum possible area of the triangle.
RandomInstance offers a second argument to generate a list of geometric objects that satifies the given concitions
Use the following line to find the area of the triangles above:
In[ ]:= (Area[Triangle[{"A","B","C"}]]/.(#["Points"]))&/@objs1
Out[ ]= {22.0393,21.6971,82.0575,18.1804,90.0952,18.8076,50.7958,25.823}
Lets try more triangles, like 20, and see where is the minimum value even we haven't analyze the problem at all:
We can graphically conclude that the minimum value is around 17.93. (Pretty good approximation comparing to the true solution later)
Next we consider adding some constraints to the GeometricScences so the illustrations of the valid triangle ABC's are aligned in regularzied manner. Well by the precision of WL and flexibility of MS OneNote, I quitck notice the following works:
I appriciate that the GeometricScene function gives a mixture of cartesian-coordinate based programming and Geogebra-style freedom. Just modify the first argument in GeometricScene and make 10 instances of valid ABC:
Looking at the the instances, you probably have some insights of the problem. The variation of the area of ABC is by
• fix the angle BAC=45 degree and its two side rays
• rotate a line $l$ that cross point D
$B$, $C$ is then determined by the intersections of rays and $l$. To find how this rotation affects the area, we can parameterize $\angle ABC = \theta$ and the area of ABC. This is valid because $\angle ABD$ is uniquely defined by $\angle ADC$, which can be interpred as a measure of the rotation of $l$.
Comparing to the objs2, we successfully sort the triangles based on the angular size of $\angle ABC$, from 10 degree to 110 degree with increment of 10. Lets plot the area vs $\angle ABC$ on a finer segmentation of angle values:
Remember that all codes that you have written is just gs1 plus some plot functions and parameter modification. Almost zero computation but finding the coordinate of $D$. This is way simpler than we used to do with Graphics function.
Let's compare our result to the analytic solution:
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- Congratulations! This post is now a Staff Pick as distinguished by a badge on your profile! Thank you, keep it coming, and consider contributing your work to the The Notebook Archive! | 2019-07-24T07:05:05 | {
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https://math.stackexchange.com/questions/3291437/the-number-of-triples-that-sum-to-a-constant?noredirect=1 | # The number of triples that sum to a constant
Problem:
How many triples are there of the form $$(x_0,x_1, x_2)$$ where $$x_0 \in I$$, $$x_1 \in I$$, $$x_2\in I$$ $$x_0 \geq 0$$, $$x_1 >= 0$$, $$x_2 >= 0$$ and $$n = x_0 + x_1 + x_2$$ where $$n \in I$$?
Let $$c(n)$$ be the number of tuples we can have for a given $$n$$. For $$n = 0$$, the only valid triple is $$(0,0,0)$$, hence $$c(0) = 1$$.
For $$c(1) = 3$$, the set of valid triples is: $$(0,0,1 ), (0,1,0), (0,0,1)$$ Hence $$c(1) = 3$$.
For $$c(2) = 6$$, the set of valid triples is: $$(1,0,1 ), (0,1,1 ), (0,0,2 ), (1,1,0), (0,2,0), (0,0,2)$$ Hence $$c(2) = 6$$.
Using the information on this URL:
How many $k-$dimensional non-negative integer arrays $(x_1,\cdots,x_k)$ satisfies $x_1+x_2+\cdots+x_k\le n$
I find the answer to be: $$c(n) = {{n+3}\choose{3}} - {{n+2}\choose{3}}$$ $$c(n) = \frac{(n+3)!}{3!n!} - \frac{(n+2)!}{3!(n-1)!}$$ $$c(n) = \frac{(n+3)(n+2)(n+1) - (n+2)(n+1)(n)}{6}$$ $$c(n) = \frac{(n+2)(n+1)(n+3 - n)}{6}$$ $$c(n) = \frac{3(n+2)(n+1)}{6}$$ $$c(n) = \frac{(n+2)(n+1)}{2}$$ Do I have that right?
Thanks,
Bob
• This is equivalent to $\frac12(n+1)(n+2)$ which agrees with the supplied values. Jul 12 '19 at 23:07
• Like this? Jul 12 '19 at 23:10
• en.wikipedia.org/wiki/Stars_and_bars_(combinatorics) Jul 12 '19 at 23:16
Yes - this is correct. This can also be found by defining a helper function $$b(n)$$, which counts the number of ways to write $$n$$ as a sum of two numbers. Clearly, $$b(n) = n+1$$, as the first value, $$v$$, can be anything from $$0$$ to $$n$$, while the second value is $$n - v$$.
