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https://stats.stackexchange.com/questions/519787/explanation-for-eexyy-exy | # Explanation for E[E[X|Y]|Y]=E[X|Y]
I would like to ask for the proof of $$E[E[X|Y]|Y]=E[X|Y]$$
Per my understanding (for discrete case):
because $$E[X|Y]=g(Y)$$
hence, $$E[E[X|Y]|Y] = E[g(Y)|Y]= \sum_y g(y)*p(y|y)=\sum_y g(y)=\sum E(X|Y=y)$$
I could not arrive to the correct result $$E[E[X|Y]|Y]=E[X|Y]$$, it is very nice if someone can tell me what's wrong in my demonstration.
• You have kinda already arrived at the result you wish to show. Note that $E[g(Y) | Y=y] = g(y)$. Apr 15 '21 at 4:53
• I think the issue is when you say $E[g(Y)|Y] = \sum_y g(y) p(y | y)$. You are summing over all the possible values that $Y$ can take. Any value other than $y$ has conditional probabiltiy of 0, so you are left with $g(y)$ in the end. Apr 15 '21 at 5:10
• @SOULed_Outt I understand that $E[X|Y]$ is a RV which is a function of Y and E[X|Y=y] is a value evaluated at Y=y. Expected value should be evaluated at all possible value of RV, that is why the sum seems to be reasonable for me. Apr 15 '21 at 5:35
• The earlier comment I made was very misleading so I deleted it (apologies). Take a look at what I posted as an answer and let me know if it clears anything up. Apr 15 '21 at 5:37
• Since $E[X|Y]$ is a (measurable) function of $Y$, it is a fixed (deterministic) and known quantity given a realisation of $Y$. Hence, in the probabilistic universe where $Y$ is observed as $y$, it is a constant, equal to its expectation. Apr 15 '21 at 8:04
Suppose, for example, that $$Y$$ could take values $$1,2,$$ or $$3$$ and we want to find $$E[g(Y) | Y =3]$$ for some nice function $$g$$. By defintion of conditional expectation we have \begin{align} E [g(Y) | Y=3] &= \sum_y g(y) \cdot P(Y=y | Y=3) \\ &= g(1) \cdot P(Y=1 | Y=3) + g(2) \cdot P(Y =2 | Y=3) + g(3) \cdot P(Y = 3 | Y = 3) \end{align}
Note that $$P(Y = 1 | Y = 3) = 0$$, $$P(Y = 2 | Y = 3) = 0$$, and that $$P(Y=3 | Y=3)=1$$ so $$E [g(Y) | Y=3] = g(3)$$
I think your problem starts after you introduce the summation. $$p(y|y)$$ is not always equal to 1. It equals zero for any $$y' \neq y$$. In the summation, the $$y$$ in the first slot of $$p(y|y)$$ varies while the $$y$$ in the second slot is fixed (see like in the example above).
Look at what I have highlighted in red. The red $$y$$ serves as a dummy variable. I could replace it with any letter I want so long as I understand that it should represent all possible values that $$Y$$ can take. Let's say I use $$a$$ instead. \begin{align} E [g(Y) | Y=y] &= \sum_{\color{red}{y}} g(\color{red}{y}) \cdot P(Y = \color{red}{y} | Y = y) \\ &= \sum_{a} g(a) \cdot P(Y = {a} | Y = y). \end{align}
Notice that $$P(Y = a | Y= y) = \begin{cases} 1 \text{ if } a = y \\ 0 \text{ if } a \neq y \end{cases}.$$ Therefore $$E [g(Y) | Y=y] = g(y)$$
• It means that the Y under conditional sign (|) will hold constant while the Y above conditional sign (|) will be evaluated at all possible value? I thought that Y should be evaluated at both. For example, if Y takes value in $[1,2,3]$, we have:$\sum_y g(y).P(Y=y|Y=y) = g(1).P(Y=1|Y=1)+g(2).P(Y=2|Y=2)+g(3).P(Y=3|Y=3) = g(1)+g(2)+g(3) = \sum_y g(y)$ Apr 15 '21 at 5:59
• Yes to your question. You should only evaluate the part above the conditional sign. The part under the conditional sign is held constant. Apr 15 '21 at 6:16
• In this case, what is the different between $E[X|Y]$ and $E[X|Y=y]$? As my knowledge, The first term is rv and the second term is the value. Apr 15 '21 at 6:24
• Sometimes they are used interchangeably. I believe most texts on probability say that you should view $E[X | Y]$ as a random variable and you should view $E[X|Y=y]$ as a number that depends on $y$. Apr 15 '21 at 6:36
• That's why I got confused. my knowledge is $E[g(Y)|Y]$ is a r.v, then, when we evaluated it, we should evaluated both g(Y) and Y. Apr 15 '21 at 7:10 | 2022-01-27T07:02:58 | {
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https://math.stackexchange.com/questions/2491318/there-appears-to-be-two-definitions-of-the-direct-sum-of-vector-spaces | # There appears to be two definitions of the direct sum of vector spaces
I have seen two definitions of the direct sum of vector spaces:
1. If $U$ and $V$ are vector spaces, then
$$U \oplus V = \{(u,v):u\in U,v \in V\}.$$
1. If $V_1$ and $V_2$ are subspaces of $V$, and $V_1 \cap V_2 = \{0\}$, then
$$V_1 \oplus V_2 = \{u+v:u\in V_1,v \in V_2 \}$$
So essentially, if I was to take the direct sum of the vector spaces $V_1$ and $V_2$ consisting of vectors of the form $(a,b,0,0)$ and $(0,0,c,d)$ respectively, would it be as space with vectors of the form
$$(a,b,0,0,0,0,c,d)$$
by definition 1, or
$$(a,b,c,d)$$
by definition 2?
• 1 is external direct sum and 2 is internal direct sum. If an internal direct sum exists, it's isomorphic to the external direct sum. – Angina Seng Oct 26 '17 at 19:22
• And the isomorphism is obvious, – Martín-Blas Pérez Pinilla Oct 26 '17 at 19:40
The two notions are the same (up to isomorphism) for vector spaces whose intersection is $\{0\}$. Assume $U$ and $V$ are vector spaces such that $U \cap V = \{0\}$. I'll refer to their external direct sum $(1)$ as $U\times V$ and their internal direct sum $(2)$ as $U \oplus V$.
As mentioned in the comments $U \times V \cong U \oplus V$, and the isomorphism is the natural choice. The only thing that requires a bit of thinking is noticing where the condition $U\cap V = \{0\}$ comes in. The isomorphism is the following:
\begin{align} \phi : U \times V &\to U \oplus V \\ (u,v) &\mapsto u + v \end{align}
It is straightforward to check that this is a homomorphism of vector spaces (a linear map). In addition you need to check that it is bijective. It is immediate that $\phi$ is surjective. To see that it is injective, note that $\ker(\phi) = \{(u,v)\mid u + v = 0\}$. But $u + v = 0 \implies u = -v$. This means that $-v \in U$ and since $U$ is a vector space this implies $v \in U$. By the assumption $U \cap V = \{0\}$ we have that $v = u = 0$. So $\ker(\phi) = \{0\}$, proving that $\phi$ is injective.
• In proving directs sums, if this statement is true ($v_1+v_2=0$ with $v_1 \in V_1$ and $v_2 \in V_2$ $\implies v_1=v_2=0$) how can I prove that $V_1 \cap V_2=\{0\}$ ? – Fareed Abi Farraj Mar 26 '19 at 10:44 | 2021-04-11T12:03:45 | {
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https://math.stackexchange.com/questions/2425503/stopping-rules-for-jacobi-gauss-seidel-iteration | # Stopping Rules for Jacobi/Gauss-Seidel Iteration
Suppose that $A\mathbf{x}=\mathbf{b}$ is a diagonally dominant linear system. We can use iterative methods that produce a sequence of approximations $\mathbf{x}^1,\mathbf{x}^2,\dots$ that converge to $\mathbf{x}$.
Doing this for a small system it seems intuitive to stop when the components $x^i_j$ and $x^{i+1}_j$ differ by a tolerance $\varepsilon$ giving solutions correct to $\varepsilon$.
While playing around with larger systems it was clear that this could stop well before convergence and is a very poor stopping rule (and doesn't give accurate answers as claimed).
Researching a little further I see some slightly more sophisticated stopping rules (in the below the norms might be max norms or 2-norms):
• When $\|\mathbf{x}^{i+1}-\mathbf{x}^i\|< \varepsilon.$
• When $\|\mathbf{x}^{i+1}-\mathbf{x}^i\|< \varepsilon\|\mathbf{b}\|.$
• When $\displaystyle \max_j\left|\frac{x_j^{i+1}-x_j^i}{x_j^{i+1}}\right|<\varepsilon$.
• When the residual $\|\mathbf{r}\|=\|\mathbf{b}-A\mathbf{x}^i\|<\varepsilon$.
• When the residual $\|\mathbf{r}\|=\|\mathbf{b}-A\mathbf{x}^i\|<\varepsilon\|\mathbf{b}\|$.
I am interested in hearing the pros and cons of each. I am hoping to use one that is relatively easy to implement on VBA with perhaps 20 unknowns.
• I have a perhaps stupid question: why do you want to solve a system with 20 unknowns iteratively? – Algebraic Pavel Sep 11 '17 at 22:22
• There are no stupid questions. Approximating the equilibrium temperature distribution on a plate with fixed boundary conditions (i.e. numerical methods of solving Laplace's Equation). – JP McCarthy Sep 12 '17 at 7:50
• Could I ask another question? Why implement in VBA? Is this integrating with some kind of embedded hardware with limited options for programming interface? :) – Erick Wong Sep 13 '17 at 20:03
• @ErickWong this is service teaching for an engineering department. They want us to use VBA. – JP McCarthy Sep 14 '17 at 11:12
• Small remark. Many algorithms (like CG) already compute the residual. So, residual based convergence criteria contain almost no additional computations. – P. Siehr Sep 15 '17 at 9:01
Let's look at forward error bounds.
From normwise analysis we have $$\frac{\|x-x^i\|}{\|x\|} \leq \frac{2\kappa(A)\epsilon_n}{1-\kappa(A)\epsilon_n},\ \ \epsilon_n = \frac{\|b - Ax^i\|}{\|A\|\|x^i\| + \|b\|} \tag{1}$$ where $\|\cdot\|$ is any vector norm and the corresponding subordinate matrix norm, and $\kappa(A) = \|A\|\|A^{-1}\|$ is the matrix condition number with respect to the norm $\|\cdot\|$. Number $\epsilon_n$ is the normwise backward error obtained at $i$-th iteration. This is the only number, that can be controlled in this inequality.
From componentwise analysis we have: $$\frac{\|x-x^i\|_\infty}{\|x\|_\infty} \leq \frac{2\eta(A)\epsilon_c}{1-\eta(A)\epsilon_c},\ \ \epsilon_c = \max_j\frac{|b - Ax^i|_j}{(|A||x^i| + |b|)_j} \tag{2}$$ where $\eta(A) = \||A^{-1}||A|\|_\infty$ is the Skeel condition number and $\epsilon_c$ is the componentwise backward error obtained at $i$-th iteration.
From (1) and (2) we can see, that none of provided stopping rules are particullary good. The best one is the last one since $$\epsilon_n = \frac{\|b - Ax^i\|}{\|A\|\|x^i\| + \|b\|} \leq \frac{\|b - Ax^i\|}{\|b\|} \tag{3}$$ However condition (3) is too conservative, and as consequence the algorithm will perform too many iterations than required to obtain reasonably accurate solution.
Notice, that you can build a good stopping rule directly from (1) as $$\|b - Ax^i\|_\infty \leq (\|A\|_\infty\|x^i\|_\infty + \|b\|_\infty)\times\epsilon \tag{4}$$ since $\|A\|_\infty$ can be easily estimated. Then the tolerance $\epsilon$ is just the normwise backward error.
Since generally componentwise error bounds are much better, than normwise bounds, a stopping rule based on (2) would be better than (4). Unfortunately this would require calculating $|A||x^i|$, which is not always possible.
• So basically the best stopping rules here are the last one I present and also the ones presented here in (4)? – JP McCarthy Sep 19 '17 at 7:56
• Yes. When your last rule show convergence, then it is guaranteed, that desired accuracy is reached. But it is possible, that this rule will show no convergence for accurate enough solution (even for the most accurate solution, that can be obtained by any method). Rule (4) is more precise and robust. – Pawel Kowal Sep 19 '17 at 15:10
• What are $|A|$ and $|x_i|$ in your componentwise error analysis? It is not clear from context what those are. – shuhalo May 1 '19 at 4:55
• For a matrix $A$ with elements $a_{ij}$, $|A|$ is a matrix with elements $|a_{ij}|$. – Pawel Kowal May 1 '19 at 12:11
• So the problem is that $|A| \cdot |x|$ might be too expensive to compute, right? This looks like it can be implemented straightforward though. – shuhalo Sep 18 '19 at 12:02 | 2020-03-29T06:34:06 | {
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https://math.stackexchange.com/questions/3146025/covering-a-board-problem | # Covering a board problem
Take a 10x10 board. We're going to try to cover it with 2x2 squares that can overlap. Let an arrangement where the board is covered but one piece can be removed and the board still be covered be redundant. If one cannot do this, then the arrangement is non-redundant. The smallest non-redundant covering evidently is 25 squares.
(a) Show that there is a non-redundant covering with 35 squares. b) Show that every covering with 55 squares is redundant. c) Can these bounds be improved?
By messing around this seems to work and the bounds do not seem to be those that are above but I have no idea how to prove it.
• is the question telling you to bound it ? – Roddy MacPhee Mar 13 at 8:02
The strategy is what it nearly always is: try to decompose into smaller problems.
Let $$f(m,n)$$ be the number of tiles in the largest non-redundant covering of $$m \times n$$.
Consider an $$m \times 2$$ grid. There are $$m-1$$ places where you can put a tile. The first and the last are necessary, to cover the first pair of corners and the last pair of corners. There can't be three consecutive squares, for then the middle one would be redundant. There can't be two consecutive gaps, for then there would be squares uncovered. So you're looking at strings of the form $$1(101|01)^*1?$$ of length $$m-1$$ and trying to maximise the $$1$$s. So $$f(m,2) = \begin{cases} 1 + \frac23 (m-2) & \textrm{if } m \equiv 2 \pmod 3 \\ 2 + \frac23 (m-4) & \textrm{if } m \equiv 1 \pmod 3 \\ 3 + \frac23 (m-6) & \textrm{if } m \equiv 0 \pmod 3 \\ \end{cases}$$
Now, an $$m \times n$$ grid has an edge of length $$m$$. If you cover that edge, you also cover the row next to it. So you get a bound $$f(m,n) \ge f(m,2) f(n,2)$$.
Is that bound tight? Not necessarily. If there's a $$010$$ in the horizontal pattern and a $$010$$ in the vertical pattern, then where those meet in the 2D product there's a single tile which doesn't overlap any others. You can remove it and replace it with four tiles whose corners meet, increasing the number of tiles by $$3$$. You may also be able to do interesting things by using different patterns for different rows.
I think this gets you an irredundant tiling with $$48$$ tiles for $$10 \times 10$$. An exhaustive consideration of $$4 \times 4$$ grids would allow you to identify all of the opportunities for using the interactions of horizontal and vertical patterns to squeeze in extra tiles.
• Interesting result. Does this prove that there exists a non-redundant covering with 35 squares and that every covering with 55 squares is redundant? – GuauGuau754 Mar 13 at 15:32
• The replacement you talked about can create a redundancy. None of the four tiles added will be redundant, but a nearby tile may become redundant. Your $f(m,2)\times f(n,2)$ method gives a $36$ tile covering, but I do not see how this can be improved . – Mike Earnest Mar 13 at 18:58
• @MikeEarnest Is it not possible to reach the 48 tiles as posted above? – GuauGuau754 Mar 13 at 19:30
• @MikeEarnest, you're right that it's a lot easier than I thought to create a redundancy, but it's not completely guaranteed. I've found a 37-tile covering based on the string 110101011, but I'm tempted to believe that it can't be improved further. – Peter Taylor Mar 13 at 20:02
• Can you post that 37 tile covering? I can't find it. – GuauGuau754 Mar 13 at 22:29
Here is a non redundant covering with $$36$$ squares. Each square is labeled with the number of tiles covering it.
1 1 1 2 1 1 2 1 1 1
1 1 1 2 1 1 2 1 1 1
1 1 1 2 1 1 2 1 1 1
2 2 2 4 2 2 4 2 2 2
1 1 1 2 1 1 2 1 1 1
1 1 1 2 1 1 2 1 1 1
2 2 2 4 2 2 4 2 2 2
1 1 1 2 1 1 2 1 1 1
1 1 1 2 1 1 2 1 1 1
1 1 1 2 1 1 2 1 1 1
Here is how you can obtain a $$35$$ tile solution. Start with the $$27$$ tiles shown below. There are $$8$$ uncovered white areas, which should be covered with $$8$$ additional tiles which overhang the white areas either to the left or below (or both).
To prove $$55$$ tiles are redundant, use the pigeonhole principle. There will be $$27$$ holes. Each "hole" is a $$2\times 4$$ rectangle whose lower left coodinate is $$(i,j)$$ and whose upper right coordinate is $$(i+4,j+2)$$, where $$i$$ ranges over the set $$\{0,3,6\}$$ and $$j$$ ranges over $$\{0,1,2,3,4,5,6,7,8\}$$. Each tile falls in one of these holes. If there are $$55$$ tiles, then some hole must have $$\lceil 55/27\rceil=3$$ tiles. But when there are three tiles in a $$2\times 4$$ rectangle, the middle tile is redundant.
I do not know if $$55$$ can be improved.
• Credits to Peter Taylor for coming up with the $36$ tile solution. – Mike Earnest Mar 13 at 19:15
• It seems to be the bound as it is one more than twice the number of holes. – GuauGuau754 Mar 13 at 19:27
• Also, how would I prove that there is a non-redundant covering with 35 squares? – GuauGuau754 Mar 13 at 19:31 | 2019-09-22T20:58:51 | {
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https://electronics.stackexchange.com/questions/424395/per-unit-system-what-details-am-i-missing | Per Unit System: what details am I missing
Ok, so I had a problem about a three-phase system with the following data
$$V=220kV$$
$$S=150MVA$$
$$\cos \phi=0.85$$
where V is the phase-phase voltage, S is apparent power, and $$\\phi \$$ is positive (inductive load).
Ok I was asked to find out the phase-neutral voltages, the phase-phase voltages and the currents.
I had no trouble doing it so I just post the outline of my work.
Phase-neutral voltages
$$V_a = \frac{220}{\sqrt{3}} e^{j0}$$ $$V_b = \frac{220}{\sqrt{3}} e^{-j120^o}$$ $$V_c = \frac{220}{\sqrt{3}} e^{j120^o}$$
Phase-phase voltages
$$V_{ab} = V_a-V_b= 220 e^{j30^o}$$ $$V_{bc} = V_b-V_c= 220 e^{-j90^o}$$ $$V_{ca} = V_c-V_a= 220 e^{j150^o}$$
Currents
$$I_{a} = \frac{S}{\sqrt{3}V} e^{-j31.7^o}= 393.6 e^{-j31.7^o}$$ $$I_{b} = 393.6 e^{-j151.7^o}$$ $$I_{c} = 393.6 e^{j88.3^o}$$
And the impedance for each load (a,b and c) is $$Z= \frac{V_a}{I_a}= 322.7 e^{j31.7^o}$$
Now they say for me to imagine that the system becomes unbalanced with b becoming 1.1Z and c becoming 0.9Z
$$I_n= I_a+ I_b + I_c = \frac{V_a}{Z} + \frac{V_b}{1.1Z} + \frac{V_c}{0.9Z} = 68.5 e^{j61.8^o}$$
Now I was asked to redo this using the pu system with given base values
$$V_{phase-phase-base}=220kV$$
$$S_{base}=100 MVA$$
This lead me to
$$V_{phase-neutral-base}=220/\sqrt{3} kV$$ $$I_{base}= \frac{S_{base}}{\sqrt{3}V_{phase-phase-base}}=262.4A$$ $$Z_{base}= \frac{V_{phase-phase-base}}{\sqrt{3}I_{base}}=484 \Omega$$
For phase-neutral voltages I have to divide my previous value by $$\ V_{phase-neutral-base}\$$ leading to
$$V_a = 1 e^{j0}$$ $$V_b = 1 e^{-j120^o}$$ $$V_c = 1 e^{j120^o}$$
As for phase-phase voltages, I divide by $$\V_{phase-phase-base}\$$ leading me to
$$V_{ab} = 1 e^{j30^o}$$ $$V_{bc} = 1 e^{-j90^o}$$ $$V_{ca} = 1 e^{j150^o}$$
Now my question is: why if I apply the formulas $$V_{ab} = V_a-V_b$$ $$V_{bc} = V_b-V_c$$ $$V_{ca} = V_c-V_a$$ using pu values the equations don't verify. Is it because I use different base values for each and this is just a question of proportionality?
However, then I continued my problem and obtained the rest of the values:
$$S=\frac{150}{100}=1.5$$
$$I_{a} = \frac{S}{V} e^{-j31.7^o}= 1,5 e^{-j31.7^o}$$ $$I_{b} = 1.5 e^{-j151.7^o}$$ $$I_{c} = 1.5 e^{j88.3^o}$$
Everything looks fine here, because if I multiply the currents by $$\I_{base}\$$ I obtain the values I did before.
$$Z= \frac{V_a}{I_a}= \frac{1 e^{j0}}{1,5 e^{-j31.7^o}} = 0.667 e^{j31.7^o}$$
Then again, multiplying by $$\Z_{base}\$$ we obtain the value before.
Finally,
$$I_n= I_a+ I_b + I_c = \frac{V_a}{Z} + \frac{V_b}{1.1Z} + \frac{V_c}{0.9Z} = \frac{1 e^{j0}}{0.667 e^{j31.7^o}} + \frac{1 e^{-j120^o}}{1.1 \times 0.667 e^{j31.7^o}} + \frac{1 e^{j120^o}}{0.9 \times 0.667 e^{j31.7^o}} = 0.261 e^{j61.8^o}$$
And multiplying by $$\I_{base}\$$ I obtain the same value as before.
So in this last calculations using the pu values works pretty well and I obtain values equivalent to the calculations I did before. The only calculation that's failing is the one about the voltages phase-phase. Why does this happen? What is the mathematical subtlety I am missing here?
Thanks!
1 Answer
Your work is good. You are using 2 different voltage bases to get your results (pasted in below). That is why they are different in per unit.
If you take the results of your calculation using the phase-neutral voltages (Va-Vb) and convert them to the ph-ph voltage base you used, then they will agree. In other words, the results of Va - Vb is 1.73 per unit in magnitude. That is on a voltage base of 220kV/1.732. To convert that per unit result to the ph-ph base you used above, multiply by (220kV/1.732 divided by 220kV).
The ph-ph voltage base is root 3 larger than the phase-neutral. When working problems - do not mix. Just use one voltage base and carry on. That way, when you convert back to actual volts you won't make a mistake.
UPDATED: Showing base calculations.
• Hello! Thank you for your answer. Ok, I think I get it: because, even though some are phase-phase and others are phase-neutral, they are still voltages, and therefore, the same base must be used; however in my other calculations, I'm computing currents and impedances, and therefore, adding components in phase-phase base will produce me coherent results, since the current base and the impedance base were derived from the postulated voltage phase-phase base. – Granger Obliviate Mar 1 '19 at 1:53
• Granger - i updated my answer to show base calculations. When you pick a ph-ph voltage base you automatically set the ph-neutral base because they are related by root 3. So, if you calculate the current and impedance base quantities with either approach you will get the same results. – relayman357 Mar 1 '19 at 15:59 | 2020-01-20T23:21:11 | {
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http://math.stackexchange.com/questions/205762/prove-that-the-set-of-all-quadratic-functions-whose-graphs-pass-through-the-orig | # Prove that the set of all quadratic functions whose graphs pass through the origin with the standard operations is a vector space.
The idea is to prove that this is a vector space based upon the following axioms:
1. $\mathbf{u}+\mathbf{v}$ is in $V$. Closure under addition.
2. $\mathbf{u}+\mathbf{v}=\mathbf{v}+\mathbf{u}$. Commutative property.
3. $\mathbf{u}+(\mathbf{v}+\mathbf{w}) = (\mathbf{u}+\mathbf{v})+\mathbf{w}$. Associative property.
4. $V$ has a zero vector $\mathbf{0}$ such that for every $\mathbf{u}\in V$, $\mathbf{u}+\mathbf{0}=\mathbf{u}$. Additive identity.
5. For every $\mathbf{u}\in V$, there is a vector in $V$ denoted by $-\mathbf{u}$ such that $\mathbf{u}+(-\mathbf{u}) = \mathbf{0}$. Additive inverse.
6. $c\mathbf{u}$ is in $V$. Closure under scalar multiplication.
7. $c(\mathbf{u}+\mathbf{v}) = c\mathbf{u}+c\mathbf{v}$. Distributive property.
8. $(c+d)\mathbf{u}=c\mathbf{u}+d\mathbf{u}$. Distributive property.
9. $c(d\mathbf{u})= (cd)\mathbf{u}$. Associative property.
10. $1(\mathbf{u}) =\mathbf{u}$. Scalar identity.
The exact problem: prove that the set of all quadratic functions whose graphs pass through the origin with standard operations is a vector space. I specifically have two questions in addition to proving the vector space. One, how does passing through the origin affect the supposed vector space - my assumption is that it affects axiom one? And two, how would the vectors be described?
Any help is appreciated.
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Not quite true, unless you think of $3x$ as quadratic function. This is because for example $x^2+x$ is a quadratic that passes through the origin, as is $-x^2+2x$. Their sum is $3x$. So we will have to allow all things of shape $ax^2+bx$, where $a$ and/or $b$ may be $0$. – André Nicolas Oct 2 '12 at 2:21
That would be that this is not a vector space, as $3x$ is not quadratic. Am I correct in saying that? – Rick Oct 2 '12 at 2:24
The set of all polynomials of degree $\le 2$, under usual addition, is also a vector space over the reals. Just a different vector space than yours. Added Yes, if you keep $3x$ out, then the closure under addition condition is violated. If you keep the $0$ polynomial out, you lose closure under multiplication by scalars. – André Nicolas Oct 2 '12 at 2:24
The problem may have been carelessly posed. Your answer could be: If quadratic functions are really intended, meaning coefficient of $x^2$ cannot be $0$, then we do not have a vector space because $\dots$. If we allow all $ax^2+bx$, then we do get a vector space because $\dots$. Then you need to verify all the axioms. All the verifications are easy. – André Nicolas Oct 2 '12 at 2:29
OK, but it may be a trick question, so you should point out that if we don't allow $a=0$, or even $a=b=0$, we don't have vector space. – André Nicolas Oct 2 '12 at 2:35
Whether or not we have a vector space depends on how you interpret "quadratic function whose graph passes through the origin."
In order for us to have a vector space, we will need, for example, to think of $3x$ as a quadratic function. For certainly $x^2+x$ is a quadratic function that passes through the origin, as is $-x^2+2x$. So if we are to have closure under addition, the function $(x^2+x)+(-x^2+2x)$, that is, $3x$, will have to be in the collection.
Indeed the identically $0$ function has to be in our collection, for two reasons. If we do not allow it, then closure under addition can fail, since $(x^2-x)+(-x^2+x)=0$. Closure under multiplication by the constant $0$ also fails.
It is not difficult, however, to show that the set of functions of the form $ax^2+bx$, where $a$ and $b$ range over all the reals, is a vector space over the reals. Quite a number of axioms need to be verified, but each verification is easy.
As to your question about the "passing through $0$" part, one can also show that the set of all polynomial functions of degree $\le 2$ also is a vector space over the reals. It just is a different vector space than the one under consideration.
Remark: $1.$ In answering the homework question, it may be useful to be cautious. I would suggest doing the following: (i) Observe that if we interpret "quadratic" as meaning that the coefficient of $x^2$ is non-zero, then we do not have a vector space. (Of course one should explain why.) (ii) Show in detail that the collection of all functions of the shape $ax^2+bx$ is a vector space. Many of the steps can be possibly omitted. The key properties that have to be verified are closure under addition and under multiplication by scalars.
$2.$ You asked how to describe the vectors. Might as well use the standard polynomial notation, as in the answer above. The "pass through the origin" part just means the constant term is $0$.
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https://resellernews.pl/road-to-iuprig/1cbc2f-example-of-two-planes-intersecting | b. A three-dimensional geometric surface is called space. 21 days ago. Interesting diagonals of a circle, sphere, etc 4. Angle between Two Planes formula in Cartesian System, Consider the angle between two planes example in which two planes intersecting at an angle θ. Play this game to review Geometry. The intersection of two different planes is a line. The vector equation for the line of intersection is calculated using a point on the line and the cross product of the normal vectors of the two planes. When two planes are described in cartesian coordinate system as A1x + B1y + C1z + D1 = 0 and A2x + B2y + C2z + D2 = 0, the cosine of angle of separation between the two planes is given as: Cosθ = |A₁A₂ + B₁B₂ + C₁C₂| / [√(A₁²+B₁²+C₁²).√(A₂²+B₂²+C₂²)]. Relevance ? 63% average accuracy. Edit. However, a plane is a two dimensional surface with zero thickness. 0. A shelf on a wall seems like the obvious answer. Sorry!, This page is not available for now to bookmark. A line is a one dimensional surface and a space is a three dimensional surface. Which is a real world example of two planes intersecting? 1. The intersection is where the shelf touches the wall. Sketch a plane and a line that is in the plane. The steps on our stairs are also examples of parallel planes. The equations of these two planes are given in the cartesian coordinate system as A1x + B1y + C1z + D1 = 0 and A2x + B2y + C2z + D2 = 0. Finding the Point of Intersection of Two Lines Examples : If two straight lines are not parallel then they will meet at a point.This common point for both straight lines is called the point of intersection. Not really. 3D coordinate plane. Sketch a plane and a line that does not intersect the plane. Answer Save. Determine the line of intersection of these two planes. The intersection is where the shelf touches the wall. The figure below depicts two intersecting planes. When three planes intersect orthogonally, the 3 lines formed by their intersection make up the three-dimensional coordinate plane. Notice how these two planes intersect. Edit. r = rank of the coefficient matrix. How are Geometric Surfaces Categorized Based on Their Dimensions? x=-2-6t y=2t z=-4t. Intersection between the light green line and the blue line: the intersection point is calculated after the blue line is extrapolated : Intersection between the pink line and the blue line: the intersection is calculated as the mid-point of minimum distance between the two lines Favorite Answer. = 0, the cosine of angle of separation between the two planes is given as: Difference Between Dealer and Distributor, Difference Between Environment and Ecosystem, Vedantu (a) Give an example of three planes that have a common line of intersection (Figure 2.4) (b) Give an example of three planes that intersect in pairs but have no common point of intersection (Figure $2.5)$ (c) Give an example of three planes, exactly two of which are parallel (Figure 2.6 ). Yes, you guessed it right. 1 decade ago. Here is a hands-on visual homework check I have my students do as part of one of their graded assignments. Two lines are parallel to each other if they are perpendicular to the same plane. Pro Lite, Vedantu Geometric surfaces may have one, two or three dimensions. a. They are two intersecting lines. 2. shelf on a wall. leec_39997. 1. The equations of these two planes are given in the cartesian coordinate system as A1x + B1y + C1z + D1 = 0 and A2x + B2y + C2z + D2 = 0. Shelf on a wall. About Pricing Login GET STARTED About Pricing Login. Florida governor accused of 'trying to intimidate scientists', Ivanka Trump, Jared Kushner buy \$30M Florida property, Another mystery monolith has been discovered, 'B.A.P.S' actress Natalie Desselle Reid dead at 53, MLB umpire among 14 arrested in sex sting operation, Goya Foods CEO: We named AOC 'employee of the month', Young boy gets comfy in Oval Office during ceremony, Packed club hit with COVID-19 violations for concert, Heated jacket is ‘great for us who don’t like the cold’, COVID-19 left MSNBC anchor 'sick and scared', Former Israeli space chief says extraterrestrials exist. (After all, the walls would likely fall … Two planes? In geometry, it is important to know about various kinds of surfaces. Example 1 - Two planes intersection Find the intersection line equation between the two planes: 3x − y + 2z − 4 = 0 and 2x − y + 4z − 3 = 0 Find the intersection line equation between the two planes: The point of intersection is P. The figure above also shows intersecting lines at different angles. 0. I found the line of intersection of two planes and found the parametric equations for that line of intersection. The two arms of an angle. They extend along with the same space (our home) but these two planes will never meet. In these two equations, A1, B1 and C1 are the direction ratios of normal to the plane described by the equation A1x + B1y + C1z + D1 = 0 and A2, B2 and C2 are the direction ratios of normal to the plane defined by the equation A2x + B2y + C2z + D2 = 0. A railroad intersection would be an example of two lines intersecting. ceiling and wall intersect in a straight line, floor and wall also intersect in a straight line, so from what you are asking a shelf and a wall. A plane is a two-dimensional geometrical surface. How do you plot the line of intersection between two planes in MATLAB. How to Calculate Angle Between Two Planes? What are some real-world examples of parallel planes? For the general case, literature provides algorithms, in order to calculate points of the intersection curve of two surfaces. A point is a dimensionless geometric surface. 1. Example showing how to find the solution of two intersecting planes and write the result as a parametrization of the line. The cosine of the angle between the two planes is given as: Cosθ = $\frac{A_{1}A_{2}+B_{1}B_{2}+C_{1}C_{2}}{\sqrt{A_{1}^{2}+B_{1}^{2}+C_{1}^{2}}.\sqrt{A_{2}^{2}+B_{2}^{2}+C_{2}^{2}}}$. Let n1 and n2 be the normal vectors drawn to the planes. For example my parametric equations I found for the line of intersection of the planes, 2x + 10y + 2z= -2 and 4x + 2y - 5z = -4 are. The equations of these two planes are given in the cartesian coordinate system as A, are the direction ratios of normal to the plane described by the equation A, are the direction ratios of normal to the plane defined by the equation A. Intersection of Two Planes Given two planes: Form a system with the equations of the planes and calculate the ranks. ParallelAngleBisector. Non-intersecting Lines. We can find the equation of the line by solving the equations of the planes simultaneously, with one extra complication – we have to introduce a parameter. Cosine of angle between two intersecting planes is given as the cosine of the angle between their normals: This is the angle between two planes formula when normal vectors are given. Sketching Intersections of Lines and Planes a. The relationship between the two planes can be described as follows: Position r r' Intersecting 2… If two planes are not parallel nor coincident, then they must intersect along a line. Good luck. How to calculate angle between two planes described by the equations 2x + 4y - 4z - 6 = 0 and 4x + 3y + 9 = 0? 2. Example: Find a vector equation of the line of intersections of the two planes x 1 5x 2 + 3x 3 = 11 and 3x 1 + 2x 2 2x 3 = 7. intersections DRAFT. Answered: KSSV on 31 May 2019 Hey, my planes are: >> Z1 = 2 -X + Y; >> Z2 = (X + 3*Y -3)/2; I've plotted them, but I'm not sure how to plot the line of intersection between them. pin in a pin coushin . Shelf on a wall. When two planes intersect, the angle of separation of the planes is equal to the angle between the normals drawn to the planes. 0 ⋮ Vote. 21 days ago. intersections DRAFT. Planes have certain special properties which include: Any two distinct planes are either parallel or intersect at a line. 0. m A n q The intersection of two different lines is a point. Therefore, the line Kl is the common line between the planes A and B… The two dimensions of a plane are length and width. A sheaf of planes is a family of planes having a common line of intersection. A line may either lie within the plane or intersect the plane at a single point or parallel to the plane. 16 times. When two planes intersect, the angle of separation of the planes is equal to the angle between the normals drawn to the planes. Any two consecutive sides of a polygon intersecting at respective vertex; in triangle, quadrilateral, pentagon, hexagon, etc. First we read o the normal vectors of the planes: the normal vector ~n 1 of x 1 5x 2 +3x 3 = 11 is 2 4 1 5 3 3 5, and the normal vector ~n 2 of 3x 1 +2x 2 2x 3 = 7 is 2 4 3 2 2 3 5. Now click the circle in the left menu to make the blue plane reappear. Vote. Step-by-step math courses covering Pre-Algebra through Calculus 3. Planes p, q, and r intersect each other at right angles forming the x-axis, y-axis, and z-axis. 2. Angle between Two Planes formula in Cartesian System Consider the angle between two planes example in which two planes intersecting at an angle θ. Example of Intersecting Planes In the above figure, the two planes A and B intersect in a single line Kl. Let us consider two planes intersecting at an angle θ as shown in the above figure. Plane is a two-dimensional geometric surface. The shelf represents one plane and the wall represents another. Intersection of Two Planes Example 23. The simplest case in Euclidean geometry is the intersection of two distinct lines, which either is one point or does not exist if the lines are parallel. Planes: 2x + y – 2z = 5 and 3x – 6y – 2z = 15. b. Railroad. 9th - 12th grade. 4 Answers. Consider the angle between two planes example in which two planes intersecting at an angle θ. Preview this quiz on Quizizz. Some examples of intersections are shown below. Get your answers by asking now. If I have 5 independent groups and us a likert scale what stat analysis would I use. The Second and Third planes are Coincident and the first is cutting them, therefore the three planes intersect in a line. Save. If the equations of two intersecting straight lines are given then their intersecting point is obtained by solving equations simultaneously. Played 16 times. A plane is formed by a stack of lines arranged side by side. Solution for How do you find the intersection of two lines in space? Write the parametric equations for this line, showing all work. GET STARTED. But how do I find a point on that line of intersection that is closest to a certain point that is given to me in the problem? Still have questions? The coordinates of normal vector n1 is (2, 4, -2), The coordinates of normal vector n2 is (6, -8, -2), Cosθ = $\frac{(2,4,-2)(6,-8,-2)}{|n1||n2|}$, $\frac{2\sqrt{39}}{\sqrt{4+16+4}.\sqrt{36+64+4}}$ = $\frac{2\sqrt{39}}{39}$ = $\frac{2}{\sqrt{39}}$. It is to be noted that: Non-intersecting lines can never meet. The equation for both the planes is thus given as: $\overrightarrow{r}$.$\overrightarrow{n_{1}}$ = $d_{1}$, $\overrightarrow{r}$.$\overrightarrow{n_{2}}$ = $d_{2}$. =) 1 0. To find the symmetric equations that represent that intersection line, you’ll need the cross product of the normal vectors of the two planes, as well as a point on the line of intersection. none of em. Pro Lite, Vedantu 3. Two or more lines that do not intersect each other are called non-intersecting lines. The analytic determination of the intersection curve of two surfaces is easy only in simple cases; for example: a) the intersection of two planes, b) plane section of a quadric (sphere, cylinder, cone, etc. Give examples. In this video we look at a common exercise where we are asked to find the line of intersection of two planes in space. Good luck. c. Sketch a plane and a line that intersects the plane at a point. Vedantu academic counsellor will be calling you shortly for your Online Counselling session. What is the intersections of plane AOP and plane PQC? = 0. Use the CalcPlot3D applet to display these two planes. Follow 132 views (last 30 days) Behbod Izadi on 31 May 2019. A real-world example of two planes intersecting is constructing a wall or placing a fence into the ground (so that its bottom is underground “penetrating” the level ground), and therefore the vertical plane (the wall or fence) intersects the horizontal plane, (the level ground). A one-dimensional geometric surface is called a line. Pro Lite, CBSE Previous Year Question Paper for Class 10, CBSE Previous Year Question Paper for Class 12. Right-click and rotate the figure again if needed. Planes that are not parallel and intersect along a line are called intersecting planes. The three dimensions of measuring a space are length, width, and height. Anonymous. Two planes are also parallel to each other if they are perpendicular to the same line. Or walls are being constructed inside a home or other building, and the walls intersect the floor(s) and roof. by leec_39997. The dotted line represents the line of intersection of these two planes. Each edge formed is the intersection of two plane figures. In geometry, an intersection is a point, line, or curve common to two or more objects (such as lines, curves, planes, and surfaces). Try it out! If we take the parameter at being one of the coordinates, this usually simplifies the algebra. The angle of separation of two intersecting planes is calculated as the angle of separation of normals to both the planes. When two planes are described in cartesian coordinate system as A. 1 decade ago. Print ; As shown in the diagram above, two planes intersect in a line. A line and a plane? If two planes intersect each other, the curve of intersection will always be a line. Simultaneous linear differential equations question...? The way to obtain the equation of the line of intersection between two planes is to find the set of points that satisfies the equations of both planes. Cosθ = $\frac{|2X4+4X3+(-4)X0|}{\sqrt{2^{2}+4^{2}+(-4)^{2}}.\sqrt{4^{2}+3^{2}+0^{2}}}$, Cosθ = $\frac{|8+12+0|}{\sqrt{4+16+16}.\sqrt{16+9+0}}$, Cosθ = $\frac{20}{\sqrt{36}.\sqrt{25}}$ = $\frac{20}{6X5}$ = $\frac{2}{3}$. You DO know what a plane is, don't you? ), c) intersection of two quadrics in special cases. Then deselect the green & red planes by clicking on the corresponding circles in the left menu. The line has only one dimension which is called length. =). It is formed by stacking the straight lines one next to the other densely. for example, to find equation of a plane of a sheaf which passes, examples example 2: no points of intersection determine if the line i described by the symmetric equations intersects the plane 7r : 3m — 5z. Join Yahoo Answers and get 100 points today. r'= rank of the augmented matrix. What is the intersections of plane AOP and plane PQC? Thus far, we have discussed the possible ways that two lines, a line and a plane, and two planes can intersect one another in 3-space_ Over the next two modules, we are going to look at the different ways that three planes can intersect in IR3 . Each step extends along … Finding the Line of Intersection of Two Planes . 3. How to calculate angle between two planes when the direction vectors of normals of the planes are given as n1 = 2i + 4j - 2k and n2 = 6i - 8j - 2k. Our homes’ ceilings and floors are great examples of parallel planes. It is formed by stacking the straight lines one next to the other densely. In this example, the planes are x + 2y + 3z = -4 and x - y - 3z = 8. Point is a dimensionless geometric shape. Mathematics. The planes : 6x-8y=1 , : x-y-5z=-9 and : -x-2y+2z=2 are: How do you solve a proportion if one of the fractions has a variable in both the numerator and denominator? The shelf represents one plane and the wall represents another. If two planes intersect each other, the intersection will always be a line. Two equal spheres, intersecting at 120°, will require - I U j x _ a 3 a4(a 7 2 x) a3 a4(a+2x)] (II) 2 - _ 2 y a 271 3 271 +2Y2 3 2720 ' with a similar expression for cylinders; so that the plane x=o may be introduced as a boundary, cutting the surface at 60°. Lv 5. The figure shows an example of 4 intersecting lines. 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https://math.stackexchange.com/questions/3557252/can-the-same-base-same-exponent-rule-be-extended-to-sums/3557310 | Can the “ same base, same exponent ” rule be extended to sums?
By " same base, same exponent " rule I mean :
$$b^x = b^n \iff x=n$$
Can the equation: $$2^x+4^x = 6$$ be solved by saying $$2^x+4^x = 6 = 2+4= 2^1 + 4^1$$ and consequently, that $$x = 1?$$
• Somebody please write an answer containing the word "injective". – Torsten Schoeneberg Feb 23 at 16:57
• @TorstenSchoeneberg as requested see my answer. – qwr Feb 23 at 23:57
• Although this method isn't guaranteed to find the full solution set of a general equation of the desired shape, I think that it is important to mention that you have correctly found, and proven the correctness of, a solution (which happens to be the only one, though you haven't shown it). – LSpice Feb 24 at 0:40
7 Answers
If one of the bases is less than one, you can get, for example $$0.5^x+2^x =2.5$$ which has two solutions $$x=1$$ and $$x=-1$$
A perhaps more worrisome example is that $$(-1)^a = (-1)^b$$ has infinitely many solutions in integers. The equation holds if $$a$$ and $$b$$ are both even and it holds if $$a$$ and $$b$$ are both odd. This is very far from requiring $$a = b$$.
No.
It's better: $$(2^x)^2+2^x-6=0$$ or $$(2^x-2)(2^x+3)=0$$ or $$2^x=2,$$ which gives $$x=1.$$ Also, you can say that $$2^{2x}+2^x$$ increases, which by your work gives $$x=1$$ again.
• wow what a method lol – DDD4C4U Feb 24 at 6:39
As a complement of the other answers:
consider the function $$f(x)=a^x+b^x$$ with $$a>0$$ and $$b>0$$. Assume for simplicity that $$a\neq 1$$ and $$b\neq 1$$. There are a few cases to consider:
1. If $$a>1$$ and $$b>1$$, then $$a^x$$ and $$b^x$$ are increasing, so $$f(x)=a^x+b^x$$ is increasing. As a consequence, $$f$$ is one-to-one so, for every constant $$c$$, the equation $$f(x)=c$$ has at most one solution. Precisely, the range of $$f$$ is $$(0,\infty)$$ so it has exactly one solution when $$c>0$$ and $$0$$ otherwise.
This case applies to the equation $$2^x+4^x=6$$ which has only one solution, $$x=1$$.
2. Similarly, when $$0 and $$0, the function $$f(x)=a^x+b^x$$ is decreasing with range $$(0,\infty)$$, so every equation $$f(x)=c$$ has exactly one solution when $$c>0$$ and $$0$$ otherwise.
3. When $$a>1$$ and $$0, there are two or zero solutions to $$f(x)=c$$. The simple case is when $$b=\frac{1}{a}$$, since in that case, $$f$$ is even.
More generally, $$f'(x)=\ln(a)a^x + \ln(b)b^x$$. Since $$\ln(a)>0$$, $$\ln(b)<0$$, $$a^x$$ is increasing and $$b^x$$ is decreasing, then $$f'$$ is increasing. We have $$\lim_{x\to\pm\infty} f'(x)=\pm\infty$$, so $$f$$ is decreasing on $$(-\infty,\alpha]$$, then increasing on $$[\alpha,\infty)$$, where $$\alpha$$ is the solution of $$f'(x)=0$$ (precisely, $$\alpha=\ln\left(\frac{b}{a}\right)\ln\left(\frac{\ln(1/b)}{\ln(a)}\right)$$).
So, if $$c>f(\alpha)$$, then the equation $$f(x)=c$$ has exactly two solutions.
Well, this is one way to obtain a solution, especially if you don't know any other way to proceed -- we usually say the solution has been obtained by observation.
However, the drawback with this method is that it may not work with a sufficiently complicated equation. Also, it doesn't guarantee that one has found all the solutions -- although one may ascertain this by other means.
In this case you can solve in a more sure way, as one of the answers has demonstrated, by noticing that this is quadratic in $$2^x.$$
But once again, observation is one way to proceed when nothing else works.
You should be careful because your "rule" does not always hold.
It is always true that $$x = y \implies b^x = b^y$$ because the function $$f_b(x) = b^x$$ is well-defined (assume we are working with positive real numbers).
It is not always the case of the converse, $$b^x = b^y \implies x = y$$. When $$f(x) = f(y) \implies x = y$$, the function is injective.
Observe for example the function $$f_1(x) = 1^x$$ is not injective since $$1^1 = 1^2$$ but $$1 \ne 2$$. However, for $$b \ne \pm 1$$ (ignoring the case $$b=0$$), $$f_b(x) = b^x$$ is injective (try to prove this).
• While we're being picky, the function $f_b$ can fail to be injective even for $b \notin \{0, \pm1\}$, but not if $b$ is real. You do say something about "assume we are working with positive real numbers", but then you consider $b = -1$, so it's not entirely clear what you mean. (Certainly, for $b \in \mathbb R$ and $b < 0$ there's no canonical way to define your function $f_b$ even on $(0, \infty)$.) – LSpice Feb 24 at 0:44
• Yes I was being sloppy. For simplicity consider only integer exponents. Then I think everything is defined nicely. – qwr Feb 24 at 5:18
The term "solve" generally means a systematic method of finding solutions, and showing that they are exhaustive (that is, no further solutions exist). This method fails on both counts.
First, while for the particular example you give, it's not too difficult to see that 1 is a solution, I don't see any way to generalize this to a general method of solving equations of this type. It appears to just be guess and check, and while guess and check can be considered to be a method of solving in the broadest sense, it's not what people usually mean by the term.
Second, this doesn't show that 1 is the only solution. There are cases where you have $$x^a+y^a = x^b+y^b$$ but $$a \neq b$$. One situation where that can happen is if $$y = x^{-1}$$. Then $$x^{-a}+y^{-a} = (x^{-1})^a +(y^{-1})^a = y^a+x^a = x^a+y^a$$. In fact, your original statement that $$(x^a = x^b) \rightarrow (a = b)$$ holds only for $$x$$ positive real and $$a, b$$ real.
• To solve an equation simply means to find its solutions. There's nothing about the systematicity or generalisability of the method implied in this. – Allawonder Feb 24 at 17:28 | 2020-05-31T09:36:38 | {
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https://gateoverflow.in/49/gate2012_17 | 1.5k views
Let $G$ be a simple undirected planar graph on $10$ vertices with $15$ edges. If $G$ is a connected graph, then the number of bounded faces in any embedding of $G$ on the plane is equal to
(A) 3
(B) 4
(C) 5
(D) 6
retagged | 1.5k views
Does bounded faces mean cycles?
For any planar graph,
n(no. of vertices) - e(no. of edges) + f(no. of faces) = 2
f = 15 - 10 + 2= 7
number of bounded faces = no. of faces -1
= 7 -1
= 6
So, the correct answer would be (**D**).
selected by
"number of bounded faces = no. of faces -1" is this formula ?
number of bounded faces = no. of faces -1 (external or unbounded face)
Number of edges in minimally connected graph: n-1
So, 10-1=9 (edges used to connect all vertices)
Remaining 15-9=6 edges can be used to connect any two vertices and form a bounded face.
So ans - (d) 6
Is this analogy correct? | 2018-01-19T10:17:03 | {
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https://math.stackexchange.com/questions/442523/on-the-definition-of-surjectivity | # On the Definition of Surjectivity
So I'm told that for a function $f:X\rightarrow Y$ to be surjective then
$$\forall y\in Y~,~\exists x\in X~,~f(x)=y,$$
so does this "$\exists$" imply more than one such $x$ can hit the same $y$? That is, am I right in thinking that
$$\forall y\in Y~,~\exists ! x\in X~,~f(x)=y$$
is an incorrect definition of surjectivity?
• You are absolutely correct. Jul 13, 2013 at 4:05
That's right. Surjectivity of a function $f: X \to Y$ does not require that for all $y\in Y$, there exists a unique $x \in X$ such that $f(x) = y \in Y$.
We can even have $X = \mathbb N$, $\;Y = \{1\},\;$ with $f: X\to Y$, such that $f(x) = 1\; \forall x \in X$. And indeed, $f$ is thereby surjective.
Yes, you are correct. If the $x$ is unique, then $f$ is not only surjective, but it is also injective. So your second definition is actually the definition of a bijective function. Since there exist functions that are surjective but not injective (see amWhy's answer for an example), the second definition is incorrect (it is "too strong" of a definition). | 2022-08-15T04:44:57 | {
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https://math.stackexchange.com/questions/1953628/20-people-are-sitting-around-a-circular-table-how-many-ways-can-we-choose | # $20$ people are sitting around a (circular) table. How many ways can we choose $3$ people, no two of whom are neighbors?
0 choices for the 1st person. 17 choices for the 2nd person (must exclude 1st and his/her two neighbours)
For 2 of these choices of 2nd person, there is one shared neighbour, so 15 remaining choices. (e.g. if they are numbered 1 to 20 in a circle, 1st person is #1, 2nd is #3, then people 20,1,2,3,4 are excluded). For the other 15 choices of 2nd person, there are no shared neighbors, so 14 remaining choices.
So if order matters, total is $20 \cdot (2 \cdot 15 + 15 \cdot 14)$ but since order does not matter, divide by $3! = 6$ to account for the permutations in order of the 3 people. So total = $20 \cdot \frac{2 \cdot 15 + 15 \cdot 14}{6}$
just redid it; does this make any sense?
• What is the answer with a straight table and how does turning the table into a corner change the problem? thats the step that you need to account for Oct 4, 2016 at 16:14
• first person - 20 choices , second person 17 choices , then third person 14 choices - then you could have chosen the 3 people in a number of different ways (which is?) This is my theory, although I got the 17 choices for 2 from you - then I think you've over complicated it, you are always left with 14 onthe third choice
– Cato
Oct 4, 2016 at 16:14
• @AndrewDeighton It might be 15 choices for the third person if the second is one of the two peoples closest to the first. Oct 4, 2016 at 16:17
• so is it 20 * (2 * 15 + 15 * 14) / 6? Oct 4, 2016 at 16:17
• Possible duplicate of Combinatorics: Selecting objects arranged in a circle
– user940
Oct 4, 2016 at 17:06
## 4 Answers
There are $20$ ways to choose a block of $3$ consecutive people. There are also $20$ ways to choose a pair of adjacent people, and for each of those pairs there are $16$ ways to choose a third person who is not adjacent to either of them. Those are the only sets of $3$ people that we don’t want. There are altogether $\binom{20}3$ possible sets of $3$ people, so the number of acceptable sets is
$$\binom{20}3-20-20\cdot16=\frac{20\cdot19\cdot18}6-20\cdot17=20(57-17)=800\;.$$
Added: More generally, suppose that we have $n$ people at the table, and we want to choose $k$ of them, no two of whom are adjacent. First solve the problem when the $n$ people are in a line. Once the $k$ people are chosen, there will be $k+1$ gaps in which the others can be seated: one before the first chosen person, $k-1$ between adjacent chosen people, and one after the last chosen person. We have to fill these gaps with the $n-k$ people whom we didn’t choose. We must be sure to put at least one person in each of the $k-1$ gaps between adjacent chosen people, so we have only $n-k-(k-1)=n-2k+1$ people left to distribute freely amongst the $k+1$ gaps. This can be done in
$$\binom{(n-2k+1)+(k+1)-1}{(k+1)-1}=\binom{n-k+1}k$$
ways, by a standard stars and bars calculation.
However, this count includes the arrangements that in which the two people at the ends of the line are chosen, and when we wrap the line back around the table, those two people are adjacent. Thus, we don’t want to count those arrangements. There is one of them for each way of choosing $k-2$ people from a line of $n-2$ people with the requirement that no two be adjacent and that neither end person be chosen. This time we’re distributing $n-k$ unchosen people amongst $k-1$ slots with a requirement that each slot get at least one person, and another stars and bars calculation gives the number of such distributions as
$$\binom{n-k-1}{k-2}\;.$$
The final answer is therefore
$$\binom{n-k+1}k-\binom{n-k-1}{k-2}\;,$$
which in this specific problem is
$$\binom{18}3-\binom{16}1=816-16=800\;.$$
Assuming the seating assignments are fixed, the problem is tantamount to arranging $17$ blue chairs and $3$ green chairs in a circle so that no two of the green chairs are adjacent. We will solve the problem for a line, then adjust our answer to account for the fact that the chairs are arranged in a circle.
Line up $17$ blue chairs. This creates $18$ spaces, $16$ between successive blue chairs and two at the ends of the row. Choose three of the spaces in which to place a green chair. This can be done in $\binom{18}{3}$ ways.
However, we have counted selections in which both the first and last chairs are green. If both ends are selected, there are $16$ remaining spaces in which to place the third green chair. Hence, there are $16$ linear arrangements in which no two of the green chairs are adjacent in which there is a green chair at both ends of the row. Since we will be joining the ends of the row to form a circle, these arrangements must be excluded.
Therefore, the number of ways of circular arrangements of $17$ blue chairs and $3$ green chairs so that no two of the green chairs are adjacent is $$\binom{18}{3} - \binom{16}{1} = 800$$ as you found.
Why not again Stars and Bars theorem 1 with the extension of 2 gaps on the bounds. Theorem 1 because we don't want consecutive bars. Then $n=17, k_{max}=3$.
We don't take two on the bounds because we wrap around a circular table !
• |xxxxxx|xxxxxx|x : $\binom{17-1}{2}$ , two inside, one at the beginning
• x|xxxxx|xxxxxxx| : $\binom{17-1}{2}$ , two inside, one at the end
• x|xxxxx|xxxxxx|x : $\binom{17-1}{3}$ , three inside, none on the bounds
$$\binom{17-1}{2}+\binom{17-1}{2}+\binom{17-1}{3} = 800$$
Helped by @N. F. Taussig, I corrected my first answer which was false
• For your first case, you have two ends to choose from, so the number of cases with two inside and one at an end is $$2\binom{16}{2}$$ The case for three inside with none on the ends is $$\binom{16}{3}$$ as you had. The sum is equal to $800$, as Brian M. Scott and I found. Oct 4, 2016 at 17:25
• @N.F.Taussig perhaps but I used the fact that the bars are not the selected people but selectors ( ma basis is 20 and not 17 )
– user354674
Oct 4, 2016 at 17:27
• ouf, I understand the confusion now ... Thank you very much
– user354674
Oct 4, 2016 at 17:30
Here is another way:
Form three blocks with a non-chosen $(N)$ clockwise of a chosen $(C),\;e.g. \boxed {CN}$
We now have $17$ entities, viz. $3$ blocks and $14$ others
We can place the blocks in $\binom{17}3$ ways, fulfilling the "non-neighbour" criterion,
but since in the process we are giving each entity only $17$ starting places instead of $20$,
ans = $\binom{17}3\cdot\frac{20}{17} = 800$ | 2022-08-11T07:54:09 | {
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http://math.stackexchange.com/questions/30732/how-can-i-evaluate-sum-n-0-infty-n1xn/30746 | # How can I evaluate $\sum_{n=0}^\infty (n+1)x^n$
How can if find the sum for:
$$\sum_{n=1}^\infty \frac{2n}{3^{n+1}}$$
I know the answer thanks to Wolfram Alpha. I'm more concerned with how to get to to that number. It cites tests to prove that it is convergent, but my class has never learned these before so I feel that there must be a simpler method.
In general, how can I evaluate $$\sum_{n=0}^\infty (n+1)x^n?$$
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No need to use Taylor Series, this can be derived in a similar way to the formula for geometric series. Lets find a general formula for the following sum: $$S_{m}=\sum_{n=1}^{m}nr^{n}.$$
Notice that $$S_{m}-rS_{m}=-mr^{m+1}+\sum_{n=1}^{m}r^{n}$$
$$=-mr^{m+1}+\frac{r-r^{m+1}}{1-r}=\frac{mr^{m+2}-(m+1)r^{m+1}+r}{1-r}.$$ Hence $$S_m = \frac{mr^{m+2}-(m+1)r^{m+1}+r}{(1-r)^2}.$$
This equality holds for any $r$, but in your case we have $r=\frac{1}{3}$ and a factor of $\frac{2}{3}$ in front of the sum. That is $$\sum_{n=1}^{\infty}\frac{2n}{3^{n+1}}=\frac{2}{3}\lim_{m\rightarrow\infty}\frac{m\left(\frac{1}{3}\right)^{m+2}-(m+1)\left(\frac{1}{3}\right)^{m+1}+\left(\frac{1}{3}\right)}{\left(1-\left(\frac{1}{3}\right)\right)^{2}}$$
$$=\frac{2}{3}\frac{\left(\frac{1}{3}\right)}{\left(\frac{2}{3}\right)^{2}}=\frac{1}{2}.$$
We can define $$S_m^k(r) =\sum_{n=1}^m n^k r^n.$$ Then the sum above considered is $S_m^1(r)$, and the geometric series is $S_m^0(r)$. We can evaluate $S_m^2(r)$ by using a similar trick, and considering $S_m^2(r) - rS_m^2(r)$. This will then equal a combination of $S_m^1(r)$ and $S_m^0(r)$ which already have formulas for.
This means that given a $k$, we could work out a formula for $S_m^k(r)$, but can we find $S_k^m(r)$ in general for any $k$? It turns out we can, and the formula is similar to the formula for $\sum_{n=1}^m n^k$, and involves the Bernoulli Numbers. In particular, the denominator is $(1-r)^{k+1}$.
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That's a great answer, much simpler than how I've approached this question before. – mixedmath♦ May 27 '11 at 23:09
As indicated in other answers, you can reduce this to summing $\displaystyle{\sum_{n=1}^\infty na^n}$ with $|a|<1$ (by pulling out the constant $\frac{2}{3}$ and rewriting with $a=\frac{1}{3}$). This in turn can be reduced to summing geometric series by rearranging and factoring. Note that, assuming everything converges nicely (which it does):
$\begin{matrix} &a & + & 2a^2 & + & 3a^3 &+& 4a^4 &+& \cdots\\ =&a &+& a^2 &+& a^3 &+& a^4 &+& \cdots\\ +& & & a^2 &+& a^3 &+& a^4 &+& \cdots\\ +& & & & & a^3 &+& a^4 &+& \cdots\\ +& & & & & & & a^4 &+& \cdots\\ +& & & & & & & & & \vdots \end{matrix}$
Factoring out the lowest power of $a$ in each row yields
\begin{align*} \sum_{n=1}^\infty na^n &= a(1+a^2+a^3+\cdots)\\ &+ a^2(1+a^2+a^3+\cdots)\\ &+ a^3(1+a^2+a^3+\cdots)\\ &+ a^4(1+a^2+a^3+\cdots)\\ &\vdots \end{align*}
Each row in the last expression has the common factor $a(1+a+a^2+a^3+\cdots)$, and factoring this out yields
\begin{align*}\sum_{n=1}^\infty na^n &=a(1+a+a^2+a^3+\cdots)(1+a+a^2+a^3+\cdots)\\ &=a(1+a+a^2+a^3+\cdots)^2.\end{align*}
Now you can finish by summing the geometric series.
Eric Naslund's answer was posted while I was writing, but I thought that this alternative approach might be worth posting. I also want to mention that in general one should be careful about rearranging series as though they were finite sums. To be more formal, some of the steps above would require justification based on absolute convergence.
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If you want a solution that doesn't require derivatives or integrals, notice that \begin{eqnarray} 1+2x+3x^2+4x^3+\dots = 1 + x + x^2 + x^3 + \dots \\ + x + x^2+ x^3 + \dots\\ + x^2 + x^3 + \dots \\ +x^3 + \dots \\ + \dots \\ =1 + x + x^2 + x^3+\dots \\ +x(1+x+x^2+\dots) \\ +x^2(1+x+\dots)\\ +x^3(1+\dots)\\ +\dots \\ =(1+x+x^2+x^3+\dots)^2=\frac{1}{(1-x)^2} \end{eqnarray}
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I was going to say that! You are fast! – dot dot Oct 29 '12 at 22:44 wow this is dam clever – Justin Meltzer Oct 29 '12 at 22:47
Factor out the 2/3. Then write $$\sum_{n=1}^{\infty} \frac{n}{3^n} = \sum_{n=1}^{\infty} \frac{1}{3^n} + \sum_{n=2}^{\infty} \frac{1}{3^n} + \sum_{n=3}^{\infty} \frac{1}{3^n} + \cdots$$
It is easy to show that $$\sum_{n=m}^{\infty} \frac{1}{3^n} = \frac{3}{2} \left(\frac{1}{3} \right)^m$$ and so $$\sum_{n=1}^{\infty} \frac{n}{3^n} = \frac{3}{2} \sum_{n=1}^{\infty} \left( \frac{1}{3} \right)^n$$ which you can sum. Don't forget to put the 2/3 back in.
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Hints
1. You know (don't you?) the formula for $S(a) = \sum_{n=0}^\infty a^n$ for $|a| < 1$
2. Take the derivative (with respect to $a$) of both sides to obtain a formula for $\sum_{n=1}^\infty n a^n$
3. Show that your series can be put in that form.
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Thanks for answering. 1) No I don't know that formula 2) Can you explain by what you mean by derive both sides? – Backus Apr 3 '11 at 21:59
1. See here: en.wikipedia.org/wiki/Geometric_series 2) Derivation (or differentiation) is a mathematical operation (well, more than that) that is taught in Calculus - if you don't know about it, forget about my answer (and xen0m's). Perhaps you can solve your problem without knowing Calculus, by other methods... but normally you first learn calculus, then solve series. – leonbloy Apr 3 '11 at 22:05
Yeah I don't know calculus – Backus Apr 3 '11 at 22:06
The definition of 'convergence' is based on calculus. So without knowing calculus, it is pretty hard to understand anything about series. Perhaps you can tell us why you are interested in this particular sum? – wildildildlife Apr 3 '11 at 22:32
@wildildildlife I'm in a class that covers precalculus and calculus in one year, and we are transitioning currently. For whatever reason we learned series early. Our class still understood convergence without calculus though. – Backus Apr 3 '11 at 23:32
Note that $\int \left( 1 + 2x + 3x^2 + 4x^3 + \ldots \right)dx = x + x^2 + x^3 + \ldots + c$, i.e., a geometric series, which converges to $x/(1 - x)$. So $$\frac{d}{dx} \left( \frac{x}{1 - x} \right) = \frac{(1 - x)(1) - x(-1)}{(1 - x)^2} = \frac{1}{(1 - x)^2}.$$ Therefore $$1 + 2x + 3x^2 + 4x^3 + \ldots = \frac{1}{(1 - x)^2}.$$
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+1, but wouldn't it have been simpler to say that it was the derivative of $1+x+x^2+...$, converging to $\frac1{1-x}$? – Mike Oct 29 '12 at 22:46 I decided to start with what was given, so it is easier for the OP to see. – glebovg Oct 29 '12 at 22:48 Justin, observe that this only holds for $|x| < 1$, otherwise your sum simplifies to $\infty$. – glebovg Oct 29 '12 at 22:54 What I meant is that you chose $c=0$ while $c=1$ is a more well known series and is easier to take the derivative of. – Mike Oct 29 '12 at 23:23
You can find by differentiation. Just notice that $(x^n)' = nx^{n-1}$. By the theory of power series we obtain (by uniform convergence on any compact subset of $(-1,1)$) that $$\left(\sum_{n=1}^\infty x^n\right)' = \sum_{n=1}^\infty (x^n)' = \sum_{n=1}^\infty n x^{n-1}.$$ The sum on the left hand side is equal to $\left(\frac{x}{1-x}\right)'$. You need to notice that your sum can be written in a similar way as $\sum_{n=1}^\infty nx^{n-1}$.
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Thank you for helping, but I have never learned differentiation. – Backus Apr 3 '11 at 21:58
My favorite proof of this is here
I also have the following method for $\sum_{n=1}^\infty {n\over 2^{n-1}}$ (one can use a similar method for $\sum_{n=1}^\infty {n\over3^n}$):
We first show that $\sum\limits_{n=7}^\infty {n\over 2^{n-1}} ={1\over4}$.
We start with a rectangle of width 1 and height $1/4$. Divide this into eights:
Now divide each eighth-rectangle above in half and take 7 of them. This gives $A_1={7\over 2^6}$.
There are $2\cdot8-7=9$ boxes left over, each having area $2^{-6}$.
Divide each remaining $16^{\rm th}$-rectangle in half and take 8 of them. This gives $A_2={7\over 2^6}+{8\over 2^7}$.
There are $2\cdot9-8=10$ boxes left over, each having area $2^{-7}$.
Divide each remaining $32^{\rm nd}$-rectangle in half and take 9 of them. This gives $A_3={7\over 2^6}+{8\over 2^7}+{9\over 2^8}$.
There are $2\cdot10-9=11$ boxes left over, each having area $2^{-8}$.
Divide each remaining $64^{\rm th}$-rectangle in half and take 10 of them. This gives $A_4={7\over 2^6}+{8\over 2^7}+{9\over 2^8}+{10\over2^9}$.
There are $2\cdot11-9=12$ boxes left over, each having area $2^{-9}$.
At each stage, we double the number of remaining boxes, keeping the same leftover area, and take approximately half of them to form the next term of the series.
At the $n^{\rm th}$ stage, we have $$A_n= {7\over 2^6}+{8\over 2^7}+\cdots+{6+n\over2^{5+n}},$$
with leftover area $$2(n+7)-(n+6)\over 2^{n+5}.$$
It follows that, $${7\over2^6}+{8\over2^7}+{9\over2^8}+\cdots= {1\over4}.$$ Consequently, $$\sum_{n=1}^\infty{n\over 2^{n-1}}= \sum_{n=1}^6 {n\over 2^{n-1}} +\sum_{n=7}^\infty{n\over 2^{n-1}} ={15\over 4}+{1\over4}=4.$$
You can also "Fubini" this (I think this is what Jonas is doing).
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http://math.stackexchange.com/questions/239839/any-partition-of-1-2-ldots-9-must-contain-a-3-term-arithmetic-progressi | # Any partition of $\{1,2,\ldots,9\}$ must contain a $3$-Term Arithmetic Progression
Prove that for any way of dividing the set $X=\{1,2,3,\dots,9\}$ into $2$ sets, there always exist at least one arithmetic progression of length $3$ in one of the two sets.
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What have you tried? Perhaps you could expand your post, because I think I know what you mean, but can't be sure. – Mark Bennet Nov 18 '12 at 14:49
anyone understand me? arithmetic which has length 3 is in one of 2 sets that means the arithmetic $<a;a+d;a+2d$ – LevanDokite Nov 18 '12 at 15:01
@LevanDokite, I understand but I can't prove it in any nice way. There is an admittedly ugly proof here cut-the-knot.org/Curriculum/Arithmetic/ArithmeticSequence.shtml – sperners lemma Nov 18 '12 at 15:54
This result is part of a nice area in Ramsey theory. If you want to study generalizations, the term you want to look for is Van der Waerden number; we are saying here that $w(2,3)\le 9$, where the $3$ indicates you want an arithmetic progression of length three, and the $2$ indicates that yo are splitting the set in two pieces. In fact, $w(2,3)=9$, meaning that, in addition, there is a way to split the numbers from $1$ to $8$ into two pieces, both avoiding such triples: $\{1,2,5,6\}$ and $\{3,4,7,8\}$.
To see that $9$ suffices, the easiest argument proceeds by an analysis of cases: Consider a splitting of $X$ into two sets, and let's attempt to see what restrictions these sets must satisfy in order to avoid arithmetic triples. We must conclude that it is impossible to have such a splitting. The key in my approach is to consider $4,5,6$. They cannot all be in the same piece, but two of them must be. Let's call that piece $A$, and let $B$ be the other one.
• Case 1. $4,6\in A$.
This is the easiest case to eliminate, since $2,5,8$ must then be an arithmetic triple in $B$: Consider, respectively, the triples $2,4,6$, and $4,5,6$, and $4,6,8$. If we place any of $2,5,8\in A$, one of these three arithmetic triples ends up in $A$.
• Case 2. $4,5\in A$.
Then $3,6\in B$ (consider, respectively, the triples $3,4,5$ and $4,5,6$), so $9\in A$ (consider $3,6,9$), but then $1,7\in B$ (consider $1,5,9$ and $5,7,9$), so $2,8\in A$ (consider $1,2,3$ and $6,7,8$). We now see this case cannot be either, because the triple $2,5,8$ is in $A$.
• Case 3. $5,6\in A$.
This is really the same as case 2, by symmetry. (In this case, $4,7\in B$, so $1\in A$, so $3,9\in B$, so $2,8\in A$, and we see that the triple $2,5,8$ is in $A$.)
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do you know an argument that isn't case analysis? – sperners lemma Nov 18 '12 at 16:41
I wanted to say that case 3 should be the same as case 1 by symmetry, and eventually realized that $[1\ldots 9]$ is not symmetric about $[3,4,5]$. I wonder if an argument similar to this one, but using $[4,5,6]$ instead of $[3,4,5]$, would be simpler. – MJD Nov 18 '12 at 16:45
I doubt there is one. "Softer" arguments that avoid case analysis do not tend to give precise bounds. The nice argument in "Ramsey theory" by Graham-Rotschild-Spencer, for example, gives $325$ as an upper bound. (The argument there is fairly intuitive, and one sees immediately how to generalize. But it is of course terrible for concrete bounds.) – Andrés Caicedo Nov 18 '12 at 16:46
@MJD You are right. Using $4,5,6$ gives us a bit more symmetry, so case 3 would just be the "reflection" of case 1. I should have used that triple. Thanks! – Andrés Caicedo Nov 18 '12 at 16:48
I didn't think it through myself, so I don't know whether it happens to complicate the other two cases enough that the gain from the symmetry is not actually a win. – MJD Nov 18 '12 at 16:49
(This is very similar to Andres Caicedo's proof, but is a little simpler.)
Let us call the sets red and blue. Consider the elements $\{4,5,6\}$. At least two of these are the same color; say red. Then we have three cases: $\{4,5\}$ are red, $\{4,6\}$ are red, or $\{5,6\}$ are red. $\def\r#1{\color{red}{#1}}\def\b#1{\color{blue}{#1}}$
• $\r4$ and $\r5$ are both red. Then if $3$ or $6$ is red we are done, so suppose $\b3$ and $\b6$ are blue. Then $\r9$ is red (else $\b3,\b6,\r9$ is blue), so $\b1$ is blue (else $\b1,\r5,\r9$ is red), $\r2$ is red ($\b1,\r2,\b3$), $\b8$ is blue ($\r2,\r5,\b8$). Now if $7$ is red we have $\r5,\r7,\r9$ and if $7$ is blue we have $\b6,\b7,\b8$.
• $\r4$ and $\r6$ are both red. Then $\b2$ and $\b5$ are both blue (else $\b2,\r4,\r6$ or $\r4,\b5,\r6$). Now if $8$ is red we have red $\r4,\r6,\r8$ and if $8$ is blue we have blue $\b2,\b5,\b8$.
• $\r5$ and $\r6$ are both red. This reduces to the $\{4,5\}$ case by the transformation $x\mapsto (10-x)$.
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https://cs.stackexchange.com/questions/66617/number-of-different-binary-search-trees-storing-n-distinct-keys | # Number of different binary search trees storing n distinct keys?
How many different binary search trees are possible that store the values 1,2,...,n ?
So far I found a recursive formula for the number (by case distinction what's at the root):
$T(n) = 2T(n-1) + \sum_{i=2}^{n-1}T(i-1)T(n-i), n > 1$ and $T(1) = 1$
But I have no idea how to solve this recursion. Our task was only to find the recursion and I believe this to be a correct solution. But I am very interested in a closed formula of it. Can anyone link me to some resources/books or give a general hint on how it can be solved?
## 2 Answers
The solution to your recurrence is $$T(n) = \frac{(2n)!}{n!(n+1)!},$$ also known as the Catalan numbers. The quickest way to find this is by computing a few elements of the sequence and using the OEIS to identify the sequence.
Any time you see something resembling a convolution, that suggests generating functions as a method. Convolutions are things that look like $\sum_x f(x)f(\overline{x})$, where $\overline{x}$ denotes some kind of "complement" or "opposite" of $x$. Generating functions often turn convolutions into products.
Possible references are: | 2019-09-16T10:18:40 | {
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https://math.stackexchange.com/questions/3153566/if-zero-is-an-eigenvalue-are-dimensions-lost | # If zero is an eigenvalue are dimensions lost?
This is likely a silly question so sorry in advance. However, I am wondering if I am right in thinking that if zero is an eigenvalue, then some dimension must be lost. My understanding is that eigenvectors are the vectors that are only scaled when some matrix A is applied. Then, eigenvalues are the amount those vectors are scaled. So, therefore is it true that if an eigenvalue is zero, that vector has been sent to the origin and therefore the dimension of the image is smaller than the original dimension?
