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# What does it mean to be proportional to something?
I am asked a physics homework question, but it is really simply a mathematical question I think, dealing with proportional reasoning.
The period of a pendulum is proportional to the square root of its length.
A pendulum of length $2$ has a period of $3$.
They give units but it really should not matter here.
What I have done is written this down, and I am not really sure the first step is the correct one, because I am unsure how to read the question:
$$3P_1 = \sqrt 2$$
$$P_1 = \frac{\sqrt 2}{3}$$
They want to know about a pendulum of length $4.5$ so,
$$xP_2 = \sqrt {4.5}$$
My reasoning then is that:
$$\frac{\sqrt 2}{3} = \frac{\sqrt{4.5}}{x}$$
$$xP_2 = \frac{3\sqrt{4.5}}{\sqrt 2} = 4.5$$
I never really deal with proportional reasoning, so I am going out on a limb here to make this connection, but it all seems very intuitive. Of course, my intuition is not always right. My question then is, am I doing this right? Is this good reasoning?
-
Hint:
Take T to be the time period of the pendulum.
$T_1 = k \cdot \sqrt {l_1}$
$T_2=k \cdot \sqrt{l_2}$
When you divide them both(Since none of the terms are $0$)
$\dfrac{T_1}{T_2}= \sqrt{\dfrac{l_1}{l_2}}$
Note: When L is proportional to M, you can write it as L=k. M, where k is a constant. | {
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Note: When L is proportional to M, you can write it as L=k. M, where k is a constant.
-
Hm that is interesting, and it confirms my answer when I use the k, that really helps too. It is truly discomforting that my brain does all that without informing me how it was done.. Many thanks! – Leonardo Apr 7 '13 at 10:56
You're Welcome. And do note that there might be many cases of proportionality, when you come across Gravitational force (or maybe coulomb's force), you come across 'inversely proportional to'. – Inceptio Apr 7 '13 at 10:59
It may be noted that here the actual value of k is 2*pi/sqrt(g) where g is accelaration due to gravity. – Sreekanth Karumanaghat Apr 7 '13 at 11:24
@SreekanthKarumanaghat: That really doesn't matter when you have two relations, cause: they just scare off each other. – Inceptio Apr 7 '13 at 11:50
Yeah I know that I was just noting for info sake :) – Sreekanth Karumanaghat Apr 7 '13 at 12:08
The correct meaning of propotional:- If x is directly propotional to y. x/y = k(a constant)
Analytically
Graph between x and y is straight line whose slope is k.
Applying the above formula here. T1 = k sqrt(L1) T2 = k sqrt(L2)
dividing 2 by 1,we get T1/T2 = sqrt(L1)/sqrt(L2)
3/T2 = sqrt(2)/sqrt(4.5)
T2 = sqrt(4.5)*3/sqrt(2)
T2 = 4.5
- | {
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# Recursion question
The first two terms of a sequence are $a_1 = 1$ and $a_2 = \frac {1}{\sqrt3}$. For $n\ge1$,
$$a_{n + 2} = \frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}}.$$ What is $|a_{2009}|$?
So I tried (thanks to Paul Wellin):
fun[n_] := Module[{prev = 1, this = 1/Sqrt[3], next},
Do[next = Simplify[(prev + this)/(1 - prev this)];
prev = this;
this = next, {n - 1}];
prev]
And I got:
fun[2009]
(* 0 *)
Which is the correct answer. Kind of amazed it went so quickly. I then tried:
Table[fun[i], {i, 1, 100}]
Which produces the first 100 terms of the sequence. Again, amazingly quick. Now, this turns out to be a periodic sequence. I counted with my finger, and determined 24.
Anyone have a good way to use Mathematica to determine if the sequence is periodic and find the period?
• mathematica.stackexchange.com/questions/80163/… – Marvin Dec 5 '15 at 9:31
• Entirely different method: Recognize the recursion as the sum of angles formula for tangent. Set $\tan \theta_{n} = a_n$ so $\theta_1 = \pi/4$, $\theta_2 = \pi/6$, and $\theta_{n+2} = \theta_n + \theta_{n+1}$. This is a much easier recurrence... – Eric Towers Dec 5 '15 at 19:08
• @EricTowers Could you add this as an answer? I'd love to see what you do. – David Dec 6 '15 at 3:33
One way of doing this is starting with your list of the first 100 terms.Then ask for
zzz = SequencePosition[lst, Take[lst, -2]]
(* {{3, 4}, {27, 28}, {51, 52}, {75, 76}, {99, 100}} *)
The length of this result is larger than 1, so your list must be periodic. The period is
zzz[[2, 1]] - zzz[[1, 1]]
(* 24 *)
• thank you introducing me to SequencePosition – ubpdqn Dec 5 '15 at 10:56
• @FredSimons I agree, really simple explanation to help students with exploration. Thanks. – David Dec 5 '15 at 18:19 | {
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Started as a comment to the OP. Turning this into an answer was requested. The OP asked for Mathematica help solving the recurrence. I suggested recognizing that the recurrence was the sum of angles formula for tangent, then going from there. Consequently, this answer uses Mathematica to solve this problem using only one "external" ingredient right at the beginning. (At the end, we explore a claim made at the beginning.)
Problem: Find the period of \begin{align} a_1 &= 1 \\ a_2 &= \frac{1}{\sqrt{3}} \\ a_{n+2} &= \frac{a_n + a_{n+1}}{1-a_n a_{n+1}}, n \geq 1 \end{align}
(It would be seriously convenient if FunctionPeriod[] just did this. We could "punt" and use the accepted answer here, applying it to the straightforward implementation of the above recursion and initial values. However, that's not what I'm here to write.)
"External" fact: The recursion is the sum of angles formula for tangent.
Claim at the beginning: We could replace the recursion formula with the analog for any other sum of angles formula and get exactly the same sequence of angles. I.e., there are at least two other recursions with this same solution.
First, we check this external fact with Mathematica and get the initial values.
Simplify[Tan[a+b] == (Tan[a] + Tan[b])/(1-Tan[a]Tan[b])
(* True *)
ArcTan[1]
(* Pi/4 *)
ArcTan[1/Sqrt[3]]
(* Pi/6 *)
So we may reformulate the question as \begin{align} \theta_1 &= \frac{\pi}{4} \\ \theta_2 &= \frac{\pi}{6} \\ \theta_{n+2} &= \theta_1 + \theta_2 \pmod{2\pi}, n \geq 1 \end{align}
Sums of multiples of $\pi/4$ and $\pi/6$ have denominators dividing $12$.
LCM[4,6]
(* 12 *)
There are $24$ twelfths of $\pi$ in a circle.
2 Pi / (Pi/12)
(* 24 *) | {
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LCM[4,6]
(* 12 *)
There are $24$ twelfths of $\pi$ in a circle.
2 Pi / (Pi/12)
(* 24 *)
Consequently, the period of this recurrence is at most $24^2$. (The recurrence is order $2$, so it is perhaps better to think of each adjacent pair of sequence members producing the next adjacent pair of sequence members.) (Although, as is frequently the case with homogeneous recurrences, if $\dots 0, 0 \dots$ occurs in our sequence, every subsequent angle is zero. So it is possible for the period to be less than $24$ and for there to be an initial aperiodic part, say if $\theta_1 = \theta_2 = \pi$, although it will turn out that this is not the case in this problem.)
So we generate the first $2 \times 24^2$ entries, which will contain two or more periods.
Remove[q];
q[1] = Pi/4;
q[2] = Pi/6;
q[n_ /; n >= 3] := q[n]= Mod[q[n-1]+q[n-2],2 Pi];
(* We memoize members of the sequence to avoid redundant recomputation. *)
Generate the $2 \times 24^2$ entries and glance at some of them to see if periodicity seems likely.
Table[ {k, q[k]} , {k, 24, 2*24^2, 24} ]
(* {{ 24, 23 Pi/12 }, { 48, 23 Pi/12 }, ... , {1152, 23 Pi/12 }} *)
Seems rather likely to be periodic. Test this with Fourier[]:
periods = Flatten[
Position[
Chop[Fourier[Table[q[k], {k, 1, 2*24^2}]]],
n_ /; n != 0
]
]
(* {1, 49, 97, 145, 193, 241, 289, 337, ... , 1009, 1057, 1105} *)
GCD @@ Select[DeleteDuplicates[Flatten[
Outer[Abs[#1 - #2] &, periods, periods]
]], # > 0 &]
(* 48 *)
So the period divides $48$. (Actually, we could know that it divides $24$. We ran to $2 \times 24^2$. Try it again with $3 \times 24^2$ and get $72$. In fact, if we run to $N \times 24^2$ for $N$ up to $24$, we'll get $24N$ for the GCD above.)
Let's have Mathematica try every divisor of $48$ as a candidate period.
{#, Simplify[q[3 + #] == q[3]]} & /@ Divisors[48]
(* {{1, False}, {2, False}, {3, False}, {4, False},
{6, False}, {8, False}, {12, False}, {16, False},
{24, True}, {48, True}} *) | {
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so the only candidate periods are $24$ and $48$. (Note that starting at q[3], we're finding out about the recursion moreso than the initial values.) Then we test the shortest candidate:
q /@ Range[24] == q /@ Range[25, 2*24]
The above method works, and is valid, but perhaps one would be happier seeing that the recurrence actually recurs after $24$ steps. Starting with the symbolic first two angles $a,b$, what is the resulting angle after (a divisor of $24$) applications of the recursion rule. We frame this as working on pairs in Nest[].
TableForm[
{#[[1]], #[[2]] === {a, b}, #[[2]]} & /@ (
{#, PolynomialMod[
Nest[
(* Thinking in pairs, the new first angle is the old second
angle and the new second angle is the sum of the two old
angles. *)
{#[[2]], #[[1]] + #[[2]]} &,
{a, b},
#],
#]} & /@
Divisors[24]),
TableDepth -> 1
]
(* { { 1, False, {0, 0}},
{ 2, False, {a + b, a}},
{ 3, False, {a + 2 b, 2 a}},
{ 4, False, {2 a + 3 b, 3 a + b}},
{ 6, False, {5 a + 2 b, 2 a + b}},
{ 8, False, {5 a + 5 b, 5 a + 2 b}},
{12, False, {5 a, 5 b}},
{24, True, {a, b}} } *)
So the recursion has a natural period of $24$. It's possible for the choice of initial conditions to cause the period to be shorter. (Recall that if the first two angles are $0,0$, then the sequence is all zeroes.) So we check for this. The resulting period still has to divide $24$, so we only check those (although if we were not certain of this, we could check every period up to $24$). | {
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TableForm[
{ #[[1]] , #[[2]] == #[[3]] , #[[2]] , #[[3]] } & /@ (
{ # , {q[1], q[2]}, {q[1 + #], q[2 + #]} } & /@
Divisors[24]
),
TableDepth -> 1]
(* { { 1, False, {Pi/4, Pi/6}, { Pi/ 6, 5 Pi/12}},
{ 2, False, {Pi/4, Pi/6}, {5 Pi/12, 7 Pi/12}},
{ 3, False, {Pi/4, Pi/6}, {7 Pi/12, Pi }},
{ 4, False, {Pi/4, Pi/6}, { Pi , 19 Pi/12}},
{ 6, False, {Pi/4, Pi/6}, {7 Pi/12, Pi/ 6}},
{ 8, False, {Pi/4, Pi/6}, {3 Pi/ 4, 11 Pi/12}},
{12, False, {Pi/4, Pi/6}, { Pi/ 4, 5 Pi/ 6}},
{24, True, {Pi/4, Pi/6}, { Pi/ 4, Pi/ 6}} } *)
And so the period is $24$. Then it's easy to compute the desired number:
Tan[q[Mod[2009,24,1]]] (*Don't want to inadvertently try to evaluate q[0]. *)
(* This is Tan[ q[17] ] = Tan[ Pi ]. *)
(* 0 *)
The original problem statement asked about $|a_{2009}|$ but since $0 = |0|$, there's no need to apply the absolute value function. | {
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Returning to the claim at the beginning. Since adding angles is pretty straightforward, we can use any other sum of angles formula. This means there are (at least) two other recursion problems looking wildly different from this one (by having a wildly different looking general recursion formula), but having the same underlying computation. The sum of angles formulas for sine and cosine are, with the corresponding recursion: \begin{align} \sin(\theta_1 + \theta_2) &= \sin(\theta_1) \cos(\theta_2) + \cos(\theta_1) \sin(\theta_2) \\ a_{n+2} &= a_n c(a_{n+1}) \sqrt{1-a_{n+1}^2} + a_{n+1} c(a_n) \sqrt{1-a_n^2} \\ \\ \cos(\theta_1 + \theta_2) &= \cos(\theta_1)\cos(\theta_2) - \sin(\theta_1)\sin(\theta_2) \\ a_{n+2} &= a_n a_{n+1} - s(a_n)s(a_{n+1}) \sqrt{1-a_n^2}\sqrt{1-a_{n+1}^2} \end{align} where the sign-tracking functions $s$ and $c$ are $s(\theta)$ is $1$ in quadrants I and II and the angle $\pi/2$ and is $-1$ otherwise, and $c$ is $1$ in quadrants I and IV and the angle $0$ and is $-1$ otherwise. We could eliminate these two sign-tracking functions by having two sequences, $q_n$ and $r_n$: $a_n \mapsto (q_n, r_n) = (\sin(\theta_n), \cos(\theta_n))$ or by switching to exponentials $a_n \mapsto \mathrm{e}^{\mathrm{i} \theta_n} = \cos(\theta_n) + \mathrm{i}\sin(\theta_n)$, which the final equality shows is really the same as switching to sines and cosines, but using real and imaginary parts of complex numbers to hold the two sequences, instead of pairs of real numbers. | {
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Taking $\theta_1 = \pi/4$ and $\theta_2 = \pi/6$ and using either the sum of angles for sine or for cosine, we get the same sequence of angles. Consequently, we will get a final, final answer of $a_{2009} = \tan^{-1}(\sin(\theta_{2009})) = 0$ for the sine recursion and similarly for the cosine recursion. (Whether you take the simple recurrence $\theta_{n+2} = \theta_n + \theta_{n+1}$ and obscure it via sines, cosines, tangents, or, I suppose, other periodic functions, like Weierstrass's $\wp$-functions, you still get the same sequence of fractions of the functions's period(s), and the same period of the recurrence.)
• +1, but the recurrence you display for the sine and cosine is in fact numerically unstable, and thus cannot be recommended in general. – J. M. will be back soon Dec 10 '15 at 5:18
• @J.M. : I agree in part. If we're going to talk about stability, the initial tangent recursion isn't stable either. Replace $a_{n+1}$ by $a_{n+1}+\varepsilon$ and expand in powers of $\varepsilon$ and find that the error is magnified by $\frac{1+a_n^2}{(1-a_n a_{n+1})^2}$. This is harder to interpret than doing the same thing to the tangent version, yielding $1+\tan^2(a_n + a_{n+1})$, which is necessarily $\geq 1$. So the instability is baked into the problem and the alternate recursions are merely reflecting it. – Eric Towers Dec 10 '15 at 17:11
You can use NestWhileList to show the same pair of consecutive elements appear and then determine the period, e.g.
res = NestWhileList[{#[[2]],
Simplify[(#[[1]] + #[[2]])/(1 - #[[1]] #[[2]])]} &, {1,
1/Sqrt[3]}, Unequal, All];
plt = (Most@(First /@ res))~Join~(Style[#, Red, Bold] & /@ res[[-1]])
Length[plt] - 2
ListPlot[plt~Join~res[[1]], Joined -> True]
I add this as a separate answer to amplify another answer,in relation to $\tan(x+y)$.
This sequence could be pursued as Fibonacci or linear recurrence of angles,e.g. | {
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This sequence could be pursued as Fibonacci or linear recurrence of angles,e.g.
mp[n_] :=
Mod[MatrixPower[{{1, 1}, {1, 0}}, n].{Pi/6, Pi/4}, Pi, -Pi/2];
an[1] := mp[0][[2]];
an[2] := mp[0][[1]];
an[n_?(# > 2 &)] := First@mp[n - 2]
os[n_] := Tan[an[n]];
SequencePosition[an /@ Range[1, 26], {Pi/4, Pi/6}]
os /@ Range[26]
I exploited Fred Simons SequencePosition (yielding: {{1, 2}, {25, 26}}) and the last line reproduces the original sequence:
$\left\{1,\frac{1}{\sqrt{3}},\sqrt{3}+2,-\sqrt{3}-2,0,-\sqrt{3}-2,-\sqrt{3}-2,\frac{1}{\sqrt{3}},-1,\sqrt{3}-2,-\sqrt{3},-\sqrt{3}-2,1,-\frac{1}{\sqrt{3}},2-\sqrt{3},\sqrt{3}-2,0,\sqrt{3}-2,\sqrt{3}-2,-\frac{1}{\sqrt{3}},-1,-\sqrt{3}-2,\sqrt{3},\sqrt{3}-2,1,\frac{1}{\sqrt{3}}\right\}$
A quick visual inspection helps to see how a pattern with period 24 pops up
n = 100;
tab = Table[fun[i], {i, 1, n}];
pos = Flatten @ Position[tab, First @ tab]
{1, 13, 25, 37, 49, 61, 73, 85, 97}
tra = Transpose[{pos, Array[First @ tab &, Length @ pos]}];
ListPlot[{tab, tra}, GridLines -> {pos[[;; ;; 2]], None}]
I rewrote the code to produce a list of the current "state" of the recursion, which is given by the last two terms $(a_n,a_{n-1}$. The system is autonomous, so the state is completely determined by this pair. Therefore if a pair ever occurs again, there is a loop.
Some utility functions:
step = (* basic recursive step *)
{Last[#], Simplify[Plus @@ #/(1 - Times @@ #)]} &;
steps[n_] := (* equivalent to fun[n], except it returns a list of all states *)
NestList[step, {1, 1/Sqrt[3]}, n];
findloop[] := (* like steps[n], except it runs until a loop is found; caveat forever *)
NestWhileList[step, {1, 1/Sqrt[3]}, UnsameQ, All];
First proof:
gr = Rule @@@ Partition[findloop[], 2, 1] // Graph;
cycle = First@FindHamiltonianCycle[gr];
Length@cycle
HighlightGraph[gr, PathGraph[cycle]]
Less fussy proof, assuming you can make a good guess (i.e. 24): | {
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Less fussy proof, assuming you can make a good guess (i.e. 24):
Rule @@@ Partition[steps[24], 2, 1] // Graph
(* graph similar to above, but without highlighting *)
Even less fussy, assuming you can make a sufficient guess (i.e. 100):
gr = Rule @@@ Partition[steps[100], 2, 1] // Graph
Length@VertexList[gr]
(*
graph similar to above, but with multiple edges
24 <-- number of vertices == period
*)
Here is a translation of Brent's algorithm for finding the length of a cycle. See Brent, doi:10.1007/BF01933190 (1980). This could be compiled when step can be compiled for cases where a loop is long or occurs a long way down the sequence.
brentcyclelength[step_, x0_, ρ_: 2, u_: 0] :=
Module[{x, y, r, j, k, done = False},
y = x0;
r = ρ^u;
k = 0;
While[! done,
x = y;
j = k;
r = ρ*r;
While[! done && k < r,
k++;
y = step[y];
done = x == y
];
];
k - j
]
brentcyclelength[step, {1, 1/Sqrt[3]}]
(* 24 *) | {
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# Picture/intuitive proof of $\cos(3 \theta) = 4 \cos^3(\theta)-3\cos(\theta)$?
Is there a nice geometric, intuitive or picture proof as to why the easily algebraically provable identity $\cos(3 \theta) = 4 \cos^3(\theta)-3\cos(\theta)$ is true?
Note I'm not looking for a computational proof like the one linked to, more a proof without words or intuitive style proof, thanks.
Enhancing my diagram for the angle-sum formula (currently featured in Wikipedia) to use three angles will get you pretty close ...