The function $$c(n)$$ is then equal to $$\sum_{i = 0}^{n}b(n-i)$$ This is because the first value, $$i$$, can be anything from $$0$$ to $$n$$. The number of ways to write the remaining two numbers so that the total sum is $$n$$ is $$b(n-i)$$. Plugging in the formula finds $$c(n) = \sum_{i = 0}^{n} (n-i+1) = n(n+1) - \frac{n(n+1)}{2} + n+1 = \frac{(n+2)(n+1)}{2}$$
which is the same formula you arrived at.
This is a prototypical “stars and bars” problem. Here we have $$n$$ stars and $$2$$ bars, with $$\binom{n+2}2 = {(n+2)(n+1)\over2}$$ ways to choose among the $$n+2$$ possible positions for the two bars.
Another way: The number of partitions of $$n$$ into three non-negative integers is equal to the coefficient of $$z^n$$ in the formal power series $$(1+z+z^2+z^3+\cdots)^3$$. This coefficient can be found using the generalized binomial theorem: $$(1+z+z^2+z^3+\cdots)^3 = \frac1{(1-z)^3} = \sum_{k=0}^\infty \binom{-3}{k} z^k = \sum_{k=0}^\infty \binom{k+2}2 z^k.$$ The number of partitions is therefore $$\binom{n+2}2$$ as before.
• All of these were mentioned in the comments, ~9h earlier. Jul 13 '19 at 8:19
• @rtybase Your point being? If you though the question was a duplicate of some other, you should’ve flagged it as such.
– amd
Jul 13 '19 at 8:21
• Something genuinly different is required. But up to you really ... I don't think the question is a duplicate. To begin with I don't know what $I$ is (in $x_0 \in I$ for example). Nobody clarrified this. Jul 13 '19 at 8:22 | 2021-10-17T13:41:26 | {
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https://math.stackexchange.com/questions/2916320/5-cards-are-chosen-from-a-standard-deck-what-is-the-probability-that-we-get-all | # 5 cards are chosen from a standard deck. What is the probability that we get all four aces, plus the king of spades?
We have $\binom{52}{5}$ ways of doing this. This will be our denominator.
We want to select all 4 aces, there are there are exactly $\binom{4}{4}$ ways of doing this.
Now we have selected 4 cards, and we need to select one more. That one card has to be the king of spades. Out of the 48 remaining cards, only one of them is our wanted one. This can be chosen in $\binom{48}{1}$ ways.
$$\dfrac{\binom{4}{4}\binom{48}{1}}{\binom{52}{5}}=\dfrac{1}{54145}?$$
Is this correct?
If we look at it another way:
Then then our probability for the aces and specific king are $(4/52)\times (3/51) \times (2/50) \times (1/49) \times (1/48)$ which is a completely different here.
Which is the correct approach?
• numerator should be 1 if you want a specific combination – Ning Wang Sep 14 '18 at 2:15
• So the second one is correct? – K Split X Sep 14 '18 at 2:17
• No, it seems like you're permuting those four aces. – Ning Wang Sep 14 '18 at 2:19
• $\binom{48}{1}$ would mean you're selecting $1$ card (any) from the remaining $48$ cards and not specifically a king of spades. – Ister Sep 14 '18 at 7:53
• $\binom44$ is correct because there are $4$ aces to choose from, but $\binom{48}1$ should be $\binom11$ because there is only one king of spades. – bof Sep 14 '18 at 23:38
Assuming you just want those 5 cards in any order:
Since you have specified all 5 specific cards that you want, you don't need to consider the aces separately from the king. The probability of selecting these 5 cards is then
(Probability choosing any one of the 5) $\times$ (Probability of choosing one of the remaining 4) $\times$ ..., i.e. $$\frac{5}{52} \times \frac{4}{51} \times \frac{3}{50} \times \frac{2}{49} \times \frac{1}{48} = \frac{1}{54145} \times \frac{1}{48},$$ just to confirm what Ross Millikan said.
If, instead, you want the 4 aces before the king, we have $$\frac{4}{52} \times \frac{3}{51} \times \frac{2}{50} \times \frac{1}{49} \times \frac{1}{48}.$$
You've fully determined the five cards you want, so the probability is $\frac{1}{{52 \choose 5}}$ since there are ${52 \choose 5}$ different ways to choose 5 cards from the deck, and only one correct one.
