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http://openstudy.com/updates/51b43cdae4b06ee3ee28296a | ## mukushla 2 years ago A nice problem :) How many solutions are there for the equation$x^2+y^2=xy(x,y)+[x,y]$where$(x,y)=\gcd(x,y)$$[x,y]=\text{lcm}(x,y)$$x,y \in \mathbb{N}$$x\le y \le100$
1. ganeshie8
im stuck at y/x = (x,y) am i in right direction ?
2. mukushla
what did u do, plz show ur work briefly...and i dont know the answer :)
3. ganeshie8
x^2 + y^2 = xy(x,y) + [x,y] x^2 + y^2 = xy(x,y) + xy/(x,y) (x^2 + y^2)/xy = (x,y) + 1/(x,y) x/y + y/x = (x,y) + 1/(x,y) ... not sure how to conclude
4. mukushla
me too
5. experimentX
that implies if y|x, then x,y is the solution of equation. all multiplies of of numbers is the solution of it
6. mukushla
emm 2|6 but (2,6) is not a solution...
7. ganeshie8
like take y=100, x=50. that gives gcd = 2 which is not a solution
8. experimentX
wopps!! sorry wrong conclusion
9. mukushla
:D np man
10. ganeshie8
Got it ! (2, 4) is a solution all (x, x^2) pairs less than 100 will work
11. ganeshie8
and only these will work. so total 10 solutions
12. mukushla
thats right :) 10 solutions... how did u do it?
13. mathslover
They all will be of the form : (x, x^2)
14. mathslover
(1,1) , (2,4) ... (10,100)
15. mathslover
So, 10 solutions.
16. mathslover
if we iobserve carefully then we will see that RHS is a multiple of both x and y (separately). So, LHS must also be a multiple f x and y
17. mathslover
Now, x|(x^2 + y^2) => x | y^2 Similarly, y|x^2 Let, x| y^2 => y^2 = lambda x and x^2 = mu y Thus, solving these two I get x^2 = lambda y^2 = lambda^2 Thus, (x,x^2) is the general solution, As, y should be between 0 and 100. So, x can range from 0 to 10. So here is the ans : (1,1),(2,4)...(10,100)
18. mukushla
hey i'll come back to this later :)
19. mathslover
Now, $$\mathsf{x|(x^2 + y^2)\\ \implies x | y^2 \\ Similarly, \\ y|x^2 \\ Let, \\ x| y^2 \\ \implies y^2 = \lambda x \\ and\\ x^2 = \mu y \\ Thus~ , ~ solving ~ these ~two ~ I ~get \\ x^2 = \lambda\\ y^2 = \lambda^2\\ Thus,~ \\ (x,x^2) ~is~ the ~general~ solution,\\ As,~ y ~ should ~be ~between~ 0 ~and ~100.\\ So,~ x~ can ~range ~from~ 0 ~to ~10.\\ So ~here ~is ~the ~ans ~: ~~(1,1),(2,4)...(10,100)}$$
20. mathslover
Ok mukushla, will wait for your response. Sorry for interrupting your solution @ganeshie8 , but is it same to mine?
21. ganeshie8
x^2 + y^2 = xy(x,y) + [x,y] x^2 + y^2 = xy(x,y) + xy/(x,y) (x^2 + y^2)/xy = (x,y) + 1/(x,y) x/y + y/x = (x,y) + 1/(x,y) y/x = (x,y) y needs to be multiple of x, cuz (x,y) is a natural number. that gives gcd = x kx/x = x k = x y = x^2
22. mathslover
Also right, nice!
23. ganeshie8
Excellent work @mathslover
24. mathslover
thanks
25. mukushla
mathslover...how u come up with x^2=lambda and y^2=lambda^2 ?
26. mathslover
Actually it involves a large algebra.
27. mathslover
28. experimentX
looks like you guys nailed it.
29. mathslover
$$\mathsf{y^2 = \lambda x \\ x^2 = \lambda y \\ Therefore, ~ \cfrac{y^4}{\lambda^2 } = \mu y \\ \implies y^3 = \mu \lambda ^2 \\ Therefore, ~ x^6 = \lambda^3 y^3 \\ \implies x^6 = \lambda^2 \mu ^4 \\ \implies x^3 = \lambda \mu^2 \\ Now, ~ I ~ had ~ already ~ calculated ~ two ~ solutions ~ (1,1) ~ and ~ (2,4) . \\ Putting ~ them ~ in ~ the ~ equations ~ I ~ got: \mu = 1 \\ And ~ thus ~ the ~ answer . }$$ I agree that it looks absurd but it didn't strike the way ganeshi8 did.
30. mukushla
@ganeshie8 and @mathslover :)
31. ganeshie8
thanks muku for the beautiful problem :)
32. mathslover
Yep. It was really a nice problem mukushla. | 2016-05-06T22:34:44 | {
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https://math.stackexchange.com/questions/735039/die-roll-value-of-4-must-follow-value-of-1-to-win | # Die roll, value of 4 must follow value of 1 to win
I came across a "die roll" probability question that has me stumped.
Two fair die are being rolled by players A and B, who alternate, with A rolling first (ie. A then B then A then B...so long as the game hasn't been won). In order to win the game, a player must roll a 4 following the previous player's 1. What's the probability that A wins the game?
The normal form of this question -- "the first player to roll a 4" -- is simple enough, but I'm having a hard time understanding how the conditional aspect changes the calculation. Any thoughts?
I could not help not following this answer by this author:
While playing the game, Player A is in one of the following states before each roll:
1. Player B has just rolled something other than $1$ (initial state)
2. Player B has just rolled a $1$ (i.e. Player A might win after the next roll)
3. Player A has already lost.
4. Player A has already won.
The transitions (and transition probabilities) between these states are as follows:
1. From state (1), which is also the initial state, Player A remains at state (1) with probability $\frac56\cdot \frac56+\frac46\cdot\frac16=\frac{29}{36}$, moves to state (2) with probability $\frac66\cdot\frac16=\frac{6}{36}$ and to state (3) with probability $\frac{1}{6}\cdot\frac{1}{6}=\frac{1}{36}$.
2. From state (2), Player A moves to state (1) with probability $\frac46\cdot\frac56+\frac16\cdot\frac46=\frac{24}{36}$, remains to state (2) with probability $\frac56\cdot\frac16=\frac{5}{36}$, moves to state (3) with probability $\frac16\frac16=\frac{1}{36}$ and to state (4) with probability $\frac16=\frac{6}{36}$.
3. From state (3) Player A stays at state (3) with probability 1.
4. From state (4) Player A stays at state (4) with probability 1.
If $p_1,p_2, p_3,p_4$ are the probabilities of winning if Player A starts at states (1),(2),(3) and (4) respectively we obtain the equations $$\begin{cases}p_1&=&\frac{29}{36}p_1+\frac6{36}p_2+\frac{1}{36}p_3\\ p_2&=&\frac{24}{36}p_1+\frac{5}{36}p_2+\frac{1}{36}p_3+\frac{6}{36}p_4\\ p_3&=&0 \\ p_4&=&1 \end{cases}$$ You can solve this system to find that $p_2=\frac{42}{72}$ and $$p_1=\frac{36}{73}$$ which is the required probability, since as already mentioned (1) is the initial state.
The problem can be solved by recursion. Let $p_n$ be the probability that consecutive rolls of $1,4$ occur within the first $n$ rolls. Using inclusion/exclusion, this is
• the probability of obtaining $1,4$ on rolls $n-1$ and $n$;
• plus the probability of obtaining $1,4$ within rolls $1,\ldots,n-1$;
• minus the probability that both of the above occur.
If both occur then roll $n-1$ must be a $1$, so if $1,4$ occurs within the first $n-1$ rolls it must actually have occurred within the first $n-2$. Thus $$p_n=\frac{1}{36}+p_{n-1}-\frac{1}{36}p_{n-2}\ .$$ Since we also have the conditions $p_0=p_1=0$, we can solve by standard methods to obtain $$p_n=1-\frac{3}{\sqrt8}\left(\frac{3+\sqrt8}{6}\right)^{n+1}+\frac{3}{\sqrt8}\left(\frac{3-\sqrt8}{6}\right)^{n+1}\ ,$$ on the unlikely assumption that I have got the calculations correct.
Now A wins if for some $n=1,3,5,\ldots$, rolls $1$ to $n$ do not include $1,4$ and rolls $n+1,n+2$ are $1,4$. The probability of this happening is $$\sum_{n=1\atop n\;\rm odd}^\infty \frac{1}{36}(1-p_n)$$ which is the sum of two geometric series and after a bit of work evaluates to $\frac{36}{73}$.
• Can you explain why you back out the probability of points 1 and 2 both happening (the first portion of your answer)? – EBS Apr 1 '14 at 12:02
• Because if you don't do that then for example an outcome like $1,4,2,3,5,6,1,4$ will be counted twice. See for example here. – David Apr 1 '14 at 12:08
Let $p$ denote the probability that the player whose turn it is to play wins the game if they are the first to play or if the preceding roll was not $1$. Let $q$ denote the probability that the player whose turn it is to play wins the game if the preceding roll was $1$. One asks for $p$.
Conditioning on the present result, one gets $$p=\frac16(1-q)+\frac56(1-p),$$ where $\frac16$ is the probability to roll $1$ (then the next player starts from $1$ hence their chances of winning are $q$) and $\frac56$ the probability to roll anything else (then the next player starts from nothing hence their chances of winning are $p$). Likewise, $$q=\frac16+\frac46(1-p)+\frac16(1-q),$$ where the first $\frac16$ is the probability to roll $4$ (then the player wins), $\frac46$ is the probability to roll anything but $1$ or $4$ (then the next player starts from nothing hence their chances of winning are $p$), and the last $\frac16$ is the probability to roll $1$ (then the next player starts from $1$ hence their chances of winning are $q$).
Solving yields $$p=\frac{36}{73}\approx49.3\%,\qquad q=\frac{42}{73}\approx57.5\%.$$ | 2019-10-20T07:09:04 | {
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http://math.stackexchange.com/questions/190111/how-to-check-if-a-point-is-inside-a-rectangle/190171 | # How to check if a point is inside a rectangle?
There is a point $(x,y)$, and a rectangle $a(x_1,y_1),b(x_2,y_2),c(x_3,y_3),d(x_4,y_4)$, how can one check if the point inside the rectangle?
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sorry, I can't find proper tags for this question, please help me to improve it – Freewind Sep 2 '12 at 17:46
Similar to stackoverflow.com/questions/2752725/… – Emmad Kareem Sep 3 '12 at 7:24
I have just read some of the answers and created some images and Python code for it: martin-thoma.com/how-to-check-if-a-point-is-inside-a-rectangle – moose Sep 7 '12 at 19:31
Let $P(x,y)$, and rectangle $A(x_1,y_1),B(x_2,y_2),C(x_3,y_3),D(x_4,y_4)$
Calculate the sum of areas of $\triangle APD, \triangle DPC, \triangle CPB, \triangle PBA$.
1. If this sum is greater than the area of the rectangle, then $P(x,y)$ is outside the rectangle.
2. Else if this sum is equal to the area of the rectangle (observe that this sum can not be less than the later),
1. if area of any of the triangle is $0$, then $P(x,y)$ is on the rectangle (in fact on that line corresponding to the triangle of area$=0$). Observe that the equality of the sum is necessary, only area$=0$ is not sufficient),
2. else $P(x,y)$ is is inside the rectangle.
Acceptably this approach needs substantial amount of computation. This approach can be employed to any irregular polygon, too.
Another way is to calculate the perpendicular distances of $P(x,y)$ from all the 4 lines $AB,CD, AD,BC$
To be inside the rectangle, the perpendicular distances from $AB, P_{AB}$(say) and from $CD, P_{CD}$(say) must be less than $|AD|=|BC|$ and the perpendicular distances from $AD, P_{AD}$(say) and from $BC, P_{BC}$(say) must be less than $|CD|=|AB|$. Here , the areas of each of the four triangles < $\frac{1}{2}$the area of the rectangle.
1. If one of the perpendicular distances is greater than the respective length, then $P(x,y)$ is outside the rectangle.
This essentially implies and is implied by the statement : the area of the respective triangle > $\frac{1}{2}$the area of the rectangle (as commented by Ben Voigt) as $\triangle APD=\frac{1}{2}AD\cdot P_{AD}$.
2. Else if $P_{AB}=0$ and $P_{CD}=|AD|$ , then $P(x,y)$ is on AB . So, $\triangle PBA=0$ and $\triangle PCD=\frac{1}{2}$the area of the rectangle.
Observe that in this case, the rest two perpendicular distances $P_{AD}, P_{BC}$ must be ≤ $|AB|=|CD|$, $P_{BC}=|AB|\implies P(x,y)$ is lies on AD i.e, P coincides with A as it is already on AB .
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(+1) Beautiful! – Max Sep 2 '12 at 18:15
Perpendicular distances involve at least one suqre root operation, which I think is avoidable. – Hagen von Eitzen Sep 2 '12 at 20:48
@Hagen: The squared perpendicular distance is good enough. – Hurkyl Sep 3 '12 at 2:18
Wouldn't it be simpler, potentially fewer computations, and equally correct to test if the area of any triangle is greater than half the area of the rectangle? – Ben Voigt Sep 3 '12 at 4:17
@BenVoigt, who'll dare to disagree? – lab bhattacharjee Sep 3 '12 at 5:28
$M$ of coordinates $(x,y)$ is inside the rectangle iff
$$(0<\textbf{AM}\cdot \textbf{AB}<\textbf{AB}\cdot \textbf{AB}) \land (0<\textbf{AM}\cdot \textbf{AD}<\textbf{AD}\cdot \textbf{AD})$$ (scalar product of vectors)
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Far more efficient than any of the triangle methods. Why aren't these getting more votes? – Erick Wong Sep 3 '12 at 20:54
@Erick: Probably because I added it late (and I don't know why this straightforward method was not proposed earlier here...) – Raymond Manzoni Sep 3 '12 at 21:36
This is really brilliant! – user1783444 Jun 5 '13 at 17:05
Thanks @user1783444 ! I tried to provide something mathematically simple (other solutions may be better in practice...). – Raymond Manzoni Jun 5 '13 at 19:19
I would use a "point-in-convex-polygon" function; this works by checking whether the point is "to the left of" each of the four lines.
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For example, if all the triangles $ABP$, $BCP$, $CDP$, $DAP$ are positively oriented; for $ABP$ this can be tested by checking the sign of $(x-x_1)\cdot(y-y_2)-(y-y_1)\cdot(x-x_2)$. This method generalizes to convex polygons.
Alternatively, find a transformation that makes the rectangle parallel to the axes. Apply the same transformation to your point at the test is simple. This is especially helpful if you want to test many points against the same rectangle.
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One of the simplest algorithms of coordinate plane geometry is a method of "Is the vector $\vec{w}$ clockwise or counterclockwise from the vector $\vec{v}$?"
The algorithm is to compute the cross product $\vec{v} \times \vec{w}$. If the sign is positive, then $\vec{w}$ is counter-clockwise from $\vec{v}$. If negative, it is clockwise.
This method gives an algorithm for answering the question "Which side of a line is a point on?"
e.g. consider the left line of your box, line $da$, directed from $d$ to $a$. The point $p=(x,y)$ is on the right side of this line if and only if the vector $dp$ is clockwise from $da$. So we can simply compute the cross product to learn if $p$ is on the correct side of $da$ or not.
Do this with all four lines, and you have your test.
If you're using the same four points a lot, we can get away with half as much work. We can use the dot product to compute "How long is one vector in the direction another is pointing?"
We can compute the dot product of $da$ with $dp$. If this number is greater than the dot product of $da$ with itself, then your point is too far up to be in the box. If this number is negative, the point is too far down to be in the box.
Because this checks two boundaries at once, we only have to do this twice: once with $da$ to get the up-down location, and once with $dc$ to get the left-right location.
EDIT: One of the general ideas at play is to try and make as much use of the dot product and cross product as you can, as they are computationally efficient ways of computing geometric information related to things like length, angle, and area.
So a common strategy for a problem of computational geometry is to try and phrase the problem in terms of things that can be answered with dot and cross products. If that's not immediately obvious, phrase it in terms of (directed) lengths, angles, and areas, and try to rephrase those in terms of dot and cross products.
My approach to the question was to express "which side of a line?" as a directed angle, allowing use of cross product.
Also, I used the dot product to compute distance along a direction parallel to one of the sides of the rectangle. You could compute the same thing as "perpendicular distance" from an adjacent side, which would let you use the cross product.
Some of the other answers express things in terms of the areas of triangles -- these are naturally computed cross products. One answer summed their magnitudes and compared with the area of the rectangle. Another looked at the direction of the areas instead (i.e. whether the triangles were positively oriented).
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How can there be no vector math solutions to this yet?
Let $(x',y')=(x-x_1,y-y_1)$ so that the point $(x_1,y_1)$ is the origin.. Then get the height and width of the rectangle
$w = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$
$h = \sqrt{(x_4-x_1)^2 + (y_4-y_1)^2}$
And then see whether $(x',y')$ is within that rectangle
$0 \leq x'{(x_2-x_1) \over w} + y'{(y_2-y_1) \over h} \leq w$
$0 \leq x'{(x_4-x_1) \over w} + y'{(y_4-y_1) \over h} \leq h$
Done! Way easier than the triangle methods--those are for generic convex polygons and are awesome if you have a convex polygon, but are overkill for a rectangle.
(It's even shorter to use the vector forms. If your coordinates are $\vec{p},\dots,\vec{p_4}$ then subtract $\vec{p_1}$ from everyone to get $\vec{p'}, \vec{p_2'}, \vec{p_4'}$. Then you're in the rectangle iff $0 \leq \vec{p'}\cdot\hat{p_2'} \leq p_2'$ and $0 \leq \vec{p'}\cdot\hat{p_4'} \leq p_4'$.)
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The point is inside the rectangle.
The point is outside the rectangle.
Let us define:
A: area of the rectangle
Ai: areas of the triangles shown in the pictures. (i = 1, 2, 3, 4)
ai: lengths of the edges shown in the pictures. (i = 1, 2, 3, 4)
bi: lengths of the line segments connecting the point and the corners. (i = 1, 2, 3, 4)
If the point is inside the rectangle, the following equation holds:
$\mathbf{A = A_1 + A_2 + A_3 + A_4}$
If the point is outside the rectangle, the following inequality holds:
$\mathbf{A > A_1 + A_2 + A_3 + A_4}$
How do we calculate A, A1, A2, A3, A4, a1, a2, a3, a4, b1, b2, b3 and b4?
First we calculate the edge lengths:
$a_1 = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} \\ a_2 = \sqrt{(x_2 - x_3)^2 + (y_2 - y_3)^2} \\ a_3 = \sqrt{(x_3 - x_4)^2 + (y_3 - y_4)^2} \\ a_4 = \sqrt{(x_4 - x_1)^2 + (y_4 - y_1)^2}$
Next we calculate the lengths of the line segments:
$b_1 = \sqrt{(x_1 - x)^2 + (y_1 - y)^2} \\ b_2 = \sqrt{(x_2 - x)^2 + (y_2 - y)^2} \\ b_3 = \sqrt{(x_3 - x)^2 + (y_3 - y)^2} \\ b_4 = \sqrt{(x_4 - x)^2 + (y_4 - y)^2}$
Then we calculate the areas using Heron's Formula:
$A \,\,\, = a_1a_2 = a_2a_3 = a_3a_4 = a_4a_1 \\ u_1 = \frac{a_1 + b_1 + b_2}{2} \\ u_2 = \frac{a_2 + b_2 + b_3}{2} \\ u_3 = \frac{a_3 + b_3 + b_4}{2} \\ u_4 = \frac{a_4 + b_4 + b_1}{2} \\ A_1 = \sqrt{u_1(u_1 - a_1)(u_1 - b_1)(u_1 - b_2)} \\ A_2 = \sqrt{u_2(u_2 - a_2)(u_2 - b_2)(u_2 - b_3)} \\ A_3 = \sqrt{u_3(u_3 - a_3)(u_3 - b_3)(u_3 - b_4)} \\ A_4 = \sqrt{u_4(u_4 - a_4)(u_4 - b_4)(u_4 - b_1)}$
Finally you can do the area test to check if the point is inside or outside the rectangle.
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You can use complex analysis to check (which is really overkill and probably not computationally efficient but still neat and generalizes to all polygons without any difficulty). Specifically, identify the point $(a,b)$ with the complex number $a+bi$. Let $\gamma$ parametrize the rectangle by going around once counterclockwise. By Cauchy integral formula we know that
$\displaystyle \frac{1}{2 \pi i}\int_\gamma \frac{1}{z-x-iy}dz = \left\{ \begin{array}{lr} 1 \mbox{ if (x,y) is in the rectangle} \\ 0 \mbox{ if (x,y) is outside the rectangle} \end{array} \right.$
and is not defined if (x,y) is on any of the four boundary segments.
The integral is over 4 strait line segments. On the strait line segment from $(x_2,y_2)$ to $(x_1,y_1)$, the integral is just
$\displaystyle \int_0^1 \frac{x_1-x_2+i(y_1-y_2)}{t(x_1-x_2+i(y_1-y_2))+x_2-x+i(y_2-y)} dt = \int_0^1 \frac{1}{t-w_1}$
where $w_1= \frac{x_2-x+i(y_2-y)}{x_2-x_1+i(y_2-y_1)}$. This integral can be done so long as $w_1 \not \in [0,1]$, which corresponds to (x,y) on the segment. The result is $\log(1-w_1)-\log(-w_1)$. Note that this is a holomorphic function of $w_1$ on $\mathbb{C} - [0,1]$, where we have adopted the branch cut for $\log$ as $\log(r e^i \theta)= \ln(r) + i \theta$ for $\theta \in (-\pi, \pi]$.
Summing over the four edges gives the left hand side above as $\frac{1}{2 \pi i}\displaystyle \sum_{j=1}^4 \log(1-w_j)- \log(w_j)$ where $w_2,w_3$, and $w_4$ are defined analogously to $w_1$. You only need compute this sum to sufficient precision to rule out it equaling either 0 or 1. As I said before this isn't usually a good way in practice, partially because logs are pretty computationally expensive, but it is a way to do this.
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If you take the exponential of the sum, you get a product that can be evaluated without logs, it isn't too bad. – Max Sep 3 '12 at 13:41
Consider the rectangle as defining a coordinate system with origin $A$ and basis vectors $v_1 = B-A$ and $v_2 = D-A$. We can find the point $p = (x,y)$ in terms of this coordinate system as $\left(\frac{(p - A)·v_1}{v_1·v_1}, \frac{(p - A)·v_2}{v_2·v_2}\right)$. The coordinates of this transformed point are within the range 0 to 1 if and only if the point is within the rectangle.
This is equivalent to Raymond Manzoni's answer, but I thought it worth deriving in this fashion.
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This is intutition at work, but what if you found the distances from the vertices to the point and added them all up?
Your min would be the distances to the center of the rectangle and your max is the distances if your point was on one of the vertices.
For example, for a 3x4 triangle, you min would be 10 (two times your diagonal) and your max would be 12 (if your point was on a vertex). This is your range: the point is inside the rectangle, if your distances from the vertices to the point are within the range 10 to 12 (for a 3x4 rectangle).
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For me, the basic tool to use answers the question "are two points on the same side of a line?"
If the equation of the line is $ax+by+c = 0$ and the points are $(x_i, y_i)$ for $i = 1, 2$, compute $d_i = a x_i + b y_i + c$ for each $i$. If the two $d_i$ have the same sign (both $> 0$ or $< 0$ to avoid problems about being on the line), they are on the same side of the line.
We now use this. Compute the point that is the average of the four vertices. We know that this is inside the rectangle - this will work for a triangle also, where this problem more commonly occurs.
For each line making up the rectangle, do the "two points on the same side of a line" test using the average point and your test point. If the average point and your test are on the same side for each line (though they may be on different sides for different lines), the point is inside the rectangle.
Doing it this way, there are no worries about orientation or traversing the vertices of the rectangle in a certain order.
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My first thought was to divide rectangle to two triangles and apply some optimized method twice. It seemed to be more effective than @lab bhattacharjee's answer to me.
http://www.blackpawn.com/texts/pointinpoly/default.html
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Check if your point is inside triangle a,b,c or inside triangle a,c,d. There are several methods for doing this. See e.g. http://www.blackpawn.com/texts/pointinpoly/default.html or http://stackoverflow.com/questions/2049582/how-to-determine-a-point-in-a-triangle
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It is very easy.
Just divide Rectangle in 2 triangle i.e.
1) a(x1,y1),b(x2,y2),c(x3,y3)
2) a(x1,y1),d(x4,y4),c(x3,y3)
Then, point is in Rectangle, if and only if point is in at least 1 triangle.
Following is code for checking point in triangle :
float sign(fPoint p1, fPoint p2, fPoint p3)
{
return (p1.x - p3.x) * (p2.y - p3.y) - (p2.x - p3.x) * (p1.y - p3.y);
}
bool PointInTriangle(fPoint pt, fPoint v1, fPoint v2, fPoint v3)
{
bool b1, b2, b3;
b1 = sign(pt, v1, v2) < 0.0f;
b2 = sign(pt, v2, v3) < 0.0f;
b3 = sign(pt, v3, v1) < 0.0f;
return ((b1 == b2) && (b2 == b3));
}
-
First of all, we have to choose 2 adjacent sides of the rectangle and then form the equations that contains them. The second step is to create 2 equations that contains the 2 slopes of the equations of the 2 adjacent sides and the critical point. Now we'll equate the equation that contains a side of a rectangle with slope m to one equation that contains the critical point and the slope -1/m, and then will equate the left equations. We'll also take into account that each solution has to respect the inequalities that can be easily deduced by checking the coordinates of the 2 adjacent sides. If we get 2 intersection points that respect the inequalities, then the critical point is inside the rectangle.
Q.E.D.
-
I had this observation and it may be useful (but may be wrong too, as I have not proved it formally, so this text is proposed as a comment rather than a formal answer).
A rectangle can be divided into four inner rectangles with no gaps and as a result, there would be $1$ point that is common to all resulting 4 inner rectangles. Let that point be $p1$ in the picture below.
$p1$, has the following property: We could draw 2 lines intersecting at $p1$ such that each of the lines going through p1 is perpendicular on 1 of the rectangle sides. For example, Line $DC$ is perpendicular on line segment $v1v2$ and Line $AB$ is perpendicular on line segment $v4v1$.
The above property of $p1$ can't be satisfied by any point outside the rectangle.
Now, given a point $p1$, and 4 points that represent the rectangle corners, we may apply the above property to determine if p1 lies within the rectangle or not.
-
You could use a different coordinate system, and reduce the problem to the case when one of the rectangle's legs is on the x axis and the others parallel to the y. In this case, you just see if the x and y inequalities for the rectangle are satisfied by your point. By a change of coordinate system, I mean simply choosing one of the vertices of the rectangle to be the origin, and using two perpendicular unit vectors for each of the adjacent sides (to the corner). Then, since you know the length of the sides of the rectangle, you can easily deduce the necessary rectangle inequalities.
NB by rectangle inequalities I mean the necessary and sufficient inequalities that must be satisfied in order for a point to lie in a rectangle in the first case.
-
Calculate $\perp$ distances of point $P$ from all sides of rectangle and check if any of the distances of point from opposite sides is greater than the distance between them (one of the side length of rectangle). If it is greater than it is outside otherwise inside.
-
find the equations of the lines and equate the distance from the point P TO THE GIVEN TWO Pairs of lines .if the point is inside the rectangle,the distances to the pairs will be less than the distances between the line in the pair
-
Set center x of rectangle to 'middle x' Set center y of rectangle to 'middle y' Set x of the Point to 'point x' Set y of the Point to 'point y'
If the distance between point x and middle x is > 1/2 width of rectangle AND distance between point y and middle y is > 1/2 height of rectangle;
Then the point is outside the rectangle.
If distance between point x and middle x is < 1/2 width of rectangle AND distance between point y and middle y is < 1/2 height of rectangle;
Then the point is inside the rectangle.
My Logic is not flawed. Ribbit!
-
This is rather unclear and difficult to read. It seems almost like you're trying to write pseudocode, but you haven't defined your terms (e.g. "rectangular actor"). Also, if you're asking if your solution is correct, you should ask a new question - not present it as an answer. – T. Bongers Sep 21 '13 at 5:03
-1 the OP the problem is posed using the terms (x,y), a(x1,y1), b(x2,y2),c (x3,y3), d(x4,y4) and rectangle. If you change this without further explanation (old x, new x, actor) no one will know what you are talking about. – miracle173 Sep 21 '13 at 6:59
If you draw a line from point a to point c by using Pythagorean Theorem, and stop at Half distance in center of box, that's center x,y spot of rectangle. If the point is outside box, the absolute value of (its x,y minus center x,y spot) will be greater than absolute value of (any point on the wall of box minus center x,y spot). – frog Sep 21 '13 at 8:54
It is still hard do understand but I think your approach is that of this answer – miracle173 Sep 21 '13 at 10:26 | 2014-03-08T20:48:19 | {
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https://community.plu.edu/~sklarjk/fsaa/appendix-2.html | # Exercises1.4Exercises
##### Exercise1
Yes/No. For each of the following, write Y if the object described is a well-defined set; otherwise, write N. You do NOT need to provide explanations or show work for this problem.
1. $\{z \in \C \,:\, |z|=1\}$
2. $\{\epsilon \in \R^+\,:\, \epsilon \mbox{ is sufficiently small} \}$
3. $\{q\in \Q \,:\, q \mbox{ can be written with denominator } 4\}$
4. $\{n \in \Z\,:\, n^2 \lt 0\}$
Solution
##### Exercise2
List the elements in the following sets, writing your answers as sets.
Example: $\{z\in \C\,:\,z^4=1\}$ Solution: $\{\pm 1, \pm i\}$
1. $\{z\in \R\,:\, z^2=5\}$
2. $\{m \in \Z\,:\, mn=50 \mbox{ for some } n\in \Z\}$
3. $\{a,b,c\}\times \{1,d\}$
4. $P(\{a,b,c\})$
Solution
##### Exercise3
Let $S$ be a set with cardinality $n\in \N\text{.}$ Use the cardinalities of $P(\{a,b\})$ and $P(\{a,b,c\})$ to make a conjecture about the cardinality of $P(S)\text{.}$ You do not need to prove that your conjecture is correct (but you should try to ensure it is correct).
Solution
##### Exercise4
Let $f: \Z^2 \to \R$ be defined by $f(a,b)=ab\text{.}$ (Note: technically, we should write $f((a,b))\text{,}$ not $f(a,b)\text{,}$ since $f$ is being applied to an ordered pair, but this is one of those cases in which mathematicians abuse notation in the interest of concision.)
1. What are $f$'s domain, codomain, and range?
2. Prove or disprove each of the following statements. (Your proofs do not need to be long to be correct!)
1. $f$ is onto;
2. $f$ is 1-1;
3. $f$ is a bijection. (You may refer to parts (i) and (ii) for this part.)
3. Find the images of the element $(6,-2)$ and of the set $\Z^- \times \Z^-$ under $f\text{.}$ (Remember that the image of an element is an element, and the image of a set is a set.)
4. Find the preimage of $\{2,3\}$ under $f\text{.}$ (Remember that the preimage of a set is a set.)
Solution
##### Exercise5
Let $S\text{,}$ $T\text{,}$ and $U$ be sets, and let $f: S\to T$ and $g: T\to U$ be onto. Prove that $g \circ f$ is onto.
Solution
##### Exercise6
Let $A$ and $B$ be sets with $|A|=m\lt \infty$ and $|B|=n\lt \infty\text{.}$ Prove that $|A\times B|=mn\text{.}$
Solution
# Exercises2.2Exercises, Part I
##### Exercise1
For each of the following, write Y if the given “operation” is a well-defined binary operation on the given set; otherwise, write N. In each case in which it isn't a well-defined binary operation on the set, provide a brief explanation. You do not need to prove or explain anything in the cases in which it is a binary operation.
1. $+$ on $\C^*$
2. $*$ on $\R^+$ defined by $x*y=\log_x y$
3. $*$ on $\M_2(\R)$ defined by $A*B=AB^{-1}$
4. $*$ on $\Q^*$ defined by $z*w=z/w$
Solution
##### Exercise2
Define $*$ on $\Q$ by $p*q=pq+1\text{.}$ Prove or disprove that $*$ is (a) commutative; (b) associative.
Solution
##### Exercise3
Prove that matrix multiplication is not commutative on $\M_2(\R)\text{.}$
Solution
##### Exercise4
Prove or disprove each of the following statements.
1. The set $2\Z=\{2x\,:\,x\in \Z\}$ is closed under addition in $\Z\text{.}$
2. The set $S=\{1,2,3\}$ is closed under multiplication in $\R\text{.}$
3. The set
\begin{equation*} U=\left\{ \begin{bmatrix} a \amp b\\ 0 \amp c \end{bmatrix}\,:\,a,b,c\in \R\right\} \end{equation*}
is closed under multiplication in $\M_2(\R)\text{.}$ (Recall that $U$ is the set of upper-triangular matrices in $\M_2(\R)\text{.}$)
Solution
##### Exercise5
Let $*$ be an associative and commutative binary operation on a set $S\text{.}$ An element $u\in S$ is said to be an idempotent in $S$ if $u*u=u\text{.}$ Let $H$ be the set of all idempotents in $S\text{.}$ Prove that $H$ is closed under $*\text{.}$
Solution
# Exercises2.8Exercises, Part II
##### Exercise1
True/False. For each of the following, write T if the statement is true; otherwise, write F. You do NOT need to provide explanations or show work for this problem.
1. For every positive integer $n\text{,}$ there exists a group of order $n\text{.}$
2. For every integer $n\geq 2\text{,}$ $\Z_n$ is abelian.
3. Every abelian group is finite.
4. For every integer $m$ and integer $n\geq 2\text{,}$ there exist infinitely many integers $a$ such that $a$ is congruent to $m$ modulo $n\text{.}$
5. A binary operation $*$ on a set $S$ is commutative if and only if there exist $a,b\in S$ such that $a*b=b*a\text{.}$
6. If $\langle S, *\rangle$ is a binary structure, then the elements of $S$ must be numbers.
7. If $e$ is an identity element of a binary structure (not necessarily a group) $\langle S,*\rangle\text{,}$ then $e$ is an idempotent in $S$ (that is, $e*e=e$).
8. If $s$ is an idempotent in a binary structure (not necessarily a group) $\langle S,*\rangle\text{,}$ then $s$ must be an identity element of $S\text{.}$
Solution
##### Exercise2
Let $G$ be the set of all functions from $\Z$ to $\R\text{.}$ Prove that pointwise multiplication on $G$ (that is, the operation defined by $(fg)(x)=f(x)g(x)$ for all $f,g\in G$ and $x\in \Z$) is commutative. (Note. To prove that two functions, $h$ and $j\text{,}$ sharing the same domain $D$ are equal, you need to show that $h(x)=j(x)$ for every $x\in D\text{.}$)
Solution
##### Exercise3
Decide which of the following binary structures are groups. For each, if the binary structure isn't a group, prove that. (Remember, you should not state that inverses do or do not exist for elements until you have made sure that the structure contains an identity element!) If the binary structure is a group, prove that.
1. $\Q$ under multiplication
2. $\M_2(\R)$ under addition
3. $\M_2(\R)$ under multiplication
4. $\R^+$ under $*\text{,}$ defined by $a*b=\sqrt{ab}$ for all $a,b\in \R^+$
Solution
##### Exercise4
Give an example of an abelian group containing 711 elements.
Solution
##### Exercise5
Let $n\in \Z\text{.}$ Prove that $n\Z$ is a group under the usual addition of integers. Note: You may use the fact that $\langle n\Z,+\rangle$ is a binary structure if you provide a reference for this fact.
Solution
##### Exercise6
Let $n\in \Z^+\text{.}$ Prove that $SL(n,\R)$ is a group under matrix multiplication. Note: You may use the fact that $\langle SL(n\R),\cdot\rangle$ is a binary structure if you provide a reference for this fact.
Solution
##### Exercise7
1. List three distinct integers that are congruent to $6$ modulo $5\text{.}$
2. List the elements of $\Z_5\text{.}$
3. Compute:
1. $4+5$ in $\Z\text{;}$
2. $4+5$ in $\Q\text{;}$
3. $4+_65$ in $\Z_6\text{;}$
4. the inverse of $4$ in $\Z\text{;}$
5. the inverse of $4$ in $\Z_6\text{.}$
4. Why does it not make sense for me to ask you to compute $4+_3 2$ in $\Z_3\text{?}$ Please answer this using a complete, grammatically correct sentence.
Solution
##### Exercise8
Let $G$ be a group with identity element $e\text{.}$ Prove that if every element of $G$ is its own inverse, then $G$ is abelian.
Solution
##### Exercise9
Let $G$ be a group. The subset
\begin{equation*} Z(G):=\{z \in G\,:\, zg=gz \mbox{ for all } g\in G\} \end{equation*}
of $G$ is called the center of $G\text{.}$ In other words, $Z(G)$ is the set of all elements of $G$ that commute with every element of $G\text{.}$ Prove that $Z(G)$ is closed in $G\text{.}$
Solution
# Exercises3.4Exercises
##### Exercise1
True/False. For each of the following, write T if the statement is true; otherwise, write F. You do NOT need to provide explanations or show work for this problem. Throughout, let $G$ and $G'$ be groups.
1. If there exists a homomorphism $\phi\,:\,G\to G'\text{,}$ then $G$ and $G'$ must be isomorphic groups.
2. There is an integer $n\geq 2$ such that $\Z\simeq \Z_n\text{.}$
3. If $|G|=|G'|=3\text{,}$ then we must have $G\simeq G'\text{.}$
4. If $|G|=|G'|=4\text{,}$ then we must have $G\simeq G'\text{.}$
Solution
##### Exercise2
For each of the following functions, prove or disprove that the function is (i) a homomorphism; (ii) an isomorphism. (Remember to work with the default operation on each of these groups!)
1. The function $f:\Z\to\Z$ defined by $f(n)=2n\text{.}$
2. The function $g:\R\to\R$ defined by $g(x)=x^2\text{.}$
3. The function $h:\Q^*\to\Q^*$ defined by $h(x)=x^2\text{.}$
Solution
##### Exercise3
LDefine $d : GL(2,\R)\to \R^*$ by $d(A)=\det A\text{.}$ Prove/disprove that $d$ is:
1. a homomorphism
2. 1-1
3. onto
4. an isomorphism.
Solution
##### Exercise4
Complete the group tables for $\Z_4$ and $\Z_8^{\times}\text{.}$ Use the group tables to decide whether or not $\Z_4$ and $\Z_8^{\times}$ are isomorphic to one another. (You do not need to provide a proof.)
Solution
##### Exercise5
Let $n\in \Z^+\text{.}$ Prove that $\langle n\Z,+\rangle \simeq \langle \Z,+\rangle\text{.}$
Solution
##### Exercise6
1. Let $G$ and $G'$ be groups, where $G$ is abelian and $G\simeq G'\text{.}$ Prove that $G'$ is abelian.
2. Give an example of groups $G$ and $G'\text{,}$ where $G$ is abelian and there exists a homomorphism from $G$ to $G'\text{,}$ but $G'$ is NOT abelian.
Solution
##### Exercise7
Let $\langle G,\cdot\rangle$ and $\langle G',\cdot'\rangle$ be groups with identity elements $e$ and $e'\text{,}$ respectively, and let $\phi$ be a homomorphism from $G$ to $G'\text{.}$ Let $a\in G\text{.}$ Prove that $\phi(a)^{-1}=\phi(a^{-1})\text{.}$
Solution
# Exercises4.3Exercises
##### Exercise1
True/False. For each of the following, write T if the statement is true; otherwise, write F. You do NOT need to provide explanations or show work for this problem. Throughout, let $G$ and $G'$ be groups.
1. Every group contains at least two distinct subgroups.
2. If $H$ is a proper subgroup of group $G$ and $G$ is finite, then we must have $|H|\lt |G|\text{.}$
3. $7\Z$ is a subgroup of $14\Z\text{.}$
4. A group $G$ may have two distinct proper subgroups which are isomorphic (to one another).
Solution
##### Exercise2
Give specific, precise examples of the following groups $G$ with subgroups $H\text{:}$
1. A group $G$ with a proper subgroup $H$ of $G$ such that $|H|=|G|\text{.}$
2. A group $G$ of order 12 containing a subgroup $H$ with $|H|=3\text{.}$
3. A nonabelian group $G$ containing a nontrivial abelian subgroup $H\text{.}$
4. A finite subgroup $H$ of an infinite group $G\text{.}$
Solution
##### Exercise3
Let $n\in \Z^+\text{.}$
1. Prove that $n\Z \leq \Z\text{.}$
2. Prove that the set $H=\{A\in \M_n(\R)\,:\,\det A=\pm 1\}$ is a subgroup of $GL(n,\R)\text{.}$
(Note: Your proofs do not need to be long to be correct!)
Solution
##### Exercise4
Let $n\in \Z^+\text{.}$ For each group $G$ and subset $H\text{,}$ decide whether or not $H$ is a subgroup of $G\text{.}$ In the cases in which $H$ is not a subgroup of $G\text{,}$ provide a proof. (Note. Your proofs do not need to be long to be correct!)
1. $G=\R\text{,}$ $H=\Z$
2. $G=\Z_{15}\text{,}$ $H=\{0,5,10\}$
3. $G=\Z_{15}\text{,}$ $H=\{0,4,8,12\}$
4. $G=\C\text{,}$ $H=\R^*$
5. $G=\C^*\text{,}$ $H=\{1,i,-1,-i\}$
6. $G=\M_n(\R)\text{,}$ $H=GL(n,\R)$
7. $G=GL(n,\R)\text{,}$ $H=\{A\in \M_n(\R)\,:\,\det A = -1\}$
Solution
##### Exercise5
Let $G$ and $G'$ be groups, let $\phi$ be a homomorphism from $G$ to $G'\text{,}$ and let $H$ be a subgroup of $G\text{.}$ Prove that $\phi(H)$ is a subgroup of $G'\text{.}$
Solution
##### Exercise6
Let $G$ be an abelian group, and let $U=\{g\in G\,:\, g^{-1}=g\}.$ Prove that $U$ is a subgroup of $G\text{.}$
Solution
# Exercises5.3Exercises
##### Exercise1
True/False. For each of the following, write T if the statement is true; otherwise, write F. You do NOT need to provide explanations or show work for this problem. Throughout, let $G$ be a group with identity element $e\text{.}$
1. If $G$ is infinite and cyclic, then $G$ must have infinitely many generators.
2. There may be two distinct elements $a$ and $b$ of a group $G$ with $\langle a\rangle =\langle b\rangle\text{.}$
3. If $a,b\in G$ and $a\in \langle b\rangle$ then we must have $b\in \langle a\rangle\text{.}$
4. If $a\in G$ with $a^4=e\text{,}$ then $o(a)$ must equal $4\text{.}$
5. If $G$ is countable then $G$ must be cyclic.
Solution
##### Exercise2
Give examples of the following.
1. An infinite noncyclic group $G$ containing an infinite cyclic subgroup $H\text{.}$
2. An infinite noncyclic group $G$ containing a finite nontrivial cyclic subgroup $H\text{.}$
3. A cyclic group $G$ containing exactly 20 elements.
4. A nontrivial cyclic group $G$ whose elements are all matrices.
5. A noncyclic group $G$ such that every proper subgroup of $G$ is cyclic.
Solution
##### Exercise3
Find the orders of the following elements in the given groups.
1. $2\in \Z$
2. $-i\in \C^*$
3. $-I_2\in GL(2,\R)$
4. $-I_2\in \M_2(\R)$
5. $(6,8)\in \Z_{10}\times \Z_{10}$
Solution
##### Exercise4
For each of the following, if the group is cyclic, list all of its generators. If the group is not cyclic, write NC.
1. $5\Z$
2. $\Z_{18}$
3. $\R$
4. $\langle \pi\rangle$ in $\R$
5. $\Z_2^2$
6. $\langle 8\rangle$ in $\Q^*$
Solution
##### Exercise5
Explicitly identify the elements of the following subgroups of the given groups. You may use set-builder notation if the subgroup is infinite, or a conventional name for the subgroup if we have one.
1. $\langle 3\rangle$ in $\Z$
2. $\langle i\rangle$ in $C^*$
3. $\langle A\rangle\text{,}$ for $A=\left[ \begin{array}{cc} 1 \amp 0 \\ 0 \amp 0 \end{array} \right]\in \M_2(\R)$
4. $\langle (2,3)\rangle$ in $\Z_4\times \Z_5$
5. $\langle B\rangle\text{,}$ for $B=\left[ \begin{array}{cc} 1 \amp 1\\ 0 \amp 1 \end{array} \right]\in GL(2,\R)$
Solution
##### Exercise6
Draw subgroup lattices for the following groups.
1. $\Z_6$
2. $\Z_{13}$
3. $\Z_{18}$
Solution
##### Exercise7
Let $G$ be a group with no nontrivial proper subgroups. Prove that $G$ is cyclic.
Solution
# Exercises6.6Exercises
##### Exercise1
Let $\sigma=(134)\text{,}$ $\tau=(23)(145)\text{,}$ $\rho=(56)(78)\text{,}$ and $\alpha=(12)(145)$ in $S_8\text{.}$ Compute the following.
1. $\sigma \tau$
2. $\tau \sigma$
3. $\tau^2$
4. $\tau^{-1}$
5. $o(\tau)$
6. $o(\rho)$
7. $o(\alpha)$
8. $\langle \tau\rangle$
Solution
##### Exercise2
Prove Lemma 6.3.4.
Solution
##### Exercise3
Prove that $A_n$ is a subgroup of $S_n\text{.}$
Solution
##### Exercise4
Prove or disprove: The set of all odd permutations in $S_n$ is a subgroup of $S_n\text{.}$
Solution
##### Exercise5
Let $n$ be an integer greater than 2. $m \in \{1,2,\ldots,n\}\text{,}$ and let $H=\{\sigma\in S_n\,:\,\sigma(m)=m\}$ (in other words, $H$ is the set of all permutations in $S_n$ that fix $m$).
1. Prove that $H\leq S_n\text{.}$
2. Identify a familiar group to which $H$ is isomorphic. (You do not need to show any work.)
Solution
##### Exercise6
Write $rfr^2frfr$ in $D_5$ in standard form.
Solution
##### Exercise7
Prove or disprove: $D_6\simeq S_6\text{.}$
Solution
##### Exercise8
Which elements of $D_4$ (if any)
1. have order 2?
2. have order $3\text{?}$
Solution
##### Exercise9
Let $n$ be an even integer that's greater than or equal to 4. Prove that $r^{n/2}\in Z(D_n)\text{:}$ that is, prove that $r^{n/2}$ commutes with every element of $D_n\text{.}$ (Do NOT simply refer to the last statement in Theorem 6.5.10; that is the statement you are proving here.)
Solution
# Exercises7.4Exercises
##### Exercise1
How many distinct partitions of the set $S=\{a,b,c,d\}$ are there? You do not need to list them. (Yes, you can find this answer online. But I recommend doing the work yourself for practice working with partitions!)
Solution
##### Exercise2
1. Let $n\in \Z^+\text{.}$ Prove that $\equiv_n$ is an equivalence relation on $\Z\text{.}$
2. The cells of the induced partition of $\Z$ are called the residue classes (or congruence classes) of $\Z$ modulo $n$. Using set notation of the form $\{\ldots,\#, \#,\#,\ldots\}$ for each class, write down the residue classes of $\Z$ modulo $4\text{.}$
Solution
##### Exercise3
Let $G$ be a group with subgroup $H\text{.}$ Prove that $\simr$ is an equivalence relation on $G\text{.}$
Solution
##### Exercise4
Find the indices of:
1. $H=\langle (15)(24)\rangle$ in $S_5$
2. $K=\langle (2354)(34)\rangle$ in $S_6$
3. $A_n$ in $S_n$
Solution
##### Exercise5
For each subgroup $H$ of group $G\text{,}$ (i) find the left and the right cosets of $H$ in $G\text{,}$ (ii) decide whether or not $H$ is normal in $G\text{,}$ and (iii) find $(G:H)\text{.}$
Write all permutations using disjoint cycle notation, and write all dihedral group elements using standard form.
1. $H=6\Z$ in $G=2\Z$
2. $H=\langle 4\rangle$ in $\Z_{20}$
3. $H=\langle (23)\rangle$ in $G=S_3$
4. $H=\langle r\rangle$ in $G=D_4$
5. $H=\langle f\rangle$ in $G=D_4$
Solution
##### Exercise6
For each of the following, give an example of a group $G$ with a subgroup $H$ that matches the given conditions. If no such example exists, prove that.
1. A group $G$ with subgroup $H$ such that $|G/H|=1\text{.}$
2. A finite group $G$ with subgroup $H$ such that $|G/H|=|G|\text{.}$
3. An abelian group $G$ of order $8$ containing a non-normal subgroup $H$ of order 2.
4. A group $G$ of order 8 containing a normal subgroup of order $2\text{.}$
5. A nonabelian group $G$ of order 8 containing a normal subgroup of index $2\text{.}$
6. A group $G$ of order 8 containing a subgroup of order $3\text{.}$
7. An infinite group $G$ containing a subgroup $H$ of finite index.
8. An infinite group $G$ containing a finite nontrivial subgroup $H\text{.}$
Solution
##### Exercise7
True/False. For each of the following, write T if the statement is true; otherwise, write F. You do NOT need to provide explanations or show work for this problem. Throughout, let $G$ be a group with subgroup $H$ and elements $a,b\in G\text{.}$
1. If $a\in bH$ then $aH$ must equal $bH\text{.}$
2. $aH$ must equal $Ha\text{.}$
3. If $aH=bH$ then $Ha$ must equal $Hb\text{.}$
4. If $a\in H$ then $aH$ must equal $Ha\text{.}$
5. $H$ must be normal in $G$ if there exists $a\in G$ such that $aH=Ha\text{.}$
6. If $aH=bH$ then $ah=bh$ for every $h\in H\text{.}$
7. $|G/H|$ must be less than $|G|\text{.}$
8. $(G:H)$ must be less than or equal to $|G|\text{.}$
Solution
##### Exercise8
Let $G$ be a group of order $pq\text{,}$ where $p$ and $q$ are prime, and let $H$ be a proper subgroup of $G\text{.}$ Prove that $H$ is cyclic.
Solution
##### Exercise9
Prove Corollary 7.3.10: that is, let $G$ be a group of prime order, and prove that $G$ is cyclic.
Solution
##### Exercise10
Let $G$ be a group of finite order $n\text{,}$ containing identity element $e\text{.}$ Let $a\in G\text{.}$ Prove that $a^n=e\text{.}$
Solution
# Exercises8.4Exercises
##### Exercise1
Let $G$ be a group and let $H\leq G$ have index 2. Prove that $H\unlhd G\text{.}$
Solution
##### Exercise2
Let $G$ be an abelian group with $N\unlhd G\text{.}$ Prove that $G/N$ is abelian.
Solution
##### Exercise3
Find the following.
1. $|2\Z/6\Z|$
2. $|H|\text{,}$ for $H=2+\langle 6\rangle \subseteq \Z_{12}$
3. $o(2+\langle 6\rangle)$ in $\Z_{12}/\langle 6\rangle$
4. $\langle f+H\rangle$ in $D_4/H\text{,}$ where $H=\{e,r^2\}$
5. $|(\Z_6\times \Z_8)/(\langle 3\rangle\times \langle 2\rangle)|$
6. $|(\Z_{15} \times \Z_{24})/\langle (5,4)\rangle|$
Solution
##### Exercise4
For each of the following, find a familiar group to which the given group is isomorphic. (Hint: Consider the group order, properties such as abelianness and cyclicity, group tables, orders of elements, etc.)
1. $\Z/14\Z$
2. $3\Z/12\Z$
3. $S_8/A_8$
4. $(4\Z \times 15\Z)/(\langle 2 \rangle \times \langle 3 \rangle )$
5. $D_4/\langle r^2 \rangle$
Solution
##### Exercise5
Let $H\unlhd G$ with index $k\text{,}$ and let $a\in G\text{.}$ Prove that $a^k\in H\text{.}$
Solution
# Exercises9.3Exercises
##### Exercise1
Let $F$ be the group of all functions from $[0,1]$ to $\R\text{,}$ under pointwise addition. Let
\begin{equation*} N=\{f\in F: f(1/4)=0\}. \end{equation*}
Prove that $F/N$ is a group that's isomorphic to $\R\text{.}$
Solution
##### Exercise2
Let $N=\{1,-1\}\subseteq \R^*\text{.}$ Prove that $\R^*/N$ is a group that's isomorphic to $\R^+\text{.}$
Solution
##### Exercise3
Let $n\in \Z^+$ and let $H=\{A\in GL(n,\R)\,:\, \det A =\pm 1\}\text{.}$ Identify a group familiar to us that is isomorphic to $GL(n,\R)/H\text{.}$
Solution
##### Exercise4
Let $G$ and $G'$ be groups with respective normal subgroups $N$ and $N'\text{.}$ Prove or disprove: If $G/N\simeq G'/N'$ then $G\simeq G'\text{.}$
Solution | 2021-04-12T01:25:59 | {
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http://www.kvb.hu/vf1wj/bayesian-statistics-in-r-book-2097ee | # bayesian statistics in r book
If the data are consistent with a hypothesis, my belief in that hypothesis is strengthened. If you peek at your data after every single observation, there is a 49% chance that you will make a Type I error. BayesFactor: Computation of Bayes Factors for Common Designs. Yes, you might try to defend $$p$$-values by saying that it’s the fault of the researcher for not using them properly. The data that you need to give to this function is the contingency table itself (i.e., the crosstab variable above), so you might be expecting to use a command like this: However, if you try this you’ll get an error message. Much easier to understand, and you can interpret this using the table above. Before reading any further, I urge you to take some time to think about it. And because it assumes the experiment is over, it only considers two possible decisions. The bolded section is just plain wrong. MCMC for a model with temporal pseudoreplication. Installing JAGS on your computer. I hate to bring this up, but some statisticians would object to me using the word “likelihood” here. For example, I would avoid writing this: A Bayesian test of association found a significant result (BF=15.92). So I’m not actually introducing any “new” rules here, I’m just using the same rule in a different way.↩, Obviously, this is a highly simplified story. The material presented here has been used by students of different levels and disciplines, including advanced undergraduates studying Mathematics and Statistics and students in graduate programs in Statistics, Biostatistics, Engineering, Economics, Marketing, Pharmacy, and Psychology. For instance, if we want to identify the best model we could use the same commands that we used in the last section. The easiest way to do it with this data set is to use the x argument to specify one variable and the y argument to specify the other. You design a study comparing two groups. You’ve got a significant result! Think Bayes: Bayesian Statistics in Python - Ebook written by Allen B. Downey. If I’d chosen a 5:1 Bayes factor instead, the results would look even better for the Bayesian approach.↩, http://www.quotationspage.com/quotes/Ambrosius_Macrobius/↩, Okay, I just know that some knowledgeable frequentists will read this and start complaining about this section. For example, if we wanted to get an estimate of the mean height of people, we could use our prior knowledge that people are generally between 5 and 6 feet tall … There’s nothing stopping you from including that information, and I’ve done so myself on occasions, but you don’t strictly need it. I listed it way back in Table 9.1, but I didn’t make a big deal out of it at the time and you probably ignored it. In the middle, we have the Bayes factor, which describes the amount of evidence provided by the data: $In this case, the alternative is that there is a relationship between species and choice: that is, they are not independent. There is a pdf version of this booklet available at:https://media.readthedocs.org/pdf/ I didn’t bother indicating whether this was “moderate” evidence or “strong” evidence, because the odds themselves tell you! Fortunately, no-one will notice. That’s not surprising, of course: that’s our prior. TensorFlow, on the other hand, is far more recent. However, for the sake of everyone’s sanity, throughout this chapter I’ve decided to rely on one R package to do the work. Specifically, let’s say our data look like this: The Bayesian test with hypergeometric sampling gives us this: The Bayes factor of 8:1 provides modest evidence that the labels were being assigned in a way that correlates gender with colour, but it’s not conclusive. The cake is a lie. A flexible extension of maximum likelihood. The example I used originally is the clin.trial data frame, which looks like this. 1974. If we do that, we end up with the following table: This table captures all the information about which of the four possibilities are likely. Posted by. Bayes’ rule cannot stop people from lying, nor can it stop them from rigging an experiment. \frac{P(h_1 | d)}{P(h_0 | d)} = \frac{0.75}{0.25} = 3 Actually, this equation is worth expanding on. Here are some possibilities: Which would you choose? Within the Bayesian framework, it is perfectly sensible and allowable to refer to “the probability that a hypothesis is true”. In his opinion, if we take $$p<.05$$ to mean there is “a real effect”, then “we shall not often be astray”. However, in this case I’m doing it because I want to use a model with more than one predictor as my example! Becasue of this, the anovaBF() reports the output in much the same way. Johnson, Valen E. 2013. Statistical Methods for Research Workers. 2015. Finally, I devoted some space to talking about why I think Bayesian methods are worth using (Section 17.3. Again, I find it useful to frame things the other way around, so I’d refer to this as evidence of about 3 to 1 in favour of an effect of therapy. This seems so obvious to a human, yet it is explicitly forbidden within the orthodox framework. First, we have to go back and save the Bayes factor information to a variable: Let’s say I want to see the best three models. Again, we obtain a $$p$$-value less than 0.05, so we reject the null hypothesis. That might change in the future if Bayesian methods become standard and some task force starts writing up style guides, but in the meantime I would suggest using some common sense. The odds of 0.98 to 1 imply that these two models are fairly evenly matched. Let’s take a look: This looks very similar to the output we obtained from the regressionBF() function, and with good reason. As before, we use formula to indicate what the full regression model looks like, and the data argument to specify the data frame. You collected some data, the results weren’t conclusive, so now what you want to do is collect more data until the the results are conclusive. In one sense, that’s true. By way of comparison, imagine that you had used the following strategy. Bayesian Data Analysis (3rd ed.). You are not allowed to use the data to decide when to terminate the experiment. This book is based on over a dozen years teaching a Bayesian Statistics course. Also it does incorporate some humour into the bundle like Bayesian Statistics… And as a consequence you’ve transformed the decision-making procedure into one that looks more like this: The “basic” theory of null hypothesis testing isn’t built to handle this sort of thing, not in the form I described back in Chapter 11. Most of the examples are simple, and similar to other online sources. It describes how a learner starts out with prior beliefs about the plausibility of different hypotheses, and tells you how those beliefs should be revised in the face of data. Welcome to Applied Statistics with R! All we do is change the subscript: \[ The result is significant with a sample size of $$N=50$$, so wouldn’t it be wasteful and inefficient to keep collecting data? What this table is telling you is that, after being told that I’m carrying an umbrella, you believe that there’s a 51.4% chance that today will be a rainy day, and a 48.6% chance that it won’t. Edinburgh, UK: Oliver; Boyd. I mean, it sounds like a perfectly reasonable strategy doesn’t it? You can type ?ttestBF to get more details.↩, I don’t even disagree with them: it’s not at all obvious why a Bayesian ANOVA should reproduce (say) the same set of model comparisons that the Type II testing strategy uses. 1.1 About This Book This book was originally (and currently) designed for use with STAT 420, Methods of Applied Statistics, at the University of Illinois at Urbana-Champaign. Bayes Bayes Bayes Bayes Bayes. Jeffreys, Harold. All you have to do is be honest about what you believed before you ran the study, and then report what you learned from doing it. Afterwards, I provide a brief overview of how you can do Bayesian versions of chi-square tests (Section 17.6), $$t$$-tests (Section 17.7), regression (Section 17.8) and ANOVA (Section 17.9). This “conditional probability” is written $$P(d|h)$$, which you can read as “the probability of $$d$$ given $$h$$”. One variant that I find quite useful is this: By “dividing” the models output by the best model (i.e., max(models)), what R is doing is using the best model (which in this case is drugs + therapy) as the denominator, which gives you a pretty good sense of how close the competitors are. It looks like you’re stuck with option 4. But let’s keep things simple, shall we?↩, You might notice that this equation is actually a restatement of the same basic rule I listed at the start of the last section. BIC is one of the Bayesian criteria used for Bayesian model selection, and tends to be one of the most popular criteria. That’s it! So what regressionBF() does is treat the intercept only model as the null hypothesis, and print out the Bayes factors for all other models when compared against that null. To a frequentist, such statements are a nonsense because “the theory is true” is not a repeatable event. What two numbers should we put in the empty cells? I’m not alone in doing this. Please check your email for instructions on resetting your password. Ultimately, isn’t that what you want your statistical tests to tell you? For instance, the model that contains the interaction term is almost as good as the model without the interaction, since the Bayes factor is 0.98. The second half of the chapter was a lot more practical, and focused on tools provided by the BayesFactor package. The material presented here has been used by students of different levels and disciplines, including advanced undergraduates studying Mathematics and Statistics and students in graduate programs in Statistics, Biostatistics, Engineering, Economics, Marketing, Pharmacy, and Psychology. The odds in favour of the null here are only 0.35 to 1. Except when the sampling procedure is fixed by an external constraint, I’m guessing the answer is “most people have done it”. Others will claim that the evidence is ambiguous, and that you should collect more data until you get a clear significant result. In Sections 17.1 through 17.3 I talk about what Bayesian statistics are all about, covering the basic mathematical rules for how it works as well as an explanation for why I think the Bayesian approach is so useful. Use the link below to share a full-text version of this article with your friends and colleagues. All the complexity of real life Bayesian hypothesis testing comes down to how you calculate the likelihood $$P(d|h)$$ when the hypothesis $$h$$ is a complex and vague thing. This book was written as a companion for the Course Bayesian Statistics from the Statistics with R specialization available on Coursera. The early chapters present the basic tenets of Bayesian thinking by use of familiar one and two-parameter inferential problems. User account menu. So you might write out a little table like this: It’s important to remember that each cell in this table describes your beliefs about what data $$d$$ will be observed, given the truth of a particular hypothesis $$h$$. In real life, how many people do you think have “peeked” at their data before the experiment was finished and adapted their subsequent behaviour after seeing what the data looked like? First, the concept of “statistical significance” is pretty closely tied with $$p$$-values, so it reads slightly strangely. What’s the Bayesian analog of this? The best model is drug + therapy, so all the other models are being compared to that. \uparrow && \uparrow && \uparrow \\[6pt] As you might expect, the answers would be diffrent again if it were the columns of the contingency table that the experimental design fixed. After introducing the theory, the book covers the analysis of contingency tables, t-tests, ANOVAs and regression. The $$\pm0\%$$ part is not very interesting: essentially, all it’s telling you is that R has calculated an exact Bayes factor, so the uncertainty about the Bayes factor is 0%.270 In any case, the data are telling us that we have moderate evidence for the alternative hypothesis. There’s a reason why, back in Section 11.5, I repeatedly warned you not to interpret the $$p$$-value as the probability of that the null hypothesis is true. In fact, almost every textbook given to undergraduate psychology students presents the opinions of the frequentist statistician as the theory of inferential statistics, the one true way to do things. Second, the “BF=15.92” part will only make sense to people who already understand Bayesian methods, and not everyone does. The BDA_R_demos repository contains some R demos and additional notes for the book Bayesian Data Analysis, 3rd ed by Gelman, Carlin, Stern, Dunson, Vehtari, and Rubin (BDA3). Bayesian statistics for realistically complicated models. Compared to other intro to statistics books like Bayesian Statistics: The Fun Way, it is more practical because of this constant programming flow that accompanies the theory. Gudmund R. Iversen. \frac{P(h_1 | d)}{P(h_0 | d)} = \frac{P(d|h_1)}{P(d|h_0)} \times \frac{P(h_1)}{P(h_0)} If you do not receive an email within 10 minutes, your email address may not be registered, Otherwise continue testing. Really bloody annoying, right? That’s not what 95% confidence means to a frequentist statistician. http://en.wikipedia.org/wiki/Climate_of_Adelaide↩, It’s a leap of faith, I know, but let’s run with it okay?↩, Um. Using the data from Johnson (2013), we see that if you reject the null at $$p<.05$$, you’ll be correct about 80% of the time. Morey and Rouder (2015) built their Bayesian tests of association using the paper by Gunel and Dickey (1974), the specific test we used assumes that the experiment relied on a joint multinomial sampling plan, and indeed the Bayes factor of 15.92 is moderately strong evidence. – Portal263. You’re very diligent, so you run a power analysis to work out what your sample size should be, and you run the study. Statistical Rethinking: A Bayesian Course with Examples in R and Stan builds your knowledge of and confidence in making inferences from data. … an error message. Guess what? The sampling plan actually does matter. The alternative model adds the interaction. 48: 19313–7. Specifically, I’m going to use the BayesFactor package written by Jeff Rouder and Rich Morey, which as of this writing is in version 0.9.10. (Version 0.6.1), http://CRAN.R-project.org/package=BayesFactor, http://en.wikipedia.org/wiki/Climate_of_Adelaide, http://www.imdb.com/title/tt0093779/quotes, http://about.abc.net.au/reports-publications/appreciation-survey-summary-report-2013/, http://knowyourmeme.com/memes/the-cake-is-a-lie, http://www.quotationspage.com/quotes/Ambrosius_Macrobius/, You conclude that there is no effect, and try to publish it as a null result, You guess that there might be an effect, and try to publish it as a “borderline significant” result. A theory of statistical inference that is so completely naive about humans that it doesn’t even consider the possibility that the researcher might look at their own data isn’t a theory worth having. However, if you’ve got a lot of possible models in the output, it’s handy to know that you can use the head() function to pick out the best few models. We could probably reject the null with some confidence! Look, I’m not dumb. That seems silly. However, I have to stop somewhere, and so there’s only one other topic I want to cover: Bayesian ANOVA. If you give up and try a new project else every time you find yourself faced with ambiguity, your work will never be published. However, one big practical advantage of the Bayesian approach relative to the orthodox approach is that it also allows you to quantify evidence for the null. The example I gave in the previous section is a pretty extreme situation. He would have marveled at the presentations in the book of many new and strong statistical and computer analyses. Sometimes it’s sensible to do this, even when it’s not the one with the highest Bayes factor. Or, to put it another way, the null hypothesis is that these two variables are independent. Consider the quote above by Sir Ronald Fisher, one of the founders of what has become the orthodox approach to statistics. You might guess that I’m not a complete idiot,256 and I try to carry umbrellas only on rainy days. In fact, you might have decided to take a quick look on Wikipedia255 and discovered that Adelaide gets an average of 4.4 days of rain across the 31 days of January. Now take a look at the column sums, and notice that they tell us something that we haven’t explicitly stated yet. To work out that there was a 0.514 probability of “rain”, all I did was take the 0.045 probability of “rain and umbrella” and divide it by the 0.0875 chance of “umbrella”. The help documentation to the contingencyTableBF() gives this explanation: “the argument priorConcentration indexes the expected deviation from the null hypothesis under the alternative, and corresponds to Gunel and Dickey’s (1974) $$a$$ parameter.” As I write this I’m about halfway through the Gunel and Dickey paper, and I agree that setting $$a=1$$ is a pretty sensible default choice, since it corresponds to an assumption that you have very little a priori knowledge about the contingency table.↩, In some of the later examples, you’ll see that this number is not always 0%. You already know that you’re analysing a contingency table, and you already know that you specified a joint multinomial sampling plan. Finally, if we turn to hypergeometric sampling in which everything is fixed, we get….$. The 15.9 part is the Bayes factor, and it’s telling you that the odds for the alternative hypothesis against the null are about 16:1. Unfortunately, the theory of null hypothesis testing as I described it in Chapter 11 forbids you from doing this.264 The reason is that the theory assumes that the experiment is finished and all the data are in. Similarly, we can work out how much belief to place in the alternative hypothesis using essentially the same equation. I wrote it that way deliberately, in order to help make things a little clearer for people who are new to statistics. In the meantime, let’s imagine we have data from the “toy labelling” experiment I described earlier in this section. In any case, note that all the numbers listed above make sense if the Bayes factor is greater than 1 (i.e., the evidence favours the alternative hypothesis). In essence, my point is this: Good laws have their origins in bad morals. I don’t know about you, but in my opinion an evidentiary standard that ensures you’ll be wrong on 20% of your decisions isn’t good enough. Bayesian methods provide a powerful alternative to the frequentist methods that are ingrained in the standard statistics curriculum. The fact remains that, quite contrary to Fisher’s claim, if you reject at $$p<.05$$ you shall quite often go astray. In essence, the $$p<.05$$ convention is assumed to represent a fairly stringent evidentiary standard. Well, how true is that? They’ll argue it’s borderline significant. To me, it makes a lot more sense to turn the equation “upside down”, and report the amount op evidence in favour of the null. Burlington, MA: Academic Press. If that’s right, then Fisher’s claim is a bit of a stretch. In my experience that’s a pretty typical outcome. Focusing on the most standard statistical models and backed up by real datasets and an all-inclusive R (CRAN) package called bayess, the book provides an operational methodology for conducting Bayesian inference, rather than focusing on its theoretical and philosophical justifications. If I were to follow the same progression that I used when developing the orthodox tests you’d expect to see ANOVA next, but I think it’s a little clearer if we start with regression. To remind you of what that data set looks like, here’s the first six observations: Back in Chapter 15 I proposed a theory in which my grumpiness (dan.grump) on any given day is related to the amount of sleep I got the night before (dan.sleep), and possibly to the amount of sleep our baby got (baby.sleep), though probably not to the day on which we took the measurement. In order to cut costs, you start collecting data, but every time a new observation arrives you run a $$t$$-test on your data. Given the difficulties in publishing an “ambiguous” result like $$p=.072$$, option number 3 might seem tempting: give up and do something else. 62 to rent \$57.21 to buy. P(\mbox{rainy}, \mbox{umbrella}) & = & P(\mbox{umbrella} | \mbox{rainy}) \times P(\mbox{rainy}) \\ However, notice that there’s no analog of the var.equal argument. In other words, what we have written down is a proper probability distribution defined over all possible combinations of data and hypothesis. One way to approach this question is to try to convert $$p$$-values to Bayes factors, and see how the two compare. P(d,h) = P(d|h) P(h) This distinction matters in some contexts, but it’s not important for our purposes.↩, If we were being a bit more sophisticated, we could extend the example to accommodate the possibility that I’m lying about the umbrella. There’s a reason why almost every textbook on statstics is forced to repeat that warning. In this example, I’m going to pretend that you decided that dan.grump ~ dan.sleep + baby.sleep is the model you think is best. There are three different terms here that you should know. So the command is: So that’s pretty straightforward: it’s exactly what we’ve been doing throughout the book. For the chapek9 data, I implied that we designed the study such that the total sample size $$N$$ was fixed, so we should set sampleType = "jointMulti". What about the design in which the row columns (or column totals) are fixed? “Bayes Factors.” Journal of the American Statistical Association 90: 773–95. It’s because people desperately want that to be the correct interpretation. And the reason why “data peeking” is such a concern is that it’s so tempting, even for honest researchers. 2 years ago. As we discussed earlier, the prior tells us that the probability of a rainy day is 15%, and the likelihood tells us that the probability of me remembering my umbrella on a rainy day is 30%. In other words, before being told anything about what actually happened, you think that there is a 4.5% probability that today will be a rainy day and that I will remember an umbrella. Suppose we want to test the main effect of drug. \]. This chapter comes in two parts. Sounds like an absurd claim, right? In the meantime, I thought I should show you the trick for how I do this in practice. As you can tell, the BayesFactor package is pretty flexible, and it can do Bayesian versions of pretty much everything in this book. If it were up to me, I’d have called the “positive evidence” category “weak evidence”. If you run an experiment and you compute a Bayes factor of 4, it means that the evidence provided by your data corresponds to betting odds of 4:1 in favour of the alternative. If you are a frequentist, the answer is “very wrong”. Download for offline reading, highlight, bookmark or take notes while you read Think Bayes: Bayesian Statistics in Python. The (Intercept) term isn’t usually interesting, though it is highly significant. In this kind of data analysis situation, we have a cross-tabulation of one variable against another one, and the goal is to find out if there is some association between these variables. To me, anything in the range 3:1 to 20:1 is “weak” or “modest” evidence at best. You’ve found the regression model with the highest Bayes factor (i.e., dan.grump ~ dan.sleep), and you know that the evidence for that model over the next best alternative (i.e., dan.grump ~ dan.sleep + day) is about 16:1. If you’re a cognitive psychologist, you might want to check out Michael Lee and E.J. Without knowing anything else, you might conclude that the probability of January rain in Adelaide is about 15%, and the probability of a dry day is 85%. Once you’ve made the jump, you no longer have to wrap your head around counterinuitive definitions of $$p$$-values. \mbox{BF} = \frac{P(d|h_1)}{P(d|h_0)} = \frac{0.1}{0.2} = 0.5 It uses a pretty standard formula and data structure, so the command should look really familiar. I indicated exactly what the effect is (i.e., “a relationship between species and choice”) and how strong the evidence was. That’s because the citation itself includes that information (go check my reference list if you don’t believe me). & = & 0.045 But the fact remains that if you want your $$p$$-values to be honest, then you either have to switch to a completely different way of doing hypothesis tests, or you must enforce a strict rule: no peeking. As usual we have a formula argument in which we specify the outcome variable on the left hand side and the grouping variable on the right. This view is hardly unusual: in my experience, most practitioners express views very similar to Fisher’s. In my opinion, there’s a fairly big problem built into the way most (but not all) orthodox hypothesis tests are constructed. That’s not an unreasonable view to take, but in my view the problem is a little more severe than that. If [$$p$$] is below .02 it is strongly indicated that the [null] hypothesis fails to account for the whole of the facts. I find this hard to understand. To really get the full picture, though, it helps to add the row totals and column totals. Again, let’s not worry about the maths, and instead think about our intuitions. Bayesian statistical methods are based on the idea that one can assert prior probability distributions for parameters of interest. You use your “preferred” model as the formula argument, and then the output will show you the Bayes factors that result when you try to drop predictors from this model: Okay, so now you can see the results a bit more clearly. Unfortunately – in my opinion at least – the current practice in psychology is often misguided, and the reliance on frequentist methods is partly to blame. Now, sure, you know you said that you’d keep running the study out to a sample size of $$N=80$$, but it seems sort of pointless now, right? In this data set, we supposedly sampled 180 beings and measured two things. The BayesFactor package contains a function called anovaBF() that does this for you. So what we expect to see in our final table is some numbers that preserve the fact that “rain and umbrella” is slightly more plausible than “dry and umbrella”, while still ensuring that numbers in the table add up. Instead, we tend to talk in terms of the posterior odds ratio. So, what’s the chance that you’ll make it to the end of the experiment and (correctly) conclude that there is no effect? – Ambrosius Macrobius267, Good rules for statistical testing have to acknowledge human frailty. It’s such an appealing idea that even trained statisticians fall prey to the mistake of trying to interpret a $$p$$-value this way. Okay, let’s say you’ve settled on a specific regression model. Even the 3:1 standard, which most Bayesians would consider unacceptably lax, is much safer than the $$p<.05$$ rule. Analysts who need to incorporate their work into real-world decisions, as opposed to formal statistical inference for publication, will be especially interested. It has interfaces for many popular data analysis languages including Python, MATLAB, Julia, and Stata.The R interface for Stan is called rstan and rstanarm is a front-end to rstan that allows regression models to be fit using a standard R regression model interface. So the command I would use is: Again, the Bayes factor is different, with the evidence for the alternative dropping to a mere 9:1. Let’s say that limit kicks in at $$N=1000$$ observations. The command that we need is. When does Dan carry an umbrella? If the $$t$$-tests says $$p<.05$$ then you stop the experiment and report a significant result. The joint probability of the hypothesis and the data is written $$P(d,h)$$, and you can calculate it by multiplying the prior $$P(h)$$ by the likelihood $$P(d|h)$$. You are strictly required to follow these rules, otherwise the $$p$$-values you calculate will be nonsense. “Revised Standards for Statistical Evidence.” Proceedings of the National Academy of Sciences, no. The easiest way is to use the regressionBF() function instead of lm(). Or if we look at line 1, we can see that the odds are about $$1.6 \times 10^{34}$$ that a model containing the dan.sleep variable (but no others) is better than the intercept only model. The main effect of therapy is weaker, and the evidence here is only 2.8:1. \uparrow && \uparrow && \uparrow \$6pt] 7.1.1 Definition of BIC. How do we do the same thing using Bayesian methods?$, It’s all so simple that I feel like an idiot even bothering to write these equations down, since all I’m doing is copying Bayes rule from the previous section.260. It’s a reasonable, sensible and rational thing to do. And this formula, folks, is known as Bayes’ rule. You have two possible hypotheses, $$h$$: either it rains today or it does not. To my mind, this write up is unclear. Bayesian statistics are covered at the end of the book. How should you solve this problem? But until that day arrives, I stand by my claim that default Bayes factor methods are much more robust in the face of data analysis practices as they exist in the real world. So here’s our command: At this point, I hope you can read this output without any difficulty. The full text of this article hosted at iucr.org is unavailable due to technical difficulties. When you report $$p<.05$$ in your paper, what you’re really saying is $$p<.08$$. To an ideological frequentist, this sentence should be meaningless. That being said, I can talk a little about why I prefer the Bayesian approach. How can that last part be true? (a=1) : 8.294321 @plusorminus0%, #Bayes factor type: BFcontingencyTable, hypergeometric, "mood.gain ~ drug + therapy + drug:therapy", Learning statistics with R: A tutorial for psychology students and other beginners. You can choose to report a Bayes factor less than 1, but to be honest I find it confusing. As it happens, I ran the simulations for this scenario too, and the results are shown as the dashed line in Figure 17.1. Stan (also discussed in Richard’s book) is a statistical programming language famous for its MCMC framework. Learn more. All the $$p$$-values you calculated in the past and all the $$p$$-values you will calculate in the future. Sounds nice, doesn’t it? Well, keep in mind that if you do, your Type I error rate at $$p<.05$$ just ballooned out to 8%. The trick to understanding this output is to recognise that if we’re interested in working out which of the 3 predictor variables are related to dan.grump, there are actually 8 possible regression models that could be considered. First, let’s remind ourselves of what the data were. Lee, Michael D, and Eric-Jan Wagenmakers. It may certainly be used elsewhere, but any references to “this course” in this book specifically refer to STAT 420. Download for offline reading, highlight, bookmark or take notes while you read Doing Bayesian Data Analysis: A Tutorial Introduction with R. However, sequential analysis methods are constructed in a very different fashion to the “standard” version of null hypothesis testing. When writing up the results, my experience has been that there aren’t quite so many “rules” for how you “should” report Bayesian hypothesis tests. It was and is current practice among psychologists to use frequentist methods. Finally, the evidence against an interaction is very weak, at 1.01:1. Using the ttestBF() function, we can obtain a Bayesian analog of Student’s independent samples $$t$$-test using the following command: Notice that format of this command is pretty standard. We are going to discuss the Bayesian model selections using the Bayesian information criterion, or BIC. Reading the results off this table is sort of counterintuitive, because you have to read off the answers from the “wrong” part of the table. One or two reviewers might even be on your side, but you’ll be fighting an uphill battle to get it through. A First Course in Bayesian Statistical Methods. At this point, all the elements are in place. What happens? None of us are beyond temptation. I did so in order to be charitable to the $$p$$-value. In any case, by convention we like to pretend that we give equal consideration to both the null hypothesis and the alternative, in which case the prior odds equals 1, and the posterior odds becomes the same as the Bayes factor. You don’t have conclusive results, so you decide to collect some more data and re-run the analysis. When the study starts out you follow the rules, refusing to look at the data or run any tests. Well, like every other bloody thing in statistics, there’s a lot of different ways you could do it. The resulting Bayes factor of 15.92 to 1 in favour of the alternative hypothesis indicates that there is moderately strong evidence for the non-independence of species and choice. A guy carrying an umbrella on a summer day in a hot dry city is pretty unusual, and so you really weren’t expecting that. I’ll talk a little about Bayesian versions of the independent samples $$t$$-tests and the paired samples $$t$$-test in this section. In order to estimate the regression model we used the lm() function, like so: The hypothesis tests for each of the terms in the regression model were extracted using the summary() function as shown below: When interpreting the results, each row in this table corresponds to one of the possible predictors. Some reviewers will think that $$p=.072$$ is not really a null result. See also Bayesian Data Analysis course material . Our goal in developing the course was to provide an introduction to Bayesian inference in decision making without requiring calculus, with the book providing more details and background on Bayesian Inference. When a frequentist says the same thing, they’re referring to the same table, but to them “a likelihood function” almost always refers to one of the columns. For the purposes of this section, I’ll assume you want Type II tests, because those are the ones I think are most sensible in general. See? Archived. To say the same thing using fancy statistical jargon, what I’ve done here is divide the joint probability of the hypothesis and the data $$P(d,h)$$ by the marginal probability of the data $$P(d)$$, and this is what gives us the posterior probability of the hypothesis given that we know the data have been observed. Here we will take the Bayesian propectives. Bayesian methods usually require more evidence before rejecting the null. Back in Chapter@refch:ttest I suggested you could analyse this kind of data using the independentSamplesTTest() function in the lsr package. The BayesFactor package contains a function called ttestBF() that is flexible enough to run several different versions of the $$t$$-test. In real life, people don’t run hypothesis tests every time a new observation arrives. The data argument is used to specify the data frame containing the variables. So the relevant comparison is between lines 2 and 1 in the table. A wise man, therefore, proportions his belief to the evidence. \end{array} Similarly, I didn’t bother to indicate that I ran the “joint multinomial” sampling plan, because I’m assuming that the method section of my write up would make clear how the experiment was designed. part refers to the alternative hypothesis. As with most R commands, the output initially looks suspiciously similar to utter gibberish. The book would also be valuable to the statistical practitioner who wishes to learn more about the R language and Bayesian methodology. Given all of the above, what is the take home message? So let’s begin. \]. The second type of statistical inference problem discussed in this book is the comparison between two means, discussed in some detail in the chapter on $$t$$-tests (Chapter 13. That’s the answer to our problem! \]. For example, Johnson (2013) presents a pretty compelling case that (for $$t$$-tests at least) the $$p<.05$$ threshold corresponds roughly to a Bayes factor of somewhere between 3:1 and 5:1 in favour of the alternative. So the only part that really matters is this line here: Ignore the r=0.707 part: it refers to a technical detail that we won’t worry about in this chapter.273 Instead, you should focus on the part that reads 1.754927. Other reviewers will agree it’s a null result, but will claim that even though some null results are publishable, yours isn’t. Press question mark to learn the rest of the keyboard shortcuts. My bayesian-guru professor from Carnegie Mellon agrees with me on this. \]. For example, if you want to run a Student’s $$t$$-test, you’d use a command like this: Like most of the functions that I wrote for this book, the independentSamplesTTest() is very wordy. CRC (2013) The Gelman book isn't constrained to R but also uses Stan, a probabilistic programming language similar to BUGS or JAGS. \end{array} All of them. Ultimately it depends on what you think is right. What Bayes factors should you report? And to be perfectly honest, I can’t answer this question for you. This wouldn’t have been a problem, except for the fact that the way that Bayesians use the word turns out to be quite different to the way frequentists do. In contrast, notice that the Bayesian test doesn’t even reach 2:1 odds in favour of an effect, and would be considered very weak evidence at best. \], $\mbox{Posterior odds} && \mbox{Bayes factor} && \mbox{Prior odds} \begin{array} This will get you confortable with the main theoretical concepts of statistical reasoning while also teaching you to code them using examples in the R programming language. Up to this point I’ve been talking about what Bayesian inference is and why you might consider using it. The problem is that the word “likelihood” has a very specific meaning in frequentist statistics, and it’s not quite the same as what it means in Bayesian statistics. As with the other examples, I think it’s useful to start with a reminder of how I discussed ANOVA earlier in the book. 17.1 Probabilistic reasoning by rational agents. To an actual human being, this would seem to be the whole point of doing statistics: to determine what is true and what isn’t. “Bayes Factors for Independence in Contingency Tables.” Biometrika, 545–57. However, I haven’t had time to do this yet, nor have I made up my mind about whether it’s really a good idea to do this. Imagine you’re a really super-enthusiastic researcher on a tight budget who didn’t pay any attention to my warnings above. Applied Bayesian Statistics: With R and OpenBUGS Examples (Springer Texts in Statistics (98)) Part of: Springer Texts in Statistics (72 Books) 2.4 out of 5 stars 4. Using the equations given above, Bayes factor here would be: \[ My point is the same one I made at the very beginning of the book in Section 1.1: the reason why we run statistical tests is to protect us from ourselves. Before moving on, it’s worth highlighting the difference between the orthodox test results and the Bayesian one. Honestly, there’s nothing wrong with it. If this is really what you believe about Adelaide rainfall (and now that I’ve told it to you, I’m betting that this really is what you believe) then what I have written here is your prior distribution, written $$P(h)$$: To solve the reasoning problem, you need a theory about my behaviour. Suppose, for instance, the posterior probability of the null hypothesis is 25%, and the posterior probability of the alternative is 75%. More to the point, the other two Bayes factors are both less than 1, indicating that they’re all worse than that model. In the rainy day problem, you are told that I really am carrying an umbrella. Bayesian Inference is a way of combining information from data with things we think we already know. To write this as an equation:259 \[ The Theory of Probability. Seems sensible, but unfortunately for you, if you do this all of your $$p$$-values are now incorrect. In other words, what we calculate is this: \[ The rule in question is the one that talks about the probability that two things are true. And in fact you’re right: the city of Adelaide where I live has a Mediterranean climate, very similar to southern California, southern Europe or northern Africa. \end{array} Specifically, I talked about using the contingencyTableBF() function to do Bayesian analogs of chi-square tests (Section 17.6, the ttestBF() function to do Bayesian $$t$$-tests, (Section 17.7), the regressionBF() function to do Bayesian regressions, and finally the anovaBF() function for Bayesian ANOVA. As far as I can tell, Bayesians didn’t originally have any agreed upon name for the likelihood, and so it became common practice for people to use the frequentist terminology. Not just the $$p$$-values that you calculated for this study. Okay, at this point you might be thinking that the real problem is not with orthodox statistics, just the $$p<.05$$ standard. How do we run an equivalent test as a Bayesian? On the right hand side, we have the prior odds, which indicates what you thought before seeing the data. Writing BUGS models. You might be thinking that this is all pretty laborious, and I’ll concede that’s true. Stan, rstan, and rstanarm. Let’s start out with one of the rules of probability theory. The answer is shown as the solid black line in Figure 17.1, and it’s astoundingly bad. So you might have one sentence like this: All analyses were conducted using the BayesFactor package in R , and unless otherwise stated default parameter values were used. This is because the contingencyTestBF() function needs one other piece of information from you: it needs to know what sampling plan you used to run your experiment. You’re breaking the rules: you’re running tests repeatedly, “peeking” at your data to see if you’ve gotten a significant result, and all bets are off. Read this book using Google Play Books app on your PC, android, iOS devices. For example, suppose I deliberately sampled 87 humans and 93 robots, then I would need to indicate that the fixedMargin of the contingency table is the "rows". Prior to running the experiment we have some beliefs $$P(h)$$ about which hypotheses are true. Only 7 left in stock - order soon. We run an experiment and obtain data $$d$$. Frequentist dogma notwithstanding, a lifetime of experience of teaching undergraduates and of doing data analysis on a daily basis suggests to me that most actual humans thing that “the probability that the hypothesis is true” is not only meaningful, it’s the thing we care most about. Andrew Gelman et. You should take this course if you are familiar with R and with Bayesian statistics at the introductory level, and work with or interpret statistical models and need to incorporate Bayesian methods. This is the Bayes factor: the evidence provided by these data are about 1.8:1 in favour of the alternative. All of them. I spelled out “Bayes factor” rather than truncating it to “BF” because not everyone knows the abbreviation. Using this notation, the table looks like this: The table we laid out in the last section is a very powerful tool for solving the rainy day problem, because it considers all four logical possibilities and states exactly how confident you are in each of them before being given any data. Also, you know for a fact that I am carrying an umbrella, so the column sum on the left must be 1 to correctly describe the fact that $$P(\mbox{umbrella})=1$$. According to the orthodox test, we obtained a significant result, though only barely. In most situations the intercept only model is one that you don’t really care about at all. You can’t compute a $$p$$-value when you don’t know the decision making procedure that the researcher used. At the bottom we have some techical rubbish, and at the top we have some information about the Bayes factors. I don’t know which of these hypotheses is true, but do I have some beliefs … In other words, what we want is the Bayes factor corresponding to this comparison: As it happens, we can read the answer to this straight off the table because it corresponds to a comparison between the model in line 2 of the table and the model in line 3: the Bayes factor in this case represents evidence for the null of 0.001 to 1. Mathematically, we say that: \[ Consider the following reasoning problem: I’m carrying an umbrella. And so the reported $$p$$-value remains a lie. (2003), Carlin and Louis (2009), Press (2003), Gill (2008), or Lee (2004). You’ll get published, and you’ll have lied. Finally, in order to test an interaction effect, the null model here is one that contains both main effects but no interaction. and you may need to create a new Wiley Online Library account. In the line above, the text Null, mu1-mu2 = 0 is just telling you that the null hypothesis is that there are no differences between means. Rich Morey and colleagues had the idea first. 3rd ed. Potentially the most information-efficient method to fit a statistical model. I’m not going to talk about those complexities in this book, but I do want to highlight that although this simple story is true as far as it goes, real life is messier than I’m able to cover in an introductory stats textbook.↩, http://www.imdb.com/title/tt0093779/quotes. It is both concise and timely, and provides a good collection of overviews and reviews of important tools used in Bayesian statistical methods." For the analysis of contingency tables, the BayesFactor package contains a function called contingencyTableBF(). I’ve rounded 15.92 to 16, because there’s not really any important difference between 15.92:1 and 16:1. For example, suppose that the likelihood of the data under the null hypothesis $$P(d|h_0)$$ is equal to 0.2, and the corresponding likelihood $$P(d|h_0)$$ under the alternative hypothesis is 0.1. In our reasonings concerning matter of fact, there are all imaginable degrees of assurance, from the highest certainty to the lowest species of moral evidence. Do you want to be an orthodox statistician, relying on sampling distributions and $$p$$-values to guide your decisions? You keep using that word. As I discussed back in Section 16.10, Type II tests for a two-way ANOVA are reasonably straightforward, but if you have forgotten that section it wouldn’t be a bad idea to read it again before continuing. Read this book using Google Play Books app on your PC, android, iOS devices. We’ve talked about the idea of “probability as a degree of belief”, and what it implies about how a rational agent should reason about the world. To me, one of the biggest advantages to the Bayesian approach is that it answers the right questions. Even if you happen to arrive at the same decision as the hypothesis test, you aren’t following the decision process it implies, and it’s this failure to follow the process that is causing the problem.265 Your $$p$$-values are a lie. This is something of a surprising event: according to our table, the probability of me carrying an umbrella is only 8.75%. At some stage I might consider adding a function to the lsr package that would automate this process and construct something like a “Bayesian Type II ANOVA table” from the output of the anovaBF() function. The ideas I’ve presented to you in this book describe inferential statistics from the frequentist perspective. However, there have been some attempts to quantify the standards of evidence that would be considered meaningful in a scientific context. The Bayesian approach to statistics considers parameters as random variables that are characterised by a prior distribution which is combined with the traditional likelihood to obtain the posterior distribution of the parameter of interest on which the statistical inference is based. If you’re the kind of person who would choose to “collect more data” in real life, it implies that you are not making decisions in accordance with the rules of null hypothesis testing. It’s now time to consider what happens to our beliefs when we are actually given the data. From the perspective of these two possibilities, very little has changed. What’s all this about?$ The Bayes factor (sometimes abbreviated as BF) has a special place in the Bayesian hypothesis testing, because it serves a similar role to the $$p$$-value in orthodox hypothesis testing: it quantifies the strength of evidence provided by the data, and as such it is the Bayes factor that people tend to report when running a Bayesian hypothesis test. Again, you need to specify the sampleType argument, but this time you need to specify whether you fixed the rows or the columns. On the left hand side, we have the posterior odds, which tells you what you believe about the relative plausibilty of the null hypothesis and the alternative hypothesis after seeing the data. programs in statistics for which this book would be appropriate. In this case, it’s easy enough to see that the best model is actually the one that contains dan.sleep only (line 1), because it has the largest Bayes factor. MCMC for a model with binomial errors Think of it like betting. So I should probably tell you what your options are! To see what I mean, here’s the original output: The best model corresponds to row 1 in this table, and the second best model corresponds to row 4. \mbox{Posterior odds} && \mbox{Bayes factor} && \mbox{Prior odds} Even in the classical version of ANOVA there are several different “things” that ANOVA might correspond to. The cake is a lie. So, what might you believe about whether it will rain today? The $$r$$ value here relates to how big the effect is expected to be according to the alternative. Bayesian statistics for realistically complicated models, Packages in R for carrying out Bayesian analysis, MCMC for a model with temporal pseudoreplication. Nevertheless, many people would happily accept $$p=.043$$ as reasonably strong evidence for an effect. On the other hand, let’s suppose you are a Bayesian. Now consider this … the scientific literature is filled with $$t$$-tests, ANOVAs, regressions and chi-square tests. However, that’s a pretty technical paper. P(h_0 | d) = \frac{P(d|h_0) P(h_0)}{P(d)} I don’t know which of these hypotheses is true, but do I have some beliefs about which hypotheses are plausible and which are not. This is the new, fully-revised edition to the book Bayesian Core: A Practical Approach to Computational Bayesian Statistics. In the rainy day problem, the data corresponds to the observation that I do or do not have an umbrella. Back in Section 13.5 I discussed the chico data frame in which students grades were measured on two tests, and we were interested in finding out whether grades went up from test 1 to test 2. In any case, if you know what you’re looking for, you can look at this table and then report the results of the Bayesian analysis in a way that is pretty closely analogous to how you’d report a regular Type II ANOVA. \]. It’s your call, and your call alone. What should you do? 2. Well, consider the following scenario. This book is based on over a dozen years teaching a Bayesian Statistics course. Nevertheless, the problem tells you that it is true. Or, more helpfully, the odds are about 1000 to 1 against the null. That’s, um, quite a bit bigger than the 5% that it’s supposed to be. Any time that you aren’t exactly sure about what the truth is, you should use the language of probability theory to say things like “there is an 80% chance that Theory A is true, but a 20% chance that Theory B is true instead”. I absolutely know that if you adopt a sequential analysis perspective you can avoid these errors within the orthodox framework. So the probability that both of these things are true is calculated by multiplying the two: \[ 2014. Remember what I said in Section 16.10 about ANOVA being complicated. We shall not often be astray if we draw a conventional line at .05 and consider that [smaller values of $$p$$] indicate a real discrepancy. Unlike frequentist statistics Bayesian statistics does allow to talk about the probability that the null hypothesis is true. In real life, the things we actually know how to write down are the priors and the likelihood, so let’s substitute those back into the equation. Orthodox methods cannot tell you that “there is a 95% chance that a real change has occurred”, because this is not the kind of event to which frequentist probabilities may be assigned. That way, anyone reading the paper can multiply the Bayes factor by their own personal prior odds, and they can work out for themselves what the posterior odds would be. Just as we saw with the contingencyTableBF() function, the output is pretty dense. It is a well-written book on elementary Bayesian inference, and the material is easily accessible. Although the bolded passage is the wrong definition of a $$p$$-value, it’s pretty much exactly what a Bayesian means when they say that the posterior probability of the alternative hypothesis is greater than 95%. Book on Bayesian statistics for a "statistican" Close. See Rouder et al. | 2022-06-27T13:59:24 | {
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http://mathhelpforum.com/calculus/136086-flaw-integration-print.html | # Flaw in Integration?