Thus,
\begin{align} \cos(\alpha+\beta+\gamma) &= \cos\alpha \cos\beta \cos\gamma - \sin\alpha \sin\beta \cos\gamma - \sin\alpha \cos\beta\sin\gamma - \cos\alpha \sin\beta\sin\gamma \\ \sin(\alpha+\beta+\gamma) &= \sin\alpha \cos\beta \cos\gamma + \cos\alpha \sin\beta \cos\gamma + \cos\alpha \cos\beta \sin\gamma - \sin\alpha \sin\beta \sin\gamma \end{align}
With $\alpha = \beta = \gamma = \theta$, these become ... \begin{align} \cos 3\theta &= \cos^3\theta - 3 \sin^2\theta \cos\theta \\ \sin 3\theta &= 3\cos^2\theta \sin\theta - \sin^3\theta \end{align} ... which the Pythagorean identity helps us rewrite as ... \begin{align} \cos 3\theta &= \cos^3\theta - 3 (1-\cos^2\theta) \cos\theta = 4\cos^3\theta - 3 \cos\theta \\ \sin 3\theta &= 3(1-\sin^2\theta) \sin\theta - \sin^3\theta = 3\sin\theta - 4\sin^3\theta \end{align}
Off-hand, I don't know of a diagram that goes directly from $\cos 3\theta$ to $4\cos^3\theta-3\cos\theta$. | {
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• That is absolutely amazing. – bobby Jun 30 '14 at 13:29
• Very colorful! =) – Arnie Bebita-Dris Sep 4 '14 at 12:36
• I just love it when someone has the time and effort to make a truly helpful and fascinating visual explanation. Sometimes when the numbers and letters just swim around, a picture can be very helpful – Asimov Sep 4 '14 at 14:17
• How do you get something featured on Wikipedia? Do you submit it somewhere and then wait for someone to place it on an article? – étale-cohomology Aug 9 '17 at 22:31
• @étale-cohomology: The "wiki" in "Wikipedia" indicates that the content is user-editable. Anybody can go in and make improvements. As for my diagrams: I'd posted them here, and then someone posted "similar" versions to Wikipedia (without credit or notice); someone from here at Math.SE edited the post so that I'd get credit. (Thanks again @ChrisSherlock!) I, myself, recently went in and edited the article's description of those diagrams and provided a link to my new website for such things in the article's list of references. – Blue Aug 9 '17 at 23:00
Assemble four congruent right triangles with hypotenuse $1$ and angle $\theta$ as shown:
In the diagram, the marked angles are all $\theta$. Since $OQ = \cos\theta$, we have $OU = \cos^2\theta$, so $OS = 2\cos^2\theta$, and so $$OS' = 2\cos^3\theta$$ On the other hand, $$OS' = OP' + P'Q' + Q'S' = \cos(3\theta) + P'Q' + \cos(\theta)$$ and $$OS' = OR' - Q'R' + Q'S' = \cos(\theta) - Q'R' + \cos(\theta)$$ Since $P'Q' = Q'R'$, adding these together yields $$\cos(3\theta) + 3\cos(\theta) = 2OS' = 4\cos^3\theta$$
I hope the diagram is reasonably self-explanatory, that it's clear what needs to be proved to justify the steps of the argument, and that it's easy to prove those things once you've identified them. But if I've misjudged this, let me know and I'll fill in some more details. | {
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I don't think it is a very good idea to try an find a geometric picture for every trig identity because that would take exceedingly long. I will show you that your formula comes from the idea that a rotation by $n\theta$ is equal to $n$ rotations by $\theta$.
Denote $\langle a,b\rangle$ as the vector to the ordered pair $(a,b)$. We will discuss rotations of these vectors. First, consider that the rotation of the vector $\langle 1,0\rangle$ by an angle of $\theta$ is given by $\langle \cos\theta,\sin\theta\rangle$. Similarly, rotation of the vector $\langle 1,0\rangle$ by $3\theta$ is given by $\langle \cos(3\theta),\sin(3\theta)\rangle$.
I will now define a new type of multiplication between vectors. Let $$\langle a,b\rangle\ltimes\langle c,d \rangle = \langle ac-bd,ac+bc\rangle$$
It so happens that if the vector $\langle a,b\rangle$ makes an angle $\phi_1$ with the horizontal and $\langle c,d \rangle$ makes an angle $\phi_2$ then $\langle a,b\rangle\ltimes\langle c,d \rangle$ makes an angle $\phi_1 + \phi_2\!$. Using this, there is another way to find the rotation of a vector by $3\theta$. We can rotate it three times by $\theta$ rather than rotating it once by $3\theta$. In other words, we will find $\left(\langle \cos\theta,\sin\theta \rangle \ltimes \cos\theta,\sin\theta \rangle\right) \ltimes \cos\theta,\sin\theta \rangle.$ Using this, one could suspect that \begin{align}\langle \cos(3\theta),\sin(3\theta)\rangle &= \left(\langle \cos\theta,\sin\theta \rangle \ltimes \cos\theta,\sin\theta \rangle\right) \ltimes \cos\theta,\sin\theta \rangle \\ &= \langle \cos^2\!\theta-\sin^2\!\theta,2\sin\theta\cos\theta\rangle \ltimes \langle \cos\theta, \sin\theta \rangle \\ &= \langle \cos^{3}\!\theta-\sin^2\!\theta\cos\theta-2\!\sin^2\!\theta\!\cos\theta, 2\!\sin\theta\!\cos^2\!\theta+\sin\theta\cos^2\!\theta-\sin^3\!\theta\rangle \\ &=\langle 4\cos^3\!\theta-3\cos\theta,3\sin\theta-4\sin^3\!\theta\rangle\end{align} | {
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Where the last step follows from the Pythagorean identities. If we then equate the first elements and the second elements we find the familiar results
\begin{align} \cos(3\theta) &= 4\cos^3\!\theta-3\cos\theta \\ \sin(3\theta) &= 3\sin\theta-4\sin^3\!\theta\end{align}
This is essentially a result of complex numbers and is called De moivre's formula. You can see it being used to prove your identity here. Our type of multiplication mirrors complex multiplication which is inherently rotational in nature. You can see why here. De moivre's formula says that a rotation by $n\theta$ is equal to $n$ rotations by $\theta$.
The best geometric proof of your formula comes from this basic idea which is so simple that no picture is needed to represent it.
This is advantageous because it can be used to show other identites such as $$\cos(5\theta) = 16 \cos^5\! \theta - 20 \cos^3 \!\theta + 5 \cos \theta$$
• This is neither geometric nor intuitive and doesn't address the question in a natural way at all. – beep-boop Jul 2 '14 at 22:14
• @alexqwx: most of the proofs, including those using De Moivre's, are geometrically based. However, it is hard to find any that I would call intuitive. – robjohn Jul 7 '14 at 0:18
This is not a direct proof but a link to the Fourier series: $4\cos^3\theta=3\cos\theta+\cos3\theta$.
Similarly, for the fifth power: $16\cos^5\theta=10\cos\theta+5\cos3\theta+\cos5\theta$.
The main lesson is that to represent the $k^{th}$ power, the odd harmonics from $1$ to $k$ suffice.
The explanation is not so difficult: the Fourier coefficients are computed from the integrals $$\int_0^{2\pi}\cos^k\theta\ e^{in\theta}d\theta=\int_0^{2\pi}\left(\frac{e^{i\theta}-e^{-i\theta}}2\right)^ke^{in\theta}d\theta.$$ When developing, the lowest exponent of $e^{i\theta}$ is $n-k$. If $n>k$, the integrand is oscillatory and the integral vanishes.
If you admit that, then $\cos^3\theta=a\cos\theta+b\cos3\theta$. | {
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If you admit that, then $\cos^3\theta=a\cos\theta+b\cos3\theta$.
From the figure, at $\theta=0$ the values add up to $1$ and at $\theta=\pi/2$ the slopes cancel out, so that $$a+b=1\\a-3b=0,$$ $$\cos^3\theta=\frac34\cos\theta+\frac14\cos3\theta.$$
Similarly, $\cos^5\theta=a\cos\theta+b\cos3\theta+c\cos5\theta$, at $\theta=0$ the values add up to $1$ and at $\theta=\pi/2$ the first and third derivatives cancel out, so that $$a+b+c=1\\ a-3b+5c=0\\ a-27b+125c=0,$$ $$\cos^5\theta=\frac{10}{16}\cos\theta+\frac{5}{16}\cos3\theta+\frac{1}{16}\cos5\theta.$$
yes.you can write: $cos(3\theta) = cos(2\theta +\theta)$ =$$cos(2\theta)cos(\theta) - sin(2\theta)sin\theta$$ =$$2cos^3(\theta)-cos \theta -sin(2\theta) sin(\theta)$$=$$2cos^3\theta -cos\theta -2sin^2\theta cos\theta$$=$$2cos^3\theta - cos \theta - 2(cos\theta -cos^3 \theta)$$ so we have : $$4cos^3\theta -3cos\theta$$
• I've already linked to an easy algebraic proof, I'm hoping for a more intuitive, geometric and less computational proof. – bobby Jun 30 '14 at 12:10
• sorry I do not see it .for picture you can use Matlab software. – user157745 Jun 30 '14 at 12:16 | {
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IntMath Home » Forum home » Methods of Integration » Find integral sqrt (x^2 + 1) using trigonometric substitution
# Find integral sqrt (x^2 + 1) using trigonometric substitution [Solved!]
### My question
Find int sqrt (x^2 + 1) dx with limits of integration from 0 to 1 using Trigonometric Substitution.
### Relevant page
8. Integration by Trigonometric Substitution
### What I've done so far
I used x = tan theta and dx = sec^2 theta d theta.
Replacing x and dx gives sqrt (tan^2 theta + 1) sec^2 theta = sqrt (sec^2 theta) sec ^2 theta = sec ^3 theta d theta.
I looked up the integral of sec ^3 theta and converted tan, sec, sin, and cos in the formula by drawing a triangle based on x = tan theta and arrived at .57155 and not 1.15.
X
Find int sqrt (x^2 + 1) dx with limits of integration from 0 to 1 using Trigonometric Substitution.
Relevant page
<a href="https://www.intmath.com/methods-integration/8-integration-trigonometric-substitution.php">8. Integration by Trigonometric Substitution</a>
What I've done so far
I used x = tan theta and dx = sec^2 theta d theta.
Replacing x and dx gives sqrt (tan^2 theta + 1) sec^2 theta = sqrt (sec^2 theta) sec ^2 theta = sec ^3 theta d theta.
I looked up the integral of sec ^3 theta and converted tan, sec, sin, and cos in the formula by drawing a triangle based on x = tan theta and arrived at .57155 and not 1.15.
## Re: Find integral sqrt (x^2 + 1) using trigonometric substitution
@phinah: Good on you for using the math entry system! (I tidied up some of the math expressions.)
Just a small (but important) point - don't miss out the "d theta" parts in your second line. It should be:
Replacing x and dx gives sqrt (tan^2 theta + 1) (sec^2 theta) d theta = sqrt (sec^2 theta) (sec ^2 theta) d theta = sec ^3 theta d theta.
Now, to give you a hint about why your final number is not correct (I'm guessing 1.15 comes from the Answers in your text book, right?). | {
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When we change x to tan theta and x goes from 0 to 1, what will theta's lower and upper values be?
X
@phinah: Good on you for using the math entry system! (I tidied up some of the math expressions.)
Just a small (but important) point - don't miss out the "d theta" parts in your second line. It should be:
Replacing x and dx gives sqrt (tan^2 theta + 1) (sec^2 theta) d theta = sqrt (sec^2 theta) (sec ^2 theta) d theta = sec ^3 theta d theta.
Now, to give you a hint about why your final number is not correct (I'm guessing 1.15 comes from the Answers in your text book, right?).
When we change x to tan theta and x goes from 0 to 1, what will theta's lower and upper values be?
## Re: Find integral sqrt (x^2 + 1) using trigonometric substitution
theta = arctan (0) = 0
theta = arctan (1) = .7853
X
theta = arctan (0) = 0
theta = arctan (1) = .7853
## Re: Find integral sqrt (x^2 + 1) using trigonometric substitution
Also this is not a textbook exercise. This is from your website, the question at the end of the Integration chapter, section 4: The Definite Integral. You state it can be solved via Trigonometric Substitution but you solved it using the Trapezoidal Rule in section 5. Thanks.
X
Also this is not a textbook exercise. This is from your website, the question at the end of the Integration chapter, section 4: The Definite Integral. You state it can be solved via Trigonometric Substitution but you solved it using the <a href="https://www.intmath.com/integration/5-trapezoidal-rule.php">Trapezoidal Rule in section 5</a>. Thanks.
## Re: Find integral sqrt (x^2 + 1) using trigonometric substitution
Ahh, I see. Thanks for giving me the context.
Your lower and upper values are correct. Now substitute those in the integral of sec^3 theta you found from the table.
X
Ahh, I see. Thanks for giving me the context.
Your lower and upper values are correct. Now substitute those in the integral of sec^3 theta you found from the table. | {
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## Re: Find integral sqrt (x^2 + 1) using trigonometric substitution
Leaving it in terms of theta:
int sec^3 theta d theta from 0 to .7853
According to Wolfram, the integral formula is
1/2[tan theta sec theta {: - ln (cos {:theta/2:} - sin {:theta /2:}) {: + ln (sin {:theta/2:} + cos {:theta/2:})]
Therefore, 1/2 [tan .7853\ sec .7853 {:- ln (cos .7853/2 - sin .7853/2) {: + ln (sin .7853/2 + cos .7853/2)] - 1/2 [0 - ln (1-0) + ln (0+1) ]
= 1/2[tan .7853\ sec .7853 {: - ln (cos .7853/2 - sin .7853/2) {: + ln (sin .7853/2 + cos .7853/2)] - [0]
=1.15
which is the web site's answer using the Trapezoidal Rule!
Thank you for the guidance.
Note: concerning how theta is written above, I wrote it out in Word.
X
Leaving it in terms of theta:
int sec^3 theta d theta from 0 to .7853
According to Wolfram, the integral formula is
1/2[tan theta sec theta {: - ln (cos {:theta/2:} - sin {:theta /2:}) {: + ln (sin {:theta/2:} + cos {:theta/2:})]
Therefore, 1/2 [tan .7853\ sec .7853 {:- ln (cos .7853/2 - sin .7853/2) {: + ln (sin .7853/2 + cos .7853/2)] - 1/2 [0 - ln (1-0) + ln (0+1) ]
= 1/2[tan .7853\ sec .7853 {: - ln (cos .7853/2 - sin .7853/2) {: + ln (sin .7853/2 + cos .7853/2)] - [0]
=1.15
which is the web site's answer using the Trapezoidal Rule!
Thank you for the guidance.
Note: concerning how theta is written above, I wrote it out in Word.
## Re: Find integral sqrt (x^2 + 1) using trigonometric substitution
OK - looks good.
BTW, the "theta" symbols didn't show because Word uses a different set of fonts. It's always best to type the math directly in the text box, for best results. I edited your answer so the thetas showed properly and also so the answer worked OK on a phone.
X
OK - looks good.
BTW, the "theta" symbols didn't show because Word uses a different set of fonts. It's always best to type the math directly in the text box, for best results. I edited your answer so the thetas showed properly and also so the answer worked OK on a phone. | {
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# Fundamental theorem of calculus on Wikipedia
In Wikipedia is written in one step that for the integral $\int_{x_1}^{x_1 + \Delta x}f(t) dt$ that by mean value theorem there exists a $c$ in $[x_1, x_1 + \Delta x]$ such that
$\int_{x_1}^{x_1 + \Delta x}f(t) dt = f(c) \Delta x$
But mean value theorem states that there exists $c$ in $(x_1, x_1 + \Delta x)$. (the open interval)
It seems to be a problem when squeeze theorem is later applied.
Is this mistake in the proof or is the step valid? If it is valid can you please explain me why? Thank you for any help.
-
The open interval is contained in the closed interval, so the condition that $c$ is in the closed interval is weaker than the conclusion of the mean value theorem. – Matthew Pressland Apr 23 '13 at 14:33
@MattPressland But if $c \in (x_1, x_1 + \Delta x)$ and by applying squeeze theorem, $c = x_1$ then this is contradiction because $\Delta x \to 0$ and $x_1 \notin (x_1, x_1)$? – blue Apr 23 '13 at 15:03
The squeeze theorem doesn't tell you $c=x_1$, it tells you that $c\to x_1$ as $\Delta x\to 0$. A sequence of points in the open interval can converge to one of the boundary points in the closed interval. – Matthew Pressland Apr 23 '13 at 15:10
@MattPressland I did not know it can converge to boundary point. You can write in answer? If you do I accept. – blue Apr 23 '13 at 15:13
@blue .Lagrage's Mean value theorem for differentiation differ's from first mean value theorem for integration(which you used!) at that point(the interval in which $c$ is expected to be found is closed in the latter and open in the former). – Fermé somme Apr 23 '13 at 15:20 | {
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As boywholived points out, a possible statement of the mean value theorem allows for $c$ to lie in the closed interval. In fact this is simply a weaker statement - if $c$ lies in the open interval, then in particular it lies in the closed interval.
This doesn't cause any problems when applying the squeeze theorem. Perhaps a change of notation is useful; for a particular choice of $\Delta x$, the mean value theorem gives you some $x<c<\Delta x$ (assuming the open interval version of the mean value theorem for now). This $c$ depends on $\Delta x$, so we should maybe write $c(\Delta x)$, making it clear that $c$ is really a function whose value depends on $\Delta x$. Then when applying the squeeze theorem, we are interested in the limit of this function as $\Delta x\to 0$. The fact that $c(\Delta x)>x$ for all $\Delta x>0$ only implies that $\lim_{\Delta x\to0}c(\Delta x)\geq x$, not that the inequality is strict. (Compare to the sequence $a_n=1/n$; every term is strictly positive, but $a_n\to 0$ as $n\to\infty$). | {
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# finding the remainder of $x^{100}-2x^{51}+1$
I have never been great with polynomials. Here's my problem.
Find the remainder of $f(x)=x^{100}-2x^{51}+1$ when $f$ is divided by $x^2-1$
This sounds easy right? Why can't I figure it out? My thought was to try and create it such that $f(x)=q(x)g(x)+r(x)$. But I can not get past getting $deg[r(x)]<deg[g(x)].$
$$f(x)=x^{100}-2x^{51}+1$$ $$=x^{100}-x^{51}-x^{51}+x^2-x^2+1$$ $$=x^{51}(x^{49}-1)-x^2(x^{49}-1)-x^2+1$$ $$=(x^{51}-x^2)(x^{49}-1)-x^2+1$$ $$=x^2(x^{49}-1)(x^{49}-1)-x^2+1$$ $$=x^2[(x^{49}-1)^2-1]+1=?.......$$ I don't see what I am missing
This is the standard approach, especially if you know the roots of the divisor.
Let $f(x) = x^{100} - 2x^{51} + 1$, and $f(x) = g(x) (x^2-1) + ax + b$ be the division
Then, $0 = f(1) = g(1) ( 1^2 - 1) + a (1) + b = a + b$,
and $4 = f(-1) =g(-1) ( (-1)^2 -1) + a(-1) + b = -a + b$.
Hence $a= -2, b = 2$.
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Hence $a= -2, b = 2$.
Thus the remainder is $-2x+2$.
• I like that. I've never seen it done like that before. That is great! – Eleven-Eleven Apr 18 '13 at 22:57
• @ChristopherErnst You should review the Remainder-Factor Theorem. This is a standard application of it. For more questions, you can check this out. – Calvin Lin Apr 18 '13 at 22:58
• So basically depending on the degree of your particular divisor, if say $deg[g(x)]=3$ you set your remainder to $ax^2+bx+c$ and take one more value and solve the system for a,b,and c. – Eleven-Eleven Apr 18 '13 at 23:25
• @ChristopherErnst Correct. This way is direct and you are simply solving linear equations, though it assumes that you can find the roots of $g(x)$. Of course, there are other ways to approach these kind of polynomial remainder problem, like the other (deleted) solution. – Calvin Lin Apr 18 '13 at 23:34
• Very nice. This way is particularly nice should the divisor factor into nice linear terms with real solutions like the one in the problem. Can this be extended when there are complex factors in our divisor? – Eleven-Eleven Apr 18 '13 at 23:42
Hint $\ \,$ Interpolate the remainder $\rm\:r(x)\:$ at the roots $\rm\:\color{#c00}{x = \pm1},\:$ where $\rm\ r(\pm1)\, =\, f(\pm1)$
$$\qquad\ \ \begin{eqnarray} &&\rm\ \ r(x) &=\,&\rm f(x) - (\color{#c00}{x^2\!-\!1})\, q(x),\ \ \ deg\ r < 2\\ \\\Rightarrow\, &&\rm 2\, r(x) &=\,&\rm f(1)\, (x\!+\!1) - f(-1)\, (x\!-\!1)\end{eqnarray}$$
Remark $\$ Or, equivalently, use Chinese Remainder (CRT) to solve
$$\begin{eqnarray}\rm r(x) &\equiv&\rm \ \ \ f(1) &&\rm\ (mod\ \color{#c00}{x-1}) \\ \rm r(x)&\,\equiv\,&\rm f(-1)&&\rm\ (mod\ \color{#c00}{x+1})\end{eqnarray}$$
Generally, as here, Lagrange interpolation is a special case of CRT. | {
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# What does .9 with a line above the 9 mean?
What does this mean? $$\Large.\overline9$$ I've never seen this notation before.