• Just to clarify, you want to get 5 cards from a set of specific 5 cards in which case your numerator should be $\binom{5}{5}$ – Ister Sep 14 '18 at 7:51
• I'm not this carl; you wording leads the casual reader to believe you are incorrect, since there are 5! ways to pick the correct 5 cards. You should stress that you started by calculating combinations, not permutations – Carl Witthoft Sep 14 '18 at 15:54
Your first computation gets the chance you get four aces and any other card. The $48 \choose 1$ factor is selecting the other card. If you want specifically the king of spades it should be $1$, so the chance is $48$ times less. Your second computation gets the chance you draw the four aces first in some order, then draw the king of spades. You should multiply by $5$ for the number of positions the king can be in. Neither is correct for the question in the title.
The first way:
There is only one set of cards, that fits the specifications (all 4 aces and the king of spades) and $\binom{52}{5}$ different sets of cards that can all be drawn with equal probability. The chance to draw your set is $$\frac{1}{\binom{52}{5}}$$
The second way:
The chance of drawing a specific card out of $n$ is $\frac{1}{n}$. So a specific sequence or order of 5 cards has the chance $\frac{1}{52}\cdot\frac{1}{51}\cdot\frac{1}{50}\cdot\frac{1}{49}\cdot\frac{1}{48}$ to be drawn. There are $5!$ different sequences or orderings of 5 cards, that countain the 5 specified cards (s. permutations). So the chance to draw the set is $$\frac{5!}{52\cdot51\cdot50\cdot49\cdot48} = \frac{5!\cdot 47!}{52!} = \frac{1}{\frac{52!}{5!\cdot 47!}} = \frac{1}{\binom{52}{5}}$$
For the last equality see Binomial coefficient.
You select 4 aces out of 4 possible aces and one king of spades out of 1 possible king of spades, so there is exactly $$\binom{4}{4}\binom{1}{1}=1$$ way of completing this set. Therefore this set is unique.
There are $\binom{52}{5}$ unique sets of five cards.
Therefore the probability of picking this set of five cards is:
$$P=\frac{\binom{4}{4}\binom{1}{1}}{\binom{52}{5}} = \frac{1}{2598960}$$ | 2019-04-21T04:22:03 | {
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https://math.stackexchange.com/questions/3013379/counting-number-of-bitstrings-that-do-not-contain-110-for-length-n-ge-4 | # Counting number of Bitstrings that do not contain $110$ for length $n \ge 4$
Question: Consider bitstrings that do not contain $$110$$. Let $$S_n$$ be the number of such strings having length $$n$$. What is $$S_n$$ for $$n \ge 4$$?
Answer: $$S_n = S_{n-1} + S_{n-2} + 1$$
Attempt: I tried writing out all the bitstrings for $$n=2$$, and $$n=3$$.
$$n=3$$ I got: $$000, 001, 011, 111, 100, 010, 101$$
$$n=2$$ I got: $$00, 01, 10, 11$$
$$n=4$$ seems to large to do this by hand. According to the answer, it should be $$S_4 = S_3 + S_2 + 1$$.
$$S_4 = 7 + 0 + 1 = 8$$? Assuming $$S_n$$ is the number of strings without $$110$$, so since $$n=3$$ has one $$110$$ I just exclude that and get $$7$$. $$n=2$$ has no $$110$$.
I'm not sure how the formula was derived. The bitstring counting logic I struggle with, any guidance on how to appriach these would help a lot.
• Your bitstring either starts with a $0$, or it starts with a $10$, or it has no $0$'s at all. Do you see how that relates to the answer's recursive formula? – Barry Cipra Nov 25 '18 at 22:15
This is not going to be an answer, but I don't have enough reputation to comment, so I apologize in advance.
My thought is to try thinking about your placement of the $$110$$. For the case of $$n=3$$, you saw that there was only one place to put it - as the whole thing.
Now consider the case of $$n=4$$. The approach will be to consider the possibilities which contain the string you are looking to avoid, find out how many such unique strings there are and subtract it from the total.
There are two places to put the string $$110$$, namely the beginning $$110\bullet$$ or the end $$\bullet110$$. There are $$2^4$$ total four-digit bitstrings, and there are four which match these placement patterns $$1100, 1101$$ and $$0110, 1110$$. Just to check, we should make sure all four of these are different, which is clear for this many (just look at the third digit) but might not be so clear on a larger scale. So as I see it there are $$2^4 - 4 = 12$$ ways to do this. This is $$16-4=12 \neq 8$$.
I agree with your logic for finding $$S_3, S_2$$, and hence $$S_4 = 8$$. Do you see any errors in my logic?
If you agree with my logic, is it possible that the question is not being asked the way you have asked it here? Alternatively are you sure the solution you have is correct?