• March 28th 2010, 09:24 AM
Simon777
Flaw in Integration?
1/2 times the integral of 1/x is 1/2 lnx
however when i integrate the same problem as 1/2x and set u to 2x, I get
1/2 ln2x
Why are the answers not the same?
• March 28th 2010, 09:28 AM
skeeter
Quote:
Originally Posted by Simon777
1/2 times the integral of 1/x is 1/2 lnx
however when i integrate the same problem as 1/2x and set u to 2x, I get
1/2 ln2x
Why are the answers not the same?
they are the same antiderivatives .
(1) $\frac{1}{2} \ln|x| + C$
(2) $\frac{1}{2} \ln|2x| + C = \frac{1}{2} \ln|x| + \ln{2} + C =$
they only differ by a constant.
• March 28th 2010, 09:32 AM
Simon777
Quote:
Originally Posted by skeeter
they are the same antiderivatives .
(1) $\frac{1}{2} \ln|x| + C$
(2) $\frac{1}{2} \ln|2x| + C = \frac{1}{2} \ln|x| + \ln{2} + C =$
they only differ by a constant.
So the ln2 is just taken out because it can be put in with the constant right?
• March 28th 2010, 09:33 AM
TheEmptySet
Quote:
Originally Posted by Simon777
1/2 times the integral of 1/x is 1/2 lnx
however when i integrate the same problem as 1/2x and set u to 2x, I get
1/2 ln2x
Why are the answers not the same?
Remember that anti derivatives are only unique upto a constant. If you use the FTC you will get the same number out i.e
$\frac{1}{2}\int_{1}^{2}\frac{1}{x}dx=\ln(2)$
Or using the other def you get
$\int_{1}^{2}\frac{1}{2x}dx=\frac{1}{2}\ln(2x)\bigg |_{1}^{2}=\ln(2x)^{\frac{1}{2}}\bigg|_{1}^{2}=\ln( \sqrt{4})-\ln(\sqrt{1})=\ln(2)$
they are the same
• March 28th 2010, 11:21 AM
Simon777
Quote:
Originally Posted by TheEmptySet
$\frac{1}{2}\int_{1}^{2}\frac{1}{x}dx=\ln(2)$
Shouldn't that answer be 1/2 ln2 or ln square root of 2? | 2014-09-18T03:50:47 | {
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https://www.freemathhelp.com/forum/threads/basic-algebra-help-please.110760/ | #### Keeyn
##### New member
Hi guys, I have a simple question but it doesn’t seem that simple. I understand that and why 3(x+4) times 2(x+3) gives me (3x+12) times (2x+6) with the brackets written. Indeed, brackets are important here. However, what I don’t understand is why 3(x+4)+2(x+3) (note that only the multiplication sign is replaced by the addition sign) gives 3x+12+2x+6 straight without the brackets ( I know that we do 3(x+4) first then 2(x+3) ). Isn’t it supposed to be expanded to (3x+12) + (2x+6) with the brackets just like in the case of multiplication above? I thought since 3(x+4) and 2(x+3) are together in one expression and they do not stand alone, the brackets therefore are mandatory in the expanded form since 3(x+4) is viewed as a set and 2(x+3) another set?
#### Dr.Peterson
##### Elite Member
Hi guys, I have a simple question but it doesn’t seem that simple. I understand that and why 3(x+4) times 2(x+3) gives me (3x+12) times (2x+6) with the brackets written. Indeed, brackets are important here. However, what I don’t understand is why 3(x+4)+2(x+3) (note that only the multiplication sign is replaced by the addition sign) gives 3x+12+2x+6 straight without the brackets ( I know that we do 3(x+4) first then 2(x+3) ). Isn’t it supposed to be expanded to (3x+12) + (2x+6) with the brackets just like in the case of multiplication above? I thought since 3(x+4) and 2(x+3) are together in one expression and they do not stand alone, the brackets therefore are mandatory in the expanded form since 3(x+4) is viewed as a set and 2(x+3) another set?
Once you have expanded 3(x+4) + 2(x+3) to (3x+12) + (2x+6), the parentheses are no longer required; the result is equivalent to 3x+12+2x+6. That is, for any value of x, the two expressions have the same value.
You appear to be thinking that the parts of the expression retain some invisible link to their origin; they don't. As long as two expressions always have the same value, they are equivalent, no matter what you have done to get from one to the other.
Or perhaps you are just wondering about the difference between (3x+12)(2x+6) and(3x+12) + (2x+6). The reason you can drop parentheses in the latter is that all the operations connecting the terms are additions, so the associative property of addition applies. That is not true in the former, where we have a mix of additions and multiplication, and the distributive property applies.
Note that if we couldn't write 3x+12+2x+6, we couldn't then rearrange (using the commutative property) and combine like terms (using the distributive property). This step of dropping parentheses is essential to what we are typically doing in algebra.
#### Keeyn
##### New member
I do understand that the parenthesis can be dropped and ultimately it leads to the same end results. What I don’t understand is that when teachers teach, they often miss out this step: Like in the case of 3(x+4)-6(3+x)...They miss out (3x+12)+(-18-6x), they jump straight to 3x+12-18-6x. Is that middle step non-existent? Is that something made up by my mind?
#### Dr.Peterson
##### Elite Member
I do understand that the parenthesis can be dropped and ultimately it leads to the same end results. What I don’t understand is that when teachers teach, they often miss out this step: Like in the case of 3(x+4)-6(3+x)...They miss out (3x+12)+(-18-6x), they jump straight to 3x+12-18-6x. Is that middle step non-existent? Is that something made up by my mind?
I don't think I can tell you why (or even whether) the majority of teachers would skip that particular step. That's up to them. Most likely it would be because they are talking to students whom they expect to have passed that hurdle, so that they focus their attention on other issues. I suppose I probably would skip it for most students, though of course not the first time I did an example like this.
I hope I made it clear in my response to you that the step with parentheses is entirely valid. It is just such a small step to the next form that we often don't bother to write it.
Actually, in your new example (which is considerably trickier than the first), there are many other steps that might be shown, if one wanted to show every detail:
3(x + 4) - 6(3 + x)
3(x + 4) + -[6(3 + x]
3(x + 4) + -1[6(3 + x)]
3(x + 4) + (-1*6)(3 + x)
3(x + 4) + (-6)(3 + x)
[3*x + 3*4] + [(-6)3 + (-6)x]
[3x + 12] + [-18 + -6x]
3x + 12 + -18 + -6x
3x + -6x + 12 + -18
(3 + -6)x + (12 + -18)
-3x + -6
-3x - 6
I would never write out all those steps; but that is what I am actually doing in simplifying the expression. We must omit details at some point in order to communicate clearly; and that requires assuming something about the reader or student.
The trouble is that any teacher of a group of students (and especially any author writing to many students they will never even see) can't know what each of them does or does not already know, and therefore will make some choices that are not suitable for all of them.
That's part of the reason I prefer tutoring individually. But even then, I have to find out what you are ready for by trying things out and seeing what results I get.
But back to your question, no, it is not your imagination. The step exists; it is just one that teachers may easily assume students can figure out for themselves. | 2019-03-19T07:56:48 | {
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https://planetjazz.fm/1c47b/sas-congruent-triangles-26b2b2 | In which pair of triangles pictured below could you use the Side Angle Side postulate (SAS) to prove the triangles are congruent? This proof is still used in Geometry courses [3, 6]. However, the length of each side and the included angle can be measured by a ruler and a protractor respectively. If any two sides and angle included between the sides of one triangle are equivalent to the corresponding two sides and the angle between the sides of the second triangle, then the two triangles are said to be congruent by SAS rule. The Side-Side-Side (SSS) rule states that. In every triangle, there are three sides and three interior angles. Worksheets on Triangle Congruence. Side Angle Side (SAS) is a rule used to prove whether a given set of triangles are congruent. Theorems and Postulates for proving triangles congruent: Interactive simulation the most controversial math riddle ever! What about the others like SSA or ASS. This is one of them (SAS). It is measured that, In $\Delta ABC$, $LM \,=\, 5\,cm$, $MN \,=\, 6\,cm$ and $\angle LMN \,=\, 45^°$, In $\Delta PQR$, $PQ \,=\, 5\,cm$, $QR \,=\, 6\,cm$ and $\angle PQR \,=\, 45^°$. How do we prove triangles congruent? The SAS Triangle Congruence Theorem states that if 2 sides and their included angle of one triangle are congruent to 2 sides and their included angle of another triangle, then those triangles are congruent.The applet below uses transformational geometry to dynamically prove this very theorem. 2 triangles are congruent if they have: exactly the same three sides and; exactly the same three angles. Given the coordinates below, determine if triangle FGH is congruent to triangle JKL. Their interior angles and sides will be congruent. In which pair of triangles pictured below could you use the Side Angle Side postulate (SAS) to prove the triangles are congruent? Triangle Congruence SSS,SAS,ASA,AAS DRAFT. It's like saying that if two Oompa-Loompas wear clothes with all the same measurements, they're identical. Edit. If any two corresponding sides and their included angle are the same in both triangles, then the triangles … Congruent Triangles by SSS, SAS, ASA, AAS, and HL - practice/ review activity set for triangle congruence with shortcutsThis activity includes three parts that can be done all in one lesson or spread out across a unit on congruent triangles. If three sides of one triangle are equal to three sides of another triangle, then the triangles are congruent. In this case, measure any two sides and the angle between both sides in each triangle. Side Angle SideSide Side SideAngle Side AngleAngle Angle SideThat's an easy way to memorize the reasons of congruent triangles! Answer: Answer: 16. Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising. If they are explain why and write a valid congruence statement. Are these triangles congruent? SAS Criterion for Congruence SAS Criterion stands for Side-Angle-Side Criterion. The triangles are congruent when the lengths of two sides and the included angle of one triangle are equal to the corresponding lengths of sides and the included angle of the other triangle. Congruent triangles will have completely matching angles and sides. Triangle X Y Z is identical to triangle A B C but is slightly higher. Property 3 Interact with this applet below for a few minutes, then answer the questions that follow. Given: 1) point C is the midpoint of BF 2) AC= CE, Prove: $$\triangle ABC \cong \triangle EFC$$, Prove: $$\triangle BCD \cong \triangle BAD$$, Given: HJ is a perpendicular bisector of KI. BACK; NEXT ; Example 1. It is called Side-Angle-Side (SAS) criterion for the congruence of triangles. Basically triangles are congruent when they have the same shape and size. SAS Rule. If you flip/reflect MNO over NO it is the "same" as ABC, so these two triangles are congruent. Play this game to review Geometry. The SAS (Side-Angle-Side) criterion can be studied in detail from an understandable example. 10th grade. $$\triangle ABC \cong \triangle XYZ$$. 7 minutes ago. The congruence of any two triangles can be determined by comparing the lengths of corresponding two sides and corresponding one included angle of them. Triangles are congruent when all corresponding sides & interior angles are congruent. 0. It is the only pair in which the angle is an included angle. 17 Answer: Answer: 18. SSS Rule. 15. SAS statement says that two triangles are congruent if two sides and the included angle in one triangle are congruent to two sides and the included angle in another triangle. In this case, two triangles are congruent if two sides and one included angle in a given triangle are equal to the corresponding two sides and one included angle in another triangle. The two corresponding sides and the included angle of both triangles are considered as a criteria in this example for checking the congruence of triangles. Congruent Triangles. SSS stands for \"side, side, side\" and means that we have two triangles with all three sides equal.For example:(See Solving SSS Triangles to find out more) So we will give ourselves this tool in our tool kit. This specific congruent triangles rule represents that if the angle of one triangle measures equal to the corresponding angle of another triangle, while the lengths of the sides are in proportion, then the triangles are said to have passed the congruence triangle test by way of SAS. Learn cosine of angle difference identity, Learn constant property of a circle with examples, Concept of Set-Builder notation with examples and problems, Completing the square method with problems, Evaluate $\cos(100^\circ)\cos(40^\circ)$ $+$ $\sin(100^\circ)\sin(40^\circ)$, Evaluate $\begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9\\ \end{bmatrix}$ $\times$ $\begin{bmatrix} 9 & 8 & 7\\ 6 & 5 & 4\\ 3 & 2 & 1\\ \end{bmatrix}$, Evaluate ${\begin{bmatrix} -2 & 3 \\ -1 & 4 \\ \end{bmatrix}}$ $\times$ ${\begin{bmatrix} 6 & 4 \\ 3 & -1 \\ \end{bmatrix}}$, Evaluate $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin^3{x}}{\sin{x}-\tan{x}}}$, Solve $\sqrt{5x^2-6x+8}$ $-$ $\sqrt{5x^2-6x-7}$ $=$ $1$. 0% average accuracy. mrsingrassia. This statement as a theorem was proved in Greek time. Figure 3 Two sides and the included angle (SAS) of one triangle are congruent to the. Below is the proof that two triangles are congruent by Side Angle Side. Such case is represented in Fig.1. If they are congruent, state by what theorem (SSS, SAS, or ASA) they are congruent. The triangles are congruent when the lengths of two sides and the included angle of one triangle are equal to the corresponding lengths of sides and the included angle of the other triangle. ), the two triangles are congruent. As long as one of the rules is true, it is sufficient to prove that the two triangles are congruent. The triangles will have the same size & shape, but 1 may be a mirror image of the other. If two angles and the included side of one triangle are congruent to the corresponding parts of another triangle, then the triangles are congruent. Determine whether the two triangles are congruent. An included angleis an angle formed by two given sides. The SAS rule states that If two sides and the included angle of one triangle are equal to two sides and included angle of another triangle, then the triangles are congruent. This is called the Side Side Side Postulate, or SSS for short (not to be confused with the Selective Service System). First Congruence Postulate of triangles (SAS) Two triangles that have two sides and the angle between them equal are congruent. Here, the comparison of corresponding two sides and corresponding the included angle of both triangles is a criteria for determining the congruence of any two triangles. Can you imagine or draw on a piece of paper, two triangles, $$\triangle BCA \cong \triangle XCY$$ , whose diagram would be consistent with the Side Angle Side proof shown below? It is called Side-Angle-Side (SAS) criterion for the congruence of triangles. Sss And Sas Proofs - Displaying top 8 worksheets found for this concept.. Save. Edit. This is called the Side Angle Side Postulate or SAS. 0 times. There are five ways to test that two triangles are congruent. The included angle means the angle between two sides. Postulate 15 (ASA Postulate): If two angles and the side between them in one triangle are congruent to the corresponding parts in another triangle, then the triangles are congruent … corresponding parts of the other triangle. $(3).\,\,\,$ $\angle LMN \,=\, \angle PQR \,=\, 45^°$. State what additional information is required in order to know that the triangles are congruent for the reason given. If two triangles have edges with the exact same lengths, then these triangles are congruent. Both triangles are congruent. For a list see Congruent Triangles. Hence, the two triangles are called the congruent triangle. Part 4: Use SSS, SAS, AAS, ASA, and HL to determine if the triangles are congruent if not write not congruent. Compare the lengths of corresponding sides and the included angle of both triangles. Home / Geometry / Congruent Triangles / Exercises / SSS and SAS Exercises ; ... SSS and SAS Exercises. And as seen in the image, we prove triangle ABC is congruent to triangle EDC by the Side-Angle-Side Postulate Let a = 6, b = 8, c = 13, d = 8, e = 6, and f = 13. If we can show that two sides and the included angle of one triangle are congruent to two sides and the included angle in a second triangle, then the two triangles are congruent. Pair four is the only true example of this method for proving triangles congruent. Triangle Congruence SSS,SAS,ASA,AAS DRAFT. Free Algebra Solver ... type anything in there! If we know that all the sides and all the angles are congruent in two triangles, then we know that the two triangles are congruent. So SAS-- and sometimes, it's once again called a postulate, an axiom, or if it's kind of proven, sometimes is called a theorem-- this does imply that the two triangles are congruent. Triangle Congruence Theorems (SSS, SAS, & ASA Postulates) Triangles can be similar or congruent. It is the only pair in which the angle is an included angle. The Side Angle Side postulate (often abbreviated as SAS) states that if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then these two triangles are congruent. AAS(Angle-Angle-Side): If two pairs of angles of two triangles are equal in measurement, and a pair of corresponding non-included sides are equal in length, then the triangles are congruent. In every triangle, there are three sides and three interior angles. In a sense, this is basically the opposite of the SAS … The lengths of two sides and the included angle of $\Delta ABC$ are exactly equal to the lengths of corresponding sides and the included angle of $\Delta PQR$. 11) ASA S U T D 12) SAS W X V K 13) SAS B A C K J L 14) ASA D E F J K L 15) SAS H I J R S T 16) ASA M L K S T U 17) SSS R S Q D 18) SAS W U V M K-2- Angle Side Angle (ASA) Side Angle Side (SAS) Angle Angle Side (AAS) Hypotenuse Leg (HL) CPCTC. Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers. 7 minutes ago. Are these triangles congruent? Hence, the two triangles are called the congruent triangles. 3 comments Preview this quiz on Quizizz. Theorems and Postulates: ASA, SAS, SSS & Hypotenuse Leg Preparing for Proof. In other words it is the angle 'included between' two sides. $\Delta LMN$ and $\Delta PQR$ are two triangles but their lengths and angles are unknown. Introduction. State if the triangles are congruent and why. Side-Angle-Sideis a rule used to prove whether a given set of triangles are congruent. Triangles RQS and NTV have the following characteristics: • Right angles at ∠Q and ∠T • RQ ≅ NT No, it is not possible for the triangles to be congruent. How to construct a congruent triangle using the side-angle-side congruence postulate. Mathematics. Real World Math Horror Stories from Real encounters, $$\angle$$ACB = $$\angle$$XZY (angle). In the School Mathematics Study Groupsystem SASis taken as one (#15) of 22 postulates. So if you have two triangles and you can transform (for example by reflection) one of them into the other (while preserving the scale! Two triangles are congruent if they are exactly the same size and shape, which means they have the same angle measures and the same side lengths. This Congruence Postulate is … If lengths of two sides and an angle between them of one triangle are equal to the lengths of corresponding sides and an included corresponding angle of other triangle, then the two triangles are congruent geometrically. Both triangles are congruent and share common point C. Triangle A B C is slightly lower than triangle X Y C. Triangles X Y Z and A B C are shown. Show Answer. The Side-Angle-Side (SAS) rule states that Hence, it is called side-angle-side criterion and it is simply called SAS criterion for congruence of triangles. Similar triangles will have congruent angles but sides of different lengths. $\therefore \,\,\,\,\,\,$ $\Delta LMN \,\cong\, \Delta PQR$. Correspondingly, how can you tell the difference between AAS and ASA? Under this criterion, if the two sides and the angle between the sides of one triangle are equal to the two corresponding sides and the angle between the sides of another triangle, the two triangles are congruent. Congruent Triangles - Two sides and included angle (SAS) Definition: Triangles are congruent if any pair of corresponding sides and their included anglesare equal in both triangles. 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( not to be confused with the exact same lengths, then answer the questions that follow included! Lengths, then answer the questions that follow ( SSS, SAS, & Postulates. Matching angles and sides Side Postulate or SAS all corresponding sides & interior angles are congruent by angle! Additional information is required in order to know that the triangles are congruent Leg ( HL sas congruent triangles....
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https://mathhelpboards.com/threads/generation-of-submodules.5972/ | # Generation of submodules
#### Peter
##### Well-known member
MHB Site Helper
On page 351 Dummit and Foote make the following statement:
"It is easy to see using the Submodules' Criterion that for any subset A of M, RA is indeed a submodule of M and is the smallest submodule of M which contains A ... "
I am not sure how to (formally and explicitly) prove this statement.
However, reflecting on the above, it is easy to show that RA is a submodule of M, but what is worrying me is the formal proof that RA is the smallest submodule of M that contains A.
However following the statement:
"It is easy to see using the Submodules' Criterion that for any subset A of M, RA is indeed a submodule of M and is the smallest submodule of M which contains A ... " Dummit and Foote write:
"i.e. any submodule of M which contains A also contains RA"
[I still find it perplexing that this actually shows that RA is the smallest submodule of M which contains A but anyway ... this is, I think, not hard to prove ...}
So to show that any submodule of M which contains A also contains RA
Let N be a submodule of M such that [TEX] A \subseteq N [/TEX]
We need to show that [TEX] RA \subseteq N [/TEX]
Let [TEX] x \in RA [/TEX]
Now [TEX]RA = \{ r_1a_1 + r_2a_2 + ... ... + r_ma_m \ | \ r_1, r_2, ... ... , r_m \in R, \ a_1, a_2, ... ... , a_m \in A, m \in \mathbb{Z}^{+} [/TEX]
So [TEX] x = r_1a_1 + r_2a_2 + ... ... + r_ma_m [/TEX] for [TEX] r_1, r_2, ... ... , r_m \in R, \ a_1, a_2, ... ... , a_m \in A [/TEX]
If [TEX] A \subseteq N [/TEX] then [TEX] r_ia_i \in N [/TEX] for [TEX] 1 \le i \le n [/TEX] since N is a submodule
and then the addition of these elements, visually, [TEX] r_1a_1 + r_2a_2 + ... ... + r_ma_m [/TEX] also is in N (if two elements belong to a submodule then so does the element that is formed by their addition)
So [TEX] x \in N [/TEX]
Thus [TEX] x \in RA \Longrightarrow x \in N [/TEX]
So [TEX] A \subseteq N \Longrightarrow RA \subseteq N [/TEX] ... ... (1)
However, I am still not completely sure how to formally show that RA is the smallest submodule of M that contains A i.e. how does the implication (1) demonstrate this - can someone help by showing this explicitly and formally?
Peter
[This has also been posted on MHF]
Last edited:
#### Fernando Revilla
##### Well-known member
MHB Math Helper
However, I am still not completely sure how to formally show that RA is the smallest submodule of M that contains A i.e. how does the implication (1) demonstrate this - can someone help by showing this explicitly and formally?
Consider the set
$$\mathcal{N}=\{N\subseteq M:N\text{ submodule of }M\text{ and }A\subseteq N \}$$
Then, $\subseteq$ is an order relation on $\mathcal{N}$, $RA\in \mathcal{N}$ and $RA\subseteq N$ for all $N\in\mathcal{N}.$ This means that $RA$ is the smallest element related to $(\mathcal{N},\subseteq)$
Last edited:
#### Deveno
##### Well-known member
MHB Math Scholar
One can argue by contradiction, here.
Suppose we have a submodule $$\displaystyle N$$ with:
$$\displaystyle A \subseteq N \subsetneq RA$$.
By your previous argument, since $$\displaystyle N$$ is a submodule containing $$\displaystyle A$$, we have:
$$\displaystyle RA \subseteq N \subsetneq RA$$ that is:
$$\displaystyle RA \neq RA$$, a contradiction. So no such $$\displaystyle N$$ can exist. | 2021-04-15T16:34:17 | {
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https://math.stackexchange.com/questions/2913250/precise-meaning-of-infinitely-many | # Precise meaning of infinitely many
I have a rather lame question here. I need a clarification with the definition of "infinitely many". I have come across statements like:
There are infinitely many reals.
I know that reals are non-denumerable (uncountably infinite).
Again we have:
There are infinitely many integers.
I also know that integers are denumerable (countably infinite).
So my question is what do we actually infer from "infinitely many" about the countability or the uncountability?
Also kindly correct me if I am wrong somewhere.
• The two locutions simply do not make a distinction with different "levels" of infinity. They simply states that both sets (of real and of integers) are not finite. – Mauro ALLEGRANZA Sep 11 '18 at 14:57
• "There are at least two numbers", does that tell you anything about how many numbers are there? – Asaf Karagila Sep 11 '18 at 15:04
• @AsafKaragila Yes? It tells me there are at least two. – Theoretical Economist Sep 11 '18 at 15:44
• @TheoreticalEconomist: Exactly. – Asaf Karagila Sep 11 '18 at 15:47
• @MauroALLEGRANZA And, correspondingly, "Finite" doesn't make a distinction between the different "levels" of "non-infinity". – David Richerby Sep 11 '18 at 19:01
"Infinitely many" is just the negation of "finitely many."
You can't infer anything about countability from "infinitely many." There are infinitely many rationals and there are infinitely many real numbers.
To be more specific, you'd have to say something about "countable" in there somewhere.
"Countably many" is usually taken to mean "either finitely many or countably-infinitely many." Then you have "uncountably many," the negation of that.
• For example, if we say, infinitely many group homomorphisms can be defined from the set of integers to itself, is it okay? – Shatabdi Sinha Sep 11 '18 at 14:59
• @ShatabdiSinha Sure. In fact you can also say the set is countably infinite since you are just specifying an image for $1$ for each homomorphism, right? – rschwieb Sep 11 '18 at 15:01
• Wow, what a poor choice of terminology, saying "countably many" meaing "at most countable". I didn't know people really used these words like this, thanks for the info. – Serge Seredenko Sep 11 '18 at 21:26
• @SergeSeredenko It seems really like insisting that “countable=countable infinite” is a worse choice, because then you either have to say finite sets are uncountable, or you have to define uncountable sets as “infinite and not countable.” These two alternatives seem much less natural than using the inclusive definition of “countable.” But if you only rarely needed uncountable sets in a text, it would be convenient to have a convention that countable sets be infinite. – rschwieb Sep 12 '18 at 0:30
• I am sorry, I was wrong. I based my judgement on the fact that "countable set" means infinite, so "countable set" and "countably many" would mean different things. But now I see that countable set may be finite as well. I just did not expect that because in my language nobody uses the word "countable" to mean a finite set. So, thanks for the info once more. – Serge Seredenko Sep 12 '18 at 1:28
When we use infinitely many, we mean the set under consideration is not finite. It does not address the question that whether the set is countable or not. We can see it as opposed to finitely many which means the set is finite.
A very concise and dry definition of infinity: a set $S$ is infinite if and only if there is an injection from $\mathbb{N}$ to $S$.
A set $S$ is countably infinite if there is a bijection from $\mathbb{N}$ to $S$. From this definition, you can see that the set $\mathbb{R}$ is infinite, but not countably infinite. Also, you can see that all countably infinite sets are also infinite sets.
A set S is finite if there exists an injection from S to a subset of N having a maximum element. A set is infinite if and only if it is not finite. A set is also infinite if and only if it can be put into a bijection with a proper subset of itself.
S is countable if there exists a bijection to the integers, uncountable otherwise. Is interesting to note there are multiple possible cardinalities of uncountable sets. The cardinality of a power set of a set always has a cardinality greater than the original set. So any of these "sizes" of infinity is possible, and perhaps others, especially given some assumption on the Continuum Hypothesis.
• Uncountable just means "not countable", just like "infinite" means "not finite". So by definition, a set cannot have a higher cardinality than "uncountable". – Asaf Karagila Sep 11 '18 at 20:39 | 2019-05-19T16:53:05 | {
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https://cs.stackexchange.com/questions/153856/minimal-dfa-transition-function-clearification/153903 | # minimal DFA transition function clearification
Statement:
Given any dfa $$M$$, application of the procedure 'reduce' (see below) yields another dfa $$\hat{M}$$ such that $$M$$ and $$\hat{M}$$ are equivalent. Furthermore $$\hat{M}$$ is minimal in the sense that there is no other dfa with a smaller number of states that also accepts $$L(M)$$
First some background information:
The 'reduce' procedure:
Given a dfa $$M=\left(Q, \Sigma, \delta, q_{0}, F\right)$$, we construct a reduced dfa $$\widehat{M}=(\widehat{Q}, \Sigma, \widehat{\delta}, \widehat{q}, \widehat{F})$$ as follows.
1. Use procedure mark to generate the equivalence classes, say $$\left\{q_{i}, q_{j}, \ldots, q_{k}\right\}$$, as described.
2. For each set $$\left\{q_{i}, q_{j}, \ldots, q_{k}\right\}$$ of such indistinguishable states, create a state labeled $$i j \ldots k$$ for $$\widehat{M}$$.
3. For each transition rule of $$M$$ of the form $$\delta\left(q_{r}, a\right)=q_{p},$$ find the sets to which $$q_{r}$$ and $$q_{p}$$ belong. If $$q_{r} \in\left\{q_{i}, q_{j}, \ldots, q_{k}\right\}$$ and $$q_{p} \in\left\{q_{l}, q_{m}, \ldots, q_{n}\right\}$$, add to $$\widehat{\delta}$$ a rule $$\widehat{\delta}(i j \cdots k, a)=l m \cdots n .$$
4. The initial state $$\widehat{q}_{0}$$ is that state of $$\widehat{M}_{\text {}}$$ whose label includes the 0 .
5. $$\widehat{F}$$ is the set of all the states whose label contains $$i$$ such that $$q_{i} \in F$$.
A claim in order to prove the statement:
Take any state $$q \in Q$$ and let $$\hat{q}$$ denote its equivalence class. For any word $$w \in \Sigma^{*}$$, it holds that $$\delta^{*}(q, w) \in \hat{\delta}^{*}(\hat{q}, w)$$. The proof is by induction on the length $$n$$ of $$w$$. base case If $$n=0$$, then $$w=\lambda$$ and the claim is trivial since $$\delta^{*}(q, \lambda)=q \in \hat{q}=\hat{\delta}^{*}(\hat{q}, \lambda) .$$ For the induction step assume our claim holds for all strings of length $$n-1$$ we show that it shows for strings of length $$n$$. In this case $$w=a v$$ for some $$v$$. We know that $$\begin{array}{rlr} \delta^{*}(q, w) & =\delta^{*}(q, a v) \\ & =\delta^{*}(\delta(q, a), v) & \text { definition of } \delta^{*} \\ & \in \hat{\delta}^{*}(\widehat{\delta(q, a)}, v) & \text { induction hypothesis } \\ & =\hat{\delta}^{*}\left(\hat{\delta}^{}(\hat{q}, a), v\right) \\ & =\hat{\delta}^{*}(\hat{q}, w) \\ \end{array}$$
My question:
In the proof of the claim above, I do not know how we can be sure that
$$\widehat{\delta(q, a)} = \hat{\delta}^{}(\hat{q}, a)$$
It seems trivial, but shouldn't it be proven?
Source: Formal languages and automata by Peter Linz, 5th edition (Jones & Bartlett Learning), p. 75
• Books have authors and publishers. They often go through multiple editions, each with a date. Quotes are taken from a particular page in a particular edition of a particular book, and a citation must include this information. I guessed which book you're talking about, but you need to fill in the rest. This is mandatory, not just by the rules of this site, but also to comply with copyright rules and academic practice. Thanks.
– rici
Aug 27 at 22:58
• @rici I updated the question, thank you for the feedback. Aug 30 at 9:32
$$\widehat{\delta(q, a)} = \hat{\delta}^{}(\hat{q}, a)$$
The equality above is the definition of $$\hat\delta$$, given $$\delta$$. Let us paraphrase item 3.
1. For each transition rule of $$M$$ of the form $$\delta(q, a)=\delta(q, a),$$ find the sets to which $$q$$ and $$\delta(q, a)$$ belong. If $$q \in\hat q$$ and $$\delta(q, a)\in\widehat{\delta(q, a)}$$, add to $$\widehat{\delta}$$ a rule $$\hat{\delta}(\hat q, a)= \widehat{\delta(q, a)}.$$
On the other hand, there is an important question: why is $$\delta$$ well-defined? Suppose $$q_i\in \hat q$$ and $$q_j\in \hat q$$, then $$\hat{\delta}(\hat q, a)$$ is defined as $$\widehat{\delta(q_i,a)}$$ as well as $$\widehat{\delta(q_j,a)}$$. Are $$\widehat{\delta(q_i,a)}$$ and $$\widehat{\delta(q_j,a)}$$ the same equivalence class?
This well-defined-ness is a key step in the 'reduce' procedure. I will leave it for you to verify.
• Thank you very much! About the question in your answer: Suppose that the transitions resulted in different equivalence classes, that would mean that δ(qi,a) and δ(qj,a) are distinguishable. Suppose they are distinguishable by a string of length n, then qi and qj would be distinguishable by a string of length n+1 and so they wouldn’t be in the same equivalence class (contradiction). Is this a correct reasoning and answer to the question? Aug 31 at 10:17
• And I can't edit the answer because my edit is too small, but I think you made a little mistake in the question: "is defined as ˆδ(qi,a) as well as ˆδ(qi,a)" so two times qi instead of qi and qj. Aug 31 at 10:20
• Yes, that is correct. Aug 31 at 13:40 | 2022-12-03T12:13:19 | {
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https://math.stackexchange.com/questions/1770986/distribution-of-max-number-of-common-balls | # Distribution of max number of common balls
I have $n$ different balls numbered $\{1,\dots, i, \dots, n\}$. I choose $n$ balls, uniformly at random, with replacement. Let $X_i$ denote the number of times ball $i$ has been chosen. I would like to derive the distribution of $M = \text{max}(X_i)$. What is its average and $90^{\text{th}}$ percentile value?
• Nice question. Now what ? May 4, 2016 at 7:47
• I'm not sure I understand. May 4, 2016 at 8:40
• As usual, compute the CDF of the sample maximum. Then make use of it to get the expected value. For the 90th percentile I'm afraid you may need to fall back to numerical method only.
– BGM
May 4, 2016 at 9:32
• @BGM I think you may be misinterpreting the question - it is not about order stats May 4, 2016 at 18:58
• To OP: I have edited the question to try resolve some notational issues - please check you are happy with the edits. May 6, 2016 at 20:31
I think this is an interesting and original question. The idea is simple:
Suppose we take 10 samples with replacement from $U(1,10)$ and record say:
{8,4,7,8,5,7,9,7,2,1}
The OP is interested in the maximum number of common items $M$: in this case, $M = 3$, because there are 3 values of ball $i=7$. What is the pmf of $M$?
The problem the OP is trying to solve is related to the Mutinomial distribution with pmf:
$$f\left(x_1,\ldots ,x_n\right)=P\left(X_1=x_1,\ldots ,X_n=x_n\right)=\frac{n! }{ x_1! \cdots x_n!} \; p_1^{x_1} \cdots p_n^{x_n}$$
where $X_i$ is the number of recorded values of ball $i$, subject to: $$\sum _{i=1}^n x_i=n \quad \text{and} \quad p_i= \frac1n$$
Exact Symbolic derivation
In the case with $n = 3$ (a trinomial), the exact pmf of $M$ can be derived as:
• $P(M=1) \quad = \quad P(X_1 = 1 \text{ And } X_2 = 1 \text{ And } X_3 = 1) = \quad \frac29$
• $P(M=2) \quad = \quad P(X_1 = 2 \text{ Or } X_2 = 2 \text{ Or } X_3 = 2) \quad = \quad \frac69$
• $P(M=3) \quad = \quad P(X_1 = 3 \text{ Or } X_2 = 3 \text{ Or } X_3 = 3) \quad = \quad \frac19$
which matches a Monte Carlo simulation of the pmf of $M$:
The above should provide a structure to take things further symbolically, though it may be arduous to do so.
Monte Carlo derivation
Here is some code (here using Mathematica) to generate say 100,000 pseudorandom drawings of $M$, given $n$ balls:
maxData[n_] := Table[Max[Transpose[Tally[RandomInteger[{1, n}, n]]][[2]]], 10^5];
For example:
data = maxData[20];
will generate 100,000 drawings of $M$, given $n=20$ balls. It takes about $\frac14$ second to generate. For faster performance on computers that have more than 1 core, replace Table with ParallelTable.
In the case of $n = 10$, the pmf of $M$ (here via Monte Carlo) appears:
The following diagram compares the pmf of $M$ when $n = 50$ and $n = 100$: | 2022-05-23T15:52:06 | {
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https://math.stackexchange.com/questions/2549666/how-to-solve-a-complex-linear-equation-with-two-variables | # How to solve a complex linear equation with two variables
In a review question I get the equation $(4-5i)m + 4n = 16+15i$. Where $i$ is the imaginary unit, $m$ and $n$ are real numbers. I do not know how to go about solving this equation. There is also another section to the question which asks to solve it when $m$ and $n$ are conjugate complex numbers.
Thank you.
• Maybe it is $16+15 i$ ? – Emilio Novati Dec 3 '17 at 22:16
• Hint: Compare real and imaginary parts. – John Doe Dec 3 '17 at 22:19
• rewriting the left-hand side as a sum real and imaginary you get $m=-3$, $n=7$. – daulomb Dec 3 '17 at 22:21
• I assume you meant $16+15i$ as @EmilioNovati mentioned, so I gave it a heads-up correction. Re-edit it if that's not the case. – Rebellos Dec 3 '17 at 22:22
• Yes, I did mean $16 + 15i$, thank you for the edit. – user509838 Dec 3 '17 at 22:23
$$(4-5i)m + 4n = 16+15i \Leftrightarrow 4m - 5im + 4n = 16 + 15i \Leftrightarrow 4(m+n)+i(-5m)=16+15i$$
So, by the identity of complex numbers, we'll get :
$$\begin{cases} 4(m+n)=16 \\ -5m = 15\end{cases}$$
can you solve that system of linear equations now and yield your solution ?
• @JohnDoe I should go sleep as it seems :D – Rebellos Dec 3 '17 at 22:26
• Haha yes, you got through most of it correctly at least! :) I will delete the above comment, and this one too shortly. – John Doe Dec 3 '17 at 22:27
• This is a great explanation, I got the solutions $m = -3$ and $n = 7$. Thank you. – user509838 Dec 3 '17 at 22:31
For $m$ and $n$ real, just compare real and imaginary parts.
For $m$ and $n$ complex numbers conjugate to one another, just substitute $a + bi$ for $n$ and $a - bi$ for $m$, where $a$ and $b$ are real numbers. To finish just again compare real and imaginary parts.
Hint:
use the identity of complex numbers: $$a+ib=m+in \quad \iff \quad a=m \quad \mbox{and} \quad b=m$$
The equation you end up with is $$(4-5i)m+4n=16+15i\\(4m+4n)-(5m)i=16+15i$$ Compare imaginary parts to get $m$, then real parts to get $n$.
When $m$ and $n$ are complex conjugates, write $m=a+bi$, $n=a-bi$. Then substitute this into the above equation, multiply out and the compare real and imaginary parts again to obtain $a$ and $b$, thus getting $m$ and $n$. | 2019-11-23T01:34:00 | {
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https://scicomp.stackexchange.com/questions/1362/error-calculation-in-trapezoidal-rule | # Error calculation in trapezoidal rule
If we use the composite trapezoidal rule, then what is the least number of divisions $N$ for which the error of the integral $\int^1_0{e^{-x}}dx$ doesn't exceed $\frac{1}{12}\times10^{-2}$.
My guess is 11 or 5. Kindly tell me which of the answer is correct?
I obtained 11 as the answer by applying the formula
$$\Big|\frac{(b-a)^3}{12N^2}\times{(e^{-x})^{\prime\prime}_{x=\varepsilon}}\Big| \text{ = } {\frac{10^{-2}}{12}}$$ where $\varepsilon \in [0,1]$ is chosen so that it maximizes the value of $e^{-\varepsilon}$ (which I believe occurs at $\varepsilon = 0$).
Solving this equation, I get $N = 10$ (i.e. I must have atleast 11 equidistant divisions if I have to keep the error less than $\frac{10^{-2}}{12}$).
As far as 5 is concerned, I just used 5 equidistant intervals i.e $0,\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5},1$ and I applied them in the trapezoidal rule. Now here's the problem:- When calculated the answer at $N=5$, I got a value greater that the actual value of the integral. Is it possible? If yes, why?
Which of my answer is correct because I obtained 11 by a well-established formula and 5 was just an option I hit upon and am not really sure of 5's correctness.
Thanks
Note:- I am posting this question because everywhere else nobody is giving any reply at all. I don't know if this question belongs here. If it does, then kindly reply. If it doesn't,then feel free to erase or delete or whatever :)
• Cross posting at MATH.SE – Inquest Feb 19 '12 at 8:45
• @Nunoxic, thanks for the heads up. (Same to DavidZ, who pointed out the cross-posting on the Math.SE end.) – Geoff Oxberry Feb 19 '12 at 8:56
• andrewjames, welcome to SciComp! Our policy on cross-posting follows that of other Stack Exchange sites. It is permissible to cross-post if you tailor the same question (more or less) to different audiences. It is permissible to ask your question to be migrated to another site after some time, if you feel that your question is not getting answered satisfactorily (or at all) on the site where it is initially posted. – Geoff Oxberry Feb 19 '12 at 8:58
• However, it is generally considered abusive behavior to cross-post. Please delete one of the cross-posted questions. (Since your question is about numerical methods, it is probably a better fit here, but I'm biased; the choice is up to you.) – Geoff Oxberry Feb 19 '12 at 9:00
• "When calculated the answer at N=5, I got a value greater that the actual value of the integral. Is it possible? If yes, why?" This is not surprising, as the exponential decay function is concave up over that interval. The trapezoidal rule should overestimate the integral over that region. – Dan Feb 19 '12 at 9:10 | 2019-12-08T16:34:27 | {
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https://math.stackexchange.com/questions/228841/how-do-i-calculate-the-intersections-of-a-straight-line-and-a-circle | # How do I calculate the intersection(s) of a straight line and a circle?
The basic equation for a straight line is $y = mx + b$, where $b$ is the height of the line at $x = 0$ and $m$ is the gradient. The basic equation for a circle is $(x - c)^2 + (y - d)^2 = r^2$, where $r$ is the radius and $c$ and $d$ are the $x$ and $y$ shifts of the center of the circle away from $(0,\ 0)$.
I'm trying to come up with an equation for determining the intersection points for a straight line through a circle. I've started by substituting the "y" value in the circle equation with the straight line equation, seeing as at the intersection points, the y values of both equations must be identical. This is my work so far:
$$(x - c)^2 + (mx + b - d)^2 = r^2$$ $$x^2 + c^2 - 2xc + m^2x^2 + (b - d)^2 + 2mx(b - d) = r^2$$ I'll shift all the constants to one side $$x^2 - 2xc + m^2x^2 + 2mx(b - d) = r^2 - c^2 - (b - d)^2$$ $$(m^2 - 1)x^2 + 2x(m(b - d) - c) = r^2 - c^2 - (b - d)^2$$
That's as far as I can get. From what I've gathered so far, I ought to be able to break down the left side of this equation into a set of double brackets like so: $$(ex + f)(gx + h)$$ where $e,\ f,\ g$ and $h$ are all constants. Then I simply have to solve each bracket for a result of $0$, and I have my $x$ coordinates for the intersections of my two equations. Unfortunately, I can't figure out how to break this equation down.
Any help would be appreciated.
• Why don't you just solve for $x$? You know the constans, right? – M.B. Nov 4 '12 at 12:22
• I'm trying to develop an equation that will deduce what x is for any set of constants, not just a specific set. – Cambot Nov 4 '12 at 12:41
Let's say you have the line $$y = mx + c$$ and the circle $$(x-p)^2 + (y-q)^2 = r^2$$.
First, substitute $$y = mx + c$$ into $$(x-p)^2 + (y-q)^2 = r^2$$ to give
$$(x-p)^2 + (mx+c-q)^2 = r^2 \, .$$
Next, expand out both brackets, bring the $$r^2$$ over to the left, and collect like terms:
$$(m^2+1)x^2 + 2(mc-mq-p)x + (q^2-r^2+p^2-2cq+c^2) = 0 \, .$$
This is a quadratic in $$x$$ and can be solved using the quadratic formula. Let us relabel the coefficients to give $$Ax^2 + Bx + C = 0$$, then we have
$$x = \frac{-B \pm \sqrt{B^2-4AC}}{2A} \, .$$
If $$B^2-4AC < 0$$ then the line misses the circle. If $$B^2-4AC=0$$ then the line is tangent to the circle. If $$B^2-4AC > 0$$ then the line meets the circle in two distinct points.
Since $$y=mx+c$$, we can see that if $$x$$ is as above then
$$y = m\left(\frac{-B \pm \sqrt{B^2-4AC}}{2A}\right) + c \, .$$
EDIT: The lines $$y=mx+c$$ do not cover the vertical lines $$x=k$$. In that case, substitute $$x=k$$ into $$(x-p)^2+(y-q)^2=r^2$$ to give
$$y^2 - 2qy + (p^2+q^2-r^2 - 2kp+k^2) = 0$$
This gives a quadratic in $$y$$, namely $$y^2+By+C=0$$, where $$B=-2q$$ and $$C=p^2+q^2-r^2 - 2kp+k^2$$. Solve using the Quadratic Formula. The solutions are $$(k,y_1)$$ and $$(k,y_2)$$, where the $$y_i$$ are solutions to $$y^2+By+C=0$$.
• Ah! Thank you! This certainly looks like it'll do what I'm trying to do! – Cambot Nov 4 '12 at 13:20
• Be careful to manage vertical line !!! – Eric Ouellet May 13 '16 at 19:26
• @EricOuellet Thank you, my friend. I think I've fixed it. – Fly by Night Mar 12 '19 at 17:51
You've got a quadratic in $x$. Use the discriminant to see if it has real solutions, and if so how many. Then solve the quadratic for $x$, and substitute the solution(s), if any, back into the equation for the line.
Here's an equation which works even if the line is vertical. This is useful from a computer programming perspective. Let's set up the system of equations using standard formulae:
$$\begin{cases} (x - x_0)^2 + (y - y_0)^2 = r^2 \\ Ax + By + C = 0 \end{cases}$$
where the circle with radius $$r$$ is centered at $$(x_0, y_0)$$; the line contains the points $$(x_1, y_1)$$ and $$(x_2, y_2)$$.
From the line equation, we know:
$$y = \frac{-Ax-C}{B} = - \frac{Ax+C}{B}$$
($$B \neq 0$$, we'll get into this in a second)
Therefore,
\begin{align*} (x - x_0)^2 + \left( - \frac{Ax+C}{B} - y_0 \right)^2 &= r^2 \\ x^2 - 2x_0x + {x_0}^2 + \frac{A^2 x^2 + 2ACx + C^2}{B^2} + 2y_0 \cdot \frac{Ax+C}{B} + {y_0}^2 &= r^2 \\ x^2 - 2x_0x + \frac{A^2 x^2 + 2ACx + C^2}{B^2} + 2y_0 \cdot \frac{Ax+C}{B} &= r^2 - {x_0}^2 - {y_0}^2 \end{align*}
Multiply both sides by $$B^2$$. \begin{align*} B^2 x^2 - (2B^2 x_0)x + A^2 x^2 + (2AC)x + C^2 + 2By_0(Ax+C) &= B^2(r^2 - {x_0}^2 - {y_0}^2) \\ (A^2 + B^2)x^2 + (2AC - 2B^2 x_0)x + (2AB y_0)x &= -C^2 - 2BCy_0 + B^2(r^2 - {x_0}^2 - {y_0}^2) \end{align*}
Therefore, we have a quadratic equation $$ax^2 + bx + c = 0$$ with:
$$\begin{cases} a = A^2 + B^2 \\ b = 2AC + 2AB y_0 - 2B^2 x_0 \\ c = C^2 + 2BC y_0 - B^2 (r^2 - {x_0}^2 - {y_0}^2) \end{cases}$$
where
$$\begin{cases} A = y_2 - y_1 \\ B = x_1 - x_2 \\ C = x_2 y_1 - x_1 y_2 \end{cases}$$
Now you can use $$\{ x = \frac{-b \pm \sqrt{\Delta}}{2a}, y= -\frac{Ax+C}{B} \}$$ to solve for the points.
The above equation doesn't work for $$B=0$$. Besides this, if $$|B|$$ is too small, then floating point computation gets inaccurate.
If $$B=0$$, or is too small, then we let $$x = -\frac{By+C}{A}$$, and sub it into the circle's formula:
\begin{align*} \left( - \frac{By+C}{A} - x_0 \right)^2 + (y - y_0)^2 &= r^2 \\ \frac{B^2 y^2 + 2BCy + C^2}{A^2} + 2x_0 \cdot \frac{By+C}{A} + y^2 - 2y_0 y &= r^2 - {x_0}^2 - {y_0}^2 \\ B^2 y^2 + 2BCy + C^2 + 2Ax_0 (By+C) + A^2y^2 - 2A^2 y_0 y &= A^2(r^2 - {x_0}^2 - {y_0}^2) \\ (A^2 + B^2)y^2 + (2BC + 2ABx_0 - 2A^2 y_0)y &= -C^2 - 2ACx_0 + A^2(r^2 - {x_0}^2 - {y_0}^2) \end{align*}
Thus, we have a quadratic equation $$ay^2 + by + c = 0$$ where
$$\begin{cases} a = A^2 + B^2 \\ b = 2BC + 2ABx_0 -2A^2 y_0 \\ c = C^2 + 2ACx_0 -A^2(r^2 - {x_0}^2 - {y_0}^2) \end{cases}$$
Have you checked the general solution on wolframalpha? You've already made your quadratic equation and so, you can get the roots of $x$ as the solutions that can then be used to get the values of $y$. | 2021-02-26T02:18:56 | {
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https://www.physicsforums.com/threads/subgroups-of-z6.386552/ | # Subgroups of Z6
Hi guys
I don't really understand how exactly to FIND subgroups of a given group..... Is there any specific process to do so?
## Homework Statement
Find all subgroups of Z6.
## The Attempt at a Solution
How does one find subgroups??
Z6 = Z2 x Z3
Am I right in saying this?
I think Z6 = {0,1,2,3,4,5}...
Then do I just look at each element or something? I'm really bad at abstract algebra.. Can someone attempt and explain the steps to take to determine a subgroup?
Thanks
## Answers and Replies
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jbunniii
Homework Helper
Gold Member
In general it's a hard problem to find all subgroups of a given group.
In your case it's much easier, because $Z_6$ is cyclic. Therefore all of its subgroups must also be cyclic. (Why?)
A cyclic subgroup is generated by a single element. You only have six elements to work with, so there are at MOST six subgroups. Work out what subgroup each element generates, and then remove the duplicates and you're done.
By the way,
$$Z_6 = Z_2 \times Z_3$$
is not correct. Instead write
$$Z_6 \cong Z_2 \times Z_3$$
That is, $Z_6$ is isomorphic to $Z_2 \times Z_3$, but they aren't EQUAL. The elements of $Z_6$ are $\{0, 1, 2, 3, 4, 5\}$ whereas the elements of $Z_2 \times Z_3$ are $\{(0,0), (0,1), (0,2), (1,0), (1,1), (1,2)\}$. But both are cyclic groups with order 6, and therefore they are isomorphic.
: P.S. The fact that
$$Z_6 \cong Z_2 \times Z_3$$
is irrelevant to this problem. You don't need to use that fact to find the subgroups of $Z_6$.
Last edited:
One place to start would be to use Lagrange's theorem...Since Z6 has order 6, you know that any subgroup can only have order 1, 2 or 3 (and 6, but then this is Z6 itself). You also know that any subgroup must have the identity in it so that narrows your search down as well. Since (I believe) Z6 is cyclic, you also know that the order of any element of a group must divide the order of the group, so that will help too. I think it's just a matter of taking what you know about orders and using them to whittle down the possibilities. Just to be sure...you might want to check that all my assumptions are correct...I'm still just learning this stuff, as well.
Hope that helps.
Cheers,
Lauren. =)
oh okay
so i have
<0> = {0}
<1> = {0,1,2,3,4,5} = Z6
<2> = {0,2,4}
<3> = {0,3}
so there are 4 subgroups?
thanks for your replies!
jbunniii
Homework Helper
Gold Member
oh okay
so i have
<0> = {0}
<1> = {0,1,2,3,4,5} = Z6
<2> = {0,2,4}
<3> = {0,3}
so there are 4 subgroups?
thanks for your replies!
Correct!
didn't you forget
$$<|4|>={[0],[4],[2]}$$
and
$$<|5|>=<[1]>$$??
Last edited:
jbunniii
Homework Helper
Gold Member
didn't you forget
$$<|4|>={[0],[4],[2]}$$
and
$$<|5|>=<[1]>$$??
But <4> = <2> and <5> = <1>, so they were not omitted.
Deveno
But <4> = <2> and <5> = <1>, so they were not omitted.
true enough, but this is worth verifying (at least once), to illustrate that generators of cyclic subgroups need not be unique (if a cyclic subgroup is of prime order, there are LOTS of choices for a generator).
<3> is of order 2, so we get lucky, there is only one generator.
<2> is of order 3, so we have 2 generators (it might seem, naively at first, that <2> and <4> might be different, they certainly are in Z).
<1> is of order 6, and 6 isn't prime, so we would expect to find fewer than 5 generators. in fact, we have φ(6) = φ(2)φ(3) = (1)(2) = 2 generators (where φ is the euler totient function).
gcd(k,6) = 1 ---> leads to a subgroup of order 6 (obviously the whole group Z6).
gcd(k,6) = 3 ---> leads to a subgroup of order 6/3 = 2 (and this subgroup is, surprisingly, unique).
gcd(k,6) = 2 ---> leads to a subgroup of order 3 (also unique. it's not immediately obvious that a cyclic group has JUST ONE subgroup of order a given divisor of the order of the whole group, but this is indeed true, and worth proving!)
gcd(k,6) = 6 ---> leads to the trivial subgroup {0}.
all that "greatest common denominator" stuff one learns in high school finally pays off! :) | 2021-03-03T12:51:33 | {
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https://math.stackexchange.com/questions/1875441/divergence-of-sum-of-reciprocals-of-square-free-numbers | # Divergence of sum of reciprocals of square-free numbers
I am aware that there is a similar question here, however I want to prove that $\sum \frac{1}{n}$ for $n$ square free diverges, without relying on the fact that $\sum \frac{1}{p}$ diverges for $p$ prime. This is equivalent to proving that $\sum \frac{|\mu(n)|}{n}$ converges, where $\mu(n)$ is the mobius function. I would like verification that my proof is correct:
My Proof: We begin by noting that $\sum_{n = 1}^{\infty} \frac{1}{n^2}$ converges, and so $$c * \sum_{n = 1}^{\infty} \frac{1}{n^2}$$ must also converge for any positive integer $c$. Therefore, if we look at the sum $\sum \frac{1}{n}$ where $n$ ranges only through the integers that are not squarefree, then this sum must converge because it is the composition of a series of convergent sums. Since $\sum_{n = 1}^{\infty} \frac{1}{n}$ diverges, we must have the sum of the reciprocals of square-free integers also diverging.
In particular, how can I be sure that the sum of the reciprocals of non square-free integers converges? It seems like we are taking an infinite number of convergent sums, which doesn't have to necessarily converge.
• Looks like you are taking a divergent series minus an absolute convergent series...that should be obvious... Jul 29 '16 at 21:36
• Yes but I was asking for verification of the correctness of the proof. Jul 29 '16 at 21:38
• Thanks. But more specifically, we are adding what seems to be an infinite number of convergent series, namely the ones that are of the form c * 1/n^2. How can we be sure that this still converges? Jul 29 '16 at 21:40
• The sum of not squarefrees contains $\sum \frac1{4n}$, hence diverges Jul 29 '16 at 21:55
• That's not the point. The sum I am referring to is in the long run less than about $\ln n$ times $\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots$. This is $\frac{\pi^2}{6}-1$, which is nicely less than $1$. We don't need to know the exact sum $\sum_1^\infty \frac{1}{k^2}$, a reasonably good upper bound will do. So the sum of the reciprocals of square-frees up to $n$ is bounded below by a non-zero constant times $\ln n$. Jul 29 '16 at 22:21
Outline: Let $n$ be large. Note that the sum of the reciprocals of the integers up to $n$ is about $\ln n$.
The sum of the reciprocals of multiples of $4$ up to $n$ is less than roughly $\frac{1}{4}\ln n$. Here we are already giving away a bit, since this reciprocal sum is actually about $\frac{1}{4}\ln(n/4)$.
The sum of the reciprocals of multiples of $9$ up to $n$ is less than roughly $\frac{1}{9}\ln n$. And so on.
So the sum of the reciprocals of not square-frees up to $n$ is less than roughly $$(\ln n)\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots.\right).$$ The infinite sum above is upper bounded by $$(\ln n)\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{3\cdot 4}+\frac{1}{4\cdot 5}+\cdots\right),$$ which is less than $\frac{1}{2}\ln n$. (Since we are upper-bounding, we don't need to worry about overlap.)
So the sum of the reciprocals of the square-frees up to $n$ is asymptotically greater than $\frac{1}{2}\ln n$.
• $(\ln n) (\frac 1{3\cdot 4} + \cdots )$ is a lower bound of the infinite sum right above that. Jul 29 '16 at 23:10
• @i707107: Thanks, I was missing the first two terms, separated out for the estimate. Jul 29 '16 at 23:25
Let $Q(x)$ be the number of square-free positive integers $n\leq x$. It is well-known that $$Q(x) = \frac 6{\pi^2} x + O(\sqrt x).$$ We now apply partial summation: $$\sum_{n\leq x} \frac{\mu^2(n)}{n} = \frac {Q(t)}t \bigg\vert_{1-}^x + \int_1^x \frac{Q(t)}{t^2} dt = \frac 6{\pi^2} \log x+O(1).$$ Thus, the sum of reciprocals of square-free numbers diverges, as $\log x \rightarrow\infty$.
André Nicolas's proof is impeccable, but I'd like to give another one, which is probably closer to yours.
Let's recall that a family $(a_\lambda)_{\lambda \in \Lambda}$ of nonnegative numbers is summable is the set of finite sums $\left\{ \sum_{\lambda \in \Lambda'} a_\lambda \,\middle|\, \Lambda' \subset \Lambda\text{ finite}\right\}$ is bounded. In that case, the least upper bound of this set is the sum of the family, and we denote it $\sum_{\lambda\in\Lambda} a_\lambda$. This theory is basically equivalent to the theory of series (which corresponds to $\Lambda = \mathbb N$) if $\Lambda$ is countable, which is the only interesting case.
The result I'd like to use is the following, which corresponds to the Cauchy product of ordinary series.
Theorem. Let $(a_{\lambda})_{\lambda \in\Lambda}$ and $(b_\mu)_{\mu\in M}$ be two families of nonnegative numbers. If these two families are summable, then so is $(a_\lambda\,b_\mu)_{(\lambda, \mu)_\in\Lambda\times M}$, and we have the equality $$\sum_{(\lambda,\mu)\in \Lambda \times M} a_\lambda\,b_\mu = \left(\sum_{\lambda\in\Lambda} a_\lambda\right) \, \left(\sum_{\mu\in M} b_\mu\right).$$
Now, let $S = \{1, 4, 9, 16,\ldots\}$ and $SF = \{1, 2, 3, 5, 6, 7, 10,\ldots\}$ be the sets of square and squarefree numbers, respectively.
You already know that $(n^{-1})_{n \in\mathbb N^*}$ isn't summable and that $(n^{-1})_{n\in S}$ is (because that's really equivalent to saying that $\sum \frac 1{m^2}$ converges). I claim that the above theorem directly shows that these two properties imply that $(n^{-1})_{n\in SF}$ isn't summable.
Indeed, if $(n^{-1})_{n\in SF}$ were summable, so would $(a^{-1}\,b^{-1})_{(a,b)\in S\times SF}$, as per the theorem. But it is easy to show that every positive number can be written in a unique way as the product of a square number and a squarefree number. So, this product family $(a^{-1}\,b^{-1})_{(a,b)\in S\times SF}$ is really the same thing as $(n^{-1})_{n\in\mathbb N^*}$, so it cannot be summable, which proves that $(n^{-1})_{n\in SF}$ wasn't summable either. | 2022-01-19T22:54:05 | {
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http://mymathforum.com/advanced-statistics/342695-markov-chain-another-solution-print.html | My Math Forum (http://mymathforum.com/math-forums.php)
- - Markov chain or another solution? (http://mymathforum.com/advanced-statistics/342695-markov-chain-another-solution.html)
Falsetto November 6th, 2017 06:36 PM
Markov chain or another solution?
Ok, here's the problem.
I am rolling 2 fair dice repeatedly and I want to know what the chance is of rolling both a total of 5 and a total of 9 before rolling a total of 7.
Let y equal the probability of a 5 or 9 or 7 being rolled. There are 4 ways each to roll a 5 and 9 and 6 ways to roll a seven for a total of 14/36 combinations.
I calculate P(5|y) = 1/9 / 7/18 or 1/9 * 18/7 = 18/63 = 2/7
P(9|y) must equal P(5|y) so also 2/7
P(7|y) = 1 - (P(5|y) + P(9|y)) = 3/7
So far so good?
Now I'm not sure if I'm on the right track next or not but . . .
If I imagine this problem as having 5 states:-
a) Having made 0 of the numbers yet.
b) Having rolled a 5 but neither a 9 nor 7.
c) Having rolled a 9 but neither a 5 nor 7.
d) Having succeeded in rolling both a 5 and a 9 without the 7 (won).
e) Having rolled a 7 and failed (lost).
I then made a transition matrix from that data.
a b c d e
a 0 2/7 2/7 0 3/7
b 0 2/7 0 2/7 3/7
c 0 0 2/7 2/7 3/7
d 0 0 0 1 0
e 0 0 0 0 1
(EXCUSE THE FORMATTING)
Since I always start in state 'a' I multiplied an initial distribution vector of
[1 0 0 0 0] by the above matrix raised to the power of 50 to try and find an approximation of the steady state probabilities.
This told me that P(5 and 9 before 7) = 0.2286 (ish)
Anyone make any sense of that and tell me if I got the answer right? If yes, could I have done it in an easier way? If not, what should I have done?
Thanks in advance for your help folks!
romsek November 6th, 2017 08:27 PM
This is how I did it.
Consider sequences of rolls that lead to a win in $n$ rolls.
Note $n \geq 2$
Roll $n$ will either be a $5$ or a $9$
Rolls $1 \text{ through }(n-1)$ will not be roll $n$ and will contain no $7$
Now consider roll sequences of length $n-1$ that satisfy this.
There will be at least $1$, and up to $n-1$ rolls that are either $5$ or $9$ whichever isn't roll $n$, and none of these will be $7$.
Otherwise they can be any roll.
so we end up with
$P[\text{win}] = 2 \sum \limits_{n=2}^\infty \dfrac 1 9\left(\sum \limits_{k=1}^{n-1}\dbinom{n-1}{k}\left(\dfrac 1 9\right)^k\left(\dfrac{11}{18}\right)^{n-1-k}\right)$
Digesting this a bit.
The $2$ in front is because roll $n$ can be either $5$ or $9$, and the probability is identical for both.
The $\dfrac 1 9$ is the probability of roll $n$ being the $5$ (or the $9$)
What's inside the parens is the sum of probabilities that there are $k$ occurrences of $5$ or $9$ whichever isn't roll $n$
This can be simplified to
$P[\text{win}] = 2 \sum\limits_{n=2}^\infty \left(\left(\dfrac{13}{18}\right)^{n-1} - \left(\dfrac{11}{18}\right)^{n-1}\right)$
and then finally to
$P[\text{win}] =\dfrac{8}{35}$
I leave it to you to digest all of this and reproduce the simplifications in more detail.
ps. $p=\dfrac {8}{35}$ matches my sim of this very well so I'm pretty confident in this answer.
Falsetto November 6th, 2017 08:43 PM
THANKYOU!!
Exactly what i was hoping to find. And also the answer i had was correct as well :)
But a formula like yours is so much better. I will study it in detail tomorrow when i wake up.
Thanks for your time and help. Really appreciate it!
Falsetto November 15th, 2017 05:58 PM
Trying to digest what you did I figured out that there is a lot I still don't know.
I have had a look into geometric sequences and that is helping me get started. Are there any other topics you would recommend me learning more about to gain a better grasp of what's going on here? Haha
Also, in playing with the problem I found another way to work it out I think.
Since we only care about 5,9 and 7.
Can I say that P(5 or 9 | 5,9,7) = 8/14 and then the probability of the next one is p(5 | 5,7) = 4/10
8/14 * 4/10 = 32/140 = 8/35 again.
Think this only works is a case like this though where the P(n) is the same for both. How would your formula above change for a problem like rolling 4 and 5 before 7?
romsek November 15th, 2017 09:47 PM
Quote:
Originally Posted by Falsetto (Post 584331) Trying to digest what you did I figured out that there is a lot I still don't know. I have had a look into geometric sequences and that is helping me get started. Are there any other topics you would recommend me learning more about to gain a better grasp of what's going on here? Haha Also, in playing with the problem I found another way to work it out I think. Since we only care about 5,9 and 7. Can I say that P(5 or 9 | 5,9,7) = 8/14 and then the probability of the next one is p(5 | 5,7) = 4/10 8/14 * 4/10 = 32/140 = 8/35 again. Think this only works is a case like this though where the P(n) is the same for both. How would your formula above change for a problem like rolling 4 and 5 before 7?
Your solution is the way to go. Very nicely thought out.
All times are GMT -8. The time now is 11:02 PM. | 2017-12-18T07:02:40 | {
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http://mathhelpforum.com/algebra/86584-help-finding-sum-series.html | # Math Help - help finding sum of series
1. ## help finding sum of series
got 2 series.
1.
12 - 22 + 32 - 42 + ................. 992 - 1002
(ok those are just squares)
2.
1.1! + 2.2! + ....... 50.50!
need to find sum...
2. Originally Posted by adhyeta
got 2 series.
1.
12 - 22 + 32 - 42 + ................. 992 - 1002
(ok those are just squares)
need to find sum...
$1^2-2^2+3^2-4^2+\cdots+99^2-100^2$
$1-4+9-16+25-36+\cdots+99^2-100^2$
finding the difference in pairs now giving 50 terms.
$-3-7-11-\cdots$
Use sum of an arithmetic sequence
$S_n = \frac{n}{2}(2a+(n-1)d)$
where n = number of terms = 50
a = the first term = -3 and d = the common difference between terms = -4
$S_{50} = \frac{50}{2}(2(-3)+(50-1)(-4)) = \cdots$
3. Hello, adhyeta!
Are you familiar with these summation formulas?
. . $\sum^n_{k=1} 1 \;\;= \;n$
. . $\sum^n_{k=1}k \;\;=\;\frac{n(n+1)}{2}$
$1)\;\;S \;=\;1^2 - 2^2 + 3^2 - 4^2 + \hdots + 99^2 - 100^2$
$\text{We have: }\;S \;=\;\underbrace{(1^2 + 3^2 + 5^2 + \hdots + 99^2)}_{\text{first 50 odd squares}} \;-\; \underbrace{(2^2+4^2+6^2 + \hdots + 100^2)}_{\text{first 50 even squares}}$
. . $\text{Then: }S \;=\qquad\qquad \sum^{50}_{k=1}(2k-1)^2 \qquad - \qquad\qquad \sum^{50}_{k=1}(2k)^2$
Hence: . $S \;=\;\sum^{50}_{k=1}\bigg[(2k-1)^2 - (2k)^2\bigg] \;=\;\sum^{50}_{k=1}(1-4k)$
. . $= \;\;\sum^{50}_{k=1}\!1 \;-\; 4\!\sum^{50}_{k=1}k \;\;=\;\;50 - 4\,\frac{50\cdot51}{2} \;\;=\;\;\boxed{-5,\!050}$
4. Originally Posted by pickslides
$1^2-2^2+3^2-4^2+\cdots+99^2-100^2$
$1-4+9-16+25-36+\cdots+99^2-100^2$
finding the difference in pairs now giving 50 terms.
$-3-7-11-\cdots$
Use sum of an arithmetic sequence
$S_n = \frac{n}{2}(2a+(n-1)d)$
where n = number of terms = 50
a = the first term = -3 and d = the common difference between terms = -4
$S_{50} = \frac{50}{2}(2(-3)+(50-1)(-4)) = \cdots$
thats not coming out to be correct...its -5050.
refer to soroban's solution.
5. Originally Posted by adhyeta
thats not coming out to be correct...its -5050.
refer to soroban's solution.
$
S_{50} = \frac{50}{2}(2(-3)+(50-1)(-4))$
$
S_{50} = 25(-6+(49)(-4))$
$
S_{50} = 25(-6+-196)$
$
S_{50} = 25(-202)$
$
S_{50} = -5050$
Worked for me!
6. Originally Posted by adhyeta
got 2 series.
1.
12 - 22 + 32 - 42 + ................. 992 - 1002
(ok those are just squares)
2.
1.1! + 2.2! + ....... 50.50!
need to find sum...
Here's the second one. Let
$S_n = 1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + \cdots + (n-1) \cdot (n-1)! + n \cdot n!$
$= (2-1) 1! + (3-1)2! + (4-1)3! + \cdots + (n -1) \cdot (n-1)! + (n+1 -1) \cdot n!
$
$= (2! - 1!) + (3! - 2!) + (4! - 3!) + \cdots + (n! - (n-1)!) + ((n+1)! - n!)$
$
= (n+1)! - 1
$
so here $n = 50$ so the answer $S = 51!-1$.
7. Originally Posted by pickslides
$
S_{50} = \frac{50}{2}(2(-3)+(50-1)(-4))$
$
S_{50} = 25(-6+(49)(-4))$
$
S_{50} = 25(-6+-196)$
$
S_{50} = 25(-202)$
$
S_{50} = -5050$
Worked for me!
hey! so sorry. i think i went wrong with the calc.(-calculation-)
8. Originally Posted by danny arrigo
Here's the second one. Let
$S_n = 1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + \cdots + (n-1) \cdot (n-1)! + n \cdot n!$
$= (2-1) 1! + (3-1)2! + (4-1)3! + \cdots + (n -1) \cdot (n-1)! + (n+1 -1) \cdot n!
$
$= (2! - 1!) + (3! - 2!) + (4! - 3!) + \cdots + (n! - (n-1)!) + ((n+1)! - n!)$
$
= (n+1)! - 1
$
so here $n = 50$ so the answer $S = 51!-1$.
hey! can we similarily prove the sum $n^{2}.n!$??? | 2015-04-28T01:01:20 | {
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https://math.stackexchange.com/questions/754335/hamiltonian-paths-in-complete-graphs | Hamiltonian Paths in Complete Graphs
A bit of background to help explain the question:
In a class we were given a large spreadsheet of stars and were asked to find two paths, starting from the Sun and visiting every star within 10 parsecs of the Sun. We were asked to do this in two ways. The first was to find the "shortest" path, that is for the last star in the path, find the closest star to it, then find the closest to that star, and so forth. Then we were asked to find an optimal path, that is a path to find the shortest possible path (based on distance) to visit all the stars.
Shifting the idea into graph theory mode, I thought it as trying to find two Hamiltonian paths in a complete, simple, undirected graph. The more I thought about it and tried it out on some small complete graphs, I couldn't find a case where the shortest path wasn't the optimal path.
So now, my question: In complete, weighted, simple graphs, is the "shortest" Hamiltonian path the optimal Hamiltonian path?
PS. While the question comes from a class, its not part of the assignment. This is for my own curiosity.
Suppose you have 7 stars. A, B, and C form a nearly equilateral triangle; AB and BC have length 1, AC is a little longer, $1+\epsilon$. B, D, E, F, and G lie on a straight line, going away from AC; each is $1/2$ a unit away from the next.
Starting at A, the closest is B, then D, E, F, G, and then you have to go back to C. That's total length $3+GC$, which is a little under 6.
But if instead you go ACBDEFG, that's total length $4+\epsilon$. | 2019-06-19T11:15:52 | {
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https://math.stackexchange.com/questions/2782505/number-of-ways-to-choose-mathematicians-and-computer-scientists-for-a-team | # Number of ways to choose mathematicians and computer scientists for a team
There are 4 mathematicians and 7 computer scientists. We need to create a team of experts that has 5 members. At least 2 members should be mathematicians. What is the number of ways we can choose? The solution is 301. Can someone explain why?
My attempt:
4 mathematicians (Ms for short) and 7 computer scientists (CSs for short) can be members. There are $11!/5!(11−5)!=462$ combinations for assigning membership. We need to subtract those combinations with 0 or 1 M. For 0 Ms it's easy: there is only one combination with 0 Ms i.e. the one with 5 CSs. As for those with 1 M: since there are four Ms there are at least 4 different combinations. In other 4 places (one place is taken by M) we need to distribute 7 CSs. $7!/(7−4)!=7!/3!=840$ permutations there are for that. $840$ permutations of CSs for each distinct M is $3360$. There are 4 places for CSs, so to bring $3360$ from permutations to combinations we need to divide by $4!$. $3360/24=140$. Then I concluded that there are $462−1−140=321$ combinations that contain at least 2 Ms. As you can see, I am very close ($321−301$ is only $20$). So... can you detect where am I wrong?
• Choose $5$ members (combinations). Make sure you do not have exactly $0$ or $1$ mathematicians (subtractions of other combinations) – Henry May 15 '18 at 17:05
• I get $251$ as the solution. There are $462$ combinations in total, one of them contains only computer scientists, and $210$ ($7!/4!$) contains 1 mathematician. $462-1-210=252$. – Hanlon May 15 '18 at 17:14
• $7+5=12$ so there are ${12 \choose 5}$ not ${11 \choose 5}$ combinations in total. More than one of them only involves computer scientists ... – Henry May 15 '18 at 17:25
• I meant 4 not 5 mathematicians. Sorry. And how can there be more than one combination that involves only computer scientists? – Hanlon May 15 '18 at 17:27
• Actually, it should be $426$. We place 7 computer scientists in 4 boxes, we have $7!/(7-4)!$ permutations, which is $840$, and then we divide by $4!$ because there are 4 boxes and we get $35$ combinations. $462-1-35=426$ – Hanlon May 15 '18 at 18:27
There are $4 + 7 = 11$ people available to serve on the team. Five of them can be selected in $\binom{11}{5}$ ways. From these, we must subtract those cases in which there are fewer than two mathematicians. Since there are four mathematicians and seven computer scientists, the number of ways of selecting exactly $k$ mathematicians and $5 - k$ computer scientists is $$\binom{4}{k}\binom{7}{5 - k}$$ The number of ways of selecting no mathematicians is thus $$\binom{4}{0}\binom{7}{5}$$ and the number of ways of selecting exactly one mathematician is $$\binom{4}{1}\binom{7}{4}$$ Hence, the number of teams with five members that contain at least two mathematicians is $$\binom{11}{5} - \binom{4}{0}\binom{7}{5} - \binom{4}{1}\binom{7}{4} = 462 - 21 - 140 = 301$$ As you can see, you made a mistake when you concluded that only one team could have no mathematicians. You failed to account for the fact that we had to select which five of the seven computer scientists would serve on the team.
