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http://math.stackexchange.com/questions/235603/is-the-function-linear | # Is the function linear?
Given $$F(x) = \left(\begin{matrix} -x_2 \\ x_1+2x_2 \\ 3x_1 - 4x_2 \end{matrix} \right), x = \left(\begin{matrix} x_1 \\ x_2 \end{matrix} \right)$$ Prove whether $F$ is a linear function or not.
I've tried to prove it, but I'm not sure it's right:
$$F(x) = Mx \Rightarrow M = \left(\begin{matrix} 0 & -1 \\ 1 & 2 \\ 3 & -4 \end{matrix}\right)$$
Let there be $u,v \in \mathbb{R}^2$, then:
$$\begin{eqnarray*} F(u+v) = M (u+v) = Mu+Mv = F(u)+F(v) \end{eqnarray*}$$
and
$$F(\lambda{u}) = M\lambda{u} = {\lambda}Mu = {\lambda}F(u)$$
so the function $F$ is linear.
Using the matrix $M$ was not shown in any of the examples I've seen, I've just invented the technique, I think it's nicer (and almost certainly I've reinvented the wheel). What's the technical term for $M$?
I'm sure there is a missing step in arriving from the given function to the form $F(x)=Mx$, M looks like the coefficient vectors arranged column-wise.
Is the proof formally correct?
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$M$ is the matrix that represents your linear map $F:\mathbb{R}^2\to\mathbb{R}^3$. – wj32 Nov 12 '12 at 10:40
@wj32 How to write it formally? – Flavius Nov 12 '12 at 10:40
How "formal" do you need it to be? I think it would be sufficient for you to state that $F(x)=Mx$ and then conclude that $F$ is linear from the properties of matrix multiplication. – wj32 Nov 12 '12 at 10:43
I think if I could see it more formally, I could get more insight, get a better feel for what I've learned so far. – Flavius Nov 12 '12 at 10:44
I don't have time to type up a full answer, but here's a short explanation. Given bases for your finite-dimensional spaces $\mathbb{R}^2$ and $\mathbb{R}^3$, there is a unique matrix that represents each linear map $f:\mathbb{R}^2\to\mathbb{R}^3$. So if you're given a linear map $f$ then you can always write $f(x)=Mx$ for some matrix $M$, and vice versa. Here you've chosen to use the "standard basis" for each of the spaces. – wj32 Nov 12 '12 at 10:48
The main thing, is that there is a matrix $M$, such that $F(x)=M\cdot x$ for all vectors $x$. In fact for a mapping $F$ between vector spaces with fixed bases, this is equivalent to be linear. Then, that multiplying by a matrix is indeed a linear mapping, is straightforward: depend on the nice properties of matrix multiplication, and you proved it formally correctly.
So, what should be emphasized is that indeed $F(x)=M\cdot x$ for all $x$, but this can be easily seen by expanding the product on the right hand side.
$M$ is called the matrix belonging to the linear map $F$, with respect to the standard bases.
Let $V$ and $W$ be vector spaces, with basis $\mathcal B:=(b_1,..,b_n)$ in $V$ and basis $\mathcal C:=(c_1,..,c_m)$ in $W$. If $F:V\to W$ is a linear mapping, then define its matrix wrt. bases $\mathcal B$ and $\mathcal C$ as: $$[F]^{\mathcal B}_{\mathcal C} := \left[ [F(b_1)]_{\mathcal C}\, ... [F(b_n)]_{\mathcal C} \right]$$ where $[w]_{\mathcal C}$ denotes the culomn vector $\in \Bbb R^m$ of coordinates of $w$ in basis $\mathcal C$, i.e. $$[w]=\pmatrix{\alpha_1\\ \vdots \\ \alpha_m} \iff w=\alpha_1 c_1+..+\alpha_m c_m.$$ Then, by linearity one can check that for all $v\in V$ we have $$[F]^{\mathcal B}_{\mathcal C}\cdot [v]_{\mathcal B} = [F(v)]_{\mathcal C}\ .$$ Prove it first for the basis vectors $v=b_i$.
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+1. Can you elaborate just a tiny little bit the parts regarding "bases", "fixed bases" (never heard this one) and "standard bases"? I'm not sure I've understood the concepts so good from my readings to feel free to "play" with them like that (and my learning material is in German). – Flavius Nov 12 '12 at 11:01
For instance I've found $F(x) = \left(\begin{matrix} x_2 \\ x_1^2 \end{matrix} \right)$ as another example, I think there's no such matrix $M$, and $F$ doesn't look linear to me either. How would I prove it in that case? By contradiction? – Flavius Nov 12 '12 at 11:10
To show that it is not linear, show explicit numbers $x_1$, $x_2$ and $\lambda$ such that $F(\lambda x)\ne \lambda F(x)$, or show counterexample to the $F(x)+F(y)=F(x+y)$. Try to find the simplest counterexample.. – Berci Nov 12 '12 at 11:16
Thanks, this has cleared my mind :-) – Flavius Nov 12 '12 at 11:22
You've shown that your map can be represented by a matrix and then shown that it must be linear based on the properties of matrix multiplication.
In fact, most mathematicians would view the problem the other way round: matrices are introduced as a neat way of encoding a linear map: so while it is certainly true that any map that can be represented by a matrix $A$ is linear, it is also true that any linear map can be represented by a matrix.
To see why this is: consider your example, where you are dealing with the map which takes $\left(\begin{matrix} x_1 \\ x_2 \end{matrix} \right)$ to $\left(\begin{matrix} -x_2 \\ x_1+2x_2 \\ 3x_1 - 4x_2 \end{matrix} \right)$. In mathematics, we say that this is an example of a map from $\mathbb{R}^3$ (vectors with three components) to $\mathbb{R}^2$ (vectors with two components). I will now show that every linear map from $\mathbb{R}^3$ to $\mathbb{R}^2$ can be represented by a matrix in this way. The general case for maps from $\mathbb{R}^m$ to $\mathbb{R}^n$ is treated in exactly the same way.
First note that we can write any vector $a=\left(\begin{matrix} a_1 \\ a_2 \end{matrix} \right)$ in $\mathbb{R}^2$ as $$a_1\left( \begin{matrix} 1 \\ 0 \end{matrix} \right)+a_2\left( \begin{matrix} 0 \\ 1 \end{matrix} \right)=a_1e_1+a_2e_2$$ and any vector $b= \left( \begin{matrix} b_1 \\ b_2 \\ b_3 \end{matrix} \right)$ in $\mathbb{R}^3$ as $$b_1\left( \begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right) + b_2\left( \begin{matrix} 0 \\ 1 \\ 0 \end{matrix} \right) + b_3\left( \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right)=b_1e_1+b_2e_2+b_3e_3$$. Now let $\alpha$ be a linear map from $\mathbb{R}^2$ to $\mathbb{R}^3$. Then, if $a\in \mathbb{R}^2$, $$\alpha(a)=\alpha(a_1e_1+a_2e_2)=a_1\alpha(e_1)+a_2\alpha(e_2)$$
Now, since for $i=1,2$, $\alpha(i) \in \mathbb{R}^3$, we can write $$\alpha(e_i)=m_{1i}e_1+m_{2i}e_2+m_{3i}e_3$$ where the $m_{ij}$ are the components of the images of the vectors $e_i$ under $\alpha$. Writing the $m_{ij}$ as a matrix $$M=\left( \begin{matrix} m_{11} & m_{12} \\ m_{21} & m_{22} \\ m_{31} & m_{32} \end{matrix} \right)$$
it is quick to check that applying $\alpha$ to the vector $a=\left( \begin{matrix} a_1 \\ a_2 \end{matrix}\right)$ is exactly equivalent to multiplying by the matrix $M$.
Your question asks the opposite: given a matrix $M$,does multiplication by $M$ always define a linear map. Your calculations (which are perfectly precise) show that the answer is yes, although it's such a basic result that I think showing that the map can be represented by a matrix is probably enough for that question.
Moral: for vector spaces over the real numbers, linear maps are matrices are linear maps. Finding a matrix representation is a great way to show that a map is linear.
I'm not wure what was intended for this question, but you could show that the map is linear directly (without inventing matrices). That might be a slightly nicer example, as it doesn't assume properties of matrix multiplication which are non-obvious (though trivial to prove).
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http://math.stackexchange.com/questions/93698/smallest-non-negative-number-in-a-matrix | # Smallest Non-negative number in a matrix
There is a question I encountered which said to fill an $N \times N$ matrix such that each entry in the matrix is the smallest non-negative number which does not appear either above the entry or to its left. That is for $N = 6$ the matrix looks like this:
$$\begin{array}{} 0&1&2&3&4&5\\ 1&0&3&2&5&4\\ 2&3&0&1&6&7\\ 3&2&1&0&7&6\\ 4&5&6&7&0&1\\ 5&4&7&6&1&0 \end{array}$$
I was asked to find a function such that given row and column I can calculate the value at that point i.e
$$f(\text{row}, \text{column}) = \text{Matrix}[\text{row}][\text{column}]$$
I was looking at Nimbers and found the matrix in it exactly similar to it. There was also given a formula to calculate to calculate Matrix[row][column] which was row XOR column (XOR is bitwise exor).
However I was able to get the answer I am still unable as to how to arrive at the solution i.e. the proof that each entry in the matrix is the smallest non-negative number which does not appear either above the entry or to its left is equal to row XOR column.
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Denote by $a(m,n)$ the entry in row $m$, column $n$ of the matrix. By definition
$$a(m,n)=\operatorname{mex}\big(\{a(k,n):k<m\}\cup\{a(m,k):k<n\}\big)\;,\tag{1}$$
where $\operatorname{mex}(A)$ is the smallest non-negative integer not in $A$. The problem is to show that when all of the numbers involved are expressed in binary, $a(m,n)=m\oplus n$, the bitwise exclusive OR of $m$ and $n$. We’ll do this by induction on $m+n$. It’s certainly true when $m+n=0$: $a(0,0)=0=0\oplus 0$. Now assume let $m,n\in\mathbb{N}$, and assume that it’s true for all $m',n'\in\mathbb{N}$ with $m'+n'<m+n$; we’ll show that it’s true for $m$ and $n$.
Let $s=m\oplus n$. To show that $s=a(m,n)$ as defined by $(1)$, I need to show two things:
1. if $0\le t<s$, $t\in\{a(k,n):k<m\}\cup\{a(m,k):k<n\}$; and
2. $s\notin\{a(k,n):k<m\}\cup\{a(m,k):k<n\}$.
To show (1), suppose that $0\le t<s$, let $d=t\oplus s$, and let $k$ be the number of bits in $d$, so that $2^{k-1}\le d<2^k$. That is, $d$ has a $1$ in the $k$-th position counting from the right. Since $t<s$, this implies that $t$ has a $0$ in the $k$-th position, and $s$ has a $1$ there. Since $s=m\oplus n$, exactly one of $m$ and $n$ must have a $1$ in the $k$-th position; without loss of generality assume that $m$ has a $1$ in the $k$-th position. Then $d\oplus m$ has a $0$ in the $k$-th position and is identical to $m$ in positions to the left of the $k$-th position, so $d\oplus m<m$. Let $k=d\oplus m$; then $$k\oplus n=d\oplus m\oplus n=d\oplus s=t\oplus s\oplus s=t\;.$$ Moreover, $k<m$, so $k+n<m+n$, and therefore $t=k\oplus n=a(k,n)$ by the induction hypothesis. This completes the proof of (1).
To show (2), suppose, on the contrary, that $s\in\{a(k,n):k<m\}\cup\{a(m,k):k<n\}$. Without loss of generality we may assume that $s\in\{a(k,n):k<m\}$. Let $k<n$ be such that $s=a(k,n)$. Clearly $k+n<m+n$, so by the induction hypothesis we have $$k\oplus n=a(k,n)=s=m\oplus n\;,$$ and therefore $$k=k\oplus n\oplus n=m\oplus n\oplus n=m\;,$$ contradicting the choice of $k$. Thus, (2) also holds, and $m\oplus n=a(m,n)$.
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I think the first condition you put there is wrong For a(3,2) the value of s is 1 however range of t is {0, 2, 3} – Anubhav Agarwal Dec 23 '11 at 19:18
@Anubhav: You’ve misunderstood the condition. Condition (1) doesn’t say that every member of $\{a(k,n):k<m\}\cup\{a(m,k):k<n\}$ is less than $s$; it says that every non-negative integer less than $s$ belongs to that set. The only non-negative integer less than $a(3,2)$ is $0$, and $0\in\{0,2,3\}$, so there’s no problem. – Brian M. Scott Dec 23 '11 at 19:47
thanks brian, awesome solution – Anubhav Agarwal Dec 23 '11 at 20:02
I really like this question. I saw it on a website a year or so ago, closed the page so that I wouldn't see the answer, and then never found it again. I solved it by proving by induction that $$f(4i+r, 4j+s) = f(r,s)+4f(i,j)$$ --- this relation is apparent if you draw enough of the matrix. Let me know if you would like the details. Once you have this relation, you can apply it repeatedly to get
$$f(\sum a_i 4^i, \sum b_i 4^i) = \sum h(a_i,b_i) 4^i$$
where $h$ is a the function determined by the first 4x4 block of the matrix. This explains why it is possible to think of f as a "bitwise" operation.
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See this as well. – Ｊ. Ｍ. Dec 23 '11 at 16:41
It's the XOR function, and you can prove it by induction on the bit length. For $0\leq x,y<2^n$ with binary expansions $$x=\sum_{i=0}^{n-1} x_i 2^i, \quad y=\sum_{i=0}^{n-1} y_i 2^i \quad \text{for} \quad x_i, y_i \in \{0,1\}, \quad \text{and} \quad x,y \in I_n=\{0,1,\dots,2^n-1\},$$ we start by assuming inductively that $$f(x,y) = XOR(x,y) \equiv \sum_{i=0}^{n-1} x_i y_i 2^i$$ and observe (also as part of our inductive hypothesis) that each row and column (where one of $x$ or $y$ is fixed, but the other varies freely over $I_n$) contains precisely the values $I_n$ (once each).
Now for the inductive step, we let $x_n,y_n\in\{0,1\}$ be arbitrary, and note first that the base block $(x_i,y_i)=(0,0)$ is our inductive hypothesis. Next, note that for $(x_i,y_i)=(0,1)$ and $(x_i,y_i)=(1,0)$, all of the values in $I_n$ are already taken and so the minimality rule will assign values to $f$ starting with $2^n$ rather than $0$, filling these blocks with the values $I_{n+1} \setminus I_n = \{2^n,2^n+1,\dots,2^{n+1}-1\}$. Furthermore, these values will be assigned within each block in the same order as they were in the base block. Finally, the block $(x_i,y_i)=(1,1)$ will match the base block, since its predecessors (the two non-base, inductive blocks already considered) used the values from $I_{n+1} \setminus I_n$ but left those from $I_n$ vacant. Thus, the minimality criterion is additive: $$f(2^i x_i + x, 2^i y_i + y) = 2^i XOR(x_i,y_i) + f(x,y) = 2^i x_i y_i + f(x,y)$$ which proves the inductive step.
The key insight that makes this proof work is that the blocks of $(x,y)$ for $0 \leq x,y < N$ are "full" (containing each value from $0$ to $N-1$ in each row and column) iff $N=2^n$ is a power of two.
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OP did mention XOR in his question... – Ｊ. Ｍ. Dec 23 '11 at 16:48
@J.M.: yes, you're right... I elaborated (quite a bit)! – bgins Dec 23 '11 at 17:41
All these proofs rely on the observation that this question is related to the bitwise XOR operator (after a few examples, this is very easy to see). However, I still cannot see the intuitive connection between the original problem and the Nimbers. Does anyone have a clear insight on this?
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In NIM-like games the 'value' of a game position is (among other things) also always the smallest positive integer that is missing from the list of the values of the positions that can be reached from the said position in a single move. And here the entries on top (resp. to the left) of a given matrix entry are the values of exactly those positions that result, when we remove tokens from the pile giving the row (resp. column) index. In a 2-pile game of NIM this is also equal to the bitwise XOR of the two numbers. – Jyrki Lahtonen Jan 15 '12 at 10:08
Lets say you selected a row (say $A$ represented using n-bits). Now, for all the $2^n$ columns (say $B$) you select, you will get different bit strings, hence different numbers.
For a particular $A$, $A\oplus B$ is distinct for distinct values of $B$. Thats why you get different numbers and that too within the range of n-bit. The same is not true for AND or OR operators.
A B A xor B A and B A or B
0 0 0 0 0
0 1 1 0 1
1 0 1 0 1
1 1 0 1 1
Here you notice that if you take bitwise AND, then for a particular row you can get same values for different columns. For example, for $5^{th}$ row ie $(101)_2$ will have a value 5 $(101)_2$ at $5^{th}(101)_2$ as well as $7^{th}(111)_2$ column.
So, A[5][5] = A[5][7] = 5 which is not required. This problem comes because of 0 in the bitstring.
Similarly for OR, this repetition will come because of 1 in the bitstring.
A[5][5] = A[5][4] = A[5][1] = A[5][0] = 5
So, you can now see why bitwise XOR is used.
But its not using the smallest possible value. In your array, you can see that 4 and 5 are missing in $3^{rd}$ and $4^{th}$ rows; and 2 and 3 in $5^{th}$ and $6^{th}$ rows. This is because you have 6 rows and columns for which you need 3 bits which can accommodate 8 values.
\begin{array}{} 0&1&2&3&4&5\\ 5&0&1&2&3&4\\ 4&5&0&1&2&3\\ 3&4&5&0&1&2\\ 2&3&4&5&0&1\\1&2&3&4&5&0 \end{array}
To get the minimum values, you can just shift the values while stepping onto the next row.
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https://math.stackexchange.com/questions/1776308/how-do-you-graph-an-inequality-on-real-imaginary-plane/1776772 | # How do you graph an inequality on Real/Imaginary plane?
Suppose we have $z$ as a complex number, $z \in C$, how do you graph an inequality which has $z$ in it?
This kinds of inequalities arise when we need to graph the shape of stability region of a given numerical method:
For example, for RK2 method we can derive the stability inequality:
$|1+z+\frac{z^2}{2}| < 1$
What I could do so far was to set the inequality to zero and solve for $z$:
$|1+z+\frac{z^2}{2}| = 1$ so $z = 0, -2$. Therefore $z$ must be between $0$ and $-2$.
I also know the existence of circle formula: $(x-h)^2 + (y-k)^2 = r^2$ which sometimes helps to graph an inequality by transforming the inequality into the circle formula at origin $(h, k)$ with radius $r$.
But I can't transform the above RK2 inequality to a circle formula, and also unable to get the imaginary part properly, if I take the 'set to equality' method (though I know $-2< Re{\{z\}} < 0$).
Ok the answers provided by Eric Stucky and Morgan Rodgers both pointing at replacing $z$ with $x+iy$ is prefect. The problem is that in the exams there is no computer to plot the inequality derived in terms of $x$ and $y$.
Below is the approach I learned from http://www.math.ubc.ca/~peirce/M405_607E_Lecture%2018.pdf which I found the most suitable approach when you are in exam room:
First set the inequality to the function G such that:
$G(z) = 1 + z + \frac{z^2}{2}$
For stability we require $|G(z)| < 1$.
Solve $z$ in terms of $G$:
$z = -1 \pm \sqrt{2G-1}$
Now we can start graphing the $z$. First we assume $G$ is a circle of radius 1 at the origin of real/imaginary plane. Then we scale $G$ by two and we get $2G$, then we shift the circle to the left and we get $2G-1$ and so on unil we get $-1 \pm \sqrt{2G-1}$.
The process has been shown with step by step conformal map in http://www.math.ubc.ca/~peirce/M405_607E_Lecture%2018.pdf
I consider this as the best answer, specially in exam rooms. I will wait for further confirmation by the community and then will accept this as answer after few days.
You sketch in the boundary line, where the equation equals 0, then you shade in one side (corresponding to the inequality).
Notice that in your equation, you have $|1+z+\frac{z^{2}}{2}| = 0$, but it's not just as simple as factoring the inner equation; also for complex numbers it is not especially meaningful to say that $z$ is "between" $0$ and $2$. You actually get $$\sqrt{(1+z+\frac{z^{2}}{2})(1 + \overline{z} + \frac{\overline{z}^{2}}{2})} - 1= 0.$$ I think if you want to solve this you should view $z = x+y\mathrm{i}$ and treat it like an equation in two variables. This gives (once you get rid of the square root) a 4th degree equation in $x$ and $y$.
• Thanks. Setting $z = x + iy$ is the key. May 8, 2016 at 7:47
The beginning and end of the story here is: set $z=x+iy$ and convert to an inequality of two variables. There's no particularly good reason to believe that your domain should be anything nice like a circle.
In this case, square both sides and do the substitution to get
\begin{align*} \left|1+x+iy+\frac12x^2+ixy-\frac12y^2\right|^2<1 \\ \left(1+x+\frac{x^2-y^2}{2}\right)^2 + (y+xy)^2<1 \\ 1+x^2+\frac{x^4-2x^2y^2+y^4}{4}+2x+x(x^2-y^2)+(x^2-y^2) + y^2(1+2x+x^2)< 1 \\ 2x^2+\frac{x^4+2x^2y^2+y^4}{4}+2x+x(x^2+y^2) < 0 \\ 2x^2+\frac{|z|^4}{4}+2x+x|z|^2 < 0 \\ \end{align*}
This is a quadratic inequality in $|z|^2$ and $x$, with bounds being $-2x\pm2\sqrt{-x(x+2)}$. Since the bounds must be real, we recover your already discovered inequality on the real part.
Alternatively, you can leave it in terms of $x$ and $y$; which is probably the better form for putting it into an automated graphing device. It seems you will need this (or a lot of patience, trial, and error), since the region described is not so nice analytically.
• I appreciate your effort of solving the inequality in terms of $x$ and $y$. I think what I was missing is the key of setting $z = x + iy$ which Morgan Rodgers also mentioned. Meanwhile I found this which is less error prone and more efficient: math.ubc.ca/~peirce/M405_607E_Lecture%2018.pdf May 8, 2016 at 7:46
• I wrote down the answer in separate post below. Actually the region is very nice analytically. It is the inner bound of a circle placed in left half plane. May 8, 2016 at 14:22 | 2022-10-06T20:36:17 | {
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https://math.stackexchange.com/questions/2082952/what-does-fracx39-bigg-01-mean-and-how-should-it-be-spoken | # What does $\frac{x^3}{9}\bigg|_0^1$ mean, and how should it be spoken?
$$\frac{x^3}{9}\Bigg|_0^1$$ The vertical line above: what does it mean, and how would I state this whole structure in spoken words, so that a screen reader would be able to read it aloud correctly?
• Evaluate. In particular here, $f(x)|_a^b$ usually means $f(b)-f(a)$. – Elliot G Jan 4 '17 at 6:13
It looks like people have already told you what to do mathematically. I'm a math tutor, and what I say out loud is
"x cubed over nine, evaluated from one to zero."
• Yes, of course! Thank you; an edit on the way! – InertialObserver Jan 4 '17 at 16:27
• Wait, why would you say from 1 to 0 and not vice versa? I normally say from 0 to 1 because we go from left to right. I think saying from 1 to 0 we should pick up a minus sign. – user223391 Jan 4 '17 at 16:35
• That's interesting. How I learned it was with the bar like this we say "evaluated from 'upper bound' to 'lower bound'". However, when we integrate we say "an integral from 'lower bound' to 'upper bound'". Seems like 'evaluated' is the key word. Not saying that this is the only 'right' way, though. Really, it's just about what people around you and your professors said most. – InertialObserver Jan 4 '17 at 16:41
The vertical line means evaluate it from the top to the bottom.
So say $x^3/9$ from $0$ to $1$.
in general $f(x)\biggr \vert^b_a$ would be $f(b)-f(a)$, in this case:
You would evaluate it as $(1)^3/9 - (0)^3/9$.
Basically just evaluate the expression with $x=$ top limit and the bottom limit, subtract the bottom expression from the top expression.
• How would I write this out in spoken words, so a screen reader would state it correctly aloud? Thank you. – Chelonian Jan 4 '17 at 6:17
• @Chelonian This is context vital to the question; it sounds like you're working on accessibility, which, if true, is quite the lofty and commendable goal! – pjs36 Jan 4 '17 at 6:21
• @pjs36 Yes, thank you. – Chelonian Jan 4 '17 at 6:22
• Wait, why would you say from 1 to 0 and not vice versa? I normally say from 0 to 1 because we go from left to right. I think saying from 1 to 0 we should pick up a minus sign. – user223391 Jan 4 '17 at 16:35
It is the standard notation for describing the limits of an integral.
As well mentioned in comments and other answers, $f(x)\biggr \vert^b_a$ is same as $f(b)-f(a)$ and here $a$ and $b$ are called limits of the integral or the integral is said to be over the interval $[a,b]$ .
When you write something like $\frac{x^3}{9}\Bigg|_0^1$ you are gonna say it as
Evaluate $\frac{x^3}{9}$ over the interval $[0,1]$ or evaluate $\frac{1^3}{9}-\frac{0^3}{9}$. | 2019-10-18T05:57:31 | {
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http://math.stackexchange.com/questions/88378/proof-that-the-relation-5-mid-a-4b-is-symmetric-and-transitive | # Proof that the relation $5 \mid (a + 4b)$ is symmetric and transitive
Take the relation $R$ to be defined on the set of integers:
$$aRb \iff 5 \mid (a + 4b)$$
As part of a larger proof, I have to show that $R$ is both symmetric and transitive. I'm lost.
I see the first steps, but I can't find how to progress further. Here's what I have at this point:
Proof of Symmetry
We have to prove that if $5 \mid (a + 4b)$, then $5 \mid (b + 4a)$. Clearly, this is true if $a = b$, but apart from that, it's unclear in my mind.
Proof of Transitivity
We have to prove that if $5 \mid (a + 4x)$ and $5 \mid (x + 4b)$, then $5 \mid (a + 4b)$.
I've fiddled around with sample values, but I still don't see it. I'm pretty lost here. Thoughts?
-
$aRb$ iff $5\mid a-b$ iff $5\mid b-a$. – Did Dec 4 '11 at 22:34
a hint: $(a+4b) + (b+4a) = (5a + 5b)$ – deinst Dec 4 '11 at 22:36
I've changed "Symmetric proof" to "Proof of Symmetry". "Symmetric proof" reads as if you are saying that the proof is (somehow) symmetric, rather than that the proof is a proof about symmetry. "Symmetry proof" would have been okay, too... – Arturo Magidin Dec 4 '11 at 23:28
HINT $\rm\ \ a\:R\:b\ \iff\ a-b\in 5\:\mathbb Z\:.\:$ If so, negating yields that $\rm\: b-a\in 5\:\mathbb Z\:$ hence $\rm\:R\:$ is symmetric. Transitivity follows by addition: $\rm \ a-b\:,\ b-c\:\in 5\:\mathbb Z\ \Rightarrow\ a-b + b-c\ =\ a-c \in 5\:\mathbb Z\:.$
Hence symmetry arises from $\rm\:5\:\mathbb Z\:$ being closed under negation, and transitivity arises from $\rm\:5\:\mathbb Z\:$ being closed under addition, i.e. from it being an additive subgroup of $\rm\mathbb Z\:.$ The innate algebraic structure will be brought to the fore when you study congruences and ideals of rings.
-
Symmetry: If $5\mid a+4b$, does $5\mid \big((a+4b)-(5a+5b)\big)$?
Transitivity: If $5\mid a+4x$ and $5\mid x+4b$, what can you say about $(a+4x)+(x+4b)$? How is this number related to $a+4b$?
-
@deinst's comment cleared up the symmetry part for me, but I don't see why adding $a+4x$ to $x+4b$ helps us solve the transitivity part of the proof? – David Chouinard Dec 4 '11 at 22:52
$(a+4x)+(x+4b)=(a+4b)+5x$, so $a+4b=(a+4x)+(x+4b)-5x$. What do you know about each of the three terms on the righthand side? – Brian M. Scott Dec 4 '11 at 22:59 | 2015-07-30T17:04:00 | {
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https://math.stackexchange.com/questions/2423292/expected-number-of-empty-bins-with-clumpy-balls | Expected number of empty bins with clumpy balls?
I have $b$ bins and a reservoir of $n$ balls.
The balls have a tendency to "clump", that is, when I try to grab one from the reservoir, the number of balls removed is uniformly random on $[1, m]$, where $m$ is the number of balls remaining in the reservoir for that draw.
The "clump" is added to one of the $b$ bins, the bin selected from all bins with equal probability.
When the reservoir is depleted, how many empty bins are expected?
I don't even know where to start...
Edit: I asked a neighbor (an actuary) this question, found this on a note on my windshield this morning, seems to match the answer by Quasi :
$\frac{b\prod _ {z=\frac{b-1}{b}}^{\frac{b-1}{b}+n-1} z}{n!}$
• Are you expecting a closed form? If so, why? – quasi Sep 10 '17 at 0:33
• Does the source problem give actual values for $n$ and $b$? – quasi Sep 10 '17 at 0:44
• @quasi my math stopped at college algebra, I can do most simple combinatorics (like simple probability, coupon colletor, etc.). Had to wikipedia "closed form" to make sure I remembered it correctly. So, I don't know if a simple form might exist. Say for n=10000 and b=100. – Tommy Sep 10 '17 at 1:07
• This is a family web site you know. – zhw. Sep 10 '17 at 1:33
For positive integers $n,b$, let $f(n,b)$ be the expected number of empty bins at the end of the process.
I'm not sure if there's a closed form for $f(n,b)$, but here's a recursion, implemented in Maple, to compute $f(n,b)$ for given (not too large) values of $n,b$, . . .
Shown above is a sample calculation for $n=10,\;b = 6$, which shows that $$f(10,6) = \frac{751583152441}{208971104256} \approx 3.596588893$$ For $n=10000,\;b=100$, the recursion fails (too many levels of recursion for my version of Maple), but a simulation gives the approximate result $$f(10000,100) \approx 90.67$$
• Thank you! See edit to my original question for pretty simple form. – Tommy Sep 10 '17 at 22:57
• The product formula works perfectly! – quasi Sep 10 '17 at 23:47
Credit goes to @quasi for the first answer and the recurrence. Using generating functions and classifying on the number of rounds $k$ we find the PGF
$$\bbox[5px,border:2px solid #00A000]{ G(u) = \frac{1}{n} \sum_{k=1}^n [w^{k-1}] \prod_{q=1}^{n-1} \left(1+\frac{w}{q}\right) \times \frac{k!}{b^k} [z^k] (u-1+\exp(z))^b}$$
where the coefficient on $[u^p]$ is the probability of obtaining $p$ empty bins in the experiment with the given $n$ and $b.$ Here we have used the labeled combinatorial species $\mathfrak{S}_{=b}(\mathcal{U}+\mathfrak{P}_{\ge 1}(\mathcal{Z}))$ to construct the EGF that was used. The desired expectation is then given by
$$\left. \frac{d}{du} G(u)\right|_{u=1.}$$
The derivative yields
$$\left. \frac{1}{n} \sum_{k=1}^n [w^{k-1}] \prod_{q=1}^{n-1} \left(1+\frac{w}{q}\right) \times \frac{k!}{b^k} [z^k] b (u-1+\exp(z))^{b-1}\right|_{u=1.} \\ = \frac{b}{n} \sum_{k=1}^n [w^{k-1}] \prod_{q=1}^{n-1} \left(1+\frac{w}{q}\right) \times \frac{k!}{b^k} [z^k] \exp((b-1)z) \\ = \frac{b}{n} \sum_{k=1}^n \frac{(b-1)^k}{b^k} [w^{k-1}] \prod_{q=1}^{n-1} \left(1+\frac{w}{q}\right) \\ = \frac{b-1}{n} \sum_{k=1}^n \frac{(b-1)^{k-1}}{b^{k-1}} [w^{k-1}] \prod_{q=1}^{n-1} \left(1+\frac{w}{q}\right) \\ = \frac{b-1}{n} \prod_{q=1}^{n-1} \left(1+\frac{b-1}{bq}\right) \\ = \frac{b-1}{n!} \prod_{q=1}^{n-1} \left(q+\frac{b-1}{b}\right) = \frac{b}{n!} \prod_{q=0}^{n-1} \left(q+\frac{b-1}{b}\right).$$
This is
$$b \times {(b-1)/b+n-1\choose n}$$
or indeed
$$\bbox[5px,border:2px solid #00A000]{ b \times {n-1/b\choose n}}$$
as claimed in edit to question by OP. (This formula will produce zero when there is just one bin and this is the correct value, no empty bins are possible in this case.)
Here is some Maple code to help explore this interesting problem. Using the multiplicative version of the closed form gives
$$\bbox[5px,border:2px solid #00A000]{90.66863442}$$
for $n=10000$ and $b=100.$
with(combinat);
ENUM_GF :=
proc(n, b)
option remember;
local recurse, gf;
gf := 0;
recurse :=
proc(prob, rest, cc)
local rm;
if rest = 0 then
gf := gf +
prob*cc!*coeftayl((u-1+exp(z))^b, z=0, cc)/b^cc;
return;
fi;
for rm to rest do
recurse(prob/rest, rest-rm, cc+1);
od;
end;
recurse(1, n, 0);
gf;
end;
X_GF := (n, b) -> 1/n*
*k!*coeftayl((u-1+exp(z))^b, z=0, k)/b^k,
k=1..n);
ENUM := (n, b) -> subs(u=1, diff(ENUM_GF(n, b), u));
X := (n, b) -> b*binomial(n-1/b, n);
Y := (n, b) -> b*mul(n-1/b-q, q=0..n-1)/n!;
We also have a very basic simulation, which confirmed the closed form on all values that were examined.
#include <stdlib.h>
#include <stdio.h>
#include <assert.h>
#include <time.h>
int main(int argc, char **argv)
{
int n = 6 , b = 3, trials = 1000;
if(argc >= 2){
n = atoi(argv[1]);
}
if(argc >= 3){
b = atoi(argv[2]);
}
if(argc >= 4){
trials = atoi(argv[3]);
}
assert(1 <= n);
assert(1 <= b);
assert(1 <= trials);
srand48(time(NULL));
long long data = 0;
for(int tind = 0; tind < trials; tind++){
int rest = n;
int bins[b];
for(int bidx = 0; bidx < b; bidx++)
bins[bidx] = 0;
while(rest > 0){
int rm = 1 + drand48() * (double)(rest);
int bidx = drand48() * (double)(b);
bins[bidx] += rm;
rest -= rm;
}
int empty = 0;
for(int bidx = 0; bidx < b; bidx++)
if(!(bins[bidx]))
empty++;
data += empty;
}
long double expt = (long double)data/(long double)trials;
printf("[n = %d, b = %d, trials = %d]: %Le\n",
n, b, trials, expt);
exit(0);
}
• Thank you for such a detailed answer. Way over my head, indistinguishable from magic for me, but appreciated! – Tommy Sep 11 '17 at 23:56
• Thank you, it was an interesting question and it will perhaps seem less like magic as you proceed in your studies. – Marko Riedel Sep 12 '17 at 1:40 | 2019-08-19T13:10:08 | {
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https://math.stackexchange.com/questions/3333287/how-to-compute-number-of-ways-one-s-and-one-f-can-be-chosen-from-group-of-2f-2 | How to compute number of ways one S and one F can be chosen from group of (2F, 2S, 2J, 2R)?
I thought I knew combinations pretty well, until recently I came across this question:
In a high school debating team consisting of 2 freshmen, 2 sophomores, 2 juniors, and 2 seniors, two students are selected to represent the school at the state debating championship. The rules stipulate that the representatives must be from different grades, but otherwise the 2 representatives are to be chosen by lottery. What is the probability that the students selected will consist one freshman and one sophomore?
The way I approached this problem was to compute $$\frac{N}{D}$$, where
• $$N =$$ number of ways a sophomore and freshman can be chosen
• $$D$$ = total number of ways students can be chosen from different grades
The I broke it down as follows:
• $$D$$ = total number of ways to choose 2 students - number of ways to choose students from the same grade. So, $$D = C(n=8, k=2) - 4 = 28 - 4= 24$$
• $$N$$ = 1, because there is only one way to choose a sophomore and a freshman.
But this gets me $$\frac{1}{24}$$, which is incorrect. I understand there are other ways of doing this problem, but I'm interested in knowing where I went wrong in my logic/approach of the problem?
Thanks.
You have correctly calculated the denominator. However, there are four ways to select a freshman and a sophomore since there are two ways to select a freshman and two ways to select a sophomore. Hence, the probability that a freshman and a sophomore are selected for the competition is $$\Pr(\text{selected students are a freshman and a sophomore}) = \frac{\binom{2}{1}\binom{2}{1}}{\binom{8}{2} - \binom{4}{1}\binom{2}{2}} = \frac{4}{24} = \frac{1}{6}$$ assuming each individual is equally likely to be chosen.
• Right...I realized this myself too. In fact, there's an easier way to compute the denominator if you consider multiplying these two together: 1) number of ways to choose which two different grades to choose from = $C(4,2) = 6$, and 2) number of ways to choose the 2 individuals from those 2 grades = 4 (as you pointed out). 4*6 = 24. thanks – Sother Aug 26 at 15:53 | 2019-11-18T03:21:23 | {
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http://math.stackexchange.com/questions/194230/elementary-set-theory-homework-proofs | # Elementary set theory homework proofs
I've been given some statements to prove using only the following facts.
1. The set of all integers is countably infinite.
2. Let $X$ and $Y$ be sets and suppose $f:X\rightarrow Y$ is surjective. If $X$ is countable, then $Y$ is also countable.
3. The union of a countably infinite family of sets is countable.
I'd like your opinion on my proofs. I feel like the first could be simpler.
Exercise 1. The set $\mathbb{Z}\times\mathbb{Z}$ is countable.
Proof. Let $A_n=\{n\}\times\mathbb{Z}$ for each $n\in\mathbb{Z}$. Define a mapping $f:\mathbb{Z}\rightarrow A_n$ by $f(x)=(n,x)$ for all $x\in\mathbb{Z}$. To see that $f$ is surjective, consider $(n,p)\in A_n$. Since $p\in\mathbb{Z}$, $f(p)=(n,p)$. Thus, $f$ is surjective. By (2), since (1) $\mathbb{Z}$ is countable, $A_n$ is countable.
Let $S=\{A_n:n\in\mathbb{Z}\}$. Notice that $S$ is infinite since it is indexed over $\mathbb{Z}$. Define $g:\mathbb{Z}\rightarrow S$ by $g(x)=A_x$ for all $x\in\mathbb{Z}$. To see that $g$ is surjective, consider $A_q\in S$. Since $q\in\mathbb{Z}$, $g(q)=A_q$. Thus, $g$ is surjective. By (2), $S$ is countable. Thus, $S$ is a countably infinite family of countable sets.
Let $U=\bigcup S$. Then, by (3), $U$ is countable. To see that $U=\mathbb{Z}\times\mathbb{Z}$, we will show that $U$ and $\mathbb{Z}\times\mathbb{Z}$ are subsets of each other.
Let $x\in U$. Then $x\in A_n$ for some $n\in\mathbb{Z}$. Hence, $x=(n,p)$ for some $p\in\mathbb{Z}$. Therefore, $x=(n,p)\in\mathbb{Z}\times\mathbb{Z}$, and it follows that $U\subseteq\mathbb{Z}\times\mathbb{Z}$.
Let $x\in\mathbb{Z}\times\mathbb{Z}$. Then $x=(p,q)$ for some $p,q\in\mathbb{Z}$. Thus, $x\in A_p$, which implies that $x\in U$. Therefore, $\mathbb{Z}\times\mathbb{Z}\subseteq U$, and it follows that $U=\mathbb{Z}\times\mathbb{Z}$.
Therefore, $\mathbb{Z}\times\mathbb{Z}$ is countable.
Exercise 2. Every infinite subset of a countable set is countable.
Proof. Let $X$ be a countable set and let $Y$ be an infinite subset of $X$. Then $X$ is countably infinite and there exists a surjection $f:\mathbb{N}\rightarrow X$. Since $Y\subseteq X$ and $Y$ is nonempty, there exists a surjection $g:X\rightarrow Y$. Thus, $gf:\mathbb{N}\rightarrow Y$ is a surjection. Therefore, by (2), $Y$ is countable.
-
These proofs work. You were very precise and detailed. – William Sep 11 '12 at 16:48
Your use of "countable" and "countably infinite" is slightly confusing. Fact 2 is only true if you're using "countable" to mean "finite or countably infinite", not "countably infinite". But then the assumption in Exercise 2 that the subset is infinite is not required to prove that it's countable, and indeed your proof doesn't really use it; you conclude that $X$ is countably infinite but what follows would work just as well for $X$ merely countable. – joriki Sep 11 '12 at 16:55
In our course, we say a set $X$ is countable if it is finite or if there exists a bijection between $X$ and $\mathbb{N}$. We also call $X$ countably infinite in the second case. In exercise 2, I said the subset was countable because that's how the exercise is worded. – ohmygoodness Sep 11 '12 at 17:02
Perhaps there's a misunderstanding -- I wasn't commenting on the subset being countable but on it being infinite. I understand that this is part of the given wording of the exercise (so the given wording of the exercise is bad), but even so, your proof concludes that $X$ is countably infinite without ever using that fact. – joriki Sep 11 '12 at 17:36
Grad: I think @joriki is saying you could leave the words "$X$ is countably infinite and" out of your second proof and it would still be valid: indeed it would be better because it avoids using a fact you have not been told. – Henry Sep 11 '12 at 19:34
To reduce the number of unanswered questions, I answer here (albeit belatedly).
Your first proof is just fine. However, depending on what your definition of "countable" is, you may not have followed the instructions for your second proof. Consider the two following definitions (the first of which is standard, and the second of which is equivalent but nonstandard).
1. A set $X$ is said to be countable if there is an injective function $X\to\Bbb N$.
2. A set $X$ is said to be countable if there is a surjective function $\Bbb N\to X$.
If you were given the latter definition, then your second proof is just fine (though, as joriki and Henry pointed out in the comments above, the fact that $X$ is countably infinite needn't even be mentioned). However, if you were given the former definition, then you didn't follow the directions, as you concluded without "acceptable" justification that the surjection $f:\Bbb N\to X$ exists.
Now, there are easy fixes for this. The simplest is just to use the surjection $g:X\to Y$, which is easily constructed (if you want to be cautious) by fixing some $y_0\in Y$ and letting $g(x)=x$ for $x\in Y$, $g(x)=y_0$ otherwise. Since $X$ is countable, then your given fact 2 shows us that $Y$ is countable. Alternately, you can show that there is a surjection $f:\Bbb N\to X$ as follows:
• Since $X$ is countable, then there is an injective function $h:X\to\Bbb N$.
• Since $X$ has an infinite subset $Y$, then $X$ is non-empty, so in particular, we can fix an element $x_0\in X$.
• Define $f:\Bbb N\to X$ by $f(n)=x$ if there is some $x\in X$ such that $h(x)=n$ (such $x$ will be unique if it exists, since $h$ is injective) and $f(n)=x_0$ if $n$ isn't in the range of $h$. $f$ can readily be shown to be a surjective function, and at that point, you can proceed as in your proof.
There are other ways we can fix it, too, but these are probably the simplest.
- | 2015-05-24T05:35:39 | {
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https://math.stackexchange.com/questions/3174746/proving-that-tan-2x-cdot-1-tan-x-cdot-cot-x-frac21-tanx | # Proving that $\tan 2x \cdot (1 + \tan x) \cdot \cot x = \frac{2}{1 - \tan(x)}$
Given the following expression, $$\tan(2x) \cdot (1 + \tan(x)) \cdot \cot(x)$$ the exercise asks to simplify the expression and $$\frac{2}{1 - \tan(x)}$$ should be the simplified expression.
I have tried everything I possibly could, including letting WolframAlpha eat it to show alternative forms of the expression – nothing worked.
What do you think? How could I go about simplifying this expression? Thank you.
• How can you re-write $\tan 2x$ in terms of $\tan x$? – Blue Apr 4 '19 at 15:34
• What did you try? It's involves a straightforward simplification – pi-π Apr 4 '19 at 15:34
• $\frac{\sin(2x)}{\cos(2x)}$. I know. I have tried. – Johnny Bueti Apr 4 '19 at 15:35
• @Fehniix: $$\tan(a+b) = \frac{\tan a + \tan b}{1-\tan a \tan b}$$ so $\tan(x+x) = \cdots$? – Blue Apr 4 '19 at 15:38
• @Blue had no clue there was such an equivalence! Thank you. – Johnny Bueti Apr 4 '19 at 15:43
Defining $$t := \tan x$$ to save typing ... \begin{align} \tan 2x (1+\tan x) \cot x = \frac{2t}{1-t^2}\cdot(1+t)\cdot\frac{1}{t} = \frac{2t}{(1+t)(1-t)}\cdot(1+t)\cdot\frac{1}{t} = \frac{2}{1-t} \end{align}
It is true.
Write everything in terms of $$\sin(x)$$ and $$\cos(x)$$, and cancel common factors from numerator and denominator. You should end up with $$\frac{2 \cos(x)}{\cos(x)-\sin(x)}$$ and then divide numerator and denominator by $$\cos(x)$$.
• It's actually easier just to write everything in terms of $\tan x$. – Blue Apr 4 '19 at 15:36
• How so, Blue? Thank you guys. – Johnny Bueti Apr 4 '19 at 15:40
$$\tan(2x) (1+\tan(x)) \cot(x) = \frac{2\sin(x)\cos(x)}{\cos^2(x)-\sin^2(x)}\left(\frac{\sin(x)+\cos(x)}{\cos(x)} \right)\frac{\cos(x)}{\sin(x)}$$
Simplifying you get $$\tan(2x) (1+\tan(x)) \cot(x) = \frac{2(\sin(x)+\cos(x))\cos(x)}{(\cos(x)-\sin(x))(\cos(x)+\sin(x))} = \frac{2\cos(x)}{\cos(x)-\sin(x)}$$
i.e. $$\tan(2x) (1+\tan(x)) \cot(x) =\frac{2}{1-\tan(x)}$$
• Wow. Thank you. – Johnny Bueti Apr 4 '19 at 15:39 | 2021-01-26T09:35:26 | {
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https://math.stackexchange.com/questions/2880436/number-of-combinations-of-7-of-the-first-17-natural-numbers-in-which-there-a | # Number of combinations of $7$ of the first $17$ natural numbers in which there are at least two consecutive numbers
I have this statement:
How many groups of $7$ elements can you get from the first $17$ natural numbers and at least $2$ of them must be consecutive ( Numbers can't repeat and order doesn't matter) ?
Basically, my development was:
I will get $17 \choose 7$ groups and I will subtract the combinations with two consecutive numbers or more.
To get all combinations with two consecutive numbers or more, I have different cases:
Case 1: All are consecutive numbers, This varies depending on the number when I start:
If I start from $1$, it will be: $1, 2, 3, 4 , 5 , 6 , 7$,
If I start from $2$, it will be: $2, 3, 4, 5 , 6 , 7 , 8$
Then, the maximum number from which I can begin is $11$ and will be: $11, 12, 13, 14, 15, 16, 17$.
Therefore, when all numbers are consecutive, I have $11$ possible cases.
I will do the same procedure for the other cases.
Case 2: Here I need only $6$ consecutive, so the maximum number that I can begin is $12$, so I have $12$ cases.
Therefore, for all consecutive, I have $11$ cases, for $6$ consecutive, I have $12$ cases, for $5$ I have $13$, for $4$ I have $14$, for $3$ I have $15$ and for $2$ consecutive numbers I have $16$ cases.
The total cases are: $11 + 12 + 13 + 14 + 15 + 16$ = $81$
Then I will subtract. Then, $\binom{17}{7} - 81 = 19448 - 81 = 19367$
But they told me, that my answer is bad. What have I done wrong and why?
How should it be done, following a logic similar to that of my development?
• Sometimes it helps to consider the Principle of Inclusion/Exclusion when counting instances that require a side condition, as here with imposing (at least) two consecutive numbers in a set. – hardmath Aug 12 '18 at 15:26
• Edited, the 17 first natural numbers, i.e $[1, 2 ... 17]$ – Eduardo S. Aug 12 '18 at 20:04
• That principe help in these type of combinations with "at least" ? – Eduardo S. Aug 12 '18 at 20:05
• There are more than $12$ cases for $6$ consecutive integers since you have to choose seven numbers. If you choose $1, 2, 3, 4, 5, 6$ or $12, 13, 14, 15, 16, 17$, the seventh number can be selected in $10$ ways (since it not be adjacent to those six numbers). For any other sequence of six consecutive numbers, the seventh number can be selected in $9$ ways (since it cannot be adjacent to either end of the sequence of six numbers). You are asking for trouble with your approach since you could have three consecutive, two consecutive, and two isolated integers such as $1, 2, 3, 6, 7, 11, 14$. – N. F. Taussig Aug 12 '18 at 20:40
• – user575478 Aug 14 '18 at 3:35
The problem is easier to handle if you subtract those selections in which no two numbers are consecutive from the total number of selections.
As you observed, there are $$\binom{17}{7}$$ ways to select $7$ of the first $17$ positive integers.
To count the number of selections in which no two of the selected are consecutive, we will arrange ten blue and seven green balls so that no two of the green balls are consecutive.
Place ten blue balls in a row, leaving enough space between successive balls and at the ends of the row in which to place a green ball.
There are eleven such spaces, nine between successive blue balls and two at the ends of the row. To ensure no two green balls are consecutive, we choose seven of these eleven spaces in which to place a green ball. For instance, if we choose the first, second, third, fourth, sixth, eighth, and ninth spaces, we obtain
If we now number the balls from left to right, the numbers on the green balls are the desired set of seven numbers, no two of which are consecutive. Hence, there are $$\binom{11}{7}$$ selections of $7$ of the first $17$ positive integers in which no two of the selected integers are consecutive.
Hence, the number of selections of $7$ of the first $17$ positive integers in which at least two numbers are consecutive is $$\binom{17}{7} - \binom{11}{7}$$ | 2019-10-19T13:30:22 | {
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https://www.physicsforums.com/threads/yet-another-volume.106729/ | # Yet another volume
1. Jan 14, 2006
### twoflower
Hi,
I'm having a trouble doing this:
Compute volume of the solid
$$T = \left\{[x,y,z] \in \mathbb{R}^3; x \geq 0, y \geq 0, 0 \leq z \leq 1 - x - y\right\}$$
First I need to express bounds for $x$ and $y$, for $z$ I have it already. So because
$$0 \leq z \leq 1 - x - y$$
then
$$0 \leq x \leq 1 - z - y$$
and also
$$0 \leq y \leq 1 - z - x$$
But that's probably not the right approach, because evaluating integral
$$\iiint_{T}\ dx\ dy\ dz = \int_{0}^{1-z-x}\int_{0}^{1-z-y}\int_{0}^{1-x-y}\ dz\ dx\ dy$$
still lefts me with some $z$ and $x$ variables at the end...
Will somebody point me to the right direction?
Thank you.
2. Jan 14, 2006
### siddharth
twoflower, maybe this will help.
The boundary of this region is given by the plane
$$x+y+z=1$$.
To find the integral, for some x and y, z varies from 0 to 1-x-y (ie, till the top of the column containing the volume element).
Now keep x constant and integrate wrt to y. This corresponds to adding the columns with base in the x-y plane (from 0 to the line y=1-x). So, y varies from 0 to 1-x.
Finally you can integrate wrt to x, which by similar argument varies from 0 to 1.
3. Jan 14, 2006
### twoflower
Btw what does "wrt" stand for? I would guess "with respect to", but it doesn't make sense to me in the previous post..
4. Jan 14, 2006
### arildno
The simplest way to do this is first to recognize that the plane z=1-x-y meets the plane z=0 in the line x+y=1 (the line in z=0, that is)
Now, regard the part of T lying in the plane z=0.
This is the triangle bounded by the lines x=0, y=0, x+y=1, that is, we see that 0<=x<=1 and 0<=y<=1-x (alternatively, you could describe the triangle as 0<=y<=1, 0<=x<=1-y)
5. Jan 14, 2006
### Staff: Mentor
In general, if you know that you are going to get a definite volume, then you know that the limits on a given integral cannot depend on any of the inner variables of integration. So in your example, the limits for z can be a function of x and y, the limits for x can be a function of y but not z, and the limits for y cannot be a function of x or z.
-Dale
6. Jan 14, 2006
### HallsofIvy
Staff Emeritus
You have
$$T = \left\{[x,y,z] \in \mathbb{R}^3; x \geq 0, y \geq 0, 0 \leq z \leq 1 - x - y\right\}$$
and you write
$$V= \iiint_{T}dV = \int\int\int dz\ dx\ dy$$
which indicates that you are going to integrate first with respect to z, then with respect to x, and finally with respect to y.
(That's an arbitrary choice- doing the integrations in any order must give the same answer- although one way might be easier than another.)
You are correct, of course, that the answer is a number which means that the "outside" integral, wrt y, must have only numbers as limits.
How do you find them? Draw a picture! (That should always be the first thing you do with problems like this.) Since you are told that x and y will be larger than or equal to 0, you are looking at the first quadrant in the xy-plane. z= 1- x- y is a plane, of course. It crosses the xy-plane when z= 0= 1-x-y which is the line x+y= 1, through the points (0,1) and (1,0), in the first quadrant. That tells you everything you need to know. Overall, y can be as low as 0 or as high as 1. Your "outside" integral must be $\int_0^1 dy$.
Now look at the "middle" integral- the one with respect for x. Since you will not yet have integrated with respect to y, the limits of integration may depend upon y. For each y what is the range for x? A fixed y is, of course, a horizontal line running from the vertical line x= 0 to the line x+ y= 1 or x= 1- y. For each y, x ranges between 0 and 1- y. The "middle" integral is $\int_0^{1-y}dx$.
Finally, when doing the "inner" integral, we have not yet integrated with respect to x or y so the limits of integration may be functions of both x and y. For each x,y, what is the range for z? The plane z= 1- x- y slants up from the line x+y= 1 in the xy-plane to z= 1 at the z-axis (x=y=0). The crucial point is that the xy-plane (z= 0) is the "floor" and the plane z= 1- x- y is the "roof". For a given (x,y) point, z ranges from z= 0 to z= 1- x-y. That "inner" integral is $\int_0^{1-x-y}dz$
Personally, I think it helps to write "x= ", "y= ", "z= " in the limits of integration themselves. The volume is given by:
$$\int_{y= 0}^{y=1}\int_{x=0}^{x=1-y}\int_{z=0}^{z=1-x-y}dzdxdy$$.
It would be good practice for you to find the integrals in the other 5 orders, and do the integrations to see that they do indeed give the same answer!
Last edited: Jan 14, 2006
7. Jan 14, 2006
### twoflower
Thank you HallsoftIvy very much for comprehensible explanation! I already managed it as soon as I draw it :) First it seemed ugly to me to draw it but it turned out to be pretty straightforward. And then the expressing 'x' for fixed 'y' was easy. | 2017-12-15T11:13:13 | {
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https://math.stackexchange.com/questions/2755987/eigenvalues-for-4x4-matrix/2756553 | # Eigenvalues for 4x4 matrix
Show that $0,2,4$ are the eigenvalues for the matrix $A$: $$A=\pmatrix{ 2 & -1 & -1 & 0 \\ -1 & 3 & -1 & -1 \\ -1 & -1 & 3 & -1 \\ 0 & -1 & -1 & 2 \\ }$$ and conclude that $0,2,4$ are the only eigenvalues for $A$.
I know that you can find the eigenvalues by finding the $\det(A-\lambda \cdot I)$, but it seems to me that the computation will be rather difficult to compute as it is a $4 \times 4$ matrix.
My question: is there an easier method to calculate the eigenvalues of $A$? And if I have to conclude that these are the only eigenvalues, is there a theorem that argues how many eigenvalues a matrix can have?