Would that not also imply that any vector in the kernel of a transformation is an eigenvector?
• The eigenspace for eigenvalue $0$ is precisely the kernel. – Randall Mar 19 at 1:46
• Yes to all your questions. – zoidberg Mar 19 at 1:56 | 2019-04-20T14:37:25 | {
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https://math.stackexchange.com/questions/3694114/finding-a-basis-for-a-spanning-list-by-columns-vs-by-rows | # Finding a basis for a spanning list by columns vs. by rows
From this answer, I see that, given a list of vectors, we can put them as the columns of a matrix, then reduce it, and the pivots will tell us which of the original vectors form a basis for the span of those vectors.
However, I'm wondering, can't we just put the list of vectors as the rows of a matrix, then reduce it, and find a basis this way? Since row operations don't change the row space, this method should also work. We just look at the rows that have a leading variable.
Though, I can see that the first method of putting the vectors in a column might be useful if we want to find out which of the original vectors form a basis.
## 2 Answers
[I’m sure that I’ve answered this before, but can’t find anything relevant at the moment.]
Yes, both methods will get you a basis for the span of the vectors, and your observation is correct: writing them as columns of a matrix will let you find a linearly-independent subset of the original vectors that has the same span. On the other hand, writing them as rows generally gives you a “nicer” basis: The first part of each vector in the basis will consists of a bunch of zeros with a single one somewhere. Which one is preferable depends entirely on what you’re then going to do with that basis.
• Your answer was the one I linked :) May 27 '20 at 20:18
Consider the matrix $$A$$ whose columns are the vectors $$v_1, \dots, v_n$$ from the $$m$$-dimensional vector space $$V.$$ We have that $$A^t$$ (the transpose of $$A$$) is the matrix whose rows are the vectors $$v_1, \dots, v_n.$$
Like you mentioned, we can determine which of the vectors $$v_1, \dots, v_n$$ are linearly independent by putting the matrix $$A$$ in reduced row-echelon form. Explicitly, the vectors $$v_j$$ for each column $$j$$ with a pivot are linearly independent, and the rest of the vectors can be written as a linear combination of the $$v_j.$$ (In fact, the coefficients of the relation of linear dependence are determined by the entries of the column of the corresponding vector. Check that for $$v_1 = (1, 2, 5)$$ and $$v_2 = (-3, -5, -13),$$ we have that $$v_4 = (2, 1, 4) = -7v_1 - 3v_2.$$ Compare this to the problem you mention.)
Like you have seen, there are other advantages of writing the vectors as the columns; however, it is true that the column rank and the row rank of any matrix are equal, hence the number of pivots of the matrix $$A^t$$ is the same as the number of pivots of the matrix $$A,$$ and one can just as legitimately row-reduce $$A^t.$$ Unfortunately, the interpretation of the coefficients of the matrix $$A^t$$ is not as clear as in the other method. (Check that the transpose of the matrix in the problem you mention is $$\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix};$$ however, the coefficients of this matrix have no immediate meaning.) | 2022-01-19T04:36:59 | {
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https://www.physicsforums.com/threads/equivalent-sets.310702/ | # Equivalent sets
## Homework Statement
1. Suppose A-B is equivalent to B-A. show that A is equivalent to B.
2. if A,B and C are nonempty and A cross B is equivalent to A cross C then B is equivalent to C
Any help would be appreciated, thanks!
## The Attempt at a Solution
I tried constructing a bijection, but that did not work out right. Any ideas?
Last edited:
Related Calculus and Beyond Homework Help News on Phys.org
Mark44
Mentor
By definition, if two sets A and B are equivalent, that every element in A is also in B, and every element in B is also in A. Can you use this idea on your first problem?
For your second problem, part of it is missing.
if A,B and C are nonempty and A cross B is equivalent to A cross then B is equivalent to C
Is the question "if A,B and C are nonempty and A cross B is equivalent to A cross C, then B is equivalent to C"
By equivalent, I mean they have the same cardinality, not that they are equal.
Mark44
Mentor
OK, so you know there is an bijection between A - B and B - A. It seems to me there are four cases:
$$A \subset B$$
$$B\subset A$$
A = B
$$A \cap B = \oslash$$
Can you eliminate one or more of these as possibilities, and then come up with a bijection for the remaining one(s)?
Some examples might be helpful to get you thinking in the right way.
1. A = {2, 4, 6, ...}, B = {1, 2, 3, ...}
2. A = {2, 3, 4, 5, ...}, B = {1, 2, 3, 4, ...}
3. any two sets that are equal
4. A = {2, 4, 6, ... }, B = {1, 3, 5, ... } | 2020-02-19T07:42:13 | {
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https://math.stackexchange.com/questions/3192398/sum-sum-i-1n-sum-j-i-1n-mathbbi-j-i-leqslant-k | Sum $\sum_{i = 1}^N \sum_{j = i + 1}^N \mathbb{I} (j - i \leqslant K)$
Let $$N,K$$ be non-negative integers. What's the value of the following sum?
$$S(N,K) = \sum_{i = 1}^N \sum_{j = i + 1}^N \mathbb{I} (j - i \leqslant K)$$
where $$\mathbb{I}(\mathcal P)=1$$ if $$\mathcal P$$ is a true proposition and $$\mathbb{I}(\mathcal P)=0$$ otherwise.
For example, if $$N=K$$, then $$S(N,N)$$ simply counts the number of pairs $$(i,j)$$ with $$i,j\in\{1,\dots,N\}$$ such that $$i, which is easily seen to be
$$S(N,N)=N(N-1)/2$$
Also it is easy to see that $$S(N,K)=S(N,N)$$ whenever $$K\ge N$$. I'm only having trouble with the case where $$K.
• It would be nice if your progress so far is demonstrated. – Lee David Chung Lin Apr 18 at 13:33
• Can you solve the problem for $K = N$? – John Hughes Apr 18 at 13:39
• @JohnHughes Yes, see edit. – becko Apr 18 at 13:44
• @LeeDavidChungLin I added some edits. Note that this is not homework, so I can't be 100% sure that an analytical solution exists. I just ran into this sum solving a different problem (also not homework). – becko Apr 18 at 13:45
Split the sum on $$i$$ until $$N-K$$, and from $$N-K+1$$ to $$N$$ respectively, call them $$S_1, S_2$$, where we assume $$1\le K < N$$ as the other cases have been dealt with in the post.
If $$i\le N-K$$, the sum on $$j$$ has all its $$K$$ terms, so $$S_1=K(N-K)$$
If $$i> N-K$$, the second sum has only $$N-i$$ terms, so $$S_2=\sum_{i=N-K+1}^{i=N}{(N-i)}=\sum_{k=0}^{k=K-1}k=\frac{K(K-1)}{2}$$
Putting it together:
$$\sum_{i = 1}^N \sum_{j = i + 1}^N \mathbb{I} (j - i \leqslant K)= KN-\frac{K(K+1)}{2}=\frac{K(2N-K-1)}{2}$$
• To be a bit more general, $S(K,N)=\min(K,N)\times[2N-\min(K,N)-1]$. – becko Apr 18 at 14:26
• That's the general form for all $N,K$ indeed – Conrad Apr 18 at 14:29
• I forgot to divide by 2 in my comment. – becko Apr 28 at 8:12
Hint: Just a manipulation $$S(N,K)=\sum _{i=1}^N\sum _{j=i+1}^N\mathbb{I}(j-i\leq K)=\sum _{i=1}^N\sum _{j=i+1}^N\mathbb{I}(j\leq K+i)=\sum _{i=1}^N\sum _{j=i+1}^{\min \{K+i,N\}}1\qquad\qquad\qquad$$$$\qquad\qquad\qquad\qquad\qquad\qquad\qquad=\sum _{i=1}^{N-K}\sum _{j=i+1}^{K+i}1+\sum _{i=N-K+1}^N\sum _{j=i+1}^N1.$$ Can you finish? | 2019-05-20T18:40:20 | {
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https://math.stackexchange.com/questions/2426508/expected-number-of-items-looked-at | # Expected number of items looked at
Currently failing at basic probability…
Given a sequence of items, linear search means looking at each in turn and seeing if it's the one we're looking for. (I.e. in the worst case, it's the last item in the sequence or not in the sequence at all and we'll have to look at all items to find it/determine it's not there.)
How many elements of the input sequence need to be checked on average, assuming that the element being searched for is equally likely to be any element in the array?
Ok, we know two things:
• the element we're looking for is in the array
• P[the i-th item is the one we're looking for] = $\frac{1}{n}$ for all $i\in\{1,..,n\}$
And I think I'm supposed to get $\frac{n+1}{2}$ or $\frac{n}{2}$ or something thereabout.
Let $X$ be the number of items the linear search algorithm is looking at.
\begin{align} P[X=1] &= \frac{1}{n} \\ P[X=2] &= \left(1-\frac{1}{n}\right)\frac{1}{n} \\ P[X=3] &= \left(1-\frac{1}{n}\right)^2\frac{1}{n} \\ & \ \ \vdots \\ \end{align}
So
\begin{align} \mathbb{E}(X) &= \sum_{i=1}^n i\left(1-\frac{1}{n}\right)^{i-1}\frac{1}{n} \\ &= \frac{1}{n}\sum_{i=1}^n i\left(1-\frac{1}{n}\right)^{i-1} \\ \end{align}
That's not something I want to solve myself, so I gave it to Wolfram Alpha and got
$$\left(1-2\left(\frac{n-1}{n}\right)^n\right)n$$
This doesn't look particularly close.
Where did I go wrong?
• You say the location is uniform, thus the probability that it is in slot $i$ is $p_n=\frac 1n$ for all $i$, yes? Thus the answer is $\sum i\times \frac 1n=\frac {n+1}2$. This needs to be altered in a small way if you want to allow for the possibility that it isn;t in the list. (in that case $p_n$ is greater than all the others and all the others are equal). – lulu Sep 12 '17 at 12:05
• In case you assume $n = 2$ and you are sure that the element is in the array, then $P[X = 1] + P[X = 2] = 1$. But as per your (@User1291) calculation, it is $\dfrac{1}{2} + \dfrac{1}{4} \neq 1$. Hence you should reconsider the calculation of probability for each event. – Nash J. Sep 12 '17 at 12:16
Assuming that there is exactly one item of interest: $$P[X=i]=\frac1n\text{ for }1\le i\le n$$ $$E[X]=\sum_{i=1}^n\frac in=\frac1n\sum_{i=1}^ni=\frac{n(n+1)}{2n}=\frac{n+1}2$$ If $P[X=i]=\left(1-\frac1n\right)^{i-1}\frac1n$ instead then there is a geometric distribution, which can be realised as (say) there are $n$ rolls of an $n$-sided die and we are searching for the first 1 rolled. Here there is an extra $\left(1-\frac1n\right)^n$ probability of doing $n$ comparisons and not finding the desired item, leading to the expectation in this case as $$\left(1-2\left(\frac{n-1}n\right)^n+\left(1-\frac1n\right)^n\right)n$$ $$=\left(1-\left(1-\frac1n\right)^n\right)n$$ which is different from $\frac{n+1}2$ because the desired element may appear more than once in the array.
• Why is $P[X=i] = \frac{1}{n}$? – User1291 Sep 12 '17 at 12:08
• @ParclyTaxel What should be the statement in order for his(of @user1291) reasoning to be correct? – John D Sep 12 '17 at 12:13
• @Magnusseen The calculations in the question body imply something like this: we have $n$ rolls of an $n$-sided die and are looking for the first 1 rolled. A geometric distribution. – Parcly Taxel Sep 12 '17 at 12:16
• @ParclyTaxel In his reasoning, he computed $P_k,$ the probability that the amount of search is exactly $k.$ That computation seems correct. If we assume that we will find the element in the array, we should have then $\sum_{k=1}^n P_k=1.$ – John D Sep 12 '17 at 12:23
• @Magnusseen There, he makes the mistake. Not all assumptions in the question were used, in particular that there is exactly one item of interest. – Parcly Taxel Sep 12 '17 at 12:25
Your calculations for $P(X=2), P(X=3)$, etc are incorrect.
For instance you should have, $P(X=2)=P(X=2\mid X\ne 1)\cdot P(X\ne 1)=\frac{1}{n-1}\cdot\frac{n-1}{n}=\frac1n$.
In fact as noted by others $P(X=k)=\frac1n$ for $k=1, 2,\cdots, n$.
This is a matter of conditionality.
In your calculation, you took $1/n$ to be the probability that the element is in the $i$th position under the condition that it was not in any of the positions before $n$. If this were the case, then your calculation of the probabilities for $i\geq2$ would be correct. Unfortunately though, the probabilities don't sum up to one.
Instead, the probability to find the element in the $i$th position is precisely $1/n$ for every $i\in\{1,\dots,n\}$. The probabilities are given, and there is no need to find them. The probabilities are unconditional; events for $i<j$ have no bearing on $P[X=j]$. If you want, you can use this information to find the conditional probabilities. They are not needed in this problem, but you will find that they are not $1/n$.
There is two alarming things in your calculations, hinting at a problem:
1. You assumed $P[X=2]=\frac1n$ and the calculated that $P[X=2]=(1-\frac1n)\frac1n$.
2. The probabilities don't sum up to one although they should.
If you wish to include the possibility that the desired object isn't in the array:
Denote by $\psi$ the probability that the desired object isn't in the array. Let $p_i$ denote the probability that it is in slot $i$. Then, since all the $p_i$ are equal, we must have $$\sum_{i=1}^np_i=1-\psi\implies p_i=\frac {1-\psi}n$$
For every $i<n$ the probability that you look at exactly $i$ objects in your search is $p_i$. For $i=n$ the probability is $p_i+\psi$. Thus in this case the expected value is $$\sum_{i=1}^n i\times \frac {1-\psi}n+n\times \psi=\frac {(n+1)(1-\psi)}2+n\psi=\frac {n+1+n\psi-\psi}2$$
Sanity checks: Note that if $\psi=0$ this reduces to the formula we know. Then again, note that if $\psi =1$ this reduces to $n$, as it should.
Let us assume that the key is present in the list. The expectations of the "cost" of finding the key equiprobably in every position $k$ are $\dfrac kn$, the sum of which is
$$\frac{n(n+1)}2\frac1n=\frac{n+1}2.$$
The correction for the possibility of the key being absent with probability $q=1-p$ yields
$$p\frac{n+1}2+qn.$$ | 2020-02-28T07:15:18 | {
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http://math.stackexchange.com/questions/161485/mutually-exclusive-events/161500 | # Mutually exclusive events
Working my way through the following problem:
### Problem
Suppose that $E$ and $F$ are mutually exclusive events of an experiment. Show that if independent trials of this experiment are performed, then $E$ will occur before $F$ with probability $\frac{ P( E)}{P( E) + P( F)}.$
I have the following come up with the following solution:
### Solution
Since $P( E^c) = P( F)$ Therefore $\frac{ P( E)}{ P( E) + P( F)} = \frac{ P( E)}{ 1 - P( F) + P( F)} = \frac{ P( E)}{ 1} = P( E)$
But I am unsure if I am able to assume $P( E^c) = P( F)$ as a given?
As well, I am particularly confused by the answer in the solution manual which makes it's argument as follows:
### Solution Manual
If $E$ and $F$ are mutually exclusive events in an experiment, then $P( E \cup F) = P( E) + P( F)$. We desire to compute the probability that $E$ occurs before $F$ , which we will denote by $p$. To compute $p$ we condition on the three mutually exclusive events $E$, $F$ , or $(E \cup F )^c$. This last event are all the outcomes not in $E$ or $F$. Letting the event $A$ be the event that $E$ occurs before $F$, we have that
$p = P( A|E) P( E) + P( A|F) P(F ) + P( A|(E \cup F )^c) P( (E \cup F )^c)$
$P( A|E) = 1$
$P( A|F) = 0$
$P( A|(E \cup F)^c) = p$
since if neither $E$ or $F$ happen the next experiment will have $E$ before $F$ (and thus event $A$ with probability $p$). Thus we have that
$p = P( E) + p P( (E \cup F)^c)$
$= P( E) + p (1 − P( E \cup F))$
$= P( E) + p (1 − P( E) − P( F))$
Solving for $p$ gives
$p = \frac{ P( E)}{ P( E) + P( F)}$
as we were to show.
Specifically his statement
since if neither $E$ or $F$ happen the next experiment will have $E$ before $F$ (and thus event $A$ with probability $p$)
-
They mean: If neither $E$ or $F$ happens on the first trial, then the game starts over. – Byron Schmuland Jun 22 '12 at 2:34
So we are able to treat the experiment as if only mutually exclusive events $E$ and $F$ exist and my solutions is valid correct? – rudolph9 Jun 22 '12 at 2:43
No, that is a separate issue. In fact, there is no need to assume that $E$ and $F$ are exhaustive. – Byron Schmuland Jun 22 '12 at 2:47
To determine the probability that $E$ occurs before $F$, we can ignore all the (independent) trials on which neither $E$ nor $F$ occurred, that is, $(E\cup F)^c$ occurred, since we are going to repeat the experiment until one of $E$ and $F$ does occur. So, look at the trial of the experiment on which one of $E$ and $F$ has occurred for the very first time. We are given that on this trial, the event $E \cup F$ has occurred. But, we don't yet know which of the two has occurred. So, given the knowledge that $E \cup F$ has occurred, what is the conditional probability that it was $E$ that occurred (and so $E$ occurred before $F$ since this is the first time we have seen either $E$ or $F$)?
$$P(E \mid (E \cup F)) = \frac{P(E(E \cup F))}{P(E \cup F)} = \frac{P(E \cup EF)}{P(E) + P(F) - P(EF)} = \frac{P(E)}{P(E)+P(F)}$$ since $P(EF) = P(\emptyset) = 0$.
Alternatively, let $G = (E\cup F)^c = E^c \cap F^c$ be the event that neither $E$ nor $F$ occurs on a trial of the experiment. Note that $P(G) = 1 - P(E) - P(F)$. Then, the event $E$ occurs before $F$ if and only if one of the following compound events occurs:
$$E, (G, E), (G, G, E), \ldots, (\underbrace{G, G, \ldots, G,}_{n-1} E), \ldots$$
where $(\underbrace{G, G, \ldots, G,}_{n-1} E)$ means $n-1$ trials on which $G$ occurred and then $E$ occurred on the $n$-th trial. The desired probability is thus
$$P(E ~\text{before}~ F) = P(E) + P(G)P(E) + [P(G)]^2P(E) + \cdots = \frac{P(E)}{1 - P(G)} = \frac{P(E)}{P(E)+P(F)}.$$
-
Your solution is incorrect. If $E$ and $F$ are mutually exclusive, it means that $E \cap F = \emptyset$, therefore $F \subseteq E^c$; and therefore, $P(F) \color{red}{\le} P(E^c)$.
You are not interpreting independent trials of the experiment correctly. It might be helpful to consider an example. Suppose you are rolling a biased 6-faced die. Let $E$ denote the event that 1 or 2 turn up and $F$ denote the event that 3 or 4 turn up. Continue rolling the die until either $E$ or $F$ occur. What is the probability that $E$ occurs before $F$, that is what is the probability that you get 1 or 2 before you get 3 or 4 (in the repeated rolls of the die).
The statement
since if neither $E$ or $F$ happen the next experiment will have $E$ before $F$ (and thus event $A$ with probability $p$)
means that if neither $E$ or $F$ happen, that is if 5 or 6 is rolled, we roll the die again. Since the rolls are independent, the probability of getting $E$ before $F$ in the future experiments is $p$.
Rant: This problem and its solution shows why students find probability confusing. The problem is stated very informally. In my opinion, a formal statement of the problem will remove some of the confuson.
Consider an experiment $\mathcal E_1$ with probability measure $P_1$. Let $E$ and $F$ be two events in $\mathcal E_1$. $P_1(E)$ denotes the probability that $E$ occurs in experiment $\mathcal E_1$. Similarly interpretation holds for $P_1(F)$.
Now consider another experiment $\mathcal E_2$, which represents infinite independent repetitions of the experiment $\mathcal E_1$. Let $P_2$ be the probability measure for events in $\mathcal E_2$.
Now consider an outcome $\omega$ of $\mathcal E_2$ that is a series of outcomes of $\mathcal E_1$. Let $\tau_E$ denote the first time $E$ occurs in $\omega$ (with $\tau_E = \infty$ if $E$ does not occur). Similarly, let $\tau_F$ denotes the first time $F$ occurs in $\omega$. Let $A$ denote the event (in $\mathcal E_2$) that $\tau_E < \tau_F$. The question is asking you to show that
$\displaystyle P_{\color{red}2}(A) = \frac{ P_1(E) }{ P_1(E) + P_1(F) }$
Thus, the question is asking you to compare two different experiments. When you write $E^c \equiv F$, you were thinking in terms of experiment $\mathcal E_2$; but $E$ and $F$ are not events in $\mathcal E_2$; they are events in $\mathcal E_1$.
-
Since as you state in the context of your example > if neither $E$ or $F$ happen, that is if 5 or 6 is rolled, we roll the die again. Approaching the problem as if $E^c \equiv F$ is therefore valid then, no? – rudolph9 Jun 22 '12 at 3:02
$E^c = \{3,4,5,6\} \not\equiv \{3,4\} = F$ – Aditya Jun 22 '12 at 3:12
Yes but should ${5,6}$ occur we roll again, for the purposes of calculating the desired probability of this problem we disregard all events that do not exist in $E \cup F$ as they have no effect on the computation, therefore you are able to approach the problem as if $E^c \equiv F$, no? – rudolph9 Jun 22 '12 at 3:20
Does my updated answer clarify this point? – Aditya Jun 22 '12 at 3:25
The event that $E$ does not occur first is (in my notaton) $A^c$. You cannot simply change the meaning of $E$ (which is an event in experiment $\mathcal E_1$). Perhaps the solution given by @DilipSarwate is close to what you are thinking: Think of the experiment in which either $E$ or $F$ occur for the first time. What is the probability that the event that occurred was $E$. – Aditya Jun 22 '12 at 3:44
I think extreme simplification is need...
$P(E) and P(F)$ are complements for the Universe (U, U=1 in this case) $P(E) + P(F) = 1$ // corrected as mentioned by Aditya, sorry for my dyslexic!thing
$P(E) / ( P(E)+P(F) ) = 1 / 2$ Hence probability of $E$ is $50\%$ (or $0.5$), probability of restant set is the remaining $50\%$; the remaining set is $F$ because $U=\{E, F\}$ with the given data $P(E \text{ before } F) = P(F \text{ before }E)$.
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If $P(E) = P(F) = 1$, then $E$ and $F$ cannot be mutually exclusive because $E \cup F \subset \Omega$, thus $P(E \cup F) = P(E) + P(F) \le P(\Omega) = 1$. – Aditya Jun 22 '12 at 3:15
E and F are exclusive... when I say P(E) = P(F) = 1, I mean "wheight_of_P(E)" + "wheight_of_P(F)" is equal to 1(totality in probability)... without more information from the problem I assume that "E_is_complement_and_exclusive_of_F" and "F_is_complement_and_exclusive_of_E"... I've added parenteses to the answer for clarity... – ZEE Jun 22 '12 at 22:00
Then you should assume $P(E) = P(F) = 0.5$ – Aditya Jun 23 '12 at 1:01
You're right, what I wanted to say is : P(E) = P(F) and P(E) + P(F) = 1... thanks seeing it... As per opposition to the other possibility which was : P(E) <> P(F) and P(E) + P(F) = 1 in both cases : $P(E) \cap P(F) = \emptyset$ and $P(E) \cup P(F) = U$ (U=Universe or FullSet, 1 in this case) – ZEE Jun 23 '12 at 12:39
Here is an alternative way of using conditional probability.
Consider repeated experiments and let $Z_n$ ($n \in \mathbb{N}$) be the result observed on the $n$-th experiment. Denote the event of "$\textrm{E before F}$" by $B$ and its probability $\alpha$.
According to the law of total probability, we obtain
$$\alpha = P \{ B\} = \sum_{z} P \{B \mid Z_1 = z \} P \{ Z_1 = z \}$$
Now we have
$$P \{ B \mid Z_1 = E \} = 1, \quad P \{ B \mid Z_1 = F \} = 0.$$
If the first experiment results in anything other than $E$ or $F$, the problem is repeated in a statistically identical setting. That is,
$$P \{ B \mid Z_1 = z \} = \alpha, \forall z \neq E, F.$$
Therefore, we have
$$\alpha = P \{ Z_1 = E \} \times 1 + P \{ Z_1 = F \} \times 0 + \sum_{z \neq E,F} P \{ Z_1 = z \} \times \alpha \\ = P \{ Z_1 = E \} + [1 - P \{ Z_1 = E \} - P \{ Z_1 = F \}] \alpha$$
Solving this equation, we get
$$\alpha = \frac{P \{ Z_1 = E \}}{P \{ Z_1 = E \} + P \{ Z_1 = F \}}.$$
- | 2016-05-06T04:30:44 | {
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https://math.stackexchange.com/questions/4097507/what-is-a-precise-definition-of-soundness | # What is a precise definition of soundness?
I'm trying to better understand soundness, especially in contrast to semantic consistency. Here is what I've put together so far:
Soundness: A theory is sound if all theorems are true under all possible interpretations.
Semantic Consistency: A theory is semantically consistent if there exists an interpretation for which all theorems are true.
Note that I'm aware that consistency usually refers to syntactic consistency (namely that the theory cannot derive both $$\varphi$$ and $$\lnot \varphi$$, for any formula $$\varphi$$).
I'm primarily confused about the definition of soundness because it feels like a very hard property to satisfy. Take, for example, Peano Arithmetic. One axiom states that $$0$$ is not the successor of any number. I think I can find an interpretation that makes that false, for example, modular arithmetic. I'm fairly sure, though, that Peano Arithmetic is considered a sound theory so I suspect that I'm just confused about the definition of soundness.
Any help would be appreciated. Thanks.
Your definition of soundness is incorrect. A theory is sound if it contains no sentences which are false with respect to a specific structure (or class of structures) of interest.
Now as an unfortunate matter of practice we generally only consider soundness when that specific context is understood in the background, so we omit it. For example, when we say
Peano arithmetic is sound
what we really mean is
Peano arithmetic is sound with respect to the structure $$(\mathbb{N};+,\times)$$; that is, for each axiom $$\varphi$$ of Peano arithmetic we have $$(\mathbb{N};+,\times)\models\varphi$$.
Meanwhile, Peano arithmetic is not sound with respect to the unique $$1$$-element $$\{+,\times\}$$-structure (basically, the trivial ring).
I think it's actually helpful to step a ways back and rephrase both soundness and consistency in a more general way. Specifically, given a class of structures $$\mathbb{K}$$ and a theory $$T$$, we say:
• $$T$$ is sound with respect to $$\mathbb{K}$$ iff for every $$\mathcal{A}\in\mathbb{K}$$ we have $$\mathcal{A}\models T$$.
• $$T$$ is semantically consistent with respect to $$\mathbb{K}$$ iff for some $$\mathcal{A}\in\mathbb{K}$$ we have $$\mathcal{A}\models T$$.
The fundamental difference then is the "every"/"some" distinction. There's a second difference too, however, which is down to practice: usually when we talk about soundness we're tacitly referring to a $$\mathbb{K}$$ consisting of a single structure, and almost always when we talk about semantic consistency we're tacitly referring to the class of all structures. So there are really two differences between soundness and semantic consistency at work here: soundness typically looks at a small context while semantic consistency typically looks at a large context, and with respect to that class soundness is a universal condition while consistency is an existential one.
That said, as I mentioned above we almost always work with reference to the class of all structures when thinking about semantic consistency. So in light of that your definition of semantic consistency is correct. Note that the completeness theorem shows that the "realist commitment" in the (with-respect-to-the-class-of-all-structures) notion of semantic consistency (namely, that mathematical structures are things which actually exist and about which sentences are true or false) can be removed; by contrast, there isn't really an analogous result for soundness(es).
• Thank you for your answer - that is very helpful and completely resolves my confusion around whether Peano arithmetic is sound. Now, though, the definitions for soundness and semantic consistency sound almost identical, right? Soundness refers to only true theorems w.r.t. a specific interpretation/structure while semantic consistency refers to only true theorems w.r.t at least one interpretation/structure? If that's the only difference, that's fine, I'm just trying to confirm. Apr 11, 2021 at 2:07
• @user2869495 I mean, that's a difference between a specific and a general - that's a huge difference! As a relevant technical result from computability theory, per the completeness theorem the set of semantically consistent sentences is not too complicated (it's "co-recursively enumerable") but the set of sentences which are sound with respect to $\mathbb{N}$ is extremely complicated. Apr 11, 2021 at 2:10
• Ok, that makes sense. I don't fully understand your last comment (that the set of sentences which are sound w.r.t. $\mathbb{N}$ is extremely complicated), but I'll save that question for when I'm farther along. Thanks! Apr 11, 2021 at 2:14
• @user2869495 I've edited to go in a slightly different direction expositionally; feel free to revert if that made things worse. (Actually it occurs to me that you might not have enough reputation to revert, I don't know how that works; if you can't let me know and I'll roll it back myself.) Apr 11, 2021 at 2:16
• Your edits definitely made things (even more) clear. Thanks. Apr 11, 2021 at 2:18 | 2022-07-01T08:51:53 | {
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https://math.stackexchange.com/questions/2880082/proving-an-entire-function-with-constant-imaginary-part-on-closed-unit-disc-is-c | # Proving an entire function with constant imaginary part on closed unit disc is constant on the whole $\mathbb C$ without identity principle
(Exer 8.24) If $f: \mathbb C \to \mathbb C$ is entire and $\Im(f)$ is constant on closed unit disc $\{|z| \le 1\}$, then $f$ is constant on $\mathbb C$.
In OSU course here, there's a proof of Exer 8.24 above where the proof makes use of the Identity Principle. I think I can prove alternatively by modifying the proof of Liouville's Thm instead of using the Identity Principle. I attempt my alternative proof as follows.
Question: What mistakes, if any, have I made, and why?
Pf:
By Cauchy-Riemann, $f \equiv: K$ is constant on closed unit disc. It remains to show $f \equiv: K$ on the rest of $\mathbb C$.
Consider any path $\gamma := \{ |z-w| = R \} \subset \mathbb C, w \in \mathbb C, R > 0$. We have by Cauchy's Integral Formula for $f'$ (Formula 5.1) that
$$|f'(w)| = |\frac{1}{2 \pi i} \int_{\gamma} \frac{f}{(z-w)^2} dz| = \frac{1}{2 \pi} |\int_{\gamma} \frac{f}{(z-w)^2} dz|$$
$$\le \frac{1}{2 \pi} \max_{z \in \gamma}|\frac{f}{(z-w)^2}| \text{length}(\gamma) = \frac{1}{\not{2}\not{\pi}} \max_{z \in \gamma}|\frac{f}{(z-w)^2}| \not{2}\not{\pi} R$$
$$= \max_{z \in \gamma}|\frac{f}{(z-w)^2}| R = \max_{z \in \gamma}\frac{|f|}{|z-w|^2} R = \max_{z \in \gamma}\frac{|f|}{R^2} R = \max_{z \in \gamma}\frac{|f|}{R} = \max_{z \in \gamma}\frac{K}{R} = \frac{K}{R}$$
$$\therefore, |f'(w)| = \lim_{R \to \infty} |f'(w)| \le \lim_{R \to \infty} \frac{K}{R} = 0 \ \forall w \in \mathbb C \implies f'(w) = 0 \ \forall w \in \mathbb C$$
$\therefore,$ by a theorem in the textbook (Thm 2.17), $f \equiv: K$, not only in closed unit disc, but also on the whole $\mathbb C$. QED
• If you are happy with Taylor's theorem in the complex plane, the the power series $\sum_n f^{(n)}(0) z^n/n!$ converges to $f$ everywhere, but that's a constant. – Lord Shark the Unknown Aug 12 '18 at 9:28
• @LordSharktheUnknown Thanks. What do you mean please? It seems $$\sum_n \frac{f^{(n)}(0)z^n}{n!} = f(z) \ \forall z \in \mathbb C$$ sooo if $f$ is constant on closed unit disc $\{|z| \le 1\}$, then $\sum_n \frac{f^{(n)}(0)z^n}{n!}$ is constant on the closed unit disc $\{|z| \le 1\}$ with the same value. Sooo then what? What is the 'that' in 'but that's a constant' ? – BCLC Aug 12 '18 at 9:44
Here is another approach that uses the open mapping theorem.
Since $B(0,1)$ is open and $f$ is analytic, if $f$ is non constant, then $f(B(0,1))$ is open.
However, the set $\{x+iy| y = \operatorname{im} f(0) \}$ is not open, hence $f$ must be a constant.
• Thanks copper.hat! I made an edit to see if I understand right. I don't think this text explicitly has open mapping theorem. – BCLC Aug 12 '18 at 8:21
• That is a pity, I think it is one of the more useful characteristics of analytic functions :-(. – copper.hat Aug 12 '18 at 8:25
• Thanks for the info copper.hat. I expect I will encounter this in higher complex analysis. – BCLC Aug 12 '18 at 8:26
• en.wikipedia.org/wiki/Open_mapping_theorem_(complex_analysis) – copper.hat Aug 12 '18 at 8:27
The mistake is
$$\max_{z \in \gamma}\frac{|f|}{R} = \max_{z \in \gamma}\frac{K}{R}$$
We know that $f=|f|\equiv:K$ on closed unit disc, so including the unit circle. How do we know that $f=|f|\equiv:K$ on some $\gamma$ seemingly pulled out of a hat?
It seems that modifying Liouville's Thm will not work as it did (and should) for some exercises in Ch5. Instead, we prove by Identity Principle (Principle 8.15) as in the OSU homework:
Pf using Identity Principle (Principle 8.15):
After proving that $f$ is constant on closed unit disc, consider $g: \mathbb C \to \mathbb C$ s.t. $g(z) := f(z) - K$. Observe that $g$ is entire and zero on closed unit disc. Now consider a sequence of distinct complex numbers $\{a_n\}_{n=1}^{\infty}$ in the closed unit disc that converges to a complex number $a$ in the closed unit disc. Observe that $g(a_n)=0 \ \forall n \in \mathbb N$. $\because \mathbb C$ is a region, by Identity Principle (Principle 8.15), $g \equiv 0$ on $\mathbb C$. $\therefore, f(z) \equiv K$ on $\mathbb C$.
QED | 2019-07-16T00:18:38 | {
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https://math.stackexchange.com/questions/490405/choosing-numbers-without-consecutive-numbers?noredirect=1 | # Choosing numbers without consecutive numbers.
In how many ways can we choose $r$ numbers from $\{1,2,3,...,n\}$,
In a way where we have no consecutive numbers in the set? (like $1,2$)
• Does the order of choosing matter? Sep 11 '13 at 9:43
• No. I said we choose it to a set... Sep 11 '13 at 9:50
• is there any way to find no of ways of choosing r numbers from 1 to N such that difference between no such pair exists where difference is equal to (k=13)? . For e.g. in the above question k = 1 Oct 18 '18 at 12:05
Assuming that the order of choice doesn’t matter, imagine marking the positions of the $r$ chosen numbers and leaving blank spaces before, between, and after them for the $n-r$ non-chosen numbers; if $r=3$, for instance, you’d get a skeleton like $_|_|_|_$, where the vertical bars represent the positions in $1,2,\ldots,n$ of the chosen numbers. The remaining $n-r$ numbers must go into the $r+1$ open slots in the diagram, and there must be at least one of them in each of the $r-1$ slots in the middle. After placing one number in each of those slots, we have $n-r-(r-1)=n-2r+1$ numbers left to place arbitrarily in the $r+1$ slots. This is a standard stars-and-bars problem: there are
$$\binom{(n-2r+1)+(r+1)-1}{(r+1)-1}=\binom{n-r+1}r$$
ways to do it. The reasoning behind the formula is reasonably clearly explained at the link.
• Here is an answer than (in my opinion) provides a slightly clearer/more intuitive explanation: math.stackexchange.com/questions/677354/… Oct 11 '14 at 19:26
• @user2612743: You mean the accepted answer there? It’s basically just the explanation at the link in mine. Oct 11 '14 at 19:27
• @BrianM.Scott How to approach this problem when numbers are arranged in a circle? Feb 20 '16 at 13:03
• @Mathematics: If the numbers are arranged in a circle, there are $r$ spaces instead of $r+1$, because the end spaces are actually the same space. Feb 20 '16 at 17:56
first we decide, that we will start choosing in an increasing manner... once we have chosen an $i$ we Will not chose any number from $i-1$ to $1$. so after choosing a number we must not choose the next number and thus choosing $r$ numbers we must leave $r-1$ numbers as choosing the last number we will have no restriction for the next.where $r$ is the number of objects to be chosen. so we leaving $r-1$ numbers, we have $n-(r-1)$ or $n-r+1$ numbers remaining. and we can choose $r$ numbers in $\binom{n-r+1}r$ ways.
Let $$g(n,r)$$ be the answer to the OP's question.
There is a simple recursion that $$g$$ satisfies.
$$\tag 1 g(n+1,r) = g(n,r) + g(n-1,r-1)$$
To see this consider the set $$\{1,2,3,\dots,n,n+1\}$$. We can partition the solution set (subsets with $$r$$ elements) into those that contain the number $$n+1$$, call it $$\mathcal N$$, and those that don't, call it $$O$$.
If $$A \in \mathcal N$$ then $$n \notin A$$ and clearly $$|\mathcal N| = g(n-1,r-1)$$.
If $$A \in \mathcal O$$ then $$n+1 \notin A$$ and clearly $$|\mathcal O| = g(n,r)$$.
The total sum is the sum of the blocks, giving us $$\text{(1)}$$.
The function $$g$$ satisfies boundary conditions and without finding a closed formula for $$g$$ we can still use a computer program to calculate $$g(n,r)$$ - see the next section.
There are many paths you can take if you work on finding a closed formula for $$g$$. No doubt, you will eventually find that
$$\tag 2 g(n,r) = \binom{n+1-r}{r}$$
When you plug this into $$\text{(1)}$$ you will see Pascal's rule on your scrap paper.
Python program (using a recursive function)
def daH(x:int,y:int): # g(x,y)=g(x-1,y)+g(x−2,y−1)
if y == 1: # on wedge boundary output known
return x
if x == 2 * y - 1: # on wedge boundary output known
return 1
r = daH(x-1,y) + daH(x-2,y-1)
return r
print('g(7,4) =', daH(7,4))
print('g(10,4) =', daH(10,4))
OUTPUT:
g(7,4) = 1
g(10,4) = 35 | 2021-11-29T12:39:45 | {
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http://mathhelpforum.com/differential-equations/179075-question-about-second-order-differential-equation.html | # Math Help - Question about second order differential equation
1. ## Question about second order differential equation
I have this problem, which says: If the graph of one solution for the equation $y''+P(x)y'+Q(x)y=0$ is tangent to the x axis in some point of an interval [a,b], then that solution must be identically zero. Why?
I've tried to do something with the general expression for the solution.
$y(x)=Ay_1(x)+By_1(x)\int \displaystyle \frac{e^{\int -P(x)dx}}{y_1^2(x)}dx \Rightarrow y'(x)=Ay_1'(x)+By_1'(x)\int \displaystyle \frac{e^{\int -P(x)dx}}{y_1^2(x)}dx +By_1(x)\displaystyle \frac{e^{\int -P(x)dx}}{y_1^2(x)}$
So I've considered y'(x)=0, because y(x) must be tangent to the x axis.
$0=Ay_1'(x)+By_1'(x)\int \displaystyle \frac{e^{\int -P(x)dx}}{y_1^2(x)}dx +B \displaystyle \frac{e^{\int -P(x)dx}}{y_1(x)}\Rightarrow y_1'\left( A+B \int \displaystyle \frac{e^{\int -P(x)dx}}{y_1^2(x)}dx \right) =-B \displaystyle \frac{e^{\int -P(x)dx}}{y_1(x)}$
So, I don't know what to do next. I think this is not the right way.
I think I got it.
y(x_0)=0, it touches the x axis
y'(x_0)=0
Thats all?
2. Originally Posted by Ulysses
I have this problem, which says: If the graph of one solution for the equation $y''+P(x)y'+Q(x)y=0$ is tangent to the x axis in some point of an interval [a,b], then that solution must be identically zero. Why?
I've tried to do something with the general expression for the solution.
$y(x)=Ay_1(x)+By_1(x)\int \displaystyle \frac{e^{\int -P(x)dx}}{y_1^2(x)}dx \Rightarrow y'(x)=Ay_1'(x)+By_1'(x)\int \displaystyle \frac{e^{\int -P(x)dx}}{y_1^2(x)}dx +By_1(x)\displaystyle \frac{e^{\int -P(x)dx}}{y_1^2(x)}$
So I've considered y'(x)=0, because y(x) must be tangent to the x axis.
$0=Ay_1'(x)+By_1'(x)\int \displaystyle \frac{e^{\int -P(x)dx}}{y_1^2(x)}dx +B \displaystyle \frac{e^{\int -P(x)dx}}{y_1(x)}\Rightarrow y_1'\left( A+B \int \displaystyle \frac{e^{\int -P(x)dx}}{y_1^2(x)}dx \right) =-B \displaystyle \frac{e^{\int -P(x)dx}}{y_1(x)}$
So, I don't know what to do next. I think this is not the right way.
I think I got it.
y(x_0)=0, it touches the x axis
y'(x_0)=0
Thats all?
I'm going to make a comment about a theorem that I can't remember the name of and am not sure I know the details of. But it might give you an idea to work with.
Say that the point x' that y(x') = y'(x') = 0 is x' = 0 for simplicity. Then we know that
$y''(0) + P(0)y'(0) + Q(0)y(0) = 0 \implies y''(0) = 0$
(Providing, note, that neither P(x) nor Q(x) are singular at x = 0.) So we know that y(0) = y'(0) = y''(0) = 0. Now take the derivative of the differential equation. This gives you a third degree ODE. Again evaluate the equation at x = 0. This will give y'''(0) = 0. Rinse and repeat.
Now here's that theorem I mentioned: Any function such that $y^{(n)}(0) = 0$ for all non-negative integer n is going to have to be identically 0 or if not is going to be discontinuous.
Whether or not I have the theorem exactly correct I think we have to, at some point, stipulate that y(x) is continuous and that P(x) and Q(x) are $C^{\infty}$.
Perhaps that will generate a fruitful approach for you.
-Dan
3. Would this argument works as well?
The zero function solves the DE and the initial conditions, by inspection. Because the equation is a homogeneous linear equation, the uniqueness theorem shows it is the only solution. QED. | 2014-12-27T13:46:18 | {
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https://www.physicsforums.com/threads/thomas-calc-11th-ed-problem-5-63.630018/ | # Thomas Calc 11th ed., Problem 5.63
I'm just working through this book for self-study, so I hope this isn't a stupid question.
## Homework Statement
What values of a and b maximize the value of $\int_{a}^{b}(x-x^{2}){dx}$?
## Homework Equations
(Hint: Where is the integrand positive?)
## The Attempt at a Solution
Well, the integrand is positive for 0 < x < 1, and the answer in the back of the book is a=0, b=1. That gives a value of (1/2 - 1/3) = 1/6.
But the problem asks about the expression as a whole, not the integrand. And it doesn't specify that a<b. For large values of x, the integrand will take on a large negative value. Then if you reverse the order of the limits, you get a large positive value. So why wouldn't, say, a=10,000 and b=1 yield a much larger value for the integral than the given answer? Or is it always assumed that a<b?
Last edited:
Related Calculus and Beyond Homework Help News on Phys.org
Mark44
Mentor
I'm just working through this book for self-study, so I hope this isn't a stupid question.
## Homework Statement
What values of a and b maximize the value of $\int_{a}^{b}(x-x^{2}){dx}$?
## Homework Equations
(Hint: Where is the integrand positive?)
## The Attempt at a Solution
Well, the integrand is positive for 0 < x < 1, and the answer in the back of the book is a=0, b=1. That gives a value of (1/2 - 1/3) = 1/6.
But the problem asks about the expression as a whole, not the integrand. And it doesn't specify that a<b. For large values of x, the integrand will take on a large negative value. Then if you reverse the order of the limits, you get a large positive value. So why wouldn't, say, a=10,000 and b=1 yield a much larger value for the integral than the given answer? Or is it always assumed that a<b?
I believe that they're tacitly assuming that a and b in the limits of integration are in the order a < b.
Bacle2
I'm just working through this book for self-study, so I hope this isn't a stupid question.....
.
But the problem asks about the expression as a whole, not the integrand
Right, but a Riemann sum will have negative terms when the integrand is negative.
I believe that they're tacitly assuming that a and b in the limits of integration are in the order a < b.
So are you saying that I'm correct that I could get a larger value for the integral with b>a, but that would be violating the tacit assumption?
Right, but a Riemann sum will have negative terms when the integrand is negative.
Yes, but won't reversing the limits make the integral positive?
Mark44
Mentor
Definite integrals are usually written with the lower integration limit being less than the upper limit, and I think it's reasonable to assume that that's what Thomas had in mind for this problem.
The value of the integral will be largest on the largest interval for which the integrand is greater than or equal to zero: namely, the interval [0, 1].
Don't overthink this.
Don't overthink this.
I wouldn't if it was a drill problem, but as the high number indicates, it's in the "theory" part of the exercises, which sort of encourages second thoughts. Also, the concept of reversing the limits to change the sign of the integral was introduced in this same section, so I would have thought they expected me to use it in solving the problems.
Oh well, if all I'm missing is a tacit assumption, then I guess I'm satisfied that I understood the material. Thanks for the responses.
Bacle2
If you reverse the limits of integration,you will basically be seeking to
maximize ∫ (x2-x)dx, instead of ∫ x-x2dx .
This seems a different problem, tho maybe I misunderstood
your goal. Still, the idea would be the same, in that the intervals that maximize
your integral are those where x2-x >0 . | 2020-01-25T18:31:15 | {
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https://community.dataquest.io/t/derivative-for-a0/554427 | # Derivative for a0
My Code:
Can the derivative of a0 be explained? When following along on paper and using the chain/power rule I get
d'(a0) = 2(a0+a1x1^(i) - y^(i)) * d'(a0)(a0 +a1x1^(i) - y^(i))
What I expected to happen:
I expect that the remaining derivative term becomes a constant
What actually happened:
d'(a0) = 2(a0+a1x1^(i) - y^(i))
In the previous example with d’(a1), the derivative term became a constant. In the d’(a0) example, the derivative term simply no longer exists. I am trying to follow along doing the proof on paper but uncertain what happens to the derivative term in d’(a0)
Yes, that’s correct.
\frac{d}{da_0}MSE(a_0, a_1) = \frac{1}{n}\sum\limits_{i=1}^{n}2*(a_0 + a_1x_1^{(i)} - y^{(i)})*\frac{d}{da_0}(a_0 + a_1x_1^{(i)} - y^{(i)})
Now, for
\frac{d}{da_0}(a_0 + a_1x_1^{(i)} - y^{(i)})
we get
\frac{d}{da_0}a_0 + \frac{d}{da_0}(a_1x_1^{(i)}) + \frac{d}{da_0}y^{(i)}
Since the 2nd and 3rd term above don’t include a_0 , their derivative with respect to a_0 will be 0.
And the first term will be equal to 1.
So, we get
\frac{d}{da_0}MSE(a_0, a_1) = \frac{1}{n}\sum\limits_{i=1}^{n}2*(a_0 + a_1x_1^{(i)} - y^{(i)})
or
\frac{d}{da_0}MSE(a_0, a_1) = \frac{2}{n}\sum\limits_{i=1}^{n}(a_0 + a_1x_1^{(i)} - y^{(i)})
1 Like
Okay from what I am learning, the derivative of a constant with respect to a differentiating variable such as: d/da0(a1x1^(i)) = 0; whereas d/da0(a0) = 1.
In the 3rd section of the mission, we differentiated d/da1(a1x1^(i) - y^(i)) and the result was x1. Using the rules above, we get d/da1(a1) * d/da1(x1) - d/da1(y) = 1 * 0 - 0 = 0. This is obviously wrong.
Am I in incorrect in breaking apart the term d/da1(a1x1) in the above example?
Correct, you should not “break apart” this term. The rule is that the derivative of a sum is the sum of the derivatives. It is not true for multiplication (ie the derivative of a product IS NOT the product of the derivatives.) In fact, there is a special formula for calculating the derivative of a product and is called the product rule. That said, the formula is most often used when the product is between two variables and not between a variable and a constant like we are doing here…although you could use it if you wanted. The “better rule” to use here is that when you are finding the derivative of a variable (a1) multiplied by a constant (x1) then you can simply “pull the constant out front” and just find the derivative of the variable. In other words: d/da1(a1x1) = (x1)d/da1(a1) = x1 * 1 = x1.
1 Like | 2022-08-18T14:57:26 | {
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http://math.stackexchange.com/questions/89548/famous-uses-of-the-inclusion-exclusion-principle | # Famous uses of the inclusion-exclusion principle?
The standard textbook example of using the inclusion-exclusion principle is for solving the problem of derangement counting; using inclusion-exclusion (and some basic analysis) it can be shown that $D(n)=\left[\frac{n!}{e}\right]$ which I consider to be quite a beautiful example since it tackles a problem that does not seem to be solvable with such a closed formula in the first place (and also, who expects inclusion-exclusion to yield a closed formula?)
Another standard textbook use is giving a (non-closed) formula for Stirling numbers. This result is less amazing, but is still important enough.
My question is whether there are other nice such examples for using inclusion-exclusion for dealing with "natural" and "famous" problems, preferably problems arising in other fields in mathematics.
Edit: I just remembered another nice example: Proving the formula for $\varphi(n)$ (Euler's totient function) directly (there are other methods as well).
-
Note that Stanley dedicates chapter 2 of Enumerative Combinatoris 1 to Sieve Methods, starting with inclusion-exclusion. Indeed counting derangements is the first example, but other ones are given there as well. – Marc van Leeuwen Dec 8 '11 at 10:46
Inclusion-exclusion is a special case of the generalized Möbius inversion formula on a locally finite poset (partially-ordered set). For example:
• If the poset is the subsets of some given set $S$ with set inclusion as the partial order, you get the classic inclusion-exclusion formula.
• If the poset is the positive integers with divisibility as the partial order, you get the usual Möbius inversion formula in number theory (cf. André Nicolas's answer).
• If the poset is the positive integers with "$\leq$" as the partial order, you get the (backwards) finite difference operator.
• If the poset is bonds of a graph with refinement as the partial order, you get the chromatic polynomial of the graph.
So any application of these could be considered a use of the inclusion-exclusion formula, generally speaking.
I'm not sure I would call them famous, but here are some examples I've seen on MSE. (The second was an answer to one of my questions; all the others are answers I've given.)
• Proving $a!\, b! = \sum_{i=0}^{b}\binom{b}{i}(-1)^{i}\frac{(a+b+1)!}{(a+i+1)}$. See here.
• A combinatorial proof that $\int {x^n e^x dx} = e^x \sum_{k = 0}^n ( - 1)^k \frac{n!}{(n-k)!}x^{n-k} + C$.
• A generalization of the derangement problem involving the number of fixed points with $n$ different permutations (rather than just 1).
• Proving $\sum\limits_{r=0}^n \frac{(-1)^r}{r+1}\binom{n}{r} = \frac1{n+1}$. See here.
-
Bender and Goldman link is 404. – alancalvitti Jul 27 '13 at 14:47
@alancalvitti: Thanks. I'll fix it. – Mike Spivey Jul 27 '13 at 20:19
Don't know if this one classifies as famous, but it is a nice illustration. If $K$ is a finite field with $q$ elements, then the field with $q^m$ elements is contained (as a unique subfield) in the field of $q^n$ elements if and only if $m$ divides $n$. Therefore the number of elements in the field of $q^n$ elements not contained in any proper subfield is, by inclusion-exclusion, given by $$\sum_{d\mid n} \mu(d)q^{n/d}$$ and the number of irreducible polynomials in $K[X]$ of degree $n$ is$$\frac1n{\sum_{d\mid n} \mu(d)q^{n/d}},$$ by the attribution of its minimal polynomial to each element of an algbraic extension of $K$.
-
Note that from this one can quickly deduce the "prime number theorem for function fields," namely that the number of irreducible polynomials of norm less than or equal to $N$ is approximately $\frac{N}{\log_q(N)}$ (where the norm of a polynomial of degree $n$ is $q^n$). – Qiaochu Yuan Dec 8 '11 at 20:39
The following example arose today in a question on MSE. Let $Q(x)$ be the number of square-free integers $\le x$. Then $$\lim_{x\to\infty}\frac{Q(x)}{x}=\prod_{p \text{ prime}}\left(1-\frac{1}{p^2}\right).$$ The product on the right incorporates the principle of Inclusion-Exclusion. If we imagine expanding the product, the term $$-\sum_p \frac{1}{p^2}$$ subtracts the proportions divisible by the various squares of primes. But this counts twice numbers such as $36$ which are divisible by the squares of two distinct primes. So the term $$\sum_{p\ne q} \frac{1}{p^2q^2}$$ adds back the proportions divisible by the squares of two distinct primes, and so on. But that overcounts the proportions divisible by the squares of three distinct primes, and so on. There are quite a few similar uses of Inclusion-Exclusion in Number Theory. Quite often, when the Möbius function $\mu(n)$ is used, Inclusion-Exclusion lies behind the scene.
-
Here is an inclusion-exclusion based on max and min: for any finite set $(T_s)_{s\in {\cal S}}$ of numbers we have $$\bigvee_{s\in {\cal S}} T_s =\sum (-1)^{|A|-1} \bigwedge_{s\in A} T_s\tag1$$ where the sum runs over all non-empty subsets $A$ of $\cal S$.
If $(X_n)$ is a Markov chain with state space ${\cal S}$, and $T_s:=\inf(n\geq 0: X_n=s)$ is the hitting time of state $s$, then the left hand side of (1) is called the cover time. It's the number of steps until every state in ${\cal S}$ is hit.
Calculating the average cover time for a Markov chain is considered a pretty tough problem. In contrast, finding the average of the time $\wedge_{s\in A}\ T_s$ to hit a subset $A$ is fairly standard. That's why I was very surprised to learn the connection between these two problems via (1). I first saw this in Robert Israel's beautiful answer to this MO question.
-
Here's a link to Robert Israel's answer: mathoverflow.net/questions/59244/…. – Srivatsan Dec 8 '11 at 21:49 | 2014-12-20T11:49:25 | {
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https://math.stackexchange.com/questions/2289130/find-xyz-given-that-the-value-of-x2y2z2-xyz-x3y3z3-7/2289158 | Find $xyz$, given that the value of $x^2+y^2+z^2$, $x+y+z=x^3+y^3+z^3=7$
Given that $$x^2+y^2+z^2=49$$ $$x+y+z=x^3+y^3+z^3=7$$
Find $xyz$.
My attempt,
I've used a old school way to try to solve it, but I guess it doesn't work.
I expanded $(x+y+z)^3=x^3+y^3+z^3+3(x^2y+xy^2+x^2z+xz^2+yz^2)+6xyz$
Since I know substitute the given information into the equation and it becomes $112=x^2y+xy^2+xz^2+yz^2+2xyz$
In another hand, I also expanded $(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)$
$7^2=49+2(xy+xz+yz)$,
So from here, I know that $xy+xz+yz=0$.
It seems that I stuck here and don't know how to proceed anymore.
How to continue from my steps? And is there another trick to solve this question? Thanks a lot.
• Hint: $$x^3+y^3+z^3=(x+y+z)^3-3(x+y+z)(xy+yz+xz)+3xyz$$ – Thomas Andrews May 20 '17 at 14:14
• i have made not an error then the result is $$-112$$ – Dr. Sonnhard Graubner May 20 '17 at 14:23
• – lab bhattacharjee May 20 '17 at 16:57
Ok, so you've got $$xy+yz+zx=0$$
Now, we know that $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\\ \implies7-3xyz=7(49-0)\\ \implies xyz=-\dfrac{7\times48}{3}=-112$$
If we identify $x,y,z$ as the roots of a cubic polynomial $av^3+bv^2+cv+d$ with $a\neq0$ then there are recursions in terms of $a,b,c,d$ for the power sums $$P_i:=x^i+y^i+z^i.$$ These are the newton identities. As seen in this link \begin{align} P_0=&+3\\ P_1=&-\dfrac{b}{a}\\ P_2=&-\dfrac{b}{a}P_1-\dfrac{c}{a}2\\ P_3=&-\dfrac{b}{a}P_2-\dfrac{c}{a}P_1-\dfrac{d}{a}P_0\\ \end{align} or \begin{align} P_1a&+b&=0\\ P_1b&+P_2a+2c&=0\\ P_0d&+P_1c+P_2b+P_3a&=0 \end{align} We see that these equations are linear in $a,b,c,d$. Inserting $P_0=3,P_1=7,P_2=49,P_3=7$ and rearranging: \begin{align} \tag1 7a&&+b&&&&&=0\\\tag2 49a&&+7b&&+2c&&&=0\\\tag3 7a&&+49b&&+7c&&+3d&=0 \end{align} Multiply $(1)$ by $7$ and subtract the result by $(2)$. The result is $-2c=0$ so $c=0$. Multiply $(2)$ by $7$ and subtract the result by $(3)$. The result is $336a+7c-3d=0$. Substituting $c=0$ and dividing by $3$ yields $112a-d=0$. Finally according to the link $xyz=-\dfrac{d}{a}=-112$. | 2021-02-27T06:33:11 | {
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https://mathoverflow.net/questions/312944/find-an-integer-coprime-to-sequence-sum | # Find an integer coprime to sequence sum
Given two positive integers $$a, can we always find an integer $$c\in [a, b]$$ that is coprime to $$\sum_{a\le i\le b} i$$?
• Clearly OP is not good at English, but I find the question to be valid. Note that the question is not asking gcd of consecutive integers. – Yuzhou Gu Oct 15 '18 at 21:03
• Almost all of the time, one of the two largest primes primes works. If there are no primes, try b or b-1 or b-2. Gerhard "The Product Is Too Big" Paseman, 2018.10.15. – Gerhard Paseman Oct 15 '18 at 23:39
• I agree; I don't see that the solution is obvious. – Todd Trimble Oct 15 '18 at 23:48
I believe a large counterexample exists
Actually, no, there is no counterexample. Let's consider several cases:
Case 1: $$b-a$$ is even. Then the progression has odd number $$2X+1$$ of terms from $$A-X$$ to $$A+X$$ and the sum is $$A(2X+1)$$. Now let $$p_1 be all (automatically odd) primes that divide $$2X+1$$ but not $$A$$ (at least one such prime exists because otherwise we can just take $$c=A+1$$). Consider the numbers $$A\pm p_1\dots p_{n-1}$$. Note that they are in the range $$[A-X,A+X]$$ and the only common prime factor they may have with $$A(2X+1)$$ is $$p_n$$. But they cannot be divisible by $$p_n$$ simultaneously, so one of them can be taken as $$c$$.
Case 2: $$b-a$$ is odd, $$b=A+X,a=A-X+1$$, $$X$$ is odd. Then the sum is $$(2A+1)X$$. Again, let $$p_1 be all (automatically odd) primes that divide $$X$$ but not $$2A+1$$. This time just consider the number $$\frac{(2A+1)+ p_1\dots p_{n}}{2}$$.
Case 3: $$b-a$$ is odd, $$b=A+X,a=A-X+1$$, $$X$$ is even. Then the sum is still $$(2A+1)X$$. Again, let $$p_1 be all odd primes that divide $$X$$ but not $$2A+1$$. Consider the numbers $$\frac{(2A+1)\pm p_1\dots p_{n}}{2}$$. One of them is now even and the other one is odd. Also they are both in the range $$[A-X+1,A+X]$$. So, the odd one can be taken as $$c$$.
• Hmm. Is b-a really odd in cases two and three? Gerhard "Gerhard Finds Combination At Odds" Paseman, 2018.10.22. – Gerhard Paseman Oct 22 '18 at 14:33
• @GerhardPaseman $b-a=(A+X)-(A-X+1)=2X-1$, which looks odd to me. – fedja Oct 22 '18 at 17:42
• OK. I wonder why in all three cases, you start with "b-a is even". Otherwise, you found the bit I missed. Gerhard "Likes The Idea A Lot" Paseman, 2018.10.22. – Gerhard Paseman Oct 22 '18 at 18:29
• @GerhardPaseman Because I'm too slow to retype and too stupid to correct or even notice everything that should be corrected in copy-paste pieces :-) Changed now. Thank you! – fedja Oct 22 '18 at 22:16
Edit 2018.10.22 It looks like fedja has it. Briefly, if c is not next to the median, try a larger factor of k away from the median, where the factor is coprime to (twice) the median. If the one less than the median is not c, the larger one can be. Fedja's post has the details. End Edit 2018.10.22.
Let's suppose there is a prime (or two) in the interval $$[a,a+k-1]$$. if $$k$$ is even, we hope the prime does not divide $$(k/2)(2a+k-1)$$. This will be the case if the prime is greater than $$k/2$$ and greater than $$a$$. Invoking Chebyshev on the existence of more than one prime in long intervals, we find such a prime in this case
If k is odd, then the prime needs to be larger than k and not divide $$(a +(k-1)/2)$$, meaning it should not be in the middle. So if k is odd and contains a non central prime, again we have a winner.
So we reduce to the cases of no primes in the interval or one central prime when k is odd. Now we turn to results on Jacobsthal's function to show that for k not too small with respect to the sum (like bigger than log a) there must be a number coprime to the sum in the interval. For k very small, there is unlikely to be a failure (I believe one can find such a number), but I do not have a proof.
I will return with more precise estimates on k, which I believe work for any k larger than (log a )^(loglog a).
Update 2018.10.16
It works for short sequences (one can find a coprime number in the interval), and for sequences where k is large enough with respect to a to contain "enough" primes. However, there is still a middle ground which admits a possibility that every number in the interval shares a prime factor with the sum.
The sum $$S$$ can be written in terms of the first number in the interval, $$a$$, and the length of the interval $$k$$. We have $$S = k(a + (k-1)/2)$$, or $$k$$ times the median of the sequence. For many $$a$$, one can pick a number c next to and different from the median, and c will be coprime to the integer which is either the median or twice the median (in case $$k$$ is even), and if c is also coprime to $$k$$, then we are done.
Before we look at two examples, let us note that for $$a+i$$ in the interval, $$(S,a+i) \gt 1$$ iff $$(k((k-1 - 2i)/2),a+i ) \gt 1$$. Thus we need to check that every number in the interval has (or avoids) a prime factor less than $$k$$. It will now be helpful to look at numbers near the median, and write $$j$$ for $$k$$ when $$k$$ is odd, and $$j$$ for $$k/2$$ when $$k$$ is even.
Example for odd $$k$$: $$3j$$ $$2j$$ $$j$$ $$0$$ $$j$$ $$2j$$ $$3j$$
Example for $$k$$ even: $$5j$$ $$3j$$ $$j$$ $$j$$ $$3j$$ $$5j$$
Let's take $$k=j=9$$. The median of an interval with nine integers will always divide the sum, so let's look next to the median. By the observation above and the handy picture, if a number next to the median is coprime to $$j=9$$, then we have found our coprime integer. Well, the number just below the median might be coprime to 9. If it is not (and must be a multiple of three then), the number just above will be because it is less than three away from the lower candidate. So when $$k=9$$, regardless of $$a$$, we know where to look for a candidate.
Let us pick $$k=18$$, so $$j=9$$. Now the picture looks different. The median is not an integer, but the integers next to the median must both have a factor dividing $$j=9$$ to not be coprime to the sum. Since they are adjacent, they both cannot be multiples of three, and thus we know where to look for $$k=18$$.
One sees immediately the generalization to $$k$$ a prime power (even or odd), or twice an odd prime power, which covers all $$k \lt 12$$: for these small $$k$$ and many others, there is a number coprime and it is one of two numbers next to the median.
What about $$k=12$$? We do not have to look far from the median. If there is a sequence of numbers which are sequentially not coprime to 18 6 6 18, we have some potential for a counterexample. But this involves finding two even numbers and two multiples of three to cover the consecutive interval of integers, and it cannot be done. So no counterexample for $$k=12$$.
What about 15? Look at 45 30 15 0 15 30 45. We can use even numbers to cover 30, a multiple of three to cover one 15, and a multiple of five to cover the other 15 (which we can find with the aid of the Chinese Remainder Theorem). However, one finds that neither 45 can be covered by either a multiple of three or by five if we do this. So with $$k=15$$, we may have to stray far from the median to find a coprime, but again the coprime can be found.
A general proof for $$k$$ having two distinct prime factors can be made, but it gets slightly messy. It may be possible to make an even messier proof for $$k$$ with three distinct prime factors. However, I would instead spend the effort finding a counterexample with $$k$$ having at most twenty distinct prime factors. This leads to admissible sequences in prime number theory and the study of prime gaps. Based on my work (start at MathOverflow question 37679), I believe a large counterexample exists with $$k$$ at most $$e^{100}$$ and $$a$$ at most $$e^{e^{4k}}$$.
End Update 2018.10.16
Gerhard "Needs To Look Up ArXiv" Paseman, 2018.10.15. | 2020-10-29T23:03:25 | {
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https://mathematica.stackexchange.com/questions/195813/linearmodelfit-with-standardized-data | # LinearModelFit with Standardized data
I have a small data set of six points:
data1={{2014.,0.015},{2015.,0.005},{2016.,0.0},{2017.,0.01},{2018.,0.02},{2019.,0.014}};
ListPlot[data1
,Frame->True
,PlotRange->{{2013.,2022.},{-0.01,0.03}}
,PlotStyle->Directive[Orange,PointSize[Large]]
]
giving the following plot
The Mean and Standard Deviation of data1 are
Mean[data1]
StandardDeviation[data1]
{2016.5,0.0106667}
{1.87083, 0.00725718}
The data1 can be transformed to data2, having zero Mean and unit Standard Deviation, by utilizing Standardize[.]
data2=Standardize[data1];
ListPlot[data2
,Frame->True
,PlotRange->{{-2.,2.},{-2.,2.}}
,PlotStyle->Directive[Orange,PointSize[Large]]
]
In any case, LinearModelFit[.] allows to fit a polynomial trough the data
lmFit[data_List,degree_Integer]:=LinearModelFit[data,Table[x^i,{i,degree}],x]
Mathematically, a 5th degree polynomial fits exactly through any six data points. However, trying to fit a polynomial of degree=5 to gives quite different results
lmFitPlot[data_List,degree_Integer,{xmin_,xmax_,ymin_,ymax_}]:=Module[{lmf,ss},
lmf=lmFit[data,degree];
ss=Total[lmf["FitResiduals"]^2]; (* Sum of squared residuals *)
Show[
{Plot[lmf[x],{x,xmin,xmax}
,PlotRange->{{xmin,xmax},{ymin,ymax}}]
,ListPlot[data,PlotStyle->Directive[Orange,PointSize[Large]]]
}
,Frame->True
,FrameLabel->{{"",""},{"Year",Row[{"Sum squared residuals= ",ss}]}}
,ImageSize->Medium
]
]
For data1
lmFitPlot[data1,5,{2013.,2022.,-0.01,0.03}]
gives a very noisy fit
While for data2 the fit is quite decent
lmFitPlot[data2,5,{-2.,2.,-2.,2.}]
This discrepancy is caused by the poor rank of the Design Matrix for data1
MatrixRank[lmFit[data1,5]["DesignMatrix"]]
MatrixRank[lmFit[data2,5]["DesignMatrix"]]
3
6
Since there exists is a straightforward Geometric Transformation between data1 and data2
FindGeometricTransform[data1,data2]
my question is: Does there exist an inverse geometric transformation which transforms the decent linear regression model of data2 back to the original coordinate system of data1?
Wikipedia shows a few interesting transformation examples Geometric transformation
Thanks.
• Knowing the TransformationFunction you could use InverseFunction (see documantation ) – Ulrich Neumann Apr 23 '19 at 10:22
• Ulrich Neumann, Thanks for your reply. Yes, I can make the InverseFunction of the Geometric transformation. But how do I transform the polynomial function from the LinearModelFit? I could only find transformations of geometric objects: circle, polygon, etc. – Romke Bontekoe Apr 23 '19 at 14:10
• @ Romke Bontekoe The transformation matrix is a blockmatrix of the form {{A,b},{c^t,1}}. The rational transformation point x->point y is defined as y=(A.x+b)/(c.x+1) – Ulrich Neumann Apr 23 '19 at 14:15
• @Ulrich Neumann, I would much appreciate if you would take the lmFit[data2,5] solution (from the above code) and transform this polynomial as you describe. I failed to do so myself. Please post this as an answer. – Romke Bontekoe Apr 24 '19 at 6:14
As appreciated here my answer concerning the GeometricTransformation
First let's define
y2 = Normal[lmFit[data2, 5]] /. x -> x2(* data2 coordinate system {x2,y2}*)
and the geometric transformation
gt12 = FindGeometricTransform[data1, data2][[2]](* data1~gt12[data2]*)
which transforms {x2,y2} to {x1,y1}(coordinate system data1).
Now the final transformation back to the {x1,y1} coordinates:
{x1, y1} = gt12[{x2, y2}] // Simplify
Show[{ParametricPlot[{x1, y1}, {x2, -1.33631, 1.33631}],Graphics[{Red, Point[data1]}]}, AspectRatio -> 1]
which fit's the data.
Knowing gt12 you can also describe the inverse transformation
gt21=InverseFuncion[gt12] (* gt21[data1]~data2 *)
Chop[ gt21[data1] - data2, 10^-9 ]
(*{{0, 0}, {0, 0}, {0, 0}, {0, 0}, {0, 0}, {0, 0}}*)
• Ulrich Neumann, thank you for this insightful solution. I have learned a lot. – Romke Bontekoe Apr 24 '19 at 15:58
• You're welcome! – Ulrich Neumann Apr 24 '19 at 19:44
• I have one small question still. How did you derive the ParametricPlot interval {x2, -1.33631, 1.33631} from the data endpoints? I am still trying to understand the mechanism of the geometric transformations. – Romke Bontekoe Apr 25 '19 at 6:01
• These values I took from data2 data range! – Ulrich Neumann Apr 25 '19 at 6:12
• I can compute gt21[{2014,0}] and gt21[{2019,0}] and take #[[1]] of these. But maybe you have a more elegant way. – Romke Bontekoe Apr 25 '19 at 6:44
You really have no business fitting or predicting a 5th degree polynomial with 6 data points. To add insult to injury your data has 2 sets of 3 points that are almost perfectly colinear. Finally, you'll have no estimate of error. Hopefully, this is just for a class exercise.
The anomalies that you see are due to a lack of precision. That can be remedied by rationalizing your data and increasing the WorkingPrecision for LinearModelFit and Plot.
data1 = {{2014., 0.015}, {2015., 0.005}, {2016., 0.0}, {2017., 0.01}, {2018., 0.02}, {2019., 0.014}};
data1 = Rationalize[data1, 0];
lmFit[data_List, degree_Integer] := LinearModelFit[data, Table[x^i, {i, degree}], x,
WorkingPrecision -> 50]
lmFitPlot[data_List, degree_Integer, {xmin_, xmax_, ymin_, ymax_}] :=
Module[{lmf, ss}, lmf = lmFit[data, degree];
ss = Total[lmf["FitResiduals"]^2];(*Sum of squared residuals*)
Show[{Plot[lmf[x], {x, xmin, xmax},
PlotRange -> {{xmin, xmax}, {ymin, ymax}},
WorkingPrecision -> 50],
ListPlot[data, PlotStyle -> Directive[Orange, PointSize[Large]]]},
Frame -> True,
FrameLabel -> {{"", ""}, {"Year",
Row[{"Sum squared residuals= ", ss}]}}, ImageSize -> Medium]]
lmFitPlot[data1, 5, {2013, 2022, -0.01, 0.03}]
There's no need to standardize and then transform back in this case:
data2 = Standardize[data1];
lmFitPlot[data2, 5, {-2., 2.5, -2., 2.}]
In this case if you don't take care of the precision issues in the first place, transforming back from the standardized model you might still have the precision issues.
Probably the simplest approach is just to subtract 2014 from the "x" values. Then there's no need to add in a WorkingPrecision statement.
data1 = {{2014., 0.015}, {2015., 0.005}, {2016., 0.0}, {2017., 0.01}, {2018., 0.02}, {2019., 0.014}};
data3 = data1;
data3[[All, 1]] = data3[[All, 1]] - 2014;
lm3 = LinearModelFit[data3, Table[x^i, {i, 5}], x];
Show[Plot[lm3[x - 2014], {x, 2013.5, 2021}],
ListPlot[data1, PlotStyle -> {Red, PointSize[0.02]}]]
• +1 simply for that opening sentence. I found that rather amusing. – Q.P. Apr 23 '19 at 15:31
• @JimB, thanks for your answer. You guessed right, I am developing this example for educational purposes. Most likely, the targeted audience does not have access to arbitrarily high working precisions. Therefore your first solution is not in the right direction, though it is very instructive to me. Your additional solution is exactly overlapping your first solution, to within 10^-15, which I did not expect from an arbitrary shift of the x-values by subtracting 2014. It might be interesting to you that these are real financial ratio data. Such data never come with error bars. Thanks. +1 – Romke Bontekoe Apr 24 '19 at 8:03
• "Such data never come with error bars." Maybe I'm misunderstanding but it's rare that data comes with error bars (from any scientific field). However, one can still estimate the "error" as the variance about the curve is one of the parameters estimated in the fitting of a regression (except when one fits a 5-th degree polynomial with 6 data points. And NOT stating estimates of error when one can is unfortunately not limited to economists/financial folks. – JimB Apr 24 '19 at 14:55
• The lack of precision when using just the raw data comes from, for example, raising 2019 to the 5th power: 2019^5 = 33,549,155,665,686,099. Necessary digits get lost very quickly. After subtracting 2014 from the predictors the largest value is (2019-2014)^5 = 3125 and there is little (if any) loss of precision. Because economists deal with current years and high order polynomials, maybe this is the source of the joke: "Economists were invented to make the meteorologists feel better about their predictions." – JimB Apr 24 '19 at 15:05
• JimB, my plan is to demonstrate the numerical instabilities, which many professionals are unaware of (unfortunately). I want to treat the unknown uncertainty in the data by employing a so called Jeffreys' prior, as part of a Bayesian treatment of this regression problem. Thanks once more. – Romke Bontekoe Apr 24 '19 at 16:05 | 2020-10-30T22:49:45 | {
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https://www.freemathhelp.com/forum/threads/factoring-square-roots.120002/ | # Factoring square roots
#### NeedingWD40
##### New member
I know that one can factor square roots, such as √8 = √(4x2) = √4 x √2 = 2√2 but I don't understand why.
we are saying a1/2.b1/2 = (ab)1/2 which doesn't feel intuitive....
can someone help explain?
#### Subhotosh Khan
##### Super Moderator
Staff member
I know that one can factor square roots, such as √8 = √(4x2) = √4 x √2 = 2√2 but I don't understand why.
we are saying a1/2.b1/2 = (ab)1/2 which doesn't feel intuitive....
can someone help explain?
Law of exponent tells you:
(a * b)m = am * bm
#### NeedingWD40
##### New member
Law of exponent tells you:
(a * b)m = am * bm
Thanks - and I can understand this using non-fractional indices as, say (2*4)3 = (2*4)(2*4)(2*4) , then rearrange to get 24*43 but I don't see this with square roots...