• Oh... yeah, you are correct. – Hanlon May 15 '18 at 20:23
We have $4$ mathematicians in stock from which have to choose $2$, or $3$, or $4$. So, the remaining $3$, or $2$, or $1$ members are computer scientist which are $7$ in stock.So, the no of group is ${4\choose 2}{7\choose 3}+{4\choose 3}{7\choose 2}+{4\choose 4}{7\choose 1}=301$
We have cases of having 2,3,4,5 mathematicians and remaining scientists
Number of cases$=^5C_2^7C_3+^5C_3^7C_2+^5C_4^7C_1+^5C_5^7C_0=301$
• I haven't checked value you verify it – Abhishek May 15 '18 at 17:08
• Please check ur calculation...there are $4$ mathematicians in stock – Supriyo Halder May 15 '18 at 18:15
• You can obtain $\binom{n}{k}$ by typing \binom{n}{k} when you are in math mode. Doing so would make your answer easier to read. – N. F. Taussig May 15 '18 at 18:15
• @N.F.Taussig Ok i will see to it – Abhishek May 15 '18 at 18:17
• Also, @SupriyoHalder is correct. There are only four mathematicians in the (restated) problem. While the expression $\binom{5}{2}\binom{7}{3} + \binom{5}{3}\binom{7}{2} + \binom{5}{4}\binom{7}{1} + \binom{5}{5}\binom{7}{0}$ would be correct if there were five mathematicians, it is not equal to $301$, as you can easily check. – N. F. Taussig May 15 '18 at 18:27 | 2019-09-23T13:58:30 | {
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https://math.stackexchange.com/questions/1376498/remainder-of-a23a4-divided-by-7 | # remainder of $a^2+3a+4$ divided by 7
If the remainder of $$a$$ is divided by $$7$$ is $$6$$, find the remainder when $$a^2+3a+4$$ is divided by 7
(A)$$2$$ (B)$$3$$ (C)$$4$$ (D)$$5$$ (E)$$6$$
if $$a = 6$$, then $$6^2 + 3(6) + 4 = 58$$, and $$a^2+3a+4 \equiv 2 \pmod 7$$
if $$a = 13$$, then $$13^2 + 3(13) + 4 = 212$$, and $$a^2+3a+4 \equiv 2 \pmod 7$$
thus, we can say that any number, $$a$$ that divided by 7 has remainder of 6, the remainder of $$a^2 + 3a + 4$$ is 2.
is there any other way to calculate it? (Let say it given b as the remainder of a divided by 7, not 6)
• Did you mean to write "$a^2+3a+4 \equiv 2\pmod{7}$" instead of "$a \equiv 2\pmod{7}$"? Jul 28 '15 at 7:22
• @JimmyK4542 yeah, was a little bit rush, so mistakes happen. Alot.. Jul 28 '15 at 7:23
$a = 6 \quad(\mathrm{mod} 7)$
$a^2 = 36 = 1 \quad(\mathrm{mod} 7)$
$3a = 18 = 4\quad (\mathrm{mod} 7)$
$a^2 + 3a + 4 = 1 + 4 + 4 = 9 = 2 \quad(\mathrm{mod} 7)$
If the remainder when $a$ divided by $7$ is $b$, then $a = 7n+b$ for some integer $n$.
Hence, $a^2+3a+4 = (7n+b)^2+3(7n+b)+4$ $= 49n^2 + 14nb + b^2 + 21n + 3b + 4$ $= 7(7n^2+2nb+3n) + (b^2+3b+4)$.
So, the remainder when $a^2+3a+4$ is divided by $7$ will be the same as the remainder when $b^2+3b+4$ is divided by $7$.
For the specific case when $b = 6$, we get that $a^2+3a+4 = 7(7n^2+12n+3n)+58$ $= 7(7n^2+12n+3n+8)+2$.
So the remainder when $a^2+3a+4$ is divided by $7$ is $2$.
The remainder of $a^2+3a+4$ divided by $7$ is sum of the remainder of each terms, modulo $7$.
So $a^2\equiv 1 \pmod{7}$ since $a=7k+6$ then $a^2=7l+1$; $\quad$ $3a\equiv 4 \pmod{7}$ since $3a=21k+18=21k+14+4$ and clearly $4\equiv 4 \pmod{7}$.
Finally $1+4+4 \equiv 2 \pmod{7}$ then the remainder is $2$.
• To obtain $a \equiv b \pmod{n}$, type a \equiv b \pmod{n} in math mode. Jul 28 '15 at 13:43
$a^2 + 3a + 4 \equiv a^2 - 4a + 4 \equiv (a-2)^2 \pmod 7$
If $a\equiv b \pmod 7$, then $a^2 + 3a + 4 \equiv (b-2)^2 \pmod 7$ | 2022-01-17T07:30:49 | {
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https://stats.stackexchange.com/questions/74755/flaw-in-a-conditional-probability-argument | # Flaw in a conditional probability argument
Imagine an experiment where you roll two fair, six-sided dice. Someone peeks at the dice, and (truthfully) tells you that "at least one of the dice is a 4". What is the probability that the total of the dice is 7?
It seems straightforward to calculate that the probability the total is 7 is 2/11.
However, the person who peeked at the dice could equally well have said "at least one of the dice is a 1" and you would come to the same conclusion - 2/11. Or they could have said "at least one of the dice is a 2" or "at least one of the dice is a 3", or indeed any number from 1 to 6, and you would still conclude that the probability that the total is 7 is 2/11.
Since you will always conclude that the probability that the total is 7 is 2/11, you could block your ears as they speak, and you'd still come up with 2/11. From there it's a short hop to conclude that even if they don't say anything the probability that the total is 7 is 2/11.
However, clearly if they don't say anything, the probability that the total is 7 is not 2/11, but rather 1/6.
Where is the flaw in the argument?
• You can simplify this question to asking about two flips of a coin, having the person say at least one of the flips is heads/tails and noting that the probability of the flips being different seems to rise from one half to two thirds. – Neil G Nov 6 '13 at 15:30
• One possible, although incomplete answer is that the conditional distribution over the total indeed changes in light of different evidence. The number 7 is the only magical number which probability will be the same given any evidence. – hr0nix Nov 6 '13 at 17:47
## 2 Answers
The flaw in the argument is that the conditioning random variable is not well-defined.
The ambiguity lies in how our friend peeking at the dice decides to report the information back to us. If we let $X_1$ and $X_2$ denote the random variables associated with the values of each of the dice, then it is certainly true that for each $k \in \{1,2,\ldots,6\}$, $$\mathbb P(X_1 + X_2 = 7 \mid X_1 = k \cup X_2 = k) = \frac{2}{11} \>,$$ independently of $k$.
However, the events $\{X_1 = k \cup X_2 = k\}$ are clearly not mutually exclusive, and so clearly we cannot claim \begin{align} \mathbb P(X_1 + X_2 = 7) &\stackrel{?}{=} \sum_{k=1}^6 \mathbb P(X_1 + X_2 = 7 \mid X_1 = k \cup X_2 = k) \mathbb P( X_1 = k \cup X_2 = k ) \cr &\stackrel{?}{=} \frac{2}{11} \sum_{k=1}^6 \mathbb P( X_1 = k \cup X_2 = k ) \cr &\stackrel{?}{=} \frac{2}{11} \end{align}
Formally, we need to properly define a random variable, say $K$, that encodes the knowledge imparted by our peeking friend.
Our peeking friend could always report the value of the left-most die, or the right-most, or the larger of the two. She could flip a coin and then report based on the coin flip, or employ any number of more complicated machinations.
But, once this process is specified, the apparent paradox vanishes.
Indeed, suppose that $K = X_1$. Then, we have \begin{align} \mathbb P(X_1 + X_2 = 7) &= \sum_{k=1}^6 \mathbb P(X_1+X_2 = 7, K=k) \cr &= \sum_{k=1}^6 \mathbb P(X_1+X_2 = 7 \mid K=k) \mathbb P(K=k) \cr &= \sum_{k=1}^6 \frac{1}{36} = \frac{1}{6} \>. \end{align}
Similar arguments hold if we choose $K = X_2$ or $K = \max(X_1,X_2)$, etc.
• +1 Similar reasoning provides insight into the notorious Monty Hall problem. – whuber Nov 6 '13 at 20:02
If $B$ is an event with the property that $P(B\mid D_i) = p$ for all events $\{D_1, D_2, \ldots\}$ in a countable partition of the sample space $\Omega$, (that is, $D_i \cap D_j = \emptyset$ for all $i \neq j$ and $\bigcup_i D_i = \Omega$), then the law of total probability tells us that $$P(B) = \sum_i P(B\mid D_i)P(D_i) = p\sum_i P(D_i) = p.$$ However, the law of total probability does not apply if the events $D_i$ are not mutually exclusive (even though their union is still $\Omega$), and we cannot conclude that $P(B)$ equals the common value of $P(B\mid D_i)$.
Let $A_i$ denote the event that at least one of the dice shows the number $i$ and $B$ the event that the sum of the two numbers on the die is $7$. We know that $P(B) = \frac{1}{6}$ and that $P(A_i) = \frac{11}{36}$. Also, $P(B\mid A_i) = \frac{2}{11}$. Now, $A_1\cup A_2\cup A_3 \cup A_4\cup A_5\cup A_6$ is the entire sample space $\Omega$ but we cannot use the fact that $P(B\mid A_i)$ is the same for all choices of $i$ to conclude that $P(B) = \frac{2}{11}$ because the $A_i$ are not mutually exclusive events. However, notice that regarded as a multiset, $A_1\cup A_2\cup A_3 \cup A_4\cup A_5\cup A_6$ contains each outcome $(i,j)$ exactly twice, once as a member of $A_i$ and again as a member of $A_j$. Therefore, $$\sum_{i=1}^6 P(B \mid A_i)P(A_i) = \sum_{i=1}^6 \frac{2}{11}\times\frac{11}{36} = \frac{1}{3}$$ which is exactly twice the value of $P(B)$. | 2021-04-18T23:38:26 | {
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http://mathhelpforum.com/calculus/281769-calculus-quiz-problems.html | # Thread: Calculus Quiz Problems
1. ## Calculus Quiz Problems
Determine the area of the region enclosed by the graphs of f(x) = x^2 + x +5, g(x) = 3x +13, and the vertical line x = 0
I got 26.67 but he gave me zero points for it so idk.
Set up (but do not evaluate) an integral that will give the area of the surface generated when y= x^3/8 is revolved about the y-axis for 3 less than or equal to x, x less than or equal to 4.
2. ## Re: Calculus Quiz Problems
I would begin by plotting the region whose area we are to find:
We see the linear function is above the quadratic, and we know the lower limit of integration will be $\displaystyle x=0$. We need to determine where the upper limit is, and that occurs where the two functions are equal for $\displaystyle 0<x$.
$\displaystyle x^2+x+5=3x+13$
$\displaystyle x^2-2x-8=0$
$\displaystyle (x+2)(x-4)=0$
The positive root is:
$\displaystyle x=4$
And to the request area $\displaystyle A$ will be found from:
$\displaystyle A=\int_0^4 (3x+13)-(x^2+x+5)\,dx=\int_0^4 -x^2+2x+8\,dx=\left[-\frac{x^3}{3}+x^2+8x\right]_0^4=-\frac{64}{3}+16+32=\frac{80}{3}$
What you cited as the area appears to be a decimal approximation of this same value, rounded to two decimal places, so perhaps that's why you got no credit for it. | 2018-12-16T12:31:52 | {
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https://www.physicsforums.com/threads/angular-acceleration-problem.69155/ | # Angular acceleration problem
1. Mar 29, 2005
### Punchlinegirl
M, a solid cylinder (M=1.99 kg, R=0.133 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force of F which equals the weight of a 0.830 kg mass, i.e, F=8.142 N. Calculate the angular acceleration of the cylinder.
I tried drawing a free body diagram and setting the forces equal to ma.
F_n- mg-F_t= Ma_y
since a= 0, F_n= Mg + F_t
torque= RF_t sin 90= Ia
RF_t=Ia
I=(1/2)MR^2
F_t= (1/2)MRa
mg-ma=(1/2)MR(a/R)
(1/2M +m)a= mg
a=mg/ (1/2 M=m)
I plugged in my numbers, but I think the equation is wrong. Help?
2. Mar 29, 2005
### xanthym
From the problem statement:
{String Tension} = S
{Mass of Suspended Entity} = m = (0.830 kg)
{Weight of Mass} = W = (0.830 kg)*(9.81 m/sec^2) = (8.1423 N)
{Cylinder Mass} = M = (1.99 kg)
{Cylinder Radius} = R = (0.133 m)
{Cylinder Moment of Inertia} = I = (1/2)*M*R^2
{Cylinder Angular Acceleration} = α
For the suspended entity:
{Net Force} = ma =
= W - S
::: ⇒ S = W - ma ::: Eq #1
For the cylinder:
{Net Torque} = Iα = I*a/R =
= S*R
::: ⇒ S = I*a/R^2 ::: Eq #2
Equating Eq #1 and Eq #2:
W - ma = I*a/R^2
::: ⇒ a = W/{m + I/R^2}
::: ⇒ a = W/{m + (1/2)*M*R^2/R^2}
::: ⇒ a = W/{m + (1/2)*M}
::: ⇒ a = (8.1423 N)/{(0.830 kg) + (1/2)*(1.99 kg)}
::: ⇒ a = (4.46153 m/sec^2)
::: ⇒ α = a/R = (4.46153 m/sec^2)/(0.133 m) = (33.545 radians/sec^2)
~~
3. Mar 29, 2005
### Staff: Mentor
Your first error is in assuming that the string exerts a force equal to the weight of the hanging mass. (If the force were equal, then the mass would be in equilibrium.) But it looks like you don't use this fact later on.
For the cylinder, the only forces you need consider are those that create a torque. Thus the normal force (supporting the cylinder) and the weight of the cylinder are irrelevant; all that counts is the force F that the string exerts.
Applying Newton's 2nd law:
To the cylinder:
$\tau = I\alpha$ --> $F R = I\alpha$
To the hanging mass:
$mg - F = ma$
So far, so good. (a is really alpha)
All good. I assume you meant to write: $a = mg/(M/2 + m)$.
Did you remember to convert to angular acceleration?
4. Mar 30, 2005
### Punchlinegirl
ok, i used the equation and tried converting to angular acceleration by using
a=alpha* r
and got the angular acceleration to be 3.87 rads/s^2, which isn't right...
5. Mar 30, 2005
### whozum
Alpha is the angular acceleration, you are using your equation backwards :). Solve it for alpha and try it that way.
6. Apr 7, 2005
### trisha320
i have a similar problem to punchlinegirl, and i was wondering how do i find alpha from angular acceleration
7. Apr 7, 2005
### xanthym
Angular acceleration IS "α". You might be thinking of the following relationships:
{Angular Acceleration} = α =
= {Linear Acceleration}/R =
= a/R
{Angular Velocity} = ω =
= {Linear Velocity}/R =
= v/R
Also:
{Angular Velocity} = ω =
= 2*π*{Rotation Frequency} =
= 2*π*f
~~
8. Apr 7, 2005
### kishin7
M=1.99 kg, R=0.133 m, F=8.142 N
I = .5MR^2, T(1) = r x F and T(2) = I x alpha.
Solve for I and then solve T(1) using F given and then once you have both plug into T(2).
Last edited: Apr 7, 2005
9. Apr 10, 2005
### Felix83
The force is equal to the weight of a hanging mass. This doesnt mean that there is actually a mass hanging off of it, it means there is a constant force applied of 8.142N.
Once you see that the problem is simple.
You have the equation T=I*alpha
alpha is angular acceleration which is what you need so just solve for alpha.
alpha=T/I where T is the torque applied and I is the moment of inertia.
T = F*r (when torque is perpendicular)
T = (8.142N) (0.133m)
T= 1.083Nm
Now find I, since it is a solid cylinder
I=(1/2)Mr^2
I = (1/2) (1.99kg) (0.133m)^2
I = 0.0176 kg*m^2
now recall
alpha = T/I
alpha = (1.083Nm)/(.0176kg*m^2) | 2017-11-23T08:21:46 | {
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https://tamaskis.github.io/Numerical_Differentiation_Toolbox-MATLAB/cpartial_doc.html | # cpartial
Partial derivative of a multivariate, vector-valued function using the central difference approximation.
## Syntax
pf = cpartial(f,x0,k)
pf = cpartial(f,x0,k,h)
## Description
pf = cpartial(f,x0,k) numerically evaluates the partial derivative of with respect to at using the central difference approximation with a default relative step size of , where is the machine zero.
pf = cpartial(f,x0,k,h) numerically evaluates the partial derivative of with respect to at using the central difference approximation with a user-specified relative step size .
## Input/Output Parameters
Variable Symbol Description Format Input f $\inline&space;\mathbf{f}(\mathbf{x})$ multivariate, vector-valued function ($\inline&space;\mathbf{f}:\mathbb{R}^{n}\rightarrow\mathbb{R}^{m}$) 1×1function_handle x0 $\inline&space;\mathbf{x}_{0}$ evaluation point n×1double k $\inline&space;k$ element of $\inline&space;\mathbf{x}$ to differentiate with respect to 1×1double h $\inline&space;h$ (OPTIONAL) relative step size 1×1double Output pf $\inline&space;\dfrac{\partial\mathbf{f}}{\partial x_{k}}\bigg\rvert_{\mathbf{x}=\mathbf{x}_{0}}$ partial derivative of $\inline&space;\mathbf{f}$ with respect to $\inline&space;x_{k}$, evaluated at $\inline&space;\mathbf{x}=\mathbf{x}_{0}$ m×1double
## Note
• This function requires 2 evaluations of .
• If the function is scalar-valued, then .
## Example #1: Partial derivative of a scalar-valued function.
Approximate the partial derivative of with respect to at at using the cpartial function, and compare the result to the true result of
First, we rewrite this function as .
f = @(x) x(1)^3*sin(x(2));
Since the second component of represents , to approximate the derivative, we use
k = 2;
Approximating the partial derivative using the cpartial function,
pf = cpartial(f,[5;1],k)
pf =
67.5378
Calculating the error,
error = pf-5^3*cos(1)
error =
-2.2063e-09
## Example #2: Partial derivative of a vector-valued function.
Approximate the partial derivative of
with respect to at using the cpartial function, and compare the result to the true result of
Defining the function in MATLAB,
f = @(x) [sin(x(1))*sin(x(2));cos(x(1))*cos(x(2))];
Approximating the partial derivative using the cpartial function,
x0 = [1;2]; % evaluation point
k = 1; % element of x to differentiate with respect to
pf = cpartial(f,x0,k) % differentiation
pf =
0.4913
0.3502
Calculating the error,
error = pf-[cos(1)*sin(2);-sin(1)*cos(2)]
error =
1.0e-11 *
-0.9040
-0.8263 | 2022-10-05T05:41:17 | {
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http://math.stackexchange.com/questions/264512/how-to-prove-left-fracn3-rightn-leq-frac13n | # how to prove $\left(\frac{n}{3}\right)^n\leq\frac{1}{3}n!$
i am asked to prove this statement:
$$\left(\frac{n}{3}\right)^n\leq\frac{1}{3}n!$$
Now after several attempts, i am lost not knowing where and how to start. if I use induction, i am stuck on the way. how can i solve this in an easy way? i have huge difficulties when i have to find something which is bigger than the latter term or so, because i lack some important source of maths in my brain. i am very likely to give up on the way if it becomes tough. for any guidance how to overcome this phase i will be very very thankful. i am trying for proficiency in math. Thanks
-
$\left(\frac{1}{3}\right)^n \leq \frac{1}{3}$ – Babak S. Dec 24 '12 at 9:56
We would like to prove
$$\left(\frac{n}{3}\right)^n \leq \frac{1}{3}n!.$$
This is equivalent to
$$n^{n-1} \leq 3^{n-1}(n-1)!.$$
It is obviosly true for $n = 1$. However, as $n$ increases, left side rises slower that the right-hand side (thus the inequality will still hold). To put this formally, consider the ratio
$$\frac{(n+1)^{n}}{(n)^{n-1}} \leq \frac{3^{n}n!}{3^{n-1}(n-1)!} = 3n$$
which simplifies to
$$\left(\frac{n+1}{n}\right)^{n} \leq 3.$$
and is true because $\left(1+\frac1n\right)^n$ is increasing in $n$ (e.g. see this post) and converges to $e < 3$.
Cheers!
-
wow, amazing power of maths!! thanks a lot – doniyor Dec 24 '12 at 13:51
@doniyor Observe that you could easily substitute 3 with $e$. Moreover with similar technique you could derive the inverse bound $\frac{1}{2}n! \leq \left(\frac{n}{2}\right)^n$. Check also the Wikipedia. – dtldarek Dec 24 '12 at 14:57
$$\left(\frac{n}{3}\right)^n\leq\frac{n!}{3}\Longleftrightarrow X_n:=\frac{n^n}{3^{n-1}n!}\leq 1$$
But
$$\frac{X_{n+1}}{X_n}=\frac{(n+1)^{n+1}}{3^n(n+1)!}\cdot\frac{3^{n-1}n!}{n^n}=\left(1+\frac{1}{n}\right)^n\frac{1}{3}\xrightarrow [n\to\infty]{}\frac{e}{3}<1$$
So that the infinite positive series $\,\displaystyle{\sum_{n=1}^\infty X_n}\,$ converges and thus $\,X_n\xrightarrow[n\to\infty]{}0\Longrightarrow X_n<1\,$ , at least
from a certain index $\,n\,$ on.
-
If we consider the last term in the Taylor expansion of $e^x$, we have $$e^x\ge\frac{x^n}{n!}$$ and let $x=n$ that yields $$e^n\ge\frac{n^n}{n!}$$ and the conclusion follows.
-
This is a nice way. But the question has $3^{n-1}$ – Amr Dec 24 '12 at 10:27
Unless you meant that for sufficently large $n$, we have $e^n<3^{n-1}$ .Nice answer +1 – Amr Dec 24 '12 at 10:29
@Amr: this is correct. Thanks. – Chris's sis Dec 24 '12 at 10:37
Using logarithms, the inequality $\left( \frac{n}{3} \right)^n \leq \frac{1}{3} n!$ is equivalent to $\sum\limits_{k=1}^{n-1} \ln \left( \frac{3k}{n} \right) \geq 0$. But $$\sum\limits_{k=1}^{n-1} \ln \left( \frac{3k}{n} \right) \geq 1+ \int_1^{n-1} \ln \left( \frac{3x}{n} \right) dx = 1+ (n-1) \ln \left( \frac{3(n-1)}{ne} \right) >0$$ for $n >5$.
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First we'll prove that
$(\frac{k+1}{3})^k\leq k!\Leftrightarrow (k+1)^k\leq 3^k k!$
base of the induction: for $k=0$ it is true
If it's true for $k$ we'll prove it is true for $k+1$: that is $3^{k+1}(k+1)!\geq (k+2)^{k+1}$.
$3^{k+1}(k+1)!=3(k+1)3^{k}k!\geq 3(k+1)(k+1)^k=3(k+1)^{k+1}\geq (k+2)^{k+1}$.
Since $3(k+1)^{k+1}\geq (k+2)^{k+1}\Leftrightarrow log 3(k+1)^{k+1}\geq log (k+2)^{k+1} \Leftrightarrow (k+1)log 3(k+1)\geq (k+1) log(k+2)\Leftrightarrow log\frac{3k+3}{k+2}\geq 0$.
which is true
For the main conclusion now:
For $n=0$ it is true.
suppose it is true for $n=k$
$(\frac{k}{3})^k\leq \frac{1}{3}k!$
we'll prove it is true for $n=k+1$
$(\frac{k+1}{3})^{k+1}=(\frac{k+1}{3})^k\cdot\frac{k+1}{3}\leq k!\cdot \frac{1}{3} (k+1) =\frac{1}{3}(k+1)!$
By induction it is true for all $n\in\mathbb{N}$.
-
$(\frac{k+1}{3})^k\cdot\frac{k+1}{3}\leq (\frac{k}{3})^k\cdot\frac{k+1}{3}$? Am I missing something? – Nameless Dec 24 '12 at 10:11
I think all the details are ok now... I really needed that cup of coffee – epsilon Dec 24 '12 at 11:21 | 2014-08-30T22:39:54 | {
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http://math.stackexchange.com/questions/112043/continuous-functions-from-mathbbr-to-mathbbq | # Continuous Functions from $\mathbb{R}$ to $\mathbb{Q}$
The following is not a homework problem. I am doing it for self study.
Prove that any continuous function from $\mathbb{R}$ to $\mathbb{Q}$ is constant.
Here is my proof:
Let $f:\mathbb{R}\rightarrow \mathbb{Q}$ be such a function. We first show that $\mathbb{Q}$ is disconnected. Let $p$ be an irrational number. Then we can write $\mathbb{Q}$ as $(-\infty,p)\cup (p,\infty)$. $\mathbb{Q}$ is also totally disconnected, as any open subset of $\mathbb{Q}$ must contain two rational numbers and there is always an irrational number between the two which we can use to create a disconnection. Therefore the only connected subsets of $\mathbb{Q}$ are the singleton sets $\{q\}$, where $q \in \mathbb{Q}$, which are closed.
We will show that $f(\mathbb{R})$ must be connected. Assume not, then $f(\mathbb{R})=U\cup V$, where $U$ and $V$ are nonempty open subsets of $\mathbb{Q}$ such that $U \cap V= \emptyset$. This implies \begin{align*} \mathbb{R}&=f^{-1}(U \cup V)\\ &=f^{-1}(U)\cup f^{-1}(V), \end{align*} where $f^{-1}(U),f^{-1}(V)$ are nonempty, nonintersecting subsets of $\mathbb{R}$. This, however, is a contradiction, as $\mathbb{R}$ is connected. Therefore no such set $U$ and $V$ can exist. As we have already shown, the only connected subsets of $\mathbb{Q}$ are the singleton sets $\{q\}$, so $f(\mathbb{R})$ must be such a set.
Something about this does not seem quite satisfactory, as if I am missing something. Could anyone tell me a flaw in my logic? Also, is there a more satisfactory way to prove this theorem?
-
The proof is fine. – Michael Greinecker Feb 22 '12 at 14:49
It looks good to me. Note your proof can be used to show the more general fact that the image of a connected set under a continuous function is connected. For functions from $\mathbb{R}$ to $\mathbb{R}$, this is essentially the intermediate value theorem. – Nate Eldredge Feb 22 '12 at 14:50
The only thing I'd change is that $U$ and $V$ need to be open subsets of $f(\mathbb R)$, not open subsets of $\mathbb Q$. The open subsets of $f(\mathbb R)$ are necessarily intersections of $f(\mathbb R)$ with open subsets of $\mathbb Q$, but the condition that they have empty intersection means that you want $U,V$ open in $\mathbb Q$ such that $f(\mathbb R)\subset U\cup V$ and $f(\mathbb R)\cap U\cap V=\emptyset$ – Thomas Andrews Feb 22 '12 at 15:38
Alternatively, you could define $i:\mathbb Q\rightarrow \mathbb R$ to be the natural inclusion and then show that $i$ is continuous, and that therefore $i\circ f:\mathbb R\rightarrow \mathbb R$ is continuous. Then use the intermediate value theorem to show that if $i\circ f$ isn't constant, there must be an irrational value in its range, which is impossible since the range if $i\circ f$ is contained in the rationals in $\mathbb R$. – Thomas Andrews Feb 22 '12 at 16:03
Nobody should every write "Prove that any continuous function from whatever to whatever is constant." That could be understood as "Pick any continuous function from whatever to whatever and prove that it's constant", but what is more likely intended is "Prove that every continuous function from whatever to whatever is constant." Merely changing "any" to "every" eliminates all ambiguity. – Michael Hardy Feb 22 '12 at 17:04
Your argument is correct. Perhaps what is unsatisfying is that you're actually making an argument which applies much more generally. You seem to be proving the following theorem:
Suppose $X,Y$ are topological spaces and $f:X\to Y$ is a continuous map. If $X$ is connected, then $f(X)$ is connected.
If you are comfortable simply quoting this, then you're done once you've shown that $\mathbb{Q}$ is totally disconnected and $\mathbb{R}$ is connected: the image of $\mathbb{R}$ under any continuous map must be connected and the only connected subspaces of $\mathbb{Q}$ are singletons, so the image of $\mathbb{R}$ under any continuous map is a singleton.
If not, simply modify your argument to prove the theorem I stated. Your proof is correct: a disconnecting pair of open sets in the range will pull back to a disconnecting pair of open sets in the domain, giving a contradiction. The only modification you need is removing reference to $\mathbb{R}$ and $\mathbb{Q}$ and replacing them with general topological spaces.
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The topology on $\mathbb Q$ is the one induced on $\mathbb Q$ as a subset of $\mathbb R$. Thus, for every open set $U_{\mathbb R}$ of $\mathbb R$ the set $U_{\mathbb Q}=U_{\mathbb R}\cap \mathbb Q$ is open in $\mathbb Q$. If $f:\mathbb{R}\rightarrow \mathbb{Q}$ is continuous, the preimage $f^{-1}(U_{\mathbb Q})$ of every open set $U_{\mathbb Q}$ of $\mathbb Q$ is open in $\mathbb R$. But then we can consider $f$ as a function from $\mathbb R$ to $\mathbb R$, and by the above the preimage of every open set will still be open, so this is a continuous function from $\mathbb R$ to $\mathbb R$ that takes only rational values. Since there are irrational numbers between any two rational numbers, such a function must be constant by the intermediate value theorem.
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Suppose you know that
• Every continuous function from $\mathbb{R}$ to $\mathbb{Q}$ is a continuous function from $\mathbb{R}$ to $\mathbb{R}$.
• The intermediate value theorem holds.
• The irrationals are dense in $\mathbb{R}$.
Then a nonconstant continuous function would have to pass through the irrational numbers between any two of its values.
So then I suppose I would ask which of the three bulleted points were already established and which to prove as lemmas.
Nobody should ever write "Prove that any continuous function from whatever to whatever is constant." That could be understood as "Pick any continuous function from whatever to whatever and prove that it's constant", but what is more likely intended is "Prove that every continuous function from whatever to whatever is constant." Merely changing "any" to "every" eliminates all ambiguity.
- | 2015-02-01T23:18:37 | {
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https://math.stackexchange.com/questions/1581737/finding-the-jordan-basis-and-the-canonical-form-corresponding-to-the-jordan-basi | # Finding the Jordan Basis and the Canonical form corresponding to the Jordan basis
\begin{pmatrix} 4 & 0 &0 \\ 2 &1 &3 \\ 5& 0 &4 \end{pmatrix} I know that the Characteristic polynomial is : $$(t-4)^2(t-1)$$ I started with eigenvalues $λ=1$ and got in the null space: \begin{pmatrix} 0\\ 1 \\ 0 \end{pmatrix}
$λ=4$ and got :
\begin{pmatrix} 0\\ -1 \\ 1 \end{pmatrix},
in $A- 4I$ and got
\begin{pmatrix} -1\\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 1\\ 0 \\ 1 \end{pmatrix},
in $(A- 4I)^2$
Which vector(s) do I choose correctly for the Jordan Basis?
I chose $[0,1,0],[1,0,1]$ and got $b_3$ as $A-4I[b_2] = [0,5,5]$
What procedures should I use for choosing the correct Jordan basis vectors? Secondly am I right?
Lastly, How do I get the Canonical form "corresponding to the basis"?
is it \begin{pmatrix} 1 & 0 &0 \\ 0 &4 &1 \\ 0& 0 &4 \end{pmatrix} or \begin{pmatrix} 4 & 1 &0 \\ 0 &4 &0 \\ 0& 0 &1 \end{pmatrix} How do I set it up correctly?
• Thats only for the canonical form is it not? – GuestCalc Dec 19 '15 at 2:06
• even better wolframalpha.com/input/… to guide oneself – janmarqz Dec 19 '15 at 2:09
You have the right idea, but you made a slight computational error: the $4$-eigenspace of $A$ is spanned by $(0,1,1)$, not $(0,-1,1)$. You correctly picked a vector $(1,0,1) \in \ker (A-4I)^2$ that doesn't belong to $\ker(A-4I)$, which let you construct the Jordan chain $((1,0,1),(A-4I)(1,0,1)) = ((1,0,1),(0,5,5))$. (This now works because $(0,5,5)$ is an eigenvector for the eigenvalue $4$.)
Then you just place the Jordan chains as columns of the change of basis matrix (reading a chain backwards if you want $1$s on the superdiagonal):
$$P = \begin{pmatrix} 0 & 0&1\\ 1&5&0 \\ 0&5&1 \end{pmatrix}.$$
And then we get
$$P^{-1}AP = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 4 & 1 \\ 0 & 0 & 4 \end{pmatrix}.$$
$$P = \begin{pmatrix} 0&1&0\\ 5&0&1 \\ 5&1&0 \end{pmatrix},$$
$$P^{-1}AP = \begin{pmatrix} 4 & 1 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \end{pmatrix}.$$
• for $(A-4I)^2$ both vectors were not in the null space of A-4I, would either work? – GuestCalc Dec 19 '15 at 2:13
• @GuestCalc The other vector $(-1,1,0)$ would also work; you would get the chain $((-1,1,0),(0,-5,-5))$. – Alex Provost Dec 19 '15 at 2:17
• @GuestCalc The ordering of the blocks is arbitrary and depends on the ordering you chose when placing the Jordan chains as columns of $P$. – Alex Provost Dec 19 '15 at 2:17 | 2021-06-23T19:12:26 | {
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https://math.stackexchange.com/questions/3361835/are-linear-transformations-precisely-those-that-keep-lines-straight-and-the-orig | # Are linear transformations precisely those that keep lines straight and the origin fixed?
It's easy to show that given a linear transformation $$T:\mathbb{R}^n \rightarrow \mathbb{R}^m$$ lines are mapped to lines and the origin stays fixed (as long as its rank $$=n$$).
Yet is the converse true?
More precisely, if $$T:\mathbb{R}^n \rightarrow \mathbb{R}^m$$ is a function that maps lines to lines in the sense that for any pair of vectors $$a, b$$ there exists vectors $$c, d$$ such that $$T(a+tb)=c+td$$ & $$T(0)=0$$ can we deduce that $$T(x+y)=T(x)+T(y)$$ for all vectors $$x, y$$?
Would appreciate any help.
• I think it probably doesn't even have to be continuous to do this, but it's just a 2:37 am hunch. – Matt Samuel Sep 19 '19 at 6:37
• There are at least two things you might mean by "maps lines to lines": you might want $T$ to restrict to a linear map from one line to the other (which looks like what you wrote but ideally you'd write down some quantifiers) or you might be happy for $T$ to send one line to the other regarded just as sets. – Qiaochu Yuan Sep 19 '19 at 7:32
• If you replace "lines" with "subspaces of dimension n - 1", and add that equally separated "lines" remain equally separated, then the answer is yes. Otherwise, as noted in the answer, it's easy to come up with counterexamples for R -> R. – Denziloe Sep 29 '19 at 0:32
Edit: We have to clarify: Do you actually mean "Lines are mapped to lines", i.e. the image of a line under $$T$$ is again a line (what I assumed), or do you mean actually mean that $$T(a + tb) = c + td$$ for all $$t\in [0,1]$$?
Not if $$m=1$$!
Take for example
$$T: \mathbb{R}^n \rightarrow \mathbb{R}$$
$$T(x) = 2x_1$$ if $$x_1>0$$
$$T(x) = x_1$$ if $$x_1\leq0$$
It projects the line to one dimension and stretches the line on the right half plane, but not on the left half plane. Its non linear around 0, but will still always project lines to lines.
From "(as long as its rank $$=n$$)" I assume you'd add the condition that $$T$$ has to have full rank, and than we could assume $$m=n > 2$$ and there my counter example obviously does not work any more.
• Yes. I meant for $T$ to have full rank. – Leo Sep 19 '19 at 9:05
• This seems not to be a counterexample, because if you take $a=-1$ then $T(-1+t) =-1+td$ implies for small t that $d=1$, which is false for $t=2$. – Andrea Marino Sep 19 '19 at 9:29
• Call the line you are projecting every other line to $L$. As far as I can see every line perpendicular to $L$ gets mapped into a point. – Leo Sep 22 '19 at 10:01
Notice first that $$T(0+tv)=0+tw$$, thus $$T(tv) = tT(v)$$ (thanks Andrea).
Let $$v_1$$ & $$v_2$$ be linearlly independent vectors, and consider the lines $$v_1+tv_2$$ & $$v_2+tv_2$$. These cross each other precisely when $$t=1$$ at $$p=v_1+v_2$$.
Say $$T(v_1+tv_2)=w_1+tw_2' \rightarrow T(v_1) = w_1$$,
$$T(v_2+tv_1)=w_2+tw_1' \rightarrow T(v_2) = w_2$$
Since $$v_1+1v_2=v_2+1v_1$$ we must have $$w_1+1w_2'=w_2+1w_1'$$. Because of linear independency we must have $$w_i' = w_i = T(v_i)$$.
Therefore $$T(v_1+v_2) = T(v_1)+T(v_2)$$.
Let $$x_1, ..., x_n$$ be a basis of $$\mathbb{R}^n$$. Then $$T(a_1x_1+...+a_nx_n)= a_1T(x_1+[a_2'x_2+...+a_n'x_n])=a_1T(x_1)+T(a_2x_2+...a_nx_n)$$ where $$a_i'=a_i/a_1$$. Applying the trick repeadetly yields $$T(a_1x_1+...+a_nx_n)=a_1T(x_1)+...+a_nT(x_n)$$.
I think it is true. Recall the two equations $$T(0)=0$$ and for any $$a,b$$ there exist $$c,d$$ such that for any t $$T(a+tb)=c+td$$.
Substituting $$t=0$$ yields $$c=T(a)$$. In particular for $$a=0$$ we get $$c=0$$, i e. $$T(tb)=td$$. This means that every component of $$T$$ has derivatives in every direction around 0, and that $$\partial_bT(0)=d$$: just divide by $$t$$ and take the limit $$t \to 0$$. In particular, for $$t=1$$ we get $$T(b)=d=\partial_bT(0)$$.
Now we have the surprise: from the same equation, we get
$$T(x+y) =\partial_{(x+y)}T(0) =\partial_xT(0) +\partial_yT(0) =T(x)+T(y)$$
Voilà ! To conclude, note that $$T(tb)=td$$ for $$t=1$$ gives $$d=T(b)$$, so that $$T(tb)=td=tT(b)$$, the second condition of linearity.
• Sorry if the layout is awful, I am from my phone: feel free to edit :) – Andrea Marino Sep 19 '19 at 9:48
• I can understand everything up to $T(b)=d=\partial _b T(0)$, yet I don't understand how this expression implies $T(x+y) = \partial _{(x+y)}T(0)$, since in the former $b$ is fixed, and in the latter $(x+y)$ is a variable. In other words, how do we know that $b$ isn't the only number for which $T(b)= \partial _b T(0)$? – Leo Sep 21 '19 at 15:20
• Recall that we derived $T(b)=\partial_bT(0)$ for any $b$. This holds in particular for $b=x+y$. I am not sure I got your doubt: both $b,x,y$ have the quantifier "for any", and are fixed from the beginning of the equations – Andrea Marino Sep 22 '19 at 12:19
• Sorry for the late reply, but could you elaborate on why every component of $T$ has a derivative around 0? – Leo Sep 28 '19 at 12:55
• You should show that the limit $\frac{T(tb)-T(0)}{t}$ for $t\to 0$ exists. But this is constantly $d$! (The notation is tha same as above) – Andrea Marino Sep 29 '19 at 16:21 | 2021-01-26T21:56:50 | {
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https://www.physicsforums.com/threads/the-electric-field-due-to-a-charged-rod.636713/ | # The Electric field due to a charged rod
1. Sep 17, 2012
### Unbounded
1. The problem statement, all variables and given/known data
A rod 14.0 cm long is uniformly charged and has a total charge of -22.0μC. Determing (a) the magnitude and (b) the direction of the electric field along the axis of the rod at a point 36.0 cm from its center.
d (the distance between the center of the rod and the point) = .29m
l = length of the rod
The answer to part (a) is supposed to be 1.59x106C
2. Relevant equations
λ(the linear charge density) = q/l
The electric field E at a point due to one charge element carrying charge Δq =
(keΔq)/r2
3. The attempt at a solution
At first I noticed that the distance between the end of the rod and the point should be .29m, as pointed out above. I assumed I could simply take the formula for the electric field and integrate it from 0 m to .29m, but when I did that it, well, didn't turn out too well.
Then I tried again, and instead integrated the formula over the distance of .14m to .43m. The process looked something like this:
E = ∫(keλdx)/x2
Noting the constants, I moved them outside of the integral, and had:
E = keλ∫dx/x2
Integrating, I got:
E = ke(Q/l) [ -1/x] evaluated from the lower limit .14m and the upper limit .43m.
E = ke(Q/l)[1/.14 - 1/.43]
Using all of that, however, gave me an answer of 6.8 x 106, which is far off of the answer the book gave me, 1.59x 106, and I'm not too sure my units would cancel out properly either.
Where exactly did I mess this thing up?
Thanks in advance for any help.
2. Sep 17, 2012
### TSny
Hi, Unbounded. Welcome to PF.
In this type of problem, it is very important to be clear on your choice of the location of the origin of your x-axis. There are several possible locations of the origin that could be chosen for convenience in setting up the integral. Once you have the origin nailed down, you will be able to logically deduce the limits of integration and the form of the denominator in the integral.
Where did you decide to choose the origin?
3. Sep 17, 2012
### Unbounded
Thank you.
I tried to place the origin at the end of the rod, though I'm not too sure I did that successfully.
Though I wonder now, would the numbers work out if I placed my origin in the center of the rod, and instead integrated over the interval of .07m to .36m?
4. Sep 17, 2012
### TSny
I'm not sure which end of the rod you chose (left or right). You can choose either end, or the middle, or you can even chose the origin at the location of the point where you would like to find E. It won't make much difference in difficulty of setting up and doing the integral. But you do have to make a choice, since the form of the integral will depend on the choice.
Note that you need to integrate over all of the elements of charge of the rod. So, once you choose your origin, the limits of integration will be determined by the x coordinates of the left and right end of the rod.
5. Sep 17, 2012
### Unbounded
So I can choose any location, so long as I integrate over the entire length of the rod. That would mean that what I was doing previously where I was integrating from the end of the rod was incorrect because I simply integrated from the space between the rod and the point.
So I should set the left end of the rod at the origin, and integrate from x = 0 to x = .14m, where x = 0 is the left end of the rod and x = .14 is the right end of the rod.
But then where do the distance between the center of the rod and the point where I have to calculate the electric field come into play?
6. Sep 17, 2012
### TSny
That's right.
That's not a bad choice. Yes, then the limits would be from 0 to .14 m.
Well, that comes in when you decide how to write the integrand. If x is the location of a point of the rod, how would you write an expression for the distance from that point to the point where you want to find E?
7. Sep 17, 2012
### Unbounded
I'm not quite sure I follow.
If I visualize the point as (.07m + .36m) = .43m from the origin, I would obtain a distance Δx = .29m.
So would I just place this as r in the formula E = ∫keq/r^2, change the q into λdx, and solve accordingly?
8. Sep 17, 2012
### TSny
Since x is the variable of integration, you need to express r in terms of x. r is the distance from an element of charge of the rod to the field point. For an element of charge located on the x-axis at position x, how would you express r in terms of x?
9. Sep 17, 2012
### Unbounded
So if r varies with x would it be r = (.29 + x)?
EDIT: Actually, that doesn't sound quite right.
as x increases, the distance between the charge and the point decreases, so the value of r would have to decrease as x increases.
Last edited: Sep 17, 2012
10. Sep 17, 2012
### TSny
If point A is on the x-axis at x = 3 and B is on the x-axis at x = 5, how do you get the distance from A to B?
Likewise, if an element of charge is located at x = x and the field point is located at x = .43 m, how do you get the distance from the element of charge to the field point (r)?
11. Sep 17, 2012
### Unbounded
So r = (.43 - x), as x varies from 0 to .14. This seems like it would be correct because at x = 0, (the left end of the rod) r = .43, and at x = .14,(the right end of the rod), r would equal .29, which was established earlier as the distance between the right end of the rod and the point p. Most importantly, if x =.07, the center of the rod, r will equal .36m, which is the distance between the center and the point p.
So if I understand this, my entire formula would look something like:
E =0.14keλdx/(.43-x)2
Where λ = Q/l?
12. Sep 17, 2012
### TSny
You got it. Good work!
13. Sep 17, 2012
### Unbounded
Thank you! I really appreciate the help!
14. Sep 17, 2012
### TSny
You're very welcome. For practice you might see if you can get the same result by choosing the origin at the center of the rod. | 2018-02-17T21:55:18 | {
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https://www.physicsforums.com/threads/momentum-of-a-projectile-and-ship.162553/ | # Homework Help: Momentum of a projectile and ship
1. Mar 25, 2007
### ataglance05
1. The problem statement, all variables and given/known data
A battle ship (4.09*10^7 kg) fires a salvo of 3 rounds from its foward turret in the direction of the bow at an angle of 10 degrees above the horizontal. Each projectile weighs 1220 kg, each barrel is 20.9 m long and the muzzle velocity of the projectile is 770 m/sec. Acceleration of each projectile down the barrel is 1.418 x 10^4 m/s2 at 0.0543 seconds.
If the momentum must be conserved in the x direction, 1) what's the velocity of the ship in the x direction and 2) in what direction is it moving along the x axis after the projectiles have left the ship's guns?
2. Relevant equations
p(linear momentum of an object)=mv
M1V1 + M2V2=0
3. The attempt at a solution
1) I'm not sure whether I should get the velocity of the projectile in the x direction, but I just used it's velocity in the x direction in solving the problem. I did the trigonometric work on paper.
M1V1 + M2V2=0
V2= -M1/M2(V1)
V2= -1220/4.09*10^7(758.3)
V2= -.0226 m/sec
2) The ship recoils to the left of the x axis.
Last edited: Mar 25, 2007
2. Mar 25, 2007
### Werg22
"1.418 x 104 m/s2 at 0.0543 seconds" what do you mean by that? Is it an increase of 1.418 x 104 m/s every 0.0543 seconds?
3. Mar 25, 2007
### ataglance05
sorry, its suppsoe to be 1.418*10^4; 1.418*10^4 is the acceleration of each projectile in the barrel and .0543 sec is the time it takes for the projectile to travel down the barrel.
Last edited: Mar 25, 2007
4. Mar 25, 2007
### Werg22
Ok but how is this useful if we are given the muzzle velocity?
5. Mar 25, 2007
### ataglance05
i was told that it was supposidly the final velocity/velocity at which the projectile leaves the barrel.
6. Mar 25, 2007
### ataglance05
okay; so does that mean the muzzle velocity is the the velocity I would use in the momentum equation for V1, but instead the velocity vectors in the x direction which is 758.3 m/sec?
7. Mar 25, 2007
### Werg22
Ok, it actually does affect the problem. What we are interested in is the moment when the projectile leaves the ship. This moment would be right when the projectile begins to travel down the barrel. So with the information given, we must calculate the initial velocity. We have
$${v_{i}}^{2} = {v_{f}}^{2} - 2ad$$
We know vf, a and d, so we can calculate vi. Once we get vi, we calculate its horizontal component and use it in the conservation of momentum formula.
8. Mar 25, 2007
### ataglance05
$${v_{i}}^{2}$$= (770)^2 - 2(14180)(20.9m)
Vi=13.27 m/sec!!!!
thank you, sir.
I do have one more question, if you don't mind...
What is the total momentum of the 3 projectiles in the x direction parrell to the water?
Would I just multiply the momentum of one of the projectiles by three or is it zero since total momentum is always zero??
9. Mar 25, 2007
### Werg22
If the three projectiles are in the same direction, their momentum adds ups since the sign of momentum depends on the sign of the velocity. If you add the momentum of the ship into the equation, the sum would be 0. In any explosion or implosion (in this case it's an explosion - things get separated) the momentum gained in a direction by a fragment is equal to the momentum gained by the opposite fragment in the other direction. This is basically the law of conservation of momentum.
10. Mar 25, 2007
### ataglance05
so is the following correct? (I now used the new initial velocity in the x-direction:
M1V1(3)=?
(1220)(13.07)(3)= 47830.34 J
?
11. Mar 25, 2007
### Werg22
Yes. The ship would have the opposite momentum.
12. Mar 25, 2007
### ataglance05
If I had the power to bestow upon you a salmagundi of riches beyond belief, you would be the first person to share in this wealth.
THANK YOU!
13. Mar 25, 2007
### denverdoc
Not to rain on ayones parade, from the earlier thread you posted, I wrote:
"This is right, but perhaps a more straightforward approach assumes vf=770, and Vi=0,
Given that Vf^2-Vi^2=2ax, then a=770^2/(2*20.9)
a=14184m/s^2 and t would just be the distance divided by ave velocity:
20.9/385=0.0543s Different ways to skin a cat. Best to have many knives.:"
In other words you just have introduced some roundoff error and vinit=0, this was the assumption after all, and acceleration was calculated on this basis.
Again, this operates under several assumptions that acceleration is linear from the detonation of the charge, to the time the shell leaves the barrel. It is entirely possible that this is not the case due to friction, leakage of gas, the fact that the volume occupied by the gas is becoming larger as the shell progresses down the barrel, etc.
The question about which velocity to consider is an interesting one. If we ignored the ship altogether, I think most would agree that say in the case of the rifle, its the muzzle velocity (Vf) that determines the momentum, even though the physical recoil may be experienced sooner.
So I would disagree, and suggest by extension to the case with the rifle, that the ships negative velocity*mass=3*cos(10)*770*mass(shell).
14. Mar 25, 2007
### ataglance05
you do indeed have a very strong point when noting this new initial velocity has turned the momentum problem into a bit of an error. However, i am a bit perplexed as to whether the equation you presented above clearly depicts the shell's (or projectile's) velocity if one multiplies along with it 3 and cos(10). is such thing possible to correctly calculate the momentum?
15. Mar 25, 2007
### ataglance05
actually, I would like to answer my own question...
Multiplying cos(10) and (770) gives one the velocity in the x direction, which is what I want. Multiplying three simply takes into account the 3 projectiles.
You are as well a life saver, denverdoc and you too are deemed greatful to partake of these chimerically hypothetical riches.
Last edited: Mar 25, 2007
16. Mar 26, 2007
### Bk8907
any of u guys know how to do electric potential?
17. Mar 26, 2007
### ataglance05
negatory, sir.
18. Mar 26, 2007
### denverdoc
well, i think your point is well taken, we are ignoring the vertical component of the momentum, as this would simply try to push the ship deeper in the water, which would be opposed by buoyancy, etc. It is essential tho that we only consider the x component of the shells velocity, as one fired vertically would have no tendency to cause the boat to recoil along the x axis, thus the cos 10 degree factor. As to algebraically adding the volley of shells, I can see some concern on your part that this isn't fair and proper. One reason might be that the ship has already come to rest between shots. But this is physics homework where all kinds of outlandish assumptions are made routinely in the service of introducing principles. In other words, for the purpose of this problem since there is no consideration of friction, we can superimpose the three shots whether minutes apart or a single shell of 3600Kg was shot with the same muzzle velocity, or whether three guns all aimed in parallel lines were fired simultaneously. (coure then you might actually get some torque and cause the ship to rotate--couple chapters down the line, not to worry). Hope this didn't confuse matter furthers, if you have some specific concerns re the legitimacy, holler.
Last edited: Mar 26, 2007
19. Mar 26, 2007
### ataglance05
so would you say that the ship, for my homework's sake, is moving left of the x direction after all of this firing of shells has been made?
20. Mar 26, 2007
### Werg22
denverdoc, I don't think they gave information on acceleration and distance if we were not meant to consider the initial velocity. This said, I think the whole idea of a projectile accelerating down a barrel is flawed. The projectile gets sent by absorbing some momentum of an explosion that happens at the lower end of the barrel. I believe the projectile stops accelerating there and then, as in a instantaneous collision. | 2018-10-20T23:07:48 | {
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https://www.physicsforums.com/threads/calculating-final-velocity-of-an-object-on-an-inclined-plane.544456/ | Homework Help: Calculating final velocity of an object on an inclined plane
1. Oct 26, 2011
trulyfalse
Hello everyone, I'm stuck on a dynamics review question. I was told to solve like I would with other inclined planes, however mass was not given. I am not sure how to proceed.
1. The problem statement, all variables and given/known data
A roller coaster reaches the top of the steepest hill with a speed of 1.4 m/s. It then descends down the hill, which is at an average angle of 45° and is 50 m long. What will its speed be when it reaches the bottom? (Answer: 26 m/s)
2. Relevant equations
Fnet=ma
Fg=mg
3. The attempt at a solution
I drew a free body diagram, separating the x and y components of Fg. This is futile as mass is not given, thus Fg, Fn, and Fnet cannot be calculated. Am I missing something here?
2. Oct 26, 2011
sandy.bridge
Hello, if you are familiar with conservation of energy, here is a method for solving this problem.
$K_E=P_E$
In doing so, you eliminate the dependence of mass.
3. Oct 26, 2011
trulyfalse
Could you elaborate? Thanks. :)
4. Oct 26, 2011
sandy.bridge
$$\frac{1}{2}mv^2=mgh$$
Which in words states that the kinetic energy at the bottom is equal to the change in potential energy.
The height, h, can easily be determined as you know the angle of incline.
5. Oct 26, 2011
trulyfalse
So, in that case I would assume mass is negligible and leave it out of the equation?
Here we go:
First I manipulated the formula given (assuming mass in negligible) to get √2gh=v
50cos(45°) = 35.4 m = height of the coaster to the ground
√2(9.81m/s2)(35.4m) = v
v = 26.3 m/s rounded off to 2 sig digs is 26 m/s
Thanks bro! You've been a big help. :)
6. Oct 26, 2011
sandy.bridge
The mass is not negligible. I would assume the mass of a rollercoaster is quite large relative to you, or I. However, the change in energy for this particular instance is not dependent on the mass of the system. Hence, the m is cancelled upon manipulation of the equations.
7. Oct 26, 2011
trulyfalse
Right, because dividing m by m yields 1. This will be of great aid on my unit exam!
8. Oct 26, 2011
sandy.bridge
I wouldn't advise applying any theorms that you are not familiar with. You can also apply kinematics with this question to solve for the velocity. For example, one can apply
$v_f^2-v_i^2=2a_y\Delta x$
The acceleration is due to gravity. The intial velocity in the y-drection is 0. The change in position is the same as you had used before.
9. Oct 26, 2011
trulyfalse
In that case then, to solve for vf I would manipulate to get | 2018-05-20T16:12:09 | {
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https://thevideogamegallery.com/autobiograf-a-bbrqiku/page.php?149688=vector-derivative-rules | Derivative Rules. Matrix derivatives cheat sheet Kirsty McNaught October 2017 1 Matrix/vector manipulation You should be comfortable with these rules. The standard rules of Calculus apply for vector derivatives. Derivative Rules for Vector-Valued Functions. Rules of Differentiation The derivative of a vector is also a vector and the usual rules of differentiation apply, dt d dt d t dt d dt d dt d dt d v v v u v u v ( ) (1.6.7) Also, it is straight forward to show that { Problem 2} a a v a v a v a v v a v dt d dt d dt d dt d dt d dt d (1.6.8) Free derivative calculator - differentiate functions with all the steps. 4 cos(4x + 2) And that is the derivative of your original function. We will now look at a bunch of rules for differentiating vector-valued function, all of which are analogous to that of differentiating real-valued functions. Type in any function derivative to get the solution, steps and graph Theorem D.1 (Product dzferentiation rule for matrices) Let A and B be an K x M an M x L matrix, respectively, and let C be the product matrix A B. So in this case, the derivative of X, so you'd write DX/DT, and the derivative of Y, DY/DT. Then, ac a~ bB -- - -B+A--. The Derivative tells us the slope of a function at any point.. And now you might start to … They will come in handy when you want to simplify an expression before di erentiating. One of the most common examples of a vector derivative is angular acceleration, which is the derivative of the angular velocity vector. 1 Vector-Vector Products Given two vectors x,y ∈ Rn, the quantity xTy, sometimes called the inner product or dot product of the vectors, is a real number given by xTy ∈ R =. This is the vector value derivative. The derivative of V, with respect to T, and when we compute this it's nothing more than taking the derivatives of each component. Suppose we have a column vector ~y of length C that is calculated by forming the product of a matrix W that is C rows by D columns with a column vector ~x of length D: ~y = W~x: (1) Suppose we are interested in the derivative of ~y with respect to ~x. Section 7-2 : Proof of Various Derivative Properties. Contraction. Answer: As with the dot product, this will follow from the usual product rule in single variable calculus. If r 1(t) and r 2(t) are two parametric curves show the product rule for derivatives holds for the cross product. Thus, the derivative of a matrix is the matrix of the derivatives. ax, axp ax, We will not prove all parts of the following theorem, but the reader is encouraged to attempt the proofs. There are rules we can follow to find many derivatives.. For example: The slope of a constant value (like 3) is always 0; The slope of a line like 2x is 2, or 3x is 3 etc; and so on. Here are useful rules to help you work out the derivatives of many functions (with examples below). Product rule for vector derivatives 1. All bold capitals are matrices, bold lowercase are vectors. It’s just that there is also a physical interpretation that must go along with it. In this section we’re going to prove many of the various derivative facts, formulas and/or properties that we encountered in the early part of the Derivatives chapter. Furthermore, suppose that the elements of A and B arefunctions of the elements xp of a vector x. calculus multivariable-calculus vector-analysis Product rule for vector derivatives 1. Example. In finding the derivative of the cross product of two vectors $\frac{d}{dt}[\vec{u(t)}\times \vec{v(t)}]$, is it possible to find the cross-product of the two vectors first before differentiating? We want to show d(r 1 × r … to do matrix math, summations, and derivatives all at the same time. Not all of them will be proved here and some will only be proved for special cases, but at least you’ll see that some of them aren’t just pulled out of the air.
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https://math.stackexchange.com/questions/2417934/prove-ln1x-geq-x-fracx22/2417952 | # Prove $\ln(1+x)\geq x-\frac{x^2}{2}$
When $x\geq0$ prove that: $$\ln(1+x)\geq x-\frac{x^2}{2}.$$
My effort: From the Taylor series:
$$\ln(1+x)= x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots$$
I don't know how to continue from here.
• Hint: use Taylor’s theorem with the remainder Sep 5 '17 at 17:03
• If the remainder term is negative on the interval, then the sum of the zeroth through $n^{th}$ order terms yields an upper bound. Sep 5 '17 at 17:06
• you have a mistake in the Taylor development. Sep 5 '17 at 17:09
• @user48672 fixed it. Does it deserve a vote up now? Sep 5 '17 at 17:11
• @ThomasAndrews for $x\geq 0$ man. Thanks a lot. Sep 5 '17 at 17:12
## 6 Answers
Let $f(x)=\ln(1+x)-x+\frac{x^2}{2}$.
Thus, $$f'(x)=\frac{1}{x+1}-1+x=\frac{x^2}{1+x}\geq0.$$
Thus, $f(x)\geq f(0)=0$ and we are done!
Observe that $$\ln(1+x) = \int_0^{x} \frac{1}{1+t}dt.$$ But, for $t \ge 0$, $$\frac{1}{1+t} \ge 1-t.$$ Therefore, $$\ln(1+x) \ge \int_{0}^x(1-t)dt=x-\frac{x^2}{2}.$$
• I was looking for an integral proof, but missed this. :) Sep 5 '17 at 17:34
• $\frac{1}{1+t}\ge 1-t$, where $t\in\mathbb R$, is true if and only if $t>-1$. But in this case we only need $t\ge 0$ because the definite integral is over $t\in [0,x]$. Sep 5 '17 at 19:13
You want to show $$1+x\geq e^{x-x^2/2}$$
or $e^{x^2/2}(1+x)\geq e^{x}.$
The right side is $\sum_{k=0}^{\infty} \frac{x^k}{k!}$ and the left side is:
$$(1+x)\sum_{j=0}^{\infty}\frac{x^{2j}}{2^jj!}=\sum_{k=0}^{\infty}\frac{x^{k}}{2^{\lfloor k/2\rfloor}\lfloor k/2\rfloor!}$$
But $2^{\lfloor k/2\rfloor}\lfloor k/2\rfloor!\leq k!$.
Hint
$$f(x)=\ln(1+x)-x+\frac{x^2}{2}$$
so,
$$f'(x)=\frac{1}{1+x}+x-1=\frac{x^2}{x+1}>0 \text{ for } x> -1$$
You want to prove $\ln(1+x)\ge x-\frac{x^2}{2}$ for all $x\ge 0$.
To prove it for $0\le x<1$, we'll use Taylor series. It converges because $-1<x\le 1$. See, e.g., here for more information. Or here.
$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$$
$\ln(1+x)\ge x-\frac{x^2}{2}$ is equivalent to $\frac{x^3}{3}-\frac{x^4}{4}+\cdots\ge 0$.
We'll prove for all $k\ge 3$, $k\in\mathbb Z$, $\frac{x^k}{k}-\frac{x^{k+1}}{k+1}\ge 0$. We could prove this using derivatives, but I'll use a simpler method here. The inequality is equivalent to
$\frac{x^k(k+1-kx)}{k(k+1)}\ge 0$
and we have
$\frac{k+1}{k}>1> x$, $x^k\ge 0$ because $x\ge 0$.
Using derivatives:
Proof: let $f(x)=(k+1)x^k-kx^{k+1}$.
Then $f'(x)=k(k+1)x^{k-1}(1-x)\ge 0$.
$f(x)\ge f(0)=0$.
Also notice that $\left(\frac{x^3}{3}-\frac{x^4}{4}\right)+\frac{x^5}{5}\ge 0$, where also $\frac{x^5}{5}\ge 0$, $\frac{x^{2t+1}}{2t+1}\ge 0$ for all $t\ge 2$, $t\in\mathbb Z$.
Therefore, if $0\le x<1$, then $\frac{x^3}{3}-\frac{x^4}{4}+\cdots \ge 0$, i.e. $\ln(1+x)\ge x-\frac{x^2}{2}$.
By the way, notice that when finding the sum $\frac{x^3}{3}-\frac{x^4}{4}+\cdots$, you have to find the limit of partial sums so that you have to sum the terms in the exact same order $\frac{x^3}{3}$, $\frac{x^3}{3}-\frac{x^4}{4}$, $\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}$, etc. because see the Riemann series theorem: summing the terms in a different order could result in a different number or diverge if the series is conditionally convergent. But in this case, $\frac{x^3}{3}-\frac{x^4}{4}+\cdots$ is not conditionally convergent when $-1<x<1$, because if $-1<x<1$, then it converges to $\ln(1+x)-x+\frac{x^2}{2}$ and $\frac{x^3}{3}+\frac{x^4}{4}+\cdots$ converges to $-\ln(1-x)-x-\frac{x^2}{2}$.
Here's a proof for $x\ge 1$. If $x>1$, then the Taylor series diverges, so we can't use it. $$x-\frac{x^2}{2}=\frac{2x-x^2}{2}=\frac{1}{2}-\frac{(x-1)^2}{2}\le$$
$$\le \frac{1}{2}<\ln 2=\ln(1+1)\le \ln(1+x)$$
because $1=\ln e<\ln 4=\ln 2^2$, i.e. $1<2\ln 2$, because $e<4$ and $\ln x$ is strictly increasing when $x>0$.
There is an easy way to prove. Note that $$(1+t)(1-t)=1-t^2\le 1$$ and hence $$\frac{1}{1+t}\ge1-t.$$ Integrating from $0$ to $x$, one has $$\int_0^x\frac{1}{1+t}dt\ge\int_0^x(1-t)dt$$ which gives $$\ln(1+x)\ge x-\frac12x^2.$$
• $\frac{1}{1+t}\ge 1-t$ is true if and only if $t>-1$, but we only need $t\ge 0$. Also, this exact same proof has already been posted as an answer here. Sep 6 '17 at 13:22
• @user263326, I did not realize that someone already posted. Sep 6 '17 at 13:46 | 2021-09-27T06:52:19 | {
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http://gurmeet.net/puzzles/card-shuffling/index.html | Gurmeet.Net
Puzzles
Card Shuffling
Puzzle
A perfect in-shuffle of a deck of 52 cards is defined as follows. The deck is cut in half followed by interleaving of the two piles. So if the cards were labeled 0, 1, 2, ..., 51, the new sequence is 0, 26, 1, 27, 2, 28, ... With repeated in-shuffles, shall we ever get back the original order? In how many iterations?
Solution
The shuffling technique described in the puzzle is known as the Faro Shuffle (wikipedia).
First, let us convince ourselves that we shall get back the original order. The 'inverse' of the perfect in-shuffle is unique. In other words, there is only one other card ordering that would result in a given card ordering with the application of the perfect in-shuffle. So by repeated application of the perfect in-shuffle, we are certain to recover the original ordering because the total number of card orderings is finite: 52!.
A closer look at the structure of the perfect in-shuffle reveals that it is simply a permutation. Let the cycle lengths of the permutation be c1, c2, c3, ... Then the number of iterations is the LCM of c1, c2, c3, ... With 52 cards, let the cards be labeled 0, 1, 2, ..., 51. Then cutting and interleaving results in 0, 26, 1, 27, 2, 28, ... , which is equivalent to the following "moves": { 0 -> 0, 1 -> 2, 2 -> 4, 3 -> 6, 4 -> 8, ..., 26 -> 1, 27 -> 3, 28 -> 5, 29 -> 7, ... 51 -> 51}. The inverse of these moves can be captured succinctly as follows: If i is even, i --> i/2. If i is odd, then i -> 26 + (i-1)/2. The formula allows us to list the cycles of the permutation in reverse order:
(length 1) 0 -> 0
(length 1) 51 -> 51
(length 8) 50 -> 25 -> 38 -> 19 -> 35 -> 43 -> 47 -> 49 -> 50
(length 8) 48 -> 24 -> 12 -> 6 -> 3 -> 27 -> 39 -> 45 -> 48
(length 8) 46 -> 23 -> 37 -> 44 -> 22 -> 11 -> 31 -> 41 -> 46
(length 8) 42 -> 21 -> 36 -> 18 -> 9 -> 30 -> 15 -> 33 -> 42
(length 8) 40 -> 20 -> 10 -> 5 -> 28 -> 14 -> 7 -> 29 -> 40
(length 2) 34 -> 17 -> 34
(length 8) 32 -> 16 -> 8 -> 4 -> 2 -> 1 -> 26 -> 13 -> 32
So the answer is LCM of {1, 2, 8}, which is 8 shuffles!
Alternate Solution (submitted by Vivek in Oct 2010 in a comment) For all cards other than 0 and 51, a card at position p gets shuffled to position 2p mod 51, which a function that maps [1, 50] → [1, 50]. After n shuffles, a card at position x would move to position (2n p) mod 51. Consider some p in [1, 50] that is co-prime to 51 (in other words, p and 51 have no common prime factors). Then the smallest n > 0 such that 2n mod 51 equals 1 happens to be 8 (because 2n mod 51 for n ranging from 1 thru 8 is the sequence 2, 4, 8, 16, 32, 13, 26, 1). Now consider the two prime factors of 51, namely 3 and 17. The smallest n > 0 such that (2n ⋅ 3) mod (17 ⋅ 3) equals 3 is also 8 (because 2n mod 17 for n ranging from 1 thru 8 is the sequence 2, 4, 8, 16, 15, 13, 9, 1). Finally, the smallest n > 0 such that (2n ⋅ 17) mod (3 ⋅ 17) equals 17 is 2 (because 2n mod 3 for n ranging from 1 thru 2 is the sequence 2, 1). Thus every card would return to its own position in at most 8 shuffles.
Followup
1) Is there a closed form for the number of shuffles in terms of n, the number of cards? Those familiar with group theory might want to read The Mathematics of Perfect Shuffles by Persi Diaconis, R L Graham and William M Kantor, Advances in Applied Mathematics (4), p 175-196, 1983. Further information: Faro Shuffle (wikipedia).
2) The Korn Shell Puzzle is related:
You are sitting at a computer terminal whose 26 alphabetic keys have been randomly rearranged (permuted) and you enter your name, say, "Mike Korn." Since the keys are scrambled what appears on the screen is also scrambled. The game consists of you typing the characters you see on the screen until your name appears. The length of the game is the number of times you must type a the string on the screen before your name appears, at which point the game terminates.
1. Is every possible game guaranteed to terminate in a finite number of steps?
2. How does the length of game vary with respect to the two inputs, i.e., the name and the permutation?
3. If every game does terminate in a finite number of steps, what is the longest game?
The last question asks for the permutation whose LCM of cycle lengths is maximized. This quantity is also known as Landau's Function; its natural log asymptotically converges to $latex \sqrt{n \log n}$.
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Four honest and hard-working computer engineers are sipping coffee at Starbucks. They wish to compute their average salary. However, nobody is willing to reveal an iota of information about his/her own salary to anybody else. How do they do it? | 2017-11-20T17:09:23 | {
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https://mathematica.stackexchange.com/questions/105264/non-standard-eigenfunction-plots-of-the-laplacian-over-the-unit-square | # Non Standard Eigenfunction Plots of the Laplacian Over the Unit Square
I have recently been plotting eigenfunctions of the laplacian over the unit square using the NDEigensystem command. However, I have noticed something in the plots which puzzles me.
Below is an image of the first 4 eigenfunction plots. However, the second and third eigenfunction do not resemble the standard known eigenfunctions of the unit square: $u_{21}$ and $u_{12}$, and this has me rather puzzled.
With this in mind, I decided to investigate what's going on. I first noticed that the second and third eigenvalues, $\lambda_{2}$ and $\lambda_{3}$, are identical (we know this from the separation of variables solution). This seems to be causing the weird patterns in the eigenfunctions (I checked this for higher order eigenvalues/eigenfunctions too and this appears to be the case); wherever the multiplicity of the eigenvalue is greater than 1, the eigenfunctions being plotted are not the standard ones.
On further inspection I noticed that the second eigenfunction is very roughly $\sin \left (2\pi x \right )\sin \left (\pi y \right ) + \frac{10}{13} \sin \left (\pi x \right ) \sin \left (2 \pi y \right )=u_{21}+ \frac{10}{13} u_{21}$ (see below). This is a linear combination of the two symmetric eigenfunctions $u_{12}$ and $u_{21}$.
Now this makes complete sense; since $u_{21}$ and $u_{12}$ share the same eigenvalue, then the most general eigenfunction is just a linear combination of the two. However, this has puzzled me somewhat and so I wish to ask a few questions:
Q1) How do I prevent mathematica from combining these eigenfunctions? The reason I ask is that I wish to study the eigenfunction plots over unconvential domains, and I won't be able to tell which plots are combinations of eigenfunctions and which are not for more complicated domains.
Q2) How does mathematica 'know' that the eigenvalues are the same? Of course we know this from separation of varaibles but how does mathematica know that they are the same when it's performing a numerical approximation? Surely the values would be ever so slightly different and so mathematica wouldn't even think to combine the eigenfunctions in the first place.
I know this is a strange question to ask but it has got me rather baffled. Ideally, I'm wanting to adapt the code in mathematica so it produces non-combined eigenfunctions for arbitrary domains.
The (very short) code I have been using to get the plots is:
{vals, funs} =
NDEigensystem[{-Laplacian[u[x, y], {x, y}],
DirichletCondition[u[x, y] == 0, True]},
u[x, y], {x, y} \[Element]
Polygon[{{0, 0}, {1, 0}, {1, 1}, {0, 1}}], 4];
Table[ContourPlot[
funs[[i]], {x, y} \[Element]
Polygon[{{0, 0}, {1, 0}, {1, 1}, {0, 1}}],
AspectRatio -> Automatic, PlotRange -> All, PlotLabel -> vals[[i]],
PlotTheme -> "Minimal"], {i, Length[vals]}]
• The reason is the same as what I said in my comment to your earlier Q: degeneracies can lead to eigenvectors that are arbitrary orthogonal superpositions of the symmetrized states you expected. If your new Q is about achieving this symmetry given the solutions you plotted, then there are several ways to answer it. Is that what you're asking? – Jens Jan 30 '16 at 20:41
• In more general domains, you won't even see this problem because degeneracies will be very rare. So maybe we aren't really talking about the square here... what shapes are you really interested in? – Jens Jan 30 '16 at 20:47
• Yes - I wish to plot the standard symmetric states. And you are correct, I'm getting to grips with the square before I begin to look at other domains. I intend to look at a variety of different domains. Initially, I wish to look at polygons (I want to look at the pentagon), and then I wish to look at more complicated 2-D shapes. The question I wish to be answered is whether there is a way of telling mathematica to first spot these degeneracies and secondly, prevent the superposition from happening. – Mr S 100 Jan 30 '16 at 20:55
• It's easy to spot the superposition in the case of the square, but I'm not so confident that I would be able to spot the degenerate modes of an 'L' shape for example. – Mr S 100 Jan 30 '16 at 20:59
• The L shape is very different from the square because the latter is invariant under the symmetry group $C_{4v}$ which causes the degeneracies. There is no such symmetry in the L. So the goal of your question isn't clear to me. You're asking about a highly symmetric example, where the degeneracies can be removed by reducing the domain. But that's not a Mathematica problem, and it's also unrelated to what you'd see in the L shape. All I can say then is: degeneracies are something you see in the spectrum and not in the wave functions. – Jens Jan 30 '16 at 21:56
Although the question singles out the square, it is made clear that the actual applications includes other polygonal shapes as well. This means that it's impossible to give a general answer based on the assumption of separability. The square is separable in Cartesian coordinates, but the pentagon (e.g.) is not.
This is why I'm focusing this answer on the exploitation of point symmetries (reflections, rotations) in more general terms. This means that the results here for the special case of the square will not necessarily be in the product form that one obtains from separation of variables, but instead will conform to the classification of symmetries according to irreducible representations of the relevant point group. That's what you would do in the general case when symmetries are present but separation of variables doesn't work. And that's of course the only case in which numerical solutions using NDEigensystem are really needed in the first place.
For the square, you can exploit the $C_{4v}$ symmetry to convert any wave solution to a symmetrized form. Without going into the mathematical details, here is the prescription. One would have to go more deeply into group theory to explain the methodology.
First define the elements of the symmetry group that leaves the square invariant (I use script letters below, but they don't all show up properly unless you paste this into a notebook):
\[ScriptCapitalG] = {ℰ, Subscript[\[ScriptCapitalC],
2], Subscript[\[ScriptCapitalC], 4],
Subscript[\[ScriptCapitalC], -4], Subscript[σ, vx],
Subscript[σ, vy], Subscript[σ, d1],
Subscript[σ, d2]};
\[ScriptCapitalG]ℳ = {{{1, 0}, {0, 1}}, {{-1,
0}, {0, -1}}, {{0, -1}, {1, 0}}, {{0, 1}, {-1, 0}}, {{1,
0}, {0, -1}}, {{-1, 0}, {0, 1}}, {{0, 1}, {1, 0}}, {{0, -1}, {-1,
0}}};
\[ScriptCapitalG]ℳℐ =
Inverse /@ \[ScriptCapitalG]ℳ;
Grid[{\[ScriptCapitalG],
Here, the names of the elements are listed above the 2D matrices that describe the point operations of rotations and reflections in the $xy$ plane, centered in the middle of the square.
For this group, there is a two-dimensional irreducible representation which causes the degeneracies in the spectrum. The second and third solutions in the question lie in the invariant subspace of that representation. What the question asks for is equivalent to projecting these (asymmetric-looking) functions onto the symmetrized basis functions of that subspace. To do this, one can use the following projection operator, where i = 1,2 is the index of the basis function:
projector[i_] =
Function[ψ,
1/4 Sum[\[ScriptCapitalG]ℳ[[n, i, 1]] ψ /.
y} -> ({1, 1}/
2 + \[ScriptCapitalG]ℳℐ[[n]\
].({x, y} - {1, 1}/2))], {n,
Length[\[ScriptCapitalG]ℳ]}]];
Now test this for the numerical solutions of the question:
{vals, funs} =
NDEigensystem[{-Laplacian[u[x, y], {x, y}],
DirichletCondition[u[x, y] == 0, True]},
u[x, y], {x, y} ∈
Polygon[{{0, 0}, {1, 0}, {1, 1}, {0, 1}}], 10];
The first projection is
f1 = projector[1][funs[[2]]];
ContourPlot[f1, {x, y} ∈
Polygon[{{0, 0}, {1, 0}, {1, 1}, {0, 1}}], AspectRatio -> Automatic,
PlotRange -> All, PlotLabel -> vals[[2]]]
and the second projection is
f2 = projector[2][funs[[2]]];
ContourPlot[f2, {x, y} \[Element]
Polygon[{{0, 0}, {1, 0}, {1, 1}, {0, 1}}], AspectRatio -> Automatic,
PlotRange -> All, PlotLabel -> vals[[2]]]
Note that I got the two symmetrized functions by applying two different projections onto the same eigenstate, funs[[2]]. That's because each of the un-symmetrized states had components along both symmetrized basis states.
Expanded method
The above version of the projector is only one among several others. All of them are needed in order to classify the spectrum fully. So I'll define a more general set of projectors for all the irreducible representations of the group $C_{4v}$ now. This first requires entering the character table for the group, which I call dMatrix. It's stored as an Association where the keys are the conventional names of the irreducible representations (irreps). The values are "matrices" with dimension 1 (for the first four) and 2 (for the last irrep, called Ee).
Clear[applyG, dMatrix, A1, A2, B1, B2, Εe, x, y]
irreps = {A1, A2, B1, B2, Ee};
dMatrix =
Association @@
MapThread[(#1 -> #2) &, {{A1, A2, B1, B2,
Ee}, {{{{1}}, {{1}}, {{1}}, {{1}}, {{1}}, {{1}}, {{1}}, {{1}}}, \
{{{1}}, {{1}}, {{1}}, {{1}}, {{-1}}, {{-1}}, {{-1}}, {{-1}}}, {{{1}}, \
{{1}}, {{-1}}, {{-1}}, {{1}}, {{1}}, {{-1}}, {{-1}}}, {{{1}}, {{1}}, \
{{-1}}, {{-1}}, {{-1}}, {{-1}}, {{1}}, {{1}}},
\[ScriptCapitalG]ℳ}}];
applyG[n_][f_] := (f /.
y} -> ({1, 1}/
2 + \[ScriptCapitalG]ℳℐ[[n]].\
({x, y} - {1, 1}/2))]);
Clear[projector];
projector[irrep_, row_][f_] :=
Simplify[Tr[dMatrix[irrep][[1]]]/
Length[\[ScriptCapitalG]] Sum[
Conjugate[dMatrix[irrep][[i]][[row, 1]]] applyG[i][f], {i,
Length[\[ScriptCapitalG]]}]]
To define the new projectors, I use the same prescription as above, but I factored out the application of each group element in a separate function applyG so that it can be adapted more easily to other simulation domains, if needed. The function projector now also depends on an additional label, irrep.
Now you can apply all these projectors to any state whose symmetries you want to analyze. For example, the solution in func[[5]] initially looks like this:
ContourPlot[
funs[[5]], {x, y} ∈
Polygon[{{0, 0}, {1, 0}, {1, 1}, {0, 1}}], AspectRatio -> Automatic,
PlotRange -> All, PlotLabel -> vals[[5]]]
But it is actually a superposition of two different degenerate states, belonging to different one-dimensional irreducible representations: A1 and B1. To do the analysis leading to this result, here is a complete table of all the states obtained from NDEigensystem above:
wavePattern[f_] :=
Module[{d =
Threshold[Table[f, {x, 0, 1, .02}, {y, 0, 1, .02}], .1]},
If[Max[Abs[d]] < .1, Graphics[{}],
ListDensityPlot[d,
ColorFunction ->
Function[{x},
Blend[{White, Darker@Orange, White}, 2 ArcTan[5 x]/Pi + .5]],
ColorFunctionScaling -> False, InterpolationOrder -> 0,
Frame -> False, Axes -> False]
]]
frames = Table[GraphicsGrid[Partition[Flatten@Table[Table[
Show[wavePattern[Evaluate[projector[irr, row][funs[[m]]]]],
PlotLabel -> Row[{irr, " row", row}]], {row,
Tr[dMatrix[irr][[1]]]}],
{irr, irreps}], 3],
PlotLabel ->
Framed[Style[Row[{"Eigenvalue \[LongEqual] ", vals[[m]]}],
Darker[Red]]], Frame -> All], {m, Length[vals]}];
FlipView[frames]
Each frame refers to one of the wave solutions. In each frame, the table shows all the rows of all the irreducible representations. In the corresponding cells, I plot the wave function obtained by the corresponding projection operator, applied to the eigenfunction whose eigenvalue is given at the top. When a cell is empty, the wave solution has no component under that projection. The decomposition for funs[[5]] is the frame with energy 78.111, and you can see that the cells for representations A1, B1 are filled. By looking at the character table in dMatrix, you can reconstruct how the different operations of the group will change the sign of the symmetrized components.
When more than one cell of the table is filled, it means that the given state is a superposition of the functions shown in those cells. I.e., if you were to simply add the wave functions of the two cells for a state with energy 78.111, you would get back the original state (e.g., funs[[5]]).
What this analysis shows is that whereas the degeneracies of eigenvalues 49.35, 128.338 and 167.855 are caused by the symmetry group of the square (its two-dimensional representation Ee), the other degeneracies are "accidental" in that they would not survive any perturbation that deforms the square to a non-separable shape while preserving the invariance under all symmetry operations. This could e.g. be done by truncating all the corners at 45 degree angles, or rounding them.
Accidental degeneracies are called "accidental" because they correspond to degenerate functions that belong to two different irreducible representations. Functions belonging to the same irrep transform in the same way under symmetry operations (e.g., they may or may not change sign under each of the possible reflections at the four symmetry lines of the square). When two such functions with different transformation behavior under symmetries still have the same eigenvalue, then that degeneracy is not caused by the known symmetry group -- this makes them accidental. The "non-accidental" degeneracies are immune to symmetry-preserving perturbations, the accidental ones aren't. However, it sometimes happens that accidental degeneracies aren't really accidental, but due to hidden symmetries that we just haven't considered. Of course in the square, what I call accidental degeneracies have perfectly simple explanations when looking at the problem with separation of variables. But as I said in the introduction, my remarks concern features that don't assume integrability.
This set of functions (projector etc.) can now in principle also be applied to other groups, by changing the definition in the first two lines, i.e., the group elements and their matrices. You may also have to change applyG if the center of symmetry is not the point $(\frac12,\frac12)$.
Finally, it should be noted that this answer assumed the wave solutions to be given as stated in the question. A cleverer use of symmetry would of course be to reduce the computational domain to a wedge shaped subset on which either Dirichlet or Neumann boundary conditions can be imposed. That will automatically lead to the properly symmetrized eigenfunctions. What I did here is "post-processing," which can be useful because the brute-force calculation for the full domain is easier to program.
• This is excellent - thank you! I've just tested it on the 2nd and 3rd eigenfunction and it works. I'll have to invest the time into finding out exactly why this works (and the group theory behind it). However, I set the eigensystem to find the first 15 functions, and tested the projection on the 5th and 6th modes (which both have eigenvalue ~98.69) and the projection isn't working for me here. Am I correct in adjusting it to f5 = projector[1][funs[[5]]]? Will this also work in the case of eigenvalue multiplicity greater than 2; i.e the 31st, 32nd, 33rd mode (all share eigenvalue 493.48)? – Mr S 100 Jan 30 '16 at 23:01
• The projector in the case funs[[5]] returns numerically zero, indicating that these states don't belong to the 2D irreducible representation onto which I project. But there are other (1D) representations, each with their own projection operators. You would have to construct them using the rules of group theory and apply them to each eigenstate to isolate the symmetrized components. You'll need a character table for the group and repeat what I did above - it can be automated, but I'll have to leave that to you or someone else for now. – Jens Jan 30 '16 at 23:56
• Amazing! Thank you for all your help – your knowledge has been invaluable. As I know nothing about group theory, it's probably best that I first read up on the mathematics behind this post in order to understand your code. From what I can see, it works – and works well. I do have one final question: you say that the analysis shows that the degeneracies at eigenvalue 98.71 are "accidental". This is obvious as the frame in the table returns irreducible representations which are not of the standard form. Is this something that just need to be spotted on sight? Thank you so much for your help. – Mr S 100 Jan 31 '16 at 12:43
• I added another paragraph on accidental degeneracies. – Jens Jan 31 '16 at 19:05
• Excellent! Thanks for clearing that up for me. I would just like to say thanks for taking the time to answer my question and having patience with me and my ignorance. This is honestly a superb answer. I now look forward to learning about the theory behind all this! – Mr S 100 Jan 31 '16 at 20:40 | 2020-10-23T03:31:56 | {
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https://math.stackexchange.com/questions/464504/if-cos4-theta-%E2%88%92-sin4-theta-x-find-cos6-theta-%E2%88%92-sin6-theta-in | # If $\cos^4 \theta −\sin^4 \theta = x$. Find $\cos^6 \theta − \sin^6 \theta$ in terms of $x$.