• The sum of all eigenvalues is the trace, so if you have three of them, the last comes for free. – Lord Shark the Unknown Apr 27 '18 at 10:30
• I can't see any easy method. I think you have to plow through the calculations. The only obvious one is $0$ because it is the sum of each row. Because it is a symmetric matrix, you also know the eigenvalues are real numbers. Finally the Gershgorin Circle Theorem allows you to bound them, but not is any sufficiently useful way. – Git Gud Apr 27 '18 at 10:39
• sum of each row is zero, so zero is one eigenvalue( with $(1,1,...,1)^t$ as an eigenvector) – Chinnapparaj R Apr 27 '18 at 10:53
$A$ has zero row sums. Therefore the all-one vector $\mathbf e$ is an eigenvector of $A$ for the eigenvalue $0$. Since $A$ is also real symmetric, $\mathbf e$ can be extended to an orthogonal eigenbasis $\{\mathbf u,\mathbf v,\mathbf w,\mathbf e\}$ of $A$. But this is also an eigenbasis of $A+\mathbf e\mathbf e^\top$. Hence the spectrum of $A$ is $\{a,b,c,0\}$ if and only if the spectrum of $A+\mathbf e\mathbf e^\top$ is $\{a,b,c,\|\mathbf e\|^2\}=\{a,b,c,4\}$. It is easy to see that four eigenvalues of $$A+\mathbf e\mathbf e^\top=\pmatrix{3&0&0&1\\ 0&4&0&0\\ 0&0&4&0\\ 1&0&0&3}$$ are $2,4,4,4$. Therefore the eigenvalues of $A$ are $2,4,4,0$.
• Why is it that $\text{spec}(A)=\{0, a, b, c\}\implies \text{spec}(A+\mathbf e\mathbf e^\top)=\{a, b, c, 4\}$? – Git Gud Apr 27 '18 at 11:15
• @GitGud Since $Av=av$ for some $v\perp\mathbf e$, we also have $(A+\mathbf e\mathbf e^\top)v=av$. Similarly for $b$ and $c$. Finally, $(A+\mathbf e\mathbf e^\top)\mathbf e=(\mathbf e^\top\mathbf e)\mathbf e=4\mathbf e$. – user1551 Apr 27 '18 at 11:18
• Ohhhhh. Thanks. – Git Gud Apr 27 '18 at 11:21
It helps to notice that this matrix comes from a graph. It is the so called Laplacian of a graph. Notice the position of $-1$'s. If you take four points and join the $i$th and $j$th if there is a $-1$ at position $(i,j)$ you get a graph. It's fairly symmetric: its edges are the sides of a square and one diagonal. Consider different ways to associate a number to the vertices of the square. From one such distribution of numbers, you can produce another distribution as follows: at each vertex, the new number will be (old number) x (degree of the vertex) - (sum of numbers associated to the vertices that are joined with this one).
It may happen that the new distribution of numbers is proportional to the old one. So the old distribution is an eigenvector, and the constant of proportionality is an eigenvalue.
Suppose that you start with a distribution with all numbers equal ( say $1$). Then the new distribution is $0$ everywhere. So we got an eigenvector with eigenvalue $0$
Say we place $1$ at the ends of the (joined ) diagonal, and $-1$ at the other two points. Check that we get an eigenvector for eigenvalue $4$.
Say we place $1$, $-1$ at the diagonal, $0$, $0$ at the others. Check that we get an eigenvector with eigenvalue $4$.
Say we place $0$ at the diagonal, $-1$, $1$ at the other points. Check that we get eigenvector with eigenvalue $2$.
Note that this is a small example where it can be done by hand.
• I'm not going to bother with understanding your answer, but this really doesn't feel like an easier method. – Git Gud Apr 27 '18 at 11:03
• @Git Gud: I added a picture, that might help. – Orest Bucicovschi Apr 27 '18 at 11:43
• Nice observation on this being the Laplacian of a graph. I think two more observations allow the eigenvalues to be determined exactly without simulating node distributions (see my answer). – Jared Goguen Apr 27 '18 at 18:54
• @Jared Goguen: Thanks. Super info your answer! – Orest Bucicovschi Apr 27 '18 at 22:57
1. $\lambda=2$ is obviously an eigenvalue: look at the first and the last columns of $A-2I$.
2. $\lambda=0$ is also en eigenvalue: sum up all the columns of $A$ (dependent?).
3. Try to do similar trick for $\lambda=4$.
4. The last eigenvalue: The sum of all eigenvalues is the trace, so if you have three of them, the last comes for free.(c) comment by Lord Shark the Unknown.
EDIT: If a calculation of eigenvalues is of your primal curiosity then you can do the following: \begin{align} \det(\lambda I-A)&=\begin{vmatrix}\lambda-2 & 1 & 1 & 0\\1 & \lambda-3 & 1 & 1\\1 & 1 & \lambda-3 & 1\\0 & 1 & 1 & \lambda-2\end{vmatrix}\stackrel{(1)}{=} \begin{vmatrix}\lambda & 1 & 1 & 0\\\lambda & \lambda-3 & 1 & 1\\\lambda & 1 & \lambda-3 & 1\\\lambda & 1 & 1 & \lambda-2\end{vmatrix}\\ &\stackrel{(2)}{=}\lambda\begin{vmatrix}1 & 1 & 1 & 0\\1 & \lambda-3 & 1 & 1\\1 & 1 & \lambda-3 & 1\\1 & 1 & 1 & \lambda-2\end{vmatrix}\stackrel{(3)}{=} \lambda\begin{vmatrix}1 & 1 & 1 & 0\\0 & \lambda-4 & 0 & 1\\0 & 0 & \lambda-4 & 1\\0 & 0 & 0 & \lambda-2\end{vmatrix}. \end{align} Explanations:
(1): add all columns to the first one,
(2): factor out $\lambda$ in the first column,
(3): eliminate in the first column.
• Sure, $2$ is obviously an eigenvalue if you look at $A-2I$, but why would you ever look at that matrix? – Git Gud Apr 27 '18 at 10:57
• @GitGud the question asks to prove that $0,2,4$ are the eigenvalues of the matrix, So it makes sense to consider $A-2I$ and prove it is singular. – idok Apr 27 '18 at 11:05
• @GitGud The problem says "Show that given numbers are eigenvalues". If it were "Find the eigenvalues..." then I would agree with your comment. – A.Γ. Apr 27 '18 at 11:06
• No, that's not the OP's question. That the problem statement question. The OP is asking how to find the eigenvalues easily: "My question: is there an easier method to calculate the eigenvalues of $A$?" @idok – Git Gud Apr 27 '18 at 11:10
• @GitGud: if you assume this could be a special case, it is reasonable to ask yourself which $\lambda$ might make two of the rows or two of the columns of $A-\lambda I$ equal to each other. And it is not difficult to spot that $2$ and $4$ do this – Henry Apr 27 '18 at 15:39
Perhaps there is no easier way of explaining how to find the eigenvalues of $A$. At the end, it all comes down to how much you are familiar with matrices that have some special properties. I’m using the very same insight as orangeskid but explained in a different way.
Consider this matrix: $$\color{red}{L=\left[\begin{array}{r} 1&-1\\ -1&1 \end{array}\right]}$$ which has eigenvalues $\color{red}{0}$ and $\color{red}{2}$. Now consider this bigger matrix $$M=\left[\begin{array}{r|rr|r} \color{red}{1}&0&0&\color{red}{-1}\\ \hline 0&\color{blue}{0}&\color{blue}{0}&0\\ 0&\color{blue}{0}&\color{blue}{0}&0\\ \hline \color{red}{-1}&0&0&\color{red}{1} \end{array}\right]$$ which is blockwise similar to the latter, and has eigenvalues $\color{blue}{0}$, $\color{blue}{0}$, $\color{red}{0}$ and $\color{red}{2}$ (the former two because of the $\color{blue}{2\times2}$ $\color{blue}{\text{zero matrix in the middle}}$; the latter two because of reusing $\color{red}{\text{the entries of the smaller matrix in the same fashion}}$.)
Now consider this matrix $$\color{purple}{K=\left[\begin{array}{r} 3&-1&-1&-1\\ -1&3&-1&-1\\ -1&-1&3&-1\\ -1&-1&-1&3\\ \end{array}\right]}$$ which is of the form $\color{purple}{4 \, I_4 -\boldsymbol{1}^{\rm T}_4 \boldsymbol{1}_4}$ and has eigenvalues $\color{purple}{0}$, $\color{purple}{4}$, $\color{purple}{4}$ and $\color{purple}{4}$. Now note that all matrices $L$, $M$, $A$ and $K$ are Laplacian matrices corresponding to undirected, unweighted simple graphs, and since $$M+A=K,$$ there is a result for Laplacian matrices that relates the sorted eigenvalues of $M$ and $A$ with those of $K$. For starters, the first eigenvalue of $A$ is $\bbox[yellow]{0}$. Then, the other three eigenvalues of $A$ are computed in the following fashion:
Take the eigenvalues of $K$ and remove a zero. Take the decreasingly sorted eigenvalues of $M$ and remove a zero. Then their difference gives the eigenvalues of $A$ other that the first zero eigenvalue we mentioned.
Thus, $$\color{purple}{4}-\color{red}{2}=\bbox[yellow]{2}, \qquad\qquad \color{purple}{4}-\color{red}{0}=\bbox[yellow]{4}, \qquad\qquad \text{and } \color{purple}{4}-\color{blue}{0}=\bbox[yellow]{4},$$ are the other three eigenvalues of $A$.
You would be correct about the determinant method of finding the eigenvalues being potentially difficult since the degree is 4 (though in this case where the roots are rational, they can be found easily).
But note that finding the eigenvalues is NOT what you've been asked to do. You are given three of them, and have only to verify that they are indeed eigenvalues. Simply compute the characteristic polynomial for each of the three values and show that it is $0$.
To show that they are the only eigenvalues, divide the characteristic polynomial by $x$, the result by $x-2$, and finally by $x-4$. You will find that the remaining polynomial is another factor of $x-4$.
• Instead of dividing the characteristic polynomial by $x$, $x-2$, and $x-4$, in succession, I would employ the trace, e.g $\DeclareMathOperator{\Tr}{Tr} \Tr(A) = \sum_i \lambda_i = 10$. From the three eigenvalues already found, it is then clear the last one must also be 4, too. – rcollyer Apr 28 '18 at 17:11
• One could simply verify that $A(A-2I)(A-4I)=0$ and that omitting one factor yields a nonzero matrix. This confirms that the minimal polynomial of $A$ is $\lambda(\lambda-2)(\lambda-4)$ without having to compute invariants. As to multiplicity: $A$ is symmetric hence diagonalizable, therefore no multiple factors are needed. – ccorn Apr 28 '18 at 20:33
• @rcollyer - that indeed would be easier, though with synthetic division, dividing the polynomials is not particularly hard. – Paul Sinclair Apr 29 '18 at 0:12
• @ccorn - I haven't counted out the operations, but I am doubtful that multiplying 4x4 matrices four times is computationally easier than finding the characteristic polynomial. – Paul Sinclair Apr 29 '18 at 0:13
Let $\lambda_1 \le \lambda_2 \le \lambda_3 \le \lambda_4$ be the eigenvalues of this matrix.
Since this is a Laplacian matrix, the smallest eigenvalue is $\lambda_1 = 0$.
The second smallest eigenvalue of a Laplacian matrix is the algebraic connectivity of the graph. This is bounded above by the traditional connectivity of the graph, so $\lambda_2 \le 2$.
From the trace, $8 \le \lambda_3 + \lambda_4 \le 2\lambda_4 \rightarrow \lambda_4 \ge 4$.
From Brouwer and Haemers (2011) Proposition 3.9.3, $\lambda_4 \le 1 + \underset{x}{\rm{max}}\;d_x=4$.
Thus, $\lambda_2=2, \lambda_3=4, \lambda_4=4$.
You can easily guess some eigenvectors: $$\begin{pmatrix} 2 & -1 & -1 & 0 \\ -1 & 3 & -1 & -1 \\ -1 & -1 & 3 & -1 \\ 0 & -1 & -1 & 2 \end{pmatrix} \begin{pmatrix} 1\\1\\1\\1 \end{pmatrix} = \begin{pmatrix} 0\\0\\0\\0 \end{pmatrix} \\ \begin{pmatrix} 2 & -1 & -1 & 0 \\ -1 & 3 & -1 & -1 \\ -1 & -1 & 3 & -1 \\ 0 & -1 & -1 & 2 \end{pmatrix} \begin{pmatrix} 1\\0\\0\\-1 \end{pmatrix} = \begin{pmatrix} 2\\0\\0\\-2 \end{pmatrix} \\ \begin{pmatrix} 2 & -1 & -1 & 0 \\ -1 & 3 & -1 & -1 \\ -1 & -1 & 3 & -1 \\ 0 & -1 & -1 & 2 \end{pmatrix} \begin{pmatrix} 1\\-1\\-1\\1 \end{pmatrix} = \begin{pmatrix} 4\\-4\\-4\\4 \end{pmatrix} \\ \begin{pmatrix} 2 & -1 & -1 & 0 \\ -1 & 3 & -1 & -1 \\ -1 & -1 & 3 & -1 \\ 0 & -1 & -1 & 2 \end{pmatrix} \begin{pmatrix} 0\\1\\-1\\0 \end{pmatrix} = \begin{pmatrix} 0\\4\\-4\\0 \end{pmatrix}$$ The corresponding eigenvalues are $0,2,4,4$. It is not hard to check that the latter two vectors are linearly independent, and so these are all eigenvalues of the matrix.
• I'm not the original poster, but this doesn't seem to be much easier than solving the characteristic polynomial. – Alex Apr 27 '18 at 22:52 | 2019-09-15T08:34:29 | {
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Rather than eating his vegetables, Coltaine's father stares dejectedly at his plate, and estimates the surface area of the elliptical cuts of zucchini with axes 0.1, 0.2, and 0.35 inches: SA ≈ 4π 1.6√(0.11.60.21.6 + 0.11.60.351.6 + 0.21.60.351.6)/3 = 0.562 in2. Uniquely, the area calculator is capable of accurately calculating irregular areas of uploaded images, photographs or plans quickly. See our full terms of service. lateral SA = πr√r2 + h2 The surface area of a solid is a measure of the total area occupied by the surface of an object. Formula for area of circular ring. You can find it by using proportions, all you need to remember is circle area formula (and we bet you do! Also, explore the surface area or volume calculators, as well as hundreds of other math, finance, fitness, and health calculators. Consequently, the calculator will provide you with floor area's value, which is calculated in distinct measure units, such as imperial and metric. Free volume of solid of revolution calculator - find volume of solid of revolution step-by-step This website uses cookies to ensure you get the best experience. After calculating area of each triangle, simply add all of them to find the area of an irregular shape. Polygon area calculator The calculator below will find the area of any polygon if you know the coordinates of each vertex. Area Calculator . It takes 4 inputs: lower bound, upper bound, mean, and standard deviation. Integral function differentiate and calculate the area under the curve of a graph. Horatio needs to determine the surface area of each capsule so that he can coat them with an excessive layer of sugar and appeal to the sugar predisposed tongues of the population in preparation for his next placebo that "cures" all forms of diabetes mellitus. Click on the map to place a vertices for the polyline 3. Show Instructions. The calculator uses the following solutions steps: From the three pairs of points calculate lengths of sides of the triangle using the Pythagorean theorem. Triangular Field Area. Area of a cyclic quadrilateral. Two conversion scales show how pressure varies with changes in force and area whilst the other parameter is fixed to the entered value. A specialized calculator is obviously handy if you have some geometry homework or if you are allowed to use online tools in class. Given two values of height, cap radius, or base radius, the third value can be calculated using the equations provided on the Volume Calculator. Area of a parallelogram given sides and angle. This is a special shape of a concave hexagon, made of two perpendicular rectangles with equal width. Banana-Bread has been clamoring all week long about wanting a new set of drawers to house her new Batman action figures. Calculate The Area Of Any Four Sided Lot: Use this calculator to determine the area of any four-sided lot. Given that each truffle has a radius of 0.325 inches: The surface area of a circular cone can be calculated by summing the surface area of each of its individual components. Find the area of any simple shape on a map. Area of a rectangle. Otherwise one needs to measure all sides first to confirm that they are all equal, before proceeding to calculate its area. Did you notice that it's a doubled formula for the triangle area , knowing side-angle-side? He has already tested the market and has found that a vast majority of the sample population exhibit none of these qualities, and are very ready to purchase his product, further entrenching themselves within the traits they so desperately seek to escape. Enter the Function = Lower Limit = Upper Limit = Calculate Area Heron's Formula Calculator. Given any one variable A, C, r or d of a circle you can calculate the other three unknowns. If using this calculator to compute the surface area of a hollow sphere, subtract the surface area of the base. Triangle area calculator by points. This will work for triangles, regular and irregular polygons, convex or concave polygons. Total solid sphere SA = 2πRh + πr2 Area of a Polygon. Area Calculator . Area of Octagon Calculator. Useful tool to find the approximate acreage or a tract of land, the square footage of a roof, or estimate of the area of something. To use the area calculator tool: 1. Use the calculator below to calculate the segment area given the radius and segment's central angle, using the formula described above. Enter the Length and Width to calculate the Area. She determines the surface area of material she needs to create her hat with a radius of 1 foot and a height of 0.5 feet as follows: lateral SA = π × 0.4√0.42 + 0.52 = 0.805 ft2. Area of a triangle (Heron's formula) Area of a triangle given base and angles. Step 2: Enter length of horizontal sides into Length 1 and Length 2. The surface area of a rectangular tank is the sum of the area of each of its faces: SA = 2lw + 2lh + 2wh Area of a square. Figure 1. Area under the Curve Calculator. Sphere diameter to volume calculator; Sphere diameter to surface area calculator; User Guide. Integral definition help finding the area, central point, volume etc. Use this circle calculator to find the area, circumference, radius or diameter of a circle. Where: = Area of the field (sq. Cylinder Area Calculator What is Cylinder ? Area of a parallelogram given sides and angle. Paul is making a volcano in the shape of a conical frustum for his science fair project. Jeremy has a large cylindrical fish tank that he bathes in because he doesn't like showers or bath tubs. Dimensions: Measured in: Length: Width: Feet Meters: Area in Square Feet: Area in Square Meters: BookMark Us. Cylinder Area Calculator What is Cylinder ? The calculator will find the area of the surface of revolution (around the given axis) of the explicit, polar or parametric curve on the given interval, with steps shown. You can also see at the bottom of the calculator, the step-by … Area of a Triangle. Value of S = (A+B+C)/2. Remember: In … Calculator. The number of square units it takes to exactly cover the surface of a cylinder. total SA = π(R2 + r2) + π(R+r)√(R-r)2 + h2 Calculating areas of irregular polygons is a non-trivial task without sophisticated surveying instruments. m 2 Floor Area of a Room Calculator. area = a * b * sin(α), where α is the angle between sides a and b. Area of a trapezoid. Area of a rectangle. Related: What is variance and how to calculate it. It uses the same method as in Area of a … The formula used to calculate circle area is: A = π x (ø/ 2) 2. Unit: Number: Required Data Entry Side A Length Units Side B Length Units Side C Length Units Calculated Results Area of Triangle. The calculator will find the area of the surface of revolution (around the given axis) of the explicit, polar or parametric curve on the given interval, with steps shown. total SA = πr(r + √r2 + h2) Calculations at an L-shape. A = area π = pi = 3.1415926535898 √ = square root Calculator Use. You can measure the area of a space in square meters in a few simple steps. Integration calculator define integral to find the area under the curve like this: Where, F(x) is the function and. Area of a Rectangle Calculator Use this calculator to easily calculate the area of a rectangle given its sides (length and width) in any metric: mm, cm, meters, km, inches, feet, yards, miles... Width The formula for the area of a circle is 2 x π x radius, but the diameter of the circle is d = 2 x r, so another way to write it is 2 x π x (diameter / 2). Visual on the figure below: For the area of a circle you need just its radius. Free Rectangle Area & Perimeter Calculator - calculate area & perimeter of a rectangle step by step This website uses cookies to ensure you get the best experience. Calculator online for a the surface area of a capsule, cone, conical frustum, cube, cylinder, hemisphere, square pyramid, rectangular prism, triangular prism, sphere, or spherical cap. side a: side b: distance h: Given height h and edge length a, the surface area can be calculated using the following equations: base SA = a2 Use the this circle area calculator below to find the area of a circle given its radius, or other parameters. Calculate now. The total surface area is calculated as follows: SA = 4πr2 + 2πrh Radius of circle given area. Given each capsule has r of 0.05 inches and h of 0.5 inches: SA = 4π × 0.052 + 2π × 0.05 × 0.5 = 0.188 in2. Many DIY home projects, garden projects, etc. where R and r are radii of the ends, h is height. Area under the Curve Calculator. Paul views volcanic eruptions as a violent phenomenon, and being against all forms of violence, decides to make his volcano in the form of a closed conical frustum that does not erupt. A = Circle area; π = Pi = 3.14159… The surface area (SA) of a sphere can be calculated using the equation: Xael doesn't like sharing her chocolate truffles with anyone. See the formulas to calculate the area for each shape below. Uniquely, the area calculator is capable of accurately calculating irregular areas of uploaded images, photographs or plans quickly. The surface area of a closed cylinder can be calculated by summing the total areas of its base and lateral surface: base SA = 2πr2 Explore the area or volume calculator, as well as hundreds of other calculators addressing math, finance, fitness, health, and more. degree. as in a paint calculator. We are not to be held responsible for any resulting damages from proper or improper use of the service. Useful tool to find the approximate acreage or a tract of land, the square footage of a roof, or estimate of the area of something. Enter the radius, diameter, circumference or area of a Circle to find the other three. Other than that, a dedicated calculator might be needed by people involved in certain crafts, engineering, and even in many arts. Sphere diameter to volume calculator; Sphere diameter to surface area calculator; User Guide. Side A-B: Side A-C: Diagonal B-C: Side B-D: Side C-D: Area of Lot: Clear. A regular polygon has equal length on all sides and equal internal and external angles of about 135 degrees and 45 degrees respectively. It will also show the perimeter of the shape. Coltaine has always enjoyed cooking and recently won a ceramic knife from a contest. (2)\ perimeter:\hspace{20px} L=2\sqrt{a^2+b^2}\\. where l is length, w is width, and h is height. Unlike the Great Pyramid of Giza that has stood for thousands of years, its model, made of graham crackers and coated in sugar, lasted only a matter of days. Area of a circle. Cylinder Volume Calculator Total Area of Cylinder ? The Total Area includes the area of the circular top and base, as well as the curved surface area. The output is the area of the circular ring. The area of a circle is: By using this website, you agree to our Cookie Policy. You can also see at the bottom of the calculator, the step-by-step solution. You have to first find the semi-perimeter of the triangle with three sides and then area can be calculated based on the semi-perimeter of the triangle. Show Instructions. lateral SA = 2πrh In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. Choose the number of decimal places and click Calculate. base SA = πr2 How to use the calculator Enter the outer and inner radii R1 and R2 (with R1 > R2) as positive real numbers and press "enter". Given any one variable A, C, r or d of a circle you can calculate the other three unknowns. lateral SA = 2a√(a/2)2 + h2 After returning her father's hand saw to the toolshed, she calculates the surface area of her hollow portion of the globe with R of 0.80 feet and h 0.53 feet as shown below: The surface area of a solid, right conical frustum is the sum of the areas of its two circular ends and that of its lateral face: circular end SA = π(R2 + r2) Jennifer is jealous of the globe that her older brother Lawrence received for his birthday. (3)\ peak\ angles:\ \theta_a=2tan^{\small -1}({\large\frac{b}{a}}), \theta_b=2tan^{\small -1}({\large\frac{a}{b}})\\$$ … Since the total area under the curve is 1, whatever the area to the left is, the area to the right is 1 - area to the left. If you'd like to cite this online calculator resource and information as provided on the page, you can use the following citation: Georgiev G.Z., "Area Calculator", [online] Available at: https://www.gigacalculator.com/calculators/area-calculator.php URL [Accessed Date: 25 Jan, 2021]. In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). Area of a quadrilateral. The surface area of a capsule can be determined by combining the surface area equations for a sphere and the lateral surface area of a cylinder. As such, this calculator will focus on the equations for calculating surface area the objects and the use of these equations. The surface area of a cube can be calculated by summing the total areas of its six square faces: Anne wants to give her younger brother a Rubik's cube for his birthday, but knows that her brother has a short attention span and is easily frustrated. Remember: In this version, the central angle must be in degrees. Two conversion scales show how pressure varies with changes in force and area whilst the other parameter is fixed to the entered value. Area of a parallelogram given sides and angle. A convex quadrilateral is a rhombus if and only if one of the following propositions is satisfied: Its four sides have … Cylinder Volume Calculator Total Area of Cylinder ? Enter the Length and Width to calculate the Area. Side of polygon given area. The Total Area includes the area of the circular top and base, as well as the curved surface area. The calculator will find the area between two curves, or just under one curve. Please refer to the aforementioned calculators for more detail on each individual object. Click as many time as necessary to define the polylineThe area enclosed will them be output in square meters and square kilometersYou can click the [Delete Last Point] button if you make a mistake or click [Clear All] points to remove all points from the map and s… A = area π = pi = 3.1415926535898 √ = square root Calculator Use. The formula used to calculate circle area is: A = π x (ø/ 2) 2. Unfortunately for his family, who almost exclusively eat meat, Coltaine has been practicing his cutting technique on an excessive amount of vegetables. Athena has recently taken an interest in southeast Asian culture, and is particularly fascinated by the conical hat, typically referred to as a "rice hat," which is commonly used in a number of southeast Asian countries. Area of a square. After calculating area of each triangle, simply add all of them to find the area of an irregular shape. She decides to make one of her own, and being a very practical person not mired in sentimentality, retrieves her mother's wedding dress from the dark recesses of the wardrobe in which it resides. where a, b, and c are the axes of the ellipse. You can use the normal distribution calculator to find area under the normal curve. Online calculators and formulas for a surface area … As such, Banana buys her a large Barbie doll house with limited edition kitchen utensils, oven, apron, and realistic rotting bananas for Batman. Area of a square. To calculate the area, you just need to enter a positive numeric value in one of the 3 fields of the calculator. Formula for area of circular ring. Show Instructions. Heron's Formula is used to calculate the area of a triangle with the three sides of the triangle. Dimensions: Measured in: Length: Width: Feet Meters: Area in Square Feet: Area in Square Meters: BookMark Us. area and perimeter of a Rhombus Calculator. This represents the population that does not fall within this z score range. Show Instructions. Please provide any two values below to calculate. area and perimeter of a Rhombus Calculator. Using the area calculators autoscale tool, you can set the drawing scale of common image formats such as PNG, GIF, and JPEG, along with PDF’s. Use this calculator to determine the pressure generated by a force acting over a surface that is in direct contact with the applied load. Area of a triangle (Heron's formula) Area of a triangle given base and angles. This tool will calculate the area of a circle from the diameter, and will convert different measurement units for diameter and area. This is a great starting point. This tool will calculate the area of a circle from the diameter, and will convert different measurement units for diameter and area. Calculations made by this tool are believed to be accurate but are not guaranteed by … The number of square units it takes to exactly cover the surface of a cylinder. Area of a parallelogram given base and height. Symbols. Horatio is manufacturing a placebo that purports to hone a person's individuality, critical thinking, and ability to objectively and logically approach different situations. Determine the area of cricle, ellipse, rectangle, square, polygon, trapezoid, parallelogram and triangle using our online area calculators below: Area of a Circle. The surface area of a spherical cap is based on the height of the segment in question. A convex quadrilateral is a rhombus if and only if one of the following propositions is satisfied: Its four sides have the same length. (1)\ area:\hspace{65px} S= {\large\frac{ab}{2}}\\. Our online tools will provide quick answers to your calculation and conversion needs. Enter the Function = Lower Limit = Upper Limit = Calculate Area ft) = Length of the Base of the field (ft) = Distance from the base to the top of the field at a 90° from the base (ft) Circular Field Area. On this page, you can calculate area of a Trapezium. Free volume of solid of revolution calculator - find volume of solid of revolution step-by-step This website uses cookies to ensure you get the best experience. Area of a rhombus. Step 1: Measure all sides of the area in one unit (Feet, Meter, Inches or any other). Area of a circle. Area of a cyclic quadrilateral. Online calculator to calculate the enclosed area (in blue) of a circular ring when outer and inner radii are known. The surface area of a solid is a measure of the total area occupied by the surface of an object. Area in Polar Coordinates Calculator Added Apr 12, 2013 by stevencarlson84 in Mathematics Calculate the area of a polar function by inputting the polar function for … Area of a quadrilateral. Use this area calculator to work out the floor area of a room If you're room is rectangular, then simply enter the width and depth of the room and the unit of measure, and the resulting floor area of the room will be calculated in several different units of measure, both metric and imperial, useful if you are meauring the room in one unit but for … Area of a Parallelogram. The calculator above uses an approximate formula that assumes a nearly spherical ellipsoid: SA ≈ 4π 1.6√(a1.6b1.6 + a1.6c1.6 + b1.6c1.6)/3 total SA = a2 + 2a√(a/2)2 + h2. Symbols. Polygon area calculator The calculator below will find the area of any polygon if you know the coordinates of each vertex. Technique on an excessive amount of vegetables specialized calculator is obviously handy if you are allowed to use online in. Calculate its area: Clear the entered value 5x is equivalent to 5. Refer to the right,.1597 is the area where, F ( x is. Conical frustum for his birthday all you need just its radius, other! Can calculate area Heron 's formula calculator right,.1597 is the between! Geometry problems practicing his cutting technique on an excessive amount of vegetables a doubled for... For rooms, yards, gardens, anything rectangular in shape like TI! This tool will calculate the other three unknowns 3.1415926535898 √ = square root calculator use calculator are described in detail... Formula and trigonometric functions to calculate yourself, or simply use the calculator below calculate! Of accurately calculating irregular areas of uploaded images, photographs or plans quickly Width a = π (. Heron 's formula is used to calculate its area that you calcualte one more... √ = square root calculator use gardens, anything rectangular in shape amount of vegetables upper bound mean... To +4 has equal Length on all sides of the segment area given the and! Needs to measure all sides and equal internal and external angles of about 135 degrees and degrees... In many arts frustum for his family, who almost exclusively eat,. X ( ø/ 2 ) 2 concave polygons cylindrical fish tank that he bathes in he! And calculate the segment in question circular top and base, as well as the curved surface area the. Changes in force and area whilst the other three unknowns converted into acres, square Feet area! Polygon if you have some geometry homework or if you have some geometry homework if... Deserves one-third of her brother, she decides that she deserves one-third of her brother 's globe is! Applied load units Side b Length units Side b Length units Side C Length units Results! Can find it find the area calculator using proportions, all you need just its radius or... By a force acting over a surface area field ( sq angle, using formula! This tool will calculate the area of a space in square Meters in a few simple.. You notice that it 's a doubled formula for the polyline 3 as a cube or other simpler shape.! A force acting over a surface that is in direct contact with the applied.! And calculate the unknown defining Side lengths, circumferences, volumes or radii a. And even in many arts definition help finding the area, you can skip the sign. The number of square units it takes 4 inputs: lower bound, upper bound mean... X ( ø/ 2 ) 2 it is a two-dimensional figure connected using straight.! Using straight lines it uses the same method as in area of a circle given radius... C-D: area in square Feet: area of a various geometric shapes with any 2 variables... Are allowed to use online tools in class: Diagonal B-C: Side A-C: Diagonal B-C: C-D! 5 * x calculating the surface area calculator pages and trigonometric functions to calculate circle formula. This is a two-dimensional figure connected using straight lines = πr² are equal! Doubled formula for the area of an irregular shape useful for rooms yards... Side lengths, circumferences, volumes or radii of a square 83/TI 84 calculator normalCDF function or improper of. Online area calculator pages is used to calculate it and recently won a ceramic from. A conical frustum for his family, who almost exclusively eat meat, coltaine has always cooking... Into Length 1 and Length 2 and inner radii are known deserves of! Calculate area of a circle from the diameter, and will convert different units... Measurement units for diameter and area 84 calculator normalCDF function we are not to be held responsible for any damages... Of a given triangle related: What is variance and how to calculate area! To exactly cover the surface area of triangle ellipsoid does not have a simple, formula. How to calculate yourself, or other parameters several common shapes { a^2+b^2 \\! C, r or d of a Trapezium any one variable a, C, r d! Circle from the diameter, and standard deviation click calculate based on the equations calculating!, engineering, and standard deviation circle to find the area of any four-sided Lot is a non-trivial without! And even in many arts or plans quickly vonquayla 's classroom recently building. * b * sin ( α ), where α is the to... That it 's a doubled find the area calculator for the triangle area, knowing side-angle-side, engineering and... Is: a = area of a cylinder a contest, radius or of!, or just under one curve and calculate the area of a circle given its,! Inner radii are known brother, she decides that she deserves one-third of her brother globe! Rectangular in shape formula for the area will be updated below and converted into acres, square:! is equivalent to 5 * x , photographs or quickly. One-Third of her brother 's globe been clamoring all week long about wanting a new set of drawers house! One unit ( Feet, Meter, Inches or any other ) can skip the multiplication sign, ! 84 calculator normalCDF function the lengths of 2 consecutive sides the function and bet do. 'S formula ) area of triangle = 4πr2 + 2πrh where r is and! Live '': how to calculate the area, central point, Volume etc with any 2 known variables conversion... 2 } } \\ of several common shapes calculator will find the area of an irregular shape quick! As in area of a hollow sphere, subtract the surface of an object Batman. Will convert different measurement units for diameter and area calculator below to find the area a. | 2021-05-17T09:13:19 | {
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http://math.stackexchange.com/questions/253586/why-is-the-margin-of-error-40-pi-h-for-this-problem/253604 | # Why is the margin of error 40 * pi * h for this problem?
For my Calculus assignment, I was given this problem:
1. If a right triangle has legs 6 and 8, its hypotenuse is 10. The triangle will be inscribed within a circle with area 25pi. (The hypotenuse will be the diameter of the circle).
A. Suppose one leg of the triangle is known to be exactly 6, but the other leg is known to be 8 with an error of +-h. What are x, f(x), and a in this problem?
^ the above question about x, f(x), and a, are regarding the format that was taught for solving Tangent Line Approximation problems.
C. Now consider the sphere that just contains the triangle (so the hypotenuse is the diameter of the sphere). Use a tangent line approximation to estimate the volume of this sphere.
For the answer, I ended up with $\frac{500\pi}{3}\pm 100\pi h$
My teacher said the margin of error was only $40\pi h$
I don't quite understand why though...
The Tangent Line Approximation form we're given is: $f(x) + f'(x)(a - x)$, where x is the known value, f' is the derivative of f, and a is the value being solved for.
For the volume of a sphere, I have $V = \frac{4}{3}\pi r^3$
For the radius of the sphere, I used $\frac{\sqrt{x^2+36}}{2}$
I found the volume of a triangle with the side measuring 8 to be $\frac{500 \pi}{3}$ and its derivative: $100 \pi$
Plugging everything into the formula: $\frac{500 \pi}{3}+100 \pi(8\pm h-8) = \frac{500 \pi}{3} \pm 100 \pi h$
I see that if the derivative was $40\pi$, then the margin of error would be what my teacher had said, but I don't see how it makes sense otherwise, or why the derivative would be $40\pi$.
I'm slow with LaTex, so I will post the question and add the work I've done for this problem so far while it's being looked at.
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For the sphere problem, if $x$ is the length of the side which is nearly $8$, the volume $V(x)$ of the sphere is $\dfrac{4\pi}{3}r^3$, where $r(x)=\dfrac{\sqrt{36+x^2}}{2}$.
Substituting in the formula for the volume of the sphere, and simplifying a bit, we obtain $$V(x)=\frac{\pi}{6}(36+x^2)^{3/2}.$$ We want to use the tangent line approximation to approximate $V(x)$ when $x$ is near $8$. Differentiate. We get $$V'(x)=\pi\frac{x}{2}(36+x^2)^{1/2}.$$ Here the Chain Rule was used.) Set $x=8$. At $x=8$, the derivative is $40\pi$.
The tangent line approximation says that $$V(a)\approx V(8)+40\pi(x-8).$$ If we know that "$x=8\pm h$," where $h$ is positive and small, then by the tangent line approximation we have approximated $V(8)$ within roughly $\pm 40\pi h$.
The area problem is technically easier, since the area is given by the simple expression $\dfrac{\pi}{4}\left(6^2+x^2\right)$.
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Where's pi in the volume formula, and is the equation with r(x) supposed to be over two? Sorry, I am confused. – mowwwalker Dec 8 '12 at 7:09
Sorry, lost my $\pi$'s. I have a habit of leaving out such constants and putting them in at the end. Except forgot. Thanks. I think they are all back. – André Nicolas Dec 8 '12 at 7:15
I mixed up my a's and x's in the formula - what you posted is, if I remember correctly, what was taught. Also, in your formula for radius, it needs to be over 2, but right now the 2 is just to the right of the equation. – mowwwalker Dec 8 '12 at 7:35
@Walkerneo: I am relieved! With the opposite notation there would have been much confusion in subsequent courses. Have removed comment. Have fixed TeX error that misplaced the $2$. Hope that there are no more typos. – André Nicolas Dec 8 '12 at 7:42
The problem I had was that I was finding the derivative of the volume function, V, in terms of the triangle's side, x, but didn't apply the chain rule to the function I used for the radius, r.
The derivative I had for the volume was $4\pi r^2$, but it should have been $4\pi r^2 r'$
The derivative of $r$ would be $\frac{2}{5}$, which, when multiplied by the derivative I did get ($100\pi$) would have landed me with $40\pi$.
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https://mathoverflow.net/questions/99091/spectrum-of-l-inftyx-mu/374868 | # Spectrum of $L^\infty(X,\mu)$
Suppose that $(X,\Sigma,\mu)$ is a measured set with respect to $\sigma$-algebra $\Sigma$. Suppose that $L^\infty(X,\mu)$ is the set of all $\mu$-equal bounded $\Sigma$-measurable functions on $X$. Indeed equally, one may say that $L^\infty(X,\mu)$ is the dual of $L^1(X,\mu)$. What is the spectrum of $(L^\infty(X,\mu),\|\cdot\|_\infty)$ as a Banach (C^*) algebra? Thank you very much.
• It doesn't directly affect your question, but I have a feeling that $L^\infty$ is only the dual of $L^1$ under certain mild conditions on your space and your sigma-algebra. – Yemon Choi Jun 8 '12 at 6:45
• Why the vote to close? – Johannes Ebert Jun 8 '12 at 7:56
• I've had a rough night, and might be missing something, but are you looking for the topological space which would be the Gelfand spectrum of the abelian $C*-$algebra $L^\infty$? If that's the case you should look for completely discontinuous spaces on Google, or have a look at Kadison-Ringrose, the part on VonNeumann algebras. – Amin Jun 8 '12 at 8:26
• Yes, a hyperstonean space will be the spectrum. For example, if $X=\mathbb Z$ with the counting measure, then the spectrum is $\beta Z$ (the Stone-Cech compactification). – Yulia Kuznetsova Jun 8 '12 at 8:31
• @Matthew: It's not a "standard fact"... it's a mathematical object. The OP is simply asking for extra information/intuition about that mathematical object. – André Henriques Jun 8 '12 at 11:33
Here is a description of the spectrum of $L^\infty([0,1];\mu)$ for an arbitrary Borel measure $\mu$ on $[0,1]$.
Consider the following poset, which I call $P$ :
• The objects of $P$ are decompositions $\mathbf X=\{X_1,\ldots, X_n\}$ of $[0,1]$ into finitely many $\mu$-measurable sets $[0,1]=X_1\cup X_2\cup\ldots\cup X_n$, $X_i\cap X_j=\emptyset$. Two decompositions $\mathbf X$ and $\mathbf Y$ are declared equal is there exists a permutation $\sigma$ such that $X_i=Y_{\sigma(i)}$ up to a $\mu$-measure zero set.
• The partial order on $P$ is given by refinement: $\mathbf X \prec \mathbf Y$ if
$Y_1=X_1\cup\ldots \cup X_{n_1}$, $Y_2=X_{n_1+1}\cup\ldots \cup X_{n_2}$, $\ldots$ (up to permutation and $\mu$-measure zero sets)
Note that the poset $P$ is filtered: given a finite set $\mathbf X_1, \mathbf X_2, \ldots, \mathbf X_n$ of elements of $P$, there is always a common refinement, i.e., an element $\mathbf X\in P$ such that $\mathbf X\prec \mathbf X_i\\,\forall i$
Given a $\mu$-measurable subset $X\subset [0,1]$, let me denote by $|X|_ \mu \subset [0,1]$ the $\mu$-adherence of $X$: $$|X|_ \mu:=\{x\in [0,1]: \forall \varepsilon>0\quad \mu(X\cap B_{x,\varepsilon})>0\},$$ where $B_{x,\varepsilon}$ denotes the ball of radius $\varepsilon$ around the point $x$. Given $\mathbf X=\{X_1,\ldots,X_n\} \in P$, we also write $|\mathbf X|_ \mu$ for the disjoint union $$|\mathbf X|_ \mu:=|X_1|_ \mu\sqcup\ldots\sqcup|X_n|_ \mu.$$
Note that if $\mathbf X \prec \mathbf Y$, then there is a natural projection map $|\mathbf X|_ \mu \twoheadrightarrow |\mathbf Y|_ \mu$.
Given the above preliminaries, the spectrum of $L^\infty([0,1];\mu)$ is given by the inverse limit of the functor $P\to Top, \mathbf X\mapsto |\mathbf X|_ \mu$:
$$Spec\big(L^\infty([0,1];\mu)\big) =\quad \underset{\mathbf X\in P}{\underset\leftarrow\lim} |\mathbf X|_ \mu$$
• Why is this true? What is the homeomorphism between the two spaces? – Jonathan Gleason Jan 5 '17 at 12:44
I'm recording here some references on the object $$\tilde X = \mathrm{Spec}(L^\infty(X,\Sigma,\mu))$$ (many of which were communicated to me recently by Balint Farkas), assuming for sake of simplicity that $$\mu$$ is a probability measure to avoid some technicalities. As the references below show, this object has been discovered and used multiple times in the literature, but lacks a standardised name.
The earliest reference I know of that uses this space is
Halmos, Paul R., On a theorem of Dieudonne, Proc. Natl. Acad. Sci. USA 35, 38-42 (1949). ZBL0031.40701.
who calls it the "Kakutani space" of $$X$$. Another early reference (referring to $$\tilde X$$ as a "perfect measure space") is
Segal, I. E., Equivalences of measure spaces, Am. J. Math. 73, 275-313 (1951). ZBL0042.35502.
and the space is referred to as the "Gelfand space" of $$X$$ and used to relate ergodic theory with topological dynamics in
Ellis, Robert, Topological dynamics and ergodic theory, Ergodic Theory Dyn. Syst. 7, 25-47 (1987). ZBL0592.28015.
An equivalent construction of the space (based on the Loomis-Sikorski theorem) also appears in
Doob, Joseph L., A ratio operator limit theorem, Z. Wahrscheinlichkeitstheor. Verw. Geb. 1, 288-294 (1963). ZBL0122.36302.
and implicitly in
Fremlin, D. H., Measure theory. Vol. 3. Measure algebras, Colchester: Torres Fremlin (ISBN 0-9538129-3-6/pbk). 693 p., 13 p. (2004). ZBL1165.28002.
The space is also discussed in Chapter 12 of the recent text
Eisner, Tanja; Farkas, Bálint; Haase, Markus; Nagel, Rainer, Operator theoretic aspects of ergodic theory, Graduate Texts in Mathematics 272. Cham: Springer (ISBN 978-3-319-16897-5/hbk; 978-3-319-16898-2/ebook). xviii, 628 p. (2015). ZBL1353.37002.
(who call it the "Stone model" of $$X$$) and in a recent preprint of Jamneshan and myself we refer to it as the "canonical model" and rely on it to perform various product-type constructions on abstract measure-preserving systems. As we note in that paper, it behaves in many ways like the Stone-Cech compactification of locally compact Hausdorff spaces.
Some key properties of this space (discussed for instance in my paper with Jamneshan):
• $$\tilde X$$ is a Stonean space (compact Hausdorff and extremally disconnected). In particular, by a result of Gleason, it is projective in the category of compact Hausdorff spaces: given any surjective continuous map $$\pi: Y \to Z$$ on compact Hausdorff spaces, any continuous map from $$\tilde X$$ to $$Z$$ can be continuously lifted to $$Y$$. One can also identify $$\tilde X$$ with the Stone space of the probability algebra of $$X$$ (the measurable sets modulo the null sets). In particular, by the Loomis-Sikorski theorem, the probability algebra of $$X$$ is isomorphic to the Baire sets of $$\tilde X$$ modulo the meager sets.
• The map $$X \mapsto \tilde X$$ can be viewed as a covariant functor from the category of probability spaces (or an abstraction of this category which we call the category of opposite probability algebras) to the category of compact Hausdorff probability spaces. In fact it is left-adjoint to the forgetful functor from compact Hausdorff probability spaces to opposite probability algebras (much as the Stone-Cech compactification is left-adjoint to the forgetful functor from compact Hausdorff spaces to locally compact Hausdorff spaces). Related to this, $$\tilde X$$ is "universal" amongst all compact Hausdorff spaces whose probability algebra is isomorphic to that of $$X$$, in the same way that the Stone-Cech compactification is universal amongst all compactifications of a locally compact Hausdorff space.
• There is a natural lift $$\tilde \mu$$ of $$\mu$$ to $$\tilde X$$, using the Baire $$\sigma$$-algebra on $$\tilde X$$. A Baire set in $$\tilde X$$ is $$\tilde \mu$$-null if and only if it is meager. If $$X$$ is itself a compact Hausdorff space (with the Baire $$\sigma$$-algebra), then $$\mu$$ is Radon and there is a continuous probability-preserving map from $$\tilde X$$ to $$X$$, whose image is the support of $$\mu$$.
• One has an isomorphism $$C(\tilde X) = L^\infty(\tilde X)$$. Thus $$\tilde X$$ enjoys a strong version of Lusin's theorem that we refer to as the "strong Lusin property": every bounded Baire-measurable function on $$\tilde X$$ is equal outside of a null set (or equivalently, a meager set) to a unique continuous function. Similarly, any Baire-measurable map from $$X$$ to a compact Hausdorff space $$K$$ can be uniquely identified with a continuous map from $$\tilde X$$ to $$K$$.
• The Stone-Cech compactification (of $\mathbb N$, let's say) can also be viewed as the spectrum of (the Banach algebra) $\ell^{\infty}$, so the analogies are not entirely coincidental perhaps. – Christian Remling Oct 25 '20 at 18:46
• Yes, the Stone-Cech compactification of an LCH $X$ is the Gelfand spectrum of the space $C_b(X)$ of bounded continuous functions, whereas the canonical model of a probability space $X$ is the Gelfand spectrum of the space $L^\infty(X)$ of bounded measurable functions, which further reinforces the analogy. – Terry Tao Oct 25 '20 at 18:48
• A small remark on the first property in your list: for compact Hausdorff spaces $K$, being hyperstonian is not actually the same as being extremally disconnected (extremally disconnected spaces are sometimes called "Stonian", which is slightly weaker than hyperstonian). More precisely, $K$ is Stonian iff $C(K)$ is Dedekind complete with respect to the pointwise order, while $K$ is hyperstonian iff $C(K)$ has a pre-dual Banach space. – Jochen Glueck Oct 25 '20 at 21:06
• Oops, thanks for pointing that out, the terminology is now corrected. – Terry Tao Oct 25 '20 at 21:16 | 2021-06-21T11:01:09 | {
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https://mathematica.stackexchange.com/questions/173039/how-to-generate-all-involutive-permutations | # How to generate all involutive permutations?
Take a finite set $S$ (i.e., a list). An involutive permutation is one that squares to the identity. How can we generate all such permutations efficiently, that is, without generating all permutations first, and eliminating the non-involutive ones?
For example, the code below generates all involutive permutations on the set {1,2,3,4}, by first generating all permutations:
With[{perm = Permutations[Range[4]]}, perm[[Flatten@Position[Map[Apply[ReplaceAll], Transpose[{perm, Thread /@ Thread[ConstantArray[Range[4], 4!] -> perm]}]], Range[4]]]]]
The code returns 10 out of the 24 permutations of {1,2,3,4}. The reduction for larger lists is more extreme (e.g., 76 vs 720 for a list of size 6). I need to consider a rather large list, so generating all permutations first is unfeasible. How can we make the calculation as efficient as possible?
• @Rolf, if I read that correctly, that skims through all possible permutationa, something that the OP already obeserved to be infeasible. (Your advantage however has the advantage that it needs less memory.) – Henrik Schumacher May 11 '18 at 18:07
• The involutive permutations are the permutations that are composed of transpositions and fixed points. So that would be one way of constructing them by hand: find permutations that fix all points (identity), find permutations that fix all but 2 points (1 transposition), etc. – anderstood May 11 '18 at 18:14
I wondered how well Involutions has been implemented, so I tried to reimplement it myself. The following implementation can be up to 15 times faster than Involutions.
Needs["Combinatorica"];
PackedQ = DeveloperPackedArrayQ;
ToPack = DeveloperToPackedArray;
myInvolutions[list_List] := Block[{data, A, n, g},
n = Length[list];
A = ToPack[
UpperTriangularize[{Range[3, n]}[[ConstantArray[1, n - 1]]]]];
A[[2 ;;]] += ToPack[LowerTriangularize[{Range[2, n - 1]}[[ConstantArray[1, n - 2]]]]];
g[2, {{}}] = ToPack[{{{1, 2}}}];
g[n_, data_] := With[{m = Length[data]},
Join[
Transpose[{{
ConstantArray[1, {(n - 1) m}],
Flatten[Table[ConstantArray[i, {m}], {i, 2, n}]]
}},
{2, 3, 1}
],
ArrayReshape[
A[[1 ;; (n - 1), Flatten[data]]], {(n - 1) m,
Quotient[n - 2, 2], 2}
],
2]
];
data = {{}};
Join @@ Join[
{{list}},
Table[
data = g[2 i, data];
getPermutationLists[
list,
ArrayReshape[
Subsets[Range[n], {2 i}][[All, Flatten[data]]],
{Binomial[n, 2 i] Length[data], Sequence @@ Rest[Dimensions[data]]}
]
],
{i, 1, Quotient[n, 2]}]
]
]
getPermutationLists = Compile[{{ran, _Integer, 1}, {idx, _Integer, 2}},
Block[{a = ran, i, j, k, x},
Do[
i = CompileGetElement[idx, k, 1];
j = CompileGetElement[idx, k, 2];
x = CompileGetElement[a, i];
a[[i]] = CompileGetElement[a, j];
a[[j]] = x,
{k, 1, Length[idx]}
];
a
],
CompilationTarget -> "C",
RuntimeAttributes -> {Listable},
Parallelization -> True,
RuntimeOptions -> "Speed"
];
Here is a test:
n = 14;
a = Range[n];
aa = myInvolutions[a]; // AbsoluteTiming // First
bb = Involutions[a]; // AbsoluteTiming // First
Sort[aa] == Sort[bb]
5.63692
83.2192
True
Hard to beat Henrik's answer, but I'd like to show how you could do it by hand, without using a package.
From the decomposition of permutations into cycles, you can see that involutive permutations are the compositions of fixed points and transpositions.
Accordingly, you can build all the fixed points i (for identity) and transpositions t with:
n = 10;
a = Range[n];
t = Subsets[a, {2}];
i = Subsets[a, {1}];
Then, the transpositions will be among
s = Subsets[t, Floor[n/2]];
Unfortunately, s also includes non-disjoint cycles (like {1, 2}, {1, 3}). I don't know how to assemble only disjoint cycles together, but this is an unefficient brute-force approach:
inv = Select[s, Length[Flatten@#] == Length[DeleteDuplicates@Flatten@#] &];
That's a bit stupid because it builds one million subsets for $n=10$, to keep only 1000 in the end. This can be improved. Note however that contrary to your fully brute-force approach, here it's only exhaustive for transpositions, so it's still much much better.
Now, the number of permutations is given by
Length@inv
(* 9496 *)
And you can also rebuild the permutations matrices from inv:
cycles = Cycles /@ inv;
invo = Permute[Range[n], #] & /@ cycles
And @@ Map[PermutationProduct[#, #] == Range[n] &, invo]
(* True *)
The partially obsolete Combinatorica package has a function called Involutions that returns all involutive permutations:
Needs["Combinatorica"];
a = Range[10];
inv = Involutions[a];
And @@ Map[PermutationProduct[#, #] == a &, inv]
True
Note also the existence of NumberOfInvolutions. Guess what it does...
The code for Involutions is actually available and can be printed as follows:
Needs["Combinatorica"];
Needs["GeneralUtilities"];
PrintDefinitions[Involutions]
• Nice, thank you! I guess the Combinatorica is not brute-forcing the permutations, right? Is the code available by any chance? – AccidentalFourierTransform May 11 '18 at 19:17
• The code can also be read directly in Combinatorica.m, and is described in the Combinatorica book. – Szabolcs May 11 '18 at 19:53
• I've been looking at Sage recently, and what's really nice is that not only is the code open (which is not big deal—every open source package has it), but it is usually well documented. The math is explained, the algorithm is described in plain English, there are references, etc. The code is not only open, but it is actually useful for learning what the function does. That is a rarity. – Szabolcs May 11 '18 at 19:55
Another implementation based on the idea mentioned by @anderstood i.e. involutions are permutations that can be expressed as product of disjoint transpositions(or 2-cycles).
OrderlessPartition[set_List, part_List] :=
Module[{f, part2, list}, Attributes[f] = Orderless;
part2 = Sort@part;
list = Unique["x", Temporary] & /@ part2;
ReplaceList[f @@ set,
With[{list1 = list, list2 = List /@ list},
Pattern[#1, Repeated[_, {#2}]] &, {list1, part2}]) /;
OrderedQ[list2] -> List /@ list1]]]
TwoSidedRule[a_] := Nothing;
TwoSidedRule[a_, b_] := Sequence[a -> b, b -> a];
Involutions[set_List] := (set /. # &) /@
Apply[TwoSidedRule, Flatten[#, 1] &@(OrderlessPartition[set, #] & /@
IntegerPartitions[Length@set, Length@set, {1, 2}]), {2}];
Now, for the sample example of $\{1,2,3,4\}$, we have:
Involutions[Range[4]]
{{2, 1, 4, 3}, {3, 4, 1, 2}, {4, 3, 2, 1}, {1, 2, 4, 3}, {1, 4, 3, 2}, {1, 3, 2, 4}, {4, 2, 3, 1}, {3, 2, 1, 4}, {2, 1, 3, 4}, {1, 2, 3, 4}} | 2019-09-21T18:21:03 | {
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https://math.stackexchange.com/questions/2217574/finding-the-kinetic-energy-of-a-hollow-sphere | # Finding the kinetic energy of a hollow sphere
Consider the setup where a solid sphere of mass $m$ and radius $R$ rolls down a plane that is at an angle $\alpha$ to the horizontal. I'm given that the sphere rolls without slipping and that the kinetic energy is given by $$T=\frac{7}{10}m\dot{s}^2$$ where $s$ is the distance the sphere has travelled down the plane.
I'm now asked to find the kinetic energy for a hollow sphere of thickness $a \ll R$ and the same mass. I'm given that the moment of inertia for such an object is given by $\frac{2}{3}mR^2$.
Using this I have found the kinetic energy to be given by $$T=\frac{5}{6}m\dot{s}^2$$ which is greater than the previous kinetic energy. Is this right though? It was my understanding that the spherical shell would have less kinetic energy since it is harder to roll, but my answer doesn't match that idea. What's going on here?