• Perhaps $0.9999...$? – ajotatxe Jun 2 '15 at 23:28
• It also means $1$, by that's a whole 'nother discussion. – PyRulez Jun 3 '15 at 0:32
• @AlbertMasclans Apparently this notation style depends on geography (en.wikipedia.org/wiki/Repeating_decimal): overline=US, overdot=China, parentheses=Europe. – user3449173 Jun 3 '15 at 8:02
• @Nordik Yes, it depends on geography, but parentheses are used only sometimes in Europe. In Germany we typically use the overline, too. Parentheses are used to denote an uncertanty in the last digit(s). – Christoph Jun 3 '15 at 10:58
• By the way, the overline is technically known as a vinculum. – Brian M. Scott Jun 3 '15 at 14:38
It is called a vinculum and it denotes a repeating decimal.
• +1 for a good word I need to remember to use when the time is right. – JoeTaxpayer Jun 2 '15 at 23:58
• Sounds like something in a gynecologist's office. – zhw. Jun 3 '15 at 0:36
• @JoeTaxpayer: Do you mean +1 or +0.9...? – mipadi Jun 3 '15 at 22:18
It means a repeating decimal. One can write $\frac 16=0.1\overline 6$, or $\frac 1{14}=0.0\overline{714852}$ for example. The repeating part is whatever is under the overline.
$0.\overline{9}=0.999999\ldots=1$
More generally, $0.\overline{n}=0.nnnnnnnnn\ldots$
For example, $\frac 13=0.333333333\ldots=0.\overline3$
As other answers have said, it stands for a repeating decimal, where the digits under the line are repeated. $$0.\overline{9} = 0.9999999\ldots$$ | {
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But if you want to be a little pedantic, you might prefer to say that both $0.\overline{9}$ and $0.9999999\ldots$ are two different forms of notation for the same number. That number is the limit of the sequence formed by repeatedly appending copies the digits covered by the line. In other words, given the notation $0.\overline{9}$, you can write out the following sequence: \begin{align} a_1 &= 0.9 \\ a_2 &= 0.99 \\ a_3 &= 0.999 \\ a_4 &= 0.9999 \\ &\vdots \end{align} As you tack on more and more repetitions, these numbers get closer and closer to some limiting value, which I'll call $A$. If you know calculus notation, $$\lim_{n\to\infty} a_n = A$$ The number represented by $0.\overline{9}$ is $A$. It happens to work out to be $1$ (or if we're being pedantic, $1$ is yet another notation for the same number).
As another example of this way of interpreting repeating decimals, consider $0.257\overline{143}$. You can write the sequence \begin{align} b_1 &= 0.257143 \\ b_2 &= 0.257143143 \\ b_3 &= 0.257143143143 \\ b_4 &= 0.257143143143143 \\ &\vdots \end{align} and similarly, the number represented by $0.257\overline{143}$ is the value that this sequence gets closer and closer to as you add more repetitions; or $$\lim_{n\to\infty} b_n$$ This one works out to $\frac{128443}{499500}$.
• Did you just randomly pick that last fraction or did you already have it in mind? – user1717828 Jun 3 '15 at 12:19
• Worth noting, since the 'overdot' notation being standard in UK and apparently China has come up in comments on OP, that in that variant we dot the first and last in a sequence (as opposed to each digit). – OJFord Jun 3 '15 at 22:13
• @user1717828 totally random, I didn't feel like going to the trouble of looking up a "significant" fraction. – David Z Jun 4 '15 at 5:08
This symbol is called a vinculum or a overline.
The number(s) under this symbol are repeated indefinitely.
$0.\overline{9} = 0.99999999999999...$
$2.52\overline{346} = 2.52346346346346...$ | {
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# Arithmetic Series Word Problem (Analysis)
#### iNET
Hi, I'm not sure if this is the right section for this question, so if not please redirect me to the correct subforum.
Q: There are k animal feeding stations arranged in a line with a supply hut. The stations are s yards apart and the nearest is t yards from the hut. An attendant carries n bags of feed, one at a time, to each feeding station. How far will he have traveled when he arrives back at the hut after servicing all stations?
Here's what I did, but I don't know if it's correct: (S.k denotes 'S of k')
S.k=(2t + 2(t+s)+...+2(t+(k-1)s))n
= n((k/2)(2(2t)+(k-1)2s))
= n((k(4t+2sk-2s))/(2))
= n((4kt+2sk^2-2sk)/(2))
= n(kt+sk^2+sk)
= nk(t+sk+s) yards
#### romsek
MHF Helper
I get
$2 n t + 2n(s+t) + 2n(2s+t) + \dots 2n((k-1) s + t) = 2 n t k+2 n s \displaystyle{\sum_{j=1}^{k-1}}j = 2 n t k + 2 n s \dfrac {k(k-1)}{2}=2 n k \left(t + \dfrac{s(k-1)}{2}\right)$
1 person
#### iNET
Thanks! I don't exactly understand the method you used to get your answer, but using your answer I was able to see the mistake in my own work that–when corrected–gave me a solution identical to yours.
#### romsek
MHF Helper
the main thing is
$\displaystyle {\sum_{j=1}^{k-1}}j =\dfrac {k(k-1)}{2}$
It looks like you set the problem up correctly. We both start from the same basic equation.
#### iNET
I see.
This may be a stupid question, but the answer could also be simplified to:
nk(2t+s(k-1))
Is one answer somehow 'more' correct than the other one, or are both answers equally acceptable?
Thanks for your help.
#### romsek
MHF Helper
I see.
This may be a stupid question, but the answer could also be simplified to:
nk(2t+s(k-1))
Is one answer somehow 'more' correct than the other one, or are both answers equally acceptable?
Thanks for your help.
they are both the same answer.
Yours is a bit simpler than mine. | {
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# Find $\lim_{n\to\infty} \frac{n^2}{2^n}$
$$\lim_{n\to\infty}\frac{n^2}{2^n}$$
Do you have some tips so I could solve this problem, without the use of L'Hôpital's rule?
Indeed, we didn't see formally L'Hôpital's rule, nor Taylor series so I'm supposed to do this without such "tools".
I've tried using the fact that $n^2 = e^{2\log(n)}$ and $2^n=e^{n\log(2)}$ but didn't manage to eliminate my indeterminate form.
• At least write down what you get after applying l'Hoptial's rule once. Then apply it again. – Alex R. Oct 23 '13 at 17:35
• It look as if $n$ may range over positive integers. In that case, use the fact that by the Binomial Theorem, $(1+1)^n=1+n+\frac{n(n-1)}{2}+\frac{n(n-1)(n-2)}{6}+\cdots\gt \frac{n(n-1)(n-2)}{6}$. – André Nicolas Oct 23 '13 at 17:40
• What, if anything, do you know about the comparative growth rates of $n$ and $\log(n)$? If you happen to know that $$\lim_{n\to\infty}\bigl[n-\log(n)\bigr]=\infty,$$ then since $$\frac{n^2}{2^n}=\frac{e^{2\log(n)}}{e^{n\log(2)}}=e^{2\log(n)-n\log(2)},$$ and since $\log(2)>0,$ it follows that the limit you're looking for is $0$, and even gives you a start on how it can be proven. – Cameron Buie Oct 23 '13 at 17:53
There are many ways to do that. For example, you can prove that for $n \ge 10$ we have $$2^n > n^3.$$ You can show that using the induction argument.
Regarding to analysis tag, you seem to be familiar to series so let us consider the following series: $$\sum(n^2/2^n)$$ it is not hard seeing that this series is convergent and so the $n-$th term approaches to zero while $n\to\infty$. Clearly, this way needn't using L'Hopital's rule. | {
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• @Amzoti: Thanks for your support my dear friend. I hope you have a good time ahead and a peaceful slumber. :-) – mrs Oct 24 '13 at 1:48
• You are welcome my friend! I appreciate all of the hard work you and amWhy put into the site helping people become better problem solvers! Time to watch a movie and relax as it has been a difficult month at work and preparing two talks! Have a great night! Regards – Amzoti Oct 24 '13 at 1:49
I noticed the OP had wanted L'Hôpital's rule off the table only after I posted my previous reply. I apologize for my inattention. Here is another answer:
Let $a_n=n^2/2^n$ for all $n\in\mathbb{Z}_+$. You can easily compute that \begin{align*} a_{n+1}-a_n=-\frac{n(n-2)-1}{2^{n+1}}, \end{align*} which is negative for $n\geq3$. Therefore, the sequence is eventually decreasing and bounded below (by zero), so it must have a limit.
Obviously, this limit cannot be negative, since $a_n>0$ for all $n$. If we show it cannot be positive, either, then we can conclude that the limit must be zero.
Suppose that the limit is positive: $\lim_{n\to\infty} a_n=c>0$. Since the sequence is decreasing (after omitting $a_1$ and $a_2$, which does not affect convergence properties), it is also true that \begin{align*} c=\lim_{n\to\infty} a_n=\inf_{n\in\mathbb{Z}_+\setminus\{1,2\}}a_n. \end{align*} Then, for $n$ large enough and beyond, we have that \begin{align*} (*)\quad c\leq a_n\leq\frac{3}{2}c. \end{align*}
Therefore, \begin{align*} (**)\quad a_{n+1}=a_{n}\times\frac{a_{n+1}}{a_n}=a_n\times\frac{(n+1)^2}{2n^2}\leq\frac{3}{2}c\times \frac{(n+1)^2}{2n^2}, \end{align*} | {
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where the last inequality follows from $(*)$. But $(n+1)^2/(2 n^2)$ converges to $1/2$ as $n\to\infty$, so that for $n$ sufficiently large and beyond, \begin{align*} \frac{(n+1)^2}{2n^2}\leq\frac{5}{8}. \end{align*} For such values of $n$, $(**)$ implies that \begin{align*} a_{n+1}\leq\frac{3}{2}c\times\frac{(n+1)^2}{2n^2}\leq\frac{15}{16}c<c, \end{align*} where the last inequality is strict because $c$ is positive. But this contradicts $c$ being the infimum of the values of the sequence. The limit thus cannot be positive.
• +1. I appreciate your effort of writing another answer after I pointed out that L'H was off-limits. :) – Lord_Farin Oct 23 '13 at 20:13
• Thanks, @Lord_Farin. :-) – triple_sec Oct 23 '13 at 20:37 | {
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# Unique combinations of $n$ non-unique elements in $k$ non-unique buckets with $c$ capacity
I have $n$ non-unique elements, and I have $k$ unordered buckets that can hold anywhere from $0$ to $c$ elements, such that $c * k \geq n$. I would like to find all possible combinations.
For example, given $n=10$, $k=4$, and $c=4$, there are 7 possible distributions:
• 3322
• 3331
• 4222
• 4321
• 4330
• 4411
• 4420
where "3322", for example, means that two buckets have three elements each and the other two buckets have two elements each.
Another way to look at it, is that I want to find unique combinations of $k$ numbers less than or equal to $c$ such that their sum is equal to $n$.
Ideally I'd like an algorithm to be able to generate a list of all acceptable combinations, but knowing a formula for finding the number of combinations given arbitrary $n$, $k$, and $c$ would be helpful. Other answers generally assume buckets have to have a minimum capacity of 1, or they deal with unique elements to some degree, which are not the case here.
First, it is important to note that all solutions can be arranged in descending sorted order. This is useful for the algorithm I will describe, such that it only searches for sorted answers.
Essentially, we start at the first bucket and assign it some item count $0\leq i\leq c$. Now, we realize that assigning the remaining items to the remaining buckets is essentially the same problem, except now we have $n-i$ items, $k-1$ buckets, and each remaining bucket can now only hold up to $i$ items, because if it were to hold anymore, the solution would be unsorted.
Only when we reach the end (no buckets and no items left) do we have a proper solution. This algorithm can also be sped up by realizing that if $\frac{n-i}{k-1}>i$, there is no continuing from this point, as the remaining items can not fit in the remaining buckets without going over $i$.
Below is a solution in Python, | {
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Below is a solution in Python,
def buckets(n,k,c,solution=[]):
total = 0
for i in range(min(c,n),-1,-1):
if k-1 != 0 and float(n-i)/(k-1) > i:
#Can't fit remaining items
break
solution.append(i)
if n-i == 0 and k-1 == 0:
#We've reached the end of the buckets and have a solution
print(solution)
solution.pop()
return 1
elif k-1 != 0:
#If there's still buckets left, try assigning them
total += buckets(n-i,k-1,i,solution)
#Done checking this bucket assignment
solution.pop()
buckets(10,4,4)
[4, 4, 2, 0]
[4, 4, 1, 1]
[4, 3, 3, 0]
[4, 3, 2, 1]
[4, 2, 2, 2]
[3, 3, 3, 1]
[3, 3, 2, 2]
• Thanks so much for this - it works completely. Jun 1, 2017 at 17:54 | {
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# Matrix exponential of a simple bidiagonal matrix
I am interested in finding an expression (closed form or recursive) for the matrix exponential of this banded matrix:
$$\begin{pmatrix} 0 & 1 & 0 & 0 & \cdots & 0 & 0 \\ 0 & a_1 & 1 & 0 & \cdots & 0 & 0 \\ 0 & 0 & a_2 & 1 & \cdots & 0 & 0 \\ 0 & 0 & 0 & a_3 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots &\ddots&\vdots&\vdots \\ 0 & 0 & 0 & 0 & \cdots & a_{n-1} & 1 \\ 0 & 0 & 0 & 0 & \cdots & 0 & a_n \end{pmatrix}$$
For simplicity, assume that $a_k>0~\forall k$. I am quite certain one must exist having played around with it for a while, and looking at the solution for small values of $n$.
Has anyone seen this structure before? Does it have a name? Do you know if there is a solution published somewhere?
If you go ahead and compute the answer for small values of $n$, you get:
$$\exp\left( \begin{array}{cc} 0 & 1 \\ 0 & a_1 \\ \end{array} \right) = \left( \begin{array}{cc} 1 & \frac{-1+e^{a_1}}{a_1} \\ 0 & e^{a_1} \\ \end{array} \right)$$
$$\exp \left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & a_1 & 1 \\ 0 & 0 & a_2 \\ \end{array} \right) = \left( \begin{array}{ccc} 1 & \frac{-1+e^{a_1}}{a_1} & \frac{-e^{a_2} a_1+a_1+e^{a_1} a_2-a_2}{a_1 \left(a_1-a_2\right) a_2} \\ 0 & e^{a_1} & \frac{e^{a_1}-e^{a_2}}{a_1-a_2} \\ 0 & 0 & e^{a_2} \\ \end{array} \right)$$
Unfortunately, $n=3$ is to large to print here, but a pattern remains.
It suffices to diagonalize $A$ (if the $(a_i)$ are pairwise distinct). I change slightly the notation $A=D+J_n$ where $D=diag(a_1,\cdots,a_n)$ and $J_n$ is the nilpotent Jordan block of dimension $n$. $A=PDP^{-1}$ where the first row of $P$ is
$1,-\dfrac{1}{a_1-a_2},\dfrac{1}{(a_1-a_3)(a_2-a_3)},-\dfrac{1}{(a_1-a_4)(a_2-a_4)(a_3-a_4)},\cdots$
We obtain the other rows of $P$ by circular permutation along the diagonals. Finally $e^A=Pdiag(e^{a_1},\cdots,e^{a_n})P^{-1}$ where the first row of $P^{-1}$ is | {
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$1,\dfrac{1}{a_1-a_2},\dfrac{1}{(a_1-a_2)(a_1-a_3)},\dfrac{1}{(a_1-a_2)(a_1-a_3)(a_1-a_4)},\cdots$
We obtain the other rows of $P^{-1}$ by circular permutation along the diagonals.
• Cool thanks very much, this looks promising. However, I'm not sure exactly what you mean by "circular permutation along the diagonals". Do you mean just "circular permutation" or (equivalently) do you mean that the diagonals, when wrapped, will be constant? When I try this I don't get $PP^{-1}=1$ – Ian Hincks May 8 '14 at 20:28
• Also, can you point me in the direction of how you arrived at this conclusion? – Ian Hincks May 8 '14 at 20:29
• Ah, okay, I figured out that "circular permutation along the diagonal" means that, for example, the second row of $P$ will be $(0,1,-(a_2-a_3)^{-1},((a_2-a_4)(a_3-a_4))^{-1},...)$. – Ian Hincks May 8 '14 at 21:12
• Yes Ian, the second row is as you write it. Let $A_n$ be the matrix of dimension $n$. Then $e^{A_n}$ is the submatrix of $e^{A_{n+1}}$ that is constituted by its first $n$ rows and its first $n$ columns. – user91684 May 9 '14 at 10:36
This response is very late but perhaps still useful to others who come across your question whilst browsing (as I did!).
Your banded (or upper bidiagonal) matrix is what Optiz termed a $\textit{Steigungsmatrix }$ (maybe ... "gradient matrix"). See McCurdy et al (1984), Mathematics of Computation, 43, 501-528. For any $n>0$, all you need is to apply $\textit{Opitz's formula}$. You will need a little knowledge of functions of matrices and divided differences to apply it but it will be worth the effort.
If we denote your matrix by A and its elements by $a_{ij}$ then, using Opitz's theorem, the matrix F defined as F = $g$(A) for a function $g$(.), has entries $f_{ij}$ as follows: $f_{ij}=0$ for $i>j$, $f_{ij} = g(a_{ii})$ for $i=j$, and $f_{ij} = g[a_{ii},a_{i+1,i+1},..,a_{jj}]$ for $i<j$ . | {
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The notation $g[x_0,x_1,...,x_n]$ is the nth divided difference defined recursively as $$g[x_0,x_1,...,x_{(n-1)},x_n] = \frac{g[x_1,...,x_n]-g[x_0,...,x_{(n-1)}]}{x_n-x_0},$$ $g[x_0,x_1] = (g[x_1]-g[x_0])/(x_1-x_0)$ and $g[x] = g(x)$.
Setting $g$(.) to be the exponential function, I get your answers for n=1 and n=2.
The above assumes distinct diagonal entries. If there are repeats in the diagonal terms then you should use the appropriate "confluent" form of the divided difference. | {
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# Prove that the sum of pythagorean triples is always even
Problem: Given $a^2 + b^2 = c^2$ show $a + b + c$ is always even
My Attempt, Case by case analysis:
Case 1: a is odd, b is odd. From the first equation,
$odd^2 + odd^2 = c^2$
$odd + odd = c^2 \implies c^2 = even$
Squaring a number does not change its congruence mod 2.
Therefore c is even
$a + b + c = odd + odd + even = even$
Case 2: a is even, b is even. Similar to above
$even^2 + even^2 = c^2 \implies c$ is even
$a + b + c = even + even + even = even$
Case 3: One of a and b is odd, the other is even Without loss of generality, we label a as odd, and b as even
$odd^2 + even^2 = c^2 \implies odd + even = c^2 = odd$
Therefore c is odd
$a + b + c = odd + even + odd = even$
We have exhausted every possible case, and each shows $a + b + c$ is even. QED
Follow Up: Is there a proof that doesn't rely on case by case analysis? Can the above be written in a simpler way?
• In fact there are no Pythagorean triples where the legs ($a$ and $b$) are both odd Feb 27, 2017 at 16:47
• Even though there are shorter proofs, there is something pleasingly straightforward about your proof. Mar 1, 2017 at 13:48
Note that $x^2\equiv x\pmod 2$ and thus $a^2+b^2=c^2$ implies $$a+b+c\equiv a^2+b^2+c^2\equiv 2c^2\equiv 0\pmod 2$$
• Well that makes it easy, props for the quick answer. Feb 26, 2017 at 21:55
• You are welcome, @spyr03. Feb 26, 2017 at 21:55
• $a^2+b^2+c^2\equiv 2(a^2+b^2)\pmod 2$ Could you explain that step? Feb 26, 2017 at 21:55
• @mrnovice $c^2 = a ^2 + b^2\implies a^2+b^2+c^2 = 2(a^2+b^2)$. Feb 26, 2017 at 21:56
• @mrnovice, not at all, the way I had written the answer was missing the context, you've asked a valid question. Feb 26, 2017 at 21:59
Also, Pythagorean triples have a well defined structure: $$a=k(m^{2}-n^{2}),\ \,b=k(2mn),\ \,c=k(m^{2}+n^{2})$$ and $$a+b+c=2k(mn+m^2)$$
Hint
Write $a+b+c=k$, so
$$a^2+b^2=(a+b)^2-2ab= (k-c)^2-2ab=c^2 → k^2-2(kc+ab)=0→k^2=2(kc+ab)$$ | {
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Hint
Write $a+b+c=k$, so
$$a^2+b^2=(a+b)^2-2ab= (k-c)^2-2ab=c^2 → k^2-2(kc+ab)=0→k^2=2(kc+ab)$$
Notice that $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)=2c^2+2(ab+bc+ac)$, so the square of $a+b+c$ is even and thus $a+b+c$ is also even.