Given $\cos^4 \theta −\sin^4 \theta = x$ , I've to find the value of $\cos^6 \theta − \sin^6 \theta$ .
Here is what I did: $\cos^4 \theta −\sin^4 \theta = x$.
($\cos^2 \theta −\sin^2 \theta)(\cos^2 \theta +\sin^2 \theta) = x$
Thus ($\cos^2 \theta −\sin^2 \theta)=x$ , so $\cos 2\theta=x$ .
Now $x^3=(\cos^2 \theta −\sin^2 \theta)^3=\cos^6 \theta-\sin^6 \theta +3 \sin^4 \theta \cos^2 \theta -3 \sin^2 \theta \cos^4 \theta$
So if I can find the value of $3 \sin^4 \theta \cos^2 \theta -3 \sin^2 \theta \cos^4 \theta$ in terms of $x$ , the question is solved. But how to do that ?
You have $$\cos^2 \theta + \sin^2 \theta = 1$$ $$\cos^2 \theta - \sin^2 \theta = x$$ Adding and subtracting the two equations gives $$\cos^2 \theta = {1 + x \over 2}$$ $$\sin^2 \theta = {1 - x \over 2}$$ Substituting you have $$\cos^6 \theta - \sin^6 \theta = \bigg({1 + x \over 2}\bigg)^3 - \bigg({1 - x \over 2}\bigg)^3$$ $$= {3 \over 4} x + {1 \over 4} x^3$$
• Very elegant solution. Even though I had those two equations , I never thought of solving them simultaneously to obtain cos^2 theta & sin^2 theta . Thanks a lot . – A Googler Aug 11 '13 at 5:13
• @AGoogler you saying you never knew half angle equations? :) – Kaster Aug 11 '13 at 5:50
• @Kaster Nope I do not mean that. I'm saying that even though I obtained the two equations (cos^2 - sin^2 =x and cos^2+sin^2=1 ) , I did not think of solving them to get cos^2 and sin^2 . – A Googler Aug 11 '13 at 7:28
• @AGoogler OK, what I meant is $\cos 2\theta = x$, so $\cos^2 \theta = \frac {1+\cos 2\theta}2 = \frac {1+x}2$; $\sin^2 \theta = \frac {1-\cos 2\theta}2 = \frac {1-x}2$ – Kaster Aug 11 '13 at 7:40
$$\cos^6\theta-\sin^6\theta = \left ( \cos^2 \theta\right )^3 - \left (\sin^2 \theta \right )^3 = \\ = \left( \cos^2 \theta - \sin^2 \theta\right ) \left(\cos^4 \theta + \sin^2\theta \cos^2\theta + \sin^4\theta \right ) = \\ = x \left ( \cos^4 \theta - 2\cos^2\theta\sin^2\theta + \sin^4 \theta + 3 \cos^2\theta\sin^2\theta\right ) = \\ = x \left( \left(\cos^2\theta - \sin^2 \theta \right )^2 + \frac 34 \sin^22\theta\right ) = x \left( x^2 + \frac 34 \left( 1-\cos^2 2\theta\right )\right ) = \\ = x \left(x^2 + \frac 34 (1-x^2) \right ) = \frac x4 \left(x^2+3 \right )$$
• I like the way you substituted ( -2cos^2sin^2 + 3cos^2sin^2 ) for (+cos^2sin^2) . How do you guys manipulate equations like this with ease ? – A Googler Aug 11 '13 at 7:36
• @AGoogler, I don't really know actually. Never thought about it. It's just the way I was taught in high school. – Kaster Aug 11 '13 at 7:44
For typing ease, let $a=\cos^2\theta$ and $b=\sin^2\theta$. Thus $a+b=1$. We are told that $a^2-b^2=x$, or equivalently that $a-b=x$.
We want $a^3-b^3$. It will be enough to find $a^2+ab+b^2$. Note the identity $$4(a^2+ab+b^2)=3(a+b)^2+(a-b)^2.$$
Remark: For other problems of a similar character, it might be preferable to extract $ab$ from the identity $4ab=(a+b)^2-(a-b)^2$, and use standard techniques for expressing symmetric functions of $a$ and $b$ in terms of the elementary symmetric functions $a+b$ and $ab$.
It would be easier if you decompose $\cos^6 \theta -\sin^6 \theta$ into $(\cos^2 \theta -\sin^2 \theta)(\cos^4 \theta +\cos^2 \theta \sin^2 \theta+\sin^4 \theta)$
As already found $\cos^2\theta-\sin^2\theta=x,\implies \cos2\theta=x$
Using $a^3-b^3=(a-b)^3+3ab(a-b),$
$$\cos^6\theta-\sin^6\theta=(\cos^2\theta)^3-(\sin^2\theta)^3$$
$$=(\cos^2\theta-\sin^2\theta)^3+3\cos^2\theta\sin^2\theta(\cos^2\theta-\sin^2\theta)$$
$$=x^3+3x\cos^2\theta\sin^2\theta$$
$$=x^3+\frac{3x}4\sin^22\theta\text{ as }\sin2A=2\sin A\cos A$$
$$=x^3+\frac{3x}4(1-\cos^22\theta)$$
$$=x^3+\frac{3x}4(1-x^2)=\frac{4x^3+3x-3x^3}4=\frac{x(x^2+3)}4$$ | 2020-02-19T05:16:15 | {
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http://kpru.fitnessnutritionshop.it/how-to-find-the-coordinates-of-a-point-on-a-circle-given-the-radius.html | # How To Find The Coordinates Of A Point On A Circle Given The Radius
x2 1 y2 5 1 10. The calculator will generate a step by step explanations and circle graph. Unit circle diagram is provided in each worksheet for reference. Learn vocabulary, terms and more with flashcards, games and other study tools. Remember that the radian measure of an angle is the length of this arc on a circle of radius 1. Investigation 2: coordinates on a circle Ferris wheels are circular and rotate about the center. Subtract the y-coordinate of the point on the circumference from the y-coordinate of the midpoint and then square the difference. Solution of exercise 5. Similarly to find the y-intercepts, set x = 0 in the equation and solve for x. This form of the equation is helpful, since you can easily find the center and the radius. The radius is any line segment from the center of the circle to any point on its circumference. The radius of the circle the circle (point O, figure 3-3) to the ends of an It should be noted that for a given intersecting angle or central angle, when using the arc definition, all the. r is (roughly) the distance from the origin to the point; is the angle between the radius vector for the point and the positive x-axis. The equation is now in the standard form of equation (2. I have points with a given latlong and a distance around them - e. Find the equations of the two circles that satisfy these conditions. Graph triangle RST and construct the perpendicular bisectors of two sides to locate the center of the circle. (x + 4)2 + (y + 2)2 = 25 Write the equation of the circle that passes through the given point and has a center at the origin. Since you have the coordinates of the center (Cx, Cy), simply add the calculated offset. What is the equation of the circle? Edit: question might be faulty since apparently there coULD be multiple answers, but it'd be cool if someone showed me how to do it when there's 3 instead of 2 circumference points given. Find the equation of a circle with center point coordinate of (3,4) and a point P (-1,3) Click input boxes to enter data. IM Commentary. _____ _____25) Sketch the graph of. Code to add this calci to your website. The centre will be given by and the radius will be. Name three more points on the circle. A circle has a radius of 5 and its circumference passes through points (3,1) and (-2,-4). The circle above has its center at point C and a radius of length r. Find the radius and then use the center and radius to write an equation. From the definition of an osculating circle, we can calculate the center of curvature which we will denote by $\vec{r_c}(t)$, by the following formula: (1). Find the line bisecting the two tangent lines. In particular, the rectangular coordinates of a point P are given by an ordered pair (x,y)\text{,} where x is the (signed) distance the point lies from the y -axis to P and y is the (signed) distance the point lies from the x -axis to P\text{. It accepts a variety of formats:. 0288632 : 20 miles, 51. This was an exercise we're required to do in my class on CS. You must also define a range or tangential points. By definition, all radii of a circle are The diameter of a circle is the segment that contains the center and whose endpoints are both on the. [AS-level Maths: Circle coordinate geometry] How do I find the center of a circle given a point on its circumference and its radius? Mathematics (A-Levels/Tertiary/Grade 11-12) Close. The great circle distance is proportional to the central angle. To use the calculator, enter x and y coordinate of a center and radius of each circle. Let r be the radius of the. Where does the point P, which has the coordinates negative six, comma, negative six, lie? We have three options. Answer : is a way to express the definition of a circle on the coordinate plane. Solution for Find the coordinates of a point on a circle with radius 30 degrees corresponding to an angle of 250 degrees. To find the area of a circle, you use the following formula: or To find the circumference of a circle, you need to use the following formula: or C = x 2r. Substitute in the values for. Functions and Graphs. Coordinate Geometry -. net turns out to be a simple application of line intersection. 9 centered at the origin. By definition, all radii of a circle are The diameter of a circle is the segment that contains the center and whose endpoints are both on the. Fortunately there is an easy way to calculate the circle center and circle radius. enter the radius of circle r=6 enter the point x1=3 to find whether the. Note that because spheres are three-dimensional, this will. ) Diameter - The distance from one edge of a circle to the other side through the center. Find the reference angle by measuring the smallest angle to the x-axis. Consider the general equation a circle is given by ${x^2} + {y^2} + 2gx + 2fy + c = 0\,\,\,{\text{ – – – }}\left( {\text{i}} \right)$. com Blog: http. Draw a line through two points. I need the formulas to get radius and x/y-coordinates of a circle, which intersects 2 given points (x1/y1 and x2/y2) and a given tangent slope (y=m*x+b). So you can substitute in (0, -5) for the center ( x0, y0) and you will get an equation involving x, y and r. Get the radius of the circle. The diameter of a circle is any straight line segment that passes through the center of the circle and whose endpoints are on the circumference of the circle. Equation of a Circle Through Three Points Calculator. Those may sound like four easy steps, but embedded within them is the knowledge to find The equation of a perpendicular line. Points of a Circle. geography ST_Buffer(geography g1, float radius_of_buffer, integer num_seg_quarter_circle); Description Returns a geometry/geography that represents all points whose distance from this Geometry/geography is less than or equal to distance. Note in Quadrant I, both X and Y Find the terminal point P(x, y) on the unit circle determined by the given value of t. Round my answers to three decimal places. Exam Style Questions Ensure you have: Pencil, pen, ruler, protractor, pair of compasses and eraser 2. Radius of circle given area. Substitute in the values for. Sal 'eye-balled' the center and was given a point on the circle, which he then used to find the radius. The formula for the unit circle relates the coordinates of any point on the unit circle to sine and cosine. Select interior, exterior, or on the circle (x - 5) 2 + (y + 3) 2 = 25 for the following point. (x 2 2) 2 1 (y 1 5) 2 5 4 15. Show Step-by-step Solutions. To create multiple rings, separate the values in the "radius" box with commas: e. Determine the appropriate signs for $x$ and $y$. The fixed distance r from the center to any point on the circle is called the radius. The radius is a line segment. Any point (x,y) on the path of the circle is x = rsin(θ), y = rcos(θ). The above formula is considered as the standard equation of a circle. Determine the appropriate signs for x and y in the given quadrant. I have points with a given latlong and a distance around them - e. Similarly to find the y-intercepts, set x = 0 in the equation and solve for x. geography ST_Buffer(geography g1, float radius_of_buffer, integer num_seg_quarter_circle); Description Returns a geometry/geography that represents all points whose distance from this Geometry/geography is less than or equal to distance. The polar coordinates are given to you in a variety of forms. Given the equation of a circle, and the coordinates of a point, how will we come to know that the point lies inside, outside or on the circle? This lesson will answer this very question. For the last part you need to think where the gradient of a radius of the circle is the negative reciprocal of 2/7. 2 Polar Coordinates The coordinate system we are most familiar with is called the Cartesian coordinate system, a rectangular plane divided into four quadrants by horizontal and vertical axes. Convention dictates that Zero degree on a circle is at 3 o'clock. But the lengths of the legs aren't just plain old x and y anymore. The radius is given in map units by default, but you can also give the value in other units by specifying a distance units abbreviation with the value that you enter. With the help of Compass & Rulers perfect circle can be drawn easily. In a unit circle, the length of the intercepted arc is equal to the radian measure of the central angle Finding Sines and Cosines of Angles on an Axis. Online geometry calculator which helps you to calculate the radius and center of a circle, from the given three points of circle coordinates. The distance between the centre and any point of the circle is called the radius of the circle. Повторите попытку позже. In the worksheet above, the $$x$$-coordinates of the blue, green and orange points are given by the angle. Find the line bisecting the two tangent lines. The task is to find minimum radius of the circle so that at-least k points lie inside the circle. Find the coordinates of a point on a circle at the opposite end of the diameter from point A (,)? The circle's formula is from the circle's formula we can see that radius ; so, diameter is , the center is at (,) now we can find equation of the line passing through points A (,) and (,):. Find the Point on a Circle Given an Angle and the Radius - Duration: 3. The fixed point is called the centre and the given constant distance is known as the radius of the circle. a Find an equation of the perpendicular bisector of AB. Substitute the values of the center of the circle in place of x and y. 1 m The radius of 08 is given. 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. The fixed point C (p, q) is the center and r is the radius of this circle, according to the definition. In other terms, it simply refers to the line drawn from the center to any point on the circle. The center of the given circle is (1, - 2) and its radius is √(98) ,so the coordinate of the vertices of the square inscribed in the given circle are ( 1 + √(98)cosθ , - 2 + √(98)sinθ) where θ = ±π/4,±3π/4 S2 square of the distance of a vertex from the origin = 5 + 98 + 2√(98)(cosθ - 2sinθ ) which is maximum when θ = - π/4. Find the radius of a circle, if the tangent segment to the circle is of 15 units long and If we draw the radius OB from the center O of the circle to the tangent point B (Figure 2), then the radius OB will be perpendicular to the tangent line AB in accordance with the lesson A tangent line to. Given the equation of a circle, and the coordinates of a point, how will we come to know that the point lies inside, outside or on the circle? This lesson will answer this very question. Solution for Find the coordinates of a point on a circle with radius 30 degrees corresponding to an angle of 250 degrees. To determine the position of a given point with respect to a circle, all we need to do is to find the distance between the point and the center of the circle. The units are in place to give. If mAOB 180, points A and B and the points of the circle in the inte-rior of AOB make up minor arcAB, written as. The radius of the circle is just the distance from its center to any point on the circle. Chord :A line segment with end points on the circle. d = 8 ft 2. Determine the ratio in which the 2x + y = 4 divides the line segment joining the points (2,-2) and (3,7). Apply the formulas to calculate diameter, circumference and area. Any circle with its center at the origin has the equation x 2 + y 2 = r 2, where r is the radius of the circle. Convention dictates that Zero degree on a circle is at 3 o'clock. The radius is a vector from the origin. EXAMPLE: Find the coordinates of the center and the radius of the circle given by the equation. Radius: the distance from the center of a circle to any point on it. 5 centered at the origin. If I pick a fixed geohash length (say equals to ~ 1x1mi. A circle can be described by defining the x,y coordinate of the centre point, as well as the the radius. area: The interior surface of a circle, given by $A = \pi r^2$. Actually, this is done surprisingly easy, by simply using the atan2 method. 6: Given the center and the radius, find the equation of a circle in the coordinate plane or given the equation of a circle in center-radius form, state the center and the radius of the circle. Re: Coordinate Geometry Formulas. The center of the circle separates the diameter into two equal segments called radii (plural for In this example, figure the radius length from the center of the circle (3,1) to the endpoint of the. I have a set of lat/long points. ) Radius - The distance from the center of a circle to its edge. The angles from 0 to n give the whole circle. create a triangle where the radius is the longest line. Points of a Circle. I think you are showing an example when the radius of the quarter circle is 8. If not on a coordinate graph use the points and the distance. This calculator can find the center and radius of a circle given its equation in standard or general form. Let the center of the circle be at (0, b) and radius of circle be r. I found what I thought was the equation on another forum, but it doesn't seem to be right. A line segment connecting two points on the circle and going through the center is called a diameter of the circle. How can I re-use this? Proof: the angle of a triangle opposite the longest side is the largest angle. 29 In the diagram below, Circle 1 has radius 4, while Circle 2 has radius 6. " The distance from the center point to a point on the circle is called the radius of the circle, shown in the diagram as r. (x + 4)2 + (y + 2)2 = 25 Write the equation of the circle that passes through the given point and has a center at the origin. Given a chord of a circle, it's possible to find the angle of the arc. This line will contain the center point of the circle. For a given circle center, we compute the optimum circle radius ˆr by solving: (5) ¶J ¶r n r=rˆ =0 )rˆ = å i=1 d i n This means that both d i and ˆr can be considered as functions of the circle center coordinates x and y which from now on will be the free parameters of our model. r = 33 ft 6. The angle unit is radian. The numerical side, the 16, is the square of the radius, so it actually indicates 16 = r 2 = 4 2, so the radius is r = 4. Since the point of tangency is given, that’s the point to use. B, and C are the corner points of the thin triangular plate of constant density shown here. Given line AB, point P, and radius R (Figure 4. !A circle has centre (5, 2) and radius 4. Site: http://mathispower4u. R = earth’s radius (mean radius = 6,371km) Digging deeper, I found Chris's Vincenty formula for distance between two Latitude/Longitude points page which includes a table on different datum models (treating Earth as an ellipsoid), it shows WGS-84 & GRS-80 having the greatest radius on an ellipsoid as 6378. To find the length of the radius of the circle you need to. Great-circle distance between two points. Find a point on the unit circle given one coordinate and the quadrant in which the point lies. Given that is the equation of a circle, state the coordinates of the centre and the radius. In fact, this was a trick people used in the past to find the area under a curve on a graph (before technology gave us. SOLUTION: Rearrange and complete the square in both x and y:. Given the start and finish coordinates stated in your Post, the start angle of your arc is either 270, or -90 deg, ending at 90, or -270 deg depending on the direction of rotation of the arc. In the worksheet above, the $$x$$-coordinates of the blue, green and orange points are given by the angle. The circle above has its center at point C and a radius of length r. Find the Equation of a Circle Given Its Center and One Point on Its Circumference. The center point of the circle is the center of the diameter, which is the midpoint between (−2,4) ( - 2. Online geometry calculator which helps you to calculate the radius and center of a circle, from the given three points of circle coordinates. State the sign of the sine or cosine value of an angle based on the quadrant in which the terminal side of an angle occurs. Any tangent to the circle will be y = mx + 5 \sqrt{1+m^2}. With this method you can restrict the generated random points to a circular region that you specify, or you can extend it to include the whole earth. I have points with a given latlong and a distance around them - e. To find the length of the radius of the circle you need to. As can be seen by the labels on the $$x$$-axis, one revolution does not correspond to 360 but to $$2\pi$$. I found what I thought was the equation on another forum, but it doesn't seem to be right. Looking at the figure above, point P is on the circle at a fixed distance r (the radius) from the center. _____ _____25) Sketch the graph of. Lesson Notes. Without loss of generality, we could transform the coordinates of the given circle (with radius R), and the given point, such that the center of the circle is at the origin, and the new coordinates of the As far as I know, you can find only the equation of pair of tangents from any point outside the circle!. A circle is the set of all points equidistant from a given point. Use this circle calculator to find the area, circumference, radius or diameter of a circle. P = (10, 6) and M = (-4, 8). Dominic thinks that angles A and B have the same radian measure. Find the coordinates of the center of the circle and length of the radius given the equation of that circle which is 5x^2+15x+5y^2-10y-10=0 Divide thru by 5 to get: x^2 + 3x + y^2 - 2y - 2 = 0. Finding the Radius as the Distance Between Two Points. The special case r = 1 is called the unit circle; its equation is x2+y2 = 1. Find the coordinates of a point on a circle with radius 15 corresponding to an angle of 355∘. Find the intersection of two lines. To use the calculator, enter x and y coordinate of a center and radius of each circle. The midpoint has much in common with the radius, for the midpoint on a diameter measures its corresponding radius since the diameter's length is twice that of its radius. So the slope of the radius is given by. To find the circle. 1 asks us to prove that all circles are similar. •Center Point that is equidistant from all points on the circle. Incorporate these simple unit circle PDFs to determine the coordinates of the terminal point for the given angle measures. Find the radius and then use the center and radius to write an equation. Explain why. The center of the circle will be (–3, 6), and the radius, which is the distance from (–3,6), will be 5. 1258957 : 100 miles}. Draw a circle of a given radius with a given center. We can use the Pythagorean Theorem to find the distance between two points on a graph. we know that the line joining two points in the circumference of a circle is its diameter and the origin lies at the midpoint of the diameter. Find the Equation of a Circle Given Its Center and One Point on Its Circumference. Find the cosine and sine of the reference angle. Example 3 Find the angle between the tangents to the circle x 2 + y 2 = 25, form the point (6, 8). As with circles, the radius of a sphere is often an essential piece of starting information for calculating the shape's diameter To begin, find the coordinates of the sphere's center point. Definition: A circle is a simple shape, consisting of those points in a plane that are a given distance from a given point - the centre. This circle is circle "A. I do not even know what a locus is. To graph a circle, visit the circle graphing. both the curves are symmetrical about the y axis and will intersect in 2 points. First, rewrite the equation so that the two x terms are next to one another, the two y terms are next to each other, and the constant is on the other side. Substitute in the values for. The radius of the circle the circle (point O, figure 3-3) to the ends of an It should be noted that for a given intersecting angle or central angle, when using the arc definition, all the. A line representing the radius appears inside the circle as you draw it. In the end, we will put all. Every circle centered at the origin will have this type of equation. What if you were given the length of the radius of a circle and the coordinates of its center? This definition can be used to find an equation of a circle in the coordinate plane. Re: Finding a point on a circle, given only anothe point, the radius, and the arc len The co-ordinates of your blue point cannot be determined with the data given. Then graph the circle. But the lengths of the legs aren't just plain old x and y anymore. SOLUTION: Rearrange and complete the square in both x and y:. To find this point, calculate the average of the x-coordinates of the endpoints, and also the average of the y. P is the point where the angle intersects the circle and is. Get angle in radians given a point on a circle. Given a circle on the coordinate plane, Sal finds its standard equation, which is an equation in the form (x-a)²+(y-b)²=r². Coordinates of a point on a circle. T must be the same point, so the radius from the center of the circle to the point of tangency is perpendicular to the tangent line, as desired. 7 Circles in the Coordinate Plane(continued) Name _____ Date _____ Work with a partner. Construct the perpendicular bisectors of two sides. x2 + y2 + 6x + 4y - 3 = 0 (x - 3)2 + (y-2) = 42, so the center is (3,2). 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. The center point of the circle is the center of the diameter, which is the midpoint between (−2,4) ( - 2. find the ratio in which the line segment joining A(2,-2)and B(-3,-5)is divided by the y axis. We can still draw a right triangle, so Pythagoras can still help us out. Destination point given distance and bearing from start point You have starting points: xcoordinate, a list of x-coordinates (longitudes) ycoordinate, a list of y-coordinates (latitudes) num_samples, the number of samples in the plane towards the destination point bearings: heading, a list of headings (degrees) and distances: range, a list of di…. The following formula gives the distance between 2 points x1,x2 and y1,y2 in the cartesian plane: sqrt(pow(x1-x2,2) - pow(y1-y2,2)) given the center and a point on the circle, you can use this formula to find the radius of the circle. The locus of all points equidistant from a single point is a circle. One point on the circle is (6,-3). Find the equation of a circle with center point coordinate of (3,4) and a point P (-1,3) Click input boxes to enter data. (ii) the radius of C. From P draw an arc with radius R, cutting line DE at C, the center of the required tangent arc. The x-coordinate tells you how many steps you have to take to the right (positive) or To find out the coordinates of a point in the coordinate system you do the opposite. Click 'reset' and note this angle initially has a measure of 40°. To support your homeschooling, we're including unlimited answers with your free account for the time being. The equation of a circle is X minus H squared plus Y minus R of course is the radius, stands for radius. Find the radius of a circle, if the tangent segment to the circle is of 15 units long and If we draw the radius OB from the center O of the circle to the tangent point B (Figure 2), then the radius OB will be perpendicular to the tangent line AB in accordance with the lesson A tangent line to. Substitute the radius and center (h, k) into (x - h) 2 + (y - k) 2 = r 2 when the center is NOT (0,0). In Cartesian coordinates, the equation of a circle is ˙(x-h)2+(y-k)2=R2. Any interval joining a point on the circle to the centre is called a radius. Examples:. The standard G-C. If the center is at (0,0) then it will look like the following:. At first, take a note on the value of the circumference value of the circle. Functions and Graphs. Equation of a Circle, given three points. Siyavula's open Mathematics Grade 12 textbook, chapter 7 on Analytical Geometry covering Equation Of A Tangent To A Circle. r = 21 em 5. 0000 Case 2) Given points are opposite ends of a diameter of the circle with center (0. asked by mwaba on March 23, 2012; Calc. Substitute the radius value into x 2 + y 2 = r 2 when the center is (0,0). Given a circle and a point on the circle, it is relatively easy to find the tangent line using coordinate geometry. Given the start and finish coordinates stated in your Post, the start angle of your arc is either 270, or -90 deg, ending at 90, or -270 deg depending on the direction of rotation of the arc. Please help. Bitmap/Midpoint circle algorithm -- Takes the center of the circle and radius, and returns the circle points v Returns points for a circle given center and radius. Straight line. Find answers to Function to find X Y coordinates of an arcs center point from the end points, circle centered at (0,0) - if not, then just apply a linear offset. ) Circle - A set of points equidistant from a given fixed point on a plane. The equation of a circle is X minus H squared plus Y minus K squared is equal to R squared. Make a conjecture about the angle between the radius and the tangent to a circle at a point on the circle. Points on a circle. Quation of a Tangent I Equation of a tangent to a circle at a given. The point A(10,7) lies on the circle. If we place the center point on the origin point, the equation of a circle with center point (0, 0) and radius r is:. 0288632 : 20 miles, 51. Below is another method to find the equation of the circle. The angle unit is radian. Determine the appropriate signs for x x and y y in the given quadrant. [AS-level Maths: Circle coordinate geometry] How do I find the center of a circle given a point on its circumference and its radius? Mathematics (A-Levels/Tertiary/Grade 11-12) Close. A line representing the radius appears inside the circle as you draw it. It works anywhere JavaScript runs. Find The Maximum Area Of A Rectangle Inscribed In A Circle Of Radius R. I drew 5 different chords with 5 perpendicular lines coming from Finding the center of circles is of great interest to me, but I'm wondering how well / easy this would work for finding the centers of much smaller diameters. Any tangent to the circle will be y = mx + 5 \sqrt{1+m^2}. Destination point given distance and bearing from start point You have starting points: xcoordinate, a list of x-coordinates (longitudes) ycoordinate, a list of y-coordinates (latitudes) num_samples, the number of samples in the plane towards the destination point bearings: heading, a list of headings (degrees) and distances: range, a list of di…. GMAT - Find the Area of the Shaded Region. This allows you to find out where they intersect, and what areas are not within the radius of any of your locations. Online geometry calculator which helps you to calculate the radius and center of a circle, from the given three points of circle coordinates. Step 1: Note the Circumference value. For a given circle center, we compute the optimum circle radius ˆr by solving: (5) ¶J ¶r n r=rˆ =0 )rˆ = å i=1 d i n This means that both d i and ˆr can be considered as functions of the circle center coordinates x and y which from now on will be the free parameters of our model. Let ∅ be the angle between Do not mix r, the polar coordinate, with the radius of the circle. Move the center of the circle a small distance toward point P (call this new center. Let us put a circle of radius 5 on a graph: Now let's work out exactly where all the points are. Google Classroom. By the definition of a circle, any two radii have the same length. The y coordinate of the point to check. Use the midpoint formula to find the midpoint of the line segment. For example, some will have a negative r value, some will have the angle in radians, and some will have the angle in degrees. Given the centre and radius of a circle, to find the equation of Circle K?. But circle equations are often given in the general format of ax2 + by2 + cx + dy + e = 0, When you are given this general form of equation and told to find the center and radius of a. The equation is of the form where (a, b) is the centre and r is the radius. If you know two points that a line passes through, this page will show you how to find the Fill in one of the points that the line passes through Calculus, Derivatives Calculus, Integration Calculus, Quotient Rule Coins, Counting Combinations, Finding all Complex Numbers, Adding of Complex. [12th July 2018] Unfortunately, due to a large price increase in back-end services, we can no longer offer some features [Zoom and Pan to find the required area on the map then click on the map to draw a circle] OR [Type a location into the text box and click Draw Radius]. Re: Coordinate Geometry Formulas. In particular, the rectangular coordinates of a point P are given by an ordered pair (x,y)\text{,} where x is the (signed) distance the point lies from the y -axis to P and y is the (signed) distance the point lies from the x -axis to P\text{. A complete lesson on equation of a circle featuring video examples, interactive practice, self-tests, worksheets and more. Find the diameter of 0B. Given a positive integer K, a circle center at (0, 0) and coordinates of some points. 6826048,-74. •Center Point that is equidistant from all points on the circle. You are told that the point (2, 3) is on the circle, which means it "solves" the equation. With this method you can restrict the generated random points to a circular region that you specify, or you can extend it to include the whole earth. Simplify the expression to find r (the radius). Find the radius. Calculating A Circle's Center and Equation From 3 Points. In this case since we are given the coordinates of the center and the fact that the circle passes thru (5,0), we can calculate the radius (needed for finding the area of the. The general equation of a circle is x 2+y +2gx+2fy +c = 0, where the centre is given by (−g,−f) and the radius by r = p g2 +f2 − c. 6: Given the center and the radius, find the equation of a circle in the coordinate plane or given the equation of a circle in center-radius form, state the center and the radius of the circle. Below is another method to find the equation of the circle. Find the coordinates of a point on a circle at the opposite end of the diameter from point A (,)? The circle's formula is from the circle's formula we can see that radius ; so, diameter is , the center is at (,) now we can find equation of the line passing through points A (,) and (,):. It uses ByRef parameters to return the coordinates of the points of intersection. (Call this point P. 9876) Case 4) Given points are farther away from each. The line segment connecting the origin to the point $$P$$ measures the distance from the origin to $$P$$ and has length $$r$$. Looking at the figure above, point P is on the circle at a fixed distance r (the radius) from the center. The standard circle is drawn with the 0 degree starting point at the intersection of the circle and the x-axis with a positive angle going in the. Coordinates of a point on a circle. Here, the point of tangency is (-13, 96) and the center of the circle is (-22, 56). This equation represents a circle with the center at (2,3) and with a radius equal to or 4. You know the angle where the two tangent lines cross, so the angle in the triangle at that point is 1/2 that. The center of the circle will be (–3, 6), and the radius, which is the distance from (–3,6), will be 5. For example, if the question states that the circle has centre (1,2) this tells us that the x coordinate of the centre is 1 and the y coordinate of the centre is 2, hence a=1 and b=2. Let be any point on the circle. Point A (9, 2) Point B (3, -4) Point C (5, -6) Let's put these points into 'x' and 'y' coordinates. 0 should be treated as never describing circles (except in the case where the points are coincident). With the help of Compass & Rulers perfect circle can be drawn easily. I already inserted a graph with the X and Y axis with the (2,0) marked. This dial should lets you mousemove anywhere in a circle to adjust the position of the dial to a point on the circle with the same angle as the click. Given the equation of a circle, and the coordinates of a point, how will we come to know that the point lies inside, outside or on the circle? This lesson will answer this very question. A radius measures the distance from a circle's middle point, or origin, to its surrounding perimeter, also known as its circumference. So now substitute (2, 3) in for ( x, y) in your equation and you will be able to find the radius r. Area of a quadrilateral. 0288632 : 20 miles, 51. Actually, this is done surprisingly easy, by simply using the atan2 method. area: The interior surface of a circle, given by $A = \pi r^2$. Sets the radius of the pen to the specified size. To find and understand this function, it must also be constructed. Given the centre and radius of a circle, to find the equation of Circle K?. Solution of exercise 5. ) The circle radius may be reduced to this value, and it will still contain all of the data points, with one point, P, lying on the circle. Given the circle with equation (x + 22) 2 + (y – 56) 2 = 1681, give an equation for the line tangent to the circle at the point (-13, 96). We then rotate a circle with fixed-radius 'R' about the point where, x and y represent the coordinates of a point and 'd' is the distance between P and Q. •Indicated by symbol – P Circle with center P. 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. The haversine formula is an equation important in navigation, giving great-circle distances between two points on a sphere from their longitudes and latitudes. The radius is a vector from the origin. (x - 2)2 + (y - 1)2 = 49 19. This form of the equation is helpful, since you can easily find the center and the radius. An arc of a circle is the part of the circle between two points on the circle. The numerical side, the 16, is the square of the radius, so it actually indicates 16 = r 2 = 4 2, so the radius is r = 4. Whenever possible it is always a good idea to draw a graph to give a better understanding of what is going on in the question. Use coordinates to prove simple geometric theorems algebraically. The fixed point C (p, q) is the center and r is the radius of this circle, according to the definition. The standard circle is drawn with the 0 degree starting point at the intersection of the circle and the x-axis with a positive angle going in the. Find the coordinates of a point on a circle at the opposite end of the diameter from point A (,)? The circle's formula is from the circle's formula we can see that radius ; so, diameter is , the center is at (,) now we can find equation of the line passing through points A (,) and (,):. To determine the position of a given point with respect to a circle, all we need to do is to find the distance between the point and the center of the circle. The hypotenuse is the radius of the circle, and the other two sides are the x and y coordinates of the point P. The slopes of two parallel lines, m1 and m2 are equal if the lines are parallel. Because we will use this value to solve the radius of the circle. One point on the circle is (6,-3). The circle has a radius r, and its center point is the vertex corresponding to the angle θ. IM Commentary. I am looking for a method to add distance to a starting GPS coordinate point to create a bounding box (to approximate a circle) which I will then use to find all matches in my MySQL database. Given 3 points which are not colinear (all on the same line) those three points uniquely define a circle. Click 'reset' and note this angle initially has a measure of 40°. Let us put a circle of radius 5 on a graph: Now let's work out exactly where all the points are. The midpoint has much in common with the radius, for the midpoint on a diameter measures its corresponding radius since the diameter's length is twice that of its radius. r = 21 em 5. In a unit circle, the length of the intercepted arc is equal to the radian measure of the central angle Finding Sines and Cosines of Angles on an Axis. This "circle generator" will plot a point on a map — given a set of coordinates or other location (an airport code, postal code, city/state pair, or coordinate pair) — and draw a circle, or circles, around that point. •Indicated by symbol – P Circle with center P. If the two lines are perpendicular, m1*m2=-1. , "10mi,50mi,100mi. As you can see, calculating a circle center based on 3 known points can be a bit complicated. Given the centre and radius of a circle, to find the equation of Circle K?. Shortcut in finding the circle's radius length Let sqrt stands for the square root. Given that is the equation of a circle, state the coordinates of the centre and the radius. 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. center (-3, 3), radius 4 Plot the center and draw a circle with the given radius. Enter the coordinates into the text boxes to try it out. Functions and Graphs. Please help. Calling this , then for , , so. We can solve these three using the method of simultaneous equations, and. number of revolutions completed by a rotating wheel in one minute. Is the radius the distance from the center to the point? The area of a circle is pi* r*r= so the circle would be 50. Coordinates of intersection of a tangent from a given point to the circle. How To: Given the angle of a point on a circle and the radius of the circle, find the $\left(x,y\right)$ coordinates of the point. Give an equation for the circle with center (107, -3. 5) In order to further reduce the radius, you must move the center position. Designate a random point on the circle (x,y). Whenever possible it is always a good idea to draw a graph to give a better understanding of what is going on in the question. Point G will be shown to be its center. Find answers to GIVEN: X-Y-Z coordinates of 3 points in 3D. I need to calculate the center point and radius of a circle that will encompass all the points. The fixed point C (p, q) is the center and r is the radius of this circle, according to the definition. SOLUTION: Rearrange and complete the square in both x and y:. Chord :A line segment with end points on the circle. Finding the Center of a Circle Given 3 Points Date: 05/25/2000 at 00:14:35 From: Alison Jaworski Subject: finding the coordinates of the center of a circle Hi, Can you help me? If I have the x and y coordinates of 3 points - i. Find the coordinates of a point on a circle with radius 15 corresponding to an angle of 355∘. (x 2 2) 2 1 (y 1 5) 2 5 4 15. Let r be the radius of the. Given the angle of a point on a circle and the radius of the circle, find the (x, y) (x, y) coordinates of the point. The center is at ( h , k ) , and the radius is r. Processing. The pen is circular, so that lines have rounded ends, and when you set the pen radius and draw a point, you get a circle of the specified radius. It works anywhere JavaScript runs. Use similar triangles to find the coordinates of the point Q that divides the segment from z,) to P2(X2, into two lengths whoseratio is p/q = r. coordinates of the center and the. Center point coordinates (3,4) ( h , k ) Any Point coordinates, for example (-1,3) on its circumference ( x , y ). The equation of circle of radius 5 and touching the coordinates axes in third quadrant, is. The task is to generate uniformly distributed numbers within a circle of radius R in the (x,y) plane. We will find the center of the circle from and then we will find the radius with the help of distance formula. The center of the circle separates the diameter into two equal segments called radii (plural for In this example, figure the radius length from the center of the circle (3,1) to the endpoint of the. Let be the smallest Integer Radius of a Circle centered at the Origin (0, 0) with Lattice Points. By the definition of a circle, any two radii have the same length. Sets the radius of the pen to the specified size. (x 2 2) 2 1 (y 2 6) 2 5 49 11. - 270 degrees b. The distance formula in coordinate geometry is used to calculate the distance between 2 points whose coordinates are given. let (x1,y1) and (x2,y2) be two points on the circumference of the circle. But can't we just estimate the no. Definition: A circle is a simple shape, consisting of those points in a plane that are a given distance from a given point - the centre. The central angle between the two points can be determined from the chord length. Coordinates of intersection of a tangent from a given point to the circle. Chapter 12 studies parametric equations in detail-here we stay with the circle. Enter a radius and address to draw a circle on a map. a Find an equation of the perpendicular bisector of AB. 6826048,-74. You know the angle where the two tangent lines cross, so the angle in the triangle at that point is 1/2 that. The hypotenuse is the radius of the circle, and the other two sides are the x and y coordinates of the point P. According to the formula, the x coordinate of a point on the unit circle is c o s ( θ) and the y coordinate of a point on the. Given a positive integer K, a circle center at (0, 0) and coordinates of some points. State the radius and center of the circle with equation 16 = (x – 2) 2 + (y – 3) 2. 0288632 : 20 miles, 51. In the worksheet above, the $$x$$-coordinates of the blue, green and orange points are given by the angle. 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. We can use the Pythagorean Theorem to find the distance between two points on a graph. Culland is finding the center of a circle whose equation is x2 + y2 + 6x + 4y - 3 = 0 by completing the squar e. d = 8 ft 2. Notice that the word ‘radius’ is being used to refer both to these intervals. Polar coordinates are an alternative to rectangular coordinates for referring to points in the plane. 7 Circles in the Coordinate Plane(continued) Name _____ Date _____ Work with a partner. Print the result. To find and understand this function, it must also be constructed. b Find the coordinates of the point where the two circles touch. " A circle is named by its center point. The general equation of a quadratic curve C(x,y)=0 passing through the given two (and exactly two) intersection points of the quadratic curves C1(x Instead of using Newton method to locate points on the curve for the marching method, a new method with dimidiate structure is proposed to trace implicit. We use it to stress test our geohash based services. The center-radius form of the circle equation is in the format (x – h) 2 + (y – k) 2 = r 2, with the center being at the point (h, k) and the radius being " r ". Name three more points on the circle. 0288632 : 20 miles, 51. It also plots them on the graph. Calculate the product of the value of PI and the value of the square of the radius. At the point of Tangency, Tangent to a circle is always perpendicular to the radius. The coordinates of the origin are (0, 0). The equation can be recognised because it is given by a quadratic expression in both x and y with no xy term, and where the coefficients of x2 and y2 are equal. Polar coordinates are an alternative to rectangular coordinates for referring to points in the plane. For this example mark up a sheet of squared paper with both & y axis then draw a circle where the coordinates of the centre. Inside the circle, on the circle, or outside the circle. Haversine Formula. The unit circle is a circle with a radius of 1. There are an infinite number of different circles that have radius 10 and contain the point (2, 3). If so any tips on how I go about finding the center? How did you get r=1? Make a right angle triangle with base of length x and height y. This is true for all angles, clockwise or anticlockwise. Then square each side of the. What is the equation of the circle? Edit: question might be faulty since apparently there coULD be multiple answers, but it'd be cool if someone showed me how to do it when there's 3 instead of 2 circumference points given. Determine the appropriate signs for x and y in the given quadrant. Find the coordinates of the vertices of the triangle. Circle C1 with center (1. Any interval joining a point on the circle to the centre is called a radius. The equation can be recognised because it is given by a quadratic expression in both x and y with no xy term, and where the coefficients of x2 and y2 are equal. Third, you need to know that = 3. The distance between the centre and any point of the circle is called the radius of the circle. By definition, all radii of a circle are The diameter of a circle is the segment that contains the center and whose endpoints are both on the. A line segment connecting two points on the circle and going through the center is called a diameter of the circle. For the last part you need to think where the gradient of a radius of the circle is the negative reciprocal of 2/7. 5 The circle C touches the y-axis at the point A (0, 3) and passes through the point B (2, 7). So now substitute (2, 3) in for ( x, y) in your equation and you will be able to find the radius r. So the slope of the radius is given by. The diameter = 2 × radius (d = 2r). Find the radius of a circle, if the tangent segment to the circle is of 15 units long and If we draw the radius OB from the center O of the circle to the tangent point B (Figure 2), then the radius OB will be perpendicular to the tangent line AB in accordance with the lesson A tangent line to. Re: Finding a point on a circle, given only anothe point, the radius, and the arc len The co-ordinates of your blue point cannot be determined with the data given. Повторите попытку позже. Step 1: You are given three points that lie on a circle. But since the points are both moving, I need to be able to determine the new coordinates based off the arc length from the point they are at, not from (0,0) Basically, given an circle with a known radius and a coordinate on that circle, how can I find a coordinate on the arc a set distance away? Thanks,-Rick. The above formula is considered as the standard equation of a circle. The great circle distance is proportional to the central angle. 135 degrees c. Any interval joining a point on the circle to the centre is called a radius. The radius of the circle the circle (point O, figure 3-3) to the ends of an It should be noted that for a given intersecting angle or central angle, when using the arc definition, all the. The tangent at a point on a circle is at We have learned that we are able to find the slope of a line, but we've never got to know how to find out. Draw a line through two points. Third, you need to know that = 3. values, giving you the coordinates of the orthocenter. Given the equation of a circle, and the coordinates of a point, how will we come to know that the point lies inside, outside or on the circle? This lesson will answer this very question. In cylindrical coordinates, we have dV=rdzdrd(theta), which is the volume of an infinitesimal sector between z and z+dz, r and r+dr, and theta and theta+d(theta). Convention dictates that Zero degree on a circle is at 3 o'clock. Note that because spheres are three-dimensional, this will. A circle has an equation 22 State the coordinates of the centre and the radius of the circle. (Call this point P. Whenever possible it is always a good idea to draw a graph to give a better understanding of what is going on in the question. A circle in the coordinate plane has a center at (3,1). The diameter of a circle is any straight line segment that passes through the center of the circle and whose endpoints are on the circumference of the circle. To find this point, calculate the average of the x-coordinates of the endpoints, and also the average of the y. It is a special case of a more general formula in spherical trigonometry, the law of haversines, relating the sides and angles of spherical "triangles". Online geometry calculator which helps you to calculate the radius and center of a circle, from the given three points of circle coordinates. Given a circle and a point on the circle, it is relatively easy to find the tangent line using coordinate geometry. The equation is of the form where (a, b) is the centre and r is the radius. (x1,y1), (x2,y2) and (x3,y3) - how do I find the coordinates of the center of a circle on whose circumference the. As with circles, the radius of a sphere is often an essential piece of starting information for calculating the shape's diameter To begin, find the coordinates of the sphere's center point. 1 (No Transcript) 2 Geometry Mini-Lesson Circle C is defined by the equation (x 4. If we place the center point on the origin point, the equation of a circle with center point (0, 0) and radius r is:. For a given point (2, 0), find the coordinates of the image point under a half-turn about the origin. Also find the co-ordinate of its centre and the length of the radius. Quation of a Tangent I Equation of a tangent to a circle at a given. Determine the appropriate signs for $x$ and $y$. EXAMPLE: Find the coordinates of the center and the radius of the circle given by the equation. Learn how to find the center and radius of a circle, either from an equation or a visual. The radius of the circle is just the distance from its center to any point on the circle. Find the coordinates of the center of the circle. Also find the co-ordinate of its centre and the length of the radius. The above formula is considered as the standard equation of a circle. You enter the latitude and longitude of a starting point and a maximum distance which is the radius of a circle centered on the starting point. Then the distance between and the center is given by. A circle is the set of all points in a plane equidistant from a given point called the center of the circle. The Web Machinist Software Program has a Circle Center 3 Known Points Calculator in it. Inside the circle, on the circle, or outside the circle. ) The circle radius may be reduced to this value, and it will still contain all of the data points, with one point, P, lying on the circle. There are an infinite number of different circles that have radius 10 and contain the point (2, 3). Then square each side of the. asked by mwaba on March 23, 2012; Calc. State whether Dominic is correct or not. Polar Coordinates and Complex Numbers. 6826048,-74. How To: Given the angle of a point on a circle and the radius of the circle, find the $\left(x,y\right)$ coordinates of the point. Section 6-2: Equations of Circles Definition of a Circle A circle is the set of all points in a plane equidistant from a fixed point called the center point. Any point (x,y) on the path of the circle is x = rsin(θ), y = rcos(θ). Similarly to find the y-intercepts, set x = 0 in the equation and solve for x. FIND: Radius of circle on those 3 points & Circle center co-ordinates from the expert community at Experts Exchange. Find the unit circle coordinates by using the simple steps that you will learn in these time-saving videos which include examples of exercises that apply to the topic. Example 3 Find the angle between the tangents to the circle x 2 + y 2 = 25, form the point (6, 8). When a circle like that modeling the Ferris wheel is placed on a rectangular coordinate grid with center at the origin (0, 0), you can. The angles from 0 to n give the whole circle. To evaluate the equation of the required circle, we must the find values of $$g,f,c$$ from the above equations (ii), (iii) and (v). " A circle is named by its center point. 6826048,-74. A circle is a shape consisting of all points in a plane that are a given distance from a given point, the centre; equivalently it is the curve traced out by a point that moves in a plane so that its distance from. Select interior, exterior, or on the circle (x - 5) 2 + (y + 3) 2 = 25 for the following point. It also plots them on the graph. Basically, the circumference is called as the perimeter of a circle. This line will contain the center point of the circle. (iii) Find the equation of each circle. Construct the perpendicular bisectors of two sides. Circles - a set of all points that are a fixed distance from a given point called the center of the circle. We use it to stress test our geohash based services. We should end up with two equations (top and bottom of circle) that can then. A bit of theory can be found below the calculator. Give an equation for the circle with center (107, -3. Also, it can find equation of a circle given its center and radius. Consider a circle with radius r and center ()hk,. The diameter = 2 × radius (d = 2r). In a unit circle, the length of the intercepted arc is equal to the radian measure of the central angle Finding Sines and Cosines of Angles on an Axis. 1258957 : 100 miles}. Polar Coordinates and Complex Numbers. Area of a circular sector. As you drag the point P around the circle, you will see that the relationship between x,y and r always holds. 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. I needed to calculate the angle between two points on the same circle. We could also know the center point and another point on the circle, which is what we are doing here. To find the x and y coordinates, we use trigonometry. Find the center and radius from the equation. A bit of theory can be found below the calculator. " The distance from the center point to a point on the circle is called the radius of the circle, shown in the diagram as r. Google Classroom. If we have the point C, which is the center of a circle, a circle of radius six, so let me draw that radius. The given end points of the diameter are. P = (10, 6) and M = (-4, 8). Important Properties: † Equation of a circle: An equation of the circle with center (h;k) and radius r is given by (x¡h)2 +(y ¡k)2 = r2:. Chapter Test A ~ i. 6: Given the center and the radius, find the equation of a circle in the coordinate plane or given the equation of a circle in center-radius form, state the center and the radius of the circle. Find the coordinates of the center of the following circle. Area of a regular polygon. Get the radius of the circle. Find the center and radius of the circle whose equation is: x 2 + y 2 + 6x — 10y — 2 = 0. If not on a coordinate graph use the points and the distance. Find the reference angle by measuring the smallest angle to the x-axis. At present, we will assume that such a model can be constructed. Given any one variable A, C, r or d of a circle you can calculate the other three unknowns. The first number in the coordinate pair, r, represents the distance from the pole and is like measuring the radius of a circle. Given 'n' points on 2-D plane, find the maximum number of points that can be enclosed by a We pick an arbitrary point P from the given set.
lgpj5mmfwgwtd1 txqy7hgp6wb f4kmy9fcj8 g6cwadndzn7 tf0941isx7 kry0vvd7h5fj rxb601hl9bkg kp51d7sl00 e4uh0qyjbatxd kvaxyf5jsm6s qy7nrpa1er7 5ai1kx18rck 00ggcye5fs70zy xgwh9cumsw6l8e 4ao7wq9m24 nxqmjml2b392we3 4ae8h4a2wvpkiu foswzekhrdqnkt u7t3p8wq2up 68pz3kvbq5w6 xg6eyjfh72t 6ns7i9wxwv 8ko2dyt0i0zrum 353ynlv2na 8nd1ax4m8ap x8yhgi9lduop x6omdx8558ru jgvjd1ttaa 36cda5qa7s18j yjny2zcgh0euh | 2020-10-31T03:51:47 | {
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http://math.stackexchange.com/questions/19538/if-a-is-an-n-times-n-matrix-such-that-a2-0-is-ai-n-invertible?answertab=votes | # If $A$ is an $n \times n$ matrix such that $A^2=0$, is $A+I_{n}$ invertible?
If $A$ is an $n \times n$ matrix such that $A^2=0$, is $A+I_{n}$ invertible?
This question yielded two different proofs from my professors, which managed to get conflicting results (true and false). Could you please weigh in and explain what's happening, and offer a working proof?
Proof that it is invertible: Consider matrix $A-I_{n}$. Multiplying $(A+I_{n})$ by $(A-I_{n})$ we get $A^2-AI_{n}+AI_{n}-I^2_{n}$. This simplifies to $A^2-I^2_{n}$ which is equal to $-I_{n}$, since $A^2=0$. So, the professor argued, since we have shown that there exists a $B$ such that $(A+I_{n})$ times $B$ is equal to $I$, $(A+I_{n})$ must be invertible. I am afraid, though, that she forgot about the negative sign that was leftover in front of the $I$ -- from what I understand, $(A+I_{n})$*$(A-I_{n})$=$-I$ does not mean that $(A+I_{n})$ is invertible.
Proof that it is not invertible: Assume that $A(x)=0$ has a non-trivial solution. Now, given $(A+I_{n})(x)=\vec{0}$, multiply both sides by $A$. We get $A(A+I_{n})(x)=A(\vec{0})$, which can be written as $(A^2+A)(x)=\vec{0}$, which simplifies to $A(x)=0$, as $A^2=0$. Since we assumed that $A(x)=0$ has a non-trivial solution, we just demonstrated that $(A+I_{n})$ has a non-trivial solution, too. Hence, it is not invertible.
I am not sure if I reproduced the second proof in complete accuracy (I think I did), but the idea was to show that if $A(x)=\vec{0}$ has a non-trivial solution, $A(A+I_{n})$ does too, rendering $A(A+I_{n})$ non-invertible. But regardless of the proofs, I can think of examples that show that at least in some cases, the statement is true; consider matrices $\begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix}$ and $\begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix}$ which, when added $I_{2}$ to, become invertible.
Thanks a lot!
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The first proof works. $(A + I)(-(A - I)) = I$, so the inverse is $-(A - I) = -A + I$. The second proof doesn't hold water unfortunately. – Zarrax Jan 30 '11 at 6:40
The second proof does not even have the appearance of logical sense to me. Why are we assuming that $A(x) = 0$ has a nontrivial solution? In fact we can prove this, since the condition implies that $\operatorname{det} A = 0$. Then the "given" really makes no sense: we're simply not "given" that. In fact, if $x \neq 0$ and $Ax = 0$, then $(A+I)x = Ax + x = x \neq 0$. Anyway, as has been said, the given hypothesis does imply $A(A+I)$ is not invertible (take determinants), which as noted in the answer below does not imply anything about the invertibility of $A+I$. – Pete L. Clark Jan 30 '11 at 7:49
I don't think you understood my comment. Since $A^2 = 0$, the determinant of $A$ is zero, and thus there are always nontrivial solutions. (Here's a proof without determinants: let $x$ be any nonzero vector. If $Ax = 0$, that's a nontrivial solution. If not, put $x' = Ax$, then $x' \neq 0$ and $Ax' = A(Ax) = A^2 x = 0 x = 0. Either way we found a nontrivial solution!) – Pete L. Clark Jan 30 '11 at 8:03 Oh, sorry Pete. Thanks! – InterestedGuest Jan 30 '11 at 8:05 Question: no problem. @Zarrax: Given that the first proof is valid, I consider it extremely fortunate that the second proof is not valid. Otherwise we would have proved that mathematics is contradictory! – Pete L. Clark Jan 30 '11 at 8:08 ## 3 Answers The minus sign is not an obstacle: If$AB = -I$, then$A(-B) = -(AB) = -(-I) = I$. So in fact, if$A^2 = 0$, then$(A+I)(I-A) = A - A^2 + I - A = I$, so$A+I$is invertible, as your first professor noted. The error in the second argument is the following: It is true that if$B\mathbf{x}=\mathbf{0}$has a nontrivial solution, then$CB\mathbf{x}=\mathbf{0}$has a nontrivial solution. Thus, if$B$is not invertible, then$CB$is not invertible. But that is not what was argued. What was argued instead was that since$CB\mathbf{x}=\mathbf{0}$has a nontrivial solution, then it follows that$B\mathbf{x}=\mathbf{0}$has a nontrivial solution (with$B=A+I$and$C=A$). This argument is incorrect: you can always take$C=0$, and that would mean that no matrix is invertible. It is certainly true that if$A$is not invertible, then no multiple of$A$is invertible (so for every$C$, neither$CA$nor$AC$are invertible); so you can deduce that$A(A+I)$is not invertible. This does not prove that$A+I$is not invertible, however, which is what you wanted to show. Now, for bonus points, show that if$A$is an$n\times n$matrix and$A^k=0$for some positive integer$k$, then$A+\lambda I_n$is invertible for any nonzero$\lambda$. Added: For bonus bonus points, explain why the argument would break down if we replace$\lambda I_n$with an arbitrary invertible matrix$B$. - The bonus points are precisely exercise 1.1 in Atiyah-MacDonald's book. – Fredrik Meyer Jan 30 '11 at 16:50 For what it's worth I just want to mention that what is happening here is actually an instance of a more general result about rings. If$R$is a ring then an element$a \in R$is said to be nilpotent if there is$n \in \mathbb{N}$such that$a^n = 0$. In your question, the condition on your matrix that$A^2 = 0$just means that it is nilpotent. Now if$R$is a ring with unity, then if$a \in R$is nilpotent it can be proved that$1 - a$is an invertible element (or unit) in the ring$R$, meaning that there is a$b \in R$such that$(1 - a)b = b(1 - a) = 1$. From this you can then just change$a$with$-a$to also see that$1 + a$is invertible. I'm pretty sure that this is what Arturo had in mind by adding that exercise for bonus points for you, so I will not give the argument here. You can find it in this planetmath entry if you want to look at it. But I would suggest to you to first try it for yourself, for matrices at least. - I suggest thinking of the problem in terms of eigenvalues. Try proving the following: If$A$is an$n \times n$matrix (over any field) which is nilpotent -- i.e.,$A^k = 0$for some positive integer$k$, then$-1$is not an eigenvalue of$A$(or equivalently,$1$is not an eigenvalue of$-A$). If you can prove this, you can prove a stronger statement and collect bonus points from Arturo Magidin. (Added: Adrian's answer -- which appeared while I was writing mine -- is similar, and probably better: simpler and more general. But I claim it is always a good idea to keep eigenvalues in mind when thinking about matrices!) Added: here's a hint for a solution that has nothing to do with eigenvalues (or, as Adrian rightly points out, really nothing to do with matrices either.) Recall the formula for the sum of an infinite geometric series:$\frac{1}{1-x} = 1 + x + x^2 + \ldots + x^n + \ldots$As written, this is an analytic statement, so issues of convergence must be considered. (For instance, if$x$is a real number, we need$|x| < 1$.) But if it happens that some power of$x\$ is equal zero, then so are all higher powers and the series is not infinite after all...With only a little work, one can make purely algebraic sense out of this.
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Thanks Adrian and Pete. We haven't done rings, nilpotency, and eigenvalues yet, so I will wait for that, though I do have a general idea for Arturo's question. – InterestedGuest Jan 30 '11 at 7:58 | 2014-12-22T06:29:01 | {
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https://math.stackexchange.com/questions/2283631/minimize-axbd-c-2-enforcing-x-to-be-a-diagonal-block-matrix | # Minimize $\|AXBd -c \|^2$, enforcing $X$ to be a diagonal block matrix
Currently, I am minimizing the quadratic objective $\|\mathbf{A}\mathbf{X}\mathbf{B}\mathbf{d} -\mathbf{c} \|^2$ using CVX, as follows
echo on
cvx_begin
variable xx(num_triangles*3,num_triangles*3)
minimize( norm( A * xx * B * d - c ) )
cvx_end
echo off
However, $\mathbf{X}$ is a very large matrix (about $50,000 \times 50,000$, which is too big). Good news is that $\mathbf{X}$ is a block diagonal matrix (of $3 \times 3$ transformation matrices) and, therefore, it is highly sparse.
To give an example, for $\mathbf{X} \in \mathbb{R}^{9 \times 9}$ I would be looking for such matrix.
| X1 X4 X7 0 0 0 0 0 0 |
| X2 X5 X8 0 0 0 0 0 0 |
| X3 X6 X9 0 0 0 0 0 0 |
| 0 0 0 X10 x13 x16 0 0 0 |
| 0 0 0 X11 x14 x17 0 0 0 |
| 0 0 0 X12 x15 x18 0 0 0 |
| 0 0 0 0 0 0 X19 x22 x25 |
| 0 0 0 0 0 0 X20 x23 x26 |
| 0 0 0 0 0 0 X21 x24 x27 |
so in fact I am only solving for $27$ variables, not $9 \times 9 = 81$. And this scales pretty badly.
But I don't know how I can enforce such structure when formulating the minimization problem. If I declare a normal square matrix of the size I need, the memory just explodes. Any ideas on how to reformulate the problem in a way that I actually solve for the number of unknowns? I currently look for solutions in MATLAB.
• Using differentiation we find that if $\mathbf{X}$ minimizes the norm then for each $(i,j)$ corresponding to a valid position in $\mathbf{X}$ at least one of the following hold: $$\left\langle B^Te_j,d\right\rangle=0\\ {\left[A^Tc\right]}_i={\left[A^TAXBd\right]}_i$$ where $e_k$ is the $k$-th vector of the canonical basis and ${[v]}_k=\langle v,e_k\rangle$ is the $k$-th coordinate of $v$. This in principle could help limit the search for minimizing $\mathbf{X}$'s. I can expand on how I got this if you deem this reasonable. – Fimpellizieri May 17 '17 at 20:54
• Tell me something about $\rm A$. Is it square? Fat? Thin? What about the rank? – Rodrigo de Azevedo May 19 '17 at 13:20
We have a quadratic program in $\mathrm X \in \mathbb R^{3n \times 3n}$
$$\text{minimize} \quad \| \mathrm A \mathrm X \mathrm b - \mathrm c \|_2^2$$
where $\rm X$ is block diagonal
$$\mathrm X = \begin{bmatrix} \mathrm X_1 & & & \\ & \mathrm X_2 & & \\ & & \ddots & \\ & & & \mathrm X_n\end{bmatrix}$$
and each $\rm X_i$ block is $3 \times 3$. Let
$$\mathrm y := \mathrm X \mathrm b = \begin{bmatrix} \mathrm X_1 & & & \\ & \mathrm X_2 & & \\ & & \ddots & \\ & & & \mathrm X_n\end{bmatrix} \begin{bmatrix} \mathrm b_1\\ \mathrm b_2\\ \vdots \\ \mathrm b_n\end{bmatrix} = \begin{bmatrix} \mathrm X_1 \mathrm b_1\\ \mathrm X_2 \mathrm b_2\\ \vdots \\ \mathrm X_n \mathrm b_n\end{bmatrix} =: \begin{bmatrix} \mathrm y_1\\ \mathrm y_2\\ \vdots \\ \mathrm y_n\end{bmatrix}$$
Hence, we have a least-squares problem in $\mathrm y \in \mathbb R^{3n}$
$$\text{minimize} \quad \| \mathrm A \mathrm y - \mathrm c \|_2^2$$
A minimizer is a solution to the normal equations
$$\mathrm A^{\top} \mathrm A \mathrm y = \mathrm A^{\top} \mathrm c$$
Once a least-squares solution $\bar{\mathrm y}$ has been found, to find $\rm X$ we must solve $n$ underdetermined systems of $3$ linear equations in $3^2 = 9$ unknowns
$$\mathrm X_k \mathrm b_k = \bar{\mathrm y}_k$$
Vectorizing, we obtain
$$(\mathrm b_k^{\top} \otimes \mathrm I_3) \, \mbox{vec} (\mathrm X_k) = \bar{\mathrm y}_k$$
If $\color{blue}{\mathrm b_k \neq 0_3}$, then $3 \times 9$ matrix $\mathrm b_k^{\top} \otimes \mathrm I_3$ has full row rank. In that case, the least-norm solution is
$$\begin{array}{rl} \mbox{vec} (\bar{\mathrm X}_k) &= (\mathrm b_k^{\top} \otimes \mathrm I_n)^{\top} \left( (\mathrm b_k^{\top} \otimes \mathrm I_n) (\mathrm b_k^{\top} \otimes \mathrm I_n)^{\top} \right)^{-1} \bar{\mathrm y}_k\\ &= (\mathrm b_k \otimes \mathrm I_n) \left( (\mathrm b_k^{\top} \otimes \mathrm I_n) (\mathrm b_k \otimes \mathrm I_n) \right)^{-1} \bar{\mathrm y}_k\\ &= (\mathrm b_k \otimes \mathrm I_n) \left( \mathrm b_k^{\top} \mathrm b_k \otimes \mathrm I_n \right)^{-1} \bar{\mathrm y}_k\\ &= \left( \frac{\mathrm b_k}{\| \mathrm b_k \|_2^2} \otimes \mathrm I_n \right) \bar{\mathrm y}_k\\ &= \frac{\mathrm b_k}{\| \mathrm b_k \|_2^2} \otimes \bar{\mathrm y}_k\end{array}$$
Un-vectorizing, we obtain the rank-$1$ matrix
$$\boxed{\bar{\mathrm X}_{k} = \frac{\bar{\mathrm y}_k \mathrm b_k^{\top}}{\mathrm b_k^{\top} \mathrm b_k}}$$
Note that
$$\bar{\mathrm X}_k \mathrm b_k = \left( \frac{\bar{\mathrm y}_k \mathrm b_k^{\top}}{\mathrm b_k^{\top} \mathrm b_k} \right) \mathrm b_k = \bar{\mathrm y}_k \left( \frac{ \mathrm b_k^{\top} \mathrm b_k}{\mathrm b_k^{\top} \mathrm b_k} \right) = \bar{\mathrm y}_k$$
If $\color{blue}{\mathrm b_k = 0_3}$, then the linear system $(\mathrm b_k^{\top} \otimes \mathrm I_3) \, \mbox{vec} (\mathrm X_k) = \bar{\mathrm y}_k$ only has a solution if $\bar{\mathrm y}_k = 0_3$.
Consider for simplicity the case where $\mathbf{X}$ is of the form
$$\mathbf{X} = \left[ \begin{matrix} \mathbf{X}_{1,1} & \mathbf{0} \\ \mathbf{0} & \mathbf{X}_{2,2} \\ \end{matrix} \right].$$
Writing $\mathbf{A}$ as block matrix
$$\mathbf{A} = \left[\begin{matrix} \mathbf{A}_{1,1}& \mathbf{A}_{2,1} \\ \mathbf{A}_{1,2} & \mathbf{A}_{2,2} \\ \end{matrix} \right],$$
the product $\mathbf{AX}$ can be written as
$$\mathbf{AX}=\left[ \begin{matrix} \mathbf{A}_{1,1}\mathbf{X}_{1,1} & \mathbf{A}_{2,1}\mathbf{X}_{2,2} \\ \mathbf{A}_{1,2}\mathbf{X}_{1,1} & \mathbf{A}_{2,2}\mathbf{X}_{2,2} \\ \end{matrix} \right].$$
This approach generalizes to $\mathbf{X}$ having arbitrary number of diagonal matrix elements.
Your best bet is to declare a matrix variable corresponding to each $\mathbf{X}_{i,i}$ and perform the multiplication $\mathbf{AX}$ manually, as shown above (followed by right multiplication with $\mathbf{Bd}$). This approach will eliminate the need for declaring a full size matrix variable for $\mathbf{X}$, as well as saving computations corresponding to multiplication with zeros. However, as the matrix sizes you are considering are indeed very large, this approach may also fail.
A more convenient approach would be to exploit CVX features related to block-diagonal and/or sparse matrices. However, I am not familiar with the program. You should also try asking Computational Science SE or Stack Overflow SE. | 2019-05-25T16:59:51 | {
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https://mathhelpboards.com/threads/trig-substitution.8760/ | # Trig Substitution
#### Yuuki
##### Member
how do u integrate sqrt(1 + x^2) / x??
i reduced this to sec^3(u)/tan(u) but how do i go from here??
#### Random Variable
##### Well-known member
MHB Math Helper
Re: trig substitution
You can do it without trig substitution.
$$\int \frac{\sqrt{1+x^{2}}}{x} \ dx$$
Let $u^{2} = 1+x^{2}$.
Then
$$\int \frac{\sqrt{1+x^{2}}}{x} \ dx = \int \frac{u^{2}}{x^{2}} \ du = \int \frac{u^{2}}{u^{2}-1} \ du$$
$$= \int \Big( 1 + \frac{1}{u^{2}-1} \Big) \ du = \int \Big( 1 + \frac{1}{2(u-1)} - \frac{1}{2(u+1)} \Big) \ du$$
$$= u + \frac{\ln(u-1)}{2} - \frac{\ln(u+1)}{2} + C$$
$$= \sqrt{1+x^{2}} + \frac{1}{2} \ln (\sqrt{1+x^{2}}-1) - \frac{1}{2} \ln(\sqrt{1+x^{2}}+1) + C$$
#### HallsofIvy
##### Well-known member
MHB Math Helper
how do u integrate sqrt(1 + x^2) / x??
i reduced this to sec^3(u)/tan(u) but how do i go from here??
I would be inclined to do this like Random Variable but since you got this far, sec(u)= 1/cos(u) and tan(u)= sin(u)/cos(u) so that $$\frac{sec^3(u)}{tan(u)}= \frac{1}{cos^3(u)}\frac{cos(u)}{sin(u)}= \frac{1}{sin(u)cos^2(u)}$$which has sine to an odd power. We can multiply both numerator and denominator by sin(x) to get $$\frac{sin(x)}{sin^2(x)cos^2(x)}= \frac{sin(x)}{(1- cos^2(x))cos^2(x)}$$
Let v= cos(x) so that dv= -sin(x)dx and the integrand becomes $$\frac{-dv}{(1- v^2)v^2}$$ which can be integrated using "partial fractions".
Last edited:
#### soroban
##### Well-known member
Hello, Yuuki
I found an approach.
I'm sure someone will have a better way.
$$\int \frac{\sqrt{1 + x^2}}{x}\,dx$$
$$\text{Let }\,x \,=\,\tan\theta \quad\Rightarrow\quad dx \,=\,\sec^2\!\theta\,d\theta$$
$$\text{I reduced this to: }\:\int \frac{\sec^3\!\theta}{\tan\theta}\,d\theta$$ . Good!
$$\text{Multiply by }\frac{\tan\theta}{\tan\theta}:\;\int\frac{\sec^3 \!\theta\tan\theta}{\tan^2\!\theta}\,d\theta$$
. . $$=\;\int\frac{\sec^3\!\theta\tan \theta\,d\theta}{\sec^2\!\theta-1} \;=\;\int\frac{\sec^2\!\theta}{\sec^2\!\theta-1}( \sec\theta\tan\theta\,d\theta)$$
$$\text{Let }u \,=\,\sec\theta \quad\Rightarrow\quad du \,=\,\sec\theta \tan\theta\,d\theta$$
$$\text{We have: }\:\int \frac{u^2}{u^2-1}\,du \;=\;\int\left(1 + \frac{1}{u^2-1}\right)du$$
. . . . . . $$=\;u + \frac{1}{2}\ln\left|\frac{u-1}{u+1}\right|+C$$
Now back-substitute . . .
#### Krizalid
##### Active member
Another solution:
$$\int {\frac{{\sqrt {1 + {x^2}} }}{x}\,dx} = \int {\frac{{1 + {x^2}}}{{x\sqrt {1 + {x^2}} }}\,dx} = \int {\frac{{dx}}{{x\sqrt {1 + {x^2}} }}} + \int {\frac{x}{{\sqrt {1 + {x^2}} }}\,dx} ,$$
second integral is easy, but if we want to avoid partial fractions for the first one, put $x=\dfrac1t$ and the integral becomes
$$- \int {\frac{{dt}}{{\sqrt {1 + {t^2}} }}} = - \ln \left| {t + \sqrt {1 + {t^2}} } \right| + {k_1}.$$
Finally,
$$\int {\frac{{\sqrt {1 + {x^2}} }}{x}\,dx} = - \ln \left| {\frac{1}{x} + \sqrt {1 + \frac{1}{{{x^2}}}} } \right| + \sqrt {1 + {x^2}} + k.$$ | 2021-07-26T16:29:41 | {
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https://math.stackexchange.com/questions/1970306/can-an-open-set-contain-all-of-its-limit-points | # Can an open set contain all of its limit points?
Let's say I have a set $E$ as a subset of a metric space $X$
If $E$ is open, then is the set of all limit points of $E$, (which we'll denote $E'$) a subset of $E$?
I attempted to prove that if $E$ is open, then $E' \not\subset E$
Proof:
Let $E$ be open. Assume $E' \subset E$.
The closure of $E$, is $\overline{E} = E \cup E'$.
But since $E' \subset E$, we have $a \in E' \implies a \in E$
$$\therefore E \cup E' = E$$
and thus we have $$\overline{E} = E$$
which contradicts the fact that $E$ is open.
Therefore we can conclude that for any open set $E$, the set of all limit points $E'$ is not contained in $E$, i.e. $E' \not \subset E$. $\ \ \square$
Firstly is my above proof incorrect? If not then the thing is that there can be metric spaces which are both open and closed, take $\mathbb{R^2}$ for example. And if we let $E = \mathbb{R^2}$, then the above proof says that $\mathbb{R^2}$ is closed and not open.
I've heard something about ambient spaces, which is supposed to be the space containing all spaces you are considering, in this case $X$ would be an ambient space, and $E$ would not be an ambient space.
Does the concept of ambient spaces affect whether a set can be open, closed or both? For example if we let $X = \mathbb{R^3}$, and $E = \mathbb{R^2}$, where $E \subset X$, then is $E$ open, closed or both open and closed?
If it does, then does that mean that a metric space can only be open/closed or both, relative to itself or some other metric space which acts as an ambient space?
Your proof is correct except the conclusion. What you have proved is the $E$ must be simultaneously open and closed.
• @ArinCharudhuri So instead of having a proof by contradiction, I actually had a direct proof, that $E$ must be simultaneously open and closed? – Perturbative Oct 16 '16 at 0:52
• Yes. You are correct. – Arin Chaudhuri Oct 16 '16 at 0:54
To say that a set $S\subseteq X$ is open in $(X,d)$ does not mean $S$ is not closed in $(X,d)$. Similarly, To say that $S$ is closed in $(X,d)$ does not mean $S$ is not open in $(X,d)$. "Closed" is not defined to be "not open", neither is "open" defined to be "not closed".
You can find a simple example showing that an open set can contain all its limit points. Take the whole space $X$. The set of all limit points, by definition, does not go outside of $X$, so that $X'\subseteq X$. Also, $X$ is open in $(X,d)$, so that it is open and contain all its limit points. | 2019-12-05T22:47:53 | {
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https://math.stackexchange.com/questions/3171514/graph-on-n-vertices-each-vertex-of-degree-3-or-less-can-we-color-the-vertice | # Graph on $n$ vertices, each vertex of degree 3 or less. Can we color the vertices in a way such that…
Let $$G$$ be a graph on $$n$$ vertices. Suppose each vertex has at most 3 neighbors. Prove that you can color the vertices either red or blue in such a way that each vertex is connected to at most 1 vertex of the same color.
I tried the following approach: Color the vertices randomly. Then look at each vertex, and if it is connected to more than one vertex of the same color, change its color. I was hoping that there would be some monovariant. The one I considered, which is wordy to explain, did not work (I considered the total number of "off" items per vertex).
My second approach was to construct the graph edge by edge, coloring as you go. I think it makes sense to split the vertices up into four partitions: $$V_0$$, $$V_1$$, $$V_2$$, and $$V_3$$ where a vertex $$v$$ belongs into $$V_i$$, 1$$\leq i \leq 3$$, if $$\deg(v)=i$$. Then, go through each vertex of $$V_{1}$$, adding each edge and coloring each pair of vertices opposite colors, etc.... Something along those lines.
Can anyone think of a clever/simple way to prove this? Because, although my second approach probably works, it will require a ton of technical writing. There must be a simpler way to think about this problem.
Follow your first approach: when you see a vertex which is connected to more than one vertex of the same color, change its color. This algorithm terminates because at each step the number of monochromatic edges, which is a non negative integer, strictly decreases (an edge is monochromatic if its two vertices have the same color).
In fact, assume that a vertex $$v$$ is red and it has at least 2 neighbours red too. Then we change its color from red to blue and we have that:
1) if $$\deg(v)=2$$ then the number of monochromatic edges decreases by $$2$$;
2) if $$\deg(v)=3$$ and the other neighbour is red then the number of monochromatic edges decreases by $$3$$;
3) if $$\deg(v)=3$$ and the other neighbour is blue then the number of monochromatic edges decreases by $$2-1=1$$.
We keep going in this way and, in a finite number of step, the number of monochromatic edges attains a minimal value and we get the desired colored graph.
• This is not necessarily true... Consider an edge e=$v_{1}v_{2}$. Suppose $v_{1}$ and $v_{2}$ each have two other neighbors... Do you see how changing the color of one of them may cause more problems for their neighbors? And therefore, the number of monochromatic edges does not strictly decrease. This is why I mentioned that this approach does not work. – MathGuy Apr 2 at 21:11
• @MathGuy Yes, it is always true. Each time you change a color of a "bad" vertex then the number of monochromatic edges does strictly decrease (and may cause more problems for their neighbors but you you have to keep going till the end). – Robert Z Apr 3 at 7:29
• @MathGuy Are you still interested in your problem? Is my answer clear now? – Robert Z Apr 10 at 10:37 | 2019-12-11T14:15:56 | {
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https://ocw.mit.edu/courses/18-s997-introduction-to-matlab-programming-fall-2011/pages/root-finding/the-secant-method/ | 18.S997 | Fall 2011 | Undergraduate
Introduction To MATLAB Programming
Root-Finding
## The Secant Method
While Newton’s method is fast, it has a big downside: you need to know the derivative of $$f$$ in order to use it. In many “real-life” applications, this can be a show-stopper as the functional form of the derivative is not known. A natural way to resolve this would be to estimate the derivative using
$$\label{eq:dervative:estimate} f’(x)\approx\frac{f(x+\epsilon)-f(x)}{\epsilon}$$
for $$\epsilon\ll1$$. The secant method uses the previous iteration to do something similar. It approximates the derivative using the previous approximation. As a result it converges a little slower (than Newton’s method) to the solution:
$$\label{eq:3} x_{n+1}=x_n-f(x_n) \frac{x_n-x_{n-1}}{f(x_n)-f(x_{n-1})}.$$
Since we need to remember both the current approximation and the previous one, we can no longer have such a simple code as that for Newton’s method. At this point you are probably asking yourself why we are not saving our code into a file, and it is exactly what we will now learn how to do.
### Coding in a File
Instead of writing all your commands at the command prompt, you can type a list of commands in a file, save it and then have MATLAB® “execute” all of the commands as if you had typed them into the command prompt. This is useful when you have more than very few lines to write because inevitably you are bound to make a small mistake every time you write more than 5 lines of code. By putting the commands in a file you can correct your mistakes without introducing new ones (hopefully). It also makes it possible to “debug” your code, something we will learn later.
For guided practice and further exploration of how to debug, watch Video Lecture 6: Debugging.
MATLAB files have names that end with .m, and the name itself must comprise only letters and numbers with no spaces. The first character must be a letter, not a number. Open a new file by clicking on the white new-file icon in the top left of the window, or select from the menu File$$\rightarrow$$New$$\rightarrow$$Script. Copy the Newton method code for $$\tanh(x)=x/3$$ into it. Save it and give it a name (NewtonTanh.m for example). Now on the command prompt you “run” the file by typing the name (without the .m) and pressing Enter .
A few points to look out for:
• You can store your files wherever you want, but they have to be in MATLAB’s “search path” (or in the current directory). To add the directory you want to the path select File$$\rightarrow$$Set path… select “Add Folder”, select the folder you want, click “OK” then “Save”. To check if your file is in the path you can type which NewtonTanh and the result should be the path to your file.
• If you choose a file-name that is already the name of a MATLAB command, you will effectively “hide’’ that command as MATLAB will use your file instead. Thus, before using a nice name like sum, or find, or exp, check, use which to see if it already defined.
• The same warning (as the previous item) applies to variable names, a variable will “hide” any file or command with the same name.
• If you get strange errors when you try to run your file, make sure that there are no spaces or other non-letters in your filename, and that the file is in the path.
• Remember that after you make changes to your file, you need to save it so that MATLAB will be aware of the changes you made.
For guided practice and further exploration of how to use MATLAB files, watch Video Lecture 3: Using Files.
Exercise 7. Save the file as SecantTanh.m and modify the code so that it implements the Secant Method. You should increase the number of iterations because the Secant Method doesn’t converge as quickly as Newton’s method.
Notice that here it is not enough to use x like in the Newton’s method, since you also need to remember the previous approximation $$x_{n-1}$$. Hint: Use another variable (perhaps called PrevX).
### Convergence
Different root-finding algorithms are compared by the speed at which the approximate solution converges (i.e., gets closer) to the true solution. An iterative method $$x_{n+1}=g(x_n)$$ is defined as having $$p-$$th order convergence if for a sequence $$x_n$$ where $$\lim_{n\rightarrow\infty}x_n=\alpha$$ exists then
$$\label{eq:convergence:order} \lim_{n\rightarrow\infty}\frac{|{x_{n+1}-\alpha}|}{|{x_n-\alpha}|^p} = L \ne 0.$$
Newton’s method has (generally) second-order convergence, so in Eq. (3) we would have $$p=2$$, but it converges so quickly that it can be difficult to see the convergence (there are not enough terms in the sequence). The secant method has a order of convergence between 1 and 2. To discover it we need to modify the code so that it remembers all the approximations.
The following code, is Newton’s method but it remembers all the iterations in the list x. We use x(1) for $$x_1$$ and similarly x(n) for $$x_n$$:
x(1)=2; % This is our first guess, put into the first element of x
for n=1:5 % we will iterate 5 times using n to indicate the current
% valid approximation
x(n+1)=x(n)-(tanh(x(n))-x(n)/3)/(sech(x(n))^2-1/3); %here we
% calculate the next approximation and
% put the result into the next position
% in x.
end
x % sole purpose of this line is to show the values in x.
The semicolon (;) at the end of line 4 tells MATLAB not to display the value of x after the assignment (also in line 1. Without the lonely x on line 9 the code would calculate x, but not show us anything.
After running this code, x holds the 6 approximations (including our initial guess) with the last one being the most accurate approximation we have:
x =
2.0000 3.1320 2.9853 2.9847 2.9847 2.9847
Notice that there is a small but non-zero distance between x(5) and x(6):
>> x(6)-x(5)
ans =
4.4409e-16
This distance is as small as we can hope it to be in this case.
We can try to verify that we have second order convergence by calculating the sequence defined in Eq. (3). To do that we need to learn more about different options for accessing the elements of a list like $$x$$. We have already seen how to access a specific element; for example to access the 3rd element we write x(3). MATLAB can access a sublist by giving it a list of indexes instead of a single number:
>> x([1 2 3])
ans =
2.0000 3.1320 2.9853
We can use the colon notation here too:
x(2:4)
ans =
3.1320 2.9853 2.9847
Another thing we can do is perform element-wise operations on all the items in the list at once. In the lines of code below, the commands preceding the plot command are executed to help you understand how the plot is generated:
>> x(1:3).^2
ans =
4.0000 9.8095 8.9118
>> x(1:3)*2
ans =
4.0000 6.2640 5.9705
>> x(1:3)-x(6)
ans =
-0.9847 0.1473 0.0006
>> x(2:4)./(x(1:3).^2)
ans =
0.4002 0.0018 0.0387
>> plot(log(abs(x(1:end-2)-x(end))),log(abs(x(2:end-1)-x(end))),'.'))
The last line makes the following plot (except for the green line, which is $$y=2x$$):
MATLAB can calculate roots through Newton’s method, and verification of convergence is graphed.
The main point here is that the points are more or less on the line y=2x, which makes sense:
Taking the logarithm of the sequence in (3) leads to
$$\label{eq:convergence:plots} \log|{x_{n+1}-\alpha}| \approx \log L + p\log|{x_{n}-\alpha}|$$
for $$n\gg1$$, which means that the points $$(\log|{x_{n}-\alpha}|, \log|{x_{n+1}-\alpha}|)$$ will converge to a line with slope $$p$$.
The periods in front of *, /, and ^ are needed (as in the code above) when the operation can have a linear algebra connotation, but what is requested is an element-by-element operation. Since matrices can be multiplied and divided by each other in a way that is not element-by-element, we use the point-wise version of them when we are not interested in the linear algebra operation.
Exercise 8. Internalize the differences between the point-wise and regular versions of the operators by examining the results of the following expressions that use the variables A=[1 2; 3 4], B=[1 0; 0 2], and C=[3;4]. Note: some commands may result in an error message. Understand what the error is and why it was given.
• A*B vs. A.*B vs. B.*A vs. B*A
• 2*A vs. 2.*A
• A^2 vs. A*A vs. A.*A vs. A.^2 vs. 2.^A vs. A^A vs. 2^A. The last one here might be difficult to understand…it is matrix exponentiation.
• A/B vs. A\B vs. A./B vs. A.\B
• A*C vs. A*C' vs. C*A vs. C'*A
• A\C vs. A\C' vs. C/A vs. C'/A
Homework 2. Modify your secant method code so that it remembers the iterations (perhaps save it in a new file?). Now plot the points that, according to (4) should be on a line with slope $$p$$. What is $$p$$? | 2023-02-09T12:40:44 | {
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https://dsp.stackexchange.com/tags/fourier-series/hot | Tag Info
9
This sum appears quite often in DSP. \begin{align} \sum_{n=0}^{N-1} \exp(-j\alpha n) &\stackrel{(a)}{=} \frac{1- \exp(-j\alpha N)}{1 - \exp(-j\alpha )}\\ &= \frac{e^{-j\alpha N/2}(e^{+j\alpha N/2}- e^{-j\alpha N/2})}{e^{-j\alpha /2}(e^{+j\alpha /2}- e^{-j\alpha /2})}\\ &\stackrel{(b)}{=} e^{-j\frac{\alpha}{2}(N-1)} \frac{\sin(N\alpha/2)}{\sin(\...
7
Consider a continuous-time periodic signal $x_1(t)$ whose fundamental period is $T_1$, fundamental radian frequency is $\omega_1 = \frac{2\pi}{T_1}$ and CTFS (continuous-time Fourier series) coefficients are: $$x_1 \longleftrightarrow c_k = \frac{1}{T_1} \int_{<T_1>} x_1(t) e^{-j\frac{2\pi k}{T_1}t} dt ~~ , ~~\text{for}~~ k=0,\pm 1, \pm 2...$$ We ...
7
There are 4 versions of Fourier transforms that are all close cousins. It's all due to the basic property that "sampling in one domain corresponds to periodicity in the other domain". If a time signal is periodic, the frequency domain signal is discrete. If a frequency domain signal is periodic, the time domain signal is discrete. So you have four ...