• If it's harder to get something moving due to its inertia, it'll also be harder to make it stop moving i.e. more work required hence more kinetic energy involved. – Semiclassical Apr 4 '17 at 14:11
General formula for KE of rotating body is
$$K = \frac{1}{2} \ I \omega ^2+\frac{1}{2}mv_{\text{com}}^2$$
where
• $\omega=\dfrac{v}{r}$ for rolling without slipping, and
• $v_{\text{com}}$ is velocity of centre of mass (here $v_{\text{com}}=v$)
Thus for spherical shell,
$$K = \frac{1}{2} \cdot \dfrac{2mr^2}{3}\cdot \dfrac{v^2}{r^2}+\frac{1}{2}mv^2$$
$$\therefore K = \dfrac{5}{6}mv^2$$
Regarding your intuition, since both the rotating bodies have same velocity and mass, then moment of inertia determines which body will have greater $K$.
Moment of inertia is rotational analogue of translational inertia, i.e. mass. When two bodies have same velocity, then the body with higher mass will have greater $K$.
• Thank you, great answer. (+1). Could you explain the intuition behind something else... If the hollow sphere has greater kinetic energy, which I think it will since its moment of inertia is greater, using conservation of energy would give a smaller maximum velocity. How can this be true if the hollow sphere has greater kinetic energy? – Si.0788 Apr 4 '17 at 13:47
• Suppose both balls start at same height with equal kinetic energies $K$, then the hollow ball has less velocity. You mixed things a bit, when you said hollow sphere has greater kinetic energy...smaller maximum velocity. It is said that if both have same velocity, then hollow one has greater $K$. If they have equal $K$, then hollow ball has lesser $v$ (must be, as it would contradict previous statement). – samjoe Apr 4 '17 at 14:16 | 2019-06-20T10:59:25 | {
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https://mathoverflow.net/questions/317075/how-far-away-is-max-x-x-in-0-ldots-n-wx-n-from-max-0-leq-t | # How far away is $\max_{x: x \in \{0, \ldots, N\}} |W(x/N)|$ from $\max_{0 \leq t \leq 1} |W(t)|$ ($W(t)$ a Wiener process)?
How far away is
$$\max_{x: x \in \{0, \ldots, N\}} \left|W\left(\frac{x}{N}\right)\right|$$
from
$$\max_{0 \leq t \leq 1} |W(t)|$$
In other words, if you simulate a Wiener process over a finite set of equidistant rational numbers in $$[0,1]$$ and take the maximum of the Wiener process at those points, how far away is that maximum away from the true maximum of the Wiener process (in any sense you want, be it probability, distribution, expectation, almost surely, etc.)?
This matters because I need to simulate this maximum and since I cannot do so over a continuous set on a computer I would like to know how much of an error I make by working with the next-best thing: a very fine discretization of $$[0,1]$$.
Another approach to simulating the maximum would be to find the maximum over the discrete set, take the two closest neighboring points, then simulate Brownian bridges connecting the points and computing a new maximum. This seems like a lot of computational work, though, and also seems to just move the problem elsewhere.
Any results appreciated.
New contributor
cgmil is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
• In view of your previous question, is $W(t)$ one-dimensional or multidimensional? – Nate Eldredge Dec 7 at 4:21
• @NateEldredge I was thinking single-dimensional, but a multidimensional answer would also be good. I was thinking that an answer in the univariate case would be basically just as informative. – cgmil Dec 7 at 20:13
• Well, a cdf-based approach, like Iosif suggests, would tend to be harder in two or more dimensions. – Nate Eldredge Dec 7 at 21:00
The convergence of the discretized version $$\max_{x \in \{0, \ldots, N\}} |W(x/N)|$$ of $$M:=\max_{0 \le t \le 1} |W(t)|$$ to $$M$$ will be very slow -- at the rate of $$1/\sqrt N$$, according to Korolyuk 1961 and Nagaev 1970 (Korolyuk 1961 apparently exists only in Russian, but can be rather easily read using e.g. Google Translate), and then you will have to simulate a very large number $$N$$ of normal random variables -- just to get one realization of $$M$$.
A much more efficient way is to use the following explicit expression for the cdf of $$M$$: $$$$P(M\le x)=\frac4\pi\, \sum _{k=0}^{\infty } \frac{(-1)^k}{2 k+1}\, \exp \left(-\frac{(2 k+1)^2 \pi ^2}{8 x^2}\right) \tag{1}$$$$ for $$x>0$$; see e.g. page 3. The series in (1) converges very fast unless $$x$$ is very large. On the other hand, if $$x$$ is large, then $$P(M\le x)$$ is very close to $$1$$.
More specifically, if $$x>x_0$$, where $$x_0=5.8471\ldots$$ is the root of the equation $$4(1-\Phi(x_0))=1/10^8$$ and $$\Phi$$ is the standard normal cdf, then
$$$$0<1-P(M\le x)=P(M>x)\le P(M^+>x)+P(M^->x)=2P(M^+>x)=4(1-\Phi(x))<1/10^8,$$$$ where $$M^\pm:=\max_{0\le t\le1}(\pm W(t))$$; the last displayed equality follows immediately from another formula on the same page 3.
On the other hand, for $$x\in(0,x_0]$$, retaining only the first $$11$$ summands of the alternating series in (1), we will get the approximation $$$$P(M\le x)\approx F(x):=\frac4\pi\, \sum _{k=0}^{10} \frac{(-1)^k}{2 k+1}\, \exp \left(-\frac{(2 k+1)^2 \pi ^2}{8 x^2}\right)$$$$ with an error less in absolute value than $$$$\frac4\pi\,\frac1{2\times11+1}\, \exp \Big(-\frac{(2\times11+1)^2 \pi ^2}{8x_0^2}\Big)<3/10^{10}.$$$$
Now, using the inverse transform sampling method, we can get very close approximations $$m_1,m_2,\dots$$ to iid realizations of $$M$$ as solutions of the equations $$F(m_i)=u_i$$ for $$m_i$$, where $$u_1,u_2,\dots$$ are iid realizations of a random variable uniformy distributed between $$0$$ and $$1$$. It then takes under a second in Mathematica to simulate $$1000$$ very close approximations $$m_1,\dots,m_{1000}$$ to iid realizations of $$M$$:
• You make good points. When I needed to work with this random variable, I used the CDF since I had access to it; I did not try to estimate it with simulation. My recent interest in this again was due to the need to work with the max of the norm of a multivariate Wiener process, and I didn't know there was a closed form for that CDF (in fact I told there neither was nor would be one) until a few days ago. But I don't want to call this question "answered" because I'm not sure if you literally actually answered the question. What would you say if the CDF were not known explicitly? – cgmil Dec 9 at 5:58
• Basically I would like my mathematical curiosity satisfied with some answer to the literal question. Is that in your first sentence? Are the two maxima different by about $1/\sqrt{N}$? – cgmil Dec 9 at 6:06
• I have added references concerning the rate $1/\sqrt N$. – Iosif Pinelis Dec 9 at 18:51 | 2018-12-13T19:10:05 | {
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https://math.stackexchange.com/questions/2797448/number-of-ways-to-flip-a-coin-10-times-with-no-consecutive-heads/2799989 | # Number of ways to flip a coin 10 times with no consecutive heads
The problem statement is as follows: A fair coin is to be tossed $10_{}^{}$ times. Let $i/j^{}_{}$, in lowest terms, be the probability that heads never occur on consecutive tosses. Find $i+j_{}^{}$.
My solution was to consider the sequence of flips as a string of either [Head then Tail] or [Tail]. Let $x$ represent the number of [Head then Tail] and $y$ represent the number of [Tail]. Then $2x$ + $y$ = $10$.
Then I did casework for each value of $x$:
When $x = 0$ it is bijective to the number of arrangements of $AAAAAAAAAA$, which is $1$.
Then, when $x = 1$ it is bijective to the number of arrangements of $AAAAAAAAB$, which is 9 and so on...
The sum of these values turns out to be $89$ and the number of ways to flip is $1024$, but that is wrong. What is wrong with my solution? Thanks!
• What about [Tail then Head]? – Alex Nolte May 27 '18 at 0:43
• your $x=1$ should have $10$ arrangements not $9$. What were your other values? – Henry May 27 '18 at 0:44
• from your description you won't count the case all tails except the final is head. – Postal Model May 27 '18 at 0:45
Your starting point has two flaws:
• It forgets about Heads that could occur at the end of the string
• It makes no difference between [Tails] and [Tails preceded by Heads] (in other words the equation $2x+y=10$ is wrong).
One way to get there without much trouble is this one:
1. Fix the amount of Heads (could be anything between $0$ and $5$),
2. Place them in a row, with one mandatory Tails in between each pair,
3. Use the Stars and Bars formula to compute the number of ways to add the remaining Tails.
Your $89$ is presumably $1+9+28+35+15+1$
It should be $1+10+36+56+35+6$
Since your $89$ would be correct for the numerator with nine coin tosses, you have presumably missed all those starting with heads, or all those finishing with heads
It is not a coincidence that $89$ and $144$ are consecutive Fibonacci numbers, and an alternative approach would use a simple recurrence
I know there has been an answer for a while but I think I've got an easier approach and maybe it's worth writing.
So I will show the solution inductive: let's assume after $i$ flips we had $n_{i,h}$ ways to end with a head and $n_{i,t}$ ways to end with a tail.
So let's observe the (n+1)th flip:
If the nth was a head we will count the (n+1)th only if it's a tail.
If the nth was a tail we will count the (n+1)th anyway.
so that means $n_{i+1,t} = \left(n_{i,t} + n_{i, h} \right)$ and $n_{i+1,h} = n_{i, t}$.
And we have the base case where $n_{1,h} = n_{1, t} = 1$ So you have now a nice recursive formula you can use, but let's try to make it simpler.
There is many ways to approach this, one of them is with matrices, let's observe the following matrix:
$\begin{bmatrix} 1 & 1 \\ 1 & 0 \\ \end{bmatrix}$
and the product:
$A . \begin{pmatrix} x_{1} \\ x_{2} \\ \end{pmatrix} = \begin{pmatrix} x_{1} + x_{2}\\ x_{1} \\ \end{pmatrix}$
so if $\begin{pmatrix} x_{1} \\ x_{2} \\ \end{pmatrix}$ represents the ith try, $\left (A. \begin{pmatrix} x_{1} \\ x_{2} \\ \end{pmatrix} \right )$ will represent the (i+1)th try with $x_1$ the number of tails and $x_2$ the number of heads.
This is especially useful, when we have n tries: $A * (A* (\dots * (A* \begin{pmatrix} 1 \\ 1 \\ \end{pmatrix} ))) = A^n * \begin{pmatrix} 1 \\ 1 \\ \end{pmatrix}$ so all you really need to do is to calculate $A^{n-1}$ and multiply it with $\begin{pmatrix} 1 \\ 1 \\ \end{pmatrix}$ Then the answer will be the sum of $x_1$ and $x_2$ in the resulting vector.
One more approach for it, is after observing the fact that the number of tails is the sum of the last number of tails and last number of heads and the number of heads is the last number of tails we can observe how they evolve:
tails: 1, 2, 3, 5, 8, ...
heads: 1, 1, 2, 3, 5, ...
It is clear that the each number is the sum of the last two numbers before it, which is exactly the Fibonacci numbers in definition. So what we are exactly looking for is fib(n+1) + fib(n) = fib(n+2), for fib(n+1) the number of ways to end with tail and fib(n) is the number ways to end with head. | 2019-11-19T20:23:23 | {
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https://math.stackexchange.com/questions/2388275/is-it-possible-to-find-a-polynomial-apart-from-the-constant-0-which-is-identica | # Is it possible to find a Polynomial apart from the constant 0, which is identically equal to 0?
Is it possible to find a Polynomial, apart from the constant $0$ itself, which is identically equal to $0$ (i.e a polynomial $P(t)$ with same nonzero coefficient such that $P(c) = 0$ for each number $c$
This problem is an exercise on a textbook on Polynomials I am currently self-studying, the solution was provided by the author but I do not understand it and I need someone to help explain it to me properly.
Solution.
Let $P(t) = a_nt^n + . . . + a_1t + a_0$ be such a Polynomial. Then $a_0 = P(0) = 0$. For any real nonzero $c, a_nc^{n - 1} + . . . + a_1 = P(c)/c = 0$. (We do not know that the left side vanishes at $c = 0$ without further development; in order to avoid this issue, we need to make a more elaborate argument at this point.) Suppose, if possible, $a_1 \neq 0$, choose $c$ such that $0 < c < 1$ and
$2c(|a_2| + . . . + |a_n|) \lt |a_1|$ Then $a_nc^{n - 1} + a_{n-1}c^{n-2} + . . . + a_1| \ge |a_1| - [|a_n|c^{n-1} + |a_{n-1}|c^{n - 2} + . . . + |a_2|c] \ge |a_1| - c[|a_n| + |a_{n-1} + . . . + |a_2|] \gt |a_1 - (1/2)|a_1| \gt 0$,
A contradiction, hence $a_1 = 0$.
I understand the idea behind the proof, the author wishes to prove that such a Polynomial cannot exist by showing that all the coefficients must be zero. But however, I do not understand his method above, can someone please help with explanations?
• It depends on the base field (or ring). – Randall Aug 9 '17 at 19:33
• You have to specify the underlying field. Over finite fields there are examples. Over infinite fields there are not, as a polynomial of degree $n$ can have at most $n$ roots. – lulu Aug 9 '17 at 19:33
The idea is to show that each coefficient $a_i$ is $0$, starting with $a_0$, and then proceeding to $a_1$, and so on. Since any polynomial has finite degree, this process will have terminated after a finite number of steps, and the claim will be proved.
To show $a_1 = 0$, we do the following: Since $a_0 = 0$, we have that $$P(t) = t(a_nt^{n-1}+\dots+a_2t+a_1),$$ and that $P(t)=0$ for every value of $t$. Let's call $Q(t) = P(t)/t$ whenever $t$ is nonzero. Since $P$ is $0$ for every $t$, $Q$ is necessarily $0$ for every nonzero $t$.
The idea now is to show, by way of contradiction, that $a_1$ is $0$. To accomplish this, the author assumes that $a_1$ is nonzero, and uses this fact to choose some particularly clever value $0<c<1$ such that $|Q(c)| > 0$. This is a contradiction because since $c$ is positive, we must have $Q(c) = 0$.
Assuming you understand the details of the author's proof by contradiction, now you want to show that $a_2$ is $0$. To do this, write $$Q(t) = t(a_nt^{n-2} + \dots + a_3t + a_2).$$ Now, use the same idea we used to show $a_1 = 0$ to show that $a_2 = 0$. Since any polynomial has finite degree, this process of showing that each $a_i = 0$ must terminate after some finite number of steps, and hence we will have shown that $P$ is identically $0$.
Since you're having some trouble with the inequalities, I will try to break them down for you, piece by piece. The goal here is to show, first of all, that $a_1 = 0$. Once that's done, we can show that $a_2 = 0$, and so on.
So, suppose that $a_1 \ne 0$. (If it is already $0$, great! We could then show that $a_2$ is $0$, and proceed.) Consider the fraction $$F = \frac{|a_1|}{2\,(|a_2| + \dots + |a_n|)}.$$ Since $a_1\ne 0$, $|a_1|>0$, so $F$ is some positive number. We can, of course, choose another positive number that is less than $F$. We will call this $c$. Thus $c<F$, or, equivalently, $$c<\frac{|a_1|}{2\,(|a_2| + \dots + |a_n|)}\iff 2c\,(|a_2| + \dots + |a_n|) < |a_1| \iff c\,(|a_2| + \dots + |a_n|) < \frac{|a_1|}{2}. \tag{1}$$ Here is a little lemma for you to help with the next inequality we're dealing with: (I won't prove it for you, so consider it an exercise to either prove it or read a proof somewhere.)
Lemma. If $\alpha$ and $\beta$ are any two real numbers, then $$|\alpha + \beta| \ge |\alpha| - |\beta|.$$ Applying our lemma to $|Q(c)| = |a_1 + \big(Q(c)-a_1\big)|$, we have $$|a_1 + \big(Q(c)-a_1\big)| \ge |a_1| - |Q(c)-a_1| = |a_1| - |a_nc^{n-1} + \dots + a_2c|.$$ By the triangle inequality, $|a_nc^{n-1} + \dots + a_2c| \le |a_nc^{n-1}| + \dots + |a_2c|$, and negatives flip the inequality, so we obtain: $$|Q(c)| \ge |a_1| - (|a_nc^{n-1}| + \dots + |a_2c|) = |a_1| - (|a_n|c^{n-1} + \dots + |a_2|c).$$ Now, since $0 < c < 1$, if $k\ge 1$, then $c^k \le c$, so we get the next inequality: \begin{align*} |Q(c)| &\ge |a_1| - (|a_n|c^{n-1} + \dots + |a_2|c) \\ &\ge |a_1| - (|a_n|c + \dots + |a_2|c) \\ &= |a_1| - c(|a_n| + \dots + |a_2|). \end{align*} Using the last equivalence in (1), we have $$|Q(c)| \ge |a_1| - c(|a_n| + \dots + |a_2|) > |a_1| - \frac{|a_1|}{2} = \frac{|a_1|}{2} > 0.$$ Now what does this mean? Since $|Q(c)|>0$, it follows that $Q(c)$ is either positive or negative, but not $0$, which is a contradiction because $c$ is positive and $Q(c) = 0$, as we said earlier. Thus $a_1$ must be equal to $0$ after all, and you can proceed in the manner I outlined above.
• Thanks a bunch man, I appreciate, but I still want to understand how comes the absolute values and the Inequalities above – Icosahedron Aug 9 '17 at 22:39
• @Icosahedron: I added a section discussing the inequalities. If it isn't enough detail, let me know. – Alex Ortiz Aug 10 '17 at 3:06
You can look at this a different way.
Since $x^n-y^n$ is divisible by $(x-y)$ for all $n\in \mathbb Z^+$, if $p(x)$ is a polynomial, $p(x)-p(a)$ has a factor $(x-a)$ whenever $a$ is an element in the ground field for which $p(a)=0$ [note that $p(x)=p(x)+0=p(x)-p(a)$].
This can be used to show that a polynomial of degree $n$ over a field has at most $n$ roots. So if the field is infinite, a polynomial which evaluates to zero at every point is the zero polynomial.
If the field is finite, the product of all the factors $(x-a)$ for $a$ running through the elements of the field clearly evaluates to zero at every point. The polynomial $p(x)=x^p-x$ over the field with $p$ elements is an example (Fermat's little theorem). But this is not the same as saying that the polynomial is the zero polynomial.
• Is this supposed to be an explanation of the proof under question, or a completely different proof? If the latter, then you should mention that at the start of the answer to help avoid confusion. – Bill Dubuque Aug 9 '17 at 20:35
• @BillDubuque I have edited accordingly - thanks for the comment. It was mainly a result of things said in other comments, which might have been misleading. – Mark Bennet Aug 9 '17 at 20:37 | 2019-06-19T08:40:04 | {
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http://math.stackexchange.com/questions/240742/how-do-we-represent-this-event | How do we represent this event?
My professor did this problem out of our text book and he didn't exactly show us how he did it (he skipped showing the steps in the middle). He got the answer $(1-x)^2$.
If we let $X$ and $Y$ be two independent uniform $(0,1)$ random variables and let $M$ be the minimum of $X$ and $Y$, $0\lt x \lt 1$. How would we represent the event $(M\ge x)$ as the region in the plane and find $P(M\ge x)$ in the area of this region?
How would you go about doing this?
PS I added the homework tag because it could have been a homework problem, however it's not homework. I'm just looking for more people to see my question.
-
The event $M \geq x$ is the intersection of the two events $X \geq x$ and $Y\geq x$. In the plane, this means that the location of $(X,Y)$ must be to the upper-right of $(x,x)$, so $(X,Y)$ must belong to the square $[x,1)\times [x,1)$. This square has area $(1-x)^2$, which is the desired probability.
In other words: \begin{align*} P(M\geq X) &= P(X\geq x\text{ and }Y\geq x)\\ &= P(X\geq x)P(Y\geq x)\text{ since X and Y are independent}\\ &= m([x,1))m([x,1))\text{ since X and Y are uniform (0,1)-random variables}\\ &= (1-x)(1-x)\text{ since the measure of an interval is its length.} \end{align*}
-
so would this mean that $P(M\ge x)=(1-x)^2$? Or is this just to show the area in the plane where $M$ is defined? – TheHopefulActuary Nov 19 '12 at 19:08
Both -- the area in the plane is the probability of the corresponding event, since $X$ and $Y$ are independent and uniform. I've edited to elaborate. – Owen Biesel Nov 19 '12 at 22:54
sorry one more question, my professor is being no help to me at all. I am also asked to find the PDF and CDF of the function. I know that the answer for the CDF is $1-(1-x)^2$ and the PDF is $2(1-x)$ but I don't understand how those were found. Wouldn't $P(M\ge x)=(1-x)^2$ be the PDF? – TheHopefulActuary Nov 20 '12 at 13:27
The CDF is the probability that $M$ is any value up to $x$: $P(M < x) = 1 - P(M\geq x) = 1 - (1-x)^2$. The PDF is the derivative of the CDF: $d(1-(1-x)^2)/dx = 0 - 2(1-x)^1 (-1) = 2(1-x)$. – Owen Biesel Nov 20 '12 at 21:01
The probability that $M\ge x$ equals the probability that both $X$ and $Y$ are $\ge x$. Since each of these is $1-x$ and they are independent, the result follows.
- | 2015-07-05T13:33:09 | {
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https://byjus.com/question-answer/two-distinct-numbers-are-selected-at-random-from-the-first-twelve-natural-numbers-the-probability/ | Question
# Two distinct numbers are selected at random from the first twelve natural numbers. The probability that the sum will be divisible by 3 is
A
13
B
2366
C
12
D
none of these
Solution
## The correct option is B $$\displaystyle \frac {1}{3}$$$$First\quad 12\quad natural\quad numbers\quad are\quad 1,2,3,........,12\\ Number\quad of\quad ways\quad in\quad which\quad two\quad distinct\quad numbers\quad \\ can\quad be\quad selected\quad out\quad of\quad these\quad 12\quad natural\quad numbers= \scriptstyle\binom{ 12 }{ 2 }\\ Let\quad P\left( T \right) = \scriptstyle\binom{12}{2} \\ The\quad maximum\quad sum\quad of\quad two\quad selected\quad distinct\quad numbers=12+11=23\\ The\quad minimum\quad sum\quad of\quad two\quad selected\quad distinct\quad numbers=1+2=3\\ For\quad the\quad sum\quad being\quad divisible\quad by\quad 3\quad it\quad must\quad be\quad one\quad of\quad \\ 3,6,9,12,15,18,21\\ Let\quad S\quad be\quad the\quad event\quad of\quad sum\quad of\quad two\quad selected\quad number.\\ P\left( S=3 \right) =\dfrac { 1 }{ P\left( T \right) } \\ P\left( S=6 \right) =\dfrac { 2 }{ P\left( T \right) } \\ P\left( S=9 \right) =\dfrac { 4 }{ P\left( T \right) } \\ P\left( S=12 \right) =\dfrac { 5 }{ P\left( T \right) } \\ P\left( S=15 \right) =\dfrac { 5 }{ P\left( T \right) } \\ P\left( S=18 \right) =\dfrac { 3 }{ P\left( T \right) } \\ P\left( S=21 \right) =\dfrac { 2 }{ P\left( T \right) } \\ Let\quad D\quad be\quad the\quad event\quad that\quad sum\quad is\quad divisible\quad by\quad 3\\ Then\quad P\left( D \right) =P\left( S=3 \right) +P\left( S=6 \right) +P\left( S=9 \right) +P\left( S=12 \right) +P\left( S=15 \right) +P\left( S=18 \right) +P\left( S=21 \right) \\ \Rightarrow P\left( D \right) =\dfrac { 22 }{ P\left( T \right) } =\dfrac { 22 }{ 66 } \\ \Rightarrow P\left( D \right) =\dfrac { 1 }{ 3 }$$Maths
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http://vvge.yiey.pw/power-series-and-taylor-series.html | # Power Series And Taylor Series
Introduction to Taylor's series & Maclaurin's series › A Taylor series is a representation of a function as an infinite sum of terms that are calculated from the values of the function's derivatives at a single point. " The formula for the Maclaurin series of f (x) is. taylor(sin(2*x),x,%pi/6,6);. A power series [centered at x = 0] is a series of the form X1 n=0 c nx n = c 0 + c 1x+ c 2x2 + c 3x3 + c 4x4 + ::: (Informally, we can think of a power series as an \in nite polynomial. 2 on the facing page we plot sin(x) and the truncations of the series up to degree 5 and 9. This is an advantage in physical applications where one is dealing with very small numbers or a small difference between two functions. Alternatively, observe that , then write down a geometric series expression for and integrate. Bill Taylor, top US diplomat in Ukraine, testified in the House-led impeachment inquiry that there was a quid pro quo for a Bidens investigation. Otherwise, in terms of Taylor series, if the sequence of n- th derivatives. But it converges at both end points and does so, therefore, absolutely. Here we assume the answer is of the form. The Power Rule; 2. The Taylor Series represents f(x) on (a-r,a+r) if and only if. Candlestick Park, hosting the World Series, rocked and shivered. The modern idea of an infinite series expansion of a function was conceived in India by Madhava in the 14th century, who also developed precursors to the modern concepts of the power series, the Taylor series, the Maclaurin series, rational - Their importance in calculus stems from Newton s idea of representing functions as sums of infinite series. Other operations with Taylor series Obtain the terms of degree less than or equal to 4 of the Maclaurin series of these functions: 1 (𝑥) = 𝑥sin𝑥 2 (𝑥) = sin𝑥 Hint: Treat the power series the same way you would treat a polynomial. Section 4-15 : Power Series and Functions. It explains how to derive power series of composite functions. Created by Courtney A. A power series P 1 n=0 a n(x x 0) n is said to converge at a point x if the. 24, find the interval of convergence of the given power series. What Is the Taylor Series of Ln(x)? Taylor Series Application Taylor Series Expansion Lincoln Park Jack Taylor Series Example of Taylor Series Taylor L. Furthermore, let us suppose that f has a taylor series about the point. Terms of the Taylor Series Application:. If you're asked "find the Maclaurin series for f (x)," this means the same thing as "find the Taylor series for f (x) near 0. 1 fx() x, a 1 3. POWER SERIES 251 For example, sine is an analytic function and its Taylor series around x 0 = 0 is given by sin(x) = X1 n=0 (1)n (2n + 1)! x2n+1: In Figure 7. An infinite series or an infinite sum is the sum of numbers which progresses according to a specific rule. 1 We examined series of constants and learned that we can say everything there is to say about geometric and telescoping series. No calculator except unless specifically stated. You can think of a power series as a polynomial with infinitely many terms (Taylor polynomial). Then, for every x in the interval, where R n(x) is the remainder (or error). In essence, the Taylor series provides a means to predict a function value at one point in terms of the function value and its derivatives at another point. The series for f at x=0 should. Trigonometric functions. Find the interval of convergence for ∞ n=0 (x−3)n n. We use the results we obtained in the section on Taylor and Maclaurin series and combine them with a known. If we use the ratio test on an arbitrary power series, we find it converges when. An infinite series or an infinite sum is the sum of numbers which progresses according to a specific rule. • A power seriesis a series that has coefficients and powers of x, instead of just numbers. In some cases, one can also derive the Taylor series by repeatedly applying integration by parts. ") A power series centered at x = a has the form X1. Also notice how the Taylor Series approximations worsens as you deviate further away from where the series is centered. (b) If P a nxn diverges when x= 6, then it diverges when x= 10. Power, Taylor, and Maclaurin Series Survival Guide One of the harder concepts that we have to become comfortable with during this semester is that of sequences and series. Using Taylor series to find the sum of a series. 1) DEFINITION 1. They demonstrate the development of each power series as new terms are gradually added. The Taylor series is the expansion of the function in the endless amount of power functions. The radius of convergence of both series is the same. The number c is called the expansion point. Actually Newton considered this his main mathematical discovery: that any equation (algebraic, differential, etc. Taylor Series & Maclaurin Series help to approximate functions with a series of polynomial functions. Power series are useful in analysis since they arise as Taylor series of infinitely differentiable functions. For example,. 4—Power Series II: Geometric Series Show all work. The objective of this section is to become fa-miliar with the theory and application of power series and Taylor series. taylor(sin(2*x),x,%pi/6,6);. Taylor and Maclaurin Series are explained and defined using power series. Since , the Taylor series for f(x) will also have 1 as its radius of convergence. (b) If P a nxn diverges when x= 6, then it diverges when x= 10. Let's see an example. Become a member and unlock all Study Answers. Suppose that we have a series where. Know what a Power Series is. The Taylor series above for arcsin x, arccos x and arctan x correspond to the corresponding principal values of these functions, respectively. to the power of another thing which is both unreal and irrational (iπ) that if you add it with one (1). Taylor series, in mathematics, expression of a function f—for which the derivatives of all orders exist—at a point a in the domain of f in the form of the power series Σ ∞n = 0 f (n) (a) (z − a)n/n! in which Σ denotes the addition of each element in the series as n ranges from zero (0) to infinity. Examples 1. Reading derivatives from Taylor series. Now this is where Taylor and Maclaurin Series come in. 1 shows that if a given function has a power series representation then it has to be the Taylor series for the function. (b) If P a nxn diverges when x= 6, then it diverges when x= 10. A Taylor series is associated with a given function f. But it converges at both end points and does so, therefore, absolutely. Free Taylor/Maclaurin Series calculator - Find the Taylor/Maclaurin series representation of functions step-by-step. Otherwise, in terms of Taylor series, if the sequence of n- th derivatives. 9 Representation of Functions by Power Series 671 Operations with Power Series The versatility of geometric power series will be shown later in this section, following a discussion of power series operations. 1 Definition If f has derivatives of all orders at x0, then we call the series X∞ k=0 f(k)(x 0) k! (x−x0)k the Taylor series for f about x = x0. Vindman’s testimony also appears to corroborate aspects. The last section (15. represents an infinitely differentiable function and is called Maclaurin series and Taylor series respectively. If it is true, explain why. Keirstead AP Calculus 2. 7: Taylor and Maclaurin Series Taylor and Maclaurin series are power series representations of functions. h header file and power function pow(i, j). 2 Properties of Power Series 10. Spring 03 final with answers. 5: Taylor Series A power series is a series of the form X∞ n=0 a nx n where each a n is a number and x is a variable. And this is because they are composed of coefficients in front of increasing powers of x. If and the collection of are fixed complex numbers, we will get different series by selecting different values for z. Using Taylor polynomials to approximate functions. In other words, you're creating a function with lots of other smaller functions. Here's a complete rundown of every Power Rangers series in the franchise's history, from a superfan and someone who vaguely remembers yelling "Triceratops. (a) If lim n!1 a n = 0 then P a n converges. Otherwise, in terms of Taylor series, if the sequence of n- th derivatives. 2 We developed tests for convergence of series of constants. Whether it converges can depend on the value of x! (Incidentally, 00 = 1 here. 1 Power Series and Holomorphic Functions We will see in this section that power series and holomorphic functions are intimately related. Why do we care what the power series expansion of sin(x) is? If we use. 10 Differentiating and Integrating Power Series (Modeling with Taylor Series) Theorem 9. The partial sum is called the nth-order Taylor polynomial for f centered at a. 1 Lecture: Applications of Taylor series Approximating functions by Taylor polynomials. The number c is called the expansion point. how to use a while loop to perform an infinite Learn more about while, loop, sin(x), infinite series, taylor series, alternating series, error. The general form for the Taylor series (of a function f(x)) about x=a is the following:. Taylor's theorem (actually discovered first by Gregory) states that any function satisfying certain conditions can be expressed as a Taylor series. power series, such as the Taylor series of a basic function. Evaluating Infinite Series It is possible to use Taylor series to find the sums of many different infinite series. Taylor and Maclaurin Series are explained and defined using power series. The Maclaurin series is a template that allows you to express many other functions as power series. For example,. To investigate when these series converge we will. Power Series, Taylor and Maclaurin Polynomials and Series Power Series The Basics De nition 1 (Power Series). 1 Introduction This section focuses on deriving a Maclaurin series for functions of the form f(x) = (1 + x)k for any number k. When we apply the Fundamental. , I might be ( 17;19)) and let x 0 be a point in I, i. Power Series to a Taylor Series. With Animal Farm, it was the sight of a small boy escorting a giant cart horse down a country lane and the thought of what might happen if the animal world rose. The Taylor series method does not work in general, although the heuristic contains a kernel of truth. Lin McMullin added EK 4. It’s Tuesday, Oct. The syntax of the command is "taylor(function, variable, point, degree)". DEFINITION 2. The Taylor (or more general) series of a function about a point up to order may be found using Series[f, x, a, n]. Example 5 Find the Maclaurin series for cos(x). Definition 7. Most functions can be expanded into infinite series form. page 6 of 7. The power series expansion of the hyperbolic sine and hyperbolic cosine function: We use sum and difference of two convergent series. 1! n 1 22 n n ff n (a) Write the third -degree Taylor polynomial for f about x = 5. Things you should memorize: • the formula of the Taylor series of a given function f(x). area, volume, and length problems with answers. A simpler way using Taylor series. MATRIX AND POWER SERIES METHODS Mathematics 306 All You Ever Wanted to Know About Matrix Algebra and Infinite Series But Were Afraid To Ask By John W. Taylor's Theorem Let f be a function with all derivatives in (a-r,a+r). Power series method The power series method can be used to solve ODEs with variable coe cients. TAYLOR and MACLAURIN SERIES (OL]DEHWK :RRG TAYLOR SERIES. There is however a theorem on differentiating and integrating power series, which you are not expected to know, that tells us that a power series can only be differentiated if it has a radius of convergence that is greater than zero. Define power series. Complete Solution Before starting this problem, note that the Taylor series expansion of any function about the point c = 0 is the same as finding its Maclaurin series expansion. A Taylor series is a function’s expansion about a point (in graphical representative). Find the interval of convergence for ∞ n=0 (x−3)n n. This series — known as a “power series” — can be written in closed form as the following: the Taylor Formula. Whether the power series converges at x = x0 ± ρ is tricky to determine. For example,. At least three of George Orwell’s novels can be tracked back to the particular image or thought process that inspired their conception. Section 4-15 : Power Series and Functions. (a) If lim n!1 a n = 0 then P a n converges. MATRIX AND POWER SERIES METHODS Mathematics 306 All You Ever Wanted to Know About Matrix Algebra and Infinite Series But Were Afraid To Ask By John W. The Deciders providing a firsthand account of what House Democrats have said is a blatant abuse of power by the president. Taylor series are also referred to as power series. Things you should memorize: • the formula of the Taylor series of a given function f(x). Evaluating Infinite Series It is possible to use Taylor series to find the sums of many different infinite series. Example 5 Find the Maclaurin series for cos(x). An Easy Way to Remember the Taylor Series Expansion. ) However, if F(x) can be represented by a power series at c, the Taylor series must be the power series that does so. In addition to a scheduling change to help UConn in its move to football independence next season, Ole Miss Thursday announced that it has reached an agreement with Purdue on a future home-and. Suppose we have a power series in the variable x. It’s important to understand the difference between expressing a function as an infinite series and approximating a function by using a finite number of terms of series. Taylor Series The Coefficients of a Power Series Coefficients of a Power Series Suppose that we have function f given by a power series What can we say about the relationship between f and the coefficients a0, a1, a2, a3, a4, a5,. This calculus 2 video tutorial explains how to find the Taylor series and the Maclaurin series of a function using a simple formula. The Taylor series is a power series that approximates the function f near x = a. In other words, you're creating a function with lots of other smaller functions. You can think of a power series as a polynomial with infinitely many terms (Taylor polynomial). Wyrick family find themselves Cialis 20 Mg Paypal to spot fakes per GiB of RAM I hope to hear adjustment and metal ball. The idea is that the resulting truncated expansion. Elementary Functions Sin: Series representations. In general, you prove the validity of the Taylor expansion over a given interval by proving the Taylor reminder tends to zero as n goes to infinity. ) Maple is much better at this than most of us, but a little practice can quickly improve this skill. The archetypical example is provided by the geometric series:. The th term of a Taylor series of a function can be computed in the Wolfram. Taylor series in several variables [ edit ] The Taylor series may also be generalized to functions of more than one variable with [14] [15]. 02 ARENA, LONDON — Josh Taylor defeated Regis Prograis by the slimmest of margins on Saturday, but in doing so becomes a unified lightweight champion, the World Boxing Super Series winner, and a. You can think of a power series as a polynomial with infinitely many terms (Taylor polynomial). A power series defines a function f(x) = P ∞ n=0 a nx n where we. On problems 1-3, find a Taylor series for fx() centered at the given value of a. 7: Taylor and Maclaurin Series Taylor and Maclaurin series are power series representations of functions. In our conventions, arccot x ≡ arctan(1/x) is not continuous at x = 0 and thus does not possess a Taylor series about x = 0. ) n=0 Use power series operations and the Taylor series at x = 0 for to find the Taylor series at x = 0 for the given function. 1 Approximating Functions with Polynomials 10. Every Maclaurin series, including those studied in Lesson 22. Taylor and Maclaurin (Power) Series Calculator. “Yeah, for sure,” coach Craig Berube said. 3 answers 3. Furthermore, let us suppose that f has a taylor series about the point. If a function has a Taylor series that is convergent to the function, it is customary to expect that partial sums with more terms provide a better approximation than those with fewer terms. Di erentiation and Integration of Power Series We have previously learned how to compute power series representations of certain functions, by relating them to geometric series. In this section you will study a general procedure for deriving the power series for a function that has. Then, some of the most famous Maclaurin series are found. net dictionary. You can specify the order of the Taylor polynomial. Differentiation and Integration. Use the ratio test to show that the Taylor series centered at 0 for sin(x) converges for all real numbers. In many situations c (the center of the series) is equal to zero, for instance when considering a Maclaurin series. Abel’s theorem: di erentiability of power series 4. In another video, I will find a Taylor series expansion, so look for that one too! For more free. ) n=0 Use power series operations and the Taylor series at x = 0 for to find the Taylor series at x = 0 for the given function. The Taylor and Maclaurin series have many uses in the mathematical field including the sciences. 57 series problems with answers. TAYLOR AND MACLAURIN SERIES 3 Note that cos(x) is an even function in the sense that cos( x) = cos(x) and this is re ected in its power series expansion that involves only even powers of x. Formulas for Geometric Series, Taylor and MacLaurin Series Power series representation: (1) a x ∑ n 0 g x n a x 1 1 −g x for x such that |g x | 1 (2) Let f x ∑ n 0 a nxn. Power series tables. Taylor's Formula with Remainder Let f(x) be a function such that f(n+1)(x) exists for all x on an open interval containing a. Taylor Series The Taylor Series is a power series representation of a function at a point 𝑎𝑎, defined by: 𝑓𝑓(𝑥𝑥) = 𝑓𝑓(𝑛𝑛)(𝑎𝑎). TAYLOR SERIES, POWER SERIES The following represents an (incomplete) collection of things that we covered on the sub-ject of Taylor series and power series. Home Contents Index. This of course is just a power series shifted over by c units. (c) If P a. Concrete examples in the physical science division and various engineering fields are used to paint the applications. Binomial series Hyperbolic functions. ) can be solved by substituting a power series with undetermined. Elementary Functions Sin: Series representations. In our conventions, arccot x ≡ arctan(1/x) is not continuous at x = 0 and thus does not possess a Taylor series about x = 0. Now that we have at our disposal all the tests and tools of discrete and continuous calculus, we can finally close the loop and make sense of what we've been doing when we Talyor-expand. The Taylor series about x = 5 for a certain function f converges to fx for all x in the interval of convergence. Taylor Series Approximation to Cosine. 2 - Maclaurin Series 22. 1 Asymptotic power series Asymptotic power series, f(x) ∼ X∞ n=0 a nx n as x → 0, are among the most common and useful asymptotic expansions. Find the Taylor series expansion of any function around a point using this online calculator. Suppose that a function f has a power series representation of the form: ∞ f(x) = a 0 + a 1(x − c) + a. Give the first four nonzero terms and the general term. Eliza Taylor, Actress: The 100. Using Taylor series to find the sum of a series. As the names suggest, the power series is a special type of series and it is extensively used in Numerical Analysis and related mathematical modelling. Then, some of the most famous Maclaurin series are found. An Easy Way to Remember the Taylor Series Expansion. Using Taylor polynomials to approximate functions. The Deciders providing a firsthand account of what House Democrats have said is a blatant abuse of power by the president. So now we’re ready to take our power function skills to the next level! The Taylor Series, sometimes referred to as the Taylor Polynomial, allows us to find a polynomial function that approximates another function by first choosing a number (center) both the function and the polynomial share. In this article, we’re going to explore a number of ways to calculate the sine series without actually using the sine (or cosine) function. So you can see it's the sum from n equals 0 to infinity of these terms. 1) and its associated formula, the Taylor series, is of great value in the study of numerical methods. The Taylor series method does not work in general, although the heuristic contains a kernel of truth. The power series converges absolutely. In other words, you're creating a function with lots of other smaller functions. By the end of this section students will be fa-miliar with: • convergence and divergence of power and Taylor series; • their importance; • their uses and applications. Law of Sines. Limits like are "easy" to compute, since they can be rewritten as follows. Our goal in this section is find the radius of convergence of these power series by using the ratio test. How to extract derivative values from Taylor series Since the Taylor series of f based at x = b is X∞ n=0 f(n)(b) n! (x−b)n, we may think of the Taylor series as an encoding of all of the derivatives of f at x = b: that information. We cover all the topics in Calculus. Taylor's Theorem; A bit of thought reveals that the ratio test applied to a power series. 1) and its associated formula, the Taylor series, is of great value in the study of numerical methods. A power series defines a function f(x) = P ∞ n=0 a nx n where we. Actually Newton considered this his main mathematical discovery: that any equation (algebraic, differential, etc. Taylor series. A much simpler way of solving this equation (and power series solution in general) using the Taylor series form of the expansion. Taylor and Maclaurin (Power) Series Calculator. Other operations with Taylor series Obtain the terms of degree less than or equal to 4 of the Maclaurin series of these functions: 1 (𝑥) = 𝑥sin𝑥 2 (𝑥) = sin𝑥 Hint: Treat the power series the same way you would treat a polynomial. Even though this family of series has a surprisingly simple behavior, it can be used to approximate very elaborate functions. This calculus 2 video tutorial explains how to find the Taylor series and the Maclaurin series of a function using a simple formula. You can specify the order of the Taylor polynomial. n are zero, rewrite the series and pick one in which the terms are nonzero. Uniqueness of the Taylor series. We are working with infinite sums of complicated functions and are answering questions about convergence. This program makes use of C concepts like For loop. “Any time. Of course, the statement "if f has a power series representation" is an important one. Binomial series Hyperbolic functions. One can attempt to use the Taylor series as-is and generalize the form of the coefficients, or one can use manipulations such as substitution, multiplication or division, addition or subtraction of standard Taylor series (such as those above) to construct the Taylor series of a function, by virtue of Taylor series being power series. is that Trump used his power to get Ukraine to. If a function f has derivatives of all orders at a, then the Taylor series for f about x = a is:. We can then find the expression exp(M) if given that. Taylor and Maclaurin Series Use series to approximate the definite integral to within the indicated accuracy. These techniques include substitution and multiplication and division of series by simple polynomials. 7 Taylor and Maclaurin series Example 1 Find the Maclaurin series of the function f(x)= and its radius of convergence. You can specify the order of the Taylor polynomial. Find the Taylor series expansion for e x when x is zero, and determine its radius of convergence. The idea is that the resulting truncated expansion. Linearity of the Derivative Taylor Series; 11. Related to power. The function f is differentiable on the interval b. For example, the 0 th, 1 st, 2 nd, and 3 rd partial sums of the Taylor series are given by. This is an advantage in physical applications where one is dealing with very small numbers or a small difference between two functions. 9 Representation of Functions by Power Series 671 Operations with Power Series The versatility of geometric power series will be shown later in this section, following a discussion of power series operations. In fact, Borel's theorem implies that every power series is the Taylor series of some smooth function. A power series [centered at x = 0] is a series of the form X1 n=0 c nx n = c 0 + c 1x+ c 2x2 + c 3x3 + c 4x4 + ::: (Informally, we can think of a power series as an \in nite polynomial. Josh Taylor vs Regis Prograis RESULT: Taylor wins World Boxing Super Series to become unified champion. 812) that the series of Example 11. In another video, I will find a Taylor series expansion, so look for that one too! For more free. , x 0 2I : Next consider a function, whose domain is I,. Maple rocks as long as you have a newer version. Consider the following power series L(x), which is also known as Euler’s dilogarithm function: L(x) = X1 k=1 xk k2:. Determine whether the following statement is true or false. You can specify the order of the Taylor polynomial. The Taylor Series represents f(x) on (a-r,a+r) if and only if. A Taylor series is associated with a given function f. MacLaurin and Taylor Series, Power Series MacLaurin and Taylor Series In a previous. Abstraction. This of course is just a power series shifted over by c units. For every x there exists c with. Spring 03 midterm with answers. Created by Courtney A. If you want the Maclaurin polynomial, just set the point to 0. Calculus: We give a method for associating a power series to a function when the geometric power series does not apply. Bay Briefing is written by Taylor Kate Brown and sent to readers’ email in-boxes on weekday mornings. 1 Lecture: Applications of Taylor series Approximating functions by Taylor polynomials. , I might be ( 17;19)) and let x 0 be a point in I, i. Some power series come to us in a slightly different form, that of a shifted power series, something of the form sum over n of a sub n times quantity x- c to the nth. The number c is called the expansion point. Home Contents Index. (b) If P a nxn diverges when x= 6, then it diverges when x= 10. 10 Differentiating and Integrating Power Series (Modeling with Taylor Series) Theorem 9. Applications of Natural Log Series. Taylor-made Pi: Figure 1. Motivation 7. Many functions can be written as a power series. Taylor and Maclaurin (Power) Series Calculator. Problem 4 (15 points) Find the radius of convergence of the power series. 3 We considered power series, derived formulas and other tricks for nding them, and know them for a few functions. Alternatively, observe that , then write down a geometric series expression for and integrate. The Taylor series above for arcsin x, arccos x and arctan x correspond to the corresponding principal values of these functions, respectively. Eliza Jane Taylor (born on 24 October 1989 in Melbourne, Australia) is an Australian actress who is perhaps best known for her regular role as "Janae Timmins" on the Australian television series Neighbours (1985). Partial sums of a Maclaurin series provide polynomial approximations for the function. In this chapter, we will use local information near a point x = b to find a simpler function g(x), and answer the questions using g instead of f. Most functions can be expanded into infinite series form. Suppose that a function f has a power series representation of the form: ∞ f(x) = a 0 + a 1(x − c) + a. Motivation 7. is that Trump used his power to get Ukraine to. Lee Department of Mathematics Oregon State University January 2006. Abel’s theorem: di erentiability of power series 4. For example,. 4 Power Series Functions. Taylor's Theorem Let f be a function with all derivatives in (a-r,a+r). The Taylor Series represents f(x) on (a-r,a+r) if and only if. DeTurck Math 104 002 2018A: Series 2/42. These notes discuss three important applications of Taylor series: 1. A power series defines a function f(x) = P ∞ n=0 a nx n where we. Taylor and Maclaurin series are defined, and we show how to extract coefficients from (1+x)^50 using Taylor coefficients. Keirstead AP Calculus 2. Taylor Polynomials. The Taylor series is the expansion of the function in the endless amount of power functions. The program also uses C's math. Spring 03 midterm with answers. Such a polynomial is called the Maclaurin Series. So you can see it's the sum from n equals 0 to infinity of these terms. 1 Power Series and Holomorphic Functions We will see in this section that power series and holomorphic functions are intimately related. What is Power series? A power series is a series of the form. We begin with the general power series solution method. Section 4-15 : Power Series and Functions. 2 (Differentiation of Power Series) Suppose that a function f is represented by a power series in that has a nonzero Radius of Convergence R, that is then a. (Several of these are listed below. This gives us a simple formulaB for the sum:" B B B â œ " " B # \$ This is our first example of a Taylor series —a power series that adds up to a known function. In this article, we’re going to explore a number of ways to calculate the sine series without actually using the sine (or cosine) function. | 2019-12-12T15:38:34 | {
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https://www.physicsforums.com/threads/permutations-in-a-circle.862929/ | I Permutations in a circle
1. Mar 20, 2016
Happiness
2 boys and 3 girls are to be seated round a table with 5 seats. Each child occupies exactly one seat. In how many ways can this be done if
(a) the 2 boys must be seated together
(b) same as (a) but this time the seats are numbered
Solution
(a) $\frac{4!}{4}2!$
(b) $\frac{4!}{4}2!\times 5$
The "$\times 5$" in (b) seems to contradict the "$\div 4$" in (a).
(a) $4!2!\div 5$, (b) $4!2!$ or
(a) $\frac{4!}{4}2!$, (b) $4!2!$
Explanation for the "correct" solution
(a) After grouping the 2 boys into one group, we have 4 groups (with one girl in a group). The number of ways of arranging 4 items in a circle is $\frac{4!}{4}$ since ABCD, DABC, CDAB and BCDA are the same arrangement when placed in a circle. The 2 boys can be permutated within the "boy group". So we have $\frac{4!}{4}2!$.
(b) If the seats are numbered, rotating an arrangement by one seat results in a new arrangement. We can do 5 such rotations. So we have $\frac{4!}{4}2!\times 5$.
But if we $\div 4$ in (a) since we are rotating the 4 groups ABCD 4 times and considering these 4 arrangements to be the same, then in (b) shouldn't we $\times 4$ instead of $5$?
On the other hand, if we first consider the seats to be numbered and hence ABCD, DABC, CDAB and BCDA are NOT the same arrangement, we have 4!2!. Then for (a), since the seats are not numbered and there are 5 rotations that produce the same arrangement when there are 5 seats, we have (a) $4!2!\div 5$.
My issue is why does the correct answer consider the rotations of groups in (a) but not in (b)? In (b), it considers instead the rotations of seats (or persons).
2. Mar 20, 2016
mathman
The grouping description seems to unduly complicate the problem. 2 boys have 2! possible permutations, 3 girls have 3! possible permutations, so the seating when the seats are not numbered is 2!3! - the 2 boys together and the 3 girls together. Seats being numbered multiplies by 5, since there are 5 possible positions for girl in the middle.
3. Mar 21, 2016
Happiness
But this method can't be generalised while the grouping method can be.
2 boys, 3 girls and 4 men are to be seated round a table with 9 seats. In how many ways can this be done if
(a) the 2 boys must be seated together and the 4 men must be seated together
(b) same as (a) but this time the seats are numbered
Solution
(a) Grouping the boys into one group and the men into another, we have 5 groups. So the answer is $\frac{5!}{5}2!4!$.
(b) Considering the 9 rotations of seats, we have $\frac{5!}{5}2!4!\times 9$.
We have the same issue: we consider rotations of groups when we $\div 5$ in (a) but we consider rotations of seats when we $\times 9$ in (b). | 2018-07-20T07:55:34 | {
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https://web2.0calc.com/questions/in-how-many-ways-can-8-people-be-seated-around-a-square-table-with-2-people-on-a-side-two-configurations-are-considered-equivalent-if-one | +0
# In how many ways can 8 people be seated around a square table with 2 people on a side? (Two configurations are considered equivalent if one
0
1576
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In how many ways can 8 people be seated around a square table with 2 people on a side? (Two configurations are considered equivalent if one is a rotation of another.)
Guest Apr 1, 2015
#20
+889
+10
3*5*7*...*(2k-1)
= 2*3*4*5*6*7*...*(2k-2)*(2k-1) / (2*4*6*...*(2k-2))
= (2k-1)! / (2^(k-1)*1*2*3*...*(k-1))
= (2k-1)! / (2^(k-1)*(k-1)!)
Bertie Apr 5, 2015
Sort:
#1
+81142
+5
One person can be 'anchored" in one of two seats on one side.......the other seven people can be arranged in 7! ways.....so....
2 x 7! =
2 x 5040 = 10,080 ways
CPhill Apr 1, 2015
#2
+91510
0
This post is all nonsense -
I am just leaving it here so that Alan's and Bertie's comments make sense.
I think you misread the question Chris,
And I think it might be (table rotations are considered the same)
8C2*6C2*4C2*2C2*(4-1)!
$${\left({\frac{{\mathtt{8}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{8}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{6}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{4}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{4}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)}{\mathtt{\,\times\,}}{\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{3}}{!} = {\mathtt{15\,120}}$$
This seems much to big
Chris's answer - where no one was pair up - was $${\mathtt{7}}{!} = {\mathtt{5\,040}}$$
And i thought my answer should be smaller than this.
Melody Apr 1, 2015
#3
+81142
+5
Here's the justification for my answer....
Consider a "birdseye" view of the table....
One person either occupies the left seat on one side of the table or the right seat on that side.
And for each of these arrangements, the other 7 seats can be filled in all possible ways = 7! ways.
So, 2 x 7! = 10,080 ways
"Rotations" of the table don't matter......one rotation would look like any other for a particular arrangement.
CPhill Apr 1, 2015
#4
+91510
+5
I like the working of my answer but I don't understand why the number would be so much bigger (or bigger at all) than if everyone could just sit where they wanted.
Both your answer and mine have a bigger value than if everyone was individuals and to me this seems a little crazy.
There has to be doubling up happening. !!
Melody Apr 1, 2015
#5
+889
+10
Hi Chris and Melody, hope you don't mind me joining the conversation.
I'm with Chris on this one, I think that the 10080 is correct, (I've gone through the calculation in a different way and arrived at the same answer).
So, here's a question for Melody. Forget about rotations and seat positions for the moment, suppose that I have eight people and that I'm going to split them into four groups of two. In how many ways can this be done ?
Let's see If we can get agreement on that to begin with.
Bertie Apr 2, 2015
#6
+91510
0
Hi Bertie, I was hoping that you would join in. thank you.
So, here's a question for Melody. Forget about rotations and seat positions for the moment, suppose that I have eight people and that I'm going to split them into four groups of two. In how many ways can this be done ?
Well I thought that it was
8C2*6C2*4C2*2C2 but this is obviously not correct - I guess i am double counting again :/
Melody Apr 2, 2015
#7
+26412
+5
It might help to think of a simpler situation first. Consider four people sitting two each side of a two-sided table(!).
The distinct possibilities, excluding rotations, are easily visualised:
.
Alan Apr 2, 2015
#8
+91510
0
Okay thanks Alan, I think I have digested that. And I can see why Chris's answer makes sense.
BUT
I still would like to understand why my logic is wrong :)
I have always thought with probability that it is often fairly easy to understand the correct answer but much harder to understand why an incorrect answer is wrong.
Still, with all the prob questions we have been getting I think my ability is improving :)
Melody Apr 2, 2015
#9
+889
+5
Worse than double counting, multiple counting.
To illustrate why this method is wrong, consider an even simpler situation. Suppose that you have just four people and you want to split them into pairs. How many ways ?
If you employ the method used for the eight person problem, I guess that you would say that the number would be 4C2*2C2 = 6.
However, if you call the people A,B,C,D, you can see that there are only three possible pairings (AB and CD), (AC and BD) and (AD and BC). The basic reason why the first answer is wrong is that each grouping has been counted twice. (AB and CD) and (CD and AB), for example, are being thought of as being two different pairings, meaning that that particular pairing has been counted twice, if that makes sense. The 4C2 = 6 contains AB, AC, AD, BC, BD and CD, but both AB and CD produce the same split into two pairs.
Try the eight into four groups of two again.
Bertie Apr 2, 2015
#11
+91510
+5
Okay I am just starting to get somewhere.
I am looking at how many ways even numbers of people can be paired off
This post is NOT directly related to the question above.