$c^2 = a^2 + b^2 = (a+b)^2 - 2ab$.
$2ab = (a+b)^2 - c^2$
$2ab = (a+b+c)(a+b-c)$
Let $n = a+b+c$, and the above becomes:
$2ab = n(n-2c)$
So the right-hand side must be even, but since $n-2c$ is odd when $n$ is odd, $n$ must be even.
• Nice explanation, I never would have thought about writing $2ab = (a + b + c)(a + b -c)$. You could have finished with $2ab = n^2 - 2cn$ and thus n^2 must be even, therefore n is even. Feb 28, 2017 at 23:29
Consider $(a+b+c)^2$
Which is $a^2 + b^2 + c^2 + 2(ab+bc+ca)$
Since $c^2 = a^2 + b^2$ (c being the hypotenuse), $(a+b+c)^2 = 2(c^2 + ab + bc + ca)$ - which is an even number.
and since squares of odd is odd and evens is even $a+b+c$ has to be even. | {
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Codecademy Python Math Symbols | interviewingthecrisis.org
# Writing mathematical expressions - MatplotlibPython plotting.
Codecademy is the easiest way to learn how to code. It's interactive, fun, and you can do it with your friends. Python Comparison Operators. These operators compare the values on either sides of them and decide the relation among them. They are also called Relational operators. Python 3.x: small greek letters are coded from 945 to 969 so,alpha is chr945, omega is chr969 so just type printchr945 the list of small greek letters in a list. :mortar_board:exercise answers. Contribute to ummahusla/Codecademy-Exercise-Answers development by creating an account on GitHub. Many programming languages include libraries to do more complicated math. You can do statistics, numerical analysis or handle big numbers. One topic many programming languages have difficulty with is symbolic math. If you use Python though, you have access to sympy, the symbolic math library. Sympy.
Learn about lists, a data structure in Python used to store ordered groups of data. :mortar_board:exercise answers. Contribute to ummahusla/Codecademy-Exercise-Answers development by creating an account on GitHub. Python 3.x: small greek letters are coded from 945 to 969 so,alpha is chr945, omega is chr969 so just type printchr945 the list of small greek letters in a list. | {
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Python Arithmetic Operators Example - Assume variable a holds 10 and variable b holds 20, then −. Numeric and Mathematical Modules¶ The modules described in this chapter provide numeric and math-related functions and data types. The numbers module defines an abstract hierarchy of numeric types. There is nothing special about multiplication in Python; we just use an asterisk in the same way we did with addition, subtraction, and division. print2 2 I’m sure it comes as no surprise to you that the above outputs 4. Note that if we multiply two integers, the result is of the data type int. Jupyter notebook recognizes LaTeX code written in markdown cells and renders the symbols in the browser using the MathJax JavaScript library. Mathematics Inline and Display Enclose LaTeX code in dollar signs $.$ to display math inline. What are operators in python? Operators are special symbols in Python that carry out arithmetic or logical computation. The value that the operator operates on is called the operand.
x = Symbol'x' y = Symbol'y' A = Matrix[[1,x], [y,1]] Once a matrix is created, you can operate on it. There are functions to do dot products, cross products or calculate determinants. Hyperbolic functions The abbreviations arcsinh, arccosh, etc., are commonly used for inverse hyperbolic trigonometric functions area hyperbolic functions, even though they are misnomers, since the prefix arc is the abbreviation for arcus, while the prefix ar stands for area. math.floor x ¶ Return the floor of x as a float, the largest integer value less than or equal to x. math.fmod x, y ¶ Return fmodx, y, as defined by the platform C library. Codecademy-Exercise-Answers / Language Skills / Python / Unit 01 Python Syntax / Edvins Antonovs Subfolder rename & ascending ordering for python Latest commit 5fc7ead Jul 14, 2016. | {
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from sympy.plotting import plot from sympy import Symbol x = Symbol 'x' p = plot 2 x3, 3 x1, legend = True, show = False p [0]. line_color = 'b' p [1]. line_color = 'r' p. show Sets Beside FiniteSet which I exemplify below, sympy also includes support for infinite sets and intervals. Ask questions, get help with an exercise, and chat about your Codecademy coursework here. “When one teaches, two learn” — Robert Heinlein.
Codecademy: Python 1. Python Syntax print "Welcome to Python!" my_int = 7 my_float = 1.23 my_bool = True my_int = 7 my_int = 3 print my_int def spam: eggs = 12 return eggs print spamjust a comment """ first comment second comment third comment """ count_to = 12 count_to = 5 - 2 ni = 2 10 ni = 20 / 4 eggs = 10 2 spam = 3 % 2 2. Strings and Console Output 'Help! Help! I\'m being. Yes, you can do symbolic math in Python! The library to take a look at is SymPy. Its official website is. This article is not a SymPy tutorial, as I only want to walk you through some examples to show you the kinds of things that it can do. Programming reference for Ruby. Codecademy is the easiest way to learn how to code. It's interactive, fun, and you can do it with your friends.
Python-based: SymPy is written entirely in Python and uses Python for its language. Lightweight: SymPy only depends on mpmath, a pure Python library for arbitrary floating point arithmetic, making it easy to use. A library: Beyond use as an interactive tool, SymPy can be embedded in other applications and extended with custom functions. Definition of codecademy in thedictionary. Meaning of codecademy. What does codecademy mean? Information and translations of codecademy in the most comprehensive dictionary definitions resource on the web. | {
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Yingxie Python notes from codecademy Chap 1 Python Syntax 13‟ Intro: Python can be used to create web apps, games, even a search engine. 1.1 Variables and Data Types. Originally Answered: is there much math in python? This is a common misconception because there are many math people who also program. In reality, programming is the process of telling a computer to do something-- like telling your friend to do something. List of all math symbols and meaning - equality, inequality, parentheses, plus, minus, times, division, power, square root, percent, per mille,. RapidTables Home › Math › Math symbols › Math symbols. SymPy is a Python library for symbolic computation. It provides computer algebra capabilities either as a standalone application, as a library to other applications, or live on the web as SymPy Live or SymPy. Math Symbols Explained with Python August 03, 2019 When working with Machine Learning projects, you will come across a wide variety of equations that you need to implement in code.
Introduction¶ SymPy is a Python library for symbolic mathematics. It aims to become a full-featured computer algebra system CAS while keeping the code as simple as possible in order to be comprehensible and easily extensible.You can find out which functions and attributesare defined in a module. import math dir mathIf you have a Python script named math.py in the samefolder as your current script, the file math.py willbe loaded instead of the built-in Python module. | {
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08.01.2013 · What is this? As many of you have guessed by now, Sandra is an A.I. meant to be a repository for your questions. Right now, her vocabulary is limited. The symbolic math in sympy is pretty good. It is not up to the capability of Maple or Mathematica, but neither is Matlab but it continues to be developed, and could be helpful in some situations. It is not up to the capability of Maple or Mathematica, but neither is Matlab but it continues to be developed, and could be helpful in some situations. Learn codecademy with free interactive flashcards. Choose from 148 different sets of codecademy flashcards on Quizlet. Symbolab: equation search and math solver - solves algebra, trigonometry and calculus problems step by step.
Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. mpmath is a free BSD licensed Python library for real and complex floating-point arithmetic with arbitrary precision. It has been developed by Fredrik Johansson. Embedding Python in LaTeX by Kjell Magne Fauske, Friday, October 24, 2008 Comments: 19 Recently, while browsing the archives of the matplotlib mailing list, I stumbled upon the small python.sty package written by Martin R. Ehmsen. Insights When One Teaches, Two Learn: Meet Codecademy’s Community. We believe “when one teaches, two learn.” To make that belief a reality for learners, we host an online community with tens of. Chapter 5 Operators and Expressions The purspose of computing is insight, not numbers. Richard Hamming Chapter Objectives Learn Python’s arithmetic, string, relational, logical, bitwise operators. | {
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Plotly's Python library is free and open source! Get started by downloading the client and reading the primer. You can set up Plotly to work in online or offline mode, or in jupyter notebooks. The Percent Sign % is a interesting beast in the Python language. It does something called string formatting and it’s a mathematic operator as well. Whatever math is. Let’s find out what the POWER of the % is first, we’ll learn by finding out what NOT to do. Unicode characters are very useful for engineers. A couple commonly used symbols in engineers include Omega and Delta. We can print these in python using unicode characters. Kenneth Tilton I am surprised you do not include the numeric character codes. kt --"The best Algebra tutorial program I have seen. in. ALT Codes - Alt Codes for Maths / Mathematics Welcome to Useful Shortcuts, THE Alt Code resource! If you are already familiar with using alt codes, simply select.
Jupyter Notebook. From the official webpage: The Jupyter Notebook is a web application that allows you to create and share documents that contain live. In this post, I am gonna show you how to write Mathematic symbols in markdown. since I am writing blog post that hosted by Github with Editor Atom, and use plugin markdown-preview-plus and mathjax-wrapper, and use mathjax Javascript display the math symbols on the web page. The type of code that best shows off the power of computers involves math. I wonder whether any coder can succeed without mastery of basic math. | {
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1. Learn Python, a powerful language used by sites like YouTube and Dropbox. Learn the fundamentals of programming to build web apps and manipulate data. Master Python loops to deepen your knowledge. Learn the fundamentals of programming to build web apps and manipulate data.
2. Any text element can use math text. You should use raw strings precede the quotes with an 'r', and surround the math text with dollar signs \$, as in TeX. Regular text and mathtext can be interleaved within the same string.
3. What is SymPy? SymPy is a Python library for symbolic mathematics. It aims to be an alternative to systems such as Mathematica or Maple while keeping the code as.
4. I want to make a dictionary in Python, but with math symbols because then i want to plot that with the symbols. So, i have something like this.
The Python Discord. News about the dynamic, interpreted, interactive, object-oriented, extensible programming language Python. If you are about to ask a "how do I do this in python" question, please try r/learnpython, the Python discord, or the python IRC channel on FreeNode. Note that wasysym also defines a \lightning symbol. The difference—other than “ ” vs. The difference—other than “ ” vs. “ ”—is that the stmaryrd version above is limited to math mode. An online LaTeX editor that's easy to use. No installation, real-time collaboration, version control, hundreds of LaTeX templates, and more.
Diese Liste mathematischer Symbole zeigt eine Auswahl der gebräuchlichsten Symbole, die in moderner mathematischer Notation innerhalb von Formeln verwendet werden. | {
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# Number of ways to distribute items (with limit of min and max at each location)
There are a total of $n$ objects in stock. You must keep at least $c_i$ objects in stock at location $i$. Assume $n \ge \sum c_i$.
a.) How many different inventories $(x_1,x_2,\dots,x_r)$ are there?
b.) Each location can store a maximum of $d_i$. Let $r = 2$ and count the number of different invetories $(x_1,x_2)$ with $c_1 \le x_1 \le d_1$ and $c_2 \le x_2 \le d_2$.
My answer for part a: We have no choice over where $c_1 + c_2 + \dots + c_r = c_t$ objects are placed. The ramaining $n-c_t$ objects can be spread across $r$ locations. So, there are $$\binom{(n-c_t) + r - 1}{r-1}$$ different inventories.
My incomplete answer for part b: After placing minimum required objects at each location, we are left with $n-c_1-c_2$ objects to distribute.
The remaining objects can be distributed in $$\binom{(n-c_1-c_2)+2-1}{2-1} = (n-c_1-c_2)+1.$$
Now from the number above, I need to subtract the number of combinations that lead to overflows in the two locations.
[---Not sure if the rest is correct---]
Number of ways location 1 overflows:
Location 1 overflows when $x_1 > d_1$ and $c_2 \le x_2 \le d_2$: $$\binom{(n-(d_1+1)-c_2)+2-1}{2-1} = n-d_1-c_2$$
Number of ways location 2 overflows:
(same reasoning as above) $$n-d_2-c_1$$
Total number of inventories is $$(n-c_1-c_2+1)-[(n-d_1-c_2)+(n-d_2-c_1)] = (d_1+d_2+1)-n$$
Total number of inventories $= (d_1+d_2+1)-n$ doesn't make sense.
-
Your answer to part (a) is fine. In (b) you’ve overlooked the possibility of overflow in both inventories simultaneously. | {
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You got off to a fine start: there are $$I_0 = n-c_1-c_2+1$$ inventories that meet the minimum requirements. Then you want get the number of inventories that overflow location $1$ but meet the minimum requirement at location $2$; your answer of $n-d_1-c_2$ is correct only if $n\ge d_1+c_1$. If $n < d_1+c_1$, it’s not possible to overflow at location $1$ and meet the minimum requirement at location $2$. Thus, the correct result for this step is $$I_1 = \max\{n-d_1-c_2,0\}$$ inventories that overflow at location $1$ and meet the minimum at location $2$. Similarly, there are $$I_2=\max\{n-d_2-c_1,0\}$$ inventories that meet the minimum at location $1$ but overflow location $2$.
However, there may be inventories that overflow at both locations. When you subtract off the inventories that overflow at location $1$ and then those that overflow at location $2$, the inventories that overflow at both locations get subtracted twice, so you have to add them back in.
The basic calculation gives $$\binom{n-(d_1+1)-(d_2+1)+2-1}{2-1} = n-d_1-d_2-1$$ of these inventories, but again this is true only if $n\ge d_1+d_2+1$; if $n < d_1+d_2+1$, it isn’t possible to overflow both locations. Thus, the correct figure is $$I_3=\max\{n-d_1-d_2-1,0\}.$$
The final result is then given by $$I_0-I_1-I_2+I_3.$$
Note that if $n \ge d_1+d_2+1$, this evaluates to $0$. This makes perfectly good sense: in that case it’s impossible to meet the requirements, and there are no acceptable inventories.
-
Unless I am reading the problem wrong, this is the same as a couple of other problems that were asked in the last couple of days. You are given a total and want to find the number of ways to make that total $\left(\displaystyle\sum_{i=1}^rx_i=n\right)$ given a range constraint on each summand $(c_i\le x_i\le d_i)$. The answer is the coefficient of $x^n$ in $$\prod_{i=1}^r\frac{x^{d_i+1}-x^{c_i}}{x-1}\tag{1}$$ We can apply this to the special cases you have listed. | {
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a) you let $d_i\to\infty$. In that case, $(1)$ becomes $\displaystyle\prod_{i=1}^r\frac{x^{c_i}}{1-x}=x^{\sum_{i=1}^rc_i}\frac{1}{(1-x)^r}$, so the answer becomes the coefficient of $x^{n-\sum_{i=1}^rc_i}$ in $\frac{1}{(1-x)^r}$. \begin{align} \frac{1}{(1-x)^r} &=\sum_{i=0}^\infty\binom{-r}{i}(-x)^i\\ &=\sum_{i=0}^\infty\binom{r+i-1}{i}x^i\\ &=\sum_{i=0}^\infty\binom{r-1+i}{r-1}x^i \end{align} Therefore, the answer is $\displaystyle\binom{r-1+n-\sum_{i=1}^rc_i}{r-1}$.
b) you set $r=2$. In this case, $(1)$ becomes $$\left(x^{c_1}+x^{c_1+1}+\dots+x^{d_1}\right)\left(x^{c_2}+x^{c_2+1}+\dots+x^{d_2}\right)\tag{2}$$ and the answer is the coefficient of $x^n$ in $(2)$.
Suppose $d_1-c_1\le d_2-c_2$, then the answer is $$\text{the coefficient of }x^n\text{ in }(2)=\left\{\begin{array}{}n-c_1-c_2+1&\text{for }c_1+c_2\le n\le d_1+c_2\\d_1-c_1+1&\text{for }d_1+c_2+1\le n\le c_1+d_2-1\\d_1+d_2-n+1&\text{for }c_1+d_2\le n\le d_1+d_2\\0&\text{otherwise}\end{array}\right.$$
-
Please show the closed form, I can't get the answer. – jamesio Oct 3 '11 at 17:04
@jamesio: I have included the gory details. – robjohn Oct 3 '11 at 17:53 | {
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# Integral substitution
## Main Question or Discussion Point
How do I evaluate the integral $$\int_a^b$$ f(x4) dx = $$\int_y^z$$ f(u) dx/du du, where u=x4
y=a4
z=b4
## Answers and Replies
HallsofIvy
Homework Helper
How do I evaluate the integral $$\int_a^b$$ f(x4) dx = $$\int_y^z$$ f(u) dx/du du, where u=x4
y=a4
z=b4
How you integrate the function f depends strongly on what f is, don't you think?
f is $$\int_-1^1$$ 1+x4 dx
f is $$\int_1^1$$ 1+x4 dx
meant to be a -1 at the bottom of the integral.
How could I do this then.
jgens
Gold Member
Well, if your integral is simply
$$\int_{-1}^1 1 + x^4\, dx$$
Then it's pretty simple. Just seperate the integrand so that,
$$\int_{-1}^1 \,dx + \int_{-1}^1 x^4 \, dx$$
But, based on the context of the question I'm not sure that's what you're looking for. Perhaps you could repost exactly what you mean clearly?
Sure. Consider the integral $$\int_{-1}^1$$ 1+x4 dx
How do I evaluate this integral using the substitution u=x4 and the formula:
$$\int_a^b$$ f(x4) dx = $$\int_{a^4}^{b^4}$$ f(u) dx/du du, where u=x4
jgens
Gold Member
Why would you use that method to evaluate the integral? The integral you have posted can easily be evaluated using the power rule for integration.
I thought this was a more interesting way but just don't know how to get started.
Also I someone could help me with this problem.
How could I evaluate the integral above using u=x4, but seperating it into 2 integrals.
Cyosis
Homework Helper
Taking detours isn't a more interesting way, but here goes.
$u=x^4 \Rightarrow du=4x^3dx$
$$\int 1+x^4 dx=\int \frac{1+u}{4(u^{3/4})} du$$
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$$\int 1+x^4 dx=\int \frac{1+u}{4(u^{3/4})} du$$
Making life a lot harder!
Last edited:
Gib Z
Homework Helper
Personally I wouldn't discourage him, I find it a good way to learn how to compute integrals and what methods to use can be supplemented by trying other methods that come to mind, and seeing why they may not be as efficient. Once he computes that integral Cyosis got him, I think he will be able to see in future why it may not be wise to compute an integral as such.
Cyosis
Homework Helper
You raise a very good point. I guess it's all too easy to see if something is going to be inefficient or not when you have some experience with the topic at hand.
Note: I didn't take the limits into account so you still have to do that yourself.
Why would you use that method to evaluate the integral? The integral you have posted can easily be evaluated using the power rule for integration.
You are too harsh. Sure, for this function there are easy ways to do the integral. But for other functions f maybe you really should do the substitution. And why not learn to do the substitution on simple problems that can be evaluated in other ways? Then, for example, you can check your answers and see when you make a mistake. | {
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# Synthetic Geometry Session Followup Questions and Problem Summary
In our first session, we spent some time on this problem: Let $ABC$ be a triangle in which $\angle A = 60^{\circ}$ Let $BE$ and $CF$ be the angle bisectors with $E$ on $AC$ and $F$ on $AB$. Let $M$ be the reflection of $A$ in the line $EF$. Prove that $M$ lies on $BC$
1. $AEIF$ is cyclic because $\angle BIC=90+\frac {\angle A}{2}=120^{\circ}\implies \angle FIE+\angle A=180^{\circ}$. The cyclic result only occurs when $\angle A=60^{\circ}$, and it gives us information about the angles of $\triangle AEF$ and consequently those of $\triangle FME$
2. $I$ is the circumcenter of $MEF$ because $IE=IF$ and $\angle FIE=2\angle FME$. This result relates $M$ with $I$ and gives us more angle relations.
3. $BMIF$ is cyclic, which we proved by using the above properties to angle chase $\angle FBI=\angle FMI$. Symmetrically $CMIE$ is cyclic, so $\angle IMB+\angle IMC=\angle AFI+\angle AEI=180^{\circ}$, which implies $B,M,C$ are collinear and we are done.
The discovery of these properties makes the problem transparent, meaning we know what makes our main result true. What this also means is that we can probably find simpler solutions. Indeed, there are many ways to solve this problem:
Using angle bisector theorem and Miquel point (Avoids tedious angle chasing):
Since $AEIF$ is a cyclic quadrilateral, we know the miqual point of its complete correspondence lies on $BC$. This means the circumcircles of $BFI,CEI, ABE,ACF$ all have a common point on $BC$, we denote the point $M'$. Because $ACM'F$ is cyclic, $\frac {FM'}{BF}=\frac {AC}{BC}=\frac {AF}{BF}$; therefore $AF=FM'$ and similarly $AE=EM'$. This is enough to establish $M=M'$ and $M$ lies on $BC$
Utilizing $EF$ as the axis of symmetry:
Let $EF\cap BC=P$. It suffices to prove $PE$ bisects $\angle APB$. We can achieve this with our first property.