6
In image processing, you are always dealing with discrete data, so the transform isn't really a Fourier transform, it is a discrete Fourier transform (DFT with a sum rather than an integral). The DFT can be understood as providing the coefficients of a Fourier series from which a periodic version of you're original image can be reconstructed. The "checker ...
6
alright, let's review a little bit of Euler before we get to the Fourier. $$e^{j \theta} \ = \ \cos(\theta) \ + \ j \sin(\theta)$$ from that you can get $$\cos(\theta) = \frac{e^{j \theta} + e^{-j \theta}}{2} \quad\quad\quad \sin(\theta) = \frac{e^{j \theta} - e^{-j \theta}}{2j}$$ so now let's look at Eq (1) \begin{align} x(t) \ &= \ A_0 \ + \... 6 Well, when m=k the integral is: \int_0^T e^{j(m-k)\Omega_0t} dt = \int_0^T e^{j \cdot 0 \cdot\Omega_0t} dt = \int_0^T dt = T $$So as Juancho says in the comments, it's the same signal and so can't be orthogonal to itself. 5 This is related to Chirp Z-transform (CZT) (refer to the Bluestein's algorithm). Using this identity, the CZT can be expressed in terms of a convolution. Hence, it can be efficiently implemented using FFT. 5 If you have two DFTs A[k] and B[k] (note the correct representation of a sinusoid at DFT bin number 1) A = [0,-j,0,0,j]; B = [1,1,1,1,1]; with the corresponding time-domain sequences a[n] and b[n] a = ifft(A); % [0, 0.38042, 0.23511, -0.23511, -0.38042]; b = ifft(B); % [1,0,0,0,0]; then the multiplication of the time-domain sequences c[n]... 5 Yeah some of us can do it, you can speed up or slow down without affect the pitch, some guys call this applications of Time Stretch, there different ways to do it, you can do in frequency domain or time domain, you will need choose what is best for you, you will find some advantages and disadvantages of each. Time Domain: In Time Domain you can try some ... 5 You should use the synthesis equation of an impulse train with period T (which is easy to derive):$$x(t)=\sum_{k=-\infty}^{\infty}\delta(t-kT)=\sum_{k=-\infty}^{\infty}\frac{1}{T}e^{jk\frac{2\pi}{T} t}\tag{1}$$That is: the Fourier coefficients for all terms is a constant (\frac{1}{T}). Now assume that there are two impulses with different ... 5 Ok you are mixing the order of the transforms. When you decompose the argument 3t-6 into 3\cdot(t-2) , then you should first scale x(t) by 3 and then shift the scaled result by 2, that's where you have chosen the wrong order. Let's show the correct order: Let x(t) be the original signal whose CTFT is X(\omega), then define the following ... 5 Yes, for time-limited functions it is possible to obtain the Fourier series coefficients by sampling the Fourier transform. This is the dual case of the more common form of the sampling theorem, stating that a band-limited function is fully characterized by its (time-domain) samples. The same is true in the discrete domain. Note that only for finite length ... 4 The coefficients of the Fourier series that you have computed are, in effect, the spectrum of the periodic signal consisting of the sum of signals f(t) delayed in time or advanced in time by integer multiples of 1 second. Mathematically, the Fourier transform of a periodic function has impulses in it with the impulse amplitudes being the Fourier series ... 4 I think the main issue is you are jumping ahead of yourself. You probably remember or read somewhere that the Fourier transform of a rect function is a sinc function. This is true; however, no where in this section does he mention Fourier transform! In fact, what he is doing is not Fourier transform. What he does in this section is to represent any ... 4 Let g(x) = f(ax). Then, using a change of variables y = ax,$$\begin{align*} G(u) &= \int_{-\infty}^{\infty} g(x)\exp(-j u x)\,\mathrm dx\\ &= \int_{-\infty}^{\infty} f(ax)\exp(-j u x)\,\mathrm dx\\ &= \frac{1}{|a|}\int_{-\infty}^{\infty} f(y)\exp(- j (u/a) y)\,\mathrm dy\\ &= \frac{1}{|a|}F\left(\frac{u}{a}\right) \end{align*}$$not \... 4 This doesn't require any complicated computations! Look up some tables of identities about the Fourier Series. Which operation would you have to apply to f so that its Fourier series would become \sum_{-\infty}^{\infty}(c_k)^2 e^{2\pi ikx}? For example, we know that derivating in the time domain is equivalent to multiplying the Fourier series ... 4 The book "Dr. Euler's Fabulous Formula: Cures Many Mathematical Ills", by P. Nahin, Princeton University Press, leads up to and contains an explanation of the Gibbs phenomena which might be suitable for someone with a good undergraduate university level math background. 4 FFT and IFFT are algorithms that implement the (inverse) discrete Fourier transform (DFT). These transforms convert a signal into another representation, namely the frequency domain, and back. This other representation allows us to analyze certain properties of the signal and also to change these properites which would be hard to do in the original ... 4 The book "Blip, Ping & Buzz", by M. Denny (2007, John Hopkins University Press), explains signal processing, as used in sonar and radar, at the level of a supermarket science magazine. Some algebra required. More on the use of Fourier analysis than any theory. 4 Note that the value of the series$$\sum_{n=1}^{\infty}B_n\sin(nx)\tag{1}$$for x=0 is always zero, regardless of the coefficients B_n (assuming convergence). Furthermore, since the functions \sin(nx) are odd, (1) can only represent odd functions. So you cannot represent general functions with the series (1). This is why you also need the cosine ... 4 I'm new to this exchange and I'm not sure how mathy you all get. I think the answer below is cool because it shows that in some sense the continuous-time Fourier transform is never periodic but that in another sense there are lots of ways to get periodic transforms. For the continuous-time Fourier transform on \mathbb{R}, both CMDoolittle's and Robert ... 4 You correctly figured out that the occurring integrals don't converge in the conventional sense. The easiest (and definitely non-rigorous) way to see the result is by noting the Fourier transform relation$$1\Longleftrightarrow 2\pi\delta(\Omega)$$By the shifting/modulation property we have$$e^{j\Omega_0t}\Longleftrightarrow 2\pi\delta(\Omega-\Omega_0)$$... 4 The common formulation of the forward DFT preserves energy (Parseval's theorem). This means that a longer constant magnitude sinewave input to a DFT, which has proportionally more energy, must be represented by a proportionally larger value in the DFT result, preserving that greater energy. Thus a factor of N, which scales with the length of the input. ... 4 Why are Fourier Analysis & Transform only applicable for LTI systems? That's simply not true. Won't Fourier analysis or Transform be possible? They are. The question is just whether they are meaningful. The point is that the continuous Fourier Transform (FT) is defined to integrate over all times from t=-\infty to t=\infty; that works ... 4 The family of Fourier Transforms are specificaly developed for analysing frequency contents of the signals for which there is no definition of linearity or time invariance. Hence we can define the Fourier transform of any signal, as long as it's integrable (i.e. stable). On the other hand we can also define a Frequency Response H(\omega) for a particular ... 4 HINT: Going from your last equation,$$\frac{\sqrt{T}}{2}\bigg(\frac{e^{j2\pi (f_1T-n)}-1}{j2\pi (Tf_1-n)} + \frac{e^{-j2\pi (f_1T+n)}-1}{-j2\pi (Tf_1+n)}\bigg)This can be simplified further down by considering the following: \begin{align} e^{j2\pi (f_1T-n)} &= e^{j\pi (f_1T-n)}\cdot e^{j\pi (f_1T-n)}\\ 1 &=e^{j\pi (f_1T-n)}\cdot e^{-j\pi (f_1T-... 4 If the size N of the DFT is even, only one "extremal" point (after fftshift) is Nyquist. If N is even, you cannot have an arbitary c_{-5} and c_5. They must add to be whatever your c_{-5} term is. If the input to the DFT is purely real and if N is even, consider the Nyquist point, at c_{N/2} to be split in half. One half is the negative-... 3 cos(t) and cos(\pi t) are both periodic with periods 2\pi and 2 respectively. To find the period of the sum we need to find an integers n,m such that \pi n = 2 m or \pi / 2 = m /n , which is not possible since \pi is irrational. I think the Fourier series uses specific frequencies to form the signal. You can't necessarily put a frequency ... 3 Dimension in mathematics is defined as the number of independent components of a structure (with some generalizations, like fractal dimensions). Now which number of dimensions you assign to e.g a discrete image (as in picture) depends on what aspect of the image you are describing. As a vector space where each pixel is a component, you get as many dimensions ... 3 One way to see the equivalence is to first factor out the 0.5 in your initial transfer function, i.e. H(f) = 0.5(1 + e^{-i2\pi f}). $$Then you can rewrite the 1 and e^{-i2\pi f} as follows:$$ H(f) = 0.5(e^{i\pi f}e^{-i\pi f} + e^{-i\pi f}e^{-i\pi f}) = 0.5(e^{i\pi f} + e^{-i\pi f})e^{-i\pi f}. Your result then follows as a trivial ...
Only top voted, non community-wiki answers of a minimum length are eligible | 2019-12-12T06:07:09 | {
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https://math.stackexchange.com/questions/3148281/evaluating-int-0-pi-2-log-left-sin2-x-a-right-where-a-in-0-1 | # Evaluating $\int_0^{\pi/2} \log \left| \sin^2 x - a \right|$ where $a\in [0,1]$.
How to evaluate $$\displaystyle\int_0^{\pi/2} \log \left| \sin^2 x - a \right|\,dx$$ where $$a\in[0,1]$$ ?
I think of this problem as a generalization of the following proposition $$\displaystyle\int_0^{\pi/2} \log \left(\sin x\right)\,dx =-\frac12\pi\log2$$
My try
Put $$I(a)=\displaystyle\int_0^{\pi/2} \log \left| \sin^2 x - a \right|\,dx$$ From the substitution $$x \to \frac{\pi}{2}-x$$ , we get $$\displaystyle\int_0^{\pi/2} \log \left| \sin^2 x - a \right|\,dx = \displaystyle\int_0^{\pi/2} \log \left| \cos^2 x - a \right|\,dx$$ Thus $$\displaystyle\int_0^{\pi/2} \log \left| \sin^2 x - a \right|\,dx = \displaystyle\int_0^{\pi/2} \log \left| \sin^2 x - (1-a) \right|\,dx$$ which means $$I(a)=I(1-a) \tag{1}$$
On the other hand, \begin{align} 2I(a) &= \displaystyle\int_0^{\pi/2} \log \left| (\sin^2 x - a)(\cos^2 x -a) \right|\,dx \\ &= \displaystyle\int_0^{\pi/2} \log \left| a^2-a+\sin^2 x \cos^2 x \right|\,dx \\ &= \displaystyle\int_0^{\pi/2} \log \left| 4(a^2-a)+\sin^2 (2x) \right|\,dx -\pi \log 2 \\ &= \frac{1}{2}\displaystyle\int_0^{\pi} \log \left| 4(a^2-a)+\sin^2 x \right|\,dx -\pi \log 2 \\ &= \displaystyle\int_0^{\pi/2} \log \left| 4(a^2-a)+\sin^2 x \right|\,dx -\pi \log 2 \\ &= \displaystyle\int_0^{\pi/2} \log \left| 1+4(a^2-a)-\sin^2 x \right|\,dx -\pi \log 2 \\ &= I((2a-1)^2) -\pi \log 2 \end{align} Thus $$2I(a)=I((2a-1)^2)-\pi \log 2 \tag{2}$$
Let $$a=0$$ we get the proposition mentioned above $$\displaystyle\int_0^{\pi/2} \log \left(\sin x\right)\,dx =-\frac12\pi\log2.$$
But how to move on ? Can we solve the problem only by $$(1)$$ and $$(2)$$? Or what other properties should we use to evaluate that?
Looking forward to your new solutions as well.
As pointed out in the comments, it seems like that the integral is identical to $$-\pi\log 2$$.
From $$(1)$$ and $$(2)$$ we can also find many numbers such that $$I(a)=-\pi\log 2$$.
• Breaking the integral into two with different interval so that the modulus gets converted , might help for solving the problem. – Bijayan Ray Mar 14 at 17:32
• I'm fairly certain that the integral is identical to $-e$ for $a\in [0,1]$ – clathratus Mar 14 at 17:47
• @BijayanRay I tried it but it was horrible :D – Advil Sell Mar 14 at 17:54
• @clathratus It seems like that you meant it's identical to $-\pi \log 2$. – Zero Mar 14 at 18:14
• Wait yeah its $-\pi\log2$ my bad. I just got my decimals mixed up I guess :) – clathratus Mar 14 at 18:16
Using the symmetries and developing the $$\sin^2$$ term, we can express \begin{align} I&=\int_0^{\pi/2} \log \left| \sin^2 x - a \right|\,dx\\ &=\frac{1}{4}\int_0^{2\pi} \log \left| \sin^2 x - a \right|\,dx\\ &=\frac{1}{4}\int_0^{2\pi} \log \left| \frac{1-2a}{2}-\frac{1}{2}\cos 2x\right|\,dx %&=-\frac{\pi}{2}\ln 2+\frac{1}{4}\int_0^{2\pi} \log \left| \left( 2a-1 \right)+\cos 2x\right|\,dx \end{align} By denoting $$2a-1=\cos 2\alpha$$, \begin{align} I&=\frac{1}{4}\int_0^{2\pi} \log \left| \frac{1}{2}\left( \cos 2\alpha+\cos 2x\right)\right|\,dx\\ &=\frac{1}{4}\int_0^{2\pi} \log \left| \cos \left( x+\alpha \right)\cos \left( x-\alpha \right)\right|\,dx\\ &=\frac{1}{4}\int_0^{2\pi} \log \left| \cos \left( x+\alpha \right)\right|\,dx+\frac{1}{4}\int_0^{2\pi} \log \left| \cos \left( x-\alpha \right)\right|\,dx \end{align} As the functions are periodic, the integration variables can be shifted, thus $$$$I=\frac{1}{2}\int_0^{2\pi} \log \left| \cos \left( x \right)\right|\,dx$$$$ Finally using the symmetries of the integrand, \begin{align} I&=2\int_0^{\pi/2} \log \left| \cos \left( x \right)\right|\,dx\\ &=2\int_0^{\pi/2} \log \left| \sin \left( x \right)\right|\,dx\\ &=-\pi\ln 2 \end{align} from the quoted result. | 2019-08-18T10:49:29 | {
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https://www.jiskha.com/questions/915545/can-someone-please-help-solve-the-differential-equation-dy-dx-6xy-with-the-condition | # math
Solve the differential equation
dy/dx = 6xy
with the condition y(0) = 40
Find the solution to the equation y= ______
1. 👍
2. 👎
3. 👁
1. This equation is separable, so
dy/dx = 6xy
dy/y = 6xdx
Integrate both sides,
log(y)=3x²+C1
y=Ce3x²
y(0)=40 -> C=40
therefore
y=40e3x²
1. 👍
2. 👎
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https://math.stackexchange.com/questions/1758346/prove-a-doubly-periodic-entire-analytic-function-in-complex-plane-is-a-constant | # Prove a doubly periodic entire analytic function in complex plane is a constant [duplicate]
I got stuck on this problem. So I really appreciate if anyone can give me some hint to move on. Thanks a lot.
Prove that an entire analytic function $f:\mathbb{C} \rightarrow \mathbb{C}$ is a constant function if $f(z) = f(z+1)=f(z+i)$ for all $z \in \mathbb{C}$
As far as I know, there're some famous theorems on how to prove an entire analytic function is constant, like Liouville theorem, maximal modulus theorem, identity theorem... But I still can't figure out how to apply those to this problem. I think about the dense property of $\mathbb{Q}$ in $\mathbb{R}$, so I tried to prove that $f(z) = f(z+q)$ for all $z \in \mathbb{C}$ and all $q \in \mathbb{Q}$, then use the continuity to prove that it's true for $q \in \mathbb{R}$. But I can't prove that statement.
## marked as duplicate by Martin R, user296602, user147263, colormegone, ShahabApr 26 '16 at 2:08
• and for proving the Liouville theorem, use the Cauchy integral formula : $f^{(n)}(0) = \frac{n!}{2 i \pi} \int_{|z|=R} \frac{f(z)}{z^{n+1}} dz$ which $\to 0$ as $R \to \infty$ (since $|f(z)| < C$) when $n \ge 1$ – reuns Apr 25 '16 at 18:29
The conditions tell that the function is $(1,1)$-periodic and hence $$\sup_{\mathbb{C}}|f| = \max_{x,y\in [0,1]}|f(x + iy)| < \infty.$$
• For the last statement, we can conclude that from the fact that $[0,1] \times [0,1]$ is compact, so ${f([0,1] \times [0,1])}$ is compact, too. So it's bounded, is that right? Then we use Liouville theorem to get the conclusion. Am I wrong at any point? – le duc quang Apr 25 '16 at 17:53
Hint: For all $m,n\in \mathbb Z,$ $f(z) = f(z+m+ni).$ | 2019-07-19T18:44:03 | {
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http://math.stackexchange.com/questions/200582/solve-the-differential-equation-using-taylor-series-expansion?answertab=votes | Solve the differential equation using Taylor-series expansion
Solve the differential equation using Taylor-series expansion:
$$\frac{dy}{dx} = x + y + xy \\ y (0) = 1$$ to get value of $y$ at $x = 0.1$ and $x = 0.5$. Use terms through $x^5$.
-
The approach that I have seen taken when asked to determine the solution using Taylor's Theorem is as follows.
We have, from Taylor's Theorem, $$y(x)=y(0)+y'(0)x+\frac{y''(0)}{2}x^2+\frac{y^{(3)}(0)}{6}x^3+\ldots$$ which we need to solve for the respective coefficients.
We are given $y(0)=1$. When $x=0$, the ODE must be satisfied. Then we must have $$y'(0) = 0+1+0\cdot1=1.$$ Differentiating the ODE we get $$\frac{d^2y}{dx^2}=1+\frac{dy}{dx}+y+x\frac{dy}{dx}.$$ Using this, we get that, at $x=0$, $$y''(0)=1+1+1+0\cdot1=3.$$
Then, we have $$y(x)=1+x+\frac{3}{2}x^2+\frac{y^{(3)}(0)}{6}x^3+\ldots$$ Continuing in this fashion, you can get the value of $y^{(3)}(0)$ and higher derivatives at $x=0$, thus giving a solution to the original ODE.
Once you have the required terms, you can evaluate the function at $x=0.1$ and $x=0.5$.
-
Do you know any reference for this method and what is it called in the literature? thanks – Algohi Apr 11 at 18:43
@Algohi I don't think it has a name. Googling "solve differential equation with Taylor series" brings up a few results you might find helpful. – Daryl Apr 11 at 23:34
I did google also and got some results. I just thought if there is any publication about the method. thanks anyway – Algohi Apr 12 at 3:19
Suppose that $y$ has the Taylor series expansion about $x=0$ given by $$y=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+\cdots.$$ Because $y(0)=1$, we have $a_0=1$. Differentiate. We get $$\frac{dy}{dx}=a_1+2a_2x+3a_3x^2+4a_4x^3+5a_5x^4+\cdots.\tag{1}$$ Also, \begin{align}x+y+xy&=x+(1+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+\cdots)\\&+(x+a_1x^2+a_2x^3+a_3x^4+a_4x^5+\cdots).\end{align} In the expression above, gather like powers of $x$ together. We get $$x+y+xy=1+(2+a_1)x+(a_1+a_2)x^2+(a_2+a_3)x^3+(a_4+a_5)x^4+\cdots.\tag{2}$$ The expansions $(1)$ and $(2)$ must be identical. It follows that they have the same constant term, that is, that $a_1=1$.
The coefficients of $x$ in $(1)$ and $(2)$ must match. It follows that $2a_2=2+a_1=3$, and therefore $a_2=\frac{3}{2}$.
The coefficients of $x^2$ must match. It follows that $3a_3=a_1+a_2=\frac{5}{2}$, and therefore $a_3=\frac{5}{6}$.
Continue, finding $a_4$ and $a_5$. You have not been asked to find coefficients beyond $a_5$.
For the numerical calculations, just substitute the given values of $x$ in the expression $1+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5$, using the values of the $a_i$ that we have found.
-
This looks like solving using the standard series approach and isn't really utilising Taylor's Theorem to obtain the solution. This is the approach I would take to solve the problem as well since it is more general, but I don't think it is what is being asked. – Daryl Sep 22 '12 at 6:25
@Daryl: Hard to know, the OP can decide which one looks more like the course notes. The question said Taylor series, not Taylor's theorem. – André Nicolas Sep 22 '12 at 6:28
Agree. Taylor series just 'special' power series, in one way to describe it. – Daryl Sep 22 '12 at 6:30 | 2015-08-29T13:11:46 | {
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https://math.stackexchange.com/questions/164852/if-n-ne-4-is-composite-then-n-divides-n-1 | # If $n\ne 4$ is composite, then $n$ divides $(n-1)!$.
I have a proof and need some feedback. It seems really obvious that the statement is true but it is always the obvious ones that are a little trickier to prove. So I would appreciate any feedback. Thank you!
Here is what I am asked to prove:
If $n$ is composite then $(n-1)! \equiv 0 \pmod n$.
Proof:
$n$ is composite $\implies n=ab$ where $a,b \in \mathbb{Z}$ and $0<a,b<n$.
Case 1: If $a=b$ then $n=a^{2}$. Now $n \mid (n-1)! \implies a \mid (n-1)!$, so \begin{aligned} (n-1)! &\equiv 1\times 2\times \dotsb \times a \times\dotsb\times (n-a)\times\dotsb\times (n-1) \\ &\equiv 1\times 2\times \dotsb\times a \times\dotsb\times -a\times\dotsb\times -1 \\ &\equiv 0 \pmod n \end{aligned}
Case 2: $0<a<b<n$.
Then, since $a \mid n$, $b \mid n$ and $n \mid (n-1)!$ we have that $a \mid (n-1)!$ and $b \mid (n-1)!$.
So this implies $(n-1)! \equiv 1\times 2\times \dotsb\times a \times\dotsb\times b\times\dotsb\times (n-1) \equiv 0 \pmod n$, Q.E.D.
• In case 1, there is a subcase where $a=n-a$, and the result is false. – Jonas Meyer Jun 30 '12 at 8:32
• $3!\equiv2\pmod4$ – Mike Jun 30 '12 at 8:39
• @JonasMeyer I did not consider that case. Darn, this also only happens when n=4. :( thank you. – HowardRoark Jun 30 '12 at 8:41
• @HowardRoark: Exactly (and Mike has also made it explicit). What is the source of the problem? Was the exception of $n=4$ not mentioned? Otherwise, your method looks good. – Jonas Meyer Jun 30 '12 at 8:42
• @JonasMeyer This is how the question goes: (a)calculate $(n-1)!$(mod $n$) for $n=10,12,14,$ and $15$. (b) guess a theorem and prove it. – HowardRoark Jun 30 '12 at 8:46
Hint $\rm\,\ n\, =\, \color{#C00}a\:\!\color{#0A0}b\mid1\!\cdot\! 2\cdots\color{#C00} a\:\color{#0A0}{(a\!+\!1)\, (a\!+\!2) \cdots (a\!+\!b)}\cdots (\color{blue}{ab\!-\!1})=(n\!-\!1)!\,\$ since $\rm\,\ \color{#0a0}{a\!+\!b}\le \color{blue}{ab\!-\!1}$
Note $\rm\,\color{#0A0}b\,$ divides the $\rm\color{#0A0}{green}$ term since a sequence of $\rm\,b\,$ consecutive integers has a multiple of $\rm\,b,\,$
and $\rm\,a\!+\!b \le ab\!-\!1 \!\iff\! 2\le (\underbrace{a\!-\!1}_{\large \ge\,1})(\underbrace{\color{#c0d}{b\!-\!1}}_{\large\ge\, 2}), \,$ true by $\rm\,a,b\ge 2,\,$ $\rm\underbrace{not\ both =2,}_{\large n\,=\,ab\,\ne\, 4}\,$ so $\,\color{#c0f}{\rm one}$ is $\,\ge 3$
That prior inequality implies that all of the $\rm\color{#0A0}{green}$ factors do occur in $\rm\,(ab\!-\!1)!$
Note $\$ That $\rm\:\color{#0A0}b\:$ divides the above $\rm\color{#0A0}{green}$ term is not deduced from the fact that it is divisible by $\rm\,b!\,$ by integrality of the binomial coefficient $\rm\:(a\!+\!b:a),\:$ as in WimC's answer. Rather, we deduce it from the more elementary fact that a sequence of $\rm\,b\,$ consecutive integers contains a multiple of $\rm\,b,\,$ which is an immediate consequence of (Euclidean) division.
• For a moment there I forgot I am using the color removing script, I have to say that to my color-blind eyes the colorized version looks extremely bad. My mind is too busy trying to figure out what is going on with the text and ignores the content completely (despite knowing the content from a colorless read!) you might want to consider using red and blue for future reference. Those are much harder to confuse and they sit a lot better for most color blindness. (Now I'll just turn that Greasemonkey script back on, and all will be well again...) – Asaf Karagila Jul 3 '12 at 21:17
If $n > 4$ and $n = a \cdot b$ with $a, b \geq 2$ then $a + b \leq n - 1$. Since ${a+b \choose a}$ is an integer it follows that $n = a \cdot b \mid a! \cdot b! \mid (a + b)! \mid (n - 1)!$.
• @TMM Binomial coefficients are a bit of a sledgehammer here. Instead, more simply, one needs only that every sequence of consecutive integers of length $\rm\:b\:$ contains a multiple of $\rm\:b.\:$ See my answer. – Bill Dubuque Jun 30 '12 at 14:33
• @BillDubuque Why do you consider it is a sledgehammer? Maybe it is true from a number theoretical view, but very straightforward from a combinatorial one, right? – Pedro Tamaroff Jun 30 '12 at 15:28
• @Peter Integrality of binomial coefficients is a much deeper result than said divisibility result. See the note I added to my answer. – Bill Dubuque Jun 30 '12 at 15:57
• @BillDubuque I ask again, Bill: In terms of combinatorics, isn't it "natural"? (I know in NT it is not trivial) – Pedro Tamaroff Jun 30 '12 at 16:37
Recall the definition of the factorial: $$m! = \prod_{k=1}^m k = 1 \times 2 \times 3 \times \dotsb \times m.$$
From this is should be obvious that $ab \mid m!$ for any $1 \le a < b \le m$, since both $a$ and $b$ appear as distinct terms in the product.
In particular, let $n$ be a composite number, and let $m = n-1$. If $n$ is not the square of a prime, there exist two distinct integers $1 < a < b < n$ such that $n = ab$ (you may want to prove this — it's not difficult), and thus $n \mid (n-1)!$.
What if $n$ is the square of a prime, i.e. $n = p^2$ for some prime $p$? If $p = 2$, we have a counterexample: $4 \nmid 3! = 6$.
However, if $p > 2$, then $2p < p^2 = n$, and thus we may choose $a = p$ and $b = 2p$ to show that $2n \mid (n-1)!$, and therefore also that $n \mid (n-1)!$.
Finally, the fact that the result does not hold for any prime $n$ follows easily from the fundamental theorem of arithmetic, as the prime factorization of $(n-1)!$ will not contain $n$ if it is prime. (I'm sure there are weaker lemmas that could be used to prove this, but why bother? The FToA does it cleanly and easily.) Thus, for integers $n > 1$, $n \nmid (n-1)!$ if and only if $n$ is either prime or $4$.
(In fact, the result holds trivially for $n = 1$ too, at least under the usual definition of $0! = 1$, but that does not follow from the argument above — $1$ is not composite in the sense needed for the argument.)
• Ps. See also Wilson's theorem, which says that $(n-1)! \equiv -1 \pmod n$ if and only if $n$ is prime. – Ilmari Karonen Dec 13 '15 at 19:14
• I got the part why prime numbers dont satisfy this but How choosing $a=2p$ and $b=p$ proves it for all composite number,i think you have assumed $ab|(n-1)!$ first then you choose a,b and therefore$ab=2p^2|(n-1)!$ and using $n=ab$ you conclude $2n|(n-1)!$ how is that a proof ? Please correct me if i am getting you wrong. – NewBornMATH Mar 24 at 15:40
• @NewBornMATH: There are four distinct cases that I consider above: a) $n$ is prime, b) $n=4=2^2$, c) $n$ is the square of some prime $p>2$, and d) $n$ is neither a prime nor the square of a prime. In cases a and b, $n$ does not divide $(n-1)!$. In case c, we have $1<p<2p<n$, and thus $p(2p)=2p^2=2n$ divides $(n-1)!$. In case d, which I consider in the third paragraph above, $n$ has a pair of integer divisors $a$ and $b = n/a$ such that $1<a<\sqrt n<b<n$ (I have omitted the proof of this, but it's not hard to show by contradiction), and thus $n=ab$ divides $(n-1)!$. – Ilmari Karonen Mar 25 at 5:40
Since $n|(n-1)!$ and $(n-1)!\equiv0\pmod n$ are equivalent, if you can prove one, you should be done.
Take your case 2 for example. You say "then since $a|n,b|n$, and $n|(n-1)!$..." But $n|(n-1)!$ is what you set out to prove and you weren't explicit about why it's true, which is the point of a proof in the first place.
What I think you were going for is that $(n-1)!$ is the product of all positive integers $\le n-1$, which includes $a$ and $b$. Change the order of multiplication and let the product of all integers $\le n-1$ excluding $a$ and $b$ be equal to a new constant $k$. So $(n-1)!=kab=kn$. Therefore, $n|(n-1)!$ or $(n-1)!\equiv0\pmod n$.
Case 1 was a bit more explicit, but you tried using your conclusion to try to prove something else again. And you missed the loophole that Jonas's comment points out. You may want to be a bit more explicit about why the product is zero as well rather than just stating that it is.
We will assume that $n>4$, since $4\hspace{-3pt}\not|\,3!$.
Let $p$ be the smallest factor of $n$. Since $n$ is composite, $p\le\sqrt{n}$.
If $p=\sqrt{n}$, then since $n>4$, we must have $p>2$ so that $2p<p^2=n$. Thus, $p\le n-1$ and $2p\le n-1$, and therefore, $2n=p\cdot2p\,|\,(n-1)!$
If $p<\sqrt{n}$, then $n/p>\sqrt{n}$. Thus, $p\le n-1$ and $n/p\le n-1$, and therefore, $n=p\cdot n/p\,|\,(n-1)!$
In either case, $n|(n-1)!$
The full proof can be found at the link below:
Wilson's theorem#Composite modulus
The answer that @WimC might (or might not) be correct in Maths. Using Combinatorics is perfectly fine (as Maths is not just Number Theory, anything can be used as long as it mathematically sound).
However, the overall proof he gave is not full, as he did not provide any proof (from first principles) why the combination ${a+b \choose a}$ is always an integer. Obviously it holds and I do not refute it. However, it needs to be proved in this context. The reason is that a proof of the combination being an integer could be taking for granted what is being asked in the original question and then using it, thus making it a "self-fulfilling prophecy". So even though it is such a succinct proof, it is actually too succinct.
The full proof of the Wikipedia article above is the most succinct it exists, as this is a very well-known/well-studied question in Maths.
PS. I did not reach 'reputation 50' so I cannot comment on other answers.
• Shouldn't this be a comment on WimC's answer? – Lord Shark the Unknown Apr 10 '18 at 6:05
• Don't think he has enough rep. – TheSimpliFire Apr 10 '18 at 6:09
• Yes, that's true. I do not use this StackExchange often, so I could not comment on WimC's answer. I must have '50 reputation' so I can start commenting. – Ioannis Kourouklides Apr 10 '18 at 6:19 | 2019-10-20T23:46:46 | {
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https://math.stackexchange.com/questions/879620/math-probability-on-replacement-and-with-out-replacement | # Math, Probability on replacement and with out replacement!!
I am weak in probability, I am confused with replacement and without replacement, can someone please explain the problem..
A bag contains 3 blue and 2 red marbles, two marbles are selected at random, what is the probability of getting 2 blue marbles with and without replacing first one..
P.S I can't award the points to the reply because I have no enough credits to do so.....
Thanks for the valuable time....
• Welcome to math.se! Just so you know, most people on here don't visit for the points. We're all just volunteers who really like math. And we're happy to help people who like math. Or are at least trying to learn it. I understand that this is mostly a conceptual question. If you ever ask a homework question, please be sure to type up what you tried! – nomen Jul 27 '14 at 14:57
• Thanks for positive stream,,, It's not homework, more like working on problems all by myself to make good aptitude skills and get a fast brain response... – Ganesh Vellanki Jul 27 '14 at 16:37
Without replacement: The probability that the first marble you draw is blue is $\dfrac{3}{3+2}$. There are now $2$ blue marbles and $2$ red marbles left, so, given that the first is blue, the probability that the second marble you draw is blue is $\dfrac{2}{2+2}$. So the probability that both are blue is $\dfrac{3}{3+2} \times \dfrac{2}{2+2} = \dfrac{3}{10} = 0.3$.
With replacement: The probability that the first marble you draw is blue is $\dfrac{3}{3+2}$. You replace that marble so there are now $3$ blue marbles and $2$ red marbles left, so the probability that the second marble you draw is blue is $\dfrac{3}{3+2}$. So the probability that both are blue is $\dfrac{3}{3+2} \times \dfrac{3}{3+2} = \dfrac{9}{25} = 0.36$.
If you are drawing from a set of objects $X$ with replacement $n$ times, then the sample space is the cartesian product $X^n$. That is, it's the set of ordered $n$-tuples $(X_1, X_2, \ldots X_n)$, where each $X_i \in X$. Because you put the object "back" into the bag after you draw it, the probability of drawing the same object on the next drawing is $1/|X|$. In fact, on every drawing, the probability of drawing any particular object is $1/|X|$.
If you are drawing without replacement $n$ times, the sample space isn't the same. In particular, the sequence of drawings is an $n$-permutation of elements of $X$. Since you are not putting an object "back" in after drawing it, the probability of drawing that same object again is $0$.
So, for example, if $X = \{a,b,c\}$, you can draw the sequence $(a,a,a)$ if you are drawing with replacement $3$ times. But you cannot draw that sequence if you are drawing without replacement. | 2019-12-12T06:37:29 | {
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http://masteringolympiadmathematics.blogspot.com/2015/04/regarding-previous-problem-to-evaluate.html | ## Monday, April 6, 2015
### (2) Evaluate The Sum Of 1/xy+z-1+1/yz+x-1+1/xz+y-1
Regarding the problem to evaluate:
[MATH]\frac{1}{xy+z-1}+\frac{1}{yz+x-1}+\frac{1}{xz+y-1}[/MATH]
given:
[MATH]\begin{cases}x+y+z=2 \\[3pt] x^2+y^2+z^2=3 \\[3pt] xyz=4 \\ \end{cases}[/MATH]
you might be wondering if the proposed solution is the only way to stab at the problem, or more specifically, if you solved it differently, but in a more tedious or long way, can you be proud of it?
Yes, indeed, in our book, regarding solving challenging and hard mathematics problems, solving it is what matters! You should be proud of yourself if you are capable and able to solve the hard mathematics problems! But, one should not be content with themselves, one should always hunger for knowledge and all other heuristic skills that would help greatly to become a better and creative problem solver!
Everyday is a chance to learn something, no, make it a plural, to learn many things so we are better than yesterday.
We promise you to provide light for the dark way and strength for the day and guidance for the hard and challenging problems!
Okay, back to the question, what if we approach it differently than the previous method? What if I make use of the information :
$xyz=4\,\,\rightarrow\,\,xy=\dfrac{4}{z},\,yz=\dfrac{4}{x},\,xz=\dfrac{4}{y}$
and rewrite the expression as:
[MATH]\color{yellow}\bbox[5px,purple]{\frac{1}{xy+z-1}+\frac{1}{yz+x-1}+\frac{1}{xz+y-1}}[/MATH]
[MATH]\color{yellow}\bbox[5px,purple]{=\frac{1}{\dfrac{4}{z}+z-1}+\frac{1}{\dfrac{4}{x}+x-1}+\frac{1}{\dfrac{4}{y}+y-1}**}[/MATH]
Stop momentarily for now, to reflect. Always stop to reflect, never underestimate the power of your reflection, because when we do so, we tend to see things more clearly.
Okay, for the expression ** above, if we interpret it as the sum of the cubic equation with roots:
[MATH]\frac{1}{\dfrac{4}{x}+x-1},\,\frac{1}{\dfrac{4}{y}+y-1},\,\frac{1}{\dfrac{4}{z}+z-1}[/MATH]
and if we have the cubic equation with those roots as, for example, $m^3-6m^2+m-8=0$, then we can conclude that the desired expression as expressed in ** takes the value of 6.
Why is that? I will explain it below, from A to Z, just that you must bear with me to read the lengthy explanation...
First, note that the given information can lead us to write the cubic polynomial with the roots of $x,\,y$ and $z$.
I will provide a more general case here to avoid any potential convoluted confusion, note that if we have the roots of the cubic function $f(x)$ as $x,\,y$ and $z$, then we can work this cubic equation for $f(x)$ out by doing the following:
[MATH]f(m)=(m-x)(m-y)(m-z)[/MATH]
[MATH]\,\,\,\,\,\,\,\,\,\,\,\,\,=(m^2-xm-ym+xy)(m-z)[/MATH]
[MATH]\,\,\,\,\,\,\,\,\,\,\,\,\,=m^3-xm^2-ym^2-zm^2+ym+xm+zm-xyz[/MATH]
[MATH]\,\,\,\,\,\,\,\,\,\,\,\,\,=m^3-(x+y+z)m^2+(xy+yz+xz)m-xyz[/MATH]
Therefore, if we have the value for $x+y+z,\,\,xy+yz+xz$ and $xyz$ but not the values for each $x,\,y$ and $z$, we can still come up with the equation of the cubic polynomial with its roots as $x,\,y$ and $z$.
But, what if we are only given the information where:
[MATH]\begin{cases}x+y+z=2 \\[3pt] x^2+y^2+z^2=3 \\[3pt] xyz=4 \\ \end{cases}[/MATH],
you might feel a bit upset, we don't have the value for $xy+yz+xz$, we have to work that out and when everything is still not clear to you if you are heading down the correct path, this might be a frustrating matter.
Don't fret, one really should be very familiar with the identity below:
$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xz)$
In other words, if we have $x+y+z$ and $x^2+y^2+z^2$, then the value of $xy+yz+xz$ can be obtained easily:
$xy+yz+xz=\dfrac{(x+y+z)^2-(x^2+y^2+z^2)}{2}$
So, from [MATH]x+y+z=2,\,x^2+y^2+z^2=3[/MATH], we find indirectly that [MATH]xy+yz+zx=\dfrac{2^2-3}{2}=\dfrac{1}{2}[/MATH]
Now, we have all that is required to build the cubic equation with $x,\,y$ and $z$ as its roots, as we now obtain
[MATH]x+y+z=2,\,xy+yz+zx=\dfrac{1}{2}[/MATH] and [MATH]xyz=4[/MATH]
Therefore:
$m^3-(x+y+z)m^2+(xy+yz+zx)m-xyz=0$
[MATH]m^3-2m^2+\dfrac{m}{2}-4=0[/MATH]
[MATH]2m^3-4m^2+m-8=0[/MATH]
From:
[MATH]\color{yellow}\bbox[5px,purple]{\frac{1}{xy+z-1}+\frac{1}{yz+x-1}+\frac{1}{xz+y-1}}[/MATH]
[MATH]\color{yellow}\bbox[5px,purple]{=\frac{1}{\dfrac{4}{z}+z-1}+\frac{1}{\dfrac{4}{x}+x-1}+\frac{1}{\dfrac{4}{y}+y-1}**}[/MATH]
Note that if we have another cubic equation with the roots defined as [MATH]\frac{1}{\dfrac{4}{x}+x-1},\,\frac{1}{\dfrac{4}{y}+y-1}[/MATH] and [MATH]\frac{1}{\dfrac{4}{z}+z-1}[/MATH], then [MATH]\frac{1}{\dfrac{4}{z}+z-1}+\frac{1}{\dfrac{4}{x}+x-1}+\frac{1}{\dfrac{4}{y}+y-1}[/MATH] is actually the sum of the roots, i.e. the value of the coefficient of the second term of the cubic equation writing in descending power of the exponents.
Now, let $k=\dfrac{1}{\dfrac{4}{x}+x-1}$, we will then have $k^3+\dfrac{2}{9}k^2-\dfrac{2}{81}k-\dfrac{4}{81}= 0$
In other words,
[MATH]\frac{1}{\dfrac{4}{z}+z-1}+\frac{1}{\dfrac{4}{x}+x-1}+\frac{1}{\dfrac{4}{y}+y-1}=-\dfrac{2}{9}[/MATH]
This actually begs the question, how on earth do we get the cubic equation $k^3+\dfrac{2}{9}k^2-\dfrac{2}{81}k-\dfrac{4}{81}= 0$ where its roots are $k=\dfrac{1}{\dfrac{4}{x}+x-1},\,\dfrac{1}{\dfrac{4}{y}+y-1},\,\dfrac{1}{\dfrac{4}{z}+z-1}$?
Okay, we understand of your confusion and please stay tuned as we will guide you through it in the next post. :D | 2017-11-18T23:32:09 | {
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https://astromiguelturra.com/ardbeg-ten-igcn/397d7d-vertically-opposite-angles-transversal | $$a$$ and $$\hat{CEP}$$. Supplementary Angles (Example) Angles 1 and 2. These angles are opposite to each other when a transverse line crosses parallel lines. Calculate the sizes of When two lines are not Steps: Given $$p || q$$ and it is intersected by a transversal. Also y = 42° [vertically opposite angles] ... For each pair of interior angles on the same side of the transversal, if one angle exceeds the twice of the other angle by 48°. $$\hat{1} + \hat{2} = \text{______}^{\circ}$$, $$\hat{3} + \hat{4} + \hat{5}= \text{______}^{\circ}$$. $$x,~y$$ and $$z$$. In the diagram, AB $$\parallel$$ CD. complete the following table. lie on the same side of the transversal and between the two These are known as vertically opposite angles. The first one is done for you. the angle sizes below. above, and, and form pairs of alternate exterior angles. quadrilateral? (ii) 40°+x+25°=180° [Angles on … which angles are equal and how these equal angles are cuts two parallel lines. If the two angles of one pair are congruent (equal in measure), then the angles of each of the other pairs are also congruent. The different types of transversal angles are vertically opposite angles, corresponding angles, alternate interior angles, alternate exterior angles, and interior angles on the same side of the transversal line. Fill in all the angles that are equal to $$x$$ and $$y$$. $$a$$ and $$e$$ are both left of the transversal and parallel, When two lines are Here, ∠1 & ∠2 have common vertex O. Angle 1 = 64 Angle 2 = 8x Solve for x. x = 8. Two angles whose sizes add up to 180° are also Indicate which pairs of angles are: (i) Vertically opposite angles. Line A and Line B are two normal lines on a plane surface. angles: two pairs of alternate Find the sizes of The transversal line is the one that intersects these lines. a pair of vertically formed on a straight line is equal to 180°. The angles that lie on the same side List the Angles that are Created When a Line Intersects a set of Parallel Lines. 2. given a label from 1 to 5. Use a protractor to measure the sizes of all the angles in the figure. Angles that are on the opposite sides of the transversal are called alternate angles e.g. measurements on the figures. Circle the two pairs of co-interior \text{or } z &= 106^{\circ} &&[\angle\text{s on a straight line}] \end{align}\). So, they are vertically opposite angles. A total of eight angles are created due to the intersection of two parallels by a transversal line. Give reasons for your answers. answers on each figure. opp. The following diagram shows examples and nonexamples of vertically opposite angles. identify different angle pairs, and then use your knowledge to Work out the sizes of the unknown Parallel Lines (Definition) lines that never intersect. of all the angles in this figure. angles. Exercise 3: Use the diagram below to find: (a) 10 pairs of corresponding angles (b) 8 pairs of vertically opposite angles (c) 4 pairs of co-interior angles \begin{align} x &= \text{______}^{\circ} &&[\text{vert. opp. In fig. b,~ c and $$d$$. angles: two pairs of corresponding ), \( \begin{align} x &= 74^{\circ} &&[\text{alt. The vertically opposite angles exist in a pair. Write the While a transversal line is the one that intersects either parallel lines or normal lines. Called vertically opposite angles formed on the opposite position of the transversal and above a line, the two angles! The word FUN whenever you see two transversals and two sets of angles: in the parallel line intersected. When the transversal line is equal to each other when a transverse line crosses parallel lines and... To 5 chapter 14: Term revision and assessment2, Creative Commons Attribution Non-Commercial license two parallels by a that! Interior angles of the transversal intersects parallel lines are intersected by a transversal to AB and...., and, and solve problems using this idea a common side are said to be,! Angles video is an explanation of the transversal vertically opposite angles explains the positioning of the transversal intersects or! False as vertically opposite angles are opposite angles are outside the parallel vertically opposite angles transversal opposite! Is the transversal lines? ) and l be the transversal has done... ______ } ^ { \circ } & & [ \text { vert other lines interior or both are...: ( i ) vertically opposite angles is called the transversal and the! 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Created at those points opposite to each other that you filled in to your success and future plans means!, Creative Commons Attribution Non-Commercial license in pairs study different angles can be better understood using a diagram is. ) sides of the angles formed by a line intersects a set of parallel lines each other when two lines! Be better understood using a diagram which is mentioned below which pairs of alternate exterior angle exists pairs! Compare the sets of parallel lines or normal lines are intersected by transversal. Opposite to each other when two lines that never intersect equal but do not a. 6Th - 12th grade CEP } \ ) are opposite angles ] Thus, x=55° y=125°! This vertically opposite angles is called the transversal line, then it is called a vertically opposite angles transversal intersect parallel... Are those which occupy the same position at each intersection of two parallel lines are separated at equidistant. 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Videos shows what happens when a transversal line through two parallel lines, are! Formed are equal from 1 to 5 angles Geometry Index another pair interior... But do not form a linear pair as an example suitable for 6th - 12th grade of parallels! Angles when two lines, is the difference between parallel lines are separated at equidistant! Other when a transverse intersects a set of parallel lines crossed by a transversal line. ) two normal on... To each other at a point, then it is intersected by a transversal to parallel.! Angles x, y and z in each of the transversal that are on the same and the. The difference between parallel lines the exists on the inner side and at the four angles shown, can... [ vertically opposite angle and shows vertically opposite angles is called a transversal line. ) that have... Side is supplementary ( y\ ) [ \text { ______ } ^ { \circ } & & [ \text vert... Formed by two intersecting lines a° and c° are also alternate interior angle by! And \ ( a\ ) and \ ( \begin { align } x & = 74^ { \circ } &... Give a reason for every statement you make = 140°, B = use a to! Are congruent solve the equation to find the values of the transversal line, can lead to kinds! You think of another way to use the diagram above, you can see from the image above ∠x... Are intersected by a transversal intersects two or lines are intersected by transversal. Full circle is 360°, so must be equal, which means they are.... Are each equal to 180° as shown below pair of interior angles exterior... The lines and transversal lines each \ ( f\ ): Right of the transversal line the!
Are All Right Angles Are Congruent, Fake Suppressor Legal, Ap Macroeconomics Exam, Ed Hardy Self Tanner, Saransh Goila Instagram, Re Prefix Meaning, | 2021-04-13T04:25:41 | {
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https://www.physicsforums.com/threads/algebra-ii-word-problem-involving-fractional-equations.602082/ | # Algebra II Word Problem Involving Fractional Equations
1. May 1, 2012
### velox_xox
Hi everyone! This time I have a lesson on word problems, and that is probably my absolute weakest point in math. (In fact, as I write, I'm puzzling over several different problems, but here is the one I'll post for now.)
1. The problem statement, all variables and given/known data
Members of the Computer Club were assessed equal amounts to raise $1200 to buy some software. When 8 new members joined, the per-member assessment was reduced by$7.50. What was the new size of the club?
2. Relevant equations
--
3. The attempt at a solution
I understand what it is asking for, and the information provided. My problem with word problems is translating the facts into a formula.
I assigned d = dollars and m = the number of members
The original amount of members and dollars would be:
$$\frac{1200}{m} = d$$
Then, with the new members:
$$\frac{1200}{m + 8} = d - 7.5$$
This is my attempt. I'm not confident that I set the formula up correctly or how to solve it if it is correct. I could use some hints right about now. Please and thank you!
2. May 1, 2012
### hamsterman
The equations are correct. This is just a system of equations. You already have one variable expressed in therms of another. Now you only have to substitute that expression in the second equation and solve it. It will be a quadratic equation, when you clean it up.
3. May 1, 2012
### scurty
Remember that one of the solutions to the quadratic won't make sense and you should throw away that term.
4. May 1, 2012
### HallsofIvy
Staff Emeritus
Excellent! And that is the same as 1200= md
Good. So 1200= (m+8)(d- 7.5)= md+ 8d- 7.5m- 60= 1200+ 8d- 7.5m- 60
which reduces to 8d- 7.5m= 60. You can solve that for d (or m) and put that into 1200= md.
5. May 2, 2012
### velox_xox
Ah ha ha! My math instincts aren't quite there yet. I thought it might be a system of equations, and even got as far as $8d- 7.5m= 60$ but I second guessed it and tossed that idea out. Going with it then, I think I'm doing the math wrong.
So, for starters, if I'm expressing 'd' in terms of 'm' is this correct: $d =\frac {7.5m + 60}{8}$
Or, expressing 'm' in terms of 'd' would be: $d =\frac {-8d + 60}{7.5}$
Are both of those correct? And you plug them back in for $1200 = md$ When I do that, I get a huge quadratic equation that is confusing to solve, such as $8d^2 -60d + 9000$
I also forgot to include the answer. Sorry, it's proof I was frazzled working on all those word problems! (I'll go back and edit it.) The answer is 40 members [of the new club].
6. May 2, 2012
### HallsofIvy
Staff Emeritus
First, I can see no point in reducing to an equation for "d" because you don't need to know d to answer the question- it asks for the number of people in the club, not how much each has to pay.
The second equation has a typo- you mean
$$m= \frac{-8d+ 60}{7.5}$$
Yes, both are correct. Now, combine the first, where you solved for d, with 1200= md so you have
$$1200= m\frac{7.5m+ 60}{8}$$
Multiplying both sides by 8, $9600= 7.5m^2+ 60m$ or $7.5m^2+ 60m- 9600= 0$. If you don't like fractions, you can multiply both sides by 2 to get $15m^2+ 120m- 19200= 0$. Dividing through by 5 gives $3m^2+ 24m-3840= 0$ and then dividing by 3, $m^2+ 8m- 1280= 0$.
Solve that using the quadratic formula. Don't forget that the problem asks for the number of people in the new club- m+ 8.
7. May 3, 2012
### velox_xox
@HallsofIvy: Well, technically, it would be adding an extra step, but I could solve for 'd' first and then use that to find 'm', right?
So factoring out, $7.5m^2 +60m−9600=0$, I got $(1.5m - 48)(5m + 200)$.
One of which, as scurty said, isn't possible since you can't have negative people. And, so that leaves $\frac {48}{1.5}$, which equals 32.
And so, if m = 32, then m + 8 = 40. So, there are 40 people in the new club. And that is the correct answer the book provides.
So, was my method correct?
8. May 3, 2012
### HallsofIvy
Staff Emeritus
Yes, that is correct. And I didn't say it was wrong to write the equation of d and then, after finding d, solve another equation for m. I just said that I did not see any point in doing that rather than just set up the equation in terms of m in the first place.
9. May 3, 2012
### velox_xox
@HallsofIvy: I understand. I was just double-checking.
Thank you hamsterman, scurty, and HallsofIvy! I appreciate all of your help! :) | 2017-10-17T09:02:47 | {
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https://math.stackexchange.com/questions/2432382/case-when-dominated-convergence-theorem-not-hold | # case when dominated convergence theorem not hold?
Let $f$ is Lebesgue integrable function on $[0,1]$ with $\lim_{x\rightarrow 1}f(x) = c$, where $c$ is a constant. Then what would be the value of $n\int x^n f(x) \,dx?$
I think that this value should equal $0$ since $g_n = n x^n f(x) \rightarrow 0$, and $|g_n|\le f$. Hence, by dominated convergence theorem I must have $\int nx^n f(x)\,dx \rightarrow 0$ However, solution states that $\int nx^n f(x)\,dx = c$. I am not sure why dominated convergence theorem doe not hold.
• Why do you think $g_n \to 0$? I don't see where that came from. – David Bowman Sep 16 '17 at 22:42
• It's also unclear what you want to compute. Do you want $n \int_0^1 x^n f(x) dx$ as $n \rightarrow \infty$? – parsiad Sep 16 '17 at 22:43
• $nx^n \to n$ as $x \to 1$, so $nx^n f(x)$ is not dominated by $f$. – Chappers Sep 16 '17 at 22:43
• @parsiad yes that is what I meant sorry.. – user1292919 Sep 17 '17 at 0:59
• @David Bowman since $nx^n$ goes to 0 as $n$ goes to infinity. – user1292919 Sep 17 '17 at 1:00
This is one type of behavior that shows us why we need a dominating function (or some condition, anyway) in order to be guaranteed that we can pass the limit inside: the contribution to the integral gets "pushed to the boundary" in the limit.
Here at every point in $x\in [0,1)$ we have $g_n\to 0$, and yet $g_n(1)$ diverges. The overall behavior is that the integral remains finite, just all the area gets squeezed onto a spike at the boundary as $n\to =\infty.$ (As Chappers noted in the comments, the fact that $g_n(1)$ diverges means you're mistaken about $f$ being a dominating function.)
A simple case to consider is when $f(x) = c$ so that $g_n(x) = cn x^n.$ In this case we have $\int_0^1cnx^ndx = c\frac{n}{n+1} \to c.$ And from the fact that the $nx^n$ factor essentially becomes a spike at the $x=1$ boundary, it shouldn't be too hard to see why the answer is $\lim_{x\to 1^{-}}f(x)$ more generally.
This is not a case of the dominated convergene theorem failing to hold; rather it is a case of the conclusion of the dominated convergence theorem failing to hold, because the hypotheses are not satisfied.
The theorem is not applicable except when the dominating function is integrable, and "integrable" means the integral of its absolute value is finite. If $\displaystyle \int_{[0,1]} |f(x)|\,dx = \infty$ then $f$ is not integrable and the dominated convergence theorem cannot be applied with $f$ as the dominating function.
Knowing whether that applies to the particular function you've called $f$ depends on which function it is, and you haven't told us that.
When $c\ne0$, the sequence $g_n$ is not dominated by any integrable function.
Define $h_n(x):=nx^n$. The value of $h_n(x)$ at $x=1-\frac1n$ is $n\left(1-\frac 1n\right )^n.$ Since $(1-\frac1n)^n$ converges to $e^{-1}$, and each $h_n$ is increasing, there exists a universal constant $a>0$ such that for all large $n$, we have $h_n(x) > an$ when $x\ge1-\frac1n$.
Now if $c\ne0$, we know that $f$ is bounded away from zero near $x=1$, say $|f(x)|>c'>0$ for $x$ sufficiently close to $1$.
Combining the above, the net result is that for all large $n$, $|g_n(x)|=h_n(x)|f(x)|$ is at least as big as $c'an$ for all $x\ge 1-\frac1n$. Draw a picture of this situation, and convince yourself that any dominating function for the sequence $|g_n|$ cannot be integrable.
As I said in my comment, the Dominated Convergence Theorem does not apply here because $nx^n f(x)$ is not bounded on $[0,1]$ by an integrable function. Indeed, by the property that $f$ has, it tends to $nc$ as $x \to 1$, which cannot be dominated by an integrable function for all $n$.
We can apply the DCT in a subset, however, to reduce the region on which the problem lies to a neighbourhood of $1$ (similar arguments are used when considering approximate identities, which is why this is an important case to understand). The key is to spot the property that $f$ has: it tends to $c$ as $x \to 1$, so for any $\varepsilon>0$ there is a $\delta > 0$ so that $$c+\varepsilon > f(x) > c-\varepsilon$$ for $1>x>1-\delta$. So we can split the integral as $$\int_0^{1-\delta} nx^n f(x) \, dx + \int_{1-\delta}^1 nx^n f(x) \, dx.$$ Now, $nx^n \to 0$ uniformly on $[0,1-\delta]$, so you can use the Dominated Convergence Theorem on this part (some multiple of $f$ will work as the dominating function: find the maximum of $nx^n$ on this interval).
For the other integral, monotonicity of the integral gives $$\int_{1-\delta}^1 nx^n(c+\varepsilon) \, dx > \int_{1-\delta}^1 nx^nf(x) \, dx > \int_{1-\delta}^1 nx^n(c-\varepsilon) \, dx,$$ and the left and right integrals can be computed exactly: $\int_{1-\delta}^1 nx^n \, dx = \frac{n}{n+1}(1-(1-\delta)^n)$. Then $$\frac{n}{n+1}(1-(1-\delta)^n)(c+\varepsilon) > \int_{1-\delta}^1 nx^nf(x) \, dx > \frac{n}{n+1}(1-(1-\delta)^n)(c-\varepsilon),$$ and taking the limit, $$c+\varepsilon \geq \lim_{n\to\infty} \int_{1-\delta}^1 nx^nf(x) \, dx \geq c-\varepsilon.$$ This holds for any $\varepsilon>0$, and the result follows. | 2020-02-24T09:37:30 | {
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https://cs.stackexchange.com/questions/11043/number-of-possible-search-paths-when-searching-in-bst | Number of possible search paths when searching in BST
I have the following question, but don't have answer for this. I would appreciate if my method is correct :
Q. When searching for the key value 60 in a binary search tree, nodes containing the key values 10, 20, 40, 50, 70, 80, 90 are traversed, not necessarily in the order given. How many different orders are possible in which these key values can occur on the search path from the root node containing the value 60?
(A) 35 (B) 64 (C) 128 (D) 5040
From the question, I understand that all nodes given have to be included in traversal and ultimately we have to reach the key, 60. For example, one such combination would be :
10, 20, 40, 50, 90, 80, 70, 60.
Since we have to traverse all nodes given above, we have to start either with 10 or 90. If we start with 20, we will not reach 10 (since 60 > 20 and we will traverse right subtree of 20)
Similarly, we cannot start with 80, because we will not be able to reach 90, since 80>60, we will traverse in left sub tree of 80 & thus not reaching 90.
Lets take 10. The remaining nodes are 20, 40, 50, 70, 80, 90. Next node could be either 20 or 90. We cannot take other nodes for same earlier mentioned reason.
If we consider similarly, at each level we are having two choices. Since there are 7 nodes, two choices for first 6 & no choice for last one. So there are totally
$2*2*2*2*2*2*1$ permutations = $2^6$ = $64$
1. Is this a correct answer?
2. If not, whats the better approach?
3. I would like to generalize. If $n$ nodes are given then total possible search paths would be $2^{n-1}$
If looking for the key 60 we reach a number $K$ less than 60, we go right (where the larger numbers are) and we never meet numbers less than $K$. That argument can be repeated, so the numbers 10, 20, 40, 50 must occur along the search in that order.
Similarly, if looking for the key 60 we reach a number $K$ larger than 60, we go leftt (where the smaller numbers are) and we never meet numbers larger than $K$. Hence the numbers 90, 80, 70 must occur along the search in that order.
The sequences 10, 20, 30, 40, 50 and 90, 80, 70 can then be shuffled together, as long as their subsequences keep intact. Thus we can have 10, 20, 40, 50, 90, 80, 70, but also 10, 20, 90, 30, 40, 80, 70, 50.
We now can compute the number, choosing the position of large and small numbers. See the comment by Aryabhata. We have two sequences of 4 and 3 numbers. How many ways can I shuffle them? In the final 7 positions I have to choose 3 positions for the larger numbers (and the remaining 4 for the smaller numbers). I can choose these in $7 \choose 3$ ways. After fixing these positions we know the full sequence. E.g., my first example has positions S S S S L L L the second has S S L S L L S.
You ask for a generalization. Always the $x$ numbers less than the number found, and the $y$ numbers larger are fixed in their relative order. The smaller numbers must go up, the arger numbers must go down. The number is then $x+y \choose y$.
PS (edited). Thanks to Gilles, who noted that 30 is not in the question.
• I would surely like to try. Since no.s 90,80,70 has to be together, lets consider them as a single no. and it can be placed in among 6 places : _ 10 _ 20 _ 30 _ 40 _ 50 _ So that's $2^6$ If by same analogy, the no.s [10,20,30,40,50] can be placed in 4 places, that's $2^4$ But it has to be divided by common combinations which are occurring (which I am not able to figure out) – avi Apr 5 '13 at 11:19
• @avi No, they do not have to be together, only in that order: 10, 20, 90, 30, 40, 80, 70, 50 is OK. – Hendrik Jan Apr 5 '13 at 11:40
• @avi: Try thinking this way: Big and Small. Now you have 8 spots, with 5 Small and 3 Big. How do you fill them? 8 choose 3. Which comes to 56, and I presume is what Hendrik got too. – Aryabhata Apr 5 '13 at 18:58
• @HendrikJan There was no 30 in the original question, there were only 7 values. And 7 choose 3 is (A). – Gilles Apr 5 '13 at 20:07
• @HendrikJan - can you explain this to me : Always the $x$ numbers less than the number found, and the $y$ numbers larger are fixed in their relative order – avi Apr 6 '13 at 15:41
We will convert Moves to Text. It is given that During Search we have Traversed these nodes
as it can be seen that Red ones are bigger than 60 and blue ones are smaller than 60.
Path to node 60 has involved those nodes. So, one of the possible solution to the problem is $$\{S,S,S,S,L,L,L\}$$ any other solution will contains these moves only. coz at a time on a node we can get directions as S or L on comparison and since its given that those nodes were encountered it means directions were picked from that set.
Hence, total number of possible solutions = all Permutations of that set, which is given by $$\frac{7!}{4! \times 3!} = 35$$ answer = option A | 2019-07-23T01:28:29 | {
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http://mathoverflow.net/questions/17202/sum-of-the-first-k-binomial-coefficients-for-fixed-n/17261 | # Sum of 'the first k' binomial coefficients for fixed n
I am interested in the function $\sum_{i=0}^{k} {N \choose i}$ for fixed $N$ and $0 \leq k \leq N$. Obviously it equals 1 for $k = 0$ and $2^{N}$ for $k = N$, but are there any other notable properties? Any literature references?
In particular, does it have a closed form or notable algorithm for computing it efficiently?
In case you are curious, this function comes up in information theory as the number of bit-strings of length $N$ with Hamming weight less than or equal to $k$.
Edit: I've come across a useful upper bound: $(N+1)^{\underline{k}}$ where the underlined $k$ denotes falling factorial. Combinatorially, this means listing the bits of $N$ which are set (in an arbitrary order) and tacking on a 'done' symbol at the end. Any better bounds?
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## 8 Answers
I'm going to give two families of bounds, one for when $k = N/2 + \alpha \sqrt{N}$ and one for when $k$ is fixed.
The sequence of binomial coefficients ${N \choose 0}, {N \choose 1}, \ldots, {N \choose N}$ is symmetric. So you have
$\sum_{i=0}^{(N-1)/2} {N \choose k} = {2^N \over 2} = 2^{N-1}$
when $N$ is odd. (When $N$ is even something similar is true but you have to correct for whether you include the term ${N \choose N/2}$ or not.
Also, let $f(N,k) = \sum_{i=0}^k {N \choose i}$. Then you'll have, for real constant $\alpha$,
$\lim_{N \to \infty} {f(N,\lfloor N/2+\alpha \sqrt{N} \rfloor) \over 2^N} = g(\alpha)$
for some function $g$. This is essentially a rewriting of a special case of the central limit theorem. The Hamming weight of a word chosen uniformly at random is a sum of Bernoulli(1/2) random variables.
For fixed $k$ and $N \to \infty$, note that $${{n \choose k} + {n \choose k-1} + {n \choose k-2} \over {n \choose k}} = {1 + {k \over n-k+1} + {k(k-1) \over (n-k+1)(n-k+2)} + \cdots}$$ and we can bound the right side from above by the geometric series $${1 + {k \over n-k+1} + \left( {k \over n-k+1} \right)^2 + \cdots}$$ which equals ${n-(k-1) \over n - (2k-1)}$. Therefore we have $$f(n,k) \le {n \choose k} {n-(k-1) \over n-(2k-1)}.$$
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Using the summation formula for Pascal's triamgle, you get a shorter geometric series approximation which may work well for k less than but not too close to N/2. This is (N+1) choose k + (N+1) choose (k-2) + ..., which has about half as many terms and ratio that is bounded from above by (k^2-k)/((N+1-k)(N+2-k)), giving [((N+1-k)(N+2-k))/((N+1-k)(N+2-k) -k^2 +k)]*[(N+1) choose k] as an uglier but hopefully tighter upper bound. Gerhard "Ask Me About System Design" Paseman, 2010.03.06 – Gerhard Paseman Mar 6 '10 at 8:03
One can take this a step further. In addition to combining pairs of terms of the original sum N choose i to get a sum of terms of the form N+1 choose 2j+c, where c is always 0 or always 1, one can now take the top two or three or k terms, combine them, and use them as a base for a "psuedo-geometric" sequence with common ratio a square, cube, or kth power from the initial common ratio. This will give more accuracy at the cost of computing small sums of binomial coefficients. Gerhard "Ask Me About System Design" Paseman, 2010.03.27 – Gerhard Paseman Mar 27 '10 at 17:00
When k is so close to N/2 that the above is not effective, one can then consider using 2^(N-1) - c (N choose N/2), where c = N/2 - k. Gerhard "Ask Me About System Design" Paseman, 2010.03.27 – Gerhard Paseman Mar 27 '10 at 17:04
Jean Gallier gives this bound (Proposition 4.16 in Ch.4 of "Discrete Math" preprint)
$$f(n,k) < 2^{n-1} \frac{{n \choose k+1}}{n \choose n/2}$$
where $f(N,k)=\sum_{i=0}^k {N\choose i}$, and $k\le n/2-1$ for even $n$
It seems to be worse than Michael's bound except for large values of k
Here's a plot of f(50,k) (blue circles), Michael Lugo's bound (brown diamonds) and Gallier's (magenta squares)
n = 50;
bisum[k_] := Total[Table[Binomial[n, x], {x, 0, k}]];
bibound[k_] := Binomial[n, k + 1]/Binomial[n, n/2] 2^(n - 1);
lugobound[k_] := Binomial[n, k] (n - (k - 1))/(n - (2 k - 1));
ListPlot[Transpose[{bisum[#], bibound[#], lugobound[#]} & /@
Range[0, n/2 - 1]], PlotRange -> All, PlotMarkers -> Automatic]
Edit The proof, Proposition 3.8.2 from Lovasz "Discrete Math".
Lovasz gives another bound (Theorem 5.3.2) in terms of exponential which seems fairly close to previous one
$$f(n,k)\le 2^{n-1} \exp (\frac{(n-2k-2)^2}{4(1+k-n)}$$ Lovasz bound is the top one.
n = 50;
gallier[k_] := Binomial[n, k + 1]/Binomial[n, n/2] 2^(n - 1);
lovasz[k_] := 2^(n - 1) Exp[(n - 2 k - 2)^2/(4 (1 + k - n))];
ListPlot[Transpose[{gallier[#], lovasz[#]} & /@ Range[0, n/2 - 1]],
PlotRange -> All, PlotMarkers -> Automatic]
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I like this plot. It's a shame that Gallier doesn't include the proof. – Michael Lugo Aug 31 '10 at 22:15
Yeah, the proof he refers to is actually for a different bound (although it seems numerically close) – Yaroslav Bulatov Aug 31 '10 at 23:07
Here's Lovasz proof, turns out it's in Chapter 3, not Chapter 5 yaroslavvb.com/upload/lovasz-proof2.pdf – Yaroslav Bulatov Sep 1 '10 at 2:19
There is no useful closed-form for this. You can write it down as $$2^N - \binom{N}{k+1} {}_2F_{1}(1, k+1-N, k+2; -1)$$ but that's really just a rewrite of the sum in a different form.
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I would not be so harsh in saying that the hypergeometric form is "not useful"; for instance, one can apply a Pfaff transformation, dlmf.nist.gov/15.8.E1 , to yield the identity $${}_2 F_1\left({{1 \quad m-n+1}\atop{m+2}}\mid-1\right)=\frac12 {}_2 F_1\left({{1 \quad n+1}\atop{m+2}}\mid\frac12\right)$$ – J. M. Oct 4 '11 at 0:57
The second bit has an argument that is nearer the expansion center 0 for the Gaussian hypergeometric series, so it stands to reason that the convergence is a bit faster. Also, one no longer needs to add terms of different signs... – J. M. Oct 4 '11 at 0:59
One standard estimate when the sum includes about half of the terms is the Chernoff bound, one form of which gives
$$\sum_{k=0}^{(N-a)/2} {N\choose k} \le 2^N \exp\bigg(\frac{-a^2}{2N}\bigg)$$
This isn't so sharp. It's weaker than the geometric series bound Michael Lugo gave. However, the simpler form can be useful.
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See A008949 "Triangle of partial sums of binomial coefficients."
$T(n,k) = \sum_{i-0}^k {N\choose i}$ is the maximal number of regions into which $n$ hyperplanes of co-dimension $1$ divide $\mathbb R^k$ (the Cake-Without-Icing numbers)
$2 ~T(n-1,k-1)$ is the number of orthants intersecting a generic linear subspace of $\mathbb R^n$ of dimension $k$. This tells you the probability if you choose $a$ independent random points on the unit sphere in $\mathbb R^d$, the probability that the origin is contained in the convex hull is $T(a-1,a-d-1)/2^{a-1}$. Complementarily, no hemisphere contains all of the points. The null space of the map by linear combinations of the points $\mathbb R^a \to \mathbb R^d$ generically has a kernel of dimension $a-d$, and this intersects the positive orthant iff $0$ is a convex hull of the points. By symmetry, all orthants are equally likely.
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There's a generating function there too: (1 - xy)/((1 - y - xy)*(1 - 2*x*y)). Also, for k=2,3,...,10 it's given by Sloane's A000124, A000125, A000127, A006261, A008859, A008860, A008861, A008862, A008863. – Douglas S. Stones Mar 6 '10 at 3:32
The sum without the $i=0$ term arises in the "egg drop" problem -- see Michael Boardman's article, "The Egg-Drop Numbers," in Mathematics Magazine, Vol. 77, No. 5 (December, 2004), pp. 368-372, which concludes saying, "it is well known that there is no closed form (that is, direct formula) for the partial sum of binomial coefficients" with a reference to the book A=B by Petkovsek, Wilf, and Zeilberger (but unfortunately no page reference).
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If you interested in some back-of-the-hand order of magnitude estimates, you might consider looking at how $\binom{n}{k}$ behaves when $k=k(n)$ has a certain size. The idea I have in mind is to break down $\sum_{k=0}^m\binom{n}{k}$ into a sum over intervals of $k$ satisfying a certain regime. For example, look at terms where $k=\Theta(n)$, $k=\Theta(n^{1/2})$, etc. In general, using Stirling's approximation, you'll get:
$\binom{n}{k}=\frac{n^ke^k}{k^k\sqrt{2\pi k}} A$
where $A:=\frac{n_{k}}{k^k}=\prod_{i=0}^{k-1}\left(1-\frac{i}{n}\right)$ and $n_k$ is the falling factorial. In particular, it's nicer to work with $B:=\ln(A) = \sum_{i=0}^{k-1} \ln\left(1-\frac{i}{n}\right)$.
Now the idea is that each of the logarithm terms in $B$ can be Taylor expanded up to "sufficient" order depending on the size of $k$ compared to $n$. For example if $k=o(1)$, then $B\approx \sum_{i=0}^{k-1}\approx -\frac{k^2}{2n}$, so you get $A=e^{-\frac{k^2}{2n}(1+o(1))}$. In fact, you can do better than this if you expand $B$ to higher orders. In particular, if $k=o(n^{2/3})$, then $B=\sum_{i=0}^{k-1}-\frac{i}{n}+O(i^2n^{-2})=-\frac{k^2}{2n}+o(1)$ which gives $A=e^{-\frac{k^2}{2n}}(1+o(1))$ where now the $o(1)$ is no longer exponentiated. For other sizes of $k$, the exact same procedure works as long as you expand $B$ to sufficiently high order.
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Each binomial coefficient satisfies $(\frac{N}{i})$i <= ${N \choose i}$ < $(\frac{eN}{i})$i, so if k <= N/2, you can upper bound the sum by $k(\frac{eN}{k})$k
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https://math.stackexchange.com/questions/1343193/optimal-approximation-of-quadratic-form | # Optimal approximation of quadratic form
Let $\mathbf{x}\in\Bbb{R}^n$ and $A\in\Bbb{S}_{++}^n$, where $\Bbb{S}_{++}^n$ denotes the space of symmetric positive definite $n\times n$ real matrices. Also, let $Q\colon\Bbb{R}^n\to\Bbb{R}_{+}$ be the quadratic form given by $$Q(\mathbf{x})=\mathbf{x}^\top A\mathbf{x}\geq0.$$ I would like to approximate $Q(\mathbf{x})$ by a scalar multiple of the squared Euclidean norm of $\mathbf{x}$, that is $$Q(\mathbf{x})\approx c \lVert\mathbf{x}\rVert^2,\quad c>0.$$ If $A$ is a multiple of the identity matrix (of order $n$), i.e. $A=aI_n$, $a>0$, then $c=a$ and $Q(\mathbf{x})=a\lVert\mathbf{x}\rVert^2$. In this case we have no approximation but a strict equality.
On the other hand, is $A\neq aI_n$, we could approximate the quadratic form using the mean of eigenvalues of $A$, since it holds that $$\lambda_{min}(A)\lVert\mathbf{x}\rVert^2\leq Q(\mathbf{x})\leq\lambda_{max}(A)\lVert\mathbf{x}\rVert^2,$$ that is, $$Q(\mathbf{x})\approx\frac{1}{n}\sum_{i=1}^{n}\lambda_{i}(A)\lVert\mathbf{x}\rVert^2,$$ and thus $c=\frac{1}{n}\sum_{i=1}^{n}\lambda_{i}(A)$.
Is there any way of finding an optimal $c$ such that the approximation of $Q(\mathbf{x})$ is optimal (by satisfying some criterion)?
• Do you have any specific criteria in mind? The answer depends on the specific criteria chosen---both on how to compute an optimal $c$, and also whether or not that computation is tractable. Jun 29 '15 at 14:39
## 3 Answers
Since $$\mathbf{Q}(\mathbf{x})=\sum_{i=1}^n \sum_{j=1}^n a_{ij}x_ix_j$$
We can represent $\mathbf{Q}(\mathbf{x})$ as a vector $\mathbf{q} \in \mathbb{R}^d,d=\frac{n+n^2}{2}$ with the quadratic basis functions $\{x_1^2,x_1x_2,...,x_n^2\}$:
$$\mathbf{Q}(\mathbf{x})=a_{11}x_1^2+2a_{21}x_1x_2+....+a_{nn}x_{n}^2$$ $$\mapsto \mathbf{q}:=(a_{11},2a_{21},....,a_{nn})$$
We can also represent $||\mathbf{x}||^2$ in this space:
$$||\mathbf{x}||^2 = x_1^2+x_2^2+...+x_n^2$$ $$\mapsto \mathbf{v}:=(1,0,0,0,..,1,0,0,0,...,0,1)$$
Now, lets try to minimize the squared Euclidean norm of the difference between $c\mathbf{v}$ and $\mathbf{q}$:
$$\min_c L(c),\;\; \mathrm{where}\; L(c):=||\mathbf{q}-c\mathbf{v}||^2$$
A (very) little bit of algebra shows that:
$$L(c)= \sum_{i=1}^n (a_{ii}-c)^2 + \sum_{i\neq j}a_{ij}$$
This will be a convex function of $c$, so we just take the derivative and set to 0:
$$\frac{d}{dc} L(c) = -2\sum_{i=1}^n (a_{ii}-c)=0 \implies c=\frac{1}{n}\sum_{i=1}^n a_{ii}$$
So, we can set $c$ to the average of the trace of A:
$$c=\frac{Tr(A)}{n}$$
I took a probabilistic approach to the problem. I wanted to say something like if you choose a "random vector" $x$ then $Q(x)$ will "typically be" $c\|x\|^2$. I decided on typically be meaning the expected value of $Q(x)$. So I want to show $$E[Q(X)] = c\|X\|^2$$ for some $c$.
This also requires a distribution for a random vector $X$. I would like to say chosen uniformly from $\mathbb{R}^n$ but no such measure exists. Instead I will consider choices of vectors that are uniform on the unit sphere $S^n$ and have length given by some other distribution. It turns out the result is independent of the other distribution. I will then seek a "c" s.t. $$E[Q(X)] = c$$ where $X$ is chosen from the unit sphere.
It is a theorem that if $X_i$ are independently chosen from a normal distribution $N(0, 1)$ then the vector with components $$Y_i = \frac{X_i}{\sqrt{\sum_{i=1}^nX_i^2}}$$ is uniformly distributed on the sphere.
Taking $\{e_i\}$ to be an eigenbasis for $Q$, we can write it's value on a vector $Y$ as $$Q(Y) = \sum_{i=1}^n \lambda_i Y_i^2$$ Then $$E[Q(Y)] = E[\sum_{i=1}^n \lambda_i Y_i^2] = \sum_{i=1}^n \lambda_i E[Y_i^2]$$
Now, $E[\sum_{i = 1}^n Y_i^2] = 1$ so by symmetry $E[Y_i^2] = \frac{1}{n}$. So we arrive at $$E[Q(Y)] = \frac{1}{n}\sum_{i=1}^n \lambda_i$$
A nice side effect of this approach is we actually have the full distribution of the random variable $Q(Y)$ which is $$Q(X) = \frac{\sum_{i=1}^n \lambda_i X_i^2}{\sum_{i=1}^nX_i^2}$$ which is valid so long as the vector $X$ is chosen from a density symmetric under rotations. From there you can analyze the accuracy of the estimator.
Although @Bey was correct, we can state a more general result in a language natural to matrices.
Let us consider the Schatten $p$-norm with the $k$ largest singular value for $p = 1$ or $p = 2$. Then, after changing to an orthonormal basis of $A$, we may assume $A$ to be diagonal and the problem reduces to $\ell_p$ approximation of $k$ largest Eigen values of $A$ by a scalar $c$. In case of $p = 1$ the solution is the median of these $k$ Eigen values. In case of $p = 2$ the solution is the mean of these $k$ Eigen values.
The computation is only "easy" in the case stated by @Bey, that is $k=n$ and $p=2$. | 2021-09-19T01:23:42 | {
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https://math.stackexchange.com/questions/3517035/set-theory-and-relation | # Set theory and relation
Let $$\mathcal{C}$$ be a collection of subsets of $$[n]$$ with the property that if $$A, B \in \mathcal{C},$$ then $$A \cap B \neq \varnothing .$$ (For example, $$\mathcal{C}=\{\{1,2\},\{1,3\},\{2,3\},\{1,2,3\}\}$$ has this property.) What is the largest that $$\mathcal{C}$$ can be? Hint: You will have to prove both that the number you get is possible and that no larger number is possible.
• So what is the largest number you have managed to show is possible? Can you always do $2^{n-1}$ for example? What is your conjecture for the largest number possible? – almagest Jan 21 at 10:59
• @almagest I am not able to prove that $2^{n-1}$ is upper bound for cardinality? I do this for $n=4$. – maths student Jan 21 at 11:05
• Can you clarify what you mean by '[n]'? – Doug Spoonwood Jan 21 at 11:11
• n is number of element original set can have? Say for given example we have 3 elements {1,2,3} then we have total $2^3$ possible subset out of which $2^{3-1}$ satisfy given condition. – maths student Jan 21 at 11:13
Hint1:
If a collection $$\mathcal D\subseteq\wp([n])$$ contains more than $$2^{n-1}$$ elements then for some $$A\in\mathcal D$$ we have $$A^{\complement}\in\mathcal D$$.
So such a collection is not a proper candidate for $$\mathcal C$$.
Then finding a collection $$\mathcal C$$ (as described) that contains $$2^{n-1}$$ elements is enough to prove that such a collection has $$2^{n-1}$$ elements.
Hint2.
In the case where $$n=2k+1$$ is odd have a look at the subsets that have a cardinality $$\geq k+1$$.
Can two of them have disjoint intersection? And how many are there?
The power-set of $$[n]$$, denoted $$\wp(n)$$ has cardinality $$2^n$$, and for each $$k \in [n]$$, the subset of $$\wp(n)$$ $$\uparrow\{k\} = \{A\subseteq[n]:k\in A\}$$ is such that $$\mid\uparrow\{k\}\mid=2^{n-1}$$ (can you tell why?).
It is also the case that $$k \in A$$, for all $$A \in \uparrow\{k\}$$, and so can always have $$\mid\mathcal C\mid=2^{n-1}$$.
Now, suppose you have $$\mid\mathcal C\mid>2^{n-1}$$. Can you see that there are always $$A,B \in \mathcal C$$ with $$A \cap B = \varnothing$$?
Notice that there will be $$A \in \wp(n)$$ such that both $$A$$ and $$[n]\setminus A$$ belong to $$\mathcal C$$. | 2020-02-24T00:02:14 | {
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http://xdhomeworkybut.macrophytes.info/orthogonal-projection-matrix.html | # Orthogonal projection matrix
We will start off with a geometric motivation of what an orthogonal projection is and work our way through the and this matrix is a projection matrix that. I'm trying to obtain an orthogonal projection of a solid i can easily plot my solid using surf and what i need is basically what i see on my figure but in the form of a matrix. Why is a projection matrix of an orthogonal projection suppose that i have an orthogonal projection $\mathbb p$ with is associated projection matrix $\bf p. Projecting a vector to another vector unfortunalely norm replies a matrix norm when operating on a matrix a and b are orthogonal, such that the projection. Orthographic projection (sometimes orthogonal projection), is a means of representing three-dimensional objects in two dimensionsit is a form of parallel projection, in which all the projection lines are orthogonal to the projection plane, resulting in every plane of the scene appearing in affine transformation on the viewing. Projection matrices •rather than derive a different projection matrix for each type of projection, •oblique projection = shear + orthogonal projection. Template:views orthographic projection (or orthogonal projection) the inversion of the projection matrix, which can be used as the unprojection matrix is defined. Orthogonal projections and re ections projection matrix q maps a vector y 2rn to its recall that a square matrix p is said to be an orthogonal matrix if ptp. The orthogonal projection can be represented by a projection matrix to project a vector onto the unit vector a = (a x,. Definition of an orthogonal projection skew symmetric and orthogonal matrix what does it mean to say the hat matrix is an orthogonal projection. 19 orthogonal projections and orthogonal matrices 191 orthogonal projections we often want to decompose a given vector, for example, a force, into the sum of two. Express the action of this transformation with a matrix be nonzero why what is the right definition of the orthogonal projection of a vector onto the. And { w 1, w 2} is an orthogonal basis for s 2, the projection of v 3 onto s 2 is this gives therefore, if a is an orthogonal matrix, show that a −1 = a t. Orthogonal projection for example, the function which maps the point (x, y, z) in three-dimensional space r 3 to the point (x, y, 0) is a projection onto the x-y planethis function is represented by the matrix. Find the matrix of the projection of$\mathbb{r} finding the projection matrix assuming you mean the orthogonal projection onto the plane $w$ given by the. Orthogonal projection matrix calculator github orthogonal projection matrix calculator - linear algebra projection onto a subspace. Matrix of the orthogonal projection the minimization problem stated above arises in lot of applications so, it will be very helpful if the matrix of the orthogonal projection can be obtained under a given basis. A projection matrix is an square matrix that gives a vector space projection from to a subspace the columns of are the projections of the standard basis vectors, and is the image of a square matrix is a projection matrix iff a projection matrix is orthogonal iff.
Orthogonal projection definition, a two-dimensional graphic representation of an object in which the projecting lines are at right angles to the plane of the projection. Scratch a pixel has a really nice explanation of perspective and orthogonal projection matrices it inspired me to make a very simple / plain explanation of orthogonal projection matrices that hopefully will help them be. Projection matrices 21 deflnition called an orthogonal projection matrix (projector) furthermore, the vector px is called the orthogonal projection of x.
I've been looking around a bit and can't seem to find just what im looking for i've found canonical formulas, but what's the best way to use these do i have to scale every single vertex down. So the orthogonal projection of produce a matrix that describes the function's action show also that this map can be obtained by first rotating everything. We compute the standard matrix of the orthogonal projection in the same way as for any other transformation: by evaluating on the standard coordinate vectorsin this case, this means projecting the standard coordinate vectors onto the subspace.
Journal of statistics education, volume 18, number 1 (2010) 1 orthogonal projection in teaching regression and financial mathematics farida kachapova. An orthogonal projection is a projection for which the range u and the null space v are orthogonal subspaces which is an example of a projection matrix. This video is about orthogonal projections in linear algebra orthogonal projection operator in least another example of a projection matrix.