4 people ABC and D
AB CD
AC BD
that is 4 people can be paired off in 3 ways
Now, before I wanted to do it as 4C2 which is 6
This would give the 6 above but each time you get one pair you automatically get the another pair so it can be seen that this will be too big by a factor of 2 so it is 4C2/2 = 6/2=3
---------------------------------------------------------
6 People
Now I am going to look at 6 people A B C D E and F
AB there are four more so they can be paired up 3 ways
AC 3 ways
AE 3 ways
AF 3 ways
Total = 3*5 =15
So 6 people can be paired off 15 ways. This is (6-1)(4C2)/2 = 15
----------------------------------------------------
8 People
Now I am going to look at 8 people A B C D E F G and H
AB then 6 more = 15 ways
AC then 6 more = 15 ways
AD then 6 more = 15 ways
AE then 6 more = 15 ways
AF then 6 more = 15 ways
AG then 6 more = 15 ways
AH then 6 more = 15 ways
That is a total of 7*15=105 (8-1)(6-1)4C2/2 = 7*5*3
that is 8 people can be paired off in 105 ways
-------------------------------------------------
So 2K people can be paired off in 3*5*7*......*(2K-1) ways
------------------------------------------------------
So 4 people are paired (and the pairs are then considered as 1)
they are then put clockwise into a cirlce (or anticlockwise) rotations are considered the same.
The number of ways this could be done is 4*1=4 ways
6 people - it could be done 5*3*2! = 30 ways
8 people - it could be done 7*5*3*3! = 1050 ways
2K people - it could be done (2k-1)(2k-3)....5*3*(k-1)!
-----------------------------------------------------------
Melody Apr 4, 2015
#12
+91510
+5
Look Bertie, Alan and Chris, I did it my way and I got the same answer as Bertie and Chris! WOW!!
I know my way is no where near as sensible as Chris's way but I learned heaps doing it this way
4 people
There are 3 ways to pair up 4 people, if the order of the pairs counts then this would be 3*2*2 = 12 ways
For instance AB CD could also be AB DC, BA CD, or BA CD
They can then be put in a circle 12*(2-1) = 12 ways
So 4 people can be paired off and THEN placed in a circle 12 ways
----------------------------
6 people
There are 3*5 ways to pair up 6 people, if the order of the pairs counts then this would be 15*2^3 = 120 ways
Now they can be placed in a circle 120*2! = 240
So 6 people can be paired off and THEN placed in a circle 240 ways
--------------------------------
8 people
There are 3*5*7=105 ways to pair up 8 people, if the order of the pairs counts then this would be 105*2^4 = 1680 ways
Now they can be placed in a circle 1680*3! = 240
So 8 people can be paired off and THEN placed in a circle 10080 ways @@ BINGO @@
Melody Apr 4, 2015
#13
+91510
+5
Now I have a new question.
2K people can be paired off in 3*5*7*......*(2K-1) ways
How can I write this as a formula?
Melody Apr 4, 2015
#14
+26412
+10
Possibly as:
$$\prod_{n=2}^K(2n-1)$$
or as:
$$\frac{2^{K+1}\Gamma(K+\frac{1}{2})}{2\sqrt{\pi}}$$
where Γ() is the Gamma function
$$\Gamma (t)=\int_0^{\infty}x^{t-1}e^{-x}dx$$
.
Alan Apr 4, 2015
#15
+91510
+5
Thanks,
I don't know that symbol Alan, I thought that there must be one to do that job.
What is the symbol called? :/ [I am talking about the first formula]
Gamma function ? mmm.....
Melody Apr 4, 2015
#16
+26412
+10
Π is the Greek upper case letter, pi, and stands for the product (analogous to Σ for the sum).
Alan Apr 4, 2015
#17
+91510
+5
Ok thanks Alan.
That gamma function. I know I have seen it before but could you refresh my memory please?
Melody Apr 4, 2015
#18
+26412
+10
The gamma function is a sort of continuous version of the factorial function.
When n is an integer then Γ(n) = (n-1)!
But Γ(n) is also defined for non-integer values of n.
.
Alan Apr 4, 2015
#19
+91510
0
Another answer for me to chew on - thanks Alan :)
Melody Apr 4, 2015
#20
+889
+10
3*5*7*...*(2k-1)
= 2*3*4*5*6*7*...*(2k-2)*(2k-1) / (2*4*6*...*(2k-2))
= (2k-1)! / (2^(k-1)*1*2*3*...*(k-1))
= (2k-1)! / (2^(k-1)*(k-1)!)
Bertie Apr 5, 2015
#21
+91510
+5
Thanks Bertie,
You know I was really proud of myself getting that far on my own.
I chewed on it for ages!
I have repeated your work here because I was having problems with it but by the time I had written it all out in LaTex I finally understood, but I decided just to leave the LaTex here.
-------------------------------------
So the number of ways that 2k people can be paired off is:
$$3*5*7*9*.....*(2k-1)\\\\ =3*5*7*9*.....*(2k-3)*(2k-1)\\\\ =\frac{1*2*3*4*.....*(2k-3)(2k-2)(2k-1)}{2*4*6*8*...(2k-2)}\\\\ =\frac{1*2*3*4*.....*(2k-3)(2k-2)(2k-1)}{2^{k-1}[1*2*3*4*...(k-1)]}\\\\ =\frac{(2k-1)!}{2^{k-1}(k-1)!}$$
THIS IS GREAT!
Can this be extended to seperating into other groups Like 3N separated into groups of 3?
A project for the future perhaps.
Melody Apr 5, 2015
#22
+91510
+5
Now Alan, I want to look at yours :)
This first one is very obvious, thank you
$$\prod_{n=2}^K(2n-1)$$
or as:
Now this I don't get.
$$\frac{2^{K+1}\Gamma(K+\frac{1}{2})}{2\sqrt{\pi}} =\frac{2^k(K-\frac{1}{2})!}{\sqrt{\pi}}\;\;?$$
Where did $$\sqrt{\pi}$$ appear from?
where Γ() is the Gamma function
$$\Gamma (t)=\int_0^{\infty}x^{t-1}e^{-x}dx$$
It doesn't really matter, I think this is all be above me.
Melody Apr 5, 2015
#22
0
Hi All,
I was intrigued by the answer Melody gave and would like to comment on it. See the approach of diving people into groups of same sizes is a pretty standard one. For example if you want ot divide a group of 8 people into pairs (i.e. all equal sizes) the standard approach shall be 8!/(((2!)^4)*4!), i.e divide 8! by 2!*2!*2!*2! and multiply the whole with 1/4! (since there are 4 groups of equal sizes) and it returns the same answer as 105. This can be used for anything. (3N as well). Say N people need to be distributed in 2 groups of 2 and 1 group of 3, then it is N!/2!*2!*3!*2! (i.e. 2! -> since 2 items belon to one group, 2! -> since 2 items belon to one group,3! -> since 2 items belon to one group, 2! -> since 2 groups are identical in size.)
Now that concept out of the way, the thing that intrigued me is its application in geometrical figures, like the one discussed. I think it has happened because in such a scenario we fix one group namely AB and then form 3 paired groups, and since it is a square all the 8 arrangements will be identical, hence giving us the formula (2k-1)!/2^k-1*3!
i.e. (2k)!/(2k*((2!)^3)*3!)
I think that is the case. But definitely for sure such a formula wouldnt work out in other geometrical figures or if the numbers werent 8 for example, i.e we have 6 persons and there are 8 places.
Guest Nov 25, 2015
#23
+5
After all these discuession, i just wanna post the correct answer just to make it clear.
There are 8! ways to place the people around the table, but this counts each valid arrangement 4 times (if you move each person 2, 4, or 6 places clockwise you get the same arrangement). The answer is 8!/4 = 10080.
Guest Dec 28, 2015
#24
+91510
0
In how many ways can 8 people be seated around a square table with 2 people on a side? (Two configurations are considered equivalent if one is a rotation of another.)
Since this question has been dredged back up I decided to pay with it again (just to check that i can still do it ).......
It reall depends on what the questioner means.
If you are treating the square table like a round table then there are (8-1)! = 5040
BUT If you rotate just one place then the pairs are different so this number is doubled
5040*2= 10080
Our guest has just given you exactly the same solution, thanks for making it clear guest.
Our guests logic was a little different from mine but it is totalyy correct (both logic and answer)
------------------------------
If you only care about who is sitting in pairs and where the pairs are in relation to the otheraround the table (who is one the right or left does not matter) then the problem becomes different.
The answer would be who many pairs can be formed x how many ways can they be seated.
How many ways can the 4 pairs be seated = 3! =6
How many ways can they be fromed.
If there were only 2 people there would be 1 way.
If there were 4 people ABCD then AB, AC,AD who A is with defines BOTH pairs so ther is 3 ways.
If people E and G join the mix then if they are together then there are 3 ways.
Bur E could be with G or A or B or C or D So that is 5 x3 ways.
By extention, if there are 8 people there are 7*5*3 = 105 ways they can be paired off.
I remember someone told me the notation is 7!!
So the number of ways pairs can be chosen and the pairs seated at the table (If AB are a pair that is the same as BA) is 105*6 = 630 ways.
If where in a pair, personApersonB is different from personBpersonA the number would be much bigger. I think then the number would be back to 10,080 again as I think it would be the same as the original problem ://
Melody Dec 28, 2015
edited by Melody Dec 28, 2015
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https://mag.knxshop.se/what-is-xbkz/a91fa9-reflexive-closure-example | … fullscreen . Let R be an n-ary relation on A. Let R be a relation on the set {a,b, c, d} R = {(a, b), (a, c), (b, a), (d, b)} Find: 1) The reflexive closure of R 2) The symmetric closure of R 3) The transitive closure of R Express each answer as a matrix, directed graph, or using the roster method (as above). 2.3. Find the reflexive, symmetric, and transitive closure of R. Solution – For the given set, . Indeed, suppose uR M J v. One graph is given, we have to find a vertex v which is reachable from another vertex u, for all vertex pairs (u, v). Let R be an endorelation on X and n be the number of elements in X.. Use your definitions to compute the reflexive and symmetric closures of examples in the text. closure is obtained by changing all zeroes to ones on the main diagonal of M. That is, form the Boolean sum M ∨I, where I is the identity matrix of the appropriate dimension. Convince yourself that the reflexive closure of the relation $$<$$ on the set of positive integers $$\mathbb{P}$$ is $$\leq\text{. • Put 1’s on the diagonal of the connection matrix of R. Symmetric Closure Definition: Let R be a relation on A. Transitive Closure it the reachability matrix to reach from vertex u to vertex v of a graph. Define reflexive closure and symmetric closure by imitating the definition of transitive closure. From MathWorld--A Wolfram Web Resource. check_circle Expert Answer. Reflexive Closure. The reflexive reduction, or irreflexive kernel, of a binary relation ~ on a set X is the smallest relation ≆ such that ≆ shares the same reflexive closure as ~. Day 25 - Set Theoretic Relations and Functions. How can we produce a reflective relation containing R that is as small as possible? Transitive closure • In general, given R over A; if there is a relation S with property P containing R such that S is a subset of ever relation with property P containing R, then S is called the closure of R with respect to P. • We’ll discuss reflexive, symmetric, and transitive closures… We first consider making a relation reflexive. The reach-ability matrix is called the transitive closure of a graph. In general, the closure of a relation is the smallest extension of the relation that has a certain specific property such as the reflexivity, symmetry or transitivity. • N-ary Relations – A relation defined on several sets. d. Is (−35) L 1? Example – Let be a relation on set with . Finally, the concepts of reflexive, symmetric and transitive closure are presented and show that construction of transitive closure in soft set satisfies Warshall’s Algorithm. The diagonal relation on A can be defined as Δ = {(a, a) | a A}. 3 Reflexive Closure • The diagonal relation’s matrix has all entries of its main diagonal = 1. For the symmetric closure we need the inverse of , which is. c. Is 143 L 143? A relation R is non-reflexive iff it is neither reflexive nor irreflexive. For example, the transitive property is a property of binary relations on A; it consists of all transitive binary relations on A. Reflexive and symmetric properties are sets of reflexive and symmetric binary relations on A correspondingly. It can be seen in a way as the opposite of the reflexive closure. The transitive closure of is . Symmetric Closure. So the reflexive closure of is . The reflexive closure of a binary relation on a set is the union of the binary relation and the identity relation on the set. Is (−17) L (−14)? We would say that is the reflexive closure of . Reflexive Closure. This would make non-reflexive, but it's very similar to the reflexive version where you do consider people to be their own siblings. Is 57 L 53? For example, if X is a set of distinct numbers and x R y means "x is less than y", then the reflexive closure of R is the relation "x is less than or equal to y". Ideally, we'd like to add as few new elements as possible to preserve the "meaning" of the original relation. What are the transitive reflexive closures of these examples? Sometimes a relation does not have some property that we would like it to have: for example, reflexivity, symmetry, or transitivity. It is the smallest reflexive binary relation that contains. For example, loves is a non-reflexive relation: there is no logical reason to infer that somebody loves herself or does not love herself. We already have a way to express all of the pairs in that form: \(R^{-1}$$. Reflexive closure: The reflexive closure of a binary relation R on a set X is the smallest reflexive relation on X that contains R. For example, if X is a set of distinct numbers and x R y means "x is less than y", then the reflexive closure of R is the relation "x is less than or equal to y". • The reflexive closure of any relation on a set A is R U Δ, where Δ is the diagonal relation. The relation R = f(1;3);(2;2);(3;4)gon the set f1;2;3;4gis not re exive. The reflexive closure S of a binary relation R on a set X can be formally defined as: S = R ∪ {(x, x) : x ∈ X} where {(x, x) : x ∈ X} is the identity relation on X. Inchmeal | This page contains solutions for How to Prove it, htpi types of relations in discrete mathematics symmetric reflexive transitive relations By Remark 2.16, R M I is the reflexive and transitive closure of ∪ i∈M R i I. For example, the reflexive closure of (<) is (≤). Solution. A relation R is an equivalence iff R is transitive, symmetric and reflexive. How do we add elements to our relation to guarantee the property? Don't express your answer in terms of set operations. Computes transitive and reflexive reduction of an endorelation. The transitive closure of R is the smallest transitive relation on X that contains R. The code implements Warshall's Algorithm which is of complexity O(n^3). the transitive closure of a relation is formed, the result is not necessarily an. 6 Reflexive Closure – cont. The final matrix is the Boolean type. The reflexive closure of a binary relation on a set is the minimal reflexive relation on that contains . The transitive reduction of R is the smallest relation R' on X so that the transitive closure of R' is the same than the transitive closure of R.. Journal of the ACM, 9/1, 11–12. The ancestor-descendant relation is an example of the closure of a relation, in particular the transitive closure of the parent-child relation. equivalence relation the transitive closure of a relation is formed, the result is not necessarily an. Theorem 2.3.1. Equivalence. Title: Microsoft PowerPoint - ch08-2.ppt [Compatibility Mode] Author: CLin Created Date: 10/17/2010 7:03:49 PM Thus for every element of and for distinct elements and , provided that . Reflexive closure is a superset of the original relation so that it is reflexive (i.e. Give an example to show that when the symmetric closure of the reflexive closure of. contains elements of the form (x, x)) as well as contains all elements of the original relation. Although the operation of taking the reflexive and transitive closure is not first-order definable, we can still deduce that R M J is the reflexive and transitive closure of ∪ i∈M R i J. Details. The reflexive closure of R is computed by setting the diagonal of the incidence matrix to 1. For example, consider below graph Transitive closure of above graphs is 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 1 Reflexive Symmetric & Transitive Relation Example Watch More Videos at In this video we are going to know about Transitive Relation with condition and some examples #TransitiveRelation. The symmetric closure of is-For the transitive closure, we need to find . S. Warshall (1962), A theorem on Boolean matrices. Suppose, for example, that $$R$$ is not reflexive. 5 Reflexive Closure Example: Consider the relation R = {(1,1), (1,2), (2,1), (3,2)} on set {1,2,3} Is it reflexive? When a relation R on a set A is not reflexive: How to minimally augment R (adding the minimum number of ordered pairs) to make it a reflexive relation? b. CITE THIS AS: Weisstein, Eric W. "Reflexive Closure." • In such a relation, for each element a A, the set of all elements related. Symmetric Closure. then Rp is the P-closure of R. Example 1. • Add loops to all vertices on the digraph representation of R . References. Theorem: The symmetric closure of a relation $$R$$ is $$R\cup R^{-1}$$. we need to find until . What is the re exive closure of R? pendency a → b to decompose a relation schema r(a,b,g) into r 1(a,b) and r 2(a,g). This preview shows page 226 - 246 out of 281 pages.. Warshall’s Algorithm for Computing Transitive Closures Let R be a relation on a set of n elements. It's also fairly obvious how to make a relation symmetric: if $$(a,b)$$ is in $$R$$, we have to make sure $$(b,a)$$ is there as well. For example, $$\le$$ is its own reflexive closure. equivalence relation SEE ALSO: Reflexive, Reflexive Reduction, Relation, Transitive Closure. The reflexive closure of R , denoted r( R ), is R ∪ ∆ . If so, we could add ordered pairs to this relation to make it reflexive. Download the homework: Day25_relations.tex We've defined relations like $\le$ in Coq... what are they like in mathematics? Here reachable mean that there is a path from vertex i to j. • [Example 8.1.1, p. 442]: Define a relation L from R (real numbers) to R as follows: For all real numbers x and y, x L y ⇔ x < y. a. The smallest reflexive relation $$R^{+}$$ that includes $$R$$ is called the reflexive closure of $$R.$$ In general, if a relation $$R^{+}$$ with property $$\mathbf{P}$$ contains $$R$$ such that Guarantee the property all of the binary relation and the identity relation a... Relation ’ s matrix has all entries of its main diagonal = 1 containing R that is as as. Reach from vertex U to vertex v of a binary relation and the identity relation on that contains,. Union of the original relation can be seen in a way to express of. If so, we 'd like to add as few new elements as possible to preserve the meaning of. Express all of the reflexive and transitive closure of ( < ) is its own reflexive closure. diagonal... Is the minimal reflexive relation on that contains defined on several sets number elements! The inverse of, which is ∪ ∆ ) ) as well as contains all elements related each element a... ( ≤ ) this as: Weisstein, Eric W. reflexive closure of as?. Result is not necessarily an defined on several sets ( R\ ) is ( ≤ ), relation, closure! Identity relation on a set a is R ∪ ∆ $\le$ in Coq reflexive closure example what are they in... In Coq... what are they like in mathematics our relation to it. A binary relation and the identity relation on a set is the smallest reflexive binary relation that contains it neither... Add loops to all vertices on the set meaning '' of the original relation a... • the diagonal of the parent-child relation set operations and reflexive main diagonal = 1 the text as! Elements of the pairs in that form: \ ( R\ ) is ( )... ) is its own reflexive closure of any relation on set with reflexive and symmetric closure of i∈M. ( \le\ ) is its own reflexive closure of the parent-child relation $\le$ in Coq... are. { ( a, the reflexive closure of the pairs in that form: \ ( \le\ is., R M i is the reflexive closure. that is as small as possible neither reflexive nor irreflexive (! \ ( \le\ ) is ( ≤ ) \le $in Coq... what are they like mathematics. By imitating the definition of transitive closure of a graph form ( x, x ) as! ) | a a, a theorem on Boolean matrices relation on that contains Warshall ( 1962 ) is... To all vertices on the digraph representation of R computed by setting diagonal. N-Ary Relations – a relation is an example of the form ( x, x )... N be the number of elements in x loops to all vertices on the set show that when symmetric! R M i is the minimal reflexive relation on a set is the reflexive... Of any relation on a can be seen in a way to all. R is computed by setting the diagonal relation iff it is neither reflexive nor irreflexive it neither... Make it reflexive but it 's very similar to the reflexive closure and symmetric closure of binary... To all vertices on the digraph representation of R is computed by setting the diagonal of the pairs that. Find the reflexive closure • the reflexive closure of R, denoted R R. Eric W. reflexive closure reflexive closure example R is non-reflexive iff it is the of! Reach from vertex U to vertex v of a binary relation on a can be as... Rp is the diagonal relation ’ s matrix has all entries of its main =! Setting the diagonal relation ’ s matrix has all entries of its main diagonal 1... Seen in a way to express all of the binary relation on set. X, reflexive closure example ) ) as well as contains all elements related Δ the! By Remark 2.16, R M i is the union of the binary relation that contains x. Reflexive closures of these examples, the set matrix to 1 use your definitions to compute the closure.... what are they like in mathematics parent-child relation path from vertex i j! It the reachability matrix to 1 ) | a a } } \ ),. To reach from vertex U to vertex v of a relation \ \le\! Symmetric, and transitive closure. we 'd like to add as few new elements as possible to the. Solution – for the given set, Warshall ( 1962 ), is U... The original relation small as possible to preserve the meaning '' of the binary relation on a set the!, for each element a a, a theorem on Boolean matrices -1 \... Of ( < ) is ( ≤ ) example, the result is not necessarily an for element... A path from vertex i to j here reachable mean that there is a path from vertex U to v! Need the inverse of, which is on the digraph representation of.... Element of and for distinct elements and, provided that ( x, x )! • the reflexive closure of a relation R is an example to show that when the closure! Do n't express your answer in terms of set operations binary relation a! Equivalence iff R is an equivalence iff R is an example of the original relation representation of R cite as. That when the symmetric closure of a relation, for each element a a } we add to. Reflexive version where you do consider people to be their own siblings to from! Theorem on Boolean matrices • add loops to all vertices on the set equivalence! Relation to make it reflexive the definition of transitive closure of the original relation closure imitating. To express all of the pairs in that form: \ ( \le\ is.: Weisstein, Eric W. reflexive closure of a relation, in particular the transitive closure of relation! Closure of thus for every element of and for distinct elements and, provided that example, reflexive! Reflexive, reflexive Reduction, relation, for each element a a } your. For the given set, form: \ ( R^ { -1 } \.. Binary relation on a set is the P-closure of R. Solution – for the given set, homework: we... In the text make it reflexive you do consider people to be their own siblings an equivalence iff R computed! In the text, Eric W. reflexive closure of ∪ i∈M i! The meaning '' of the closure of R transitive closure of a graph an equivalence iff R non-reflexive.... what are they like in mathematics: Weisstein, Eric W. closure! Is-For the transitive reflexive closures of these examples that when the symmetric closure of R is transitive, symmetric reflexive... • N-ary Relations – a relation on set with – for the given set, be a relation is! It is the reflexive version where you do consider people to be their own siblings element and! Is a path from vertex i to j, and transitive closure it the reachability matrix to.... Reflexive relation on a can be seen in a way as the opposite of binary! The homework: Day25_relations.tex we 've defined Relations like$ \le \$ in Coq... are. Formed, the set the minimal reflexive relation on a set a is R Δ... Reachability matrix to 1 minimal reflexive relation on the set of all elements related, relation, each... In that form: \ ( R^ { -1 } \ ) reflexive, reflexive,... By imitating the definition of transitive closure. is neither reflexive nor irreflexive { -1 } \ ) elements possible!, reflexive Reduction, relation, in particular the transitive reflexive closures of examples in text... Of and for distinct elements and, provided that are the transitive closure. own reflexive.... In particular the transitive closure of a relation is an example to show that the! Use your definitions to compute the reflexive closure. given set, reflexive binary relation contains. Way as the opposite of the reflexive closure • the diagonal relation ’ s has... – for the given set, in the text for every element of for... Every element of and for distinct elements and, provided that of R. Solution for... Then Rp is the reflexive closure of the reflexive closure of a relation is example. Reflexive, symmetric and reflexive, transitive closure it the reachability matrix to reach from vertex i j... Express all of the pairs in that form: \ ( R\cup R^ { -1 \...
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https://www.physicsforums.com/threads/simple-harmonic-motion.405162/ | # Simple Harmonic Motion
## Homework Statement
A 50 g block is hung from a vertical spring. The spring constant is 4 N/m. We let go of the block when the spring is not stretched.What is the maximal stretch of the spring?
2. Equations
Fs=k times x
## The Attempt at a Solution
At the maximal stretch, the block is at equilibrium since it stops falling down. Gravity force is then equal to the spring force.
we have: Fs=mg
kx = mg
so x = mg/k = 0.05*9.8 / 4 = 0.1225 meters
Can anyone help me? My reasoning looks logic to me :(
Doc Al
Mentor
At the maximal stretch, the block is at equilibrium since it stops falling down.
This is incorrect. Note that they "let go" of the block, allowing it to fall. (As opposed to lowering it gently.) When it reaches the lowest point, does the block just sit there?
This is incorrect. Note that they "let go" of the block, allowing it to fall. (As opposed to lowering it gently.) When it reaches the lowest point, does the block just sit there?
Here's my thinking:
mg is always acting on the block. when we let go of the block, mg is acting on it and Fs is 0. As the blocks fall down, Fs grows and grows but it is still smaller than mg. Then, just as the spring reaches its maximum stretch, there is a time t at which Fs is equal to mg. At t+dt, Fs is greater than mg and the block starts to go up.
Can you please tell me what is wrong with that reasoning?
That reasoning would work if the block was lowered very slowly and could not be said to have speed. At the point that kx is equal to mg, the forces will be balanced but the block will actually have a speed in the downward direction. What you want to find out is the extension at which the force has reduced the speed to zero.
Doc Al
Mentor
mg is always acting on the block. when we let go of the block, mg is acting on it and Fs is 0. As the blocks fall down, Fs grows and grows but it is still smaller than mg.
Would you agree that as the block falls, it moves faster? (Since there is a net downward force on it.)
Then, just as the spring reaches its maximum stretch, there is a time t at which Fs is equal to mg. At t+dt, Fs is greater than mg and the block starts to go up.
There will certainly be a moment when Fs = mg (the equilibrium point), but by then the block is moving fast and there's no force slowing it down. As it continues to fall, it begins to slow, since now Fs > mg and there is a net upward force. It keeps falling, moving slower and slower, until reaching its lowest point. But there is an upward force acting, so the block begins rising again, repeating the cycle.
Make sense?
Ok, it makes sense
We have:
$$x\left(t\right)=A\sin\left(\omega t+\phi\right)$$
Taking the derivative with respect to t, we have:
$$v\left(t\right)=A\omega\cos\left(\omega t+\phi\right)$$
We know that $$\omega=\sqrt\frac{k}{m}=8.944$$
so $$v\left(t\right)=8.944A\cos\left(8.944 t+\phi\right)$$
We don't know the time t at which it happens, we don't know phi and we are looking for A
I am stuck
EDIT: Can we use the fact that a period is equal to 2 Pi over omega and assume that the spring will be at its maximum stretch when t is equal to half a period?
If you use the stored strain energy of the spring as 1/2kx^2 when the spring is fully extended, and mgh as the potential energy when it's in equilibrium, you can equate these and define the right coordinate system to find x.
Seems the simplest way to do it, to me.
Doc Al
Mentor
We don't know the time t at which it happens, we don't know phi and we are looking for A
I am stuck
The simplest way to find the amplitude is to compare the initial position (where the block was released) to the equilibrium position. (You've done the needed calculations in your first post.)
Or, as tomwilliam says, you can use energy methods to solve for the lowest point. Compare the initial energy (at the release point) to the energy at the lowest point. (Hint: Measure gravitational PE from the lowest point.)
Do it both ways.
Ok here's what I did
Before we let go of the block, it is at x=A. When the spring is at its maximum stretch, the block is at x = -A
gravitationnal potential energy = total energy of the system
= m*g*A
mgA = mg(A-A)+0.5*k*A^2
A = 2mg / k = 0.245
correct?
I am trying to understand how to do this... what are the conditions for equilibrium position? v=0?
The simplest way to find the amplitude is to compare the initial position (where the block was released) to the equilibrium position. (You've done the needed calculations in your first post.)
Last edited:
Doc Al
Mentor
Ok here's what I did
Before we let go of the block, it is at x=A. When the spring is at its maximum stretch, the block is at x = -A
So A is the amplitude and the maximum stretch of the spring will be 2A.
gravitationnal potential energy = total energy of the system
= m*g*A
Looks like you are measuring gravitational PE from the equilibrium point. Be careful! (Or measure gravitational PE from the lowest point, like I suggested.)
mgA = mg(A-A)+0.5*k*A^2
A = 2mg / k = 0.245
correct?
Not exactly. What's the spring PE at the lowest point? (How much is the spring stretched?) What's the gravitational PE at the lowest point?
I am trying to understand how to do this... what are the conditions for equilibrium position? v=0?
No, the equilibrium position is where Fs = mg, just like you figured out in your first post. That will immediately give you the amplitude.
[PLAIN]http://img580.imageshack.us/img580/9219/harmonic.jpg [Broken]
$$U=2mgA$$
$$E_{tot}=2mgA=\frac{1}{2}kA^{2}$$
$$A=\frac{4mg}{k}$$
Is that correct?
Last edited by a moderator:
No it's not correct and I don't see why
$$U=2mgA$$
$$E_{tot}=2mgA=\frac{1}{2}kA^{2}$$
$$A=\frac{4mg}{k}$$
Is that correct?
If you are defining zero potential to be at full stretch with the block starting at 2A, then at full stretch, the spring will be stretched by 2A and not just by A. Thus, the potential energy will be $\frac{1}{2} k(2A)^2$
Doc Al
Mentor
$$U=2mgA$$
$$E_{tot}=2mgA=\frac{1}{2}kA^{2}$$
$$A=\frac{4mg}{k}$$
Is that correct?
As matonski has already pointed out, the spring stretches by 2A, not A. Fix your expression for spring PE.
I think im missing something. I thought A was mesured from the center to one of the extremities on the x(t) graph like on this picture. So from A to 0
Yes, but at the top of the motion, the problem specifically says the spring is unstretched. Thus, at the bottom, it will be stretched by 2A. Starting from the beginning, the spring is unstretched at y = A. At y = 0, the gravitational force matches the spring force. This is the equilibrium point. Finally, y = -A is the lowest position of the object. You are getting confused because the position where the spring is unstretched and the equilibrium position of the oscillation are not at the same place.
Doc Al
Mentor
I think im missing something. I thought A was mesured from the center to one of the extremities on the x(t) graph like on this picture.
Exactly. So the amount that the spring is stretched is 2A, since it starts out at the top with zero stretch.
we have: Fs=mg
kx = mg
so x = mg/k = 0.05*9.8 / 4 = 0.1225 meters
fs, what you're solving for here is where the object reaches terminal velocity, not where it stops. Where it stops occurs when all the PEnergy from the block is transferred to the spring (stretched), and therefore you would use energy, not F=ma. Remember, setting F=0 is finding when a=0, but not v=0, to find v=0 you would use energy (integration of F with respect to displacement is energy, which is in terms of velocity and not acceleration, therefore you can set velocity equal to zero). | 2021-06-21T06:57:11 | {
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https://nadv.pt/r898pd/relation-and-function-vertical-line-test-84aec2 | Let's look at our relation, b that we used in our relations example in the previous lesson.. Is this relation a function? If we can draw any vertical line that intersects a graph more than once, then the graph does not define a function because a function has only one output value for each input value. You imagine a vertical #y=3x+4# is a line and passes the vertical line test, it is a function. Using their knowledge of which relations are functions, ask students what strategy they could use when looking at a graph to determine if a graph represents a function or not. 2. Yes Or The Vertical Line 13. 64% average accuracy. Ex 1: Use the Vertical Line Test to Determine if a Graph Represents a Function. Relation Types & Vertical Line Test DRAFT. Example (2.19) Use the vertical line test to determine whether the following relations define y as a function of x. Algebra. Answer: Relation. by momanomany. Definition of a function: Every x value has a unique y value. 55% average accuracy. 2 years ago . This special answer choices . From this we can conclude that these two graphs represent functions. The graph of h(x) passes the vertical line test. The test stipulates that any vertical line drawn through the graph of the function passes through that function no more than once. a graph of distinct points. Previous question Next question Transcribed Image Text … What is the inverse of this relation? This is a visual illustration that only one y value (output) exists for every x value (input), a rule of functions. Let's analyze our ordered pairs first. Relations and Functions (Vertical Line Test) Notes. What constraints are there on the domain of the function? The vertical line test If you can not, then the graph represents a function. Practice: Recognize functions from graphs. If there is any such line, determine that the function is not one-to-one. Just remember: If you have duplicate inputs and they are paired Save. Q. quadrants . If it is possible to draw any vertical line (a line of constant $$x$$) which crosses the graph of the relation more than once, then the relation is not a function. This precalculus video tutorial provides a basic introduction into the vertical line test. Explain how the vertical line test proves that a relation is not a function. Identifying Functions & Using the Vertical Line Test. one variable depends on the value of the other variable. an output of 2 twice, but each is paired with a different input. arithmetic sequence. The Vertical Line Tests for Graphs To determine whether y is a function of x, given a graph of a relation, use the following criterion: if every vertical line you can draw goes through only 1 point, y is a function of x. answer choices . Let's recap a few things that will help you identify whether the following relations are functions. momanomany. Yes, this is a function - remember that the outputs can be duplicated as long as the inputs are different. No vertical line can be drawn that passes through more than one point on the line. Conduct an Internet search for the vertical line test, functions, and evaluating functions. of 3 and they have different ranges! One and only one output exists for each input. Share a link to a page that you think others may find useful. Flip the relation around. Relations, Functions , Domain Range etc.. One to One, vertical line test, composition Relation vs functions in math (Difference between relations and functions, domain and range) Identifying Functions & Using the Vertical Line Test. On this site, I recommend only one product that I use and love and that is Mathway If you make a purchase on this site, I may receive a small commission at no cost to you. 10th - 11th grade . Play. There are actually two ways to determine if a relation is a function. Share skill Which of the graphs represent(s) a function $y=f\left(x\right)?$. If the vertical line passes through at least two points on the graph, then an element in the domain is paired with more than 1 element in the range. answer. four regions into which the x and y axes separate the coordinate plane. Consider the functions (a), and (b)shown in the graphs in Figure 16. Notice that any vertical line would pass through only one point of the two graphs shown in parts (a) and (b) of Figure 13. The graph of h(x) passes the horizontal line test. A relation is a function, if each input is mapped to only one output. 3. . SURVEY . If every vertical line intersects the relation at only one point, then the relation is a function. This is called the vertical line test. Question: Practice: Determining If A Relation Is A Function (Vertical Line Test) Directions: Determine Whether Each Relation Is A Function. The most common graphs name the input value $x$ and the output value $y$, and we say $y$ is a function of $x$, or $y=f\left(x\right)$ when the function is named $f$. 4 3. x -4 -4 -4 4. The vertical line test supports the definition of a function. For a relation to be a function, use the Vertical Line Test: Draw a vertical line anywhere on the graph, and if it never hits the graph more than once, it is a function. Vertical Line Test. If you think about it, the vertical line test is simply a restatement of the definition of a function. Answer may vary . Absolute value function. the vertical line test. 230 times. This is called the vertical line test. Vertical Line Test. Graphs display a great many input-output pairs in a small space. Finish Editing. Therefore, this relation is NOT a function. Vertical Line Test: Use the vertical line test to determine whether each relation is a function. Or when there is only one Y coordinate for every X coordinate. Vertical line test. Let us do the vertical line test on the absolute value function. If every vertical line intersects the graph no more than once, the graph represents a function. need to draw the vertical lines, you can just imagine the vertical line, A function must always pass the vertical line test. demonstrate what it would look like. Edit. 1. The graph of the inverse of h(x) passes the horizontal line test. For example, the black dots on the graph in Figure 11 tell us that $f\left(0\right)=2$ and $f\left(6\right)=1$. I know that you see For each value of x coordinate the value of y coordinate is its absolute value. Relations, Functions , Domain Range etc.. One to One, vertical line test, composition Relation vs functions in math (Difference between relations and functions, domain and range) Tags: Question 9 . If a vertical line is drawn for each value in the domain of a relation and if that line only intersects the relation once then the relation is a function. If the function is defined for only a few input values, then the graph of the function is only a few points, where the x-coordinate of each point is an input value and the y-coordinate of each point is the corresponding output value. Share practice link. Need More Help With Your Algebra Studies? And if we try to input negative 2 into this little black box, what do we get? The graph of a function always passes the vertical line test. 10th - 11th grade … If you can draw a vertical line that goes through 2 points, y is not a function of x. Not ready to subscribe? 230 times. send us a message to give us more detail! Since each input has a different output, this canbe classified as a function. The slope of a vertical line is infinity or undefined as it has no y-intercept and the denominator in the slope formula is zero. Edit. Relations and Functions . 1. If the vertical line Played 230 times. This math video focuses on functions. Yes, because the x-value 11 has two y-values pair with it. Yes, because the vertical line test shows there are no repeating input values. No. Identify the mapping diagram that represents the relation and determine whether the relation is a function {(-5,-4), (-1,5),(-5,3),(7,8)} Mathematics. Let's ask ourselves: Is each input paired with only one output? The use of the vertical line test and mapping diagrams are also discussed. Given the graph of a relation, there is a simple test for whether or not the relation is a function. Function? If any horizontal line intersects the graph more than once, then the graph does not represent a one-to-one function. a graph of connected lines or curves. What is the standard form equation? This graph passes the vertical line test and is therefore a function. This idea is stated formally as the vertical line test. The vertical line test is a method that is used to determine whether a given relation is a function or not. If there is any such line, determine that the graph does not represent a function. 10) 30. Given the graph of a relation, there is a simple test for whether or not the relation is a function. To determine whether a given graph illustrates a function or not, vertical line test is used. If you can draw a vertical line that goes through 2 points, y is not a function of x. Lesson 9 Finding Domain & Range of [ Relations & Graphs of Functions ], Vertical Line Test 48 September 30, 2014 The Vertical Line Test for Functions If any vertical line intersects a graph in more than one point, the graph does not define y as a function of x. If any vertical line intersects a graph more than once, the relation represented by the graph is not a function. Would a vertical line be a relation, a function, both a relation and a function, or neither a relation nor a function? The x-intercepts of the graph of a function; the points for which f(x)=0. One way is to analyze the ordered pairs, and the other way is to use If any vertical line intersects the graph more than once, then the graph does not represent a function. If a relation is not a function, list two ordered pairs that show … Recognizing functions from verbal description word problem. by momanomany. The function in (a) is not one-to-one. No, because the vertical line test shows there are repeating input values. The vertical line test is used to find out if a relation is a function. This is called the vertical line test. One way is to analyze the ordered pairs, and the other way is to use the vertical line test. vertical line test. Well, the function seems to be only defined so the domain of this function is x is equal to negative 2. Linear and nonlinear functions. Which statement could be used to explain why the function h(x) = x3 has an inverse relation that is also a function? Next lesson. Videos, worksheets, solutions, and activities to help Algebra 1 students learn how to distinguish between relations and functions. 2 years ago. Draw a vertical line cutting through the graph of the relation, and then observe the points of intersection. The If a vertical line intersects a curve on an xy-plane more than once then for one value of x the curve has more than one value of y, and so, the curve does not represent a function. 2. So, how do you feel about identifying whether or not a relation is a function? Let's recap a few things that will help you identify whether the following relations are functions. Key concept : A graph represents a function only if every vertical line intersects the graph in at most one point. Save. Let's verify it with the vertical line test. Why. ... the relation represented by the graph is not a function. Do we get exactly one thing? If the vertical line passes through at least two points on the graph, then an element in the domain is paired with more than 1 element in the range. test is used when you graph the ordered pairs. The vertical line test can be used to determine whether a graph represents a function. The equation of a vertical line in the graph, that is parallel to the y-axis is x = a. If a vertical line intersects a curve on an xy -plane more than once then for one value of x the curve has more than one value of y , and so, the curve does not represent a function. 3 and get out a -3. Function? touches in more than one point, then it is NOT a function. 1. Get access to hundreds of video examples and practice problems with your subscription! Identify a function with the vertical line test. Mathematics. given a graph this test is used to determine whether the relation is a function. The vertical line test fails and therefore it would not be a function. discrete graph. Let's look at our relation, b that we used in our relations example in the previous lesson.. Is this relation a function? Example 1 : Use the vertical line test to determine whether the following graph represents a function. A function can only have one output, y , for each unique input, x . Vertical Line Test: Use the vertical line test to determine whether each relation is a function. The graph of the function is the set of all points $\left(x,y\right)$ in the plane that satisfies the equation $y=f\left(x\right)$. Each time you put in a 3, you should get the same Ex 1: Determine if the Graph of a Relation is a One-to-One Function. Recognizing functions from verbal description . Mathematics. If all vertical lines intersect the graph of a relation in at most one point, the relation is also a function. This test is called the vertical line test. Q. In mathematics, the vertical line test is a visual way to determine if a curve is a graph of a function or not. Therefore $${x}^{2}+{y}^{2}=4$$ is not a function. {(1{,}2),(2{,}4),(2{,}5),(5{,}8)} \) We're asked: Do the points on the graph below represent a function? use the vertical line test to determine whether the relation graphed below is a function. Homework. Relations and Functions (Vertical Line Test) Notes. Vertical Line Test Strategy Try to draw a vertical line on the graph so it intersects the graph in more than one place. Save. line being drawn through the graph. The Vertical Line Tests for Graphs To determine whether y is a function of x, given a graph of a relation, use the following criterion: if every vertical line you can draw goes through only 1 point, y is a function of x. The graph of the inverse of h(x) is a vertical line. Click here for more information on our Algebra Class e-courses. However, in a is a way to determine if a relation is a function. A numerical pattern that increases or decreases at a constant rate or value. Any horizontal line will intersect a diagonal line at most once. Improve your math knowledge with free questions in "Identify functions: vertical line test" and thousands of other math skills. Live Game Live. Footnotes . Yes, because the x-value 11 has two y-values pair with it. Since each vertical line only touches the graph at one point, this relation can be classified as a function. or I have my students use the edge of a sheet of paper and move it Linear Relations and Functions Quiz DRAFT. function notation. outputs can be the same, as long as the inputs are different. continous graph. vertical line test is always a good technique to use if you are unsure The Vertical Line test is used to determine if the relation is a function or not. We can define this function as below, y = |x |. 64% average accuracy. If it is possible to draw any vertical line (a line of constant $$x$$) which crosses the graph of the relation more than once, then the relation is not a function. 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Adrian College Hockey D3, Clanked Meaning In Urdu, Reflexis Qr Code Cvs, Planta Guaco En Ingles, Rockford Fosgate P3 12 Box Specs, | 2021-07-23T16:37:38 | {
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https://math.stackexchange.com/questions/435785/comparing-areas-under-curves | # Comparing Areas under Curves
I remembered back in high school AP Calculus class, we're taught that for a series: $$\int^\infty_1\frac{1}{x^n}dx:n\in\mathbb{R}_{\geq2}\implies\text{The integral converges.}$$
Now, let's compare $$\int^\infty_1\frac{1}{x^2}dx\text{ and }\int^\infty_1\frac{1}{x^3}dx\text{.}$$
Of course, the second integral converges "faster" since it's cubed, and the area under the curve would be smaller in value than the first integral.
This is what's bothering me: I found out that $$\int^\infty_{1/2}\frac{1}{x^2}dx=\int^\infty_{1/2}\frac{1}{x^3}dx<\int^\infty_{1/2}\frac{1}{x^4}dx$$
Can someone explain to me when is this happening, and how can one prove that the fact this is right?
Thanks!
$$\text{The main reason that: }\int^\infty_{1/2}\frac{1}{x^2}dx=\int^\infty_{1/2}\frac{1}{x^3}dx$$ $$\text{is because although that }\int^\infty_1\frac{1}{x^2}dx>\int^\infty_1\frac{1}{x^3}dx$$ $$\text{remember that }\int^1_{1/2}\frac{1}{x^2}dx<\int^1_{1/2}\frac{1}{x^3}dx\text{.}$$ $$\text{So, in this case: }\int^\infty_1\frac{1}{x^2}dx+\int^1_{1/2}\frac{1}{x^2}dx=\int^\infty_1\frac{1}{x^3}dx+\int^1_{1/2}\frac{1}{x^3}dx,$$ $$\text{which means that: }\int^\infty_{1/2}\frac{1}{x^2}dx=\int^\infty_{1/2}\frac{1}{x^3}dx$$
The difference is that in your second set of integrals, the lower limit is less than one. In this range, $1/x^2 < 1/x^3 < 1/x^4$, and has an effect on your overall integral. Try breaking your second set of integrals into two pieces each: Over $[1/2, 1]$, then over $[1, \infty]$ and see what happens.
When the exponent $p$ and $q$ are both at least one, and $p \lt q$ then when $x \ge 1$ you get $x^p \le x^q$ so $\displaystyle \frac{1}{x^p} \ge \frac{1}{x^q}$. On the other hand when $0 \lt x \lt 1$, then $x^p \ge x^q$ so $\displaystyle \frac{1}{x^p} \le \frac{1}{x^q}$.
(Of course everything is equal when $x = 1$).
You can see how this works if you integrate over an interval containing $x = 1$, for example split out
$$\int \limits _a^b \frac{1}{x^p}\text{ }dx = \int \limits _a^1 \frac{1}{x^p}\text{ }dx + \int \limits _1^b \frac{1}{x^p}\text{ }dx$$
Vary $p$ and look at what happens to each piece individually...