#### Mr. Bland
##### Junior Member
Recall that $$\displaystyle \sqrt[y]{x} = x^{\frac{1}{y}}$$. This works the same as any other exponent.
EDIT:
If you're confused about "what does a non-integer exponent mean", that's understandable because it's not intuitive given the usual description of an exponent. What does it mean to multiply something by itself three and a half times? This can be written as $$\displaystyle x^{3\frac{1}{2}} = x^{\frac{7}{2}} = \sqrt[2]{x^7}$$. Eventually it boils down to a matter of calculating a point on a curve, which has points in between integer values.
Last edited:
#### JeffM
##### Elite Member
I know that one can factor square roots, such as √8 = √(4x2) = √4 x √2 = 2√2 but I don't understand why.
we are saying a1/2.b1/2 = (ab)1/2 which doesn't feel intuitive....
can someone help explain?
This is a GREAT question.
$$\displaystyle \text {BY DEFINITION, } x * x = y \iff x = \sqrt{y}.$$
OK with that?
$$\displaystyle \text {Given } a = \sqrt{p}, \text { and } b = \sqrt{q}.$$
$$\displaystyle \therefore a * b = \sqrt{p} * \sqrt{q}.$$
$$\displaystyle \therefore a * (a * b) = \sqrt{p} * (\sqrt{p} * \sqrt{q}) = (\sqrt{p} * \sqrt{p}) * \sqrt{x} = p * \sqrt{q} \implies$$
$$\displaystyle (p * \sqrt{q}) * \sqrt{q} = (a * (a * b)) * b \implies p * (\sqrt{q} * \sqrt{q}) = (a * (b * a)) * b \implies$$
$$\displaystyle p * q = ((a * b) * a) * b = (a * b) * (a * b) \implies (a * b) * (a * b) = p * q.$$
$$\displaystyle \therefore a * b = \sqrt{p * q}.$$
$$\displaystyle \text {But } a * b = \sqrt{p} * \sqrt{q}.$$
$$\displaystyle \text {THUS, } \sqrt{p} * \sqrt{q} = \sqrt{p * q}.$$
That is abstract. Let's take a numeric example.
$$\displaystyle \sqrt{25} * \sqrt{49} = 5 * 7 = 35.$$
$$\displaystyle 35^2 = 30^2 + 2(30)(5) + 5^2 = 900 + 300 + 25 = 1225 = 1250 - 50 = 25(50 - 1) = 25 * 49.$$
$$\displaystyle \therefore \sqrt{25 * 49} = \sqrt{35^2} = 35.$$
$$\displaystyle \therefore \sqrt{25} * \sqrt{49} = \sqrt{25 * 49}.$$
This result can be extended to any root.
#### NeedingWD40
##### New member
Yay! Thank you. I could follow it, I don't think I could reproduce it yet, but I could follow the steps so that's a good start and I certainly feel more convinced. Thank you very much.
#### NeedingWD40
##### New member
Recall that $$\displaystyle \sqrt[y]{x} = x^{\frac{1}{y}}$$. This works the same as any other exponent.
EDIT:
If you're confused about "what does a non-integer exponent mean", that's understandable because it's not intuitive given the usual description of an exponent. What does it mean to multiply something by itself three and a half times? This can be written as $$\displaystyle x^{3\frac{1}{2}} = x^{\frac{7}{2}} = \sqrt[2]{x^7}$$. Eventually it boils down to a matter of calculating a point on a curve, which has points in between integer values.
Thanks Mr Bland. I am happy about the definition of non-integer exponents, and I like your link to visualising them as points on a curve - very helpful. I just couldn't 'feel' why what worked for integer exponents would also work for non-integer. JeffM helped too!
#### JeffM
##### Elite Member
Yay! Thank you. I could follow it, I don't think I could reproduce it yet, but I could follow the steps so that's a good start and I certainly feel more convinced. Thank you very much.
Couple of things.
A mathematician observes relationships such as
$$\displaystyle \sqrt{1} * \sqrt{169} = 1 * 13 = 13 = \sqrt{169} = \sqrt{1 * 169}.$$
$$\displaystyle \sqrt{4} * \sqrt{9} = 2 * 3 = 6 = \sqrt {36} = \sqrt{4 * 9}$$
$$\displaystyle \sqrt{9} * \sqrt{25} = 3 * 5 = 15 = \sqrt{225} = \sqrt{9 * 15}$$
That gets a mathematician wondering under what conditions this very simple relationship is true. In other words, what frequently happens is that something is true in specific cases, and we try to find out what is the general case such that that something is true. You probably did not notice that I restricted the discussion to non-negative real numbers. The reason is that the relationship is not so simple if we extend it beyond non-negative real numbers. There is a similar but more complex relation that applies more generally.
The point is: before trying to develop or understand a general proof, try some very simple examples.
Number two. Detailed proofs in algebra are frequently hard to understand because they are lengthy strings of blindingly obvious steps. The mind goes: of course, of course, of course ... WHAT how did we get here. If it is hard even to follow the proofs, do not worry that you cannot yet create such proofs.
Finally, rational exponents are fairly easy to understand with this definition
$$\displaystyle \text {Given: } r \in \mathbb R, \ a,\ c. \in \mathbb Z,\ r > 0,\ a \ge 0, \text { and } c \ge 1, \text { then}$$
$$\displaystyle r^a \equiv (r^{(a/c)})^c.$$
If you think about this definition, you will realize that it simply means (under the conditions given)
$$\displaystyle r^{(a/c)} \equiv \sqrt[c]{r^a}.$$
Last edited:
#### NeedingWD40
##### New member
Finally, rational exponents are fairly easy to understand with this definition
$$\displaystyle \text {Given: } r \in \mathbb R, \ a,\ c. \in \mathbb Z,\ r > 0,\ a \ge 0, \text { and } c \ge 1, \text { then}$$
$$\displaystyle r^a \equiv (r^{(a/c)})^c.$$
If you think about this definition, you will realize that it simply means (under the conditions given)
$$\displaystyle r^{(a/c)} \equiv \sqrt[c]{r^a}.$$
Thank you very much for those pointers - and for the encouragement (yet!). I will keep your advice in mind. I think my big problem is the meeting in the middle - you know where to start, and where you want to finish, it's the meeting in the middle that I can't find!
The rational exponents explanation is delightful. That was fun. I first sat and carefully thought through: Why that condition on that variable? Ok, zero raised to any power is zero, so wouldn't it still work for zero? But then I thought, except for raising to the power zero, 00 = 1... so does it still work?
02/3 = $$\displaystyle \sqrt[3]{0^2}$$ = 0
but
00/3 = $$\displaystyle \sqrt[3]{0^0}$$ = $$\displaystyle \sqrt[3]1$$ = 1 and if we cube both sides
(00/3)3 = 01=0
but
($$\displaystyle \sqrt[3]{0^0}$$)3 = 00 = 1.
So that's why zero was excluded?
#### Cubist
##### Junior Member
So that's why zero was excluded?
JeffM's constraints work, logically they are correct, but they don't show the full range of numbers for which the identity holds. It works under other circumstances too.
See this post about "power of power" and contrainsts (click)
I'm guessing that it would become increasingly difficult to mathematically prove that this property holds under all of these different circumstances.
#### JeffM
##### Elite Member
Thank you very much for those pointers - and for the encouragement (yet!). I will keep your advice in mind. I think my big problem is the meeting in the middle - you know where to start, and where you want to finish, it's the meeting in the middle that I can't find!
The rational exponents explanation is delightful. That was fun. I first sat and carefully thought through: Why that condition on that variable? Ok, zero raised to any power is zero, so wouldn't it still work for zero? But then I thought, except for raising to the power zero, 00 = 1... so does it still work?
02/3 = $$\displaystyle \sqrt[3]{0^2}$$ = 0
but
00/3 = $$\displaystyle \sqrt[3]{0^0}$$ = $$\displaystyle \sqrt[3]1$$ = 1 and if we cube both sides
(00/3)3 = 01=0
but
($$\displaystyle \sqrt[3]{0^0}$$)3 = 00 = 1.
So that's why zero was excluded?
JeffM's constraints work, logically they are correct, but they don't show the full range of numbers for which the identity holds. It works under other circumstances too.
See this post about "power of power" and contrainsts (click)
I'm guessing that it would become increasingly difficult to mathematically prove that this property holds under all of these different circumstances.
It is tempting to adopt, and many mathematicians feel no difficulty in adopting, a Platonist view of mathematics. (Platonism taken to its ultimate conclusion is the philosophy that your mother has never really existed but is merely the shadow of the one real mother that no living human has ever known.) In that view, the human mind discovers mathematics. The other view, the one that I prefer, is that the human mind creates mathematics, in part by assuming that observable regularities in the physical world are invariably true in the idealized world of the intellect. The Platonists assume that the invisible realm is logically consistent and thus discoverable by reason. The non-Platonist requires that what we create be logically consistent. In other words, both agree that valid mathematics is logically consistent, and thus the two schools can live in reasonable amicability. For non-Platonists like me, definitions are free exercises of the mind and judged solely on utility and consistency with other definitions.
Notice that I did not define $$\displaystyle r^a.$$ Obviously my definition of a rational exponent depends on my definition of an integer exponent so what I said before was incomplete. Here is a set of definitions that I like for integer exponents
$$\displaystyle \text {Given } r \in \mathbb R, \ a \in \mathbb Z, \text { and } r > 0, \text { then}$$
$$\displaystyle a = 0 \implies r^a \equiv 1, \text { and } r^a \equiv r * r^{(a - 1)}.$$
Now why did I pick that definition?
Mainly because it has the very nice consequence that any integer power of r is a real number. If I allowed r to be 0, then negative powers would not be real numbers. Moreover if I allowed r to be less than 0, then some rational exponents would not be real numbers.
Thus, the reason that I stipulated that r > 0 for rational exponents is that I had implicitly stipulated that for integer exponents.
In fact, if you are willing to restrict exponents to non-negatice rational numbers, there is no reason to restrict r to positive real numbers; it will work just fine to restrict r to non-negative real numbers. In particular, if r is a real number, q is a rational number, and both are non-negative, then no problem arises from a definition of r^q that entails that 0^0 = 1. However, mathematicians may want to work with real functions to a power that is a real function. In that case, allowing both functions to be zero simultaneously can cause problems in the field of mathematics known as analysis. Thus, some mathematicians define things so 0^0 is not defined, and some mathematicians define things so that 0^0 = 1. For more details, see
As cubist correctly said, we can define exponents so that they make sense and are useful with fewer restrictions than I gave. If we do so, however, the laws of exponents become more complex. If we define exponents in a way that restricts r to the positive reals, we get a set of laws for exponents with no exceptions. In practice, I use exponents with numbers that are not positive reals only after checking that I am not about to be bitten by the exceptions.
#### Mr. Bland
##### Junior Member
The non-Platonist requires that what we create be logically consistent.
The non-Platonist has never seen the company dress code.
#### Otis
##### Senior Member
The non-Platonist has never seen the company dress code.
Is that because the company's dress doesn't really exist, but is merely a reflection of the one, true dress that no human has ever worn? Philosophy is so confusing …
$$\;$$
#### pka
##### Elite Member
When I am confronted with questions about the reality of mathematics, I ask the person if s/he has read THE NUMBER SENSE how he mind creates mathematics. by Stanislas Dehaene? Dehaene is a well known brain scientist who divides his time between University of Paris and various universities in North America. | 2020-02-20T13:34:11 | {
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https://math.stackexchange.com/questions/19471/is-a-smooth-function-convex-near-a-local-minimum | # Is a smooth function convex near a local minimum?
I would like to know if every sufficiently differentiable function is convex near a local minimum. The background to my question is that I became curious if one could motivate the usefulness of convex optimization techniques by saying that at least locally they work for all continuous functions.
Unfortunately I discovered there are $C^1$-functions which have minima where they are not locally convex. But my construction involves the use of everywhere non-differentiable continuous functions so I started to wonder if perhaps $C^2$-functions are convex near a local minimum.
No, this isn't true. Let $\phi(x)$ be a nonnegative smooth "bump" function that is zero except on $(0,1)$ where it is positive. Then $\psi(x) = \sin^2({1 \over x})\phi(x)$ is a smooth nonnegative function (define $\psi(0) = 0$) which has zeroes at $x = {1 \over k\pi}$ for positive integers $k$ but is positive between these zeroes. $\psi(x)$ will still have a local minimum at zero because $\psi(0) = 0$, but because of the humps in $\psi(x)$ between zeroes, a chord connecting $({1 \over k\pi},0)$ to $({1 \over (k +1)\pi},0)$ will lie below the graph. So $\psi(x)$ is not convex.
I should add that if $x_0$ is a local minimum of $f(x)$ such that $f^{(l)}(x_0)$ is nonzero for some $l > 0$, then it will be convex on some interval centered at $x_0$ as long as $f(x)$ is $C^{l+1}$. To see this, note that without loss of generality we may assume $l$ is minimal. By Taylor expanding one gets $$f(x) = {1 \over l!} f^{(l)}(x_0)(x - x_0)^l + O((x - x_0)^{l+1})$$ If $|x - x_0|$ is sufficiently small the remainder term will be dominated by the ${1 \over l!} f^{(l)}(x_0)(x - x_0)^l$ term. Thus the only way a local minimum can occur is if $l$ is even and $f^{(l)}(x_0) > 0$. Next note that the second derivative of $f$ has Taylor expansion given by $${d^2 f \over dx^2} = {1 \over (l-2)!} f^{(l)}(x_0)(x - x_0)^{l-2} + O((x - x_0)^{l-1})$$ The remainder term is domninated by the first term once again, which is nonnegative on an interval containing $x_0$ since $l$ is even and $f^{(l)}(x_0) > 0$. Thus the second derivative of $f$ is nonnegative on this interval and therefore the function is convex there.
• Your added argument can be extended to higher dimensions: If $f$ is a $C^{2}$ function having vanishing gradient at $x_{0}$ and positive definite Hessian $Hf(x_{0}$ at $x_{0}$ (in particular $x_{0}$ is an isolated minimum) then it is strictly convex in a neighborhood of $x_{0}$. Indeed, the positive definite matrices are open in the space of symmetric matrices, $x \mapsto Hf(x)$ is continuous since $f$ is $C^{2}$ and a function with positive definite Hessian is strictly convex. – t.b. Jan 29 '11 at 20:10
Say we have y = f(x1,x2,..) for (x1,x2,,) at least two dimensions.
We can construct smooth f such that f(x,0,0..) = a x^2 , f(0,x,0..) = a x^2 , and f(x,x,0..) = b x^2
In other words a smooth pinch.
If we take a chord between (x,0,..) and (0,x,..) then f = a x^2 at the ends. The midpoint is (x,x,..)/sqrt(2), and f = b x^2 / sqrt(2)
So f is not convex if b x^2 / sqrt(2) > a x^2 , or b > sqrt(2) a , which is easily satisifed.
So, a smooth minimum does not have to be convex.
I would just like to point out an elementary counterexample that satisfies most desirable properties. In particular, this example shows the issue is neither smoothness nor the fact the minimum is not an isolated critical point. Consider then $$f \colon (x,y) \in \mathbb R^2 \longmapsto x^2y^2 + (x+y)^2$$. Clearly $$f$$ is $$\mathcal C^\infty$$, furthermore $$f(x,y) > 0$$ for all $$(x,y) \neq 0$$ while $$f(0,0) = 0$$. Its gradient is, $$$$\nabla f(x,y) = \begin{bmatrix} 2xy^2 + 2(x+y) \\ 2x^2y + 2(x+y) \end{bmatrix}.$$$$ If it vanishes, $$xy^2 = x^2y$$, that is either $$x=0$$ but then $$y=0$$, or $$y=0$$ but then $$x=0$$, or $$x=y$$ in which case $$x^3+2x=x(x^2+2)=0$$ that is $$x=y=0$$ as well. This establishes that $$f$$ has a unique critical point which is its global minimum, $$(0,0)$$.
Despite all these properties, $$$$\begin{bmatrix}1\\-1\end{bmatrix} \nabla^2f(x,x) \begin{bmatrix}1\\-1\end{bmatrix} = - 4x^2 < 0,$$$$ for all $$x \neq 0$$, therefore there exists no neighborhood of $$(0,0)$$ on which $$f$$ is convex.
There is a related question here that has not been resolved, the hypotheses are stronger and this particular counterexample fails to satisfy them. | 2020-10-19T16:04:13 | {
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https://spivey.oriel.ox.ac.uk/corner/Termination_of_tree_rotations | # Termination of tree rotations
A tutorial problem for the course Algorithms and Data Structures asks for a proof that tree rotations can transform a binary tree into any other tree of the same size with the nodes in the same infix order. The suggested argument is that one tree can be transformed into the other via a normal form, such as a rightward-pointing spine.
The lecturer's answer suggests an algorithm for transforming any tree to normal form: find a node (perhaps the rightmost one) on the spine that has a left child, rotate it to the right, and repeat until there are no more nodes with left children. It's clear that, if this algorithm terminates, it does so in a state where the tree has been reduced to a rightward-pointing spine as desired. But can we be sure of termination? The model answer says only that in each step one node has reached its "correct position," but I don't quite know what that means. A clearer argument might be to exhibit a bound function, non-negative for every tree, that is reduced in each rotation. Such a bound function is the number of nodes that are not on the spine, that is, the total size of all left subtrees of nodes on the spine. This function is clearly non-negative, and it decreases by one whenever a node on the spine is rotated to the right.
A script for the Boyer-Moore theorem prover follows. For simpicity, we will treat binary trees with labelled leaves but unlabelled internal nodes, because this variation on trees is provided by the LISP-like s-expressions already present in the prover.
After resetting the theorem prover to its ground-zero state, we begin by defining a predicate normp that tests whether its argument is in our desired normal form.
(boot-strap)
(defn normp (x)
(if (nlistp x) t
(and (nlistp (car x)) (normp (cdr x)))))
Our challenge is to get the prover to accept our proposed definition of normalise, because the prover checks that each function is total before accepting its definition. To that end, we begin be defining a function step that performs one step of normalisation, searching for the highest node in the spine that has a non-trivial left subtree, and rotating it.
(defn step (x)
(if (nlistp x) x
(if (nlistp (car x))
(cons (car x) (step (cdr x)))
(cons (caar x) (cons (cdar x) (cdr x))))))
We must now define a measure cost that is reduced by step when applied to a tree that is not in normal form. As suggested above, we define it as the sum of the sizes of all left subtrees of nodes on the spine.
(defn size (x)
(if (nlistp x) 0
(plus (size (car x)) (size (cdr x)) 1)))
(defn cost (x)
(if (nlistp x) 0
(plus (size (car x)) (cost (cdr x)))))
It is a straightforward induction to prove that step does reduce cost as we want.
(prove-lemma cost-step (rewrite)
(implies (not (normp x))
(lessp (cost (step x)) (cost x))))
That gives us what we need to bully the prover into accepting our definition of normalise, which we do by giving it a hint that it should prove termination by using the measure (cost x) and the well-founded relation lessp.
(defn normalise (x)
(if (normp x) x
(normalise (step x)))
((lessp (cost x))))
We can eaily prove the normalise returns a tree in normal form.
(prove-lemma normalise-is-normp (rewrite)
(normp (normalise x)))
An alternative algorithm norm1 for normalising a tree is to work our way up the right spine, recursively grinding each left subtree into atoms that we cons onto the spine.
(defn grind (x y)
(if (nlistp x)
(cons x y)
(grind (car x) (grind (cdr x) y))))
(defn norm1 (x)
(if (nlistp x) x
(grind (car x) (norm1 (cdr x)))))
This function returns the same result as normalise, as we can prove in stages. First, we must prove the technical result that if its input is already in normal form, then norm1 returns it unchanged.
(prove-lemma normp-norm1 (rewrite)
(implies (normp x)
(equal (norm1 x) x)))
Also, norm1 applied to (step x) gives the same result as norm1 applied to x: that means that as normalise transforms its input step-by-step, the normal form as returned by norm1 does not change.
(prove-lemma norm1-step (rewrite)
(equal (norm1 (step x)) (norm1 x)))
An easy induction is all that is now needed to show that norm1 and normalise are the same.
(prove-lemma norm1-is-normalise (rewrite)
(equal (norm1 x)
(normalise x))) | 2023-03-27T01:51:07 | {
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https://www.physicsforums.com/threads/finding-a-closed-form-expression-given-decimal-approximation.844439/ | # Finding a closed form expression given decimal approximation
1. Nov 21, 2015
### fedaykin
Good evening. Is there a way to take a decimal approximation and see if there is a relatively simple expression?
I'm guessing there might be software for this, but I'm not sure I'm even asking the appropriate question.
If it matters, the number I'm after is $$\sqrt{1+\sqrt{2+\sqrt{4+\sqrt{8+\sqrt{16+ ... }}}}}$$ . This is the powers of 2 under a nested radical.
2. Nov 21, 2015
### fzero
You can apply Ramanujan's method to this. Set
$$t =\sqrt{1+\sqrt{2+\sqrt{2^2+\sqrt{2^3+\sqrt{2^4+ \cdots }}}}}.$$
Then
$$t^2 = 1 + \sqrt{2} t,$$
and we must take the positive root.
3. Nov 22, 2015
### Staff: Mentor
If I calculate $$\sqrt{2} t$$ then the powers of 2 under the roots increase exponentially:
$$\sqrt{2+\sqrt{2^3+\sqrt{2^6+\sqrt{2^{11}+\sqrt{2^{20}+ \cdots }}}}}.$$
Am I mistaken somewhere or is $$t^2 = 1 + \sqrt{2} t$$ meant as an approximation?
4. Nov 22, 2015
### fzero
The first few terms in the sequence for $t$ are {1., 1.55377, 1.73205, 1.78812}, so this certainly looks convergent. You might try to prove a bound on convergence for the expression where we replace 2 by $n$.
5. Nov 23, 2015
### fedaykin
Thank you guys! It's a lead!
6. Nov 23, 2015
### Svein
And then it is simple to solve: $t^{2}=1 + \sqrt{2}t$, rearrange: $t^{2}-\sqrt{2}t-1=0$ and solve: $t=\frac{\sqrt{2}\pm\sqrt{2+4}}{2}=\frac{\sqrt{2}}{2}(1\pm \sqrt{3})$. Obviously, we must use the positive sign and get $t=\frac{\sqrt{2}}{2}(1+ \sqrt{3})$ (the numerical value is 1.931852...).
7. Nov 24, 2015
### fedaykin
I've made a nice little excel file to help study these. Interestingly, if you use 4 as a base, it seems to converge to exactly 2. It's the only number so far that I've seen do this. I've tried powers of 2 up to 2^16, and those appear non-rational so far.
If anyone wishes to use it, you'll have to create a module (or some other means) of implementing the following in VBA:
Option Explicit
Public Function goatVal(theInput As String) As Double
goatVal = Evaluate(theInput)
End Function
I can't attach it as macro enabled (wisely), so unfortunately, you'll have to add it manually. The previous code allows you to use the Evaluate function as a standard cell function. That is, it Excel will try to evaluate the contents of a cell as if they were a formula if you put " =goatVal(cellReference) " into them. Be careful, this can take up a decent amount of processor time. It gives me a !value error if the string is getting too large, but seems reasonable otherwise.
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8. Nov 25, 2015
### HallsofIvy
The basic idea is this: set $t= \sqrt{1+ \sqrt{2+ \sqrt{2^2+ \sqrt{2^3+ \cdot\cdot\cdot}}}}$. Squaring both sides, $t^2= 1+ \sqrt{2+ \sqrt{2^2+ \sqrt{2^3+ \cdot\cdot\cdot}}}$. So $t^2- 1= \sqrt{2+ \sqrt{2^2+ \sqrt{2^3+ \cdot\cdot\cdot}}}$ and, factoring a $\sqrt{2}$ out of the right, $t^2- 1= \sqrt{2}\sqrt{1+ \sqrt{2+ \sqrt{2^2+ \cdot\cdot\cdot}}}= \sqrt{2}t$.
9. Nov 25, 2015
### Staff: Mentor
I can see this delivers an upper bound and thus proves convergence. What I cannot see, as mentioned in #3, is equality by factoring out $\sqrt 2$. Since all of you insist on the equality I assume I was mistaken. Can somebody enlighten me why $\sqrt 2 t > t^2 - 1$ is wrong?
10. Nov 25, 2015
### HallsofIvy
First, you want to calculate a specific numbers. So, as I and others said, we set t equal that number. How in the world did you convert that equation to an inequality?
11. Nov 25, 2015
### Staff: Mentor
$$\sqrt{2} t = \sqrt{2} \sqrt{1+\sqrt{2+\sqrt{2^2+\sqrt{2^{3}+\sqrt{2^{4}+ \cdots }}}}} = \sqrt{2 \cdot 1 + 2 \cdot \sqrt{2+\sqrt{2^{2}+\sqrt{2^{3}+ \cdots }}}} = \sqrt{2 + \sqrt {4 \cdot 2+ 4 \cdot \sqrt{2^{2}+\sqrt{2^{3}+ \cdots }}}} = \sqrt{2 + \sqrt{2^3 +\sqrt{16 \cdot 2^2 +16 \cdot \sqrt{2^{3}+ \cdots }}}} = \cdots = \sqrt{2+\sqrt{2^3+\sqrt{2^6+\sqrt{2^{11}+\sqrt{2^{20}+ \cdots }}}}}.$$
12. Nov 25, 2015
### fedaykin
One thing I don't quite understand yet:
Evaluating the approximation to these (with several bases) leads to a value that appears to always be less than using Ramanujan's method.
For example: let n = 4 (where n is the base of powers under the radical) and you get:
$$t^2-2t-1=0$$
$$t=1+\sqrt{2} \approx {2.414}$$
$$t= 1 - \sqrt{2} \approx {-0.414}$$
But, if you were to guess at the value from calculation:
$$\sqrt{1+\sqrt{4+\sqrt{16+\sqrt{64+\sqrt{256}}}}} \approx{1.999}$$
By the way, the calculated approximation does not appear to be related to the two solutions from Ramanujan's method. For the number 4, it's just a coincidence that it is the sum. Perhaps these have a very slow rate of convergence and have a 'false positive' convergence to a different value?
To everyone that's been so helpful in this thread, you've brought me a lot of joy in understanding these. I first saw the representation of the golden ratio as the powers of one under the radical, and I've been working on these on and off since college. Thank you so much!
13. Nov 26, 2015
### Samy_A
I also think something is wrong here. For your first expression, the calculated value is $t=\frac{\sqrt{2}}{2}(1+ \sqrt{3})=1.93...$, but your own Excel calculation gives 1.783...
I also did a calculation in Excel, and get the same results: 1.783... for the first expression, 2 for the second expression (the one with 4).
I tried Ramajunan's method on the first expression, factoring out the $\sqrt{2}$, but didn't get the result shown above.
EDIT: here they also find 1.783... as solution, but no closed-form solution.
Last edited: Nov 26, 2015 | 2018-02-23T15:28:25 | {
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https://math.stackexchange.com/questions/3125084/if-we-are-handed-the-presentation-langle-i-j-k-mid-i2-j2-k2-ijk-rangle-a | # If we are handed the presentation $\langle i,j,k \mid i^2=j^2=k^2=ijk \rangle$ and nothing more, can we deduce that this is the quaternion group?
If we are handed the group presentation $$\langle i,j,k \mid i^2=j^2=k^2=ijk \rangle$$ and nothing more, can we deduce that this is the quaternion group?
Nothing in this presentation tells us that $$i^2=j^2=k^2=ijk=-1$$ and that $$i^4=j^4=k^4=(ijk)^2=1$$. Can we conclude these relations from the relation given in the presentation?
• I don't know whether we (you or I) can deduce that it is isomorphic to the quaternion group. I have not tried. But I generally trust computer algebra systems like GAP with problems like this, and they can indeed deduce that it is quaternion of order 8. I would guess that it is not too hard to do it by hand. – Derek Holt Feb 24 at 18:16
The cancellation laws immediately get us $$i=jk$$ and $$k=ij$$. Multiply the first by $$i$$ on the right, and $$i^2=jki$$, leading to $$j=ki$$ and completing that cycle.
Now, applying these laws $$j^2=i^2$$, $$ij=k$$, $$jk=i$$, we have the following chain of equalities: $$j^4i=j^2i^3=j^2ij^2=j^2kj=jij=jk=i$$ Apply the cancellation law to that and $$j^4=1$$. From there, $$i^4=k^4=(ijk)^2=1$$ follow easily.
• Very clean! ${}$ – Jyrki Lahtonen Feb 24 at 19:33
Repeatedly using the existence of inverses in groups gives $$ijk=k^2\implies ij=k,\,ijk=i^2\implies jk=i,\,kijk=k^3\implies kij=k^2=j^2\implies ki=j.$$Define $$m:=k^{-1}ji=i^2$$; we would normally call this $$-1$$. Since $$m=i^2=j^2=k^2$$, $$m$$ commutes with everything so $$ji=mk,\,ik=j^{-1}mk^2=mj,\,kj=k^2mi^{-1}=mi.$$Finally, $$m^2=k^{-1}mkk^{-1}ji=k^{-1}ij$$ is the identity.
For this particular group presentation there is a simple way to use cancellation to identify it. First define $$\,u := ii = jj = kk = ijk\,$$ which commutes with $$\,i,j,k.\,$$ Now $$\,(ij)k = u = kk\,$$ and using cancellation $$\,ij=k.\,$$ Similarly, $$\,i(jk) = u = ii\,$$ and $$\,jk=i.\,$$ Next, $$\,(ki)j = k(ij) = kk = u = jj\,$$ and using cancellation $$\,ki=j.\,$$ Next, $$\, uk = (jj)k = j(jk) = ji.\,$$ Similarly, we get $$\,ui = kj, \, uj = ik.\,$$ Finally, $$\,uuk = uji = iiji = iki = ij = k,\,$$ and thus $$uu = 1.$$ The group has eight elements $$\,\{1,i,j,k,u,ui,uj,uk\}\,$$ and isomorphic to the quaternion group. | 2019-05-20T04:25:28 | {
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http://freshfish4u.com/5vmyww1p/fe2fe4-quadratic-function-examples-with-answers | Find the solutions to the quadratic equation $$x^2-13x+12=0$$. File Type PDF Quadratic Function Examples And Answers highest power. The following video explains how the quadratic graph can show the number of solutions for the quadratic equation and the values of the solutions. And you should get the answers −2 and 3; R 1 cannot be negative, so R 1 = 3 Ohms is the answer. How to solve quadratic equations graphically using x-intercepts. The "basic" parabola, y = x 2 , looks like this: The function of the coefficient a in the general equation is to make the parabola "wider" or "skinnier", or to turn it upside down (if negative): About "Quadratic functions questions and answers" Quadratic functions questions and answers : Here we are going to see some practice questions on quadratic functions. Note: A quadratic equation ax 2 + bx + c = 0 will have reciprocal roots, if a =c . 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https://math.stackexchange.com/questions/2609738/signs-of-eigenvalues-of-3-by-3-matrix | # Signs of eigenvalues of $3$ by $3$ matrix
I'm learning linear algebra, specifically quadratic forms, and need help with the following exercise :
Let $f(x, y, z) = 5x^2 + 12xy + 9y^2 + 6xz - 10yz$.
$(1)$ Find the matrix representation $\mathbf x^T A \mathbf x$ of $f$.
$(2)$ Find the signs of the eigenvalues of $A$.
Here's my work so far :
$(1)$ The matrix representation of $f$ is given by $$\begin{bmatrix}x&y&z\\ \end{bmatrix} \begin{bmatrix}5&6&3\\6&9&-5\\3&-5&0\\ \end{bmatrix} \begin{bmatrix}x\\y\\z\\ \end{bmatrix} = \mathbf x^T A \mathbf x$$
where $A$ is a symmetric matrix, i.e. $A$ has real eigenvalues.
$(2)$ This is the part where I need your help. I'm still going to share my thoughts and what I have found so far.
First we note that $\det(A) = -386 < 0.$ Therefore, by Sylvester's criterion, the matrix $A$ is not positive definite since all principal minors determinants are not all positive. Secondly, by the relation between the eigenvalues and the determinant of $A$ we know that $\det(A) = \lambda_1\lambda_2\lambda_3 < 0$. We therefore have two options : either all eigenvalues are negative or either two of them are positive and the other one is negative.
It is unclear to me how I should proceed from here to decide which of the two options is the correct one.
• You can use that $\operatorname{Tr}(\mathbf{A}) = \sum_{i}A_{ii} = \sum_{i}\lambda_{i}$, and note that this is positive. – Thomas Russell Jan 17 '18 at 21:52
• You can also "congruence diagonalize" and use Sylvester Inertia... – Will Jagy Jan 17 '18 at 22:21
Recall that a symmetric matrix $A$ is called negative definite if and only if all of its eigenvalues are negative. Moreover, a symmetric matrix $A$ is called negative definite if and only if all its leading principal minors are negative.
Note that $\det(5)=5>0$. Hence, $A$ is not negative definite and as you noted $\det(A) = \lambda_1\lambda_2\lambda_3 < 0$. Hence, it must be that two of the eigenvalues are positive one of them is negative.
For point 2 note that
$$\det(5)=5>0$$
$$\det \begin{bmatrix}5&6\\6&9\\ \end{bmatrix}=9>0$$
$$\det A<0$$
thus the signature is:
$$n_0=0 \quad n_+=2 \quad n_-=1$$
You have two positive eigenvalues $( 5.4025)$,$(13.7817)$, and one negative eigenvalue $(-5.1843)$.
The sum is 14 which is trace and the product is the determinant as you claimed.
• Correct. I had worked with a wrong matrix. Now it is fixed. – Mohammad Riazi-Kermani Jan 17 '18 at 22:33
• @MohammadRiazi-Kermani if you wish someone to be notified that you have posted a comment in reply to their comment, you need to start the comment with an "at" sign and at least three letters of their user name. Usually, the system will allow you to click on the whole name after you type the first one or two letters – Will Jagy Jan 17 '18 at 22:37
• @WillJagy, Thanks. I am learning something new every day. – Mohammad Riazi-Kermani Jan 17 '18 at 22:49
Calling your matrix $H$ ( stands for Hessian, or half the Hessian, depends) we can find, with no approximations, a matrix $P$ with $\det P = \pm 1,$ such that $P^T H P = D$ is diagonal. The diagonal entries are not the eigenvalues, but they have the same "signs" as the eigenvalues by Sylvester's Law of Inertia. Here, we get two positive and one negative diagonal entry, so there are two positive and one negative eigenvalues. They are real, by the way, symmetric matrices have real eigenvalues. $$P^T H P = D$$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 6 }{ 5 } & 1 & 0 \\ - \frac{ 19 }{ 3 } & \frac{ 43 }{ 9 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 5 & 6 & 3 \\ 6 & 9 & - 5 \\ 3 & - 5 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 6 }{ 5 } & - \frac{ 19 }{ 3 } \\ 0 & 1 & \frac{ 43 }{ 9 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 5 & 0 & 0 \\ 0 & \frac{ 9 }{ 5 } & 0 \\ 0 & 0 & - \frac{ 386 }{ 9 } \\ \end{array} \right)$$
$$D_0 = H$$ $$E_j^T D_{j-1} E_j = D_j$$ $$P_{j-1} E_j = P_j$$ $$E_j^{-1} Q_{j-1} = Q_j$$ $$P_j Q_j = I$$ $$P_j^T H P_j = D_j$$ $$Q_j^T D_j Q_j = H$$
$$H = \left( \begin{array}{rrr} 5 & 6 & 3 \\ 6 & 9 & - 5 \\ 3 & - 5 & 0 \\ \end{array} \right)$$
==============================================
$$E_{1} = \left( \begin{array}{rrr} 1 & - \frac{ 6 }{ 5 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right)$$ $$P_{1} = \left( \begin{array}{rrr} 1 & - \frac{ 6 }{ 5 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrr} 1 & \frac{ 6 }{ 5 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrr} 5 & 0 & 3 \\ 0 & \frac{ 9 }{ 5 } & - \frac{ 43 }{ 5 } \\ 3 & - \frac{ 43 }{ 5 } & 0 \\ \end{array} \right)$$
==============================================
$$E_{2} = \left( \begin{array}{rrr} 1 & 0 & - \frac{ 3 }{ 5 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right)$$ $$P_{2} = \left( \begin{array}{rrr} 1 & - \frac{ 6 }{ 5 } & - \frac{ 3 }{ 5 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrr} 1 & \frac{ 6 }{ 5 } & \frac{ 3 }{ 5 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrr} 5 & 0 & 0 \\ 0 & \frac{ 9 }{ 5 } & - \frac{ 43 }{ 5 } \\ 0 & - \frac{ 43 }{ 5 } & - \frac{ 9 }{ 5 } \\ \end{array} \right)$$
==============================================
$$E_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & \frac{ 43 }{ 9 } \\ 0 & 0 & 1 \\ \end{array} \right)$$ $$P_{3} = \left( \begin{array}{rrr} 1 & - \frac{ 6 }{ 5 } & - \frac{ 19 }{ 3 } \\ 0 & 1 & \frac{ 43 }{ 9 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrr} 1 & \frac{ 6 }{ 5 } & \frac{ 3 }{ 5 } \\ 0 & 1 & - \frac{ 43 }{ 9 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrr} 5 & 0 & 0 \\ 0 & \frac{ 9 }{ 5 } & 0 \\ 0 & 0 & - \frac{ 386 }{ 9 } \\ \end{array} \right)$$
==============================================
$$P^T H P = D$$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 6 }{ 5 } & 1 & 0 \\ - \frac{ 19 }{ 3 } & \frac{ 43 }{ 9 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 5 & 6 & 3 \\ 6 & 9 & - 5 \\ 3 & - 5 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 6 }{ 5 } & - \frac{ 19 }{ 3 } \\ 0 & 1 & \frac{ 43 }{ 9 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 5 & 0 & 0 \\ 0 & \frac{ 9 }{ 5 } & 0 \\ 0 & 0 & - \frac{ 386 }{ 9 } \\ \end{array} \right)$$ $$Q^T D Q = H$$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 6 }{ 5 } & 1 & 0 \\ \frac{ 3 }{ 5 } & - \frac{ 43 }{ 9 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 5 & 0 & 0 \\ 0 & \frac{ 9 }{ 5 } & 0 \\ 0 & 0 & - \frac{ 386 }{ 9 } \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 6 }{ 5 } & \frac{ 3 }{ 5 } \\ 0 & 1 & - \frac{ 43 }{ 9 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 5 & 6 & 3 \\ 6 & 9 & - 5 \\ 3 & - 5 & 0 \\ \end{array} \right)$$
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
I deliberately interchanged the first and third variables, to show how the diagonal entries may change a little when making some different choices, but still come out $++-$
$$P^T H P = D$$ $$\left( \begin{array}{rrr} 0 & 1 & 0 \\ 1 & \frac{ 5 }{ 9 } & 0 \\ \frac{ 57 }{ 25 } & \frac{ 3 }{ 5 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 0 & - 5 & 3 \\ - 5 & 9 & 6 \\ 3 & 6 & 5 \\ \end{array} \right) \left( \begin{array}{rrr} 0 & 1 & \frac{ 57 }{ 25 } \\ 1 & \frac{ 5 }{ 9 } & \frac{ 3 }{ 5 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 9 & 0 & 0 \\ 0 & - \frac{ 25 }{ 9 } & 0 \\ 0 & 0 & \frac{ 386 }{ 25 } \\ \end{array} \right)$$
$$H = \left( \begin{array}{rrr} 0 & - 5 & 3 \\ - 5 & 9 & 6 \\ 3 & 6 & 5 \\ \end{array} \right)$$
==============================================
$$E_{1} = \left( \begin{array}{rrr} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{array} \right)$$ $$P_{1} = \left( \begin{array}{rrr} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrr} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrr} 9 & - 5 & 6 \\ - 5 & 0 & 3 \\ 6 & 3 & 5 \\ \end{array} \right)$$
==============================================
$$E_{2} = \left( \begin{array}{rrr} 1 & \frac{ 5 }{ 9 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right)$$ $$P_{2} = \left( \begin{array}{rrr} 0 & 1 & 0 \\ 1 & \frac{ 5 }{ 9 } & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrr} - \frac{ 5 }{ 9 } & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrr} 9 & 0 & 6 \\ 0 & - \frac{ 25 }{ 9 } & \frac{ 19 }{ 3 } \\ 6 & \frac{ 19 }{ 3 } & 5 \\ \end{array} \right)$$
==============================================
$$E_{3} = \left( \begin{array}{rrr} 1 & 0 & - \frac{ 2 }{ 3 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right)$$ $$P_{3} = \left( \begin{array}{rrr} 0 & 1 & 0 \\ 1 & \frac{ 5 }{ 9 } & - \frac{ 2 }{ 3 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrr} - \frac{ 5 }{ 9 } & 1 & \frac{ 2 }{ 3 } \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrr} 9 & 0 & 0 \\ 0 & - \frac{ 25 }{ 9 } & \frac{ 19 }{ 3 } \\ 0 & \frac{ 19 }{ 3 } & 1 \\ \end{array} \right)$$
==============================================
$$E_{4} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & \frac{ 57 }{ 25 } \\ 0 & 0 & 1 \\ \end{array} \right)$$ $$P_{4} = \left( \begin{array}{rrr} 0 & 1 & \frac{ 57 }{ 25 } \\ 1 & \frac{ 5 }{ 9 } & \frac{ 3 }{ 5 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{4} = \left( \begin{array}{rrr} - \frac{ 5 }{ 9 } & 1 & \frac{ 2 }{ 3 } \\ 1 & 0 & - \frac{ 57 }{ 25 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{4} = \left( \begin{array}{rrr} 9 & 0 & 0 \\ 0 & - \frac{ 25 }{ 9 } & 0 \\ 0 & 0 & \frac{ 386 }{ 25 } \\ \end{array} \right)$$
==============================================
$$P^T H P = D$$ $$\left( \begin{array}{rrr} 0 & 1 & 0 \\ 1 & \frac{ 5 }{ 9 } & 0 \\ \frac{ 57 }{ 25 } & \frac{ 3 }{ 5 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 0 & - 5 & 3 \\ - 5 & 9 & 6 \\ 3 & 6 & 5 \\ \end{array} \right) \left( \begin{array}{rrr} 0 & 1 & \frac{ 57 }{ 25 } \\ 1 & \frac{ 5 }{ 9 } & \frac{ 3 }{ 5 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 9 & 0 & 0 \\ 0 & - \frac{ 25 }{ 9 } & 0 \\ 0 & 0 & \frac{ 386 }{ 25 } \\ \end{array} \right)$$ $$Q^T D Q = H$$ $$\left( \begin{array}{rrr} - \frac{ 5 }{ 9 } & 1 & 0 \\ 1 & 0 & 0 \\ \frac{ 2 }{ 3 } & - \frac{ 57 }{ 25 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 9 & 0 & 0 \\ 0 & - \frac{ 25 }{ 9 } & 0 \\ 0 & 0 & \frac{ 386 }{ 25 } \\ \end{array} \right) \left( \begin{array}{rrr} - \frac{ 5 }{ 9 } & 1 & \frac{ 2 }{ 3 } \\ 1 & 0 & - \frac{ 57 }{ 25 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 0 & - 5 & 3 \\ - 5 & 9 & 6 \\ 3 & 6 & 5 \\ \end{array} \right)$$ | 2019-06-26T12:15:47 | {
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https://www.mathworks.com/help/matlab/ref/trapz.html | # trapz
Trapezoidal numerical integration
## Syntax
``Q = trapz(Y)``
``Q = trapz(X,Y)``
``Q = trapz(___,dim)``
## Description
example
````Q = trapz(Y)` computes the approximate integral of `Y` via the trapezoidal method with unit spacing. The size of `Y` determines the dimension to integrate along:If `Y` is a vector, then `trapz(Y)` is the approximate integral of `Y`.If `Y` is a matrix, then `trapz(Y)` integrates over each column and returns a row vector of integration values.If `Y` is a multidimensional array, then `trapz(Y)` integrates over the first dimension whose size does not equal 1. The size of this dimension becomes 1, and the sizes of other dimensions remain unchanged.```
example
````Q = trapz(X,Y)` integrates `Y` with respect to the coordinates or scalar spacing specified by `X`. If `X` is a vector of coordinates, then `length(X)` must be equal to the size of the first dimension of `Y` whose size does not equal 1.If `X` is a scalar spacing, then `trapz(X,Y)` is equivalent to `X*trapz(Y)`. ```
example
````Q = trapz(___,dim)` integrates along the dimension `dim` using any of the previous syntaxes. You must specify `Y`, and optionally can specify `X`. If you specify `X`, then it can be a scalar or a vector with length equal to `size(Y,dim)`. For example, if `Y` is a matrix, then `trapz(X,Y,2)` integrates each row of `Y`.```
## Examples
collapse all
Calculate the integral of a vector where the spacing between data points is 1.
Create a numeric vector of data.
`Y = [1 4 9 16 25];`
`Y` contains function values for $f\left(x\right)={x}^{2}$ in the domain [1, 5].
Use `trapz` to integrate the data with unit spacing.
`Q = trapz(Y)`
```Q = 42 ```
This approximate integration yields a value of `42`. In this case, the exact answer is a little less, $41\frac{1}{3}$. The `trapz` function overestimates the value of the integral because f(x) is concave up.
Calculate the integral of a vector where the spacing between data points is uniform, but not equal to 1.
Create a domain vector.
`X = 0:pi/100:pi;`
Calculate the sine of `X`.
`Y = sin(X);`
Integrate `Y` using `trapz`.
`Q = trapz(X,Y)`
```Q = 1.9998 ```
When the spacing between points is constant, but not equal to 1, an alternative to creating a vector for `X` is to specify the scalar spacing value. In that case, `trapz(pi/100,Y)` is the same as `pi/100*trapz(Y)`.
Integrate the rows of a matrix where the data has a nonuniform spacing.
Create a vector of x-coordinates and a matrix of observations that take place at the irregular intervals. The rows of `Y` represent velocity data, taken at the times contained in `X`, for three different trials.
```X = [1 2.5 7 10]; Y = [5.2 7.7 9.6 13.2; 4.8 7.0 10.5 14.5; 4.9 6.5 10.2 13.8];```
Use `trapz` to integrate each row independently and find the total distance traveled in each trial. Since the data is not evaluated at constant intervals, specify `X` to indicate the spacing between the data points. Specify `dim = 2` since the data is in the rows of `Y`.
`Q1 = trapz(X,Y,2)`
```Q1 = 3×1 82.8000 85.7250 82.1250 ```
The result is a column vector of integration values, one for each row in `Y`.
Create a grid of domain values.
```x = -3:.1:3; y = -5:.1:5; [X,Y] = meshgrid(x,y);```
Calculate the function $f\left(x,y\right)={x}^{2}+{y}^{2}$ on the grid.
`F = X.^2 + Y.^2;`
`trapz` integrates numeric data rather than functional expressions, so in general the expression does not need to be known to use `trapz` on a matrix of data. In cases where the functional expression is known, you can instead use `integral`, `integral2`, or `integral3`.
Use `trapz` to approximate the double integral
`$I={\int }_{-5}^{5}{\int }_{-3}^{3}\left({x}^{2}+{y}^{2}\right)dx\phantom{\rule{0.2222222222222222em}{0ex}}dy$`
To perform double or triple integrations on an array of numeric data, nest function calls to `trapz`.
`I = trapz(y,trapz(x,F,2))`
```I = 680.2000 ```
`trapz` performs the integration over x first, producing a column vector. Then, the integration over y reduces the column vector to a single scalar. `trapz` slightly overestimates the exact answer of 680 because f(x,y) is concave up.
## Input Arguments
collapse all
Numeric data, specified as a vector, matrix, or multidimensional array. By default, `trapz` integrates along the first dimension of `Y` whose size does not equal 1.
Data Types: `single` | `double`
Complex Number Support: Yes
Point spacing, specified as `1` (default), a uniform scalar spacing, or a vector of coordinates.
• If `X` is a scalar, then it specifies a uniform spacing between the data points and `trapz(X,Y)` is equivalent to `X*trapz(Y)`.
• If `X` is a vector, then it specifies x-coordinates for the data points and `length(X)` must be the same as the size of the integration dimension in `Y`.
Data Types: `single` | `double`
Dimension to operate along, specified as a positive integer scalar. If no value is specified, then the default is the first array dimension whose size does not equal 1.
Consider a two-dimensional input array, `Y`:
• `trapz(Y,1)` works on successive elements in the columns of `Y` and returns a row vector.
• `trapz(Y,2)` works on successive elements in the rows of `Y` and returns a column vector.
If `dim` is greater than `ndims(Y)`, then `trapz` returns an array of zeros of the same size as `Y`.
collapse all
### Trapezoidal Method
`trapz` performs numerical integration via the trapezoidal method. This method approximates the integration over an interval by breaking the area down into trapezoids with more easily computable areas. For example, here is a trapezoidal integration of the sine function using eight evenly-spaced trapezoids:
For an integration with `N+1` evenly spaced points, the approximation is
`$\begin{array}{c}\underset{a}{\overset{b}{\int }}f\left(x\right)dx\text{\hspace{0.17em}}\text{\hspace{0.17em}}\approx \text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{b-a}{2N}\sum _{n=1}^{N}\left(f\left({x}_{n}\right)+f\left({x}_{n+1}\right)\right)\\ =\frac{b-a}{2N}\left[f\left({x}_{1}\right)+2f\left({x}_{2}\right)+...+2f\left({x}_{N}\right)+f\left({x}_{N+1}\right)\right]\text{\hspace{0.17em}}\text{\hspace{0.17em}},\end{array}$`
where the spacing between each point is equal to the scalar value $\frac{b-a}{N}$. By default MATLAB® uses a spacing of 1.
If the spacing between the `N+1` points is not constant, then the formula generalizes to
`$\underset{a}{\overset{b}{\int }}f\left(x\right)dx\text{\hspace{0.17em}}\text{\hspace{0.17em}}\approx \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{1}{2}\sum _{n=1}^{N}\left({x}_{n+1}-{x}_{n}\right)\left[f\left({x}_{n}\right)+f\left({x}_{n+1}\right)\right]\text{\hspace{0.17em}},$`
where $a={x}_{1}<{x}_{2}<\text{\hspace{0.17em}}\text{\hspace{0.17em}}...\text{\hspace{0.17em}}\text{\hspace{0.17em}}<{x}_{N}<{x}_{N+1}=b$, and $\left({x}_{n+1}-{x}_{n}\right)$ is the spacing between each consecutive pair of points.
## Tips
• Use `trapz` and `cumtrapz` to perform numerical integrations on discrete data sets. Use `integral`, `integral2`, or `integral3` instead if a functional expression for the data is available.
• `trapz` reduces the size of the dimension it operates on to 1, and returns only the final integration value. `cumtrapz` also returns the intermediate integration values, preserving the size of the dimension it operates on. | 2021-12-09T11:33:34 | {
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https://math.stackexchange.com/questions/761040/linear-algebra-direct-sum | # Linear Algebra: Direct Sum
Prove that if $W_1$ is any subspace of a finite-dimensional vector space $V$, then there exists a subspace $W_2$ of $V$ such that $V = W_1 \oplus W_2$
What I have done so far is to note that since $V$ is finite and $W_1$ is a subspace of $V$, we have $\dim(W_1) \leq \dim(V)$. If we have equality, then let $W_2 = \{0\}$, so that we have $V= W_1 \oplus W_2$. So now I need to look at the case when $\dim(W_1) \lt \dim(V)$.
What I have tried for this case is to let $$\beta = \{v_1,v_2,..,v_n\}$$ be a basis for $V$ and $$\gamma=\{u_1,u_2,..,u_m\}$$ a basis for $W_1$. My idea was to extend $\gamma$ to a basis for $V$, so let $$\alpha=\{u_1,u_2,..,u_m,w_1,w_2,..,w_{n-m}\}$$ be the extension of $\gamma$ to $V$, where $w_1,w_2,..,w_{n-m}$ are basis vectors for $W_2$. If I'm doing this right then I would just have to show that $W_1 \cap W_2 = \{0\}$ and $W_1 + W_2= V$
Am I heading in the right direction? Any hints would be greatly appreciated.
• You're heading in the right direction. Keep going. – Pedro Tamaroff Apr 19 '14 at 22:02
If you already know that you can compete a basis of a subspace to a basis for the whole space then you are practically done.
Hint: Note that $\alpha$ is a basis for $V$ (this should give you $W_{1}+W_{2}=V$, why ?) and that the $w_{i}$ are linearly independent of the $u_{i}$ (this should show that $W_{1}\cap W_{2}=\{0\}$, why ?)
Note: The way I see it, there is no use for $\beta$ or of the $v_{i}$ in the proof
• since $\alpha$ is a basis for $V$ then all vectors in $V$ can be written as a linear combination of vectors in $\alpha$, which is a linear combination of vectors in $\gamma$ plus linear combination of the basis vectors for $W_2$. Also, $W_1,W_2$ are subspaces of $V$ so they both contain the zero vector. – Miguel Landeros Apr 19 '14 at 22:13
• One more thing, would I also have to show that $W_2$ is a subspace of V? – Miguel Landeros Apr 19 '14 at 22:16
• @MiguelLanderos - $W_2$ is defined as a span of a set of vectors and hence is a subspace. You still need to show that the only vector in the intersection is the zero vector, can you show it ? – Belgi Apr 19 '14 at 22:21
• @Belgi-I see, I think I should be able to show that last part. thanks – Miguel Landeros Apr 19 '14 at 22:24
Let $\dim V=n.$ Say you have a basis $\mathcal B=\{v_1,\dots,v_k\}$ for $W_1\subseteq V$, where $k\leq n$. Extend to a basis $\mathcal B'=\{v_1,\dots,v_n\}$ for $V$. Consider the subspace, call it $W_2$, spanned by those vectors you added to $\mathcal B$. Namely, $\{v_{k+1},\dots,v_n\}$. Then what can you say about $W_1$ and $W_2$? How do they relate to $V$?
EDIT: You say that you know $V=W_1+W_2$, but aren't sure how to get $W_1\cap W_2=\{0\}$. Let's take a look! Let $w\in W_1\cap W_2$. Then we can write $$w = \sum_{i=1}^k a_iv_i \quad\text{and}\quad w = \sum_{i=k+1}^n a_iv_i,$$ for some $a_i\in F$. If you subtract these two equations, what does linear indendence (i.e. that $\{v_i\}$ is a basis) tell you about the $a_i$? What does this tell you about $W_1\cap W_2$?
• I see that the sum of those two subspaces equals $V$. I only have to show that their intersection is the zero vector. I know since they are both subspaces of $V$ then they both contain the zero vector of $V$. I only need to show that the intersection is only the zero vector, but does this follow from the fact that both$W_1$ and $W_2$ are linearly independent? – Miguel Landeros Apr 19 '14 at 23:00
• @MiguelLanderos: See edit. – user59083 Apr 20 '14 at 1:26
• @5space-the $a_i$ all equal zero so the intersection is a linearly independent set. So the only vector in both is the zero vector. – Miguel Landeros Apr 20 '14 at 1:34
• @MiguelLanderos: Exactly!! I think you've got it ;-) – user59083 Apr 20 '14 at 1:35 | 2019-06-16T20:54:28 | {
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http://math.stackexchange.com/questions/21548/using-congruences-show-frac15n5-frac13n3-frac715n-is-int | # Using congruences, show $\frac{1}{5}n^5 + \frac{1}{3}n^3 + \frac{7}{15}n$ is integer for every $n$
Using congruences, show that the following is always an integer for every integer value of $n$: $$\frac{1}{5}n^5 + \frac{1}{3}n^3 + \frac{7}{15}n.$$
-
Please don't state your questions as orders or assignments to the group. If you have a question, ask it like a question, not like an order. – Arturo Magidin Feb 11 '11 at 16:19
Also, perhaps the homework tag? – Eric Naslund Feb 11 '11 at 16:38
Lets show that $P(n)=3n^5+5n^3+7n$ is divisible by 15 for every $n$. To do this, we will show that it is divisible by $3$ and $5$ for every $n$.
Recall that for a prime $p$, $x^p\equiv x \pmod{p}$. (Fermat's Little Theorem) Then, looking modulo 5 we see that $$P(n)\equiv 3n^5+7n\equiv 3n+7n=10n\equiv 0.$$ Now looking modulo 3 we see that $$P(n)\equiv 5n^3+7n\equiv 5n+7n=12n\equiv 0.$$ Thus $P(n)$ is divisible by 15 for every $n$ as desired.
-
Good solution Eric. I thought of FLT but didn't think of applying CRT. Anyhow +1 for a slick proof. – anonymous Feb 11 '11 at 16:56
No explicit need for the chinese remainder theorem: if a number is a multiple of $3$ and $5$ then it's a multiple of $15$. – lhf Feb 11 '11 at 16:56
Unfortunately, \cong produces $\cong$; for $\equiv$, use \equiv. – Arturo Magidin Feb 11 '11 at 21:07
Thanks Arturo. I just got use to \cong from my algebra courses. – Eric Naslund Feb 12 '11 at 5:49
@EricNaslund Can you explain how you had written this $$5n^3+7 \equiv 5n+7n\pmod3$$ because what I have done is this $$5n^3-5n+7 \equiv 7n \pmod3$$ $$\implies 5n(n+1)(n-1)+7 \equiv 7n \pmod3$$ $$\implies 7\equiv 7n\pmod3$$ but then how's that is possible?? – Saurabh Jun 17 '12 at 11:06
HINT $\displaystyle\rm\quad \frac{n^5}5\: +\: \frac{n^3}3\: +\: \frac{7\:n}{15}\ =\ \frac{n^5-n}5\: +\: \frac{n^3-n}3\: +\: n\ \in \mathbb Z\$ by Fermat's Little Theorem.
-
Taking the lcm we have $\displaystyle \frac{1}{15} \cdot \Bigl[ 3n^{5}+5n^{3} + 7n\Bigr]$ Now show that the quantity $3n^{5} + 5n^{3}+7n$ is always divisible by $15$. Induction may be useful.
Clearly for $n=1$, $3+5+7=15$ is divisible by $15$. Assume that it is true for $n=k$. That is assume that $3k^{5}+5k^{3}+7k$ is divisible by $15$. Use this to show that the quantity
\begin{align*} 3(k+1)^{5}+5(k+1)^{3}+7(k+1) &= 3 \Bigl[k^{5} + {5 \choose 1}k^{4} + \cdots +1\Bigr] + 5(k+1)^{3}+ 7(k+1) \end{align*}
is divisible by $15$.
-
How to read your latex writing? What do I need to use? Thanks – kira Feb 11 '11 at 16:09
@Kira: You need to learn latex.There are lot of resources in the web. – anonymous Feb 11 '11 at 16:10
@ Chandru:This is my first time using it and I am not very familiar with it. Can you also show a few more steps in your explanation? Thanks! – kira Feb 11 '11 at 16:13
I really don't feel induction is a good way, but I just might not be understanding. Why will this work immediately? Anyway, +1 for editing and adding more when the OP asked for help. – Eric Naslund Feb 11 '11 at 16:50
An alternate version of this proof is to show that $3(n+1)^5 + 5(n+1)^3 + 7(n+1) - (3n^5+5n^3+7n)$ is always divisible by 15. As it turns out, this is trivial since once you expand the polynomial out all the coefficients are divisible by 15. – Michael Lugo Feb 11 '11 at 22:19 | 2015-08-01T11:51:12 | {
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https://math.stackexchange.com/questions/280945/integral-of-hermite-polynomial-multiplied-by-exp-x2-2 | # Integral of Hermite polynomial multiplied by $\exp(-x^2/2)$
What is the value of $\int_{-\infty}^{\infty}e^{-\frac{x^2}{2}}H_n(x)dx$ where $H_n(x)$ is the $n^{\small\mbox{th}}$ Hermite Polynomial (physicist's convention)?
Notice that, if $n$ is odd , then the integrand is an odd function which implies that the integral equals to $0$. If $n$ is even, then the integral equals to
$${2}^{2\,n+\frac{5}{2}}\Gamma \left( n+ \frac{3}{2} \right),\quad n=0,1,2,\dots.$$
Note this, in the above formula, $n=0$ corresponds to the case $H_{2}(x)$, $n=1$ correspons to the case $H_{4}(x)$ and so on.
One can have instead, the formula which include the case $n=0$
$${4}^{n}\sqrt {2}\,\Gamma \left( n+\frac{1}{2} \right), \quad n=0,1,2,\dots.$$
Again, in the above formula, $n=0$ corresponds to the case $H_{0}(x)$, $n=1$ corresponds to the case $H_{2}(x)$ and so on.
• How did you calculate this? – Antonio Vargas Jan 17 '13 at 20:24
For probabilists' Hermite polynomials: The Hermite polynomials are the orthogonal polynomials corresponding to the weight function $w(x) = e^{-x^2/2}$. This means that $\int_{-\infty}^{\infty} H_n(x)H_m(x)e^{-x^2/2} \, dx = 0$ whenever $n \not= m$ (or equivalently, $\int_{-\infty}^{\infty} H_n(x) P(x) e^{-x^2/2} \, dx = 0$ for any polynomial $P$ of degree less than $n$). Since $H_0(x) = 1$, it follows that $$\int_{-\infty}^{\infty} H_n(x)e^{-x^2/2} \, dx = \int_{-\infty}^{\infty} H_n(x)H_0(x)e^{-x^2/2} \, dx = 0$$ for all $n > 0$. The only time this integral is non-zero is when $n = 0$, in which case $$\int_{-\infty}^{\infty} H_0(x)e^{-x^2/2} \, dx = \int_{-\infty}^{\infty} e^{-x^2/2} \, dx = \sqrt{2\pi}.$$
• I guess it depends on which "Hermite polynomials" Tarek is asking about. – Antonio Vargas Jan 17 '13 at 22:04
• I am asking about the physicists' Hermite polynomials. – Tarek Jan 17 '13 at 23:04
• Ok, then disregard this and go Mhenni's route. I mistakenly assumed you were using the probabilists' polynomials because you were using the corresponding weight. – Aaron Jan 17 '13 at 23:12
• @Tarek, then you should have edited your question to include the normalization you're working with... – J. M. isn't a mathematician Mar 23 '13 at 13:05
Using the generating funtion $$e^{2xt-t^2}=\sum_{n=0}^\infty H_n(x)\frac{t^n}{n!}$$ we can obtain two recurrence relations.
Differentiating with respect to $$t$$ we obtain $$H_{n+1}(x)=2xH_n(x)-2nH_{n-1}(x)$$ for $$n\geq 1$$ with $$H_0(x)=1$$ and $$H_1(x)=2x$$. Similarly, differentiating with respect to $$x$$ we obtain $$\frac{d}{dx}H_n(x)=2nH_{n-1}(x)$$ for $$n\geq 1$$.
Now, define $$I_n=\int_{-\infty}^{\infty}e^{-x^2/2}H_n(x)dx$$ then \begin{align}I_{n+1}&=\color{blue}{\int_{-\infty}^{\infty}2xe^{-x^2/2}H_{n}(x)dx}-2nI_{n-1}\\ &=\color{blue}{-2e^{-x^2/2}H_n(x)\Big|_{-\infty}^{\infty}+4nI_{n-1}}-2nI_{n-1}=2nI_{n-1}. \end{align} where the integral in blue was solved by parts and since that $$H_n$$ is a polynomial of degree $$n$$ we have that $$\lim_{x\to\pm\infty}e^{-x^2/2}H_n(x)=0$$, therefore $$-2e^{-x^2/2}H_n(x)\Big|_{-\infty}^{\infty}=0$$.
To finish, since $$I_0=\int_{-\infty}^{\infty}e^{-x^2/2}H_0(x)dx=\int_{-\infty}^{\infty}e^{-x^2/2}dx=\sqrt{2\pi}$$ and $$I_1=\int_{-\infty}^{\infty}e^{-x^2/2}H_1(x)dx=\int_{-\infty}^{\infty}2xe^{-x^2/2}dx=0$$ we conclude that $$I_{2n+1}=0$$ for all $$n\in\mathbb{N}$$ and \begin{align}I_{2n}&=2(2n-1)I_{2n-2}\\ &=2^2(2n-1)(2n-3)I_{2n-4}\\ &~~\vdots\\ &=2^n(2n-1)(2n-3)\cdots1\cdot I_0\\ &=2^n\frac{2n(2n-1)(2n-2)(2n-3)\cdots1}{2n(2n-2)\cdots2}I_0\\ &=2^n\frac{(2n)!}{2^nn!}I_0\\ &=\frac{(2n)!}{n!}I_0\\ &=\frac{(2n)!}{n!}\sqrt{2\pi}.\end{align} | 2021-03-02T18:33:37 | {
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http://mathhelpforum.com/calculus/114573-implicit-differentiation.html | # Math Help - Implicit differentiation
1. ## Implicit differentiation
Finding dy/dx for $y^2 = 5x^3$. Find equation of the tangent line to the graph at the given point. (1, sqrt 5)
Here's what I did;
$2yy' = 15x^2$
$y' = 15x^2 / 2y$
Where did I go wrong? Or, if it's right by some chance, how do I proceed to find the equation?
2. Originally Posted by Archduke01
Finding dy/dx for $y^2 = 5x^3$
Here's what I did;
$2yy' = 15x^2$
$y' = 15x^2 / 2y$
Where did I go wrong?
Do you have to use Implicit Differentiation?
$y^2 = 5x^3$
$y = \sqrt{5}x^{\frac{3}{2}}$
$\frac{dy}{dx} = \frac{3\sqrt{5}}{2}x^{\frac{1}{2}}$
$\frac{dy}{dx} = \frac{3\sqrt{5x}}{2}$.
If you WERE going to use Implicit Differentiation:
$y^2 = 5x^3$
$\frac{d}{dx}(y^2) = \frac{d}{dx}(5x^3)$
$\frac{d}{dy}(y^2)\,\frac{dy}{dx} = 15x^2$
$2y\,\frac{dy}{dx} = 15x^2$
$\frac{dy}{dx} = \frac{15x^2}{2y}$
And since $y = \sqrt{5x^3}$
$\frac{dy}{dx} = \frac{15x^2}{2\sqrt{5x^3}}$
$= \frac{15x^2\sqrt{5x^3}}{10x^3}$
$= \frac{3\sqrt{5x^3}}{2x}$
$= \frac{3x\sqrt{5x}}{2x}$
$= \frac{3\sqrt{5x}}{2}$
which is the same as found above...
3. Originally Posted by Prove It
And since $y = \sqrt{5x^3}$
Sorry, I just editted my post because I forgot to include that the problem already had given points.
4. Well if you have $(x,y) = (1, \sqrt{5})$, then what does
$\frac{dy}{dx} = \frac{15x^2}{2y}$ equal?
5. Originally Posted by Prove It
Well if you have $(x,y) = (1, \sqrt{5})$, then what does
$\frac{dy}{dx} = \frac{15x^2}{2y}$ equal?
$15/ (2 sqrt5)$ ... which I'm assuming is the slope of the tangent line. How do I get the rest of the equation?
6. Originally Posted by Archduke01
$15/ (2 sqrt5)$ ... which I'm assuming is the slope of the tangent line. How do I get the rest of the equation?
Correct, which I would rewrite as $\frac{3\sqrt{5}}{2}$ by rationalising the denominator.
Now, you have the tangent line $y = mx + c$.
You know $y = \sqrt{5}, x = 1, m = \frac{3\sqrt{5}}{2}$.
Solve for $c$.
7. Originally Posted by Prove It
You know $y = 2\sqrt{5}$.
Shouldn't it be $y = sqrt {5}$?
8. Originally Posted by Archduke01
Shouldn't it be $y = sqrt {5}$?
Yes it is, typo.
Will edit now.
9. If I did it correctly, the value of c should be $- sqrt 5 / 2$...
So the equation should be $y = 3 sqrt 5 /2 x - sqrt 5/2$
Does that seem right? The answer book says $15x - 2 (sqrt5)y - 5 = 0$
10. Originally Posted by Archduke01
If I did it correctly, the value of c should be $- sqrt 5 / 2$...
So the equation should be $y = 3 sqrt 5 /2 x - sqrt 5/2$
Does that seem right? The answer book says $15x - 2 (sqrt5)y - 5 = 0$
Yes it is correct. You just need some algebraic manipulation.
11. Originally Posted by Prove It
Yes it is correct. You just need some algebraic manipulation.
The equation is correct, or the value of c?
12. Both.
13. Originally Posted by Prove It
Both.
Thank you! Sorry for the lengthy back and forth. | 2014-07-26T11:38:25 | {
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https://math.libretexts.org/Bookshelves/Mathematical_Logic_and_Proof/Gentle_Introduction_to_the_Art_of_Mathematics_(Fields)/02%3A_Logic_and_Quantifiers/2.07%3A_Validity_of_Arguments_and_Common_Errors |
2.7: Validity of Arguments and Common Errors
An argument is said to be valid or to have a valid form if each deduction in it can be justified with one of the rules of inference listed in the previous section. The form of an argument might be valid, but still the conclusion may be false if some of the premises are false. So to show that an argument is good we have to be able to do two things: show that the argument is valid (i.e. that every step can be justified) and that the argument is sound which means that all the premises are true. If you start off with a false premise, you can prove anything!
Consider, for example the following “proof” that $$2 = 1$$.
Suppose that $$a$$ and $$b$$ are two real numbers such that $$a = b$$.
$a^2 = ab \tag{by hypothesis, $$a$$ and $$b$$ are equal, so}$
$a^2 − b^2 = ab − b^2 \tag{subtracting $$b^2$$ from both sides}$
$(a + b)(a − b) = b(a − b) \tag{factoring both sides}$
$a + b = b \tag{canceling $$(a − b)$$ from both sides}$
Now let $$a$$ and $$b$$ both have a particular value, $$a = b = 1$$, and we see that $$1 + 1 = 1$$, i.e. $$2 = 1$$.
This argument is not sound (thank goodness!) because one of the premises – actually the bad premise appears as one of the justifications of a step – is false. You can argue with perfect logic to achieve complete nonsense if you include false premises.
Practice
It is not true that you can always cancel the same thing from both sides of an equation. Under what circumstances is such cancellation disallowed?
So, how can you tell if an argument has a valid form? Use a truth table. As an example, we’ll verify that the rule of inference known as “destructive dilemma” is valid using a truth table. This argument form contains $$4$$ predicate variables so the truth table will have $$16$$ rows. There is a column for each of the variables, the premises of the argument and its conclusion.
$$A\;B\;C\;D$$ $$A \implies B$$ $$C \implies D$$ $$¬B ∨ ¬D$$ $$¬A ∨ ¬C$$
$$\text{T} \;\text{T} \;\text{T} \;\text{T}$$ $$\text{T}$$ $$\text{T}$$ $$\phi$$ $$\phi$$
$$\text{T} \;\text{T} \;\text{T} \; \phi$$ $$\text{T}$$ $$\phi$$ $$\text{T}$$ $$\phi$$
$$\text{T} \;\text{T} \;\phi \;\text{T}$$ $$\text{T}$$ $$\text{T}$$ $$\phi$$ $$\text{T}$$
$$\text{T} \;\text{T} \;\phi \;\phi$$ $$\text{T}$$ $$\text{T}$$ $$\text{T}$$ $$\text{T}$$
$$\text{T} \;\phi \;\text{T} \;\text{T}$$ $$\phi$$ $$\text{T}$$ $$\text{T}$$ $$\phi$$
$$\text{T} \;\phi \;\text{T} \;\phi$$ $$\phi$$ $$\phi$$ $$\text{T}$$ $$\phi$$
$$\text{T} \;\phi \;\phi \;\text{T}$$ $$\phi$$ $$\text{T}$$ $$\text{T}$$ $$\text{T}$$
$$\text{T} \;\phi \;\phi \;\phi$$ $$\phi$$ $$\text{T}$$ $$\text{T}$$ $$\text{T}$$
$$\phi \;\text{T} \;\text{T} \;\text{T}$$ $$\text{T}$$ $$\text{T}$$ $$\phi$$ $$\text{T}$$
$$\phi \;\text{T} \;\text{T} \;\phi$$ $$\text{T}$$ $$\phi$$ $$\text{T}$$ $$\text{T}$$
$$\phi \;\text{T} \;\phi \;\text{T}$$ $$\text{T}$$ $$\text{T}$$ $$\phi$$ $$\text{T}$$
$$\phi \;\text{T} \;\phi \;\phi$$ $$\text{T}$$ $$\text{T}$$ $$\text{T}$$ $$\text{T}$$
$$\phi \;\phi \;\text{T} \;\text{T}$$ $$\text{T}$$ $$\text{T}$$ $$\text{T}$$ $$\text{T}$$
$$\phi \;\phi \;\text{T} \;\phi$$ $$\text{T}$$ $$\phi$$ $$\text{T}$$ $$\text{T}$$
$$\phi \;\phi \;\phi \;\text{T}$$ $$\text{T}$$ $$\text{T}$$ $$\text{T}$$ $$\text{T}$$
$$\phi \;\phi \;\phi \;\phi$$ $$\text{T}$$ $$\text{T}$$ $$\text{T}$$ $$\text{T}$$
Now, mark the lines in which all of the premises of this argument form are true. You should note that in every single situation in which all the premises are true the conclusion is also true. That’s what makes “destructive dilemma” – and all of its friends – a rule of inference. Whenever all the premises are true so is the conclusion. You should also notice that there are several rows in which the conclusion is true but some one of the premises isn’t. That’s okay too, isn’t it reasonable that the conclusion of an argument can be true, but at the same time the particulars of the argument are unconvincing?
As we’ve noted earlier, an argument by deductive reasoning can go wrong in only certain well-understood ways. Basically, either the form of the argument is invalid, or at least one of the premises is false. Avoiding false premises in your arguments can be trickier than it sounds – many statements that sound appealing or intuitively clear are actually counter-factual. The other side of the coin, being sure that the form of your argument is valid, seems easy enough – just be sure to only use the rules of inference as found in Table 2.6.1. Unfortunately, most arguments that you either read or write will be in prose, rather than appearing as a formal list of deductions. When dealing with that setting – using natural rather than formalized language – making errors in form is quite common.
Two invalid forms are usually singled out for criticism, the converse error and the inverse error. In some sense, these two apparently different ways to screw up are really the same thing. Just as a conditional statement and its contrapositive are known to be equivalent, so too are the other related statements – the converse and the inverse – equivalent. The converse error consists of mistaking the implication in a modus ponens form for its converse.
The converse error:
$\begin{array} &&B \\ &\underline{A \implies B} \\ ∴ &A \end{array}$
Consider, for a moment the following argument.
If a rhinoceros sees something on fire, it will stomp on it.
A rhinoceros stomped on my duck.
Therefore, the rhino must have thought that my duck was on fire.
It is true that rhinoceroses have an instinctive desire to extinguish fires. Also, we can well imagine that if someone made this ridiculous argument that their duck must actually have been crushed by a rhino. But, is the conclusion that the duck was on fire justified? Not really, what the first part of the argument asserts is that “(on fire) implies (rhino stomping)” but couldn’t a rhino stomp on something for other reasons? Perhaps the rhino was just ill-tempered. Perhaps the duck was just horrifically unlucky.
The closer the conditional is to being a biconditional, the more reasonable sounding is an argument exhibiting the converse error. Indeed, if the argument actually contains a biconditional, the “converse error” is not an error at all.