Now I will propose a few follow up questions which you guys should now easily answer: | {
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Now I will propose a few follow up questions which you guys should now easily answer:
We keep the notations of our main problem:
1. $DBC$ is equilateral such that $D,A$ are on opposite sides of $BC$. Prove that $DE=DF$
2. Through $I$ construct the perpendicular to $EF$ which intersects $EF,BC$ at $X,Y$. Prove that $2IX=IY$
3. $O$ is the circumcenter of $AEF$, Prove that $OM\perp BC$.
Note by Xuming Liang
5 years, 5 months ago
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For follow up problem 1 :
To prove $DE=DF$ , it suffices to establish congruence between $\Delta DIF$ and $\Delta DIE$ (not dead yet) . We have $DI=DI$ and we have already established that $IF=IE$ by having $\angle IFE= \angle IEF=30$. So we need to prove a pair of angles congruent. Since we want a good connection between the angles and the discovered stuff , we prefer proving $\angle DIF=\angle DIE$. By cyclicity we have $m\angle BIF=m\angle CIE=60$. So we need to prove $\angle DIB=\angle DIC=\dfrac{\angle BIC}{2}=\dfrac{120}{2}=60$. Now how to achieve this? We have not yet utilized the equilateral triangle yet!
So, consider $\Box BICD$ Can we prove it cyclic? Yes! We have to prove that $\angle EBD+\angle ECD=180$ and its easy.
$\angle IBD+\angle ICD =\angle IBC+\angle CBD+\angle ICD+\angle BCD=\dfrac{B}{2}+60+\dfrac{C}{2} + 60=120+60=180 \\ \Rightarrow \Box BICD \ is \ cyclic!$
So by cyclicity we have $\angle DIB=\angle DIC=60$ which we had to prove.
- 5 years, 5 months ago
Very nice reasoning using congruence! For the cyclic part, I believe you meant to say $B,I,C,D$ are concyclic, which is true because $\angle BIC+\angle BDC=180$. Since $BD=DC$, $ID$ bisects $\angle BIC$, which is what you wanted to prove.
- 5 years, 5 months ago
Sorry , yes it is that $B,I,C,D$ are concyclic. I have edited it and few angles too.
- 5 years, 5 months ago
A fast way to prove that $BICD$ is cyclic is to notice that $\angle BIC = 120^{\circ}$ and $\angle BDC = 60^{\circ}$.
- 5 years, 5 months ago | {
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- 5 years, 5 months ago
First we have $\Delta IXF$ as 30-60-90 triangle. So $FI=2IX$. So we have to prove that $FI=IY$. So consider $\Delta BIF$ and $\Delta BIY$ , where we have $\angle FBI=YBI$ , $BI=BI$ , $\angle BIF=\angle BIY=60$ (by angle chase). So they are congruent by ASA test and hence $FI=IY$ is proved.
- 5 years, 5 months ago
I pretty much proved it the same as you. Nice solution.
- 5 years, 5 months ago
Yesterday we did find that $\angle EMC=60$ So to prove our result , we must prove that $\angle OME=30$. Using the fact that $O$ is circumcentre , we have $\angle EOF=\angle EIF=120$ and by some angle chase we have $\angle OEI=\angle OFI=60$ where $\angle OFE=\angle IFE=\angle OEF=\angle IEF=30$. We get a beautioful result that $\Box FOEI$ is a rhombus which has one of its diagonals that is $OI$ equal to its side. Using these facts and that $I$ is circumcenter of $\Delta EMF$ we have $EI=OI=IF=IM$ and thus $\Box EOFM$ is cyclic giving $\angle OFE= \angle OME=30$ which was to be proved :)
- 5 years, 5 months ago
When will be the next discussion?
- 5 years, 5 months ago
Problem 3: Since $M$ is the Miquel point, there exists a spiral similarity centered at $M$ that maps $FI \rightarrow AE$. Denote $M_1, M_2$ as the midpoint of $AF$ and $EI$, respectively. Thus, the spiral similarity maps $FI \rightarrow M_1M_2$. Using the circumcircles of $\triangle BFI$ and $BM_1M_2$, this implies that $M_1M_2BM$ is cyclic. Since $M_1OM_2B$ is cyclic, this implies that $M_1OM_2MB$ is cyclic with diameter $OB$. Thus, $\angle OMB = 90^{\circ}$ and we are done.
- 5 years, 5 months ago
Great solution. This fact is actually true as long as $BICD$ is cyclic.
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A bacteria colony is made of billions of organisms. “If you can't explain it simply, you don't understand it well enough.” —Einstein (, Understanding Discrete vs. Asking for help, clarification, or responding to other answers. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. But if you're asking if it's spatially continuous then the answer is no. Hence, the temperature variable placed against the time variable would be represented by a continuous curve. It … Now that we know the continuous rate is -47.9% per year, we can work out how long until we're at 50%: This is a tricky one: the stock market changes every day, so it seems like it'd continuous, but there isn't an underlying predictable rate. Random variables represent quantities or qualities that randomly change within a population. This can be done regardless of whether the sample size is specified by a discrete unit or a continuous measure. Measurement Scale and Context For example, 8 bits have 256 possible values, and 16 bits have 65536. The natural log works on the ratio between the new and old value: $\frac{\text{new}}{\text{old}}$. Learn more about how features and surfaces can be represented as either discrete or continuous in ArcGIS. It is continuous if you consider the values. What to do to speed up the paper publication process? The measurements themselves are discrete, but how you represent them need not be. What's that Professor? For example, you could represent them as a continuous density surface: Or, as discrete 3D bar charts extruded from the census tracts (in this case a grid): A population count is a point measure. We can't have half a student! Why is population density a continuous data type when it is typically measured for aggregate areas such as census tracts or districts/neighbourhoods (ie, it can't be measured at any point on a surface like gradient or temperature). The population of a city is always a | {
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at any point on a surface like gradient or temperature). The population of a city is always a continuous variable as people are always moving to and from the city as well as births & deaths. To simulate a weather system, for example, the tracking occurs continuously as all elements are constantly changing. How to say "garlic", "garlic clove" and "garlic bulb" in Japanese? "12% interest per day" is different than "12% interest per year".). BetterExplained helps 450k monthly readers with friendly, intuitive math lessons (more). (Notice how the rate must be scaled to match the time period. Something areal data is not. If the population is discrete, this distribution is a stick (or bar) graph. ), How To Think With Exponents And Logarithms, Q: Why is e special? It is areal data as @Radar have said. Continuous Growth. Updated my answer. Sample Data Representing a Population Distributions. In Star Trek TNG Episode 11 "The Big Goodbye", why would the people inside of the holodeck "vanish" if the program aborts? The function itself need not be continuous. Ah, but the caveat is the application of a kernel function on the point data. A surface for which each location has a specified or derivable value. For a census tract you can ask the question: How many people exist per unit area? Why was the name of Discovery's most recent episode "Unification III"? To subscribe to this RSS feed, copy and paste this URL into your RSS reader. The ratio between new and old was 37/53, so ln(37/53) = -.359 = -35.9% continuous growth over our time period. What's the best way to produce a density map from weighted points in QGIS? For a discrete distribution, probabilities can be assigned to the values in the distribution - for example, "the probability that the web page will have 12 clicks in an hour is 0.15." Argh! Enjoy the article? For a spatial application the consideration of interpolation is important. (Continuous growth requires a smaller rate because of compounding.). Discrete and | {
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is important. (Continuous growth requires a smaller rate because of compounding.). Discrete and Continuous Data. We can take them, split them into smaller, more frequent changes, and spread them out. In our case, we grew from 1 to 2, which means our continuous growth rate was ln(2/1) = .693 = 69.3%. $2^n$ (where n is an integer) models discrete scenarios like coin flips or binary digits. Are you telling me the bacteria colony just happens to have a continuous rate of precisely ln(2) over the course of a day? elevation). There are two definitions of continuous data ( at least that I could find online). find the population density of some points in qgis/postgis, Calculating population density from urban raster reclasiffy, How to create a population density heatmap from attributes, Find average density value within a polygon in ArcMap, Sources for population density map of Europe. For example, the half-life of Carbon-14 is 5700 years. These practice problems focus on distinguishing discrete versus continuous random variables. A clear understanding of the difference between discrete and continuous data is critical to the success of any Six Sigma practitioner. +1 This answer addresses both spatial and non-spatial forms of data continuity, which is important considering the forum in which the question was asked. To learn more, see our tips on writing great answers. What are the possible outcomes? (The model gets more complex as you account for how long it takes for cubs to have children of their own.). We see a lot of jumpy changes, and sample them at yearly intervals to see how we're doing. I'd prefer you told me the colony doubled while a grad student stared at a petri dish for 24 hours straight. None of this "wait until we decay by 50% so humans can count it easier" nonsense. Let’s see the definition: Continuous data is information that could be meaningfully divided into finer levels. A continuous distribution is one in which data can take on any value within a | {
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into finer levels. A continuous distribution is one in which data can take on any value within a specified range (which may be infinite). What does “blaring YMCA — the song” mean? Population ecology is the study of population and how they change over time in relation to the environment including the environmental influences on population density and distribution, age structure and variations in population size. A discrete variable is a number that can be counted. Breeding seasons introduce some delay in the regulative process. The natural log finds the continuous rate behind a result. One of my pet peeves were problems like "A bacteria colony doubles after 24 hours...". Discrete and Continuous Data. Density is an areal measure and in itself implies that there exists a container variable (e.g. All data that are the result of counting are called quantitative discrete data. I visualize change as events along a timeline: Discrete changes happen as distinct green blobs. With discrete growth, we can see change happening after a specific event. Occasionally when analyzing data between groups or variables there are limitations on the validity of measured tests due to the lack of evidence, data, or information because a sample size is too small or has low value. Example: Material X decayed from 53kg to 37kg over 9 months. We can't point to an event and say "It changed here". The radioactive material is changing every instant. Automating calculation of Population Density in Census Tracts? jar that holds water, or in this case a census tract that holds people). Discrete Data. (1.98kg... 1.99kg... 2.00kg. If your system does change continuously, why not provide the continuous rate and write $e^{\ln(2) x}$? Should live sessions be recorded for students when teaching a math course online? Show Answer. This happened over 9 months, so the monthly continuous rate is -35.9/9 = -3.98%. Using $e$ as a base ($e^{\text{rate} \cdot \text{time}}$) implies you want people to think about | {
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Using $e$ as a base ($e^{\text{rate} \cdot \text{time}}$) implies you want people to think about change that happens at every moment. (Brush up on the number e and the natural logarithm.). I wrote this post because my video on e had questions about how $2^x$ represents "staircase growth". This is similar to saying the average family has 2.3 kids.). That is, the function's domain is an uncountable set. | {
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## is population discrete or continuous
Backyard System Of Poultry Rearing, Anti American Synonym, Sm Megamall Cyberzone Store Directory, Dollar Tree Pharmacy, Gilchrist County Map, Haier Window Air Conditioner, Roasted Eggplant Baby Led Weaning, How To Make Mcdonald's Habanero Sauce, | {
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The chromatic number of the union of two graphs
Let $$G_n$$ be the graph on the set of all binary strings of length $$n$$ with two strings adjacent whenever they are Hamming distance $$2$$ away from each other, or one of them lies below another one; thus, for instance, $$G_2=K_4$$, and $$G_3$$ has $$25$$ edges. What is the chromatic number of $$G_n$$?
There is a simple, but not quite obvious construction showing that $$\chi(G_n)\le n(n+1)/2+1$$, and I am interested in a matching lower bound. Computations give $$\chi(G_1)=2,\ \chi(G_2)=4,\ \chi(G_3)=5,\ \chi(G_4)=9,$$ $$\chi(G_5)=12,\ \chi(G_6)=16,\ \text{and}\ \chi(G_7)\le 17.$$ (The first five values are easy to compute, the last two are reported by Gordon Royle in the comments.)
An update. The colorings I am interesting in are of a special nature: if the strings $$s$$ and $$t$$ are same-colored, then (identifying strings with sets) also $$s\setminus t$$ and $$t\setminus s$$ are same-colored. Let $$\chi^*(G)$$ denote the smallest number of colors needed to properly color the graph $$G$$ in this special way. Is it true that $$\chi^*(G_n)=(1/2+o(1))n^2$$? That, say, $$\chi^*(G_n)>(1/4+o(1))n^2$$? | {
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• @JoshuaZ: sure; color each vertex with the scalar product of the corresponding string and the vector $(1,2,\dotsc,n)$.
– Seva
Nov 20 at 16:36
• what do you mean by lying below? Nov 21 at 8:48
• @FedorPetrov: one of the corresponding subsets of the $n$-element set being contained in another one.
– Seva
Nov 21 at 9:03
• Here is a nice $17$-colouring of $G_7$ where I am using subsets of $\{0,1,\ldots,6\}$ to represent the vertices. Let $\sigma = (0,1,2,3,4,5,6)$ be a cyclic permutation and let $\tau$ be the complement map on the subsets. Let $F= \{013, 124, 235, 346, 450, 561, 602\}$ be the Fano plane and set $C = F \cup F^\tau$. Then take $D = \{0, 16, 25, 34, 124, 135, 236, 456\}$ and its seven rotations under powers of $\sigma$, and $D^\tau$ and its seven rotations under powers of $\sigma$. Then one colour class for each of $\emptyset$ and the entire set, making a total of 17 classes. Nov 22 at 7:04
• So $G_6$ has chromatic number $16$ - this took quite a long time to compute, almost all of which was spent searching in vain for a $15$-colouring. I think it will be hard to get close to an exact answer. Nov 23 at 2:29
This is a comment, not an answer, but it will be more convenient to post it as an answer. Consider the vertices of $$G_n$$ as subsets of $$[n]=\{1,\dots,n\}$$.
Observation 1. $$\chi(G_n)\ge\left\lfloor\frac{3n}2\right\rfloor+1$$.
This is because $$G_n$$ contains a clique of that size, namely,
$$\varnothing,\{1\},\{2\},\{1,2\},\{1,2,3\},\{1,2,4\},\{1,2,3,4,\},\{1,2,3,4,5\},\dots$$.
Observation 2. The OP noted that $$\chi(G_n)\le\binom{n+1}2+1=\frac{n^2+n+2}2$$. For odd $$n$$ this bound can be improved to $$\chi(G_n)\le\frac{n^2+3}2$$.
The vertices $$\varnothing$$ and $$[n]$$ get their own colors. If $$1\le|X|\le n-1$$, color $$X$$ with the ordered pair $$\left(\sum_{x\in X}x\mod(n+1),\ \left\lceil\frac{|X|}2\right\rceil\right).$$ This is a proper coloring, so (assuming $$n$$ is odd) $$\chi(G_n)\le2+(n+1)\cdot\frac{n-1}2=\frac{n^2+3}2.$$ | {
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Observation 3. The clique number $$\omega(G_n)$$ is exactly $$\left\lfloor\frac{3n}2\right\rfloor+1$$.
We showed in Observation 1 that $$\omega(G_n)\ge\left\lfloor\frac{3n}2\right\rfloor+1$$. Let us prove by induction that $$\omega(G_n)\le\left\lfloor\frac{3n}2\right\rfloor+1$$.
We may assume $$n\ge2$$. Let $$\mathcal C$$ be a maximal clique in $$G_n$$. Let $$\mathcal C_h=\{X\in\mathcal C:|X|=h\}$$. Choose $$h$$ so that $$0\lt h\lt n$$ and $$\mathcal C_h\ne\varnothing$$.
If $$\bigcup\mathcal C_h=[n]$$ and $$\bigcap\mathcal C_h=\varnothing$$, then $$|\mathcal C|=|\mathcal C_h|+2\le n+2\le\left\lfloor\frac{3n}2\right\rfloor+1$$. Otherwise, choose $$X\in\{\bigcup\mathcal C_h,\ \bigcap\mathcal C_h\}$$ so that $$0\lt|X|\lt n$$, and let $$k=|X|$$. Then every element of $$\mathcal C$$ is a subset or superset of $$X$$. Since $$0\lt k\lt n$$, by induction we have $$|\mathcal C|\le\omega(G_k)+\omega(G_{n-k})-1\le\left\lfloor\frac{3k}2\right\rfloor+\left\lfloor\frac{3(n-k)}2\right\rfloor+1\le\left\lfloor\frac{3n}2\right\rfloor+1.$$
• Some more computational snippets: the cliques you found are the unique maximum clique (up to equivalence under the automorphism group) for $G_4$, $G_6$ and $G_8$. For $G_5$, $G_7$ and $G_9$ there are other classes of equally-large cliques, but none larger. Nov 25 at 2:56
• So I liked your original proof that if $|\mathcal{C}_h| = 1$ for some $h$, then the result follows. Did you decide that it was not so obvious to prove that there must be a layer containing just one vertex of the clique? Nov 26 at 5:11 | {
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# How to solve for x/z and y/z here?
I got stuck solving these two equations: $$a_1(x/z) + b_1 (y/z) + c_1 = 0$$ and, $$a_2(x/z) + b_2 (y/z) + c_2 = 0$$ for $$x/z$$ and $$y/z$$.
The desired result would be: $${x \over z} = {b_1c_2 - b_2c_1\over a_1b_2 - a_2b_1}$$,
$${y \over z} = {c_1a_2 - c_2a_1\over a_1b_2 - a_2b_1}$$
How do I get there? I keep getting dead ends.
Edit: I basically came this far: $$u = v{-b_1 \over a_1} - {c_1 \over a_1}$$ and, $$u = v{-b_2 \over a_2} - {c_2 \over a_2}$$
And ofcourse $$u = x/z$$, $$v = y/z$$
Now i got here, $$v{-b_1 \over a_1} - {c_1 \over a_1} = v{-b_2 \over a_2} - {c_2 \over a_2}$$
Then I took all the v's to one side and factored v outside: $$v({-b_2 \over a_2} + {b_1 \over a_1}) = {c_1 \over a_1} - {c_2 \over a_2}$$
now divide both sides by that b and a thingy on the left, but now it gets messy...
• Can you show which dead end you run into? It should be simple enough to substitute your proposed solutions into the equations and simplify to show that they work. Jan 30 '16 at 16:08
• Rename $\frac xz=u$ and $\frac yz=v$ and rewrite the initial equations. Jan 30 '16 at 16:08
• Hint: take the last thing, check that it is correct (the second term on the left should be $\frac{b_2}{a_2}$, and the plus on the right should be a minus), then multiply by $a_1a_2$ on both sides then solve for $v$. You just obtained $\frac yz$ as in the solution you give :). Edit You just fixed the sign while I typed. Jan 30 '16 at 16:44
• I suppose I am stuck on how to divide c1/a1 - c2/a2 by that thing on the left now. Any clever way to do that? Jan 30 '16 at 16:56
• See my answer below. I started from a little earlier than that equation, but to make the division, I suggest you multiply both sides by $a_1a_2$, and then divide. The multiplication eliminates all denominators. This is why it is advisable. Jan 30 '16 at 17:04
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## 2 Answers
HINT: with $$\frac xz=u$$ and $$\frac yz=v$$ we get $$a_1u+b_1v+c_1=0$$ $$a_2u+b_2v+c_2=0$$ if $$b_1\ne 0$$ we have $$v=-\frac{c_1}{b_1}-\frac{a_1}{b_1}u$$ and we get an equation for $$u$$ $$a_2u+b_2\left(-\frac{c_1}{b_1}-\frac{a_1}{b_1}u\right)+c_2=0$$ can you proceed?
• it seems we get a different thing, did I do something wrong in my original post? I edited it in. Jan 30 '16 at 16:42
• I urge you to decide if you mean to use slash fractions $a/b$ or normal fractions $\frac ab$, because mixing the two can be quite confusing for readers. @TheProgramMAN123 I think you just followed a different path than he did. You solved both for $u$ and then equalled the two expressions, he solved one for $v$ and then substituted the expression in the other one. Jan 30 '16 at 16:48
• If you solve the last equation in this post for $u$, you get the solution in the question, with signs changed both in numerator and in the denominator. Jan 30 '16 at 16:51
Let me just continue your attempt. You got to:
$$\left\{\begin{array}{c} u=-\frac{b_1}{a_1}v-\frac{c_1}{a_1} \\ u=-\frac{b_2}{a_2}v-\frac{c_2}{a_2} \\ \end{array}\right.$$
If we equal these two expressions for $u$ and multiply both sides by $a_1a_2$, we get:
$$-b_1a_2v-c_1a_2=-b_2a_1v-c_2a_1.$$
Bringing the RHS's $v$ term to the left and the LHS's "constant" term to the right, this becomes:
$$v(a_1b_2-a_2b_1)v=a_2c_1-a_1c_1.$$
Yes, I also factored out $v$ from the LHS and reordered the factors in the various $a,b,c$ products. Is this not precisely the solution you gave, that is:
$$\frac yz=v=\frac{a_2c_1-a_1c_2}{a_1b_2-a_2b_1}?$$
Wonderful! To complete, we substitute this into the equation for $u$, one of the two, I mean. To simplify the calculations, we multiply by $a_1$ the first equation of the system at the start of the answer, and then substitute, getting:
$$a_1u=-b_1\frac{a_2c_1-a_1c_2}{a_1b_2-a_2b_1}-c_1.$$
Let us make that RHS into a single fraction: | {
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$$a_1u=-b_1\frac{a_2c_1-a_1c_2}{a_1b_2-a_2b_1}-c_1.$$
Let us make that RHS into a single fraction:
$$a_1u=\frac{\overline{-b_1a_2c_1}+b_1a_1c_2-c_1a_1b_2+\overline{c_1a_2b_1}}{a_1b_2-b_1a_2}=\frac{a_1(b_1c_2-c_1b_2)}{a_1b_2-b_1a_2}.$$
The overlines simply indicate those two terms cancel. Oh, but if we divide both sides by $a_1$, we get the desired result, don't we? Very good. We are done.