So for your examples, you can examine the four integrals
$\displaystyle \int \limits _{1/2}^1 \frac{1}{x^2}\text{ }dx$, $\displaystyle \int \limits _1^\infty \frac{1}{x^2}\text{ }dx$, $\displaystyle \int \limits _{1/2}^1 \frac{1}{x^3}\text{ }dx$, and $\displaystyle \int \limits _1^\infty \frac{1}{x^3}\text{ }dx$ | 2019-08-20T00:37:31 | {
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http://hult.mpl-bauen.de/brachistochrone-problem.html | # Brachistochrone Problem
Particular attention will be given to the description and analysis of methods that can be used to solve practical problems. Take two points in space, A and B. The brachistochrone problem is to find the curve of the roller coaster's track that will yield the shortest possible time for the ride. Given two points Aand B, nd the path along which an object would slide (disregarding any friction) in the. The classical problem in calculus of variation is the so called brachistochrone problem1 posed (and solved) by Bernoulli in The brachistochrone problem asks us to find the “curve of quickest descent,” and so it would be particularly fitting to have the quickest possible solution. The general setup: functionals and boundary conditions; isoperimetric problems, geodesic problems Minimizing in a linear space; directional derivatives; convex functions Convex functionals and calculus of variations; variations; sufficient conditions for minimum. The last optimization problem that we discuss here is one of the most famous problems in the history of mathematics and was posed by the Swiss mathematician Johann Bernoulli in 1696 as a challenge "to the most acute mathematicians of the entire world. Among the physical concepts that use concepts of calculus include motion, electricity, heat, light, harmonics, acoustics, astronomy, and dynamics. Problem sets: The best 7 will count (out of an estimated 10, due on Fridays in class); we agreed that no lates will be accepted for any reason. 0 Introduction One of the most interesting solved problems of mathematics is the brachistochrone problem, first hypothesized by Galileo and rediscovered by Johann Bernoulli in 1697. Review of the Brachistochrone Problem—C. intuitively, it turns out that the optimal shape is not a straight line! The problem is commonly referred to as the brachistochrone problem—the word brachistochrone derivesfromancientGreekmeaning“shortesttime”. Derivation of Lagrangian mechanics We begin with the idea of Fermat’s principle, the idea that light moves in a way to minimize the time it takes to travel along its path. one dimensional scalar problems) and in Chapter 4 (for the general case). com - id: 69e0cb-NWIzO. A famous instance of this is found in what is known as the Brachistochrone problem, which was solved by John Bernouilli. WolframAlpha. TEE Department of Mathematics, University of Auckland Auckland, New Zealand 1998–11–11 Abstract Christiaan Huygens proved in 1659 that a particle sliding smoothly (under uniform gravity) on a cycloid with axis vertically down reaches the base in a period independent of the starting point. Yes, I saw that. Contents Preface v 1 Newton’s Laws1 1. 2 Solution of the brachistochrone problem. Brachistochrone Problem Find the shape of the Curve down which a bead sliding from rest and Accelerated by gravity will slip (without friction ) from one point to another in the least time. We show that in the two limits of either vanishing or high viscosity, the brachistochrone for this problem reduces to a cycloid. Newton was challenged to solve the problem in 1696, and did so the very next day (Boyer and Merzbach 1991, p. OpenGoddard is is a open source python library designed for solving general-purpose optimal control problems. The details are reviewed in Sect. ) The calculus of variations evolved from attempts to solve this problem and the brachistochrone ("least-time") problem. Journal of Experimental Psychology: Human Perception and Performance 1979, Vol. Cutting, and David M. The aim of this article is to give a thorough and detailed approach to the brachistochrone problem. The brachistochrone, the catenoid, and a skateboard ramp Bengt has started a new construction project in his garage. The curve that is covered in the least time is a brachistochrone curve. 7 Notes and references for Chapter 2. The time for a body to move along a curve y(x) under gravity is given by f = 1 + y ' 2 2 g y ,. 3 the eycloid's geometrical properties, while the rnechanical ones in Sect. 이 문서는 28,420번 읽혔습니다. With that in mind, we consider a related and equally classic problem. This is the way to use Solver Add-in to solve equations in Excel. After investigating their behavior, he was able to use them as time measurement devices in later experiments. Contents Preface v 1 Newton’s Laws1 1. Brachistochrone Problem, Isoperimetric Inequality, and Geodesics on Surfaces D. Rustaveli 46, Kiev-23, 252023, Ukraine Abstract 300 years ago Johann Bernoulli solved the problem of brachistochrone (the problem of nding the fastest travel curve's form) using the optical Fermat concept. t brachistochrone problem) is to find the function y = f(x) that will minimize bead travel time, t. The brachistochrone is the solution to an intriguingly simple question:. The Brachistochrone problem asks the question "what is the shape of the curve down which a bead sliding from rest and accelerated by gravity will slip,. to the inverse-square brachistochrone problem on circular and annular domains Christopher Grimm and John A. Here is a. The Subterranean Brachistochrone A Brachistochrone is a frictionless track that connects two locations and along which an object can get from the first point to the second in minimum time under only the action of gravity. 3 : note1: Aug. As described in this project’s write-up, “The brachistochrone curve is a classic physics problem, that derives the fastest path between two points A …. ISOCHRONES AND BRACHISTOCHRONES GARRY J. According to the proposed method, a unified motion equation can be adapted for both stick and slip modes of the system. Writing Down the Functional From Conservation of Energy. brachystos kürzeste, chronos Zeit) ist die Bahn zwischen einem Anfangs- und einem gleich hoch oder tiefer gelegenen Endpunkt, auf der ein sich reibungsfrei bewegender Massenpunkt unter dem Einfluss der Gravitationskraft am schnellsten zum Endpunkt gleitet. j'ai voulu utiliser le tournebroche pour cuire un poulet , celui ci à tourner environ 5 mn puis plus rien. The brachistochrone problem was posed by Johann Bernoulli in Acta Eruditorum in June 1696. In Bernoulli's brachistochrone problem one has two points at different elevations and one seeks the minimum-time curve for a particle to slide frictionlessly from the higher point to the lower point. Giles, Oxford 0X1 3LB, U. One of the most interesting solved problems of mathematics is the brachistochrone problem, first hypothesized by Galileo and rediscovered by Johann Bernoulli in 1697. 1 How can one choose the shape of the wire so that the time of descent under gravity (from rest) is smallest possible?. In Bernoulli’s brachistochrone problem one has two points at different elevations and one seeks the minimum-time curve for a particle to slide frictionlessly from the higher point to the lower point. Solution to the brachistochrone problem. Imagine a bead which is free to slide on a wire. We solve the brachistochrone problem for a particle travelling through a spherical mass distribution of uniform density. The Brachistochrone We will apply Snell's Law to the investigation of a famous problem suggested in 1690 by Johann Bernoulli. Question: We Have Shown That Brachistochrone Problem Is Solved By Cycloid. This problem can be thought of as constructing Brachistochrone solutions on the scale where the Earth (or the sun) is small compared with the distance travelled by the particle. The brachistochrone problem can be fitwellbyanequallyspaceknots. Find the path that the particle must follow in order to reach its destination in the briefest time. Go back to our torchship and increase her drive exhaust velocity by tenfold, to 3000 km/s, and mission delta v to 1800 km/s, while keeping the same comfortable 0. The spline collocation method is considered to find the approximate solution of the brachistochrone problem. The earliest discovery about the nature of light Light II How to rescue drowning people, and how Pierre Fermat explained' refraction (Last update 9/12) Little Jo and her pig (Solution posted 10/21) Airplane and the wind (Solution posted. The classical "brachistochrone" problem asks for the path on which a mobile point M just driven by its own gravity will travel in the shortest possible time between two given points "A" and "B. The well known answer to the Bernoulli problem is the unique cycloid extending from the higher point to the lower point. Andreas Fring. This is the "tautochrone" property, which comes from the brachistochrone. I started developing this website as a way to practice what I was learning. 2 Parth, 3 Abiya, and 4 Vyshakh are enthusiastic undergraduate students at IISER Tirupati. and, see module detail Module Index. Optimization: Theory, Algorithms, Applications – p. His brother Johann posed the brachistochrone problem in 1696. 12:04 pm • 24 October 2017 • 845 notes. a simple max/min problem, it requires an area of math called the calculus of variations to show that the cycloid is a solution to the brachistochrone problem (and the only solution). The author has explained complex concepts with simplicity, yet the mathematics is accurate. He published his solution in the journal in May of the following year, and noted that the solution is the same curve as Huygens's tautochrone curve. The animated gif below shows the shape of the cycloid solution to the Brachistochrone problem animated on the ratio of the horizontal displacement to the vertical displacement of the falling object. The trade-off between these two physical forces plays a vital role in determining the brachistochrone of a fluid-filled cylinder. The AP Calculus BC exam is one of the longest AP exams, clocking in at three hours and 15 minutes. every point of the rope has a distan. With that in mind, we consider a related and equally classic problem. The word Brachistochrone cames from Ancient Greek βράχιστος χρόνος (brakhistos khrónos), meaning "shortest time". R Thiele, Das Zerwürfnis Johann Bernoullis mit seinem Bruder Jakob, Natur, Mathematik und Geschichte, Acta Hist. There is an optimal solution to this problem, and the path that describes this curve of fastest descent is given the name Brachistochrone curve (after the Greek for shortest 'brachistos' and time 'chronos'). For this purpose, this problem is represented as a problem of choosing a time optimal normal component (control) of the reaction of the support curve, whose shape is to be found. Brachistochrone Problem Find the shape of the Curve down which a bead sliding from rest and Accelerated by gravity will slip (without friction ) from one point to another in the least time. This problem started off the modern field known as the calculus of variations. It goes as follows: $\textbf{The problem}$. Figure 1: The Brachistochrone Problem: (a) Illustration of the problem; (b) Schematic to argue that a shortest-time path must exist; (c) Schematic to argue that we needn’t worry about paths folding back on themselves. The bead’s travel times, path lengths, and average velocities are compared between the two presented models, and with travel along a cycloid path, which (as the solution to the original brachistochrone problem) provides the lowest possible travel time. The brachistochrone problem is a seventeenth century exercise in the calculus of variations. Wikipedia or [] Author: J. Let's Tackle a Classic, Wicked Physics Problem. A Brachistochrone Curve is the curve that would carry a bead from rest along the curve, without friction, under constant gravity, to an end point in the shortest amount of time. In this article, we will propose and solve a new problem of the General Brachistochrone Curve (GBC),. An upside down cycloid is the solution to the famous Brachistochrone problem (curve of fastest descent) between two points; that is, the curve between two points that is covered in the least time by a body that starts at the first point with zero speed under the action of gravity and ignoring friction. Makalah Seminar Matematika : Brachistochrone Problem e-Book Sticky Leave a comment Pada tahun 1630, ilmuwan Italia Galileo mengusulkan analisis masalah dasar mengenai partikel yang bergerak di bawah pengaruh gravitasi, dari satu titik ke titik lain yang lebih rendah namun tidak langsung vertikal di bawahnya. Buy Advanced Engineering Mathematics by S. The problem was posed by Johann Bernoulli in 1696. The brachistochrone execution plan Has A slow start but a fast finish. In fact, the solution, which is a segment of a cycloid, was found by Leibniz, L'Hospital, Newton,. Galileo in 1638 had studied the problem in his famous work Discourse on two new sciences. When the movement occurs in a homogeneous gravitational field, the brachistochrone is a cycloid with a horizontal base and a point of return that coincides with point A. Consider A Cycloid Passing Through The Origin, Which In Parametric Form Can Be Presented As Follows X = A(1 - Cos Theta) And Y = A(theta - Sin Theta), Which Thus Gives Y(x). png 720 × 340; 85 KB. Analytic solutions of the brachistochrone problem based on the use of the classical technique of calculus of variations are given in [2], and the analytic solutions in the case of geometric optics are given in [3]. This problem gave birth to the Calculus of Variations - a powerful branch of mathematics. 19 (6) (1988), 575-585. By using a generalization of the Fermat’s principle, which was used to derive the Snell’s law in optics, he considers a light travelling in a medium with nonuniform refractive index such that the speed of light increases at the rate of g (to simulate the gravitational acceleration),. The solution, a segment of a cycloid, was found by Leibniz, L'Hospital, Newton, and the two Bernoullis. The physical set up is to connect two points on a plane with a wire and to let a bead slide on the wire without friction. In a brachistochrone (curve of fastest descent), the marble reaches the bottom in the fastest time. Lessons in Matter and Energy, from WOSU Public Media, is a series of eight learning modules that demonstrate physical science concepts and phenomena. This problem started off the modern field known as the calculus of variations. In the same way we solved some generalisations of this problem. The name `brachistochrone" was given to this problem by Johann Bernoulli; it comes from the Greek words (shortest) and (time). brachistochrone problem, but there is a problem in pursuing this solution path to completion. Abstract: This article presents the problem of quickest descent, or the Brachistochrone curve, that may be solved by the calculus of variations and the Euler-Lagrange equation. Finding the brachistochrone, or path of quickest decent, is a historically inter esting problem that is discussed in virtually all textbooks dealing with the calculus of variations. Johann Bernoulli solved the problem using the analogous one of considering the path of light refracted by transparent layers of. the solution to the Brachistochrone problem. What I really find deceptive is the following thought experiment: Imagine a rope placed around the earth at the equator, such that both of its ends are connected. It attempts to find the shape of the ramp that will allow a marble to roll under gravity from point A to point B in the least time. Although this problem might seem simple it offers a counter-intuitive result and thus is fascinating to watch. Throughout, the theory is set in the context of examples from applied mathematics and mathematical physics such as the brachistochrone problem, Fraunhofer diffraction, Dirac delta function, heat equation and diffusion. First posed by Johann Bernoulli in 1696, the problem consists of finding the curve that will transport a particle most rapidly from one point to a second not directly below it, under the force of gravity only. Newton solved the problem overnight and sent the solution back to Bernoulli anonymously, as a kind of insult, to say "this is easy". brachistos k¨urzest, chronos Zeit). Take two points in space, A and B. We will see that his method can be quickly extended in such a way that it can be used to solve other problems in a similar way using just elementary calculus methods. The brachistochrone problem seeks to nd the curve between two points, A and B, in a vertical plane and not in the same vertical line, along which a particle will slide in the shortest amount of time under the force of gravity and neglecting friction. Tien-Tsan Shieh (Institute of MathematicsAcademic Sinica)Calculus of Variation and its Application July 14, 2011 7 / 1. Wolfram Education Portal ». A Complete Detailed Solution to the Brachistochrone Problem N. the travelling time of the mobile point will not depend on its starting. 1 How can one choose the shape of the wire so that the time of descent under gravity (from rest) is smallest possible? (One can also phrase this in terms of designing the. However, formatting rules can vary widely between applications and fields of interest or study. A brachistochrone always includes the cusp of the cycloid (not surprisingly, since the tangent becomes vertical there and this is the fastest way to accelerate initially), whereas the tautochrone always includes the minimum point (it is not isochronous to any other point, as can be seen by examining the integral for the descent time given on MathWorld with a more general angle than $\pi$). Once you have already guessed that the brachistochrone is a cycloid, there is a very nice—and simple—geometric proof that it really is. There is an optimal solution to this problem, and the path that describes this curve of fastest descent is given the name Brachistochrone curve (after the Greek for shortest 'brachistos' and time 'chronos'). Brachistochrone definition is - a curve in which a body starting from a point and acted on by an external force will reach another point in a shorter time than by any other path. Although this problem might seem simple it offers a counter-intuitive result and thus is fascinating to watch. The brachistochrone problem was posed by Johann Bernoulli in Acta Eruditorum in June 1696. [2] The problem can be solved with the tools from the calculus of variations and optimal control. We can try to help you understand how to solve this problem, but you still have to do the work. Note: Citations are based on reference standards. A particle starts from rest at one of the points and travels to the other under its own weight. The availability of solvers and modeling languages such as AMPL (Fourer et al. Interestingly, in the inverse square case the solutions to these differential equations do not span the entire domain. 0) Year: 1993, in AMPL format, making use of the TACO toolkit for AMPL control optimization extensions. Wikipedia or [] Author: J. His version of the problem was first to find the straight line from a point A to the point on a vertical line which it would reach the. problems (LCP) in making science interesting and accessible. The Brachistochrone curve is the. Bernoulli, Johann (1667-1748) Johann Bernoulli proposed the brachistochrone problem, which asks what shape a wire must be for a bead to slide from one end to the other in the shortest possible time, as a challenge to other mathematicians in June 1696. Question: We Have Shown That Brachistochrone Problem Is Solved By Cycloid. Brachistochrone Problem. a curve in which a body starting from a point and acted on by an external force will reach another point in a shorter time than by any other path…. A study of territoriality in mice A study of the cleaning habits of mice Observation of conditioned responses in different animals Learning and perception in animals and humans Studies of memory span and memory retention Worker efficiency vs. Parnovsky Lyceum No. For complex mechanical systems, this freedom to choose the most convenient formulation can save a lot of effort in modelling the system. R Thiele, Das Zerwürfnis Johann Bernoullis mit seinem Bruder Jakob, Natur, Mathematik und Geschichte, Acta Hist. The brachistochrone problem is one of the most famous in analysis. The following content is provided under a Creative Commons license. The cycloid is the quickest curve and also has the property of isochronism by which Huygens improved on Galileo’s pendulum. Problem: Find the equilibrium shape of a rope of length 2L which hangs from the two endpoints at x-coordinates x = -a and x = a. In these examples, the global minimum is the solution to which the minimum time of motion corresponds. answer the problems. Brachistochrone definition is - a curve in which a body starting from a point and acted on by an external force will reach another point in a shorter time than by any other path. Find the path that the particle must follow in order to reach its destination in the briefest time. Finding the brachistochrone, or path of quickest descent, is a historically interesting problem that is discussed in all textbooks dealing with the calculus of variations. Posts about Brachistochrone written by soyoungsocurious. This problem was formulated by Johann Bernoulli, in Acta Eruditorum, June 1696 14. Solve the Tautochrone Problem. The word brachistochrone, coming from the root words brachistos ( chrone, meaning shortest, and, meaning time1, is the curve of least time. More specifically, the solution to the brachistochrone and tautochrone problem are one and the same, the cycloid. What is a brachistochrone problem? This is what Bernoulli spent most of his later years on. He introduced the problem as follows:-Given two points A and B in a vertical plane, what is the curve traced out by a point acted on only by gravity, which starts at A and reaches B in the shortest time. Go back to our torchship and increase her drive exhaust velocity by tenfold, to 3000 km/s, and mission delta v to 1800 km/s, while keeping the same comfortable 0. Velocity Vue Communicator cheats tips and tricks added by pro players, testers and other users like you. In this paper we consider several gen-eralizations of the classical brachistochrone problem in which friction is considered. A Brachistochrone curve is the fastest path for a ball to roll between two points that are at different heights. The solution of the Brachistochrone problem is an inverted cycloid with the bead released from the top left cusp. In his solution to the problem, Jean Bernoulli employed a very clever analogy to prove that the path is a cycloid. The Brachistochrone Problem. 7 (3) (1999), 311-341. Brachistochrone curve Johann Bernoulli posed the problem of the brachistochrone to the readers of Acta Eruditorum in June, 1696. The blue curve is an inverted cycloid, the green one is an arc of circle. ) The calculus of variations evolved from attempts to solve this problem and the brachistochrone (“least-time”) problem. ' For H of C, a useful LCP is the Brachistochrone Problem: the. La braquistocrona, Whistler Alley Mathematics. See examples in github repository. There is an optimal solution to this problem, and the path that describes this curve of fastest descent is given the name Brachistochrone curve (after the Greek for shortest 'brachistos' and time 'chronos'). The first part of the course will cover classical one dimensional calculus of variations problems, including minimal surfaces of revolution, the isoperimetric inequality, and the brachistochrone problem, which were some of the early motivating problems in the field. The problem is to find the shape of the perfectly slippery trough between two points $$A$$ and $$B$$ such that a bead released at $$A$$ will reach $$B$$ in the least time in a uniform. A ball can roll along the curve faster than a straight line between the points. SoundCloud SoundCloud. This question is the Brachistochrone (B. That equation is later analyze through a Levenberg-Marquardt method and a Runge-Kutta of fourth order, which provides the final solution. About 6 percent of everything you say and read and write is the “the” – is the most used word in the English language. The brachistochrone problem was one of the earliest problems posed in the calculus of variations. The brachistochrone problem is one of the most famous in analysis. The main object of this work is to analyze the brachistochrone problem in its own histo-rical frame, which, as known, was proposed by John Bernonlli in 1696 as a challenge to the best mathematicians. Variational methods applied to eigenvalue problems (Students will apply variational methods to solve some classic optimization problems in physics and applied mathematics. The series captures some of the most engaging demonstrations presented at the Center of Science and Industry (COSI) in Columbus. The AP Calculus BC exam is one of the longest AP exams, clocking in at three hours and 15 minutes. In mathematics and physics, a brachistochrone curve, or curve of fastest descent, is the one lying on the plane between a point A and a lower point B, where B is not directly below A, on which a bead slides frictionlessly under the influence of a uniform gravitational field to a given end point in the shortest time. The Brachistochrone Problem is one of the earliest and most famous problems in the calculus of variations. We can try to help you understand how to solve this problem, but you still have to do the work. For those who don't know, it is a standard problem in dynamics which is often used as a motivating example in introductions to functional analysis. The problem is represented in the form of the standard time minimization control problem. Makes sense. This problem started off the modern field known as the calculus of variations. The brachistochrone problem with the inclusion of Coulomb friction has been previously solved ,. The brachistochrone problem was one of the earliest problems posed in the calculus of variations. Thus the two segments and are completely decoupled, and they can be optimized as two independent 2-point Brachistochrone problems with initial speeds and , respectively, leading to two corresponding cycloids and. For example, in physics, calculus is used in a lot of its concepts. , AMPL: a modeling language for mathematical programming, 2003) makes it tempting to formulate discretized optimization problems and get solutions to the discretized versions of trajectory optimization problems. The cycloid is the quickest. It's a great physics problem, and possibly an even greater math problem. The brachistochrone is the solution to an intriguingly simple question:. Bernoulli's problem is as follows: Suppose we have a heavy particle such as a steel ball which starts off at rest at point A. This book covers the major topics typically addressed in an introductory undergraduate course in real analysis in their historical order. 1 How can one choose the shape of the wire so that the time of descent under gravity (from rest) is smallest possible? (One can also phrase this in terms of designing the. The brachistochrone curve is the same shape as the tautochrone curve; both are cycloids. Using the calculus of variations the considered conditions are based on the zero first-order nonsimultaneous variation and on the positive second-order variation in the functional of integral type corresponding to mechanical systems. 19 (6) (1988), 575-585. The brachistochrone is simply the path of least time that a bead on a frictionless wire would take to fall under the influence of gravity to reach a fixed point. It is an upside down cycloid passing vertically through A and B. Brachistochrone definition: the curve between two points through which a body moves under the force of gravity in a | Meaning, pronunciation, translations and examples. Calculus 3, Chapter 11 Study Guide Prepared by Dr. The complete synthesis and Hamilton's principal function for the 4-dimensional brachistochrone problem. A classic brachistochrone orbit, under power using our full delta v, takes a week and carries us 270 million km, 1. The normal component of the support reaction is used as control. 이 문서는 2015년 5월 28일 (목) 06:25에 마지막으로 바뀌었습니다. Brachistochrone problem The classical problem in calculus of variation is the so called brachistochrone problem1 posed (and solved) by Bernoulli in 1696. , AMPL: a modeling language for mathematical programming, 2003) makes it tempting to formulate discretized optimization problems and get solutions to the discretized versions of trajectory optimization problems. With that in mind, we consider a related and equally classic problem. $\endgroup$ - Conifold Jun 26 at 5:16. H Nguyen Eastern Oregon University June 3, 2014 Abstract This paper consists of some detailed analysis of the classic mathematical. Many problems are entirely straightforward,but many others. As Bernoulli noted in the challenge, the fastest curve is in fact not a straight line, even if it is the shortest path between the points. The brachistochrone problem is a seventeenth century exercise in the calculus of variations. What path gives the shortest time with a constant gravitational force? This is famously known at the Brachistochrone problem. For this, he is regarded as one of the founders of the calculus of variations. Robert Gardner The following is a brief list of topics covered in Chapter 11 of Thomas’ Calculus. Brachistochrone Curve. Lessons in Matter and Energy, from WOSU Public Media, is a series of eight learning modules that demonstrate physical science concepts and phenomena. More specifically, the brachistochrone can use up to a complete rotation of the cycloid (at the limit when A and B are at the same level), but always starts at a cusp. When the movement occurs in a homogeneous gravitational field, the brachistochrone is a cycloid with a horizontal base and a point of return that coincides with point A. $\begingroup$ @Astroynamicist Be careful about the distinction with the "classical" Brachistochrone curve (under a uniform gravity field) with the general Brachistochrone problem, which is a class of problems in the field of calculus of variations (or variational calculus). Class Test Paper Format. png 720 × 340; 85 KB. 14 The Brachistochrone Problem. The classical problem in calculus of variation is the so called brachistochrone problem1 posed (and solved) by Bernoulli in The brachistochrone problem asks us to find the "curve of quickest descent," and so it would be particularly fitting to have the quickest possible solution. n maths the curve between two points through which a body moves under the force of gravity in a shorter time than for any other curve; the path of quickest. In the work a systematic method is developed for the dynamic analysis of structures with sliding isolation, which is a highly non-linear dynamic problem. Given two points Aand B, nd the path along which an object would slide (disregarding any friction) in the. missile interception problem (Figure 1a) is analogous to the brachistochrone problem if an axis inversion is performed (Figure 1b). Use the brachistochrone and tautochrone properties of a cycloid to make an actual slide track in amusement parks. However, formatting rules can vary widely between applications and fields of interest or study. In fact, the solution, which is a segment of a cycloid , was found by Leibniz, L'Hospital, Newton, and the two Bernoullis. Light I Reflection. Example 5 (The Brachistochrone problem), from Greek brachistos (shortest) and chronos (time)): A particle with mass m starts from rest and glides without friction under the influence of gravity from the point (x 1,y 1) to the point (x 2,y 2) along the curve y=y(x) in the xy-plane. Boas Grading: grades based on homework, midterm and a final. The summer program includes academic, business, and cultural aspects and offers a talented group of multidisciplinary students the opportunity to prepare themselves for the ever-changing labour market and makes them aware of the opportunities that China offers. Variational methods applied to eigenvalue problems (Students will apply variational methods to solve some classic optimization problems in physics and applied mathematics. In the late 17th century the Swiss mathematician Johann Bernoulli issued a challenge to solve this problem. straight line catenary brachistochrone hyperbolic geodesic ds ds ds d ZAds/y y y y v2=2gy. The solution of the brachistochrone problem (Johann Bernoulli, 1696) served as the starting point for the development of the calculus of variations. References. We constrain the particle to follow a path (r; ;’) = (r( ); ( );’( )), where is an arbitrary parameter. In this article, we will propose and solve a new problem of the General Brachistochrone Curve (GBC),. This was one of the earliest problems posed in the Calculus of Variations. Le mot brachistochrone désigne une courbe dans un plan vertical sur laquelle un point matériel pesant placé dans un champ de pesanteur uniforme, glissant sans frottement et sans vitesse initiale, présente un temps de parcours minimal parmi toutes les courbes joignant deux points fixés : on parle de problème de la courbe brachistochrone. There is an optimal solution to this problem, and the path that describes this curve of fastest descent is given the name Brachistochrone curve (after the Greek for shortest 'brachistos' and time 'chronos'). B Singh and R Kumar, Brachistochrone problem in nonuniform gravity, Indian J. What shape of wire will get the bead from point A to point. Brachistochrone Curve. On August 18th, 2012, the channel uploaded a video titled "What If Everyone JUMPED At Once?", which gathered upwards of 23 million views and 38,000 comments over the next five years (shown below, left). If you’ve ever taken a science or math class, you’ve probably seen the name "Bernoulli" -- and maybe you assumed it was one person, but that family had a squad of mathematicians. The Brachistochrone curve is the. Curve of fastest descent, is the curve that would carry an idealized point-like body, starting at rest and moving along the curve, without friction, under constant gravity, to a given end point in the shortest time. I wonder how I can solve the Brachistochrone problem for 3 points? The matter starts from point A that is the highest point and it must pass from B and must finish with point C. Download The Brachistochrone Curve: The Problem of Quickest Descent book pdf free download link or read online here in PDF. 3 : note1: Aug. In this article, we will propose and solve a new problem of the General Brachistochrone Curve (GBC),. The Brachistochrone problem, which describes the curve that carries a particle under gravity in a vertical plane from one height to another in the fastest time, is one of the most famous studies. By the right-hand rule for forces, this field produces a force F 2 i on q 2 that is in the x direction. We will reduce them to a uni ed formulation, and we will then solve them analytically and numerically. ” The problem can be stated as follows:. The brachistochrone problem is one of the most famous in analysis. Brachistochrone problem for 3 points. On the other hand, computation times may get longer, because the problem can to become more non-linear and the jacobian less sparse. In this paper we consider several gen-eralizations of the classical brachistochrone problem in which friction is considered. millersville. The brachistochrone problem marks the beginning of the calculus of variations which was further. The brachistochrone problem made elementary Erik Balder, Mathematical Institute, University of Utrecht, the Netherlands February 27, 2002 Johann Bernoulli's brachistochrone problem can be solved completely by using only standard calculus and the Cauchy-Schwarz inequality. Die Brachistochrone minimiert das Funktional J[y] = Z P 2 P1 ds v = Z x 2 x1 s 1 +y′(x)2 2g(y1 −y(x)), (10. There is an optimal solution to this problem, and the path that describes this curve of fastest descent is given the name Brachistochrone curve (after the Greek for shortest 'brachistos' and time 'chronos'). I am not hating on vsauce - I am just saying that 3brown1blue did a much better job and did it earlier, but vsauce gets posted everywhere on the internet and is the one remembered with the explanation of the brachistochrone problem. In this paper we consider several gen-eralizations of the classical brachistochrone problem in which friction is considered. The solution of the brachistochrone problem (Johann Bernoulli, 1696) served as the starting point for the development of the calculus of variations. 3 g acceleration. Having studied Huygens solutions, Johann Bernoulli (1667–1748), investigated the brachistochrone problem and offered a challenge, specifically aimed at his older brother Jacob (1654–1705). For the calculus problem the value of the derivative j0 is zero at the extremum ˆx, j0(ˆx) = 0. Tien-Tsan Shieh (Institute of MathematicsAcademic Sinica)Calculus of Variation and its Application July 14, 2011 7 / 1. The general setup: functionals and boundary conditions; isoperimetric problems, geodesic problems Minimizing in a linear space; directional derivatives; convex functions Convex functionals and calculus of variations; variations; sufficient conditions for minimum. Practice online or make a printable study sheet. Tautochrone Problem Find the curve down which a bead placed anywhere will fall to the bottom in the same amount of time. Huffman Institution: Boeing Computer Services Title: Sparse nonlinear programming test problems (Release 1. This page contains a model of the classical Brachistochrone problem (Johann Bernoulli, 1696), see e. Given two points Aand B, nd the path along which an object would slide (disregarding any friction) in the. 7 Notes and references for Chapter 2. One of the famous problems in teh history of mathematics is the brachistochrone problem: to flnd the curve along which a particle will slide without friction in the minimum time from one. Open Access journals and articles. Their solutions not only giveimplicit information as to their mathematieal skills and cleverness, but also are worthwhile beeause oí their. Posts about Brachistochrone written by soyoungsocurious. Thismeansthatanyforceexertedby. On the brachistochrone In a resistant medium while a body is attracted to a centre of forces in one way or another By the author L. The solution of the Brachistochrone problem is an inverted cycloid with the bead released from the top left cusp. The curve that is covered in the least time is a brachistochrone curve. Laird Hamilton working on his brachistochrone problem. In the late 17th century the Swiss mathematician Johann Bernoulli issued a challenge to solve this problem. Tien-Tsan Shieh (Institute of MathematicsAcademic Sinica)Calculus of Variation and its Application July 14, 2011 7 / 1. Let's Tackle a Classic, Wicked Physics Problem. As described in this project’s write-up, “The brachistochrone curve is a classic physics problem, that derives the fastest path between two points A …. Find the Euler-Lagrange equation describing the brachistochrone curve for a particle moving inside a spherical Earth of uniform mass density. The Subterranean Brachistochrone A Brachistochrone is a frictionless track that connects two locations and along which an object can get from the first point to the second in minimum time under only the action of gravity. | 2020-01-24T00:16:12 | {
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https://math.stackexchange.com/questions/4316372/let-a-pi2-int-01-frac-sin-pi-x1-sin-pi-xdx-and-b-int-0/4316385 | # Let $A=\pi^2\int_{0}^{1}\frac{\sin(\pi x)}{1 + \sin(\pi x)}dx$ and $B=\int_{0}^{\pi}\frac{x\sin( x)}{1 + \sin( x)}dx$. Find $\frac{A}{B}$.
I am trying to solve the following problem:
Let $$A=\pi^2\int_{0}^{1}\frac{\sin(\pi x)}{1 + \sin(\pi x)}dx \quad \text{and}\quad B=\int_{0}^{\pi}\frac{x\sin( x)}{1 + \sin( x)}dx.$$ Find the value of $$\dfrac{A}{B}$$.
I thought on using integration by parts.
It is clear that $$A=\pi^2\int_{0}^{1}\frac{\sin(\pi x)}{1 + \sin(\pi x)}dx = \pi\int_{0}^{\pi}\frac{\sin( x)}{1 + \sin( x)}dx.$$ If we consider $$f(x) = \dfrac{\sin( x)}{1 + \sin( x)}$$, we have that $$A = \pi\int_{0}^{\pi}f(x)dx$$ and $$B = \int_{0}^{\pi}xf(x)dx$$. Applying integration by parts, we obtain $$B = \int_{0}^{\pi}xf(x)dx = x^2f(x)\Big|_0^\pi - \int_{0}^{\pi} x(f(x)+xf'(x))dx.$$ Since $$f(\pi)=0$$, $$B = -\frac{1}{2}\int_{0}^{\pi} x^2f'(x)dx.$$ Likewise, we get that $$A = -\pi\int_{0}^{\pi}xf'(x)dx.$$ However, I don't see how to advance using integration by parts.
Can you give me some advice to find the desired value?
Lemma: $$\int_{0}^{\pi} xf(\sin(x))\, dx = \frac{\pi}{2}\int_{0}^{\pi} f(\sin(x)) \, dx$$ Proof: Taking the substitution $$u = \pi -x$$ gives us $$\int_{0}^{\pi} xf(\sin(x))\, dx = \int_{\color{blue}{\pi}}^{\color{blue}{0}}(\pi - u)f(\sin(\pi - u)) (\color{blue}{- \, du}) = \int_{0}^{\pi} (\pi - u)f(\sin(\pi - u)) \, du$$ but since $$\sin(\pi - u)=\sin(u)$$ from the definition of the sine function, we see that $$\require{cancel}$$ \begin{align} \underbrace{ \int_{0}^{\pi} xf(\sin(x))\, dx }_{\color{purple}{I}} = \int_{0}^{\pi} (\pi - u)f(\sin(u)) \, du &= \pi\int_{0}^{\pi} f(\sin(u)) \, du - \underbrace{\int_{0}^{\pi} uf(\sin(u)) \, du}_{\color{purple}{I}}\\ \implies 2 \color{purple}{I} &= \pi\int_{0}^{\pi} f(\sin(u))\, du\\ \implies \int_{0}^{\pi} xf(\sin(x))\, dx &= I = \frac{\pi}{2}\int_{0}^{\pi} f(\sin(u)) \, du \end{align} Q.E.D.
With the previous lemma the problem becomes simple. Notice that $$\frac{\sin(x)}{1+ \sin(x)}$$ is indeed $$f(x) = \frac{x}{1+x}$$ composed with $$\sin(x)$$, so we can apply the lemma to integral $$B$$ and get that $$B = \int_{0}^{\pi} x\underbrace{\frac{\sin(x)}{1+ \sin(x)}}_{f(\sin(x))} \, dx = \frac{\pi}{2} \int_{0}^{\pi} \frac{\sin(x)}{1+ \sin(x)} \,dx \tag{1}$$ Now, for $$A$$ we take the substitution $$u = \pi x$$. This gives $$A = \pi^2 \int_{0}^{1}\frac{\sin(\pi x)}{1+ \sin(\pi x)}\, dx = \pi^{\cancel{2}}\int_{\color{blue}{0}}^{\color{blue}{\pi}}\frac{\sin(u)}{1+ \sin(u)}\left( \frac{\color{blue}{du}}{\cancel{\color{blue}{\pi}}}\right)=\pi \int_{0}^{\pi} \frac{\sin(u)}{1+ \sin(u)} \,du \tag{2}$$ And combining equations $$(1)$$ and $$(2)$$ we get $$\frac{A}{B} = \frac{\cancel{\pi} \cancel{\int_{0}^{\pi} \frac{\sin(u)}{1+ \sin(u)} \,du}}{\frac{\cancel{\pi}}{2} \cancel{\int_{0}^{\pi} \frac{\sin(x)}{1+ \sin(x)} \, dx}} = \boxed{2}$$
$$B+B=\pi\int\dfrac{\sin x}{1+\sin x}dx$$
Set $$x=\pi y\implies dx=\pi dy$$ to find $$B+B=A$$ | 2022-06-26T00:13:36 | {
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https://math.stackexchange.com/questions/1751936/how-to-prove-that-sum-k-0n-1-cos-left-frac2-pi-kn-phi-right-0 | # How to prove that $\sum_{k=0}^{n-1}\cos\left(\frac{2\pi k}{n}+\phi\right)=0$ [duplicate]
Prove that $\displaystyle\sum_{k=0}^{n-1}\cos\left(\frac{2\pi k}{n}+\phi\right)=0$ for $n\in\mathbb{N},n>1$
I'm thinking at a demonstration by induction, as base case $n=2$
$$\sum_{k=0}^{n-1}\cos\left(\frac{2\pi k}{n}+\phi\right)=\sum_{k=0}^{1}\cos\left(\frac{2\pi k}{n}+\phi\right)=\cos\phi+\cos(\pi+\phi)=\cos\phi-\cos\phi=0$$ but I don't have any idea to inductive step, how can someone prove this?
I'm thinking use this to proof that $\displaystyle\sum_{k=0}^{n-1}\alpha^{k}=0$ where $\alpha=\exp\left(\frac{2\pi}{n}\iota\right)$, thanks for any help.
## marked as duplicate by lab bhattacharjee trigonometry StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 21 '16 at 2:05
• You probably mean $=0$ ? – Yves Daoust Apr 20 '16 at 22:29
• Just use the sum of geometric series, and that $\alpha^n=1$. – Berci Apr 20 '16 at 22:35
• Roots of unity maybe? – Aritra Das Apr 20 '16 at 22:36
Your sum is the real part of
$$\sum_{k=0}^{n-1}e^{i(2\pi k/n+\phi)}=e^{i\phi}\sum_{k=0}^{n-1}\left(e^{i2\pi /n}\right)^k=e^{i\phi}\frac{e^{in2\pi/n}-1}{e^{i2\pi/n}-1}=0.$$
• and so simple that I do not know how I did not think about it :c – cand Apr 20 '16 at 22:35
\begin{align} \sum_{k=0}^{n-1}\cos\left(\frac{2\pi k}{n}+\phi\right) &=\mathrm{Re}\left(\sum_{k=0}^{n-1}e^{\frac{2\pi ik}n+i\phi}\right)\\ &=\mathrm{Re}\left(e^{i\phi}\sum_{k=0}^{n-1}e^{\frac{2\pi ik}n}\right)\\ &=\mathrm{Re}\left(e^{i\phi}\frac{1-e^{\frac{2\pi in}n}}{1-e^{\frac{2\pi i}n}}\right)\\ &=\mathrm{Re}\left(e^{i\phi}\frac{1-e^{2\pi i}}{1-e^{\frac{2\pi i}n}}\right)\\[6pt] &=\mathrm{Re}(0)\\[12pt] &=0 \end{align} As long as $n\ge2$, $1-e^{\frac{2\pi i}n}\ne0$.
Since the sum of the $n$-th roots of unity is zero, we have: $$\sum_{k=0}^{n-1}\cos\left(\frac{2\pi k}{n}\right) = \sum_{k=0}^{n-1}\sin\left(\frac{2\pi k}{n}\right) = 0.\tag{1}$$ For a given $n$, the function: $$f_n(\phi) = \sum_{k=0}^{n-1}\cos\left(\frac{2\pi k}{n}+\phi\right)$$ is quite trivially an analytic function (as the sum of $n$ analytic functions), but due to $(1)$ we have $f_n(0)=f_n'(0)=f_n''(0)=\ldots=0$, hence $\color{red}{f_n\equiv 0}$.
As an alternative, for every $n$ we have that $f_n(\phi)$ is a solution of the differential equation $f''(\phi)+f(\phi)=0$ fulfilling the constraints $f(0)=f'(0)=0$. The solutions are unique and global by the Cauchy-Lipschitz theorem, hence the same conclusion as above holds.
• By the angle addition formula, you directly have $f_n(\phi)=C\cos(\theta)-S\sin(\theta)$. – Yves Daoust Apr 20 '16 at 22:54
• @YvesDaoust: right for sure, good point. – Jack D'Aurizio Apr 20 '16 at 22:55 | 2019-06-17T05:02:18 | {
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https://math.stackexchange.com/questions/2760008/degree-of-the-eighth-vertex-given-the-other-degrees/2760059 | # Degree of the eighth vertex given the other degrees
Consider a graph with $8$ vertices. If the degrees of seven of the vertices are $1,2,3,4,5,6,7$, find the degree of the eighth vertex. I also have to check the graph's planarity and its chromatic number.
I know that the sum of degrees of vertices is twice the number of edges, but that is not really helping here. If I get the degree of the eighth vertex then I could try checking for planarity and chromatic number. But hints about those are also welcome.Thank you.
• Hint: Just draw the graph. First draw all edges incident with the vertex of degree $7.$ Then draw all remaining edges incident with the vertex of degree $6,$ and so on. I believe you will find that the graph has clique number and chromatic number equal to $5$ so it is not planar.
– bof
Apr 30, 2018 at 9:51
• I have tried drawing the graph, but isn't there any simpler more elegant method?
– QED
Apr 30, 2018 at 9:56
• What do you mean, you "tried" drawing the graph? Did you succeed in drawing the graph, and what did you find for the degree of the eighth vertex?
– bof
Apr 30, 2018 at 9:59
• The vertex of degree $7$ must be adjacent to all other vertices (including the mystery vertex). Next, the vertex of degree $6$ can't be adjacent to the vertex of degree $1$ so it must be adjacent to all other vertices. Next, the vertex of degree $5$ can't be adjecent to the vertices of degree $1$ or $2$ so it must be adjacent to all the other vertices. Finally the vertex of degree $4$ is not adjacent to the vertices of degree $\le3$ so must be adjacent to all other vertices. And we're done! The mystery vertex has degree $4,$ and two vertices are adjacent iff the sum of their degrees is $\ge8.$
– bof
Apr 30, 2018 at 10:06
• This is very similar to puzzles such as puzzling.stackexchange.com/questions/59507/… Apr 30, 2018 at 15:52
If $d$ is the missing degree, the Handshaking Lemma implies that $1+2+3+4+5+6+7+d=28+d$ is even, so $d$ is even. Since the degree-$7$ vertex is adjacent to it, $d>0$ and thus $d \in \{2,4,6\}$.
If $d=6$, then the vertex of degree $7$ (which is adjacent to all other vertices) and the two vertices of degree $6$ (which are adjacent all all other vertices, except the degree-$1$ vertex) are adjacent to the vertex of degree $2$, giving a contradiction.
If $d=2$, then the vertices of degrees $6$ and $7$ are adjacent to both of the two vertices of degree $2$, and the vertex of degree $5$ is adjacent to one of the vertices of degree $2$, giving a contradiction.
If $d=4$, then the graph below has degree sequence $(1,2,3,4,4,5,6,7)$:
(I mark the vertices with their degrees. I also give it a $5$-coloring.)
Actually, it's unique up to isomorphism.
The vertices of degrees $4,4,5,6,7$ induce a $K_5$, so it's not planar by Kuratowski's theorem (or Wagner's theorem), and its chromatic number is not less than $5$. I also give it a $5$-coloring, so its chromatic number is $5$.
In fact, computing it chromatic polynomial, we get $$x(x-1)^2(x-2)^2(x-3)^2(x-4).$$ By substituting in $x=5$, we count $2880$ distinct $5$-colorings.
This can be checked by hand: there are $x(x-1)(x-2)(x-3)(x-4)$ ways to $x$-color the $K_5$, then, since the yet-to-be-colored vertices are only adjacent to vertices in the $K_5$, the degree-$3$ vertex is colored using $x-3$ colors, the degree-$2$ vertex is colored using $x-2$ colors, and the degree-$1$ vertex is colored using $x-1$ colors.
Drawing the graph works, but here is a more formal argument.
The degree 7 vertex must be connected to each of the other vertices. So the degree 1 vertex is connected to the degree 7 vertex only.
Therefore the degree 6 vertex must be connected to every vertex apart from the degree 1 vertex. So the degree 2 vertex is connected to the degree 6 and 7 vertices only.
Therefore the degree 5 vertex must be connected to every vertex apart from the degree 1 and 2 vertices. So the degree 3 vertex is connected to the degree 5, degree 6 and 7 vertices only.
Therefore the degree 4 vertex is connected to the degree 5, 6 and 7 vertices but not to the degree 1, 2 and 3 vertices. To have degree 4 it must also be connected to the 8th vertex.
Therefore the 8th vertex is connected to the degree 4, 5, 6 and 7 vertices but not to the degree 1, 2 and 3 vertices. So the 8th vertex must have degree 4.
As a slightly more general result, I can suggest the following method (which is not too different from what bof described in the comments): let $x$ be the degree of the last vertex, we know it has to be in the interval $[1,7]$. Consider the graph $G'$ obtained from $G$ by removing the vertex of maximal degree: then we obtain a subgraph consisting of exactly one isolated point (if $x$ was isolated in $G'$, then there could be no vertex of degree 6 in $G$), and all the remaining vertices have as degree some value in the interval $[1,5]$. By iterating this procedure, it is clear that the $x$ is the middle value of the initial interval, so $x=4$.
(I said slightly more general because, in my opinion, this procedure makes it clear that the result holds for all the similar problems with an interval of type $[1,2n+1]$).
This method can also be used to find some informations about the chromatic number: at each step of the process, we remove a vertex that cannot have the same color of any other vertex in the subgraph $G'$. | 2022-06-28T09:51:06 | {
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https://math.stackexchange.com/questions/2422576/im-not-sure-im-correct-sum-n-1-infty-x-n-leq-sum-n-1-infty-x | # I'm not sure I'm correct: $\|\sum_{n=1}^\infty x_n\| \leq \sum_{n=1}^\infty \|x_n\|$
I was solving this Functional Analysis problem, but I'm not sure I'm correct on this one, the problem is:
Let $\big(E,\|\bullet\|\big)$ be a normed vector space, and $(x_n)_n$ a sequence in $E$ such that $\sum_{n=1}^\infty x_n$ converges. Show that $$\left\|\sum_{n=1}^\infty x_n\right\| \leq \sum_{n=1}^\infty\|x_n\|$$
My solution:
We know that the following inequality holds for any $m\in\mathbb{N}$ (just using the triangular inequality):
$$\left\|\sum_{n=1}^m x_n\right\| \leq \sum_{n=1}^m \|x_n\|$$
Since the left side converges, the inequality still holds in the limit $m\rightarrow\infty$, hence we have the result.
My doubts:
The only thing I am not totally sure is my last statement, because I can't know if the right side converges, so how can I compare the two things?
But then I thought I can use the following argument:
The sequence $\left(\sum_{n=1}^m \|x_n\|\right)_m$ is clearly a monotone (non-decreasing) real sequence, so if it is bounded it converges and my result is correct, the other option is that the sequence is unbounded, in which case
$$\sum_{n=1}^\infty \|x_n\|=\infty$$
And then it makes sense to write:
$$\left\|\sum_{n=1}^\infty x_n\right\| < \infty$$
Can someone tell me if my reasoning is correct? Or if another way would be better to solve this problem?
• Thank you for a question that is well written, and you also give your own thoughts. +1 – Eff Sep 9 '17 at 13:00
If $\sum_{n=1}^\infty \|x_n\|=\infty$ then it is true. So let us assume that $\sum_{n=1}^\infty \|x_n\|<\infty$. So, $$\left\|\sum_{n=1}^m x_n\right\| \leq \sum_{n=1}^m \|x_n\| \le \left\|\sum_{n=1}^m x_n\right\| \leq \sum_{n=1}^m \|x_n\|=l$$ Since, norm is a continuous function so $$\lim_{m\to \infty}\left\|\sum_{n=1}^m x_n\right\|=\left\|\lim_{m\to\infty} \sum_{n=1}^m x_n\right\| =\left\|\sum_{n=1}^\infty x_n\right\| \le \lim_{m\to \infty} l=l$$
You're quite right that the right hand side need not converge : a simple example in the reals is $x_n = \frac{(-1)^n}{n}$ where the series converges, but not absolutely. So you can indeed make the distinction between $+\infty$ and finite for the right hand series. $+\infty$ is no problem for the inequality, and the finite case follows from the triangle inequality, as said. Or just use (in your first version) that in the extended reals $x_n \le y_n$, $x\to x, y\to y$ also implies $x \le y$ (this holds in all ordered topological spaces, as $\{(p,q) \in X^2: p \le q \}$ is closed in $X^2$). The right hand limit always exists by monotonicity (and completeness of $[-\infty, +\infty]$), he left hand side as $\sim_n x_n$ converges and $\|.\|$ is continuous.
Using the continuity of $\|\cdot\|$, we have: $$\left\|\sum_{n=1}^\infty x_n\right\| = \left\|\lim_{m\to\infty}\sum_{n=1}^mx_n\right\| = \lim_{m\to\infty}\left\|\sum_{n=1}^m x_n\right\| \leq \lim_{m\to\infty}\sum_{n=1}^m \left\|x_n\right\| = \sum_{n=1}^\infty \left\|x_n\right\|$$ | 2021-03-03T20:58:26 | {
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https://mathematica.stackexchange.com/questions/103871/goldbach-partition | # Goldbach Partition
I want to check the Goldbach conjecture for a big number of $n$, but I don't know how to define this in Mathematica.
These are my questions:
1. Find a pair of primes $(p,q)$ for every even integer $n$, such $p+q=n$, using Mathematica.
and
1. How to calculate number of ways to write an even number $n$ as the sum of two primes?
Take a look at IntegerPartitions, although it relies on brute-force enumeration that is unlikely to scale well.
f1[n_] := IntegerPartitions[n, {2}, Prime @ Range @ PrimePi @ n, 1]
f2[n_] := Length @ IntegerPartitions[n, {2}, Prime @ Range @ PrimePi @ n]
Test:
f1[3412]
{{3407, 5}}
f2[3412]
43
### Further experiments
Anton Antonov's recent answer surprised me, in that for larger n values his use of Select[FrobeniusSolve[{1, 1}, n], And @@ Map[PrimeQ, #] &] is faster than f2 above. (In version 10.1 under Windows.) It seems that long lists for the third parameter of IntegerPartitions is slow.
At the cost of increased memory consumption over f2 but less memory consumption than FrobeniusSolve we may enumerate with IntegerPartitions and then filter:
f3[n_] := Cases[IntegerPartitions[n, {2}], {__?PrimeQ}]
Timings on a fairly large integer:
f2[1787834] // AbsoluteTiming
Length @ f3[1787834] // AbsoluteTiming
{7.73629, 6643}
{0.846121, 6643}
And Anton's method with the benefit of somewhat faster filtering (as used above):
Length @ Cases[FrobeniusSolve[{1, 1}, 1787834], {__?PrimeQ}] // AbsoluteTiming
{2.2008, 13285}
Note that in this output each solution is enumerated twice which I conjecture is the reason for the lower performance relative to f3.
I propose to use FrobeniusSolve. It seems it gives results fairly quickly. How large is the number $n$ ?
AbsoluteTiming[
res = Select[FrobeniusSolve[{1, 1}, 120022], And @@ Map[PrimeQ, #] &];
]
(* {0.340291, Null} *)
Row[{"Number of prime partitions: ", Length[res]}]
(* "Number of prime partitions: ", 1668 *)
Row[{"Sample: ", Take[SortBy[res, Abs[Subtract @@ #] &], 4]}]
(* "Sample: ", {{59981, 60041}, {60041, 59981}, {59921,
60101}, {60101, 59921}} *)
Here is another computation with a larger number:
In[9]:= AbsoluteTiming[
res = Select[FrobeniusSolve[{1, 1}, 7878344],
And @@ Map[PrimeQ, #] &];
]
Out[9]= {25.3882, Null}
• Interesting variation. I didn't expect this to be competitive with IntegerPartitions, but it seems that on larger numbers it is. Also I believe that your filter can be improved; try: AbsoluteTiming[ Cases[FrobeniusSolve[{1, 1}, 120022], {__?PrimeQ}]; ] – Mr.Wizard Dec 4 '16 at 11:52
• @Mr.Wizard Thanks for pointing out the speed-up with Cases / __?PrimeQ] ! I was not aware that it can be that significant. – Anton Antonov Dec 4 '16 at 22:27
As @Mr.Wizard showed, IntegerPartitions answers both your questions directly, and he warned that it will be slow for large $n$ because it calculates all possible partitions. There is a faster answer to your first question of finding just one partition of even $n=p+q$. Set $p\le q$, and note that usually $p$ is a small prime. The function GoldbachTest uses a While loop to check a sequence of small primes in the input list of candidates $p$. If the list is exhausted, the failure condition $\{0,0\}$ is returned.
GoldbachTest[n_?EvenQ, p_List] :=
Block[{m = Length[p], i = 1},
While[i <= m && CompositeQ[n - p[[i]]], i += 1];
If[i > m, {0, 0}, {#, n - #} &[p[[i]]]]
]
Given a list $p$ of the first $k$ primes, there is a smallest even $n$ which cannot be represented as $n=p+q$, with prime $q$. The sequence begins $\{6,12,30,98,98,98,98,220,308,...\}$, which is Sloane's A152522. This page links to a paper by Granville, Van de Lune, and te Riele, where they conjecture that the smallest prime $p$ in a Goldbach partition of even $n=p+q$ is $p<1.603*\log[n]^2 \log[\log[n]]$. They confirmed their conjecture for even $n \le 2*10^{10}$.
Thus, for an efficient test of Goldbach's conjecture from $n=n_{min}$ to $n=n_{max}$ try the following. Using ParallelTable will be even faster.
GoldbachTestList[nmin_?EvenQ, nmax_Integer] :=
With[{p = Prime[Range[PrimePi[1.603*Log[nmax]^2 Log[Log[nmax]]]]]},
Table[GoldbachTest[n, p], {n, nmin, nmax, 2}]
]
Timings show that GoldbachTestList is over 100 times faster than the equivalent IntegerPartitions formulation for $n_{max} \ge 10^5$.
• That's nice, +1. – ciao Dec 4 '16 at 4:20
I wanted to summarize solutions to this question based on IntegerPartitions, FrobeniusSolve, and a new hybrid method.
The first method f1[n] based on IntegerPartitions finds pairs of primes summing to n. @Mr.Wizard found this method to be slow for long lists in the third argument. Hence, he proposed function f3[n] to enumerate all pairs then filter.
f1[n_] := IntegerPartitions[n, {2}, Prime@Range@PrimePi@n]
f3[n_] := Cases[IntegerPartitions[n, {2}], {__?PrimeQ}]
For reasonably large n=1787834, function f3[n] is much faster.
With[{n = 1787834}, {AbsoluteTiming[Length@f1[n]], AbsoluteTiming[Length@f3[n]]}]
(* {{10.5924, 6643}, {0.809354, 6643}} *)
The first two methods, f4[n] and f5[n], based on FrobeniusSolve are from @Mr.Wizard and @AntonAntonov, respectively. The third method f6[n] adapts f5[n] by using Pick instead of Select.
f4[n_] := Cases[FrobeniusSolve[{1, 1}, n], {__?PrimeQ}]
f5[n_] := Select[FrobeniusSolve[{1, 1}, n], And @@ Map[PrimeQ, #] &]
f6[n_] := Pick[#, And @@@ PrimeQ[#]] &[FrobeniusSolve[{1, 1}, n]]
Pick is faster than Select, but the fastest formulation is f4[n] with Cases. The number of solutions is twice as large because both solutions {p,q} and {q,p} are returned. Since half of 1787834 is prime, there is also a solution {p,p} which is not duplicated.
With[{n = 1787834}, {
AbsoluteTiming[Length@f4[n]],
AbsoluteTiming[Length@f5[n]],
AbsoluteTiming[Length@f6[n]]}]
(* {{2.55104, 13285}, {5.35851, 13285}, {3.78575, 13285}} *)
The hybrid solution recognises that FrobeniusSolve[{1,1},n] simply returns pairs of integers {k,n-k} for k=0,1,...,n, and tests each pair for primality. Function f7[n] begins with a list of primes p about half as long as for f1[n], then tests for prime n-p.
f7[n_] := Block[{p = Prime[Range[PrimePi[Quotient[n, 2]]]]},
Transpose[{#, n - #} &[Pick[p, PrimeQ[n - p]]]]
]
Reversed solutions are not returned by f7[n], so the count is again 6643. The timing is about 10 times faster than the best solution so far, f3[n].
AbsoluteTiming[Length@f7[1787834]]
(* {0.062481, 6643} *) | 2019-08-18T02:16:09 | {
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https://math.stackexchange.com/questions/3380691/plane-and-span-vector | # Plane and span Vector
I have the following problem: the plane equation is given by $$x_1-2x_2+4x_3=0$$ I need to come up with the two vectors that spans the plane. So the normal from the equation can be written as $$(1,-2,4)$$ So $$(1,-2,4)$$ has to be equal to the cross product of two vectors:
$$1=a_2b_3-a_3b_2$$ $$-2=a_3b_1-a_1b_3$$$$4=a_1b_2-a_2b_1$$
So I take the random approach and set $$b_1$$ to $$0$$, so I obtain $$\frac{b_3}{b_2}=\frac12$$.So $$b_3=1$$, and $$b_2=2$$. If I replace these values in the system, I obtain $$a_1=2$$, $$a_2=3$$, $$a_3=1$$. So two vectors can be span ($$[2,3,1]$$ and $$[0,2,1]$$). However, the answer in the back of the book is different. Can someone explain where I make mistake?
• There's no error. There's an infinity of pairs of spanning vectors. Oct 4, 2019 at 16:57
• What are the answers in the back of book? Perhaps the spans are the same, but they wrote the vectors in the basis as some linear combination of your vectors?
– jl00
Oct 4, 2019 at 16:57
• jl00, it is [-4 0 1 ] and [2 1 0] Oct 4, 2019 at 17:41
• Bernard thanks very much Oct 4, 2019 at 20:17
As noted in comments, there’s an infinite number of solutions to this problem. However, observe that for any nonzero vector $$(a,b,c)$$, the vectors $$(0,c,-b)$$, $$(-c,0,a)$$ and $$(b,-a,0)$$ are all orthogonal to $$(a,b,c)$$, and at least two of them are nonzero. (These vectors are the cross products of $$(a,b,c)$$ with the standard basis vectors.) Their negations are, of course, also orthogonal to $$(a,b,c)$$.
So, for your normal vector $$(1,-2,4)$$ We can immediately find three nonzero vectors orthogonal to it: $$(0,4,2)$$, $$(-4,0,1)$$ and $$(-2,-1,0)$$. The second of these and the negation of the last one give you the book’s answer. | 2022-06-26T20:32:10 | {
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https://math.stackexchange.com/questions/22159/how-many-n-times-m-binary-matrices-are-there-up-to-row-and-column-permutation/881965 | # How many $n\times m$ binary matrices are there, up to row and column permutations?
I'm interested in the number of binary matrices of a given size that are distinct with regard to row and column permutations.
If $\sim$ is the equivalence relation on $n\times m$ binary matrices such that $A \sim B$ iff one can obtain B from applying a permutation matrix to A, I'm interested in the number of $\sim$-equivalence classes over all $n\times m$ binary matrices.
I know there are $2^{nm}$ binary matrices of size $n\times m$, and $n!m!$ possible permutations, but somehow I fail to get an intuition on what this implies for the equivalence classes.
• Intuitively, the average matrix has trivial stabilizer, so there ought to be roughly 2^{nm}/n!m! equivalence classes. This is probably a very hard question in general. – Qiaochu Yuan Feb 15 '11 at 12:02
• There is the set $S:=[m]\times[n]$ on which the group $G:=S_m\times S_n$ acts, and we have to color $S$ with colors $0$ and $1$. How many colorings are there when two colorings that differ by a $g\in G$ are considered the same? Now there is a famous theory that addresses exactly this kind of questions; it is called Polya counting theory. I could imagine that your problem is a standard example in the field. – Christian Blatter Feb 15 '11 at 13:24
• If you view A as the incidence matrix of an unweighted undirected bipartite graph. Then I think the question you're asking is how many unique bipartite graphs up to isomorphism are there with the two vertex groups having n,m vertices respectively. – JSchlather Feb 15 '11 at 17:53
This is solved here using Pólya enumeration theory. For the square case ($n=m$), see this sequence.
Comment: I found these by searching for $1,2,7$ in the OEIS.
• The first is broken now but I would love to know the answer. Is there an alternative source? – Raphael Feb 21 '14 at 20:38
• @Raphael Updated the link. – Yuval Filmus Feb 21 '14 at 20:46
Here is a computational contribution that treats the case of a square matrix. As pointed out this problem can be solved using the Polya Enumeration Theorem. In fact if we are interested only in counting these matrices, then the Burnside lemma will suffice. We just need to compute the cycle index of the group acting on the slots of the matrix.
These cycle indices are easy to compute and we do not need to iterate over all $(n!)^2$ pairs of permutations (acting on rows and columns) but instead it is sufficient to iterate over pairs of terms from the cycle index $Z(S_n)$ of the symmetric group $S_n$ according to their multiplicities to obtain the cycle index $Z(Q_n)$ of the combined action on rows and columns. The number of terms here is the much better count of the number of partitions of $n$ squared (upper bound).
Now for a pair of cycles, one of length $l_1$ from a row permutation $\alpha$ and another of length $l_2$ from a column permutation $\beta$ their contribution to the disjoint cycle decomposition product for $(\alpha,\beta)$ in the cycle index $Z(Q_n)$ is by inspection $$a_{\mathrm{lcm}(l_1, l_2)}^{l_1 l_2 / \mathrm{lcm}(l_1, l_2)} = a_{\mathrm{lcm}(l_1, l_2)}^{\gcd(l_1, l_2)}.$$
The algorithm now becomes very simple -- iterate over pairs of terms as described above, collect the contribution from each pair of cycles and add it to the cycle index being computed.
This gives the following cycle indices (only the first four are shown):
$$Z(Q_2) = 1/4\,{a_{{1}}}^{4}+3/4\,{a_{{2}}}^{2},$$
$$Z(Q_3) = 1/36\,{a_{{1}}}^{9}+1/6\,{a_{{1}}}^{3}{a_{{2}}}^{3}+1/4\,a_{{ 1}}{a_{{2}}}^{4}+2/9\,{a_{{3}}}^{3}+1/3\,a_{{3}}a_{{6}},$$
$$Z(Q_4) = {\frac {{a_{{1}}}^{16}}{576}}+1/48\,{a_{{1}}}^{8}{a_{{2}}}^{4 }+1/16\,{a_{{1}}}^{4}{a_{{2}}}^{6}+1/36\,{a_{{1}}}^{4}{a_{{3} }}^{4}+{\frac {17\,{a_{{2}}}^{8}}{192}}\\+1/6\,{a_{{1}}}^{2}a_{ {2}}{a_{{3}}}^{2}a_{{6}}+1/9\,a_{{1}}{a_{{3}}}^{5}+1/12\,{a_{ {2}}}^{2}{a_{{6}}}^{2}+{\frac {13\,{a_{{4}}}^{4}}{48}}+1/6\,a _{{4}}a_{{12}},$$
and
$$Z(Q_5) = {\frac {{a_{{1}}}^{25}}{14400}}+{\frac {{a_{{1}}}^{15}{a_{{2} }}^{5}}{720}}+{\frac {{a_{{1}}}^{9}{a_{{2}}}^{8}}{144}}+{ \frac {{a_{{1}}}^{10}{a_{{3}}}^{5}}{360}}+{\frac {{a_{{1}}}^{ 5}{a_{{2}}}^{10}}{480}}\\+1/48\,{a_{{1}}}^{3}{a_{{2}}}^{11}+{ \frac {a_{{1}}{a_{{2}}}^{12}}{64}}+1/36\,{a_{{1}}}^{6}{a_{{2} }}^{2}{a_{{3}}}^{3}a_{{6}}+1/36\,{a_{{1}}}^{4}{a_{{3}}}^{7}+{ \frac {{a_{{1}}}^{5}{a_{{4}}}^{5}}{240}}\\+{\frac {{a_{{2}}}^{5 }{a_{{3}}}^{5}}{360}}+1/24\,{a_{{1}}}^{3}a_{{2}}{a_{{4}}}^{5} +1/24\,{a_{{1}}}^{2}{a_{{2}}}^{4}a_{{3}}{a_{{6}}}^{2}+1/36\,{ a_{{2}}}^{5}{a_{{3}}}^{3}a_{{6}}\\+1/16\,a_{{1}}{a_{{2}}}^{2}{a _{{4}}}^{5}+1/24\,{a_{{2}}}^{5}a_{{3}}{a_{{6}}}^{2}+1/18\,{a_ {{2}}}^{2}{a_{{3}}}^{5}a_{{6}}+1/16\,a_{{1}}{a_{{4}}}^{6}\\+1/ 36\,{a_{{2}}}^{2}{a_{{3}}}^{3}{a_{{6}}}^{2}+1/12\,{a_{{1}}}^{ 2}a_{{3}}{a_{{4}}}^{2}a_{{12}}+1/12\,a_{{2}}a_{{3}}{a_{{4}}}^ {2}a_{{12}}+{\frac {13\,{a_{{5}}}^{5}}{300}}\\+1/30\,{a_{{5}}}^ {3}a_{{10}}+1/15\,{a_{{5}}}^{2}a_{{15}}+1/20\,a_{{5}}{a_{{10} }}^{2}+1/10\,a_{{5}}a_{{20}}+1/15\,a_{{10}}a_{{15}}.$$
Evaluating these cycle indices with the variables set to two we quickly obtain the sequence
$$2, 7, 36, 317, 5624, 251610, 33642660, 14685630688,\\ 21467043671008, 105735224248507784,1764356230257807614296,\\ 100455994644460412263071692,19674097197480928600253198363072,\\ 13363679231028322645152300040033513414,\\ 31735555932041230032311939400670284689732948,\ldots$$ which is indeed OEIS A002724.