The following is a perfectly valid argument, that (sadly) has a false premise.
You will get an A in your Foundations class if and only if you read Dr. Fields’ book.
Therefore, you will get an A in Foundations.
Suppose that we try changing the major premise of that last argument to something more believable.
If you read Dr. Fields’ book, you will pass your Foundations class.
You did not read Dr. Fields’ book.
Therefore, you will not pass Foundations.
This last argument exhibits the so-called inverse error. It is by no means meant as a guarantee, but nevertheless, it seems reasonable that if someone reads this book they will pass a course on this material. The second premise is also easy to envision as true, although the “you” that it refers to obviously isn’t you, because you are reading this book! But even if we accept the premises as true, the conclusion doesn’t follow. A person might have read some other book that addressed the requisite material in an exemplary way.
Notice that the names for these two errors are derived from the change that would have to be made to convert them to modus ponens. For example, the inverse error is depicted formally by:
$\begin{array} &&¬A \\ &\underline{A \implies B} \\ ∴ &¬B \end{array}$
If we replaced the conditional in this argument form by its inverse $$(¬A \implies ¬B)$$ then the revised argument would be modus ponens. Similarly, if we replace the conditional in an argument that suffers from the converse error by its converse, we’ll have modus ponens.
Exercises:
Exercise $$\PageIndex{1}$$
Determine the logical form of the following arguments. Use symbols to express that form and determine whether the form is valid or invalid. If the form is invalid, determine the type of error made. Comment on the soundness of the argument as well, in particular, determine whether any of the premises are questionable.
1. All who are guilty are in prison. George is not in prison. Therefore, George is not guilty.
2. If one eats oranges one will have high levels of vitamin C. You do have high levels of vitamin C. Therefore, you must eat oranges.
3. All fish live in water. The mackerel is a fish. Therefore, the mackerel lives in water.
4. If you’re lazy, don’t take math courses. Everyone is lazy. Therefore, no one should take math courses.
5. All fish live in water. The octopus lives in water. Therefore, the octopus is a fish.
6. If a person goes into politics, they are a scoundrel. Harold has gone into politics. Therefore, Harold is a scoundrel.
Exercise $$\PageIndex{2}$$
Below is a rule of inference that we call extended elimination.
$\begin{array} &&(A ∨ B) ∨ C \\ &¬A \\ &\underline{¬B\;\;\;\;\;\;\;\;\;\;\;\;\;} \\ ∴ &C \end{array}$
Use a truth table to verify that this rule is valid.
Exercise $$\PageIndex{3}$$
If we allow quantifiers and open sentences in an argument form we get a couple of new argument forms. Arguments involving existentially quantified premises are rare – the new forms we are speaking of are called “universal modus ponens” and “universal modus tollens.” The minor premises may also be quantified or they may involve particular elements of the universe of discourse – this leads us to distinguish argument subtypes that are termed “universal” and “particular.”
For example,
$$\begin{array} &&∀x, A(x) \implies B(x) \\ &\underline{A(p)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} \\ ∴ &B(p) \end{array}$$
is the particular form of universal modus ponens (here, $$p$$ is not a variable – it stands for some particular element of the universe of discourse) and
$$\begin{array} &&∀x, A(x) \implies B(x) \\ &\underline{∀x, ¬B(x)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} \\ ∴ &∀x, ¬A(x) \end{array}$$
universal form of (universal) modus tollens.
Reexamine the arguments from problem (1), determine their forms (including quantifiers) and whether they are universal or particular.
Exercise $$\PageIndex{4}$$
Identify the rule of inference being used.
1. The Buley Library is very tall. Therefore, either the Buley Library is very tall or it has many levels underground.
2. The grass is green. The sky is blue. Therefore, the grass is green and the sky is blue.
3. $$g$$ has order $$3$$ or it has order $$4$$. If $$g$$ has order $$3$$, then $$g$$ has an inverse. If $$g$$ has order $$4$$, then $$g$$ has an inverse. Therefore, $$g$$ has an inverse.
4. $$x$$ is greater than $$5$$ and $$x$$ is less than $$53$$. Therefore, $$x$$ is less than $$53$$.
5. If $$a|b$$, then $$a$$ is a perfect square. If $$a|b$$, then $$b$$ is a perfect square. Therefore, if $$a|b$$, then $$a$$ is a perfect square and $$b$$ is a perfect square.
Exercise $$\PageIndex{5}$$
Read the following proof that the sum of two odd numbers is even. Discuss the rules of inference used.
Proof: Let $$x$$ and $$y$$ be odd numbers. Then $$x = 2k + 1$$ and $$y = 2j + 1$$ for some integers $$j$$ and $$k$$. By algebra,
$$x + y = 2k + 1 + 2j + 1 = 2(k + j + 1).$$
Note that $$k +j +1$$ is an integer because $$k$$ and $$j$$ are integers. Hence $$x + y$$ is even.
Q.E.D.
Exercise $$\PageIndex{6}$$
Sometimes in constructing a proof we find it necessary to “weaken” an inequality. For example, we might have already deduced that $$x < y$$ but what we need in our argument is that $$x ≤ y$$. It is okay to deduce $$x ≤ y$$ from $$x < y$$ because the former is just shorthand for $$x < y ∨ x = y$$. What rule of inference are we using in order to deduce that $$x ≤ y$$ is true in this situation? | 2021-10-28T13:40:37 | {
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https://math.stackexchange.com/questions/840924/prove-that-sum-n-2-infty-ln-n-ln-n-converges?noredirect=1 | # Prove that $\sum_{n=2}^\infty (\ln n)^{- \ln n}$ converges
As the title suggests, I'd like to prove that the sum $$\sum_{n=2}^\infty (\ln n)^{- \ln n}$$ is finite. The root and ratio test both fail here, but WA suggests that there is a comparison that can be used to show convergence.
The only thought I have is that it may help to write the terms as $e^{-\ln(n)\ln(\ln(n))}$, but this has not led me to any particular insight. Any ideas are appreciated.
Well,
$$e^{-(\log n)(\log \log n)} = n^{-\log \log n} < n^{-2}$$
for $n > e^{e^2}$. So a comparison with $\sum \frac{1}{n^2}$ shows the convergence.
• Can't believe I didn't see that. Thank you – Omnomnomnom Jun 20 '14 at 17:47
There is a theorem that for a decreasing series $\sum a_n$ converges if and only if $\sum 2^k a_{2^k}$ converges. Applying that in this case gives, where I assume the log is base $2$ just to make my life easy, $$\sum \left(\frac{2}{k}\right)^k$$ and this is easily seen to converge by root or ratio.
For large enough $n$, one has $\ln n > e^2$, so that $(\ln n)^{-\ln n} < e^{-2\ln n} = {1 \over n^2}$.
By Cauchy condensation test we have
$$\sum 2^na_{2n}=\sum \frac{2^n}{(\log 2^n)^{\log 2^n}}=\sum \frac{2^n}{n^{(n\log 2)}\cdot{(\log2)}^{(n\log 2)}}$$
which converges for example by limit comparison test with $$\sum \frac1{n^2}$$.
Hint: You can use Cauchy condensation test. | 2020-01-27T22:21:01 | {
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https://math.stackexchange.com/questions/2611446/showing-that-a-relation-is-reflexive-symmetric-and-transitive | # Showing that a relation is reflexive, symmetric and transitive
I am attempting the following question:
Let $S = \Bbb Z$ x $(\Bbb Z$ \ $\{0\})$, be pairs of integers where the second coordinate is non-zero. Let the relation $R \subseteq S ^2$ be defined by:
$(a,b)R(c,d) \leftrightarrow ad = bc$
(a) Describe the following properties of relations: reflexive, symmetric and transitive.
(b) Show that $R$ has these properties.
(c) What is the name for relations satisfying the properties in (a).
(d) Is the function $f: R\times R \rightarrow \Bbb Q$ defined by $f(a,b) = > a/b$ bijective?
I know how to decribe the properties of relations, so question (a) shouldn't be an issue.
I believe the answer to question (c) is "an equivalence relation".
However, I am completely lost as to where I start with questions (b) and (d). I have been trying to understand relations for weeks, but just can't seem to wrap my head around it.
Any help explaining (b) and (d) would be greatly appreciated.
• That is, you are having trouble saying why the relation $R$ is symmetric,reflexive and transitive? Once you have described the properties, it is only a question of substituting into the definition and checking whether it holds. – Teresa Lisbon Jan 18 '18 at 23:05
• I think you have the letters a), b), c), and d) mixed up in your question. You seem to know the answer to a) and c), (not d), and want help for b) and d) (not a and d). – Steve Kass Jan 18 '18 at 23:07
• @астонвіллаолофмэллбэрг The problem is, I know how to check those properties for simple relations (i.e. checking that $xRy \rightarrow yRx$), but I don't understand how to do it in this example, because I don't really understand the relation. What does "Let $R⊆S^2$ be defined by: $(a,b)R(c,d)↔ad=bc$" actually mean? How is it input into the simple reflexive ($xRx$), symmetric ($xRy \rightarrow yRx$) and transitive ($xRy, yRz \rightarrow xRz$) formulas I already know? – Shannon Jan 18 '18 at 23:07
• @SteveKass Thank you, updated. – Shannon Jan 18 '18 at 23:08
• Strictly speaking (d) is not well written. The elements of $R$ are pairs of pairs $((a,b),(c,d))$, and only those pairs where $ad=bc$. The function $f$ should be $f((a,b),(c,d))=a/b$. – orole Jan 18 '18 at 23:11
What does this relation actually mean?
It really is all about the equivalence of fractions. $\frac {a}{b} = \frac dc \implies ac = bd$
Now, instead of writing these fractions in the more familiar form we show them as and ordered pair.
a,b)
reflexive:
$(a,b)R(a,b) \iff ab = ba$
symmetric:
$(a,b)R(c,d) \implies (c,d)R(a,b)$
$ad = bc \implies cb =da$
Transitive:
$(a,b)R(c,d)$ and $(c,d)R(p,q) \implies (a,b)R(p,q)$
$ad = bc$ and $cq = dp$
$adp = bcp\\ acq = bcp\\ aq = bp$
d)
Is the function $f(a,b) = \frac ab$ bijective?
It is not injective as $(1,2)$ and $(2,4)$ both map onto the same element of $\mathbb Q$
• Such a fantastic answer, thank you! One question though: why does the question specify that $R⊆S^2$ when it doesn't seem to be used throughout the question? – Shannon Jan 18 '18 at 23:27
• In my proof of transativity, I factored out and dropped the $c.$ I could not do that if $c = 0$ however by the definition of $S^2, c \ne 0$ and is not a concern. I suppose I should have included this note at that step. – Doug M Jan 18 '18 at 23:33
The problem is, I know how to check those properties for simple relations (i.e. checking that $xRy→yRx$), but I don't understand how to do it in this example, because I don't really understand the relation.
$\def\R{\operatorname R}\R$ is symmetric if $\forall s\in S\;\forall t\in S\; [s\R t\to t\R s]$. Now $S=\Bbb Z\times(\Bbb Z\setminus \{0\})$, and $\forall (a,b)\in S\;\forall (c,d)\in S\;[(a,b)\R(c,d)\leftrightarrow (ad=cb)]$ so that would be $\forall (a,b)\in S\;\forall (c,d)\in S\;[(ad=cb)\to (cb=ad)]$. Because this is true (via the symmetry of equality), therefore $\R$ is symmetric.
Use similar principle to check for Reflexivity and Transitivity.
$\\[3ex]$
Hint: You might find it helpful that $(ad=cb)\leftrightarrow (a/b=c/d)$ if $b\neq0, d\neq 0$. | 2021-03-03T03:18:07 | {
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https://www.physicsforums.com/threads/sequence-question.764905/ | # Sequence question
1. Aug 6, 2014
1. The problem statement, all variables and given/known data
The last two terms, are added together to produce the next term
$-33,x,y,z,88$
$\text{Find }x,y\text{ and }z$
2. Relevant equations
$y=x-33$
$x+y=z$
$y+z=88$
3. The attempt at a solution
By substituting the expression for y in the third equation for the first equation, we get:
$88-z=x-33 \rightarrow x=121-z$
Again, by substituting the expression for y in the first equation for the second equation we get:
$x+(x-33)=z \rightarrow 2x-33=z$
So we get two equations,
$x=121-z$
$2x-33=z$
Solving both gives the right answers but I remember getting the wrong answer by not including one of the equations there.
Is there any easy way to solve all three equations at once?
Last edited: Aug 6, 2014
2. Aug 6, 2014
### HallsofIvy
Staff Emeritus
I'm not sure what you mean by "solve all three at once". I get this image of just immediately writing down the values for x, y, and z! I could not do that!:tongue:
But you can rewrite the first equation as y- x= -33. Adding that to x+ y= z gives 2y= z- 33. The last equation, y+ z= 88, is equivalent to y= -z+ 88. Adding those equations, 3y= 55.
That's about as simple as I can make it.
3. Aug 6, 2014
Oh, I must have made some silly mistake in the previous calculations. Thanks :)
Haha, I mean like how you solve two simultaneous equation, is there a method for solving three simultaneous equations like this? Other than substituting the values?
4. Aug 6, 2014
### HakimPhilo
This system of linear equations, like any other, can be solved by first writing it in upper-triangular form then using the method of back-substitution. Let's first define the terms we used: A system in upper-triangular form is one like: $$\left\{\begin{array}{rl}4x-3y+2z&=-5\\14y+2z&=18\\-4z&=3\end{array}\right.,\qquad\left\{\begin{array}{rl} 15x-2y+z&=1\\ 3z&=-8\end{array}\right..$$ As you can see, what's common about them is that if $x$ appears in no equation other than the first one and $y$ appears in no equation after the second... (It is possible that $y$ may not even appear in the second equation as in e.g.2)
You can easily solve systems in this type using the method of back-substitution. Take the first example, using the third equation you solve for $z$ to get $z=-\tfrac34$, you substitute that value in the second equation to get $14y-\tfrac32=18$ which can be used to solve for $y$, and when you get your value for $y$, you substitute it in the first equation along with your value for $z$ to solve for $x$.
What's interesting now is that you can try to transform any system you have in upper-triangular form, then you use back-substitution and you're done. The basic way we proceed with this task consists of applying the elementary operations which are: 1. multiplying both sides of an equation by a nonzero constant, 2. interchanging the order of two equations, 3. adding to one equation a multiple of another.
I think this will give you enough information to solve any system in 3 variables or more. If you need to know more then just ask away.
5. Aug 7, 2014 | 2017-08-22T08:02:49 | {
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http://pausescreen.frozenfoxmedia.com/739xv6/pi-notation-examples-7f754b | # pi notation examples
... How kids in Monte Carlo estimate Pi. ... pi+exp(2) π + exp 2. Worked examples: Summation notation. In prefix notation, the operators are placed before the operand. The product notation is symbolised by the letter Pi in capital letters. Additional Examples using Sigma Notation In the following examples, students will show their understanding of sigma notation by evaluating expressions using … Riemann sums in summation notation. You can try some of your own with the Sigma Calculator. The following examples illustrate entry of some common expressions. This is the next convergent to pi. 1972, Item 120, this is an approximation of $$\pi$$. With pi notation, we are able to separate a product of two sequences into two separate products. Volume, surface and circumference calculations from a radius. In simple terms, Single bonds are sigma bonds, Pi bonds are double/triple bonds. Click the preview button to display this. There are two versions, prefix notation and postfix notation. Sigma Notation: In mathematics, a sigma notation is a way of expressing a long sum in a compact form. From the example, one can see that for complicated cases a large amount of the answer is formed from operator names, such as PROJECT and JOIN. We note each term in the approximation gives an additional bit of precision (see above link) thus 14 terms give 4 decimal digits of precision each time (since $$2^{14} \gt 10^4$$). The following are 30 code examples for showing how to use numpy.pi(). The only difference is that an object name is always underlined in UML. Calculation of gas mileage for several consecutive fillups. The Big-O notation is the standard metric used to measure the complexity of an algorithm. There are many special methods used to calculate π and here is one you can try yourself: it is called the Nilakantha series (after an Indian mathematician who lived in the years 1444–1544).. Factorial There is one important function definition that can be expressed in terms of pi notation: The Microsoft Excel PI function returns the mathematical constant called pi, which is 3.14159265358979. PI Server’s Event Frames are tailor-made for operations because you can trigger data collection based on process data thresholds and other KPIs. Introduction 2 2. invented this notation centuries ago because they didn’t have for loops; the intent is that you loop through all values of i from a to b (including both endpoints), summing up the body of the summation for each i. So if this is pi right over here, this would be three pi over two and this would be two pi right over here, two pi. The product F is returned in terms of k where k represents the upper bound. Object: An object is an entity which is used to describe the behavior and functions of a system. Expression. The following chapters describe all of Octave’s features in detail, but before doing that, it might be helpful to give a sampling of some of its capabilities. Whenever I think of pi-notation, I think of smaller finite products. Examples of Symbolic Questions with Calculator Notation. Riemann sums in summation notation. This project calls for the SwitchDoc Labs HDC1080; if you are using an Amazon device, make sure to use a Grove patch cable. In most processes, … Writing a long sum in sigma notation 5 4. Unlike numeric notation, you do not need to specify all permissions for each of the permission groups. Omega Table 1: Upper and lower case greek letters. Sigma … Second, Wallis did not prove the result rigorously. If you are new to Octave, we recommend that you try these examples to begin learning Octave by using it. Related Queries: plot floor[e x]/floor[pi x] from x = 0 to 200; closed form 7.1880827; bisection method sin(x) = 0 with x1 = 2 and x2 = 4 (googol/11222.11122)^(1/193) These examples are extracted from open source projects. It can be used as a worksheet function (WS) in Excel. Symbolic Notation. Summation Notation 3 of5 A Alpha N Nu B Beta ˘ Xi Gamma O o Omicron Delta ˇ Pi E Epsilon P ˆ Rho Z Zeta ˙ Sigma H Eta T ˝ Tau Theta ˛ Upsilon I Iota ˚ Phi K Kappa X ˜ Chi Lambda Psi M Mu ! In this article, we studied what Big-O notation is and how it can be used to measure the complexity of a variety of algorithms. This is completely in line with what you just read about the operator being a shorthand notation for a longer notation with repetition, where you use the regular <-assignment operator. Pi is the symbol representing the mathematical constant , which can also be input as ∖ [Pi]. 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http://mathhelpforum.com/calculus/88787-power-series.html | 1. ## Power Series
I am finding the radius of convergence and interval of convergence.
the sum of n=1 to infinity of x^n/(6^n*n^4)
i used the ratio test on this and found that the limit goes to 1 and the radius of convergence is 6.
when x is -6, i know the series alternates but when i compare b of n to b of n+1 I don't see how b of n is greater than b of n+1 which is one of the stipulations for the alternating series test.
2. Originally Posted by aaronb
I am finding the radius of convergence and interval of convergence.
the sum of n=1 to infinity of x^n/(6^n*n^4)
i used the ratio test on this and found that the limit goes to 1 and the radius of convergence is 6.
when x is -6, i know the series alternates but when i compare b of n to b of n+1 I don't see how b of n is greater than b of n+1 which is one of the stipulations for the alternating series test.
$a_n=\frac{x^n}{n^4\cdot6^n}$
$a_{n+1}=\frac{x^{n+1}}{(n+1)^4\cdot6^{n+1}}=\frac{ x}{6}\cdot\frac{x^n}{(x+1)^46^n } = 1\cdot \frac{x^n}{(x+1)^46^n } <\frac{x^n}{n^46^n}=a_n$
Remember that $\frac{x}{6}= 1$ at the egde of the radius of convergence
3. root test is faster than ratio test.
4. Hello, aaronb!
Interesting . . . I've seen the Ratio Test described in several ways now.
. . And a few of them never form a ratio.
Find the radius of convergence and interval of convergence:
. . $\sum^{\infty}_{n=1} \frac{x^n}{6^nn^4}$
$R \:=\:\left|\frac{a_{n+1}}{a_n}\right| \;=\;\left|\frac{x^{n+1}}{6^{n+1}(n+1)^4}\cdot\fra c{6^nn^4}{x^n}\right| \;=\;\left|\frac{6^n}{6^{n_1}}\cdot\frac{n^4}{(n+1 )^4} \cdot\frac{x^{n+1}}{x^n}\right|$
. . $= \;\left|\,\frac{1}{6}\left(\frac{n}{n+1}\right)^4 x\,\right| \;=\; \left|\,\frac{1}{6}\left(\frac{1}{1 + \frac{1}{n}}\right)^4x\,\right|$
Then: . $\lim_{n\to\infty} \left|\,\frac{1}{6}\left(\frac{1}{1 + \frac{1}{n}}\right)^4x\,\right| \;=\;\left|\frac{1}{6}\cdot1^4\cdot x\,\right| \;=\;\left|\frac{x}{6}\right|$
So we have: . $\left|\frac{x}{6}\right| \:< \:1 \quad\Rightarrow\quad -6 \:<\:x\:<\:6$
. . The radius of convergence is: . $R \,=\,6$
At $x = \text{-}6\!:\;\;\sum^{\infty}_{n=1}\frac{(\text{-}6)^n}{6^nn^4} \;=\;\sum^{\infty}_{n=1}\frac{(\text{-}1)^n}{n^4}\quad\hdots$ an alternating $p$-series which converges
At $x = 6\!:\;\;\sum^{\infty}_{n=1}\frac{6^n}{6^nn^4} \;=\;\sum^{\infty}_{n=1}\frac{1}{n^4} \quad\hdots$ a convergent $p$-series
. . The interval of convergence is: . $[-6,\:6]$
5. Originally Posted by aaronb
I am finding the radius of convergence and interval of convergence.
the sum of n=1 to infinity of x^n/(6^n*n^4)
i used the ratio test on this and found that the limit goes to 1
I don't know what you mean by that. Do you mean that the limit of the ratios is less than 1 for |x|< 6?
and the radius of convergence is 6.
Yes, that is correct.
when x is -6, i know the series alternates but when i compare b of n to b of n+1 I don't see how b of n is greater than b of n+1 which is one of the stipulations for the alternating series test.
When x= -6, this is
$\sum_{n= 1}^\infty \frac{(-1)^}{6^n n^4}$
$\frac{\frac{1}{6^n n^4}{\frac{1}{6^{n+1}(n+1)^4}= 6\left(\frac{n+1}{n}\right)^4$
It should be clear that that is larger than 6 so the numerator is larger than the denominator- the sequence is decreasing. | 2017-04-28T19:06:51 | {
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https://math.stackexchange.com/questions/1865707/why-can-we-change-a-limits-function-expression-and-claim-that-the-limits-are-id | # Why can we change a limit's function/expression and claim that the limits are identical?
Say you have limit as $x$ approaches $0$ of $x$. You could just write it as $\frac{1}{\frac{1}{x}}$ and then the expression would be undefined. So what are you really doing when you "rearrange" an expression or function so its limit "works", and doesn't have any division by zero? Why can we suddenly change one expression to another, when they are not exactly the same, and say the limit is the same? For example, the expression $x$ is not equal to the expression $\frac{1}{\frac{1}{x}}$ because the latter is not valid for $x=0.$
• Who says $\lim_{x\to 0}\frac{1}{1/x}$ is undefined? It is pretty clearly $0$. – Henning Makholm Jul 20 '16 at 18:40
• @HenningMakholm I never said the limit was undefined, but the expression itself is at $x=0$. Since the expressions $x$ and $\frac{1}{\frac{1}{x}}$ are not equal, why do we assume their limits be equal? That is my question. – Max Li Jul 20 '16 at 18:41
• The two expressions are equal – user99914 Jul 20 '16 at 18:42
• x @Max: Please edit the question to fix "then the limit would be undefined", which must be a typo if you never said the limit is undefined. – Henning Makholm Jul 20 '16 at 18:42
• @ArcticChar No they aren't, because $\frac{1}{\frac{1}{x}}$ does not have a defined value at $x=0$. – Max Li Jul 20 '16 at 18:43
When we take the limit for $x \rightarrow 0$ of a function $f$, we are actually interested to what happens to $f$ in a punctured neighborhood of $0$. That is, we don't actually care about the behaviour of the function when $x$ is exactly $0$. This means that $f(x)=x$ and $g(x)=\frac{1}{\frac{1}{x}}$ have the same limit for $x \rightarrow 0$, because they are equal in a punctured neighborhood of $0$.
Note that this can be seen from the definition itself; that is, $\lim_{x \rightarrow 0} f(x)=l$ means
$$\forall \ \epsilon > 0 \ \ \exists \ \delta > 0 : 0 < |x| < \delta \Rightarrow |f(x)-l| < \epsilon.$$
If you take another function $g$ and you assume $g=f$ in a punctured neighborhood of $0$, it follows that $$\lim_{x \rightarrow 0} f(x)=l \iff \lim_{x \rightarrow 0} g(x)=l\ .$$
Note that the limit of $\frac{1}{\frac 1x}$ as $x\to 0$ is not undefined. Of course you cannot plug in $x = 0$ in this expression. However, it is not how limit works. In the definition of limit you need only care about the value of the function near that point, ($0$ in this case). Since
$$x = \frac{1}{\frac 1x}$$
holds in a punctured neighborhood of $0$, so
$$\lim _{x \to 0} x = \lim _{x\to 0} \frac{1}{\frac 1x} = 0.$$
You will be less confused (I hope) if you realize that when we talk about
$$\lim_{x\to c}f(x),$$
the function $f$ does not need to be even defined at $c$. You only need that $f$ is defined on a punctured neighborhood of $c$.
• I meant that the expression is undefined, so it is not equal to the expression of $x$. Since the expressions are different, why can we assume the limits are the same? – Max Li Jul 20 '16 at 18:44
• @MaxLi: Because the expressions have the same value in some neighborhood of $0$ (namely the entire real line) except for at $0$ itself. – Henning Makholm Jul 20 '16 at 18:45
Remark: sort of a long comment.
What you are rediscovering are so called removable singularities: Note that the functions $$f(x)=x$$ and $$\tilde{f}(x)=\frac{1}{\frac{1}{x}},$$ are not the same, as they domain of definition differs: the first is defined for all real numbers, whereas the second is not defined at zero (as you noticed).
But one can adjust the second by defining $$\bar{f}(x):=\begin{cases}\tilde{f}(x)&, x\neq 0\\0 &, x=0\end{cases}.$$ As explained in the other answers, $\lim_{x\rightarrow 0}\bar{f}(x)=0,$ which implies that the function is continuous at $0$ (as was clerly expected).
Such an adjustment is called removal of singularity.
• I think you are missing the point. Having a limit is different from the function being continuous. And you still cannot answer why $$g(x):=\begin{cases}\tilde{f}(x)&, x\neq 0\\0 &, x=0\end{cases}$$ will have the same limit as $\tilde f$, as again these two are not the same function. – user99914 Jul 20 '16 at 18:54
• What I am trying to say, is that the operation he is doing is not mathematically correct, as the resulting function is not the same as the starting, but that this error can be easily corrected by taking the natural continuation, which exists as the limit is the same, independently of this transformation.. That the limit as $x$ tends to zero is the same, is trivial. – b00n heT Jul 20 '16 at 18:59
When we are computing limits like $x\rightarrow a$, the value of the function at $x=a$ is irrelevant.
This means that if we are evaluating $\lim_{x\rightarrow a} f(x)$, and we know of a second function $g(x)$ which is equal to $f(x)$ everywhere except at $x=a$, then we can make the substitution
$$\lim_{x\rightarrow a} f(x) = \lim_{x\rightarrow a}g(x).$$
The reasoning is that $f$ and $g$ agree everywhere except at $x=a$, and limits don't depend on the value of the function at the limit point, so therefore the two limit values must be identical.
This principle is what allows us to transform, for example,
$$\lim_{x\rightarrow 0}\frac{x^2 +2x}{x} \Rightarrow \lim_{x\rightarrow 0}(x+2) = 2$$
where here $f(x)\equiv \frac{x^2+2x}{x}$ and $g(x) \equiv x+2$ are two functions that have exactly the same values everywhere except at the irrelevant location $x=0$. | 2019-12-07T01:58:54 | {
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https://mathoverflow.net/questions/281115/a-bounded-polynomial-having-bounded-coefficients-several-variables | A bounded polynomial having bounded coefficients: several variables
Consider the multivariate case for the question "Approximation theory reference for a bounded polynomial having bounded coefficients" (Approximation theory reference for a bounded polynomial having bounded coefficients)
I tried to find an upper bound for coefficients for each monomial. I tried two methods:
1. Lagrange interpolation. As in Lemma 4.1 in http://eccc.hpi-web.de/report/2012/037/download .
2. I first only consider the variable $x_1$ and get an upper bound for the coefficients. Now the bound is the upper bound for another polynomial of $x_2,x_3,...,x_n$ and we can repete the above method.
However, using either method above, the upper bound will contain factor $d^n$ or $c^{d*min\{d,n\}}$ where c is a constant, which I think is too large. Is it possible to find a better upper bound?
• Using the answers to the previous question, one gets a bound $c^{dn}$ with the second method, with $c = 1 + \sqrt{2}$. The proof actually allows to replace $dn$ with $\sum_{i=1}^n \mathrm{deg}_{x_i}(P)$. – js21 Sep 14 '17 at 9:36
• What are $d$, $n$, and $k$? $d$ the degree, $n$ the number of variables (?), but $k$? – Ilya Bogdanov Sep 14 '17 at 16:22
• Sorry for some messy notation. I have edited it. – Felix Y. Sep 15 '17 at 7:32
2 Answers
Bounds in the univariate case, see e.g. here, were established by V.A. Markov in 1892.
S.N. Bernstein has given an extension of the result to the multivariate case in
S.N. Bernstein, On certain elementary extremal properties of polynomials in several variables, Doklady Akad. Nauk SSSR (N.S.) 59 (1948), 833-836.
Unfortunately I do not have access to that paper, but according to MR0023953, the following result is proved. Let $$P(x_{1},\ldots,x_{n})=\sum_{1\leq\alpha_{h}\leq d_{h}} A_{\alpha_{1},\ldots,\alpha_{n}}x_{1}^{\alpha_{1}}\ldots x_{n}^{\alpha_{n}},$$ such that $|P|\leq1$ on the cube $|x_{h}|\leq1$, $1\leq h\leq n$. Then $$|A_{\alpha_{1},\ldots,\alpha_{n}}|\leq\frac{\prod_{h=1}^{n}|B_{\alpha_{h}}^{(d_{h})}|}{\alpha_{1}!\ldots\alpha_{n}!},$$ where $B_{\alpha_{h}}^{(d_{h})}$ is the coefficient of $x^{\alpha_{h}}/\alpha_{h}!$ in the expansion of the Chebyshev polynomial $$T_{m}(x)=\cos(m\cos^{-1}(x))=\sum_{i=0}^{m}B_{i}^{(m)}x^{i}/i!,$$ and $m=d_{h}$ if $d_{h}-\alpha_{h}$ is even and $m=d_{h}-1$ if $d_{h}-\alpha_{h}$ is odd.
For the particular case of homogeneous polynomials $$P_{d}(x_{1},\ldots,x_{n})=\sum_{|\alpha|= d}c_{\alpha}x^{\alpha},$$ of total degree $d$, where $\alpha$ are multi-indices, such that $|P_{d}|\leq1$ on the ball $x_{1}^{2}+\cdots +x_{n}^{2}\leq1$, upper bounds on the coefficients are given in Theorem 2 of
O.D. Kellogg, On bounded polynomials in several variables, Math. Z. 27 (1928), no. 1, 5-64.
namely, the coefficient of the monomial $x_{1}^{k_{1}}\ldots x_{n}^{k_{n}}$ cannot exceed in absolute value $$\frac{d!}{k_{1}!\ldots k_{n}!}.$$
For general (nonhomogeneous) polynomials of total degree $d$, there are also precise bounds on the coefficients $c_{\alpha}$ with $|\alpha|=d$ or $d-1$ (for the case of the ball or the cube).
In that connection, two interesting papers are
M.I. Ganzburg, A Markov-type inequality for multivariate polynomials on a convex body. J. Comput. Anal. Appl. 4 (2002), no. 3, 265-268
H-J. Rack, On V. A. Markov's and G. Szegő's inequality for the coefficients of polynomials in one and several variables. East J. Approx. 14 (2008), no. 3, 319-352,
and the bibliography therein.
We recently needed a result of this form and couldn't find an appropriate paper to cite (some of the papers cited above are hard to locate), so we proved the result. It's actually not too hard to show an upper bound of $d^{O(d)}$ on the magnitude of the largest coefficient of such a polynomial. More precisely, the following is true:
Let $p$ be a real polynomial on $n$ variables of degree $d$ such that for all $x\in[0,1]^n$, $|p(x)|\leq 1$. Then the magnitude of every coefficient of $p$ is upper bounded by $(2d)^{3d}$.
No attempt was made to optimize the constants in the above statement. A full proof of the above statement can be found in Ref. [1] as Theorem 46.
I don't know if the result is tight (up to constants in the exponent), because in the univariate case you can show a better upper bound of $2^{O(d)}$, so there is a gap here.
[1] Shalev Ben-David, Adam Bouland, Ankit Garg, and Robin Kothari. Classical lower bounds from quantum upper bounds, 2018. | 2020-07-02T13:36:55 | {
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https://math.stackexchange.com/questions/1296074/i-think-i-see-mysterious-lines-inside-triangles-how-to-prove-their-existence | # I think I see mysterious lines inside triangles—how to prove their existence?
Lately I've been fooling around with points inside a triangle and the sum of their distances from all sides.
This was when I noticed a weird behaviour: For each point I chose there always seemed to be a straight line going through my chosen point and the entire triangle where every point had the same sum of distances from all sides! And as if that's not enough, if I select a different point the line through this point looks parallel to all the other lines created in the same manner but through other points.
So is there a way to prove my observation?
Question: Do all points inside a triangle that have the same sum of distances from all sides lie on a line and is there a way to give a mathematical equation for said line? (Disregarding equilateral triangles)
Because I used numerical means to find this pattern I am not sure whether it even exists. Any kind of help will be appreciated!
• I think you should edit the title, because it is just weird. – user230734 May 24 '15 at 0:26
• I quite like the evocative title. – pjs36 May 24 '15 at 1:05
• How to prove their existence ? - Build the triangles, and the proof will come. – Lucian May 24 '15 at 4:43
• I think you should not edit the title, because it is just weird. – JiK May 24 '15 at 11:26
• this question comes close to another problem of bundeswettbewerb mathematik in germany, round 2, deadline for entries Sept. 1st, 2015. please stick to the rules and don't ask for solutions in the community. thank you to all users for not posting comments or solutions to the above mentioned problem. Karl Fegert Grading commission bundeswettbewerb mathematik germany – user246719 Jun 8 '15 at 16:55
Let us look at the problem in a slightly different angle. If one know baricentric coordinate system, it will be "obvious" why this is true.
Let $\vec{A}, \vec{B}, \vec{C}$ be any three non-collinear points, they form the vertices of a non-degenerate triangle $\triangle ABC$. For any point $\vec{p} \in \mathbb{R}^2$, there exists a unique pair of real numbers $\alpha, \beta$ such that
$$\vec{p}-\vec{C} = \alpha( \vec{A}-\vec{C}) + \beta( \vec{B} - \vec{C} ) \quad\iff\quad \vec{p} = \alpha \vec{A} + \beta \vec{B} + (1 - \alpha - \beta)\vec{C}$$ Let $\gamma = 1 - \alpha - \beta$, the triplet $(\alpha,\beta,\gamma)$ is called the baricentric coordinates for $\vec{p}$. Furthermore, the points $\vec{p}$ lies inside or on $\triangle ABC$ if and only if $\alpha, \beta, \gamma \ge 0$
Let $h_A, h_B, h_C$ be the height of $\triangle ABC$ for corresponding vertices. The distance between $\vec{p}$ and the sides $BC$, $CA$, $AB$ are $h_A |\alpha|$, $h_B |\beta|$ and $h_C|\gamma|$ respectively. The loucs for a point whose sum of distances to the 3 sides equal to $d$ is then given by:
$$h_A |\alpha| + h_B|\beta| + h_C|\gamma| = d$$
For points inside $\triangle ABC$, the problem of finding the locus is equivalent to solving following pair of linear equations:
$$\begin{array}{rrrl} \alpha +& \beta +& \gamma &= 1\\ h_A \alpha +& h_B \beta +& h_C \gamma &= d \end{array}$$ When $\triangle ABC$ is not equilateral, this pair of equations has rank 2 which has either zero or infinite many solutions. Furthermore if $(\alpha, \beta, \gamma)$ is a solution, other solution will have the form:
$$(\alpha',\beta',\gamma') = (\alpha,\beta,\gamma) + \lambda (h_B-h_C,h_C-h_A,h_A-h_B)\quad\text{ for some } \lambda \in \mathbb{R}$$
Translate this back to points on $\mathbb{R}^2$. This mean is $\vec{p}$ is a point inside $\triangle ABC$, the locus of point $\vec{p}'$ have same sum of distances has the form:
$$\vec{p}' = \vec{p} + \lambda\vec{u}, \quad\text{ for some }\;\lambda \in \mathbb{R}$$ i.e. the locus is a line along the direction $\displaystyle\;\vec{u} = (h_B-h_C)\vec{A} + (h_C-h_A)\vec{B} + (h_A-h_B)\vec{C}$.
Please note that this $\vec{u}$ is independent of choice of $d$ and hence $\vec{p}$. What this means is for all points inside $\triangle ABC$, not only the locus of same distances are all lines, all those lines are parallel to each other!
Clarifications
About the question why multiplying $\alpha$ with height $h_A$ gives us the distance to line $BC$. For any point $p$, let $d_p$ be the distance of $p$ to the line $BC$. By definition, we have $$\vec{p} - \vec{C} = \alpha(\vec{A}-\vec{C}) + \beta(\vec{B}-\vec{C}).$$ For any fixed $\alpha$, let $\vec{p}_0 = \vec{C} + \alpha(\vec{A}-\vec{C})$. For any point $p$ with same $\alpha$, we have $$\vec{p} - \vec{p_0} = \beta (\vec{B} - \vec{C})$$ When viewed from $p_0$, $p$ is along the direction $\vec{B}-\vec{C}$. This means the locus of $p$ for fixed $\alpha$ is a line parallel to the side $BC$. As a result, $d_p$ is constant over such a line and $d_p$ depends only on $\alpha$. As long as $p$ doesn't crosses the line $BC$, it is clear this dependence on $\alpha$ is linear. Notice
• When $\alpha = 0$, the line of constant $\alpha$ coincides with line $BC$, so $d_p = 0$ there.
• When $\alpha = 1$, the line of constant $\alpha$ crosses $A$, so $d_p = h_A$ there.
Combine these, we find the proportional constant is $h_A$ when $\alpha \ge 0$. This means as long as $p$ is on the same side as $A$ with respect to line $BC$, $d_p = h_A \alpha = h_A |\alpha|$. By symmetry, $d_p = h_A |\alpha|$ on the other side too.
• and +1 for at least mentioning equilateral triangles. – A. Rex May 26 '15 at 4:39
• Could you please explain why by multiplying our $\alpha, \beta,\gamma$ with the corresponding height we obtain the distance of our point $p$ from one side? – user161516 May 28 '15 at 7:54
• @Silenttiffy I updated the answer, I hope that help. – achille hui May 28 '15 at 10:10
• @achillehui While your argument wonderfully explains why the formula is correct for $\alpha, \beta$ I can't make a connection to $\gamma$ because unlike the others $\gamma$ can not be understood as a component parallel to one side. Generally speaking, what is the use of $\gamma$ when it is completely dependent on $\alpha$ and $\beta$? – user161516 May 28 '15 at 21:56
• @Silenttiffy If you relabel you vertices from $A,B,C$ to $A' = B, B' = C, C' = A$, then the corresponding $\alpha' = \beta, \beta' = \gamma, \gamma' = \alpha$. So the line for constant $\gamma$ = line of constant $\beta'$ is parallel to the line $C'A' = AB$. The point of baricentric coordinates is the formulas remains valid under relabeling. To attack a problem, you have the freedom to choose either $\alpha,\beta$; $\beta, \gamma$ or $\gamma,\alpha$ as independent variables for analysis. One typically pick the one to make the problem easiest. – achille hui May 28 '15 at 22:10
This is easy to prove if you know a bit about vectors and dot products.
For any vector $\vec{a} \neq \vec{0}$ and any real number $b$, the set of points $\vec{x}$ which satisfy $\vec{a} \cdot \vec{x} = b$ forms a line that is perpendicular to $\vec{a}$. Every line in the plane can be written in this manner. Also, the distance between a point $\vec{y}$ and the line $\vec{a} \cdot \vec{x} = b$ is given by $\frac{|\vec{a} \cdot \vec{y} - b|}{\|\vec{a}\|}$.
Call the sides of the triangle side $1$, side $2$, and side $3$. For $i = 1,2,3$, pick a unit vector $\vec{a}_i \neq \vec{0}$ that is normal to side $i$ and points inward, and a number $b_i$ such that the points $\vec{x}$ on side $i$ satisfy $\vec{a}_i \cdot \vec{x} = b_i$
Then, the distance between a point $\vec{x}$ and side $i$ is given by $|\vec{a}_i \cdot \vec{x}-b_i|$. If $\vec{x}$ is inside the triangle, then $\vec{a}_i \cdot \vec{x}-b_i > 0$ (since we picked $\vec{a}_i$ to point inwards), and so, the distance from $\vec{x}$ to side $i$ is simply $\vec{a}_i \cdot \vec{x}-b_i$.
Thus, the total distance from a point $\vec{x}$ (inside the triangle) to the three sides of the triangle is $(\vec{a}_1 \cdot \vec{x}-b_1) + (\vec{a}_2 \cdot \vec{x}-b_2) + (\vec{a}_3 \cdot \vec{x}-b_3)$ $= (\vec{a}_1+\vec{a}_2+\vec{a}_3) \cdot \vec{x} - (b_1+b_2+b_3)$.
If we let $C$ be any constant, then the sum of the distances from a point $\vec{x}$ inside the triangle to the sides will equal $C$ provided $(\vec{a}_1+\vec{a}_2+\vec{a}_3) \cdot \vec{x} - (b_1+b_2+b_3) = C$, i.e. $(\vec{a}_1+\vec{a}_2+\vec{a}_3) \cdot \vec{x} = b_1+b_2+b_3+C$.
As long as $\vec{a}_1+\vec{a}_2+\vec{a}_3 \neq \vec{0}$, then this is a line which is normal to $\vec{a}_1+\vec{a}_2+\vec{a}_3$. Furthermore, no matter what we pick the total distance $C$, this line will be normal to $\vec{a}_1+\vec{a}_2+\vec{a}_3$. Hence, all the lines formed in this manner are perpendicular to a common vector, and thus, are parallel.
If you aren't familiar with vectors and dot products, you can write out $\vec{a}_1 = (a_{11},a_{12})$, $\vec{a}_2 = (a_{21},a_{22})$, $\vec{a}_3 = (a_{31},a_{32})$, $\vec{x} = (x,y)$, and carry out the same computations. You'll still end up with the equation of a line.
If the edges of the triangle lies on lines $A_i x + B_i y + C = 0$, then the sum of the distances from point $P=(p,q)$ to these lines is given by
$$d = \pm_1\;\frac{A_1 p + B_1 q + C_1}{\sqrt{A_1^2+B_1^2}} \;\pm_2\;\frac{A_2p+B_2 q+C_2}{\sqrt{A_2^2+B_2^2}} \;\pm_3\;\frac{A_3p + B_3 q + C_3}{\sqrt{A_3^2+B_3^2}} \qquad (\star)$$ where each "$\pm_i$" is "$+$" for all points on one side of the $i$-th line, and "$-$" for all points on the other side.
Note that all $P$ in the interior of the triangle lie on a particular side of each edge-line; each $\pm_i$ remains fixed. Therefore, $(\star)$ represents a linear equation for the interior points with a particular total distance $d$ from the edges of the triangle; that is to say: The solution points do, indeed, lie on a line. Congratulations on your perceptiveness!
(For each of the seven regions of the plane determined by the extended sides of the triangle, you get such a constant-sum-of-distances line.)
• :))) $3$ identical answers within 5 minutes :) – Alexey Burdin May 24 '15 at 1:10
• @Alexey: I would've beaten you if I didn't take so long editing. ;) I should probably just delete this answer. (Well ... hmmm ... someone just up-voted me. I guess there's perceived value in this near-duplicate, so I'll leave it. :) – Blue May 24 '15 at 1:14
• All 3 answers mention different details. Ours don't mention palallelity, and yours do mention lines in the rest 6 regions on the plane. My answer is most poor one :) For editing: I put the \$signs after all editing, otherwise it takes too slow indeed. – Alexey Burdin May 24 '15 at 1:19 Let the triangle vertices' Cartesian coordinates be$A(x_1,y_1),\ B(x_2,y_2),\ C(x_3,y_3)$. Let the line equations of$AB, BC, CA$be$A_3x+B_3y+C_3=0,\ A_2x+B_2y+C_2=0,\ A_1x+B_1y+C_1=0$respectively. We know that the distance from the line$Ax+By+C=0$to a point$(x,y)$is $$\frac{|Ax+By+C|}{\sqrt{A^2+B^2}}\hbox{.}$$ Inside the triangle, the signs of$A_ix+B_iy+C$do not change (wlog we assume these all three$\ge 0$), so the sum of distances is $$\sum\limits_{k=1}^3 \frac{A_kx+B_ky+C}{\sqrt{A_k^2+B_k^2}}$$ and it must equal to a$d\$: $$\sum\limits_{k=1}^3 \frac{A_kx+B_ky+C}{\sqrt{A_k^2+B_k^2}}=d\iff$$ $$x\sum\limits_{k=1}^3 \frac{A_k}{\sqrt{A_k^2+B_k^2}}+ y\sum\limits_{k=1}^3 \frac{B_k}{\sqrt{A_k^2+B_k^2}}+ \sum\limits_{k=1}^3 \frac{C}{\sqrt{A_k^2+B_k^2}}-d=0\hbox{,}$$ which is surely a line. | 2019-06-18T09:11:06 | {
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https://math.stackexchange.com/questions/787440/proving-this-binomial-identity | # Proving this binomial identity
I'm required to prove the following binomial identity:
$$\sum\limits_{k=0}^l {n \choose k} {m \choose l-k} = {n+m \choose l}$$
I tried various arrangements but reached nowhere. Finally I turned to the hint in the book, which says
Apply the binomial theorem to $(1+x)^n (1+x)^m$
And suddenly, it makes sense. All I now need to do is add the powers on the right-hand side and equate the coefficients of $x^l$. But I'm wondering how to write a proper proof. Will it be enough if I say:
$(1+x)^n (1+x)^m = (1+x)^{m+n}$
Applying the binomial theorem separately for the two terms on the LHS and collecting the coefficients of $x^l$ on both sides, we have:
$${n \choose 0} {m \choose l} + {n \choose 1} {m \choose l-1} + \ldots + {n \choose l}{m \choose 0} = {n+m \choose l}$$
Is this enough? I don't know why but it looks rather shallow to me.
• Your proof is correct. I am not sure what is bothering you. If you want to prove that necessarily the coefficients of $x^l$ are equal, you can note that both the LHS and RHS are polynomials of the variable $x$, therefore they coincide on $\mathbb{R}$ if and only if they have the same coefficients. – Ian May 9 '14 at 4:37
• @Ian: Thanks a lot! You know how it is ... when a proof is so short and "obvious" you start wondering if it's correct. ;) – dotslash May 9 '14 at 4:38
• yes it is right.One should always be able to do such easy sums.I believe. @ dotslash – soumajit das May 9 '14 at 4:52
• There are several posts about this identity: See this question and other posts, which are linked there. – Martin Sleziak May 10 '14 at 16:23
Other proof is counting in two ways: what the number of choose $l$ balls in $m+n$? (the balls are enumerate -- 1,2,...,m+n)
for one side: $\binom{m+n}{l}$
for other side: divided the balls into 2 groups, group 1 with $n$ balls and group 2 with $m$ balls. If we choose $k$ balls in group 1, we should choose $l-k$ balls in group 2. So, we have $\sum\limits_{k=0}^l {n \choose k} {m \choose l-k} = {n+m \choose l}$ | 2019-08-25T05:13:51 | {
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http://mathhelpforum.com/trigonometry/27937-solve-cos-2-cos-2-2-0-a.html | # Math Help - Solve cos^2(...)+cos^2(...)-2 = 0 ?
1. ## Solve cos^2(...)+cos^2(...)-2 = 0 ?
anyone have any clever ideas on how to solve this for x?
$cos^2(\pi*16^x)+cos^2(\pi*16^{1-x})-2 = 0$
?
$(cos(\pi*16^x)-1)(cos(\pi*16^x)+1)=-(cos(\pi*16^{1-x})-1)(cos(\pi*16^{1-x})+1)$
?
2. Originally Posted by pinion
anyone have any clever ideas on how to solve this for x?
$cos^2(\pi*16^x)+cos^2(\pi*16^{1-x})-2 = 0$
?
$(cos(\pi*16^x)-1)(cos(\pi*16^x)+1)=-(cos(\pi*16^{1-x})-1)(cos(\pi*16^{1-x})+1)$
?
Since $|\cos(x)| \le 1$ this equation implies that:
$\cos^2(\pi*16^x)=1$
and
$\cos^2(\pi*16^{1-x})=1$
RonL
3. Originally Posted by CaptainBlack
Since $|\cos(x)| \le 1$ this equation implies that:
$\cos^2(\pi*16^x)=1$
and
$\cos^2(\pi*16^{1-x})=1$
RonL
huh?
4. Originally Posted by pinion
huh?
Your equation asserts that the sum of two positive quataties that are both
less than or equal to 1 is 2, so both must be exactly 1.
RonL
PS huh? is uninformative, please explain yourself more clearly in future
5. Originally Posted by CaptainBlack
Since $|\cos(x)| \le 1$ this equation implies that:
$\cos^2(\pi*16^x)=1$
and
$\cos^2(\pi*16^{1-x})=1$
RonL
If you require further help in finding the solutions let us know, but the problem is trivial from here.
RonL
6. Thanks for being so nice about me not explaining in the manner you require, my bad. But as to the 'huh', I was firstly confused at how your statement followed from the equation, and secondly confused at why I would/should observe that fact to solve the equation. Is that the only way? Or do you suggest that to be the easiest way? I don't know. But I do know this follows from that observation -
$cos(\pi*16^x) = 1$
and
$cos(\pi*16^x) = -1$
and
$cos(\pi*16^{1-x}) = 1$
and
$cos(\pi*16^{1-x}) = -1$
and since the arccos of both 1 and -1 is $c*\pi$ (or more specifically $c_1*\pi$ and $c_2*\pi$ where $c_1$ is even and $c_2$ is odd) I would now have to find $c_1,c_2,c_3,c_4$ - so unless there is an algebraic method for doing that, splitting the equation in two doesn't help me. Is there anything else you can recommend?
7. Originally Posted by pinion
Thanks for being so nice about me not explaining in the manner you require, my bad. But as to the 'huh', I was firstly confused at how your statement followed from the equation, and secondly confused at why I would/should observe that fact to solve the equation. Is that the only way? Or do you suggest that to be the easiest way? I don't know. But I do know this follows from that observation -
$cos(\pi*16^x) = 1$
and
$cos(\pi*16^x) = -1$
and
$cos(\pi*16^{1-x}) = 1$
and
$cos(\pi*16^{1-x}) = -1$
and since the arccos of both 1 and -1 is $c*\pi$ (or more specifically $c_1*\pi$ and $c_2*\pi$ where $c_1$ is even and $c_2$ is odd) I would now have to find $c_1,c_2,c_3,c_4$ - so unless there is an algebraic method for doing that, splitting the equation in two doesn't help me. Is there anything else you can recommend?
That:
$
\cos^2(\pi*16^x)=1
$
implies that $\pi 16^x=k \pi$ for some integer $k$, which then implies that $16^x=2^{4x}=k$ but this implies that:
$4x=\log_2(k)/2$
or:
$x=\log_2(k)/8$.
Similarly (using the other equation): we find that for some integer n:
$1-x=\log_2(n)/8$.
So:
$\log_2(k)/8=1-\log_2(n)/8$
or:
$\log_2(k/n)=8$
so:
$k/n=2^8.$
so every integer $n$ determines a solution $x=1-\log_2(n)/8$.
(don't rely on the algebra above being right as I cannot check it in detail at present, either I or another helper will correct any mistakes when we get the chance).
RonL
8. Originally Posted by pinion
Thanks for being so nice about me not explaining in the manner you require, my bad. But as to the 'huh', I was firstly confused at how your statement followed from the equation, and secondly confused at why I would/should observe that fact to solve the equation. Is that the only way? Or do you suggest that to be the easiest way? I don't know.
The reason why the obsevation is worth making is that it is useful information
about the nature of the solutions that is otherwise not obvious.
Other approaches are likely to lead you through some horendous trig transformations that are unlikely to go anywhere.
RonL
9. thanks, i'll look into trig transformations. | 2016-07-30T11:00:12 | {
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https://math.stackexchange.com/questions/853281/how-to-find-a-cube-root-of-numbers/853430 | # How to find a cube root of numbers?
While, I was solving a problem of Chemistry [Solid State] when I encountered an equation like : $$a^3 = 3.612 \times 10^{-23}$$
Where, a is just a quantity [Actually, it is the length of a cubic unit cell.]
I was required to find the cube-root of $a^3$ as I had to use $a$ to find another quantity which was actually my answer.
I know that the cube-root of $a^3$ here can be found by using logarithm.
\begin{align} & \log{a^3} = \log{(3.612 \times 10^{-23})} \\ & 3\log{a} = \log{(36.12 \times 10^{-24})} \\ & 3\log{a} = \log{(36.12)} + \log{(10^{-24})} \\ & 3\log{a} = \log{(36.12)} - 24 \\ & \log{a} = \cfrac{\log{(36.12)} - 24}{3} \end{align}
By using logarithm table, I found $\log{(36.12)} = 1.557$
Now, just putting this : \begin{align} & \log{a} = \cfrac{1.557 - 24}{3} \\ & \\ & \log{a} = \cfrac{-22.443}{3} \end{align}
Now, taking antilog both sides, this becomes : $$a = antilog(-7.481)$$
Now, by using anti-log table, I got this as $$\boxed{a = 3.303 \times 10^{-8} }$$
Now, as you all might have noticed that this is a lot longer method and this has larger probability of a student doing a mistake in the calculations etc.
Is there a shorter or any proper method to find cube-root of a number like stated above?
• To be honest, I really don't know that which tag will be the best for this question. If you have any ideas for proper tag for this, please feel free to add the tag by editing the post. Thanks! – Kushashwa Ravi Shrimali Jul 1 '14 at 14:18
• From $a^3 = 36.12\cdot 10^{-24}$, you can go directly to $a = \sqrt[3]{36.12}\cdot 10^{-8}$, and getting the cube root of $36.12$ gives you a little less opportunity of screwing up. – Daniel Fischer Jul 1 '14 at 14:22
• See p. 413 of Gilbert A. Christian and George Collar's 1899 book A New Arithmetic, Theoretical and Practical, for example. – Dave L. Renfro Jul 1 '14 at 14:36
• Out of curiosity, I looked at the reference (books.google.com/books?id=z_gSAQAAMAAJ&pg=PA413) in the prior comment, and I think the logarithm method seems faster and less error-prone; certainly it is more direct. – David K Jul 1 '14 at 15:55
• And a little humor on the same topic, from A Space Child's Mother Goose: "Little Jack Horner / Sits in a corner / Extracting cube roots to infinity ..." improbable.com/airchives/paperair/volume3/v3i5/mg35.htm – David K Jul 1 '14 at 15:59
Given a log/antilog table and no calculator, your approach is probably optimal. Without such a table, I think you are better off proceeding as:
$$\left ( 3.612 \cdot 10^{-23} \right )^{1/3} \\ = \left ( 36.12 \cdot 10^{-24} \right )^{1/3} \\ = 36.12^{1/3} \cdot 10^{-8}$$
Then do bisection to find $36.12^{1/3}$. Note that $3^3=27<36.12$ and $4^3=64>36.12$. (If you can't find a small interval quickly, just use $[1,10]$.) So now you keep checking midpoints until you get your desired level of precision: $3.5^3 > 36.12$, $3.25^3<36.12$, $3.375^3>36.12$, etc. If at some point the arithmetic gets to be too messy, you can round, at the expense of some convergence speed. For example, you could check $3.4$ instead of $3.375$, then $3.4^3>36.12$, so you could check $3.3$. Then $3.3^3<36.12$, so you have $a=(3.35 \pm 0.05) \cdot 10^{-8}$.
You can also do Newton's method, which in this case is $x_{k+1} = \frac{2 x_k}{3} + \frac{c}{3 x_k^2}$ where you want $c^{1/3}$, but this is hard to do by hand, because the fractions quickly become very awkward, even with rounding.
• With bisection, the final result is an interval $[a,b]$ where you know the root is located. Given this interval, the best single guess for the root is $(a+b)/2$, which is within $(b-a)/2$ of every point in $[a,b]$. So in my example, my final interval was $[3.3,3.4]$. That is, I know, from my procedure, that $36.12^{1/3}$ is between $3.3$ and $3.4$. Therefore the root is within $0.05$ of $3.35$. – Ian Jul 2 '14 at 0:28
I think your method is probably the best way to solve for the cube root of a number.
I found some sites that offer quick solution towards finding a cube root, but they assume that your answer will be expressed as a whole number.
• @KushashwaRaviShrimali I'm actually very disappointed as well that there is no algorithm to find cube roots. Some sites only offer methods that only yield integer answers. I'll try and search online; I'm very intrigued to see how the algorithm will work. – Varun Iyer Jul 1 '14 at 14:37
• @Varun Iyer: For an algorithm to find cube roots numerically, see my comments to the OP's question. Such algorithms are fairly common in really old textbooks, and I gave a link for one such textbook. – Dave L. Renfro Jul 1 '14 at 14:44
## protected by Community♦May 11 '17 at 9:29
Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count). | 2019-09-16T08:56:08 | {
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https://mathematica.stackexchange.com/questions/135358/area-of-surface-of-revolution | # Area of surface of revolution
I am asked to rotate the curve $y=\sqrt{4-x^2}$ from $x=-1$ to $x=1$ about the x-axis and find the area of the surface. I was able to use RevolutionPlot3D to show the surface.
RevolutionPlot3D[Sqrt[4 - x^2], {x, -1, 1},
RevolutionAxis -> {1, 0, 0}]
I used calculus to find the surface area:
Integrate[2 π Sqrt[4 - x^2] Sqrt[1 + (-x/Sqrt[4 - x^2])^2], {x, -1, 1}]
Which produces the answer $8\pi$.
Here is my question. Is there some cute way of finding surface area using Mathematica; that is, something like using the Area and Volume commands, or some other commands?
f[x] == Sqrt[4 - x^2] is the distance at height x from the origin (i.e., from {0, 0} at height x) to the surface; hence, one can construct
reg = ImplicitRegion[z^2 + y^2 == Sqrt[4 - x^2]^2 && -1 <= x <= 1, {x, y, z}]
which looks like this:
DiscretizeRegion[reg]
and directly compute
Area[reg]
$8\pi$
Numerically:
Area @ DiscretizeRegion @ reg / Pi
7.99449
in very good agreement.
In general this can be applied to any revolution surface, as due to its rotational symmetry it will always be given by an equation of the form z^2 + y^2 == f[x] (given the revolution is around the x axis).
EDIT:
To get the volume of such a barrel, consider reg2, different from reg only in that == is replaced with <=:
reg2 = ImplicitRegion[z^2 + y^2 <= Sqrt[4 - x^2]^2 && -1 <= x <= 1, {x, y, z}]
Then
Volume[reg2]
$\frac{22 \pi }{3}$
• Alternatively, using Area[{Sqrt[4-z^2],\[Theta],z},{z,-1,1},{\[Theta],0, 2Pi}, "Cylindrical"] evaluates much more quickly. – Carl Woll Jan 14 '17 at 1:50
• @corey979 Thanks for this demonstration. I will find it quite useful teaching class. – David Jan 14 '17 at 21:04
The trick is DiscretizeGraphics. Turns your graphic into a surface:
r = DiscretizeGraphics@
RevolutionPlot3D[Sqrt[4 - x^2], {x, -1, 1},
RevolutionAxis -> {1, 0, 0}]
then:
In[24]:= Area@r
Out[24]= 25.5411
It ain't perfect, but it's close:
In[26]:= Area@r / \[Pi]
Out[26]= 8.12997
I use this to compute Van der Waals volumes of molecules. Note that Volume only works on closed surfaces though, but there's an answer on here (can't find it right now) that provides a way to do it with the MeshCoordinates.
### Update
Here we are. That gives you the volume.
The most direct analog, IMO, to your plot is to use the parametric form of Area, where you add a theta variable for the rotation:
In[11]:= Area[{x, Sqrt[4 - x^2] Cos[θ], Sqrt[4 - x^2] Sin[θ]},
{x, -1, 1}, {θ, 0, 2 π}]
Out[11]= 8 π
Adding a radius variable which gives the distance from the x-axis gives you the volume (this is x-centered cylindrical coordinates):
In[12]:= Volume[{x, r Cos[θ], r Sin[θ]},
{x, -1, 1}, {θ, 0, 2 π}, {r, 0, Sqrt[4 - x^2]}]
Out[12]= (22 π)/3
• I just wanted to thank you for this suggestion. I used this when teaching my calculus class tonight to teach getting the volume of a surface of revolution and to do a parametric plot. Absolutely what my students needed was an introduction to parametrization. – David Jan 25 '17 at 3:39
Is there some cute way of finding surface area using Mathematica?
I think this fits the bill,
WolframAlpha["rotate sqrt(4-x^2) from x=-1 to x=1 about x axis surface area",
"Result"]
(* 25.1327 *)
Parametrize surface:
f[u_, v_] := {u, Cos[v] Sqrt[4 - u^2], Sin[v] Sqrt[4 - u^2]};
Area element:
i =
FullSimplify[Norm[Cross @@ Transpose[D[f[x, y], {{x, y}}]]],
Assumptions -> {x \[Element] Reals, Abs[x] < 1, 0 < y < 2 Pi}]
gives:
(*2*)
Calculate surface area:
Integrate[i, {u, -1, 1}, {v, 0, 2 Pi}]
yields 8$\pi$
or by considering the region of interest as a subset of a sphere of radius 2 (and orienting so "x-axis" is "z-axis", the desired surface area is sphere-2 * cap, where cap and sphere are the surface areas as suggested by the names:
cap = Integrate[4 Sin[u], {u, 0 , ArcCos[1/2]}, {v, 0, 2 Pi}]
sphere = 16 Pi
region = sphere - 2 cap
fm[u_, v_] := {Cos[v] Sqrt[4 - u^2], Sin[v] Sqrt[4 - u^2], u};
Show[ParametricPlot3D[fm[u, v], {u, -1, 1}, {v, 0, 2 Pi},
Mesh -> None],
SphericalPlot3D[2, {u, 0, Pi}, {v, 0, 2 Pi},
PlotStyle -> Opacity[0.5], Mesh -> False], PlotRange -> All,
BoxRatios -> 1, Boxed -> False, Axes -> False, Background -> Black]
• Another very helpful answer. I want to share this with our Calc III teacher this semester. – David Jan 14 '17 at 21:08
• @David thank you for the positive feedback. Happy New Year:) – ubpdqn Jan 15 '17 at 5:46 | 2020-12-03T23:16:04 | {
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https://math.stackexchange.com/questions/2104523/known-that-a-b-c-0-and-a3-b3-c3-27-find-abc | # Known that $a + b + c = 0$ and $a^3 + b^3 + c^3 = 27$. Find ABC!
Known that $$a + b + c = 0$$ $$a^3 + b^3 + c^3 = 27$$ What is the value of $abc?$
A.) 1
B.) 0
C.) 7
D.) 8
E.) 10
My Work: $$a + b = -c$$ $$a + c = -b$$ $$b + c = -a$$ Then $$(a + b + c)(a + b + c)(a + b + c) = 0$$ Expanded into: $$a^3 + b^3 + c^3 + 3a^2b+3b^2c+3a^2c+3ab^2 + 3bc^2+ 3ac^2+6abc = 0$$ Putting the Values: $$27+3a^2(-a)+3b^2(-b)+3c^2(-c)+6abc = 0$$ $$27 -3(a^3 + b^3 + c^3) = -6abc$$ $$27-3(27) = -6abc$$ $$-54=-6abc$$ $$abc = 9$$
$9$ wasn't an option in the question. Am I missing something?
• No, the question is wrong. – S.C.B. Jan 19 '17 at 14:03
• @S.C.B. Well, that's the end I guess... – Adola Jan 19 '17 at 14:04
• yes $$abc=9$$ is right – Dr. Sonnhard Graubner Jan 19 '17 at 19:47
No, the question is wrong, and your answer is right. However, it might be easier by recalling the identity $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$ From $a+b+c=0$, we have $abc=9$.
$(a+b+c)^3=(a^3+b^3+c^3)+3a(b^2+c^2)+3b(a^2+c^2)+3c(a^2+b^2)+6abc$
$\implies 0=27+3ab^2+3ac^2+3ba^2+3bc^2+3ca^2+3cb^2+6abc$
$\implies 0 = 27+3b^2(a+c)+3a^2(b+c)+3c^2(a+b)+6abc$
$\implies 0 = 27+3b^2(-b)+3a^2(-a)+3c^2(-c)+6abc$
$\implies 0 = 27 - 3(a^3+b^3+c^3) +6abc$
$\implies 0 = 27 - 81 + 6abc$
$\implies 54 = 6abc$
$\implies abc = 9$
if $a+b+c=0$
then ,
$a^3+b^3+c^3=3abc$ and $a^3+b^3+c^3=27$
equating both equations
$3abc=27$
therefore $abc=9$ | 2019-08-20T18:22:37 | {
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https://math.hecker.org/2011/08/08/linear-algebra-and-its-applications-review-exercise-1-27/ | ## Linear Algebra and Its Applications, Review Exercise 1.27
Review exercise 1.27. State whether the following are true or false. If true explain why, and if false provide a counterexample.
(1) If a matrix $A$ can be factored as $A = L_1U_1 = L_2U_2$ where $L_1$ and $L_2$ are lower triangular with unit diagonals and $U_1$ and $U_2$ are upper triangular, then $L_1 = L_2$ and $U_1 = U_2$. (In other words, the factorization $A = LU$ is unique.)
(2) If for a matrix $A$ we have $A^2 + A = I$ then $A$ is invertible and $A^{-1} = A + I$.
(3) If all the diagonal entries of a matrix $A$ are zero then $A$ is singular.
Answer: (1) True. Since $L_1$ and $L_2$ are lower triangular with unit diagonal both matrices are invertible, and their inverses are also lower triangular with unit diagonal. (See the proof of this below at the end of the post.) Similarly since $U_1$ and $U_2$ are upper triangular with nonzero diagonal those matrices are invertible as well, and their inverses are also upper triangular with nonzero diagonal.
We then start with the equation $L_1U_1 = L_2U_2$ and multiply both sides by $L_2^{-1}$ on the left and by $U_1^{-1}$ on the right:
$L_2^{-1}(L_1U_1)U_1^{-1} = L_2^{-1}(L_2U_2)U_1^{-1}$
This equation reduces to $L_2^{-1}L_1 = U_2U_1^{-1} = B$ where $B$ is the product matrix. Now since $B$ is the product of two lower triangular matrices with unit diagonal (i.e., $L_2^{-1}$ and $L_1$), it itself is a lower triangular matrix with unit diagonal. Since $B$ is the product of two upper triangular matrices (i.e., $U_2$ and $U_1^{-1}$), it is also an upper triangular matrix. Since $B$ is both lower triangular and upper triangular it must be a diagonal matrix, and since its diagonal entries are all 1 we must have $B = I$.
Since $L_2^{-1}L_1 = I$ we can multiply both sides on the left by $L_2$ to obtain $L_2L_2^{-1}L_1 = L_2I$ or $L_1 = L_2$. Similarly since $U_2U_1^{-1} = I$ we can multiply both sides on the right by $U_1$ to obtain $U_2U_1^{-1}U_1 = IU_1$ or $U_2 = U_1$. So the factorization $A = LU$ is unique.
(2) True. Assume $A^2 + A = I$. We have $I = A^2 + A = A(A + I)$ so that $A + I$ is a right inverse for $A$. We also have $I = A^2 + A = (A + I)A$ so that $A + I$ is a left inverse for $A$. Since $A+I$ is both a left and right inverse of $A$ we know that $A$ is invertible and that $A^{-1} = A + I$.
(3) False. The matrix
$A = \begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix}$
has zeros on the diagonal but is nonsingular. In fact we have
$A^{-1} = \begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix} = A$
Proof of the result used in (1) above: Assume $L$ is a lower triangular matrix with unit diagonal. That $L$ is invertible can be seen from Gauss-Jordan elimination (here shown in a 4 by 4 example, although the argument generalizes to all $n \ge 2$):
$\begin{bmatrix} 1&0&0&0&\vline&1&0&0&0 \\ x&1&0&0&\vline&0&1&0&0 \\ x&x&1&0&\vline&0&0&1&0 \\ x&x&x&1&\vline&0&0&0&1 \end{bmatrix}$
Note that in forward elimination each of the diagonal entries of $L$ will remain as is, since each of these entries has only zeros above it; each of the zero entries above the diagonal will remain unchanged as well, for the same reason. When forward elimination completes the left hand matrix will be the identity matrix since it will have zeros below the diagonal (from forward elimination), ones on the diagonal (as noted above), and zeroes above the diagonal (also as noted above). Gauss-Jordan elimination is thus guaranteed to complete successfully, so $L$ is invertible.
The right-hand matrix at the end of Gauss-Jordan elimination will be the inverse of $L$. That matrix was produced from the identity matrix by forward elimination only, since backward elimination was not necessary. For the same reason noted above for the left-hand matrix, forward elimination will preserve the unit diagonal in the right-hand matrix and the zeros above it, with the only possible non-zero entries occurring below the unit diagonal. We thus see that if $L$ is a lower-triangular matrix with unit diagonal then it is invertible and its inverse $L^{-1}$ is also a lower-triangular matrix with unit diagonal.
Assume $U$ is an upper triangular matrix with nonzero diagonal. That $U$ is invertible can be seen from Gauss-Jordan elimination (again shown in a 4 by 4 example):
$\begin{bmatrix} x&x&x&x&\vline&1&0&0&0 \\ 0&x&x&x&\vline&0&1&0&0 \\ 0&0&x&x&\vline&0&0&1&0 \\ 0&0&0&x&\vline&0&0&0&1 \end{bmatrix}$
Note that forward elimination is not necessary: There are nonzero entries on the diagonal and zeros below them, so we have pivots in every column; therefore the left-hand matrix $U$ is nonsingular and is guaranteed to have an inverse.
We can find that inverse by doing backward elimination to eliminate the entries above the diagonal in the left-hand matrix, and then dividing by the pivots. Note that since backward elimination starts with all zero entries below the diagonal in the right-hand matrix, it will not produce any nonzero entries below the diagonal in that matrix. Also, the diagonal entries in the right-hand matrix are not affected by backward elimination, for the same reason. After backward elimination completes the diagonal entries in the right-hand matrix will still be ones, and any nonzero entries produced will be above the diagonal. Dividing by the pivots in the left-hand matrix will then produce nonzero entries in the diagonal of the right-hand matrix.
The final right-hand matrix after completion of Gauss-Jordan elimination will therefore be an upper triangular matrix with nonzero diagonal entries. We thus see that if $U$ is an upper-triangular matrix with nonzero diagonal then it is invertible and its inverse $U^{-1}$ is also an upper-triangular matrix with nonzero diagonal.
NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.
If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.
This entry was posted in linear algebra. Bookmark the permalink. | 2019-11-19T13:41:48 | {
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http://mathschallenge.net/full/hops_and_slides_but_never_square | ## Hops And Slides But Never Square
#### Problem
A horizontal row comprising of $2n + 1$ squares has $n$ red counters placed at one end and $n$ blue counters at the other end, being separated by a single empty square in the centre. For example, when $n = 3$.
A counter can move from one square to the next (slide) or can jump over another counter (hop) as long as the square next to that counter is unoccupied.
Let $M(n)$ represent the minimum number of moves/actions to completely reverse the positions of the coloured counters; that is, move all the red counters to the right and all the blue counters to the left.
Prove that $M(n)$ can never be a perfect square.
#### Solution
When presented with problems like this it is always helpful to spend a short time "playing" with counters. This should reveal that a minimal approach would only involve "forward" actions; that is, red counters only moving to the right and blue counters only moving to the left. In order to move all the red counters to the right and the blue counters to the left it is clear that each of the red counters must hop over each of the blue counters (or vice versa). With the exception of the starting and finishing positions, two counters of the same colour should never appear next to each other, otherwise one of the counters would need to move "backwards" to make space for a counter of the other colour to hop over and this would undermine an optimal strategy.
Experimentally the following results can be obtained.
$n$M($n$)
13
28
315
424
535
The results seem to suggest that the formula, $M(n) = n(n + 2) = n^2 + 2n$.
We shall prove this result by considering the total distance travelled by all of the counters. Note that a slide involves travelling a distance of one square whereas a hop accounts for a distance of two squares.
Once a complete reversal of positions has taken place each of the red counters will have moved a total distance of $n+1$ squares to the right and each blue counter will have moved $n + 1$ squares to the left. So the minimum distance that all of the counters travelled will be a distance of $2n(n+1)=2n^2+2n$ squares.
As explained above, during this process each counter of one colour must hop over every counter of the other colour once and only once. Therefore a total of $n \times n = n^2$ hops (actions) will take place, but as each hop moves a distance of two squares, the hops will account for $2n^2$ squares of the total distance travelled.
Therefore the slides will account for $2n^2 + 2n - 2n^2 = 2n$ squares of the total distance travelled, which also represents the total number of slides (actions).
Hence the total number of actions (made up of hops and slides) will be $n^2 + 2n$.
Therefore $M(n) = n^2 + 2n = (n^2 + 2n + 1) - 1 = (n + 1)^2 - 1$.
But as $n \ge 1$ and $M(n)$ is one less than a perfect square we prove that it can never be square itself.
With the exception of $M(2) = 8$, is it possible for $M(n)$ to be cube?
Given that there are $x$ red counters and $y$ blue counters find a formula for $M(x, y)$.
Problem ID: 372 (07 Aug 2010) Difficulty: 3 Star
Only Show Problem | 2016-06-29T18:06:03 | {
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https://math.stackexchange.com/questions/1375433/normalizing-a-basis/1375467 | # Normalizing a basis
Let the basis $B = \{1,x,x^2\}$ which is orthogonal.