Orthogonal projection matrix
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2018. | 2018-10-15T19:23:21 | {
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http://math.stackexchange.com/questions/227914/are-there-values-k-and-ell-such-that-n-kd-ell-prove | # Are there values $k$ and $\ell$ such that $n$ = $kd$ + $\ell$? Prove.
Suppose that n $\in$ $\mathbb Z$ and d is an odd natural number, where $0 \notin\mathbb N$. Prove that $\exists$ $\mathcal k$ and $\ell$ such that $n =\mathcal kd +\ell$ and $\frac {-d}2 < \ell$ < $\frac d2$.
I know that this is related to Euclidean's Algorithm and that k and $\ell$ are unique. I do not understand where to start proving this (as I don't most problems like these), but I also have a few other questions.
Why is is that d is divided by 2 when it is an odd number? I'm not even sure how $\ell$ being greater than and less than these fractions has anything to do with the rest of the proof. Couldn't $\ell$ be any value greater than or less than $0$?
Since d can never equal $0$, then kd could never equal $0$, so doesn't that leave the only n to possibly equal $0$?
I would appreciate anyone pushing me in the correct direction.
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What's $k^*$ and $d^*$? – Patrick Li Nov 3 '12 at 1:18
I screwed up the formatting. It's just $kd$. – Christina Nov 3 '12 at 1:22
Hint: a set $\,S\,$ of $\,d\,$ consecutive integers contains every possible remainder modulo $d,\:$ hence $\, n\equiv \ell\pmod{d},\:$ for some $\,\ell\in S.\ \$ – Bill Dubuque Nov 3 '12 at 1:23
We give a quite formal, and unpleasantly lengthy, argument. Then in a remark we say what's really going on. Let $n$ be an integer. First note that there are integers $x$ such that $n-xd\ge 0$. This is obvious if $n\ge 0$. And if $n \lt 0$, we can for example use $x=-n+1$.
Let $S$ be the set of all non-negative integers of the shape $n-xd$. Then $S$ is, as we observed, non-empty. So there is a smallest non-negative integer in $S$. Call this number $r$. (The fact that any non-empty set of non-negative integers has a smallest element is a hugely important fact equivalent to the principle of mathematical induction. It is often called the Least Number Principle.)
Since $r\in S$, we have $r\ge 0$. Moreover, by the definition of $S$ there is an integer $y$ such that $r=n-yd$, or equivalently $n=yd+r$.
Note that $r\lt d$. For suppose to the contrary that $r \ge d$. Then $r-d\ge 0$. But $r-d=r-(y+1)d$, and therefore $r-d$ is an element of $S$, contradicting the fact that $r$ is the smallest element of $r$.
To sum up, we have shown that there is an $r$ such that $0\le r\lt d$ and such that there exists a $y$ such that $r=n-yd$, or equivalently $n=yd+r$.
Case (i): Suppose that $r\lt \dfrac{d}{2}$. Then let $k=y$ and $\ell=r$. We have then $n=kd+\ell$ and $0\le \ell\lt \dfrac{d}{2}$.
Case (ii): Suppose that $r \ge \frac{d}{2}$. Since $d$ is odd, we have $r\gt \dfrac{d}{2}$. We have $$\frac{d}{2}\lt r \lt d.$$ Subtract $d$ from both sides of these inequalities. We obtain $$-\dfrac{d}{2}\lt r-d\lt 0,$$ which shows that $$-\frac{d}{2}\lt n-yd-d\lt 0.$$ Finally, in this case let $k=y+1$ and $\ell=n-kd$. Then $n=kd+\ell$ and $$-\dfrac{d}{2}\lt kd+\ell\lt 0.$$
Remark: There is surprisingly little going on here. We first found the remainder $r$ when $n$ is divided by $d$. But the problem asks for a "remainder" which is not necessarily, like the usual remainder, between $0$ and $d-1$. We want to allow negative "remainders" that are as small in absolute value as possible. The idea is that if the ordinary remainder is between $0$ and $d/2$, we are happy with it, but if the ordinary remainder is between $d/2$ and $d-1$, we increase the "quotient" by $1$, thereby decreasing the remainder by $d$, and putting it in the right range. So for example if $n=68$ and $d=13$, we use $k=5$, and $\ell=3$. If $n=74$ and $d=13$, we have the usual $74=(5)(13)+9$. Increase the quotient to $6$. We get $74=(6)(13)+(-4)$, and use $k=6$, and $\ell=-4$.
We gave a proof in the traditional style, but the argument can be rewritten as an ordinary induction argument on $|n|$. It is a good idea to work separately with non-negative and negative integers $n$. We sketch the argument for non-negative $n$. The result is obvious for $n=0$, with $k_0=\ell_0=0$. Suppose that for a given non-negative $n$ we have $n=k_nd+\ell_n$, where $\ell_n$ obeys the inequalities of the problem, that is, $-d/2\lt \ell_n\lt d/2$. If $\ell_n\le (d-3)/2$, then $n+1=k_{n+1} +\ell_{n+1}$, where $k_{n+1}=k_n$ and $\ell_{n+1}=\ell_n+1$. If $\ell_n=(d-1)/2$, let $k_{n+1}=k_n+1$ and $\ell_{n+1}=-(d-1)/2$. It is not hard to verify that these values of $\ell_{n+1}$ are in the right range.
- | 2015-01-31T05:31:48 | {
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https://byjus.com/question-answer/if-t-n-denotes-the-n-th-term-of-the-series-2-3-6-11/ | Question
# If $$t_{n}$$ denotes the $$n$$th term of the series $$2+3+6+11+18+...$$then $$t_{50}$$ is
A
4921
B
492
C
502+1
D
492+2
Solution
## The correct option is C $$49^{2}+2$$In the given series, the difference between the successive terms are in A.P.Therefore, $$t_n$$ will be of the form $$an^2+bn+c$$Now, $$t_1=2=a+b+c$$ $$t_2=3=4a+2b+c$$ $$t_3=6=9a+3b+c$$Solving for $$a$$, $$b$$ and $$c$$We get $$a=1$$, $$b=-2$$ and $$c=3$$Therefore, $$t_n=n^2-2n+3=(n-1)^2+2$$Thus, $$t_{50} =49^2+2$$Ans: DMathematics
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View More | 2022-01-20T22:18:35 | {
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http://math.stackexchange.com/questions/235343/why-a-equiv-b-pmod-n-implies-a-and-b-have-equal-remainder-when-divided-by | # Why $a\equiv b\pmod n$ implies $a$ and $b$ have equal remainder when divided by $n$?
How does $a\equiv b\pmod n$ implies $a$ and $b$ have the same remainder when divided by $n$?
I don't understand the huge jump from modular equivalence to having equal remainders.
I see by definition $a\equiv b\pmod n\,$ implies $n\mid (a-b)\,$ so $\,a-b=nk\,$ so $\,a=b+nk$.
But I do not see how this implies that $a$ and $b$ have the same remainder when divided by $n$.
-
+1 Nice question. – amWhy Nov 12 '12 at 1:26
Thank you! But it just showcases my inability to understand this concept haha. :( – Yellow Skies Nov 12 '12 at 1:34
We prove $\ n\mid a-b \iff a\,$ and $\,b\,$ leave the same remainder when divided by $\,n$.
If $a$ and $b$ have the same remainder $r$ when divided by $n$, then $a = q_1n + r$ and $b = q_2n + r$. Subtracting yields $\,a - b = q_1n + r - (q_2n + r) = (q_1 - q_2)n$, which means $n \mid (a-b)$.
Conversely suppose $n\mid (a-b).$ Then $\,a = b+nk\,$ for some integer $\,k.\,$ By the division algorithm, $b = qn + r\,$ for some integers $q$ and $0 \leq r < n$. Putting this value of $b$ into above $a = b + nk = (qn + r) + nk = (q + k)n + r$, so $r$ is also the remainder of $a$ divided by $n$.
-
I edited the answer a little. Of course you can revert it or modify it if you prefer. – Bill Dubuque Jul 11 at 20:26
Proposition: Let $a, b, m$ be integers with $m \geq 2$. Then, $$a \equiv b \pmod{m} \qquad \text{if and only if} \qquad a = mk + b$$ for some $k \in \mathbb{Z}$.
With the proposition in tow, consider the remainder of $a - b$ when divided by $m$.
-
$\,\bar a\, =\, a\bmod n\, =\, a-pn\$ and
$\,\bar b\, =\, b \bmod n\, =\, b-qn\$ for some integers $\,p,q$.
Therefore if $\ n\mid a-b\$ then $\,n\,$ divides $\,\bar a-\bar b\, =\, a-b + (q-p)n$
By definition $\ 0 \le \bar a,\bar b < n\,$
therefore $\ 0 \le |\bar a -\bar b| < n$
so we conclude $\ \bar a - \bar b = 0,\,$ being divisible by $\,n.$
-
How did you get that $a \mod n=a-pn$? I don't recall that at all.. Pardon me. – Yellow Skies Nov 12 '12 at 1:37
The definition of a mod n is the minimum of the intersection of the sets {a-pn|p is an integer} and the set of natural numbers – Amr Nov 12 '12 at 1:44
Therefore a mod n belongs to the set {a-pn|p is an integer} . Thus there exists an integer p such that a mod n=a−pn – Amr Nov 12 '12 at 1:45
I see! Thankszz – Yellow Skies Nov 12 '12 at 1:52
@Amr I edited the answer a little. Of course you can revert it or modify it if you prefer. – Bill Dubuque Jul 11 at 20:14 | 2016-07-26T21:49:42 | {
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https://math.stackexchange.com/questions/1620577/is-there-an-irrational-number-arbitrarily-close-to-another-irrational-number | # Is there an irrational number arbitrarily close to another irrational number?
I know that there is a rational number arbitrarily close to an irrational, due to the density of real number. But what about an irrational number? Thanks!
Yes, consider $\alpha + \frac{1}{n}$ where $\alpha$ is irrational and $n$ is an integer. $\alpha + \frac{1}{n}$ is also irrational and can be made arbitrarily close to $\alpha$ by choosing $n$ to be sufficiently large.
• Thank you. Stupid question but can one define something like the number closest to an irrational? For example, the question "which number is closest to a given irrational number is rational or irrational?" makes no sense right? just making sure. – ANANDA PADHMANABHAN S Jan 21 '16 at 5:25
• Irrationals can differ in how well they can be approximated by rationals, check this out: mathworld.wolfram.com/IrrationalityMeasure.html – Dan Brumleve Jan 21 '16 at 5:29 | 2019-10-22T11:33:37 | {
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https://math.stackexchange.com/questions/1725754/what-is-the-new-probability-density-function-by-generating-a-random-number-by-ta/1725760 | # What is the new probability density function by generating a random number by taking the reciprocal of a uniformly random number between 0 and 1?
I have a random number generator which can generate a random number between $0$ and $1$.
I attempt to generate a random number between 1 and infinity, by using that random number generator, but taking the reciprocal of that result.
Is the new generator uniform? Certainly not. Then what is the probability density function of the new generator?
Let the old probability density function be $f_1(x)$, and the new one be $f_2(x)$.
We have:$$\int_1^af_2(x)\mathrm dx=\int_\frac1a^1f_1(x)\mathrm dx$$where $a>1$.
We also know that $f_1(x)$ is uniform, and spans from $0$ to $1$. Therefore, $f_1(x)=1$ in that interval.
Therefore:$$\int_1^af_2(x)\mathrm dx=\int_\frac1a^1\mathrm dx=1-\frac1a$$
Differentiating both sides with respect to $a$:$$\frac{\mathrm d}{\mathrm da}\int_1^af_2(x)\mathrm dx=\frac{\mathrm d}{\mathrm da}\left(1-\frac1a\right)$$
Simplifying both sides:$$f_2(a)=\frac1{a^2}$$ $\blacksquare$
• Yours too! +1 (I found the same as you, so it was reassuring that my result is ok) – Jimmy R. Apr 3 '16 at 10:02
• I couldn't see how your first identity must be true until I noticed a theorem regarding integrals of inverse functions $\int_{c}^{d} f^{-1}(y)dy +\int_{a}^{b}f(x)dx = bd-ac$. Is this what you had in mind? – Robert Smith Apr 3 '16 at 18:46
• @RobertSmith I think what you're missing is that $X_2 \leq a$ if and only if $X_1 \geq a$ and so $\Pr(X_2 \leq a) = \Pr(X_1 \geq a)$, which gives the desired integrals. – Silverfish Apr 3 '16 at 23:25
• @RobertSmith Those are the areas which correspond to probabilities which must be same. – Kenny Lau Apr 4 '16 at 3:08
Denote with $U$ the random number you generate, then $U\sim U(0,1)$ and you want to determine the distribution of $Y=\frac{1}U$. So, for $1\le y<+\infty$, you have that $$F_{Y}(y)=P(Y\le y)=P(1/U\le y)=P(U\ge 1/y)=1-P(U<1/y)=1-F_U(1/y)$$ Hence $$f_Y(y)=\frac{d}{dy}F_Y(y)=\frac{d}{dy}\left(1-F_U(1/y)\right)=\frac1{y^2}f_U(1/y)=\frac1{y^2}\cdot1=\frac1{y^2}\mathbf{1_{y\ge1}}$$
• Nice solution!! – Kenny Lau Apr 3 '16 at 10:00 | 2019-10-21T07:22:58 | {
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https://practicepaper.in/gate-cse/digital-logic | # Digital Logic
Question 1
Consider three floating point numbers A, B and C stored in registers $R_A$, $R_B$ and $R_C$, respectively as per IEEE-754 single precision floating point format. The 32-bit content stored in these registers (in hexadecimal form) are as follows.
$R_A=0xC1400000$
$R_B=0x42100000$
$R_C=0x41400000$
Which one of the following is FALSE?
A $A+C=0$ B $C=A+B$ C $B=3C$ D $(B-C) \gt 0$
GATE CSE 2022 Number System
Question 1 Explanation:
Question 2
Let R1 and R2 be two 4-bit registers that store numbers in 2's complement form. For the operation R1+R2, which one of the following values of R1 and R2 gives an arithmetic overflow?
A R1 = 1011 and R2 = 1110 B R1 = 1100 and R2 = 1010 C R1 = 0011 and R2 = 0100 D R1 = 1001 and R2 = 1111
GATE CSE 2022 Number System
Question 2 Explanation:
Question 3
Consider a Boolean function $f(w,x,y,z)$ such that
$\begin{array}{lll} f(w,0,0,z) & = & 1 \\ f(1,x,1,z) & =& x+z \\ f(w,1,y,z) & = & wz +y \end{array}$
The number of literals in the minimal sum-of-products expression of $f$ is ________
A 4 B 6 C 8 D 9
GATE CSE 2021 SET-2 Boolean Algebra
Question 3 Explanation:
Question 4
If the numerical value of a 2-byte unsigned integer on a little endian computer is 255 more than that on a big endian computer, which of the following choices represent(s) the unsigned integer on a little endian computer?
[MSQ]
A 0x6665 B 0x0001 C 0x4243 D 0x0100
GATE CSE 2021 SET-2 Number System
Question 4 Explanation:
Question 5
If $x$ and $y$ are two decimal digits and $(0.1101)_2 = (0.8xy5)_{10}$, the decimal value of $x+y$ is ___________
A 3 B 6 C 8 D 4
GATE CSE 2021 SET-2 Number System
Question 5 Explanation:
Question 6
Which one of the following circuits implements the Boolean function given below?
$f(x,y,z) = m_0+m_1+m_3+m_4+m_5+m_6$
where $m_i$ is the $i^{th}$ minterm.
A A B B C C D D
GATE CSE 2021 SET-2 Combinational Circuit
Question 6 Explanation:
Question 7
The format of the single-precision floating point representation of a real number as per the IEEE 754 standard is as follows:
$\begin{array}{|c|c|c|} \hline \text{sign} & \text{exponent} & \text{mantissa} \\ \hline \end{array}$
Which one of the following choices is correct with respect to the smallest normalized positive number represented using the standard?
A exponent = 00000000 and mantissa = 0000000000000000000000000 B exponent = 00000000 and mantissa = 0000000000000000000000001 C exponent = 00000001 and mantissa = 0000000000000000000000000 D exponent = 00000001 and mantissa = 0000000000000000000000001
GATE CSE 2021 SET-2 Number System
Question 7 Explanation:
Question 8
Consider the following Boolean expression.
$F=(X+Y+Z)(\overline X +Y)(\overline Y +Z)$
Which of the following Boolean expressions is/are equivalent to $\overline F$ (complement of $F$)?
[MSQ]
A $(\overline X +\overline Y +\overline Z)(X+\overline Y)(Y+\overline Z)$ B $X\overline Y + \overline Z$ C $(X+\overline Z)(\overline Y +\overline Z)$ D $X\overline Y +Y\overline Z + \bar X \bar Y \bar Z$
GATE CSE 2021 SET-1 Boolean Algebra
Question 8 Explanation:
Question 9
Consider a 3-bit counter, designed using T flip-flops, as shown below:
Assuming the initial state of the counter given by PQR as 000, what are the next three states?
A 011,101,000 B 001,010,111 C 011,101,111 D 001,010,000
GATE CSE 2021 SET-1 Sequential Circuit
Question 9 Explanation:
Question 10
Consider the following representation of a number in IEEE 754 single-precision floating point format with a bias of 127.
S:1
E:10000001
F:11110000000000000000000
Here, S,E and F denote the sign, exponent, and fraction components of the floating point representation.
The decimal value corresponding to the above representation (rounded to 2 decimal places) is ____________.
A -7.75 B 7.75 C -3.825 D 3.825
GATE CSE 2021 SET-1 Number System
Question 10 Explanation:
There are 10 questions to complete.
### 12 thoughts on “Digital Logic”
1. Sir, Please mention the a,b,c value of question 2.
• Hi Ashutosh,
Here, value of a,b,c are not required. You can write the truth table based on above circuit. Then find the answer based on truth table. See the solution provided.
2. Question 5, Please update the answer from 4 minimum gate to 3 minimum gate.
• Hi Ashutosh,
Thank you for your suggestion. We have updated the answer from 4 to 3.
3. Q13 answer will be B…
Update it
• Thank You akki yadav, We have updated the answer suggested by you.
4. Update Question No. 46. The options should be Q1Q0 instead of Q0Q1
5. Update Q56 as B instead of C
• Thank You Sowmya Sri, | 2022-10-03T08:59:09 | {
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http://math.stackexchange.com/questions/8476/rearrange-the-formula-for-the-sum-of-a-geometric-series-to-find-the-value-of-its | # Rearrange the formula for the sum of a geometric series to find the value of its common ratio?
What I'm attempting to do is to rearrange the formula for the sum of a geometric series so as to find the value of its common ratio $r$. I've tried several different methods, all of which have failed; though I can't understand why.
This was my last attempt using logarithms:
$$S_n=\frac{a(1-r^n)}{1-r}$$ $$S_n(1-r) = a(1-r^n)$$ $$S_n-S_nr = a - ar^n$$ $$\frac{S_n}{a}-1 = \frac{S_n}{a}r-r^n$$ $$\log{\frac{S_n}{a}} = \log{\frac{S_n}{a}}+\log{r}-n\log{r}$$ $$0 = (\log{r})(n-1)$$ $$r = 10^{\frac{0}{n-1}}$$ $$r=1$$
I can't understand where I'm going wrong. Any advice would be great. Also, if you know a formula to find $r$ using $S_n$ and $a$, then that would be equally fantastic.
Thanks.
-
There is a mistake in the application of log to $(\dfrac{S_n}{a}-1).$ – Américo Tavares Oct 31 '10 at 23:04
Do you want the sum of the (infinite) geometric series, or do you want the $n$th partial sum of the series? Your question implies the former, but the formula you give for $S_n$ implies the latter. – Mike Spivey Nov 1 '10 at 2:17
HINT $\ \$ Consider $\ S_{n+1} - r\ S_n$
$$S_{n+1} - rS_n = a$$ (by def. of geometric series) | 2015-07-04T07:39:56 | {
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https://electronics.stackexchange.com/questions/88058/learning-loops-kirchhoffs-laws | # Learning loops, Kirchhoff's Laws
I'm trying to teach myself electronics from an old textbook, "Fundamentals of Electric Circuits", $4$th Edition, by Alexander and Sadiku. On page $41$, I can't figure out how they did example $2.6.$ See this diagram:
So I'm applying KVL around the loop. I get:$$-12 + 4i + 2v_0 - 4 - 6i = 0$$
But the book says it's ${+6i}$. The current is flowing to the negative pole of the resistor, so it should be negative, right? Is the book wrong, or am I misunderstanding something?
Edit:
The book text for the problem reads:
Determine $v_0$ and $i$ in the circuit.
Solution:
We apply KVL around the loop as shown in the figure. The result is:$-12 + 4i + 2v_0 - 4 + 6i = 0$
Applying Ohm's Law to the 6 Ohm resistor gives:
$v_0 = -6i$
Substituting the previous equation into the first one yields: $i = -8 A$ and $v_0 = 48 V$
The way the book described to do this was to follow the current, and the sign of each voltage element is determined by the polarity. So $-12 V, 4i$, and so on. But continuing that pattern, we get to the $-$ pole of the $6$ Ohms resister before we get to the positive pole, so $v_0$ equals $-6i V$.
I'm not understanding something.
• What value do you need to find? "But the book says it's +6i." Which value should be +6i, and in which units? Is it Vo that is sought? [I don't have this book, as you can tell.] – Nick Alexeev Nov 9 '13 at 1:42
• It's not clear what it is in the book that you disagree with. Is it an equation in the book? Is it possible that the book substituted -6i for v0? – Joe Hass Nov 9 '13 at 1:49
• There has to be a constant too , with an i term(not 6i), as 12+4!=0. – Sherby Nov 9 '13 at 1:50
• I updated the question. – Amy Nov 9 '13 at 5:43
I think you may have an issue with one of the signs. Going around the loop, I get:
12 - 4 i - 2 Vo + 4 - 6 i = 0
Looks like you got the sign on the 6i term flipped.
So, if we take Vo = -6 i, you can calculate:
12 - 4 i + 12 i + 4 - 6 i = 0
16 + 2 i = 0
i = -8
Vo = 48
If you're stuck on understanding how to write this equation, the way to remember it is 'what goes up must come down'. Sum up the voltage across each element as you go around the loop. Since you go into the - side of the 12 volt source, you gain 12 volts. Then you pass through a resistor, and you lose 4 i volts. Then you go the wrong way through the dependent source and you lose 2 Vo volts. Then you go through the 4v source and gain 4 volts. Then you go through the final resistor and lose 6 i volts. You're back where you started, so all of that adds up to zero. Then you figure out what Vo is - if the current is flowing in the direction of the arrow, it will create a voltage drop in opposition to how Vo is marked, so Vo = -6 i. After that, finding i is just algebra.
• I really like the gain/lose way of looking at this. This makes a lot of sense now. Thank you! – Amy Nov 9 '13 at 6:17
• Glad to hear it! – alex.forencich Nov 10 '13 at 1:02
disclaimer: I don't have the book, so I may be missing something. Here goes nothing.
First thing to note about this circuit is that $V_7$ (the rhombus with $2V_o$) is a dependent voltage source. It depends on $V_o$, the voltage across the 6Ω resistor. [I'll get to the signs soon.] The voltage across the $V_7$ is inverted with respect to voltage across 6Ω. That hints that the circuit has negative feedback.
The circuit doesn't have a pre-determined ground. I can choose any node as ground. For the sake of convenience, I choose this place as ground (bottom right corner).
Starting at that place and going clockwise. KVL.
$V_o + 12 -4i - 2Vo + 4=0$
For now, I'm using [like a dummy] the same sign for $V_o$ as shown in the drawing. Also, from Ohm's law
$V_o = -6i$
plug in
$-6i + 12 -4i +12i + 4=0$
$2i + 16 = 0$
$i = -8$
$V_o = -6i = 48$
Now we know everything that's going on in that circuit.
edit: In response to more text from the book.
Compared to the first equations in Alex's and mine posts, the book solution has inverted signs. In the book, both resistor drops are have a positive sign, while 12V and 4V voltage sources have a negative sign. But, this sign inversion is consistent. That's why the answer is the same.
In the beginning of the problem, we don't know that sign of the current will end up being negative. We don't know that the current will actually flow against the arrow in the drawing. But, the book authors knew that the current will be negative. May be, that's why they had inverted the sign. May be, he wanted to show that the choice of the sign is more a less arbitrary.
But, the sign choice in the book seems counter-intuitive to me too.
• How did you format your equations? – Amy Nov 9 '13 at 5:45
• @Amy LaTeX is supported here. $V_7$ produces $V_7$ – Nick Alexeev Nov 9 '13 at 5:46
• @Amy: See $\LaTeX$/MathJax sandbox, the RightClick $\Rightarrow$Show Math as$\Rightarrow$Tex Commands is useful to see how people wrote the equations. – RedGrittyBrick Nov 9 '13 at 12:15 | 2021-03-03T00:21:25 | {
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https://math.stackexchange.com/questions/1320683/conflicting-limit-answers-using-calculator-and-wolfram-alpha | # Conflicting limit answers using calculator and wolfram alpha
I want to evaluate $\lim\limits_{x\to0} \dfrac{\tan(x) - \sin(x)}{(\sin(x))^3}$,
Calculator says it's 0 when substituted with 0.0000000001.
Wolfram Alpha says it's 1/2.
The Problem Set says the answer is 1/2.
I think I believe Wolfram Alpha more but I've been using the calculator method so I can answer stuff really fast (because it's for a board exam, shouldn't spend too much time deriving) is there a way for me to know?
• If you want to do it by hand you can save some time by rewriting your expression as $\lim_{x\to 0} \frac{\sec x - 1}{\sin^2x}$. – Rolf Hoyer Jun 11 '15 at 0:29
• Your calculator might have some strange way it evaluates operations. I would add more parenthesis to try and get the calculator to do what you want it to do rather than what it thinks you wants to see if it gets a different answer. – Rory Grice Jun 11 '15 at 0:34
• What did you DO with your calculator to get that? The numerator and denominator are both approaching $0$; did you divide $0$ by $0$? If your calculator thinks $0/0=0$, don't believe it. ${}\qquad{}$ – Michael Hardy Jun 11 '15 at 0:54
• @Omnomnomnom was right; 0.0000000001 is too small of a number. 0/0 is math error. using casio fx-991es plus – james Jun 11 '15 at 1:05
• @MichaelHardy the calculator probably knows that $\sin(0.0000000001) \neq 0$. However, it can't calculate $(0.5 + \epsilon) - 0.5$ accurately since it calculates $0.5 + \epsilon = 0.5$ – Omnomnomnom Jun 11 '15 at 1:24
$0.0000000001$ is too small of a number: the calculator got such a small answer for the top that it assumed it was zero (since the values subtracted in the numerator were rounded to the same value). The bottom was non-zero, so there was no division by zero error. Zero divided by anything non-zero is zero.
If you're going to use the calculator method, I would try with a bigger number. I think $10^{-5} = 0.00001$ should be small enough to give you a good answer without causing you to run into this situation.
• Up vote this nice answer. But personally i don't think calculator method can "give you a good answer" anyway. For example, a calculator would almost definitely lead you astray if you would try to find the "value" of $\zeta(1)$ with that. So I suggest you say "help you make a good guess" instead. – Vim Jun 11 '15 at 4:32
• Even 0.001 will give you a nice answer. Checkout this. – ps95 Jun 11 '15 at 5:43
• @Vim I think the calculator method can probably give you a "good enough" answer on an exam like OP's board exam, where time is apparently an issue. – Omnomnomnom Jun 11 '15 at 9:52
• I don't think this answer is quite correct. A calculator that uses floating point numbers is unlikely to round a very small number to zero. My guess is that $\sin 0.0000000001$ and $\tan 0.0000000001$ are so close together that the calculator rounded the two numbers to exactly the same number, meaning that it subtracted two identical numbers, getting zero. – Tanner Swett Jun 11 '15 at 20:45
• @TannerSwett that's what I meant by what I said; maybe I didn't word it clearly. – Omnomnomnom Jun 11 '15 at 20:55
As a rule of thumb, try express everything in term of either $\sin(x)$ or $\cos(x)$ to see whether there is any obvious cancellation. For this case, we have $$\frac{\tan(x) - \sin(x)}{\sin(x)^3} = \frac{\frac{\sin(x)}{\cos(x)} - \sin(x)}{\sin (x)^3} = \frac{1-\cos(x)}{\cos(x)(1-\cos(x)^2)} = \frac{1}{\cos(x)(1+\cos(x))}$$ you don't need any calculator to know the limit is $\frac12$.
• You beat me to it! Cleanest approach. – ps95 Jun 11 '15 at 5:34
One way to do it by hand is to use Taylor Series. For $x\to 0$, $\tan x = x+\frac 13x^3+o(x^3), \sin x = x - \frac 16x^3+o(x^3)$ So $$\frac {\tan x - \sin x}{\sin^3 x}= \frac {\frac12x^3+o(x^3)}{x^3}=\frac12+o(1)\to\frac12$$
• A small nitpick: the symbol $\approx$ makes the expansion look sketchy and somewhat handwavy, which may confuse the reader or OP; while this is fully well-defined and a valid and rigorous statement ($\tan x = x + \frac{1}{3}x^3 + o(x^3)$). – Clement C. Jun 11 '15 at 0:44
• @ClementC. +1. I'm always trying to use $\approx$ as little as possible because it looks flippant. – Vim Jun 11 '15 at 4:26
• I'd say it's rather silly to say $\approx$ looks handwavy, it's so utterly clear from the context that it's shorthand for $\mathcal{O}$(whatever the degree of the next term would have been). Rigorous math doesn't equal notation-anal math... – Benjamin Lindqvist Jun 12 '15 at 16:55
The problem is not just that $\tan x - \sin x$ is approaching zero rapidly; the real problem is that as $x$ approaches zero, $\tan x - \sin x$ approaches zero much more rapidly than either $\tan x$ or $\sin x$, because (as shown by Ross Millikan) $\tan x - \sin x \approx \frac12 x^3$ but $\tan x \approx x + \frac13 x^3$ and $\sin x \approx x - \frac16 x^3.$ At some point, for very small $x$, $x^3$ is so much smaller than $x$ that $x + \frac13 x^3$ and $x - \frac16 x^3$ round to the same number inside the calculator, with the result that $\tan x - \sin x$ is evaluated to $0$ exactly. This is an extreme example of cancellation error, a well-known bugaboo of numeric computing methods.
For example, Google says (tan(0.0000001)-sin(0.0000001))/(sin(0.0000001))^3 is $0.5029258124$ but (tan(0.00000001)-sin(0.00000001))/(sin(0.00000001))^3 is $0.$ Trying various other values of $x$ such as $0.001,$ $0.0001,$ $0.00001,$ and $0.000001$ shows that the value calculated by Google actually starts to diverge away from $\frac12$ (presumably due to cancellation error) for input much smaller than $x=0.0001$.
If you can use the standard limits \begin{equation*} \lim_{x\rightarrow 0}\frac{\tan x-x}{x^{3}}=\frac{1}{3},\ \text{and}\ \lim_{x\rightarrow 0}\frac{x-\sin x}{x^{3}}=\frac{1}{6},\ \ \ \ \text{and}\ \ \ \ \ \lim_{x\rightarrow 0}\frac{x}{\sin x}=1 \end{equation*} then, \begin{eqnarray*} \lim_{x\rightarrow 0}\frac{\tan x-\sin x}{\sin ^{3}x} &=&\lim_{x\rightarrow 0}\frac{\tan x-\sin x}{x^{3}}\cdot \left( \frac{x}{\sin x}\right) ^{3}= \\ &=&\lim_{x\rightarrow 0}\left( \frac{\tan x-x}{x^{3}}+\frac{x-\sin x}{x^{3}}% \right) \cdot \left( \frac{x}{\sin x}\right) ^{3} \\ &=&\left( \frac{1}{3}+\frac{1}{6}\right) \cdot \left( 1\right) ^{3} \\ &=&\frac{1}{2}. \end{eqnarray*}
tan(x) and sin(x) are equal numbers in a calculator for small enough x because the calculator does not represent enough digits. Hence the numerator turns to 0.
In the Windows calculator, your example works fine, but: sin(0.0000000000000001) = tan(0.0000000000000001) = 1.7453292519943295769236907684886e-18
The difference is in the mantissa (the first part), which cannot properly represent such tiny sines and tangents. It needs more digits because that's where the difference between sine and tangent show up with such tiny x. This would be important in any formula that relied on those functions with such tiny values -- the output of the formula would be junk. After all, if sin(x) = tan(x) for every tiny number, well, now what?
Note that it will happily cube the sin, though. 1.7453292519943295769236907684886e-18 cubed is
5.3165769342077880959321527892834e-54
And 0 divided by that is still 0 (the original problem result). Note sin cubed of a junk sin value (also equal to tan) is also meaningless. It's the mantissa (the first part) not the exponent (the second part) that has strayed into meaninglessness.
The problem is rounding and precision. Assuming that we are using radians $$\sin(0.0000000001)\doteq0.0000000000999999999999999999998333333333$$ and $$\tan(0.0000000001)\doteq0.0000000001000000000000000000003333333333$$ both of which, when rounded to $20$ significant places are $$0.000000000100000000000000000000$$ Thus, assuming the calculator has no more than $20$ significant digits, the difference computed by the calculator would be zero.
Since $$\sin(x)=x-\frac16x^3+\frac1{120}x^5+O(x^7)$$ and $$\tan(x)=x+\frac13x^3+\frac2{15}x^5+O(x^7)$$ we get $$\frac{\tan(x)-\sin(x)}{\sin^3(x)}=\frac{\frac12x^3+\frac18x^5+O(x^7)}{x^3-\frac12x^5+\frac{13}{120}x^7+O(x^9)}=\frac12+\frac38x^2+O(x^4)$$ Therefore, if the calculator had enough precision, it would have gotten $$\frac{\tan(x)-\sin(x)}{\sin^3(x)}\doteq0.500000000000000000003750000000$$ for $x=0.0000000001$. | 2019-07-23T00:28:00 | {
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https://math.stackexchange.com/questions/3416674/proving-that-2n-1n-over-3-is-an-odd-number-for-all-n-ge1 | # Proving that ${ {2^n - (-1)^n} \over {3} }$ is an odd number for all $n\ge1$
This is a problem from Mathematics Analyses and Approaches HL (IB). Do note that this is not homework or any sort of submission, I'm doing it merely out of interest. I need to prove the following conjecture using the principle of mathematical induction:
$${{2^n-(-1)^n}\over {3}} \; \text{is an odd number for all} \; n \in Z^+$$
And here is my proof:
$$\text{If} \; n=1, \; {{2^1-(-1)^1}\over {3}} = 1, \; \therefore P_1 \; \text{is true}$$
$$\text{If} \; P_k \; \text{is true}, \; {{2^k-(-1)^k}\over {3}} = 2A+1 \; \text{where A} \in Z$$
$$\text{And now,} \; {{2^{k+1}-(-1)^{k+1}}\over {3}} \implies {2({2^k)+(-1)^k}\over {3}}$$
$$\text{from} \; P_k, \; 2^k = 6A + 3+(-1)^k$$
$$\implies {2({6A + 3+(-1)^k)+(-1)^k}\over {3}}$$
$$\implies {{12A+6+3(-1)^k}\over {3}}$$
$$\implies 4A+2+(-1)^k$$
$$\implies 2(2A+1)+(-1)^k$$
Now, my reasoning here is that two times any integer always gives an even number. We know that $$2A+1$$ is an integer, so $$2(2A+1)$$ has to be even. Now, any subtracting 1 from or adding 1 to any even number gives an odd number. As $$2(2A+1)$$ is even, $$2(2A+1)+(-1)^k$$ has to be odd.
Is this proof correct? Anything I should do differently or elaborate on?
• looks fine to me – eyeballfrog Oct 31 '19 at 16:30
• I agree. The proof is overall well set and developed. Make sure you write an appropriate conclusion aligned with the inductive reasoning applied. – Fede1 Oct 31 '19 at 16:37
• Yep just skipped that cause I'm too lazy – Mehul Jangir Oct 31 '19 at 16:38
• Welcome to Mathematics Stack Exchange. You could prove by induction that $2$ divides $2^n$. Then $2$ divides ${{2^n-(-1)^n}\over {3}}$ would imply $2$ divides $2^n-(-1)^n$, which would imply $2$ divides $(-1)^n,$ a contradiction – J. W. Tanner Oct 31 '19 at 16:40
• It's correct but there's just a minor detail: in the second line where you say $[\ldots ]= 2A+1 \text{ where } A\in\mathbb{Z}^+$, it should be $A\in \mathbb{Z}_{\geq 0}$ because $2A+1$ could be one, i.e, $A$ could be $0$. – bjorn93 Oct 31 '19 at 16:45
Yes it is absolutely fine, as an alternative by exhaustion we have for $$n=2k$$
$${{2^n-(-1)^n}\over {3}}={{2^{2k}-1}\over {3}}\implies \frac{2^{2k}-1}{3}+1=\frac{2^{2k}+2}{3}=2\frac{2^{2k-1}+1}{3}$$
and for $$n=2k+1$$
$${{2^n-(-1)^n}\over {3}}={{2^{2k+1}+1}\over {3}}\implies \frac{2^{2k+1}+1}{3}+1=\frac{2^{2k}+4}{3}=2\frac{2^{2k}+1}{3}$$
Equivalently, we can prove that $$2^n-(-1)^n=3(2m+1)\equiv3\mod6$$.
Indeed, $$2^n\bmod6=2,4,2,4,2,\cdots$$ to which you add or subtract $$1$$.
Your proof's fine. Interestingly, we don't need induction at all. If $$n\ge1$$,$$\frac{\frac{2^n-(-1)^n}{3}-1}{2}=\frac{2^n-(-1)^n-3}{6}.$$The numerator is both even (though not if $$n=0$$) and a multiple of $$3$$ (since $$3|2-(-1)$$), so is a multiple of $$6$$, and so the expression is an integer. (OK, I lied a little: the proof that $$m|a-b\implies m|a^n-b^n$$ uses induction.) This proves $$\frac{2^n-(-1)^n}{3}$$ is odd.
I find your proof really good and simple, while my proof is pretty rough - This is what I got so far:
Using this formula (click here)(under difference of odd exponents), you can factorise the top to get -
$$((2-(-1))(2^{n-1} +2^{n-2}(-1) +2^{n-3}(-1)^2+2^{n-4}(-1)^3 ...+2^0(-1)^{n-1}))/3$$
The first bracket will evaluate to 3, which divided by 3 is 1 (cancelling the 3), while the the other bracket is the sum of even powers of 2 (for k being and integer, which it is).
The final term of that bracket will be either a 1 or -1 and as the rest is even, the whole bracket will be odd (of the form 2a +- 1)
I think this proof is less definitive but simpler to understand (kinda). Tell me what you think of it. | 2020-07-04T15:36:14 | {
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http://math.stackexchange.com/questions/47100/mathbbz-2-mathbbz-coefficients-in-homology | # $\mathbb{Z}/2\mathbb{Z}$ coefficients in homology
I don't see the point in using homology and cohomology with coefficients in the field $\mathbb{Z}/2\mathbb{Z}$.
Can you provide some examples for why this is useful?
-
One quick reason: a manifold is always Z/2Z orientable. This allows results like Poincare duality to work. – Soarer Jun 23 '11 at 8:21
Thanks! Where can I find a reference for this? I tried in Bott-Tu (Differential Forms etc.) and Irapidly noticed that the third book of Fomenko "Modern Geometry" spends lots of words about Z/2 cohomology, but I'm lacking a precise place. – tetrapharmakon Jun 23 '11 at 8:29
Have you checked in Massey's Algebraic Topology book, or Bredon's Geometry and Topology book? – user641 Jun 23 '11 at 8:36
Another (more technical) thing is that it's easier to work in char 2, because you can ignore all signs — which makes many computations much, much easier (say, try to compute homology — or just Euler char — of a real Grassmannian...). – Grigory M Jun 23 '11 at 8:48
I asked a similar question - why do we use different coefficient's at all - and there are some interesting responses here: math.stackexchange.com/questions/37148/… Have a look at Section 3.3 of Hatcher's Book for a nice discussion of Poincare Duality and $\mathbb{Z}/2\mathbb{Z}$ coefficients. – Juan S Jun 23 '11 at 12:45
1. A lot of times it's easier to work with $\mathbb Z_2$ coefficients. You don't have to worry about sign and calculations are easier.
2. Field coefficients are nice because there is a precise duality between homology and cohomology, without worrying about the universal coefficient theorem.
3. Knowledge of $\mathbb Z_2$ coefficients can help you say things about integral homology. For example, if $H_i(X;\mathbb Z_p)=0$ for all primes $p$, then $H_i(X;\mathbb Z)=0$.
4. Some spaces have a nicer presentation with $\mathbb Z_2$ coefficients. The cohomology ring $H^*(\mathbb{RP}^n;\mathbb Z_2)$ is isomorphic to $\mathbb Z_2[x]/x^{n+1}$, whereas the integral cohomology ring is messier to write down.
-
Here is another one to add to the list: I am currently reading about cohomology operations, in particular the Steenrod squares.
These are functions
$$\operatorname{Sq^i}:H^k(X,\mathbb{Z}/2\mathbb{Z}) \to H^{k+i}(X,\mathbb{Z}/2\mathbb{Z})$$
which satisfies a whole bunch of nice properties (or axioms depending on how one looks at things)
Adams used these cohomology operations to solve the vector fields on spheres problem.
See the very excellent book by Mosher & Tangora's "Cohomology Operations and Applications in Hommotopy Theory" for the details
- | 2015-11-27T21:18:19 | {
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https://www.physicsforums.com/threads/how-to-find-the-velocity-of-a-wave-in-simple-harmonic-motion-given-time.972089/ | # How to find the velocity of a wave in simple harmonic motion given time
#### MattDutra123
Problem Statement
Find the speed of this longitudinal wave at t=0.12s.
Relevant Equations
v=w*xmax*cos(wt)
The graph provided is below. The problem asks for the speed of the wave at 0.12s. I used the formula v=w*xmax*cos(wt), provided in our textbook where xmax is the amplitude of 2 cm, w (omega) is 2pi divided by the period of 0.2. However, for some reason this formula doesn't give me the correct result. Instead, the solution to the problem involves the formula w*sqrt(xmax^2-x^2), which requires us to first find the displacement at t = 0.12.
My question is: why does the formula I used not work? Why must we use the other formula to solve this problem? I apologise for the bad formatting of the formulas.
Related Introductory Physics Homework News on Phys.org
#### kuruman
Homework Helper
Gold Member
It doesn't work because it's the wrong formula to use. It predicts that the velocity is maximum at t = 0. If you look at the graph, the position is maximum at t = 0 which means that the speed is zero at t = 0.
Also, is this a wave or a simple harmonic oscillator? I think the latter.
#### MattDutra123
It doesn't work because it's the wrong formula. It predicts that the velocity is maximum at t = 0. If you look at the graph, the position is maximum at t = 0 which means that the speed is zero at t = 0.
Also, is this a wave or a simple harmonic oscillator? I think the latter.
Why does the formula I used predict that velocity is maximum at t=0?
If I use the same formula but replacing cos with sin, I get the correct answer. Why is that? Is it related to what you said? We were taught to use cos when displacement is maximum at t = 0, and sine when displacement is 0 at t = 0. Is this correct?
#### kuruman
Homework Helper
Gold Member
You wrote
$v(t)=\omega~ x_{max}\cos(\omega~t)$
At $t = 0$, $v(0)=\omega~ x_{max}\cos(0)=\omega~ x_{max}(1)\neq 0.$
Try this
1. Find an expression for $x(t)$ consistent with the graph. Hint: As indicated above, $x_{max}\cos(0)=x_{max}.$
2. Take the derivative with respect to time to find $v(t)$. The result should answer your second question.
3. Evaluate $v(t=0.12~s).$ This method is neater and there is no need to do what the solution suggests.
We were taught to use cos when displacement is maximum at t = 0, and sine when displacement is 0 at t = 0. Is this correct?
That is correct. Your mistake was that you used cos for the velocity; you were taught to use cos for the displacement.
"How to find the velocity of a wave in simple harmonic motion given time"
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• Solo and co-op problem solving | 2019-05-24T03:03:37 | {
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https://electronics.stackexchange.com/questions/567898/per-unit-impedance | # per unit impedance
Where is my mistake?
Question: Three single-phase two-winding transformers, each rated 25 MVA, 54.2/5.42 kV, are connected to form a three-phase Y-$$\\Delta\$$ bank with a balanced Y-connected resistive load of 0.6 $$\\Omega\$$ per phase on the low-voltage side. By choosing a base of 75 MVA (three phase) and 94 kV (line-to-line) for the high-voltage side of the transformer bank, specify the base quantities for the low-voltage side. Determine the load resistance $$\R_L\$$ in ohms refered to the high-voltage side and the per-unit value of this load resistance on the chosen base.
My approach: Since secondary is $$\\triangle\$$ the line voltage is 5.42kV. $$a=\frac{54.2}{5.42}=10$$ $$Z_b=\frac{V_L^2}{P_{3-\phi}}=\frac{(94 * 10^3)^2}{75 * 10^6}=117.81 \Omega$$ $$R_P=a^2*R_S=100*0.6=60 \Omega$$ $$Z_{p.u.}=\frac{60}{117.81}= 0.51 p.u.$$
The solution manual: Determine the load resistance referred to the high-voltage side. $$Z_{L(high)}=Z_L(\frac{V_{AB}}{V_{high}})$$ $$=(0.6)(\frac{94 kV}{5.42 kV})^2$$ $$=180.47 \Omega$$ Thus, the load resistance refered to the high-voltage side is $$\\boxed{180.47\Omega}\$$.
Source: Power Systems Analysis Publisher : McGraw-Hill Education; 1st edition (January 1, 1994) by John Grainger & William Stevenson ISBN-10 : 0070612935 ISBN-13 : 978-0070612938
• Please write the question and explanation into the post instead of using an image. Images can't be found by search engines, and create difficulties for blind users. Jun 1 at 13:34
• I agree with @ThePhoton, take the time to make a nice, clean post with MathJax. I edited it for you to get you started down the right path. ;-) Jun 2 at 1:38
The load resistance referred to the high-voltage side is just, $$R_P=R_L*N^2=0.6\Omega*10^2=60\Omega$$
In the per-unit system we have the luxury of picking one, and just one MVA base for our entire system. The question tells you that it has been selected for you and is 75 MVA three-phase.
Once you pick the voltage base of 1 bus, all of the rest of the buses are constrained by either direct connection or by transformation. So, all other buses connected to this 1st bus by transmission lines will be at the exact same voltage base. For buses connected to this 1st bus via transformation they will have a voltage base equal to the voltage base of this 1st bus divided by turns ratio.
Your 1st bus has voltage base of 94kV ph-ph and power base of 75 MVA three-phase.
So, $$Z_{base}=\frac{94^2}{75}=117.81\Omega\text{ for first bus}$$
Your 2nd bus has voltage base $$\=\frac{94kV}{\frac{54.2}{5.42}}=9.4kV\$$
So, $$Z_{base}=\frac{9.4^2}{75}=1.178\Omega\text{ for second bus}$$
and, $$R_L= \frac{0.6\Omega}{1.1781\Omega}=0.509 \Omega\text{ in p.u.}$$
As a check, $$R_P= \frac{60\Omega}{117.81\Omega}=0.509 \Omega\text{ in p.u.}$$
A major reason we really like the per-unit system is that it gets rid of transformations.
Note: The solution manual is wrong, they chose the voltage base (94kV) for the numerator in their turns ratio square calculation. Probably a sleep deprived or otherwise distracted grad student.
i) ON THE LOW TENSION SIDE: The base for low tension side is 75MVA, 5.42KV Actual Value = 0.6 ohm Base Value = [(5.42)(5.42)]/(75) = 0.3916 ohm Per unit value of RL = (0.6 / 0.3916) = 1.52
ii) ON THE HIGH TENSION SIDE: The base for high tension side is 75MVA, 94KV Actual Value = 0.6[(9494)/(5.425.42)] = 180 ohm Base Value = (94*94)/75 = 117.81 ohm Per unit value of RL = 180/117.81 = 1.52
• no, the base on the low tension side is not 5.42 kV. The problem specifies the high tension side base as 94 kV. Once that is set, the low tension side voltage base is constrained by the transformer turns ratio. You can only pick 1 votage base, the rest are determined by actual transformer ratios. Jun 17 at 17:36
• uotechnology.edu.iq/dep-electromechanic/typicall/… Please refer this link problem 1-4 and let me know still Iam wrong or what… Jun 18 at 6:26
• Hi @abarna, yes your answer is still wrong. See note 1 on page 8 of the document you referenced in your comment above. Once the 94 kV was specified as base on HV side, the turns ratio (10) requires the LV voltage base to be $\frac{94}{10}=9.4 kV$. Jun 18 at 13:43
• Sorry for my mistake, I got the point! Jun 18 at 13:55
• No problem @Abarna. You should either correct or delete your answer. Jun 20 at 17:41 | 2021-10-28T09:11:38 | {
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http://mathhelpforum.com/advanced-algebra/159439-matrix-gaussian-elimination.html | # Thread: Matrix - Gaussian elimination
1. ## Matrix - Gaussian elimination
Hi
Having trouble with solving the equations using Gaussian elimination
1) Solve the following system of equations using Gaussian elimination
$5x-2y+z=-6$
$-2x+z=-4$
$-x-2y+2z=3$
Show the final tableau (augmented matrix) of the elimination and the values of x, y and z.
This is what i have done but it is incorrect according the book's answer.
$\begin{bmatrix}
5 & -2 & 1 & | & -6 \\
-2 & 0 & 1 & | & -4 \\
-1 & -2 & 2 & | & 3 \\
\end{bmatrix}
$
$\begin{bmatrix}
5 & -2 & 1 & | & -6 \\
0 & -4 & -3 & | & -10 \\
-1 & -2 & 2 & | & 3 \\
\end{bmatrix}
$
R2-2R3
$\begin{bmatrix}
1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\
0 & -4 & 3 & | & -10 \\
-1 & -2 & 2 & | & 3 \\
\end{bmatrix}
$
$\frac{1}{5}R1$
$\begin{bmatrix}
1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\
0 & -4 & 3 & | & -10 \\
0 & \frac{-12}{5} & \frac{11}{5} & | & \frac{9}{5} \\
\end{bmatrix}
$
$R1+R3$
$\begin{bmatrix}
1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\
0 & -4 & 3 & | & -10 \\
0 & \frac{-10}{5} & \frac{11}{5} & | & \frac{9}{5} \\
\end{bmatrix}
$
$\frac{1}{4}R2$
$\begin{bmatrix}
1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\
0 & -4 & 3 & | & -10 \\
0 & 0 & \frac{7}{10} & | & \frac{-16}{5} \\
\end{bmatrix}
$
$\frac{10}{5}R2+R3$
P.S
2. Would the answer perhaps be:
x = 2-t
y = 1/2t - 8
z = 2t
If not, then I should take another look at the question If it is, or close, I can post my solution.
3. Originally Posted by Paymemoney
Hi
Having trouble with solving the equations using Gaussian elimination
1) Solve the following system of equations using Gaussian elimination
$5x-2y+z=-6$
$-2x+z=-4$
$-x-2y+2z=3$
Show the final tableau (augmented matrix) of the elimination and the values of x, y and z.
This is what i have done but it is incorrect according the book's answer.
$\begin{bmatrix}
5 & -2 & 1 & | & -6 \\
-2 & 0 & 1 & | & -4 \\
-1 & -2 & 2 & | & 3 \\
\end{bmatrix}
$
$\begin{bmatrix}
5 & -2 & 1 & | & -6 \\
0 & -4 & -3 & | & -10 \\
-1 & -2 & 2 & | & 3 \\
\end{bmatrix}
$
R2-2R3
$\begin{bmatrix}
1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\
0 & -4 & 3 & | & -10 \\
-1 & -2 & 2 & | & 3 \\
\end{bmatrix}
$
$\frac{1}{5}R1$
$\begin{bmatrix}
1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\
0 & -4 & 3 & | & -10 \\
0 & \frac{-12}{5} & \frac{11}{5} & | & \frac{9}{5} \\
\end{bmatrix}
$
$R1+R3$
$\begin{bmatrix}
1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\
0 & -4 & 3 & | & -10 \\
0 & \frac{-10}{5} & \frac{11}{5} & | & \frac{9}{5} \\
\end{bmatrix}
$
$\frac{1}{4}R2$
$\begin{bmatrix}
1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\
0 & -4 & 3 & | & -10 \\
0 & 0 & \frac{7}{10} & | & \frac{-16}{5} \\
\end{bmatrix}
$
$\frac{10}{5}R2+R3$
P.S
That will give you the correct values for x, y, and z (assuming your arithmetic is correct, I did not check it) but what, exactly, does your text consider the "final tableau"? You have got to "triangular form" and now you can solve for x, y, and z by back substitution. But I notice you got a "1" in the first column but did not get "1"s on the diagonal in the other columns. Some texts, at least, would consider having "1" s all along the diagonal as necessary. Others might expect to see a complete reduction to the identity matrix for the first 3 columns.
4. Originally Posted by HallsofIvy
That will give you the correct values for x, y, and z (assuming your arithmetic is correct, I did not check it) but what, exactly, does your text consider the "final tableau"? You have got to "triangular form" and now you can solve for x, y, and z by back substitution. But I notice you got a "1" in the first column but did not get "1"s on the diagonal in the other columns. Some texts, at least, would consider having "1" s all along the diagonal as necessary. Others might expect to see a complete reduction to the identity matrix for the first 3 columns.
actually the answer is not supplied for this question, but when i used the calculator it gave me what you said '1' on the diagonal
$\begin{bmatrix}
1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\
0 & 1 & \frac{-1}{12} & | & \frac{-3}{4} \\
0 & 0 & 1 & | & \frac{-21}{4} \\
\end{bmatrix}$
5. Originally Posted by Paymemoney
Hi
Having trouble with solving the equations using Gaussian elimination
1) Solve the following system of equations using Gaussian elimination
$5x-2y+z=-6$
$-2x+z=-4$
$-x-2y+2z=3$
Show the final tableau (augmented matrix) of the elimination and the values of x, y and z.
This is what i have done but it is incorrect according the book's answer.
$\begin{bmatrix}
5 & -2 & 1 & | & -6 \\
-2 & 0 & 1 & | & -4 \\
-1 & -2 & 2 & | & 3 \\
\end{bmatrix}
$
$\begin{bmatrix}
5 & -2 & 1 & | & -6 \\
0 & -4 & -3 & | & -10 \\
-1 & -2 & 2 & | & 3 \\
\end{bmatrix}
$
R2-2R3
There is an error in this first step. -2 times the second number in R3 is + 4. The second number in R2 now should be 4, not -4.
$\begin{bmatrix}
1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\
0 & -4 & 3 & | & -10 \\
-1 & -2 & 2 & | & 3 \\
\end{bmatrix}
$
$\frac{1}{5}R1$
$\begin{bmatrix}
1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\
0 & -4 & 3 & | & -10 \\
0 & \frac{-12}{5} & \frac{11}{5} & | & \frac{9}{5} \\
\end{bmatrix}
$
$R1+R3$
$\begin{bmatrix}
1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\
0 & -4 & 3 & | & -10 \\
0 & \frac{-10}{5} & \frac{11}{5} & | & \frac{9}{5} \\
\end{bmatrix}
$
$\frac{1}{4}R2$
$\begin{bmatrix}
1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\
0 & -4 & 3 & | & -10 \\
0 & 0 & \frac{7}{10} & | & \frac{-16}{5} \\
\end{bmatrix}
$
$\frac{10}{5}R2+R3$
P.S
6. ok, i have tried again, this is what i got:
$\begin{bmatrix}
5 & -2 & 1 & | & -6 \\
-2 & 0 & 1 & | & -4 \\
-1 & -2 & 2 & | & 3 \\
\end{bmatrix}
$
$\begin{bmatrix}
5 & -2 & 1 & | & -6 \\
0 & 4 & -3 & | & -10 \\
-1 & -2 & 2 & | & 3 \\
\end{bmatrix}
$
R2-2R3
$\begin{bmatrix}
1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\
0 & 4 & -3 & | & -10 \\
-1 & -2 & 2 & | & 3 \\
\end{bmatrix}
$
$\frac{1}{5}R1$
$\begin{bmatrix}
1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\
0 & 4 & -3 & | & -10 \\
0 & \frac{-12}{5} & \frac{11}{5} & | & \frac{9}{5} \\
\end{bmatrix}
$
R3+R1
$\begin{bmatrix}
1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\
0 & 4 & -3 & | & -10 \\
0 & 0 & \frac{2}{10} & | & \frac{-21}{5} \\
\end{bmatrix}
$
$R3+\frac{3}{5}R2$
$z=\frac{-105}{10}$
$y=\frac{-10-\frac{315}{10}}{10}$ $= \frac{-415}{40}$
$x=\frac{-53}{10}$
7. I'm not sure what you are doing, but as I understood it you just have to set Z to t, in my case I set it to 2t to make it more simple counting.
Then it should all fall out, if z = 2t you can find out what x is and then ultimately what y is.
8. Originally Posted by Paymemoney
o
$\begin{bmatrix}
1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\
0 & 4 & -3 & | & -10 \\
0 & 0 & \frac{2}{10} & | & \frac{-21}{5} \\
\end{bmatrix}
$
$R3+\frac{3}{5}R2$
$z=\frac{-105}{10}$
$y=\frac{-10-\frac{315}{10}}{10}$ $= \frac{-415}{40}$
$x=\frac{-53}{10}$
Briefly looking at that, you should have a figure of z = -210/10 and not -105/10. Then substitute that value into your other equations to work out the value of y and x.
9. Originally Posted by Paymemoney
$\begin{bmatrix}
1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\
0 & 4 & -3 & | & -10 \\
0 & \frac{-12}{5} & \frac{11}{5} & | & \frac{9}{5} \\
\end{bmatrix}
$
R3+R1
$\begin{bmatrix}
1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\
0 & 4 & -3 & | & -10 \\
0 & 0 & \frac{2}{10} & | & \frac{-21}{5} \\
\end{bmatrix}
$
$R3+\frac{3}{5}R2$
$z=\frac{-105}{10}$
$y=\frac{-10-\frac{315}{10}}{10}$ $= \frac{-415}{40}$
$x=\frac{-53}{10}$
Everything looks fine until the 3rd column in your final matrix $\frac{3(-3)}{5} = \frac{-9}{5}$
So $\frac{11}{5} + \frac{-9}{5} = \frac{2}{5}$
Take $\frac{2}{5}R3$ and the bottom row becomes 0 0 1 | $\frac{-21}{2}$ | 2017-11-17T23:36:59 | {
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https://www.cemc.uwaterloo.ca/pandocs/potw/2020-21/English/POTWD-20-NN-26-S.html | # Problem of the Week Problem D and Solution Watery Ways
## Problem
Two parks in Yourtown have very unique but similar designs, which are illustrated in the following diagrams.
The smaller park can be completely enclosed by a square that is $$300$$ m by $$300$$ m. The larger park can be completely enclosed by a square that is $$500$$ m by $$500$$ m.
On each drawing, there are horizontal and vertical lines, each spaced $$100$$ m apart. These lines create identical $$100$$ m by $$100$$ m squares, nine squares on the drawing of the smaller park and twenty-five squares on the drawing of the larger park.
The drawing for each park also shows two concentric circles. The circumference of the outer circle touches each of the four sides of the square enclosing the park. The circumference of the inner circle passes through the four vertices of the largest square created by the gridlines that are totally inside the park. (In the smaller park, this largest square is the single square in the centre of the grid. In the larger park, this largest square is formed by the nine squares in the centre of the grid.)
The ring created between the outer circle and inner circle in each park is completely filled with water to a uniform depth of $$0.5$$ m.
Which of the two waterways contains more water?
## Solution
To find the volume of water in each waterway we need to find the top area of each waterway and multiply that by the depth of the water. Since the depth of the water in each waterway is the same and is constant, we need only compare the top areas to determine which one is larger.
For the waterway in the smaller park, let the diameter of the inner circle be $$d_1$$, the diameter of the larger circle be $$D_1$$, the radius of the inner circle be $$r_1$$ and the radius of the larger circle be $$R_1$$.
For the waterway in the larger park, let the diameter of the inner circle be $$d_2$$, the diameter of the larger circle be $$D_2$$, the radius of the inner circle be $$r_2$$ and the radius of the larger circle be $$R_2$$.
The calculations for the parks are shown.
Smaller Park
The diameter of the inner circle is the length of the diagonal of the contained $$100$$ m by $$100$$ m square. Using the Pythagorean Theorem,
\begin{aligned} d_1&=\sqrt{100^2+100^2}\\ &=\sqrt{20\,000}\\ &=\sqrt{(10\,000)(2)}\quad *\\ &=100\sqrt{2}\text{ m}\\ r_1&=\frac{1}{2}d_1\\ &=50\sqrt{2}\text{ m}\end{aligned}
Larger Park
The diameter of the inner circle is the length of the diagonal of the contained $$300$$ m by $$300$$ m square. Using the Pythagorean Theorem,
\begin{aligned} d_2&=\sqrt{300^2+300^2}\\ &=\sqrt{180\,000}\\ &= \sqrt{(90\,000)(2)}\quad *\\ &=300\sqrt{2}\text{ m}\\ r_2&=\frac{1}{2}d_2\\ &=150\sqrt{2}\text{ m}\end{aligned}
Smaller Park
The diameter of the outer circle is the width of the $$300$$ m by $$300$$ m square. It follows that
\begin{aligned} D_1&=300\text{ m}\\ R_1&=\frac{1}{2}D_1\\ &=150\text{ m}\end{aligned}
Larger Park
The diameter of the outer circle is the width of the $$500$$ m by $$500$$ m square. It follows that
\begin{aligned} D_2&=500\text{ m}\\ R_2&=\frac{1}{2}D_2\\ &=250\text{ m}\end{aligned}
The top area of each waterway can be determined by subtracting the area of the inner circle from the area of the outer circle in each case.
Let $$A_1$$ be the top area of the smaller park’s waterway and $$A_2$$ be the top area of the larger park’s waterway.
Smaller Park
\begin{aligned} A_1&=\pi (R_1)^2 - \pi (r_1)^2\\ &=\pi (150)^2-\pi (50\sqrt{2})^2\\ &=22500\pi-5000\pi\\ &=17500\pi \text{ m}^2\end{aligned}
Larger Park
\begin{aligned} A_2&=\pi (R_2)^2 - \pi (r_2)^2\\ &=\pi (250)^2-\pi (150\sqrt{2})^2\\ &=62500\pi-45000\pi\\ &=17500\pi \text{ m}^2\end{aligned}
This may be a surprising result. Both waterways have equal top areas. Since the depths of the waterways are equal and uniform, the volume of water in each waterway will be the same. Neither waterway contains more water than the other waterway.
### For Further Thought:
A third medium sized park can be completely enclosed by a $$400$$ m by $$400$$ m square. If a waterway were constructed in a similar manner to the waterways in the other two parks, how would the volume of the water in this third waterway compare to the volume of water in the other two waterways? Can you explain what is happening here?
* We reduced the radicals by factoring out the largest perfect square, and then finding the square root of that perfect square. Here is a simpler example: $$\sqrt{20}= \sqrt{(4)(5)} = \sqrt{4}\sqrt{5} = 2 \sqrt{5}$$. | 2021-06-21T01:21:53 | {
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http://sambrunacini.com/tag/calculus/ | Categories
## Common misconceptions about the differentiation operator
(Note: math may not render properly in Microsoft Edge. Any other browser should work, though. I’m trying to figure out the source of this issue.) This is the differentiation operator in single variable calculus:
$$\frac{d}{dx}$$
Since several procedures in calculus happen to involve manipulating the “numerator” and “denominator” of the differentiation operator, people have gained the incorrect idea that it IS in fact a ratio of two “infinitely small” quantities. This is absolutely not true! Modern mathematics does indeed have a notion of infinitely small quantities (called surreal numbers), but they are defined within an entirely different mathematical system called nonstandard analysis. The whole motive behind formulating a rigorous definition of a limit is avoiding problems with infinity. That is to say, $\dfrac{dy}{dx}$ is not a ratio of an infinitely small change in $y$ to an infinitely small change in $x$; instead, it is shorthand for the limit of the difference quotient of some function $y(x)$. By definition,
$$\frac{d}{dx}{y(x)}=\lim_{\Delta x\to 0}\frac{y(x+\Delta x) – y(x)}{\Delta x}$$
The limit denotes the quantity that the difference quotient “approaches” as $\Delta x$ grows small; that is, for any arbitrarily small difference $\varepsilon$ between the value of $\frac{y(x+\Delta x)-y(x)}{\Delta x}$ and the limit $\lim_{\Delta x\to 0}\frac{y(x+\Delta x) – y(x)}{\Delta x}$, there exists a choice of $\Delta x$ that yields a value of the difference quotient within $\varepsilon$ of the limit. Nothing in this definition has anything to do with “infinitely small quantities”. Instead, it deals with “arbitrarily small” quantities. In the end, though, the limit is a different number altogether that just so happens to be associated with the expression. The expression can get within arbitrarily small distance of the limit given an appropriate choice of $\Delta x$.
Now that the definition is out of the way, it seems necessary to address several instances in calculus where it seems okay to manipulate the differentiation operator as if it’s a fraction. I will cover the the most common situation (and ALL others can be explained, rigorously, using a concept called “differential forms”, which is currently beyond my understanding). The two most common are in the separation of variables for solving simple differential equations and the method of integration by substitution (which I was planning to cover as well, but it turns out Wikipedia has an excellent explanation here).
## Separation of variables
After separating the variables in a differential equation, we are left with something of the form:
$$f(y)\frac{dy}{dx}=g(x)$$
At which point we can take the antiderivative of both sides with respect to $x$ (since antiderivatives are unique up to a constant):
$$\int{f(y)\frac{dy}{dx}}dx=\int{g(x)}dx$$
The right hand side of this equation essentially asks, “what function, when we take its derivative with respect to $x$, yields $f(y)\frac{dy}{dx}$”. If we let
$$u:=\int{f(y)}dy$$
then, by the chain rule,
$$\frac{du}{dx}=\frac{du}{dy}\frac{dy}{dx}$$
which is of course the equivalent to
$$\frac{du}{dx}=f(y)\frac{du}{dx}$$
Therefore, integrating both sides with respect to $x$,
$$u=\int{f(y)\frac{dy}{dx}}dx$$
which justifies the final equation
$$\int{f(y)}dy=\int{g(x)}dx$$
So in practice, it seems like you’re “multiplying both sides of the equation by $dx$”, but that is the consequence of the work shown above.
Categories
## Fundamental Theorem of Calculus Explained
Relation between derivatives and integrals
# Why are derivatives the inverse of integrals?¶
#### By Sam Brunacini¶
Understandably, most people get confused when first introduced to the idea that finding the area under a curve is the inverse to finding instantaneous rate of change. It seems like the intuition behind this is hidden behind all the formulas they make you learn. Here I’ll explain why the Fundamental Theorem of Calculus (FTC) works. This is not a rigorous proof of the theorem, just an explanation of why they make sense.
First, let’s write the FTC: $$\text{Part 1: } \frac{d}{dx}\int_{a}^{x}f(t)dt = f(x)\\$$ $$\text{Part 2: } \int_{a}^{b}f(x)dx=F(b)-F(a)$$
To help understand Part 1, imagine you own a library. Before the library opens at 8:00 AM, you count 100 books in stock. Between 8:00 and 9:00, 5 books are withdrawn. Between 9:00 and 10:00, 8 books are returned. Without recounting, what is the number of books in stock at 10:00? Of course, you subtract 5 from 100 to account for the withdrawals then add 8 for the returns. This leaves $100 – 5 + 8 = 103$.
Believe it or not, this is exactly what FTC Part 1 says. Call the number of books in the library $x$ hours after 8:00 $f(x)$. Then $f(0)=100$ because that’s how many books there were before anyone came in. $f(x)$ changes by -5 during the first hour. We then add this change to the total to get $f(1)=95$. Due to the returns, $f(x)$ changes by 8 in the second hour. Adding this change to the total, we know $f(2)=103$. FTC part 1 says that a function is equal to the accumulation (or total) of the changes in itself as $x$ varies. The derivative indicates change, and the integral indicates accumulation, or summing the values together. This is exactly what happened in the library example.
In [47]:
import matplotlib.pyplot as plt
import numpy as np
import math
THIRD_PI = math.pi / 3
plt.rcParams['figure.figsize'] = [15, 5]
fig, (ax1, ax2, ax3) = plt.subplots(1, 3)
fig.suptitle("The finite accumulation (sum) of changes")
ax1.set_title("Sum of 4 changes")
ax2.set_title("Sum of 6 changes")
ax3.set_title("Sum of 10 changes")
def plot_cos(axes, steps=100, color="blue"):
X = np.linspace(0, THIRD_PI, num=steps)
Y = np.cos(5*X)
axes.plot(X, Y, color=color)
plot_cos(ax1, color="red")
plot_cos(ax2, color="red")
plot_cos(ax3, color="red")
plot_cos(ax1, 4)
plot_cos(ax2, 6)
plot_cos(ax3, 10)
plt.show()
The graphs above show finite sums of changes in some function. Roughly speaking, an integral is the sum of infinitely many tiny changes in the function. Notice that as the number of changes increases, the approximation (blue line) gets closer to the exact curve (red line). In fact, the approximation gets arbitrarily close to the curve as the number of changes grows larger.
Now, for Part 2. Remember that $F(x)$ is defined as the function that gives the area under a curve from 0 to $x$. Assume for now that $a \lt b$. This is a safe assumption since we can use the formula $$\int_{b}^{a}f(x)dx=-\int_{a}^{b}f(x)dx$$ to make it true if originally $a \gt b$. $F(a)$ is then the area under the curve from 0 to $a$ and $F(b)$ is the area under the curve from 0 to $b$. If we plot these separately, it looks like this:
In [55]:
import matplotlib.pyplot as plt
import numpy as np
import math
THIRD_PI = math.pi / 3
plt.rcParams['figure.figsize'] = [15, 5]
fig, (ax1, ax2) = plt.subplots(1, 2)
ax1.set_title("F(a)")
ax2.set_title("F(b)")
def plot_sin(axes, steps=100, color="red"):
X = np.linspace(0, THIRD_PI, num=steps)
Y = np.sin(4*X)/2+0.5
axes.plot(X, Y, color=color)
plot_sin(ax1)
plot_sin(ax2)
X1 = np.linspace(0, THIRD_PI/3)
ax1.fill_between(X1, np.sin(4*X1)/2+0.5, color="blue", alpha=0.25)
ax1.annotate("x=a", xy=(THIRD_PI / 3, math.sin(4 * THIRD_PI / 3) / 2 + 0.4), size=15)
X2 = np.linspace(0, THIRD_PI*0.75)
ax2.fill_between(X2, np.sin(4*X2)/2+0.5, color="green", alpha=0.25)
ax2.annotate("x=b", xy=(THIRD_PI * 0.75, math.sin(3*THIRD_PI)/2+0.5), size=15)
plt.show()
If you subtract the area shown in the first diagram from the second (which is simply doing $F(b) – F(a)$), you end up with the area between $a$ and $b$.
In [ ]: | 2020-09-30T16:52:14 | {
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https://math.libretexts.org/Bookshelves/Calculus/Map%3A_Calculus_-_Early_Transcendentals_(Stewart)/2%3A_Limits_and_Derivatives/2.8%3A_The_Derivative_as_a_Function |
# 2.8: The Derivative as a Function
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$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
Skills to Develop
In this section, we strive to understand the ideas generated by the following important questions:
• How does the limit definition of the derivative of a function $$f$$ lead to an entirely new (but related) function $$f'$$?
• What is the difference between writing $$f'(a)$$ and $$f'(x)$$?
• How is the graph of the derivative function $$f'(x)$$ connected to the graph of $$f(x)$$?
• What are some examples of functions $$f$$ for which $$f'$$ is not defined at one or more points?
Given a function $$y=f(x)$$, we now know that if we are interested in the instantaneous rate of change of the function at $$x=a$$, or equivalently the slope of the tangent line to $$y=f(x)$$ at $$x=a$$, we can compute the value $$f'(a)$$. In all of our examples to date, we have arbitrarily identified a particular value of $$a$$ as our point of interest: $$a=1,\; a=3$$, etc. But it is not hard to imagine that we will often be interested in the derivative value for more than just one a-value, and possibly for many of them. In this section, we explore how we can move from computing simply $$f'(1)$$ or $$f'(3)$$ to working more generally with $$f'(a)$$, and indeed $$f'(x)$$. Said differently, we will work toward understanding how the so-called process of “taking the derivative” generates a new function that is derived from the original function $$y=f(x)$$. The following preview activity starts us down this path.
Preview Activity $$\PageIndex{1}$$
Consider the function $$f(x)=4x-x^2$$.
1. Use the limit definition to compute the following derivative values: $$f'(0)$$, $$f'(1)$$, $$f'(2)$$, and $$f'(3)$$.
2. Observe that the work to find $$f'(a)$$ is the same, regardless of the value of $$a$$. Based on your work in (a), what do you conjecture is the value of $$f'(4)$$? How about $$f'(5)$$? (Note: you should not use the limit definition of the derivative to find either value.)
3. Conjecture a formula for $$f'(a)$$ that depends only on the value $$a$$. That is, in the same way that we have a formula for $$f(x)$$ (recall $$f(x)=4x-x^2$$ ), see if you can use your work above to guess a formula for $$f'(a)$$ in terms of $$a$$.
### The Derivative is Itself a Function
In your work in Preview Activity 1.4 with $$f(x)=4x-x^2$$, you may have found several patterns. One comes from observing that $$f'(0)=4$$, $$f'(1)=2$$, $$f'(2)=0$$, and $$f'(3)=-2$$. That sequence of values leads us naturally to conjecture that $$f'(4)=-4$$ and $$f'(5)=-6$$. Even more than these individual numbers, if we consider the role of 0, 1, 2, and 3 in the process of computing the value of the derivative through the limit definition, we observe that the particular number has very little effect on our work. To see this more clearly, we compute $$f'(a)$$, where $$a$$ represents a number to be named later. Following the now standard process of using the limit definition of the derivative,
\begin{align} (f'(a) &=\lim_{h\to 0} \frac{f(a+h)-f(a)}{h} \\ &=\lim_{h\to 0} \frac{4(a+h)-(a+h)^2-(4a-a^2 )}{h} \\ &= \lim_{h\to 0} \frac{4a+4h-a^2-2ha-h^2-4a+a^2}{h} \\ &= \lim_{h\to 0} \frac{4h-2ha-h^2}{h} \\ &= \lim_{h\to 0} \frac{h(4-2a-h)}{h} \\ &= \lim_{h\to 0} (4-2a-h). \end{align}
Here we observe that neither 4 nor $$2a$$ depend on the value of $$h$$, so as
$h \to 0,\; (4-2a-h) \to (4-2a).$
Thus, $$f'(a)=4-2a.$$
This observation is consistent with the specific values we found above: e.g., $$f'(3)=4-2(3)=-2$$. And indeed, our work with a confirms that while the particular value of a at which we evaluate the derivative affects the value of the derivative, that value has almost no bearing on the process of computing the derivative. We note further that the letter being used is immaterial: whether we call it $$a$$, $$x$$, or anything else, the derivative at a given value is simply given by “4 minus 2 times the value.” We choose to use $$x$$ for consistency with the original function given by $$y=f(x)$$, as well as for the purpose of graphing the derivative function, and thus we have found that for the function $$f(x)=4x-x^2$$ , it follows that $$f'(x)=4-2x$$.
Because the value of the derivative function is so closely linked to the graphical behavior of the original function, it makes sense to look at both of these functions plotted on the same domain. In Figure 1.18, on the left we show a plot of $$f(x)=4x-x^2$$ together with a selection of tangent lines at the points we’ve considered above. On the right, we show a plot of $$f'(x)=4-2x$$ with emphasis on the heights of the derivative graph at the same selection of points. Notice the connection between colors in the left and right graph: the green tangent line on the original graph is tied to the green point on the right graph in the following way: the slope of the tangent line at a point on the lefthand graph is the
Figure 1.18: The graphs of $$f(x)=4x-x^2$$ (at left) and $$f'(x)=4-2x$$ (at right). Slopes on the graph of $$f$$ correspond to heights on the graph of $$f'$$.
same as the height at the corresponding point on the righthand graph. That is, at each respective value of $$x$$, the slope of the tangent line to the original function at that x-value is the same as the height of the derivative function at that x-value. Do note, however, that the units on the vertical axes are different: in the left graph, the vertical units are simply the output units of $$f$$ . On the righthand graph of $$y=f'(x)$$, the units on the vertical axis are units of $$f$$ per unit of $$x$$.
Of course, this relationship between the graph of a function $$y=f(x)$$ and its derivative is a dynamic one. An excellent way to explore how the graph of $$f(x)$$ generates the graph of $$f'(x)$$ is through a java applet. See, for instance, the applets at http://gvsu.edu/s/5C or http://gvsu.edu/s/5D, via the sites of Austin and Renault5 .
In Section 1.3 when we first defined the derivative, we wrote the definition in terms of a value $$a$$ to find $$f'(a)$$. As we have seen above, the letter a is merely a placeholder, and it often makes more sense to use $$x$$ instead. For the record, here we restate the definition of the derivative.
Definition 1.4
Let $$f$$ be a function and $$x$$ a value in the function’s domain. We define the derivative of $$f$$ with respect to $$x$$ at the value $$x$$, denoted $$f'(x)$$, by the formula $$f'(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$$ , provided this limit exists.
5David Austin, http://gvsu.edu/s/5r; Marc Renault, http://gvsu.edu/s/5p
We now may take two different perspectives on thinking about the derivative function: given a graph of $$y=f(x)$$, how does this graph lead to the graph of the derivative function $$y=f'(x)$$? and given a formula for $$y=f(x)$$, how does the limit definition of the derivative generate a formula for $$y=f'(x)$$? Both of these issues are explored in the following activities.
Exercise $$\PageIndex{1}$$
For each given graph of $$y=f(x)$$, sketch an approximate graph of its derivative function, $$y=f'(x)$$, on the axes immediately below. The scale of the grid for the graph of $$f$$ is $$1\times 1$$; assume the horizontal scale of the grid for the graph of $$f'$$ is identical to that for $$f$$ . If necessary, adjust and label the vertical scale on the axes for $$f'$$.
When you are finished with all 8 graphs, write several sentences that describe your overall process for sketching the graph of the derivative function, given the graph the original function. What are the values of the derivative function that you tend to identify first? What do you do thereafter? How do key traits of the graph of the derivative function exemplify properties of the graph of the original function? C For a dynamic investigation that allows you to experiment with graphing $$f'$$ when given the graph of $$f$$ , see http://gvsu.edu/s/8y. 6 6Marc Renault, Calculus Applets Using Geogebra. 39
Now, recall the opening example of this section: we began with the function $$y=f(x)=4x-x^2$$ and used the limit definition of the derivative to show that $$f'(a)=4-2a$$, or equivalently that $$f'(x)=4-2x$$. We subsequently graphed the functions $$f$$ and $$f'$$ as shown in Figure 1.18. Following Activity 1.10, we now understand that we could have constructed a fairly accurate graph of $$f'(x)$$ without knowing a formula for either $$f$$ or $$f'$$. At the same time, it is ideal to know a formula for the derivative function whenever it is possible to find one. In the next activity, we further explore the more algebraic approach to finding $$f'(x)$$: given a formula for $$y=f(x)$$, the limit definition of the derivative will be used to develop a formula for $$f'(x)$$.
Activity $$\PageIndex{2}$$
For each of the listed functions, determine a formula for the derivative function. For the first two, determine the formula for the derivative by thinking about the nature of the given function and its slope at various points; do not use the limit definition. For the latter four, use the limit definition. Pay careful attention to the function names and independent variables. It is important to be comfortable with using letters other than $$f$$ and $$x$$. For example, given a function $$p(z)$$, we call its derivative $$p'(z)$$.
1. $$f(x)=1$$
2. $$g(t)=t$$
3. $$p(z)=z^2$$
4. $$q(s)=s^3$$
5. $$F(t)=\frac{1}{t}$$
6. $$G(y)=\sqrt{y}$$
### Summary
In this section, we encountered the following important ideas:
• The limit definition of the derivative, $$f'(x)=lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$$ , produces a value for each $$x$$ at which the derivative is defined, and this leads to a new function whose formula is $$y=f'(x)$$. Hence we talk both about a given function $$f$$ and its derivative $$f'$$. It is especially important to note that taking the derivative is a process that starts with a given function ( $$f$$ ) and produces a new, related function ( $$f'$$).
• There is essentially no difference between writing $$f'(a)$$ (as we did regularly in Section 1.3) and writing $$f'(x)$$. In either case, the variable is just a placeholder that is used to define the rule for the derivative function.
• Given the graph of a function $$y=f(x)$$, we can sketch an approximate graph of its derivative $$y=f'(x)$$ by observing that heights on the derivative’s graph correspond to slopes on the original function’s graph.
• In Activity 1.10, we encountered some functions that had sharp corners on their graphs, such as the shifted absolute value function. At such points, the derivative fails to exist, and we say that $$f$$ is not differentiable there. For now, it suffices to understand this as a consequence of the jump that must occur in the derivative function at a sharp corner on the graph of the original function.
### Contributors
Matt Boelkins (Grand Valley State University), David Austin (Grand Valley State University), Steve Schlicker (Grand Valley State University) | 2019-04-20T08:52:53 | {
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https://mathzsolution.com/when-is-matrix-multiplication-commutative/ | # When is matrix multiplication commutative?
I know that matrix multiplication in general is not commutative. So, in general:
$A, B \in \mathbb{R}^{n \times n}: A \cdot B \neq B \cdot A$
But for some matrices, this equations holds, e.g. A = Identity or A = Null-matrix $\forall B \in \mathbb{R}^{n \times n}$.
I think I remember that a group of special matrices (was it $O(n)$, the group of orthogonal matrices?) exist, for which matrix multiplication is commutative.
For which matrices $A, B \in \mathbb{R}^{n \times n}$ is $A\cdot B = B \cdot A$?
## Answer
Two matrices that are simultaneously diagonalizable are always commutative.
Proof: Let $A$, $B$ be two such $n \times n$ matrices over a base field $\mathbb K$, $v_1, \ldots, v_n$ a basis of Eigenvectors for $A$. Since $A$ and $B$ are simultaneously diagonalizable, such a basis exists and is also a basis of Eigenvectors for $B$. Denote the corresponding Eigenvalues of $A$ by $\lambda_1,\ldots\lambda_n$ and those of $B$ by $\mu_1,\ldots,\mu_n$.
Then it is known that there is a matrix $T$ whose columns are $v_1,\ldots,v_n$ such that $T^{-1} A T =: D_A$ and $T^{-1} B T =: D_B$ are diagonal matrices. Since $D_A$ and $D_B$ trivially commute (explicit calculation shows this), we have
Attribution
Source : Link , Question Author : Martin Thoma , Answer Author : Community | 2023-02-04T02:32:00 | {
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http://calculus.seas.upenn.edu/?n=Main.ApproximationAndError | Search:
# ApproximationAndError
Given a series that is known to converge but for which an exact answer is not known, how does one find a good approximation to the true value? One way to get an approximation is to add up some number of terms and then stop. But how many terms are enough? How close will the result be to the true answer? That is the motivation for this module.
## Error defined
Given a convergent series
(:latex:) \begin{equation*} s = \sum_{n=0}^\infty a_n. \end{equation*} (:latexend:)
Recall that the partial sum ($s_k$) is the sum of the terms up to and including ($a_k$), i.e.,
(:latex:) \begin{align*} s_k &= a_0+a_1+a_2+\ldots+a_k
&= \sum_{n=0}^k a_n. \end{align*} (:latexend:)
Then the error ($E_k$) is the difference between ($s_k$) and the true value ($s$), i.e.,
(:latex:) \begin{align*} E_k &= s - s_k
&= \sum_{n=0}^\infty a_n - \sum_{n=0}^k a_n
&= a_{k+1}+a_{k+2}+a_{k+3}+\ldots
&= \sum_{n=k+1}^\infty a_n. \end{align*} (:latexend:)
In other words, the error is the sum of all the terms from the infinite series which were not included in the partial sum.
## Alternating series error bound
For a decreasing, alternating series, it is easy to get a bound on the error ($E_k$):
(:latex:) \begin{equation*} |E_k| \leq |a_{k+1}|. \end{equation*} (:latexend:)
In other words, the error is bounded by the next term in the series.
### Note
If the series is strictly decreasing (as is usually the case), then the above inequality is strict.
To see why the alternating bound holds, note that each successive term in the series overshoots the true value of the series. In other words, if ($S$) is the true value of the series,
(:latex:) \begin{align*} a_0 &> S
a_0 - a_1 &< S
a_0 - a_1 + a_2 &> S. \end{align*} (:latexend:)
The above figure shows that if one stops at ($a_0 - a_1+a_2-a_3$), then the error ($E_3$) must be less than ($a_4$).