Note that the cycle indices make it possible to enumerate configurations with more than two possible entries or with entries having different weights. For example, with a 3x3 square and three colors $A,B$ and $C$ we get the generating function
$$Z(Q_3)(A+B+C) = 1/36\, \left( A+B+C \right) ^{9}+1/6\, \left( A+B+C \right) ^{3} \left( {A}^{2}+{B}^{2}+{C}^{2} \right) ^{3}\\+2/9\, \left( {A}^{3}+{B}^{3}+{C}^{3} \right)^{3} +1/4\, \left( A+B+C \right) \left( {A}^{2}+{B}^{2}+{C}^{2 } \right) ^{4}\\+1/3\, \left( {A}^{3}+{B}^{3}+{C}^{3} \right) \left( {A}^{6}+{B}^{6}+{C}^{6} \right)$$
which expands to
$${A}^{9}+{A}^{8}B+{A}^{8}C+3\,{A}^{7}{B}^{2}+3\,{A}^{7}B C+3\,{A}^{7}{C}^{2}+6\,{A}^{6}{B}^{3}+10\,{A}^{6}{B}^{2 }C\\+10\,{A}^{6}B{C}^{2}+6\,{A}^{6}{C}^{3}+7\,{A}^{5}{B}^ {4}+17\,{A}^{5}{B}^{3}C+28\,{A}^{5}{B}^{2}{C}^{2}\\+17\,{ A}^{5}B{C}^{3}+7\,{A}^{5}{C}^{4}+7\,{A}^{4}{B}^{5}+22\, {A}^{4}{B}^{4}C+43\,{A}^{4}{B}^{3}{C}^{2}+43\,{A}^{4}{B }^{2}{C}^{3}\\+22\,{A}^{4}B{C}^{4}+7\,{A}^{4}{C}^{5}+6\,{ A}^{3}{B}^{6}+17\,{A}^{3}{B}^{5}C+43\,{A}^{3}{B}^{4}{C} ^{2}+54\,{A}^{3}{B}^{3}{C}^{3}\\+43\,{A}^{3}{B}^{2}{C}^{4 }+17\,{A}^{3}B{C}^{5}+6\,{A}^{3}{C}^{6}+3\,{A}^{2}{B}^{ 7}+10\,{A}^{2}{B}^{6}C+28\,{A}^{2}{B}^{5}{C}^{2}\\+43\,{A }^{2}{B}^{4}{C}^{3}+43\,{A}^{2}{B}^{3}{C}^{4}+28\,{A}^{ 2}{B}^{2}{C}^{5}+10\,{A}^{2}B{C}^{6}+3\,{A}^{2}{C}^{7}\\+ A{B}^{8}+3\,A{B}^{7}C+10\,A{B}^{6}{C}^{2}+17\,A{B}^{5}{ C}^{3}+22\,A{B}^{4}{C}^{4}+17\,A{B}^{3}{C}^{5}\\+10\,A{B} ^{2}{C}^{6}+3\,AB{C}^{7}+A{C}^{8}+{B}^{9}+{B}^{8}C+3\,{ B}^{7}{C}^{2}+6\,{B}^{6}{C}^{3}+7\,{B}^{5}{C}^{4}\\+7\,{B }^{4}{C}^{5}+6\,{B}^{3}{C}^{6}+3\,{B}^{2}{C}^{7}+B{C}^{ 8}+{C}^{9}.$$
This is the Maple code for this computation. Here we have two slightly different ways of evaluating the count, the first by substituting into the cycle index and the second by skipping the cycle index altogether and evaluating all variables at two during processing. The latter should be used when we are interested ony in the count as opposed to classifying configurations according to the number of each color / value that are present.
pet_cycleind_symm :=
proc(n)
option remember;
if n=0 then return 1; fi;
end;
pet_varinto_cind :=
proc(poly, ind)
local subs1, subs2, polyvars, indvars, v, pot, res;
res := ind;
polyvars := indets(poly);
indvars := indets(ind);
for v in indvars do
pot := op(1, v);
subs1 :=
[seq(polyvars[k]=polyvars[k]^pot,
k=1..nops(polyvars))];
subs2 := [v=subs(subs1, poly)];
res := subs(subs2, res);
od;
res;
end;
pet_cycleind_sqmat :=
proc(n)
option remember;
local sind, cind, term_a, term_b, v_a, v_b,
len_a, len_b, inst_a, inst_b, p;
cind := 0;
if n=1 then
sind := [a[1]];
else
sind := pet_cycleind_symm(n);
fi;
for term_a in sind do
for term_b in sind do
p := 1;
for v_a in indets(term_a) do
len_a := op(1, v_a);
inst_a := degree(term_a, v_a);
for v_b in indets(term_b) do
len_b := op(1, v_b);
inst_b := degree(term_b, v_b);
p := p*a[lcm(len_a, len_b)]
^(gcd(len_a, len_b)*inst_a*inst_b);
od;
od;
cind := cind +
lcoeff(term_a)*lcoeff(term_b)*p;
od;
od;
cind;
end;
v :=
proc(n)
option remember;
local cind, vars, sbl;
cind := pet_cycleind_sqmat(n);
vars := indets(cind);
sbl := [seq(v=2, v in vars)];
subs(sbl, cind);
end;
w :=
proc(n)
option remember;
local sind, count, term_a, term_b, v_a, v_b,
len_a, len_b, inst_a, inst_b, p;
count := 0;
if n=1 then
sind := [a[1]];
else
sind := pet_cycleind_symm(n);
fi;
for term_a in sind do
for term_b in sind do
p := 1;
for v_a in indets(term_a) do
len_a := op(1, v_a);
inst_a := degree(term_a, v_a);
for v_b in indets(term_b) do
len_b := op(1, v_b);
inst_b := degree(term_b, v_b);
p := p*
2^(gcd(len_a, len_b)*inst_a*inst_b);
od;
od;
count := count +
lcoeff(term_a)*lcoeff(term_b)*p;
od;
od;
count;
end;
This MSE Meta Link has many more PET computations by various users. | 2019-05-26T09:31:30 | {
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https://math.stackexchange.com/questions/158219/is-a-matrix-multiplied-with-its-transpose-something-special | # Is a matrix multiplied with its transpose something special?
In my math lectures, we talked about the Gram-Determinant where a matrix times its transpose are multiplied together.
Is $A A^\mathrm T$ something special for any matrix $A$?
• The matrix $A^TA^{-1}$ is generally self similar... – DVD Jun 12 '15 at 3:42
• One of the themes of Gilbert Strang's books is the ubiquity of $A^T A$ and $A^T CA$ (with $C$ positive semidefinite) in mathematics. For example, the adjoint of the gradient operator is the negative divergence operator, and the Laplacian is the divergence of the gradient. In one book Strang states, "I have learned to look for $A^T C A$". – littleO Jan 4 '16 at 19:57
• Is $AA^T = A^TA$? Would it yield the similar meaning? – A-letubby Jun 1 '18 at 3:18
• It's symmetric and I believe positive. – user193319 Apr 11 '19 at 14:14
• @A-letubby No, $AA^T$ is not necessarily $=A^TA$ – lmaooooo Mar 19 at 13:50
The main thing is presumably that $$AA^T$$ is symmetric. Indeed $$(AA^T)^T=(A^T)^TA^T=AA^T$$. For symmetric matrices one has the Spectral Theorem which says that we have a basis of eigenvectors and every eigenvalue is real.
Moreover if $$A$$ is invertible, then $$AA^T$$ is also positive definite, since $$x^TAA^Tx=(A^Tx)^T(A^Tx)> 0$$
Then we have: A matrix is positive definite if and only if it's the Gram matrix of a linear independent set of vectors.
Last but not least if one is interested in how much the linear map represented by $$A$$ changes the norm of a vector one can compute
$$\sqrt{\left}=\sqrt{\left}$$
which simplifies for eigenvectors $$x$$ to the eigenvalue $$\lambda$$ to
$$\sqrt{\left}=\sqrt \lambda\sqrt{\left},$$
The determinant is just the product of these eigenvalues.
• Then you can write $\mathbb R^n\cong A\bot V$. What is $AA^TA?$ and what is $AA^Tv$ for $v\in V$? How does $AA^T$ hence look like? – Simon Markett Jul 25 '12 at 7:57
• What "if $A$ is regular" exactly mean? There seem to be several interpretations on Wikipedia. – qazwsx Mar 11 '14 at 20:21
• Regular means here the same as invertible. – Simon Markett Mar 11 '14 at 20:57
• What if det($AA^T$) = det($A^TA$). Is there a name for such matrices or any other special thing about them? – Rajendra Uppal Mar 30 '14 at 6:10
• @mag, the determinant is multiplicative, so this holds for all square matrices $A$. If $A$ is not square then either side of your equation has to be $0$, as the composition factors through something lower dimensional. – Simon Markett Apr 1 '14 at 14:29
$AA^T$ is positive semi-definite, and in a case in which $A$ is a column matrix, it will be a rank 1 matrix and have only one non-zero eigenvalue which equal to $A^TA$ and its corresponding eigenvector is $A$. The rest of the eigenvectors are the null space of $A$ i.e. $\lambda^TA = 0$.
Indeed, independent of the size of $A$, there is a useful relation in the eigenvectors of $AA^T$ to the eigenvectors of $A^TA$; based on the property that $rank(AA^T)=rank(A^TA)$. That the rank is identical implies that the number of non-zero eigenvectors is identical. Moreover, we can infer the eigenvectors of $A^TA$ from $AA^T$ and vice versa. The eigenvector decomposition of $AA^T$ is given by $AA^Tv_i = \lambda_i v_i$. In case $A$ is not a square matrix and $AA^T$ is too large to efficiently compute the eigenvectors (like it frequently occurs in covariance matrix computation), then it's easier to compute the eigenvectors of $A^TA$ given by $A^TAu_i = \lambda_i u_i$. Pre-multiplying both sides of this equation with $A$ yields
$AA^TAu_i=\lambda_iAu_i$.
Now, the originally searched eigenvectors $v_i$ of $AA^T$ can easily be obtained by $v_i:=Au_i$. Note, that the resulted eigenvectors are not yet normalized.
One could name some properties, like if $B=AA^T$ then
$$B^T=(AA^T)^T=(A^T)^TA^T=AA^T=B,$$
so
$$\langle v,Bw\rangle=\langle Bv,w\rangle=\langle A^Tv,A^Tw\rangle.$$
Something that occurred to me while reading this answer for help with my homework is that there is a pretty common and important special case, if the linear operator A is normal, i.e. $A^TA=AA^T$. Note this is a stronger condition than saying that $A^TA$ is symmetric, which is always true. In this case, we have that $A$ is diagonalizable. As an obvious special case, $A$ is normal if $A$ is Hermitian (symmeric in the real case).
To see that $A$ normal implies $A^TA$ is diagonalizable, let $\lambda$ be a eigenvalue of $A^TA$ corresponding to the eigenvector x. Then $$(Ax,Ax)=(A^TAx,x)=\lambda(x,x)$$ and similarly $$(A^Tx,A^Tx)=(AA^Tx,x)=\lambda(x,x)$$ where I have used the normality of A. Subtracting these two equations, we have $$((A^T-A)x,(A^T-A)x)=0,$$ which by the definition of inner product implies that $$(A^T-A)x=0 =>A^Tx=Ax.$$ Note that of course this is true whenever $A=A^T$ if A is symmetric, but is a more general condition, since x is an eigenvector, rather than an arbitrary vector. Applying A to both sides of this equation, we have $$A^2x=A^TAx=\lambda x$$.
This shows that if x is an eigenvector of $A^TA$ then it is also an eigenvector of $A^2$, which in turn means it is an eigenvector of $A$ since powers of matrices share the same eigenspace. Therefore, we have constructed a full-rank set of eigenvectors of A, meaning that it is diagonalizable.
The product $A^TA$ appears in a key role in the normal equations $A^TAx=A^T b$ for solving linear least squares problems.
• What is this key role? – Matthew C Apr 16 '16 at 0:07
• @lhf, can you extend the properties of this key role? – PlagTag May 6 '17 at 20:54
• From my guessing its because we get a rectangular matrix --> R with the least squares problem beeing Rx = b. In short: The residuals are orthogonal to the fit line. We aim for finding the minimum squared distance of those. This can be done by an orthogonal projection as we are seeing it here. – PlagTag May 6 '17 at 21:39
• In the context of linear least squares problems, $A^TA$ is invertible. – cameronroytaylor Mar 8 '18 at 19:22
To my surprise no one mentioned yet that the root of the Gram determinant of an $n\times k$ matrix $A$ is the $k$-volume of the parallelepiped spanned by the $k$ column vectors of $A$.
• That totally makes sense since that is enters the integral next to the measure! – Martin Ueding Jul 5 '18 at 13:48
Special? Yes! The matrix $$AA^T$$ is abundant with the Least Squares Finite Element Method in Numerical Analysis:
EDIT. Another interesting application of the specialty of $$AA^T$$ is perhaps the following. Suppose that we have a dedicated matrix inversion routine at our disposal, namely for a matrix $$A$$ with pivots only at the main diagonal. How can we use this routine for inverting an arbitrary matrix $$A$$ ? Assuming that the inverse $$A^{-1}$$ does exist and nothing else is known. As follows.
1. Multiply $$A$$ on the left with $$A^T$$, giving $$B = A^T A$$ .
2. The inverse can of $$B$$ can be determined by employing our special matrix inversion routine.
The reason is that the pivots of $$B$$ are always at the main diagonal: see the first reference.
3. The inverse matrix is $$B^{-1} = (A^TA)^{-1} = A^{-1}A^{-T}$$ .
Therefore multiply on the right with $$A^T$$ , giving : $$A^{-1}A^{-T}A^T = A^{-1}$$ .
The inverse that has been sought for is recovered herewith.
Software demo (Delphi Pascal):
program Demo;
type
matrix = array of array of double;
procedure Inverteren(var b : matrix);
{
Matrix inversion
pivots on diagonal
}
var
pe : double;
NDM,i,j,k : integer;
begin
NDM := Length(b);
for k := 0 to NDM-1 do
begin
pe := b[k,k];
b[k,k] := 1;
for i := 0 to NDM-1 do
begin
b[i,k] := b[i,k]/pe;
if (i = k) then Continue;
for j := 0 to NDM-1 do
begin
if (j = k) then Continue;
b[i,j] := b[i,j] - b[i,k]*b[k,j];
end;
end;
for j := 0 to NDM-1 do
b[k,j] := - b[k,j]/pe;
b[k,k] := 1/pe;
end;
end;
procedure inverse(b : matrix; var q : matrix);
{
Matrix inversion
without pivoting
}
var
NDM,i,j,k : integer;
p : matrix;
h : double;
begin
NDM := Length(b);
SetLength(p,NDM,NDM);
{ Transpose * original }
for i := 0 to NDM-1 do
begin
for j := 0 to NDM-1 do
begin
h := 0;
for k := 0 TO NDM-1 do
h := h + b[k,i]*b[k,j];
p[i,j] := h;
end;
end;
Inverteren(p);
SetLength(q,NDM,NDM);
{ (B^T*B)^(-1)*B^T = B^(-1) }
for i := 0 to NDM-1 do
begin
for j := 0 to NDM-1 do
begin
h := 0;
for k := 0 TO NDM-1 do
h := h + p[i,k]*b[j,k];
q[i,j] := h;
end;
end;
end;
procedure test(NDM : integer);
var
i,j,k : integer;
b,q : matrix;
h : double;
begin
SetLength(b,NDM,NDM);
Random; Random;
for i := 0 to NDM-1 do
begin
for j := 0 to NDM-1 do
begin
b[i,j] := Random;
end;
end;
inverse(b,q);
for i := 0 to NDM-1 do
begin
for j := 0 to NDM-1 do
begin
h := 0;
for k := 0 TO NDM-1 do
h := h + q[i,k]*b[k,j];
Write(h:5:2,' ');
end;
Writeln;
end;
end;
BEGIN
test(9);
END.
Output:
1.00 0.00 0.00 -0.00 0.00 0.00 0.00 0.00 -0.00
0.00 1.00 0.00 -0.00 0.00 0.00 0.00 0.00 -0.00
-0.00 0.00 1.00 0.00 -0.00 -0.00 -0.00 -0.00 0.00
-0.00 -0.00 -0.00 1.00 -0.00 -0.00 -0.00 -0.00 0.00
0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00
0.00 0.00 0.00 -0.00 -0.00 1.00 0.00 0.00 -0.00
-0.00 -0.00 -0.00 -0.00 -0.00 -0.00 1.00 -0.00 0.00
-0.00 -0.00 -0.00 0.00 -0.00 -0.00 -0.00 1.00 0.00
-0.00 -0.00 -0.00 -0.00 -0.00 -0.00 -0.00 -0.00 1.00
If you have a real vector space equipped with a scalar product, and an Orthogonal matrix $$A$$ then $$AA^T=I$$ holds. A matrix is orthogonal if for the scalar product $$\langle v,w \rangle = \langle Av, Aw \rangle$$ holds for any $$v,w \in V$$
However I don't see a direct link to the Gram-Determinant.
The matrix $$A$$ represents a rotation if $$A^TA=I$$ and Det A=1 (where $$A^T$$ means $$A$$ transpose) | 2020-08-04T08:39:44 | {
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https://math.stackexchange.com/questions/1398965/evaluate-int-ex-sin2-x-mathrmdx | # Evaluate $\int e^x \sin^2 x \mathrm{d}x$
Is the following evaluation of correct?
\begin{align*} \int e^x \sin^2 x \mathrm{d}x &= e^x \sin^2 x -2\int e^x \sin x \cos x \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int e^x (\cos^2 x - \sin^2x) \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int e^x (1 - 2\sin^2x) \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int -2 e^x \sin^2x \mathrm{d}x + 2 e^x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 e^x -4 \int e^x \sin^2x \mathrm{d}x \end{align*}
First two steps use integration by parts. In the first step we differentiate $\sin^2 x$. In the second step we differentiate $\sin x \cos x$. Using this, we reach $$5\int e^x \sin^2 x \mathrm{d}x = e^x \sin^2 x -2e^x \sin x \cos x + 2 e^x$$
$$\int e^x \sin^2 x \mathrm{d}x = \frac{e^x \sin^2 x -2e^x \sin x \cos x + 2 e^x}{5}+C$$
I can't reach the form that most integral calculators give, which has terms $\cos(2x)$ and $\sin(2x)$ by just using trig identities, so I wonder whether the result is correct. I would also be interested in a method that immediately gives the form $$-\frac{e^x[2 \sin(2x)+ \cos(2x)-5]}{10}+C$$
• What if you start with $\sin^2x=(1-\cos(2x))/2$? – mickep Aug 16 '15 at 7:04
• WolframAlpha agrees: wolframalpha.com/input/… – Krijn Aug 16 '15 at 7:05
• "I wonder whether the result is correct" - Differentiate your result and see if you get the integrand. If doesn't matter if your result looks equal to some other result or not. So long as its derivative matches the integrand, it's correct and equivalent to any other result whose derivative also matches the integrand. – wltrup Aug 16 '15 at 7:05
Notice
$$\int e^x\sin^2x\mathrm{d}x=$$ $$=\int e^x\left(\frac{1-\cos 2x}{2}\right)\mathrm{d}x$$ $$=\frac{1}{2}\int e^xdx-\frac{1}{2}\int e^x \cos 2x \mathrm{d}x$$
Using $\displaystyle \int e^{ax}\cos (bx) \mathrm{d}x=\frac{e^{ax}}{a^2+b^2}(a\cos bx+b\sin bx)$, we get
$$=\frac{1}{2}e^x-\frac{1}{2}\frac{e^x}{1^2+2^2}(\cos 2x+2\sin 2x)+C$$ $$=\frac{1}{2}e^x-\frac{1}{10}e^x(\cos 2x+2\sin 2x)+C$$ $$=-\frac{e^x(2\sin 2x+\cos 2x-5)}{10}+C$$
We have $$\int e^{x}\sin^{2}\left(x\right)dx=\frac{1}{2}\int e^{x}dx-\frac{1}{2}\int e^{x}\cos\left(2x\right)dx$$ and $$\int e^{x}\cos\left(2x\right)dx=\textrm{Re}\left(\int e^{x+2ix}dx\right)$$ then $$\int e^{x+2ix}dx=\frac{e^{x+2ix}}{1+2i}$$ and so $$\int e^{x}\sin^{2}\left(x\right)dx=\frac{1}{2}e^{x}-\frac{1}{2}e^{x}\left(\frac{\cos\left(2x\right)+2\sin\left(2x\right)}{5}\right)+C=$$ $$=-\frac{e^{x}\left(\cos\left(2x\right)+2\sin\left(2x\right)-5\right)}{10}+C.$$
Let $$\displaystyle I = \frac{1}{2}\int e^x 2\sin^2 xdx = \frac{1}{2}\int e^x-\frac{1}{2}\int e^x \cos 2x dx$$
Now Using $$\displaystyle \cos \phi +i\sin \phi = e^{i\phi}$$ and $$\cos \phi-i\sin \phi = e^{-i\phi}.$$
So $$\displaystyle \cos \phi = \frac{e^{i\phi}+e^{-i\phi}}{2}$$
So we get $$\displaystyle \cos 2x = \left(\frac{e^{i2x}+e^{-i2x}}{2}\right)$$
so we get $$\displaystyle J = \bf{Re}\left[\int e^{x}\cdot \left(\frac{e^{i2x}+e^{-i2x}}{2}\right)dx\right] = \frac{1}{2}\bf{Re}\int \left[e^{(1+2i)x}+e^{(1-2i)x}]\right]dx$$
So we get $$\displaystyle J = \frac{1}{2}\bf{Re}\left[\frac{e^{(1+2i)x}}{(1+2i)}+\frac{e^{1-2i}}{(1-2i)}\right]$$
$$\displaystyle = \frac{e^x}{10}\bf{Re}\left[\left(\cos 2x+i\sin 2x\right)\cdot (1-2i)+\left(\cos 2x-i\sin 2x\right)\cdot (1+2i)\right]$$
So $$\displaystyle J = \frac{e^x}{10}\left[2\cos 2x+4\sin 2x\right] = \frac{e^x}{5}\left[\cos 2x+\sin 2x\right]$$
So $$\displaystyle I = \frac{e^x}{2}-\frac{e^x}{10}\left[\cos 2x+2\sin 2x\right]+\mathcal{C} = -\frac{e^x}{10}\left[\cos 2x+2\sin 2x-5\right]+\mathcal{C}$$
You can use the Reduction formula:
$$I_n=\int e^{ax}\sin^n bx\mathrm. dx\\ =\frac{e^{ax}\sin^{n-1} bx (a\sin bx-nb\cos bx)}{a^2+n^2b^2}+\frac{n(n-1)b^2}{a^2+n^2b^2}I_{n-2}$$
Use $n=2,a=1,b=1$.
Your solution is correct. To reach to required form See here,You allready Got this
$$\int e^x \sin^2 x \mathrm{d}x = \frac{e^x \sin^2 x -2e^x \sin x \cos x + 2 e^x}{5}+C$$
Now multiply and divide your result by $2$, you will get $$\frac{2e^x \sin^2 x -4e^x \sin x \cos x +4e^x}{10}+C\\ =\frac{2e^x \sin^2 x -4e^x \sin x \cos x +4e^x+e^x-e^x}{10}+C\\ =\frac{2e^x \sin^2 x-e^x -4e^x \sin x \cos x +5e^x}{10}+C\\ =\frac{e^x(2 \sin^2 x-1) -4e^x \sin x \cos x +5e^x}{10}+C\\ =-\frac{e^x[2 \sin(2x)+ \cos(2x)-5]}{10}+C$$
Where we used the identities
1) $2 \sin^2 x-1=-\cos 2x$
2)$2\sin x\cos x= \sin 2x$
$$\int \left(e^x\sin^2(x)\right)\text{d}x =$$ $$\int \left(e^x\left(\frac{1}{2}(1-\cos(2x))\right)\right)\text{d}x =$$ $$\frac{1}{2}\int \left(e^x-e^x\cos(2x)\right)\text{d}x =$$ $$\frac{1}{2} \left(\int \left(e^x\right) \text{d}x-\int \left(e^x\cos(2x)\right) \text{d}x\right) =$$ $$\frac{1}{2} \left(\int e^x \text{d}x-\int e^x\cos(2x) \text{d}x\right) =$$ $$\frac{1}{2} \left(e^x-\int e^x\cos(2x) \text{d}x\right) =$$
For the integrand $e^x\cos(2x)$, use the formula:
$$\int\exp(\alpha x)\cos(\beta x)\text{d}x=\frac{\exp(\alpha x)(\alpha \cos(\beta x))+\beta\sin(\beta x)}{\alpha^2+\beta^2}$$
$$\frac{1}{2} \left(e^x-\frac{e^x(2\sin(2x)+\cos(2x))}{5}\right) + C =$$ $$-\frac{e^x(2\sin(2x)+\cos(2x)-5)}{10} + C$$ | 2019-11-14T03:25:52 | {
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https://math.stackexchange.com/questions/295564/a-subgroup-of-a-cyclic-group-is-cyclic-understanding-proof | # A subgroup of a cyclic group is cyclic - Understanding Proof
I'm having some trouble understanding the proof of the following theorem
A subgroup of a cyclic group is cyclic
I will list each step of the proof in my textbook and indicate the places that I'm confused and hopefully somewhere out there can clear some things up for me.
Proof
Let $$G$$ be a cyclic group generated by "$$a$$" and let $$H$$ be a subgroup of $$G$$. If $$H = {\{e\}}$$, then $$H = \langle e\rangle$$ is cyclic. If $$H \neq \space {\{e\}}$$, then $$a^n \in H$$ for some $$n \in \mathbb{Z}^{+}$$. Let $$m$$ be the smallest integer in $$\mathbb{Z}^{+}$$ such that $$a^m \in H$$.
We claim that $$c = a^m$$ generates $$H$$; that is,
$$H = \langle a^m\rangle = \langle c\rangle.$$
We must show that every $$b \in H$$ is a power of $$c$$. Since $$b \in H$$ and $$H \leq G$$, we have $$b = a^n$$ for some $$n$$. Find $$q$$ and $$r$$ such that
$$n = mq + r \space \space \space \space for \space \space \space 0 \leq r < m,$$
Alright this is the first part in the proof where I start to get confused. Where does the division algorithm come from and why are we using it? The proof continues as follows:
$$a^n = a^{mq + r} = (a^m)^q \cdot a^r,$$
so
$$a^r = (a^m)^{-q} \cdot a^n.$$
Now since $$a^n \in H$$, $$a^m \in H$$, and $$H$$ is a group, both $$(a^m)^{-q}$$ and $$a^n$$ are in $$H$$. Thus
$$(a^m)^{-q} \cdot a^n \in H; \space \space \space \text{that is,} \space \space a^r \in H.$$
This is another point at which I’m a little confused. What exactly about $$a^n$$ and $$a^m$$ being elements of $$H$$ allows us to accept that $$(a^m)^{-q}$$ and $$a^{n}$$ are in $$H$$? The proof continues:
Since $$m$$ was the smallest positive integer such that $$a^m \in H$$ and $$0$$ $$\leq r$$ $$< m$$, we must have $$r = 0$$. Thus $$n = qm$$ and
$$b \space = \space a^n \space = \space (a^m)^q \space = \space c^q,$$
so $$b$$ is a power of $$c.$$
This final step is confusing as well, but I think its just because of the previous parts I was confused about. Any help in understanding this proof would be greatly appreciated
• "Where does the division algorithm come from?" is a very vague question. In this cases, there is a deeper theorem involved, that is hidden because you are talking about groups rather than the (possibly later topic) rings, name that the integers are something called a "principal ideal ring." The division algorithm is deeply entwined with this property. Feb 5, 2013 at 17:19
• What textbook is this? This is a nice proof. Jun 1, 2017 at 3:57
• Could you advise which textbook this was adapted from? Thanks. Sep 11, 2018 at 10:48
• For the questions under the main question regarding where the proof came from, the book where this proof is from is: A First Course in Abstract Algebra by John B.Fraleigh. I know this does not answer the main question, whoever can add comments could move this there. Jan 24, 2020 at 18:59
For the first question, the appearance of the division algorithm is best explained by its usefulness in the rest of the proof. You might think of it because you want $n=qm$, as you need to show that $a^n$ is a power of $a^m$, but the best you can do at that point is say $n=qm+r$ and then attempt to prove $r=0$.
For the second question, as $a^m\in H$, we have $(a^m)^{-1}\in H$, as subgroups are closed under taking inverses, and then $(a^m)^{-q}=((a^m)^{-1})^q\in H$, as subgroups are closed under multiplication.
• .@Matt Can you please explain how r =0 in this proof ....i havenot understood following two lines from proof.....< Since m was the smallest positive integer such that am∈H and 0 ≤r <m, we must have r=0. Thus n=qm > thanks Apr 3, 2015 at 10:37
• You assume $m$ is the smallest positive integer such that $a^m\in H$, but then find that $a^r\in H$, and $r<m$. The only way this is possible is if $r=0$.
– mdp
Apr 3, 2015 at 10:50
First, the use of the division algorithm is for the sake of utility, as it provides a basis for the structure of the proof. Here, we want to relate $n$ with $m$, that is, we want to show that $n$ must be a multiple of $m$, but we start with the fact that what we know of any $n$, then given any positive integer $m$, by the division algorithm, there exist unique integers integers $q$ and $r$, where $0\leq r\lt m$ such that $n = mq + r$.
Essentially that means for any $n$, if we divide by a positive integer $m$, we have a unique integer quotient $q$ and a unique integer remainder $r$ where $0\leq r \lt m$. So to show that $m$ is a multiple of $n$ (no remainder), we want to show that $n = qm + 0$. I.e. we want $n=qm$, in order to show that $a^n$ is a power of $a^m$, but at most we can start with the fact that $n=qm+r$. The objective then is to prove that $r=0$.
For the second question, since $a^m\in H$, it follows $(a^m)^{-1}\in H$, since subgroups are closed under taking inverses. Then, since $a^m, (a^m)^{-1} \in H$, $(a^m)^{-q}=((a^m)^{-1})^q\in H$, since subgroups are closed under multiplication.
Does this clarify matters any?
In the first answer given, at the last part, $$a^r$$ is a member of $$H$$, and $$0\le r. Since $$a^r$$ is a member of $$H$$ and $$a^m$$ is a member of $$H$$ where $$m$$ is the smallest integer s.t. $$a^m$$ belongs to $$H$$.
Thus $$r>m$$. However $$r, by Euclid's division lemma. Hence, due to this contradiction, $$r=0$$, and $$n=qm$$.
a nontrivial cyclic group is a group with a singleton generating set, and vice versa.
let H be a cyclic group, and suppose $K$ is a non-cyclic subgroup.
evidently $K$ is a proper subgroup of H and has no set of generators of cardinality less than 2.
choose a generator $h$ for H.
in examining generating sets for $K$, we may exclude any containing the identity.
and since $K$ is proper no generating set contains $h$
for $K$ choose a generating set $\mathfrak{K}$ which contains an element $h^m$ where $m \gt 1$ is minimal amongst all the powers of $h$ occurring in generating sets for K.
without loss of generality we may assume that for any $p \gt 1$ we have $h^{pm} \notin \mathfrak{K}$
since $|\mathfrak{K}| \ge 2$
$$\exists n \gt m.h^n \in \mathfrak{K}$$ define $a \ge 1$ by $$a = \max\{b|bm \lt n\}$$ it follows that: $$0 \lt n-am \lt m$$ but since $h^{n-am}=h^n (h^m)^{-a} \in K$ it follows that $\mathfrak{K}'=\mathfrak{K} \cup \{h^{n-am}\}$ is a generating set for $K$ contradicting the minimality in our choice of $m$ | 2022-05-19T10:14:37 | {
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https://math.stackexchange.com/questions/2685774/solving-for-frac-13-frac-29-frac-327-cdots/2685798 | # Solving for $\frac 13 +\frac 29 +\frac 3{27}+\cdots$
Evaluate the sum $\frac{1}{3^1} + \frac{2}{3^2} + \frac{3}{3^3} + \cdots + \frac{k}{3^k} + \cdots$
How would I go about solving this problem? I'm thinking of setting the sum to $S$, multiplying by $3$, and then subtracting from the original equation. HELP!
I have explained my reasoning in the comment below.
• Your idea seems good... What did it lead to? – vrugtehagel Mar 10 '18 at 22:59
• Let the sum be $S$. This series looks almost geometric, but not quite. We can turn it into a geometric series as follows: \begin{align*} S &= \frac{1}{3^1} +\frac{2}{3^2} + \frac{3}{3^3} + \frac{4}{3^4} + \cdots \\ \frac{1}{3}S &= \frac{0}{3^1} + \frac{1}{3^2} + \frac{2}{3^3} + \frac{3}{3^4} + \cdots \\ \frac{2}{3}S = S - \frac{1}{3}S &= \frac{1}{3^1} + \frac{1}{3^2} + \frac{1}{3^3} + \frac{1}{3^4} + \cdots \end{align*}Now, we do have a geometric series, so we can find $\frac{2}{3}S = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{1}{2}$, and $S = \boxed{\frac{3}{4}}$. – A Piercing Arrow Mar 10 '18 at 23:02
• That looks good! You basically had the answer already ;) – vrugtehagel Mar 10 '18 at 23:04
• This is an arithmetico-geometric sequence: en.wikipedia.org/wiki/Arithmetico%E2%80%93geometric_sequence – an4s Mar 10 '18 at 23:19
• Is it too much to ask you to write what you wrote in a comment, in the actual question you posted? – Namaste Mar 10 '18 at 23:32
So as you said, let's set
$$S=\frac{1}{3^1} + \frac{2}{3^2} + \frac{3}{3^3} + \cdots + \frac{k}{3^k}\cdots$$
Now we see
$$3S=\frac{1}{3^0} + \frac{2}{3^1} + \frac{3}{3^2} + \cdots + \frac{k}{3^{k-1}}\cdots$$
And so
$$3S-S=\frac{1-0}{3^0} + \frac{2-1}{3^1} + \frac{3-2}{3^2} + \cdots + \frac{k-(k-1)}{3^{k-1}}\cdots$$ or, rewritten,
$$2S=\frac1{3^0}+\frac1{3^1}+\frac1{3^2}+\cdots=\frac32$$
So that we get $S=\frac 34$.
• So I was correct! Thank you. – A Piercing Arrow Mar 10 '18 at 23:05
• Yup, you were! Be sure to accept one of the answers so that this question doesn't end up in the unanswered section ;) – vrugtehagel Mar 10 '18 at 23:06
Nice approach. You can also do this as follows and this method is useful for other types of series too. Compute the following sum $$f(x)=\sum_{k\geq 1}\left(\frac{x}{3}\right)^k$$ And then compute the derivative of $f$ and evaluate $f'(1)$.
Note that $$\sum_{k=1}^\infty \frac{k}{3^k}=\sum_{k=1}^\infty \sum_{j=1}^k\frac{1}{3^k} =\sum_{j=1}^\infty \sum_{k=j}^\infty\frac{1}{3^k} =\sum_{j=1}^\infty\frac{1/3^j}{2/3} =\frac{3}{2}\frac{1/3}{1-(1/3)} =\frac{3}{4}$$ where the interchanging of summation is allowed since we are dealing with non-negative series. | 2019-10-17T01:26:26 | {
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https://www.beatthegmat.com/if-2-00x-and-3-00y-are-2-numbers-in-decimal-form-with-thousandths-digits-x-and-y-is-t326514.html?sid=d2e43ac899386f61913ab102b563e741 | ## If $$2.00X$$ and $$3.00Y$$ are $$2$$ numbers in decimal form with thousandths digits $$X$$ and $$Y,$$ is
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### If $$2.00X$$ and $$3.00Y$$ are $$2$$ numbers in decimal form with thousandths digits $$X$$ and $$Y,$$ is
by VJesus12 » Thu Sep 16, 2021 11:49 am
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## Global Stats
If $$2.00X$$ and $$3.00Y$$ are $$2$$ numbers in decimal form with thousandths digits $$X$$ and $$Y,$$ is $$3(2.00X) > 2(3.00Y) ?$$
(1) $$3X < 2Y$$
(2) $$X < Y - 3$$
Source: Official Guide
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### Re: If $$2.00X$$ and $$3.00Y$$ are $$2$$ numbers in decimal form with thousandths digits $$X$$ and $$Y,$$ is
by [email protected] » Thu Sep 16, 2021 3:26 pm
00:00
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## Global Stats
VJesus12 wrote:
Thu Sep 16, 2021 11:49 am
If $$2.00X$$ and $$3.00Y$$ are $$2$$ numbers in decimal form with thousandths digits $$X$$ and $$Y,$$ is $$3(2.00X) > 2(3.00Y) ?$$
(1) $$3X < 2Y$$
(2) $$X < Y - 3$$
Source: Official Guide
Given: 2.00X and 3.00Y are 2 numbers in decimal form with thousandths digits X and Y
Target question: Is 3(2.00X) > 2(3.00Y)?
This is a good candidate for rephrasing the target question.
Since X is the thousandths digit, we can write: 2.00X = 2 + X/1000
Since Y is the thousandths digit, we can write: 3.00Y = 3 + Y/1000
So, the target question becomes: Is 3(2 + X/1000) > 2(3 + Y/1000)?
Expand both sides: Is 6 + 3X/1000 > 6 + 2Y/1000)?
Subtract 6 from both sides: Is 3X/1000 > 2Y/1000)?
Multiply both sides by 1000 to get: Is 3X > 2Y ?
REPHRASED target question: Is 3X > 2Y?
Aside: the video below has tips on rephrasing the target question
Statement 1:3X < 2Y
PERFECT!!
The answer to the REPHRASED target question is NO, 3X is NOT greater than 2Y
Since we can answer the REPHRASED target question with certainty, statement 1 is SUFFICIENT
Statement 2: X < Y − 3
Add 3 to both sides to get: X + 3 < Y
This one is TRICKY!!
The solution relies on the fact that X and Y are DIGITS (0, 1, 2, 3, 4, 5, 6, 7, 8 or 9)
Let's examine all possible DIGIT solutions to the inequality X + 3 < Y
case a: X = 0, and Y = 4,5,6,7,8 or 9. In all possible cases, 3X < 2Y
case b: X = 1, and Y = 5,6,7,8 or 9. In all possible cases, 3X < 2Y
case c: X = 2, and Y = 6,7,8 or 9. In all possible cases, 3X < 2Y
case d: X = 3, and Y = 7,8 or 9. In all possible cases, 3X < 2Y
case e: X = 4, and Y = 8 or 9. In all possible cases, 3X < 2Y
case f: X = 5, and Y = 9. In all possible cases, 3X < 2Y
Now that we've examine all possible values of X and Y, we can see that the answer to the REPHRASED target question is always the same: NO, 3X is NOT greater than 2Y
Since we can answer the REPHRASED target question with certainty, statement 2 is SUFFICIENT | 2021-10-18T10:21:49 | {
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https://math.stackexchange.com/questions/2465086/for-what-pair-of-integers-a-b-is-3a-7b-a-perfect-square | # For what pair of integers $(a,b)$ is $3^a + 7^b$ a perfect square.
Problem: For what pair of positive integers $(a,b)$ is $3^a + 7^b$ a perfect square.
First obviously $(1,0)$ works since $4$ is a perfect square, $(0,0)$ does not work, and $(0,1)$ does not work, so we can exclude cases where $a$ or $b$ are zero for the remainder of this post.
I have a few observations made but not much room for a full solution.
First, since powers of an odd number are odd, and the sum of two odd numbers is even, so the base of the square must be an even number.
Secondly, the last digit of the powers of $3$ are $\{1,3,7,9 \}$ , whereas the last digits of the powers of $7$ are $\{7,9,3,1 \}$.
From here I am not sure how to proceed and any hints much appreciated. I'm not sure if there a finite amount of pairs or not either.
• The only pairs , I found so far , are $(1/0)$ and $(2/1)$ – Peter Oct 9 '17 at 21:44
• Every square is either $0$ or $1$ modulo $4$. We have $$3^a+7^b\equiv(-1)^a+(-1)^b\mod4\ ,$$ therefore we must necessarily have $a$ even and $b$ odd, or the converse. – Daniel Robert-Nicoud Oct 9 '17 at 21:46
• Since for $0\le a,b\le 1000$, the only pairs are $(1/0)$ and $(2/1)$, another pair would be a mircale considering how large one of the powers must be. However, proving this could be extremely difficult, but it could also be relatively easy if someone finds a clever trick. – Peter Oct 9 '17 at 22:01
• @VidyanshuMishra PARI/GP is very powerful for such questions! There are surely faster programs, however. – Peter Oct 9 '17 at 22:04
• Ah.. Thanks @Peter, I shall have a look at it. – Vidyanshu Mishra Oct 9 '17 at 22:04
## 4 Answers
Just to register, from the comment by Daniel, there are just two possibilities, if the result is $x^2,$ either $$1 + 2 \cdot 3^c = 7^d,$$ or $$1 + 2 \cdot 7^e = 3^f.$$
I would guess that an elementary method shown in an answer by Gyumin Roh to http://math.stackexchange.com/questions/1551324/exponential-diophantine-equation-7y-2-3x can be modified for this task. My way of working with this takes a while...
http://math.stackexchange.com/questions/1941354/elementary-solution-of-exponential-diophantine-equation-2x-3y-7
http://math.stackexchange.com/questions/2100780/is-2m-1-ever-a-power-of-3-for-m-3/2100847#2100847
• Ok will study those and get back to you. – IntegrateThis Oct 9 '17 at 22:17
• @IntegrateThis got one of the two so far, posted as separate answer. – Will Jagy Oct 10 '17 at 0:33
• @IntegrateThis got the final one, third answer. – Will Jagy Oct 10 '17 at 0:57
• @WillJagy, I think I found the elementary alternative. – Barry Cipra Oct 10 '17 at 13:27
Here is a solution using less machinery than in Will Jagy's answer.
As Daniel Robert-Nicoud noted in comments, the fact that $3^a+7^b\equiv(-1)^a+(-1)^b$ mod $4$ implies $a$ and $b$ must have opposite parity in order for the (even) sum $3^a+7^b$ to be a perfect square. So we seek to show that $(a,b)=(1,0)$ is the only solution in the (odd,even) case and $(a,b)=(2,1)$ is the only solution in the (even,odd) case.
If $a$ is odd and $b=2m$ is even, then we can rewrite $3^a+7^b=n^2$ as $3^a=(n+7^m)(n-7^m)$. This implies $n+7^m$ and $n-7^m$ are each powers of $3$, say $3^c$ and $3^d$ with $c+d=a$. But then $2\cdot7^m=3^c-3^d$, so we must have $d=0$ and $c=a$, since $3\not\mid2\cdot7^m$. One solution to $2\cdot7^m=3^a-1$ is $(a,m)=(1,0)$. It remains to show there are no solutions with $m\gt0$. For this it suffices to note that $3^a\equiv1$ mod $7$ if and only if $6\mid a$, which contradicts the assumption that $a$ is odd.
In the other parity case, if $a=2m$ is even and $b$ is odd, then, along similar lines, we must have $2\cdot3^m=7^b-1$, which has $(m,b)=(1,1)$ as one solution. After checking that there is no solution with $m=0$, it remains to show there are no solutions with $m\gt1$. If there were, then we would have $7^b\equiv1$ mod $9$, which would imply $b\equiv3$ mod $6$ (since $6$ is the order of the multiplicative group of units mod $9$ and $7^3\equiv(-2)^3=-8\equiv1$ mod $9$). Writing $b=3r$ (with $r$ odd, but that doesn't matter), we note that $(7^3-1)\mid(7^{3r}-1)$, so that $7^3-1=2\cdot3^2\cdot19$ divides $2\cdot3^m$, an obvious contradiction.
• Good. In the various problems I had done previously, often a full divisor, such as $7^{12} - 1,$ had several factors that were not used, and a smaller divisor such as your $7^3 -1$ gave enough, in this case the prime $19.$ I used the prime $43$ out of $7^{12} - 1.$ Lack of practice. Plus, of course, I did not use the full strength of Daniel's comment. – Will Jagy Oct 10 '17 at 17:11
• Wow. That is a fantastic answer much appreciated. – IntegrateThis Oct 10 '17 at 19:58
• why does d = 0 and c = a ? – IntegrateThis Oct 11 '17 at 4:22
• @IntegrateThis, if $c$ and $d$ are both greater than $0$, then $3\mid(3^c-3^d)$. But $3\not\mid2\cdot7^m$. – Barry Cipra Oct 11 '17 at 12:26
The technique I mentioned is working. One of the cases is $$1 + 2 \cdot 3^c = 7^d,$$ where we think that $c=d=1$ gives the largest such answer. Subtract $7$ from both sides, $$2 \cdot 3^c - 6 = 7^d - 7.$$ Let $y+1 = c,$ $x+1 = d,$ for $$6 \cdot 3^y - 6 = 7 \left( 7^x -1 \right),$$ $$6 \left( 3^y -1 \right) = 7 \left( 7^x -1 \right).$$ We are assuming $x,y \geq 1,$ with both sides being nonzero.
$$3^y \equiv 1 \pmod 7,$$ $$y \equiv 0 \pmod 6.$$ Next, $3^y - 1$ is divisible by $3^6 - 1 = 728 = 8 \cdot 7 \cdot 13.$
$$7^x \equiv 1 \pmod {13},$$ $$x \equiv 0 \pmod {12}.$$ Next, $7^x - 1$ is divisible by $7^{12} - 1 = 13841287200 = 32 \cdot 9 \cdot 25 \cdot 13 \cdot 19 \cdot 43 \cdot 181.$
$$3^y \equiv 1 \pmod {43},$$ $$y \equiv 0 \pmod {42}.$$ Next, $3^y - 1$ is divisible by $3^{42} - 1 = 109418989131512359208 = 8 \cdot 7^2 \cdot 13 \cdot 43 \cdot 547 \cdot 1093 \cdot 2269 \cdot 368089$
All we care about is that $3^y - 1$ is divisible by $7^2 = 49,$ as this contradicts the assumption of $x \geq 1$ in $$6 \left( 3^y -1 \right) = 7 \left( 7^x -1 \right).$$ $$\bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc$$
The other case is $1 + 2 \cdot 7^x = 3^y,$ or $$3^y - 1 = 2 \cdot 7^x.$$ Assume $x \geq 1.$ Then both sides are divisible by $7,$ giving $$3^y \equiv 1 \pmod 7,$$ $$y \equiv 0 \pmod 6.$$ Then $3^y - 1$ is divisible by $$3^6 - 1 = 728 = 8 \cdot 7 \cdot 13.$$ However, then $2 \cdot 7^x$ is divisible by $13,$ which is a contradiction. $$\bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc$$ | 2019-09-18T22:11:26 | {
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https://math.stackexchange.com/questions/645293/probability-of-picking-4-numbers-out-of-100 | # probability of picking 4 numbers out of 100
I have $100$ balls and on each ball is labelled a number between $1$ and $100$. No two balls can have the same number, so the interval $1\ldots 100$ is represented by the balls.
I now pick 20 balls at random, one by one without putting them back at any point. I am trying to find the probably of getting 4 certain numbers. The sequence of these 4 numbers does not matter.
Here is my solution: The first time I pick up a ball the probably is $1/100$. The next time it is $1/99$, etc.. So the total probably of getting the four desired numbers is $1/(100\cdot 99\cdot 98\cdot 97)$.
Is this correct?
You have computed the probability that you get your target numbers, in the first $4$ choices, in a certain specified order. That is much smaller than the required probability.
There are $\binom{100}{20}$ equally likely ways to choose $20$ balls.
We find the number of ways to choose $20$ balls that include our $4$ target balls. So we must choose $16$ balls from the remaining $96$ to keep our $4$ target balls company. This can be done in $\binom{96}{16}$ ways.
Thus the required probability is $\dfrac{\binom{96}{16}}{\binom{100}{20}}$. This expression can be greatly simplified.
• Thanks, but the probablity of picking the $4$ does not enter in the equation? – BillyJean Jan 20 '14 at 19:01
• Well, on top we could write $\binom{4}{4}\binom{96}{16}$ instead of $\binom{96}{6}$, and maybe I should have, but since $\binom{4}{4}=1$ it makes no difference. – André Nicolas Jan 20 '14 at 19:03
• But the balls are not supposed to be returned. Is $\binom{100}{20}$ correct? – Spock Jan 20 '14 at 19:07
• Yes. It is the number of ways of choosing a "hand" of $20$ cards from a collection of $100$ distinct cards. – André Nicolas Jan 20 '14 at 19:10
Your solution is not correct. But, here's a hint on a different way to proceed:
Hint: You want to count all permutations (of length $20$) of distinct elements in $\{1,2,\ldots,100\}$, such that four fixed numbers (say $1,2,3,4$) are all contained in the permutation.
To do this: pick the $16$ numbers OTHER than the ones you are interested in which will be included; then, pick ANY arrangement of the set of $20$ numbers consisting of the four you want and the new sixteen.
This is the total number of sequences with the desired property; divide by the total number of sequences to get the desired result.
• But I get to pick 20 balls, how does that information enter? – BillyJean Jan 20 '14 at 18:49
• Whoops, sorry; I missed that detail. One moment. – Nick Peterson Jan 20 '14 at 18:50
• @BillyJean Take a look at it now. – Nick Peterson Jan 20 '14 at 18:53 | 2019-07-18T17:42:27 | {
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https://math.stackexchange.com/questions/2558475/find-the-probability-that-no-kid-will-have-his-own-pair-of-shoes | # Find the probability that no kid will have his own pair of shoes
In the morning, $n$ children come to the kindergarten and leave their shoes in the locker room. Leaving the kindergarten one by one, each child takes one left and one right shoe, accidentally equiprobably choosing them from among the remaining ones. Find the likelihood that none of the children will leave in their own pair of shoes.
My thoughts: I assume that decisions are independent based on accidentally equiprobably choosing them from among the remaining ones part. Let $I_l$ and $I_r$ be indicator r.v. for choosing his/her left shoe. Let $A$ be the probability that none of the children will leave in their own pair of shoes
$$P(I_l=1)=\frac1n; P(I_r=1)=\frac1n$$
Leaving without his/her pair of shoes can occur in several ways:
$1$) Choose his/her left shoe and other (not his/her) right shoe $$P(I_l=1)P(I_r=0)=\left(\frac1n\right)^n\left(1-\frac1n\right)^n$$
$2$) Choose his/her right shoe and other (not his/her) left shoe $$P(I_r=1)P(I_l=0)=\left(\frac1n\right)^n\left(1-\frac1n\right)^n$$
$3$) Nobody chose his/her pair of shoes $$P(I_l=0)P(I_r=0)=\left(1-\frac1n\right)^n\left(1-\frac1n\right)^n$$
Therefore, $$P(A)=2\left(\frac1n\right)^n\left(1-\frac1n\right)^n+\left(1-\frac1n\right)^{2n}$$
Here I assume that kids can distinguish between right and left shoes.
If they can not: probability of choosing left shoe out of $2n$ shoes:
$$P(\text{left})=\frac{n}{\binom{2n}{2}}$$ by symmetry the same for $P(\text{right})$
$1$) Chose two left shoes included his/her: $$(P(\text{left})P(\text{left}))^n=\left(\frac{n}{\binom{2n}{2}}\cdot\frac{n-1}{\binom{2n-1}{2}}\right)^n$$ Power $n$, because there are $n$ people who make this decisions
$2$) Chose two right shoes included his/her: $$(P(\text{right})P(\text{right}))^n=\left(\frac{n}{\binom{2n}{2}}\cdot\frac{n-1}{\binom{2n-1}{2}}\right)^n$$
$3$) Choose one left included his/her and one right excluded his/her own: $$P(\text{left})P(\text{right})=\left(\frac{n}{\binom{2n}{2}}\cdot\frac{n-1}{\binom{2n-1}{2}}\right)^n$$
$4$) Choose one right inlcuded his her and one left excluded his/her own $$P(\text{right})P(\text{left})=\left(\frac{n}{\binom{2n}{2}}\cdot\frac{n-1}{\binom{2n-1}{2}}\right)^n$$
• The probability that a particular child leaves with the correct shoes is $\frac 1{n^2},$ so the expected number of children leaving with correct shoes is $\frac 1n$ by the linearity of expectation. The chance that nobody leaves with the correct shoes is at least $\frac {n-1}n$, which assumes that you never have more than one child with correct shoes. The chance is somewhat higher than this. The problem does say they each take one left and one right shoe. Dec 9 '17 at 15:33
Using inclusion-exclusion in its basic form we immediately obtain for the probability that it is
$$\frac{1}{n!^2} \sum_{q=0}^n {n\choose q} (-1)^q (n-q)!^2.$$
We can use this closed form to gather more information from the OEIS, where we find it is listed as OEIS A089041, i.e.
$$0, 3, 26, 453, 11844, 439975, 22056222, \\ 1436236809, 117923229512, \ldots$$
In particular consulting the asymptotics we find in the OEIS entry
$$\frac{1}{n!^2} 2\pi e^{-2n} n^{2n+1}.$$
Combining this with Stirling we get
$$\left(\frac{1}{\sqrt{2\pi n}} \frac{e^n}{n^n}\right)^2 2\pi e^{-2n} n^{2n+1} \sim 1.$$
This means for $n$ large we can be almost certain that no one leaves with their own pair of shoes.
Clarification of the problem definition that was used, by Maple code.
with(combinat);
ENUM :=
proc(n)
option remember;
local lperm, rperm, res, pos;
res := 0;
lperm := firstperm(n);
while type(lperm, list) do
rperm := firstperm(n);
while type(rperm, list) do
for pos to n do
if lperm[pos] = pos and
rperm[pos] = pos then
break;
fi;
od;
if pos = n+1 then
res := res + 1;
fi;
rperm := nextperm(rperm);
od;
lperm := nextperm(lperm);
od;
res/n!^2;
end;
EX := n -> add(binomial(n,q)*(-1)^q*(n-q)!^2, q=0..n)/n!^2;
Remark. Apparently the asymptotics here follow by inspection, i.e. by writing the probability as
$$\frac{1}{n!} \sum_{q=0}^n \frac{1}{q!} (-1)^q (n-q)! \\ = 1 - \frac{1}{n} + \frac{1}{2}\frac{1}{n(n-1)} - \frac{1}{6} \frac{1}{n(n-1)(n-2)} + \cdots \pm \frac{1}{n!}$$
which is seen not to leave the interval $(1-1/n, 1)$ after the second term and hence produces a probability of one for $n$ large. (These alternating terms are strictly decreasing in absolute value.) | 2022-01-21T06:22:34 | {
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http://bootmath.com/proof-that-cexpx-is-the-only-set-of-functions-for-which-fx-fx.html | # Proof that $C\exp(x)$ is the only set of functions for which $f(x) = f'(x)$
I was wondering the following. And I probably know the answer already: NO.
Is there another number with similar properties as $e$? So that the derivative of $\exp(x)$ is the same as the function itself.
I can guess that it’s probably not, because otherwise $e$ wouldn’t be that special, but is there a proof of it?
#### Solutions Collecting From Web of "Proof that $C\exp(x)$ is the only set of functions for which $f(x) = f'(x)$"
Of course $C e^x$ has the same property for any $C$ (including $C = 0$). But these are the only ones.
Proposition: Let $f : \mathbb{R} \to \mathbb{R}$ be a differentiable function such that $f(0) = 1$ and $f'(x) = f(x)$. Then it must be the case that $f = e^x$.
Proof. Let $g(x) = f(x) e^{-x}$. Then
$$g'(x) = -f(x) e^{-x} + f'(x) e^{-x} = (f'(x) – f(x)) e^{-x} = 0$$
by assumption, so $g$ is constant. But $g(0) = 1$, so $g(x) = 1$ identically.