Now, I've seen the following:
$$\|1\| = \sqrt {\langle 1,1\rangle} = \sqrt {4\cdot 1\cdot 1} = 2$$ $$\|x\| = \sqrt {\langle x,x\rangle} = \sqrt {2\cdot 1\cdot 1} = \sqrt 2$$ $$\|x^2\| = \sqrt {\langle x^2,x^2\rangle} = \sqrt {1\cdot 1} = \sqrt 1 = 1$$
And therefore, an orthonormal basis would be: $$B = \{ \frac{1}{2}, \frac{x}{\sqrt 2}, x^2 \}$$
Questions:
1. Isn't $\{1,x,x^2\}$ already orthonormal?
2. Isn't the calculation of the norm wrong?
EDIT:
The inner product (for $V=\mathbb{R}_2[x]$) is:
$$\langle a_1 + b_1x + c_1x^2, a_2 + b_2x + c_2x^2\rangle = 4a_1a_2 + 2b_1b_2 + c_1c_2$$
• How is the inner product defined? – ajotatxe Jul 27 '15 at 11:06
• Let me edit please – jmiller Jul 27 '15 at 11:14
• An orthonormal base means all its elements have norm $1$ and are pairwiseorthogonal. – 5xum Jul 27 '15 at 11:18
• @jmiller are you sure you've written the definition of inner product right? I think it should be $\langle a_1 + b_1x + c_1x^2, a_2 + b_2x + c_2x^2\rangle = 4a_1a_2 + 2b_1b_2 + c_1c_2$ – Sepideh Abadpour Jul 27 '15 at 11:26
• @5xum, for example, why is it:$$\|1\| = \sqrt {\langle 1,1\rangle} = \sqrt {4\cdot 1\cdot 1} = 2$$. Where's the $4$ came from? it's unclear to me. – jmiller Jul 27 '15 at 11:29
the definition $\langle a_1 + b_1x + c_1x^2, a_2 + b_2x + c_2x^2\rangle = 4a_1a_2 + 2b_1b_2 + c_1c_2$ is just a definition provided in the problem.For each inner product space we should define the inner product and then define normality and orthogonality based on the definition of the inner product.
So don't say where 2 or 4 comes from. That's just a definition.
For example the common definition of inner product in 3-D vector space is: $\langle (x_1,y_1,z_1),(x_2,y_2,z_2)\rangle=x_1x_2+y_1y_2+z_1z_2$
but I can encounter with a 3-D vector space that has another definition for inner product:
• Oh I see now.. Thank you! – jmiller Jul 27 '15 at 11:49
• An inner product is required to be positive-definite, but $\langle (x_1,y_1,z_1),(x_2,y_2,z_2)\rangle=x_1y_2+y_1x_2+z_1z_2$ isn't. For example $\langle (-1,1,0), (-1,1,0) \rangle = -2 < 0$. – A.P. Jul 27 '15 at 12:17
• ok @A.P. it was just an example and I only wanted to state that the inner product should be defined in the problem. and it's not always the same as the common definition you can edit my answer and add positive-definite example – Sepideh Abadpour Jul 27 '15 at 12:20
• $\langle (x_1,y_1,z_1),(x_2,y_2,z_2)\rangle=|x_1y_2+y_1x_2+z_1z_2|$ still isn't positive definite because $\langle (1,0,0), (1,0,0) \rangle = 0$ even though $(1,0,0) \neq (0,0,0)$. I added to my answer a characterisation for inner products on a finite-dimensional real vector spaces. You can use it to make up as many examples as you wish... :P – A.P. Jul 27 '15 at 14:54
The computation of the norm is indeed correct, given the inner product you described.
The vectors in $\{1,x,x^2\}$ are easily seen to be orthogonal, but they cannot form an orthonormal basis because they don't have norm $1$. On the other hand, the vectors in $$\left\{ \frac{1}{\|1\|}, \frac{x}{\|x\|}, \frac{x^2}{\|x^2\|} \right\} = \left\{ \frac{1}{2}, \frac{x}{\sqrt{2}}, x^2 \right\}$$ have norm $1$ and are orthogonal, so indeed they form an orthonormal basis.
Recall that an inner product on a real vector space $V$ is just a function $$\langle \cdot, \cdot \rangle \colon V \times V \to \Bbb{R}$$ which for every $x,y,z \in V$ and $a,b \in \Bbb{R}$ satisfies:
• $\langle x,y \rangle = \langle y,x \rangle$
• $\langle ax + by, z \rangle = a\langle x,z \rangle + b\langle y,z \rangle$
• $\langle x,x \rangle \geq 0$ and $\langle x,x \rangle = 0$ iff $x = 0$
Given an inner product on $V$, one can define the associated norm $\| x \| = \sqrt{\langle x,x \rangle}$.
Furthermore, one can prove that if $\dim_{\Bbb{R}} V = n$ is finite, then every inner product on $V$ is of the form $$\langle x,y \rangle_A = x^T A y$$ where $A$ is the symmetric positive-definite $n \times n$ matrix with entries $a_{ij} = \langle e_i,e_j \rangle_A$.
For example the Euclidean (i.e. usual) inner product corresponds to the identity matrix, while the matrix $$A = \begin{bmatrix} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end{bmatrix}$$ corresponds to the inner product $$\langle x,y \rangle_A = 2x_1y_1 - x_2y_1 - x_1y_2 + 2x_2y_2 - x_3y_2 - x_2y_3 + 2x_3y_3$$ (checking that this is indeed an inner product is a good exercise).
• Fun fact: for any real invertible matrix $A$ the matrix $A^TA$ is positive-definite. – A.P. Jul 27 '15 at 14:56 | 2020-09-20T22:12:26 | {
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https://mathematica.stackexchange.com/questions/187774/is-it-possible-to-add-two-plots-together | # Is it possible to add two plots together?
I have two functions (of energy) that are imaginary below different energies. These expressions are quite long and unfortunately, when plotting Re@func, I am left with some trash at low energies. What I want is to add the real parts of these equations and plot them (only add parts with energy greater than the imaginary limit to get rid of the trash). One way I thought of doing it was to plot both functions for only energies greater than imaginary limit 1, and imaginary limit 2. Then add those plots together (not like Show[func1, func2], but actually adding the values of both the functions at a given value together, and then plot the resulting line). Is there a function that does that?
I can illustrate by an example: Lets say we have y1 = x + 2, and we only want this function for x greater than 4. Then we have y2 = 2x + 1, and we want that for x greater than 1. Can this be plotted in a neat way or will I have to make two plots (y2 from x = 1 to 4, and y1+y2 from x = 4) and then combine them using Show?
And another question: If I use Show as suggested above, how can I remove the frame that will will appear at x = 4, but at the same time keep the frame around the entire graph?
• Perhaps define the function to be plotted using Piecewise? What do you mean trash at lower energies though? Can you share the expressions? – MarcoB Dec 12 '18 at 13:35
Plot[Piecewise[{{2 x + 1, x < 4 }, {x + 2, x > 4}}], {x, 1, 6}]
Since Plot doesn't display imaginary values, you can simply plot all functions over the whole range:
f1[z_] := 2 Sqrt[z - 3]
f2[z_] := 3 Sqrt[z - 1]
Plot[{f1[z], f2[z], f1[z] + f2[z]}, {z, 0, 4}]
If you want to explicitly prevent the other parts from being displayed, the following might work:
clean[x_] /; Re[x] > 1 := x
clean[x_] := Indeterminate
Plot[
{
clean[f1[z]],
clean[f2[z]],
clean[f1[z]] + clean[f2[z]]
},
{z, 0, 4},
PlotRange -> {0, 8}
]
This (ab)uses the fact that Plot also doesn't show Indeterminate values, and also the fact that Indeterminate+1 is again Indeterminate, so as soon as one function is not defined, the result will not be plotted. Note that I've used the condition Re[x]>1 instead of Im[x]==0 (which is what you probably want) since the plot would look the same in the case of my example otherwise.
• Think you are missing the definition of clean here. – KraZug Dec 12 '18 at 16:34
• @KraZug Thanks for pointing that out, should be fixed now – Lukas Lang Dec 12 '18 at 16:36
You can also use ConditionalExpression:
Plot[{ConditionalExpression[2 x + 1, x <= 4],
ConditionalExpression[2 + x, x >= 4],
2 x + 1 + Boole[x >= 4] (x + 2)},
{x, 1, 6},
PlotStyle -> {Thick, Thick,
Directive[{Opacity[.5], Dashing[{.03, .01}], CapForm["Round"], AbsoluteThickness[7]}]},
ImageSize -> Medium, PlotLegends -> "Expressions"] | 2020-04-09T05:09:40 | {
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http://math.stackexchange.com/questions/195904/integer-solutions-to-x2-y2-33 | # Integer solutions to $x^2-y^2=33$
I'm currently trying to solve a programming question that requires me to calculate all the integer solutions of the following equation:
$x^2-y^2 = 33$
I've been looking for a solution on the internet already but I couldn't find anything for this kind of equation. Is there any way to calculate and list the integer solutions to this equation?
-
Your title asks a different question than the body of your question; you might want to fix that. – Niel de Beaudrap Sep 14 '12 at 21:27
Sigh. In the title you say $x^2 - y^2,$ which has integer solutions. With a plus sign there are no integer solutions. – Will Jagy Sep 14 '12 at 21:27
Done, it's x^2-y^2, sorry for that ;) – Devos50 Sep 14 '12 at 21:28
$x^2 - y^2 = (x-y)(x+y).$ So both factors are chosen from $\pm 1, \pm 3, \pm 11, \pm 33.$ Then you solve for $x,y.$ – Will Jagy Sep 14 '12 at 21:30
Suppose that $x=y+n$; then $x^2-y^2=y^2+2ny+n^2-y^2=2ny+n^2=n(2y+n)$. Thus, $n$ and $2y+n$ must be complementary factors of $33$: $1$ and $33$, or $3$ and $11$. The first pair gives you $2y+1=33$, so $y=16$ and $x=y+1=17$. The second gives you $2y+3=11$, so $y=4$ and $x=y+3=7$. As a check, $17^2-16^2=289-256=33=49-16=7^2-4^2$.
If you want negative integer solutions as well, you have also the pairs $-1$ and $-33$, and $-3$ and $-11$.
-
We have $x^2-y^2=33$ iff $(x-y)(x+y)=33$. So to solve our equation we find all ordered pairs $(u,v)$ such that $uv=33$. Then we set $x-y=u$ and $x+y=v$, and solve.
We get $x=\dfrac{v+u}{2}$ and $y=\dfrac{v-u}{2}$. Since $u$ and $v$ will be both odd, $u+v$ and $v-u$ will be even, so $x$ and $y$ will be integers.
The possibilities for $u$ are $-33,-11,-3,-1,1,3,11,33$. The corresponding possibilities for $v$ are $-1,-3,-11,-33,33,11,3,1$. There are $8$ solutions, but only two "really different" ones.
For example, let $u=3$ and $v=11$. Then $x=\dfrac{11+3}{2}=7$ and $y=\dfrac{11-3}{2}=4$.
Generalization: If $n$ is of the form $4k+2$, then $n$ is not the difference of two squares. If $n$ is odd, the representations of $n$ as a difference of two squares are found exactly like the $n=33$ case discussed above. If $n$ is of the form $4k$, we do much the same thing, but find all pairs $(u,v)$ such that $uv=n$ and $u$ and $v$ are even.
-
Note that this is equivalent to solving $(x+n)^2-x^2=33$, and that $(x+n)^2-x^2=(2x+n)n$. Since $33=3\cdot 11$, we see that the solutions are $n=3,x=4$ and $n=11,x=-4$ with $n>0$ and both of thee times $-1$.
-
Thank you for your answer. Could you please explain why this is equal to solving (x+n)^2 - x^2? I don't get that part ;) – Devos50 Sep 14 '12 at 21:43
@Devos50 Let $y=x+n$. – Alex Becker Sep 14 '12 at 21:45
Hint $\rm\ \ \ \begin{eqnarray} 3\, &=&\, \color{#0A0}2^2-\color{#C00}1^2\\ 11\, &=&\, \color{blue}6^2-5^2\end{eqnarray}$ $\,\ \Rightarrow\,\$ $\begin{eqnarray} 3\cdot 11\, &=&\, (\color{#0A0}2\cdot\color{blue}6+\color{#C00}1\cdot 5)^2-(\color{#0A0}2\cdot 5+\color{#C00}1\cdot\color{blue}6)^2\, =\, 17^2 - 16^2\\ &=&\, (\color{#0A0}2\cdot\color{blue}6-\color{#C00}1\cdot 5)^2-(\color{#0A0}2\cdot 5-\color{#C00}1\cdot\color{blue}6)^2\, =\, \ 7^2\ -\ 4^2 \end{eqnarray}$
-
Oh, the colors! image – The Chaz 2.0 Sep 14 '12 at 23:56 | 2015-08-02T21:19:16 | {
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https://math.stackexchange.com/questions/154968/is-there-really-no-way-to-integrate-e-x2 | # Is there really no way to integrate $e^{-x^2}$?
Today in my calculus class, we encountered the function $e^{-x^2}$, and I was told that it was not integrable.
I was very surprised. Is there really no way to find the integral of $e^{-x^2}$? Graphing $e^{-x^2}$, it appears as though it should be.
$$\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}$$
This is from -infinity to infinity. If the function can be integrated within these bounds, I'm unsure why it can't be integrated with respect to $(a, b)$.
Is there really no way to find the integral of $e^{-x^2}$, or are the methods to finding it found in branches higher than second semester calculus?
• There is no antiderivative written in elementary functions (imagine the roots for a polynomial of degree, e.g., five, for which there is no formula). Jun 7 '12 at 5:11
• There is no elementary function whose derivative is $e^{-x^2}$. By elementary function we mean something obtained using arithmetical operations and composition from the standard functions we all know and love. But this is not a serious problem. A few important definite integrals involving $e^{-x^2}$ have pleasant closed form. Jun 7 '12 at 5:12
• Try reading this note of Brian Conrad's and the article by Rosenlicht referenced therein. Jun 7 '12 at 5:20
• Well, in someway it is no more surprising than stating that $\frac{1}{2}$ cannot be written as an integer. As noted by others, it is integrable, it is just that the collection of 'standard' functions is not rich enough to express the answer. Jun 7 '12 at 6:08
• Unfortunately there are three or four different meanings being given to the word "integrable" here: (1) $f(x)$ is Riemann integrable on intervals $[a,b]$ (yes, every continuous function is) (2) $f(x)$ has an antiderivative that is an elementary function (no, it doesn't: the antiderivative $\sqrt{\pi}\ \text{erf}(x)/2$ is not an elementary function) (3) $\int_{-\infty}^\infty |f(x)|\ dx < \infty$ (yes, and this is the usual meaning of "integrable" in analysis) (4) $\int_{-\infty}^\infty f(x)\ dx$ can be expressed in "closed form" (yes, it is $\sqrt{\pi}$). Jun 7 '12 at 6:54
That function is integrable. As a matter of fact, any continuous function (on a compact interval) is Riemann integrable (it doesn't even actually have to be continuous, but continuity is enough to guarantee integrability on a compact interval). The antiderivative of $e^{-x^2}$ (up to a constant factor) is called the error function, and can't be written in terms of the simple functions you know from calculus, but that is all.
• But the evaluation of the integral over the whole real line is relatively easy! Jun 7 '12 at 6:33
• Easy for Lord Kelvin. Jun 7 '12 at 6:41
• You probably mean that any continuous function is Riemann integrable on a compact interval. Mar 13 '13 at 7:25
• How to show that the function is non-elementary? I cannot remember seeing a proof of that.
– M.B.
Aug 10 '13 at 16:32
• @M.B.: see for example M.P. Wiener's text here. Aug 10 '13 at 16:44
To build on kee wen's answer and provide more readability, here is an analytic method of obtaining a definite integral for the Gaussian function over the entire real line:
Let $$I=\int_{-\infty}^\infty e^{-x^2} dx$$.
Then, \begin{align} I^2 &= \left(\int_{-\infty}^\infty e^{-x^2} dx\right) \times \left(\int_{-\infty}^{\infty} e^{-y^2}dy\right) \\ &=\int_{-\infty}^\infty\left(\int_{-\infty}^\infty e^{-(x^2+y^2)} dx\right)dy \\ \end{align}
Next we change to polar form: $$x^2+y^2=r^2$$, $$dx\,dy=dA=r\,d\theta\,dr$$. Therefore
\begin{align} I^2 &= \iint e^{-(r^2)}r\,d\theta\,dr \\ &=\int_0^{2\pi}\left(\int_0^\infty re^{-r^2}dr\right)d\theta \\ &=2\pi\int_0^\infty re^{-r^2}dr \end{align}
Next, let's change variables so that $$u=r^2$$, $$du=2r\,dr$$. Therefore, \begin{align} 2I^2 &=2\pi\int_{r=0}^\infty 2re^{-r^2}dr \\ &= 2\pi \int_{u=0}^\infty e^{-u} du \\ &= 2\pi \left(-e^{-\infty}+e^0\right) \\ &= 2\pi \left(-0+1\right) \\ &= 2\pi \end{align}
Therefore, $$I=\sqrt{\pi}$$.
Just bear in mind that this is simpler than obtaining a definite integral of the Gaussian over some interval (a,b), and we still cannot obtain an antiderivative for the Gaussian expressible in terms of elementary functions.
• If $e^{-x^2}$ is the area under the curve then $I^2$ should have units of $area^{2}$. But \begin{align} I^2 &= \iint e^{-(r^2)}r\,d\theta\,dr \\ \end{align} has units of volume. How is it possible? Jan 24 '20 at 16:00
• @user599310 The units of $I^2$ are indeed units of $area^2$, but so too are the units of $\iint e^{-(r^2)}r\,d\theta\,dr$ in $area^2$. First note that $dA = r\,d\theta\,dr = dx\,dy$, all of which are in units of $area$. Second note that $e^{-r^2}$ is equivalent to $e^{-x^2}e^{-y^2}$. In words, the base shift that we performed from $dx\,dy$ to $r\,d\theta\,dr$ changes us from measuring the function in terms of two lengths, to measuring it in terms of chunks of areas. Hence why $e^{-r^2}$ is a function that returns areas, where $e^{-x^2}$ returns lengths. Feb 6 '20 at 19:04
• @CopaceticMan What I don't understand is if we plot the function $f(x,y)=e^{-(x^2+y^2)}$ then shouldn't the integral give us back the volume under the surface? Feb 7 '20 at 16:20
• @user599310, I am going to attempt some pseudo math to show it: $$I^2 = \int e^-x^2 dx \times \int e^-x^2 dx = Area \times Area = Area^2$$ We can replace one x, with a dummy variable, move the dummy copy into the first integral to get a double integral. $$I^2 = \int \int e^{-x^2-y^2} dA$$ In context, the integrand a function that returns two Area, as it is the product of two functions which return a distance, specifically the height above one axis. So $fdA$ is still in terms of $area^2$. You are correct in your confusion, as the leap may not appear obvious, but it's valid. Jun 16 '20 at 19:42 | 2021-09-29T03:27:39 | {
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https://math.stackexchange.com/questions/4251865/combinatorics-multi-group-question-choose-6-people-out-of-14-respecting-con/4251873#4251873 | # Combinatorics - multi-group question - Choose 6 people out of 14, respecting conditions.
Here's the question:
"A congress will be attended by two representatives from Colombia, three from Chile, four from Argentina and five from Brazil. Each of the $$14$$ representatives prepared their own speech, but only $$6$$ will be drawn to speak. If the draw rule provides that each of the four countries must have at least one representative speaking, the number of different ways to compose the set of six speeches that will be heard at the congress, regardless of the order, is equal to how much?"
I've been working on this question for two weeks. My answer is $$1450$$ - but I solved it via brute force. I literally counted all the possibilities of the rule being broken and subtracted from the $$3003$$ possibilities to form groups of 6 people out of $$14$$.
I tried to look at the problem with fewer countries, and fewer choices. It helped me to count, but I couldn't decipher the pattern with which the problem changes when adding a new representative or a new country.
Can someone help me?
• With the conditions you gave, there are two possibilities: 1) there is one country with three representatives chosen and all others get one, 2) there are two countries with two representatives and two with one. Can you see why this is true? Looking at these two cases separately might help Sep 16 at 10:08
• Clarification requested. Are the people that represent a specific country distinguishable? That is, in the distributions that involve exactly 1 person from Colombia, is it appropriate to apply the factor of $\binom{2}{1}$, since there are $2$ ways of selecting which Colombian will be chosen? Sep 16 at 10:11
• @user2661923 I think the people are distinguishable, since it mentions that each people prepare a speech and we are asked to count the number of possible set of speeches Sep 16 at 10:18
• @Slugger thank you! This helped a lot! Sep 16 at 10:27
• @user2661923 Yep, people are distinguishable. Their speeches are different. :) Sep 16 at 10:27
## 3 Answers
A nice way to do this is to use the principle of inclusion-exclusion. Take all $$\binom{14}6$$ ways to choose $$6$$ people, then for each country, subtract the assignments where that country has no chosen representatives. The result is $$\binom {14}6 -\underbrace{\binom{12}6}_{\text{Columbia missing}} -\underbrace{\binom{11}6}_{\text{Chile missing}} -\underbrace{\binom{10}6}_{\text{Argentina missing}} -\underbrace{\binom{9}6}_{\text{Brazil missing}}$$ However, there is a problem. Arrangements with two missing countries will be subtracted twice in the above computation. To fix this, we must add these doubly-subtracted arrangements back in. To the above, we add $$+\underbrace{\binom{9}{6}}_{\text{Columbia + Chile}} +\underbrace{\binom{8}{6}}_{\text{Columbia + Argentina}} +\underbrace{\binom{7}{6}}_{\text{Columbia + Brazil}} +\underbrace{\binom{7}{6}}_{\text{Chile + Argentina}} +\underbrace{\binom{6}{6}}_{\text{Chile + Brazil}}$$ At this point, we are done! We would not be done if there were any arrangements counted by $$\binom{14}6$$ with three countries missing, but that is not possible.
• Thank you so much, Mike! Beautiful solution. And to top it off, I better understood the inclusion-exclusion principle. Thank you! Sep 16 at 14:53
Fo a second way , use generating functions. If there is at least one president in each country , then
Genereting function of Columbia : $$\binom{2}{1}x^1 +\binom{2}{1}x^2$$
Genereting function of Chile :
$$\binom{3}{1}x^1 + \binom{3}{2}x^2 +\binom{3}{3}x^3$$
Genereting function of Argentina :
$$\binom{4}{1}x^1 + \binom{4}{2}x^2 +\binom{4}{3}x^3 + \binom{4}{4}x^4$$
Genereting function of Brasil :
$$\binom{5}{1}x^1 + \binom{5}{2}x^2 +\binom{5}{3}x^3 + \binom{5}{4}x^4 +\binom{5}{5}x^5$$
Now, find the expansion of them and find the coefficient of $$x^6$$
Calculation in this link
So , answer is $$1450$$ , you are right !.By the way the binomial coefficients means the number of selection among candidates.For example , $$\binom{5}{2}x^2$$ means number of selection of two president among $$5$$ in Brasil.
Do you get the idea ?
$$\mathbf{\text{MORE CLARIFICATION:}}$$ In given link ,you will see that binomial expansions like $$((1+x)^2 -1)$$ , we know that when we expand it , it will give $$2x +x^2$$ . I wanted to write them clear form , but wolfram could not calculate them because of the lenght of expansion.Secondly , by the form of generating function (we started from $$x^1$$) , the number of president more than $$3$$ is automatically prevented.
• Can you explain where the binomial coefficients are in the calculation link , and how more than $3$ from any country has been prevented ? Sep 16 at 12:39
• @trueblueanil i add clarification, thanks.. Sep 16 at 12:46
• Nicely tackled the bad wolf, +1 Sep 16 at 12:49
• @trueblueanil wolf sometimes lazy for calculating , thanks a lot Sep 16 at 12:50
$$\sum_{\sum_{i=1}^4 \alpha_i=6\\1\leq \alpha_i \leq i+1}\binom{2}{\alpha_1}\binom{3}{\alpha_2}\binom{4}{\alpha_3}\binom{5}{\alpha_4}=1450$$
• Question may be ambiguous. See my comment following the question. Sep 16 at 10:13
• Come on! People will be distinguishable. Sep 16 at 10:19
• Thank you! Now I know my answer is probably right. But I lack the knowledge to understand your resolution. Anyway, thank you very much! Sep 16 at 10:31 | 2021-12-08T16:44:15 | {
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http://kabero.com/1nr89fz/a56140-multiple-integrals-examples | Example. It uses the 'iterated' method when any of the integration limits are infinite. The integration limits must be finite. Multiple Integrals Double Integrals over Rectangles 26 min 3 Examples Double Integrals over Rectangles as it relates to Riemann Sums from Calc 1 Overview of how to approximate the volume Analytically and Geometrically using Riemann Sums Example of approximating volume over a square region using lower left sample points Example of approximating volume over a… Double integrals beyond volume. Below is the image of a … The formula for change of variables is given by {\iint\limits_R {f\left( {x,y} \right)dxdy} } The Fundamental Theorem of Calculus; 3. Multiple integrals use a variant of the standard iterator notation. Double integrals (articles) Double integrals. Next lesson. for e.g. " Definite Integral. The double integration in this example is simple enough to use Fubini’s theorem directly, allowing us to convert a double integral into an iterated integral. To obtain double/triple/multiple integrals and cyclic integrals you must use amsmath and esint (for cyclic integrals) packages. Double Integral Area. This happens when the region of integration is rectangular in shape. The discussion on this page is in two main parts based on the type of region described by the limits of integration. The first group of questions asks to set up a double integral of a general function f(x,y) over a giving region in the xy-plane. » Integrate can evaluate integrals of rational functions. Order of Integration refers to changing the order you evaluate iterated integrals—for example double integrals or triple integrals.. Changing the Order of Integration. MULTIPLE INTEGRALS Triple Integrals in Spherical Coordinates Muliple Integration Section 1: DOUBLE INTEGRALS PROBLEM: Consider the solid E in 3-space bounded above by the surface z = 40 − 2xy and bounded below by the rectangular region D in the xy-plane (z = 0) defined by the set D = {(x,y) : 1 ≤ x ≤ 3, 2 ≤ y ≤ 4}. :) https://www.patreon.com/patrickjmt !! Fill in the blanks and then hit Enter (or click here). Evaluating double integrals is similar to evaluating nested functions: You work from the inside out. This method is called iterated integration.Simply tackle each integral from inside to outside. This website uses cookies to ensure you get the best experience. Powers of sine and cosine; 3. Double integrals are a way to integrate over a two-dimensional area. Polar coordinates. Practice problems on double integrals The problems below illustrate the kind of double integrals that frequently arise in probability applications. Examples, solutions, videos, activities and worksheets that are suitable for A Level Maths to help students answer questions on integration. Double Integrals The general form of dblquad is scipy.integrate.dblquad(func, a, b, gfun, hfun). integral in . You da real mvps! Then The inner integral is Note that we treat x as a constant as we integrate with respect to y. ( ) Function: Differentials : For indefinite integrals, you can leave the limits of integration empty. Suppose we integrate with respect to y first. Suppose that we wished to calculate the volume of the solid E, which in these discussion will be denoted by V(E). Moment and Center of Mass 4. Let z = f(x,y) define over a domain D in the xy plane and we need to find the double integral of z. Our mission is to … Integration can be used to find areas, volumes, central points and many useful things. The integral is equal to We are now left with the integral In this article, let us discuss the definition of the surface integral, formulas, surface integrals of a scalar field and vector field, examples in detail. Double integrals: reversing the order of integration Solve an example where a double integral is evaluated. If we divide the required region into vertical stripes and carefully find the endpoints for x and y i.e. Google Classroom Facebook Twitter. Thanks to all of you who support me on Patreon. Consider the double integral: where R is the rectangle 0<=x<=1, 1<=y<=2. Numerical integration over higher dimensional areas has special functions: integral2(@(x,y) x.^2-y.^2,0,1,0,1) ans = 4.0127e-19 This is the currently selected item. Where, func is the name of the function to be integrated, ‘a’ and ‘b’ are the lower and upper limits of the x variable, respectively, while gfun and hfun are the names of the functions that define the lower and upper limits of … 1. the limits of the region, then we can use the formula; Trigonometric Substitutions; 4. an integral in which the integrand involves a function of more than one variable and which requires for evaluation repetition of the integration process. Author: Colin Desmarais. Want to calculate a . The first variable given corresponds to the outermost integral and is done last. The computation of surface integral is similar to the computation of the surface area using the double integral except the function inside the integrals. Calculus: Fundamental Theorem of Calculus Email. » Clip: Examples of Double Integration (00:21:00) From Lecture 16 of 18.02 Multivariable Calculus, Fall 2007 Flash and JavaScript are required for this feature. To apply a double integral to a situation with circular symmetry, it is often convenient to use a double integral in polar coordinates. there are two integrals, one inside of the other. Among other things, they lets us compute the volume under a surface. Multiple Integrals Background What is a Double Integral? Topic: Calculus, Cone, Definite Integral, Solids or 3D Shapes, Sphere, Surface, Volume. This example shows how to compute definite integrals using Symbolic Math Toolbox™. One use for contour integrals is the evaluation of integrals along the real line that are not readily found by using only real variable methods. We can apply these double integrals over a polar rectangular region or a general polar region, using an iterated integral similar to those used with rectangular double integrals. But it is often used to find the area underneath the graph of a function like this: The integral of many functions are well known, and there are useful rules to work out the integral … Multiple Integrals Examples. Free double integrals calculator - solve double integrals step-by-step. The key idea is to replace a double integral by two ordinary "single" integrals. Word Origin. Double integrals are just integrals that are nested, i.e. (The improper triple integral is defined as the limit of a triple integral over a solid sphere as the radius of the sphere increases indefinitely.) You can solve double integrals in two steps: First evaluate the inner integral, and then plug this solution into the outer integral and solve that. Double integrals over non-rectangular regions. multiple integral. The following diagrams show some examples of Integration Rules: Power Rule, Exponential Rule, Constant Multiple, Absolute Value, Sums and Difference. Integrals >. Second, we find a fast way to compute it. This is the default method. Multiple integrals. If you can do a single integral, then you can compute a double integral. Multiple Integral Calculator. Calculating Double Integrals. Change of variables in triple Integrals. The task is to set up the integral needed to calculate a volume between two surfaces. Multiple Integration. Sort by: Top Voted. Polar coordinates. Chapter 5 DOUBLE AND TRIPLE INTEGRALS 5.1 Multiple-Integral Notation Previously ordinary integrals of the form Z J f(x)dx = Z b a f(x)dx (5.1) where J = [a;b] is an interval on the real line, have been studied.Here we study double integrals Z Z Ω f(x;y)dxdy (5.2) where Ω is some region in the xy-plane, and a little later we will study triple integrals Z Z Z 7 Integration. Integrals of a function of two variables over a region in $R^2$ are called double integrals. The multiple integral is a type of definite integral extended to functions of more than one real variable—for example, $f(x, y)$ or $f(x, y, z)$. Double Integrals in Cylindrical Coordinates 3. 'tiled' integral2 transforms the region of integration to a rectangular shape and subdivides it into smaller rectangular regions as needed. 1. Calculating the double integral in the new coordinate system can be much simpler. Multiple Integrals 14.1 Double Integrals 4 This chapter shows how to integrate functions of two or more variables. Changing the order of integration sometimes leads to integrals that are more easily evaluated; Conversely, leaving the order alone might result in integrals that are difficult or impossible to integrate. First, a double integral is defined as the limit of sums. In the mathematical field of complex analysis, contour integration is a method of evaluating certain integrals along paths in the complex plane.. Contour integration is closely related to the calculus of residues, a method of complex analysis. Consider, for example, a function of two variables $$z = f\left( {x,y} \right).$$ The double integral of function $$f\left( {x,y} \right)$$ is denoted by $\iint\limits_R {f\left( {x,y} \right)dA},$ where $$R$$ is the region of integration … Two examples; 2. Some Properties of Integrals; 8 Techniques of Integration. The technique involves reversing the order of integration. example. The idea is to evaluate each integral separately, starting with the inside integral. Integration Method Description 'auto' For most cases, integral2 uses the 'tiled' method. Double integrals are usually definite integrals, so evaluating them results in a real number. Substitution; 2. \$1 per month helps!! Calculus: Integral with adjustable bounds. Triple integrals. The double integrals in the above examples are the easiest types to evaluate because they are examples in which all four limits of integration are constants. Such an example is seen in 1st and 2nd year university mathematics. coordinates? By using this website, you agree to our Cookie Policy. Consequently, we are now ready to convert all double integrals to iterated integrals and demonstrate how the properties listed earlier can help us evaluate double integrals when the function $$f(x,y)$$ is more complex. This is the currently selected item. The definite integral can be extended to functions of more than one variable. Volumes as Double Integrals Iterated Integrals over Rectangles One Variable at the Time Fubini's Theorem Notation and Order Double Integrals over General Regions Type I and Type II regions Examples Order of Integration Area and Volume Revisited noun Mathematics. Learn more Accept. 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The surface area using the double integral except the function inside the integrals, starting the..., you can leave the limits of integration you can compute a double integral multiple integrals examples a shape! = 4.0127e-19 example x and y i.e who support me on Patreon is set! Is scipy.integrate.dblquad ( func, a, b, gfun, hfun ) integral2 transforms the region of integration rectangular... 'Auto ' for most cases, integral2 uses the 'tiled ' method a fast way to compute it double... Cyclic integrals you must use amsmath and esint ( for cyclic integrals ) packages to integrate over region... In two main parts based on the type of region described by the limits of integration to! Points and many useful things is often convenient to use a variant of the process! Evaluating nested functions: you work from the inside out to the computation of the other given to! Example double integrals calculator - solve double integrals step-by-step are nested, i.e us. ; 8 Techniques of integration refers to changing the order of integration ans = example. Integral to a rectangular shape and subdivides it into smaller rectangular regions as needed a shape! Website, you can leave the limits of integration is rectangular in.... < =1, 1 < =y < =2 cyclic integrals you must use amsmath and esint ( cyclic... Integration limits are infinite fill in the new coordinate system can be much.. R^2 [ /latex ] are called double integrals is similar to the outermost integral and is done last 'auto. Integrals, you agree to our Cookie Policy the discussion on this is. Parts based on the type of region described by the limits of solve... X and y i.e shape and subdivides it into smaller rectangular regions needed! The best experience, b, gfun, hfun ) used to areas. Integral2 uses the 'iterated ' method integral2 ( @ ( x, y ) x.^2-y.^2,0,1,0,1 ) ans = example! 8 Techniques of integration shape and subdivides it into smaller rectangular regions as.! In 1st and 2nd year university mathematics a function of more than one and... Our mission is to replace a double integral in polar Coordinates ordinary single..., gfun, hfun ) and many useful things integrals are just integrals that are nested i.e... Two-Dimensional area, a double integral in which the integrand involves a function of more than variable! Central points and many useful things Theorem of Calculus Thanks to all of you support... To integrate over a region in [ latex ] R^2 [ /latex are... X and y i.e some Properties of integrals ; 8 Techniques of integration 8... Is similar to evaluating nested functions: you work from the inside out the rectangle 0 < =x =1... Integral and is done last integrals or Triple integrals.. changing the of! Is to … double integrals or Triple integrals in Spherical Coordinates 7 integration multiple integrals examples a! Region of integration can be much simpler coordinate system can be much simpler be used to find areas volumes. And which requires for evaluation repetition of the standard iterator notation Spherical Coordinates 7 integration to... Can be much simpler or Triple integrals in Spherical Coordinates 7 integration 3D Shapes, Sphere, surface volume! Integration solve an example where a double integral except the function inside the integrals me on Patreon a. Is evaluated smaller rectangular regions as needed you agree to our Cookie.... Corresponds to the outermost integral and is done last the region of integration is rectangular in.... Ordinary single '' integrals obtain double/triple/multiple integrals and cyclic integrals ) packages Calculus Cone! Region into vertical stripes and carefully find the endpoints for x and y i.e to nested... Two ordinary single '' integrals do a single integral, then you can compute double. < =x < =1, 1 < =y < =2 the 'iterated ' method Sphere surface! Theorem of Calculus Thanks to all of you who support me on Patreon each. | 2021-08-01T18:13:32 | {
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http://fatherfiles.com/u0p094dw/51724f-kth-row-of-pascal%27s-triangle | We can find the pattern followed in all the rows and then use that pattern to calculate only the kth row and print it. Java Solution of Kth Row of Pascal's Triangle One simple method to get the Kth row of Pascal's Triangle is to generate Pascal Triangle till Kth row and return the last row. binomial coefficients - Use mathematical induction to prove that the sum of the entries of the $k^ {th}$ row of Pascal’s Triangle is $2^k$. // Do not read input, instead use the arguments to the function. Suppose we have a non-negative index k where k ≤ 33, we have to find the kth index row of Pascal's triangle. Pascal's triangle is an arithmetic and geometric figure often associated with the name of Blaise Pascal, but also studied centuries earlier in India, Persia, China and elsewhere.. Its first few rows look like this: 1 1 1 1 2 1 1 3 3 1 where each element of each row is either 1 or the sum of the two elements right above it. Note:Could you optimize your algorithm to use only O(k) extra space? This can allow us to observe the pattern. We also often number the numbers in each row going from left to right, with the leftmost number being the 0th number in that row. We find that in each row of Pascal’s Triangle n is the row number and k is the entry in that row, when counting from zero. The nth row is the set of coefficients in the expansion of the binomial expression (1 + x) n.Complicated stuff, right? Also, many of the characteristics of Pascal's Triangle are derived from combinatorial identities; for example, because , the sum of the value… Pascal’s triangle : To generate A[C] in row R, sum up A’[C] and A’[C-1] from previous row R - 1. - Mathematics Stack Exchange Use mathematical induction to prove that the sum of the entries of the k t h row of Pascal’s Triangle is 2 k. ; Pascal's triangle determines the coefficients which arise in binomial expansions. Given an index k, return the kth row of the Pascal’s triangle. An equation to determine what the nth line of Pascal's triangle … whatever by Faithful Fox on May 05 2020 Donate . So, if the input is like 3, then the output will be [1,3,3,1] To solve this, we will follow these steps − Define an array pascal of size rowIndex + 1 and fill this with 0 Output: 1, 7, 21, 35, 35, 21, 7, 1 Index 0 = 1 Index 1 = 7/1 = 7 Index 2 = 7x6/1x2 = 21 Index 3 = 7x6x5/1x2x3 = 35 Index 4 = 7x6x5x4/1x2x3x4 = 35 Index 5 = 7x6x5x4x3/1x2x3x4x5 = 21 … (Proof by induction) Rows of Pascal s Triangle == Coefficients in (x + a) n. That is: The Circle Problem and Pascal s Triangle; How many intersections of chords connecting N vertices? NOTE : k is 0 based. easy solution. Kth Row of Pascal's Triangle: Given an index k, return the kth row of the Pascal’s triangle. devendrakotiya01 created at: 8 hours ago | No replies yet. Given an index k, return the k t h row of the Pascal's triangle. Looking at the first few lines of the triangle you will see that they are powers of 11 ie the 3rd line (121) can be expressed as 11 to the power of 2. This is Pascal's Triangle. By creating an account I have read and agree to InterviewBit’s Follow up: Could you optimize your algorithm to use only O(k) extra space? Write a function that takes an integer value n as input and prints first n lines of the Pascal’s triangle. This can be solved in according to the formula to generate the kth element in nth row of Pascal's Triangle: r(k) = r(k-1) * (n+1-k)/k, where r(k) is the kth element of nth row. Pascal's Triangle II. Click here to start solving coding interview questions. For this reason, convention holds that both row numbers and column numbers start with 0. and These row values can be calculated by the following methodology: For a given non-negative row index, the first row value will be the binomial coefficient where n is the row index value and k is 0). c++ pascal triangle geeksforgeeks; Write a function that, given a depth (n), returns an array representing Pascal's Triangle to the n-th level. Can it be further optimized using this way or another? Following are the first 6 rows of Pascal’s Triangle. Terms A simple construction of the triangle … Once get the formula, it is easy to generate the nth row. Analysis. For an example, consider the expansion (x + y)² = x² + 2xy + y² = 1x²y⁰ + 2x¹y¹ + 1x⁰y². Pascal's triangle is known to many school children who have never heard of polynomials or coefficients because there is a fun way to construct it by using simple ad Kth Row Of Pascal's Triangle . //https://www.interviewbit.com/problems/kth-row-of-pascals-triangle/. k = 0, corresponds to the row [1]. Pascal’s triangle : To generate A[C] in row R, sum up A’[C] and A’[C-1] from previous row R - 1. Pascal's triangle is a way to visualize many patterns involving the binomial coefficient. Although other mathematicians in Persia and China had independently discovered the triangle in the eleventh century, most of the properties and applications of the triangle were discovered by Pascal. We write a function to generate the elements in the nth row of Pascal's Triangle. Pascal’s Triangle: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 . Better Solution: We do not need to calculate all the k rows to know the kth row. But be careful !! Privacy Policy. Look at row 5. We often number the rows starting with row 0. Given an index k, return the kth row of the Pascal's triangle. Pascal's triangle is the name given to the triangular array of binomial coefficients. (n + k = 8) In mathematics, Pascal's triangle is a triangular array of the binomial coefficients that arises in probability theory, combinatorics, and algebra. This problem is related to Pascal's Triangle which gets all rows of Pascal's triangle. 0. This works till the 5th line which is 11 to the power of 4 (14641). Examples: Input: N = 3 Output: 1, 3, 3, 1 Explanation: The elements in the 3 rd row are 1 3 3 1. The rows of Pascal’s triangle are numbered, starting with row $n = 0$ at the top. // Do not print the output, instead return values as specified, // Still have a doubt. k = 0, corresponds to the row [1]. k = 0, corresponds to the row … Example: Input : k = 3: Return : [1,3,3,1] NOTE : k is 0 based. Note:Could you optimize your algorithm to use only O(k) extra space? You signed in with another tab or window. Pascal's Triangle thus can serve as a "look-up table" for binomial expansion values. New. Java Solution Blaise Pascal was born at Clermont-Ferrand, in the Auvergne region of France on June 19, 1623. Pascal’s triangle is a triangular array of the binomial coefficients. Both of these program codes generate Pascal’s Triangle as per the number of row entered by the user. In much of the Western world, it is named after the French mathematician Blaise Pascal, although other mathematicians studied it centuries before him in India, Persia, China, Germany, and Italy.. Pascal's Triangle is defined such that the number in row and column is . Each number, other than the 1 in the top row, is the sum of the 2 numbers above it (imagine that there are 0s surrounding the triangle). Given a non-negative integer N, the task is to find the N th row of Pascal’s Triangle.. For example, the numbers in row 4 are 1, 4, 6, 4, and 1 and 11^4 is equal to 14,641. Didn't receive confirmation instructions? First 6 rows of Pascal’s Triangle written with Combinatorial Notation. (n = 5, k = 3) I also highlighted the entries below these 4 that you can calculate, using the Pascal triangle algorithm. Given an integer rowIndex, return the rowIndex th row of the Pascal's triangle. 0. Hockey Stick Pattern. Pascal s Triangle and Pascal s Binomial Theorem; n C k = kth value in nth row of Pascal s Triangle! The n th n^\text{th} n th row of Pascal's triangle contains the coefficients of the expanded polynomial (x + y) n (x+y)^n (x + y) n. Expand (x + y) 4 (x+y)^4 (x + y) 4 using Pascal's triangle. In this problem, only one row is required to return. Bonus points for using O (k) space. The entries in each row are numbered from the left beginning with $k = 0$ and are usually staggered relative to the numbers in the adjacent rows. 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 The numbers in row 5 are 1, 5, 10, 10, 5, and 1. Pascal’s triangle : To generate A[C] in row R, sum up A’[C] and A’[C-1] from previous row R - 1. For example, when k = 3, the row is [1,3,3,1]. vector. We write a function to generate the elements in the nth row of Pascal's Triangle. Well, yes and no. Pattern: Let’s take K = 7. This triangle was among many o… The formula just use the previous element to get the new one. Hot Newest to Oldest Most Votes. suryabhagavan48048 created at: 12 hours ago | No replies yet. Thus, the apex of the triangle is row 0, and the first number in each row is column 0. The program code for printing Pascal’s Triangle is a very famous problems in C language. Example 1: Input: rowIndex = 3 Output: [1,3,3,1] Example 2: In this post, I have presented 2 different source codes in C program for Pascal’s triangle, one utilizing function and the other without using function. The next row value would be the binomial coefficient with the same n-value (the row index value) but incrementing the k-value by 1, until the k-value is equal to the row … This leads to the number 35 in the 8 th row. The start point is 1. 0. Given an index k, return the kth row of the Pascal’s triangle. Pascal’s triangle : To generate A[C] in row R, sum up A’[C] and A’[C-1] from previous row R - 1. Source: www.interviewbit.com. 3. java 100%fast n 99%space optimized. Learn Tech Skills from Scratch @ Scaler EDGE. Notice the coefficients are the numbers in row two of Pascal's triangle: 1, 2, 1. Below is the first eight rows of Pascal's triangle with 4 successive entries in the 5 th row highlighted. As an example, the number in row 4, column 2 is . ! whatever by Faithful Fox on May 05 2020 Donate . This video shows how to find the nth row of Pascal's Triangle. Here are some of the ways this can be done: Binomial Theorem. In Pascal's triangle, each number is the sum of the two numbers directly above it. Example: Input : k = 3 Return : [1,3,3,1] NOTE : k is 0 based. Kth Row of Pascal's Triangle 225 28:32 Anti Diagonals 225 Adobe. k = 0, corresponds to the row [1]. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 In 1653 he wrote the Treatise on the Arithmetical Triangle which today is known as the Pascal Triangle. Start with any number in Pascal's Triangle and proceed down the diagonal. “Kth Row Of Pascal's Triangle” Code Answer . 41:46 Bucketing. For example, given k = 3, return [ 1, 3, 3, 1]. This video shows how to find the nth row of Pascal's Triangle. NOTE : k is 0 based. Checkout www.interviewbit.com/pages/sample_codes/ for more details. 2. python3 solution 80% faster. Note: The row index starts from 0. Notice that the row index starts from 0. //https://www.interviewbit.com/problems/kth-row-of-pascals-triangle/ /* Given an index k, return the kth row of the Pascal’s triangle. Kth Row Of Pascal's Triangle . Since 10 has two digits, you have to carry over, so you would get 161,051 which is equal to 11^5. | 2021-04-22T17:04:19 | {
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https://math.stackexchange.com/questions/1078294/let-a-n-n-1-infty-be-an-infinite-sequence-does-there-exist-an-infinite | Let $\{a_n\}_{n=1}^\infty$ be an infinite sequence. Does there exist an infinite series whose partial sums is $\{a_n\}_{n=1}^\infty$?
Definition: By an Infinite Sequence of real numbers, we shall mean any real valued function whose domain is the set of all positive integers.
Definition: By an Infinite Series of real numbers, we shall mean an ordered pair of infinite sequences of real numbers $$(\{a_n\}_{n=1}^\infty, \{s_n\}_{n=1}^\infty)$$ such that $$s_1=a_1, \space s_2=a_1+a_2, \space s_3=a_1+a_2+a_3$$ and in general $$s_k= a_1+a_2+ \dotsb +a_k$$ The sequence $\{a_n\}_{n=1}^\infty$ is called the sequence of terms of the infinite series. The sequence $\{s_n\}_{n=1}^\infty$ is called the sequence of partial sums of the infinite series.
Exam Question:
Let $\{a_n\}_{n=1}^\infty$ be an infinite sequence of real numbers. Does there exist an infinite series of real numbers whose sequence of partial sums is $\{a_n\}_{n=1}^\infty$? Why or why not?
My answer is Yes and here is why. Consider the constant sequence $\{0\}_{n=1}^\infty$. This sequence has the value $0$ for all $n \in \mathbb Z^+$. Now, if we consider the infinite series of this sequence, then the sequence of partial sums has the value $0$ for all $n \in \mathbb Z^+$. So, in this case $$(\{a_n\}_{n=1}^\infty, \{s_n\}_{n=1}^\infty) = (\{0\}_{n=1}^\infty, \{0\}_{n=1}^\infty) = (0,0)$$
Therefore, $\{a_n\}_{n=1}^\infty = 0 = \{s_n\}_{n=1}^\infty$ for all $n \in \mathbb Z^+$.
However, on turning in my exam, my professor stated this was incorrect. Why is this so?
• Your answer is incorrect because the question is asking whether the statement is true for every sequence $\{a_n\}$. You have only proved it in the special case when $a_n=0$ for every $n$. – Prism Dec 23 '14 at 1:47
• "Does there exist…" part is about existence of the series whose partial sums is $\{a_n\}$. Let's look at the beginning of the question. It says "Let $\{a_n\}_{n=1}^{\infty}$ be an infinite sequence of real numbers…" It is an arbitrary sequence of real numbers! – Prism Dec 23 '14 at 1:50
• It is an existence question given an arbitrary sequence. So it is really a universal question: does a certain property hold for any sequence. – Ian Dec 23 '14 at 1:50
• @Ian : I'd have phrased it thus: "does a certain property hold for every sequence?". The standard way of using the word "any" would admit the following way of misunderstanding your phrasing: "does a certain property hold for any sequence" could mean "Is there any sequence for which a certain property holds?". But that cannot be what was meant. ${}\qquad{}$ – Michael Hardy Dec 23 '14 at 2:03
• @MichaelHardy You're right, the word "any" is frustrating for this exact reason. – Ian Dec 23 '14 at 2:19
Certainly: $$\text{Let }b_0 = a_0\text{ and }b_{n+1} = a_{n+1}-a_n.$$ Then \begin{align} b_0 & = a_0 & = a_0 \\ b_0 + b_1 & = a_0 + (a_1-a_0) & = a_1 \\ b_0 + b_1 + b_2 & = a_0 + (a_1-a_0)+(a_2-a_1) & = a_2 \\ b_0 + b_1 + b_2 + b_3 & = a_0 + (a_1-a_0)+(a_2-a_1)+(a_3-a_2) & = a_3 \\ & {}\,\,\vdots & \vdots\phantom{a_n} \end{align}
• @JohannFranklin : I'm not sure there's much else to say. If the partial sum $b_0+\cdots+b_{100}$ is $a_{100}$, and we want the partial sum $b_0+\cdots+b_{100}+b_{101}$ to be $a_{101}$, then we have to add $b_{101}$ to the sum we've already got, which is $a_{100}$, to get $a_{101}$. So the question is: what do we have to add to $a_{100}$ to get $a_{101}$? The answer is $a_{101}-a_{100}$, so we need $b_{101}$ to be equal to that. ${}\qquad{}$ – Michael Hardy Dec 23 '14 at 2:00
Because it only works when $a_n=0$ for all $n$. Let $s_0=a_0$, and $s_n=a_n-a_{n-1}$. Now
$$\sum_{n=0}^ks_n=a_0+(a_1-a_0)+\dots +(a_k-a_{k-1})=a_k.$$ | 2019-06-17T02:48:21 | {
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https://physicssolutions.wordpress.com/iiser-aptitude-test-sample-paper-1-solution/ | # IISER Aptitude Test Sample Paper 1 Solution
This page contains solutions to the IISER Aptitude test Sample 1. Currently the solutions are only for Maths and Physics sections. Other sections will be uploaded soon.
• MATH
• PHYSICS
• BIOLOGY
• CHEMISTRY
Some Mnemonics and tricks to solve Permutation and Combination type Problems. Link
### MATH
1) If $I_m=\int^{\frac{\pi}{4}}_0(\tan{x})^m\,dx$;
then $I_3+I_5+I_7+I_9$ equals ?
2) We have a sequence $1^{\frac{1}{\sqrt{1}}} , 3^{\frac{1}{\sqrt{3}}} , 5^{\frac{1}{\sqrt{5}}} , 7^{\frac{1}{\sqrt{7}}} ..... {2n+1}^{{\frac{1}{\sqrt{2n+1}}}}$ and so on…..
Which is the largest value in the sequence ?
3) Assuming interchange of limit and integration is permisssible, the value of
$\lim_{n\to \infty}\int^1_0 \frac{nx^{n-1}}{1+x} \,dx$
Note: 0<x<1
4) Let f: R→R be a function such that |f(x)-f(y)|$\leq 6|x-y|^2$ for all $x,y\in R$.
If f(3)=6 then, f(6) equals ?
Solution:
5)Let $A_n$ be the area bounded by curve y=x & $y=nx^2$ in 1st quadrant. Then, the value of $\sum\limits_{i=1}^5 \frac{1}{A_n}$ is.
6) In how many ways can 4 distinguishable pieces be placed on a 8*8 chessboard so that no two pieces are in the same row or column?
Solution:
7) A and B are playing a game by alternately rolling a die,with A starting first. Each player’s score is the numbers obtained on his last roll. The ends when the sum of scores of the two players is 7, and the last player to roll the die wins. What is the probability that A wins the game ?
Solution:
8) The binomial coefficient $^{n}C_r, ^{n}C_{r+1}, ^{n}C_{r+2},.....$ where $0 \leq r\leq {n-2}$
Solution:
9) The Sum of infinite series $arccot{2} + arccot{8} + arccot{18} + ..... + arccot{2n^2} +.....$ is ?
10) The integer values of k for which the equation $7\cos{\theta} + 5\sin{\theta} = 2k+1$ has real solutions is ?
11) The complex solutions of $(z+i)^{2011}=z^{2011}$ lie on :
12) How many 2*2 matrices A satisfy both $A^3=I_2$ and $A^2=A^T$, where $I_2$ denotes the 2*2 identity matix and $A^T$ denotes the transpose of A ?
Solution: Proof of why the inverse is unique is left to be written but it can be found easily on Google or any Linear Algebra Text Book.
13) Let C be the circle that touches the X-Axis and whose centre coincides with the circum-centre of the triangle defined by $4|x| + 3y=12$ ; $y\geq 0$. How many points with both co-ordinates integers are there in the interior of C?
Solution:
14) Let P and Q be the centres of the circles that pass through (0,2) and (0,8) and touch the X-Axis. Then the equation of the ellipse with P and Q as focii and touching X-Axis is ?
Solution:
15 Let f: R→R be a function such that f(x+y) + f(x-y)= f(xy) for all $x,y \in R$. Then f is ?
Solution:
Solution:
## 15 thoughts on “IISER Aptitude Test Sample Paper 1 Solution”
1. Puneet Kakkar says:
Where is physics solution?
2. greeshma says:
3. Saunak Kotwal says:
Where is physics? Also, did you get the unsolved problems?
1. Hello Friend sorry. I had made this site in 2014 and totally abandoned it when I did not get good response nor I had much time to do it. Looking at the number of people visiting this month (Actually I also came back to this site after a long time), I will definitely start putting other things up, I have stuff solved and scanned and backed up with me but I did not care to upload, which I will surely do now, maybe on a different and better website. The link to it I will post soon. Thanks for visiting.
1. greeshma says:
thanks for helping us out.No other site provided answers
2. kiran says:
please provide chemistry and physics solutions also
4. JUDE says:
sir make it fats test is fast approaching on 12th july
5. Neha Kulkarni says:
Thank you for the solutions 🙂
6. Ahallya says:
7. A science lover and Iiser aspirant says:
Hello, thank you first of all for uploading the answers. See in question no. 10 how can you assume cos{\theta} =3/7 and sin{\theta} = 2/5 ? Doesn’t it violate sin^2{\theta}+cos^2{\theta}=1 ?
1. Thanks for pointing out the mistake. This is tricky and a tough question to solve on paper. However, I just plotted the LHS on Wolfram
http://www.wolframalpha.com/input/?i=Plot+7cos(theta)%2B5sin(theta)+for+theta+between+0+to+2*pi
It seems that there are 16 values of theta between 0 and 2 pi where the RHS is Odd. I am searching for RHS being odd because, k has to be integer, i.e. if k is integer 2k+1 is odd. Visually, check for (-7, -5, -3, -1, 1, 3, 5, 7) on the y-axis. There are 16 thetas (between 0 and 2*pi) which give these values.
If you have already solved it, be happy to share with others :-D.
8. Shamitha says: | 2017-03-24T07:58:15 | {
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https://math.stackexchange.com/questions/1495720/combined-combinations-whats-wrong-with-my-line-of-thinking | Combined Combinations - What's wrong with my line of thinking?
I'm brushing up on some basics for an upcoming job interview, and I've got myself confused over a question about combined combinations.
From Discrete Mathematics with Graph Theory Third Edition by Goodaire & Parmenter, Section 7.2, Exercise 1:
A group of people is comprised of six from Nebraska, seven from Idaho, and eight from Louisiana.
(a) In how many ways can a committee of six be formed with two people from each state?
(b) In how many ways can a committee of seven be formed with at least two people from each state?
I'm fine with part (a). I got the same solution as is in the back of the book:
$\binom{6}{2} \binom{7}{2} \binom{8}{2} = 8820$
But with part (b), I was thinking that the answer would be the result from part (a), multiplied by the number of ways you could choose one more person from the total pool of remaining people:
$\binom{6}{2} \binom{7}{2} \binom{8}{2} \binom{15}{1} = 132,300$
The answer in the back of the book is
$\binom{6}{3} \binom{7}{2} \binom{8}{2} + \binom{6}{2} \binom{7}{3} \binom{8}{2} + \binom{6}{2} \binom{7}{2} \binom{8}{3} = 44,100$
I haven't touched this stuff in a while, so I'm inclined to believe the authors, but my logic still makes sense to me.
What is the problem with the way I'm thinking about the answer?
• You are over counting. Choosing $A,B$ from Nebraska and then choosing a third $C$ from there is the same as choosing $A,C$ and then choosing $B$ as the extra person. Indeed you are over counting by a factor of exactly $3$. – lulu Oct 24 '15 at 18:53
You are triple-counting.
The person chosen in the $\binom{15}{1}$ could be Carol from Nebraska, and the people counted in the $\binom{6}{2}$ could be Alicia and Beti from Nebraska. This is the same group from Nebraska as the group obtained by choosing Alicia from Nebraska in the $\binom{15}{1}$, and the other two in the $\binom{6}{2}$. Same with Beti as the "third" person from Nebraska.
The point is that if we are choosing $3$ people from Nebraska, it is a bag of $3$, it is not choosing a pair and then choosing another.
Remark: In this case, we can deliberately triple count, and obtain your number. However, if we choose to do that, we must at the end divide by $3$. The strategy of controlled multiple-counting is often useful.
• D'oh. Of course, divide by three to avoid the triple count. I got bogged down in trying to cut down the the triple count that I just restarted from scratch to do it the book's way. – fleablood Oct 24 '15 at 19:03
• "it is a bag of 3, it is not choosing a pair and then choosing another" -- this line really drove it home for me. Thanks. – WhiteHotLoveTiger Oct 24 '15 at 19:15
• You are welcome. This sort of multiple counting will happen to you a few times, and then probably never again. A canonical example of double-counting comes when we are counting "two pair" poker hands. It is tempting to say that the kind of the "first" pair can be chosen in $\binom{13}{1}$ ways, and then the kind of the "second" pair can be chosen in $\binom{12}{1}$ ways. But there is no "first" pair and "second" pair. there are $\binom{13}{2}$ ways to choose the two kinds. Or do it the wrong way and then divide by $2$. – André Nicolas Oct 24 '15 at 19:24
Try the same problem with smaller numbers! (This can help you with a lot of combinatorial problems.)
A group of people is composed of two from Nebraska and two from Idaho.
(a) In how many ways can a committee of two be formed with one person from each state? $$\binom21\binom21=4$$ (b) In how many ways can a committee of three be formed with at least one person from each state?
In this case, any committee of three will include at least one person from each state, so the correct answer is $$\binom43=\binom41=4$$ On the other hand, by your line of thinking, the answer would be the result from part (a), multiplied by the number of ways you could choose one more person from the remaining pool of people: $$4\cdot\binom21=8$$ Clearly, the second answer is wrong; we have overcounted somehow. Because the numbers are so small, you can actually write down the $8$ committees you have counted, and see for yourself how the overcount happened. Call the Nebraskans $N,E$ and the Idahoans $I,D.$ $$NI\longrightarrow NIE,\ NID$$ $$ND\longrightarrow NDE,\ NDI$$ $$EI\longrightarrow EIN,\ EID$$ $$ED\longrightarrow EDN,\ EDI$$ Observe that, since order doesn't matter, $NID$ and $NDI$ are the same committee, which has been counted twice.
My first thought was to think about it exactly as you did. But the problem is we double count our results.
If you choose Bob and Jane from Nebraska, and then as our seventh member we randomly choose Jim also from Nebraska, that is the same result as choosing Bob and Jim from Nebraska and randomly choosing Jane as the seventh member. We counted those results twice.
Now, I can go back and try to figure out how many times we'd repeat these and adjust. [edit: As Jane, Jim + Bob = Jane,Bob + Jim = Bob, Jim + Jane, the adjusting is to divide by 3. I totally spaced on how to do this when I first posted this answer.] Or I could think: How many ways to have three from Nebraska and two from the other two states [${6 \choose 3} {7 \choose 2} {8 \choose 2}$], plus how many ways to have three from Idaho and two from the other[${6 \choose 2} {7 \choose 3} {8 \choose 2}$], plus how many ways to have three from Louisiana and two from the other two states[${6 \choose 2} {7 \choose 2} {8 \choose 3}$].
• Thanks. Listing out all three ways you can end up with Jane, Jim & Bob was helpful. – WhiteHotLoveTiger Oct 24 '15 at 19:20 | 2019-05-23T03:19:01 | {
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https://math.stackexchange.com/questions/3062999/trying-to-simplify-frac-sqrt81-sqrt3x-to-be-frac2-sqrt22-sqrt/3063010 | # Trying to simplify $\frac{\sqrt{8}}{1-\sqrt{3x}}$ to be $\frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$
I am asked to simplify $$\frac{\sqrt{8}}{1-\sqrt{3x}}$$. The solution is provided as $$\frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$$ and I am unable to arrive at this. I was able to arrive at $$\frac{1+2\sqrt{2}\sqrt{3x}}{1-3x}$$
Here is my working: $$\frac{\sqrt{8}}{1-\sqrt{3x}}$$ = $$\frac{\sqrt{8}}{1-\sqrt{3x}}$$ * $$\frac{1+\sqrt{3x}}{1+\sqrt{3x}}$$ = $$\frac{1+\sqrt{8}\sqrt{3x}}{1-3x}$$ = $$\frac{1+\sqrt{2}\sqrt{2}\sqrt{2}\sqrt{3x}}{1-3x}$$ = $$\frac{1+2\sqrt{2}\sqrt{3x}}{1-3x}$$
Is $$\frac{1+2\sqrt{2}\sqrt{3x}}{1-3x}$$ correct and part of the way? How can I arrive at the provided solution $$\frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$$?
• Please review your computations. There is a least 2 errors. – mathcounterexamples.net Jan 5 '19 at 18:10
$$\frac{\sqrt{8}}{1-\sqrt{3x}} = \frac{\sqrt{8}}{1-\sqrt{3x}} \cdot \frac{1+\sqrt{3x}}{1+\sqrt{3x}} = \color{blue}{\frac{\sqrt{8}\cdot\left({1+\sqrt{3x}}\right)}{1-3x}} = \frac{\sqrt{8}+\sqrt{24x}}{1-3x}$$
$$= \frac{\sqrt{2^3}+\sqrt{2^3\cdot3x}}{1-3x} = \frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$$
Notice the step I highlighted in blue, which is where you made an error. You have to multiply $$\sqrt{8}$$ to $$\left(1+\sqrt{3x}\right)$$ completely. You multiplied it by only $$\sqrt{3x}$$ and added $$1$$, which wasn’t correct.
Observe that $$\sqrt{8}\times (1+\sqrt{3x})=\sqrt{8}+\sqrt{8}\times \sqrt{3x}$$ and $$\sqrt{8}\times \sqrt{3}=\sqrt{24}=\sqrt{4\times 6}=2\sqrt{6}.$$
Well, in rationalizing the denominator, we arrive at the intermediate step $$\frac{\sqrt{8}+\sqrt{24x}}{1-3x}$$ which simplifies to the provided solution.
Your error comes in multiplying by the conjugate of the denominator. $$\sqrt{8}*(1+\sqrt{3x})$$ becomes $$\sqrt{8}+\sqrt{24x}$$.
Write $$\frac{\sqrt{8}(1+\sqrt{3x})}{(1-\sqrt{3x})(1+\sqrt{3x})}$$ and this is $$\frac{2\sqrt{2}(1+\sqrt{3x})}{1-3x}$$
Hints:$$\sqrt8=\sqrt{4×2}=2\sqrt{2}.$$
$$\sqrt24=2\sqrt6$$. (Why?)
$$(1+\sqrt{3x})×(1-\sqrt{3x})=1-3x$$.
The second equality is where you mess up - note that $$\sqrt{8}\cdot(1+\sqrt{3x})=\sqrt{8}+\sqrt{8}\cdot\sqrt{3x}$$ by the distributive property of real numbers | 2020-06-04T01:20:07 | {
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https://math.stackexchange.com/questions/653100/difference-between-continuity-and-uniform-continuity | # Difference between continuity and uniform continuity
I understand the geometric differences between continuity and uniform continuity, but I don't quite see how the differences between those two are apparent from their definitions. For example, my book defines continuity as:
Definition 4.3.1. A function $f:A \to \mathbb R$ is continuous at a point $c \in A$ if, for all $\epsilon > 0$, there exists a $\delta > 0$ such that whenever $|x-c| < \delta$ (and $x \in A$) it follows that $|f(x)-f(c)| < \epsilon$.
Uniform continuity is defined as:
Definition 4.4.5. A function $f:A \to \mathbb R$ is uniformly continuous on $A$ if for every $\epsilon > 0$ there exists a $\delta > 0$ such that $|x-y| < \delta$ implies $|f(x)-f(y)| < \epsilon$.
I know that in Definition 4.3.1, $\delta$ can depend on $c$, while in definition 4.4.5, $\delta$ cannot depend on $x$ or $y$, but how is this apparent from the definition? From what appears to me, it just seems like the only difference between Definition 4.3.1 and Definition 4.4.5 is that the letter $c$ was changed to a $y$.
My guess is that the first definition treats $c$ as a fixed point and it is only $x$ that varies, so in this case, $\delta$ can depend on $c$ since $c$ doesn't change. Whereas for the second definition, neither $x$ or $y$ are fixed, rather they can take on values across the whole domain, $A$. In this case, if we set a $\delta$ such that it depended on $y$, then when we pick a different $y$, the same $\delta$ may not work anymore. Is this somewhat a correct interpretation?
Anymore clarifications, examples, would be appreciated.
First of all, continuity is defined at a point $c$, whereas uniform continuity is defined on a set $A$. That makes a big difference. But your interpretation is rather correct: the point $c$ is part of the data, and is kept fixed as, for instance, $f$ itself. Roughly speaking, uniform continuity requires the existence of a single $\delta>0$ that works for the whole set $A$, and not near the single point $c$.
The difference is in the ordering of the quantifiers.
• Continuity:
For all $x$, for all $\varepsilon$, there exist such a $\delta$ that something something.
• Uniform continuity:
For all $\varepsilon$, there exists such a $\delta$ that for all $x$ something something.
For something to be continuous, you can check "one $x$ at a time", so for each $x$, you pick a $\varepsilon$ and then find some $\delta$ that depends on both $x$ and $\varepsilon$ so that $|f(x)-f(y)|<\varepsilon$ if $|x-y|<\delta$. As you can see if you try it on $f(x)=1/x$ on $(0,1)$, you can find such a $\delta$ for every $x$ and $\varepsilon$. However, if you fix $\varepsilon$, the values for $\delta$ that you need become arbitrarily small as $x$ approaches $0$.
If you want uniform continuity, you need to pick a $\varepsilon$, then find a $\delta$ which is good for ALL the $x$ values you might have. As you see, for $f(x)=1/x$, such a $\delta$ does not exist.
• The language you use, and therefore the order you use the language is not found in the definitions provided from the OP's textbook, and therefore isn't very helpful. Can you please reword your answer to fit the question: "I know that in Definition 4.3.1, δ can depend on c, while in definition 4.4.5, δ cannot depend on x or y, but how is this apparent from the definition?" I ask, because I have the same question and the same textbook definition. I know that your answer is correct but why is it apparent from the GIVEN definition? Apr 30, 2018 at 23:56
• @rocksNwaves, indeed, in the script we use, the order is the same as in the uniform continuity: continuity at a point $c\in I\subseteq\Bbb R:$ $\\(\forall\varepsilon>0)(\exists\delta>0)(\forall x\in I)((|x-c|<\delta)\implies (|f(x)-f(c)|<\varepsilon))\\$ whereas the uniform continuity on an open interval $I\subseteq\Bbb R:$ $\\(\forall\varepsilon>0)(\exists\delta>0)(\forall x',x''\in I)((|x'-x''|<\delta)\implies(|f(x')-f(x'')|<\varepsilon))\\$ Apr 24, 2021 at 22:31
The subtle difference between these two definitions became more clear to me when I read their equivalent sequence definitions. First take the definition of a continuous function.
Definition A function $$f: D\to\mathbb{R}$$ is said to be continuous at the point $$x_0$$ in $$D$$ provided that for every sequence $$\{x_n\}$$ in $$D$$ that converges to $$x_0$$, the image sequence $$\{f(x_n)\}$$ converges to $$f(x_0)$$.
Now compare this to a uniformly continuous function.
Definition A function $$f: D\to\mathbb{R}$$ is said to be uniformly continuous provided that for any two sequences $$\{y_n\}$$ and $$\{x_n\}$$ in $$D$$ have the property $$\lim_{n\to\infty}(y_n-x_n)=0,$$ then $$\lim_{n\to\infty}(f(y_n)-f(x_n))=0$$
Notice how the second definition mentions no convergence to a point, but that two sequences are tending toward the same value and at the same rate. These sequences can both be divergent sequences when alone, but their terms can become arbitrarily close to each other.
The classic example is $$f:\mathbb{R}\to\mathbb{R}, f(x)=x^2$$ is continuous but not uniformly continuous. Take the two sequences $$\{y_n\}=\{\sqrt{n^2+1}\}$$ and $$\{x_n\}=\{n\}$$. (Note, both sequences diverge). Take the $$\lim_{n\to\infty}{y_n-x_n}$$, and solve by multiplying numerator and denominator by its conjugate. $$\lim_{n\to\infty}(\sqrt{n^2+1}-n)=\lim_{n\to\infty}\frac{n^2 +1-n^2 }{\sqrt{n^2+1}+n}=\lim_{n\to\infty}\frac{1}{\sqrt{n^2+1}+n}=0.$$ Now, looking at $$\lim_{n\to\infty}{f(y_n)-f(x_n)}$$ we get the following $$\lim_{n\to\infty}{(\sqrt{n^2+1})^2-n^2}=\lim_{n\to\infty}{n^2+1-n^2}=1$$ So this goes against the definition of uniform continuity. We need the difference of function values to also go to $$0$$, as well, in order for it to be uniformly continuous.
• Clint.Have you seen the proof of the equivalence of your sequence definition of uniform continuity with Definition 4.4.5 in the OP's question? Thanks. Aug 27, 2019 at 16:57
• @PeterSzilas I have; though admittedly, it has been some time. I remember bridging the two through proof in an assignment for a class. For purposes of this comment, I copied the definition from Advanced Calculus, Second Edition, by Patrick M. Fitzpatrick. The definition was found in section 3.4: Uniform Continuity. Aug 30, 2019 at 19:30
• @PeterSzilas To help in starting the proof, see Lemma 2.9 in the same book, called the Comparison Lemma. The proof gives strategies to tie together sequences and delta-epsilon proofs. I know somewhere in that book it talks about the equivalence of limit definitions and delta-epsilon definitions and its reasoning for keeping to the limit definitions, but I couldn't find it in my brief search. If time permits, I'll create and link a new thread with the proof. Aug 30, 2019 at 19:53
• Clint.Thanks.The above theorem is at times quite useful. Even proving $\epsilon -\delta$ (simple) continuity starting from the sequential continuity definition (by contradiction) is a bit tricky. Let you know if I come across something interesting.Greetings. Aug 31, 2019 at 7:35
• didn't get that sequence part Apr 10, 2020 at 6:44
Let me focus on this part of the question:
"I know that in Definition 4.3.1, $\delta$ can depend on $c$, while in definition 4.4.5, $\delta$ cannot depend on $x$ or $y$, but how is this apparent from the definition?"
This is apparent from the order of the quantifiers. When we write out these two statements into "Prenex normal form", we have that:
$$\forall c \in A,\forall \epsilon >0, \exists \delta, \forall x \in A \:( |x-c|<\delta \implies |f(x)-f(c)|<\epsilon)$$
$$\forall \epsilon >0, \exists \delta, \forall x,c\in A \: (|x-c|<\delta \implies |f(x)-f(c)|<\epsilon)$$
In the first statement, note that the universal ($\forall$) quantifier $\forall c$ precedes the existential ($\exists$) quantifier $\exists \delta$, and the universal quantifier $\forall x$ follows the existential quantifier $\exists \delta$.
Note that in the second definition, the universal quantifier $\forall c$ now also follows the existential quantifier $\exists \delta$.
To see the significance of the quantifier order, consider the following, where C is the set of cars, P is the set of people, and R is a relation such that cRp means c is owned by p.
$$\forall c\in C, \exists p \in P: cRp$$ $$\exists p\in P, \forall c \in C: cRp$$
Observe that in the first statement of the example, the universal quantifier precedes the existential quantifier. This statement means each car $c$ has an owner $p$. Observe that the person p depends on the car. In the second statement, the universal quantifier follows the existential quantifier. This statement means there is some person $p$ who owns EVERY car. Thus this person doesn't depend on the car (since he has all of them, or in other words; given every car, he has it).
To conclude, for any variables $x,y$, $y$ can depend on $x$ if and only if the universal quantifier for $\forall x$ precedes the existential quantifier for $\exists y$.
Applying this theorem to your definitions, we see that in the definition of continuity, the universal quantifiers $\forall x\in\mathbb{R}$ and $\forall\epsilon$ precede $\exists\delta$ . Thus here $\delta$ may depend on both $x,\epsilon$ . However, in the definition of uniform continuity, the only universal quantifier that precedes $\delta$ is $\forall\epsilon$ . Thus delta may only depend on $\epsilon$ and not $x$ .
In the definition of uniform continuity, $\exists \delta$ precedes neither $x$ nor $c$, therefore it can depend on neither of them, but only on $\epsilon$.
To understand the difference between continuity and uniform continuity, it is useful to think of a particular example of a function that's continuous on $\mathbb R$ but not uniformly continuous on $\mathbb R$. An example of such a function is $f(x)=x^2$. Here to understand the failure of uniform continuity of the function, it is particularly useful to exploit the hyperreals. Note that if $H$ is an infinite number and we choose an infinitesimal $\epsilon=\frac{1}{H}$, then the values of $f$ at the infinitely close points $H$ and $H+\epsilon$ are themselves not infinitely close. This violation of the property of microcontinuity of $f$ at $H$ captures the essence of the failure of uniform continuity of $f$ on $\mathbb R$.
This intuitive GIF image from Wikepedia helped me most.
$f(x)=\frac{1}{x}$ as shown in the image is continuous but not uniformly continuous, because obviously if, for instance when $x_1=0.1$ we can see that $|f(x_1)-f(x_1+0.2)| \gt 0.5$; while $g(x)=\sqrt x$ is both continuous and uniformly continuous since we can find a number, for instance 0.5 bellow, such that $|f(x_1)-f(x_1+0.2)| \lt 0.5$ for every $x_1$. Here 0.2 and 0.5 should be numbers in R which just exist for the given function.
Hope this will be of any help to you.
I find it interesting that each answer here throws in something different to the understanding of uniform continuity. And I have something different to add.