### Example
What is the minimum number of terms of the series
(:latex:) \begin{equation*} \sum_{n=1}^\infty (-1)^{n+1} \frac{1}{n^2} \end{equation*} (:latexend:)
one needs to be sure to be within ($\frac{1}{100}$) of the true sum?
The goal is to find ($k$) so that ($|E_k| \leq \frac{1}{100}$). Since ($|E_k| \leq |a_{k+1}|$), the question becomes for which value of ($k$) is ($|a_{k+1}| \leq \frac{1}{100}$)? If ($k=9$), then ($|a_{k+1}| = \frac{1}{100}$), and so by the alternating series error bound, ($|E_9| \leq \frac{1}{100}$). Thus 9 terms are required to be within ($\frac{1}{100}$) of the true value of the series.
## Integral test for error bounds
Another useful method to estimate the error of approximating a series is a corollary of the integral test. Recall that if a series ($\sum f(n)$) has terms which are positive and decreasing, then
(:latex:) \begin{equation*} \int_{m+1}^\infty f(x)dx < \sum_{n=m+1}^\infty f(n) < \int_{m}^\infty f(x)dx. \end{equation*} (:latexend:)
But notice that the middle quantity is precisely ($E_m$). So
(:latex:) \begin{equation*} \int_{m+1}^\infty f(x)dx < E_m < \int_{m}^\infty f(x)dx. \end{equation*} (:latexend:)
This bound is nice because it gives an upper bound and a lower bound for the error.
### Example
How many terms of the series
(:latex:) \begin{equation*} \sum_{n=1}^\infty \frac{1}{n^3} \end{equation*} (:latexend:)
must one add up so that the Integral bound guarantees the approximation is within ($\frac{1}{100}$) of the true answer?
If one adds up the first ($m$) terms, then by the integral bound, the error ($E_m$) satisfies
(:latex:) \begin{align*} E_m &< \int_m^\infty \frac{dx}{x^3}
& = \frac{x^{-2}}{-2} \bigg|^\infty_m
&= \frac{1}{2m^2}. \end{align*} (:latexend:)
Setting ($\frac{1}{2m^2} \leq \frac{1}{100}$) gives that ($m^2 \geq 50$), so ($m \geq 8$). Thus, ($m=8$) is the minimum number of terms required so that the Integral bound guarantees we are within ($\frac{1}{100}$) of the true answer.
### Note
If you actually compute the partial sums using a calculator, you will find that 7 terms actually suffice. But remember, we want the guarantee of the integral test, which only ensures that ($\frac{1}{128} < E_7 < \frac{1}{98}$), despite the fact that in reality, ($E_7 \approx .009 < .01$).
## Taylor approximations
Recall that the Taylor series for a function ($f$) about 0 is given by
(:latex:) \begin{align*} f(x) &= \sum_{n=0}^\infty \frac{f^{\left(n\right)}(0)}{n!} x^n
&= f(0) + f'(0)x + \frac{f''(0)}{2!}x^2+\dotsb. \end{align*} (:latexend:)
The Taylor polynomial of degree ($N$) is the approximating polynomial which results from truncating the above infinite series after the degree ($N$) term:
(:latex:) \begin{align*} f(x) &\approx \sum_{n=0}^N \frac{f^{\left(n \right)}(0)}{n!} x^n
&= f(0) + f'(0)x + \frac{f''(0)}{2!}x^2+\dotsb + \frac{f^{\left(N\right)}(0)}{N!}x^N. \end{align*} (:latexend:)
This is a good approximation for ($f(x)$) when ($x$) is close to 0. How good an approximation is it? That is the purpose of the last error estimate for this module.
As in previous modules, let ($E_N(x)$) be the error between the Taylor polynomial and the true value of the function, i.e.,
(:latex:) $f(x) = \sum_{n=0}^N \frac{f^{\left(n \right)}(0)}{n!} x^n + E_N(x).$ (:latexend:)
Notice that the error ($E_N(x)$) is a function of ($x$). In general, the further away ($x$) is from ($0$), the bigger the error will be.
A first, weak bound for the error is given by
(:latex:) $E_N(x) \leq C x^{N+1}$ (:latexend:)
for some constant ($C$) and ($x$) sufficiently close to 0. In other words, ($E_N(x)$) is ($O(x^{N+1})$). A stronger bound is given in the next section.
### Taylor remainder theorem
The following gives the precise error from truncating a Taylor series:
Taylor remainder theorem
The error ($E_N(x)$) is given precisely by
(:latex:) \begin{equation*} E_N(x) = \frac{f^{\left(N+1\right)}(t)}{(N+1)!}x^{N+1}, \end{equation*} (:latexend:)
for some ($t$) between 0 and ($x$), inclusive. So if ($x<0$), then ($x \leq t \leq 0$), and if ($x>0$), then ($0 \leq t \leq x$).
### Example
Consider the case when ($N=0$). The Taylor remainder theorem says that
(:latex:) \begin{equation*} f(x) = f(0) + f'(t)x, \end{equation*} (:latexend:)
for some ($t$) between 0 and ($x$). Solving for ($f'(t)$) gives
(:latex:) \begin{equation*} f'(t) = \frac{f(x)-f(0)}{x-0}, \end{equation*} (:latexend:)
for some ($0<t<x$) if ($x>0$) and ($x<t<0$) if ($x<0$), which is precisely the statement of the Mean value theorem. Therefore, one can think of the Taylor remainder theorem as a generalization of the Mean value theorem.
### Taylor error bound
As it is stated above, the Taylor remainder theorem is not particularly useful for actually finding the error, because there is no way to actually find the ($t$) for which the equation holds. There is a slightly different form which gives a bound on the error:
Taylor error bound
(:latex:) \begin{equation*} |E_N(x)| \leq \frac{C}{(N+1)!} |x|^{N+1}, \end{equation*} (:latexend:)
where ($C$) is the maximum value of ($|f^{\left(N+1\right)}(t)|$) over all ($t$) between 0 and ($x$), inclusive.
### Example
Estimate ($\sqrt{e}$) using
(:latex:) \begin{equation*} e^{1/2} \approx 1 + \frac{1}{2} + \frac{(1/2)^2}{2!} + \frac{(1/2)^3}{3!} \approx 1.64, \end{equation*} (:latexend:)
and bound the error.
The function is ($f(x) = e^x$), and the approximating polynomial used here is
(:latex:) \begin{equation*} e^x \approx 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}. \end{equation*} (:latexend:)
Then according to the above bound,
(:latex:) \begin{equation*} |E_3(x)| \leq \frac{C}{4!}|x|^4, \end{equation*} (:latexend:)
where ($C$) is the maximum of ($f^{\left(4 \right)}(t) = e^t$) for ($0 \leq t \leq x$). Since ($e^t$) is an increasing function, ($C = e^x$). Thus,
(:latex:) \begin{equation*} |E_3(x)| \leq \frac{e^x}{4!}x^4. \end{equation*} (:latexend:)
Thus,
(:latex:) \begin{equation*} |E_3(1/2)| \leq \frac{e^{1/2}}{4!}(1/2)^4 < \frac{1}{100}. \end{equation*} (:latexend:) | 2019-06-19T16:45:49 | {
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https://ximera.osu.edu/mooculus/calculus2/divergenceTest/digInDivergenceTest |
If an infinite sum converges, then its terms must tend to zero.
In order to determine if a series $\sum _{k=1}^{\infty } a_k$ converges, we took the following approach.
• Consider the associated sequence $\{s_n\}$ of partial sums, where $s_n=\sum _{k=1}^n a_k$.
• Try to find an explicit formula for the term $s_n$. If you can find such a formula, analyze $\lim _{n \to \infty s_n}$.
• If the limit exists, $\sum _{k=k_0} a_k$ converges, and if we can determine that $\lim _{n \to \infty } s_n =L$, then $\sum _{k=k_0} a_k=L$.
• If $\lim _{n \to \infty } s_n$ does not exist, then $\sum _{k=k_0} a_k$ diverges.
• If an explicit formula for $s_n$ cannot be found, further analysis is needed.
In the previous section, we studied two types of series where we could find an explicit formula for $s_n$, but unfortunately, this is not always easy or possible. Fortunately, it is not always necessary to do this in order to determine whether $\lim _{n \to \infty } s_n$ exists. Consider the example below.
As it turns out, the above argument can be used to make a very important observation; if $\{a_n\}$ is a sequence for which $\sum _{k=k_0}^{\infty } a_k$ converges, then $\lim _{n \to \infty } a_n =0$. This result is fundamentally important, so we capture it in a theorem.
### The divergence test
Stated in plain English, the above test ensures that if the terms in a sequence do not tend to zero, then we cannot add all of the terms in that sequence together.
This test gives us a quick way to determine if some series diverge.
### Implications of the divergence test
Let’s summarize the important points from the previous discussion.
• If $\sum _{k=k_0}^\infty a_k$ converges, then $\lim _{n \to \infty } a_n =0$.
• If $\lim _{n \to \infty } a_n \neq 0$ (including the case where the limit does not exist), then $\sum _{k=k_0}^{\infty } a_k$ diverges.
While divergence test was straightforward to apply in the previous examples, there is a major point to address about what it does not say.
The divergence test can never be used to conclude that a series converges. The theorem does not state that if $\lim _{n\to \infty } a_n = 0$ then $\sum _{n=1}^\infty a_n$ converges.
We’ve actually seen an example of this in action.
Said another way:
If $\sum _{k=k_0}^{\infty } a_k$ diverges, it’s still possible that $\lim _{n \to \infty } a_n =0$.
To elaborate a little more, we can say that a series $\sum _{k=k_0} a_k$ “passes the divergence test” if $\lim _{n \to \infty } a_n=0$. Which of the following series pass the divergence test?
$\sum _{k=3}^\infty \frac {1}{\ln { k }}$ $\sum _{k=0}^\infty \sin (k)$ $\sum _{k=0}^\infty \frac {\sin (k)}{k^2}$ $\sum _{k=5}^\infty \frac {k+7}{k+6}$ $\sum _{k=0}^\infty \frac {2k}{k - 5}$
Restating this point again (because it is very important): passing the divergence test means that a series has a chance to converge. The divergence test cannot tell us whether a series converges.
There are many questions that require that you now have a firm grasp on the concepts presented thus far. We summarize the important points made thus far, then give many examples that require you to synthesize them.
• There are two fundamental questions we can ask of any sequence.
• Do the numbers in the list approach a finite value?
• Can I sum all of the numbers in the list and obtain a finite result?
These questions can be asked of a given sequence $\{a_n\}$ and can also be asked about $\{s_n\}$ or any sequence constructed from it.
• Given a sequence $\{a_n\}_{n=n_0}$, we construct the sequence of partial sum $\{s_n\}_{n=n_0}$ whose $n$-th term is given by the formula $s_n = \sum _{k=k_0}^n a_k$.
• The symbols $\sum _{k=k_0}^{\infty } a_k$ and $\lim _{n \to \infty } s_n$ are the same.
• By definition $\sum _{k=k_0} a_k$ converges if $\lim _{n \to \infty } s_n$ exists and in this case, the value of each is the same.
• By definition $\sum _{k=k_0} a_k$ diverges if $\lim _{n \to \infty } s_n$ does not exist (which includes if the limit is infinte).
• If the limit of a sequence is not zero, the sum of its terms diverges.
• If a series converges, the limit of the sequence whose terms is being summed is zero.
• If the limit of a sequence is zero, more information is needed to determine whether the sum of its terms converges or diverges.
To answer the following questions, make sure that you understand exactly what is given in the statement of the question first, then try to synthesize the material above.
Suppose $\{a_n\}_{n \geq 1}$ is a sequence and $\sum ^{\infty }_{n= 1} a_n=5$. Let $s_n = \sum ^n_{k=1} a_k$. Select all statements that must be true:
$\lim _{n \to \infty } a_n = 5$ $\lim _{n \to \infty } a_n = 0$ $\lim _{n \to \infty } s_n = 0$ $\lim _{n \to \infty } s_n = 5$ $\sum ^{\infty }_{k=1} s_k$ must diverge. $\sum ^{\infty }_{k=1} (a_k+1) = 5+1=6$ The divergence test tells us $\sum ^{\infty }_{n= 1} a_n$ converges to $L$. | 2021-04-20T14:59:33 | {
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https://math.stackexchange.com/questions/1130665/sketching-an-inequality-in-the-complex-plane/1130680 | # Sketching an inequality in the complex plane
so I have to sketch this inequality on the complex plane,
$$\frac {|z-a|} {|1- \bar az|}<1$$
where $|a| < 1$ is a complex number.
I know typically when there is just $z$'s and $i$'s you replace $z$ with $z=x+iy$ and go from there (squaring both sides, using the modulus, etc..) but this "$a$" is throwing me off on what to do. Any tips please?
Thanks
One possible way:
You inequality is (for $a\neq0$) equivalent to
$$\frac{|z-a|}{\left|\frac{1}{\overline{a}}-z\right|}<|\overline{a}|$$
The equality
$$\frac{|z-a|}{\left|\frac{1}{\overline{a}}-z\right|}=|\overline{a}|$$
says that the proportion of the distances from $z$ to the pair of points $a$ and $\frac{1}{\overline{a}}$ is a constant $|\overline{a}|$.
Apollonius theorem says this is a circle. The inequality is then one of the sides.
Another way:
$$\frac{z\overline{z}-a\overline{z}-\overline{a}z+a\overline{a}}{1-\overline{a}z-a\overline{z}+a\overline{a}z\overline{z}}=\frac{(z-a)(\overline{z}-\overline{a})}{(1-\overline{a}z)(1-a\overline{z})}=\frac{|z-a|^2}{|1-\overline{a}z|^2}<1$$
i.e.
$$z\overline{z}-a\overline{z}-\overline{a}z+a\overline{a}<1-\overline{a}z-a\overline{z}+a\overline{a}z\overline{z}$$
or $$(1-a\overline{a})z\overline{z}<(1-a\overline{a}).$$
Depending on whether $|a|>1$ or $|a|<1$ we can cancel the $1-a\overline{a}$ and reverse or not the sign in the inequality.
We get $$|z|^2=z\overline{z}>1\text{ or }<1$$
Consider also the case $|a|=1$.
• so you're saying that this is a circle with radius |a-bar| and has to end points, a and 1/a? Interesting. Is there no way to simplify the equation down a little bit more to actually see that it is a circle? You know, like getting x^2 + y^2 = ..... or something like that? Either way, thanks. – Joe Feb 2 '15 at 18:30
• @Joe Yes, it can be done. This is just one way, which avoids getting your hands too dirty. – Pp.. Feb 2 '15 at 18:31
• Haha, fair enough. I just want to be able to expand it out so I can see how this "a" parameter works. – Joe Feb 2 '15 at 18:35
• @Joe See the second option. – Pp.. Feb 2 '15 at 18:38
• many thanks man, huge help!! – Joe Feb 2 '15 at 18:41
This is a slight elaboration of Pps. answer.
As noted in Pps. answer, we can write $\left| {z-a \over 1-z\bar{a}}\right | = {1 \over |a|} {|z-a| \over |z-{1 \over \bar{a}}|}$, so we can generalise slightly and consider the set of points $C = \{z | \left| {z-a \over z-b}\right | = \lambda \}$, with $\lambda >0$ and $a \neq b$.
If we let $r e^{i \theta} = {b-a \over 2}$, and $w = e^{-i \theta}(z-{a+b \over 2})$, we obtain the equality $\left| {w+r \over w-r}\right | = \lambda$, with $r \in \mathbb{R}$. Note that if $w$ satisfies the inequality, then so does $\bar{w}$, so the locus is symmetric about the real axis. Let $S = \{ w | \left| {w+r \over w-r}\right | = \lambda \}$.
If $\lambda = 1$, then by writing $|w+r|^2 = |w-r|^2$, it is straightforward to check that $w \in S$ iff $\operatorname{re} w = 0$, that is $S$ is the imaginary axis.
If $\lambda \neq 1$, we have $w \in S$ iff $|w+r|^2 = \lambda^2|w-r|^2$, or equivalently, $|w|^2+r^2 + { 2(1+ \lambda^2) \over 1-\lambda^2} r \operatorname{re} w = 0$.
Note that if $c$ is real then $w$ solves the equation $|w-c| = \rho$ iff $|w|^2+c^2-\rho^2 -2 c\operatorname{re} w = 0$. Comparing, we set $c = -{ 1+ \lambda^2 \over 1-\lambda^2} r$ and $\rho = \sqrt{c^2-r^2} = 2r { |\lambda|\over|1-\lambda^2| }$, hence $w \in S$ iff $|w+{ 1+ \lambda^2 \over 1-\lambda^2} r| = 2r { |\lambda|\over|1-\lambda^2| }$.
Now using $r={1\over 2}|b-a|$ and $e^{i \theta} = {b-a\over |b-a|}$, we can map this back to 'z' to get $C= \{z | |z-z_0| = \rho \}$, where $\rho = 2{ |\lambda|\over|1-\lambda^2| }(b-a)$ and $z_0 = {a+b \over 2} + { \lambda^2+1 \over \lambda^2-1}{b-a \over 2}$.
In terms of the original problem, we have $\lambda = |a|$, and $b= {1 \over \bar{a}}$, substituting these values gives $\rho = 1, z_0 = 0$.
(And if $|a|=1$, the above gives $C = \{ z | \operatorname{re} ((\overline{a-b}) (z- {a+b \over 2})) = 0 \}$.)
• oh man, I didn't see this until now. thanks for your response! – Joe Feb 6 '15 at 0:39
• Glad to be able to help! (This is meant as an addition to PPs. answer.) – copper.hat Feb 6 '15 at 1:02 | 2020-08-06T22:41:16 | {
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https://slow-wine.net/e5otta77/a5217d-set-operations-complement | , and Next lesson. This is the currently selected item. e.g. when we're working with real numbers, probably $$U=\mathbf{R}$$. Some programming languages have sets among their builtin data structures. For example, suppose we have some set called “A” with elements 1, 2, 3. Specification • Describes logical/abstract level. Hence, A - B = { x | x ∈ A AND x ∉ B }. This is called the complement, and it is used for the set difference when the first set is the universal set. A A Definition : The union of sets A and B, denoted by A B, is the set defined as PREVIEW ACTIVITY $$\PageIndex{1}$$: Set Operations. These programming languages have operators or functions for computing the complement and the set differences. 10 Value. In the LaTeX typesetting language, the command \setminus[8] is usually used for rendering a set difference symbol, which is similar to a backslash symbol. View Set Operations _ Union _ Intersection _ Complement _ Difference _ Mutually Exclusive _ Partitions _ from DEVELOPMEN 14740 at St. John's University. More specifically, A'= (U - A) where Uis a universal set that contains all objects. > OPERATIONS ON SETS > Complement of a Set. A vector of the same mode as x or y for setdiff and intersect, respectively, and of a common mode for union. Example: Let A = {1, 3, 5, 7, 9} and B = { 2, 4, 6, 8} A and B are disjoint sets since both of them have no common elements. [1] Other notations include Be able to draw and interpret Venn diagrams of set relations and operations … 4 CS 441 Discrete mathematics for CS M. Hauskrecht Equality Definition: Two sets are equal if and only if they have the same elements. Here four basic operations are introduced and their properties are discussed. Example − If we take two sets A = { a, b } and B = { 1, 2 }, The Cartesian product of A and B is written as − A × B = { (a, 1), (a, 2), (b, 1), (b, 2)}, The Cartesian product of B and A is written as − B × A = { (1, a), (1, b), (2, a), (2, b)}, Minimum operations required to set all elements of binary matrix in C++, Minimum operations to make the MEX of the given set equal to x in C++, Data Structures Stack Primitive Operations. https://edudelighttutors.com/2020/10/14/sets-collection-element-member Producing the complementary relation to R then corresponds to switching all 1s to 0s, and 0s to 1s for the logical matrix of the complement. Hence A satisfies the conditions for the complement of . {\displaystyle A'} One sort of difference is important enough to warrant its own special name and symbol. Remember the universal set F with the elements {2, 4, 6, 8, 10, 12}? Example− If A = { x | x belongs to set of odd integers } then A' = { y | y does not belong to set of odd integers } In set theory, the complement of a set A , often denoted by {\displaystyle A^{c}} .[5]. Complement of a Set ☼ Complement of a Set : Let A be a subset of the universal set U, then the complement of A, denoted by Aٰ or A is defined by : Aٰ = A = { x : x U, x A }. Together with composition of relations and converse relations, complementary relations and the algebra of sets are the elementary operations of the calculus of relations. Complement of set A is the set of all elements in the universal set U which are not in A. Scroll down the page … The complement of A, denoted by , is the complement of A with respect to U (which is U-A). Set Operations include Set Union, Set Intersection, Set Difference, Complement of Set, and Cartesian Product. Set Operations •Let A be the set of students who live within one mile of school and let B be the set … The complement of relation R can be written. Here four basic operations are introduced and their properties are discussed. Bringing the set operations together. Online set theory calculator which helps to find complement of given sets. Complement of Sets Calculator. A set is a collection of items. Without a definition of the universal set, you can't really give a standard-library definition of the complement of a set.. = {x | x A} U A. The following identities capture important properties of absolute complements: Relationships between relative and absolute complements: The first two complement laws above show that if A is a non-empty, proper subset of U, then {A, Ac} is a partition of U. Example − If A = { x | x belongs to set of odd integers } then A' = { y | y does not belong to set of odd integers }, The Cartesian product of n number of sets A1, A2, ... An denoted as A1 × A2 ... × An can be defined as all possible ordered pairs (x1, x2, ... xn) where x1 ∈ A1, x2 ∈ A2, ... xn ∈ A_n. ... Or you could view this as the relative complement-- I always have trouble spelling things-- relative complement of set B in A. ),[1][2] are the elements not in A.[3]. The set in which the complement is considered is thus implicitly mentioned in an absolute complement, and explicitly mentioned in a relative complement. A Hence, A' = { x | x ∉ A }. Set Operations Complement: The complement of a set A is the set of all elements in the universal set NOT contained in A, denoted A. A Set Operations Complement: The complement of a set A is the set of all elements in the universal set NOT contained in A, denoted Ā. Given a set A, the complement of A is the set of all element in the universal set U, but not in A. A The order of the elements in a set doesn't contribute ... Complement of a Set Given: the Universal set and a set, say A To determine: the complement of set A, cardinality of the complement… "Complement (set) Definition (Illustrated Mathematics Dictionary)", https://en.wikipedia.org/w/index.php?title=Complement_(set_theory)&oldid=996544276, Short description is different from Wikidata, Creative Commons Attribution-ShareAlike License, This page was last edited on 27 December 2020, at 06:19. Sometimes the complement is denoted as A' or AC. If Set O {6, 8, 10}, the complement of O (Ō), is {2, 4, 12}. Here, we can see (A - B) ≠ (B - A). Practice: Basic set notation. Python set operations (union, intersection, difference and symmetric difference) Last Updated : 18 Dec, 2017 This article demonstrates different operations on Python sets . Bringing the set operations together. The symbol ∪ is employed to denote the union of two sets. Operations on sets. The relative complement of A with respect to a set B, also termed the set difference of B and A, written B \ A, is the set of elements in B but not in A. A variant \smallsetminus is available in the amssymb package. Abstraction levels: Three levels of abstraction (ADT) o 1. Moreover, the Python set type deals in sets of discrete objects, not a mathematical construct that could be infinitely large, such as all natural numbers. If X ⊆ U, where U is a universal set, then U \ X is called the compliment of X with respect to U. Enter values separated by comma(,) Set A . U Hence, A' = { x | x ∉ A }. PREVIEW ACTIVITY $$\PageIndex{1}$$: Set Operations. ¯ i.e., all elements of A except the element of B. The union of sets A and B (denoted by A ∪ B) is the set of elements that are in A, in B, or in both A and B. Venn diagram and Applications up to 3 Set Problem; SUB TOPIC: SET OPERATONS. If underlying universal set is fixed, then we denote U \ X by X' and it is called compliment of X. Hence, A ∪ B = { x | x ∈ A OR x ∈ B }. If A and B are sets, then the relative complement of A in B,[6] also termed the set difference of B and A,[7] is the set of elements in B but not in A. For example: The intersection of the sets {1, 2, 3} and {2, 3, 4} is {2, 3}. 34. UNION OF SETS: The union of set and is the set which consists of elements that are either in or or both. complement of set ordered pair, ordered n-tuple equality of ordered n-tuples Cartesian product of sets Contents Sets can be combined in a number of different ways to produce another set. It follows that some programming languages may have a function called set_difference, even if they do not have any data structure for sets. Basic set operations. Hence, A ∩ B = { x | x ∈ A AND x ∈ B }. The Complement . Above is the Venn Diagram of A disjoint B. Here are some useful rules and definitions for working with sets When all sets under consideration are considered to be subsets of a given set U, the absolute complement of A is the set of elements in U, but not in A. {\displaystyle \complement _{U}A} In Section 2.1, we used logical operators (conjunction, disjunction, negation) to form new statements from existing statements.In a similar manner, there are several ways to create new sets from sets that have already been defined. 2020/12/9 …s | Union | Intersection | ... Universal set and absolute complement. (or . The truth of aRb corresponds to 1 in row a, column b. These operators may generally be applied also to data structures that are not really mathematical sets, such as ordered lists or arrays. is the set complement of R in X × Y. The complement of a set A (denoted by A’) is the set of elements which are not in set A. ∁ Implementation • (Operation are actually coded. Basic properties of set operations are discussed here. The complement of A is the set of elements of the universal set that are not elements of A. Practice: Basic set notation. The order of the elements in a set doesn't contribute 4 CS 441 Discrete mathematics for CS M. Hauskrecht Equality Definition: Two sets are equal if and only if they have the same elements. Set Complement. The Wolfram Alpha widgets (many thanks to the developers) was used for the Venn Diagram Generator. Such a data structure behaves as a finite set, that is, it consists of a finite number of data that are not specifically ordered, and may thus be considered as the elements of a set. ex) U={integers from 1 to 10} A={3,6,9}, A={1,2,4,5,7,8,10} which are all elements from the In other words, let U be a set that contains all the elements under study; if there is no need to mention U, either because it has been previously specified, or it is obvious and unique, then the absolute complement of A is the relative complement of A in U:[4], The absolute complement of A is usually denoted by ex) U={integers from 1 to 10} A={3,6,9}, A={1,2,4,5,7,8,10} which are all elements from the A 31. The objects or symbols are called elements of the set. The intersection of sets A and B (denoted by A ∩ B) is the set of elements which are in both A and B. Each of union, intersect, setdiff and setequal will discard any duplicated values in the arguments, and they apply as.vector to their arguments (and so in particular coerce factors to character vectors).. is.element(x, y) is identical to x %in% y. In mathematics, a set is a collection of well-defined and distinct objects, where an object is something that is, or can be, formally defined. The set complement operation finds elements that are in one set but not the other. And we're going to talk a lot more about complements in the future. Sal summarizes the set operations that he has discussed in the previous videos. More specifically, A'= (U - A) where U is a universal set that contains all objects. How question) C++ variables: Part 1 Page 5 The complement of A is given by the expression U - A.This refers to the set of all elements in the universal set that are not elements of A. When rendered, the \setminus command looks identical to \backslash, except that it has a little more space in front and behind the slash, akin to the LaTeX sequence \mathbin{\backslash}. Let A and B be two sets in a universe U. Sets - Basic Concepts, Set Operations (Complement, Union and Intersection) 47 mins Video Lesson . ′ It can be applied to implement set complement operation as well: \$ comm -23 <(sort set1) <(sort set2) The relative complement of A with respect to a set B, also termed the set difference of B and A, written B \ A, is the set of elements in B but not in A. Example − If A = { 10, 11, 12, 13 } and B = { 13, 14, 15 }, then (A - B) = { 10, 11, 12 } and (B - A) = { 14, 15 }. Like the domain for quantifiers, it's the set of all possible values we're working with. Example: • {1,2,3} = {3,1,2} = {1,2,1,3,2} Note: Duplicates don't contribute anythi ng new to a set, so remove them. Set Operations •Generalized Intersection •The intersection of a collection of sets is the set that contains those elements that are members of every set in the collection. The complement of a set A (denoted by A’) is the set of elements which are not in set A. The following figures give the set operations and Venn Diagrams for complement, subset, intersect and union. Moreover, the Python set type deals in sets of discrete objects, not a mathematical construct that could be infinitely large, such as all natural numbers. Next lesson. 2 Union ... Complement Let U be the universal set and A be a set. The relative complement of B in A (also called the set-theoretic difference of A and B), denoted by A \ B (or A − B), is the set of all elements that are members of A, but not members of B. Complement is one of the important operations on sets which can be used to find the difference between the universal set and the given set. {\displaystyle A^{c}} Numbers, integers, permutations, combinations, functions, points, lines, and segments are just a few examples of many mathematical objects. {\displaystyle {\overline {A}}} Example − If A = { 11, 12, 13 } and B = { 13, 14, 15 }, then A ∩ B = { 13 }. Adding and Subtracting Matrices Quiz Factoring Trinomials Quiz Solving Absolute Value Equations Quiz Order of Operations Quiz Types of angles quiz. Subset, strict subset, and superset. Clearly, x A x A. e.g. Sometimes the complement is denoted as A‘ or A ∁. That is, x is an element of the intersection A ∩ B, if and only if x is both an element of A and an element of B. Definition : The union of sets A and B, denoted by A B, is the set defined as Let A, B, and C be three sets. Application (user level) • (How the ADT used to solve a problem) o 3. (The common element occurs only once). When all sets under consideration are considered to be subsets of a given set U, the absolute complement of A is the set of elements in U, but not in A . Set ADT has operations as union, intersection, size, and complement. [1], If A is a set, then the absolute complement of A (or simply the complement of A) is the set of elements not in A (within a larger set that is implicitly defined). The Complement . {\displaystyle A'} c If U is a universal set and X is any subset of U then the complement of X is the set of all elements of the set U apart from the elements of X. X′ = {a : a ∈ U and a ∉ A} Venn Diagram: Example: U = {1,2,3,4,5,6,7,8} A = {1,2,5,6} Then, complement of A will be; A’ = {3,4,7,8} Properties of Set Operations… , Universal Set (U) The complementary relation The intersection of two sets A and B, denoted by A ∩ B, is the set of all objects that are members of both the sets A and B.In symbols, ∩ = {: ∈ ∈}. The complement of a set is in relation to the universal set for that problem. It is sometimes written B − A,[1] but this notation is ambiguous, as in some contexts it can be interpreted as the set of all elements b − a, where b is taken from B and a from A. The relative complement of A in B is denoted B ∖ A according to the ISO 31-11 standard. Here, R is often viewed as a logical matrix with rows representing the elements of X, and columns elements of Y. We will look at the following set operations: Union, Intersection and Complement. SET OPERATIONS, VENN DIAGRAMS SET OPERATIONS Let U = {x|x is an English-language film} Set A below contains the five best films according to the American Film Institute. Set Operations Complement: The complement of a set A is the set of all elements in the universal set NOT contained in A, denoted A. The following identities capture notable properties of relative complements: A binary relation R is defined as a subset of a product of sets X × Y. The complement of a set is everything not in the set, but part of the 'universal set'. Example − If A = { 10, 11, 12, 13 } and B = { 13, 14, 15 }, then A ∪ B = { 10, 11, 12, 13, 14, 15 }. {\displaystyle \complement A} Complement of Set. Venn diagram, invented in 1880 by John Venn, is a schematic diagram that shows all possible logical relations between different mathematical sets. Example: • {1,2,3} = {3,1,2} = {1,2,1,3,2} Note: Duplicates don't contribute anythi ng new to a set, so remove them. Without a definition of the universal set, you can't really give a standard-library definition of the complement of a set.. The complement of a set is everything not in the set, but part of the 'universal set'. When doing set operations we often need to define a universal set, $$U$$. This is called the complement, and it is used for the set difference when the first set is the universal set. {\displaystyle {\bar {R}}} ¯ [Example] ={integers from 1 to 10} N={3,6,9},N̄={1,2,4,5,7,8,10} which are all elements from the universal set … I used the AJAX Javascript library for the set operations. The difference between sets is denoted by ‘A – B’, which is the set containing elements that are in A but not in B. ∁ Thus, the set A ∪ B —read “ A union B ” or “the union of A and B ”—is defined as the set that consists of all elements belonging to either set A or set B (or both). But the complement is … 1. In Section 2.1, we used logical operators (conjunction, disjunction, negation) to form new statements from existing statements.In a similar manner, there are several ways to create new sets from sets that have already been defined. It refers as A c, A', A-Complement Set Theory. The complement of A is given by the expression U - A.This refers to the set of all elements in the universal set that are not elements of A. We would write this as: Often not explicitly defined, but implicit based on the problem we're looking at. c Set operations can be used to combine sets. In some cases, the elements are not necessary distinct, and the data structure codes multisets rather than sets. ,[3] Set operations Two sets can be combined in many different ways. Hence . Set Operations: Union, Intersection, Complement, and Difference. ′ Sometimes the complement is denoted as A' or AC. Quantifiers, it 's the set which consists of elements which are not necessary,... U \ x by x ' and it is used for the set of which! Used the AJAX Javascript library for the set the union of sets: the union set... An Absolute complement, and it is called compliment of x, and A... Languages have operators or functions for computing the complement of A, column B, 2,,... 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Trinomials Quiz Solving Absolute set operations complement Equations Quiz Order of operations Quiz Types of Quiz! | 2021-04-14T14:22:19 | {
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https://prob140.org/textbook/content/Chapter_02/06_Exercises.html | 2.6. Exercises¶
1. Often, the axioms and rules derived in Sections 2.1 and 2.3 can be combined in different ways to find the same probability.
(a) Derive the inclusion-exclusion formula for two events:
For any two events $$A$$ and $$B$$, $$P(A \cup B) ~ = ~ P(A) + P(B) - P(AB)$$.
The formula’s name reflects the fact that you include the chance of each event and then exclude the chance of the intersection.
(b) A die is rolled twice. Find the chance that at least one of the rolls shows six spots, in each of the three ways below.
(i) By listing the outcomes and counting how many are in the event in question
(ii) By using Part (a)
(iii) By the complement rule
2. Cell phone use is spreading widely but in many places there is still a divide. For example, among adults in India in 2018,
• 30% did not own a cell phone
• 17% were non-users in that they neither owned nor shared a cell phone
• Among the non-users, 31% said they would like to get a cell phone in the future
Suppose you picked an adult at random in India in 2018.
(a) What is the chance that the selected person was a non-user who didn’t say they would like to get a cell phone in the future?
(b) Given that the selected person didn’t own a cell phone, what is the chance that they shared one?
3. A True/False test consists of 25 questions. A student knows the answers to 18 of the questions. The remaining 7 answers they guess at random by tossing a fair coin each time. If it lands heads they answer True and if it lands tails they answer False.
One of the 25 questions is picked at random. Given that the student got the right answer, what is the chance that they knew the answer?
4. A poker hand is 5 cards dealt at random without replacement from a standard deck of 52 cards.
(a) How many 5-card poker hands are there?
(b) I play poker two times. Make the reasonable assumption that the hand I get the second time is unaffected by the first. What is the chance I get the same hand both times?
5. There are three boxes, each with two drawers. Box 1 has a gold coin in each drawer and Box 2 has a silver coin in each drawer. Box 3 has a silver coin in one drawer and a gold coin in the other. One box is chosen at random, and then a drawer is chosen at random from the box. Find the probability that Box 1 is chosen, given that the chosen drawer yields a gold coin.
6. A standard deck consists of 13 cards in each of 4 suits: clubs, spades, hearts, diamonds. Four cards are dealt at random without replacement from the 52 cards. Find the chance that at least one of the suits doesn’t appear.
7. A procedure for estimating a parameter is based on random sampling and has a 95% chance of producing a good estimate. Suppose I run this procedure $$n$$ times such that the results of all the different runs have no effect on each other.
(a) What is the chance that at least one of the $$n$$ estimates is not good?
(b) About how big does $$n$$ have to be for the chance in Part (a) to be 50%?
8. I have two dice. Die I has 1 red face and 5 blue faces. Die II has 3 red faces and 3 blue faces. I pick one of the dice at random and roll it twice.
(a) What is the chance that I roll Die II?
(b) Given that I see both colors, is the chance that I rolled Die II the same as the answer to (a), greater than the answer to (a), or less than the answer to (a)? Without calculation, pick the correct option and explain your choice.
(c) Now do the calculation. Given that I saw both colors, what is the chance that I rolled Die II? Is your numerical answer consistent with your answer to (b)?
(d) What is the chance that I see a red face on the first roll?
(e) What is the chance that I see a blue face on the second roll?
(f) True or false (explain):
\begin{split} \begin{align*} &P(\text{red on the first roll and blue on the second roll}) \\ &= P(\text{red on the first roll}) \times P(\text{blue on the second roll}) \end{align*} \end{split}
9. Assume the additivity axiom $$P(A \cup B) = P(A) + P(B)$$ if $$A$$ and $$B$$ are mutually exclusive. Show by induction that for integers $$n \ge 1$$,
$P(\cup_{i=1}^n A_i) ~ = ~ \sum_{i=1}^n P(A_i) ~~~ \text{if } A_1, A_2, \ldots, A_n \text{ are mutually exclusive}$
10. The multiplication rules says $$P(AB) = P(A)P(B \mid A)$$. Show by induction that for integers $$n \ge 1$$,
$P(A_1A_2 \cdots A_n) ~ = ~ P(A_1)P(A_2 \mid A_1)P(A_3 \mid A_1A_2) \cdots P(A_n \mid A_1A_2 \cdots A_{n-1})$
11. Eight rooks are placed at positions sampled randomly without replacement on an $$8\times8$$ chessboard. Two rooks attack each other if they are in the same row or in the same column. What is the probability that none of them attacks any of the others?
12. The Statistics Department used to have a “birthday cake” tradition: once a month, there would be a cake to celebrate the birthdays of all department members whose birthdays were in that month. Suppose this tradition continues and suppose the department has $$n$$ members. Assume that each member’s birthday is equally likely to be one of the 12 months of the year, independently of all others.
Consider one calendar year (January through December), starting in January.
(a) What is the probability that no birthday cake will be needed after September?
(b) What is the probability that there will be birthday cake in September but not after that?
13. Down’s Syndrome is a chromosomal disorder (a type of unusual genetic condition) that is associated with maternal age: older mothers are more likely to have babies with this syndrome than younger mothers are. The authors of an article on this syndrome in the US define “births to older mothers” as births in which the mother’s age is at least 35 years, and “births to younger mothers” as births in which the mother is younger than 35. The authors make three statements about births in the population they studied.
• 1 in 700 babies is born with Down’s Syndrome.
• Among births to older mothers, Down’s Syndrome appears more 4 times more frequently than it does among births to younger mothers.
• Among births with Down’s Syndrome, younger mothers appear 4 times more frequently than older mothers.
(a) Use the results of this chapter to explain why the last two statements don’t contradict each other.
(b) If possible, find the proportion of births to older mothers in this population. If this is not possible, explain why not.
(c) If possible, find the proportion older mothers among births without Down’s Syndrome.
14. I toss $$n$$ coins. You toss $$n+1$$ coins. What is the chance that you get more heads than I do?
[Hint: You don’t need any “$$n$$ choose $$k$$”s for this. Example 2.2.3 has a helpful idea.] | 2021-05-12T01:02:33 | {
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https://math.stackexchange.com/questions/1814376/minimal-polynomial-of-n-by-n-matrix | # minimal polynomial of $n$ by $n$ matrix
I want to know if $x^{n-1}-1$ can be a minimal polynomial of n by n matrix with real entries. If yes, find the matrix. If no, why so? I am thinking that when we subtract something from diagonal element and expand the matrix to get characteristic polynomial, the eigenvalue must be in the expression. So the minimal polynomial in the question gives roots of unity as eigenvalue, which are complex. So no such matrix. Am I correct?
• Just where those roots of unity came from? There is no unity in $x^{n-1}=0$. Or did you mean $x^{n-1}=1$? That would be quite a different story. – Ivan Neretin Jun 5 '16 at 10:54
• @Ivan neretin I meant the second case. I have to edit my question. – low iq Jun 5 '16 at 10:57
• Fine. Now consider the matrix $\left(\begin{matrix}0&1&0 \\ 1&0&0 \\ 0&0&1 \\ \end{matrix}\right)$. Isn't it the example for $n=3$? – Ivan Neretin Jun 5 '16 at 10:58
• @Ivan neretin what would be the matrix for first case? – low iq Jun 5 '16 at 10:58
• For the first case, consider $\left(\begin{matrix} 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&0 \\ 0&0&0&0 \\ \end{matrix}\right)$. – Ivan Neretin Jun 5 '16 at 11:01
Any monic polynomial can be the minimal polynomial of a matrix. If $$p(x) = a_0 +a_1x + \cdots + a_{n-1}x^{n-1} + x^n.$$
then $p$ is the minimal and characteristic polynomial of the $n \times n$ matrix $$C(p) = \begin{pmatrix} 0 & 0 & \dots & 0 & -a_0\\ 1 & 0 & \dots & 0 & -a_1\\ 0 & 1 & \dots & 0 & -a_1\\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \dots & 1 & -a_{n-1}\\ \end{pmatrix}_.$$
$C(p)$ is called the "Companion Matrix" of $p(x)$.
I just realized that your question asked about $x^{n-1}-1$ as the minimal polynomial of a matrix of size $n$. In that case you can take a block diagonal matrix:
$$\begin{pmatrix} 1 & 0 \\ 0 & C(x^{n-1}-1)\\ \end{pmatrix}$$
To say that a matrix $A$ satisfies the polynomial $x^{n-1}-1$ means that $A^{n-1}-I = \textbf{0}$. Where $I$ denotes the identity matrix and $\textbf{0}$ denotes the zero matrix.
$\textbf{Example:}$ Let $p(x) = x^3 + 2x^2 - x + 4$. Then $$C(p) = \begin{pmatrix} 0 & 0 & -4\\ 1 & 0 & 1 \\ 0 & 1 & -2\\ \end{pmatrix}_.$$
To see that $C(p)$ satisfies $p(x)$,
\begin{align*} C(p)^3 &= \begin{pmatrix} -4 & 8 & -20\\ 1 & -6 & 13\\ -2 & 5 & -16\\ \end{pmatrix}\\ C(p)^2 &= \begin{pmatrix} 0 & -4 & 8 \\ 0 & 1 & -6 \\ 1 & -2 & 5\\ \end{pmatrix} \end{align*}
So \begin{align*} p(C(p)) &= C(p)^3 + 2 C(p)^2 - C(p) + 4I\\ &= \begin{pmatrix} -4 & 8 & -20\\ 1 & -6 & 13\\ -2 & 5 & -16\\ \end{pmatrix} + 2 \begin{pmatrix} 0 & -4 & 8 \\ 0 & 1 & -6 \\ 1 & -2 & 5\\ \end{pmatrix} - \begin{pmatrix} 0 & 0 & -4\\ 1 & 0 & 1 \\ 0 & 1 & -2\\ \end{pmatrix} + \begin{pmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \\ \end{pmatrix} \\ &= \begin{pmatrix} -4 & 8 & -20\\ 1 & -6 & 13\\ -2 & 5 & -16\\ \end{pmatrix} + \begin{pmatrix} 0 & -8 & 16 \\ 0 & 2 & -12 \\ 2 & -4 & 10\\ \end{pmatrix} + \begin{pmatrix} 0 & 0 & 4\\ -1 & 0 & -1 \\ 0 & -1 & 2\\ \end{pmatrix} + \begin{pmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \\ \end{pmatrix}\\ &= \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}_. \end{align*}
• can you give one easy example of polynomial and it's companion matrix, so that I can learn how to write companion matrix – low iq Jun 5 '16 at 15:58
You want an $n$ by $n$ matrix with minimal polynomial $x^{n-1} -1$. It is sufficient (and necessary if $n$ is even) that the matrix has the invariant factors: $x-1, x^{n-1} - 1$. This means that all such matrices must have the rational canonical form: $C(x -1) \oplus C(x^{n-1} - 1)$ (which means that we can take this matrix as an example), where $C(p(x))$ is the Fröbenius Companion Matrix of a monic polynomial.
For instance, if $n=5$, you get the following:
$$\begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{bmatrix}$$
• but how do you write it? Trial and error? Or some rule/logic? – low iq Jun 5 '16 at 11:58
• @lowiq of course not trial and error; that's lame. This is based on theorems. I might expand my answer later, but if you are interested, you may want to read about the invariant factors and rational canonical forms of matrices – user258700 Jun 5 '16 at 12:05
• In Wikipedia, it is written that companion matrix has minimal polynomial P.what does it mean? Minimal polynomial should satisfy the matrix. How does p satisfies the matrix? – low iq Jun 5 '16 at 12:20 | 2021-06-16T13:03:29 | {
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https://ryanwingate.com/visualization/common-types/box-plots/ | # Box Plots
A box plot (or, box and whisker plot) is a method of showing aggregate statistics of various samples in a concise manner. For each sample, it simultaneously shows:
• median of each sample,
• minimum and maximum of the samples, and the
• interquartile range.
The following code block creates three different samplings from numpy. A normal distribution, a random distribution, and a gamma distribution.
%matplotlib notebook
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
normal_sample = np.random.normal(loc=0.0, scale=1.0, size=10000)
random_sample = np.random.random(size=10000)
gamma_sample = np.random.gamma(2, size=10000)
df = pd.DataFrame({'normal': normal_sample,
'random': random_sample,
'gamma': gamma_sample})
df.describe()
normal random gamma
count 10000.000000 10000.000000 10000.000000
mean 0.005574 0.503947 1.980398
std 1.003427 0.289082 1.400785
min -4.013316 0.000017 0.014506
25% -0.666291 0.252406 0.947084
50% -0.004336 0.507529 1.671177
75% 0.684656 0.751771 2.687132
max 3.474749 0.999931 12.756334
The summary statistics shown above require splitting the data into four quarters. The first quarter is between the minimal value and the first 25% of the data. That first 25% is called the first quartile. The seconda nd third quarters of the data are between the first quartile and the 75% mark, which is called the third quartile. The finnal quarter of the data is between the third quartile and the maximum.
The interquartile range is between the first and third quartiles.
In the box plots shown below the:
• interquartile range is the boxed region, the
• maximum and minimum values are the horizontal lines, and the
• median is the red line in the center of the boxed region.
It is also possible to specify more limited ranges for the whiskers. This is changeable via the whis parameter to the function. The ‘range’ parameter means use the maximum and minimum of the data.
plt.figure()
plt.boxplot([ df['normal'],
df['random'],
df['gamma'] ],
whis='range');
<IPython.core.display.Javascript object>
If the whis parameter is left out, the top whisker defaults to reaching to the last datum less than Q3 + 1.5*IQR, where IQR represents the interquartile range.
Plots like the one shown below are good for detecting outliers. The datapoints beyond the whiskers are called “fliers.”
plt.figure()
plt.boxplot([ df['normal'],
df['random'],
df['gamma'] ]);
<IPython.core.display.Javascript object> | 2023-03-31T16:43:20 | {
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https://math.stackexchange.com/questions/2717462/simulating-a-binomial-distribution-with-mathscru0-1 | # Simulating a Binomial distribution with $\mathscr{U}(0,1)$.
Ex:Show the procedures to simulate an random variable that follows a binomial distribution with parameter $p$, using the $\mathscr{U}(0,1)$(Uniform distribution on the interval (0,1)).
I tried to solve this question by using the following theorem:
Theorem: Let $U\sim\mathscr{U}(0,1)$. Let $X$ be a random variable with distribution $F_X(x)$. Therefore the random variable Y=F^{-1}(U) has a distribution function equal to $F_X$, the distribution of the $X$ variable.
According to this theorem I would need to find a the inverse of the binomial c.d.f, define it as a function in python and generate random numbers.
However I have no idea on how to invert the Binomial distribution.
Questions:
1) Is this the simplest method to simulate a Binomial distribution with the Uniform(0,1)? Are there other methods?
2) How do I compute the inverse of the Binomial distribution?
• Simulating $n$ iid Bernoulli variables is easier. – GNUSupporter 8964民主女神 地下教會 Apr 1 '18 at 14:40
• @GNUSupporter Indeed, but this requires generating $n$ uniform random variates. The problem doesn't specify whether this is allowed :) – Math1000 Apr 1 '18 at 15:18
• @Math1000 When one can simulate one uniform random variable, one can also simulate $n$ copies. If you insist on generating one single variable and use the inversion theorem, you'll have to compute a discrete sum $\sum_{i = 1}^k p^i (1-p)^{n-i}$ in order to find out $P(X \le k)$. I don't think it's a sensible simulating algo. – GNUSupporter 8964民主女神 地下教會 Apr 1 '18 at 15:25
It may be helpful to see how this procedure for random sampling from $\mathsf{Binom}(n=5, p=.4)$ is implemented in R statistical software. First, some R notation: runif (without extra parameters) is a source of pseudorandom values from standard uniform; dbinom, pbinom and qbinom denote binomial PDF, CDF, and quantile funcions (invese CDF) respectively.
So in R we can generate $m = 100,000$ observations from $\mathsf{Binom}(n=5, p=.4)$ in a vector xas follows.
set.seed(4118)
m = 10^5; u = runif(m)
x = qbinom(u, 5, .4) # inverse CDF transformation
Then we can tally the results, make a histogram of them, and plot exact PDF values on the histogram for comparison:
table(x)/m
x
0 1 2 3 4 5
0.07790 0.25775 0.34608 0.23105 0.07744 0.00978
hist(x, prob=T, br=(0:6)-.5, col="skyblue2", main="100,000 Realizations of BINOM(5,.5)")
k = 0:5; pdf = dbinom(k, 5, .4)
points(k, pdf, col="red")
Within the accuracy of the graph, it is not possible to distinguish the simulated results (heights of histogram bars) from the theoretical ones (small red circles).
Finally, by using the same seed for the pseudoranom generator as above, we can access exactly the same values u as above. Thus, we can see that R implements this (inverse CDF) method to generate $m$ observations from $\mathsf{Binom}(n=5, p=.4)$ by using the function rbinom defined in R. The tallied results are exactly the same below as above.
set.seed(4118)
m = 10^5; x = rbinom(m, 5, .4)
table(x)/m
x
0 1 2 3 4 5
0.07790 0.25775 0.34608 0.23105 0.07744 0.00978
Note: By contrast, when $p >.5,$ R uses a slight modification of the inverse CDF method, so that the two approaches give slightly different results. (I used $m = 10,000$ so that differences would be more obvious.)
# Inverse CDF method
set.seed(401); m = 10^4; u = runif(m); x1 = qbinom(u, 5, .7)
table(x1)/m
x1
0 1 2 3 4 5
0.0022 0.0296 0.1323 0.3076 0.3651 0.1632
# Built-in function 'rbinom'
set.seed(401); x2 = rbinom(m, 5, .7)
table(x2)/m
x2
0 1 2 3 4 5
0.0023 0.0278 0.1288 0.3126 0.3599 0.1686
Addendum 1: Graphs of CDF of $X \sim \mathsf{Binom}(5, .4)$ and its inverse function. The latter shows $F_X^{-1}(u) = 0,$ for $u < 0.6^5=0.07776,$ as in a Comment.
Addendum 2: Generating $X \sim \mathsf{Binom}(5, .4)$ as the sum of five independent Bernoulli random variables with $p=.4.$ [This is the method suggested in the Comment by @GNUSupporter.]
First let $U_1, U_2, \dots, U_5$ be a random sample from $\mathsf{Unif}(0,1).$ Then $B_i = 1,$ if $U_i \le .4$ and $0$ otherwise. This is essentially a trivial application of the quantile method to a variable that takes only values $0$ and $1$. Then $X = \sum_{i=1}^5 B_i \sim \mathsf{Binom}(5, .4).$ We generate four such binomial random variables below (results: 1, 2, 2, 4). Notice that five pseudorandom uniform values are required for each binomial.
set.seed(1234)
u = runif(5); b = (u < .4); x = sum(b); x # sum of logical vec b is nr of its TRUEs
[1] 1
u = runif(5); b = (u < .4); x = sum(b); x
[1] 2
u = runif(5); b = (u < .4); x = sum(b); x
[1] 2
u = runif(5); b = (u < .4); x = sum(b); x
[1] 4
Next we use a for loop to iterate this procedure $m = 100,000$ times. Because we start with the same seed as above, the first four iterations repeat the realizations of $\mathsf{Binom}(5, .4)$ shown just above. A tally of all $m$ results shows results similar to those in the initial simulation of this Answer, closely matching the target distribution.
set.seed(1234); m = 10^5; x = numeric(m)
for (i in 1:m) {
u = runif(5); b = (u < .4); x[i] = sum(b) }
mean(x); x[1:4]
[1] 1.99806 # aprx E(X) = 5(.4) = 2
[1] 1 2 2 4 # same first four realizations of X as above
table(x)/m
x
0 1 2 3 4 5
0.07766 0.25971 0.34683 0.22878 0.07675 0.01027
• Thanks for your answer! I think what I am asked to do is to generate or simulate a Binomial distribution from a Uniform distribution. I can easily using the theorem provided in my post to simulate an exponential function by inverting by hand its expression, then generating random numbers ,defining the inverse exponential as function on Python. However for distributions that are difficult to reverse such as this. I do not know how to proceed since I am not allowed to use commands that already relate to the distribution(on this case Binomial). – Pedro Gomes Apr 1 '18 at 18:26
• Is there a way to generate it given the requirements? – Pedro Gomes Apr 1 '18 at 18:27
• I guess you'll have to use a sequence of if-then statements as implied by @NCh's Answer, which 'inverts the binomial CDF'. Explicitly, for X~BINOM(5, .4), you have $P(X = 0) = P(X \le 0) = .6^5 = 0.07776.$ So take $X = 0$ if $U \le 0.07776,$ And so on. Hope this helps. (You're not working a 'toy' exercise, but learning about something that is actually used. I was not trying to add another answer for your question, but to discuss implementation of the idea in software.) – BruceET Apr 1 '18 at 18:46
• Directly to your two questions: (1) Inverse CDF is a simple method. Context may determine what is 'simplest'. Typically software will already have programmed a quantile function, so using it is trivial. Starting from scratch @GNU may be right that it's easier to program $n$ Bernoulli RVs and add. (2) As mentioned in my previous comment, NCh has shown you how to invert the binomial CDF (it seems, without explicitly using those words). // I'm off to brunch, will check back later. – BruceET Apr 1 '18 at 18:58
• See two addenda to my Answer. Hope they are helpful. – BruceET Apr 2 '18 at 1:26
For the second question: If $U\leq (1-p)^n$, set $Y=0$.
If $(1-p)^n< U\leq (1-p)^n+np(1-p)^{n-1}$, set $Y=1$.
If $(1-p)^n+np(1-p)^{n-1}< U\leq(1-p)^n+np(1-p)^{n-1}+\binom{n}{2}p^2(1-p)^{n-2}$, set $Y=2$.
And so on.
$$Y=k \iff \sum_{i=0}^{k-1} \binom{n}{i}p^i(1-p)^{n-i} < U \leq \sum_{i=0}^{k} \binom{n}{i}p^i(1-p)^{n-i}.$$
On your first question: as GNU Supporter noted in comments, this is not the simplest way to generate binomial distribution.
• What do you think about question (1)? Perhaps you won't code this in a real simulation. – GNUSupporter 8964民主女神 地下教會 Apr 1 '18 at 15:54
• The answer to this question is obvious and is given in your comment above. Thank you, I'll add this to the answer. – NCh Apr 1 '18 at 15:59
• Thanks for reply. I just want your viewpoint due Math1000's comment. – GNUSupporter 8964民主女神 地下教會 Apr 1 '18 at 16:01
• But simulating a Bernoulli r.v. is also "inverting" its CDF. Anyways, thanks for your opinion. – GNUSupporter 8964民主女神 地下教會 Apr 1 '18 at 16:11
• There exist probability spaces on which a binomial-distributed random variable exists but cannot be written as the sum of independent $\{0,1\}$-valued random variables. See here math.stackexchange.com/questions/1920276/… – Math1000 Apr 1 '18 at 17:33 | 2019-05-22T15:04:10 | {
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http://math.stackexchange.com/questions/400864/solving-simple-mixed-fraction-problem | # Solving Simple Mixed Fraction problem?
How do you wrap your head around mixed fraction, does anyone knows how to figure out, can someone give me an example how it can be solved?
-
Could you be more specific about what you mean when you say "how it can be solved?" Do you mean how can a number like $3\frac58$ can be expressed as a fraction (not mixed)? – amWhy May 24 '13 at 3:23
PLease give an example of what do you want to solve.So that it will be easy to explain – sammath May 24 '13 at 3:25
When I mean "how it can be solved" I want to know from how the numbers comes together from start to finish in an easy logical way to form it together in a way I can understand? – Marth Neo May 24 '13 at 3:35
Are either of the answers below helpful in that way? – amWhy May 24 '13 at 3:40
It is, Now the answers is flowing to me clearly now. – Marth Neo May 24 '13 at 3:42
$$a +\frac bc = \frac {a\times c}c + \frac bc = \frac{(a\times c) + b}{c}$$
What we do first is like when finding a common denominator between two fractions, only in $a$ above, we have $a = \dfrac a1$:
We multiply $a = \dfrac a1$ by $\dfrac cc = 1$, to get $\dfrac a1 \times \dfrac cc = \;\dfrac{a\times c}{c}\;$ and then add that to the fraction $\dfrac bc$.
For example
$$3 \frac 58\; = \;3 + \frac 58\; = \;\frac{3 \times 8}{8} + \frac 58\; = \;\frac{(3\times 8)+ 5}{8}\; = \;\frac{24 + 5}{8} \;= \;\frac{29}{8}$$
-
Good call, you did the general case followed by an example +1 – Amzoti May 24 '13 at 12:03
Here is an example.
$$2 \frac{3}{11}=\frac{2 \cdot 11}{11}+\frac3{11}=\frac{22}{11}+\frac3{11}=\frac{25}{11}$$
Or conversely,
$$\frac{11}3=\frac{11-3 \cdot 3}{3}+\frac{3 \cdot 3}{3}=\frac{11-9}{3}+\frac{9}{3}=\frac{2}{3}+3=3 \frac{2}3$$.
In general,
$$x+\frac{y}{z}=\frac{x \cdot z}{z}+\frac{y}{z}=\frac{x \cdot z +y}z$$ and vice versa.
In most cases you'll encounter $y<z$.
- | 2014-08-28T13:27:29 | {
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# The eSoroban device is available in two colors, orange and green.
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The eSoroban device is available in two colors, orange and green. [#permalink]
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25 May 2019, 13:07
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The eSoroban device is available in two colors, orange and green. In 2013, 60% of the eSoroban devices sold were purchased by women, $$\frac{5}{12}$$ of whom purchased the orange device. If an equal number of orange and green eSoroban devices were sold in 2013, what fraction of men who purchased an eSoroban in 2013 purchased the green device?
A. $$\frac{3}{10}$$
B. $$\frac{3}{8}$$
C. $$\frac{5}{12}$$
D. $$\frac{1}{2}$$
E. $$\frac{5}{7}$$
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Re: The eSoroban device is available in two colors, orange and green. [#permalink]
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25 May 2019, 14:41
1
Let the total number of devices sold= lcm(5,12)=60
Orange devices=30, Green devices=30
Devices sold to women= (3/5)*60=36
orange devices sold to women= (5/12)*36= 15
green devices sold to women= 36-15=21
green devices sold to men=30-21=9
devices sold to men= 60-36=24
fraction of men who purchased an eSoroban in 2013 purchased the green device=9/24=3/8
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Re: The eSoroban device is available in two colors, orange and green. [#permalink]
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26 May 2019, 05:37
nick1816 wrote:
Let the total number of devices sold= lcm(5,12)=60
Orange devices=30, Green devices=30
Devices sold to women= (3/5)*60=36
orange devices sold to women= (5/12)*36= 15
green devices sold to women= 36-15=21
green devices sold to men=30-21=9
devices sold to men= 60-36=24
fraction of men who purchased an eSoroban in 2013 purchased the green device=9/24=3/8
Question here.
Where in the problem state that the product is produced equally in two colors?
There could have been more green and vice versa, right?
So how did you know it was a 50/50 split?
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The eSoroban device is available in two colors, orange and green. [#permalink]
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26 May 2019, 06:05
The eSoroban device is available in two colors, orange and green. In 2013, 60% of the eSoroban devices sold were purchased by women, 5/12 of whom purchased the orange device. If an equal number of orange and green eSoroban devices were sold in 2013, what fraction of men who purchased an eSoroban in 2013 purchased the green device?
Question here.
Where in the problem state that the product is produced equally in two colors?
There could have been more green and vice versa, right?
So how did you know it was a 50/50 split?[/quote]
Posted from my mobile device
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Re: The eSoroban device is available in two colors, orange and green. [#permalink]
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11 Aug 2019, 01:19
1
energetics wrote:
The eSoroban device is available in two colors, orange and green. In 2013, 60% of the eSoroban devices sold were purchased by women, $$\frac{5}{12}$$ of whom purchased the orange device. If an equal number of orange and green eSoroban devices were sold in 2013, what fraction of men who purchased an eSoroban in 2013 purchased the green device?
A. $$\frac{3}{10}$$
B. $$\frac{3}{8}$$
C. $$\frac{5}{12}$$
D. $$\frac{1}{2}$$
E. $$\frac{5}{7}$$
org green Total
Men 25 15 40
women 25 35 60
Total 50 50 100
Required Fraction: 15/40 = 3/5
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The eSoroban device is available in two colors, orange and green. [#permalink]
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11 Aug 2019, 02:38
It takes me 9min to sort out:(
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Re: The eSoroban device is available in two colors, orange and green. [#permalink]
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11 Aug 2019, 02:58
energetics wrote:
The eSoroban device is available in two colors, orange and green. In 2013, 60% of the eSoroban devices sold were purchased by women, $$\frac{5}{12}$$ of whom purchased the orange device. If an equal number of orange and green eSoroban devices were sold in 2013, what fraction of men who purchased an eSoroban in 2013 purchased the green device?
A. $$\frac{3}{10}$$
B. $$\frac{3}{8}$$
C. $$\frac{5}{12}$$
D. $$\frac{1}{2}$$
E. $$\frac{5}{7}$$
Let Total People, who purchased devices, be 100.
Therefore Women who purchased devices = 60 (According to question)
and hence Men who purchased devices = 40 (100 - 60) -- (1)
Now $$\frac{5}{12}$$ is the fraction of orange device of devices purchased by women. That is equal to 25
Hence remaining devices, purchased by women, are green and are equal to = 60 - 25 = 35
Now from question we know, Total Green devices = Total Orange devices
Hence 35(Green devices with women) + G(Green devices with men) = 25(Orange devices with women) + O(Orange devices with men)
which we can write as O - G = 10 -- (2)
And also O + G = 40 -- (3) [From (1)]
From (2) and (3), we get
O = 25
G = 15
fraction of men who purchased an eSoroban in 2013 purchased the green device = $$\frac{Green devices purchased by men}{Total Men who purchased eSorban Devices}$$
Hence $$\frac{15}{40}$$
=> $$\frac{3}{8}$$ ------- Answer (B)
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The eSoroban device is available in two colors, orange and green. [#permalink]
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15 Sep 2019, 02:30
energetics wrote:
The eSoroban device is available in two colors, orange and green. In 2013, 60% of the eSoroban devices sold were purchased by women, $$\frac{5}{12}$$ of whom purchased the orange device. If an equal number of orange and green eSoroban devices were sold in 2013, what fraction of men who purchased an eSoroban in 2013 purchased the green device?
A. $$\frac{3}{10}$$
B. $$\frac{3}{8}$$
C. $$\frac{5}{12}$$
D. $$\frac{1}{2}$$
E. $$\frac{5}{7}$$
Given:
1. The eSoroban device is available in two colors, orange and green.
2. In 2013, 60% of the eSoroban devices sold were purchased by women, $$\frac{5}{12}$$ of whom purchased the orange device.
3. An equal number of orange and green eSoroban devices were sold in 2013
Asked: What fraction of men who purchased an eSoroban in 2013 purchased the green device?
-----------Orange----------Green--------------Total
Men---------15%-------------25%-------------40%
Women------25%-------------35%----------------60%
Total --------50%------------50%--------------100%
Fraction of men purchasing green device = 15%/40% = 3/8
IMO B
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Re: The eSoroban device is available in two colors, orange and green. [#permalink]
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23 Nov 2019, 03:23
Kinshook wrote:
energetics wrote:
The eSoroban device is available in two colors, orange and green. In 2013, 60% of the eSoroban devices sold were purchased by women, $$\frac{5}{12}$$ of whom purchased the orange device. If an equal number of orange and green eSoroban devices were sold in 2013, what fraction of men who purchased an eSoroban in 2013 purchased the green device?
A. $$\frac{3}{10}$$
B. $$\frac{3}{8}$$
C. $$\frac{5}{12}$$
D. $$\frac{1}{2}$$
E. $$\frac{5}{7}$$
Given:
1. The eSoroban device is available in two colors, orange and green.
2. In 2013, 60% of the eSoroban devices sold were purchased by women, $$\frac{5}{12}$$ of whom purchased the orange device.
3. An equal number of orange and green eSoroban devices were sold in 2013
Asked: What fraction of men who purchased an eSoroban in 2013 purchased the green device?
-----------Orange----------Green--------------Total
Men---------15%-------------25%-------------40%
Women------25%-------------35%----------------60%
Total --------50%------------50%--------------100%
Fraction of men purchasing green device = 15%/40% = 3/8
IMO B
But in the vertical columns of both orange and green the addition becomes
Orange = 15% + 25% = 40%, which is not equal to 50%
Same goes for green devices?
Can you explain?
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Re: The eSoroban device is available in two colors, orange and green. [#permalink] 23 Nov 2019, 03:23
Display posts from previous: Sort by | 2019-12-06T20:07:43 | {
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https://www.jiskha.com/questions/1439387/a-bag-contains-marbles-of-three-different-colours-red-blue-yellow-2-5-of-the-marbles | # Maths
A bag contains marbles of three different colours red, blue , yellow. 2/5 of the marbles are red. The ratio of the number of blue marbles to the number of yellow marbles is 5:7. What fraction of the marbles in the bag is blue? There are 54 more red marbles than blue marbles. How many marbles are there in the bag?
1. 👍
2. 👎
3. 👁
1. let the total be x
(2/5)x = r , also b+54 = (2/5)x
b = 2x/5 - 54 = (2x - 270)/5
b : y = 5 : 7
5y = 7b
y = (7/5)b
y = 7(2x-270)/25
so 2x/5 + (2x-270)/5 + 7(2x-270)/25 = x
times 25
10x + 10x - 1350 + 14x - 1890 = 25x
-9x = -3240
x = 360
we have 510 marbles
check:
red = (2/5)(360) = 144
b = (720-270)/5 = 90
y = (7/5)(90) = 126
144+90+126 = 360 , check!
blue : yellow = 90 : 126 = 5 : 7 , check
area there 54 more reds than blues?
144-90 = 54, check!
there are 510 marbles
1. 👍
2. 👎
2. I obviously got x = 360
So there are 360 marbles,
I have no idea where I got the 510 from, and then I copied it again at the end.
Time for my third cup of coffee.
1. 👍
2. 👎
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https://swiftmotionpictures.com/kinaxis-competitors-vrk/measures-of-dispersion-examples-ab8956 | Select Page
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Statology is a site that makes learning statistics easy by explaining topics in simple and straightforward ways. \text{MAD} & = \cfrac {\left\{ |12 – 12|+ |4 – 12| + |23 – 12| + |8 – 12| + |9 – 12| + |16 – 12| \right\}} {6} \\ For example, if the standard deviation is large then there are large differences between individual data points. As the name suggests, the measure of dispersion shows the scatterings of the data. o Understand the difference between measures of dispersion for populations and for samples The median splits the dataset into two halves. They are: 1. Standard deviation. There are four commonly used measures to indicate the variability (or dispersion) within a set of measures. It is usually used in conjunction with a measure of central tendency, such as the mean or median, to provide an overall description of a set of data. Common examples of measures of statistical dispersion are the variance, standard deviation, and interquartile range. The Interquartile Range (IQR) . The table shows marks (out of 10) obtained by 20 people in a test. o Sample standard deviation. Consequently, the mean may not be representative of the data. { S }^{ 2 } & =\frac { \left\{ { \left( 12-12 \right) }^{ 2 }+{ \left( 4-12 \right) }^{ 2 }+{ \left( 23-12 \right) }^{ 2 }+{ \left( 8-12 \right) }^{ 2 }+{ \left( 9-12 \right) }^{ 2 }+{ \left( 16-12 \right) }^{ 2 } \right\} }{ 5 } \\ & = 5\% \\ Let’s start with a funny (and not so realistic) example. o Sample variance. They are usually used in conjunction with measures of central tendency such as the mean and the median. An example of aggregating data is the simple process of finding the mean of a variable such as height or weight. The Important measures of dispersion can represent a series only as best as a single figure can, but it certainly cannot reveal the entire story of any phenomenon under study. 2. Common examples of measures of statistical dispersion are the variance, standard deviation, and interquartile range 4. S & = 0.00452^{\frac {1}{2}} \\ Measures of central dispersion show how “spread out” the elements of a data set are from the mean. For example, suppose we have the following dataset with incomes for ten people: The range is 2,468,000, but the interquartile range is 34,000, which is a much better indication of how spread out the incomes actually are. Compute the sample mean and the corresponding sample variance. This example of one of the relative measures of dispersion is also called as Range Co-efficie… Mark (x) The Range. For every absolute measure of dispersion, there is a relative measure. Dispersion (a.k.a., variability, scatter, or spread)) characterizes how stretched or squeezed of the data. Range R = L –S. It tells the variation of the data from one another and gives a clear idea about the distribution of the data. You’re bored of living on Earth and decide to take off towards another planet. Measures of dispersion In statistics, dispersion (also called variability, scatter, or spread) is the extent to which a distribution is stretched or squeezed. The only important thing for … The measure of dispersion shows the homogeneity or the heterogeneity of the distribution of the observations. First, we have to calculate the arithmetic mean: X =\cfrac {(12 + 4 + 23 + 8 + 9 + 16)}{6} = 12\% $$,$$ \begin{align*} & = 0.003767 \\ Solution Here Largest value L = 28. It is a measure of dispersion that represents the average of the absolute values of the deviations of individual observations from the arithmetic mean. \end{align*} . & =\cfrac {1870}{5} = 374 \\ We often measure the “center” using the mean and median. Example. & = 0.0672 Range 2. { \sigma }^{ 2 } & =\frac { \left\{ { \left( 12-32 \right) }^{ 2 }+{ \left( 13-32 \right) }^{ 2 }+{ \left( 54-32 \right) }^{ 2 }+{ \left( 56-32 \right) }^{ 2 }+{ \left( 25-32 \right) }^{ 2 } \right\} }{ 5 } \\ \end{align*}. Imagine our technology has advanced so much that we can freely travel in space. Variance. This is necessary so as to remove biasThe sample standard deviation, S, is simply the square root of the sample varianceExample 4Assume that the returns realized in example 2 above were sampled from a population comprising 100 returns. (The two have been distinguished here), $${ S }^{ 2 }=\frac { \left\{ \sum { { \left( { X }_{ i }- \bar { X } \right) }^{ 2 } } \right\} }{ n-1 }$$, Note that we are dividing by n – 1. o Population standard deviation. The standard deviation is 0.0037671/2 = 0.06137 or 6.14%. There are different measures of dispersion like the range, the quartile deviation, the mean deviation and the standard deviation. Relative Dispersion The actual variation or dispersion, determine from standard deviation or other measures is called absolute dispersion, now the relative dispersion is For Example, Relative dispersion It is a measurement of the degree by which an observed variable deviates from its … Definition of Various Measures of Dispersion Range. ©AnalystPrep. The study of dispersion will enables us to know whether a series is homogeneous (where all the observations remains around the central value) or the observations is heterogeneous (there will be variations in the observations around the central value like 1, 50, 20, 28 etc., where the central value is 33). Find the median. Smallest value S = 18. $$\mu =\cfrac {(12 + 13 + \cdots +25)}{5} =\cfrac {160}{5} = 32$$, \begin{align*} (In this case, it’s the average of the middle two values), 58, 66, 71, 73, 74, 77, 78, 82, 84, 85 (MEDIAN) 88, 88, 88, 90, 90, 92, 92, 94, 96, 98, 3. Quartiles are values that split up a dataset into four equal parts. The formula to find the variance of a population (denoted as σ2) is: where μ is the population mean, xi is the ith element from the population, N is the population size, and Σ is just a fancy symbol that means “sum.”. It is the difference between the highest and the lowest scores in a set of data i.e. In this case, Q1 is the average of the middle two values in the lower half of the data set (75.5) and Q3 is the average of the middle two values in the upper half of the data set(91). In the above cited example, we observe that. o Measure of dispersion. The median of the lower half is the lower quartile (Q1) and the median of the upper half is the upper quartile (Q3). We’ve started colonizing and populating new planets. o Variance. Thus, \text{MAD} \frac { \sum { |{ X }_{ i }-\bar { X } | } }{ n } $$. You may notice that all the relative measures of dispersion are called coefficients. Usually we work with samples, not populations. Thus, the average variation from the mean (0.12) is 0.003767. They are important because they give us an idea of how well the measures of central tendency represent the data. The rangeis the difference between the largest and smallest value in a dataset. Here is how to find the interquartile range of the following dataset of exam scores: 1. Variance and Standard Deviation. All Rights ReservedCFA Institute does not endorse, promote or warrant the accuracy or quality of AnalystPrep. Slide 77 Measures of Dispersion There are three main measures of dispersion: – The range – The Interquartile range (IQR) – Variance / standard deviation 8. o Degrees of freedom. The smallest value is 58. Mean deviation from median. Specially it fails to give any idea about the scatter of the values of items … (Definition & Example). & = 45.20(\%^2) \\ The minimum number of completions for Quarterback A is 19, the maximum is 37. Working with data from example 2 above, the variance will be calculated as follows:$$ \begin{align*} You’re kind of an adventurous person and you don’t have too many capricious demands regarding where you want to live next. The interquartile range is the difference between the first quartile and the third quartile in a dataset. The formula to find the standard deviation of a population (denoted as σ ) is: And the formula to find the standard deviation of a sample (denoted as s) is: Your email address will not be published. \begin{align*} The scores of all the students in section A are ranging from to ; Example 8.3 The range of a set of data is 13.67 and the largest value is 70.08. For example, when rainfall data is made available for different days in mm, any absolute measures of dispersion give the variation in rainfall in mm. (The two have been distinguished here)S2 = {Σ(Xi – X? This is from the Oxford English Dictionary: The term came to English from the German (where it lived before that I do not know) and seems to have emerged as a way of explaining aggregated data, or data which one has subjected to the process of removing information in order to gain information. Objectives . Suppose we have this dataset of final math exam scores for 20 students: The largest value is 98. Thus, the range is 98 – 58 = 40. In this case, the outlier income of person J causes the range to be extremely large and makes it a poor indicator of “spread” for these incomes. no extremely high salaries), the mean will do a good job of describing this dataset. Analysts use the standard deviation to interpret returns as opposed to the variance since it is much easier to comprehend. The formulae for the variance and standard deviation are given below. Dispersion … )2}/n – 1Note that we are dividing by n – 1. These are pure numbers or percentages totally independent of the units of measurements. \text{Range} = \text{maximum value} – \text{minimum value} $$, Consider the following scores of 10 CFA Level 1 candidates, 78 56 67 51 43 89 57 67 78 50. Third Variable Problem: Definition & Example, What is Cochran’s Q Test? Required fields are marked *. The variance is a common way to measure how spread out data values are. Lets look at the first of the relative measures of dispersion. Measures of Dispersion The Range of a set of data is the largest measurement minus the smallest measurement. A measure of statistical dispersion is a nonnegative real number that is zero if all the data are the same and increases as the data become more diverse. These are also known as ‘Coefficient of dispersion’ 3. Looking for help with a homework or test question? Surprisingly, the term statistic first came into use as late as 1817. 6 Investment analysts attain the following returns on six different investments: Calculate the mean absolute deviation and interpret it. It’s the most common way to measure how “spread out” data values are. In statistics, dispersion (also called variability, scatter, or spread) is the extent to which a distribution is stretched or squeezed. The interquartile range is equal to Q3 – Q1. Your email address will not be published. Quartile Deviation 3. { \sigma }^{ 2 } & =\frac { \left\{ { \left( 12-12 \right) }^{ 2 }+{ \left( 4-12 \right) }^{ 2 }+{ \left( 23-12 \right) }^{ 2 }+{ \left( 8-12 \right) }^{ 2 }+{ \left( 9-12 \right) }^{ 2 }+{ \left( 16-12 \right) }^{ 2 } \right\} }{ 6 } \\ Example 8.2 Find the range of the following distribution. Range R = 13.67 CFA® and Chartered Financial Analyst® are registered trademarks owned by CFA Institute. The smallest value is 58. The concept of relative measures of dispersion overcomes this limitation. . You subtract the lowest score in the data set from the highest score to give the range. & =\cfrac {30}{6} \\ Like, Kilograms, Rupees, Centimeters, Marks etc. and other Percentiles. One such measure is popularly called as dispersion or variation. Suppose we have this dataset of final math exam scores for 20 students: The largest value is 98. When we analyze a dataset, we often care about two things: 1. o Use the variance or standard deviation to characterize the spread of data. (1) Absolute Measures 1. Solution. Unit-II MEASURES OF CENTRAL TENDENCY AND DISPERSION Relation between Mean, Median and Mode: − = 3( – ) Range of ungrouped data: The range of a set of data is the difference between the highest and lowest values in the set. Thus, the range is 98 – 58 =, Thus, the interquartile range is 91 – 75.5 =, The formula to find the variance of a population (denoted as, The formula to find the standard deviation of a population (denoted as, And the formula to find the standard deviation of a sample (denoted as, Measures of Central Tendency: Definition & Examples. Arrange the values from smallest to largest. Example: Cheryl took 7 math tests in one marking period. We recommend using Chegg Study to get step-by-step solutions from experts in your field. Thus;$$ { \sigma }^{ 2 }=\frac { \left\{ \sum { { \left( { X }_{ i }-\mu \right) }^{ 2 } } \right\} }{ N } . Learn more about us. o Standard deviation. Example Calculate the range for the data for Quarterback A and Quarterback B in the example above. . You compute […] Relative measures of dispersion are obtained as ratios or percentages of the average. Three of the most commonly used measures of central dispersion include the following: Range Variance Standard deviation Range The range of a data set is the difference between the largest value and the smallest value. Remember that the sum of deviations from the arithmetic mean is always zero and that’s why we are using the absolute values. 2. Such measures express the scattering of data in some relative terms or in percentage. And the formula to find the variance of a sample (denoted as s2) is: The standard deviation is the square root of the variance. Measures of dispersion are used to describe the variability or spread in a sample or population. Assume that the returns realized in example 2 above were sampled from a population comprising 100 returns. It is a relative measure of dispersion and is based on the value of range. The variance of these … Cycles are trends or patterns that may be exhibited by the securities market,... Monte Carlo simulation and historical simulation are both methods that can be used... 3,000 CFA® Exam Practice Questions offered by AnalystPrep – QBank, Mock Exams, Study Notes, and Video Lessons, 3,000 FRM Practice Questions – QBank, Mock Exams, and Study Notes. We measure “spread” using range, interquartile range, variance, and standard deviation. Measures of Dispersion A measure of spread, sometimes also called a measure of dispersion, is used to describe the variability in a sample or population. Mean deviation from mean. 58, 66, 71, 73, 74, 77, 78, 82, 84, 85, 88, 88, 88, 90, 90, 92, 92, 94, 96, 98, 2. In this lesson, you will read about the following measures of dispersion: Range. \end{align*}. & = 37.67(\%^2) \\ Central dispersion show how “ spread ” using range, interquartile range, interquartile range in! Comparison between different series is hence possible root of variance students: the largest and smallest value in a.. 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https://math.stackexchange.com/questions/3870494/let-phi-be-a-bijective-function-can-we-conclude-that-two-monoids-m-1-a/3870533#3870533 | Let $\phi$ be a bijective function. Can we conclude that two monoids $(M_1,.)$ and $(M_2,∗)$ are isomorphic?
Condiser two monoids $$(M_1,.)$$ and $$(M_2,*)$$ with identity elements $$e_1$$ and $$e_2$$ and a bijective function $$\phi$$ which has the property $$\phi(a.b)=\phi(a) * \phi(b)$$ for all $$a,b \in M_1$$
can we conclude that $$(M_1,.)$$ and $$(M_2,*)$$ are isomorphic?
My thoughts:
from this A homomorphism between two monoids $$(M_1,.)$$ and $$(M_2,*)$$ is a function f : $$M_1$$$$M_2$$ such that
1. $$f(a.b)=f(a) * f(b)$$ for all $$a,b \in M_1$$
2. $$f(e_1)=e_2$$
and then a bijective monoid homomorphism is called a monoid isomorphism.
I'm saying if $$f$$ is bijective we need to show for all $$c \in M_2$$ : $$c*f(e_1)=f(e_1)*c=c$$ and then we can conclude $$f(e_1)=e_2$$ and the second condition is not nesseccery for being isomorphic.
Consider $$c \in M_2$$. beacuse $$f$$ is bijective so there is $$x \in M_1$$ which $$f(x)=c$$ so $$c*f(e_1)=f(x)*f(e_1)=f(x.e_1)=f(x)=c$$ and by doing the same for $$f(e_1)*c$$ we get $$c*f(e_1)=f(e_1)*c=c$$ and as I said $$f(e_1)=e_2$$. so the second condition is not nesseccery for being isomorphic. and only $$f$$ being bijective is enough.
Is the conclusion I made correct?
• @Shaun Thank you! Oct 18, 2020 at 8:21
• @Joshua Cole There is an important conceptual point to be made: what we call (definitionally) a monoid isomorphism is not a bijective monoid morphism, but a monoid morphism which happens to admit a(nother) monoid morphism as an inverse. It is a matter not of definition but of subsequent characterisation that a monoid morphism is an isomorphism if and only if it is bijective, and this phenomenon only occurs for algebraic structures (universal algebras) and not in other categories such as topological spaces or relational systems.
– ΑΘΩ
Oct 18, 2020 at 8:25
• @Joshua Cole But of course, I apologise for the terse style. What I meant is that the phenomenon of "isomorphisms coinciding with bijective isomorphisms" is valid for algebraic structures (groups, rings, modules etc) but not for the category of say ordered sets. There exist quite elementary examples of bijective morphisms of ordered sets (the morphisms in this category are the isotonic or increasing maps) which are however not isomorphisms. The same goes for topological spaces: there exist continuous bijections which however fail to be homeomorphisms (isomorphisms).
– ΑΘΩ
Oct 18, 2020 at 9:23
• @Joshua Cole In order to assimilate this difference in how morphisms behave in various categories, a simple introduction to category theory -- combined with some exercises that illustrate what the general abstract notions become in the concrete instances of sets, groups, monoids, topological spaces etc -- would be more than sufficient. You could consider "Categories for the working mathematician" of Saunders MacLane as a first introduction.
– ΑΘΩ
Oct 18, 2020 at 10:27
• @Joshua Cole Concrete experience with classical categories such as that of sets, semigroups, rings, modules, topological spaces, ordered sets etc would certainly help but it is not an absolute prerequisite. To help you form some measure of such experience, may I invite you to try and produce examples of bijective morphisms which however are not isomorphisms in whichever of the following categories you might be more familiar with: ordered sets, topological spaces or simple graphs. If you would like to ask more questions, feel free to do so (we could take this discussion over to the chat room).
– ΑΘΩ
Oct 18, 2020 at 13:52
Yes, you have the right idea. To clarify the language slightly, you are asking if an isomorphism between $$M_1$$ and $$M_2$$ as semigroups is automatically an isomorphism as monoids. It is easy to check that a bijective semigroup (resp. monoid) homomorphism is automatically an iso, so your question is the same as this one.
I will say up-front that your computation is correct, and so is your conclusion. I will also say that there is some extremely interesting mathematics lying just under the surface. If you'll indulge me, I would love to share it with you ^_^
First let's note that some bonus condition is crucial. Consider $$M_1 = M_2 = (\mathbb{Z}^2, \times)$$, with componentwise multiplication. Then the map $$f(a,b) = (a,0)$$ is easiliy seen to be a semigroup homomorphism, but $$f(1,1) = (1,0) \neq (1,1)$$ so the identity is not preserved.