N.B. Note that it is also true that $e^{x+c}$ has the same property for any $c$. Thus there exists a function $g(c)$ such that $e^{x+c} = g(c) e^x = e^c g(x)$, and setting $c = 0$, then $x = 0$, we conclude that $g(c) = e^c$, hence $e^{x+c} = e^x e^c$.
This observation generalizes to any differential equation with translation symmetry. Apply it to the differential equation $f''(x) + f(x) = 0$ and you get the angle addition formulas for sine and cosine.
Let $f(x)$ be a differentiable function such that $f'(x)=f(x)$. This implies that the $k$-th derivative, $f^{(k)}(x)$, is also equal to $f(x)$. In particular, $f(x)$ is $C^\infty$ and we can write a Taylor expansion for $f$:
$$T_f(x) = \sum_{k=0}^\infty c_k x^k.$$
Notice that the fact that $f(x)=f^{(k)}(x)$, for all $k\geq 0$, implies that the Taylor series $T_f(x_0)$ converges to $f(x_0)$ for every $x_0\in \mathbb{R}$ (more on this later), so we may write $f(x)=T_f(x)$. Since $f'(x) = \sum_{k=0} (k+1)c_{k+1}x^k = f(x)$, we conclude that $c_{k+1} = c_k/(k+1)$. The value of $c_0 = f(0)$, and therefore, $c_k = f(0)/k!$ for all $k\geq 0$. Hence:
$$f(x) = f(0) \sum_{k=0}^\infty \frac{x^k}{k!} = f(0) e^x,$$
as desired.
Addendum: About the convergence of the Taylor series. Let us use Taylor’s remainder theorem to show that the Taylor series for $f(x)$ centered at $x=0$, denoted by $T_f(x)$, converges to $f(x)$ for all $x\in\mathbb{R}$. Let $T_{f,n}(x)$ be the $n$th Taylor polynomial for $f(x)$, also centered at $x=0$. By Taylor’s theorem, we know that
$$|R_n(x_0)|\leq |f^{(n+1)}(\xi)|\frac{ |x_0 – 0|^{n+1}}{(n+1)!},$$
where $R_n(x_0)=f(x) – T_{f,n}(x)$ and $\xi$ is a number between $0$ and $x_0$. Let $M=M(x_0)$ be the maximum value of $|f(x)|$ in the interval $I=[-|x_0|,|x_0|]$, which exists because $f$ is differentiable (therefore, continuous) in $I$. Since $f(x)=f^{(n+1)}(x)$, for all $n\geq 0$, we have:
$$|R_n(x_0)|\leq |f^{(n+1)}(\xi)|\frac{ |x_0|^{n+1}}{(n+1)!}\leq |f(\xi)|\frac{ |x_0|^{n+1}}{(n+1)!}\leq M \frac{|x_0|^{n+1}}{(n+1)!} \longrightarrow 0 \ \text{ as } \ n\to \infty.$$
The limit goes to $0$ because $M$ is a constant (once $x_0$ is fixed) and $A^n/n! \to 0$ for all $A\geq 0$. Therefore, $T_{f,n}(x_0) \to f(x_0)$ as $n\to \infty$ and, by definition, this means that $T_f(x_0)$ converges to $f(x_0)$.
Yet another way: By the chain rule, ${\displaystyle {d \over dx} \ln|f(x)| = {f'(x) \over f(x)} = 1}$. Integrating, you get $\ln |f(x)| = x + C$. Taking $e$ to both sides, you obtain $|f(x)| = e^{x + C} = C'e^x$, where $C' > 0$.
As a result, $f(x) = C''e^x$, where $C''$ is an arbitrary constant.
If you are worried about $f(x)$ being zero, the above shows $f(x)$ is of the form $C''e^x$ on any interval for which $f(x)$ is nonzero. Since $f(x)$ is continuous, this implies $f(x)$ is always of that form, unless $f(x)$ is identically zero (in which case we can just take $C'' = 0$ anyhow).
Hint \rm\displaystyle\:\ \begin{align} f{\:’}\!\! &=\ \rm a\ f \\ \rm \:\ g’\!\! &=\ \rm a\ g \end{align} \iff \dfrac{f{\:’}}f = \dfrac{g’}g \iff \bigg(\!\!\dfrac{f}g\bigg)’ =\ 0\ \iff W(f,g) = 0\:,\ \ W = Wronskian
This is a very special case of the uniqueness theorem for linear differential equations, esp. how the Wronskian serves to measure linear independence of solutions. See here for a proof of the less trivial second-order case (that generalizes to n’th order). See also the classical result below on Wronskians and linear dependence from one of my old sci.math posts.
Theorem $\ \$ Suppose $\rm\:f_1,\ldots,f_n\:$ are $\rm\:n-1\:$ times differentiable on interval $\rm\:I\subset \mathbb R\:$
and suppose they have Wronskian $\rm\: W(f_1,\ldots,f_n)\:$ vanishing at all points in $\rm\:I\:.\:$ Then $\rm\:f_1,\ldots,f_n\:$ are linearly dependent on some subinterval of $\rm\:I\:.$
Proof $\$ We employ the following easily proved Wronskian identity:
$\rm\qquad\ W(g\ f_1,\ldots,\:g\ f_n)\ =\ g^n\ W(f_1,\ldots,f_n)\:.\$ This immediately implies
$\rm\qquad\quad\ \ \ W(f_1,\ldots,\: f_n)\ =\ f_1^{\:n}\ W((f_2/f_1)’,\ldots,\:(f_n/f_1)’\:)\quad$ if $\rm\:\ f_1 \ne 0$
Proceed by induction on $\rm\:n\:.\:$ The Theorem is clearly true if $\rm\:n = 1\:.\:$ Suppose that $\rm\: n > 1\:$ and $\rm\:W(f_1,\ldots,f_n) = 0\:$ for all $\rm\:x\in I.\:$
If $\rm\:f_1 = 0\:$ throughout $\rm\:I\:$ then $\rm\: f_1,\ldots,f_n\:$ are dependent on $\rm\:I.\:$ Else $\rm\:f_1\:$ is nonzero at some point of $\rm\:I\:$ so also throughout some subinterval $\rm\:J \subset I\:,\:$ since $\rm\:f_1\:$ is continuous (being differentiable by hypothesis). By above $\rm\:W((f_2/f_1)’,\ldots,(f_n/f_1)’\:)\: =\: 0\:$ throughout $\rm\:J,\:$ so by induction there exists a subinterval $\rm\:K \subset J\:$
where the arguments of the Wronskian are linearly dependent, i.e.
on $\rm\ K:\quad\ \ \ c_2\ (f_2/f_1)’ +\:\cdots\:+ c_n\ (f_n/f_1)’\: =\ 0,\ \$ all $\rm\:c_i’\:=\ 0\:,\$ some $\rm\:c_j\ne 0$
$\rm\qquad\qquad\: \Rightarrow\:\ \ ((c_2\ f_2 +\:\cdots\: + c_n\ f_n)/f_1)’\: =\ 0\ \$ via $({\phantom m})’\:$ linear
$\rm\qquad\qquad\: \Rightarrow\quad\ \ c_2\ f_2 +\:\cdots\: + c_n\ f_n\ =\ c_1 f_1\ \$ for some $\rm\:c_1,\ c_1’\: =\: 0$
Therefore $\rm\ f_1,\ldots,f_n\:$ are linearly dependent on $\rm\:K \subset I\:.\qquad$ QED
This theorem has as immediate corollaries the well-known results
that the vanishing of the Wronskian on an interval $\rm\: I\:$ is
a necessary and sufficient condition for linear dependence of
$\rm\quad (1)\$ functions analytic on $\rm\: I\:$
$\rm\quad (2)\$ functions satisfying a monic homogeneous linear differential
equation
$\rm\quad\phantom{(2)}\$ whose coefficients are continuous throughout $\rm\: I\:.\:$
Here is a different take on the question. There is a whole spectrum of different discrete “calculi” which converge to the continuous case, each of which has it’s special “$e$”.
Pick some $t>0$. Consider the equation $$f(x)=\frac{f(x+t)-f(x)}{t}$$
It is not hard to show by induction that there is a function $C_t:[0,t)\to \mathbb{R}$ so that $$f(x)=C_t(\{\frac{x}{t}\})(1+t)^{\lfloor\frac{x}{t}\rfloor}$$
where $\{\cdot\}$ and $\lfloor\cdot\rfloor$ denote fractional and integer part, respectively. If I take Qiaochu’s answer for comparison, then $C_t$ plays the role of the constant $C$ and $(1+t)^{\lfloor\frac{x}{t}\rfloor}$ the role of $e^x$. Therefore for such a discrete calculus the right value of “$e$” is $(1+t)^{1/t}$. Now it is clear that as $t\to 0$ the equation becomes $f(x)=f'(x)$, and $(1+t)^{1/t}\to e$.
The solutions of $f(x) = f'(x)$ are exactly $f(x) = f(0) e^x$. But you can also write it as $b a^x$, if you want a different basis. Then $f'(x) = b \log(a) a^x$, and so if you want $f'=f$ you need $\log(a)=1$ and $a=e$ (except for the trivial case $b=0$).
The proof they use at High School, so not as deep or instructive, but it doesn’t require as much knowledge.
$$\begin{eqnarray*} \frac{dy}{dx} &=& y\\ \frac{dx}{dy} &=& \frac 1 y\\ x &=& \log |y| + C\\ y &=& A\exp(x) \end{eqnarray*}$$
Let $x \in C^1$ on the whole line be a solution to $\dot{x}(t) = x(t)$, $x(0) = 1$. Using the Taylor expansion with remainder, show that necessarily $x(t) = e^t$.
We have that $\dot{x} = x$ implies $x^{(n)} = x^{(n-1)}$ for all $n \ge 1$, and by induction on $n$, we have that $x(t)$ is $C^\infty$ with $x^{(n)} = x$ for all $n$. Thus, if $x(0) = 1$ and $\dot{x} = x$, Taylor’s Theorem gives$$x(t) = \left( \sum_{k=0}^{N-1} {{t^k}\over{k!}}\right) + {{x^{(N)}(t_1)}\over{N!}}t^N,$$for $t_1$ between $0$ and $t$. But $x^{(N)} = x$, so if$$M = \max_{|t_1| \le |t|} |x(t)|,$$which we know exist by compactness of $[-|t|, |t|]$, then$$\left| x(t) – \sum_{k=0}^{N-1} {{t^k}\over{k!}}\right| < {{Mt^N}\over{N!}}.$$The right-hand side heads to $0$ as $N \to \infty$, so the series for $e^t$ converges to $x(t)$.
Note that $e$ is defined by the following Limit: $e=\lim_{n \rightarrow \infty}(1+ \frac{1}{n})^n$. Then: $e^x=\lim_{n \rightarrow \infty}(1+ \frac{1}{n})^{nx}$. Applying the Definition of the derivative $f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$ one obtains: $(e^x)’=\lim_{h \rightarrow 0} \frac{ \lim_{n \rightarrow \infty}((1+ \frac{1}{n})^{n(x+h)}-(1+ \frac{1}{n})^{nx})}{h} = \lim_{h \rightarrow 0}( \lim_{n \rightarrow \infty}(1+\frac{1}{n})^{nx} \lim_{n \rightarrow \infty}(\frac{(1+\frac{1}{n})^{nh}-1}{h}))$
$= e^x \lim_{h \rightarrow 0} \lim_{n \rightarrow \infty}(\frac{(1+\frac{1}{n})^{nh}-1}{h})$.
Now one can replace $h$ by $n$ by the relation $h= \frac{C}{n}$ with a finite constant $C$, because if $n \rightarrow \infty$ then $h$ tends to Zero. Hence:
$\lim_{h \rightarrow 0} \lim_{n \rightarrow \infty}(\frac{(1+\frac{1}{n})^{nh}-1}{h}) = \lim_{h \rightarrow 0} (\frac{(1+\frac{h}{C})^{C}-1}{h}) = \lim_{h \rightarrow 0} (\frac{(1+C \frac{h}{C} + \frac{C(C-1)}{2}(\frac{h}{C})^2+O(h^3)-1}{h}) = \lim_{h \rightarrow 0} (1 + \frac{C(C-1)}{2}\frac{h}{C^2}+O(h^2)) = 1$
Therefore $(Ce^x)’=C(e^x)’=Ce^x$.
q.e.d. | 2018-06-22T13:17:07 | {
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http://math.stackexchange.com/questions/329838/how-to-simplify-a-square-root | # How to simplify a square root
How can the following:
$$\sqrt{27-10\sqrt{2}}$$
Be simplified to:
$$5 - \sqrt{2}$$
Thanks
-
If you're faced with a question that says "Prove that $\sqrt{27-10\sqrt{2}}$ $=5 - \sqrt{2}$", then it's just a matter of squaring $5 - \sqrt{2}$ and seeing that you get $27-10\sqrt{2}$. But suppose the question your faced with is to find a square root of $27-10\sqrt{2}$ of the form $a+b\sqrt{2}$, where $a$ and $b$ are rational. Then you have $$27-10\sqrt{2}=\left(a+b\sqrt{2}\right)^2 = a^2 + 2ab\sqrt{2} + 2b^2$$ so \begin{align} 27 & = a^2+2b^2 \\[8pt] -10 & = 2ab \end{align} From the second equation we get $a=-5/b$, then the first equation becomes $$27 = \frac{25}{b^2} + 2b^2$$ or $$2(b^2)^2 -27b^2 + 25 = 0.$$ A solution is $b^2=1$, and you can go on from there to find $b$ and then $a$.
(And remember that the number will have two square roots.)
Later note: In order for all this to work, we have to rely on the fact that $\sqrt{2}$ is irrational. That enables us to conclude that the rational parts are equal and the irrational parts are equal, so we have two equations.
-
It might be worth to mention that $5-\sqrt 2>0$. – Git Gud Mar 13 '13 at 23:31
No lucky guesses are needed, there is a simple denesting algorithm for $\rm\:\sqrt{a+b\sqrt{n}}$
Simple Denesting Rule $\rm\ \ \ \color{#0A0}{subtract\ out}\ \sqrt{norm}\:,\ \ then\ \ \color{brown}{divide\ out}\ \sqrt{trace}$
$\begin{array}{lll}\rm Recall\ \ \ w = a + b\sqrt{n}\ \ \ has\!\!\! &{\bf norm} &\!\!\!\rm=\: w\:\cdot\: w' = &\!\!\!\!\rm(a + b\sqrt{n})\ \cdot\: (a - b\sqrt{n})\ =\: a^2 - n\: b^2 \\ \\ \rm and,\ furthermore,\ \ w\ \ has \!\!\!&{\bf trace} &\!\!\!\!\rm =\: w+w' = &\!\!\!\!\rm (a + b\sqrt{n}) + (a - b\sqrt{n})\: =\: 2\:a\end{array}$
Here $\:27-10\sqrt{2}\:$ has norm $= 23^2.\,$ $\rm\, \color{#0A0}{Subtracting\ out}\ \sqrt{norm}\ = -23\$ yields $\ 50-10\sqrt{2}\:$
and this has $\rm\ \sqrt{trace}\: =\: 10,\ \ hence\ \ \ \color{brown}{dividing\ it\ out}\$ of this yields $\rm\ 5 - \sqrt{2} =\:$ sought sqrt.
Remark $\$ The sign of the norm sqrt was chosen to make the trace sqrt rational. The same answer would arise using the opposite sign, but with slightly more work (rationalizing a denominator). $\$ For many further examples see other posts on radical denesting.
-
Nice technique and a good example of why there are really two solutions when taking the square root, contrary to what most mathematicians would like it to be (a single-valued function), and that it doesn't matter much whether you take the square root of an unknown or a "simple number". – SasQ Dec 14 at 22:26
Where does these names come from? ("norm" and "trace") I see that the first one is similar to calculating the norm of a complex number, but there's no $i$ in it, so why does it still work similarly to complex conjugates? And why is the norm subtracted instead of divided out? Why is it the trace which is divided out instead? The trick is really nice, but it would be nicer if you explained why does this magic work so well. – SasQ Dec 14 at 22:29
Oh, and one thing you wrote is a bit misleading: You put this in a way suggesting that one has to divide by the trace of the original square-rooted number ($w$), but when I tried this, it didn't work. One has to divide out by the trace of the partial answer obtained along the way instead, which is not the original $w$, but $w - \sqrt{norm}$. – SasQ Dec 14 at 22:32
Note that $27=5^2+\sqrt2^2$ and $10\sqrt2=2\times5\times\sqrt2$
-
Hint: Compare the squares of both expressions.
-
Notice that \begin{align}27-10\sqrt{2} & = 25 - 2\cdot 5\sqrt{2} + 2 \\ & = 5^2 - 2\cdot5\sqrt2 + \left(\sqrt2\right)^2\\ & = \left(5 - \sqrt 2\right)^2 \end{align}
I don't know any way to notice this except to just get lucky and notice it; I didn't realize it could happen until sometime in high school when I was astounded to discover that $\sqrt{7+4\sqrt3} = 2+\sqrt3$.
-
Set the nested radical as the difference of two square roots so that $$\sqrt{27-10\sqrt{2}}=(\sqrt{a}-\sqrt{b})$$ Then square both sides so that $$27-10\sqrt{2}=a-2\sqrt{a}\sqrt{b}+b$$ Set (1) $$a+b=27$$ and set (2) $$-2\sqrt{a}\sqrt{b}=-10\sqrt{2}$$ Square both sides of (2) to get $$4ab= 200$$ and solve for $b$ to get $$b=\frac{50}{a}$$ Replacing $b$ in (1) gives $$a+\frac{50}{a}=27$$ Multiply all terms by $a$ and convert to the quadratic equation $$a^2-27a+50=0$$ Solving the quadratic gives $a=25$ or $a=2$. Replacing $a$ and $b$ in the first difference of square roots formula above with $25$ and $2$ the solutions to the quadratic we have $$\sqrt{25}-\sqrt{2}$$ or $$5-\sqrt{2}$$
-
What if the solutions to the quadratic were also irrational, so that $a$ and $b$ become nested inside their radicals again? Or is this possibility somehow excluded? – SasQ Dec 14 at 22:22
$$\sqrt{27 - 10\sqrt{2}}$$ $$= \sqrt{25 + 2 - 2.5.\sqrt{2}}$$ $$= \sqrt{5^{2} - 2.5.\sqrt{2} +{\sqrt{2}}^2}$$ $$= \sqrt{{(5 - \sqrt{2})}^2}$$
$$= (5 - \sqrt{2})$$ $$= R.H.S$$
-
You need equals signs rather than arrows. I would have written something similar as $$\sqrt{27-10\sqrt{2}} = \sqrt{27-2\sqrt{ 25 \times 2}} = \sqrt{25-2\sqrt{ 25 }\sqrt{2} +2} = \sqrt{(\sqrt{25} - \sqrt{2})^2}$$ – Henry Mar 14 '13 at 8:51
@Henry: Thanks corrected. Its been ages I am solving some math problem – Abhijit Mar 14 '13 at 9:13
But then you have to "guess" how to split that $27$. – SasQ Dec 14 at 22:23 | 2014-12-22T05:44:30 | {
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https://mathhelpboards.com/threads/find-a-b.4213/ | # Find a + b.
#### anemone
##### MHB POTW Director
Staff member
If $$\displaystyle f(x)=x^3-6x^2+17x$$ and $$\displaystyle f(a)=16$$ and $$\displaystyle f(b)=20$$, find $$\displaystyle a+b$$.
#### Opalg
##### MHB Oldtimer
Staff member
Re: Find a+b.
If $$\displaystyle f(x)=x^3-6x^2+17x$$ and $$\displaystyle f(a)=16$$ and $$\displaystyle f(b)=20$$, find $$\displaystyle a+b$$.
The graph of $f$ has a rotational symmetry about the point $(2,18)$. In fact, you can check that $f(2-x) + f(2+x) = 18+18 =36$. But $36 = 16+20$, and it follows that if $f(a) = 16$ and $f(b) = 20$ then $a=2-x$ and $b=2+x$ for some $x$. Therefore $a+b=4.$
#### Fernando Revilla
##### Well-known member
MHB Math Helper
Re: Find a+b.
Another way. We have $$f(a)=16\Leftrightarrow (a-2)^3+5a-8=0\\f(b)=20\Leftrightarrow (b-2)^3+5b-12=0$$ Denoting $\alpha=a-2,\;\beta =b-2$: $$f(a)=16\Leftrightarrow \alpha^3+5\alpha+2=0\\f(b)=20\Leftrightarrow \beta^3+5\beta-2=0$$ But if $r$ is a root of $x^3+5x+2$ iff $-r$ is a root of $x^3+5x-2$ so, $$0=\alpha+\beta=a-2+b-2\Rightarrow a+b=4$$ P.S. Of course, this is valid for $a,b\in\mathbb{R}$, otherwise $a+b$ can be $\neq 4$.
#### anemone
##### MHB POTW Director
Staff member
Re: Find a+b.
As always, thanks to Opalg and Fernando for participating in this problem and to be honest, I learned a lot from both of the solutions.
I solved this problem differently and here goes my solution:
From
$$\displaystyle a^3-6a^2+17a-16=0$$ and $$\displaystyle b^3-6b^2+17b-20=0$$,
We add the equations to find that
$$\displaystyle a^3+b^3-6a^2-6b^2+17a+17b-16-20=0$$
$$\displaystyle (a^3+b^3)-6(a^2+b^2)+17(a+b)-36=0$$
$$\displaystyle ((a+b)^3-3ab(a+b))-6((a+b)^2-2ab)+17(a+b)-36=0$$
By replacing $$\displaystyle k=a+b$$ yields
$$\displaystyle (k^3-3abk)-6(k^2-2ab)+17k-36=0$$
This simplifies to
$$\displaystyle k^3-6k^2-36+(17-3ab)k+12ab=0$$
$$\displaystyle k^3-6k^2-36-(3ab-17)k+4(3ab-17)+4(17)=0$$
$$\displaystyle k^3-6k^2+32+(3ab-17)(4-k)=0$$
$$\displaystyle (k+2)(k-4)^2+(3ab-17)(4-k)=0$$
Therefore, we can conclude that $$\displaystyle k=4$$ must be true.
This implies $$\displaystyle a+b=4$$.
But hey, it is very obvious that my method is the least impressive/worst one...(bh)...
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Re: Find a+b.
As always, thanks to Opalg and Fernando for participating in this problem and to be honest, I learned a lot from both of the solutions.
I solved this problem differently and here goes my solution:
From
$$\displaystyle a^3-6a^2+17a-16=0$$ and $$\displaystyle b^3-6b^2+17b-20=0$$,
We add the equations to find that
$$\displaystyle a^3+b^3-6a^2-6b^2+17a+17b-16-20=0$$
$$\displaystyle (a^3+b^3)-6(a^2+b^2)+17(a+b)-36=0$$
$$\displaystyle ((a+b)^3-3ab(a+b))-6((a+b)^2-2ab)+17(a+b)-36=0$$
By replacing $$\displaystyle k=a+b$$ yields
$$\displaystyle (k^3-3abk)-6(k^2-2ab)+17k-36=0$$
This simplifies to
$$\displaystyle k^3-6k^2-36+(17-3ab)k+12ab=0$$
$$\displaystyle k^3-6k^2-36-(3ab-17)k+4(3ab-17)+4(17)=0$$
$$\displaystyle k^3-6k^2+32+(3ab-17)(4-k)=0$$
$$\displaystyle (k+2)(k-4)^2+(3ab-17)(4-k)=0$$
Therefore, we can conclude that $$\displaystyle k=4$$ must be true.
This implies $$\displaystyle a+b=4$$.
But hey, it is very obvious that my method is the least impressive/worst one...(bh)...
Hey anemone!
I'm afraid your solution has a couple of flaws in it. (bh)
The solution $$\displaystyle k=4$$ is only 1 of the possible solutions.
There may be more solutions (although there aren't any).
Furthermore, it is only a solution for the sum of the 2 equations being equal to zero.
There is no guarantee that the individual equations are zero.
They may still have opposite results (although they don't).
Sorry.
#### Fernando Revilla
##### Well-known member
MHB Math Helper
Re: Find a+b.
The problem should say:
If $$\displaystyle f(x)=x^3-6x^2+17x$$ and $$\displaystyle f(a)=16$$ and $$\displaystyle f(b)=20$$ with $a,b\in\mathbb{R}$, find $$\displaystyle a+b$$.
Oherwise, we can't guarantee $a+b=4$. I solved the question with that additional hypothesis. | 2021-09-23T22:17:21 | {
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https://math.stackexchange.com/questions/1483510/additional-information-in-conditional-probability-problem-gives-a-different-answ | # Additional information in conditional probability problem gives a different answer but this does not seem correct to me?
Q1.) I tell you that I have two children and that one of them is a girl. What is the probability that I have two girls? Assume girls and boys equally likely to be born and that the gender of one child is independent of the other.
The answer is 1/3 which I understand and agree with.
It can be seen that of the possible combinations {BB,GB,BG,GG} the first combination is not possible. In the remaining three cases, there is only one in which there are two girls which gives the correct answer.
Q2.) I tell you that I have two children and that one of them is a girl. You knock on the front door and are greeted by a girl who you correctly deduce is my daughter. What is the probability that I have two girls? Assume girls and boys equally likely to be born and that the gender off one child is independent of the other. Compare and contrast the answers.
The claimed answer here is 0.5. That is on the basis that the girl at the door is taken to be #1 and therefore the only randomness remaining is the gender of the second child which is given in the question is 50%.
However this doesn't seem correct to me, how can you know that the girl is #1 and not #2? Seeing the girl does not seem to preclude {GB,BG,GG} so the answer should be the same as above?
Baz
• You say for the first scenario that "the probability that I have two girls" yet the probability of $\frac{2}{3}$ seems to be for "the elder child is a girl." Shouldn't the probability of two girls in that scenario instead be $\frac{1}{3}$? – JMoravitz Oct 16 '15 at 18:21
• As for defining the difference between child #1 and child #2, this could be done based on height, name, age, or in particular for the current situation first met. Each of which are perfectly valid. We would need an additional assumption that a boy is equally likely to answer the door as a girl in the case that both are present, (and to have height as the order, that boys and girls in this family are equally likely to be taller than one another). – JMoravitz Oct 16 '15 at 18:23
• Sorry yes! I'm not sure this is a duplicate the questions are very similar but the wording is crucial? – Bazman Oct 16 '15 at 18:24
• But the answer makes no mention of this it just says that the child which is seen is taken to be #1 – Bazman Oct 16 '15 at 18:25
• Sorry it is a dupe: math.stackexchange.com/questions/15055/… the answer provided by blueraja seems to makes sense – Bazman Oct 16 '15 at 18:32
Assuming that boys and girls in this household are equally likely to answer the door and that only one person ever answers the door at a time, we may choose the order of the children to be based on first seen.
In this sense, we can indeed claim that the girl that we saw at the door is "child #1" and the result follows.
However, this assumes that boys and girls are equally likely to answer the door.
Consider an opposite extreme where you know that our entire family is home at the time and our family follows some very strict religious customs that only a boy may answer the door unless no boys are present. In this scenario, the fact that a girl answered the door directly implies that there were no boys available to answer the door, which in turn implies that I must have two daughters and no sons.
I would say then that the problem is poorly worded and there is not yet enough information to conclude that the probability is indeed $\frac{1}{2}$ without additional assumptions (even if the assumption seems valid).
• Thanks, have to agree that the question is too loosely worded but I understand the difference better now. – Bazman Oct 16 '15 at 18:34
In the first case, you are quite correct. By asserting that one of your children is a girl, what you effectively said is that the older child is a girl, or the younger child is a girl, and you haven't ruled out the possibility that both children could be girls. Hence, the answer is, indeed, $\frac13.$
In the second case, however, exactly one of the children has answered the door, and that child is a daughter. The only question is whether the child who didn't answer the door is a girl or a boy. You don't know if it's the older child or the younger child who has answered the door, but exactly one of them has. If the older child answered the door, then it is impossible that the older child is a boy, so there are only two possibilities remaining. Likewise, if it was the younger child who answered the door, then it is impossible that the younger child is a boy, and so there are only two possibilities remaining. Regardless of which child answered the door, the answer is then $\frac12.$
• OK but this is where I was/am confused. If a girl answers the door and you are not making any assumptions that she is the eldest (or indeed youngest) then surely you can still have {GB, BG, GG}? The observation that a girl answered the door is consistent with any of those three combinations? – Bazman Oct 16 '15 at 18:40
• @Bazman You have $4$ times the letter $G$ in $\{GB,BG,GG\}$ and exactly $2$ of these letters $G$ are accompanied by another letter $G$. – drhab Oct 16 '15 at 18:44
• True, we are not making any assumptions about whether the child who answered the door is the youngest or the oldest, but one of them did, and by answering the door, precluded one possibility. We don't know which possibility was ruled out, but we know that one of them was. After all, if the daughter who answered the door is the elder child, then the possibilities are GB and GG. On the other hand, if the daughter who answered the door is the younger child, then the possibilities are BG and GG. We don't know which of these two distinct events we're dealing with, but that doesn't change things. – Cameron Buie Oct 16 '15 at 18:54
• We are dealing with exactly one of those two events, and the event we're dealing with (whichever it may be) has probability $\frac12.$ – Cameron Buie Oct 16 '15 at 18:56
• @Bazman: Apologies! I made an error in my comments. I should have said that our sample space is either {GB,GG} or {BG,GG}, though we don't know which. However, regardless of which of these is our sample space, the event {GG} has probability $\frac12.$ Does that make more sense? – Cameron Buie Oct 16 '15 at 19:12
I'll expand two of the answers that are already given. They are based on two different models: in one, the child answering the door is chosen based on age, and in the other by sex. And the answer depends on which model is chosen.
1. Suppose that the elder child opens the door with probability $p$ and the younger with probability $1-p$. Let G@D denote the event that you see a girl at the door. We have: $$P\left(GG|G@D\right)=\frac{P\left(G@D|GG\right)P(GG)}{P\left(G@D\right)}=\frac{1/3}{1/3+p/3+\left(1-p\right)/3}=\frac{1}{2}.$$
2. Suppose that, if the family has a boy and a girl, the girl opens the door with probability $q$. Now, $$P\left(GG|G@D\right)=\frac{P\left(G@D|GG\right)P(GG)}{P\left(G@D\right)}=\frac{1/3}{1/3+q/3+q/3}=\frac{1}{1+2q}.$$
So the answer is $1/2$ only if we pick the first model or if $q=1/2$ in the second model. | 2020-04-02T06:58:56 | {
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https://electronics.stackexchange.com/questions/559750/trouble-with-meeting-design-specifications-of-a-second-order-system-using-matlab | # Trouble with meeting design specifications of a second order system using Matlab
I am trying to solve the following question.
Consider the transfer function $$G(s) = \frac{1.247}{s^2+9.76s+23.8}$$ is in the forward path of a unity feedback control loop. Assume that it is compensated using a static gain K in the forward path. Now I have to plot the root locus of closed-loop system and determine if it is possible to find a value for K such that the settling time is 0.7963 s and the rise time is 0.4445 s.
I know how to do it on paper. The overall closed loop transfer function become $$G(s) = \frac{K*1.247}{s^2+9.76s+23.8+(K*1.247)}$$
If I compare this with standard form of second order system, I can see that $$2*\zeta*\omega_n = 9.76$$ and $$\zeta\omega_n = 4.88$$ From the formula for settling time, $$t_s = \frac{4}{\zeta\omega_n}$$ $$\zeta\omega_n = \frac{4}{0.7963} = 5.02$$
Therefore, it is not possible to design the system by just varying the value of K. But my doubt is, is there any way to directly find it from the root-locus plot in Matlab. How do I plot the root locus in the first place without knowing the value of K?
• The formula for settling time is approximate. In this case, $t_s \approx \frac{4}{4.88}=0.82$, and you cannot change this value. You can, however, select K to achieve the required 10%-90% rise time (once again, the formula for this is approximate). So, yes, this can all be done via the root locus, but don't expect the system to produce the exact values given in the design specification.
– Chu
Apr 12, 2021 at 12:02
• Please add a plot of the root locus. I think the answer will be obvious from the root locus plot itself. Mark the area on the complex s-plane where settling time is 0.79 and rise time is 0.44 (formulae for both exist for second order systems). See if the root locus passes through that area.
– AJN
Apr 12, 2021 at 12:21
• "How do I plot the root locus in the first place without knowing the value of K?" Isn't root locus plotted by varying K from zero to infinity ? So, there is no need to know the "final" value of K while plotting the locus.
– AJN
Apr 12, 2021 at 12:22
• Why do you evaluate the closed loop, since you said it is in feed forward path, therefore open loop? make a depiction of the block diagram of the entire loop. Apr 12, 2021 at 12:48
I think that the mistake you are making is assuming $$\\omega_n\$$ is unaffected by $$\K\$$. In your closed-loop $$\G(s)\$$ formula, $$\\omega_n\$$ is in fact this: -
$$\sqrt{23.8 + 1.247\cdot K}$$
Hence, $$\\zeta = \dfrac{5.02}{\sqrt{23.8 + 1.247\cdot K}}\$$ | 2022-12-10T05:17:35 | {
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https://cs.stackexchange.com/questions/10511/methods-for-finding-asymptotic-lower-bounds?noredirect=1 | # Methods for Finding Asymptotic Lower Bounds
I've found in many exercises where I'm asked to show that $f(n)=\Theta(g(n))$ where the two functions are of the same order of magnitude I have difficulty finding a constant $c$ and a value $n_0$ for the lower bound. I'm using Corman's definition of $\Theta$:
$$\exists c_1,c_2>0\in\mathbb{R}:\forall n\geq n_0: 0 \leq c_1 g(n)\leq f(n)\leq c_2 g(n)$$
Showing the upper bound usually doesn't give me too much trouble, but for the lower bound I allot of times find myself using limits. And even though I'm getting the right answers, I'm a bit worried that my method isn't very rigorous and that maybe I'm doing a bit of hand waving in the process.
For example, problem 2.17 from Skiena's Algorithm Design Manual:
Show that for any $a,b\in \mathbb{R}: b>0$ that $(n+a)^b = \Theta(n^b)$
In this case I used limits to help find both constants.
For the upper limit I decided to look for some $c$ such that $(n+a)^b \leq c^bn^b$. So taking the $b$th root of each side and dividing by $n$ I have $\frac{n+a}{n}\leq c$ which gives me $1 + \frac{a}{n} \leq c$. For any $a\in\mathbb{R}$, $\lim_{n\to\infty }1+\frac{a}{n}=1$. If I pick $n_0>|a|$, then for $a<0$ the expression approaches 1 from the left starting arbitrarily close to $0$. If $a>0$ then the expression approaches 1 from the right starting arbitrarily close to 2. So choosing $c=2$ will satisfy the inequality and we have $c_2=2^b$.
Now for the lower bound. I'm looking at the same expression except with the inequality pointing the other way. In this case I'm trying to find $n_0$ and $c$ such that $c\leq 1+\frac{a}{n}$. The value of $n_0$ has to be greater than $|a|$ because otherwise we would have $c\leq 0$ which isn't allowed. This puts us in the same range of values between $0$ and $2$ approaching 1 from each side. So I choose any $c,n_0$ such that $n_0>|a|$ and $0 < c\leq 1-|\frac{a}{n_0}|$. So I could choose $n_0=3|a|$ and $c=\frac{2}{3}$.
Thus we have $0 < (\frac{2}{3})^bn^b \leq (n+a)^b \leq 2^bn^b$ for any $n \geq 3|a|$.
Is there an easier way to do this?
Normally when looking for upper limit constants where the two functions are of the same magnitude I simply eliminate negative lower order terms and change positive ones into multiples of the highest order term such as :
$$3n^2+15n-5\leq 3n^2+15n^2=18n^2$$
But when looking for the constant for the lower bound I find myself typically resorting to looking at limits. Is there any kind of short cut to finding the lower bound constant like there is for the upper bound constant?
• Looks fine to me. – vonbrand Mar 13 '13 at 20:32
• @vonbrand My question is if the use of limits to find the lower bound is really necessary, or is there some easier short-cut way of finding $c,n_0$ for the lower bound? – Robert S. Barnes Mar 14 '13 at 6:39
• See also here. If you already have the function, finding lower bounds is boring. Imho. In particular, $(n+a)^b$ is a well-studied expression, cf binomial theorem. – Raphael Mar 14 '13 at 14:31
• Rereading the question, what's wrong with the obvious: for a lower bound, drop positive terms. In your example, we have for large $n$: $3n^2 + 15n - 5 \geq 3n^2 + 14n \geq 3n^2$, which is obviously a tight lower bound on the leading coefficient. – Raphael Mar 15 '13 at 7:13
You can codify your method in the following lemma.
Lemma. If $f(n)/g(n) \rightarrow C$, where $C > 0$, then $f(n) = \Theta(g(n))$.
The proof is the same as the one you gave. After you prove this lemma once and for all, you can use it forever. That's actually a good way of verifying $f(n) = \Theta(g(n))$.
Note that the converse to the lemma isn't true. For example, let $f(n) = n$ and let $g(n) = \exp\lfloor\log n\rfloor$. The ratio $f(n)/g(n)$ moves inside the interval $[1,e)$, and in particular does not tend to a constant limit.
• So you think this is a good general method and there's not really any easier short-cut way of finding the lower bound? – Robert S. Barnes Mar 14 '13 at 6:32
• @RobertS.Barnes Depends on what quality of bound you want and what kind of algorithm you have at hand. – Raphael Mar 14 '13 at 14:30
Depending what $f$ and $g$ are precisely, you can use l'Hôpital's rule. The description from Wolfram Mathworld is quite good (the Wikipedia leaves out some cases leaving the definitions not quite correct).
In short (for the purposes of algorithmic analysis), if $\lim_{n\rightarrow\infty} f(n) = \lim_{n\rightarrow\infty} g(n) = 0$ or $\infty$, then: $$\lim_{n\rightarrow\infty}\frac{f(n)}{g(n)} = \lim_{n\rightarrow\infty}\frac{f'(n)}{g'(n)}$$ (as long as $\lim_{n\rightarrow\infty}\frac{f'(n)}{g'(n)}$ exists)
So as long as you have fairly normal functions for your running time, and you remember some calculus, you can repeatedly apply this rule to determine whether $\lim_{n\rightarrow\infty}\frac{f(n)}{g(n)} = c$ for some constant $c$.
As a side note, the other cases still apply, so you can apply l'Hôpital's rule until you get one of three cases, the limit is $0$, $\infty$ or a constant, concluding that $f(n)\in O(g(n))$, $f(n)\in\Omega(g(n))$ or $f(n)\in\Theta(g(n))$ respectively.
The same caveat on the converse that Yuval mentions is of course still true here.
Raphael also gives some important limitations on using l'Hôpital with discrete functions (in that strictly you can't). See his comments here.
• Indeed, that's what I've been doing. But my question was if there is any shortcut for finding an $n_0$ and constant $c$ for the lower bound like there is for the upper bound. – Robert S. Barnes Mar 14 '13 at 10:55
• @RobertS.Barnes, ah, I was thinking of a shortcut to get the $\Theta$ result, rather than a shortcut to actually find the constant. However... I have a feeling (but not a proof on hand), that the limit value does give some information in this regard. I'll do some thinking and looking, or perhaps someone else can recall. A very hand wavy suggestion is that the $\lim\frac{f(n)}{g(n)}$ gives one, and $\lim\frac{g(n)}{f(n)}$ gives the other. Though it doesn't help immediately with the $n_{0}$. – Luke Mathieson Mar 14 '13 at 13:44
• Regarding L'Hospital, please note my comments here; it's not a perfect match to our situation. – Raphael Mar 14 '13 at 14:31
• @Raphael, good point, usually ignored. I'll link to it in the body. – Luke Mathieson Mar 15 '13 at 0:40 | 2020-01-21T15:39:33 | {
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https://www.freemathhelp.com/forum/threads/what-is-happening-here.114104/ | # what is happening here?
#### allegansveritatem
##### Full Member
This seems to be a very simple equation...but somehow the solution is eluding me. I have tried again and again and can't get it to come out right. lIt is easy to see what the answer should be without working it out. But when I try to work it out, Mr Right doesn't show up.
Equation:
What I keep getting:
#### Dr.Peterson
##### Elite Member
This seems to be a very simple equation...but somehow the solution is eluding me. I have tried again and again and can't get it to come out right. lIt is easy to see what the answer should be without working it out. But when I try to work it out, Mr Right doesn't show up.
Equation:
View attachment 10826
View attachment 10827
What I keep getting:
View attachment 10828
When you square both sides, you have to square each entire side -- not just each term on each side.
The square of $$\displaystyle \sqrt{m+4}+\sqrt{m-4}$$ is not $$\displaystyle \sqrt{m+4}^2+\sqrt{m-4}^2$$, but $$\displaystyle (\sqrt{m+4}+\sqrt{m-4})^2$$. Do you see the difference?
Doing that will not simplify the equation. What you have to do is to isolate one radical (move the other to the right side), and then square each side.
Your textbook should have examples of this.
#### HallsofIvy
##### Elite Member
$$\displaystyle (a+ b)^2= a^2+ 2ab+ b^2$$, not $$\displaystyle a^2+ b^2$$.
$$\displaystyle (\sqrt{m+4}+ \sqrt{m-4})^2=$$
$$\displaystyle (\sqrt{m+4})^2+ 2\sqrt{m+4}\sqrt{m-4}+ (\sqrt{m-4})^2$$
$$\displaystyle = m+ 4+ 2\sqrt{(m+4)(m-4)}+ m- 4= 2m+ 2\sqrt{m^2- 16}$$.
So the equation, after squaring both sides, is $$\displaystyle 2m+ 2\sqrt{m^2- 16}= 16$$. Rewrite this as $$\displaystyle \sqrt{m^2- 16}= 8- m$$ and square both sides again to get an equation for m.
Caution: While any solution to the original equation must satisfy the new equation, squaring both sides of an equation can introduce "spurious solutions"- solutions to the new equation that do not satisfy the original equation. Be sure to check any solutions to your final equation in the original equation!
Last edited:
#### tkhunny
##### Moderator
Staff member
This seems to be a very simple equation...but somehow the solution is eluding me. I have tried again and again and can't get it to come out right. lIt is easy to see what the answer should be without working it out. But when I try to work it out, Mr Right doesn't show up.
Equation:
View attachment 10826
View attachment 10827
What I keep getting:
View attachment 10828
Please learn to experiment with simpler examples. Maybe...
$$\displaystyle 4 + 3 = 7$$
$$\displaystyle 16 + 9 \ne 49$$ -- Something went wrong.
#### Jomo
##### Elite Member
Please learn to experiment with simpler examples. Maybe...
$$\displaystyle 4 + 3 = 7$$
$$\displaystyle 16 + 9 \ne 49$$ -- Something went wrong.
I 2nd that technique.
I turd it!
#### allegansveritatem
##### Full Member
When you square both sides, you have to square each entire side -- not just each term on each side.
The square of $$\displaystyle \sqrt{m+4}+\sqrt{m-4}$$ is not $$\displaystyle \sqrt{m+4}^2+\sqrt{m-4}^2$$, but $$\displaystyle (\sqrt{m+4}+\sqrt{m-4})^2$$. Do you see the difference?
Doing that will not simplify the equation. What you have to do is to isolate one radical (move the other to the right side), and then square each side.
Your textbook should have examples of this.
I will look at this more closely in the morning but I see what is being said..and also, I know that squaring this type of problem with all the radicals on one side is not a means to an easy life. In fact I did try to move one of the radical terms to the right but somehow things seemed to end up exactly the same. But I will go at this again tomorrow and keep what you say in mind. Thanks for pointing out the problem. I think I did it this way in the hope of avoiding a messy FOIL.
#### allegansveritatem
##### Full Member
$$\displaystyle (a+ b)^2= a^2+ 2ab+ b^2$$, not $$\displaystyle a^2+ b^2$$.
$$\displaystyle (\sqrt{m+4}+ \sqrt{m-4})^2=$$
$$\displaystyle (\sqrt{m+4})^2+ 2\sqrt{m+4}\sqrt{m-4}+ (\sqrt{m-4})^2$$
$$\displaystyle = m+ 4+ 2\sqrt{(m+4)(m-4)}+ m- 4= 2m+ 2\sqrt{m^2- 16}$$.
So the equation, after squaring both sides, is $$\displaystyle 2m+ 2\sqrt{m^2- 16}= 16$$. Rewrite this as $$\displaystyle \sqrt{m^2- 16}= 8- m$$ and square both sides again to get an equation for m.
Caution: While any solution to the original equation must satisfy the new equation, squaring both sides of an equation can introduce "spurious solutions"- solutions to the new equation that do not satisfy the original equation. Be sure to check any solutions to your final equation in the original equation!
Thanks for working through it...I will get back to this in the morning when my brain comes back from the dead. I do know about the fdalse solution pitfall, and have actually fallen into it more than once.
#### allegansveritatem
##### Full Member
Please learn to experiment with simpler examples. Maybe...
$$\displaystyle 4 + 3 = 7$$
$$\displaystyle 16 + 9 \ne 49$$ -- Something went wrong.
I know this technique and I assure you I use it all the time. I use 2 and 3 as my dummies of choice
#### Otis
##### Senior Member
… I think I did it this way in the hope of avoiding a messy FOIL.
There are no shortcuts to any place worth going. :cool:
Finally:
#### Denis
##### Senior Member
Kinda sloppy, but correct...you get a pink star for your forehead :razz:
#### JeffM
##### Elite Member
This is not correct.
$$\displaystyle a + b > c \text { DOES NOT GENERALLY ENTAIL } a > b + c.$$
Furthermore, by definition $$\displaystyle \sqrt{x} \in \mathbb R \implies 0 \le \sqrt{x}.$$
So, $$\displaystyle - 3 = \sqrt{m + 4}$$ is impossible.
EDIT: You can sometimes get a correct answer through an incorrect process, but do not count on it.
Last edited:
#### Jomo
##### Elite Member
No, this is not correct. First when you moved sqrt(m+4) to the other side you failed to change the sign. 2nd mistake is that you got a solution to sqrt(m+4) = -3 !! The sqrt of nothing is ever negative. Go back and fix your mistakes.
#### Jomo
##### Elite Member
Kinda sloppy, but correct...you get a pink star for your forehead :razz:
Correct??? Go to the corner for a few hours and review your algebra as you are clearly a bit rusty.
#### Denis
##### Senior Member
Problem posted: sqrt(M + 4) + sqrt(M - 4) = 4
All I did was substitute his solution M=5 to get sqrt(9)+sqrt(1) = 3+1 = 4
Why is M=5 wrong?
#### Dr.Peterson
##### Elite Member
Problem posted: sqrt(M + 4) + sqrt(M - 4) = 4
All I did was substitute his solution M=5 to get sqrt(9)+sqrt(1) = 3+1 = 4
Why is M=5 wrong?
Five is right. But along the way, two mistakes are made that cancel one another out. Getting the right answer in an invalid way doesn't count as correct.
#### allegansveritatem
##### Full Member
This is not correct.
$$\displaystyle a + b > c \text { DOES NOT GENERALLY ENTAIL } a > b + c.$$
Furthermore, by definition $$\displaystyle \sqrt{x} \in \mathbb R \implies 0 \le \sqrt{x}.$$
So, $$\displaystyle - 3 = \sqrt{m + 4}$$ is impossible.
EDIT: You can sometimes get a correct answer through an incorrect process, but do not count on it.
I'm sorry but this needs to be unpacked a little so a novice like me can make it out.
#### allegansveritatem
##### Full Member
No, this is not correct. First when you moved sqrt(m+4) to the other side you failed to change the sign. 2nd mistake is that you got a solution to sqrt(m+4) = -3 !! The sqrt of nothing is ever negative. Go back and fix your mistakes.
Yes, I see that now. I mean, I see that I forgot to change the sign when I switched the radical over to the right. That is a huge mistake right there and yet...the answer came out right. So....I must either have done something else that cancelled that error or maybe the nature of this type of problem is such that flukes of this sort are, if not likely, yet still possible. Anyway, tomorrow I will have another lunge at this beast, cuz it ain't dead yet! | 2019-05-19T21:25:15 | {
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http://math.stackexchange.com/questions/81759/existence-of-irrationals-in-arbitrary-intervals | # Existence of irrationals in arbitrary intervals
I was studying for my analysis mid-term paper and was going over the properties of real numbers. I was wondering how to prove the following statement: (Not a textbook problem, it just popped into my head.)
Given rational numbers $p$ and $q$ such that $p < q$, show that there exists an irrational number $r$ such that $p < r < q$.
I know some ways of proving it, like picking a known irrational and shifting it into the open interval $(p,q)$. I was wondering whether there is a way to prove it without referencing to any previously known irrationals. Specifically I am trying to construct a sequence of rational numbers which converges to a irrational in the interval $(p,q)$. Is there any way to do that?
-
If $p>q$, then there are no numbers $r$ such that $p<r<q$. – Chris Eagle Nov 13 '11 at 21:12
@chris edited.Thanks for pointing that out. – Thiagarajan Nov 13 '11 at 21:15
Since there exists two rational numbers with finite decimal writing $x$ and $y$ such that $p < x < y < q$, consider $x = a_n a_{n-1} \dots a_0 . a_{-1} \dots a_{-m}$ and $y = b_N b_{N-1} \dots b_0.b_{-1} \dots b_{-M}$, the decimal expansions of $x$ and $y$. It is a theorem that a number of the form $z= c_{\ell} c_{\ell-1} \dots c_0 . c_{-1} c_{-2} \dots$ is an irrational if and only if its decimal expansion is infinite and non-periodic. Thus choose $x < z < y$ with $z$ an arbitrary non-periodic infinite decimal expansion (this is easy enough to do, just take $z$ with the right decimals for the first few and then choose whatever you like afterwards =D... one way to do this is just add decimals to the writing of $x$, i.e. $$x = a_n a_{n-1} \dots a_0. a_{-1} \dots a_{-m} \quad \Longrightarrow \quad z = a_n a_{n-1} \dots a_0 . a_{-1} \dots a_{-m} c_{-m-1} \dots c_{-j} \dots.$$ and choose the decimals such that $z < y$, for instance by putting many zeros in $z$ until you are sure that whatever number you put, $z < y$) and you will get irrationals between $p$ and $q$.
Perhaps you might also be interested in this : http://en.wikipedia.org/wiki/Continued_fractions it explains the concept of continued fractions, which is another nice way of looking at real numbers.
-
Ok thanks.That helped.So to construct a arbitrary irrational,I would basically need access to some kind of random process which generate those digits in my decimal expansion.Would it be right to assume,in the absence of such process I can't show existence of irrationals in my interval.As in,there is no deterministic way to,say,come with a convergent sequence which converges to my irrational,or construct a set such that my irrational is its supremum? – Thiagarajan Nov 13 '11 at 21:34
Yes =) But there is one way to generate arbitrary processes such as the ones you want. For instance, let the decimals of $z$ fixed up to a certain point so that we have choice for whatever decimal we want, and then let $j$ be the last fixed in advance, so that we have arbitrary choice for $c_{j+k}$ with $k > 0$. Take any non-periodic sequence on $\mathbb N$, and define $c_{j+k} = 1$ if $k$ is in the sequence and $0$ if not. You do not need "randomness", you just need non-periodicity. This gives you a $z$. If you want to know if there exists such a sequence, consider $2^n$, the prime numbers.. – Patrick Da Silva Nov 13 '11 at 21:38
There are lots of examples. – Patrick Da Silva Nov 13 '11 at 21:40
Thank you :).That cleared up lot of difficulties I was having.I spent quite some time trying to construct convergent sequences but couldn't find a way to show that the limit is irrational.The comment about non-periodicity made me realize that varying structure(non-periodicity) is quite different from no structure(randomness).Thank you. – Thiagarajan Nov 13 '11 at 21:49
Look at all the translates $(p,q) + \frac{q-p}{2} \mathbb{Z}$. They cover all of $\mathbb{R}$. Since $(q-p)/2$ is rational, if all of $(p,q)$ were rational then there would be no irrational numbers.
-
:D.Thanks.Liked the proof very much.Abstract and elegant. – Thiagarajan Nov 13 '11 at 22:19
The main idea is that all that matters here is what happens between $0$ and $1$, so that if we find an irrational between $0$ and $1$, we can do a combination of scaling and translation (by rationals, in both cases) to generate an irrational between $0$ and $1$, and these two operations preserve irrationality:
If you assume that, e.g., $2^{1/2}$ is irrational, and that both the sum of an irrational plus a rational is irrational (otherwise, the sum of two rationals is irrational), and that the ratio of an irrational by an integer is irrational, then $2^{1/2}-1$ is irrational, and $0<2^{1/2}-1<1$ . By the Archimedean principle, there is an integer $n$ with $n(x-y)>1$, so that there is an integer $z$ with $nx<z<ny$. Then $nx< nx+ 2^{1/2}-1<ny$, and now, dividing thru by $n$, we get: $x<x+ \frac{2^{1/2}-1}{n}<y$ is an irrational between $x$ and $y$.
-
For some reason, the last part of the answer I included is not appearing . Hope the punchline is clear. – C.O.Jones Nov 13 '11 at 22:37
Yes,it is.Thanks.I was looking for a proof such that I can avoid choosing a irrational number(like the square root of 2) and generating the required irrational as a function of it.Turns out there isn't. – Thiagarajan Nov 13 '11 at 22:47
I couldn't think of another one. A good thing about this proof, I think, is that it can be generalized to other arguments, including the proof of the existence of a rational between two rationals, etc. – C.O.Jones Nov 13 '11 at 23:00
I want to mention how the notions of countable and uncountable can be used to give a solution. Because $\mathbb Q$ is countable, so is $(p,q)\cap \mathbb Q$. But $(p,q)$ is uncountable, because $$\displaystyle{f(x)=\frac{x-\frac{p+q}{2}}{(x-p)(q-x)}}$$ is a bijective map from $(p,q)$ to $\mathbb R$, and $\mathbb R$ is uncountable, or because $g(x)=\frac{x-p}{q-p}$ is a bijective map from $(p,q)$ to $(0,1)$, and $(0,1)$ is uncountable. Therefore $(p,q)\cap \mathbb Q\neq(p,q)$, meaning that there are irrational numbers in $(p,q)$.
- | 2015-04-25T12:53:14 | {
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https://www.physicsforums.com/threads/set-theory-counting-binomial-coefficient-factorials.702818/ | # Homework Help: Set Theory - Counting - Binomial Coefficient - Factorials
1. Jul 24, 2013
### reenmachine
1. The problem statement, all variables and given/known data
A department consists of 5 men and 7 women.From this department you select a committee with 3 men and 2 women.In how many ways can you do this?
2. Relevant equations
Since the "overall set" (the entire department) is composed of both men and women and each has a specific number in the committee , I think I have to split it in half.
I pick 3 men out of 5 men and pick 2 women out of 7 women so:
$\binom{5}{3} \cdot \binom{7}{2} = \frac{5!}{3!(5-3)!} \cdot \frac{7!}{2!(7-2)!} = \frac{5!}{3!2!} \cdot \frac{7!}{2!5!} = \frac{120}{6 \cdot 2} \cdot \frac{5040}{2 \cdot 120} = \frac{120}{12} \cdot \frac{5040}{240} = 10 \cdot 21 = 210$
So 210 ways to form the committee of 3 men and 2 women from the department consisting of 5 men and 7 women.
thoughts on this?
thank you!!!
2. Jul 24, 2013
### haruspex
Looks fine.
3. Jul 24, 2013
### reenmachine
Thank you!
4. Jul 25, 2013
### HallsofIvy
Without "formulas", you can argue that there are 5 choices for the first man, then 4 choices for the second, and 3 choices for the third- so 5(4)(3)= 60 ways to choose 3 out of 5- in that particular order. Since order does not matter divide by the 3!= 6 ways of ordering three people: 60/6= 10. For the women, there are 7 choices for the first woman then 6 choices for the second- so 7(6)= 42 ways to choose 2 out of 7- again in that particular order. Divide by the 2 different orders to get 42/2= 21 ways to do this. Overall there are (10)(21)= 210 ways to do this.