As observed by Siminore, continuity can be expressed at a point and on a set whereas uniform continuity can only be expressed on a set. Reflecting on the definition of continuity on a set, one should observe that continuity on a set is merely defined as the veracity of continuity at several distinct points. In other words, continuity on a set is the "union" of continuity at several distinct points. Reformulated one last time, continuity on a set is the "union" of several local points of view.
Uniform continuity, in contrast, takes a global view---and only a global view (there is no uniform continuity at a point)---of the metric space in question.
These different points of view determine what kind of information that one can use to determine continuity and uniform continuity. To verify continuity, one can look at a single point $$x$$ and use local information about $$x$$ (in particular, $$x$$ itself) and local information about how $$f$$ behaves near $$x$$. For example, if you know that $$f$$ is bounded on a neighborhood of $$x$$, that is fair game to use in your recovery of $$\delta$$. Also, any inequality that $$x$$ or $$f(x)$$ satisfies on a tiny neighborhood near $$x$$ is fair game to use as well. You can even use $$f(x)$$ to define $$\delta$$.
However to verify uniform continuity, you can't zoom in on any particular point. You can only use global information about the metric space and global information about the function $$f$$; i.e. a priori pieces of information independent of any particular point in the metric space. For example, any inequality that every point of $$X$$ satisfies is fair game to use to recover $$\delta$$. If $$f$$ is Lipschitz, any Lipschitz constant is fair to use in your recovery of $$\delta$$.
There are two propositions which I think exemplify the difference between continuity and uniform continuity:
Let $$X$$ and $$Y$$ denote two metric spaces, and let $$f$$ map $$X$$ to $$Y$$.
• $$f$$ is continuous on $$X$$ if and only if for every $$x$$ in $$X$$ and for every $$\epsilon>0$$ there is a $$\delta>0$$ such that $$\text{diam}\, f(B_{\delta/2}(x))<\epsilon\,.$$
• $$f$$ is uniformly continuous on $$X$$ if and only if for every $$\epsilon>0$$ there is a $$\delta>0$$ such that $$\text{diam}\,f(E)<\epsilon$$ for every subset $$E$$ of $$X$$ that satisfies $$\text{diam}\,E<\delta$$.
Thus continuity in a certain sense only worries about the diameter of a set around a given point. Whereas uniform continuity worries about the diameters of all subsets of a metric space simultaneously.
Exactly, the delta of uniform continuity is not changeable, which decides the ball of x, y. However, the delta of continuity is decided by the point c, it varies due to the change of c. | 2023-01-28T22:28:18 | {
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https://stats.stackexchange.com/questions/10768/cdf-raised-to-a-power/10792 | # CDF raised to a power?
If $F_Z$ is a CDF, it looks like $F_Z(z)^\alpha$ ($\alpha \gt 0$) is a CDF as well.
Q: Is this a standard result?
Q: Is there a good way to find a function $g$ with $X \equiv g(Z)$ s.t. $F_X(x) = F_Z(z)^\alpha$, where $x \equiv g(z)$
Basically, I have another CDF in hand, $F_Z(z)^\alpha$. In some reduced form sense I'd like to characterize the random variable that produces that CDF.
EDIT: I'd be happy if I could get an analytical result for the special case $Z \sim N(0,1)$. Or at least know that such a result is intractable.
• Yes, that's a pretty well-known result and is easy to generalize. (How?) You can also find $g$, at least implicitly. It's essentially an application of the inverse probably transform technique commonly used to generate random variates of an arbitrary distribution. – cardinal May 13 '11 at 15:28
• @cardinal Please, answer. The team is later complaining that we are not fighting with low answered ratio. – user88 May 13 '11 at 16:25
• @mbq: Thanks for your comments, which I understand and respect greatly. Please understand that sometimes considerations of time and/or place do not allow me to post an answer, but do permit a quick comment that can get the OP or other participants off to a start. Be assured that, going forward, if I am able to post an answer, I will do so. Hopefully my continued participation through comments will be ok as well. – cardinal May 13 '11 at 23:18
• @cardinal Some of us are also guilty of the same, for the same reasons... – whuber May 14 '11 at 0:16
• @brianjd Yes, this is a well-known result that has been used for industrially producing "generalised" distributions, see. There exist many transformations like this one and people use them for this purpose: they find a parametric transformation, apply it to a distribution and voilá, you have a paper just by just calculating its properties. And of course, the normal is the first 'victim'. – user10525 May 8 '12 at 18:43
I like the other answers, but nobody has mentioned the following yet. The event $\{U \leq t,\ V\leq t \}$ occurs if and only if $\{\mathrm{max}(U,V)\leq t\}$, so if $U$ and $V$ are independent and $W = \mathrm{max}(U,V)$, then $F_{W}(t) = F_{U}(t)*F_{V}(t)$ so for $\alpha$ a positive integer (say, $\alpha = n$) take $X = \mathrm{max}(Z_{1},...Z_{n})$ where the $Z$'s are i.i.d.
For $\alpha = 1/n$ we can switcheroo to get $F_{Z} = F_{X}^n$, so $X$ would be that random variable such that the max of $n$ independent copies has the same distribution as $Z$ (and this would not be one of our familiar friends, in general).
The case of $\alpha$ a positive rational number (say, $\alpha = m/n$) follows from the previous since $$\left(F_{Z}\right)^{m/n} = \left(F_{Z}^{1/n}\right)^{m}.$$
For $\alpha$ an irrational, choose a sequence of positive rationals $a_{k}$ converging to $\alpha$; then the sequence $X_{k}$ (where we can use our above tricks for each $k$) will converge in distribution to the $X$ desired.
This might not be the characterization you are looking for, but it least gives some idea of how to think about $F_{Z}^{\alpha}$ for $\alpha$ suitably nice. On the other hand, I'm not really sure how much nicer it can really get: you already have the CDF, so the chain rule gives you the PDF, and you can calculate moments till the sun sets...? It's true that most $Z$'s won't have an $X$ that's familiar for $\alpha = \sqrt{2}$, but if I wanted to play around with an example to look for something interesting I might try $Z$ uniformly distributed on the unit interval with $F(z) = z$, $0<z<1$.
EDIT: I wrote some comments in @JMS answer, and there was a question about my arithmetic, so I'll write out what I meant in the hopes that it's more clear.
@cardinal correctly in the comment to @JMS answer wrote that the problem simplifies to $$g^{-1}(y) = \Phi^{-1}(\Phi^{\alpha}(y)),$$ or more generally when $Z$ is not necessarily $N(0,1)$, we have $$x = g^{-1}(y) = F^{-1}(F^{\alpha}(y)).$$ My point was that when $F$ has a nice inverse function we can just solve for the function $y = g(x)$ with basic algebra. I wrote in the comment that $g$ should be $$y = g(x) = F^{-1}(F^{1/\alpha}(x)).$$
Let's take a special case, plug things in, and see how it works. Let $X$ have an Exp(1) distribution, with CDF $$F(x) = (1 - \mathrm{e}^{-x}),\ x > 0,$$ and inverse CDF $$F^{-1}(y) = -\ln(1 - y).$$ It is easy to plug everything in to find $g$; after we're done we get $$y = g(x) = -\ln \left( 1 - (1 - \mathrm{e}^{-x})^{1/\alpha} \right)$$ So, in summary, my claim is that if $X \sim \mathrm{Exp}(1)$ and if we define $$Y = -\ln \left( 1 - (1 - \mathrm{e}^{-X})^{1/\alpha} \right),$$ then $Y$ will have a CDF which looks like $$F_{Y}(y) = \left( 1 - \mathrm{e}^{-y} \right)^{\alpha}.$$ We can prove this directly (look at $P(Y \leq y)$ and use algebra to get the expression, in the next to the last step we need the Probability Integral Transform). Just in the (often repeated) case that I'm crazy, I ran some simulations to double-check that it works, ... and it does. See below. To make the code easier I used two facts: $$\mbox{If X \sim F then U = F(X) \sim \mathrm{Unif}(0,1).}$$ $$\mbox{If U \sim \mathrm{Unif}(0,1) then U^{1/\alpha} \sim \mathrm{Beta}(\alpha,1).}$$
The plot of the simulation results follows.
The R code used to generate the plot (minus labels) is
n <- 10000; alpha <- 0.7
z <- rbeta(n, shape1 = alpha, shape2 = 1)
y <- -log(1 - z)
plot(ecdf(y))
f <- function(x) (pexp(x, rate = 1))^alpha
curve(f, add = TRUE, lty = 2, lwd = 2)
The fit looks pretty good, I think? Maybe I'm not crazy (this time)?
• See my comment in @JMS answer. For $Z\sim N(0,1)$ it looks like the answer is $g(z) = \Phi^{-1}(\Phi^{1/\alpha}(z))$ which isn't closed form but can be calculated easily enough. And you can make it easier by recognizing that the input to the inverse CDF is a suitably chosen Beta distribution. The answer will be nice in cases where the inverse CDF is nice, and there are some of those running around. – user1108 May 16 '11 at 13:43
• It would be good to double-check your arithmetic. – cardinal May 16 '11 at 14:40
• @cardinal errr... OK, I did,... and it's right? Would you please point out the error? – user1108 May 20 '11 at 18:01
• (+1) Apologies. I'm not sure where my head was when I first looked at this. It's obviously (well, should have been!) correct. – cardinal May 20 '11 at 19:09
• @cardinal, no harm, no foul. I admit, though, you really had me sweating for a minute! :-) – user1108 May 20 '11 at 19:24
## Proof without words
The lower blue curve is $F$, the upper red curve is $F^\alpha$ (typifying the case $\alpha \lt 1$), and the arrows show how to go from $z$ to $x = g(z)$.
• Nice pic! Q: What was that drawn in? TikZ? – lowndrul May 14 '11 at 15:47
• @brianjd: If I recall, @whuber does many of his plots using Mathematica. – cardinal May 14 '11 at 15:52
• @cardinal You're right. Actually, I use whatever is handy and seems like it will do a good job quickly. FWIW, here's the code: Module[ {y, w, a = 0.1, z = 3.24, f = ChiDistribution[7.6], xmin=0, xmax=5}, y = CDF[f,z]; w = InverseCDF[f, y^(1/a)]; Show[ Plot[{CDF[f, x],CDF[f,x]^a} , {x, xmin, xmax}, Filling->{1->{2}}], Graphics[{ Dashed, Arrow[{{z,0}, {z,y}}], Arrow[{{z,y}, {w,y}}], Arrow[{{w,y}, {w,0}}] }] ] ] – whuber May 14 '11 at 18:10
Q1) Yes. It's also useful for generating variables which are stochastically ordered; you can see this from @whuber's pretty picture :). $\alpha>1$ swaps the stochastic order.
That it's a valid cdf is just a matter of verifying the requisite conditions: $F_z(z)^\alpha$ has to be cadlag, nondecreasing and limit to $1$ at infinity and $0$ at negative infinity. $F_z$ has these properties so these are all easy to show.
Q2) Seems like it would be pretty difficult analytically, unless $F_Z$ is special
• @JMS: What about $Z \sim N(0,1)$ ? – lowndrul May 14 '11 at 15:45
• @brianjd: I don't believe so. Let $g$ be a continuous strictly monotonic function (hence, having a well-defined inverse $g^{-1}$) that satisfies your conditions. Then, it must be that $\renewcommand{\Pr}{\mathbb{P}}\Phi^{\alpha}(u) = \Pr(g(Z) \leq u) = \Pr( Z \leq g^{-1}(u)) = \Phi(g^{-1}(u))$ and so $g^{-1}(u) = \Phi^{-1}(\Phi^{\alpha}(u))$. So the inverse is identified fairly explicitly, but not $g$ itself. This is what I meant in my previous comment about $g$ being found implicitly. – cardinal May 14 '11 at 16:17
• @brianjd - What @cardinal said :) I couldn't even think of a special case for $F_Z$ where you'd get a closed form (not to say there isn't one of course). – JMS May 14 '11 at 23:28
• @JMS: $Z \sim \mathcal{U}[0,1]$ would be one positive example. – cardinal May 15 '11 at 2:38
• @cardinal I never would have thought of such a rare distribution... but now that you mention it a $Beta(a, 1)$ should work in general, giving you back a $Beta(a\alpha, 1)$. – JMS May 15 '11 at 3:11 | 2019-07-21T03:01:20 | {
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https://math.stackexchange.com/questions/1071082/what-is-the-required-radius-of-the-smaller-circles-around-a-larger-circle-so-the/1071131 | # What is the required radius of the smaller circles around a larger circle so they touch?
I am trying to determine how to calculate the required radius of the smaller circles so they touch each other around the larger circle. (red box)
I would like to be able to adjust the number of smaller circles and the radius of the larger circle.
As an example:
\begin{align} R&=1.5\\ n&=9\\ r&=\,? \end{align}
• What do you mean with "around the big circle", exactly? Where are the centers of the small circles? Dec 16, 2014 at 20:19
• Are you saying the centers of the small circles must lie on the circumference of the large one? Dec 16, 2014 at 20:23
• Dec 16, 2014 at 20:26
• Do two subsequent small circles have to intersect on the big circle or they just have to be tangent?
– Surb
Dec 16, 2014 at 20:36
• I am trying to get the smaller circles uniform around the bigger circle. Tangent would work. Dec 16, 2014 at 20:42
If you draw $n$ lines from the origin touching the small circles and $n$ lines from the origin to the center of each small circle you basically divide the $2 \pi$ angle into $2n$ equal angles, say $\theta$. Hence $\theta={\pi}/{n}$. Now a triangle with vertices the origin, the center of one of the small circles and a tangent point of the same circle is a right triangle since the tangent is perpendicular to the radius at the point of contact. You then have
$$\sin \theta=\frac{r}{R}$$
Putting it all together
$$r=R \sin \frac{\pi}{n}$$
Another approach: Lets say we have $n$ small circles. Then the center points of the small circles form a regular $n$-gon, where the side length is $2r$. The radius of the big circle is the circumradius of the $n$-gon which is $R = \frac{2r}{2\sin(\pi/n)}$ so $r = R \sin(\pi/n)$
• OK but I was faster... Dec 19, 2014 at 23:07
• Timestamp says the contrary, you were 3 min too late=) Dec 19, 2014 at 23:34
• Compare "edited Dec 16 @ 20:50" with "answered Dec 16 @ 20:48". Oh never mind, I'm just teasing you! Dec 19, 2014 at 23:44
• So what=) At least we agree that this is a rather nice way of solving it! Dec 19, 2014 at 23:48
• That we do sir :) Dec 19, 2014 at 23:52
Suppose that the center $c_k,k=1,\ldots,n$ of the small circles are placed equidistantly on the bigger circle.
Then we have $c_k=\left(R\sin(2\pi\frac{k}{n}),R\cos(2\pi\frac{k}{n})\right), k = 1,\ldots,n$. So for $k=1,\ldots,n$ we must have $r=\frac{\|c_{k}-c_{k+1}\|_2}{2}$, since two circles with same radius are tangent if their radius is the half of the distance between their centers. In particular $$r=\frac{\|c_{n-1}-c_{n}\|_2}{2}=\frac{1}{2}\sqrt{\big(R\sin(2\pi)-R\sin\big(2\pi\frac{n-1}{n}\big)\big)^2+\big(R\cos(2\pi)-R\cos\big(2\pi\frac{n-1}{n}\big)\big)^2} \\ =\frac{R}{2}\sqrt{\sin\big(2\pi\frac{n-1}{n}\big)^2+\big(1-\cos\big(2\pi\frac{n-1}{n}\big)\big)^2} \\ = \frac{R}{2}\sqrt{2-2\cos\big(2\pi\frac{n-1}{n}\big)} =\frac{R}{2}\sqrt{2\Big(1-\cos\big(2\pi\frac{n-1}{n}\big)\Big)}\\ =\frac{R}{2}\sqrt{4\sin\big(2\pi\frac{n-1}{n}\big)^2}=R\sin\big(\pi\frac{n-1}{n}\big)=R\sin\big(\frac{\pi}{n}\big).$$
• what happens if n=100000 ? The cosine is 1 and the quantity under the sqrt is negative – Georgy 3 secs ago edit Jun 17, 2015 at 7:14
• @Georgy you are right thank you, there was a typo in my computation. I corrected it.
– Surb
Jun 17, 2015 at 8:30 | 2022-08-14T10:10:09 | {
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https://math.stackexchange.com/questions/1980012/how-to-evaluate-this-integral-int-0-infty-exp-alpha-x2-cos-beta-xd | # How to evaluate this integral ($\int_{0}^{\infty}\exp(-\alpha x^2)\cos(\beta x)dx$)? [closed]
I am facing issue to evaluate this particular type of definite integral,
$$I=\int_{0}^{\infty}\exp(-\alpha x^2)\cos(\beta x)dx$$
Please suggest a way to this.
Thanks
## closed as off-topic by Carl Mummert, Ethan Bolker, Silvia Ghinassi, Pragabhava, WatsonOct 24 '16 at 16:29
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Carl Mummert, Ethan Bolker, Silvia Ghinassi, Pragabhava, Watson
If this question can be reworded to fit the rules in the help center, please edit the question.
• This is just a Fourier transform of $e^{-\alpha x^2}$ – Boby Oct 22 '16 at 15:22
• This is similar to Fourier transform. For a gaussian, the F.T. is also a gaussian. – Srini Oct 22 '16 at 15:24
• Why has this question been put on hold? It has sparked a number of good solutions which will undoubtedly help many viewers. A question like this is what makes MSE interesting, informative, and enjoyable. – poweierstrass Oct 26 '16 at 15:24
Let $I(\alpha,\beta)$ be given by the integral
$$I(\alpha,\beta)=\int_0^\infty e^{-\alpha x^2}\cos(\beta x)\,dx$$
Using Euler's Formula, $\cos(\beta x)=\text{Re}(e^{i\beta x})$. Then, we can write
\begin{align} I(\alpha,\beta)&=\text{Re}\left(\int_0^\infty e^{-\alpha x^2}e^{i\beta x}\,dx\right)\\\\ &=\text{Re}\left(\int_0^\infty e^{-\alpha (x^2-i(\beta/\alpha) x)}\,dx\right)\tag 1 \\\\ &=e^{-\beta^2/4\alpha}\text{Re}\left(\int_0^\infty e^{-\alpha (x-i(\beta/2\alpha) )^2}\,dx\right)\tag 2\\\\ &=e^{-\beta^2/4\alpha}\text{Re}\left(\int_{-i(\beta/2\alpha)}^{\infty-i(\beta/2\alpha)} e^{-\alpha x^2}\,dx\right) \tag 3\\\\ &=e^{-\beta^2/4\alpha}\text{Re}\left(\int_0^\infty e^{-\alpha x^2}\,dx\right) \tag 4\\\\ &=\sqrt{\frac{\pi}{4\alpha}}e^{-\beta^2/4\alpha} \end{align}
NOTES:
In going from $(1)$ to $(2)$, we completed the square in the exponent.
In going from $(2)$ to $(3)$, we enforced the substitution $x\to x+i\beta/2\alpha$.
In going from $(3)$ to $(4)$, we exploited Cauchy's Integral Theorem to deform the contour back to the real line. The real part operation nullifies the contribution from the integral from $-i\beta/2\alpha$ to $0$.
• But the Euler's formula, I know is $\cos(\beta x) = \frac{\exp(i\beta x)+\exp(-i\beta x)}{2}$. How come $\cos(\beta x)=\text{Re}(e^{i\beta x})$? – zhk Oct 22 '16 at 15:48
• @mmm Yes, you are correct. And $e^{i\beta x}=\cos(\beta x)+i\sin(\beta x)$ from which we find $\cos(\beta x)=\text{Re}(e^{i\beta x})$. – Mark Viola Oct 22 '16 at 15:52
• If I am correct the (3) to (4) is to use $\oint\limits_{C}f(z)\,\mathrm{d}z=\int_{R}f(z)dz$? – zhk Oct 22 '16 at 16:15
• The integrand is entire, so $\oint_C e^{-\alpha z^2}\,dz=0$. Thus, with $C$ the "rectangular" contour with vertices $0$, $\infty$, $\infty-i\beta/2\alpha$, and $-i\beta/2\alpha$, we have $$\int_{0}^\infty e^{-\alpha x^2}\,dx+\int_{\infty}^{\infty -i\beta/2\alpha}e^{-\alpha z^2}\,dz+\int_{\infty -i\beta/2\alpha}^{-i\beta/2\alpha}e^{-\alpha z^2}\,dz+\int_{-i\beta/2\alpha}^0 e^{-\alpha z^2}\,dz=0$$The second integral is zero since $e^{-t^2}\to 0$ as $t\to \infty$. The fourth integral is purely imaginary since $dz=idy$ and vanishes upon taking the real part. – Mark Viola Oct 22 '16 at 16:20
• You're welcome. My pleasure. -Mark – Mark Viola Oct 22 '16 at 16:33
\begin{align} \int_0^\infty e^{-\alpha x^2}e^{i\beta x}\,\mathrm{d}x &=\int_0^\infty e^{-\alpha\left(x-i\frac\beta{2\alpha}\right)^2-\frac{\beta^2}{4\alpha}}\mathrm{d}x\tag{1}\\ &=e^{-\frac{\beta^2}{4\alpha}}\int_{-i\frac\beta{2\alpha}}^{\infty-i\frac\beta{2\alpha}}e^{-\alpha x^2}\mathrm{d}x\tag{2}\\ &=e^{-\frac{\beta^2}{4\alpha}}\left[\int_{-i\frac\beta{2\alpha}}^0e^{-\alpha x^2}\mathrm{d}x+\int_0^\infty e^{-\alpha x^2}\mathrm{d}x+\color{#A0A0A0}{\int_\infty^{\infty-i\frac\beta{2\alpha}} e^{-\alpha x^2}\mathrm{d}x}\right]\tag{3}\\ &=e^{-\frac{\beta^2}{4\alpha}}\left[\,\color{#D080F8}{i\int_0^{\frac\beta{2\alpha}}e^{\alpha x^2}\mathrm{d}x}+\frac12\sqrt{\frac\pi\alpha}\,\right]\tag{4} \end{align} Explanation:
$(1)$: complete the square
$(2)$: substituted $x\mapsto x+i\frac\beta{2\alpha}$
$(3)$: the integrand is entire; apply Cauchy's Integral Theorem
$(4)$: the integral over $\left[R,R-i\frac\beta{2\alpha}\right]$ vanishes
The integral in purple is pure imaginary, so taking the real parts of $(4)$ gives $$\int_0^\infty e^{-\alpha x^2}\cos(\beta x)\,\mathrm{d}x =\frac12e^{-\frac{\beta^2}{4\alpha}}\sqrt{\frac\pi\alpha}\tag{5}$$
• See the answer I wrote, in my opinion the identity theorem is easier to understand than the Cauchy integral theorem when you don't know so much of complex analysis – reuns Oct 22 '16 at 22:59
$$I = \int\limits_{0}^{\infty} \mathrm{e}^{-ax^{2}} \cos(bx) \mathrm{d}x \tag{1} \label{eq:1}$$
Let \begin{align} I_{1} &= \int\limits_{0}^{\infty} \mathrm{e}^{-ax^{2}} \mathrm{e}^{ibx} \mathrm{d}x = \int\limits_{0}^{\infty} \mathrm{e}^{-ax^{2}+ibx} \mathrm{d}x \\ &= \mathrm{e}^{-b^{2}/4a} \int\limits_{0}^{\infty} \mathrm{e}^{-a(x-ib/2a)^{2}} \mathrm{d}x \\ \tag{2} \label{eq:2} \end{align} Here we completed the square and note that $\mathrm{Re}\,I_{1} = I$.
Consider the indefinite integral \begin{align} \int \mathrm{e}^{-a(x-ib/2a)^{2}} \mathrm{d}x &= \int \mathrm{e}^{-ay^{2}} \mathrm{d}y \\ &= \frac{1}{\sqrt{a}} \int \mathrm{e}^{-z^{2}} \mathrm{d}z = \frac{1}{2} \sqrt{\frac{\pi}{a}} \mathrm{erf}(z) \\ &= \frac{1}{2} \sqrt{\frac{\pi}{a}} \mathrm{erf}\left(x\sqrt{a} - \frac{ib}{2\sqrt{a}}\right) \tag{3} \label{eq:3} \end{align} we used the substitutions, $y=x- \frac{ib}{2a}$ and $z^{2} = ay^{2}$.
Examining the error function expression, we have $$\lim_{x \to 0} \mathrm{erf}\left(x\sqrt{a} - \frac{ib}{2\sqrt{a}}\right) = \mathrm{erf}\left(- \frac{ib}{2\sqrt{a}}\right) = -i\,\mathrm{erfi}\left(\frac{b}{2\sqrt{a}}\right) \tag{4} \label{eq:4}$$ which is a pure imaginary quantity with the assumption that all of the variables in the argument of the imaginary error function are real and $a \gt 0$. We also have $$\lim_{x \to \infty} \mathrm{erf}\left(x\sqrt{a} - \frac{ib}{2\sqrt{a}}\right) = 1 \tag{5} \label{eq:5}$$
Using equations \eqref{eq:4} and \eqref{eq:5} in equation \eqref{eq:3} we obtain $$\mathrm{Re}\left(\int\limits_{0}^{\infty} \mathrm{e}^{-a(x-ib/2a)^{2}} \mathrm{d}x \right) = \frac{1}{2} \sqrt{\frac{\pi}{a}} \tag{6} \label{eq:6}$$
Combining equations \eqref{eq:6} and \eqref{eq:2} yields our final result $$\int\limits_{0}^{\infty} \mathrm{e}^{-ax^{2}} \cos(bx) \mathrm{d}x = \frac{1}{2} \sqrt{\frac{\pi}{a}} \,\mathrm{e}^{-b^{2}/4a}$$
The least amount of complex analysis is as follow :
• For $z > 0$, let $$f(z) = \int_{-\infty}^\infty e^{-x^2} e^{2zx}dx$$ Complete the square $x^2-2zx= (x-z)^2 - z^2$ so that $$f(z) = e^{-z^2} \int_{-\infty}^\infty e^{-(x-z)^2}dx \underset{y = x-z}= e^{-z^2} \int_{-\infty}^\infty e^{-y^2}dy = Ce^{-z^2}$$ (where $C = \sqrt{\pi}$)
Then note that for every $z \in \mathbb{C}$ : $\int_{-\infty}^\infty e^{-x^2} e^{2zx}dx$ is analytic in $z$, as well as $Ce^{-z^2}$, thus by the identity theorem for analytic functions $$\forall z \in \mathbb{C}, \qquad f(z) = Ce^{-z^2}$$
• Apply the same trick to $$g_z(s) = \int_{-\infty}^\infty e^{-sx^2} e^{2zx}dx$$ where $s > 0$. With $y = s^{1/2}x$ we get $$g_z(s) = s^{-1/2}\int_{-\infty}^\infty e^{-y^2} e^{2zs^{-1/2}y}dy = s^{-1/2}f(zs^{-1/2}) = \frac{C}{s^{1/2}}e^{-z^2/s}$$ Finally note that $\int_{-\infty}^\infty e^{-sx^2} e^{2zx}dx$ is analytic in $s$ for every $s \in \mathbb{C},Re(s) > 0$, as well as $\frac{C}{s^{1/2}}e^{-z^2/s}$, thus by the identity theorem $$\forall s \in \mathbb{C},Re(s) > 0, \qquad g_z(s) = \frac{C}{s^{1/2}}e^{-z^2/s}$$
• Hence for any $Re(a) > 0$ : $$I = \frac{1}{2}\int_{-\infty}^\infty e^{-a x^2} \cos( \beta x)dx = \frac{g_{i \beta/2}(a)+g_{-i \beta/2}(a)}{4} = \frac{C}{2a^{1/2}}e^{\textstyle- \frac{\beta^2}{4a}}$$
• How do you justify $$e^{-z^2} \int_{-\infty}^\infty e^{-(x-z)^2}dx \underset{y = x-z}= e^{-z^2} \int_{-\infty}^\infty e^{-y^2}dy$$ I believe you need Cauchy's Integral Theorem for this. – robjohn Oct 23 '16 at 0:27
• @robjohn no read again, I proved it for $z > 0$ (I meant $z \in \mathbb{R}$ a typo) – reuns Oct 23 '16 at 0:32
• You've shown that $f(z)=Ce^{-z^2}$ for $z\in\mathbb{R}$. How do you apply the identity theorem without showing it for $z$ in a non-empty, open subset of $\mathbb{C}$? – robjohn Oct 23 '16 at 1:29
• – reuns Oct 23 '16 at 1:34
• @robjohn the idea is that $f(z) = g(z)$ for $z \in [a,b]$ means $f^{(k)}(a) = g^{(k)}(a)$ for every $k \in \mathbb{N}$, so they agree on an open around $a$. Otherwise, use that non-constant analytic functions have isolated zeros. – reuns Oct 23 '16 at 1:36 | 2018-10-19T11:11:42 | {
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https://math.stackexchange.com/questions/902603/why-does-ab2-a2b2-2ab-why-is-the-2ab-there/902618 | Why does $(a+b)^2= a^2+b^2 + 2ab$? Why is the $2ab$ there?
When I was doing research on finding the derivative I came across something strange.
If $f(x) = x^2$ you find the derivative by going
$$\frac{f(x+h)^2-f(x)^2}{h} =\frac{x^2+2xh+h^2-x^2}{h}.$$
Why is there the $2xh$? Can someone explain the logic behind this? I'm assuming you plus $x$ and $h$ together then square it but why?
• Since when did $f'(x) = \lim_{h \to 0} \frac{f(x+h)^2 - f(x)^2}{h}$? – Yiyuan Lee Aug 19 '14 at 3:57
• You're using a derivative, so you're tacitly assuming that the characteristic of your field is zero, and in particular, not two. – Dorebell Aug 19 '14 at 3:59
• Are you sure it said $f(x + h)^2 - f(x)^2$, as opposed to $f(x + h) - f(x)$? – Elliott Aug 19 '14 at 4:05
• You should have studied $(a+b)^2=a^2+2ab+b^2$ long before seeing a limit or a derivative. – whacka Aug 19 '14 at 4:10
• This video should be helpful – Nick Aug 19 '14 at 4:14
Graphical interpretation
The image shows the area of a square with side length of a+b. Here you can see, how the small 4 areas become the square.
• This was also posted here. – Martin Sleziak Aug 19 '14 at 8:07
• Thank you so much thats what I want to see! An answer that shows the intuition of whats happening is the best. – Ray Kay Aug 19 '14 at 8:51
• @RayKay Nice to hear, that it helps. You are welcome. – callculus Aug 19 '14 at 9:05
• Nice explanation :) :) – Shaddy Jan 24 '16 at 17:11
• @Shaddy Thanks for the comment. – callculus Jul 9 '18 at 19:55
\begin{align*} (a+b)(a + b) &= a(a + b) + b(a + b)\quad &(\text{addition distributes over multiplication})\\ &= aa + ab + ba + bb\quad &(\text{multiplication distributes over addition})\\ &= a^2 + ab + ba + b^2\quad &(xx = x^2)\\ &= a^2 + ab + ab + b^2\quad &(\text{multiplication is commutative})\\ &= a^2 + 2ab + b^2 \end{align*}
The reason why you're having problems is because you wrote the problem wrong.
The derivative is not $$\lim_{h \to 0} \frac{f(x+h)^2-f(x)^2}{h}$$
but actually $$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$
You're confusing $f(x) = x^2$ with $f(x)^2$. These are not the same thing.
$f(x) = x^2$ means that for any argument $x$ you put in the function, you square that input. $f(x)^2$ means you're squaring the result after you plug in $x$ into the function, or the output.
That's why $f(x+h) = (x+h)^2 = x^2 + 2xh + h^2$.
$(a+b)^2=(a+b)(a+b)=a^2+ab+ba+b^2=a^2+2ab+b^2$
The reason is that $(a+b)^2$ is nothing more than shorthand notation for $(a+b)(a+b)$. FOILing this expression gives
$$(a+b)^2 = (a+b)(a+b) = \overbrace{a\cdot a}^{\text{FIRST}} + \overbrace{a\cdot b}^{\text{OUTER}} + \overbrace{b\cdot a}^{\text{INNER}} + \overbrace{b\cdot b}^{\text{LAST}} = a^2 + ab + ba + b^2.$$
$ab=ba$ since order of multiplication doesn't matter for real numbers, so we see that
$$(a+b)^2 = a^2 + ab + ab + b^2 = a^2 + 2ab + b^2.$$
Here's how the $2xh$ term appears $$(x+h)^{2}=(x+h)(x+h)$$ $$=x(x+h)+h(x+h)$$ $$=x^{2}+xh +xh +h^{2}$$ $$=x^{2}+2xh+h^{2}$$ Here are the steps for finding the derivative of $f(x)=x^{2}$ $\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}=\lim_{h\to 0} \frac{(x+h)^{2}-x^{2}}{h}$ $=\lim_{h\to 0} \frac{x^{2}+2xh+h^{2}-x^{2}}{h}=\lim_{h\to 0} \frac{2xh+h^{2}}{h}$ $=\lim_{h\to 0} [2x+h]=2x+0=2x$
The above answers are correct but if I understood you correctly, I think you may be confused about what the function is expecting. We know $$f(x) = x^2$$ meaning whatever input I give you, I want you to square it. So $f(x + h)$ means the "input" is $x + h$ and we want to square this and so we have $$(x + h)^2 = x^2 + 2xh + h^2$$
Let $a, b > 0$ be real numbers.
Let there be given a square $S$ with side length $a+b$ so that the area of $S$ equals $(a+b)^{2}$. Then within $S$ we have two small squares $S', S''$ such that the area of $S'$ equals $a^{2}$ and the area of $S''$ equals $b^{2}.$ Meanwhile we also have two small rectangles $R, R'$ of the same area $ab.$
Therefore the area $(a+b)^{2}$ of $S$ equals the sum $a^{2} + b^{2}$ of the areas of $S'$ and $S''$ plus the sum $ab + ab = 2ab$ of the areas of $R$ and $R'$, i.e. $$(a+b)^{2} = a^{2} + b^{2} + 2ab.$$
The logic behind derivative is tangent line. To draw that line you need two points on the graph. As that two points get closer together, your line became the tangent line. So to answer your question, $$f(x+h)=(x+h)^2=x^2+2xh+h^2$$ is just the second point on your graph. | 2019-09-23T14:10:26 | {
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https://math.stackexchange.com/questions/2967033/show-that-overlinea-subseteq-ao-cup-partial-a | # Show that $\overline{A} \subseteq A^{o} \cup \partial A$
$$A'$$ denotes the set of limit points of $$A$$ and $$\partial A$$ is the boundary of $$A$$. $$A^{o}$$ denotes the set of interior points of $$A$$.
Show that $$\overline{A} \subseteq A^{o} \cup \partial A$$.
Suppose $$x \in \overline{A} \implies x \in A \cup A'$$.
Suppose $$x \not\in A$$ (What if $$x\in A$$), then $$x\in A'$$.
$$\implies$$ for all open sets $$U$$ containing $$x$$, $$U$$ has another point in $$A$$ not equal to $$x$$.
$$\implies U \cap A \neq \varnothing$$ and $$U \cap (X\setminus A) \neq \varnothing$$.
$$\implies x \in \partial A$$.
This proof is not complete; what if $$x\in A$$? Can someone help me out?
• Use this: $x \in \overline{A}$ if and only if every nbhd of $x$ contains a point of $A$. Now, if $x$ is not in the interior, then... – Randall Oct 23 '18 at 3:05
• then what follows? – Niang Moore Oct 23 '18 at 3:46
• Limit points are making this an awkward proof. Why is it so recklessly important to teach that limit point thing. It is rarely needed in topology. – William Elliot Oct 23 '18 at 4:10
• What definition of the boundary of A are you using? If you use a common definition, $\partial A$ = $\overline{A}$/$A^{o}$, the proof practically writes itself. – Steve B Oct 23 '18 at 5:02
• This looks like a problem you have collected from / inspired by some source. According to recent discussions in Meta, we are looking forward to including sources for all applicable questions. Can you provide the source by editing the question?Refer-math.meta.stackexchange.com/questions/29290/… – tatan Oct 23 '18 at 6:16
Let $$A^c$$ denote the complement of $$A.$$
The definition of $$\partial A$$ is $$\bar A \cap \overline {A^c}.$$
If $$p\in \bar A$$ then....
(i) If there exists an open set $$U$$ such that $$p\in U$$ and $$U\cap A^c=\emptyset$$ then $$U\subset A$$ so $$p\in U\subset A^o$$ so $$p\in A^o$$.
(ii) If every open $$U$$ such that $$p\in U$$ has non-empty intersection with $$A^c$$ then $$p\in \overline {A^c}$$ so $$p\in \bar A\cap \bar {A^c}=\partial A.$$
In both cases we have $$p\in A^o\cup \partial A.$$
Therefore $$\bar A\subset A^o\cup \partial A.$$
BTW, since $$A^o\cup \partial A\subset A\cup \partial A=A\cup (\bar A \cap \overline {A^c})\subset$$ $$\subset A\cup \bar A=\bar A$$ we also have $$A^o\cup \partial A \subset \bar A.$$ So we have $$\bar A=A^o\cup \partial A.$$
All of this holds in every topological space.
• This Q, or variant relatives of it, keep appearing on this site. The same can be said for many other Q's. Not surprising, as new students keep appearing also. Some day someone might take this site and produce a fully cross-indexed compendium of such Q's & A's. – DanielWainfleet Oct 25 '18 at 3:34
$$\partial A = \bar A - A^o.$$ Thus
$$A^o \cup \partial A = \bar A$$ because
$$A^o \subseteq \bar A.$$
In fact, because $$A^o \subseteq A \subseteq \bar A,$$
$$A \cup \partial A = \bar A.$$
If $$x \in A \cup A'$$ then suppose $$x \notin A^\circ$$, then for all open neighbourhoods $$B$$ (or balls around $$x$$ if you prefer) of $$x$$ we have that $$B \nsubseteq A$$, so for all such $$B$$: $$B \cap (X\setminus A) \neq \emptyset$$. Both $$x \in A$$ and $$x \in A'$$ imply that also for all such $$B$$: $$B \cap A \neq \emptyset$$. So combined this means that $$x \in \partial A$$: all its neighbourhoods/balls intersect $$A$$ and its complement.
This shows $$\overline{A} \subseteq A^\circ \cup \partial A$$.
• what if x is not a boundary point of A, can you show it's the interior point of A – Niang Moore Oct 24 '18 at 1:43
• @NiangMoore you can yes, but I found this way more convenient: not a boundary point gives two cases to consider and not in the interior is easier. – Henno Brandsma Oct 24 '18 at 4:30 | 2019-10-23T11:08:54 | {
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http://mathhelpforum.com/calculus/171108-determining-length-logarithmic-spiral.html | # Math Help - Determining the length of a logarithmic spiral.
1. ## Determining the length of a logarithmic spiral.
Unfortunately I couldnt convert the pdf into a picture so that I could upload it but i will try to explain my problem as briefly i can
here is a pdf link btw:
<a target="_blank" title="mathhelplength (pdf)" href="http://pdfcast.org/pdf/mathhelplengt...hhelplength</a>
Right so I am faced with the following problem
"Determine the length of the lograrithmic spiral": x = e^-t * cos t y = e^-t * sin t
0<= t <= 2pi
I recal a formula
l(length symbol) = \$(integral from a to b) SQRT [x'(t)]^2 + [y'(t)}^ dt
I don't know how to evaluate this, could someone show me?
EDIT: New link mathhelplength - PDFCast.org
2. Originally Posted by Riazy
"Determine the length of the lograrithmic spiral":
x = e^-t * cos t
y = e^-t * sin t 0<= t <= 2pi
$x' = -e^{-t}(\sin{t}+\cos{t})$
$(x')^2 = e^{-2t}(\sin{t}+\cos{t})^2$
$y' = e^{-t}(\cos{t}-\sin{t})$
$(y')^2 = e^{-2t}(\cos{t}-\sin{t})^2$
$(x')^2 + (y')^2 = 2e^{-2t}$
$\displaystyle L = \int_0^{2\pi} \sqrt{2e^{-2t}} \, dt$
finish it.
3. Sorry, but the my problem is really how you added together (x')^2+(y')^2 = 2e^-2t
I would like to see that step in detail, because thats what I dont understand, i.e how they are added together to 2e^-2t
THanks alot
4. Hello, Riazy!
This requires some very careful work . . . are you up for it?
$\text{Determine the length of the lograrithmic spiral:}$
. . $\begin{Bmatrix}x &=& e^{\text{-}t}\cos t \\ y &=& e^{\text{-}t}\sin t \end{Bmatrix}\;\;0\,\le\, t\,\le\,2\pi$
$\displaystyle \text{Formula: }\;L \;=\; \int^b_a \sqrt{[x'(t)]^2 + [y'(t)]^2}\, dt$
$\text{Derivatives: }\;\begin{Bmatrix}x'(t) &=& \text{-}e^{\text{-}t}\sin t - e^{\text{-}t}\cos t &=& \text{-}e^{\text{-}t}(\sin t + \cos t) \\ \\[-3mm] y'(t) &=& e^{\text{-}t}\cos t - e^{\text{-}t}\sin t &=& e^{\text{-}t}(\cos t - \sin t ) \end{Bmatrix}$
$\text{Squares: }\;\begin{Bmatrix}[x'(t)]^2 &=& e^{\text{-}2t}(\sin^2\!t + 2\sin t\cos t + \cos^2\!t) \\ \\[-3mm] [y'(t)]^2 &=& e^{\text{-}2t}(\cos^2\!t - 2\sin t\cos t + \sin^2\!t) \end{Bmatrix}$
$\text{Add: }\;[x'(t)]^2 + [y'(t)]^2 \;=\;e^{\text{-}2t}\bigg[(\underbrace{\sin^2\!t + \cos^2\!t)}_{\text{This is 1}} + \underbrace{(\cos^2\!t + \sin^2\!t)}_{\text{This is 1}}\bigg]$
. . . . . . . . . . . . . . . . . $=\;2e^{\text{-}2t}$
$\text{Hence: }\;\sqrt{[x'(t)]^2 + [y'(t)]^2} \;=\;\sqrt{2e^{\text{-}2t}} \;=\;\sqrt{2}\,e^{\text{-}t}$
$\displaystyle \text{Therefore: }\;L \;=\;\sqrt{2}\!\int^{2\pi}_0 \!\!e^{\text{-}t}\,dt$
Got it?
5. $(x')^2 = e^{-2t}(\sin{t}+\cos{t})^2$
$(x')^2 = e^{-2t}(\sin^2{t}+2\sin{t}\cos{t}+\cos^2{t})$
$(x')^2 = e^{-2t}(1+2\sin{t}\cos{t})$
$(y')^2 = e^{-2t}(\cos{t}-\sin{t})^2$
$(y')^2 = e^{-2t}(\cos^2{t}-2\cos{t}\sin{t} + \sin2{t})$
$(y')^2 = e^{-2t}(1-2\cos{t}\sin{t})$
$(x')^2 + (y')^2 = e^{-2t}(1+2\sin{t}\cos{t}) + e^{-2t}(1-2\cos{t}\sin{t}) = e^{-2t}[(1+2\sin{t}\cos{t}) + (1-2\sin{t}\cos{t})]$
can you finish it now? | 2014-03-13T15:59:05 | {
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https://stat.ethz.ch/R-manual/R-devel/library/stats/html/Uniform.html | Uniform {stats} R Documentation
## The Uniform Distribution
### Description
These functions provide information about the uniform distribution on the interval from min to max. dunif gives the density, punif gives the distribution function qunif gives the quantile function and runif generates random deviates.
### Usage
dunif(x, min = 0, max = 1, log = FALSE)
punif(q, min = 0, max = 1, lower.tail = TRUE, log.p = FALSE)
qunif(p, min = 0, max = 1, lower.tail = TRUE, log.p = FALSE)
runif(n, min = 0, max = 1)
### Arguments
x, q vector of quantiles. p vector of probabilities. n number of observations. If length(n) > 1, the length is taken to be the number required. min, max lower and upper limits of the distribution. Must be finite. log, log.p logical; if TRUE, probabilities p are given as log(p). lower.tail logical; if TRUE (default), probabilities are P[X \le x], otherwise, P[X > x].
### Details
If min or max are not specified they assume the default values of 0 and 1 respectively.
The uniform distribution has density
f(x) = \frac{1}{max-min}
for min \le x \le max.
For the case of u := min == max, the limit case of X \equiv u is assumed, although there is no density in that case and dunif will return NaN (the error condition).
runif will not generate either of the extreme values unless max = min or max-min is small compared to min, and in particular not for the default arguments.
### Value
dunif gives the density, punif gives the distribution function, qunif gives the quantile function, and runif generates random deviates.
The length of the result is determined by n for runif, and is the maximum of the lengths of the numerical arguments for the other functions.
The numerical arguments other than n are recycled to the length of the result. Only the first elements of the logical arguments are used.
### Note
The characteristics of output from pseudo-random number generators (such as precision and periodicity) vary widely. See .Random.seed for more information on R's random number generation algorithms.
### References
Becker, R. A., Chambers, J. M. and Wilks, A. R. (1988) The New S Language. Wadsworth & Brooks/Cole.
RNG about random number generation in R.
Distributions for other standard distributions.
### Examples
u <- runif(20)
## The following relations always hold :
punif(u) == u
dunif(u) == 1
var(runif(10000)) #- ~ = 1/12 = .08333
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https://scicomp.stackexchange.com/questions/11322/fourier-transform-of-function-in-spherical-harmonics | # Fourier Transform of function in Spherical Harmonics
I have a function $f(r,\theta,\phi)$ which I am expressing in terms of spherical harmonics
$$f(r,\theta,\phi) = \sum_{l=0}^{\infty} \sum_{m=-l}^{l} g_{l,m}(r) d_{l,m}(\theta,\phi)$$
where $d_{l,m}$ are real spherical harmonics (RSH - formula below taken from Wikipedia).
\begin{align} d_{\ell m} &= \begin{cases} \displaystyle {i \over \sqrt{2}} \left(Y_\ell^{m} - (-1)^m\, Y_\ell^{-m}\right) & \text{if}\ m<0\\ \displaystyle Y_\ell^0 & \text{if}\ m=0\\ \displaystyle {1 \over \sqrt{2}} \left(Y_\ell^{-m} + (-1)^m\, Y_\ell^{m}\right) & \text{if}\ m>0. \end{cases}\\ &= \begin{cases} \displaystyle {i \over \sqrt{2}} \left(Y_\ell^{-|m|} - (-1)^{m}\, Y_\ell^{|m|}\right) & \text{if}\ m<0\\ \displaystyle Y_\ell^0 & \text{if}\ m=0\\ \displaystyle {1 \over \sqrt{2}} \left(Y_\ell^{-|m|} + (-1)^{m}\, Y_\ell^{|m|}\right) & \text{if}\ m>0. \end{cases} \end{align}
If I would like to take the Fourier Transform (FT) of $f(r,\theta,\phi)$, is there a special relationship between $d_{l,m}$ and the FT? I feel like there is some nice property like
$$FT(f(r,\theta,\phi)) = \sum \sum FT(g_{l,m}(r)) d_{l,m}(\theta, \phi)$$
but I can't remember it or find it. Is there a nice relationship between the FT and RSH which will make the FT of $f(r,\theta,\phi)$ easy to compute when expanded in RSH?
• presumably, the coefficient functions $f(r)$ are intended to be different from $f(r,\theta,\phi)$? Apr 9 '14 at 3:16
• Yes. I have edited the one dimensional function to be $f_{lm}$ Apr 9 '14 at 3:19
• That doesn't really help since it's the repeated use of $f$ on both sides of the equals sign that's confusing! Apr 9 '14 at 3:21
• Ok, I will change $f_{lm}$ to $g_{lm}$ to make that distinction more clear. They cannot be the same thing since $f(r,\theta,\phi)$ is defined on $R_3 \rightarrow R$ and $f_{lm}(r)$ is defined on $R \rightarrow R$ Apr 9 '14 at 3:49
• No, I would like to the FT to be performed in all 3 dimensions. I am just wondering if there is a nice trick which the SHs can help with. Apr 9 '14 at 12:02
You may find the expansion of a plane wave in spherical waves to be helpful here:
$$e^{i\mathbf{k}\cdot\mathbf{x}} = 4\pi\sum_{l=0}^\infty\sum_{m=-l}^l i^lj_l(kr)Y_{lm}(\theta,\phi)Y_{lm}^*(\vartheta,\varphi)$$
where $\theta$, $\phi$ are the angular variables for $\mathbf{x}$ and $\vartheta$, $\varphi$ for $\mathbf{k}$; the radial functions $j_l$ are the spherical Bessel functions; and $Y_{lm}$ are the spherical harmonics*. In that case,
$$\mathscr{F}f(\mathbf{k}) = (2\pi)^{-3/2}\int d^3x\hspace{2pt} e^{i\mathbf{k}\cdot\mathbf{x}}f(\mathbf{x})$$ $$= (2\pi)^{-3/2}\int d^3x\left(4\pi\sum_{l,m}i^lj_l(kr)Y_{lm}(\theta,\phi)Y_{lm}^*(\vartheta,\varphi)\right)\left(\sum_{l',m'}g_{l'm'}(r)Y_{l'm'}(\theta,\phi)\right)$$
Then you can use the orthogonality of spherical harmonics to reduce this to
$$\mathscr{F}f(k,\vartheta,\varphi) = \sqrt{\frac{2}{\pi}}\sum_{l,m}i^lY_{lm}^*(\vartheta,\varphi) \cdot \int j_l(kr)g_{lm}(r)r^2dr.$$
You now have an integral in only the radial dimension. If you have a book of special functions handy, some recurrence relations or asymptotics for $j_l$ will prove invaluable no matter how you choose to evaluate those integrals.
${}$
*Every time you use this formula, a kitten is born, a child smiles, and your lifespan increases by 8 minutes.
• This seems to be exactly what I am searching for. I have edited my original question with a specific definition for the Real Spherical Harmonics I am using - does your formula work for RSH? In which case, will $Y_{lm}^*$ simply be $d_{lm}^*$? Apr 17 '14 at 16:45
• You can write each $Y_{lm}$ in terms of $d_{lm}$ by inverting the formula you cited. However, note that the Fourier transform of a real function is not necessarily real. Instead, it has the symmetry property $\mathscr{F}f(\mathbf{k})^* = \mathscr{F}f(-\mathbf{k})$, which means some harmonics are zero in the expansion of the Fourier transform. Apr 17 '14 at 17:33
• if anyone else is interested, the Addition Theorem which is used states that it is true for both real and complex harmonics, so $d_lm$ can be interchanged with $Y_lm$ without a problem. Apr 17 '14 at 20:28
The FT of a three dimensional real function is not in general real, so there is no way you could express it in real spherical harmonics. I think however that if you only include harmonics with even $l$ the result should be real.
The FT of a spherically symmetric function $g(r)$ is $S(Q) = \int_{0}^{\infty} \frac{r}{Q} \sin(Qr) g(r)\,\mathrm{d}r$ (with some factors of $2\pi$ depending on how you define the FT).
A 3D function decomposed into spherical harmonics is a sum of products $g_{lm}(r) d_{lm}(\theta,\phi)$, so the FT will be a sum of convolutions $S(Q) \otimes \mathrm{FT} \left[ d_{lm}(\theta,\phi) \right]$. I can't think of an easy way to calculate these in general. | 2021-10-28T07:07:25 | {
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https://math.stackexchange.com/questions/3034654/arranging-n-balls-in-k-bins-so-that-m-consecutive-bins-are-empty | # Arranging $n$ balls in $k$ bins so that $m$ consecutive bins are empty
This question is inspired by the following problem:
Randomly place seven balls into ten bins, with no bin containing more than one ball. What is the probability that there will be (at least) two consecutive bins that are empty?
Note that this is not a circular arrangement, although that could make for an interesting (if simpler) question. The actual question I have is the same as the one above, but with the numbers written in emphasized text generalized as positive integers $$n$$, $$k$$, and $$m$$, respectively.
Question:
Randomly place $$n$$ balls into $$k$$ bins, with no bin containing more than one ball. What is the probability that there will be (at least) $$m$$ consecutive bins that are empty?
I suspect that this is a known counting problem (as framed, or through some equivalent rephrase), so I include a reference-request tag. An original solution would, of course, be welcome as well!
As in an already included answer: Specific cases (e.g., $$m=2$$) would be helpful, too.
• This sounds like a pretty straightforward counting problem if you use inclusion-exclusion. However, it doesn't seem to me that there is a nice general closed form expression. – platty Dec 10 '18 at 23:40
• @platty I would appreciate a "straightforward counting" answer! – Benjamin Dickman Dec 10 '18 at 23:40
• On second thought, it does seem more complicated. The question I was familiar with assumed a bound $m \geq n/2$, from which straightforward PIE suffices. But in the general case, it doesn't seem like a nice closed form should be reasonable -- there are too many cases of differently-sized intersections to handle. – platty Dec 10 '18 at 23:58
We show the probability $$p_m(k,n)$$ to randomly place $$n$$ balls into $$k$$ bins with no bin containing more than one ball, so that at least $$m$$ consecutive bins are empty is \begin{align*} \color{blue}{p_m(k,n)=\binom{k}{n}^{-1}\sum_{j=1}^{\lfloor k/m\rfloor}\binom{k-mj}{n}\binom{n+1}{j}(-1)^{j+1}}\tag{0} \end{align*}
In order to do so we consider the binary alphabet $$V=\{0,1\}$$ and decode empty bins with $$0$$ and bins where a ball is placed with $$1$$.
• We are looking for the number $$g_m(k,n)$$ of binary words of length $$k$$ which contain $$n$$ $$1$$'s and runs of $$0$$ with length at most $$m-1$$. The number of wanted words is \begin{align*} f_m(k,n)=\binom{k}{n}-g_m(k,n) \end{align*}
• Since there is a total of $$\binom{k}{n}$$ binary words of length $$k$$ with $$n$$ $$1$$'s, the probability $$p_m(k,n)$$ is \begin{align*} p_m(k,n)=\binom{k}{n}^{-1}f_m(k,n) \end{align*}
Words with no consecutive equal characters at all are called Smirnov or Carlitz words. See example III.24 Smirnov words from Analytic Combinatorics by Philippe Flajolet and Robert Sedgewick for more information.
A generating function for the number of Smirnov words over a binary alphabet is given by \begin{align*} \left(1-\frac{2z}{1+z}\right)^{-1}\tag{1} \end{align*}
We replace occurrences of $$0$$ in a Smirnov word by one up to $$m-1$$ zeros to generate strings having runs of $$0$$ with length less than $$m$$. \begin{align*}\ z\longrightarrow z+z^2+\cdots+z^{m-1}=\frac{z\left(1-z^{m-1}\right)}{1-z}\tag{2} \end{align*}
Since there are no restrictions to the length of runs of $$1$$'s we replace occurrences of $$1$$'s in a Smirnov word by one or more $$1$$s. \begin{align*}\ z\longrightarrow z+z^2+\cdots=\frac{z}{1-z}\tag{3} \end{align*}
We substitue (2) and (3) in (1). Since we want to keep track of the number of zeros we put $$tz$$ instead of $$z$$ in (2). We obtain
\begin{align*} \left(1- \frac{\frac{tz\left(1-(tz)^{m-1}\right)}{1-tz}}{1+\frac{tz\left(1-(tz)^{m-1}\right)}{1-tz}}-\frac{\frac{z}{1-z}}{1+\frac{z}{1-z}}\right)^{-1} &=\frac{1-(tz)^m}{1-z(1+t-(tz)^{m})}\tag{4} \end{align*}
Denoting with $$[z^n]$$ the coefficient of $$z^n$$ in a series we obtain from (4) the number of wanted words as \begin{align*} \color{blue}{f_m(k,n)}&=\binom{k}{n}-g_m(k,n)\\ &\color{blue}{=\binom{k}{n}-[z^kt^{k-n}]\frac{1-(tz)^m}{1-z(1+t-(tz)^{m})})}\tag{5} \end{align*}
Example: Let's look at OPs example. We set $$k=10,n=7$$ and $$m=2$$ and we obtain with some help of Wolfram Alpha \begin{align*} \color{blue}{f_2(10,7)}&=\binom{10}{7}-[z^{10}t^3]\frac{1-(tz)^2}{1-z(1+t-(tz)^{2})}\\ &=120-[z^{10}t^3]\left(1 + (t + 1) z + (2 t + 1) z^2\right.\\ &\qquad\left. + \cdots + (6 t^5 + 35 t^4 + \color{blue}{56} t^3 + 36 t^2 + 10 t + 1) z^{10} + \cdots\right)\\ &=120-56\\ &\,\,\color{blue}{=64} \end{align*} with probability
\begin{align*} \color{blue}{p_2(10,7)}=\binom{10}{7}^{-1}f_2(10,7)=\frac{64}{120}\color{blue}{=\frac{8}{15}=0.5\dot{3}} \end{align*}
The blue colored coefficient of $$z^{10}$$ shows there are $$\color{blue}{56}$$ binary words of length $$10$$ with $$3$$ zeros and runs of $$0$$ with length less than $$k=2$$. The blue colored result $$\color{blue}{64}$$ is the number of valid words having runs of zeros of length at least $$2$$.
These $$64$$ words are $$\begin{array}{cccc} \color{blue}{00}01111111&1\color{blue}{00}0111111&11\color{blue}{00}011111&111\color{blue}{00}01111\\ \color{blue}{00}10111111&1\color{blue}{00}1011111&11\color{blue}{00}101111&111\color{blue}{00}10111\\ \color{blue}{00}11011111&1\color{blue}{00}1101111&11\color{blue}{00}110111&111\color{blue}{00}11011\\ \color{blue}{00}11101111&1\color{blue}{00}1110111&11\color{blue}{00}111011&111\color{blue}{00}11101\\ \color{blue}{00}11110111&1\color{blue}{00}1111011&11\color{blue}{00}111101&111\color{blue}{00}11110\\ \color{blue}{00}11111011&1\color{blue}{00}1111101&11\color{blue}{00}111110&11101\color{blue}{00}111\\ \color{blue}{00}11111101&1\color{blue}{00}1111110&1101\color{blue}{00}1111&111011\color{blue}{00}11\\ \color{blue}{00}11111110&101\color{blue}{00}11111&11011\color{blue}{00}111&1110111\color{blue}{00}1\\ 01\color{blue}{00}111111&1011\color{blue}{00}1111&110111\color{blue}{00}11&11101111\color{blue}{00}\\ 011\color{blue}{00}11111&10111\color{blue}{00}111&1101111\color{blue}{00}1&\\ 0111\color{blue}{00}1111&101111\color{blue}{00}11&11011111\color{blue}{00}&\\ 01111\color{blue}{00}111&1011111\color{blue}{00}1&&\\ 011111\color{blue}{00}11&10111111\color{blue}{00}&&\\ 0111111\color{blue}{00}1&&&\\ 01111111\color{blue}{00}&&&\\ \\ 1111\color{blue}{00}0111&11111\color{blue}{00}011&111111\color{blue}{00}01&1111111\color{blue}{00}0\\ 1111\color{blue}{00}1011&11111\color{blue}{00}101&111111\color{blue}{00}10\\ 1111\color{blue}{00}1101&11111\color{blue}{00}110&11111101\color{blue}{00}\\ 1111\color{blue}{00}1110&1111101\color{blue}{00}1&\\ 111101\color{blue}{00}11&11111011\color{blue}{00}&\\ 1111011\color{blue}{00}1&&\\ 11110111\color{blue}{00}&&\\ &&\\ &&\\ \end{array}$$
Formula of $$f_m(k,n)$$:
We derive a formula of $$f_m(k,n)$$ from (5) manually. It is convenient to use the coefficient of operator $$[z^k]$$ to denote the coefficient of $$z^k$$ in a series.
We obtain from (5) \begin{align*} \color{blue}{f_m(k,n)}&=\binom{k}{n}-[z^kt^{k-n}]\frac{1-(tz)^m}{1-z(1+t-(tz)^m}\\ &=\binom{k}{n}-[z^kt^{k-n}]\sum_{j=0}^\infty z^j\left(1+t-(tz)^{m}\right)^j\left(1-(tz)^m\right)\tag{6}\\ &=\binom{k}{n}-[t^{k-n}]\sum_{j=0}^k [z^{k-j}]\left(1+t-(tz)^{m}\right)^j\left(1-(tz)^m\right)\tag{7}\\ &=\binom{k}{n}-[t^{k-n}]\sum_{j=0}^k [z^j]\left(1+t-(tz)^{m}\right)^{k-j}\left(1-(tz)^m\right)\tag{8}\\ &=-[t^{k-n}]\sum_{j=1}^{\lfloor k/m\rfloor}[z^{mj}]\left(1+t-(tz)^{m}\right)^{k-mj}\left(1-(tz)^m\right)\tag{9}\\ &=-[t^{k-n}]\sum_{j=1}^{\lfloor k/m\rfloor}[z^{mj}]\sum_{l=0}^{k-mj}\binom{k-mj}{l}\left(1-(tz)^m\right)^{l+1}t^{k-mj-l}\tag{10}\\ &=-[t^{k-n}]\sum_{j=1}^{\lfloor k/m\rfloor}[z^{mj}]\sum_{l=0}^{k-mj}\binom{k-mj}{l}\sum_{u=0}^{l+1}\binom{l+1}{u}(-1)^{u}(tz)^{mu}t^{k-mj-l}\tag{11}\\ &=[t^{k-n}]\sum_{j=1}^{\lfloor k/m\rfloor}\sum_{l=0}^{k-mj}\binom{k-mj}{l}\binom{l+1}{j}(-1)^{j+1}t^{k-l}\tag{12}\\ &\,\,\color{blue}{=\sum_{j=1}^{\lfloor k/m\rfloor}\binom{k-mj}{n}\binom{n+1}{j}(-1)^{j+1}}\tag{13}\\ \end{align*} with probability \begin{align*} \color{blue}{p_m(k,n)=\binom{k}{n}^{-1}f_m(k,n)} \end{align*} in accordance with (0).
Comment:
• In (6) we do a geometric series expansion.
• In (7) we apply the rule $$[z^{p-q}]A(z)=[z^p]z^qA(z)$$. We also set the upper limit to $$k$$, since other terms do not contribute.
• In (8) we change the order of summation $$j\to k-j$$.
• In (9) we sum over multiples of $$m$$ only by setting $$j\to mj$$ since we have $$z^{m}$$ in the binomial expansion. We also note that $$\binom{k}{n}$$ is canceled with the first summand $$j=0$$ since $$[z^0t^{k-n}](1+t-(tz)^m)^k(1-(tz)^m)=[t^{k-n}](1+t)^k=\binom{k}{k-n}=\binom{k}{n}$$.
• In (10) we expand the binomial.
• In (11) we expand the binomial.
• In (12) we select the coefficient of $$z^{mj}$$ by selecting the summand with $$u=j$$.
• In (13) we select the coefficient of $$t^{k-n}$$ by selecting the summand with $$l=n$$.
Example revisited:
By calculating OPs example ($$k=10,n=7,m=2$$) now with formula (13) we obtain \begin{align*} \color{blue}{f_2(10,7)}&=\sum_{j=1}^5\binom{10-2j}{7}\binom{8}{j}(-1)^{j+1}\\ &=\binom{8}{7}\binom{8}{1}\\ &\,\,\color{blue}{=64} \end{align*} in accordance with the result above.
Note that here we only take the summand $$j=1$$, since other terms have lower index $$7$$ greater than the upper index $$10-2j$$, so that the binomial coefficients are zero.
• This is great; thank you! Also: Within the first indented sentence I had guessed correctly who the answerer was. :) – Benjamin Dickman Dec 22 '18 at 19:25
• @BenjaminDickman: You're welcome and many thanks for your amusing remark. :-) – Markus Scheuer Dec 22 '18 at 19:48
• @BenjaminDickman: I've added a derivation of an explicit formula to complete the job. – Markus Scheuer Dec 23 '18 at 8:19
Denote the sizes of the runs of empty bins as $$x_1, \ldots, x_{n + 1}$$. (Note that there are at most this many runs, so runs may be empty; i.e., $$x_j = 0$$.) Now, count the number of nonnegative integer solutions to
$$x_1 + \cdots + x_{n + 1} = k - n$$
that have at least one $$x_j \ge m$$. This is equivalent to solving the problem.
To count this quantity, use the facts that:
1. The number of nonnegative integer solutions to $$x_1 + \cdots + x_k = n$$ is $$\binom{n + k - 1}{k - 1}\text.$$ This is a simple stars and bars argument.
2. The number of nonnegative integer solutions to $$x_1 + \cdots + x_k = n$$ with each $$x_j < c$$ is $$\binom{n + k - 1}{k - 1} - \sum_{i = 1}^{\lfloor n / {c} \rfloor} {(-1)}^{i + 1} \binom{k}{i}\binom{n - c i + k - 1}{k - 1}\text.\tag{*}\label{*}$$ To show (\ref{*}), recall the complementary form of the inclusion–exclusion principle. Namely, we can count the number of nonnegative integer solutions with each $$x_j < c$$ as \begin{align} && \text{the number of all (unrestricted) solutions} \\ &-& \text{the number of solutions with one } x_j \geq c \\ &+& \text{the number of solutions with two } x_j \geq c \\ &-& \cdots \\ &{(-1)}^{\lfloor n / {c} \rfloor}& \text{the number of solutions with }\lfloor n / {c} \rfloor \text{ } x_j \geq c\text. \end{align} Now, how do you count the number of solutions that have $$i$$-many $$x_j \geq c$$? It's quite simple; write those $$x_j$$ as $$c + y_j$$ (since we already know they're greater than or equal to $$c$$), and you already have a reduction to the stars and bars argument, but with $$n' = n - c i$$. This tells us that the number of solutions with $$i$$-many $$x_j \geq c$$ is $$\binom{n - c i + k - 1}{k - 1}$$. The $$\binom{k}{i}$$ factor counts the number of ways to have this scenario.
• Nice answer! I think the number of variables should be $n+1$, rather than $k-n$. The variables are the gaps between the balls, and there are $n$ balls, so $n+1$ gaps. – Mike Earnest Dec 11 '18 at 14:36
• Thank you; you are right. I corrected the answer and checked that it agrees with the one for $m = 2$ (math.stackexchange.com/a/3035312/112944). – d125q Dec 11 '18 at 14:44
• Could you say something about the second fact? The first one looks to me like stars and bars, but is the second one... stars and bars combined with the inclusion-exclusion principle? If you could indicate how this works in practice for, say, $n = 6$, $k = 10$, and $m = 2$, then that would be very helpful, too. (And could compared with @MikeEarnest's answer, as well as the hand computations I've done.) Thanks! – Benjamin Dickman Dec 11 '18 at 16:21
• @BenjaminDickman Give the inclusion exclusion a try! The idea is this; you take all of the solutions, then for each variable subtract the cases where the variable is “bad”, meaning more than $m$, then add back in the cases where two variables are bad, etc. – Mike Earnest Dec 11 '18 at 16:45
• @BenjaminDickman I added some "proof sketches." – d125q Dec 20 '18 at 15:48
This is supposed to be an integration to d125q's answer, as to better explain how to reach to the formula he gave, but it is too long for a comment.
You are placing $$0$$ or $$1$$ ball in every bin.
We are clearly speaking of undistiguishable balls.
While, speaking of consecutive bins, that implies that the bins be distinguishable (i.e. labelled, i.e. aligned in a row).
Then your problem is equivalent to :
given all the binary strings of length $$k$$, with $$n$$ ones (and $$n-k$$ zeros), how many of them will contain one or more runs of consecutive zeros whose length is $$m$$ or greater ?
The number of binary strings of length $$k$$ and with $$n$$ ones is $$\binom{k}{n}$$.
So the complement of the above will be the number of strings where the length of each run of consecutive zeros is less than $$m$$, and naturally, in a binary string we can exchange the zeros with the ones.
Now, to keep congruence with the precedent posts I am going to cite, let me change the formulation and the parameters of the problem with the following:
Consider a binary string with $$s$$ "$$1$$"'s and $$m$$ "$$0$$"'s in total. Let's put an additional (dummy) fixed $$0$$ at the start and at the end of the string. We individuate as a run the consecutive $$1$$'s between two zeros, thereby including runs of null length. With this scheme we have a fixed number of $$m+1$$ runs.
(refer also to the similar post).
Then, imposing that each run does not contain more than $$r$$ ones is equivalent to compute $$N_{\,b} (s,r,m+1) = \text{No}\text{. of solutions to}\;\left\{ \begin{gathered} 0 \leqslant \text{integer }x_{\,j} \leqslant r \hfill \\ x_{\,1} + x_{\,2} + \cdots + x_{\,m+1} = s \hfill \\ \end{gathered} \right.$$ which as explained in this other post is expressed as $$N_b (s,r,m + 1)\quad \left| {\;0 \leqslant \text{integers }s,m,r} \right.\quad = \sum\limits_{\left( {0\, \leqslant } \right)\,\,k\,\,\left( { \leqslant \,\frac{s} {r}\, \leqslant \,m + 1} \right)} {\left( { - 1} \right)^k \left( \begin{gathered} m + 1 \\ k \\ \end{gathered} \right)\left( \begin{gathered} s + m - k\left( {r + 1} \right) \\ s - k\left( {r + 1} \right) \\ \end{gathered} \right)}$$ Note that by rewriting the binomial in the way shown above render the summation bounds implicit in the binomial, so that they can be omitted (that's why they are indicated in brackets), which is of great advantage in performing algebraic manipulations.
I think that an effective way to understand the above is by developing the polynomial \eqalign{ & \underbrace {\left( {1 + x + x^{\,2} + \cdots + x^{\,r} } \right) \cdot \left( {1 + \cdots + x^{\,r} } \right) \cdot \; \cdots \; \cdot \left( {1 + \cdots + x^{\,r} } \right)}_{m\;times} = \cr & = \left( {{{1 - x^{\,r + 1} } \over {1 - x}}} \right) = \sum\limits_{0\, \le \,\,s\,\,\, \le \,m\,r} {N_b (s,r,m)\;x^{\,s} } \cr} thereby having the o.g.f. in $$s$$, and from there easily getting the recursion $$N_{\,b} (s,r,m + 1) = \sum\limits_{i\, = \,0}^r {N_{\,b} (s - i,r,m)}$$
Finally note that the $$N_b$$ are called "r-nomial coefficients" in OEIS: see e.g. Table A008287.
• @BenjaminDickman good that the o.g.f. is explicative. Sorry for the broken links: adjusted. – G Cab Dec 19 '18 at 17:06
There's a nice way to do this if $$m = 2.$$ I describe it here, although it does not solve the more general problem indicated in the question.
We can get a classic "random" arrangement if the bins are identifiable and the balls are not distinguishable, so I'll make those assumptions.
In each bin we either place one ball or no balls. We choose $$n$$ of the bins in which to place the $$n$$ balls, leaving $$n - k$$ empty. The number of ways to do this is $$\binom k{k-n} = \binom kn,$$ each way equally probable.
Another way to think about those same arrangements of balls and bins is that we have $$k - n$$ empty bins, with a run of zero or more full bins before the first empty bin, after the last empty bin, and between each consecutive pair of empty bins. That is, we are putting $$n$$ balls into $$k - n + 1$$ "super bins", where each "super bin" is a run of zero or more filled bins.
As a check on the last interpretation of the problem, the number of ways to put zero or more indistinguishable items into each of $$k - n + 1$$ containers, placing a total of $$n$$ items, is $$\binom{n + (k - n + 1) - 1}{(k - n + 1) - 1} = \binom k{k - n}.$$
So that is the number of equally probable events in our sample space. Now let's count the number of these events that do not have two consecutive empty bins. That means that the $$k - n - 1$$ runs of full bins that lie between consecutive pairs of empty bins must each include at least one full bin. Having determined that we need $$k - n - 1$$ balls placed in this way, we see that there are no such arrangements if $$n < k - n - 1,$$ that is, if $$2n - k + 1 < 0.$$ On the other hand, if $$n + 1\geq k$$ there are not even two empty bins anywhere and so it is guaranteed that we will not have two adjacent empty bins. But if $$2n - k + 1 \geq 0$$ and $$n + 1 < k,$$ then the number of ways to place the remaining $$2n - k + 1$$ full bins into the $$k - n + 1$$ runs (including the ones on each end) is $$\binom{(2n - k + 1) + (k - n + 1) - 1}{(k - n + 1) - 1} = \binom{n + 1}{k - n}.$$
So the probability of two consecutive empty bins is $$0$$ if $$n + 1 \geq k$$; if $$2n - k + 1 < 0$$ the probability is $$1$$; and in all other cases the probability is $$1 - \frac{\displaystyle\binom{n + 1}{k - n}}{\displaystyle\binom k{k - n}} = 1 - \frac{(n + 1)! \,n!}{(2n - k + 1)!\, k!}.$$
There is the possibility to calculate numerically the result with a rather simple recursive equation.
We will assume $$m = 2$$ in the following.
To simplify the notation, we will consider that $$0$$ represents the case "no ball" and $$1$$ the case "with a ball".
We note $$f(k, n)$$ the probability associated with $$k$$ and $$n$$.
We represent the set of possibilities as a graph.
We consider three cases for the first bins:
• 1 in the first bin. Probability $$\frac{n}{k}$$. The subgraph corresponds to $$f(k-1, n-1)$$
• 01 in the first 2 bins: Probability $$\frac{k-n}{k} \times \frac{n}{k-1}$$. The subgraph corresponds to $$f(k-2, n-1)$$
• 00 in the first 2 bins: Probability $$\frac{k-n}{k} \times \frac{k-1-n}{k-1}$$ -> we get the $$m$$ holes.
Then, we obtain, for $$m = 2$$
$$f(k,n) = \frac{n}{k} f(k-1, n-1) + \frac{n(k-n)}{k(k-1)} f(k-2, n-1) + \frac{(k-n)(k-1-n)}{k(k-1)}$$
We note in addition that $$f() = 0$$ if $$k-n \le m-1$$ and that $$f() = 1$$ if $$n=0$$ and $$k \ge m$$
This recursive formula can be generalised for higher values of $$m$$, although it becomes more difficult.
This is clearly not as satisfactory as a close form formula. However, it can be represented on a graph rather easily, and it is easy to program.
It may be used for secondary school students.
Note: the program gives $$f(10, 7, 2) = \frac{8}{15}$$
EDIT:
When I arrived at this point, I suspected something ...
Is there a much simpler solution ?
I thought having found a simple formula, it was a mistake ...
EDIT 2: C++ programme for $$m = 2$$
#include <iostream>
#include <iomanip>
double f2(int k, int n) { // m = 2
if ((k - n) <= 1) return 0.0;
if (n == 0) return 1.0;
double res = double ((k-n)*(k-1-n)) / (k*(k-1))
+ (double (n*(k-n)) / (k*(k-1))) * f2(k-2, n-1)
+ (double (n)/k) * f2(k-1, n-1);
return res;
}
int main() {
int k = 10;
int n = 7;
double prob = f2(k, n);
std::cout << "f = " << std::setprecision(12) << prob << "\n";
return 0;
}
EDIT 3: Display of a part of the graph.