Bottom line: when you have coefficients with fractions, the trick is often to sum all fractions and see if you can multiply/divide to simplify numerators or denominators. Don't be afraid of complicated coefficients: just have patience and do the algebra to the end, and you will get to the desired result :).
• thanks a bunch! I guess the lesson for me here is to remove those fractions as soon as you can. They make things confusing and messy. i guess multiplying both sides by a1a2 is a clever way to do that. the solution really is painfully simple then. Jan 30 '16 at 18:29 | {
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After we have found the second moment of inertia about an axis, we can find it about another parallel axis using the parallel axis theorem. The matrix of the values is known as the moment of inertia tensor. - The formula for moment of inertia is - If there are 3 particles of mass 'm' placed at each of the vertex of this equilateral triangle then we consider three times m. Equation 18) also holds for polar moments of inertia i. The moment of inertia (I) of a body is a measure of its ability to resist change in its rotational state of motion. Similar to the centroid, the area moment of inertia can be found by either integration or by parts. EHE-08): Where: Mf = Mcrk = Nominal cracking moment of the cross section. Doing the same procedure like above, and below is the work. 3/2MR^2 The center of mass of the original triangle (the part that has been cut out) was at the center of the circle, at a distance R from the pivot. Now based on symmetry you can apply the definition of the moment of inertia to calculate the moment of inertia about the y axis which equals the cendroidal y axis. Find the moment of inertia of the system of particles about a perpendicular bisector of the meter scale. dI = r2dm (1) (1) d I. Considering an element DE parallel to y-axis at a distance x from origin and width dx. Various such parameters include centre of gravity, moment of inertia, centroid , first and second moment of inertias of a line or a rigid body. So if the moment of inertia of the rectangle is, about its centroid, is bh cubed over 12, and the moment of inertia of the hole, the circle, from the previous tables is pi r to the 4th, over 4. Beam Deflection Equations are easy to apply and allow engineers to make simple and quick calculations for deflection. 000965387 kg*m^2. A piece of thin uniform wire of mass m and length 3b is bent into an equilateral triangle. • Th t fi ti fth hdd iThe moment of inertia of the shaded area is obtained by subtracting the moment of inertia of the | {
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iThe moment of inertia of the shaded area is obtained by subtracting the moment of inertia of the half-circle from the moment of inertia of the rectangle Determine the moment of inertia of the shaded area with respect to the x axis. This is a simulation of five objects on an inclined plane. Moments of Inertia of Geometric Areas Frame 28-1 * Introduction This unit will deal with the computation of second moments, or moments of inertia, of The general expression for the moment of inertia of a right triangle about a centroidal axis parallel to a side is. In order to continue, we will need to find an expression for dm. Central axis of hallow cylinder. T 1 – the instantaneous value of load torque, referred to a motor shaft, N-m. A framework, in the shape of an equilateral triangle ABC, is formed by rigidly joining three uniform rods, each of mass m and length 2a. Table 8-2 Gross and Cracked Moment of Inertia of Rectangular and Flanged Section b d nA s kd n. Rectangle Triangle. Moment of Inertia of Mass, Mass Moment inertia of Common Shapes page Sideway Output on 28/4. The angle in between the masses is 60 degrees. A: area of the shape. calculate the moment of inertia when the plate is rotating about an axis perpendicular to the plate and passing through the vertex tip. The calculator has been provided with educational purposes in mind and should be used accordingly. University. Angular momentum. The moment of inertia of an area with respect to any given axis is equal to the moment of inertia with respect to the centroidal axis plus the product of the area and the square of the distance between the 2 axes. Rolling without slipping problems. The equation of the line is $y = \dfrac{a}{b} x + a$. 8680 rad/s^2 α_down -0. Determine polar moment of inertia of an isosceles triangle 1 answer below » Polar Moments of Inertia Determine the polar moment of inertia I P of an isosceles triangle of base b and altitude h with respect to its apex (see Case 5, Appendix D). If you're | {
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triangle of base b and altitude h with respect to its apex (see Case 5, Appendix D). If you're unsure about what deflection actually is, click here for a deflection definition Below is a concise beam deflection table that shows how to calculate the maximum deflection in a beam. 025kg) g = gravity (9. The Poisson's ratio of the shaft material is , the moment of inertia about the y axis is , and the applied force at the end of the shaft is P. The SI unit of moment of inertia is kg m2. 2 x 10⁻³ kg*(0. Various such parameters include centre of gravity, moment of inertia, centroid , first and second moment of inertias of a line or a rigid body. Mass Moment of Inertia, I G (cont’d) I G for a body depends on the body’s mass and the location of the mass. Axis through center. Mass Moment of Inertia - Mass Moment of Inertia (Moment of Inertia) depends on the mass of the object, its shape and its relative point of rotation - Radius of Gyration Pipe Formulas - Pipe and Tube Equations - moment of inertia, section modulus, traverse metal area, external pipe surface and traverse internal area - imperial units. The following is a list of second moments of area of some shapes. m = point mass. 17/12mL2 2. We found the moment of inertia of the apparatus alone to be 0. Presented here is a table of formulas which permit direct solution for required moment of inertia for several simple loading cases, for the two most common deflection criteria, L/240 and L/360. ) is the moment of inertia about the centroid of the area about an x axis and d y is the y distance between the parallel axes Similarly 2 y I y Ad x Moment of inertia about a y axis J Ad 2 o c Polar moment of Inertia 2r 2 d 2 o c Polar radius of gyration 2 r 2 d 2 Radius of gyration. To find the perimeter of the triangle, you simply need to add together the lengths of the base and the two sides. The moment of inertia matrix is referred to the principal axes, again frame O 2 and the products of inertia are zero. This engineering | {
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to the principal axes, again frame O 2 and the products of inertia are zero. This engineering data is often used in the design of structural beams or structural flexural members. In mathematical notation, the moment of inertia is often symbolized by I, and the radius is symbolized by r. And I was wondering whether someone could give me some more information/examples on first and second moment of area (tech calculus, wouldnt let me. It is measured by the mass of the body. Now to calculate the moment of inertia of the strip about z-axis, we use the parallel axis theorem. Derivation of moment of inertia of triangle and cone. Some examples of simple moments of inertia Let's try an easy calculation: what's the moment of inertia of these three balls? Each ball has mass m = 3 kg, and they are arranged in an equilateral triangle with sides of length L = 10 m. We will use the parallel axis theorem and we will take the centroid as a reference in this case. 0 kg per leg. d ' (n -1)A 's Gross Section Cracked Transformed Section Gross and Cracked Moment of Inertia b h A's As b h As b bw hf h b h hf y t A's As b d nA s kd n. If you're unsure about what deflection actually is, click here for a deflection definition Below is a concise beam deflection table that shows how to calculate the maximum deflection in a beam. I need to calculate the change in moment of intertia due to modifing a simple angled beam from 120 x 120 x 10 to 120 x 112 x 10. Considering an element DE parallel to y-axis at a distance x from origin and width dx. A piece of thin uniform wire of mass m and length 3b is bent into an equilateral triangle. Equation 18) also holds for polar moments of inertia i. Moment of inertia, denoted by I, measures the extent to which an object resists rotational acceleration about a particular axis, and is the rotational analogue to mass. half the value of the moment of inertia about the central axis to the value of the moment of inertia about the base plane. About the Moment of | {
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the central axis to the value of the moment of inertia about the base plane. About the Moment of Inertia Calculator. University. Answer this question and win exciting prizes. The moment of inertia of a body is its tendency to resist rolling motions and angular accelerations. How to calculate the moment of inertia of a triangular plate rotating about the apex. Kinetic Energy is the energy possessed by an object because it is in motion. Let the mass of the triangle be M. What is the moment of inertia of this triangle for rotation about an axis that is perpendicular to the plane of the triangle and through one of vertices of the triangle? The moment of inertia of a rod rotated about its center of mass is Irod, cm =1/12mL2. ): 6391 Instructor: E-mail: [email protected] Topic - Moment of Inertia ,Ans - (Mh^2)/6. 9803 rad/s^2. 3 (4) 3 Determine the AP whose fourth term is 15 and the difference of 6th term from 10th term is 16 Prove that ratio of area of two triangle is equal to the square of the corresponding sides. Center of mass, moments of inertia, volume of a body of rotation. A framework, in the shape of an equilateral triangle ABC, is formed by rigidly joining three uniform rods, each of mass m and length 2a. Calculate the 2nd moment of area for each element about the reference axes. ANSWER: Right angled triangle. Second moment of area for triangle trough x-axis = (ah3)/36. Calculate the moment of inertia of strait angle triangle about its $$y$$ axis as shown in the Figure on the right. Moment of Inertia of a Triangular Lamina about its Base. It appears in the relationships for the dynamics of rotational motion. Two conditions may be considered. The moment of the large triangle, with side $$2L$$, is $$I_z(2L)$$. The mass moment of inertia of an object about an axis (𝑎) is equal to the moment about an axis (𝑏) through the. l : moment of inertia about the axis parallel to x-axis. The moment of inertia with respect to central longitudinal axis would be m r2/2 m r2/3 m | {
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to x-axis. The moment of inertia with respect to central longitudinal axis would be m r2/2 m r2/3 m r2/6 m r2/12 The ratio of moment of inertia of a rectangle and that of a triangle, having same base and height, with respect to their bases would be 2 : 1 3 : 1 4 : 1 6 : 1 The ratio of the moment of inertia of a triangle of base width b and height h with. Find the moment of inertia of the wire triangle about an axis perpendicular to the plane of the triangle and passing through one of its vertices. They are; Axis passing through the centroid. The units of the product of inertia are the same as for moment of inertia. In addition to the moment of inertia, the product of inertia is commonly used. this apparently involves integrals but we havent even touched integrals in calculus need help please. Four leg-loading conditions were employed: 1) no load (NL) on the legs; 2) a baseline load (BSLN) condition, with a mean of 2. dI y (dy)x 3 = 3 5. suppose the mean speed of such molecule in a gas 500 m per second and its kinetic energy of rotation is 2/3 of its kinetic energy of translation. Ix = b h3 / 36 (4a) Iy =h b3 / 36 (4b) Area Moment of Inertia for typical Cross Sections I. Calculating Moment Of Inertia Of A Triangle. 8·10-2 Kg·m2 Submit Figure < 1of1 Incorrect; Try Again: 3 Attempts Remaining Part B What Is The Triangle's. The mass and moment of inertia through the sphere's center of mass are given. Each "typical" rectangle indicated has width dx and height y 2 − y 1, so its area is (y 2 − y 1)dx. When that happens equation 4 and 5 would be used to calculate the stress and. Inertia is a property of a body to resist the change in linear state of motion. Polar Moment of Inertia for Circular Cross-section. In particular, the same object can have different moments of inertia when rotating about different axes. Angular momentum. Weld design Moment of inertia of fillet weld J [mm 4 , in 4 ] Position of center of gravity of weld group section J = π a (r + a / 2) 3 - Meaning | {
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J [mm 4 , in 4 ] Position of center of gravity of weld group section J = π a (r + a / 2) 3 - Meaning of used variables: a fillet weld height [mm, in] B width of weld group [mm, in] H height of weld group [mm, in] L weld length [mm, in] r weld radius [mm, in] s web thickness [mm, in] t flange thickness [mm, in]. Moment of inertia can be defined by the equation The moment of inertia is the sum of the masses of the particles making up the object multiplied by their respective distances squared from the axis of rotation. (8), derived in the moment of inertia example, the moment of inertia of the disk is = at 5 digits Therefore, the moment of inertia of the disk is 12. The product moment of an area A of a right angle triangle about the axes xy is Product Moment of Inertia of a Right Angle Triangle by Parallel-axis Theorem. Question: Three point masses, each of mass {eq}m {/eq}, are placed at the corners of an equilateral triangle of side {eq}L {/eq}. If the polar moment of inertia is calculated at the centroid of the area, it is denoted. The second moment of area, also known as moment of inertia of plane area, area moment of inertia, polar moment of area or second area moment, is a geometrical property of an area which reflects how its points are distributed with regard to an arbitrary axis. Answer this question and win exciting prizes. What would the moment of inertia of a thin equilateral triangular sheet of mass M and sides S be with respect to an axis through one vertex perpendicular to the sheet? I got 3Ms^2/4 but I'd like to know if I'm right. Table of Selected Moments of Inertia Note: All formulas shown assume objects of uniform mass density. Let M represent the mass of the triangle and L the length of the base of the triangle. More on moment of inertia. The moment of inertia of the triangle about the point O is I = M*r^2 , where r is the distance of the center of mass from O. Calculate the moment of inertia of a thin plate, in the shape of a right triangle, about | {
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from O. Calculate the moment of inertia of a thin plate, in the shape of a right triangle, about an axis that passes through one end of the hypotenuse and is parallel to the opposite leg of the triangle, as in Figure P10. The polar moment of inertia JO of an area about O and the polar moment of inertia JC of the area about its c d o centroid are related to the distance d between points C and O by the relationship J O = J C + Ad 2 The parallel-axis theorem is used very effectively to compute the moment of inertia of a composite area with respect to a given axis. Solve: The moment of inertia of the triangle is I mr= × = =3 3(0. The mass moment of inertia of an object about an axis through the center of mass is smaller than that about any other axis in the same direction. In mathematical notation, the moment of inertia is often symbolized by I, and the radius is symbolized by r. DIY Brick Rocket Stove - Cooking Without Electrical Power - Duration: 23:40. 0580 kg, an inner radius of 0. Center of mass, moments of inertia, volume of a body of rotation. Moment of Inertia and Polar Moment of Inertia are both the quantities expressing a body’s tendency to resist changes when certain torque is being applied. \end{equation*} If the object is made of a number of parts, each of whose moment of inertia is known, the total moment of inertia is the sum of the moments of inertia of the pieces. In order to find the moment of inertia of the triangle we must use the parallel axis theorem which ius as follows: The moment of inertia about any axis parallel to that axis through the center. All the formulas for a triangle are in any handbook. Area Moment of Inertia - Filled Right Triangle Solve. Rolling without slipping problems. When that happens equation 4 and 5 would be used to calculate the stress and. We are concerned here with area only and the area multiplied by a distance twice is the second moment of area. The inertia of both systems can be found using the equation: m = mass of | {
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the second moment of area. The inertia of both systems can be found using the equation: m = mass of hanging mass (0. 0 revolutions per s or 10 rad/s. Let be the position vector of the th mass element, whose mass is. Explanation: No explanation is available for this question! 2) What is the C. Look up I for a triangle in your table if you have forgotten. If k is the mass per unit area, then each typical rectangle has mass k(y 2 − y 1)dx. From the diagram below, we have:. Moment of inertia of this disc about the diameter of the rod is, Moment of inertia of the disc about axis is given by parallel axes theorem is, Hence, the moment of inertia of the cylinder is given as, Solid Sphere a) About its diameter Let us consider a solid sphere of radius and mass. Now that we have determined the moments of inertia of regular and truncated equilateral triangles, it is time to calculate them for the corresponding right prisms. This works especially well when the general shape of the area can be decomposed into simpler shapes for which the moment of inertia is calculated for. Further explanation. We have step-by-step solutions for your textbooks written by Bartleby experts!. Moment of inertia, also called mass moment of inertia or the angular mass, (SI units kg m 2) is a measure of an object’s resistance to changes in its rotation rate. Data: 23 d'abril de 2006 (original upload date) Font: No machine-readable source provided. 94 into 10 to the power of minsis 46 kg metre square bout an Axis through its Centre perpendicular to the lines joining the two atoms. Related Questions. The units. See how the eigenvectors of the inertia tensor change as you change a configuration of point masses, or the shape of a solid plate of material. Moment of inertia of an equilateral triangular lamina ABC, about the axis passing through its centre O and perpendicular to its plane is Io as shown in the figure. Radius and elevation of the semi-circle can be changed with the blue point. 2) A long rod | {
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figure. Radius and elevation of the semi-circle can be changed with the blue point. 2) A long rod with mass has a moment of inertia , for rotation around an axis near one. For example, given the axis O-O and the shaded area shown, one calculates the second moment of the area by adding together for all the elements of area dA in the shaded area. 4)and the second moment of the area about the y. Find the moment of inertia for the following about the y axis and x axis of a right triangle whose base is on the +x axis and whose height is on the +yaxis Source(s): moment inertia triangle: https://shortly. The more far away from the axis, the more moment of inertia the object has. The coordinate variables are x and y, respectively. Area A = 200 mm x 100 mm = 20000 mm2 I x. In other words, the centroid will always be 2/3 of the way along. Whatever kind you are trying to compute I would suggest breaking up the cross section into triangles with two vertices on successive points of your boundary and the third at the center about which the moment of inertia is to be taken. Asked in Algebra, Geometry. \end{equation*} If the object is made of a number of parts, each of whose moment of inertia is known, the total moment of inertia is the sum of the moments of inertia of the pieces. The mass and moment of inertia through the sphere's center of mass are given. Going to the division, we get. When the axes are such that the tensor of inertia is diagonal, then these axes are called the principal axes of inertia. svg 512 × 569; 4 KB. The moment of inertia of a uniform object depends not only on the size and shape of that object but on the location of the axis about which the object is rotating. 91, b < 10a. Angular acceleration of the system + triangle (long base) α_up 0. Q: Moment of Inertia of a thin spherical shell of mass m and radius r about its diameter is a) mr²/3 b) 2mr²/3 c) 2mr²/5 d) 3mr²/5 Q: Moment of inertia of a triangular section of base b and height h about an axis passing | {
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d) 3mr²/5 Q: Moment of inertia of a triangular section of base b and height h about an axis passing through its. The force of attraction is proportional to mass of the body. calculate the moment of inertia when the plate is rotating about an axis perpendicular to the plate and passing through the vertex tip. Radius of Gyration for a equilateral triangle can be calculated as. 3 (4) 3 Determine the AP whose fourth term is 15 and the difference of 6th term from 10th term is 16 Prove that ratio of area of two triangle is equal to the square of the corresponding sides. Weld design Moment of inertia of fillet weld J [mm 4 , in 4 ] Position of center of gravity of weld group section J = π a (r + a / 2) 3 - Meaning of used variables: a fillet weld height [mm, in] B width of weld group [mm, in] H height of weld group [mm, in] L weld length [mm, in] r weld radius [mm, in] s web thickness [mm, in] t flange thickness [mm, in]. We are concerned here with area only and the area multiplied by a distance twice is the second moment of area. 3) Three particles each of mass 100 g are placed at the vertices of an equilateral triangle of side length 10 cm. The mass moment of inertia of an object about an axis (𝑎) is equal to the moment about an axis (𝑏) through the. How you find moment of inertia of isosceles triangle? Wiki User 2014-05-12 13:36:50. After we have found the second moment of inertia about an axis, we can find it about another parallel axis using the parallel axis theorem. 5626 x 10⁻³ initial inertia * angular velocity = new inertia * angular velocity. 2012/2013. Sometimes, we need to find the moment of inertia of an object about the origin, which is known as the polar moment of inertia. It is the rotational analog of mass. The moment of inertia of the rod is simply $$\frac{1}{3} m_rL^2$$, but we have to use the parallel-axis theorem to find the moment of inertia of the disk about the axis shown. We want to find the moment of inertia, I y of the given area, which is | {