• The mistake in Toby's computation is that $S_2 = 4$, not $0$. The given solution ($S_n = S_{n-1} + S_{n-2} + 1$) is correct. (See Barry Cipria's comment above.) – Fabio Somenzi Nov 25 '18 at 22:41 | 2019-05-23T01:28:22 | {
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https://math.stackexchange.com/questions/1969668/depiction-of-all-graphs-not-necessarily-connected-with-6-vertices?noredirect=1 | # Depiction of all graphs (not necessarily connected) with 6 vertices
I am searching (with no unsuccess) the WWW for a depiction of all non-isomorphic graphs with 6 vertices.
On the site http://www.graphclasses.org/smallgraphs.html I found an incomplete list, but all graphs with 5 vertices (34 ones). According to the integer sequence A000088, there should be 156 of them. A paper by Cvetkovic and Petric http://www.sciencedirect.com/science/article/pii/0012365X84900335 gives all connected graphs with 6 vertices (112 ones).
If I take all connected graphs (112) and take all graphs with 5 vertices adding a separate node (34), then I have 146 graphs. Which 10 graphs are missing?
Thank you very much!
• By all complete graphs did you perhaps mean all connected graphs? – hardmath Oct 15 '16 at 15:54
• Sorry. Yes, thanks. – user36124 Oct 15 '16 at 15:56
• Given your existing work, you may find the earlier Question Why there are 11 non-isomorphic graphs of order 4? a useful reference. These include non-connected possibilities. – hardmath Oct 15 '16 at 16:03
• You need all connected graphs on 4 vertices (6 of them) with two separate vertices and the pair of 3-line/triangle with 3 disconnected vertices, etc. 6 + 2 + 1 + 1 = 10. To put it another way, the number of graphs on N are the sum of the number of connected graphs on N plus the number of connected graphs on N-1 ... 1. – gilleain Oct 15 '16 at 16:08
• You may be interested to look at findstat.org/StatisticFinder/Graphs by the FindStat project – Alexander Konovalov Oct 15 '16 at 17:05
We can partition the graphs by the number of connected components.
One connected component: There are $112$ connected graphs with $6$ vertices
Two connected components: There can be
• A connected component of size $1$ and a connected component of size $5$ ($1 \cdot 21 = 21$ possibilities)
• A connected component of size $2$ and a connected component of size $4$ ($\color{red}{1 \cdot 6 = 6}$ possibilities)
• A connected component of size $3$ and another connected component of size $3$ ($\color{red}{3}$ possibilities - $P_3 \sqcup P_3$, $P_3 \sqcup K_3$, or $K_3 \sqcup K_3$)
Three connected components: There can be
• Two connected components of size $1$ and a connected component of size $4$ ($1 \cdot 1 \cdot 6=6$ possibilities)
• A connected component of size $1$, one of size $2$, and one of size $3$ ($1 \cdot 1 \cdot 2 = 2$ possibilities)
• Three connected components of size $2$ ($\color{red}{1}$ possibility - $P_3 \sqcup P_2 \sqcup P_2$)
Four connected components: There can be only be three connected components of size $1$ and one of size $3$, for $1 \cdot 1 \cdot 1 \cdot 2 = 2$ possibilities
Five connected components: There can only be four connected components of size $1$ and one of size $2$, for $1 \cdot 1 \cdot 1 \cdot 1 \cdot 2 = 2$ possibilities
Six connected components: One possibility
Add these up and you get $112+21+6+3+6+2+1+2+2+1=156$ graphs.
The ones in red are the ones you are missing - they are not obtained from adding an isolated vertex to a graph with $5$ vertices.
Perhaps the easiest way to systematically count is to begin with the integer partitions of six as denoting the sizes of components in your (not necessarily connected) graphs. There are eleven possibilities.
Some of these give rise to a single (up to isomorphism) graph:
$1 + 1 + 1 + 1 + 1 + 1$
$2 + 1 + 1 + 1 + 1$
$2 + 2 + 1 + 1$
$2 + 2 + 2$
Some give rise to two possibilities each as three nodes can form a connected component in exactly two ways:
$3 + 1 + 1 + 1$
$3 + 2 + 1$
Then there are three possibilities for this partition:
$3 + 3$
because the two components can be non-isomorphic in one way but isomorphic in two ways.
From the earlier Question we see that there are six connected graphs on four vertices, so each of these gives six possibilities:
$4 + 1 + 1$
$4 + 2$
Then we get down to the cases you previously considered:
$5 + 1$ (a connected component of five vertices and one extra node)
$6$ (a connected component of six vertices)
There are respectively $21$ and $112$ of these cases (see OEIS A001349, Number of connected graphs with n nodes). The previous small cases numbered $23$, and $23+21+112= 156$. | 2019-11-20T02:56:56 | {
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