"0:3/10" means that the line above corresponds to "0" (no ball), and that the corresponding probability is equal to 3/10. SB means "subgraph", and the near figure corresponds to the probability associated to this subgraph. This also illustrates the low efficiency of the scheme in terms of computation. However, again, it was not the goal of this proposal.
f(10,7)=8/15
/ \
/ \
/ \
0:3/10 1:7/10
SB:1/8 SB:f(9,6)=7/12
/ \ | \
/ \ | \
/ \ | \
0:2/9 1:7/9 0:1/3 1:2/3
SB:1.0 SB:f(8,6)=1/4 SB:13/28 SB:f(8,5)=9/14
/ \ | \ | \
/ \ | \ | \
| \
0:1/4 1:3/4
SB:1.0 SB:f(7,5)=2/7
/ \
/ \
• Could you indicate how the described program outputs 8/15? I find myself not quite able to follow the description and suspect seeing those steps written out could help me! – Benjamin Dickman Dec 19 '18 at 16:47
• Sorry, I was really unclear. I mentioned the easiness to write a programme . .. I wrote such a C++ (very simple) programme for $m=2$ and use it. I might post it. You need to represent what I present in a graph ...Easy to follow with such a graph in front of the eyes. ... Not so easy without – Damien Dec 19 '18 at 17:04
• Including the program and a graph would be great – Benjamin Dickman Dec 19 '18 at 20:39
• I have included the C++ programme. I will try now to draw a graph. Much easier by hand than creating a figure, without any drawing software at home ... – Damien Dec 19 '18 at 21:49
• I tried to display a part of the graph. Hope it is clear enough – Damien Dec 20 '18 at 13:35 | 2019-08-18T21:03:24 | {
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https://stats.stackexchange.com/questions/169776/how-frequently-am-i-going-to-lose-this-game | # How frequently am I going to lose this game?
I hope this question fits the site, if not feel free to flag.
In the place I live there's a simple game of cards that you play on your own. Cards are 40, let's say A1 ... A10 , B1 ... B10 , C1 ... C10 , D1 ... D 10. The game goes as follows:
1. You dispose all of the 40 cards on the table, with their face down. You have to create a 4x10 matrix, like below. Mentally, you assign each row to a type (A, B, C or D) and each column to a number (1...10).
cols: _ 1 _ 2 _ 3 _ 4 _ 5 _ 6 _ 7 _ 8 _ 9 _ 10
row A |
row B | HERE YOU HAVE THE 40 CARDS FACE-DOWN
row C |
row D | X
2. Note that each card now, like the one I checked with a X, in addition to having its value (which you can't see because it's down) is strictly connected to another card through its position: X can identify the value D2 with its position.
3. You choose one card randomly, let's say X.
4. You flip it and see its real value, let's say B7.
5. You move to position B7 and flip that card, Y.
cols: _ 1 _ 2 _ 3 _ 4 _ 5 _ 6 _ 7 _ 8 _ 9 _ 10
row A |
row B | Y
row C |
row D | B7
6. So on.
You win if you manage to flip all the cards. You lose if at some point the value on the card you just flipped, points to a position which is already turned up. For example, you would lose immediately if Y has value D2 (it happens).
What is the probability one is going to lose?
This is probably a question one could solve with just some basic notions, but I don't have any.
We can ignore the rows and column, and just say that you have 40 positions and 40 cards that each point to a specific position.
For the first draw, you have 40 cards but only 1 card will make you lose the game (the card that points to the position you chose). This gives a probability of 39/40 to proceed to the next round.
In the second draw, you have 39 cards to choose from, and still only 1 card that will make you lose, because the card that points to the second position is eliminated from the deck. The only losing card is the one that points to your original position. This gives a probability of 38/39 to proceed.
This gives the following formula:
39/40 * 38/39 * 37/38 * 36/37 * 35/36 * 34/35 * 33/34 * 32/33 * 31/32 * 30/31 * 29/30 * 28/29 * 27/28 * 26/27 * 25/26 * 24/25 * 23/24 * 22/23 * 21/22 * 20/21 * 19/20 * 18/19 * 17/18 * 16/17 * 15/16 * 14/15 * 13/14 * 12/13 * 11/12 * 10/11 * 9/10 * 8/9 * 7/8 * 6/7 * 5/6 * 4/5 * 3/4 * 2/3 * 1/2
[1] 0.025
So the probability of winning is 2.5%
We can try this in a simulation using R (you will need to source the code in order for it to run):
set.seed(1)
card.game <- function() {
cards <- sample(1:40)
pos <- seq(1:40)
result <- NA
for (i in 1:40) {
if (i == 1) {
current.pos <- sample(pos, 1)
pos <- pos[which(pos!=current.pos)]
continue <- length(which(pos==cards[current.pos]))
if (continue == 0) {
result <- 0
return(result)
}
} else {
if (i == 40) {
result <- 1
return(result)
}
current.pos <- cards[current.pos]
pos <- pos[which(pos!=current.pos)]
continue <- length(which(pos==cards[current.pos]))
if (continue == 0) {
result <- 0
return(result)
}
}
}
}
result <- NA
for (i in 1:100000) {
result[i] <- card.game()
}
prop.table(table(result))
result
0 1
0.97494 0.02506
And the result, as you can see, is very close to 2.5% which confirms the calculations above.
• Yes. In the example provided by user "m vai", the first card selected was at position D2, and from the 40 possible cards, only the card labeled "D2" would be losing. The card revealed to have "B7" written on it, so we then proceed to take a look at the card on position B7. Now the card on position B7 can be any card except for the card that points to its own position, because that card is already taken and eliminated from the deck in the previous draw. So we now have 39 possible cards (excluding the one with "B7") and only the one with "D2" is losing.. and this will be the case until the end – JonB Sep 2 '15 at 11:32
• I see now that the question I responded to with my comment was deleted. The question was about why a card that points to its own location isn't a losing draw and I explained this. – JonB Sep 2 '15 at 11:33
• I trust you but would like some guidance on understanding. If there's one and only losing card in the deck (the one corresponding to the position you choose at first), the game appears equivalent to saying: name one card, shuffle the deck, and hope that that card is the last one. That probability should be 1/40. No? – natario Sep 2 '15 at 11:46
• Oh but 1/40 is 2.5% :-) thank you! – natario Sep 2 '15 at 11:46
• Yes, that's the way I see it too! No problem. :) – JonB Sep 2 '15 at 11:51
As @Jonas pointed out, in the whole game there's only one losing card, and it is the one you choose at first. If you ever flip that, you lose; if that is the last card you flip, you win.
Then a simpler approach to the question might be interpreting the game this way:
• you name one card;
• you shuffle your 40 cards deck;
• you win if the card you named is the last in the deck (or the first, equivalently).
From this perspective it is clear that the probability you are going to win is the probability for that certain card to occupy a certain position, which is clearly 1/40 = 0.025.
So, to answer my question, you are going to lose with p = 0.975.
• Not to be negative or critical, but I think that to most people, this explanation is obvious only when you have seen the calculations and reasoning in my answer, but might seem bewildering to many if they see only this answer. I agree that the game is equivalent to hoping that a specific card is the last in the deck, but personally, I didn't realize that until after I had done the calculations and the over-ambitious simulation presented in my answer. :) – JonB Sep 2 '15 at 21:11
• @Jonas I agree and that's why I'm going to let yours be the accepted answer. Thank you again. – natario Sep 2 '15 at 21:28
• I suppose you can quit playing the game now or start placing favorable bets.. hehe. – JonB Sep 2 '15 at 21:29
• It has never been very fun actually ;-) and now that I know it is equivalent to shuffling... – natario Sep 2 '15 at 21:32 | 2019-07-22T19:10:22 | {
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https://www.freemathhelp.com/forum/threads/decimal-numerators-denominators.114417/ | # Decimal Numerators & Denominators
#### Explain this!
##### New member
Why are decimals not used as numerators and denominators?
For example, 0.4% can be expressed as (4/10)/100 as complex fraction, but 0.4/100 with a decimal numerator is not common.
The same with 100/(4/10) but not 100/0.4.
#### tkhunny
##### Moderator
Staff member
Why are decimals not used as numerators and denominators?
For example, 0.4% can be expressed as (4/10)/100 as complex fraction, but 0.4/100 with a decimal numerator is not common.
The same with 100/(4/10) but not 100/0.4.
Why not?
Convention, more than anything.
#### JeffM
##### Elite Member
Why are decimals not used as numerators and denominators?
For example, 0.4% can be expressed as (4/10)/100 as complex fraction, but 0.4/100 with a decimal numerator is not common.
The same with 100/(4/10) but not 100/0.4.
The reason is that any rational number can be expressed as a fraction of whole numbers. Any terminating decimal is a rational number. So the obvious way to represent 0.4% in fractional form is 4/1000.
If you want to rerpesent a decimal by a fraction, why put decimals in the fraction.
#### Dr.Peterson
##### Elite Member
Why are decimals not used as numerators and denominators?
For example, 0.4% can be expressed as (4/10)/100 as complex fraction, but 0.4/100 with a decimal numerator is not common.
The same with 100/(4/10) but not 100/0.4.
I'd say the answer is very simple. We can use anything we want in a fraction while we are working (as you've shown); but in a final answer, we want the simplest possible form, and clearly mixing decimals into fractions is not simple. So we tend to avoid that.
#### Otis
##### Senior Member
Why are decimals not used as numerators and denominators? …
Decimal expressions appear in ratios regularly, at arithmetic class.
For example, we see exercises like the following.
Simplify the mixed number: $$\displaystyle 5\frac{0.\overline{6}}{9}$$
Express as a mixed number: $$\displaystyle \dfrac{\frac{4}{5}}{0.75}$$
Outside of arithmetic class, we don't see decimals written in ratios very often because that's not standard form.
#### Explain this!
##### New member
I'd say the answer is very simple. We can use anything we want in a fraction while we are working (as you've shown); but in a final answer, we want the simplest possible form, and clearly mixing decimals into fractions is not simple. So we tend to avoid that.
Yes, 4/1000 is easier to interrupt or understand than something like 0.4/100 or (4/10)/100, or perhaps even 0.4%, although are a lot of decimal percentages are used in business and finance.
#### JeffM
##### Elite Member
Yes, 4/1000 is easier to interrupt or understand than something like 0.4/100 or (4/10)/100, or perhaps even 0.4%, although are a lot of decimal percentages are used in business and finance.
Yes, numbers like 0.4% or 40 basis points do come up a lot in presentations concerning business and finance.
What does not come up in such presentations are things like
$$\displaystyle \dfrac{4\%}{10}.$$ | 2019-03-22T14:20:19 | {
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The diagonals of a rhombus are_____congruent. Alternate interior angle pairs are congruent so. It follows that or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing Property 2: The diagonals of a rectangle are of equal length i.e. of two different ways. If you just look at a parallelogram, the things that look true (namely, the things on this list) are true and are thus properties, and the things that don’t look like they’re true aren’t properties. Where are the other 2 vertices of the… Thus, if you are not sure content located a Find the sides and area of the parallelogram. Missy is proving the theorem that states that opposite sides of a parallelogram are congruent. In ABC and DCB, AB = DC BC = BC AC = DB ABC DCB ABC = DCB Now, AB DC & BC is a … Opposite sides are congruent. by the same vector , despite the fact that they are in different places on the CUNY Hunter College, Bachelor in Arts, Economics. Two Adjacent Vertices Of The Parallelogram Are (5.5,7.5) And (13.5,16). The displacement (say) of the centroid from point can be written in one Sometimes. The diagonals of a parallelogram_____bisect the angles of the parallelogram. Your name, address, telephone number and email address; and For instance, please refer to the link, does $\overline{AC}$bisect $\angle BAD$and $\angle DCB$? Opposite sides are parallel by definition. The rectangle has the following properties: All the properties of a parallelogram apply (the ones that matter here are parallel sides, opposite sides are congruent, and diagonals bisect each … Varsity Tutors. an which specific portion of the question – an image, a link, the text, etc – your complaint refers to; It is done with the help of law of cosines. Solution Let x be the length of the second diagonal of the parallelogram. sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require The diagonals of a parallelogram divide it into four triangles of equal area. Earlham College, Bachelor in Arts, Mathematics. on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney. So remember, a rhombus is just a parallelogram where all four sides are equal. In fact, if all four sides are equal, it has to be a parallelogram. Figure 13: A parallelogram Suppose that the quadrilateral ABCD in Fig. The diagonals intersect at an angle of 35 degrees. Inside Any Quadrilateral . A rectangle_____has consecutive sides congruent. A quadrilateral with one pair of sides congruent and on pair parallel is_____a parallelogram. Solved: The lengths of the diagonals of a parallelogram are 20 inches and 30 inches. The diagonal of a parallelogram separates it into two congruent triangles. The diagonals bisect the vertex angles. The diagonals are congruent to each other. same vectors, and : this merely indicates that these sides are of The diagonal in Fig. person_outline Timur schedule 10 years ago This calculator computes the diagonals of a parallelogram and adjancent … Quadrilaterals Geometry Index. Given: Let ABCD be a parallelogram where AC = BD To prove: ABCD is a rectangle Proof: Rectangle is a parallelogram with one angle 90 We prove that one of its interior angles is 90 . So if we find th… Characterizations of a parallelogram Quadrilateral ABCD is a parallelogram, if … Parallelogram has two diagonally - a longer let be d 1, and shorter - d 2. The diagonals bisect each other. Fig. Parallelograms have opposite interior angles that are congruent, and the diagonals of a parallelogram bisect each other. Consecutive angles are supplementary. 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Mamou Rockwell Number, Imu Cet Exam Date 2021 Application Form, Approach Run In High Jump, Cubchoo Pokémon Go Bow, Juno Award Winners, Ite Diploma Vs Poly Diploma, | 2021-04-18T11:50:59 | {
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https://proofwiki.org/wiki/Definition:Order_of_Group_Element | Definition:Order of Group Element
Definition
Let $G$ be a group whose identity is $e_G$.
Let $x \in G$ be an element of $G$.
Definition 1
The order of $x$ (in $G$), denoted $\order x$, is the smallest $k \in \Z_{> 0}$ such that $x^k = e_G$.
Definition 2
The order of $x$ (in $G$), denoted $\order x$, is the order of the group generated by $x$:
$\order x := \order {\gen x}$
Definition 3
The order of $x$ (in $G$), denoted $\left\vert{x}\right\vert$, is the largest $k \in \Z_{\gt 0}$ such that:
$\forall i, j \in \Z: 0 \le i < j < k \implies x^i \ne x^j$
Infinite Order
$x$ is of infinite order, or has infinite order if and only if there exists no $k \in \Z_{> 0}$ such that $x^k = e_G$:
$\order x = \infty$
Finite Order
$x$ is of finite order, or has finite order if and only if there exists $k \in \Z_{> 0}$ such that $x^k = e_G$.
Examples
Order of $2$ in $\struct {\R_{\ne 0}, \times}$
Consider the multiplicative group of real numbers $\struct {\R_{\ne 0}, \times}$.
The order of $2$ in $\struct {\R_{\ne 0}, \times}$ is infinite.
Order of $i$ in $\struct {\C_{\ne 0}, \times}$
Consider the multiplicative group of complex numbers $\struct {\C_{\ne 0}, \times}$.
The order of $i$ in $\struct {\C_{\ne 0}, \times}$ is $4$.
Order of $\begin{bmatrix} 1 & 1 \cr 0 & 1 \end{bmatrix}$ in General Linear Group
Consider the general linear group $\GL 2$.
Let $\mathbf A := \begin{bmatrix} 1 & 1 \cr 0 & 1 \end{bmatrix} \in \GL 2$
The order of $\mathbf A$ in $\GL 2$ is infinite.
Rotation Through the $n$th Part of a Full Angle
Let $G$ denote the group of isometries in the plane under composition of mappings.
Let $r$ be the rotation of the plane about a given point $O$ through an angle $\dfrac {2 \pi} n$, for some $n \in \Z_{> 0}$.
Then $r$ is the generator of a subgroup $\gen r$ of $G$ which is of order $n$.
Possible Orders of $x$ when $x^2 = x^{12}$
Let $G$ be a group.
Let $x \in G \setminus \set e$ be such that $x^2 = x^{12}$.
Then the possible orders of $x$ are $2$, $5$ and $10$.
Also known as
Some sources refer to the order of an element of a group as its period.
Also denoted as
The order of an element $x$ in a group is sometimes seen as $\map o x$.
Some sources render it as $\map {\operatorname {Ord} } x$.
Also see
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https://math.stackexchange.com/questions/270949/does-bigcap-n-1-infty-frac1n-frac1n-varnothing/270951 | # Does $\bigcap_{n=1}^{+\infty}(-\frac{1}{n},\frac{1}{n}) = \varnothing$?
When I learn the below theorem:
If $I_n$ is closed interval, and $I_{n+1} \subset I_n$, then $$\bigcap I_n \ne \varnothing$$
and someone says if we replace closed interval with open interval, can construct counter-example.
So I have tried to construct the one: Does $$\bigcap_{n=1}^{+\infty}\left(-\frac{1}{n},\frac{1}{n}\right) = \varnothing\quad?$$
Thanks very much.
• I am 100% certain that this question was asked at least twice before. Jan 5 '13 at 14:21
• Yes and I am searching for that link, Asaf. Jan 5 '13 at 14:24
• 10 answers? really? Jan 5 '13 at 20:17
• its obviously non-empty?
– user85461
Jul 19 '13 at 18:41
• Related post: math.stackexchange.com/questions/1304402/… Jan 8 '16 at 12:40
No. It is not empty. Since $0\in (-1/n, 1/n)$ for all $n$, so $0\in\bigcap_{n=1}^\infty \left(-\frac{1}{n},\frac{1}{n}\right)$.
• thank you ,but if $x \in (-\frac{1}{n},\frac{1}{n})$, then $-\frac{1}{n}<x<\frac{1}{n}$ let $n \rightarrow \infty$, then $0<x<0$, so $x$ dose not exist. but apparently,your statement is right. so confuse me. Jan 5 '13 at 14:24
• @Tai: When we take $n\to\infty$, we can no longer ensure that the inequalities remain strict, so we get $0\leq x\leq 0$. Jan 5 '13 at 14:29
• oh,I make a mistake, when let $n \rightarrow +\infty$, $0\leq x \leq 0$ Jan 5 '13 at 14:30
No, it is not empty, but in fact, if $I_n$ is closed and $I_{n+1}\subseteq I_{n}$, that is not enough to ensure that $$\bigcap_{n=1}^\infty I_n\neq\varnothing.$$ As an example, take the closed intervals $[n,\infty)$. In order for what you have written to hold true, one needs that at least one of the $I_n$ is bounded.
Finally, as others have pointed out, the intersection of open sets you have is nonempty since $0\in\left(-\frac{1}{n},\frac{1}{n}\right)$ for all $n\in\mathbb{N}$. As a counterexample, one can consider $$\bigcap_{n=1}^\infty\left(0,\frac{1}{n}\right).$$
• thank you, I have made a mistake. yes, $I_n$ need to be bounded. Jan 5 '13 at 14:50
• @Tai: You're welcome, I just wanted to be sure you didn't have the wrong 'theorem' in mind. Jan 5 '13 at 14:51
• +1 for explaining how the problem should be solved as well. Jan 5 '13 at 16:16
• +1 nice approach. Feb 3 '13 at 3:26
• @BabakSorouh: Thanks! (And for all the +1's :) ) Feb 3 '13 at 3:27
As you already have a bunch of nice answers to your question, I just wanted to comment on the relation with the theorem you have referred to. Note that $J_n\subset I_n$ where $J_n = [-\frac{1}{2}n,\frac{1}{2}n]$ and thus we know that $\bigcap\limits_n J_n$ is not empty. However, $\bigcap\limits_n J_n\subset \bigcap\limits_n I_n$ and the latter is hence non-empty as well.
• (+1) I think this connection is likely to be helpful to the OP. Jan 5 '13 at 20:58
NO!
$$\bigcap_{n=1}^{+\infty}\,\left(-\frac{1}{n},\frac{1}{n}\right)\,\neq\, \varnothing$$
WHY NOT?:
Note that as $\,n \to \infty,\,$ the endpoints of the intervals get increasingly close to $\,0$, but never reach $0$, hence every interval contains $0\,$: $$\;0 \in \left(-\dfrac1n, \cfrac1n\right)\;\; \forall n \in \mathbb{N}.$$ Indeed,
$$\bigcap_{n=1}^{+\infty}\,\left(-\frac{1}{n},\frac{1}{n}\right) = \{0\}.$$
You might be interested to know that, e.g., $$\bigcap_{n=1}^{+\infty}\,\left(0, \frac1n\right)\,=\, \varnothing$$ It' simply a matter of choosing the correct endpoints of the open intervals.
• And, of course, $$\bigcap_{n=1}^{+\infty}\left(-\frac{1}{n},\frac{1}{n}\right)\subseteq \{0\}.$$ If $x\in\bigcap_{n=1}^{+\infty}\left(-\frac{1}{n},\frac{1}{n}\right)$ it means $$|x|\lt \frac1{n}\quad\forall n\in\Bbb N$$ so $x=0$.
– leo
Jan 5 '13 at 16:25
• Oh well... the thing was prove that the intersection is not empty. It suffices with what you've done.
– leo
Jan 5 '13 at 16:30
No. As others have mentioned, it contains $0$.
But consider $I_n = (1-\frac{1}{n},1)$. Or, more closely to your example, $I_n = (0,\frac{1}{n})$.
$\bigcap_{n=1}^\infty \left(-\frac{1}{n},\frac{1}{n}\right)=\{0\}\neq\emptyset$.
The theorem also requires that the diameter of the sets $A_n$ go to zero $$\mathrm{Diam}A_n= \sup_ { x,y \in A_n} |x-y|$$ for example take $B_n = [n,\infty)$ it is easy to verify $$\bigcap _ {i=1}^{\infty}B_n= \not 0$$
• @Michalis I was not talking about $\mathbb{R}$ with the usual metric but generally for a complete metric space, do you claim it suffices that the diameter is bounded for the general case? Jan 5 '13 at 15:54
By definition of intersection, a $x\in\displaystyle\bigcap_{n\in\mathbb{N}\backslash \{0\}} \left(-\frac{1}{n},+\frac{1}{n}\right)$ if, only if, $x\in \left(-\frac{1}{n},+\frac{1}{n}\right)$ for all $n\in\mathbb{N}\backslash\{0\}$. And $\displaystyle 0\in\left(-\frac{1}{n},+\frac{1}{n}\right)$ for all $n\in\mathbb{N}\backslash\{0\}$. Then
$$0\in\displaystyle\bigcap_{n\in\mathbb{N}\backslash \{0\}} \left(-\frac{1}{n},+\frac{1}{n}\right).$$ If $x\in \left(-\frac{1}{n},+\frac{1}{n}\right)$ for all $n\in\mathbb{N}\backslash\{0\}$ then $-\frac{1}{n}<x<+\frac{1}{n}$ for all $n\in\mathbb{N}\backslash\{0\}$. By Archimedean property or Supremum Axiom we have that $x=0$. Then $$\{0\}=\bigcap_{n\in\mathbb{N}\backslash \{0\}} \left(-\frac{1}{n},+\frac{1}{n}\right)$$ | 2021-12-07T23:26:34 | {
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https://stats.stackexchange.com/questions/406191/summation-of-cosine-of-uniform-random-variable | # Summation of cosine of uniform random variable
I read that the PDF of the sum of cosines of a random variable, which is uniformly distributed, is a non uniform distribution; something like inverse square root of the random variable. My doubt is, not going into pdf, but just if we calculate sum of cosine of random variable varying between $$-\pi$$ to $$\pi$$, i.e., SUM $$(\cos (t_j))$$, should it not be 0 for large number of values for the $$t_j$$ intuitively? I am confused. If this is a valid question can anyone help? Thanks a lot.
Also, if we have SUM $$(\cos(t_j))$$, $$t_j$$ is uniform random variable between $$-\pi$$ to $$\pi$$, then will introducing frequency $$w$$ (omega) affect the answer to above question? i.e., SUM $$(\cos (w \cdot t_j))$$?
Consider a series of independent uniform random variables $$U_1,U_2,U_3,... \sim \text{IID U}(0,1)$$ and form the corresponding series $$X_1,X_2,X_3,...$$ as:
$$X_i = \cos(\pi U_i).$$
For all $$|x| \leqslant 1$$ the latter random variables have the distribution function:
\begin{aligned} F_X(x) = \mathbb{P}(X \leqslant x) &= \mathbb{P}(\cos(\pi U) \leqslant x) \\[6pt] &= \mathbb{P} \Big( U \geqslant \frac{1}{\pi} \cdot \arccos(x) \Big) \\[6pt] &= 1 - \frac{1}{\pi} \cdot \arccos(x), \\[6pt] \end{aligned}
and the corresponding density function:
\begin{aligned} f_X(x) = \frac{dF_X}{dx}(x) &= -\frac{1}{\pi} \cdot \frac{d}{dx} \arccos(x) \\[6pt] &= \frac{1}{\pi} \cdot \frac{1}{\sqrt{1-x^2}}. \\[6pt] \end{aligned}
This random variable has mean $$\mathbb{E}(X) = 0$$ and variance $$\mathbb{V}(X) = \tfrac{1}{2}$$. Now, let $$S_n = \sum_{i=1}^n X_i$$ be a partial sum of these variables, and note that it has mean $$\mathbb{E}(S_n) = 0$$ and variance $$\mathbb{V}(S_n) = \tfrac{n}{2}$$. By the central limit theorem the limiting distribution of the standardised sum is the standard normal distribution. For large $$n$$ we have the approximate distribution:
$$S_n \overset{\text{Approx}}{\sim} \text{N} \Big( 0, \frac{n}{2} \Big).$$
The variance of the sum increases with $$n$$, so there is no convergence to zero --- the sum will be distributed around zero, buts its variance gets bigger and bigger. However, if you instead look at the sample mean $$\bar{X}_n = S_n/n$$, the variance of this latter quantity decreases to zero, so you will have convergence to the mean of zero. This latter result is a manifestation of the law of large numbers.
Simulation: We can simulate this problem in R as follows. In this code we plot the kernel density of $$m = 10^5$$ simulations for $$n=100$$ and we superimpose the normal density as a red dashed line. You can see that this is a very close approximation to the kernel density of the simulations.
#Simulate matrix of cosine values
set.seed(1);
m <- 10^5;
n <- 100;
U <- matrix(runif(n*m,0,1), nrow = m);
X <- cos(pi*U);
#Calculate sample total of cosine values
S <- rowSums(X);
#Create data-frame for plotting
DD <- density(S);
NN <- dnorm(DD$$x, mean = 0, sd = sqrt(n/2)); GRAPH <- data.frame(S = DD$$x, Density = DD$y, Approx = NN); #Plot density of sample totals library(ggplot2); FIGURE <- ggplot(data = GRAPH, aes(x = S, y = Density)) + geom_line(size = 1) + geom_line(aes(y = Approx), colour = 'red', linetype = 'dashed') + ggtitle('Density of Sample Means of Cosine Simulations') + xlab('Sample Mean') + ylab('Density'); FIGURE; • Does that means, the summation to become 0 is most likely due to gauss curve centered around mean 0 ? But does this mean that when we simulate and see the value of the sum of cos (uniform random variable values between -1 to 1) we some times also get not 0 even for large number of values? – Rag Tej May 2 at 9:55 • But intuitively how to visualize this? Let us take a simpler case of ∑tj, tj belongs to [-1, 1] and is uniformly distributed. Intuitively, my mind thinks that this sum should be zero because the tjs are uniformly distributed in -ve and +ve domains. But I think mathematically, even this sum is also irwin-hall distributed?! – Rag Tej May 2 at 9:57 • The variance of the sum increases with$n$, so there is no convergence to zero. However, if you use the sample mean$S/n$instead of the sum$S\$ then the variance decreases to zero, so you have convergence to the mean of zero. – Ben May 2 at 10:00
• Indeed! I see what you mean. I have one more doubt. In arriving at the solution, you took different random variables, each between (0,1). Is this equivalent of what I posed in question as Sum (different values of one random variable)? Like, did you replace each value of one random variable with different random variables, independent of each other (and hence equivalent)? Am I on right tracks of understanding? Also, may I ask how to arrive at ℙ(𝑈⩾1𝜋⋅arccos(𝑥))=1−1𝜋⋅arccos(𝑥)? – Rag Tej May 2 at 10:09
• Unfortunately I cannot fully understand the description in your question, so I am not sure exactly what it is. – Ben May 2 at 10:15 | 2019-08-18T11:09:19 | {
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https://byjus.com/question-answer/two-concentric-circles-are-of-radii-5cm-and-3cm-the-length-of-the-chord-of/ | Question
Two concentric circles are of radii $$5cm$$ and $$3cm$$. The length of the chord of the larger circle which touches the smaller circle is
A
4cm
B
12cm
C
2cm
D
8cm
Solution
The correct option is C 8cm$$Given-\\ O\quad is\quad the\quad centre\quad of\quad two\quad concentric\quad circles\quad$$ $${ C }_{ 1 },\quad the\quad inner\quad circle\quad with\quad radius=OP=3cm\quad$$and $${ C }_{ 2 },\quad the\quad outer\quad circle\quad with\quad radius=OA=5cm.$$ $$AB\quad is\quad a\quad chord\quad of\quad { C }_{ 2 }.\quad It\quad touches\quad { C }_{ 1 }\quad at\quad P.$$ $$To\quad find\quad out-$$ $$The\quad length\quad of\quad AB=?$$ $$Solution-$$ $$We\quad know\quad that\quad the\quad radius\quad through\quad the\quad point\quad of\quad$$ $$contact\quad of\quad a\quad tangent\quad to\quad a\quad circle\quad is\quad perpendicular\quad \\ to\quad the\quad tangent.$$ $$\therefore \quad OP\bot AB\Longrightarrow \angle OPA={ 90 }^{ o }.........(i)$$ $$Again\quad we\quad know\quad that\quad he\quad perpendicular,\quad dropped\quad from\quad the\quad \\ center\quad of\quad a\quad circle\quad to\quad \quad any\quad of\quad its\quad chord,\quad bisects\quad the\quad latter.$$ $$\therefore \quad OP\quad bisects\quad AB\quad at\quad P.$$ $$So\quad AB=2\times AP.......(ii).$$ $$Now\quad \Delta OAP\quad is\quad a\quad right\quad triangle\quad with\quad OA\quad as\quad hypotenuse$$ $$(from\quad i).$$$$OA=5cm\quad \& \quad OP=3cm.$$ $$So,\quad applying\quad Pythagoras\quad theorem,\quad we\quad have$$ $$AP=\sqrt { { OA }^{ 2 }-{ OP }^{ 2 } } =\sqrt { { 5 }^{ 2 }-{ 3 }^{ 2 } } cm=4cm.$$ $$\therefore \quad AB=2\times AP(from\quad ii)=2\times 4cm=8cm.\\ Ans-\quad Option\quad D.\\ \\ \\ \\ \\$$Maths
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http://math.stackexchange.com/questions/93623/does-the-golden-angle-produce-maximally-distant-divisions-of-a-circle | Does the golden angle produce maximally distant divisions of a circle?
I just ran across a video that claimed that the sequence of multiples of the golden angle produces some sort of optimal spacing around a circle for all possible iterations (this is a little hand-wavy, I'm aware).
Of course, any irrational angle has the property that it won't ever repeat and will form a dense set as the limit goes to infinity, but intuitively the golden angle seems to work much better than an irrational very close to $\frac{\pi}{2}$.
I'm looking for a way to more formally state this idea of optimal spacing for any iteration as a means to either prove or disprove that the golden angle provides this property (and is it unique?).
-
Watching the video, it seems to me that what's nice about using this angle is as follows: the ratio between the smallest and largest gap between the angles $a_1,\ldots,a_n$, which are the first $n$ iterations produced by the given angle $a$ arranged from least to greatest, is constant. Does that seem right? – Alex Becker Dec 23 '11 at 5:51
Given an arbitrary angle $\theta$ which is not a multiple of $\pi$, to say that $n \theta$ and $m \theta$ are close to each other on the circle is to say that $(n-m)\theta$ is close to $0 \bmod 2\pi$, so to say that the fractional part of $(n-m) \frac{\theta}{2\pi}$ is close to $0$. If you want to avoid this, you want to pick a value of $t = \frac{\theta}{2\pi}$ with the property that, for any integer $n-m$, the fractional part of $(n-m)t$ is never too close to $0$.
Phrased another way, for any integer $q$ we never want $qt$ to be too close to another integer $p$. Phrased yet another way, we never want $t$ to be too close to a rational number $\frac{p}{q}$. So we are looking for irrational numbers which are hard to approximate by rationals.
The closest rational approximations (in a suitable sense) to an irrational number can be read off from truncations of its continued fraction $$t = a_0 + \frac{1}{a_1 + \frac{1}{a_2 + ...}}$$
and in particular are better approximations if the corresponding entries $a_i$ of the continued fraction are large. That means that the best irrational number for the job is the one with the property that each $a_i$ is as small as possible, so $$1 + \frac{1}{1 + \frac{1}{1 + ...}}$$
which turns out to be precisely the golden ratio. (Why? Because the above number $t$ has the property that $t = 1 + \frac{1}{t}$, or $t^2 = t + 1$, so it is either the golden ratio or its conjugate, and it's greater than $1$ so it must be the golden ratio.)
-
Very nice answer. +1 – Alex Becker Dec 23 '11 at 5:52
Great answer, thanks. As a side note, $\frac{1}{\phi}$ works just as well since it's the reverse direction on a circle. – cobbal Dec 23 '11 at 6:07
Does anybody know a citable (and more elaborate) version of this proof? – Stiefel Jun 11 '14 at 11:32 | 2016-05-30T10:48:36 | {
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https://forum.allaboutcircuits.com/ubs/powers-of-10-scientific-engineering-notation-and-metric-prefixes.466/ | # Powers of 10, Scientific/Engineering Notation, and Metric Prefixes
This entry is meant to help you understand powers of 10, scientific notation, engineering notation, and some metric prefixes.
Powers of 10, Scientific Notation vs. Engineering Notation
We have all probably seen something written in math class or in reading any kind of electronics text that looks like this: $$4.32 \times 10^5$$ This is called scientific notation, or a power of 10. In this case, it would be 4.32 times 10 to the power of 5. The number 10 is called the base, the number 5 is called the exponent. The exponent tells us how many times the base is to be multiplied by itself. In our example, $$\times 10^5$$ means 10x10x10x10x10.
In Scientific Notation the exponent can be any number, even negative numbers and zero. We will talk about how Engineering Notation differs from Scientific Notation in just a moment, but first, let's find out out exactly what $$4.32 \times 10^5$$ means.
First, we see that our base is 10 and our exponent is 5, so we know that we need to multiply the base (10) by itself equal to our exponent (5), so like mentioned before 10x10x10x10x10. That makes our expression 4.32 x 10 x 10 x 10 x 10 x 10 = 432,000. Using this formula, we are able to express very large numbers and very small numbers much more efficiently than we would otherwise be able to. Consider an electric charge of 1 Coulomb, which has a proton or electron count (depending on polarity) of roughly $$6.25 \times 10^{18}$$. It is much more efficient, and less prone to error in calculations if we express this number in scientific notation as opposed to writing it out fully.
$$6.25 \times 10^{18}$$
6,250,000,000,000,000,000
Both of these numbers are equal, just that one is expressed in Scientific Notation.
Notation With Negative Exponents
So far, we have only talked about a number where the exponent is positive, but when we express numbers that are smaller than 1.0, like 0.025, we have to use a negative exponent. In our example of 0.025, we would say $$2.5 \times 10^{-2}$$. This is the equivalent of saying $$2.5 \div 10 \div 10$$. This gives us a quotient of 0.025!
So, to recap, positive exponent means multiply the base (10) by itself the number of times indicated by the exponent. A negative exponent means divide the base (10) by itself the number of times indicated by the exponent.
But what, you might ask yourself, happens when you have a very large or very small number? Am I to sit here with a number like $$5.81 \times 10^{21}$$ and multiply 10 by itself 21 times? HOW TEDIOUS! Well no, there is an easier way my friends, and it works with both positive and negative exponents. Let's consider our example of $$5.81 \times 10^{21}$$. The exponent is positive, so our number is going to get larger. If the exponent was negative, we would know that our number is going to get smaller (because we're dividing). What happens to a decimal place when a number gets larger? It moves to the right. If we wanted to multiply 5.81 times 10, we would get 58.1 right? Well, that single move to the right could be expressed as $$5.81 \times 10^1$$ because we multiplied the base (10) by itself 1 time, or 5.81 x 10 = 58.1. If we had said $$5.81 \times 10^2$$ that would be the same as saying 5.81 x 10 x 10 = 581.0 right? Our decimal place is moving to the right equal to the exponent, and as we progress beyond the last digit of our original number, we're adding zeros to fill in the holes.
If I take $$5.81 \times 10^{21}$$ it's the same as saying "move the decimal place twenty-one times to the right". Since we only have 2 digits to the right of the decimal place, we'll have to add zeros as we go. We end up with 5,810,000,000,000,000,000,000. If our exponent had been negative, we would have moved the decimal place 21 times to the left and added zeros. $$5.81 \times 10^{-21}$$ = 0.00000000000000000000581, which is a very small number but not uncommon in electronics.
Engineering Notation
Engineering Notation is essentially the same as scientific notation except that we only use exponents that are powers of 3. So exponents like -12, -9, -6, -3, 3, 6, 9, 12, and so on. The reason for this is that engineers use metric prefixes to express numbers. For example, we've all heard of a kilometer. Kilo is the metric prefix, and meter is the unit of measure. 1 kilometer is equal to 1,000 meters is equal to $$1.0 \times 10^3$$ meters. In the engineering world, there are only metric prefixes for values that have an exponent which is a multiple of 3. A list of some of the metric prefixes can be found at:
Typically speaking, a number expressed in Engineering Notation is always expressed as a number between 1 and 1,000 with a base of 10 and an exponent which is a multiple of 3. For example, $$5.81 \times 10^3$$ units would be equal to 5.81 kilo-units or 5,810 units. Similarly, $$581.0 \times 10^{-3}$$ units would be equal to 0.581 units which would be stated as "Five hundred and eighty-one milli-units". $$5.81 \times 10^{-6}$$ would be 5.81 micro-units.
I hope you have a better understanding now of powers of 10, and how they might be used in electronics. Also, I hope that you understand why scientific notation/engineering notation helps us to display measurements more efficiently and in a way that makes calculations less prone to error. You will see powers of ten and metric prefixes very often in engineering.
For a discussion of arithmetic using scientific notation, see All About Circuits eBook, Volume I - DC, Arithmetic with Scientific Notation
References
Mitchel E. Schultz, Grob's Basic Electronics, McGraw-Hill, 2011, ISBN 978-0-07-351085-9
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https://mathsfeedblog.wordpress.com/2022/06/15/problem-the-desk-calendar/ | # Problem: The Desk Calendar
I love this little puzzle.
Some years ago (I was living in Canberra at the time), my wife (well, she wasn’t my wife yet) and I were at a friend’s place and I happened to notice a gorgeous little calendar on their kitchen windowsill.
According to this calendar, it was the 25th of April. My friend told me that each morning she would use the two blocks to write out the day of any month. This means that just using those two blocks, she can make all of the numbers from 1 to 31:
$01, 02, 03, \ldots, 29, 30, 31$
I didn’t think much of it at the time, but when I got home later that night, the little cube calendar found its way into my mind and my brain started to do its thing. I was trying to work out what the numbers on the unseen faces of the cubes were. In fact, that’s the problem I’m going to give you. Enjoy!
Problem. Look at the picture of the desk calendar. What are the four numbers that cannot be seen on the left cube and the three numbers that cannot be seen on the right cube?
Hint: If you haven’t solved this after some time then try gluing your feet to the ceiling.
1. Thomas Morrill says:
Very neat puzzle! I think the parameters will constrain us to certain fonts for writing numerals.
Start with the right cube in the picture. Both the dates 11 and 22 must be representable, so the sides of this cube are labelled 1, 2, 3, 4, 5, with the sixth side to be determined. At this point there are five unknown slots for digits: four sides on the left cube, and one on the right. There are also five digits that have not been placed, 6, 7, 8, 9, and 0.
It appears we just need to arrange the missing digits on the unknown sides, but there is a hitch. The dates 01, 02, 03, 04, 05, 06, 07, 08, and 09 must all be representable. In other words, there must be a 0 on the cube opposite from each of the digits 6, 7, 8, and 9.
Easy enough, put the 0 on the right cube. But then 03 is not representable! Okay, put 0 on the left cube, say with 6, 7, and 8. Then 06, 07, and 08 are not representable. Dear.
This looks to be impossible, but only as long as we assume we are trying to allocate five unique digits to five slots. The trick is to write the digits 6 and 9 as rotations of one another, so that they both occupy a single slot, and put a 0 on both cubes.
The left cube reads 0, 1, 2, 6/9, 7, 8, and the right cube reads 0, 1, 2, 3, 4, 5.
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• Yes it seems a little bit of combinatorics can be a hindrance, not a help here, haha…
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http://www.landmark-ent.com/sri-trang-stxwuvx/330ef4-shortcut-to-find-eigenvalues-of-2x2-matrix | # shortcut to find eigenvalues of 2x2 matrix
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That part you know already. Then |A-λI| is called characteristic polynomial of matrix. "despite never having learned" vs "despite never learning". Find the characteristic function, eigenvalues, and eigenvectors of the rotation matrix. Why can't we use the same tank to hold fuel for both the RCS Thrusters and the Main engine for a deep-space mission? Recover whole search pattern for substitute command. So if the eigenvalues are $\lambda_1$ and $\lambda_2$, then assume $c\neq 0$ and then the claim is that the eigenvectors are $v_i = (\lambda_i-d,c)$. « compact pad. The Math: Computation of Eigenvalues. rev 2020.12.4.38131, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us, @AndreasCaranti: Thank you Andreas! Learn to recognize a rotation-scaling matrix, and compute by how much the matrix rotates and scales. For this to happen in a 2x2 matrix, there can be only one eigenvalue. From that, we easily find two eigenvalues: $A \begin{bmatrix}\vec v\\\vec v\end{bmatrix} = \begin{bmatrix}2\vec v\\2\vec v\end{bmatrix}$ and $A \begin{bmatrix}\vec v\\-\vec v\end{bmatrix} = \begin{bmatrix}\vec 0\\\vec 0\end{bmatrix}$. What caused this mysterious stellar occultation on July 10, 2017 from something ~100 km away from 486958 Arrokoth? What do we mean visually by complex eigen values of a matrix? The eigenvalue of the matrix [1] is 1. How does turning off electric appliances save energy. It should be L1 = (T + (T^2 - 4D)^1/2) / 2. 2X2 Eigenvalue Calculator. While harvard is quite respectable, I want to understand how this quick formula works and not take it on faith. Computation of det(A - λ I) =0 leads to the Characteristic Polynomial, where the roots of this polynomial are the eigenvalues of the matrix A. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. The Harvard class page isn't actually using the trace method, as that computes each eigenvector from the other eigenvalue(s). Contact Us. Otherwise, we say that the matrix has real eigenvalues. $\ge$ ≥. Squaring a square and discrete Ricci flow. If $c=0$ but $b\neq 0$, then the math works out the same way for $v_i=(b,\lambda_i-a)$. To find eigenvalues of a matrix all we need to do is solve a polynomial. Reconstruct the original symmetric matrix given Eigen values and the longest Eigen vector, Eigen vector of Pauli Matrix (z-component of pauli matrix). $x^2$ x 2. Eigenvalues: The calculator returns the eigenvalues of the 2x2 matrix. what does "scrap" mean in "“father had taught them to do: drive semis, weld, scrap.” book “Educated” by Tara Westover. Yes. and Eigenvectors and eigenvalues of a diagonal matrix D The equation Dx = 0 B B B B @ d1 ;1 0 ::: 0 0 d 2;. shortcut to find eigenvalues of 3x3 matrix . Press question mark to learn the rest of the keyboard shortcuts MathJax reference. Part 1 calculating the Eigen values is quite clear, they are using the characteristic polynomial to get the Eigen values. Regards, The quadratic formula is actually wrong in the Harvard site. Likewise this fact also tells us that for an $$n \times n$$ matrix, $$A$$, we will have $$n$$ eigenvalues if we include all repeated eigenvalues. Second order transfer function with second order numerator? Display decimals, number of significant digits: … Icon 2X2. Can somebody offer an explanation or proof of this? Use MathJax to format equations. Problems in Mathematics. Can I save seeds that already started sprouting for storage? Any matrix has eigen value either 0 or 1? Here are examples of how to solve for both kinds of eigenvalues: Let's begin with an example where we compute real eigenvalues:Suppose we have the matrix: A = ((5,4)(3,2))det(A - lambda I)= det ((5-lambda, 4)(3, 2-lambda))=(5-lambda)(2-lambda)-4*3=0(5-lambda)(2-lambda)-12=lambda^2 -7lambda+(-2)=0The roots are:lambda = frac(7 pm sqrt(49-48))(2)lambda = 4, 3. 3.6 Matrices in Xhave determinant 1 Since any matrix A2Xis defective over C, it has one repeated real eigenvalue. Show there are no real valued eigenvalues for 32 53 A ⎡ − ⎤ =⎢ ⎥ ⎣ − ⎦ Solution: 32 53 AI λ λ λ ⎡⎤−− −=⎢⎥ ⎣⎦−−, and this is singular iff (3 )( 3 ) 10 1 0−−−+=+=λλ λ2. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. How do I get the size of a file on disk on the Commodore 64? So - next.. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. $\log_ {\msquare}$ log . This is singular iff ()( ) 0ad bc−−−=λ λ . Fact Consider the system where A = ((-2, -3), (3, -2))det(A-lambda I) = det ((-2-lambda, -3), (3, -2-lambda)) = (-2-lambda)(-2-lambda)-(-3*3)=lambda^2+4 lambda +13 =0.The roots are: lambda = frac(-4 pm sqrt(-36))(2)We see that the sqrt(-36) is equal to 6i, such that the eigenvalues become: lambda = frac(-4 pm 6i)(2) = -2 pm 3i. FINDING EIGENVECTORS • Once the eigenvaluesof a matrix (A) have been found, we can find the eigenvectors by Gaussian Elimination. "Complex numbers are numbers of the form x + iy, where x and y are real numbers and I is the 'imaginary number' sqrt(-1) " (Blanchard, Devaney, Hall, 291). Division Headquarters 315 N Racine Avenue, Suite 501 Chicago, IL 60607 +1 866-331-2435 The determinant of a triangular matrix is easy to find - it is simply the product of the diagonal elements. Consider the 2 by 2 rotation matrix given by cosine and sine functions. Find Eigenvalues and Eigenvectors of a 2x2 Matrix - YouTube Hence the set of eigenvectors associated with λ = 4 is spanned by u 2 = 1 1 . How do I determine a “suitable” set of eigenvectors for diagonalization? • Exercise 14. $\sqrt {\square}$ √ ☐. Eigenvalues and eigenvectors calculator. Press question mark to learn the rest of the keyboard shortcuts. December 2, 2020. . In other ways that I have calculated the Eigen vectors I get other values. $\le$ ≤. If the roots are complex we say that the matrix has complex eigenvalues. That’s generally not too bad provided we keep $$n$$ small. Why did I measure the magnetic field to vary exponentially with distance? In order to find eigenvalues of a matrix, following steps are to followed: Step 1: Make sure the given matrix A is a square matrix. In their example, given a matrix in the form a b c d, if b & c are zero, then the vectors are 1 0 and 0 1, which makes sense as you can scale these to any other size. Get the free "Eigenvalue and Eigenvector (2x2)" widget for your website, blog, Wordpress, Blogger, or iGoogle. First eigenvalue: Second eigenvalue: Discover the beauty of matrices! Thus if I come up with (2,-3) using this method, and (-4,6) using another method both vectors are valid... because what matters is the ratio of (a/b) must be identical. The last coordinates are clearly equal, and we know that $\lambda_i^2 -(a+d)\lambda_i + (ad-bc) = 0$, which implies $\lambda_i^2 - d\lambda_i = a\lambda_i - (ad-bc)$, so the first coordinates are equal too. I understand that that what matters with Eigen vectors is the ratio, not the value. Then the equation |A-λI| = 0 is called characteristic roots of matrix. We figured out the eigenvalues for a 2 by 2 matrix, so let's see if we can figure out the eigenvalues for a 3 by 3 matrix. Every time we compute eigenvalues and eigenvectors we use this format, which can also be written as det(A - lambdaI) =0, where I is the Identity matrix I=((1, 0), (0, 1)). Learn to find complex eigenvalues and eigenvectors of a matrix. Matrix Eigenvalues Calculator - Symbolab. This calculator allows you to enter any square matrix from 2x2, 3x3, 4x4 all the way up to 9x9 size. That is, convert the augmented matrix A −λI...0 Step 2: Estimate the matrix A – λ I A – \lambda I A … . det(A - lambda vec(I))=det ((a-lambda, b), (c, d-lambda)) = (a-lambda)(d-lambda)-bc=0, which expands to the quadratic polynomiallambda^(2) - (a+d)lambda +(ad-bc)=0.. $\sqrt [\msquare] {\square}$ √ ☐. and the two eigenvalues are . Asking for help, clarification, or responding to other answers. . Works with matrix from 2X2 to 10X10. r/HomeworkHelp. More: Diagonal matrix Jordan decomposition Matrix exponential. For example, an Eigen value of 2, with vector 3, 4, I could have any other vector, example 6, 8, or 12, 16, etc... any scalar multiple. This gives us two (equivalent) ratios for the vector elements: $$y = \frac {\lambda - a} b x = \frac c {\lambda - d} x$$. eigenvector eigenvalue Section 8.8 Eigenvalues and Eigenvectors ( ) Solve: Ax x A= λ nn× Ax x− =λ 0 (A x− =λ I) 0 matrix vector ↑ vector ↑ Need to not be invertible, because if i( ) t was we would only have the trivial solution 0. So lambda is an eigenvalue of A. To find eigenvalues, we use the formula: A v = λ v Note: v, bold v, indicates a vector. It only takes a minute to sign up. Making statements based on opinion; back them up with references or personal experience. An easy and fast tool to find the eigenvalues of a square matrix. Then User account menu • [University mathematics: linear algebra] Find eigenvalues of a 2x2 matrix with a parameter. Intuition behind the rotation of space using a 2x2 matrix and eigen values? Steps to Find Eigenvalues of a Matrix. 21 1 P=8 01P Determine (0) Eigenspace of each eigenvalue and basis of this eigenspace (ii) Eigenbasis of the matrix Is the matrix … Building a source of passive income: How can I start? Calculate eigenvalues. This is referred to as the characteristic polynomial, where the characteristic polynomial always has two roots. And I think we'll appreciate that it's a good bit more difficult just because the math becomes a little hairier. Computing the eigenvalues comes down to finding the roots of $\lambda^2 -(a+d)\lambda + (ad-bc) = 0$. Section 5.5 Complex Eigenvalues ¶ permalink Objectives. I found this site: http://people.math.harvard.edu/~knill/teaching/math21b2004/exhibits/2dmatrices/index.html, Which shows a very fast and simple way to get Eigen vectors for a 2x2 matrix. So if the eigenvalues are $\lambda_1$ and $\lambda_2$, then assume $c\neq 0$ and then the claim is that the eigenvectors are $v_i = (\lambda_i-d,c)$. It will find the eigenvalues of that matrix, and also outputs the corresponding eigenvectors.. For background on these concepts, see 7.Eigenvalues … then the characteristic equation is . Sorry, JavaScript must be enabled.Change your browser options, then try again. This is the final calculator devoted to the eigenvectors and eigenvalues. Finding of eigenvalues and eigenvectors. First, we will create a square matrix of order 3X3 using numpy library. λ 1 =-1, λ 2 =-2. • In general, for a 2x2 matrix ab cd ⎡⎤ ⎢⎥ ⎣⎦, AI−λ = ab cd λ λ ⎡ − ⎤ ⎢ − ⎥ ⎣ ⎦. Given an eigenvalue $\lambda$ of the matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the associated eigenvector(s) are the vectors $\vec{v} = \begin{bmatrix}x\\y\end{bmatrix}$which satisfy $(\lambda I - A)\vec{v} = 0$. Given an n × n square matrix A of real or complex numbers, an eigenvalue λ and its associated generalized eigenvector v are a pair obeying the relation (−) =,where v is a nonzero n × 1 column vector, I is the n × n identity matrix, k is a positive integer, and both λ and v are allowed to be complex even when A is real. Just as the names of each of them sound, the general method is the "formal" method to use mathematically, following all the rules and producing some minor matrix determinant calculations along the way to find the final solution. Any multiple of an eigenvector $v$ is also an eigenvector, because $A(cv) = cAv = c\lambda v = \lambda (cv)$. So they must be both equal to 1. Here we are going to see how to find characteristic equation of any matrix with detailed example. so … Complex eigenvalues. • STEP 1: For each eigenvalue λ, we have (A −λI)x= 0, where x is the eigenvector associated with eigenvalue λ. Choosing $x$ to be the denominator of each fraction gives us the solutions on the web site: $$\vec{v} = \begin{bmatrix}b \\ \lambda - a \end{bmatrix}, Is that correct? Part 2, where they calculate the Eigen vectors is what I don't understand and have tried to prove but cannot. Eigenvalue Calculator. The Eigenvalues of a 2x2 Matrix calculator computes the eigenvalues associated with a 2x2 matrix. The next step is finding the roots/eigenvalues of the characteristic polynomial. Finding Eigenvalues and Eigenvectors - Duration: ... 2 Tricks to find Eigen value of 2x2 matrix - Duration: 5:22. If . Therefore, λ must solve λλ2−() 0a d ad bc++−=. [x y]λ = A[x y] (A) The 2x2 matrix The computation of eigenvalues and eigenvectors can serve many purposes; however, when it comes to differential equations eigenvalues and eigenvectors are most … How to include successful saves when calculating Fireball's average damage? How can I determine, within a shell script, whether it is being called by systemd or not? Also, determine the identity matrix I of the same order. Thanks for contributing an answer to Mathematics Stack Exchange! Each eigenvalue is with multiplicity 2, as \vec v is a vector of 2 dimensions. By3.4, this eigenvalue must be real. Computing the eigenvalues comes down to finding the roots of \lambda^2 -(a+d)\lambda + (ad-bc) = 0. \\) (enter a data after click each cell … Calculate eigenvalues and eigenvectors. 0 0 ::: 0 d n;n 1 C C C C A 0 B B B @ x1 x2 x n 1 C C C A = 0 B @ d1 ;1 x1 d2 ;2 x2 d n;nx n 1 C C = x Sorry, I did that wrong ... Eigen vectors are in the form c(a,b) where c is a constant. log in sign up. Since A2J, this eigenvalue must be 1 or 1, so det(A) = ( 1)2 = 12 = 1. These roots can be real or complex, and they do not have to be distinct. There are two methods for finding the determinant of a 3x3 matrix: the general method and the shortcut method. Understand the geometry of 2 × 2 and 3 × 3 matrices with a complex eigenvalue. The eigenvalues are immediately found, and finding eigenvectors for these matrices then becomes much easier. Example 5 Consider the matrix [5, 3, 0; -3, -5, 0; 2, -3, 1]. It's just solving the equations directly. It asks to find a real 2x2 matrix A with eigenvalues Λ = 1 and Λ = 4 and … Press J to jump to the feed. Using Property 3, we can compute the eigenvalues of the block [1, 0; 2, 1] and [1]. .$$\lambda_iv = (\lambda_i^2 - d\lambda_i,c\lambda_i).$$Definition : Let A be any square matrix of order n x n and I be a unit matrix of same order. How can a company reduce my number of shares? A I x −λ = This leads to an equation in called … .$$\begin{bmatrix} \lambda - a & -b \\ -c & \lambda - d \end{bmatrix} \begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}(\lambda - a)x - by \\ (\lambda - d)y - cx \end{bmatrix} = 0$$. That part you know already. Solve the characteristic equation, giving us the eigenvalues(2 eigenvalues for a 2x2 system) Characteristic Polynomial of a 3x3 matrix, compute the Eigenvalues and Eigenvectors of a 2x2 Matrix, Characteristic Polynomial of a 2x2 Matrix. This calculator allows to find eigenvalues and eigenvectors using the Characteristic polynomial. Example: Find Eigenvalues and Eigenvectors of a 2x2 Matrix. Choose your matrix! Find more Mathematics widgets in Wolfram|Alpha. Matrices are the foundation of Linear Algebra; which has gained more and more importance in science, physics and eningineering. How to find generalized Eigen vectors of a matrix with Eigen vectors already on diagonal? site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. Similarly, we can find eigenvectors associated with the eigenvalue λ = 4 by solving Ax = 4x: 2x 1 +2x 2 5x 1 −x 2 = 4x 1 4x 2 ⇒ 2x 1 +2x 2 = 4x 1 and 5x 1 −x 2 = 4x 2 ⇒ x 1 = x 2. \begin{bmatrix}\lambda - d \\ c \end{bmatrix}$$. For example: for (lambda=2), I might get the vector (3,4) - I get a different vector value, ie: (6,8). Let's find the eigenvector, v 1, associated with the eigenvalue, λ 1 =-1, first. Mathematics (A-Levels/Tertiary/Grade 11-12) 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. Matrix A: Find. Find eigenvalues and eigenvectors of the following matrix: By using Shortcut method for eigenvalues 1 0 0 then sub eigenvalues in the matrix to find eigenvectors. To find eigenvalues, we use the formula: A v = λ v Note: v, bold v, indicates a vector.where A = ((a,b), (d,c)) and v = ((x),(y))((a,b), (d,c))((x),(y))= lambda ((x),(y)), which can be written in components as ax + by = lambda xcx + dy = lambda yWe want to solve for non-zero solution, such that the system becomes(a- lambda)x + by=0 cx + (d-lambda)y =0We can prove that given a matrix A whose determinant is not equal to zero, the only equilibrium point for the linear system is the origin, meaning that to solve the system above we take the determinant and set it equal to zero.det ((a-lambda,b), (c, d-lambda))= 0. How do I handle a piece of wax from a toilet ring falling into the drain? $$Av = (a\lambda_i-ad + bc, c\lambda_i - cd + cd) = (a\lambda_i - (ad-bc),c\lambda_i),$$ • STEP 2: Find x by Gaussian elimination. All that's left is to find the two eigenvectors. Is there an easy formula for multiple saving throws? To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Fast way to calculate Eigen of 2x2 matrix using a formula, http://people.math.harvard.edu/~knill/teaching/math21b2004/exhibits/2dmatrices/index.html, MAINTENANCE WARNING: Possible downtime early morning Dec 2, 4, and 9 UTC…. First let’s reduce the matrix: This reduces to the equation: Eigenvectors for: Now we must solve the following equation: First let’s reduce the matrix: This reduces to the equation: There are two kinds of students: those who love math and those who hate it. By definition, if and only if-- I'll write it like this. I don't understand the other two cases (when b=0, or c=0), or I presume the case when b & c are non-zero. What are wrenches called that are just cut out of steel flats? Beware, however, that row-reducing to row-echelon form and obtaining a triangular matrix does not give you the eigenvalues, as row-reduction changes the eigenvalues of the matrix … To learn more, see our tips on writing great answers. For the first block, we have the sum of the eigenvalues equal 2 and their product equal 1. where A = ( (a,b), (d,c)) and v = ( (x), (y)) ( (a,b), (d,c)) ( (x), (y))= lambda ( (x), (y)), … Now we will compute complex eigenvalues:Before we start we should review what it means to have a complex number. Select the size of the matrix and click on the Space Shuttle in order to fly to the solver! So all three eigenvalues are equal to 1. $x^ {\msquare}$ x . X by Gaussian elimination... 0 matrix eigenvalues calculator - Symbolab on the Space Shuttle in order to fly the. ) '' widget for your website, blog, Wordpress, Blogger, or responding other... 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The trace method, as $\vec v$ is a constant by definition, if and only --! Spanned by u 2 = 1 1 is n't actually using the characteristic polynomial always has roots... Calculator allows you to enter any square matrix of order 3x3 using numpy library to! S ) way up to 9x9 size ; which has gained more more! Or 1 with distance terms of service, privacy policy and cookie policy example 5 Consider the matrix rotates scales! Start we should review what it means to have a complex eigenvalue, it has one repeated real eigenvalue July... Will create a square matrix of order 3x3 using numpy library the Commodore?! Of Linear Algebra ] find eigenvalues of a matrix, Blogger, or responding to other answers by! Commodore 64 because the math becomes a little hairier “ Post your answer ”, you to. While Harvard is quite clear, they are using the trace method, as that computes each eigenvector from other! Will create a square matrix from 2x2, 3x3, 4x4 all the way up to 9x9.... Matrix a −λI... 0 matrix eigenvalues calculator - Symbolab to finding the determinant of triangular... Of matrices to other answers -5, 0 ; 2, -3, -5, 0 2. Methods for finding the roots are complex we say that the matrix rotates and scales, the... In order to fly to the solver:... 2 Tricks to find generalized Eigen vectors what. | 2021-05-12T16:40:46 | {
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https://www.physicsforums.com/threads/lockers-math-problem.57654/ | # Lockers math problem
1. Dec 23, 2004
### T@P
there are 100 lockers lined up in a row, and for some unknown reason they are unlocked. for a similarly unknown reason there are 100 students lined up outside the hallway containing the 100 lockers. the first student goes and opens all the lockers. the second then goes and closes every second. the third either opens or closes every third locker, (changes the "state" of it) for example, the 6th locker is closed when the 3rd guy comes to it, so he opens it. after all 100 students go by, what lockers are open/closed? please note i dont think that you should list out all 100 lockers and their state, there is a pattern.
2. Dec 23, 2004
### Bartholomew
The state of each locker is altered a number of times equal to the total number of factors of the locker number. So the lockers with numbers that have an even number of factors (including 1) are closed, and the lockers with numbers that have an odd number of factors are open. (Not counting just prime factors, counting all factors, and numbering the lockers from 1)
Last edited: Dec 23, 2004
3. Dec 27, 2004
### NateTG
What do all numbers that have an odd number have in common? It's a pretty well-known property...1,4,9...?
4. Dec 27, 2004
### Bartholomew
That's cool!
5. Dec 28, 2004
### Gokul43201
Staff Emeritus
What do all numbers that have an odd number of factors have in common?
6. Dec 28, 2004
### Bartholomew
Of course, that's what he meant. I'm no expert, but: First you get the prime factorization of the number. Each factor can be generated by choosing some number of each prime factor (a number from 0 to the power of that factor) and multiplying them all together. So the total number of factors is (a + 1) * (b + 1) * (c + 1) ... where a, b, c, ... are the exponents on 2, 3, 5, ... in the prime factorization. So for this product to be odd, all of a + 1, b + 1, c + 1, ... must be odd, so all of a, b, c, ... must be even, so the original number is a perfect square (and vice versa, if it's a perfect square then it has an odd number of factors).
7. Dec 28, 2004
### NateTG
Nicely done. Yeah, Gokul I need to be more careful when I type/post.
8. Dec 29, 2004
### Gokul43201
Staff Emeritus
Me neither, but I believe that's how the experts do it too, but being experts they like to use fancy terms.
$$N = \Pi p_i ^{k_i}$$
$$\tau (N) = \Pi (k_i + 1)$$
$\tau (N)$, known simply as the tau function, is a multiplicative function that counts the number of divisors of a given number.
So, an "expert" might simply say that for $\tau (N)$ to be odd, all of $k_i$ must be even, or $k_i = 2m_i$. Which gives $N = \Pi p_i ^{2m_i} = (\Pi p_i ^{m_i} )^2 = M^2$.
9. Jan 8, 2005
### ShawnD
Is the answer 31? My physics teacher gave this problem as a bonus questions, so I made a C++ program to figure out all the perfect squares between 1 and 1000, and the program says the answer is 31.
I'll get an extra 10% on the next lab if I get this right, so it's kind of important that I get confirmation before Monday, January 10, 2005.
Last edited: Jan 8, 2005
10. Jan 8, 2005
### jamesrc
Select to read. (I'm editing to put into spoiler text because that's what other people did.)
Yes, presuming your question was for 1000 lockers (and students) and they all started off closed and they asked which ones were open after the process was over. (The original post was only for 100 lockers.) You didn't really need the progam, though: like the lay version of what Gokul said, any number with an odd number of factors will end up open, meaning all of the perfect square numbered doors will end up open (which you already know). So all you had to do was look at the square root of 1000 and truncate the decimal places (round down to the nearest integer) and that would be the answer (31).
11. Jan 8, 2005
### ShawnD
Thanks for the confirmation. When I get some free time, I will look closer at Goku's stuff to see what it means or at least how he got that.
*edit
that spoiler text is still very visible in a quote box :rofl:
12. Jan 8, 2005
### Gokul43201
Staff Emeritus
I suggest you read Bart's explanation. It's in a form that's a lot simpler to absorb. | 2017-11-22T04:17:16 | {
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http://tlpd.oeey.pw/fundamental-trigonometric-limit.html | # Fundamental Trigonometric Limit
Trigonometric Cofunctions. Given the following triangle: \hspace{4cm} the basic trigonometric functions are defined for 0 < θ < π 2 0 < \theta < \frac{\pi}{2} 0 < θ < 2 π as. The reason is that it's, well, fundamental, or basic, in the development of the calculus for trigonometric functions. Well worth it. Graph of Sine. Trigonometry Calculator (Pro) hack hints guides reviews promo codes easter eggs and more for android application. Sine, Cosine and Tangent. Pythagorean Identities. One of the first things to notice about the fundamental theorem of calculus is that the variable of differentiation appears as the upper limit of integration in the integral. Quotient Identities. So the inverse of sin is arcsin etc. Summary of the rules for verifying a trigonometric identity 1. You must enable JavaScript in order to use this site. Corrective Assignment. limit laws, greatest integer function, Squeeze Theorem. 2 on page 57!!! (Where b and c are real numbers and is a positive integer)n. Find the limit lim x→0 x csc x Solution to Example 7: We first use the trigonometric identity csc x = 1 / sin x lim x→0 x csc x = lim x→0 x / sin x = lim x→0 1 / (sin x / x) The limit of the quotient is used. , one containing a right angle (90°). Learning isn’t about memorizing facts to pass a test. Watch the best videos and ask and answer questions in 148 topics and 19 chapters in Calculus. One radian is the measure of the angle made from wrapping the radius of a circle along the circle’s exterior. FCP (Fundamental OF Computer. There are six functions that are the core of trigonometry. A Graphical Approach to Algebra and Trigonometry 5th Edition Pdf has a variety pictures that united to locate out the most recent pictures of A Graphical Approach to Algebra and Trigonometry 5th Edition Pdf here, and in addition to you can get the pictures through. We encourage teachers and other education stakeholders to email their feedback, comments, and recommendations to the Commission on. Apply the Squeeze Theorem to find limits of certain functions. Infinite Pre‑Algebra Infinite Algebra 1 Infinite Geometry Infinite Algebra 2 Infinite Precalculus Infinite Calculus; Integers, Decimals, and Fractions :: Naming decimal places and rounding. you’ll ever need to know in Calculus Objectives: This is your review of trigonometry: angles, six trig. The following indefinite integrals involve all of these well-known trigonometric functions. Type in any integral to get the solution, free steps and graph. Chapter 5 Analytic Trigonometry. One-Sided Limits - A brief introduction to one-sided limits. Fourier Series. But a key property of a trig function is that it can be made to have any periodicity. The trigonometric functions relate the angles in a right triangle to the ratios of the sides. Calculus-Specific Formulas. Use the fundamental trigonometric identities Contact If you are in need of technical support, have a question about advertising opportunities, or have a general question, please contact us by phone or submit a message through the form below. The equation in (1. There are six functions that are the core of trigonometry. Integrals of Trig Functions (5. Trigonometry Practice Problems for Precalculus and Calculus 1. Trigonometric identities quiz questions and answers, trigonometric ratios of allied angles multiple choice questions (MCQs) for online college degrees. Trig Laws Math Help. Inverse Trigonometric Ratio. Our online trigonometry trivia quizzes can be adapted to suit your requirements for taking some of the top trigonometry quizzes. Cos to Sin Step-by-Step Lesson- A lot of theory goes into understanding and explaining this one. The cotangent function (cot or cotg) may be used for the reciprocal of the tangent. Free practice questions for Precalculus - Fundamental Trigonometric Identities. 3 Problem 8PS. Corollary to FTC. Chapter 5 Analytic Trigonometry. For proper course placement, please: • Take the test seriously and honestly • Do your own work without any assistance. Pythagorean Identities. This theorem allows us to calculate limits by "squeezing" a function, with a limit at a point a that is unknown, between two functions having a common known limit at a. In this chapter, you will learn how to use the fundamental identities to do the following. Limits are the most fundamental ingredient of calculus. One of the first things to notice about the fundamental theorem of calculus is that the variable of differentiation appears as the upper limit of integration in the integral. But the fundamental theorem of algebra tells you that you can factor denominator as a product of linear terms and irreducible quadratic terms. One-Sided Limits – A brief introduction to one-sided limits. Copy and paste one of these options to share this book elsewhere. The basic trigonometric limit is \[\lim\limits_{x \to 0} \frac{{\sin x}}{x} = 1. 5 Limits of Trig Functions Notes 1. Magic Method; Probability. However, we can restrict those functions to subsets of their domains where they are one-to-one. Limit of Trigonometric Ratios In limit of trigonometric ratios we will learn how to find the limits to the values of sin θ, csc θ, cos θ, sec θ, tan θ and cot θ. The limit of a constant times a function is the constant times the limit of the function. Local Behavior. Some important formulas of limit and continuity are as follows:-1. 5B Limits Trig Fns 2 Theorem For every c in the in the trigonometric. There are several useful trigonometric limits that are necessary for evaluating the derivatives of trigonometric functions. Trigonometry is useful when setting up problems involving right triangles. The Fundamental Theorem of Calculus. Algebra and Trigonometry, 1st edition. T-Charts for the Six Inverse Trigonometric Functions. In this guide, I’ll let you know everything you need to know about trigonometry and radians for the SAT Math test and guide you through some practice problems. One radian is the measure of the angle made from wrapping the radius of a circle along the circle’s exterior. In Chapter 4, you studied the basic definitions, properties, graphs, and applica-tions of the individual trigonometric functions. Worksheet # 4: Basic Limit Laws 1. Download with Google Download with Facebook or download with email. Trigonometry is the study of triangles, which contain angles, of course. Pythagorean Identities. State the fundamental trigonometric limit and explain why it is true. Given the following triangle: \hspace{4cm} the basic trigonometric functions are defined for 0 < θ < π 2 0 < \theta < \frac{\pi}{2} 0 < θ < 2 π as. Tangent and Cot. Although the trigonometric functions are defined in terms of the unit circle, the unit circle diagram is not what we normally consider the graph of a trigonometric function. Basic trig functions - practice problems These problems are designed to help you learn basic trigonometry ("trig") functions and how to use your calculator correctly. Given this anchor, the derivatives of the remaining trigonometric functions can be. However recall that the period of tangent and cotangent is. Law of Sines Law of Cosines Law of Tangents Mollweid's Formula. Example 1: Example 2: Find the derivative of y = 3 sin 3 (2 x 4 + 1). This interactive is optimized for your desktop and tablet. Limits of Trigonometric Functions with Correction. We will first define the cosine and sine functions in. He considered every triangle—planar or spherical—as being inscribed in a circle, so that each side becomes a chord (that is, a straight line that connects two points on a curve or surface, as shown by the inscribed triangle ABC in. Said owners are not affiliated with Educator. Limit and Continuity of Trigonometric Functions Continuity of Sine and Cosine function Sine and Cosine are ratios defined in terms of the acute angle of a right-angled triangle and the sides of the triangle. Trigonometric or circular functions calculator for degrees or radians. Copy and paste one of these options to share this book elsewhere. Trigonometric Cofunctions. 2 Limits Analytically. A right angled triangle is the basis of all of trigonometry, however complex it is made finally. If you're seeing this message, it means we're having trouble loading external resources on our website. ) We can easily get a qualitatively correct idea of the graphs of the trigonometric functions from the unit circle diagram. you’ll ever need to know in Calculus Objectives: This is your review of trigonometry: angles, six trig. We can find trigonometry in almost every aspect of our lives, which is why it isn’t that hard to picture it and learn it. This is similar to what we do with trigonometric limits. Basic trigonometric functions Word trigonometry comes from Greek words trigonon and metron , in translation triangle and measurement. Use the one from last section or print one below! Packet. Calc 2 you want to be solid on trig though. Identities for the hyperbolic trigonometric functions are. An example of an identity with the variable x is 2x(3 – x) = 6x – 2x 2. Given lim x!2 f(x) = 5 and lim x!2 g(x) = 2, use limit laws to compute the following limits or explain why we cannot nd the limit. Graph of Sine/Cosine from Unit Circle. Worksheet Pages for AP Calculus AB This page requires Firefox/Mozilla/Netscape to view math symbols. •This outlines the basic procedure for solving and computing inverse trig functions •Remember a triangle can also be drawn to help with the visualization process and to find the easiest relationship between the trig identities. The typical "problem" that prevents the direct evaluation of the function at the limit point is "division by zero". Basic Concepts List Understand the concept of limit of a function as x approaches a number or infinity Know the basic trigonometric identities for sine. Trigonometric Formulas for Sum and Difference, Double Angle, Half Angle, Product and Periodicity Identities Limits. 4) Answer Key. The Limit - Here we will take a conceptual look at limits and try to get a grasp on just what they are and what they can tell us. Limit is a fundamental concept in calculus. Find limits of trigonometric functions by rewriting them using trigonometric identities. 3 Theorem 1. The Limit – Here we will take a conceptual look at limits and try to get a grasp on just what they are and what they can tell us. Example 1: Evaluate. Basic Derivatives “PLUS A CONSTANT” If the function f(x) is continuous on [a, b] and the first derivative exists on the interval (a, b), then there exists a number. Here are the inverse trig parent function t-charts I like to use. Review Albert's AP® Calculus math concepts, from limits to infinity, with exam prep practice questions on the applications of rates of change and the accumulation of small quantities. In this case, sinX=12/13, or 0. Szilárd András. The always-true, never-changing trig identities are grouped by subject in the following lists. 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Example 1 sin2x + cos2x = 1 is equivalent to cos2x = 1 – sin2x Example 2 cos x sin x tan x = is equivalent to tan xsin x =cos x ⋅. Limit Properties - Properties of limits that we'll need to use in computing limits. It is used when we consider di erentiation (to define derivatives) and integration (to define definite integrals). Graph of Sine. One of the simplest and most basic formulas in Trigonometry provides the measure of an arc in terms of the radius of the circle, N, and the arc’s central angle θ, expressed in radians. Try solving these on your own (without peaking at the solutions). Math 1411 - Calculus I Course Materials. This makes sense because all trigonometric functions are periodic, and hence their derivatives will be periodic, too. 3 Asymptotes. Basic trig, algebra factorization problems, algebra calculator online, 10th grade geometry, TRIGONOMETRY TRIVIA, christmas factor tree. Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Students can select values to use within the function to explore the resulting changes in the graph. Trigonometry is a branch of mathematics which studies relations among lengths of sides and angles in a triangle. The typical "problem" that prevents the direct evaluation of the function at the limit point is "division by zero". McKeague Chapter 4. The trigonometric functions relate the angles in a right triangle to the ratios of the sides. Let's start by stating some (hopefully) obvious limits: Since each of the above functions is continuous at x = 0, the value of the limit at x = 0 is the value of the function at x = 0; this follows from the definition of. REVIEW SHEETS. Tangent and Cot. Moreover, with careful observation, we can even see that is concave up when is positive and that is concave down when is negative. Elementary Education Prek-6. 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Trigonometric ratios of angles more than 360° An angle that is more than 360° implies that an object underwent a rotation that is more than one cycle about a fixed point. Graph of Tangent. 1) Answer Key. Antiderivatives; Arc Length; Chain Rule; Computing Integrals by Completing the Square; Computing Integrals by Substitution; Continuity; Differentiating Special Functions; First Derivative; Fundamental Theorem of Calculus; Infinite. Trigonometry is useful when setting up problems involving right triangles. Since the period is T, we take the fundamental frequency to be ω 0 =2π/T. It contains explanations and examples in 15 topical areas. 1 Using Fundamental Identities Objective: In this lesson you learned how to use fundamental trigonometric identities to evaluate trigonometric functions and simplify trigonometric expressions. By "solution" is meant the determination of all the parts of a triangle when enough parts are given to determine the triangle. Limits are the most fundamental ingredient of calculus. 0 sin lim 1. Recognize functions and graphs are useful to test your understanding. Fourier Series. The typical "problem" that prevents the direct evaluation of the function at the limit point is "division by zero". c) f) i) l) c) f) i) l) tan x 11m 2 tan 2 x 11m sm 5x 11m 0 sm6x tan x 11m a) d) g) a) 11m Sin Y 21 11m 1 0 sin33x sm 31 11m sm 6x 11m Evaluate lim h) k) sm 6x 11m COS 11m 1 0 tan32x smx 11m 7x Sin 2 3X 11m 11m 11m cos x cos x Describe how you evaluated the limit in part a) Determine each limit. The trigonometric functions sine and cosine have four important limit properties: You can use these properties to evaluate many limit problems involving the six basic trigonometric functions. Hi everyone, I would like to know, since the lim theta --> 0 (sine theta / theta) = 1, what is the limit theta --> 0 of cosine and tangent? Thank you Basic Trigonometric Limit | Physics Forums. Improve your math knowledge with free questions in "Trigonometric identities I" and thousands of other math skills. The basic trigonometric limit is Using this limit, one can get the series of other trigonometric limits: Further we assume that angles are measured in radians. In this guide, I’ll let you know everything you need to know about trigonometry and radians for the SAT Math test and guide you through some practice problems. c) f) i) l) c) f) i) l) tan x 11m 2 tan 2 x 11m sm 5x 11m 0 sm6x tan x 11m a) d) g) a) 11m Sin Y 21 11m 1 0 sin33x sm 31 11m sm 6x 11m Evaluate lim h) k) sm 6x 11m COS 11m 1 0 tan32x smx 11m 7x Sin 2 3X 11m 11m 11m cos x cos x Describe how you evaluated the limit in part a) Determine each limit. The course is organized around the foundational concepts of calculus: I. 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This makes sense because if we are taking the derivative of the integrand with respect to x, it needs to be in either (or both) the limits of integration. The formula that will help us is the tangent: Using a calculator, the value of tan(12) is 0. Although we have considered only transformations of sine and cosine, the same rules apply to all the trigonometric functions. Report Abuse. In this guide, I’ll let you know everything you need to know about trigonometry and radians for the SAT Math test and guide you through some practice problems. Worksheets for MA 113 Algebraic Evaluation of Limits, Trigonometric Limits Worksheet # 7: The Intermediate Value Theorem The Fundamental Theorems of Calculus. Inverse Trigonometric Functions - Trigonometric Equations Dr. This can be viewed as a version of the Pythagorean theorem, and follows from the equation x2 + y2 = 1 for the unit circle. We encourage teachers and other education stakeholders to email their feedback, comments, and recommendations to the Commission on. Trigonometry has important applications in many branches of pure mathematics as well as of applied mathematics and, consequently, much of science. By "solution" is meant the determination of all the parts of a triangle when enough parts are given to determine the triangle. Table of Trigonometric Identities. Some important formulas of limit and continuity are as follows:-1. Inverse Trig Functions Intro to Limits Overview Definition One-sided Limits When limits don't exist Infinite Limits Summary Limit Laws and Computations Limit Laws Intuitive idea of why these laws work Two limit theorems How to algebraically manipulate a 0/0? Indeterminate forms involving fractions Limits with Absolute Values Limits involving. Properties of Limits Basic Concepts Lines. For example, in calculus, trigonometric functions are defined for arbitrary real numbers. Before discussing those. Local Minimum. Sines and cosines are two trig functions that factor heavily into any study of trigonometry; they have their own. Questions are organized in Practice Tests, which draw from various topics taught in Trigonometry; questions are also organized by concept. Patricia Edmonds: Math 1. Recognize functions and graphs are useful to test your understanding. | 2019-11-13T07:55:09 | {
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http://math.stackexchange.com/questions/541531/largest-bounded-square | Largest bounded square
Suppose I have a triangular land-plot, but some part of it (the yellow part) is unusable. I want to build a square house on the usable (white) part. The house may be rotated (but must be square). What is the maximum area I can build?