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disk about the axis shown. We want to find the moment of inertia, I y of the given area, which is rotating around the y-axis. In order to continue, we will need to find an expression for dm. A = bh ¸ 2 Ic = bh 3 ¸ 36 Base on x-axis, centroidal axis parallel to x-axis: x = h ¸ 3 Ax = bh 2 ¸ 6 Ix = bh 3 ¸ 12 x-axis through vertex, Base and centroidal axis parallel to x-axis: x = 2h ¸ 3 Ax = bh 2 ¸ 3 Ix = bh 3 ¸ 4. 16-24 From: Rabiei. The z2A term is the moment of inertia that area A would have about the y axis if all of the area were to be concentrated at the centroid. The moment of inertia block, which is a table containing the results of the moment of inertia calculation, is displayed and can be inserted anywhere in the drawing. £20 £200 £40. This works especially well when the general shape of the area can be decomposed into simpler shapes for which the moment of inertia is calculated for. The equilateral triangle actually makes the strongest column for a given area, but not by much (12% stronger than the circle). Rotational kinetic energy. Ix = b h3 / 36 (4a) Iy =h b3 / 36 (4b) Area Moment of Inertia for typical Cross Sections I. Relative to principal axes of inertia, the product of inertia of a figure is zero. • Th t fi ti fth hdd iThe moment of inertia of the shaded area is obtained by subtracting the moment of inertia of the half-circle from the moment of inertia of the rectangle Determine the moment of inertia of the shaded area with respect to the x axis. Determine polar moment of inertia of an isosceles triangle 1 answer below » Polar Moments of Inertia Determine the polar moment of inertia I P of an isosceles triangle of base b and altitude h with respect to its apex (see Case 5, Appendix D). DIY Brick Rocket Stove - Cooking Without Electrical Power - Duration: 23:40. Area, center of mass, moments of inertia. This theorem is really powerful because the moment of inertia about any set of axes can be found by finding the moment of inertia about the centroidal | {
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of inertia about any set of axes can be found by finding the moment of inertia about the centroidal axes and adding the distance-area term to it. The right triangle comes along frequently in geometry. When that happens equation 4 and 5 would be used to calculate the stress and. Mass and moment of inertia properties of accessory components. This calculates the Area Moment of Inertia of a semi-circle about various axes. The following effective moment of inertia expression was originally proposed by Branson [9] and was adopted by ACI [10] and presented as (2). The help tool instructs me to click on the inspect menu and choose AREA. 1 DefinitionsThe second moment of the area about the x axis (IX) is defined as:I X = ∫ y 2 dA (11. EHE-08): Where: Mf = Mcrk = Nominal cracking moment of the cross section. The second moment of area is also known as the moment of inertia of a shape. PARALLEL-AXIS THEOREM, RADIUS OF GYRATION & MOMENT OF INERTIA FOR COMPOSITE AREAS. Synchronised Similar Triangles for Three-Body Orbit with L = 0 4 where (i,j,k) runs for the cyclic permutations of (1,2,3). Relative to principal axes of inertia, the product of inertia of a figure is zero. I = Second moment of area, in 4 or mm 4; J i = Polar Moment of Inertia, in 4 or mm 4; K = Radius of Gyration, in or mm; P = Perimeter of shape, in or mm; S = Plastic Section Modulus, in 3 or mm 3; Z = Elastic Section Modulus, in 3 or mm 3; Online Parabolic Half Property Calculator. Table of Selected Moments of Inertia Note: All formulas shown assume objects of uniform mass density. Is there a way to calculate this to X-X? The instructions o. 3 (4) 3 Determine the AP whose fourth term is 15 and the difference of 6th term from 10th term is 16 Prove that ratio of area of two triangle is equal to the square of the corresponding sides. Determine polar moment of inertia of an isosceles triangle 1 answer below » Polar Moments of Inertia Determine the polar moment of inertia I P of an isosceles triangle of base b and | {
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Moments of Inertia Determine the polar moment of inertia I P of an isosceles triangle of base b and altitude h with respect to its apex (see Case 5, Appendix D). \] The moment of inertia of the area about the center can be found using in equation (40) can be done in two steps first calculate the moment of inertia in this coordinate system and then move the coordinate system to center. Figure 2: Deriving an equation for moment of inertia of the triangle rotating around its base. The moment of inertia of an area with respect to any given axis is equal to the moment of inertia with respect to the centroidal axis plus the product of the area and the square of the distance between the 2 axes. Product of inertia for triangle. (6) Theorems of Moment of Inertia. Moment of inertia of the remaining part of lamina about the same axis is :. The mass moment of inertia of an object about an axis through the center of mass is smaller than that about any other axis in the same direction. Centroids & Moment of Inertia. o The moment of inertia of a triangular section of height h about its base is given as, I = bh 3 /12. Actual physical properties may vary due to tolerances which occur in the manufacturing process. The mass moment of inertia equation for a point mass is simply: I = mr 2. (8), derived in the moment of inertia example, the moment of inertia of the disk is = at 5 digits Therefore, the moment of inertia of the disk is 12. The moment of inertia must be specified with respect to a chosen axis of rotation. The product of inertia of triangle (a) with respect to its centroid is I ¯ x y = b 2 h 2 / 72. Right Triangle The output of this equation is the I x and I y components of the area moment of inertia when the triangle is defined to be in the x/y plane. It is formed by the intersection of the medians. This table provides formula for calculating section Area, Moment of inertia, Polar moment of inertia, Section modulus, Radius of gyration, and Centroidal distance, for various | {
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Polar moment of inertia, Section modulus, Radius of gyration, and Centroidal distance, for various cross section shapes. Solve: The moment of inertia of the triangle is I mr= × = =3 3(0. 91, b < 10a. Question: Three point masses, each of mass {eq}m {/eq}, are placed at the corners of an equilateral triangle of side {eq}L {/eq}. (ii) Moment of inertia about new axes which is turned through an angle of 30 0 anticlockwise to the old axis. We see it in action all the time. Equation 18) also holds for polar moments of inertia i. It will help in deciding whether the failure will be on the compression face or on the tension face of the beam. When that happens equation 4 and 5 would be used to calculate the stress and. 18) I s = I c + Ad 2. In order to continue, we will need to find an expression for dm. The moment of inertia is ∑mi*ri²; all the m are the same = 0. The radius of gyration is the radius at which we consider the mass to rotate such that the moment of inertia is given by I = M k2. Four leg-loading conditions were employed: 1) no load (NL) on the legs; 2) a baseline load (BSLN) condition, with a mean of 2. With great regard for economy of words we call the expression written above "the moment of inertia of the area about the x axis" or I x for short. l : moment of inertia about the axis parallel to x-axis. 728(1) 30 (2) 0. Determine the moment of inertia of the area about the x and y. 9 µC, are located at the corners of an equilateral triangle as in the figure above. Going to the division, we get. Figure to illustrate the area moment of a triangle at the list of moments of inertia. Live Simple, Live Free - Tinyhouse Prepper Recommended for you. Figure to illustrate the area moment of a triangle at the list of moments of inertia. For the sake of one more bit of integration practice, we shall now use the same argument to show that the moment of inertia of a uniform circular disc about a. 2 Second Moment of Area11. The equation of the line is $y = \dfrac{a}{b} x + | {
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circular disc about a. 2 Second Moment of Area11. The equation of the line is $y = \dfrac{a}{b} x + a$. Moment of Inertia of a Triangular Lamina about its Base. This simple, easy-to-use moment of inertia calculator will find moment of inertia for a circle, rectangle, hollow rectangular section (HSS), hollow circular section, triangle, I-Beam, T-Beam, L-Sections (angles) and channel sections, as well as centroid, section modulus and many more results. Moment of Inertia 5 An example of this is the concrete T-beam shown. The second moment of area is also known as the moment of inertia of a shape. Explanation: No explanation is available for this question! 2) What is the C. Mass Moment of Inertia Calculator in Excel, Pt. The distance of the center of mass of the triangle in its new position from the pivot is the same. Moment of inertia, also called mass moment of inertia or the angular mass, (SI units kg m 2) is a measure of an object’s resistance to changes in its rotation rate. Right: Triangles with centroidal axes re-positioned with respect to the x-axis. Moment of inertia can be defined by the equation The moment of inertia is the sum of the masses of the particles making up the object multiplied by their respective distances squared from the axis of rotation. It is the rotational analog of mass. Uniform circular lamina about a diameter. We are concerned here with area only and the area multiplied by a distance twice is the second moment of area. 707(h) to get the actual I, h being the weld size. It is required in the design of machines, bridges, and other engineering systems. Find the polar moment of inertia. Using these, the moment of inertia for the parallel axis can be calculated using the formula: The moment of inertia for rotation around the axis at the surface of the sphere is. Moment of Inertia of Isosceles Triangle Jalal Afsar October 25, 2013 Uncategorized No Comments Moment of Inertia of Isosceles triangle can be easily find out by using formulas with | {
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No Comments Moment of Inertia of Isosceles triangle can be easily find out by using formulas with reference to x-axis and y-axis. The moment of inertia of the triangle about the point O is I = M*r^2 , where r is the distance of the center of mass from O. Moment of Inertia Tensor Consider a rigid body rotating with fixed angular velocity about an axis which passes through the origin--see Figure 28. The total moment of inertia is due to the sum of masses at a distance from the axis of rotation. If the triangle were cut out of some uniformly dense material, such as sturdy cardboard, sheet metal, or plywood, the centroid would be the spot where the triangle would balance on the tip of your finger. The Area Moment of Inertia equation, I = (b•h 3)/12 , (b 3 •h)/4 , computes the Area Moment of Inertia for a right triangle with right angle on right of the base. Question: What Is The Triangle's Moment Of Inertia About The Axis Through The Center? Express Your Answer To Two Significant Figures And Include The Appropriate Units. It is also popular as angular mass or rotational inertia of the given rigid body. A point mass does not have a moment of inertia around its own axis, but using the parallel axis theorem a moment of inertia around a distant axis of rotation is achieved. Angular acceleration of the system + triangle (long base) α_up 0. Worthy of note, in order to solve for the moment of inertia of the right triangular thin plate, we first had to measure the the triangle's mass, base length, and height. How you find moment of inertia of isosceles triangle? Wiki User 2014-05-12 13:36:50. Since this term is always zero or positive, the centroidal moment of inertia is the minimum moment of inertia with respect to all parallel axes. This table provides formula for calculating section Area, Moment of inertia, Polar moment of inertia, Section modulus, Radius of gyration, and Centroidal distance, for various cross section shapes. Four leg-loading conditions were employed: 1) no | {
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distance, for various cross section shapes. Four leg-loading conditions were employed: 1) no load (NL) on the legs; 2) a baseline load (BSLN) condition, with a mean of 2. Moment of Inertia of a Triangular Lamina about its Base. PARALLEL-AXIS THEOREM, RADIUS OF GYRATION & MOMENT OF INERTIA FOR COMPOSITE AREAS. The Questions and Answers of The three point masses each of mass m are placed corner of equilateral triangle offside of side l then moment of inertia of the system about an axis along one side of triangle is? are solved by group of students and teacher of Class 11, which is also the largest student community of Class 11. It is also known as rotational inertia. Moment of Inertia is defined as: $$I={\sum}mr^2$$ which in this case can be rewritten into an integral: $$I=\rho\int_A{r^2dA}$$ Since the shape of the triangle can't be described by one formula, you would have to split the integral into multiple sections. Moment of Inertia. Centroid, Area, Moments of Inertia, Polar Moments of Inertia, & Radius of Gyration of a Triangular Cross-Section. For a point mass, the moment of inertia is just the mass times the square of perpendicular distance to the rotation axis, I = mr 2. Axis through center. “The poles of inertia” is another way of saying “mass concentration centers”. Undeniable momentum, on any stage - anywhere. The second moment of inertia rectangle is the product of height and cube of width divided by 12. Write (but do not evaluate) an integral expressing the moment of inertia of the region between y= sinxand y= sinx(for 0 x ˇ) rotated around: (a)the xaxis; (b)the yaxis; (c)the zaxis. The computation of moments of inertia can often be. Chapter-3 Moment of Inertia and Centroid Page- 1 3. “twisting”) about a given axis due to an applied torque. The distance of the center of mass of the triangle in its new position from the pivot is the same. J z' = I x' + I y'. 3 × 10⁻⁵ kg. The plane figures (like triangle, quadrilateral, circle, trapezoid, etc. d' (n -1)A 's | {
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3 × 10⁻⁵ kg. The plane figures (like triangle, quadrilateral, circle, trapezoid, etc. d' (n -1)A 's Without compression steel With. of inertia of the rectangle. Note that each component of the moment of inertia tensor can be written as either a sum over separate mass elements, or as an integral over infinitesimal mass elements. automatic weight calculator for rectangular, square, round, or hexagonal, plate, tube, bar, beams, sheet, rod and other engineering material shapes. Hallow cylinder. It is analogous to mass in that it is a measure of the resistance a body offers to torque or rotational motion. 2012/2013. We will look at each expression below. The domain of the triangle is defined by. Moment of Inertia of Triangle about its Base | Very Important. Let R be the triangle with vertices (0, 0), (1, 0), (1, √ 3) and density δ = 1. 1 RADIUS OF GYRATION k All rotating machinery such as pumps, engines and turbines have a moment of inertia. Purpose: Determine the moment of inertia of a right triangular thin plate around its center of mass, for two perpendicular orientations of the triangle. Finding the Centroid. Angular momentum. = Polar Moment of Inertia. Description. dV = dxdydz. Add to Solver. The number of revolutions that the shaft will make during this time is. Worthy of note, in order to solve for the moment of inertia of the right triangular thin plate, we first had to measure the the triangle's mass, base length, and height. Determine the moment of inertia of the triangle for rotation about an axis that bisects one of its angles. Derivation of the Moment of Inertia Formula Suppose a particle of mass m is attached to a pivot by a thin rod of length r. Rotations in 2D are about the axis perpendicular to the 2D plane, i. Moment of inertia of three uniform rods of mass M and length l joined to form an equilateral triangle, about an axis passing through one of its sides. It is also de ned as I= Z r2 dm (3) for a continuous distribution of mass. In either case, use | {
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It is also de ned as I= Z r2 dm (3) for a continuous distribution of mass. In either case, use of the formulas is cumbersome and prone to error, especially in converting to consistent units. Today we will see here the method to determine the moment of inertia for the triangular section about a line passing through the center of gravity and parallel to the base of the triangular section with the help of this post. Integrate to derive a formula for the moment of inertia for a general triangle. 025kg) g = gravity (9. Evaluation of Moments of Inertia 2008 Waterloo Maple Inc. But I don't know how to do that. Moment of Inertia of Isosceles Triangle Jalal Afsar October 25, 2013 Uncategorized No Comments Moment of Inertia of Isosceles triangle can be easily find out by using formulas with reference to x-axis and y-axis. The force of attraction is proportional to mass of the body. 91, b < 10a. The moment of inertia of the triangular shaped area is 3. What is the moment of inertia of ball about the axis of rotation AB? Ignore cord’s mass. 8680 rad/s^2 α_down -0. The inertia matrix (aka inertia tensor) of a sphere should be diagonal with principal moments of inertia of 2/5 mass since radius = 1. Since moment of inertia is to be determined about an axis of rotation and from the provided diagram, it seems that the student is interested in finding the moment of inertia about the side (AB) along y-axis. 9 µC, are located at the corners of an equilateral triangle as in the figure above. Calculate the mass moment of inertia of the triangular plate about the y-axis. 2 comments. Find the moment of inertia of a plate cut in shape of a right angled triangle of mass M side AC=BC=a about an axis perpendicular to the plane of the plate and passing. Units 9 to 17,are assigned to:-Estimation of the Moment of inertia for Right-angled triangle (about X,Y) &Product of inertia &Polar Moment of Inertia, the radius of gyrations, by using two ways of Estimations, for the two cases of a right-angle | {
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the radius of gyrations, by using two ways of Estimations, for the two cases of a right-angle triangle. When the axes are such that the tensor of inertia is diagonal, then these axes are called the principal axes of inertia. The mass moments of inertia of an object about any parallel axes are identical. If you are consitent about which way you go around the triangle, the orgin can be anywhere, as it will subtract tnegative areas automatically. Central axis of hallow cylinder. The formula calculates the Moment of Inertia of a right triangle of base b and height h in respect to an axis collinear with the base of the triangle (one of the sides. Just add up area, centroid and self I of a series of triangles comprised of the origin, the ith point, and the i+1th point. Two circular loops of radii R and nR are made of same wire. (by the parallel axis theorem). For each segment defined by two consecutive points of the polygon, consider a triangle with two. Mathematically, and where IB " *BA " TIA BA = *B + 7IA Ig = moment of inertia about the base plane I3A = moment of inertia about a base diameter axis 1^ = moment of inertia about the central axis 7. Using the parallel axis theorem, you can find the moment of inertia about the center by subtracting Mr^2, where r is 2/3h. One can define the moment of inertia as the ratio of the angular moment to the angular velocity of the particular object moving at its principal axis. Area Moment of inertia. The moment of inertia about the X-axis and Y-axis are bending moments, and the moment about the Z-axis is a polar moment of inertia(J). 91, b < 10a. Now to calculate the moment of inertia of the strip about z-axis, we use the parallel axis theorem. 7899 Working Ic. Then the periphery of the rectangle is 2x (10+20)=60 mm. In[4]:= Out[4]= You compute the moment of inertia about the y axis using the function SectionInertialMoments from the SymCrossSectionProperties package. Let the lengths of sides $AB$ and $BC$ be $a$ and $b$ respect. | {
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SymCrossSectionProperties package. Let the lengths of sides $AB$ and $BC$ be $a$ and $b$ respect. This is the currently selected item. The SI unit of moment of inertia is kg m2. The moment of inertia of the disk about its center is $$\frac{1}{2} m_dR^2$$ and we apply the parallel-axis theorem (Equation \ref{10. dm = M A dA (2) (2) d m. Let the mass of the triangle be M. moment of inertia with respect to x, Ix I x Ab 2 7. Where "dM" are small mass in the body and "y" is the distance of each on of them from the axis O-O. 32075h^4M/AL, where h is the height of the triangle and L is the area. Determine the moments of inertia about the centroid of the shape. Check the basic shapes at the bottom of. Bending of Beams with Unsymmetrical Sections C = centroid of section Assume that CZ is a neutral axis. We can relate these two parameters in two ways: For a given shape and surface mass density, the moment of inertia scales as the size to the fourth power, on dimensional grounds. Inventor has a function for moments but it rotates the center plane to something like VxV in the link below. The 2nd moment of area, also known as moment of inertia of plane area, area moment of inertia, or second area moment, is a geometrical property of an area which reflects how its points are distributed with regard to an arbitrary axis. Some examples of simple moments of inertia Let's try an easy calculation: what's the moment of inertia of these three balls? Each ball has mass m = 3 kg, and they are arranged in an equilateral triangle with sides of length L = 10 m. 1 Centre of Gravity Everybody is attracted towards the centre of the earth due gravity. 3×10^ - 26 kg and a moment of inertia of 1. An 800g steel plate has the shape of an isosceles triangle. Our thin right triangular plate. The moment of inertia of this system about an axis along one side. Answered by Expert 5th October 2017, 8:56 PM. Ball hits rod angular momentum example. This is given by the table above which indicates that the | {
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Ball hits rod angular momentum example. This is given by the table above which indicates that the centroid of a triangle is located, from the corner that is opposite of the hypotenuse (the longest side of the triangle), one-third of the length of the base in the y direction and one-third of the length of the height in the x direction in this case. Second, finding the moment of inertia when the triangle rotates around its base (shorter leg). Below is the list of moments of inertia for common shapes. Beam Deflection Equations are easy to apply and allow engineers to make simple and quick calculations for deflection. Solution: The mass moment of inertia about the y-axis is given by. A cavity DEF is cut out from the lamina, where D, E, F are the mid points of the sides. It is analogous to mass in that it is a measure of the resistance a body offers to torque or rotational motion. Axis on surface. Although it is a simple matter to determine the moment of inertia of each rectangular section that makes up the beam, they will not reference the same axis, thus cannot be added. Moments of Inertia of Geometric Areas Frame 28-1 * Introduction This unit will deal with the computation of second moments, or moments of inertia, of The general expression for the moment of inertia of a right triangle about a centroidal axis parallel to a side is. Once you have done this, run the "massprop" command and click. 1 DefinitionsThe second moment of the area about the x axis (IX) is defined as:I X = ∫ y 2 dA (11. Find the moment of inertia for the following about the y axis and x axis of a right triangle whose base is on the +x axis and whose height is on the +yaxis Source(s): moment inertia triangle: https://shortly. The moment of inertia of the triangle about the point O is I = M*r^2 , where r is the distance of the center of mass from O. 2nd moment of an area or moment of inertia is the moment of all small areas dA about any axis. Area Moment of inertia. An isosceles triangle is a | {