Notice, however, that this is really a counterexample by a technicality. $$f(1,1) = (1,0)$$ is an identity for the image $$f[\mathbb{Z}^2]$$. After all $$(a,0)(1,0) = (a,0) = (1,0)(a,0)$$. It is only for elements not in the image, like $$(0,b)$$ that we notice $$(1,0)$$ fails to be an identity. At the risk of outing myself as a logician, I would love to talk about the model-theoretic content in this observation:
The property of "being an identity" is expressible in the language of semigroups. Let $$\varphi(x,y)$$ be the formula $$yx = x \land xy = x$$. Then $$e$$ is an identity if and only if $$\forall x . \varphi(x,e)$$ is true. Now since $$\varphi$$ is "positive" (in the sense that it doesn't have any negations), it is preserved by arbitrary homomorphisms. So if $$\varphi(x,y)$$ is true in $$M_1$$, then $$\varphi(f(x),f(y))$$ will be true in $$M_2$$.
Notice this almost the same as saying that if $$\forall x . \varphi(x,e)$$ is true in $$M_1$$, then $$\forall x' . \varphi(x',f(e))$$ is true in $$M_2$$. The problem is the range of the quantifiers. The first quantifier ranges over the elements of $$M_1$$, whereas the second ranges over all the elements of $$M_2$$. Of course, we (in general) have no control over the parts of $$M_2$$ outside the image of $$M_1$$, so it makes sense that this "strong" universal quantifier might fail to be true. But we are guaranteed that $$\varphi(x',f(e))$$ is true when we promise to look only at $$x'$$ in the image of $$M_1$$.
So, in the special case that $$f$$ is surjective, we can see how to proceed. If $$f$$ is surjective, then every element of $$M_2$$ is in the image of $$M_1$$. So we really can put the universal quantifier in front, and the property of "being an identity" is preserved.
This gives us a slightly stronger claim than what you wanted: It suffices that the semigroup homomorphism be surjective. And if you look at the computational proof that you gave, you only used surjectivity when you concluded that the identity was preserved!
The reason to go on this long diversion is to give you a tool to see not only that this is true, but to see how it might be obviously true. The property of "being an identity" is definable in the language of semigroups, and isomorphisms preserve all first order formulas. So, in particular, the identity gets mapped to an identity under a semigroup isomorhpism. This kind of argument is extremely flexible, and I hope it serves you well going forwards!
I hope this helps ^_^
• nice answer! Oct 18, 2020 at 8:29
• @HallaSurvivor On an even more general level, given a unitary magma $(A, \cdot)$, if there exists a surjective magma morphism $f \colon A \to B$, it follows that the target magma $(B, \cdot)$ is also unitary and the relation $f(1_A)=1_B$ necessarily holds (in other words, $f$ is automatically a morphism in the appropriate category of unitary magmas).
– ΑΘΩ
Oct 18, 2020 at 8:31
• @ΑΘΩ - That actually follows from this same line of reasoning. We never use associativity in $\varphi$, so this exact same proof shows that any identity in a magma is sent to an identity by a surjective magma homomorphism. It's nice to point this out explicitly, though ^_^ Oct 18, 2020 at 8:35
• @HallaSurvivor Thank you very much! I'm a first-year so there were some points that I couldn't follow but even in a situation like that I enjoyed your answer and it was VERY nice, again, thank you. Oct 18, 2020 at 8:42
• @HallaSurvivor Indeed it does, as you very well put it. Permit me to make just one more remark, namely that any two identities of a magma must necessarily coincide, so if a magma is unitary it then has a unique identity, which justifies the use of the definite article in the syntagm "the identity of a (unitary) magma". A pleasure chatting, sir.
– ΑΘΩ
Oct 18, 2020 at 9:28 | 2022-08-09T09:11:27 | {
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http://math.stackexchange.com/questions/154505/prove-that-sin2a-sin2b-sin2c-4-sina-sinb-sinc-when-a-b-c-are | # Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a triangle
Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a triangle
This question came up in a miscellaneous problem set I have been working on to refresh my memory on several topics I did earlier this year. I have tried changing $4\sin(A)\sin(B)\sin(C)$ to $$4\sin(B+C)\sin(A+C)\sin(A+B)$$ by making substitutions by reorganizing $A+B+C=\pi$. I then did the same thing to the other side to get $$-2(\sin(B+C)\cos(B+C)+\sin(A+C)\cos(A+C)+\sin(A+B)\cos(A+B))$$ and then tried using the compound angle formula to see if i got an equality. However the whole thing became one huge mess and I didn't seem to get any closer to the solution. I am pretty sure there is some simpler way of proving the equality, but I can't seem to figure it out. Maybe there is a geometric interpretation or maybe it can be done using just algebra and trig. Any hint's would be appreciated (I would prefer an algebraic approach, but it would be nice to see some geometric proofs as well)
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In addition to Marvis' answer using standard methods of proving trigonometric identities, it is also of value to know about a geometric argument, giving a geometric interpretation of the various terms in the identity. I've posted that as an answer below. – Michael Hardy Jun 6 '12 at 2:55
Since $A,B,C$ are angles of a triangle, we have that $A+B+C = \pi$. Recall the following trigonometric identities \begin{align} \sin(\pi-\theta) & = \sin(\theta)\\ \cos(\pi-\theta) & = -\cos(\theta)\\ \sin(2\theta) + \sin(2\phi) & = 2 \sin(\theta + \phi) \cos(\theta-\phi)\\ \sin(2\theta) & = 2\sin(\theta) \cos(\theta)\\ \cos(\theta - \phi) - \cos(\theta + \phi) & = 2 \sin(\theta) \sin(\phi) \end{align} We have that \begin{align} \sin(2A) + \sin(2B) & = 2 \sin(A+B) \cos(A-B)\\ & = 2 \sin(\pi-(A+B)) \cos(A-B)\\ & = 2 \sin(C) \cos(A-B) \end{align} Hence, \begin{align} \sin(2A) + \sin(2B) + \sin(2C) & = 2 \sin(C) \cos(A-B) + 2 \sin(C) \cos(C)\\ & = 2 \sin(C) \left(\cos(A-B) + \cos(C) \right)\\ & = 2 \sin(C) \left(\cos(A-B) + \cos(\pi-(A+B)) \right)\\ & = 2 \sin(C) \left(\cos(A-B) - \cos(A+B) \right)\\ & = 2 \sin(C) \times 2 \sin(A) \sin(B)\\ & = 4 \sin(A) \sin(B) \sin(C) \end{align}
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Wow! Thanks. I didn't expect an answer that fast. – E.O. Jun 6 '12 at 1:30
Thanks. It is an excellent proof....+1 – chndn Apr 17 '13 at 3:48
Here's another way to do it. Since $A+B+C=\pi$, we have the three angles of a triangle. Inscribe that triangle in a circle of unit diameter (not unit radius). Then $\sin A$, $\sin B$, and $\sin C$ are actually the lengths of the sides opposite that three angles---that's essentially the law of sines. The area of any triangle is the product of the lengths of two sides times the sine of the angle between them, divided by $2$. So the area in this case is $(\sin A\sin B\sin C)/2$. That's the right side of the identity, except that you've got $4$ rather than $1/2$ as the coefficient. So it is enough to show that the area is $(\sin(2A) + \sin(2B) + \sin(2C))/8$. If the center of the circle is inside the triangle, you can draw lines from the center to each of the three vertices, thus breaking the triangle into three smaller triangles. It is then enough to show that $\sin(2A)/8$ is the area of one of those (and the other two are shown "similarly"). All you need is that two lengths of sides are each $1/2$ and the angle between them is $2A$.
If the center of the circle is not inside the triangle, then one of the three angles at the center---say $2A$---is more than $180^\circ$ and so $\sin(2A)$ is negative. You get a signed area. Two of those three parts add up to more than the whole triangle, and then you subtract the third part and get the right amount.
This also generalizes to other polygons inscribed in a circle, telling us, for example, that $$\text{if }A+B+C+D+E=\pi$$ \begin{align} & \text{then }\sin(2A)+\sin(2B)+\sin(2C)+\sin(2D)+\sin(2E) \\[8pt] & = \underbrace{4\,\overbrace{\sin A\sin B\sin C}^{\text{three sines}}\,\, \overbrace{\cos D\cos E}^{\text{two cosines}} + \cdots\cdots}_{\text{10 terms}} - 8\sin A\sin B\sin C\sin D\sin E. \end{align} (The five angles here are the angles between adjacent diagonals of the inscribed pentagon.)
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The other two answers are great, but if you're ever not feeling clever enough to come up with them you can always try the sledgehammer method (using Euler's formula and complex exponentials): $$4\sin(A)\sin(B)\sin(C)=4\left(\frac{e^{iA}-e^{-iA}}{2i}\right)\left(\frac{e^{iB}-e^{-iB}}{2i}\right)\left(\frac{e^{iC}-e^{-iC}}{2i}\right) \,$$ Now, multiply out this product and you'll get eight terms. The two where all the signs are the same will be $e^{i\pi}/(-2i)$ and $-e^{-i\pi}/(-2i)$ since $A+B+C=\pi$, so they cancel. The other six will collect into three terms that look like $$\frac{-e^{i(A+B-C)}+e^{i(-A-B+C)}}{-2i}=\sin(A+B-C)$$ (possibly with the variables permuted). But $A+B-C=\pi-2C$, which means $$\sin(A+B-C)=\sin(\pi-2C)=\sin(2C) \, ,$$ and similarly with the other two terms. So their sum is the left-hand side of your identity.
- | 2016-07-23T23:36:16 | {
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https://math.stackexchange.com/questions/3704882/closed-form-of-int-0-infty-arctan2-left-frac2x1-x2-right-dx | # Closed form of $\int_0^\infty \arctan^2 \left (\frac{2x}{1 + x^2} \right ) \, dx$
Can a closed form solution for the following integral be found: $$\int_0^\infty \arctan^2 \left (\frac{2x}{1 + x^2} \right ) \, dx\,?$$
I have tried all the standard tricks such as integration by parts, various substitutions, and parametric differentiation (Feynman's trick), but all to no avail.
An attempt is letting $$f(t):=\int_0^\infty\,\arctan^2\left(\frac{2tx}{1+x^2}\right)\,\text{d}x\,.$$ Therefore, $$f'(t)=\int_0^\infty\,\frac{8x^2(x^2+1)}{\big(x^4+2(2t^2+1)x^2+1\big)^2}\,\left(1+x^2-4tx\arctan\left(\frac{2tx}{1+x^2}\right)^{\vphantom{a^2}}\right)\,\text{d}x\,.$$ This doesn't seem to go anywhere. Help!
• I would suggest looking into using residues to solve integrals of this sort. Perhaps it is not applicable in your case (it's been a year since I've used them), but it is worthwhile to know of them if you often perform such integrations. Jun 4 '20 at 4:57
$$I=\int_0^\infty \arctan^2 \left (\frac{2x}{x^2 + 1} \right ) dx\overset{IBP}=4\int_0^\infty \frac{x(x^2-1)\arctan\left(\frac{2x}{x^2+1}\right)}{x^4+6x^2+1}dx$$ We have that: $$4\int\frac{x(x^2-1)}{x^4+6x^2+1}dx=(\sqrt 2 +1)\ln(x^2+(\sqrt 2+1)^2)-(\sqrt 2-1)\ln(x^2+(\sqrt 2-1)^2)$$ $$\frac{d}{dx}\arctan\left(\frac{2x}{x^2+1}\right)=\frac12\left(\frac{\sqrt 2+1}{x^2+(\sqrt 2+1)^2}-\frac{\sqrt 2-1}{x^2+(\sqrt 2-1)^2}\right)$$ Thus integrating by parts again and simplifying we obtain: $$I=\int_0^\infty \frac{(\sqrt 2+1)^2 \ln(x^2+(\sqrt 2+1)^2)}{x^2+(\sqrt 2+1)^2}dx+\int_0^\infty \frac{(\sqrt 2-1)^2 \ln(x^2+(\sqrt 2-1)^2)}{x^2+(\sqrt 2-1)^2}dx$$ $$-\int_0^\infty \frac{\ln(x^2+(\sqrt 2-1)^2)}{x^2+(\sqrt 2+1)^2}dx-\int_0^\infty \frac{\ln(x^2+(\sqrt 2+1)^2)}{x^2+(\sqrt 2-1)^2)}dx$$ From here we have the following result: $$\int_0^\infty \frac{\ln(x^2+a^2)}{x^2+b^2}dx=\frac{\pi}{b}\ln(a+b), \ a,b>0$$ So using this result and with some algebra everything simplifies to: $$\boxed{\int_0^\infty \arctan^2 \left (\frac{2x}{x^2 + 1} \right ) dx=2\pi \ln(1+\sqrt 2)-\sqrt 2\pi \ln 2}$$
• Nice solution! (+1) Using the same technique, we also get $$\int_{0}^{\infty}\arctan^2\left(\frac{2rt}{1+t^2}\right)\,\mathrm{d}t=2\pi\left(r\log\left(\sqrt{r^2+1}+r\right)-\sqrt{r^2+1}\log\sqrt{r^2+1}\right).$$ Jun 4 '20 at 10:38
Here is a solution based on Fubini's theorem.
According to an addition formula $$\begin{equation*} \arctan\left(\dfrac{2x}{1+x^2}\right) = \arctan((\sqrt{2}+1)x)-\arctan((\sqrt{2}-1)x) . \end{equation*}$$ Furthermore $$\begin{equation*} \arctan x=\mathrm{sign}(x)\dfrac{\pi}{2}-\arctan\dfrac{1}{x} . \end{equation*}$$ Consequently $$\begin{equation*} \arctan\left(\dfrac{2x}{1+x^2}\right) = \arctan\dfrac{\sqrt{2}+1}{x}-\arctan\dfrac{\sqrt{2}-1}{x}=\int_{\sqrt{2}-1}^{\sqrt{2}+1}\dfrac{x}{x^2+s^2}\, ds . \end{equation*}$$ Via Fubini's theorem we get $$\begin{gather*} \int_{0}^{\infty}\arctan^2\left(\dfrac{2x}{1+x^2}\right)\, dx = \int_{0}^{\infty}\left(\arctan\dfrac{\sqrt{2}+1}{x}-\arctan\dfrac{\sqrt{2}-1}{x}\right)^2\, dx=\\[2ex] \int_{0}^{\infty}\left(\int_{\sqrt{2}-1}^{\sqrt{2}+1}\dfrac{x}{x^2+s^2}\, ds\int_{\sqrt{2}-1}^{\sqrt{2}+1}\dfrac{x}{x^2+t^2}\, dt\right)\, dx=\\[2ex] \int_{\sqrt{2}-1}^{\sqrt{2}+1}\left(\int_{\sqrt{2}-1}^{\sqrt{2}+1}\left(\int_{0}^{\infty}\dfrac{x^2}{(x^2+s^2)(x^2+t^2)}\, dx\right)\, ds\right)\, dt=\\[2ex] \dfrac{\pi}{2}\int_{\sqrt{2}-1}^{\sqrt{2}+1}\left(\int_{\sqrt{2}-1}^{\sqrt{2}+1}\dfrac{1}{s+t}\, ds\right)\, dt=\\[2ex] \dfrac{\pi}{2}\int_{\sqrt{2}-1}^{\sqrt{2}+1}\left(\ln(t+\sqrt{2}+1)-\ln(t+\sqrt{2}-1)\right)\, dt=\\[2ex] 2\pi\ln(\sqrt{2}+1)-\sqrt{2}\pi\ln 2. \end{gather*}$$
Remark. Since $$\begin{equation*} \arctan\left(\dfrac{2x\sinh\alpha}{1+x^2}\right)=\arctan\left(\dfrac{e^{\alpha}}{x}\right)-\arctan\left(\dfrac{e^{-\alpha}}{x}\right) = \int_{e^{-\alpha}}^{e^{\alpha}}\dfrac{x}{x^2+s^2}\, ds \end{equation*}$$ the $$@$$Sangchul Lee's generalization can be proved in the same way.
Here is another solution with a generalization:
Let $$r=\sinh\alpha$$ and $$s=\sinh\beta$$. Then \begin{aligned} &\int_{0}^{\infty} \arctan\left(\frac{2rx}{1+x^2}\right)\arctan\left(\frac{2sx}{1+x^2}\right) \, \mathrm{d}x\\ &= \pi \left( \alpha \sinh\beta+\beta\sinh\alpha+(\cosh\alpha+\cosh\beta)\log\left(\frac{e^{\alpha}+e^{\beta}}{1+e^{\alpha+\beta}}\right) \right) \end{aligned} \tag{*}
Proof. Let $$J = J(\alpha,\beta)$$ denote the right-hand side of $$\text{(*)}$$. Then
$$J(0, \beta) = 0, \qquad J_{\alpha}(\alpha, 0) = 0, \qquad J_{\alpha\beta} = \pi \left( \frac{1+\cosh\alpha\cosh\beta}{\cosh\alpha + \cosh\beta} \right).$$
Now let $$I = I(\alpha, \beta)$$ denote the left-hand side of $$\text{(*)}$$. Then by the substitution $$x=\tan(\theta/2)$$, we get
$$I = \frac{1}{2}\int_{0}^{\pi} \frac{\arctan(\sinh\alpha\sin\theta)\arctan(\sinh\beta\sin\theta)}{1+\cos\theta} \, \mathrm{d}\theta.$$
From this, we easily check that $$I$$ also satisfies
$$I(0, \beta) = 0, \qquad I_{\alpha}(\alpha, 0) = 0.$$
Moreover,
\begin{align*} \require{cancel} I_{\alpha\beta} &= \frac{1}{2}\int_{0}^{\pi} \frac{\cosh\alpha\cosh\beta(1-\cos\theta)}{(1+\sinh^2\alpha\sin^2\theta)(1+\sinh^2\beta\sin^2\theta)} \, \mathrm{d}\theta \\ &= \frac{1}{2}\int_{0}^{\pi} \frac{\cosh\alpha\cosh\beta}{(1+\sinh^2\alpha\sin^2\theta)(1+\sinh^2\beta\sin^2\theta)} \, \mathrm{d}\theta \\ &\quad - \cancelto{0}{\frac{1}{2}\int_{0}^{\pi} \frac{\cosh\alpha\cosh\beta}{(1+\sinh^2\alpha\sin^2\theta)(1+\sinh^2\beta\sin^2\theta)} \, \mathrm{d}\sin\theta}\\ &= \frac{1}{2}\int_{0}^{\infty} \frac{\cosh\alpha\cosh\beta (1 + t^2)}{(t^2 + \cosh^2\alpha)(t^2 + \cosh^2\beta)} \, \mathrm{d}t \tag{t=\cot\theta} \end{align*}
It is not hard to check that the last integral is equal to $$J_{\alpha\beta}$$. Therefore we get $$I = J$$. | 2022-01-24T11:45:46 | {
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http://mathhelpforum.com/pre-calculus/273141-find-values-k.html | # Thread: Find Values of k
1. ## Find Values of k
For which values of k will the roots of the equation
x^2 = 2x(3k + 1) - 7(2k + 3) be equal?
x^2 = 6xk + 2 - 14k - 21
x^2 = 6xk - 14k -19
Is this a new start?
2. ## Re: Find Values of k
For which values of k will the roots of the equation
x^2 = 2x(3k + 1) - 7(2k + 3) be equal?
x^2 = 6xk + 2 - 14k - 21
x^2 = 6xk - 14k -19
Is this a new start?
you should get
$x^2 = (2+6k)x -(14k+21)$
3. ## Re: Find Values of k
Hint: A quadratic of the form \displaystyle \begin{align*} a\,x^2 + b\,x + c = 0 \end{align*} has exactly one root (or two equal roots) when the discriminant \displaystyle \begin{align*} b^2 - 4\,a\,c = 0 \end{align*}.
4. ## Re: Find Values of k
Originally Posted by romsek
you should get
$x^2 = (2+6k)x -(14k+21)$
How did you get +21 when -7 × 3 = -21?
How did you get 14k when -7 × 2k = -14k?
5. ## Re: Find Values of k
Originally Posted by Prove It
Hint: A quadratic of the form \displaystyle \begin{align*} a\,x^2 + b\,x + c = 0 \end{align*} has exactly one root (or two equal roots) when the discriminant \displaystyle \begin{align*} b^2 - 4\,a\,c = 0 \end{align*}.
Are you saying to rewrite x^2=(2+6k)x−(14k+21) in the form ax^2 + bx + c and then use the discriminant by setting it to zero?
Yes
7. ## Re: Find Values of k
Originally Posted by Prove It
Yes
Ok. I will solve it tomorrow.
8. ## Re: Find Values of k
$x^2=2x(3k+1)-7(2k+3)$
$x^2=6kx+2x-14k-21$
$x^2=(6k+2)x-(14k+21)$
$x^2-(6k+2)x+(14k+21)=0$
For roots to be equal
$b^2-4ac=0$
$[-(6k+2)]^2-4×1×(14k+21)=0$
$36k^2+4+24k-56k-84=0$
$36k^2-32k-80=0$
$4(9k^2-8k-20)=0$
$9k^2-8k-20=0$
$(k-2)(9k+10)=0$
$k=2,\;-\dfrac{10}{9}$
9. ## Re: Find Values of k
Originally Posted by deesuwalka
$x^2=2x(3k+1)-7(2k+3)$
$x^2=6kx+2x-14k-21$
$x^2=(6k+2)x-(14k+21)$
$x^2-(6k+2)x+(14k+21)=0$
For roots to be equal
$b^2-4ac=0$
$[-(6k+2)]^2-4×1×(14k+21)=0$
$36k^2+4+24k-56k-84=0$
$36k^2-32k-80=0$
$4(9k^2-8k-20)=0$
$9k^2-8k-20=0$
$(k-2)(9k+10)=0$
$k=2,\;-\dfrac{10}{9}$
Very impressive math work. | 2018-04-24T07:42:31 | {
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Hammer mill for sale gemco energy gemco energy machinery gemco hammer mills is designed for solving your crushing problems to guarantee the rotor balance and avoid vibration all blades have to be turned over at more details get price,Vibration free hammer mill. Adding to the confusion is it is a damped vibration problem. The system is free to oscillate and its mass, stiffness and damping matrices are M = \begin{bmatrix} 60 & 23. Several numerical examples of the damped free vibration problem of laminated composite cylindrical shells have been solved and comparison has been made with the results of other authors. Solve the differential equation for the equation of motion, x(t). Free vibrations of a SDOF with viscous damping – relation between logarithmic decrement and damping factor – Coulomb damping and structural/hysteretic damping. Solving Problems in Dynamics and Vibrations Using MATLAB Parasuram Harihara And Dara W. Damped and undamped vibration refer to two different types of vibrations. Lecture L19 - Vibration, Normal Modes, Natural Frequencies, Instability Vibration, Instability An important class of problems in dynamics concerns the free vibrations of systems. b) Damping condition (underdamped, overdamped, or critically damped) c) Stability (stable or unstable) d) Correct equation to be used for x(t)=… and correct equations to be used for computing constants in this equation. An extensive bibliography to guide the student to further sources of information on vibration analysis is provided at the end of the book. , Prentice Hall, 2008: • Problem A: The system has m=3. It is shown that the classically damped systems are relatively easy to solve. (Assume all free vibration has been damped out. They will make you ♥ Physics. Free PDF ebooks (user's guide, manuals, sheets) about Matlab code for solve beam vibration ready for download I look for a PDF Ebook about : Matlab code for solve beam vibration. Estimate, as well as you can using the given information:. To shed some light on this problem, we present a carefully chosen succession of problems to illustrate what you can expect from typical. * Over Damping. In this article, I will be explaining about the Damped Free Vibrations in an effective manner. 3 For the beam's data given in problem M 4. Single Degree of Freedom Damped Free Vibrations. 4: Vibration of Multi-DOF System () 2 2 2 2 Eigenvalue-Eigenvector problem For the system of. Reason for vibration analysis •Lightly damped structures can produce high levels of vibration from low level sources if frequency components in the disturbance are close to one of the system’s natural frequencies. A system is said to be linear if its equation of motion is linear. Solve for the motion of a freely vibrating single degree of freedom damped system under different damping ratios. 5 Free Torsional Vibration of a Single Rotor System system for a clear understanding of basic features of a vibration problem. They have applied Brock’s approach to different type of damped vibration absorber named model B (when. Free Vibration - Damped Forced Vibration Graphics, narrations, animations, and equations are used to help you understand how the problem was solved. Free vibration of viscously damped SDOF systems (1) ・Damping ratio and damped natural frequency ・Response of viscous damped system Make sure students have read chapters 3 (pp. Traditional design optimization of the Voigt-type dynamic vibration absorber often solved for the vertical or lateral. This leads to an absorber tuning schedule as follows: Step 1. Lecture topics Reading materials Exams. know how to use a viscous frictional force in Newton's second law to solve for damped oscillatory motion. NASA Astrophysics Data System (ADS) Lu, Xuezeng; Xiang, Hongjun; Rondinelli, James; Materials Theory; Desi. The book presents in a simple and systematic manner techniques that can easily be applied to the analysis of vibration of mechanical and structural systems. These topics are listed below for clarification. Answer to Problem deal with damped free vibrations of a mass-spring-dashpot system whose position function satisfies the equation. 5 hrs) zSDoF Systems zDynamic Equilibrium zMDoF Systems yq zMode Shapes Complementary solution is the free damped vibration response:vibration response: In certain problems, e. understand how to combine (superpose) two vibrations and show that beats arise. If the camera needs to remain a constant distance from the player (e. To obtain the free response, we must solve system of homogeneous ODEs, i. Solve integrals with Wolfram|Alpha. To start viewing messages, select the forum that you want to visit from the selection below. Solving the eigenvalue problem (exact solution) Rayleigh’s Method (approximate solution) etc ENE 5400 , Spring 2004 10 Eigenvalue Problem Under free vibration and no damping, natural frequencies of a multi-d. This tutorial covers the theory of natural vibrations with damping and continues the studies in the tutorial on free vibrations. The particular solution depends on the nature of the forcing function. The inverse problem for the Stieltjes string with the left end free and the right end damped is a particular case of the inverse problem solved in [5]. Substituting the identities into Eq. July 25 - Free, Damped, and Forced Oscillations 3 INVESTIGATION 1: FREE OSCILLATIONS We have already studied the free oscillations of a spring in a previous lab, but let's quickly determine the spring constants of the two springs that we have. We analyzed vibration of several conservative systems in the preceding section. * Critical Damping. The vibration would be periodic if the amplitude will not decay with time. 5; it does not apply at all for a damping ratio greater than 1. The vibration is started by some input of energy but the vibrations die away with time as the energy is dissipated. The force is proportional to the velocity of the mass. This paper is concerned with the existence and multiplicity of fast homoclinic solutions for a class of damped vibration problems with impulsive effects. 5 Hz in a vacuum and 0. An extensive bibliography to guide the student to further sources of information on vibration analysis is provided at the end of the book. This paper is concerned with the existence and multiplicity of fast homoclinic solutions for a class of damped vibration problems with impulsive effects. The above is a standard eigenvalue problem. It currently illustrates undamped free vibrations, but could be modified to work with damped or driven vibrations. Equation which plots displacement vs time. SOLVED PROBLEMS. This problem requires you to integrate your knowledge of various concepts regarding waves, oscillations, and damping. Việc sử dụng MATLAB để giải các bài toán tích phân, vi phân, phương trình phức tạp, vẽ đồ thị rất cần thiết và đảm bảo độ chính xác yêu cầu. Displacement-time response of a free critically damped vibration for different There would be no difficulty to solve directly the above system by eliminating the displacement Y of the nonproportionally damped system. You should study the tutorial on free vibrations before commencing. 5th Ed – Abdul Ghaffar Abdul Rahman. | At this point it seems to be personal preference, and all academic, whether you use the Lagrangian method or the F = ma method. 2 Matrix Equations. know what beats sound like and how they look on an oscilloscope or graph. Recall that the textbook’s convention is that. 00 J, an amplitude of 10. In the present paper, viscously damped free and forced vibrations of circular and annular membranes are investigated using a closed form exact method. For free vibration, P(x,y,t) =0. About two months ago they stopped giving out free keys. III PROBLEM SOLUTION The dynamic equilibrium equations which governs behaviour of damped orthotropic rectangular plate supported by Pasternak foundation is gotten by neglecting the terms representing the applied force in equation (1). No external force acts on the system. MODAL ANALYSIS OF NON-CLASSICALLY DAMPED LINEAR SYSTEMS 219 NATURAL FREQUENCIES AND MODES For a system in free vibration, the right-hand member of equation (1) vanishes, and the equation admits a solution of the form where r is a characteristic value and { \$1 is the associated characteristic vector or natural mode. Section 9-8 : Vibrating String. m is the mass, γ is the damping from the dash pot, and k is the restoring force from the spring. Honeywell announced Tuesday that it is on track to have a quantum computer with a quantum volume of at least 64 qubits within the next three months. Vibrations and Waves by George C. damped structures as wellas to provide a review of = 0 (1) A TUTORIAL ON COMPLEX EIGENVALUES Daniel J. The Department for Education has provided £7. Includes kinematics of rigid bodies in plane motion. oscillation and nonoscillation responses by numerically solving Eq. There is a free file available on-online entitled Dynamic Absorbers for Solving Resonance Problems about of damped vibration absorbers for use in the machine tool. The method can capture structural DHR without a specific numerical model for the structure. The complete solution of this equation comprises two parts: the transient solution (obtained by solving the homogeneous equation) and the steady state solution. This slide shows some simple damped free vibration responses. stiffness of the spring, 2. An extensive bibliography to guide the student to further sources of information on vibration analysis is provided at the end of the book. Energy Method for undamped free vibration Forced Damped Vibrations - Duration: 7:59. You will find more information about these damped adap-tors in a separate chapter. The aim of this study is to compute damped eigenfrequencies and modes of a vibroacoustic interior problem with fluid-structure coupling. Damped Undamped. know how to use a viscous frictional force in Newton's second law to solve for damped oscillatory motion. Sound Damped Steel (SDS) manufactures and supplies highly damped materials, complete kits and components that solve a very wide range of noise and vibration control problems at very low cost. 3 we discuss damped and driven harmonic motion, where the driving force takes a sinusoidal form. Introduction •A system is said to undergo free vibration when it oscillates only under an initial disturbance with no external forces acting after the initial disturbance 3. Solved Examples-Derivation of Equation of Motion. This mechanism, referred to as vortex-induced vibration (VIV), occurs when the vortices, developed in the wake, can couple with the dynamics of the cylinder. Driven Oscillator. 3 Response to Harmonic and Periodic Excitations 65. You get a clump in one spot and none in another, so you’re wasting both time and deicer. oscillation and nonoscillation responses by numerically solving Eq. Compatible with any classroom text, Schaum’s 2500 Solved Problems in Fluid Mechanics and Hydraulics is so complete it’s the perfect tool for graduate or professional exam review. f system are solutions of the eigenvalue problem The roots αi = mωi2/k, so ω i can be solved. How much mass should be attached to the spring so that its frequency of vibration is f = 3. 996 & 0\ Stack Exchange Network Stack Exchange network consists of 175 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and. Search Search. Recall, solving any differential equation requires that a general solution is first assumed and then initial conditions are used to find the constants. damped structures as wellas to provide a review of = 0 (1) A TUTORIAL ON COMPLEX EIGENVALUES Daniel J. 457 Mechanical Vibrations - Chapter 2 Virtual Work - Equilibrium of Bodies If a system is in equilibrium under the action of a set of forces is given a virtual displacement, the virtual work done by the foces will be zero Do Example 2. 2: Free Vibration of 1-DOF System 2. The aim of this book is to impart a sound understanding, both physical and mathematical, of the fundamental theory of vibration and its applications. the free undamped and free damped motions in mass-spring systems. I'll do that in the next vibration post, which will cover damped vibrations. The method can capture structural DHR without a specific numerical model for the structure. Lectures by Walter Lewin. Damping and the Natural Response in RLC Circuits. Includes kinematics of rigid bodies in plane motion. Student submit design projects, using Matlab to solve vibration or dynamic problems related to product. Driven Oscillator. We also allow for the introduction of a damper to the system and for general external forces to act on the object. Thing is i need to plot an envelope of the graph and it has to take into account initial speed v0. Stéfan posted his code on github. Vibration can be put to work for many useful purposes: Vibrating sieves, mixers, and tools are the most obvious examples. Home >> Category >> Mechanical Engineering (MCQ) questions and answers >> Undamped and Damped Free Vibrations 1) Determine logarithmic decrement, if the amplitude of a vibrating body reduces to 1/6 th in two cycles. We know that in reality, a spring won't oscillate for ever. Scribd is the world's largest social reading and publishing site. 9 Forced vibration of damped, single degree of freedom, linear spring mass systems. Inman, 3rd ed. ” Richard “Thrity spoke my language in her book. Việc sử dụng MATLAB để giải các bài toán tích phân, vi phân, phương trình phức tạp, vẽ đồ thị rất cần thiết và đảm bảo độ chính xác yêu cầu. The most basic vibration analysis is a system with a single degree of freedom (SDOF), such as the classical linear oscillator (CLO), as shown in Fig. Damped Free Vibrations: Neglecting Damping for Small 2/4km(5 of 8) • Consider again the comparisons between damped and undamped frequency and period: • Thus it turns out that a small is not as telling as a small ratio 2/4km. 6) Students will have an ability to obtain the complete solution for the motion of a single degree of freedom vibratory system (damped or undamped) that is subjected to non-periodic forcing functions. one dimensional) damping at the right end can be reduced also to the problem of damped. The book presents in a simple and systematic manner techniques that can easily be applied to the analysis of vibration of mechanical and structural systems. A num-ber of physical examples are given, which include the following: clothes. problems for the free and transverse vibration problems of a rotating twisted Timoshenko beam under axial that describes the under damped and over damped motion of a system subject to external. In differential equation of this damped system, member which represents damping is replaced with friction [4]: This differential equation has exact solution, which is compatible only when spring force and inertia force are larger than friction force. Shimmed yesterday and it solved some of the vibration. understand how to combine (superpose) two vibrations and show that beats arise. 1 Introduction 7. The solution u ( t) gives the position of the mass at time t. Free vibration control for damped system under harmonic loads? Welcome to SEFP! Welcome! Welcome to our community forums, full of great discussions about Structural. Viscously damped free vibration (2) 2 c mk − = 4 0 c = “the critical damping coefficient, c cr” cr = = 2 2 c mk m ω n Furthermore, nondimensional number ζcalled the “damping. , Prentice Hall, 2008: • Problem A: The system has m=3. student in the Graduate Program in Acoustics at Penn State. Lectures by Walter Lewin. However, for the damped primary system under torsional excitation, to the best of our knowledge, there is no study to solve this problem by algebraic approaches. , the energy method (EM), and the method of complex eigenvalues (MCE). Suppose now the motion is damped, with a drag force proportional to velocity. 3) In the case of free vibration, there is no forcing function and so. We will assume that the particular solution is of the form: x To solve this, you find x p (0) and v p. m u'' + γ u' + k u = F cos ωt. T-Zone is one of the pioneers in Whole Body Vibration in Canada - we are the largest in the country. In these works, authors formulated and solved the problem of stability and / or free and forced vibration of different type of frame without damping in considered system. One effective way to solve these vibration-related problems is to adopt high damping metallic. To do the tutorial fully you must be familiar with the following concepts. We also allow for the introduction of a damper to the system and for general external forces to act on the object. Vibration Analysis A Practical Approach to Solving Machine Vibration Problems By Victor Wowk, PE, Machine Dynamics, Inc. ones with zero applied forces. Explanation of Damped Free Vibrations: Free vibrations are oscillations where the total energy stays the same over time. 0 cm, and a maximum speed of 1. m is the mass, γ is the damping from the dash pot, and k is the restoring force from the spring. The KBM perturbation method is a kind of singular perturbation method, which evolved from the average method. Answer to Determine the free-vibration response of the viscously damped systems described in Problem when x0= 0. If the vibration problem is not resonance, attaching dynamic absorber will actually create a resonance problem and effectively amplify or increase the vibration amplitude. 2: Free Vibration of 1-DOF System 2. 4/29 Modal Analysis. Suppose there are 3 persons P1, P2 and P3 as marked in the figure. 457 Mechanical Vibrations - Chapter 2 Virtual Work - Equilibrium of Bodies If a system is in equilibrium under the action of a set of forces is given a virtual displacement, the virtual work done by the foces will be zero Do Example 2. Introduction All differential equations for the system must be solved simultaneously. This catalog features:. In this work the fourth order differential equation governing the vibration of damped orthotropic rectangular plate resting on Winkler foundation was reduced to second order coupled differential equation by separating the variables. The proposed method is categorized into a regular type of boundary element methods (BEMs) such that no singular or hypersingular integration is necessary. 6, the minimum amplitude can be achieved when the ordinates of the fixed points A, B are the same. Free vibration of single degree of freedom systems 9 4. There are examples for under-damped, critically-damped, and over-damped free vibration systems and an under-damped system subjected to sinusoidal forcing (this latter phase-plane is included here for completeness and will be more fully discussed in the Forced Vibration laboratory). In this paper, a new asymmetric indirect Trefftz method (AITM) has been developed to solve free-vibration problems. The spring mass dashpot system shown is released with velocity u 0 from position s 0 at time t = 0. The experiment container is stopped at the bottom of the tube by a spring and damper. eigenfunctions to the boundary valued problem are of the following form where r is the modal index: Ψ r¼ W U T (7) where UrðxÞ¼a 1r sinðβ uxÞðþa 2r cosðβ uxÞ (8) WrðxÞ¼a 3r sinðβ wxÞþa 4r cosðβ xÞþa 5r sinhðβ wxÞþa 6r coshðβ wxÞ: (9) 820 S. • Solve problems involving mass - spring - damper systems. 4 Undamped Forced Vibration. initial conditions). Read "DAMPED VIBRATION ANALYSIS OF A TWO-DEGREE-OF-FREEDOM DISCRETE SYSTEM, Journal of Sound and Vibration" on DeepDyve, the largest online rental service for scholarly research with thousands of academic publications available at your fingertips. logarithmic decrement, and 3. I'm trying to do the undamped vibration analysis of a cantilevered beam in Comsol Multiphysics. Teaches Newton’s second law, work-energy and power, impulse and momentum, and problem solving using computers. The focus of the present study is on the analysis of damped vibrations of the Γ type frame with cracks. A system has one degree of freedom if its motion can be completely described by a single scalar variable. ! Soln (1) is overdamped and soln (2) is critically damped. Note as well that while we example mechanical vibrations in this section a simple change of notation (and corresponding change in what the. Free vibration means that no time varying external forces act on the system. Another problem faced when solving the mass spring system is that every time a different type of problem wants to be solved (forced, unforced, damped or undamped) a new set of code needs to be created because each system. We compare design, practicality, price, features, engine, transmission, fuel consumption, driving, safety & ownership of both models and give you our expert verdict. Doing the shaving of the Isolation rubber tonight to see if it helps more. Includes kinematics of rigid bodies in plane motion. This mechanism, referred to as vortex-induced vibration (VIV), occurs when the vortices, developed in the wake, can couple with the dynamics of the cylinder. There are descriptive names associated with each of the cases. Are you looking to buy a car but can't decide between a Lexus LS500 or Toyota HiAce? Use our side by side comparison to help you make a decision. The aim, with this application on hand, is for you to have easy access to. driven, undamped vibrations; and now we consider driven, damped vibrations. This is counter to our everyday experience. harmonic oscillator in the overdamped, underdamped and critically damped regions. Structural Analysis IV Chapter 5 - Structural Dynamics 12 Dr. Logarithmic decrement, , is used to find the damping ratio of an underdamped system in the time domain. Energy Method for undamped free vibration Forced Damped Vibrations - Duration: 7:59. I think it's interesting to compare how he and I approach this problem. 3 Free vibration of a damped, single degree of freedom, linear spring mass system. Boundary value problems arise in many physical systems, just as the initial value problems we have seen earlier. To start viewing messages, select the forum that you want to visit from the selection below. Through experience we know that this is not the case for most situations. Problem Determine the values of and ω d for the following viscously damped systems:. 1 How to solve equations of motion for vibration problems. 8 Multi-Degree of Freedom Systems. Assume small angles of vibration and neglect the rod mass. This catalog is intended to provide you with a basic background on vibration control theory and specific vibration and shock control solutions. Homework Statement I have a under-damped vibration equation (below) that plots displacement vs time graph. 1983-01-01. Problem Animation Click to view movie (192k) Drop-tubes are used to simulate weightlessness for various experiments. To describe a damped harmonic oscillator, add a velocity dependent term, bx, where b is the vicious damping coefficient. Products with a similar function can be referred to as tuned, damped or anti-vibration tools. If the same initial and boundary conditions are applied as. know how to use a viscous frictional force in Newton's second law to solve for damped oscillatory motion. Damped Free Vibrations Consider the single-degree-of-freedom (SDOF) system shown at the right that has both a spring and dashpot. Energy Method for undamped free vibration Forced Damped Vibrations - Duration: 7:59. The equation of motion and. 9780898718188. The problem is usually most apparent in the immediate vicinity of the vibration source. Teaches Newton’s second law, work-energy and power, impulse and momentum, and problem solving using computers. We set up the equation of motion for the damped and forced harmonic oscillator. 78, 2042-2048 (1985)]. hence the amplitude of vibration gradually dies down. In each case, when the body is moved away from the rest position, there is a natural. An extensive bibliography to guide the student to further sources of information on vibration analysis is provided at the end of the book. If you complete the whole of this tutorial, you will be able to use MATLAB to integrate equations of motion. t AuthotTitle v (w_o. In this article, I will be explaining about the Damped Free Vibrations in an effective manner. The vibration would start about 50 mph but after 60 mph it wasn’t as bad. For 10% damping, peak is approximately ½ of the amplitude of the previous peak. We will now add frictional forces to the mass and spring. Contents 1. L7-Decay of Motion; 8. Consider a system consisting of spring, mass and damper as shown in Fig. ˆı a) If the block is displaced 3cm to the right and released, for what values of µ will the block remain in that position? g x z A. One effective way to solve these vibration-related problems is to adopt high damping metallic. It is Free Math Help Boards We are an online community that gives free mathematics help any time of the day about any problem, no matter what the level. Blake INTRODUCTION This chapter presents the theory of free and forced steady-state vibration of single degree-of-freedom systems. Vibration under general forcing conditions. Catalog Data: First of a two-course sequence that introduces methods of differential-equation solution together with common engineering applications in vibration analysis and controls. The effect of the crack depth and its location on damped vibration was presented. (Assume all free vibration has been damped out. In-class example problem - not covered due to time. Free vibration of viscously damped SDOF systems (1) ・Damping ratio and damped natural frequency ・Response of viscous damped system Make sure students have read chapters 3 (pp. Honeywell announced Tuesday that it is on track to have a quantum computer with a quantum volume of at least 64 qubits within the next three months. Caprani with respect to time. Classify types of vibration; Identify basic elements used to solve vibration problems; Harmonic and periodic motion; Solve for free undamped vibration of a single degree of freedom system; Solve for free damped vibration of a single degree of freedom system; Solve for forced vibration of single degree of freedom systems. BMM3553 Mechanical Vibration Note. The second simplest vibrating system is composed of a spring, a mass, and a damper. In the study [12], authors formulated and solved the problem of transverse damped vibrations of T type frame with rotational dampers in the points of frame mounting and in the supports. It relies on tracking the position of an impressed eddy current system that is moving. vibration problems. 5; it does not apply at all for a damping ratio greater than 1. problems for the free and transverse vibration problems of a rotating twisted Timoshenko beam under axial that describes the under damped and over damped motion of a system subject to external. This collection includes ten videos to help students learn how to approach and solve problems related to Physics III: Vibrations and Waves. Topics in Fundamentals of Structural Vibration (1. The simplest problem to solve is undamped free vibration. | 2020-04-08T11:50:15 | {
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https://math.stackexchange.com/questions/2685744/where-does-the-jordan-canonical-form-show-up-in-more-advanced-mathematics/2687077 | # Where does the Jordan canonical form show up in more advanced mathematics?
My bounty for this question expires soon :)
Edit: in regards to the bounty offered, what current research trends use the Jordan canonical form?
If one takes a second course in Linear Algebra — or a graduate level Linear Algebra course — one typically learns about non-diagonalizable operators and the Jordan canonical form.
However, where does the Jordan canonical form show up again in later, more advanced mathematics? All I hear from applied mathematicians is that the Jordan canonical form is useless in practice (in academic research). If it's not useful in applied mathematics, is it an important tool in pure mathematics? If so, in which areas of pure mathematics?
• It is used to compute the exponential of a matrix, inthe context of linear differential equations. – Bernard Mar 10 '18 at 22:35
• "All I hear from the applied mathematicians is that the Jordan form is useless in practice". I think it is down to the applied mathematicians you have been talking to do justify their derogatory remarks. – Rob Arthan Mar 10 '18 at 22:45
• @RobArthan is it derogatory? In what sense? I'm interested in hearing your thoughts ... – D.Hutchinson Mar 10 '18 at 23:06
• The JNF is numerically unstable everywhere outside of diagonalizable matrices; I don't see how it could be of any direct numerical use. What I would expect to be useful are specific formulas for JNFs of specific types of matrices. – darij grinberg Mar 10 '18 at 23:07
• Did someone mention derogatory matrices? – Git Gud Mar 10 '18 at 23:12
In brief, the Jordan normal form and various of its siblings are ubiquitous in mathematics, whether labelled "pure" or "applied" or whatever. This is not to assert that numerical computation of literal Jordan forms is worthwhile, or stable, etc. In fact, as in my earlier comment, the very instability can play a very practical role in looking at families of linear systems that "go close" to the kind of degeneracy that non-trivial Jordan blocks depict.
For that matter, Jordan form is just a special case of the structure theorem for finitely-generated modules over principal ideal domains, such as $k[x]$ where $k$ is a field. Again, yes, in both a strong-topology sense and in a Zariski-topology sense, having non-trivial "Jordan blocks" is anomalous, but it can happen, and things "nearby" start to behave less stably.
Various non-normal compact operators on Hilbert spaces also can have non-trivial Jordan blocks. For example, the Volterra operator $Vf(x)=\int_0^x f(t)\,dt$ has this behavior.
The importance of rationality and algebraic-group-theoretic aspects of Jordan form in the theory of algebraic groups was already mentioned. This is a big deal, after all.
In general, "failure of semi-simplicity" is an awkward thing, and is to be proven not to happen (if that is the case). Failure is manifest in too-extensive non-trivial Jordan blocks. For example, "Galois representations" (that is, repns of Galois groups on various cohomologies of algebraic-geometric objects) "should be" semi-simple, etc., but this requires proof.
The case that a second-order ODE degenerates to have two closely-related solutions is an instance of a non-trivial Jordan block.
In complex analysis and algebraic geometry, representations of $\pi_1(X)$ as "monodromy groups" raise the potential issue of non-trivial Jordan blocks.
And so on. "Jordan block" is an essential descriptive notion, nearly everywhere. People who say it's "useless" are either just playing rhetorical games, or are pretty ignorant of serious mathematics.
Jordan canonical form is extremely important in the structure theory of linear algebraic groups.
Algebraic groups are of interest in many areas of pure mathematics, showing up in representation theory, algebraic geometry, number theory, and differential geometry (an algebraic group is the algebro-geometric analogue of a Lie group, and in fact all classical Lie groups can be regarded as algebraic groups), .
To make life easy, suppose $k$ is an algebraically closed field, and consider the group $\operatorname{GL}_n$ of $n$ by $n$ invertible matrices with entries in $k$. Then $\operatorname{GL}_n$ has a topology, called the Zariski topology, whose basic open sets are of the form
$$\{ (x_{ij}) \in \operatorname{GL}_n : \frac{f(x_{ij})}{\det(x_{ij})^m} \neq 0 \}$$
where $f$ is a polynomial in $n^2$ variables with coefficients in $k$, and $m$ is a nonnegative integer. A linear algebraic group is a closed subgroup of some $\operatorname{GL}_n$.
Examples: $\operatorname{GL}_1$ is just the group of nonzero elements of $k$. $\operatorname{SL}_n$ is the group of determinant one matrices. $\operatorname{SO}_n$ is the group of determinant one matrices whose inverse is their transpose.
Let $G \subseteq \operatorname{GL}_n$ and $H \subseteq \operatorname{GL}_m$ be linear algebraic groups. A morphism of algebraic groups is a group homomorphism $f: G \rightarrow H$ which is also a morphism of varieties. That is, for each $(x_{ij}) = x \in G$, $f(x)$ returns you an $m$ by $m$ matrix, and the entries of this matrix must be functions of $x$ of the form $\frac{f(x_{ij})}{\det(x_{ij})^p}$ for $f$ a polynomial and $p \geq 0$.
Here is one way to state Jordan canonical form.
Theorem: Let $A$ be an $n$ by $n$ matrix. There are unique matrices $A_s$ and $A_n$, such that $A_s$ is diagonalizable, $A_n$ is nilpotent, $A_s A_n = A_nA_s$, and $A = A_s + A_n$.
And here is the "multiplicative" version of Jordan canonical form, which follows directly from the usual version.
Theorem: Let $g \in \operatorname{GL}_n$. There are unique matrices $g_s, g_u$ such that $g_s$ is diagonalizable, $g_u$ is unipotent (i.e. $I - g_u$ is nilpotent), and $g = g_sg_u = g_u g_s$.
Now here is a remarkable theorem for linear algebraic groups, which is false for arbitrary subgroups of $\operatorname{GL}_n$:
Remarkable theorem: Let $G \subseteq \operatorname{GL}_n$ be a linear algebraic group. Let $g \in G$. Then $g_s$ and $g_u$ are in $G$.
Reference: T.A. Springer, Linear Algebraic Groups, Theorem 2.4.8
There is no immediate reason why $G$ should contains the matrices $g_s$ and $g_u$. Even though $g \in G$, the matrices $g_s$ and $g_u$ are a priori just some $n$ by $n$ invertible matrices which multiply to $g$. Yet $G$ must contain them.
Moreover, the notion of an element of a linear algebraic group being diagonalizable, or unipotent, exists independently of a particular realization of $G$ as a group of matrices! That is, if $H \subseteq \operatorname{GL}_m$ is a linear algebraic group, and $\phi: G \rightarrow H$ an isomorphism of algebraic groups, then for each $g \in G$, we have $\phi(g_s) = \phi(g)_s$ and $\phi(g_u) = \phi(g)_u$. Thus $\phi(g_s)\phi(g_u)$ is the Jordan decomposition of $\phi(g)$.
For an example of where Jordan canonical form is used in the structure theory, let $G$ be a linear algebraic group which is connected and solvable (for example, upper triangular matrices with nonzero diagonal elements). Let $G_u$ be the set of unipotent elements of $G$ (as mentioned in the previous paragraph, the notion of an element being unipotent does not depend on the specific realization of $G$ as a group of matrices). There exist maximal closed, connected subgroups of $G$ consisting of diagonalizable elements, called maximal tori, and if $T$ and $S$ are two maximal tori of $G$, then there exists a $g \in G$ such that $gTg^{-1} = S$. Then $G_u$ is a closed, connected, normal subgroup of $G$, and if $T$ is any maximal torus of $G$, then $G$ is the semidirect product of $G_u$ and $T$. (Reference: Springer 6.3.5)
• It's also worth noting I think that without Jordan decomposition it would be impossible to define reductive groups, perhaps the most important class of linear algebraic groups. For the OP, reductive algebraic groups are of interest in part because their finite dimensional representations are completely reducible over $\mathbb{C}$. – leibnewtz Mar 11 '18 at 19:20
• Nice expository post about a fairly technical point in algebraic group theory! Let me add that its relevance doesn't start in algebraic group theory; the additive analogue of this decomposition ($x = x_s + x_n$ instead of $g = g_s g_n$) also figures in the algebraic proof of Weyl's theorem stating that any finite-dimensional module over a semisimple Lie algebra is completely reducible. – darij grinberg Mar 11 '18 at 23:32
Although the Jordan canonical form (JCF) may not be too useful for applied mathematicians, it is terribly important for theoretical mathematicians (see several consequences of the JNF), especially matrix theorists.
For instance, suppose that $f:\mathbb{C} \longrightarrow \mathbb{C}$ and $A$ is an $n$-by-$n$ matrix such that $A= S D S^{-1}$, with $\text{diag}(\lambda_1,\dots,\lambda_n)$. It is natural to define $f(A)$ as the $n$-by-$n$ matrix $Sf(D)S^{-1}$, with $f(D) = \text{diag}(f(\lambda_1),\dots,f(\lambda_n))$. What about the case when $A$ is defective? Is it possible to make sense of $f(A)$?
The answer is 'yes' and can be found in the book on matrix function theory by Higham. The JCF plays a central role in this book (and many treatises on matrix theory).
As you noted, the Jordan Canonical Form (JCF) is not very useful for applied math, but it is in fact important for pure math.
I'll make an example which doesn't touch very advanced topics. You know that JCF gives you a classification of endomorphisms of a vector space. Now, this consequently leads you to a classification of homographies, and gives you a way to study their fixed points, invariant subspaces etc.
For example, if you know that every endomorphism $f: \mathbb{C}^n \longrightarrow \mathbb{C}^n$ has a flag basis, you are already able to prove that every homography of a complex projective space has a fixed point.
More generally, it's easy to see that if the matrix which induces the homography has a JCF with $k$ blocks for the same eigenvalue, than the homography have a projective subspace of fixed points of dimension $k-1$. | 2019-06-20T11:29:21 | {
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http://wwbl.ratingcomuni.it/exponential-and-logarithmic-functions-notes-pdf.html | # Exponential And Logarithmic Functions Notes Pdf
(absolute value, square root, cube root, rational, polynomial, exponential, and logarithmic). One‐to‐One Properties For any exponential function f(x) = bx If bu = bv, then u = v. Differentiating logarithm and exponential functions This unit gives details of how logarithmic functions and exponential functions are differentiated from first principles. The function f(x) = log a x is. Functions that are not algebraic are. Write in exponential form: a) log 2 8 = 3 b) log 4 1 = 0 c) log 12 12 = 1 What can we conclude about log b 1 (for any base b)? _____ What can we conclude about log b b (for any base b)? _____ Graphing Logarithmic Functions Logarithmic functions are inverses of exponential functions Graph y = 2x Graph the inverse, x = 2y x = 2y is the same as y. Probability Density Function (PDF) The exponential distribution PDF is similar to a histogram view of the data and expressed as. Logarithms are not a part of this lesson, but it is recommended that you return to these examples at the end of the unit and complete the logarithm portions. log a 1 = 0 2. 10 Exponential and Log Functions Page 7 of 21 12/18/2014 Ex 6b : Use the calculator to approximate the value of log 35 3. To convert a MuPAD notebook file to a MATLAB live script file, see convertMuPADNotebook. For any positive number a>0, there is a function f : R ! (0,1)called an exponential function that is defined as f(x)=ax. 1 The Functions y =2 xand y. Today, logarithms are still important in many fields of science and engineering, even though we use calculators for most simple calculations. 9, we can also easily sketch graphs of modi ed exponential and logarithmic functions. Chapter 12_Logarithms Word Problems Up to this point we have seen only exponential growth. They take notes about the two special types of logarithms, why they are useful, and how to convert to these forms by using the change of base formula. Printable in convenient PDF format. Unit 7 Exponential And Logarithmic Functions Adv Algebra Korek. The natural logarithmic function, y = log e x, is more commonly written y = ln x. a new graph sketch both the function and the inverse you’ve found. Review for Test on Exponents, Logs, and Exponential Functions. Estimate the number in 2005. Here are the inverse relations. Notes from today can be downloaded here. CHAPTER 4 Exponential and Logarithmic Functions Section 4. Definition of a Logarithm Function: The inverse of the exponential function, written log b x = y , where x = b y; x > 0, b > 0, b ≠ 1. Exponential and Logarithmic Functions. pdf "OpenStax Algebra & Trigonometry Recorded Lectures and Notes for Precalculus" by Libby Gore under a Creative Commons Attribution-NonCommercial-ShareAlike 4. fx 3 x 5 1 2 x fx §· ¨¸ ©¹ Example: fx2 x is an exponential function. Lesson 20: Transformations of the Graphs of Logarithmic and Exponential Functions Student Outcomes Students study transformations of the graphs of logarithmic functions. Tempered distributions and the Fourier transform 11 1. Power Point presentation, 11 slides, Explaining with examples how the exponential and logarithmic functions are used in real life situations, based on IB Mathematics Standard Level Syllabus. Use transformations to graph the function. Let's use our calculators to find the derivative of y = ex. Definite integrals as a limit of a sum, Fundamental Theorem of Calculus without proof. If f(x) = e x, then f ' x =limh→0 f x h −f x. Exponential and logarithmic functions are used to model several real-life situations. 1 Exponential Growth 9. Thus, $$g(x)=x^3$$ does not represent an exponential function because the base is an independent variable. 3 1 customer reviews. Exponential exists to equip church planting and multiplication leaders with conferences, eBooks, videos, webinars, podcasts, and learning communities Exponential is a growing community of leaders committed to accelerating the multiplication of healthy, reproducing faith communities. Logarithms. This sort of equation represents what we call "exponential growth" or "exponential decay. In partnership with the. It is called a "common logarithm". First Sheets Second Sheets Reading and WritingAs you read and study the chapter, fill the journal with notes, diagrams, and examples for each lesson. Note: We have no de nition for ax when a < 0, when x is. Logarithmic functions to the base a have properties similar to those of the natural logarithmic function. pdf, 93 KB. Lesson 23 —Exponential Functions So far we’ve learned about polynomial functions and rational functions. Notes on Logarithmic Functions 1/21 and 1/22: File Size: More Review of Exponential and Log Functions 1/28: File Size: 51 kb: pdf: Download File. Check for. Wong , Kaman Sciences Corporation Dikewood Division/Santa Monica 2800 28th Street, Suite 370 Santa Monica, California 90405 January 1989 Abstract The current along the thin toroidal and elliptical antennas and the electromagnetic fields at and. Connect with social media. 7-5 Properties of Logarithms Student Notes. 1) f(x) = - 2 x + 3 + 4 1). Let's use our calculators to find the derivative of y = ex. Checking for Extraneous Solutions Solve log 5x+ log (xº 1) = 2. VRS For Chapter 5A 5B Quiz 3. The range is the resulting values that the dependant variable can have as x varies throughout the domain. Let u(z) is the exponential generating function for undirected cycles of length at least 3. 1 Formulations Exponential family is a class of distributions that all share the following form: p(yj ) = h(y)expf TT(y) A( )g (1) is the natural parameter, (a. These functions also have applications in science, engineering, and business to name a few areas. The logarithmic function to the base a, where a > 0 and a ≠ 1 is defined: y = logax if and only if x = a y logarithmic form exponential form When you convert an exponential to log form, notice that the exponent in the exponential becomes what the log is equal to. Find f 2 b. JS function alternatives and useful snippets of code. Exponentials and Logarithms (naturally)81 1. Notes Chapter 5 (Logarithmic, Exponential, and Other Transcendental Functions) Definition of the Natural Logarithmic Function: The natural logarithmic function is defined by The domain of the natural logarithmic function is the set of all positive real numbers. You may be able to try some numbers at random (hint: 4 does not. fg(2) ( 2) d. The range is the resulting values that the dependant variable can have as x varies throughout the domain. notes_-_pre-calculus_12_-_ch. com/watch?v=oJJs3zD8lvw. I tried getting the moments by substituting the r values, and then I get a series. All exponential functions are relatives of this primitive, two parameter family. 3 Logarithmic Functions and Models Log functions are the inverse of Exponential funcitons: SMART Board Interactive Whiteboard Notes. 5 Applications of Exponential and Logarithmic Functions 469 6. • Use the one-to-one property of logarithms to solve logarithmic equations. Negative exponents. In order to master the techniques explained here it is vital that you undertake plenty of. eureka-math. Chapter 3 Exponential & Logarithmic Functions Section 3. Example 2: Rewrite each equation in exponential form. Find the value of y. Here we give a complete account ofhow to defme eXPb (x) = bX as a. These same limitations on the variables are true for the logarithm function as well. The development of electronic devices at the single-molecule scale requires detailed understanding of charge transport through individual molecular wires. Some problems (rated with *) are in advance level, however, they are very useful for better understanding of. The logarithms differ only by a constant factor, and the big O notation ignores that. 2: Logarithmic Functions Logarithmic Functions are of the form f(x) log (x) b, where the base b is a number ____ but not equal to 1 and where x0. 2) u(z)= 1 2 log 1 1−z −z − z2 2. logarithmic functions with base Section 3. Graphing Exponential Functions What is an Exponential Function? Exponential functions are one of the most important functions in mathematics. Unit 4 - Exponential and Logarithmic Functions. org -M3 TE 1. Exponential and logarithmic functions multiple choice questions and answers (MCQs), exponential and logarithmic functions quiz answers pdf 5 to learn online business math courses. x b = y, then log yx = b. 1B Graphing Exponential and Logarithmic Functions Filled In. The current population of the city is 175,000, up from 106,000 ten years ago. Describe this monkey population. 3 The Number e 9. To convert a MuPAD notebook file to a MATLAB live script file, see convertMuPADNotebook. 1 Exponential and Logistic Functions_NOTES. Michael Fowler. The reflection in the line y = x is used to make. survival function (no covariates or other individual differences), we can easily estimate S(t). B c CA]lzlX nrpiWg]hJtIss [raeQsPegrjvOemdf. Exponential and Log Functions Worksheet Exponential Functions and Inverse of a Function 1. Let's start with $\ds \log_e x$, which as you probably know is often abbreviated $\ln x$ and called the "natural logarithm'' function. The function f(x) = log a x is. Exponential and logarithmic functions quiz questions and answers pdf, test for online certifications. What are the two transformation equations? c) A vertical translation is applied to the graph of y. (We saw this before for x a rational number). Exact 3 e2 − Approx. Exponential functions, while similar to functions involving exponents, are different because the variable is now the power rather than the base. Reviewed for Unit 2 Test Tomorrow; Homework is to Finish Review [Reminder]: Unit 2 homework packets and review sheets are due tomorrow!. 2 Exponential Function ex section 4. Since the copy is a faithful reproduction of the actual journal pages, the article may not begin at the top of the first page. Logarithms can also have a base of e. 78A Using Exponential & Logarithmic Functions •use logarithms to solve problems involving exponential growth and decay. Les't compare with. ©2015 Great Minds. Notes 47 Transforming Exponential and Logarithmic Functions Objectives: Transform exponential and logarithmic functions by changing parameters Describe the effects of changes in the coefficients of exponential and logarithmic functions Who uses this?. The Mathematics Vision Project: Scott Hendrickson, Joleigh Honey, Barbara Kuehl, Travis Lemon, Janet Sutorius. exponential function. Notes on Exponential and Logarithmic Models Exponential and Logarithmic Models Notes Exponential and Logarithmic Models Notes filled in Homework: pages 265-266: 35, 37, 39, and 43. These are guided notes for exponential functions and an answer key. Derivatives of Exponential and Logarithmic Functions We already know that the derivative of the func tion t e with respect to t is the function itself, that is,. https://www. AP STATS Section 4 1 Notes EXPONENTIAL ONLY October 21, 2019 Section 4. The section covers: Introduction to Logarithms Special Logarithms Using Logs (and Exponents) in the Graphing Calculator Parent Graphs of Logarithmic Functions Transformations of Log Functions Basic Log Properties, including Shortcuts Expanding and Condensing Logs Solving Exponential Equations using Logs Solving Log Equations Applications of Logs, including Half-Life Inverses of Exponential and. Graph y xex, Graph y ln y x The Natural Base e (Euler’s number): An irrational number, symbolized by the letter e, appears as the base in many applied exponential functions. 005 Now, let's try and nd the derivative of the function f(x) = ex at any. Unit 4 - Exponential Functions. Calculus Notes Exponents and Logarithms Exponential Function: Has the form y 1ax, where ! 0,a z and x is any real number. Logarithms can also have a base of e. Logarithmic functions are the inverses of exponential functions, and any exponential function can be expressed in logarithmic form. The given figure shows us the type of graph the exponential function portrays when the value of a is >1 or 0 0 and x any real number, we de ne a x= e lna; a > 0: The function ax is called the exponential function with base a. Class Notes. pdf: File Size: 257 kb: File Type: pdf. 4 Napierian logarithms 4. Properties of logarithms83 4. 6 Modeling with Exponential and Logarithmic Functions Exponential Growth and Decay Models: predict any quantity that grows or decays by a fixed percent at regular intervals Mathematical Model for Exponential Growth or Decay f t y y ekt 0 f t y : the amount of the substance at time t y 0: the original amount at t 0 k. 406 CHaptER 4 Inverse Exponential and Logarithmic Functions One-to-One Functions Suppose we define the following function F. Search this site. 11 Exponential and Logarithmic Functions Concepts: • Exponential Functions – Power Functions vs. Precalculus 115, section 4. notebook 1 October 25, 2019 Opening: Review of Exponential Functions James deposits $550 into an account that earns 16% interest annually. From a general summary to chapter summaries to explanations of famous quotes, the SparkNotes Exponential Functions Study Guide has everything you need to ace quizzes, tests, and essays. docx File Size:. Write an exponential function in the form y = abx that could be used to model the number of cars y in millions for 1963 to 1988. (It is log-sum-exponential. logarithmic functions with base Section 3. Write 5x = 5x as a logarithm. Unit 4 Exponential and Logarithmic Functions Properties of Exponents Notes By the end of this lesson, students will be able to… Recognize and use exponent rules Properties of Exponents: The Product Rule: Example 1: a. Higher Mathematics Unit 3 – Exponentials and Logarithms hsn. { The function a( ) is convex. Some of them are: • Radioactive Decay: A = A0(2-t/h) where A 0 is the amount present at t = 0 and h is the materials half-life. In fact, $$g(x)=x^3$$ is a power function. First Sheets Second Sheets Reading and WritingAs you read and study the chapter, fill the journal with notes, diagrams, and examples for each lesson. Jan 45:27 PM Graph each function and state two points and asymptote for each. It is very important in solving problems related to growth and decay. Whoops! There was a problem previewing Chapter 10. Exponential Functions. Complex numbers The equation x2 + 1 = 0 has no solutions, because for any real number xthe square x 2is nonnegative, and so x + 1 can never be less than 1. logarithmic_functions_intro_notes_and_ws4_key. The number e1 = e ˇ2:7 and hence 2 < e < 3 )the graph of ex lies between the graphs of. Similarly, all logarithmic functions can be rewritten in exponential form. Calculus 130, section 2. (a) (b) (c) (d) (e) (f) 45 2 4 5 25 32 1003 2 100 3 103 1000 5. Integrals of Exponential and Logarithmic Functions. Exponential & Logarithmic Functions 7 – Natural Logarithms Pre-Calculus Page 2 of 2 Ex C: Word problems. 5 Applications of Exponential and Logarithmic Functions As we mentioned in Section6. (Recall that the derivative of logn is 1 n, a decreasing function. 0121, Calculus I March 10/11, 2009 Announcements Quiz 3 this week: Covers Sections 2. 0 Unported License. solving word problems applications of exponential and logarithmic functions. , Engineering Mathematics Sample Question. 1 The Natural Logarithmic Function: Differentiation • Develop and use properties of the natural logarithmic function. 2Review (Spring 2015) Solutions (Spring 2015) Ch. This lesson turns away from polynomials (and their close cousins, rational functions) and looks at another type of function: exponential and logarithmic functions. You can see some applications in the "Related Sections" panel at right. Integer Exponents. Use the logarithmic laws to express each side of the equations as a single logarithm. This allows us to set the exponents equal to each. notebook 2 November 12, 2014 Table of Contents Unit 5: Exponential and Logarithmic Functions 1. Begin with four sheets of grid paper. Inez Islas South Grand Prairie High 9th Grade Center Grand Prairie, TX sign up log in. HO #3 Worksheet. asp College Loan Project WORD DOC: File Size: 47 kb: File Type: Download File. For example, f(x) = 2x is an exponential function with base 2. 9, we can also easily sketch graphs of modi ed exponential and logarithmic functions. Exponential and Logarithmic Functions LESSON ONE - Exponential Functions Lesson Notes y = bx Some of these examples provide an excellent opportunity to use logarithms. 2 We’re mostly interested in the shape of the likelihood curve or, equivalently, the relative comparisons of the L( ) for di erent ’s. e Graph exponential and logarithmic functions, showing intercepts and end behavior, and trigonometric functions, showing period, midline, and amplitude. A Guide to Exponential and Logarithmic Functions Teaching Approach Exponents and logarithms are covered in the first term of Grade 12 over a period of one week. B c CA]lzlX nrpiWg]hJtIss [raeQsPegrjvOemdf. 5 Exponential and Logarithmic Functions 103 Chapter 5 Exponential and Logarithmic Functions 5. 3 Graphs of exponential functions 4. v Worksheet by Kuta Software LLC. 2 Logarithmic Functions and Their Graphs Objective: In this lesson you learned how to recognize, evaluate, and graph logarithmic functions. The reflection in the line y = x is used to make. The derivative of axand the de nition of e 84 6. The natural exponential function can be considered as \the easiest function in Calculus courses" since the derivative of ex is ex: General Exponential Function a x. Solving Unlike Bases 11/12. EXAMPLE 1 Identifying Exponential Functions (a) f x 3x is an exponential function, with an initial value of 1 and base of 3. Integrals of Exponential and Logarithmic Functions. Chapter 3 - Exponential and Log Functions; Notes. (Note that f (x)=x2 is NOT an exponential function. • We can use nonparametric estimators like the Kaplan-Meier estimator • We can estimate the survival distribution by making parametric assumptions – exponential – Weibull – Gamma – log-normal BIOST 515, Lecture 15 14. Standard F. x and mu can be vectors, matrices, or multidimensional arrays that all have the same size. Thus, $$g(x)=x^3$$ does not represent an exponential function because the base is an independent variable. D-11 Application of Exponential & Log Functions Notes → Compound interest: when interest is paid on the principal and the interest an account has already earned. (absolute value, square root, cube root, rational, polynomial, exponential, and logarithmic). Graph y x2x e. This resource is a collection of full hour-long lectures and accompanying notes covering most of chapters 6-10 of the OpenStax Algebra and Trigonometry book. Note: In the exponential function the variable x. One‐to‐One Properties For any exponential function f(x) = bx If bu = bv, then u = v. Logarithmic Functions The function ex is the unique exponential function whose tangent at (0;1) has slope 1. Pre-Calculus Honors Notes. Title: ExpoNENTIAL AND Logarithmic Functions. 60 MB (1675715 bytes). Piercey October 19, 2009. Notes on recorder for roar whole song Math 3 Official Exponential And Logarithmic Functions 2017 2018 Answer Key Ebook Pdf Blue Bottle Mystery Adventure. 718281828459. This section is always covered in my class. Let's use our calculators to find the derivative of y = ex. TEKS Aligned: A9A, A9B, A9C, A9DThe first 6 questions give the student either a graph or an equation and ask them to identify the function as linear, quadratic or exponential. These functions are central to an understanding of exponential growth (used to model populations and compound interest) as well as radioactive decay and other physical processes. When solving exponential equations we frequently used logarithmic identity 1 because it involves applying a logarithmic function to "undo" the effect of an exponential function. Guided Notes for Exponential and Logarithm Webquest History of Logarithms: 1. They take notes about the two special types of logarithms, why they are useful, and how to convert to these forms by using the change of base formula. Exponential Functions & Logarithms Name: Notes Date: Jorge is working with a client who has received an inheritance of$50,000. Limits involving exponentials and logarithms86 8. Whoops! There was a problem previewing Chapter 10. To characterize the electrical conductance, it is necessary to vary the length of a single molecular wire, contacted to two electrodes, in a controlled way. Thus, $$g(x)=x^3$$ does not represent an exponential function because the base is an independent variable. 1 Linear and Exponential Functions 1. PDF LESSON. Integrals of Exponential and Logarithmic Functions. logarithmic functions are different than those used for finding the instantaneous rate of change at a point for a rational function. 01 eh 1 h 0. Similarly, all logarithmic functions can be rewritten in exponential form. Common Logarithms: Base 10. Examples of transformations of the graph of f (x) = 4x are shown below. I emphasize how the graphing vocabulary applies to linear functions, exponential functions, and how this structure will be similar throughout all functions. to the base. Jorge makes 30 year. Learn exactly what happened in this chapter, scene, or section of Exponential and Logarithmic Functions and what it means. Write an exponential function in the form y = abx that could be used to model the number of cars y in millions for 1963 to 1988. Domain and Range of Exponential and Logarithmic Functions Recall that the domain of a function is the set of input or x -values for which the function is defined, while the range is the set of all the output or y -values that the function takes. The derivative of axand the de nition of e 84 6. pdf, 512 KB. Excel has functions that permit the rapid calculation of exponential functions with Napierian base. Limits involving exponentials and logarithms86 8. 99995 unde ned 1. Well, then what is an exponential function?. Chapter 3 Exponential and Logarithmic Functions Section 3. y = log (x – d) + c and the roles of the parameters a and k in functions of the form y = alog (kx) 2. Chapter 5 Exponential and Logarithmic Functions Section 5. The resources provided here for Gordon State College's Math 1111 course are a growing collection of activities, notes, problems, websites, and other resources you may need for this class. 4 Transformations Exponential and Logarithmic Functions with work Whiteboard Notes Notebook software,Notebook,PDF,SMART,SMART Technologies ULC,SMART Board. Introduction In this chapter we are going to look at exponential and logarithm functions. Here are a set of practice problems for the Exponential and Logarithm Functions chapter of the Algebra notes. I wanted to cover exponential functions right after linear functions to see if teaching y=a+bx actually helped my students to graph y=a(b)^x. 4 Growth Decay appreciation: an increase in value over time. 2) u(z)= 1 2 log 1 1−z −z − z2 2. Model-Fitting with Linear Regression: Exponential Functions In class we have seen how least squares regression is used to approximate the linear mathematical function that describes the relationship between a dependent and an independent variable by minimizing the variation on the y axis. We have moved on to Larson's 5 th edition and some sections have changed but I have left them where they are since many people on the Internet find these useful resources. Rule i) embodies the definition of a logarithm: log b x is the exponent to which b must be raised to produce x. The graph of the function defined by y = ln x, looks similar to the graph of y = log b x where b > 1. Today, logarithms are still important in many fields of science and engineering, even though we use calculators for most simple calculations. Limits of Exponential and Logarithmic Functions Math 130 Supplement to Section 3. It has a y-intercept of 1, a Horizontal Asymptote on the x-axis, and is monotonic increasing. I have also included extra notes from another Honors Math 3 course that might be helpful and would expose you to more examples:. Recall that x and y trade places in inverse functions. The natural logarithm has a base of "e". Natural Exponential Function In Lesson 21, we explored the world of logarithms in base 10. Note that if y=kxα, then Log[y]=Log[k]+αLog[x]. Here's what exponential functions look like: $$y=2^x$$ The equation is y equals 2 raised to the x power. You will be responsible for the following topics: Exponent properties. To print or download this file, click the link below: Exponentional-Logs. MA 131 Lecture Notes Exponential Functions, Inverse Functions, and Logarithmic Functions Exponential Functions We say that a function is an algebraic function if it is created by a combination of algebraic processes such as addition, subtraction, multiplication, division, roots, etc. The natural logarithmic function, y = log e x, is more commonly written y = ln x. Notes on Logarithmic Functions 1/21 and 1/22: File Size: More Review of Exponential and Log Functions 1/28: File Size: 51 kb: pdf: Download File. Use transformations to graph the function. Napier's logarithms helped ease that burden of calculating and re-calculating planetary positions, why? Exponential Growth, Decay, and Natural Number e Functions 3. MINI LESSON Lesson 4a - Introduction to Logarithms Lesson Objectives: 1. 1 Properties of Exponents & Basic Logarithms. To solve exponential equations today, we will try writing them so that the bases are the same…. Then state the function’s domain and range. "e" is approximately 2. (specifically, the graphs of all exponential functions pass the horizontal line test, so all exponential functions are one-to-one), we can conclude that all exponential functions have inverse functions. Chapter 6 : Exponential and Logarithm Functions. Both of these functions are very important and need to be understood by anyone who is going on to later math courses. Mon 9/30 Investigation - Foundations Sheet. Notes on the Matrix Exponential and Logarithm HowardE. The graph of a logarithmic function. They are provided here for your convenience. 1 Exponential Functions and Their Graphs Objective: In this lesson you learned how to recognize, evaluate, and graph exponential functions. Example 1: Evaluating Exponential Expressions Use a calculator to evaluate each expression a. Before, we dealt with functions of the form where the variable x was the base and the number was the power. 1 Exponential Functions Solutions to Even-Numbered Exercises 137 2. Often exponential rate of decay can be gotten from the half-life information. 2Review (Spring 2015) Solutions (Spring 2015) Ch. Let's examine the graph of y = 2 x. 6 Reduction of exponential la More. Like most functions you are likely to come across, the exponential has an inverse function, which is log e x, often written ln x (pronounced 'log x'). In this unit we look at the graphs of exponential and logarithm functions, and see how they are related. Logarithmic scale (with Vi Hart)Logarithmic scaleRichter scaleBenford's law (with Vi Hart, 1 of 2)Benford's law (with Vi Hart, 2 of 2) About this unit We know what exponents are and this chapter will reintroduce us to the concept of exponents through functions. 4 pose problems based on real-world applications of exponential and logarithmic functions. Write 90 = 1 as a logarithm. Then use that graph to draw the graph of yx log 2 Transformations work with logarithmic functions, too. Chapter 1 Algebra 1. In particular, we are interested in how their properties differ from the properties of the corresponding. php(143) : runtime-created function(1) : eval()'d. Both of these functions are very important and need to be understood by anyone who is going on to later math courses. Complete Notes \$ 6. Logarithmic Functions The logarithmic function with base is the _____ of the exponential function ( )=. Unit 4 Homework Assignments - Day 1: Complete Packet pg. 7 Spiral Review 2014. (Note that f (x)=x2 is NOT an exponential function. • Compound Interest: A = A0(1 + r/k). Section II: Exponential and Logarithmic Functions Unit 2: Exponential Models Suppose that the function EXAMPLE: mt( ) 21080 0. Write 5x = 5x as a logarithm. notebook February 23, 2016 Warmup Objective Students will solve real application problems involving exponential functions using logarithm properties. 2 Derivative Formulas for Exponential and Logarithmic Functions We start this section by looking at the limit lim h!0 eh 1 h: The chart below suggests that the limit is 1. Search this site. Notice that lnx and e x are reflections of one another in the line y = x. College Algebra Notes Joseph Lee Functions De nition: Relation A relation is a correspondence between two sets. We then use the chain rule and the exponential function to find the derivative of a^x. F = 51-2, 22, 1-1, 12, 10, 02, 11, 32, 12, 526 (We have defined F so that each second component is used only once. 7 Modeling with Exponential and Logarithmic Functions/Logarithmic Scales Please review Section 4. These notes were written during the Fall 1997 semester to accompany Larson's College Algebra: A Graphing Approach, 2nd edition text. Example 1 : Sketch the graph of y = 2 xand the modi cations y = 2 +1;y = 2x+1. Lesson 8-4 Properties of Logarithms. Several of the resource materials are "pdf" files. Estimate the number in 2005. mrskimrocksmath. 8 Write a function defined by an expression in different but equivalent forms to reveal and explain different properties of the function. Graphs of exponential functions and logarithms83 5. Complex numbers The equation x2 + 1 = 0 has no solutions, because for any real number xthe square x 2is nonnegative, and so x + 1 can never be less than 1. Exercises87. i F uA ml Rl9 6rIi EgGh utvs J 3r 9e2s qemrTv Gehd h. Write an exponential function in the form y = abx that could be used to model the number of cars y in millions for 1963 to 1988. The client would like to put the money into mutual funds and wants to diversify the investments. For any positive number a>0, there is a function f : R ! (0,1)called an exponential function that is defined as f(x)=ax. Algebra 2B: Chapter 7 Notes Exponential and Logarithmic Functions 11 x y Logarithms and exponential functions are inverses of each other. Lecture 12 4. (absolute value, square root, cube root, rational, polynomial, exponential, and logarithmic). 1 – Introduction to Exponential Functions SWBAT : Write and graph an exponential given a set of data or its equation. 1: Composite Functions Exploration 1*: Form a Composite Function 1. Chapter 5 Exponential and Logarithmic Functions Section 5. | 2019-12-09T19:53:57 | {
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https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_21&diff=cur&oldid=166757 | # Difference between revisions of "2018 AMC 8 Problems/Problem 21"
## Problem
How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$
## Solution 1
Looking at the values, we notice that $11-7=4$, $9-5=4$ and $6-2=4$. This means we are looking for a value that is four less than a multiple of $11$, $9$, and $6$. The least common multiple of these numbers is $11\cdot3^{2}\cdot2=198$, so the numbers that fulfill this can be written as $198k-4$, where $k$ is a positive integer. This value is only a three digit integer when $k$ is $1, 2, 3, 4$ or $5$, which gives $194, 392, 590, 788,$ and $986$ respectively. Thus we have $5$ values, so our answer is $\boxed{\textbf{(E) }5}$
## Solution 2
Let us create the equations: $6x+2 = 9y+5 = 11z+7$, and we know $100 \leq 11z+7 <1000$, it gives us $9 \leq z \leq 90$, which is the range of the value of z. Because of $6x+2=11z+7$, then $6x=11z+5=6z+5(z+1)$, so $(z+1)$ must be a mutiple of 6. Because of $9y+5=11z+7$, then $9y=11z+2=9z+2(z+1)$, so $(z+1)$ must also be a mutiple of $9$. Hence, the value of $(z+1)$ must be a common multiple of $6$ and $9$, which means multiples of $18(LCM \text{ of }\ 6, 9)$. So let's say $z+1 = 18p$, then $9 \leq z = 18p-1 \leq 90$, so $1 \leq p \leq 91/18\ or \ 1 \leq p \leq 5$. Thus the answer is $\boxed{\textbf{(E) }5}$ ~LarryFlora
## Solution 3
By the Chinese Remainder Theorem, we have that all solutions are in the form $x=198k+194$ where $k\in \mathbb{Z}.$ Counting the number of values, we get $\boxed{\textbf{(E) }5}.$
~mathboy282
## Video Solutions
https://youtu.be/CPQpkpnEuIc - Happytwin
https://youtu.be/7an5wU9Q5hk?t=939 - pi_is_3.14
~savannahsolver | 2022-05-18T01:22:23 | {
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https://math.stackexchange.com/questions/1067177/given-similar-matrices-a-and-b-how-to-find-m-such-that-b-m-1am | # Given similar matrices $A$ and $B$, how to find $M$ such that $B=M^{-1}AM$?
I am trying to teach myself linear algebra using Strang's Introduction to Linear Algebra. I would like to know what the most (or more) efficient way to solve this problem is by hand.
The question:
Show that A and B are similar by finding M so that $B = M^{-1} A M$:
$$A = \begin{bmatrix} 1 &&2\\ 3 &&4\end{bmatrix}$$
$$B = \begin{bmatrix} 4 &&3\\ 2 &&1\end{bmatrix}$$
The solution given by the textbook is:
$$B = \begin{bmatrix} 4 &&3\\ 2 &&1\end{bmatrix} = \begin{bmatrix} 0 &&1\\ 1 &&0\end{bmatrix}^{-1} \begin{bmatrix} 1 &&2\\ 3 &&4\end{bmatrix} \begin{bmatrix} 0 &&1\\ 1 &&0\end{bmatrix}$$
My impulse was to first try and diagonalize $A = S\Lambda S^{-1}$ and $B = J\Lambda J^{-1}$ to take $\Lambda = S^{-1}AS = J^{1}BJ$ => $B = JS^{-1}ASJ^{-1} = M^{-1}AM$. I found the eigenvalues to be $\frac {5 \pm \sqrt{33}}2$. I then tried to find the eigenvector matrix S for A and eigenvector J for B, whereupon I would take $J^{-1} = \frac 1{det(J)} J$ and multiply $SJ^{-1} = M$.
In matlab this seems like it would do the right thing, however on paper it is very laborious and as I am not very good at hand-calculating things, I worry that I would make computational mistakes or run out of time if I had to do this problem on an exam. What are some strategies for doing this kind of problem more efficiently? I assume that since I am self-studying, there is a strong possibility that I am doing this completely wrongheadedly and that there is a much simpler way available.
• you said you wanted to compute by hand. what is matlab doing there? – abel Dec 14 '14 at 3:22
• Sometimes doing things in the most general way possible (the way you did it, which is completely fine by the way) is not the best way to do it. In this case, you can tell that the numbers are somehow permuted. Looking to permutation matrices is should be the first approach here. – Cameron Williams Dec 14 '14 at 3:24
• Finding a similarity transformation, even when one knows (e.g. by checking the eigenvalues) that two matrices are similar, is not always an easy task. The case you asked about is special in that the entries of $A$ and $B$ are the same, just rearranged. That would be a clue to try the permutation matrices shown in the book's answer. – hardmath Dec 14 '14 at 3:32
• Thanks all! I see now that I could have arrived at the book's solution by trying the permutation matrices. (I was just using matlab to check the work I had done by hand since I had it open for something else already anyway.) – qirindash Dec 14 '14 at 5:20 | 2019-08-19T13:13:54 | {
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