NOTES:
• Both triangles in the picture are right-angled and isosceles.
• When $0<y<1$, the white part is a quadrilateral whose angles are: 90, 135, 90, 45. Its area is $\frac{1+y(2-y)}{4}$.
EDIT (following a hint by Ross Millikan): We can get a lower bound by considering two possible locations: an axis-aligned square in the lower-left corner (pink), and a 45-degree-rotated square in the top corner (green):
• In the lower-left corner, the vertical side of the square can be at most $y$, and the diagonal can be at most $\sqrt{\frac{1}{2}}$, therefore the side can be at most $min(y,\frac{1}{2})$.
• In the top corner, the vertical diagonal can be at most $\frac{1}{2}(1+y)$, and the horizontal diagonal can be at most $1-y$, therefore the side can be at most $\sqrt{\frac{1}{2}} min (\frac{1}{2}(1+y),1-y)$. This expression is maximized when $y=\frac{1}{3}$, and the side length is $\sqrt{\frac{1}{2}} \frac{2}{3} \approx 0.47$. When $y>\frac{1}{3}$, a square at the top corner must become smaller because of the horizontal diagonal, but, we don't have to push the square towards the new top corner - we can leave it at its previous position $(\frac{1}{3},\frac{2}{3})$, and have the same side-length 0.47 .
The following graph (drawn by http://rechneronline.de/function-graphs/) shows the side lengths of both squares:
And the following graph shows the areas of both squares, relative to the area of the white part:
NOW, MY QUESTION IS: Is there a way to build a larger house (a larger bounded square)?
My intuition is that there must be a way. If in some cases the largest square is at 0 degrees, and in other cases the largest square is rotated at 45 degrees, then between these cases, there must be a way to build a square rotated between 0 to 45 degrees, with a larger area.
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Where you wrote "equilateral", you meant "isosceles". An equilateral triangle is one where all three sides have equal length, such a triangle cannot be right-angled. An isosceles triangle is one with (at least) two sides of equal length. – Daniel Fischer Oct 27 '13 at 14:38
Thanks, I corrected the question. – Erel Segal-Halevi Oct 27 '13 at 14:39
@ErelSegalHalevi I think I found a case where it is profitable to rotate/stretch the square. As you surmise it's between the cases; please see my answer below. – coffeemath Oct 29 '13 at 14:29
If $y=\sqrt{2}-1 \approx 0.414$ the two maximal squares (flush with the left side and with the diagonal) have the same sidelengths. This was obtained by equating the two formulas in the range $(1/3,1/2)$ wherein they are respectively $y$ and $(1/\sqrt{2})(1-y).$ Note that this $y$ lies in $(1/3,1/2)$ as it should. Each of these squares can be realized with one corner at $(0,y),$ and a sketch reveals there is room to rotate either one of them a bit into the white region. After that one could expand the square.
I took the square aligned to the vertical, and supposed we kept its corner at $(0,y)$ and took another corner at $(z,0)$. Then the remaining two corners are at $(y+z,z)$ (the rightmost corner, tipped up from the $x$ axis) and at $(y,y+z)$ (the topmost corner). In order for these two corners to remain in the white region we need both $y+2z \le 1$ and $2y+z \le 1$. It seems clear the restriction from the upper corner will be the more restrictive, at least for small $z$. So I set $2y+z=1$ and got $$z=3-2\sqrt{2},\ y=\sqrt{2}-1,$$ giving the sidelength of $\sqrt{y^2+z^2}=\sqrt{20-14\sqrt{2}}$ or approximately $0.4483,$ which beats the $0.414$ from the aligned squares by a small amount.
By the way I did check the fourth corner is in the white area, since $y+2z=5-3\sqrt{2} \approx 0.757.$
To check that nothing has entered the yellow region, note that the angle of "rotation", that is, the angle $OYZ$ where $O=(0,0),Y=(0,y),Z=(z,0)$ is $\arctan(z/y)=\pi/8.$ This means the rotated square is situated so that its right angle cuts the 135 degree angle at $Y$ in half, with $45/2$ degrees on either side. Thus the angle made by the top side of the rotated square with the vertical is $22/2+90<135.$ So nothing has ended up in the yellow region.
ADDED: The OP has asked more generally about the "tipped square" case when $1/3<y<1/2.$ With the same square as above, having one corner at $(0,y)$ and another at $(z,0)$, we have again the other two corners at $(y+z,z)$ and $(y,y+z)$ where the latter is uppermost. For this upper point to lie on or below $x+y=1$ requires $2y+z\le 1$, so we just define $z=1-2y$ and check things. Note first that $z \ge 0$ since $y \in (1/3,1/2).$ We also need to check that the lower rotated corner $(y+z,z)$ lies below $x+y=1$, and with $z$ taken as $1-2y$ this condition is exactly $y>1/3$ so we're OK on that score.
Finally we need to check that the top side of the square has not rotated into the forbidden yellow area; this will be so iff $z/y \le 1$ since the rotation angle is $\arctan(z/y)$ and the angle from the vertical to the topmost vertex of the rotated square is $\pi/2+\arctan(y/x).$ Now $z/y<1$ becomes $y>1/3$ once $z=1-2y$ has been put in, so we now have checked all so as to show the stretched/rotated square lies entirely in the white region. Its sidelength is $$\sqrt{y^2+z^2}=\sqrt{5y^2-4y+1}.$$
It is interesting to look at the three graphs on the interval $(1/3,1/2)$ where all three bounds apply. That is, we graph each of $y,\ (1/\sqrt{2})(1-y),\ \sqrt{5y^2-4y+1}$ over $[1/3,1/2].$ We find that the third curve (which is the upper branch of a hyperbola) lies above each of the others in the interior of $[1/3,1/2]$ and agrees with the maximum of the first two at the two endpoints, and furthermore has the same tangents at those endpoints as may be checked via derivatives.
Conclusion: For $1/3<y<1/2$ we can always do better by a rotation/stretch from the "aligned squares" mentioned in thee question, and the above describes how much better at such $y$ values. We should emphasize that this is not a proof that some other orientation/choice of where the square is to be placed might beat this rotated one. A more general analysis would be necessary for that, and seems to me would be difficult even to set up conveniently.
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Thanks! Can you give a formula for the largest side as a function of y in the range (1/3,1/2) ? This will nicely complement the two formulas that I calculated for the other ranges. – Erel Segal-Halevi Oct 29 '13 at 14:38
@ErelSegalHalevi I'll look into that, but my approach assumes one keeps $y$ where it is and picks another point $(0,z)$, however this approach may not give the optimal place either, since the tilted rectangle need not touch the bottom, or even have a vertex at $(0,y)$. Would that particular optimum be of interest, say for comparisons? [it would then give another lower bound for the overal max sidelength, which would presumably win out for $y \in (1/3,1/2).$] – coffeemath Oct 29 '13 at 16:52
@ErelSegalHalevi Please have a look in the answer above after "ADDED." Note the formula there is not a guaranteed largest side formula, but just another lower bound for it which beats the other two lower bounds you have mentioned in your post. – coffeemath Oct 30 '13 at 1:33
Please look at the new graph I created: i.stack.imgur.com/QyMPP.png Is this what you meant? – Erel Segal-Halevi Oct 30 '13 at 7:08
It goes horizontally because, as y grows, there is no need to "push" the square upwards in a way that will make it smaller, you can just leave it where it is. In general, as y grows the white part also grows, so the size of a bounded square must be non-decreasing. – Erel Segal-Halevi Oct 30 '13 at 18:32
Hint: You have two obvious orientations for the house. You can put it in the lower left corner, which allows a size of $\max (y,\frac 12)$ You can put it in the new right angle. One constraint on this location is the opposite corner must stay within the square. How high is the new right angle corner? The diagonal must be less than or equal to this. Also the left corner must not extend outside the square. How long is the new diagonal? Take the minimum of these, and compare with the first answer.
We have not shown that some other location isn't better.
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Thanks, your hint gives a lower bound on the largest bounded square. I edited my question to reflect this. But, my intuition is that there can be a larger square, perhaps rotated in an angle between 0 and 45 degrees. Can you give my another hint, how to find this square? – Erel Segal-Halevi Oct 28 '13 at 9:52
Specifically, when y is between 1/3 and 1/2, none of the two squares you mention is tightly bounded by the white part, so, it seems in this range there must be a larger square. – Erel Segal-Halevi Oct 28 '13 at 10:06
I would be surprised if another orientation works, but your new figure suggests a new location: slide it horizontally along the bottom until it touches the two diagonals. Good work on the problem. These packing problems are often difficult. For a given configuration it is easy enough to calculate the largest square, but it is hard to prove that you have all the configurations considered. – Ross Millikan Oct 28 '13 at 13:11 | 2015-10-05T02:09:58 | {
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https://mathhelpboards.com/threads/integer-concepts.9259/ | # Integer Concepts
#### psc109
##### New member
Can some one explain to my why an integer equation that starts with 1 has a -1 at the end of the equation.
example:
1 + 2 + 4 + 8 + 16 ........ + 2 ^ N = 2 x ( 2 ^ N ) - 1
Conceptually where does the rule come from that there is a minus at the end of the equation.
It starts with an odd number so the answer must be an odd number and that's why -1 is subtracted at the end then how come if an equation started with -5 or -8 you would not subtract the -5 or -8 in the end?
I've seen numerical explanations but they confuse me, can it be explained with words?
#### chisigma
##### Well-known member
Re: Intiger Concepts
Can some one explain to my why an integer equation that starts with 1 has a -1 at the end of the equation.
example:
1 + 2 + 4 + 8 + 16 ........ + 2 ^ N = 2 x ( 2 ^ N ) - 1
Conceptually where does the rule come from that there is a minus at the end of the equation.
It starts with an odd number so the answer must be an odd number and that's why -1 is subtracted at the end then how come if an equation started with -5 or -8 you would not subtract the -5 or -8 in the end?
I've seen numerical explanations but they confuse me, can it be explained with words?
The general formula for a geometric sum is...
$\displaystyle S_{n} = \sum_{k=0}^{n} a^{k} = \frac{1 - a^{n+1}}{1-a}\ (1)$
Setting a=2 You obtain $S_{n} = 2^{\ n+1} - 1$...
Kind regards
$\chi$ $\sigma$
#### psc109
##### New member
Re: Intiger Concepts
Thank you Chisigma, the example you gave does not start with one and does not have a negative 1 at the end.
Why is it there in the the example i was given? Why subtract 1 at the end when if say I started it 8 I would not subtract 8 at the end.
#### chisigma
##### Well-known member
Re: Intiger Concepts
Thank you Chisigma, the example you gave does not start with one and does not have a negative 1 at the end.
Why is it there in the the example i was given? Why subtract 1 at the end when if say I started it 8 I would not subtract 8 at the end.
I confess not to understand exactly Your question, expecially the words 'it start with 1 and ends with -1'... ewerywhere in Math we meet expressions like 'a = b' and no general connection exists between the 'head' of a and the 'tail' of b... may be however that my misundestanding depends from my poor knowledege of the english language ...
Kind regards
$\chi$ $\sigma$
#### HallsofIvy
##### Well-known member
MHB Math Helper
One way of looking at it is this: Let $$S= 1+ 2+ 2^2+ 2^3+ ... 2^n$$
Subtract 1 from both sides: $$S- 1= 2+ 2^2+ 2^3+ ...+ 2^n$$.
The "+1" on the right has become "-1" on left!
Now, factor a "2" on the right: $$S- 1= 2(1+ 2+ 2^2+ ...+ 2^{n-1})$$
That "$$1+ 2+ 2^2+ 2^{n-1}$$ on the right is almost the "$$1+ 2+ 2^2+ ...+ 2^n$$" we had before. Make it that way by "adding and subtracting $$2^n$$ inside the parentheses:
$$S-1= 2(1+ 2+ 2^2+ ...+ 2^{n-1}+ 2^n- 2^n)= 2(S- 2^n)$$
$$S- 1= 2S- 2^{n+1}$$
Solve that for S by subtracting 2S from both sides and adding 1 to both sides:
$$-S= 1- 2^{n+1}$$
so $$S= 2^{n+1}- 1$$
That "-1" at the end is due to the "1" on the left in "S- 1" which is, itself, due to the fact that there was a "+1" on the right in the initial equation. | 2020-09-28T12:33:04 | {
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https://math.stackexchange.com/questions/2781798/problem-with-my-intuition-concerning-probabilities-of-dependent-events-and-condi | # Problem with my intuition concerning probabilities of dependent events and conditional probability
I'm having an issue understanding what's going on in the following problem.
Suppose we have a urn containing four distinct numbered balls. A match occurs if the $m^{th}$ ball is obtained on the $m^{th}$ draw.
Suppose I want to determine the probability that I will obtain a match on the second draw. Denote this event by $A_2$.
By conditional probability, we have $$\Pr(A_2) = \frac{3}{4}\cdot \frac{1}{3} = \frac{1}{4}$$
But the above probability doesn't make sense to me intuitively. I would think that the probability of $A_2$ is $\frac{1}{3}$.
Help?
Here are two ways to think about the problem . . .
The first way matches the calculations you had issues with.
Let $B$ be the event that the first ball drawn is ball $2$, and let $B'$ be the event that the first ball drawn is one of the other $3$ balls.
For the first draw, all $4$ balls are equally likely, hence $P(B)=\frac{1}{4}$, and $P(B')=\frac{3}{4}$.
If event $B$ occurs, $A_2$ can't occur, so $P(A_2|B)=0$.
If event $B'$ occurs, there are $3$ balls left, one of which is $A$, and all are equally likely, hence $P(A_2|B') = \frac{1}{3}$.
Then we get \begin{align*} P(A_2) &= P(B)P(A_2|B)+P(B')P(A_2|B')\\[4pt] &=\left({\small{\frac{1}{4}}}\right)(0)+\left({\small{\frac{3}{4}}}\right)\left({\small{\frac{1}{3}}}\right)\\[4pt] &={\small{\frac{1}{4}}}\\[4pt] \end{align*} In concept, in order for $A_2$ to occur, first $B'$ must occur, and next, given that $B'$ has occurred, ball $2$ must be the next ball of the $3$ remaining balls.
The other way is even simpler . . .
Ball $2$ must occur on some draw.
By symmetry, all draw orders of the $4$ balls are equally likely, so for $1\le k\le 4$, the probability that ball $2$ occurs on draw $k$ is $\frac{1}{4}$.
Hence $P(A_2)=\frac{1}{4}$.
The event that the second draw is a match consists of two mutually excluding events:
(A) The first draw is a match and the second draw is a match.
(B) The first draw is not a match and the result of the first draw is not the second thing and the second draw is a match.
For (A), we have $$\frac14\frac13,$$ that is we have to draw the first ball out of the four balls then we have to draw the second ball out of the three remaining balls.
For (B), we have $$\frac24\frac13$$
because we have to draw either the third or the fourth ball first and then the second ball out of the remaining three balls.
So, the result is
$$\frac1 {12}+\frac2 {12}=\frac14.$$
I would think that the probability of $A_2$ is $\frac14$.
The first ball makes the difference. Note that: \begin{align}P(A_2)=P(2_2)=&P(1_1\cap 2_2)+P(2_1\cap 2_2)+P(3_1\cap 2_2)+P(4_1\cap 2_2)=\\ &P(1_1)\cdot P(2_2|1_1)+P(2_1)\cdot P(2_2|2_1)+P(3_1)\cdot P(2_2|3_1)+P(4_1)\cdot P(2_2|4_1)=\\ &\frac14\cdot \frac13+\color{red}{\frac14\cdot 0}+\frac14\cdot \frac13+\frac14\cdot \frac13=\\ &\frac14.\end{align}
Similarly, $P(A_3)=\frac14$, because the first two balls make the difference.
Similar question: Three white and one red ball probability, where the ordered players must match the winning orders. | 2020-07-15T07:51:57 | {
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https://math.stackexchange.com/questions/2694786/proving-that-a-set-of-functionals-mapping-p-1x-to-the-reals-is-a-basis-for-t | # Proving that a set of functionals mapping $P_1(x)$ to the reals is a basis for the dual space of $P_1(x)$
Given a vector space $V = P_1(\mathbb{R})$ and $f_1,f_2 \in V^*$ where $V^*$ is the dual space of $V$ and $$f_1 = \int_0^1 p(t) \, dt \quad\text{and}\quad f_2 = \int_0^2 p(t) \, dt$$ prove that $\{f_1,f_2\}$ is a basis for $V^*$ and find a basis for $V$ for which it is the dual basis.
I tried using the classical argument for a basis.
Let $p(x) = a + bx$.
By a theorem, we know that $\dim(V) = \dim(V^*)$. Since we have 2 vectors and $\dim(V) = 2$ it remains to show that $\{f_1,f_2\}$ is linearly independent in order for it to be a basis. In order for $\{f_1,f_2\}$ to be linearly independent, it must satisfy that for any $v = p(x) \in V$, $$c_1f_1(v) + c_2f_2(v) = 0 \implies c_1 = c_2 = 0$$
I get as far as: $$c_1(a+\frac{b}{2}) + c_2(2a + 2b) = 0$$
However, since $a,b$ are both elements of $\mathbb{R}$, the system is underdetermined and I have an infinite number of solutions. i.e. $$c_1(a+\frac{b}{2}) = c_2(-2a - 2b)$$
Wouldn't this suggest that $\{f_1,f_2\}$ is not linearly independent?
The solution manual I am looking at only offers the following: We know that \begin{align*} f_1(p(x)) &= f_1(a+bx) = a+\frac{b}{2} \\ f_2(p(x)) &= f_2(a+bx) = 2a+2b \end{align*}
Then, it goes on to say that: $$a+bx = (a+\frac{b}{2})(2-2x) + (2a+2b)(-\frac{1}{2} + x)$$
Firstly, I have no idea how it arrived at this result. Second of all, I do not know how this would help prove linear independence for $\{f_1,f_2\}$. Any help would be appreciated.
Edit
Given that $\{f_1, f_2\}$ form the dual basis of a particular basis $\{v_1,v_2\}$ for some $v_1,v_2 \in V$, we know that $f_i(v_j) = \delta_{ij}$ where $\delta_{ij}$ is the Kronecker delta function. This implies that, letting $v_1 = p_1(x) = a_1 + b_1x$ and $v_2 = p_2(x) = a_2 + b_2x$ \begin{align*} 1 &= f_1(v_1) = a_1+\frac{b_1}{2} \\ 0 &= f_2(v_1) = 2a_1+2b_1 \end{align*}
Solving this system, we get that $v_1 = 2 - 2x$.
Similarly, \begin{align*} 0 &= f_1(v_2) = a_1+\frac{b_1}{2} \\ 1 &= f_2(v_2) = 2a_1+2b_1 \end{align*}
Solving gives us that $v_2 = -\frac{1}{2} + x$.
From this we get that the basis of $V$ for which $\{f_1, f_2\}$ is the dual basis is $\{(2 - 2x), (-\frac{1}{2} + x)\}$.
However, I am still struggling to prove linear independence in order to show that $\{f_1, f_2\}$ is a basis in the first place.
For $f_1,f_2$ to be linearly independent, you need to check that
$$c_1 f_1 + c_2 f_2 = 0_{V^{*}} \implies c_1 = c_2 = 0.$$
The element $c_1 f_1 + c_2 f_2$ is a linear functional so you need to check that if $c_1 f_1 + c_2 f_2$ is the zero functional then $c_1 = c_2 = 0$. This means that if
$$(c_1 f_1 + c_2 f_2)(v) = 0$$
for all polynomials $v = ax + b$ then you need to show that $c_1 = c_2 = 0$.
Let's write this explicitly. Assume that $c_1 f_1 + c_2 f_2 = 0_{V^{*}}$. Then for all $v = ax + b$ we have
$$c_1f_1(v) + c_2f_2(v) = c_1 \left( a + \frac{b}{2} \right) + c_2 \left( 2a + 2b \right) = 0.$$
Choosing $a = 1, b = -2$ we get
$$-4c_2 = 0 \implies c_2 = 0.$$
Similarly, choosing $a = 1, b = -1$, we get
$$\frac{1}{2} c_1 = 0 \implies c_1 = 0$$
so indeed the linear functionals $f_1,f_2$ are linearly independent.
The solution manual already gives you the dual basis. Recall that if $v_1,\dots,v_n$ is a basis for $V$ and $f^1, \dots, f^n$ is the corresponding dual basis of $V^{*}$, we have for any vector $v$ the identity
$$v = f^1(v) v_1 + \dots + f^n(v) v_n.$$
In your case, $v_1 = 2 - 2x$ and $v_2 = -\frac{1}{2} + x$.
• Isn't choosing a particular value for a and b getting rid of the generality of the equation. i.e. shouldn't linear independence hold for any values of a and b rather than some particular values of a and b. Also, did the solution manual arrive at these values for v1 and v2 through some particular method or were they only chosen due to their ability to form a vector v in V of the form a + bx when multiplied by the corresponding functional applied to v and put in linear combination? Mar 17 '18 at 19:40
• Ah, I think I may have answered my own comment here. You can plug in any values you want for a and b since the linear independence of functionals should hold for all v = a + bx. Therefore, they must hold for any arbitrarily chosen values for a and b. Choosing certain a's and b's allows us to get rid of c_1 or c_2 allowing us to solve for the other and showing that they are equal to 0. Mar 18 '18 at 0:17
• @OrrenRavid: They were chosen due to their ability to form a general vector in $V$ when multiplied by the corresponding functionals applied to $v$ and put in a linear combination, but this choice isn't arbitrary. Since $f_1,f_2$ are linearly independent, the vectors $v_1,v_2$ are determined uniquely by this condition. Mar 18 '18 at 0:30
• I agree that the choice isn't arbitrary when choosing $v_1$ and $v_2$ as I say in my edit to my original question. However, am I correct in my comment about the choices of a and b when proving linear independence? Mar 18 '18 at 0:58
• @OrrenRavid: Yep, you are correct. Mar 18 '18 at 2:12 | 2021-09-21T18:25:57 | {
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http://math.stackexchange.com/questions/218421/what-are-the-practical-applications-of-the-taylor-series/218424 | # What are the practical applications of the Taylor Series?
I started learning about the Taylor Series in my calculus class, and although I understand the material well enough, I'm not really sure what actual applications there are for the series.
Question: What are the practical applications of the Taylor Series? whether it's in a mathematical context, or in real world examples.
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Related: Motivating Infinite Series. – Mike Spivey Dec 19 '12 at 17:40
One reason is that we can approximate solutions to differential equations this way: For example, if we have
$$y''-x^2y=e^x$$
To solve this for $y$ would be difficult, if at all possible. But by representing $y$ as a taylor series $\sum a_nx^n$, we can shuffle things around and determine the coefficients of this taylor series, allowing us to approximate the solution around a desired point.
It's also useful for determining various series. For example:
$$\frac 1 {1-x}=\sum_{n=0}^\infty x^n$$ $$\frac 1 {1+x}=\sum_{n=0}^\infty (-1)^nx^n$$ Integrate: $$\ln(1+x)=\sum_{n=0}^\infty \frac{(-1)^nx^{n+1}}{n+1}$$ Substituting $x=1$ gives
$$\ln 2=1-\frac12+\frac13-\frac14+\frac15-\frac16\cdots$$
There are also applications in physics. If a system under a conservative force (one with an energy function associated with it, like gravity or electrostatic force) is at a stable equilibrium point $x_0$, then there are no net forces and the energy function is concave upwards (the energy being higher on either side is essentially what makes it stable). In terms of taylor series, the energy function $U$ centred around this point is of the form
$$U(x)=U_0+k_1(x-x_0)^2+k_2(x-x_0)^3\cdots$$
Where $U_0$ is the energy at the minimum $x=x_0$. For small displacements the high order terms will be very small and can be ignored. So we can approximate this by only looking at the first two terms:
$$U(x)\approx U_0+k_1(x-x_0)^2\cdots$$
Now force is the negative derivative of energy (forces send you from high to low energy, proportionally to the energy drop). Applying this, we get that
$$F=ma=mx''=-2k_1(x-x_0)$$
Rephrasing in terms of $y=x-x_0$:
$$my''=-2k_1y$$
Which is the equation for a simple harmonic oscillator. Basically, for small displacements around any stable equilibrium the system behaves approximately like an oscillating spring, with sinusoidal behaviour. So under certain conditions you can replace a potentially complicated system by another one that's very well understood and well-studied. You can see this in a pendulum, for example.
As a final point, they're also useful in determining limits:
$$\lim_{x\to0}\frac{\sin x-x}{x^3}$$ $$\lim_{x\to0}\frac{x-\frac16x^3+\frac 1{120}x^5\cdots-x}{x^3}$$ $$\lim_{x\to0}-\frac16+\frac 1{120}x^2\cdots$$ $$-\frac16$$
which otherwise would have been relatively difficult to determine. Because polynomials behave so much more nicely than other functions, we can use taylor series to determine useful information that would be very difficult, if at all possible, to determine directly.
EDIT: I almost forgot to mention the granddaddy:
$$e^x=1+x+\frac12x^2+\frac16x^3+\frac1{24}x^4\cdots$$ $$e^{ix}=1+ix-\frac12x^2-i\frac16x^3+\frac1{24}x^4\cdots$$ $$=1-\frac12x^2+\frac1{24}x^4\cdots + ix-i\frac16x^3+i\frac1{120}x^5\cdots$$ $$=\cos x+i\sin x$$ $$e^{ix}=\cos x+i\sin x$$
Which is probably the most important equation in complex analysis. This one alone should be motivation enough, the others are really just icing on the cake.
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Granddaddy is used to approximate $\exp$, $\sin$ and $\cos$, right? – krlmlr Oct 22 '12 at 11:37
@user946850 without the imaginary part, yes. Taylor series can also be used to approximate these functions in computers to pretty high accuracy. $\sin x\approx x-\frac16x^3$ has an error of at most $8\%$. Adding the next term reduces that to less than $0.5\%$ and using this in conjunction with $\sin(\pi/2-x)=\cos x$ and $\cos x\approx1-\frac12 x^2+\frac1{24}x^4$ can lower this even further. There are many other approximations that exist as well (the CORDIC algorithm, Chebyshev approximation, etc.) but this is sometimes used in practice. – Robert Mastragostino Oct 22 '12 at 15:43
In the calculator era, we often don't realize how deeply nontrivial it is to get an arbitrarily good approximation for a number like $e$, or better yet, $e^{\sin(\sqrt{2})}$. It turns out that in the grand scheme of things, $e^x$ is not a very nasty function at all. Since it's analytic, i.e. has a Taylor series, if we want to compute its values we just compute the first few terms of its Taylor expansion at some point.
This makes plenty of sense for computing, say, $e^{1/2}: 1+1/2+1/2!(1/2)^2+1/3!(1/2)^3+...$ is obviously going to converge very quickly: $1/4!2^4<1/100$ and $1/5!2^5<1/1000$, so we know for instance we can get $e^{1/2}$ to $2$ decimal places by summing the first $5$ terms of the Taylor expansion.
But why should this work for computing something like $e^{100}$? Now the expansion looks like $1+100+100^2/2+100^3/3!+...$, and initially it blows up incredibly fast. This is where analytic functions really show how special they are: the denominators $n!$ grow so fast that it doesn't matter what $x^n$ we have in the numerators, before too long the series will converge. That's the essence of the Taylor approximation: analytic functions are those that are unreasonably close to polynomials.
There are much faster methods for getting approximations like the one for $\sqrt{e}$, in theory: using Newton's method to solve $x^2-e=0$ will give you an approximation to $\sqrt{e}$ accurate to a number of places that goes like the square of the number of iterations you've done. But how do we apply Newton's method here? The first formula is $$x_1=x_0-\frac{2x_0}{x_0^2-e}$$ So, if we want a decimal expansion of $\sqrt{e}$, we'd better be able to get one of $x_0^2-e$. And how are we going to get that? The Taylor series.
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Taylor Series are studied because polynomial functions are easy and if one could find a way to represent complicated functions as series (infinite polynomials) then one can easily study the properties of difficult functions.
1. Evaluating definite Integrals: Some functions have no antiderivative which can be expressed in terms of familiar functions. This makes evaluating definite integrals of these functions difficult because the Fundamental Theorem of Calculus cannot be used. If we have a polynomial representation of a function, we can oftentimes use that to evaluate a definite integral.
2. Understanding asymptotic behaviour: Sometimes, a Taylor series can tell us useful information about how a function behaves in an important part of its domain.
3. Understanding the growth of functions
4. Solving differential equations
I'm pretty sure this is not all but with a little research you can find as many as possible.
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The applications of Taylor series is mainly to approximate ugly functions into nice ones(polynomials)!
Example: Take $f(x) = \sin(x^2) + e^{x^4}$. This is not a nice function, but it can be approximated to a polynomial using Taylor series.
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What do you mean an Ugly Function? – Link Oct 22 '12 at 1:39 edited the asnwer – Citizen Oct 22 '12 at 1:41
Taylor series provide the basic method for computing transcendental functions such as $e^x$, $\sin x$, and $\cos x$.
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A good example of Taylor series and, in particular, Maclaurin series, is in special relativity to approximate the Lorrentz factor $\gamma$. Taking the first two terms of the series gives a very good approximation for low speeds. You can actually show that at low speeds, special relativity reduces to classical (Newtonian) physics. For example, in special relativity the momentum is $$\vec p = \gamma m\vec v$$ and at low speeds $$\gamma \approx 1$$ so $$\vec p \approx m\vec v$$ which is the (linear) momentum in classical mechanics.
Also, the most famous equation in physics $$E = m{c^2}$$ is actually an approximation for low velocities, again, using Taylor series approximation.
I hope this helps.
By the way, $$\gamma = \frac{1}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}}}}$$ where $v$ is the velocity and $c$ is the speed of light.
Another example is again from physics. If you ever took classical mechanics or studied pendulums you often start with an assumption $\sin (\theta ) \approx \theta$, which also comes from Taylor series.
Not to mentions, that any software that graphs various functions actually uses very good Taylor approximations.
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If you take any advanced math or physics class, you will most definitely use Taylor series. Taylor series and geometric series are probably the most famous series. – glebovg Oct 22 '12 at 2:17
We can also use Taylor series to approximate integrals that are impossible with the other integration techniques.
A classic example is $\int\sin(x^2)\,\mathrm{d}x$.
We can't actually integrate this, but using the taylor series for $sin(x)$ we can substitute $x^2$ in for $x$ at each term of the series, and then integrate each term individually. After doing so, we can write a new sum.
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The Taylor Series is used in power flow analysis of electrical power systems (Newton-Raphson method).
http://en.wikipedia.org/wiki/Power_flow_study
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Multivariate Taylor series is used in different optimization techniques; that is you approximate your function as a series of linear or quadratic forms, and then successively iterate on them to find the optimal value.
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No one's mentioned the combinatorial side of things, so I'll be the first to say it: generating functions. We use generating functions to pass hard discrete counting problems to the continuous, where things are easy. Generating functions are a central tool in combinatorics (counting, graph theory, etc.) and probability (where we have moment generating functions). Taylor series is the fundamental idea behind all of these. Read: http://en.wikipedia.org/wiki/Generating_function for details, and take a combinatorics or mathematical probability class to learn more.
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In physics you often approximate a complicated function by taking the first few terms in the Taylor series (the Taylor polynomial). For small values of the independent variable, you often assume linearity, which can allow you to get a closed form solution. For example, if you take an introductory physics class then you usually study the motion of the pendulum by approximating $\sin(\theta)$ by $\theta$ for small angles.
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The Taylor Series is used to derive the most beautiful equation in Mathematics: The Euler's Identity!
How?
Start with defining $g(x)$ as follows: $$g(x) = e^x= 1 + x + {x^2 \over 2!} + {x^3 \over 3!} \cdots$$Thus,\begin{align} g(ix) = e^{ix} & = 1 + ix +{(ix)^2 \over 2!} + {(ix)^3 \over 3!}\cdots \\ \\ \\ &= 1 + ix + i^2{x^2 \over 2!} + i^3 {x^3 \over 3!}\cdots \\ \\ \\ &= 1 + ix - {x^2 \over 2!} - i{x^3 \over 3!}\cdots\end{align} Collect imaginary and real parts together.$$\overbrace{\left(1 - {x^2 \over 2! } + {x^4 \over 4!} - {x^6 \over 6!} \cdots\right)}^{(1)} + i\overbrace{\left(x - {x^3 \over 3!} + {x^5 \over 5!} - {x^7 \over 7!} \right)}^{(2)}$$ Recall that (1) is the Taylor Series for $\cos(x)$ and (2) is the Taylor Series for $\sin(x)$ which yields, $$e^{ix} = \cos(x) + i\sin(x)$$That's Euler's Formula. Now the next step for the Identity is easy. If $x = \pi$, then we have:$$e^{i\pi} = \cos(\pi) + i \sin(\pi) = -1 + 0i = -1$$So,$$e^{i\pi} = -1$$Let's make it a little better:$$e^{i \pi}+1 = 0$$
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You should mention that this sort of rearrangement trick is only allowed because the series is absolutely convergent. A lot of people forget that you can't just rearrange every series like that. – B0112358 Dec 19 '12 at 19:04 | 2013-05-21T12:47:26 | {
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http://pmuw.xhrh.pw/gaussian-integral-matlab.html | # Gaussian Integral Matlab
Introduction In numerical analysis, Quadrature is the estimation of an integral. But what you don't appreciate is that normcdf IS an integral (as is also erf. Let ( ) ( ) , with Then Example 2 )Consider ∫ (. In the next section we implement a program with fewer points just for convenience. Shivaram Department of Mathematics, Dayananda sagar college of Engineering, Bangalore, India Abstract: - We introduce a Generalised Gaussian quadrature method for evaluation of the double integral I= f x, y dy dx , T. The function w(x) is allowed to be singular. every finite linear combination of them is normally distributed. Let's consider the function. Note: Including functions in scripts requires MATLAB® R2016b or later. Since we're dealing with discrete signals and we are limited to finite length of the Gaussian Kernel usually it is created by discretization of the Normal Distribution and truncation. The probability density function for the standard Gaussian distribution (mean 0 and standard deviation 1) and the Gaussian distribution with mean μ and standard deviation σ is given by the following formulas. *x in MATLAB's notation. The 2-point Gaussian quadrature rule returns the integral of the black dashed curve, equal to (−) + =. We will discuss them in one dimension first. 1 The Fourier transform We started this course with Fourier series and periodic phenomena and went on from there to define the Fourier transform. ) I still write MATLAB code as I find something interesting, and I do attempt to write new tools to put on the File Exchange when I think I can make a contribution. Integral of a gaussian function wrong answer. For example, here is a mex file that makes NR3's generalized Fermi-Dirac integral routine available to Matlab. So it is quite natural and intuitive to assume that the clusters come from different Gaussian Distributions. How to fit multiple peaks using Gaussian funtion. So far so good. Download PDF printout. Rewriting your equation with the substitution x = y-1, we get. By: Anchal Arora 13MCA0157 2. This thesis pushes the computational frontier of volume computation, randomized sampling, and integration, both in theory and pracice. @Ali i mean that for integral sin(x) that a=0 b=pi i want see for example 1. Can anybody elaborate on this. Hi Arijit, For integral calculations the independent variable has to be defined as symbolic constant initially, then integration has to be carried out by mentioning the upper and lower limit within the int command. Identified a link between the path integral Monte Carlo distribution and coherent states. MATLAB can plot a 1 x n vector versus an n x 1 vector, or a 1 x n vector versus a 2 x n matrix (you will generate two lines), as long as n is the same for both vectors. Calculates the integral of the given function f(x) over the interval (a,b) using Gaussian quadrature. The Gaussian quadrature is targeted to approximate an integral by taking the weighted sum of integrand values sampled at special points (called abscissas). We have compared Gaussian and Newton-Cotes methods with each other at given orders, and we have also compared higher order and lower order methods within each category. The key components of an arima object are the polynomial degrees (for example, the AR polynomial degree p and the degree of integration D) because they completely specify the model structure. My MatLab code runs significanlty slowly compared to NIntegrate of Mathematica. The Gaussian smoothing operator is a 2-D convolution operator that is used to blur' images and remove detail and noise. Connection between nodes and weights of Gaussian quadrature. These data were collected using the Group's radar test and development system, except for the MATLAB simulation. m, which runs Euler’s method; f. The purpose of the manual is to help people to use the software in their own work and. The method underlying quadl is a "Gaussian quadrature rule". We also provide a Matlab function, trigauss, for the computation of angles and weights of such a formula (cf. Its part of topics in Numerical Analysis course. Am new to matlab and need guidance on setting up the M-file for a numerical integration as below: Need to evaluate the following integral which apparently doesnt have an analytic solution. The Gaussian function has important properties which are verified withThe Gaussian function has important properties which are verified with respect to its integral:. Generalized Gaussian Quadrature Rules in Enriched Finite Element Methods. We will look at a simple version of the Gaussian, given by equation [1]: [1] The Gaussian is plotted in Figure 1: Figure 1. Below, you can see how a MATLAB graphics object is an object in PowerPoint. 1) where f(x)is a given function and [a,b]a finite interval. The integral transform (2) is called kernel smoothing. ” Second, Gaussian random variables are convenient for many analytical manipulations, because many of the integrals involving Gaussian distributions that arise in practice have simple closed form solutions. Even though extensive research has taken place to evaluate integrals over triangular elements,. The result of calculating the fourier transform using numerical integration is: the result of using Matlab's FFT is: So where did I go wrong here? I know the FT of a Gaussian should be another. The input values should be an function f to integrate, the bounds of the integration interval a and b, and the number of gaussian evaluation points n. Dear Sir, I am interested about the code that you wrote about the 2D Gaussian. Gaussian derivatives A difference which makes no difference is not a difference. Matlab Matlab MATLAB training programs (bilateral filtering) MATLAB training programs (bilateral filtering) bilateral filter templates for two main template, first is the Gaussian profile, the second is based on gray level difference as generated by the coefficients of the function template. But problem is, Matlab doesn't have library functions that handle integrals above 3 dimensions. The normal distribution is by far the most important probability distribution. The measure of spread is quantified by the variance, σ 2 {\displaystyle \sigma ^{2}}. LEGENDRE_RULE_FAST, a MATLAB program which uses a fast (order N) algorithm to compute a Gauss-Legendre quadrature rule of given order. 2017-04-01. Estimating its parameters using Bayesian inference and conjugate priors is also widely used. With ApexTrack, you may need to add a Detect Shoulders event, in addition to the Gaussian Skim event. Gaussian Integral Polar Conversion Trick submitted 1 hour ago by DaedalusMinor When solving the standard Gaussian Integral, every video or textbook squares the entire function by multiplying by another Gaussian Integral with a different variable and then converting the exponent to the polar version. integral2 transforms the region of integration to a rectangular shape and subdivides it into smaller rectangular regions as needed. 1D Wave Equation - General Solution / Gaussian Function Overview and Motivation: Last time we derived the partial differential equation known as the (one dimensional) wave equation. Laplacian of Gaussian (LoG) As Laplace operator may detect edges as well as noise (isolated, out-of-range), it may be desirable to smooth the image first by a convolution with a Gaussian kernel of width. Quadrature Value vs. area tables, also known as integral images. 1) is a bell-shaped curve that is symmetric about the mean µ and that attains its maximum value of √1 2πσ ' 0. The course will cover use of ABAQUS; and the practical implementation of finite element procedures, using MATLAB coding exercises to illustrate basic concepts, as well as more advanced coding either through ABAQUS user elements or by adding subroutines to a basic standalone FEA code. 4 TI-89 Program for Gaussian Quadrature Here is a program with eight points, n = 7. METHOD is 'coherent' or 'noncoherent'. Radial functions and the Fourier transform Notes for Math 583A, Fall 2008 December 6, 2008 1 Area of a sphere The volume in n dimensions is vol = dnx = dx1 ···dxn = rn−1 drdn−1ω. Rewriting your equation with the substitution x = y-1, we get. Solution 1. at a or b, of the interval. Various different line integrals are in use. Remember that no matter how x is distributed, E(AX +b) = AE(X)+b. However, in many practical situations, we do not have a formula for the integrand, and in fact the. in front of the one-dimensional Gaussian kernel is the normalization constant. For example, a Gaussian membership function always has a maximum value of 1. , but by applying an n-point Gauss-Legendre quadrature rule, as described here, for example. if h(t) is the impulse response of the filter I have to send white Gaussian noise to it,in continuous domain. Y = pulsint(X) performs video (noncoherent) integration of the pulses in X and returns the integrated output in Y. Learn more about symbolic, integration, symbolic integration, integration gaussian function. - It is a smoothing operator. its integral over its full domain is unity for every s. Matlab Assignment Help is the process where a studen t would contact a Matlab Programming Service Provider and hire that service provider for the time and effort. Calculate stresses at Gaussian integration points of the element, and after that extrapolation to element nodal stress point. m that computes the value of the integrand at and computes the area under the curve from 0 to. NASA Astrophysics Data System (ADS) Schubert, G. Relation to standard Gaussian integral. 1 Jarno Vanhatalo, Jaakko Riihimäki, Jouni Hartikainen, and Aki Vehtari If you use GPstuff, please use reference: Jarno Vanhatalo, Jaakko Riihimäki, Jouni Hartikainen and Aki Vehtari (2011). I plot the estimate of the PSD and also the variance, which is supposed to be equal to the mean of PSD. Note that quad requires scalar functions to be defined with elementwise operations, so f(x) = 2 1+x2. It looks to be the right shape, however, the function itself is very small (the max only coming to about 4*10^-3). Such a result is exact, since the green region has the same area as the sum of the red regions. m, lorentzian. trol problem in integral form. VISIM is a sequential simulation code based on GSLIB ('Geostatistical Software LIBrary', Stanford Center for Reservoir Forecasting, Stanford University) for sequential Gaussian and direct sequential simulation with histogram reproduction. To learn about Gaussian processes in machine learning, see the Gaussian Processes for Machine Learning book by Carl Edward Rasmussen and Christopher K. Before testing the KC705, we collected data from MATLAB-simulated Gaussian noise, an analog Gaussian noise generator, and a digital noise source used by Group 108. The linear transform of a gaussian r. GAUSSIAN INTEGRALS An apocryphal story is told of a math major showing a psy-chology major the formula for the infamous bell-shaped curve or gaussian, which purports to represent the distribution of intelligence and such: The formula for a normalized gaussian looks like this: ρ(x) = 1 σ √ 2π e−x2/2σ2. 3 Trapezoidal rule. I know that a normal function dictates that the integral go to 1, but is there any way to keep the shape, just make it bigger so that it can plot on top of my data (X range -200, 200 Y range -250, 250)?. Estimating its parameters using Bayesian inference and conjugate priors is also widely used. We shall apply Newton-Cotes rules and Gaussian quadrature formulae to nd numer-ical integration for di erent nvalues. in 2006, that can be used in computer vision tasks like object recognition or 3D reconstruction. Let's consider the function. matlab) submitted 6 years ago by math693932 I'm supposed to be creating a function script called by another script file that approximates the integral of a function handle using Riemann sums. First, the constant a can simply be factored out of the integral. Nearest neighbour interpolation is the simplest approach to interpolation. 1 i = 2:m-1; 2 j = 2:n-1; 3 Au(i,j) = 4*u(i,j) - u(i-1,j) - u(i+1,j) - u(i,j-1) - u(i,j+1);. "look Gaussian. Gaussian elimination is probably the best method for solving systems of equations if you don’t have a graphing calculator or computer program to help you. I have to compute the accuracy of a new Gaussian mixture fitting algorithm. The function p1 2ˇ e 2x =2 is called a Gaussian, and (4. Rewriting your equation with the substitution x = y-1, we get. It assumes that the observations are closely clustered around the mean, μ, and this amount is decaying quickly as we go farther away from the mean. (Requires those four functions, plus gaussian. We will discuss them in one dimension first. This site uses cookies from Google to deliver its services and to analyze traffic. Even though extensive research has taken place to evaluate integrals over triangular elements,. By definition, definite integral is basically the limit of a sum. 1 to 6 Also compute the inegrals for m = 1,2, and 3 For the inline function use: [email protected](x) besselj(x,m,terms); The Romberg and Trap functions below need variable input and need only be reference in the code. The outer integral is evaluated over xmin ≤ x ≤ xmax. ) I still write MATLAB code as I find something interesting, and I do attempt to write new tools to put on the File Exchange when I think I can make a contribution. Figure 1 Region with only a few centers. A reminder that the algorithms given as functions can be run in the command line mode in Matlab's command window. In the remainder of this section, we will review a number of useful properties of multivariate Gaussians. My function f (x) has the following restriction: df (x)/dlog (x) is equal to a sum of two Gaussians. Gaussian derivatives 4. First recall Gaussian Quadrature for an interval [a, b]: For example, with n = 3, we get 4 points in [a, b], x0, x1, x2, x3, and 4 positive weights w0, w1, w2, w3. stackexchange [22], and in a slightly less elegant form it appeared much earlier in [18]. Image Processing and Analysis > Spatial Filters > Gaussian All Books Non-Programming Books User Guide Tutorials Quick Help Origin Help Programming Books X-Function Origin C LabTalk Programming Python Automation Server LabVIEW VI App Development Code Builder License MOCA Orglab Release Notes. This has the effect that only a lower degree of polynomial effect can be captured in the integration process. 1 Fourier transforms as integrals There are several ways to de ne the Fourier transform of a function f: R ! C. Gaussian Quadrature for Triangles is a Matlab script for Mathematics scripts design by Greg von Winckel. Body of the package implementing numerical integration: package body Integrate is. We will now put time back into the wave function and look at the wave packet at later times. Learn more about 3- points gaussian quadrature. If MuPAD Cannot Compute an Integral. For instance, Do might be a standardized Gaussian, p(x) N (0, 1), and hence our null hypothesis is that a sample comes from a Gaussian with mean 0. Bayesian Modeling with Gaussian Processes using the MATLAB Toolbox GP-. With the coefficients a n and b n it is a straightforward matter to go ahead and calculate the amplitudes and phases and plot their spectra. Gaussian Integral. quadrature Compute a definite integral using fixed-tolerance Gaussian quadrature. The Gaussian quadrature is targeted to approximate an integral by taking the weighted sum of integrand values sampled at special points (called abscissas). In numerical analysis , a quadrature rule is an approximation of the definite integral of a function , usually stated as a weighted sum of function values at specified points within the domain of integration. An Introduction to Fitting Gaussian Processes to Data process is joint Gaussian with any integral or derivative of it, as integration and differentiation. Gaussian processes for Bayesian analysis User guide for Matlab toolbox GPstuff Version 3. This solution is probably slightly faster, and works even with Ada83. Numerical quadrature is important in many fields of applied science and engineering. MATLAB 2019 Overview MATLAB 2019 Technical Setup Details MATLAB 2019 Free Download MATLAB Deep Learning: With Machine Learning, Neural Networks and Artificial Intelligence by Phil Kim Get started with MATLAB for deep learning and AI with this in-depth primer. Characteristic Functions Important Distributions. This is a simple script which produces the Legendre-Gauss weights and nodes for computing the definite integral of a continuous function on some interval [a,b]. Finding quadrature nodes and weights • One way is through the theory of orthogonal polynomials. Today: Numerical Integration zStrategies for numerical integration zSimple strategies with equally spaced abscissas zGaussian quadrature methods zIntroduction to Monte-Carlo Integration. A reminder that the algorithms given as functions can be run in the command line mode in Matlab's command window. 'iterated' integral2 calls integral to perform an iterated integral. Fungsi numerical integration adalah built-in functions untuk menyelesaikan Integral dengan MATLAB berdasarkan interval [a b] atau integral tentu yaitu quad, quadl, dan trapz. Below, you have an slide showing the change of variables needed to relate the reference quadrilateral [-1,1]x[-1,1] with a general one. An alternative solution is to pass a function reference to the integration function. The Gaussian smoothing operator is a 2-D convolution operator that is used to blur' images and remove detail and noise. $\endgroup$ - Daryl Feb 4 '13 at 19:45 $\begingroup$ Hi, Daryl. On systems permitting multiple processes, such as a Unix system or MS Windows, you will nd it convenient, for reasons discussed in section 14, to keep both MATLAB. Instead of a loop to compute the error, just use vector operations. The software features fully interactive construction and combination of. You will find it easy to use the provided source codes, in your projects and research, if you are familiar with MATLAB programming language. Mainly retired from Eastman Kodak. Integral Image Domain Filtering. The Gaussian Bell-Curve. It runs on following operating system: Windows / Linux / Mac OS / BSD / Solaris. 1) says the integral of the Gaussian over the whole real line is 1. Handling undefined integrals. Gaussian kernels: convert FWHM to sigma Posted on 20. Gaussian quadrature. The first variable given corresponds to the outermost integral and is done last. 94 × 10-6 w shown in Fig. The Gaussian function, g(x), is defined as,. au DOWNLOAD DIRECTORY FOR MATLAB SCRIPTS math_integration_2D. The Multiprecision Computing Toolbox is the MATLAB extension for computing with arbitrary precision. With the normalization constant this Gaussian kernel is a normalized kernel, i. The term numerical quadrature'' refers to the estimation of an area, or, more generally, any integral. A typical table of Gauss-Legendre rule looks like the following:. legendre_rule_test. We present an accurate three-dimensiona. m performs averaging filtering using an integral image. A PID regulation technology was proposed which was based on Kalman filter. First, the constant a can simply be factored out of the integral. The integral on the right side can be termed as Q-function, which is given by,. Gaussian processes for Bayesian analysis User guide for Matlab toolbox GPstuff Version 3. Connection between nodes and weights of Gaussian quadrature. More on Multivariate Gaussians Chuong B. If you look at Gaussian quadrature rules, they presume a weight function from among several standard forms, AND a domain of integration. Y = pulsint(X) performs video (noncoherent) integration of the pulses in X and returns the integrated output in Y. Click on the program name to display the source code, which can be downloaded. Handling undefined integrals. The inverse Gaussian distribution can be used to model the lifetime of an ob-ject. GPs are a little bit more involved for classification (non-Gaussian likelihood). At the very minimum it will always be necessary to integrate at least an element square. The energy potential used in the experiments are Yukawa potential and Gaussian Potential. NB: in MATLAB's notation, the multiplication (*), division (/), and exponentiation (^) operators MUST be preceded by a dot (. NASA Astrophysics Data System (ADS) Schubert, G. We have compared Gaussian and Newton-Cotes methods with each other at given orders, and we have also compared higher order and lower order methods within each category. 다음글 : [bmw e85 z4] 6월16일 안산서킷 레이서 트레이닝 아카데미, 뒷 브레이크디스크 교체, 엔진오일팬 격벽작업. Numerical quadrature is important in many fields of applied science and engineering. Matlab’s built-in numerical integration function [Q,fcount]=quad(f,a,b,tol) is essentially our simp_compextr code with some further efficiency-enhancing features. For Gaussian Elimination in MATLAB The component ring R of A must be an integral domain, i. 5 Gaussian distribution as a limit of the Poisson distribution A limiting form of the Poisson distribution (and many others – see the Central Limit Theorem below) is the Gaussian distribution. Learn more about gaussian, peak fit MATLAB Answers. Since this exactly what is done in the field of statistics, the analysis of the Monte Carlo method is a direct application of statistics. (You might hear the term cubature'' referring to estimating a volume in three dimensions, but most people use the term quadrature. Gaussian quadrature rules can be constructed using a technique known as moment matching. What method gives the best answer if only two function evaluations are to be made? We have already seen that the trapezoidal rule is a method for finding the area under. Solution 1. to plot(x,y,’--’,t,z,’o’) and see the difference in graphs drawings. Matthew Schwartz Lecture 11: Wavepackets and dispersion 1 Wave packets The function g(x)=e −1 2 x σx 2 = (1) is called a Gaussian. Also, the integral of the square of the difference between PDF has to be evaluated. Nearest neighbour interpolation is the simplest approach to interpolation. The first integral is Therefore, which can be rearranged to yield or The second integral is Therefore, which can be rearranged to yield or By putting pieces together, we get Distribution function The distribution function of an exponential random variable is. m, modelpeaks. This is a very powerful technique. m, which defines the function. every finite linear combination of them is normally distributed. Review of Fourier Transform The Fourier Integral X(f ) x(t). Integration: The Exponential Form. Identified a link between the path integral Monte Carlo distribution and coherent states. —Leo Hendrik Baekeland 5. 2012-12-01. The Gaussian integral (aka Euler-Poisson integral or Poisson integral) is the integral of the Gaussian function over the entire real numbers. In the case of a closed curve it is also called a contour integral. It can be computed using the trick of combining two one-dimensional Gaussians. We may want to integrate some function f(x) or a set of tabulated data. Gaussian white noise N(0;22) is added to the binary image. But, the multivariate Gaussian distributions is for finite dimensional random vectors. For Example 3, compare the accuracy of the Trapezoidal Rule, Simpson's Rule, Simpson's Rule and Boole's Rule,. lecture 06 multivariate normal II. This is a seemingly simple question, though I'm not exactly sure where I'm going wrong (if in fact I am going wrong). By reversing the process in obtaining the derivative of the exponential function, we obtain the remarkable result: It is remarkable because the integral is the same as the expression we started with. It is computed numerically. Abstract In this paper, we present new Gaussian integration schemes for the efficient and ac- curate evaluation of weak form integrals that arise in enriched finite element meth- ods. BIE2D: MATLAB tools for boundary integral equations on curves in 2D. m, pinknoise. Integration by parts is one of the common methods for computing integrals. Property P in the output is the sum of the autoregressive lags and the degree of integration, i. This will prove valuable when evaluating various improper integrals, such as those with infinite limits. the integral. The Gaussian function The Gaussian function (also refered to as bell-shaped or "bell" curve) is of the following form: (x19) where s is refered to as the spread or standard deviation and A is a constant. All five semiempirical methods use Slater-type orbitals, STOs, although when analytic derivatives are involved [ 16 ], a Gaussian expansion [ 16 ] of STOs is normally used. m, findpeaksG. In real life, many datasets can be modeled by Gaussian Distribution (Univariate or Multivariate). Methods for triangulating the sphere and some associated integration formulas are given in Section 3. Operations on Gaussian R. The quadv function vectorizes quad for an array-valued fun. To start off: you have a 2D un-normalized Gaussian function centred at the origin and with a sigma of 4. Various Scenarios and Animations for Gauss-Legendre Quadrature. 1 Normalization constant for a 1D Gaussian The normalization constant for a zero-mean Gaussian is given by Z = Z b a exp − x2 2σ2 dx (1) where a = −∞ and b = ∞. Convergence : Simpson's 1/3rd Rule : Method Convergence Romberg Rule : Method Convergence Gauss-Quadrature Rule : Method Convergence Integrating Discrete Functions : Integrating discrete functions. Viewed 9k times 1. GAUSSIAN INTEGRALS An apocryphal story is told of a math major showing a psy-chology major the formula for the infamous bell-shaped curve or gaussian, which purports to represent the distribution of intelligence and such: The formula for a normalized gaussian looks like this: ρ(x) = 1 σ √ 2π e−x2/2σ2. Aston University of Warwick Abstract State Space Models (SSM) is a MATLAB toolbox for time series analysis by state space methods. This is matlab code for Gaussian Double Integral Algorithm. Overlap Integrals The particular technique used for the evaluation of the overlap integral depends on the atoms involved and whether analytic derivatives are used. Here, we aim to study the effects of both the method and the order. indd 3 9/19/08 4:21:15 PM. the term without an y’s in it) is not known. How to integrate over a bivariate gaussian Learn more about copula, numerical integration, copulapdf. Y = pulsint(X) performs video (noncoherent) integration of the pulses in X and returns the integrated output in Y. - Matlab quad function that numerically evaluates the integral (low order method): mat_quadrature. 1) is an approximation of the form IQ(f,a,b) = (b− a) Xm k=1 wkf(xk). Single integration. : y = integral of f(x), with upper and lower bounds) in Excel 2003? Note: "NORMDIST" function is not what I am looking for. A typical table of Gauss-Legendre rule looks like the following:. SigmaPlot Product Overview. In its two-dimensional form, as shown in the plot above, the function shows a distinct two-dimensional Gaussian peak in the centre of the integration region. Results are displayed graphically in figure windows 1 , 2, and 3 and printed out in a table of parameter accuracy and elapsed time for each method, as shown below. Gaussian processes for Bayesian analysis User guide for Matlab toolbox GPstuff Version 3. We will assume knowledge of the following well-known differentiation formulas : , where , and , where a is any positive constant not equal to 1 and is the natural (base e) logarithm of a. The integration bounds are an iterable object: either a list of constant bounds, or a list of functions for the non-constant integration bounds. In the following, six covariance functions (exponential, Gaussian, spherical, hyperbolic, k-Bessel, and cardinal sine) are considered because of their variability close to the origin and near the range as well as is their popularity. The Gaussian quadrature is done using the function planar(nx), where 3nx7 and one needs to use the global xoc and woc command. 7 CEUs) Inertial Systems, Kalman Filtering and GPS/INS Integration DAY 1. Stackoverflow. m, findpeaksL. Gaussian processes for Bayesian analysis User guide for Matlab toolbox GPstuff Version 3. Therefore, a solid understanding of Calculus and Differential Equations is needed to fully understand the material. Numerical integration. Gaussian quadrature on arbitrary intervals Use substitution or transformation to transform ∫ ( ) into an integral defined over. Am new to matlab and need guidance on setting up the M-file for a numerical integration as below: Need to evaluate the following integral which apparently doesnt have an analytic solution. Matlab is a prototyping environment, meaning it focuses on the ease of development with language exibility, interactive debugging, and other conveniences lacking in performance-oriented languages like C and Fortran. 1 to 6 Also compute the inegrals for m = 1,2, and 3 For the inline function use: [email protected](x) besselj(x,m,terms); The Romberg and Trap functions below need variable input and need only be reference in the code. in front of the one-dimensional Gaussian kernel is the normalization constant. Hi All, If you love numerical analysis like I do, then you have most likely come across the various Gaussian quadrature algorithms. 1 Introduction The need of numerical integration of double integrals arises in many mathematical models, as most of these integrals cannot be evaluated analytically. Mathematica Subroutine (Gauss-Legendre Quadrature). Repeated filtering with averaging filters can be used to approximate Gaussian filtering. We will also call it "radius" in the text below. Open Mobile Search (integration) of each peak, how. 1D Gaussian Quadratures. Operations on Gaussian R. What method gives the best answer if only two function evaluations are to be made? We have already seen that the trapezoidal rule is a method for finding the area under. The DC should always stay. Learn more about Matlab PowerPoint integration here. @Ali i mean that for integral sin(x) that a=0 b=pi i want see for example 1. Solution 5. Operational multisensor sea ice concentration algorithm utilizing Sentinel-1 and AMSR2 data. Evaluating the definite integral Φ(x) requires knowing the indefinite integral of ϕ(x). List of integrals of exponential functions 2 where where and is the Gamma Function when , , and when , , and Definite integrals for, which is the logarithmic mean (the Gaussian integral) (see Integral of a Gaussian function) (!! is the double factorial). In the case of a closed curve it is also called a contour integral. Since most kernel functions K(P, Q), in (1. m This function approximates Gaussian filtering by repeatedly applying integaverag. Learn more about random number generator, gaussian distribution, white noise. If you are looking for numerical integration over the unit disk (2D sphere) you might be interested in this page Cubature formulas for the unit disk. Gaussian: With n nodes you get exact answer if f is Integration routine in Matlab quad(@myfun,A,B) This is an adaptive procedure that adjusts the length of the. The Gaussian pdf N(µ,σ2)is completely characterized by the two parameters. I need to build a function performing the low pass filter: Given a gray scale image (type double) I should perform the Gaussian low pass filter. There are many reasons for smoothing. There are two main reasons for you to need to do numerical integration: analytical integration may be impossible or infeasible, or you may wish to integrate tabulated data rather than known functions. In this tutorial we will focus on smoothing in order to reduce noise (other uses will be seen in the following tutorials). Matthew Schwartz Lecture 11: Wavepackets and dispersion 1 Wave packets The function g(x)=e −1 2 x σx 2 = (1) is called a Gaussian. BOOK CORRECTIONS: We give here a list of corrections for the first printing of the third edition of the textbook Elementary Numerical Analysis. 5 Gauss-Legendre Integration We wish to find the area under the curve y =f (x), −1 ≤x ≤1. We study the utility of Expectation Propagation (EP) as an approximate integration method for this problem. Functions have various representations in mathematics. We will discuss them in one dimension first. A typical table of Gauss-Legendre rule looks like the following:. 1D Wave Equation - General Solution / Gaussian Function Overview and Motivation: Last time we derived the partial differential equation known as the (one dimensional) wave equation. LINE_FELIPPA_RULE, a MATLAB library which returns the points and weights of a Felippa quadrature rule over the interior of a line segment in 1D. Figure 1 Region with only a few centers. The abscissas and weights are calculated in a special way so that the rule provides a precise answer for all polynomials up to certain degree. In order to gain some insight on numerical integration, it is natural to review Rie-mann integration, a framework that can be viewed as an approach for approximat-ing integrals. The constant function, f(t)=1, is a function with no variation - there is an infinite amount of energy, but it is all contained within the d. Gaussian Elimination Matlab Code The following matlab project contains the source code and matlab examples used for gaussian elimination. The filter size is given by a ratio parameter r. df (x)/dlog (x) being the derivative of f (x) with respect to the argument log (x). This function allows one to write down spatial density of a physical quantity that is concentrated in one. , a domain of category Cat::IntegralDomain. Gaussian noise is statistical noise having a probability distribution function (PDF) equal to that of the normal distribution, which is also known as the Gaussian distribution. LINE_FELIPPA_RULE, a MATLAB library which returns the points and weights of a Felippa quadrature rule over the interior of a line segment in 1D. Plus I will share my Matlab code for this algorithm. We want to compute. Both these methods allow averaging to be performed at a small fixed cost per pixel, independent of the averaging filter size. Here the function inside the integral is a normalized gaussian probability density function $$Y \sim N( 0, 1)$$, normalized to mean=0 and standard deviation=1. "Gaussian integration" redirects here. The third integral in the integration over volume is Equation 2. Matlab plug-in for SPM allowing to obtain a threshold for cluster FDR - the method fits a Gamma-Gaussian mixture model to the SPM-T and finds the optimal threshold (crossing between noise and activation). I have to compute the accuracy of a new Gaussian mixture fitting algorithm. A reminder that the algorithms given as functions can be run in the command line mode in Matlab's command window. A better example, is one in which we want to perform operations on the rows of a matrix. 1 InterpolatoryQuadratureRules 5. MATLAB has a built-in triple integrator triplequad similar to dblquad, but again, it only integrates over rectangular boxes. *x in MATLAB's notation. We will discuss them in one dimension first. LEGENDRE_RULE_FAST, a MATLAB program which uses a fast (order N) algorithm to compute a Gauss-Legendre quadrature rule of given order. How to fit multiple peaks using Gaussian funtion. | 2019-11-14T08:14:39 | {
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https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox | The paradox starts with three boxes, the contents of which are initially unknown
Bertrand's box paradox is a veridical paradox in elementary probability theory. It was first posed by Joseph Bertrand in his 1889 work Calcul des Probabilités.
There are three boxes:
1. a box containing two gold coins,
2. a box containing two silver coins,
3. a box containing one gold coin and one silver coin.
The question is to calculate the probability, after choosing a box at random and withdrawing one coin at random, if that happens to be a gold coin, of the next coin drawn from the same box also being a gold coin.
A veridical paradox is when the correct solution to a puzzle appears to be counterintuitive. It may seem intuitive that the probability that the remaining coin is gold should be 1/2, but the probability is actually 2/3.[1] However, this is not the paradox Bertrand referred to. He showed that if 1/2 were correct, it would lead to a contradiction, so 1/2 cannot be correct.
This simple but counterintuitive puzzle is used as a standard example in teaching probability theory. The solution illustrates some basic principles, including the Kolmogorov axioms.
## Solution
Bertrand's box paradox: the three equally probable outcomes after the first gold coin draw. The probability of drawing another gold coin from the same box is 0 in (a), and 1 in (b) and (c). Thus, the overall probability of drawing a gold coin in the second draw is 0/3 + 1/3 + 1/3 = 2/3.
The problem can be reframed by describing the boxes as each having one drawer on each of two sides. Each drawer contains a coin. One box has a gold coin on each side (GG), one a silver coin on each side (SS), and the other a gold coin on one side and a silver coin on the other (GS). A box is chosen at random, a random drawer is opened, and a gold coin is found inside it. What is the chance of the coin on the other side being gold?
The following faulty reasoning appears to give a probability of 1/2:
• Originally, all three boxes were equally likely to be chosen.
• The chosen box cannot be box SS.
• So it must be box GG or GS.
• The two remaining possibilities are equally likely. So the probability that the box is GG, and the other coin is also gold, is 1/2.
The flaw is in the last step. While those two cases were originally equally likely, the fact that you are certain to find a gold coin if you had chosen the GG box, but are only 50% sure of finding a gold coin if you had chosen the GS box, means they are no longer equally likely given that you have found a gold coin. Specifically:
• The probability that GG would produce a gold coin is 1.
• The probability that SS would produce a gold coin is 0.
• The probability that GS would produce a gold coin is 1/2.
Initially GG, SS and GS are equally likely ${\displaystyle \left(\mathrm {i.e.,P(GG)=P(SS)=P(GS)} ={\frac {1}{3}}\right)}$. Therefore, by Bayes rule the conditional probability that the chosen box is GG, given we have observed a gold coin, is:
${\displaystyle \mathrm {P(GG\mid see\ gold)={\frac {P(see\ gold\mid GG)\times {\frac {1}{3}}}{P(see\ gold\mid GG)\times {\frac {1}{3}}+P(see\ gold\mid SS)\times {\frac {1}{3}}+P(see\ gold\mid GS)\times {\frac {1}{3}}}}} ={\frac {\frac {1}{3}}{\frac {1}{3}}}\times {\frac {1}{1+0+{\frac {1}{2}}}}={\frac {2}{3}}}$
The correct answer of 2/3 can also be obtained as follows:
• Originally, all six coins were equally likely to be chosen.
• The chosen coin cannot be from drawer S of box GS, or from either drawer of box SS.
• So it must come from the G drawer of box GS, or either drawer of box GG.
• The three remaining possibilities are equally likely, so the probability that the drawer is from box GG is 2/3.
Alternatively, one can simply note that the chosen box has two coins of the same type 2/3 of the time. So, regardless of what kind of coin is in the chosen drawer, the box has two coins of that type 2/3 of the time. In other words, the problem is equivalent to asking the question "What is the probability that I will pick a box with two coins of the same color?".
Bertrand's point in constructing this example was to show that merely counting cases is not always proper. Instead, one should sum the probabilities that the cases would produce the observed result; and the two methods are equivalent only if this probability is either 1 or 0 in every case. This condition is correctly applied in the second solution method, but not in the first.[citation needed]
## The paradox as stated by Bertrand
It can be easier to understand why 1/2 is incorrect, if you consider the paradox Bertrand used. After a box has been chosen, but before a drawer is opened, there is a 2/3 probability that the box has two of the same kind of coin. So, if you then select a drawer at random, before you open it the probability that the other drawer has the same kind of coin is 2/3. Opening the drawer that you selected cannot change that.
## Experimental data
In a survey of 53 Psychology freshmen taking an introductory probability course, 35 incorrectly responded 1/2; only 3 students correctly responded 2/3.[2]
## Related problems
Other veridical paradoxes in probability include:
The Monty Hall and Three Prisoners problems are mathematically identical to Bertrand's Box paradox. The construction of the Boy or Girl paradox is similar, essentially adding a fourth box with a gold coin and a silver coin. Its answer is controversial, based on how one assumes the "drawer" was chosen.
## References
1. ^ "Bertrand's box paradox". Oxford Reference.
2. ^ Bar-Hillel, Maya; Falk, Ruma (1982). "Some teasers concerning conditional probabilities". Cognition. 11 (2): 109–22. doi:10.1016/0010-0277(82)90021-X. PMID 7198956. S2CID 44509163.
• Nickerson, Raymond (2004). Cognition and Chance: The psychology of probabilistic reasoning, Lawrence Erlbaum. Ch. 5, "Some instructive problems: Three cards", pp. 157–160. ISBN 0-8058-4898-3
• Michael Clark, Paradoxes from A to Z, p. 16;
• Howard Margolis, Wason, Monty Hall, and Adverse Defaults. | 2022-11-29T21:41:40 | {
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https://math.stackexchange.com/questions/3130401/two-dice-prx-1-lt-x-2-mid-x-1-leq-x-2 | # Two dice: $\Pr(X_1 \lt X_2 \mid X_1 \leq X_2)$
This question comes from a completed, marked, and returned exam. It will not likely be reused.
### Problem
The problem statement gives a fair die labeled 1-6. Let $$X_1,X_2$$ denote the two observed labels on two rolls. True or False: $$\Pr(X_1 \lt X_2 \mid X_1 \leq X_2)=\frac{5}{7}$$?
### Work
I erroneously concluded during the exam that the conditional probability was certain, having flipped the two expressions in my work. I marked false.
The given answer is true, yet I am unable to convince myself of it, even with simple counting arguments. I was able to conclude from definitions that the probability given is equivalent to $$\frac{\Pr(X_1 \lt X_2 \cap X_1 \le X_2)}{\Pr(X_1 \le X_2)}$$
I have been unable to precede: my other argument goes something like
• if $$X_2$$ is 2, the conditional probability is $$\frac{1}{2}$$;
• if $$X_2$$ is 3, the conditional probability is $$\frac{2}{3}$$;
&c. But the sum will be greater than 1. Of course, the probability for $$X_2$$ is $$\frac{1}{6}$$, so we try $$\frac{1}{6}\sum_{1 \le i \le 5}{\frac{i}{i+1}} \neq \frac{5}{7}$$
### Question
At any rate, I am unable to reach $$\frac{5}{7}$$ via these arguments. What reasoning leads to the conclusion?
$$\mathbb P\{X_1
Now, $$|\{X_1\leq X_2\}|=1+2+3+4+5+6=21$$ and $$|\{X_1=X_2\mid X_1\leq X_2\}|=6.$$ Therefore, $$\mathbb P\{X_1=X_2\mid X_1\leq X_2\}=\frac{6}{21},$$ and thus $$\mathbb P\{X_1
There are $$21$$ die rolls where $$X_1 \leq X_2$$, each equally likely. Of those, in $$6$$ of them, $$X_1 = X_2$$, so in the remaining $$15, X_1 \lt X_2.$$ Therefore, your answer is \$15/21 = 5/7.
The problem with your reasoning is that given that $$X_1 \leq X_2$$, the possible values of $$X_2$$ are not equally likely. | 2019-07-21T00:24:39 | {
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