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the moment of all small areas dA about any axis. Area Moment of inertia. An isosceles triangle is a triangle with two equal sides. A higher moment of inertia is an indication that you need to apply more force if you want to cause the object to rotate. 5 • The moment of inertia of the shaded area is obtained by subtracting the moment of inertia of the half-circle from the moment of inertia of the rectangle. 2 Second Moment of Area11. Kinetic Energy is the energy possessed by an object because it is in motion. I Average value of a function. Therefore Moment of Inertia of Rectangle about its center = m 1 2 a 2 + b 2 Distance of P point from center of rectangle is 2 a 2 + b 2 Therefore Moment of Inertia of Rectangle about P, I= m 1 2 a 2 + b 2 + m 4 a 2 + b 2 = m 3 a 2 + b 2 Mass of triangle PQR = 2 m = 2 ρ a b Moment of Inertia of Triangle PQR about its centroid = ρ 1 2 a b 3 + b. Moment of Inertia is strictly the second moment of mass, just like torque is the first moment of force. How to calculate the moment of inertia of a triangular plate rotating about the apex. T 1 – the instantaneous value of load torque, referred to a motor shaft, N-m. The moment of inertia of the shaded area is obtained by subtracting the moment of inertia of the half-circle from the moment of inertia of the rectangle. A 100-gram ball connected to one end of a cord with a length of 30 cm. 4)and the second moment of the area about the y. The area moment of inertia about the X and the Y axis are calculated by subtracting the second moment of inertia values of the inner rectangular area from that of the outer rectangular area. Some examples of simple moments of inertia Let's try an easy calculation: what's the moment of inertia of these three balls? Each ball has mass m = 3 kg, and they are arranged in an equilateral triangle with sides of length L = 10 m. Figure 2: Deriving an equation for moment of inertia of the triangle rotating around its base. Let ‘h’ be the distance between the two axes i. | {
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of inertia of the triangle rotating around its base. Let ‘h’ be the distance between the two axes i. Moment of inertia of a same object will change against different axis. We will take the case where we have to determine the moment of inertia about the centroid y. Calculate The Moment Of Inertia Of The Triangle With Respect To The X Axis. The moment of inertia is ∑mi*ri²; all the m are the same = 0. In either case, use of the formulas is cumbersome and prone to error, especially in converting to consistent units. dm = M A dA (2) (2) d m. If the triangle were cut out of some uniformly dense material, such as sturdy cardboard, sheet metal, or plywood, the centroid would be the spot where the triangle would balance on the tip of your finger. It is based not only on the physical shape of the object and its distribution of mass but also the specific configuration of how the object is rotating. Today we will see here the method to determine the moment of inertia for the triangular section about a line passing through the center of gravity and parallel to the base of the triangular section with the help of this post. 0mm and n = 15. The inertia of both systems can be found using the equation: m = mass of hanging mass (0. Central axis of disk. Note the dy is assigned the value 1 so that the Maple integrator does not confuse it as a mathematical variable. Journal of Graphics Tools: Vol. Chapter-3 Moment of Inertia and Centroid Page- 1 3. The axis perpendicular to its base. It is the measure of an object’s resistance against the angular acceleration. The moment of inertia of the triangle about the point O is I = M*r^2 , where r is the distance of the center of mass from O. 0 kg per leg. 0 , calculate the moment of inertia of the shaded area shown (Part B figure) about the x axis. More on moment of inertia. Calculate the mass moment of inertia of the triangular plate about the y-axis. Physics moment of inertia help? Three 210 g masses are connected to form an equilateral | {
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the y-axis. Physics moment of inertia help? Three 210 g masses are connected to form an equilateral triangle with side lengths of 40 cm. However, this is only true for uniform or ordinary objects, such as an orb attached to a string whirling around at a certain angular velocity. Multiply the Area of each element by the square of the distance from the centroid of each element to the centroid of the cross-section(x1 and y1). Polar moment of inertia is equal to the sum of inertia about X-axis and Y-axis. Principle Axes of Axes of Inertia of a Mass. Three point charges, A = 2. Moment of inertia, denoted by I, measures the extent to which an object resists rotational acceleration about a particular axis, and is the rotational analogue to mass. The domain of the triangle is defined by. Description Mass (kg) I (kg m2) 648-07627 9-Inch Disc 1. Area Moments of Inertia by Integration • Second moments or moments of inertia of an area with respect to the x and y axes, x ³ yI y ³ xdA 2 2 • Evaluation of the integrals is simplified by choosing dA to be a thin strip parallel to one of the coordinate axes. Remark: The moment of inertia of an object is a measure of the resistance of the object to changes in its rotation. The moment of inertia of an area with respect to any given axis is equal to the moment of inertia with respect to the centroidal axis plus the product of the area and the square of the distance between the 2 axes. Conversely, a lower moment of inertia means that you only need to apply a minimal amount of force to cause a rotation. I = mass moment of inertia. Hemmingsen assumed (based on copyright claims). 28 Rectangle Area, in 2, in. I Average value of a function. 15 Centroid and Moment of Inertia Calculations An Example ! Now we will calculate the distance to the local centroids from the y-axis (we are calculating an x-centroid) 1 1 n ii i n i i xA x A = = = ∑ ∑ ID Area x i (in2) (in) A 1 2 0. The moment of inertia is defined as the quantity expressed by the body | {
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Area x i (in2) (in) A 1 2 0. The moment of inertia is defined as the quantity expressed by the body resisting angular acceleration which is the sum of the product of the mass of every particle with its square of a distance from the axis of rotation. Second, finding the moment of inertia when the triangle rotates around its base (shorter leg). Determination of the center of gravity. The angle in between the masses is 60 degrees. (Eq 2) Φ = T GJ. 3/2MR^2 The center of mass of the original triangle (the part that has been cut out) was at the center of the circle, at a distance R from the pivot. ld parallel increases in mechanical power, and that EMG amplitude would increase with greater limb mass or limb moment of inertia. All the equations given below contain I, the moment of inertia of a beam, which is a constant determined by the beam's cross-sectional shape and thickness. Therefore Moment of Inertia of Rectangle about its center = m 1 2 a 2 + b 2 Distance of P point from center of rectangle is 2 a 2 + b 2 Therefore Moment of Inertia of Rectangle about P, I= m 1 2 a 2 + b 2 + m 4 a 2 + b 2 = m 3 a 2 + b 2 Mass of triangle PQR = 2 m = 2 ρ a b Moment of Inertia of Triangle PQR about its centroid = ρ 1 2 a b 3 + b. Determine the moment of inertia of the cross section about the x axis. - The formula for moment of inertia is - If there are 3 particles of mass 'm' placed at each of the vertex of this equilateral triangle then we consider three times m. J z' = I x' + I y'. "now, if the axis is passing through A, then sphere B, C, and D each rotate around this axismy attempt was this: m1(r^2) for sphere B (perpendicular to A) = 0. 1st moment of area is area multiplied by the perpendicular distance from the point of line of action. Annulus Moment of Inertia M5 revision thread Surface integrals of scalar fields show 10 more Urgent physics angular motion problem Angular momentum/moment of inertia Intuition question about Stokes' theorem. Going to the division, we get. | {
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momentum/moment of inertia Intuition question about Stokes' theorem. Going to the division, we get. base=20cm, height=30cm. 707(h) to get the actual I, h being the weld size. Find the moment of inertia and radius of gyration in each of the following cases when axis of rotation is. I = Second moment of area, in 4 or mm 4; J i = Polar Moment of Inertia, in 4 or mm 4; K = Radius of Gyration, in or mm; P = Perimeter of shape, in or mm; S = Plastic Section Modulus, in 3 or mm 3; Z = Elastic Section Modulus, in 3 or mm 3; Online Parabolic Half Property Calculator. Author: No machine-readable author provided. Mechanical Engineering: Ch 12: Moment of Inertia (27 of 97) Moment Centroid, Area, Moments of Inertia, Polar Moments of Inertia EMech full notes. The moment of inertia of a body is its tendency to resist rolling motions and angular accelerations. However, if we found the moment of inertia of each section about some. We see it in action all the time. the Z-axis. In particular, the same object can have different moments of inertia when rotating about different axes. They are; Axis passing through the centroid. Cross product and torque. In mathematical notation, the moment of inertia is often symbolized by I, and the radius is symbolized by r. dI = r2dm (1) (1) d I. 94 into 10 to the power of minsis 46 kg metre square bout an Axis through its Centre perpendicular to the lines joining the two atoms. Topic - Moment of Inertia ,Ans - (Mh^2)/6. of inertia of the rectangle. Calculating the second moment of area of geometric figures can be confusing and time consuming by hand, so let this calculator do all the work for you. Area Moment of Inertia of a Triangle. The product of inertia of area A relative to the indicated XY rectangular axes is IXY = ∫ xy dA (see illustration). Physics - Moment of Inertia Consider an equilateral triangle cut from a thin board. Calculating Moment of Inertia of a Uniform Thin Rod. The moment of inertia of two or more particles about an axis of | {
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of Inertia of a Uniform Thin Rod. The moment of inertia of two or more particles about an axis of rotation is given by the sum of the moment of inertia of the individual particles about the same axis of rotation. Let's divide the triangle into strips along y-axis, each of width dx. ANSWER: Right angled triangle. When the axes are such that the tensor of inertia is diagonal, then these axes are called the principal axes of inertia. Considering an element DE parallel to y-axis at a distance x from origin and width dx. This is given by the table above which indicates that the centroid of a triangle is located, from the corner that is opposite of the hypotenuse (the longest side of the triangle), one-third of the length of the base in the y direction and one-third of the length of the height in the x direction in this case. Is there a way to calculate this to X-X? The instructions o. 20}) to find. The tensor of inertia will take different forms when expressed in different axes. "now, if the axis is passing through A, then sphere B, C, and D each rotate around this axismy attempt was this: m1(r^2) for sphere B (perpendicular to A) = 0. These bodies, with mass density $$\rho$$, can be seen as stacks of infinitesimally thin triangles of thickness $$\text{d}h$$ and surface density $$\text{d} \mu = \rho \text{d}h$$ (we preserve the notations from the previous posts. The moments of inertia of an angle can be found, if the total area is divided into three, smaller ones, A, B, C, as shown in figure below. one rectangle; one square; one triangle; At this stage, we calculate their surface area, the moment of inertia and the moment of deviation. Cross product and torque. Matt Anderson 18,225 views. The polar moment of inertia of the area A is calculated as. Three point charges are located at the corners of an equilateral triangle(q1=2microC,q2=-4microC). Annulus Moment of Inertia M5 revision thread Surface integrals of scalar fields show 10 more Urgent physics angular motion problem | {
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thread Surface integrals of scalar fields show 10 more Urgent physics angular motion problem Angular momentum/moment of inertia Intuition question about Stokes' theorem. Area Moments of Inertia Parallel Axis Theorem • Moment of inertia IT of a circular area with respect to a tangent to the circle, ( ) 4 4 5 4 2 2 4 2 1 r IT I Ad r r r π π π = = + = + • Moment of inertia of a triangle with respect to a. See how the eigenvectors of the inertia tensor change as you change a configuration of point masses, or the shape of a solid plate of material. If you're unsure about what deflection actually is, click here for a deflection definition Below is a concise beam deflection table that shows how to calculate the maximum deflection in a beam. Considering an element DE parallel to y-axis at a distance x from origin and width dx. This table provides formula for calculating section Area, Moment of inertia, Polar moment of inertia, Section modulus, Radius of gyration, and Centroidal distance, for various cross section shapes. Solution 3. - The formula for moment of inertia is - If there are 3 particles of mass 'm' placed at each of the vertex of this equilateral triangle then we consider three times m. 1 DefinitionsThe second moment of the area about the x axis (IX) is defined as:I X = ∫ y 2 dA (11. , in 4 ¦ xyA III II I x y xyA Apply the parallel axis theorem to each rectangle, xy ¦ I xcyc xyA Note that the product of inertia with respect to. Inventor has a function for moments but it rotates the center plane to something like VxV in the link below. Question: Part A - Moment Of Inertia Of A Triangle With Respect To The X Axis A Composite Area Consisting Of The Rectangle, Semicircle, And A Triangular Cutout Is Shown(Figure 1). Moment of Inertia for body about an axis Say O-O is defined as ∑dM*y n 2. The situation is this: I know the moment of inertia with respect to the x axis and with respect to the centroidal x axis because its in the table. 2 An Example: Moment of Inertia of | {
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with respect to the centroidal x axis because its in the table. 2 An Example: Moment of Inertia of a Right Circular Cone For a right circular cone of uniform density we can calculate the moment. The formula calculates the Moment of Inertia of a right triangle of base b and height h in respect to an axis collinear with the base of the triangle (one of the sides. ) is the moment of inertia about the centroid of the area about an x axis and d y is the y distance between the parallel axes Similarly 2 y I y Ad x Moment of inertia about a y axis J Ad 2 o c Polar moment of Inertia 2r 2 d 2 o c Polar radius of gyration 2 r 2 d 2 Radius of gyration. Moment of Inertia Tensor Consider a rigid body rotating with fixed angular velocity about an axis which passes through the origin--see Figure 28. Bending of Beams with Unsymmetrical Sections C = centroid of section Assume that CZ is a neutral axis. Please enter the "Input Values" in the form. Similar to the centroid, the area moment of inertia can be found by either integration or by parts. This calculates the Area Moment of Inertia of a semi-circle about various axes. A piece of thin uniform wire of mass m and length 3b is bent into an equilateral triangle. Whatever kind you are trying to compute I would suggest breaking up the cross section into triangles with two vertices on successive points of your boundary and the third at the center about which the moment of inertia is to be taken. Open Section Properties Case 11 Calculator. Check the basic shapes at the bottom of. = Two point masses, m 1 and m 2, with reduced mass μ and separated by a distance x, about an axis passing through the center of. EXAMPLE 2: MASS MOMENT OF INERTIA Calculate the mass moment of inertia of the triangular plate about the y-axis. This engineering data is often used in the design of structural beams or structural flexural members. find the average angular velocity of. Just add up area, centroid and self I of a series of triangles comprised of the | {
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angular velocity of. Just add up area, centroid and self I of a series of triangles comprised of the origin, the ith point, and the i+1th point. Moment of Inertia Question (Edexcel M5) Moment Of Inertia Kinematics. ) is the moment of inertia about the centroid of the area about an x axis and d y is the y distance between the parallel axes Similarly 2 y I y Ad x Moment of inertia about a y axis J Ad 2 o c Polar moment of Inertia 2r 2 d 2 o c Polar radius of gyration 2 r 2 d 2 Radius of gyration. calculate the moment of inertia when the plate is rotating about an axis perpendicular to the plate and passing through the vertex tip. 0 , calculate the moment of inertia of the shaded area shown (Part B figure) about the x axis. The centroid is 8" above the base. 1st lesson free! 1st lesson free! 1st lesson free! 1st lesson free! 1st lesson free! 1st lesson free! 1st lesson free! 1st lesson free!. Mass Moment of Inertia Calculator in Excel, Pt. 8680 rad/s^2 α_down -0. The centroidal moments of inertia and the product of inertia are determined using the table below Product of inertia = Ixy = A (dx)(dy) = 0 8" 3"-3" Part Area Ix dy1 d 2 y 1 (A) Ix. This simple, easy-to-use moment of inertia calculator will find moment of inertia for a circle, rectangle, hollow rectangular section (HSS), hollow circular section, triangle, I-Beam, T-Beam, L-Sections (angles) and channel sections, as well as centroid, section modulus and many more results. The ratio of length to radius is 1) 2: 1 2) 3:1 3) 3: 1 4) 2:1 28. Purpose: Determine the moment of inertia of a right triangular thin plate around its center of mass, for two perpendicular orientations of the triangle. 12 Moment of Inertia With Respect to an Arbitrary Axis Ellipsoid of Inertia. Get an answer for 'Q. See how the eigenvectors of the inertia tensor change as you change a configuration of point masses, or the shape of a solid plate of material. new inertia = 1. Along the height it is hb^3/48 and along base it is bh^3/36. Let be | {
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of material. new inertia = 1. Along the height it is hb^3/48 and along base it is bh^3/36. Let be the position vector of the th mass element, whose mass is. Sorry to see that you are blocking ads on The Engineering ToolBox! If you find this website valuable and appreciate it is open and free for everybody - please contribute by. 001472 Kg*m^2. The moment of inertia about an axis through a vertex is 0. 4 Find the moment of inertia of a plate cut in shape of a right angled triangle of mass M, side AC = BC = a about an axis perpendicular to the plane of the plate and passing through the mid point of side AB. Now to calculate the moment of inertia of the strip about z-axis, we use the parallel axis theorem. Moment of Inertia 5 An example of this is the concrete T-beam shown. 8·10-2 Kg·m2 Submit Figure < 1of1 Incorrect; Try Again: 3 Attempts Remaining Part B What Is The Triangle's. Transfer of Axis Theorem. J s = J g + Ad 2. This theorem is really powerful because the moment of inertia about any set of axes can be found by finding the moment of inertia about the centroidal axes and adding the distance-area term to it. It is the measure of an object’s resistance against the angular acceleration. | {
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IEVref: 103-01-11 ID: Language: en Status: Standard Term: system of orthogonal functions Synonym1: orthogonal system Synonym2: Synonym3: Symbol: Definition: set of functions, such that each of them is orthogonal to any otherNote 1 to entry: Examples: Legendre polynomials P constitute a system of orthogonal functions on the interval $\left[-1,\text{\hspace{0.17em}}+1\right]$ because ${\int }_{\text{ }-1}^{\text{ }+1}{P}_{k}\left(x\right){P}_{l}\left(x\right)\text{d}x=0$ for any integers $k\ne l$. Laguerre polynomials L constitute a system of orthogonal functions on the interval $\left[0,\text{\hspace{0.17em}}+\infty \right]$ with the weight $\text{e}\text{x}\text{p}\left(-x\right)$ because ${\int }_{\text{ }0}^{\text{ }+\infty }{L}_{k}\left(x\right){L}_{l}\left(x\right)\text{exp}\left(-x\right)\text{d}x=0$ for any integers $k\ne l$. Trigonometric functions sine and cosine constitute a system of orthogonal functions on the interval $\left[0,\text{\hspace{0.17em}}2\pi \right]$ because ${\int }_{\text{ }0}^{\text{ }2\pi }\mathrm{sin}\left(kx\right)\text{sin}\left(lx\right)\text{d}x=0$ and ${\int }_{\text{ }0}^{\text{ }2\pi }\mathrm{cos}\left(kx\right)\mathrm{cos}\left(lx\right)\text{d}x=0$ for any integers $k\ne l$, and ${\int }_{\text{ }0}^{\text{ }2\pi }\text{sin}\left(kx\right)\mathrm{cos}\left(lx\right)\text{d}x=0$ for any integer k and l. Publication date: 2009-12 Source: Replaces: Internal notes: 2017-02-20: Editorial revisions in accordance with the information provided in C00020 (IEV 103) - evaluation. JGO CO remarks: TC/SC remarks: VT remarks: Domain1: Domain2: Domain3: Domain4: Domain5: | {
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# Birthday Problem for 3 people
I know that, in a room of 23 people, there is a 50-50 chance that two people have the same birthday. However, what I want to know is: How many people do you need to have a 50-50 chance that 3 people share the same birthday?
Note: Assume for this question, that birthdays are equally distributed
• math.stackexchange.com/questions/25876/… – user940 Sep 6 '13 at 4:38
• BirthDay Problem ---> en.wikipedia.org/wiki/Birthday_problem – Felix Marin Sep 6 '13 at 4:47
• The linked question asks for the probability for 30 people. What I want is the amount of people needed for the probability to be above 50%. – Thomas Sep 6 '13 at 5:04
• Notice that the number 23 uses the assumption that birthdays are equally distributed, i.e., that every birthday is equally-likely, with probability 1/365 (or 1/366). There is data that puts this into doubt, at least in the U.S. – DBFdalwayse Sep 6 '13 at 5:10
• My question is assuming that the distribution is uniform. – Thomas Sep 6 '13 at 5:31
The question linked in the comments, while not your exact question, is very relevant. In particular, reading the answers posted there will tell you that an exact computation for triples is gross. However, Byron Schmuland's answer gives us a useful estimate: $$P(\text{at least one triple with } N \text{ people}) \approx 1 - e^{-{N \choose 3}/365^2}$$ From this we can simply solve for what $N$ makes this value $\frac12$. We need \begin{align*} 1 - e^{-{N \choose 3}/365^2} &= \frac12 \\ e^{-{N \choose 3}/365^2} &= \frac12 \\ e^{{N \choose 3}/365^2} &= 2 \\ {N \choose 3} &= 365^2 \ln 2 \\ N(N-1)(N-2) &= 6 \cdot 365^2 \ln 2 \\ N^3 - 3N^2 + 2N - 6 \cdot 365^2 \ln 2 &= 0 \end{align*}
A computer calculation reveals that the only real root of this cubic polynomial is $N \approx 83.13$. Hence we would expect to need $83$ or $84$ people in the room before having a 50-50 chance of getting three people with the same birthday. | {
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