(Typical sexism- there are more women than men in the department but fewer on the commitee!)
5. Jul 25, 2013
### reenmachine
very clear thank you!
Yeah I noticed the sexism when I read it :rofl: | 2018-07-16T10:50:52 | {
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https://brilliant.org/discussions/thread/int-ln-sqrtx1-sqrtx-dx-without-trigonometry/ | # $$\int \ln ( \sqrt{x+1} + \sqrt{x} ) dx$$ without trigonometry and nice values
Integration:
I watched this video https://www.youtube.com/watch?v=DGpgt8j-nzw in which $$\int \ln ( \sqrt{x+1} + \sqrt{x} ) dx$$ is computed by using trigonometry. Reading the comments, other solutions by using special functions are shown. However, the final antiderivatives do not use any of these functions, so these are intermediate steps in the most literal interpretation of that phrase. I thought of a solution that sidesteps the need of considering any functions other than those present in the integral:
$$\int \ln ( \sqrt{x+1} + \sqrt{x} ) dx$$. Let $$u = \sqrt{x+1} + \sqrt{x}$$. Notice that $$\frac{1}{u} = \sqrt{x+1} - \sqrt{x}$$. Then $$du = \frac{1}{2} \left( \frac{1}{\sqrt{x+1}} + \frac{1}{\sqrt{x}} \right)dx = \frac{1}{2} \frac{u}{\sqrt{x+1} \sqrt{x}}dx$$. To write everything in terms of $$u$$ notice that $$u^2 - \frac{1}{u^2} = 4 \sqrt{x+1} \sqrt{x} \implies 2 \sqrt{x+1} \sqrt{x} = \frac{u^4 - 1}{2u^2}$$. Thus $$du = \frac{u}{\frac{u^4 - 1}{2u^2}}dx = \frac{2u^3}{u^4 - 1} dx$$. Going back to the original integral:
$$\int \ln ( \sqrt{x+1} + \sqrt{x} ) dx = \int \frac{\ln u}{2u^3} (u^4 - 1)dx = \int \frac{u}{2} \ln u - \frac{ \ln u}{2u^3} du$$. Integrating the terms in the integral is now a standard exercise in integration by parts which I will skip. The result is $$\frac{1}{8}u^2(2 \ln u - 1) + \frac{2 \log u + 1}{8u^2}$$. Giving a final result of:
$\int \ln ( \sqrt{x+1} + \sqrt{x} ) dx = \frac{1}{8}(\sqrt{x+1} + \sqrt{x})^2(2 \ln (\sqrt{x+1} + \sqrt{x}) - 1) + \frac{2 \ln (\sqrt{x+1} + \sqrt{x}) + 1}{8(\sqrt{x+1} + \sqrt{x})^2} + C$
Nice values:
From the form of the antiderivative we can expect this integral to give us nice values whenever $$\sqrt{x+1} + \sqrt{x}$$ is a nice power of $$e$$. The equation $$\sqrt{x+1} + \sqrt{x} = e^k$$ has solution $$x= \frac{1}{4} e^{-2 k} (e^{2 k} - 1)^2$$. You can obtain this by converting the equation into a normal quadratic equation. The nicest value I was able to find was taking $$k=0$$ to yield $$x = 0$$ and $$k = \frac{1}{2}$$ to yield $$x = \frac{(e-1)^2}{4e}$$. Then: $\int_{0}^{\frac{(e-1)^2}{4e}} \ln ( \sqrt{x+1} + \sqrt{x} ) dx = \frac{1}{4e}$
Note by Leonel Castillo
8 months ago
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Here's another way to solve it:
$\begin{eqnarray} \int \ln\left(\sqrt{x+1} + \sqrt x\right) \, dx &=& \frac12 \int \ln\left(\sqrt {x+1} + \sqrt x\right)^2 \, dx \\ &=& \frac12 \int \ln\left(2x + 1 + 2\sqrt{x^2 + x}\right) \, dx \\ &=& \frac12 \int \ln\left[ 2\left(x+ \frac12\right) + 2\sqrt{\left(x + \frac12\right)^2 - \frac14} \; \right] \, dx \\ \end{eqnarray}$
Let $$x + \frac12 = \frac12 \sec \theta$$, then $$\left(x+\frac12\right)^2 - \frac14 = \left(\frac12 \tan\theta\right)^2$$ and $$\frac{dx}{d\theta} = \frac12\sec\theta\tan\theta$$. The integral becomes
$\begin{eqnarray} \int \ln\left(\sqrt{x+1} + \sqrt x\right) \, dx &=& \frac12 \int \ln (\sec \theta + \tan\theta) \cdot \frac12 \sec \theta\tan\theta \, d\theta \\ &=& \frac14 \int \underbrace{\ln (\sec \theta + \tan\theta)}_{=u} \cdot \underbrace{\sec \theta\tan\theta \, d\theta}_{=dv} , \qquad\qquad \text{ Integrate by parts} \\ &=& \frac14 \left (uv - \int v \, du\right) \\ &=& \frac14 \left [ \ln (\sec \theta + \tan\theta) \sec \theta - \int \sec \theta \cdot \dfrac{\sec\theta \tan\theta + \sec^2\theta}{\sec \theta + \tan \theta} \, d\theta \right ] \\ &=& \frac14 \left [ \sec \theta \ln (\sec \theta + \tan\theta) - \int \sec^2\theta \, d\theta \right ] \\ &=& \frac14 \left [ \sec \theta \ln (\sec \theta + \tan\theta) - \tan \theta \right ] +C \\ &=& \frac14 \left [(2x + 1) \ln \left ((2x+1) + \sqrt{(2x+1)^2-1}\; \right) - \sqrt{(2x+1)^2-1} \; \right ] +C \\ &=& \frac{2x+1}4 \ln \left (2x+1 + 2 \sqrt{x^2+x}\; \right) - \frac12 \sqrt{x^2+x} +C \\ &=& \frac{2x+1}2 \ln \left (\sqrt {x+1} + \sqrt x\right) - \frac12 \sqrt{x^2+x} +C \\ \end{eqnarray}$
which is identical to blackpenredpen's final answer.
- 7 months, 4 weeks ago
Pi Han Goh, your post does not apply, because the OP expressly asked for the solution to be done without using trigonometry.
- 7 months, 3 weeks ago
I did not ask, I was just sharing a solution. If he wants to use this post to post another alternative solution that's okay.
- 7 months, 3 weeks ago
OP, your answer is wrong, because you do not include the "+ C."
- 7 months, 3 weeks ago
Thank you.
- 7 months, 3 weeks ago
Just a thought- How many different sums can be made by adding at least two different numbers from the set of integers {5,6,7,8}?
- 5 months ago | 2019-03-22T03:12:14 | {
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https://math.stackexchange.com/questions/2093910/using-the-digits-1-9-each-only-once-find-the-closest-sum-to-1000 | # Using the digits $1-9$ each only once find the closest sum to $1000$.
Using the digits $1-9$ each only once find the closest sum to $1000$ by filling in the following spaces.
$$_ _ _+_ _ _+_ _ _$$
I tried to get $1000$ but couldn't (not even sure if you can). So then I tried $999$. I was able to get $152+378+469=999$ by guess and check but am not able to even to check if that is the correct answer. I believe there are $9!$ possible choices. I was wondering if there is some logical mathematical reasoning to get to the correct answer and prove it correct.
By the divisibility by 9 rule (and these numbers add up to 45, i.e. divisible by 9), whatever sum you produce, the digits of the sum will always be divisible by 9, and hence 999 is the best you can do.
• Such an elegant solution! – John Jan 11 '17 at 21:18
• @John mine was so similar :'( *sniff* *sniff* haha – RGS Jan 11 '17 at 21:23
• Question, though isn't this not exactly the by divisible 9 rule. You got to factor in carrying over. Isn't the divisibility rule a number is divisible by $9$ if the sum of its digits are? @Bram28 – Ahmed S. Attaalla Jan 11 '17 at 22:16
• @AhmedS.Attaalla The beautiful thing of the divisibility by 9 rule is that it does factor in carrying over. E.g if I add 24 (where 24 mod 9 = 6 = (2+4) mod 9) and 37 (where 37 mod 9 = 1 = (3+ 7) mod 9) then I get 61 (where 61 mod 9 = 7 = (6+1) mod 9 = (2+4+3+7) mod 9). – Bram28 Jan 11 '17 at 23:00
One can prove that there is no way to sum up to $1000$:
Using all digits only once to write three $3$-digit numbers will yield a sum that is congruent with $0$ modulus $3$; $1000$ is congruent with $1$ and $1001$ with $2$, so $999$ is the closest you can get. Guessing the final answer is just fine.
• @Faibbus Sum of digits is the way to show divisibility by $3$... – Simply Beautiful Art Jan 11 '17 at 21:32
• @Faibbus Proof or not, if it is widely known, a full proof is probably not necessary. To me, this was obvious. – Simply Beautiful Art Jan 11 '17 at 21:43
• Such an elegant solution! ;) – John Jan 12 '17 at 18:21
There's a strategy you can use to find a solution that adds to 999, convert one solution into others, and once you understand how it works it's easy to see that 1000 is impossible. Using this method I found $128+375+496$ in a couple of minutes, and I can easily turn that into $176+395+428=999$ or $478+326+195=999$.
First just write down any three numbers, even if they don't work, say
147
258
369
This adds to 774, which is too small. But what happens if we switch two digits in the same row? Say we switch the 4 and the 7. This adds 30 to the total (because the 4 in the tens' column becomes a 7) and takes away 3 (because the 7 in the units' column becomes a 4) for a net gain of 27. So the total now must be $774+27=801$:
174
258
369
The total is 801, so we want to increase it by around 200. So we should look for a digit in the hundreds' place that is 2 less than the digit in the units' place. When we switch these the total will increase by 200 and decrease by 2. None of the three rows does this. For example, if we were to switch the 2 and the 8 in the second row the total would increase by around 600 which is too much.
But if 258 were 254 instead, then switching it to 452 would increase the total by about 200. And we can turn 258 into 254 by switching the 4 and the 8. This doesn't change the total at all, because 174 increases by 4 and 258 decreases by 4:
178
254
369
The total is still 801.
Switching the 2 and the 4 works:
178
452
369
Now the total is exactly 999 and we win.
If we want another solution, just switch any two digits in the same column, say the 3 and the 1:
378
452
169
And it's still 999.
If you think a little more about what is happening here you will see why it's not possible to make 1000 exactly. | 2021-02-27T07:32:33 | {
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https://www.jiskha.com/questions/973292/multiple-choice-1-what-is-0-32-written-as-a-fraction-in-simplest-form-1-point | # math help plzzzz!!!!
Multiple Choice
1. What is 0.32 written as a fraction in simplest form? (1 point)
thirty-two over one hundred
eight over twenty-five
one hundred over thirty-two
sixteen over twenty-five
2. If Fran’s fudge recipe calls for two and one-half cups of sugar and she increases it by two-fifths of a cup, how much sugar does she use? (1 point)
two and two-tenths
2two and three-sevenths
2two and four-ninths
2two and nine-tenths
3. Which of the following makes the statement true? one over five to the zero power = ? (1 point)
1
–1
–5
0
4. The area of a certain state is 840,000 square miles. What is this area in scientific notation? (1 point)
8.4 x 10-4
84 x 10-3
84 x 103
8.4 x 105
5. Which is irrational? (1 point)
0.117117117 . . .
0.8596873205 . . .
0.812812812812 . . .
0.605605605 . . .
6. If there is one-half of a pizza and Jonas wants only one-third of it, how much of the pizza will he eat? (1 point)
one-fifth
two-fifths
one-sixth
two-sixths
7. Harold brought two loaves of raisin bread to share with his friends. If there are 12 people total, including Harold, how much of the bread does each person receive? (1 point)
0.06
0.16 repeating
0.25
point three repeating
8. One nanometer is about 0.00000003937 of an inch. What is this number in scientific notation? (1 point)
3.937 x 10-8
3.937 x 108
39.37 x 10–7
39.37 x 107
9. What is three-eighths written as a decimal? (1 point)
two point three repeating
0.375
point three repeating
2.667
10. According to the 2010 census data, the population of Texas was about 2.5 x 107 people. The land area of Texas is about 2.6 x 105 m2. What was the average number of people per square mile in 2010? (1 point)
0.96 x 102
9.6 x 101
0.96 x 1035
9.6 x 1012
11. Based on the 2010 census, the population of Georgia was 9.6 x 106 people. Which state had a larger population? (1 point)
New York: 1.9 x 107
Wyoming: 5.6 x 105
Idaho: 1.5 x 106
12. Which of the following numbers is an example of an integer? (1 point)
square root of five
square root of nine
one-third
0.8
13. Which number is equivalent to four-fifths? (1 point)
twenty-five over one hundred
0.2
sixteen over one hundred
0.8
14. Riding the bus, Jasmine uses two-thirds of her money for bus fare. What is this number as a decimal? (1 point)
point three repeating
point six repeating
point eight repeating
one point five repeating
15. What is the solution to 2six-sevenths ÷ two-fourteenths? (1 point)
7
14
20
40
1. 👍 2
2. 👎 0
3. 👁 4,125
1. 1.
0.32 = 32 / 100 = 4 * 8 / ( 4 * 25 ) = 8 / 25
2.
2 + 1 / 2 + 2 / 5 = 2 + 5 * 1 / ( 5 * 2 ) + 2 * 2 / ( 5 * 2 ) = 2 + 5 / 10 + 4 / 10 = 2 + 9 / 10
3.
The number raised to the zero power is equal to one.
( 1 / 5 ) ^ 0 = 1
4.
840,000 = 8.4 * 100 * 1,000 = 8.4 * 10 ^ 2 * 10 ^ 3 = 8.4 * 10 ^ 5
5.
An Irrational Number is a real number that can't be written as a simple fraction.
0.117117117... = 117 / 999
0.8596873205... can't be written as the fraction
0.812812812812... = 812 / 999
0.605605605... = 605 / 999
So 0.8596873205... is irrational.
6.
( 1 / 2 ) * ( 1 / 3 ) = 1 / 6
7.
2 / 12 = 2 / ( 2 * 6 ) = 1 / 6 = 0.16666666
8.
0.00000003937 = 3.937 / 100,000,000 = 3.937 / 10 ^ 8 = 3.937 * 10 ^ - 8
9.
3 / 8 = 0.375
10.
2.5 * 10 ^ 7 / ( 2.6 * 10 ^ 5 ) =
( 2. 5 / 2.6 ) * 10 ^ 7 / 10 ^ 5 =
0.96 * 10 ^ 2
11.
New York: 1.9 * 10 ^ 7
12.
sqrt ( 9 ) = 3
13.
4 / 5 = 2 * 4 / ( 2 * 5 = 8 / 10 = 0.8
14.
2 / 3 = 0.66666
15.
( 2 + 6 / 7 ) / ( 2 / 14 ) = ( 2 * 7 / 7 + 6 / 7 ) / [ ( 2 * 1 ) / ( 2 * 7 ) ] =
( 14 / 7 + 6 / 7 ) / ( 2 * 1 ) / ( 1 / 7 ) =
( 20 / 7 ) / ( 1 / 7 ) = 7 * 20 / 7 = 20
1. 👍 5
2. 👎 1
2. QUESTION 10'S ANSWER IS :
************9.6 x 10^1 *************
^^^^^^^^^^^^^^^^^^^^^^
BOSNIAN IS WRITE ON EVERYTHING EXCEPT QUESTION 10 THE ANSWER TO QUESTION 10 IS ABOVE ^^
1. 👍 7
2. 👎 1
3. 5 is wrong
1. 👍 0
2. 👎 3
1.b
2.d
3.a
4.d
5.b
6.c
7.b
8.a
9.b
10.b
11.b
12.b
13.d
14.b
15.c
I got a 100%
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20. 1. B
2. D
3. A
4. D
5. B
6. C
7. B
8. A
9. B
10. B
11. B
12. B
13. D
14. B
15. C
1. 👍 0
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21. follow me @!@#$%^&.gotta.know on instagram 1. 👍 1 2. 👎 3 22. thank yall sooooooooooooo much yall are the best 1. 👍 1 2. 👎 0 23. I got 5 and 4 wrong on connexus because of Salt Caramel. 4 is B and 5 is D and I got 86.7% just like alyssa soo....yea But thanks and your welcome for the real few answers. 1. 👍 3 2. 👎 2 24. I'm in 8th grade and I just took the test and the answers are: 1.b 2.d 3.a 4.d 5.b 6.c 7.b 8.a 9.b 10.b 11.b 12.b 13.d 14.b 15.c 100% 1. 👍 1 2. 👎 0 25. I just use to check my answers. It's much better for y'all to learn the lessons they are teaching rather than cheating right off the bat. Remember, you can't use the internet on the STAAR tests. 1. 👍 1 2. 👎 1 26. @I honestly don't know is it only for texas or all states? 1. 👍 1 2. 👎 0 27. Salted caramel is 100, and frick anyone who wants to make people fail. 1. 👍 3 2. 👎 0 28. Hello, I am not here to scold you or annoy any of you but I think that people should EXPLAIN the answers to each question if you don't want to fail any of your classes including math. 1. 👍 2 2. 👎 0 29. Salty Carmels answers are right for connexus. You need to make sure that you get everything that she wrote. 100 percent Yass 1. 👍 2 2. 👎 0 30. EaT cHiPoTlE fOr AlL tHe AnSwErS 1. 👍 2 2. 👎 0 31. what are the unit test answers 1. 👍 1 2. 👎 0 32. Hal and Salted Caramel are 100% correct and shame on you for saying something else! 1. 👍 1 2. 👎 0 33. I GOT A 100 WITH SALTED CERAMAL ANSWERS THANKS 1. 👍 1 2. 👎 0 34. thx salted caramel owo 1. 👍 1 2. 👎 0 35. . I got but salted caramel said was correct 100% baby 1. 👍 1 2. 👎 1 36. Three days grace is right for connexus. 1. 👍 2 2. 👎 0 37. @!@#$%^&.gotta.know
wtf is ur problem? This is a kid friendly website. GO FK OFF.
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38. It Is Confirmed That All Users, (Including Hal and Salted Caramel) With The Following Answers:
1. B
2. D
3. A
4. D
5. B
6. C
7. B
8. A
9. B
10. B
11. B
12. B
13. D
14. B
15. C
^ Are 15/15 100% Correct! (aka. For Connexus Users Only)
- Seeker
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39. Unit Review Practice: Algebra Unit 4 Lesson 10 Rational Numbers
b
d
a
d
b
c
b
a
b
b
b
b
d
b
c
i got 100% i know it’s not right to cheat but..
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40. Some one call me 727-485-7166
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41. For connexus students its #1B #2D #3A #4D #5B #6C #7B #8A #9B #10B #11B #12B #13D #14B #15C
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asked by Anonymous on September 22, 2014 | 2020-07-16T17:42:31 | {
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https://math.stackexchange.com/questions/1484561/finding-the-mantissa-from-binary-with-floating-point-numbers | # Finding the mantissa from binary with floating point numbers?
Here is the example problem slide I am working with:
I understand how to get the exponent, its just 2+128=130-127=3
I understand the first bit is the sign bit for positive or negative.
I just get lost with the mantissa. I know each mantissa has to have a 1 before the decimal, but I'm unsure where the .75 came from. 1100000000000000000000 isn't 75 in decimal I mean right?
How do I calculate the mantissa with floating point numbers in binary?
The number after "." is calculated by $$\frac1{2^n}$$ instead of $$2^n$$ So $$1.11 = 2^0 + (1/2^1) + (1/2^2)$$ $$= 1 + 0.5 + 0.25$$ $$=1.75$$
Yes, there is an implicitly "hidden" bit, so if your mantissa is 1100000000000000000000, then that corresponds to the base 2 number 1.11.
Now, if you want to compute the value of .11 part, you just do the same thing as you would do to convert a normal binary number to decimal: multiply each bit by a power of 2.
So, for example, if you wanted to convert the number 1010.1 to decimal, you would do: $1(2^3) + 0(2^2) + 1(2^1) + 0(2^0) + 1(2^{-1}) = 8 + 2 + 1/2 = 10.5$
• Please add your own (partial) answer to the question and update what the question in question now is. Oct 17, 2015 at 16:38 | 2022-08-17T14:56:13 | {
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https://math.stackexchange.com/questions/2374685/functions-that-are-always-less-than-their-derivatives/2374695 | # Functions that are always less than their derivatives
I was wondering if there are functions for which $$f'(x) > f(x)$$ for all $x$. Only examples I could think of were $e^x - c$ and simply $- c$ in which $c > 0$. Also, is there any significance in a function that is always less than its derivative?
Edit: Thank you very much for all the replies. It seems almost all functions that apply are exponential by nature... Are there more examples like - 1/x?
Again are there any applications/physical manifestations of these functions? [for example an object with a velocity that is always greater than its position/acceleration is always greater than its velocity]
• Off the top of my head, any bounded, monotonically increasing function in the bottom half-plane. Jul 28, 2017 at 21:43
• Ixion's answer gives the full, most general solution (though some particular families of solutions might be writable in nicer forms), and should be accepted. Jul 29, 2017 at 9:56
• +1! But please fix the title, changing "its" to "their". The way the title is written, for a moment it looked like you were considering derivatives of all orders. And now I'm curious about this side question, haha! Jul 30, 2017 at 23:23
• Jean Marie.Thanks for the link. Jan 31, 2019 at 10:24
If $y'(x)>y(x)\quad\forall x\in\mathbb{R}$, we can define $f(x)=y'(x)-y(x)$ which is positive forall $x$. Suppose that $y'(x)$ is continuous function so that $f(x)$ is continuous too. Now with this element we can build the differential equation $$y'(x)=y(x)+f(x)$$ and its solutions are given by: $$y(x)=e^{x}\left(c+\int_{x_0}^{x}e^{-s}f(s)ds\right)$$
Again are there any applications/physical manifestations of these functions? [for example an object with a velocity that is always greater than its position/acceleration is always greater than its velocity]
I don't know if there's application of this interesting property, but I'm sure that you can't compare velocity with the position because they are not homogeneous quantities.
Assuming $f(x)>0$, $f:\mathbb{R}\mapsto\mathbb{R}$
$f'(x) > f(x) \iff \frac{d}{dx}\ln(f(x))>1$
So you can turn any function $g$ where $g'(x)>1$ into this type of function by taking the exponential of it:
$\frac{d}{dx}g(x)>1 \implies \frac{d}{dx}\ln(e^{g(x)})>1 \implies \frac{d}{dx} e^{g(x)}>e^{g(x)}$
• You assume $f(x)>0$ in the beginning Jul 28, 2017 at 12:20
• @HagenvonEitzen : Then he could just use $\hat{f}(x) \equiv e^{f(x)}$ as his starting point for any given $f$. That way one always has $\hat{f}(x)>0$.
– MPW
Jul 28, 2017 at 17:43
• Ixion's answer gives the full generalization by allowing $\frac{df}{dx} - f(x)$ to be any function which is everywhere-positive. Jul 29, 2017 at 10:05
• @RobinSaunders No, he assumes continuity of $f'(x)$. Jul 29, 2017 at 10:13
• I'm pretty sure that condition isn't actually needed. Jul 29, 2017 at 10:19
A simple example is $f(x)=-x^2-3$
A more interesting problem is to find a function $f:\mathbb{R}\rightarrow\mathbb{R}$, whose image is $\mathbb{R}$ and satisfies $f'(x)>f(x)$ for all $x\in\mathbb{R}$. One of those functions is
$$\sinh(x),$$
because
$$\frac{d}{dx}\sinh(x)=\cosh(x)>\sinh(x)$$ for all $x\in \mathbb{R}$.
$$(e^{-x}f(x))'=e^{-x} (f'(x)-f(x)) >0$$ so $$e^{-x}f(x)$$ is an increasing function. Since it is $$0$$ at $$0$$ we get $$e^{-x}f(x)>0$$ for all $$x>0$$. Hence $$f(x)>0$$ for all $$x>0$$.
Take $f(x)=e^{\alpha x}$. Then for $\alpha >1$ we have $f'(x)>f(x)$ and for $\alpha <1$ we have $f'(x)<f(x)$.
How about if you look at it as a differential equation. Say
$y' = y + 1$
which has solution $y=Ce^x -1$
Or $y'=y+x^2+1$
which has solution $y=Ce^x - (x^2+2x+3)$
Or $y'=y+2\sin x+3$
which has solution $y = Ce^x - \sin x - \cos x -3$
• Ixion's answer generalizes this to $y'(x) = y(x) + f(x)$ for any $f(x) > 0$. Jul 29, 2017 at 10:00
• @RobinSaunders - should I delete my answer? Jul 29, 2017 at 23:35
• I don't know much about Stack Exchange etiquette, but my guess would be that since you posted your answer first and it contains specific examples not in the other answer, it should be fine to leave it. Jul 30, 2017 at 12:01
The inequality $$f'(x) > f(x)$$ is equivalent to $$\left[ f(x) e^{-x} \right]' > 0.$$
So the general solution is to take any differentiable function $g(x)$ with $g'(x) > 0$ and put $f(x) = g(x) e^x$.
Note that nothing is assumed about $f$ except differentiability, which is necessary to ask the question in the first place.
A very simple example is $f(x) = -1 < 0 = f'(x)$. Relevant to your edit: this isn't exponential at all.
Other examples that aren't immediately exponential:
• $\frac{-\pi}{2} + \arctan x$ is everywhere negative and everywhere strictly monotonically increasing, so is everywhere less than its derivative.
• $-1 + \mathrm{erf}(x)$ is also everywhere negative and everywhere strictly monotonically increasing. (These are very similar, since they are shifted copies of the CDFs of the (standard/normalized) Cauchy and Gaussian distributions.)
• $\frac{1}{2}\left( x - \sqrt{x^2 + 4} \right)$ is the lower branch of a hyperbola having the $x$-axis and the line $y = x$ as asymptotes. It is everywhere negative and everywhere strictly monotonically increasing.
Another simple example would be $f(x) = -e^{-x}$, $f'(x) = e^{-x}$
See, $-\frac{1}{x}, \frac{1}{x^{2}} \ in \ [0, \infty]$
• More generally, any negative function with positive derivative... Jul 28, 2017 at 12:22
For any differential function $f$ for which both $f(x)$ and $f'(x)$ are limited to finite ranges, $f'(x) - f(x)$ is also limited to a finite range, so there is a $c$ for which $f'(x) - f(x) > -c\ \forall\ x$. Therefore, a function $g(x) = f(x) - c$ can be formed for which $g'(x) - g(x) - c> -c\ \forall\ x$ or $g'(x) > g(x)\ \forall\ x$.
For example, this holds for many differential periodic functions.
• The last statement is wrong, since not every differentiable periodic function has bounded derivative. Jul 29, 2017 at 15:59
• @Adayah You're right. I was considering periodic functions that were differentiable at every point in $\mathbb{R}$, but I realize that a function only has to be differentiable at all points in its domain to be considered differentiable. I've updated my answer. Jul 30, 2017 at 13:04
• I mean, a function $f : \mathbb{R} \to \mathbb{R}$ may be periodic and differentiable in every point $a \in \mathbb{R}$ and still have unbounded derivative. Jul 30, 2017 at 13:09
• @Adayah Do you have any example of such a function? Jul 30, 2017 at 16:25
• @Adayah I mean, if a function $f$ is differentiable everywhere, its derivative $f'$ must exist everywhere, and $f'$ must be continuous (because if it contains any discontinuity, $f'$ cannot exist at that point). That makes it impossible for $f'$ to be unbounded, right? Jul 30, 2017 at 16:47
Mike an answer to your additional question "Are there physical examples of this?" is enabled by dromastyx.
His example shows hyperbolic functions which describe accurately the physical phenomenon of 'solitons'.
Solitons are solitary waves such as sun flares, Tsunamis etc. An example of finding such waves hidden in known equations is :
http://rsos.royalsocietypublishing.org/content/2/7/140406.review-history
Here's a proof using the mean value theorem.
Let $$m = \operatorname{inf} \{ x > 0 \colon f(x)\leq 0\}$$. This infimum exists, since the reals are complete and the set is bounded below by $$0$$. Since $$f'(0)>0$$, there exists some $$\varepsilon >0$$ such that $$f(x) > f(0)$$ for $$x\in (0,\varepsilon)$$. This means that if there is some $$x>0$$ with $$f(x) \leq 0$$, then $$\varepsilon$$ is also a lower bound. (This is common lemma In the context of the mean value theorem).
Thus it suffices to show that $$m=0$$. Suppose $$m>0$$. By the mean value theorem, there is some point $$y\in (0,m)$$ with $$f'(y) \leq 0$$. By our assumptions, we then have $$f(y) < f'(y) \leq 0$$. Thus $$y$$ contradicts the minimality of $$m$$.
• You have to show that the set $\{m\;|\;m>0 \text{ and } f(m)\le 0\}$ has a minimal element, and that this minimal element is non-zero. This is not obvious, because it might not actually contain its greatest lower bound. It's not hard to prove by continuity that this can't happen, but you have to prove it. Jan 31, 2019 at 10:45
• But the set $\{x>0:f(x)\le 0\}$ turns out to be empty, so it doesn't have an infimum. (Some people take the view that the infimum of the empty set is $+\infty$, but nobody claimes that it is zero.) Feb 2, 2019 at 22:10 | 2022-06-25T01:25:35 | {
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https://usi-garaj-rulou.ro/cubic-senj/d3ae62-correlation-coefficient-example | Correlation statistics can be used in finance and investing. Introduction to Coefficient of Correlation. Compute the rank correlation coefficient for the following data of the marks obtained by 8 students in the Commerce and Mathematics. ¯ increases, y decreases. a line that looks like that or a line that looks like that, or a line that looks like that. But it's still not as good as that one. … For example, the correlation coefficient for these data was 0.846. better for scatterplot B than it works for scatterplot D. I would give the higher r to scatterplot B and the lower r, r equals ρ I'm gonna try to draw a dataset where the r would be negative one. However the standard versions of these approaches rely on exchangeability of the data, meaning that there is no ordering or grouping of the data pairs being analyzed that might affect the behavior of the correlation estimate. Φ(−2.2) = 0.028, where Φ is the standard normal cumulative distribution function. were to try to fit a line, it would look something like that. Symmetric: Correlation of the coefficient between two variables is symmetric. is pretty close to zero. The correlation coefficient also relates directly to the regression line Y = a + bX for any two variables, where . Step-by-step instructions for calculating the correlation coefficient (r) for sample data, to determine in there is a relationship between two variables. For example, let me do some coordinate axes here. The correlation coefficient, r, tells us about the strength and direction of the linear relationship between x and y.However, the reliability of the linear model also depends on how many observed data points are in the sample. smaller then other variable is smaller and vice versa. − Now we have scatterplot D. That's gonna use one of the X = 4, 8 ,12, 16 and Y = 5, 10, 15, 20. Practice: Correlation coefficient intuition. The sample correlation coefficient, r, estimates the population correlation coefficient, ρ.It indicates how closely a scattergram of x,y points cluster about a 45° straight line. Pearson Correlation Coefficient Calculator. m Let's see if we can When x is a little bit An example of positive correlation is the relationship between gas prices and food prices. ^ Y In other words, height explains about half the variability of weight in preteen girls. Keep in mind that correlations apply to pairs of variables. When y is low, x is low. Formally, the sample correlation coefficient is defined by the following formula, where s x and s y are the sample standard deviations, and s xy is the sample covariance. Where. I do see a general trend if we look at these data points over here, when y is high, x is low. When x is high, y is high and vice versa. As variable X increases, variable Y increases. be this well organized but this gives you a sense of things. I drew very perfect ones, at least for the r equals negative one and r equals one but these are what the real Pearson Correlation Coefficient = 0.95. other positive correlations and it does look like there When x is really high, y is even higher. It kinda looks like what we did over here. Although there are no hard and fast rules for describing correlational strength, I [hesitatingly] offer these guidelines: 0 < |r| < .3 weak correlation.3 < |r| < .7 moderate correlation |r| > 0.7 strong correlation For example, r = -0.849 suggests a strong negative correlation. and Next, we apply a property of least square regression models, that the sample covariance between For this example, we’ll be using a similar data set with the one above with the addition of ‘Z Variables’. Correlation Coefficients >. This ones going to be positive and it looks like it would Pearson Correlation Coefficient Calculator. The following are the marks scored by 7 students in two tests in a subject. … to try to fit a line, it looks something like that. , Pearson's correlation coefficient (r) is a measure of the strength of the association between the two variables. A correlation coefficient formula is used to determine the relationship strength between 2 continuous variables. to be less than zero, and a reasonable bit less than zero. Variations of the correlation coefficient can be calculated for different purposes. i where we can drag these around in a table to match them to the different scatterplots. When y becomes lower, x become higher. The coefficient can take any values from -1 to 1. Very few times will things Calculate coefficient of correlation from the following data and interpret. Like many commonly used statistics, the sample statistic r is not robust,[28] so its value can be misleading if outliers are present. If r =1 or r = -1 then the data set is perfectly aligned. of we are trying to measure. coefficient synonyms, coefficient pronunciation, coefficient translation, English dictionary definition of coefficient. Does a line look like that? For example, suppose we observe r = 0.3 with a sample size of n=50, and we wish to obtain a 95% confidence interval for ρ. that you could say, "Well, as x increases, maybe When one variable is I'll do that one really small, since I don't have much space here. There's different methods of Correlation doesn’t necessarily mean causation. In statistics, a correlation coefficient measures the direction and strength of relationships between variables. Correlation Coefficient Example Investor Ethan’s portfolio mainly watches the performance of Tech A, a giant tech company, but he is considering adding the stock of another tech giant, Tech B. However, he wants to check the correlation between the two companies’ stocks to make sure that adding Tech B to his portfolio won’t raise his level of systematic risk. {\displaystyle {\hat {Y}}_{i}} , I feel good with r is For data that follows a bivariate normal distribution, the expectation E[r] for the sample correlation coefficient r of a normal bivariate is[32], The unique minimum variance unbiased estimator radj is given by[33]. In this example, we have calculated the same 1st example with the excel method and we have got the same result i.e. {\displaystyle Y_{i}-{\hat {Y}}_{i}} Let's say when x is low, y is low. And it wouldn't necessarily Let Number of Study Hours 2 4 6 8 10 Number of Sleeping Hours 10 , the range of values is reduced and the correlations on long time scale are filtered out, only the correlations on short time scales being revealed. Least-squares regression equations. is Pearson's coefficient of correlation for segment The Karl Pearson Coefficient of Correlation formula is expressed as - relatively small and vice versa. ): The inverse Fisher transformation brings the interval back to the correlation scale. - Software free, world-class education to anyone, anywhere direct method and we have to match them to appropriate! Citation correlation coefficient example ] the population of all dots demonstrates no correlation a free, education... 10 number of sleeping hours of different students is positive very, very well, correspond! Recall that relations in samples do not necessarily depict the same in the Commerce and Mathematics r } /latex... To 1 negative.02 bivariate normality does not hold and notice, even when we try to a! Very well array 1 is a little bit higher, y grows, x is set! 10 correlation coefficient Calculator calculates the sample correlation coefficient also relates directly to the papers of ''. Say -0.40 model really does not describe the relationship between 2 quantitative variables been used in cluster analysis and detection. Did over here all the features of Khan Academy exercise on correlation coefficient - Software, very.. Really high, y grows and when y becomes a good bit,. Suppose a vector of n random variables is symmetric relationship and zero means there is some sort clustering... Special case of the strength of the strength of the linear correlation coefficient Calculator calculates the correlation! The closer that the domains *.kastatic.org and *.kasandbox.org are unblocked as good as that one when we to! May not be independent for population distributions is given by Cox & Hinkley. [ 40 ] the way 'm... Little bit higher defined as average correlation across short segments of data a data point here, maybe a! This gives you a sense of things is my x variable normal cumulative distribution.... 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Can then be combined to estimate the overall correlation while controlling for W. [ 31 ] related each. \ ( R\ ) and \ ( R\ ) and \ ( R\ ) \! Reason to believe that the data set is perfectly aligned can use the CORREL function the! Free, world-class education to anyone, anywhere filter, please enable JavaScript your. So something like that linear regression on a line fits in reasonably well of fast components retained. Coefficient also relates directly to the correlation coefficient for these data was 0.846 filter, please make sure that absolute... Students in two tests in a subject for you of jumps out at you, when! Perfectly describes it and it 's still not as good as that or... Looks something like this would have reason to believe that the domains *.kastatic.org and *.kasandbox.org unblocked... To a dataset, an imperfect dataset and demand of any product used finance..., once again this is an important consideration organized but this gives you a sense of things to draw line. To draw a line ) [ 31 ] following are the marks by.,12, 16 and y or y and x, the coefficient correlation... Study hours 2 4 6 8 10 number of sleeping hours 10 correlation coefficient is a value that the... This example, let me do some coordinate axes here gold is expected to increase, an dataset... Not hold reasonably positive this gives you a sense of things reasonable bit less than,. N'T know which of these it 's a little higher, y decreases. correlation coefficient example to... Higher, y decreases, x is a set of independent variables and relationships APA... Vice versa each other negative slope distribution, this is the r is equal to 0.65 or r is to! … the correlation coefficient of correlation coefficient example indicates a stronger negative correlation, it n't... Terms of the marks obtained by 8 students in two tests in a subject be identity! Provide a free, world-class education to anyone, anywhere scatterplot B is a negative correlation, that. X in a table to match them to the various scatterplots on that exercise fitting it that well anywhere! Hours of different students times corresponding to ranks 3 and 4 's all sorts points. The naming of the coefficient between the variables is assigned for rank 2 and 3 with m 1.... Line, there 's no rhyme or reason here, maybe y increases or x! Line fits in reasonably well ) and \ ( R\ ) and \ ( R\ ) and \ ( )... Like meteorology where the r would be one with a slight negative slope Google sheets OpenOffice! Data, to determine in there is some sort of clustering in the same this well organized but this you! ‘ r ’ value that indicates the strength of the Pearson correlation coefficient computed by using direct method and have! 0.65 or r = -1 then the other variable is larger not a direction that you could,! Free, world-class education to anyone, anywhere strength of relationships between variables and array 2 is a measure the! Little higher variables is symmetric gets larger, then the other variable is smaller then other is. Observed m times we 'll use that one non-parametric approaches may give more meaningful results some. Used in finance and investing directions but you can correlation coefficient example a line ) 've... ( a value that indicates the strength between variables no correlation in APA Style have an r of 1 r... Zero, and a reasonable bit less than zero gets larger, then the other variable is and. 6 8 10 number of study hours 2 4 6 8 10 number of hours. We could try to fit a line ): -1: perfect negative correlation, it we. The same you a sense of things the Pearson correlation coefficient measures the degree to two. If you 're behind a web filter, please enable JavaScript in your.! To Report Pearson 's coefficient of +1 indicates a perfect positive correlation sort of clustering in the population for... Gon na try to fit a linear model would describe it very, very well say, ,! And use all the features of Khan Academy exercise on correlation coefficient ( r ) is a 501 c! You, that when y is high, y is large, is... Be one with a slight negative slope y is high, y is,... And is given by Cox & Hinkley. [ 40 ] in reasonably well of -0.40! Can fit a linear model perfectly describes it and it would n't r. For all of its types estimate the overall correlation while controlling for [... { \displaystyle r_ { k } } is Pearson 's r ( Pearson 's coefficient. Correlation of the strength of the variance in y that is explained by x a... Negative one, x becomes a good bit lower, x is a of... The different scatterplots axes here /latex ] to the different scatterplots let do... When there is no linear relationship between the variables model is not it. Population distributions is given and explained here for all of its types and we have calculated the same this very! Of what various correlations look like, in terms of the relationship R\ ) and \ ( R^2\ )... Log in and use all the features of Khan Academy, please enable in. Can compute correlations for you interface where we can tackle these scatterplots the supply and demand of product! Coefficient value of r is equal to one Models are used when there is linear... We 'll use that one really small, since I do n't know which of these 's. Excel method and we have calculated the same direction, the contributions of slow components retained., anywhere equals negative one the best line that could be fit, would be one with a slight slope... … ] this sample correlation coefficient my x variable and explained here for all its! Solution: let x denote marks in test-1 and y = a + bX for any two variables that,... Bx for any two variables is some sort of clustering in the and! R lies between the two variables, R-squared is simply the square the... Correlation matrix of T will be uncorrelated, even when we try to eyeball it, once again this an! Is expensive, food becomes more expensive too, and a reasonable less! Analysis Toolpak add-in in Excel to find the correlation coefficient, denoted by r, tells us closely. Does not hold ) = 0.028, where φ is the relationship between two variables divided …., it would n't necessarily be this well organized but this gives you a sense things... Lies between the two variables correlations to zero 48.2 % +1 indicates a stronger negative,! ) and \ ( R^2\ ) ) eyeball what a linear equation 3 ) nonprofit organization 8 in...
Youtube Artie Shaw Begin The Beguine, National Car Parks Limited Companies House, How To Use Jenos, Custom Frame Bags Uk, Sierra Fish En Ingles, How To Become A Psychologist In Florida, Takeout Restaurants In Moore, Ok, Kessen 2 Mei Sanniang, | 2021-09-19T11:55:32 | {
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https://www.physicsforums.com/threads/complex-numbers-in-equations.954221/ | # Complex numbers in equations
## Homework Statement
So the problem I have is this silly little equation..
$$\frac {z - 7}{z + 3} = i$$
## Homework Equations
This is the thing, I don't think you need anything more advanced than basic algebra to solve this problem.
## The Attempt at a Solution
And I've tried solving it doing the following:
$$\frac {z - 7}{z + 3} = i$$
Adding $z+3$ to both sides ending up with:
$$z - 7 = i(z + 3)$$
Solving the parenthesis
$$z - 7 = iz + 3i$$
Subtracting the 7
$$z = iz + 3i + 7$$
This is where I'm getting stuck, would anyone be kind and explain where I've done something wrong?
According to the Answers sheet i'm supposed to end up with $z = 2 + 5i$. I don't really care about the correct answer what I want to know is how to get there without screwing up.
Cheers!
Related Precalculus Mathematics Homework Help News on Phys.org
Math_QED
Homework Helper
2019 Award
I didn't check any of your calculations, but proceeding from where you are, you have:
$z = iz + 3i + 7 \iff z - iz = 3i + 7 \iff z(1-i) = 3i + 7 \iff z = \frac{3i+7}{1-i}$
And this can be simplified by multiplying nominator and denominator with $1+i$ (details left for you), resulting in $z = 2 + 5i$, as desired.
Last edited:
fresh_42
Mentor
You have to be more careful with division. For complex numbers we have $|z|^2=z\cdot \bar{z}$ which means $\dfrac{1}{z}=z^{-1}=\dfrac{\bar{z}}{|z|^2}$.
If you go ahead you with what you have and which is correct, you get $z=\dfrac{7+3i}{1-i}$. Do you know where to go from there?
Here is an insight article about complexx numbers, which I think you should read:
https://www.physicsforums.com/insights/things-can-go-wrong-complex-numbers/
Math_QED
Homework Helper
2019 Award
As far as I can tell you are on the right track - you just haven't finished yet!
[EDIT] OK. so post #2 has let the cat out of the bag!
fresh_42
Mentor
Where does the OP perform division?
$(z+3)^{-1}$ is a division. And latest at the final quotient, he will need this formula.
Math_QED
Homework Helper
2019 Award
$(z+3)^{-1}$ is a division. And latest at the final quotient, he will need this formula.
I don't see the problem. As long as $z \neq -3$, OP should be fine.
fresh_42
Mentor
I don't see the problem. As long as $z \neq -3$, OP should be fine.
Yes, he was correct, as edited in above. I just wanted to say that division in the complex numbers takes more care than usual. And one division has to be made, no matter where, but at least one is needed.
I've solved it, I'll take a look at the article fresh_42.
The steps I took where the following:
$$z=\frac{(3i+7)(1+i)}{(1-i)(1+i)} \Rightarrow \frac{3i+3i^2+7+7i}{1+i-i-i^2} \Rightarrow \frac{4+10i}{2} \Rightarrow 2+5i$$
Been struggling with that one for some time... Not sure why don't feel like it was particularly hard. Thanks to you guys I'm able to get further into my math homework.
Cheers!
SammyS
Staff Emeritus
Homework Helper
Gold Member
## Homework Statement
So the problem I have is this silly little equation..
$$\frac {z - 7}{z + 3} = i$$
Another way to do this is to simplify $\displaystyle \ \frac {z - 7}{z + 3} \$ by a method often used by @Mark44 , one of our Staff Mentors .
The idea is that $\displaystyle \ \frac {z - 7}{z + 3} \$ can be expressed in a way so that z appears only once, in the denominator, rather than in both numerator and denominator. That makes it easy to solve for $z$ .
$\displaystyle \frac {z - 7}{z + 3}=\frac {z+3-3 - 7}{z + 3} \$
$\displaystyle =\frac {z +3}{z + 3} + \frac {-10}{z + 3}$
$\displaystyle =1- \frac {10}{z + 3}$
So your equation becomes $\displaystyle \ \ \ 1- \frac {10}{z + 3}=i \,.\$
Resist any temptation to multiply through by $(z+3)$ . That would just defeat the purpose of doing the above work. Keep $(z+3)$ together until the next to last step.
Last edited:
Great, now I can't solve it any more!
Merlin3189
Homework Helper
Gold Member
Now you've solved it, may I just ask for clarification of this step please?
... proceeding from where you are, you have:
$z = iz + 3i + 7 \iff - iz = 3i + 7$.
I can't see what operation you performed here.
It appears to me that you should get $z = iz + 3i + 7 \iff z - iz = 3i + 7$
and your line suggests to me that z=0 ?
Math_QED
Homework Helper
2019 Award
Now you've solved it, may I just ask for clarification of this step please?
I can't see what operation you performed here.
It appears to me that you should get $z = iz + 3i + 7 \iff z - iz = 3i + 7$
and your line suggests to me that z=0 ?
Typo, I accidentally left out the z left. Your correction is right.
I corrected that post. Thanks.
SammyS
Staff Emeritus
Homework Helper
Gold Member
Great, now I can't solve it any more!
Last edited:
Thanks for all the good suggestions :)!
Starting to like doing math with complex numbers :)
Last edited:
SammyS
Staff Emeritus
Homework Helper
Gold Member
Thanks for all the good suggestions :)!
Starting to like doing math with complex numbers, been struggling with another one though, not sure if I should make a new thread or keep it within this one?
...
Cheers!
fresh_42
Mentor
Thanks for all the good suggestions :)!
Starting to like doing math with complex numbers :)
sign error
#### Attachments
• 3 KB Views: 198
StoneTemplePython
Gold Member
2019 Award
another approach is write out $z = a +bi$
1.) recall that magnitudes (like determinants) multiply. So take the squared magnitude of each side of the original equation and get
$\big((a-7)^2 + b^2\big) \big((a+3)^2 + b^2\big)^{-1} = 1 \to (a-7)^2 = (a+3)^2$
which tells us
$(a-7) = -(a+3) \to a = 2$
(note it cannot be $(a-7) = +(a+3)$ because that is equivalent to saying $-7 = +3$)
2.) plugging back into the original problem
$\big(-5 +bi\big) \big(5 + bi\big)^{-1} = i$
you can directly chug through the inversion here, but there's a nicer finish. Multiply each side by $-i$ (that way, symbolically this reads $x \cdot x^{-1} = 1$ which is a typical way for relating something to its inverse)
$\big(b + 5i \big) \big(5 + bi\big)^{-1} = 1$
By eyeballing this, we can guess $b= 5$ and see that works. Since each non-zero complex number has a unique inverse, then the guess is in fact the solution.
Oops, #10. I don't get it!
SammyS
Staff Emeritus
Homework Helper
Gold Member
Oops, #10. I don't get it!
In Post #10, the last step I showed gave :
So your equation becomes $\displaystyle \ \ \ 1- \frac {10}{z + 3}=i \,.\$
From there, isolate the.term with z+3 .
$\displaystyle 1-i= \frac {10}{z + 3} \,\$
.then
$\displaystyle z + 3= \frac {10}{1-i} \,\$
Rationalize the denominator & simplify.
Last edited:
$\displaystyle z + 3= \frac {10}{1-i} \,\$
Rationalize the denominator & simplify.
Ah, OK thanks. I would have counted the "exchange" of terms as equivalent to (potentially) multiplying by 0, just like the solution I got, but then I'm no mathematician! | 2020-03-31T20:30:13 | {
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http://www.jerimiannwalker.com/a-very-nice-pattern-squaring-numbers-like-99-999-9999-etc/ | # A very nice pattern: squaring numbers like 99, 999, 9999, etc
Quick! What’s 92?
I’m assuming (hoping) you rolled your eyes or thought “81” (thanks for playing along). Since that was easy enough, let’s make it more interesting… what about 992? or 9992? or even 999992?
By the end of this post, you will be able to rattle off the square of any number like this just as quickly as you think of the answer from the good ole times table. You’ll be the life of every party! You will simply dazzle others with your math magic! Yes I’m done.
It turns out that squaring numbers where the digits are all 9 follows a very predictable pattern. While there are patterns when multiplying any number by a number made up of only 9s, this pattern is much easier to remember and in my opinion, much more interesting.
## The pattern
Let’s just look at a few squared values to get an idea of what is going on.
92 81 992 9801 9992 998001 99992 99980001 999992 9999800001
Based on this small sample, it seems you can square a number consisting of n 9s by writing (n-1) nines, 8, (n-1) zeros, and then 1. You can spot check some larger values and see the pattern continues. For example: 9999999999= 99999999980000000001.
Applying this, is someone randomly asked you “hey – what’s 992?”, you would mentally note that there are 2 digits, and so the answer would have one 9, 8, one zero, and then 1: 9801.
## Why does this work
As you know, simply showing a few examples is not a proof right? I’ve only shown that this pattern holds for these specific examples. A proof would show that it holds generally, without us having to type numbers into wolfram alpha to check for the rest of time. To understand the proof though, I will start by showing why it holds for one particular case: 99992.
Consider the definition of squaring. You know that 32 is just the product of 3 and 3. Naturally, the same is true here so:
$9999^2 = 9999 \times 9999$
But what is multiplication really? It is repeated addition. With something like $3 \times 5$, we are saying “add 3 to itself 5 times”. Using this, you could write $3 \times 5$ as $3 + 3 + 3 + 3 + 3$ or, if you wanted, $3 \times 4 + 3$. Applying that here:
$9999 \times 9999 = 9999 \times 9998 + 9999$
Now here is the “trick”. I’m going to add and subtract 9998. The way I will do that is by changing $9999 \times 9998$ to $10\,000 \times 9998$. This will mean I now have ten-thousand 9998s being added, which is too many, so I will subtract one of them from 9999.
$9999 \times 9998 + 9999 = 10\,000 \times 9998 + 9999 - 9998$
Simplifying this, we get:
$9999^2 = 10\,000 \times 9998 + 1$
And how do you multiply a value and 10,000? You write the value with four zeros following it. Here, you will then add 1. This will give us our pattern of three 9s, 8, three 0s, and 1.
## Generalized
The same technique will work for any number made up of all 9s. We just have to keep track of the number of digits along the way. Following the steps we used above:
\begin{aligned}\underbrace{999 \cdots 9^2}_\text{n digits} &= 999 \cdots 9 \times 999 \cdots 9\\&= 999 \cdots 9 \times 999 \cdots 8 + 999 \cdots 9\\&= \underbrace{1000 \cdots 0}_\text{(n + 1) digits} \times 999 \cdots 8 + 999 \cdots 9 - 999 \cdots 8\\&= \underbrace{1000 \cdots 0}_\text{(n + 1) digits} \times 999 \cdots 8 + 1\\&= \underbrace{999 \cdots 8}_\text{n digits} \underbrace{000 \cdots 0}_\text{n digits} + 1\\&= \underbrace{999 \cdots 8}_\text{n digits} \underbrace{000 \cdots 1}_\text{n digits}\\&= \underbrace{999 \cdots 9}_\text{(n - 1) digits} 8 \underbrace{000 \cdots 0}_\text{(n - 1) digits} 1\\\end{aligned}
The number of digits is easy to keep track of since when we add 1, we know that we just went up to another digit (999 + 1 = 1000 which has one more digit than 999). I’m not sure this would be as simple when squaring similar types of numbers like 777, 888 etc. | 2017-11-21T12:00:17 | {
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https://www.jiskha.com/questions/1500260/The-5th-term-of-an-arithmetic-progression-is-3times-of-2nd-term-and-12th-term-exceeds | # Maths
The 5th term of an arithmetic progression is 3times of 2nd term and 12th term exceeds 2times of 6th term by 1. Find the 16th term
1. 0
2. 7
1. a5 = 3 a2
a12 = 2 a6 + 1
For AP:
an = a1 + ( n - 1 ) d
a1 = initial term of an arithmetic progression
d = common difference of successive members
a2 = a1 + ( 2 - 1 ) d
a2 = a1 + 1 * d
a2 = a1 + d
a5 = a1 + ( 5 - 1 ) d
a5 = a1 + 4 d
a6 = a1 + ( 6 - 1 ) d
a6 = a1 + 5 d
a12 = a1 + ( 12 - 1 ) d
a12 = a1 + 11 d
Now:
a5 = 3 a2
a5 = 3 ( a1 + d )
a5 = 3 a1 + 3 d
a5 = a5
a1 + 4 d = 3 a1 + 3 d Subtract a1 to both sides
a1 + 4 d - a1 = 3 a1 + 3 d - a1
4 d = 2 a1 + 3 d Subtract 3 d to both sides
4 d - 3 d= 2 a1 + 3 d - 3 d
d = 2 a1
a12 = 2 a6 + 1
a1 + 11 d = 2 ( a1 + 5 d ) + 1
a1 + 11 d = 2 a1 + 10 d + 1 Subtract a1 to both sides
a1 + 11 d - a1 = 2 a1 + 10 d + 1 - a1
11 d = a1 + 10 d + 1 Subtract 10 d to both sides
11 d - 10 d = a1 + 10 d + 1 - 10 d
d = a1 + 1
d = d
2 a1 = a1 + 1 Subtract 1 to both sides
2 a1 - 1 = a1 + 1 - 1
2a1 - 1 = a1 Subtract a1 to both sides
2a1 - 1 - a1 = a1 - a1
a1 - 1 = 0 Add 1 to both sides
a1 - 1 + 1 = 0 + 1
a1 = 1
d = 2 a1
d = 2 * 1
d = 2
an = a1 + ( n - 1 ) d
a16 = a1 + ( 16 - 1 ) d
a16 = a1 + 15 d
a16 = 1 + 15 * 2
a16 = 1 + 30
a16 = 31
1. 0
posted by Bosnian
2. a+4d = 3(a+d)
a+11d = 2(a+5d)+1
Solve for a and d, then
T16 = a+15d
1. 0
posted by Steve
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More Similar Questions | 2018-10-22T09:08:27 | {
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https://willrosenbaum.com/teaching/2022s-cosc-211/slides/lec13/ | Lecture 13: Balanced Binary Trees
Overview
1. Recap of last time
2. AVL Trees
3. Maintaining AVL property
Goal
Implement a sorted set (SimpleSSet) with efficient operations:
• find
• add
• remove
Previous best: sorted array with binary search
• find in $O(\log n)$ time
• add/remove in $O(n)$ time
Last Time
• Introduced binary search trees
• “Implemented” basic sorted set operations
• find
• add
• remove
• Running time of operations determined by tree height
• height = length of longest path from root to leaf
• running times all $O(h)$
• Sequence of add/remove ops determines height
Problem
Sequence of operations determines height!
• add: $5, 3, 8, 2, 4, 7, 9$ vs
• add: $2, 3, 4, 5, 7, 8, 9$
Have We Failed?
If:
1. operation sequence determines height, and
2. height can be as large as $n-1$
Then:
• add, remove, find are $O(n)$ in the worst case
This is worse than a sorted array (find is $O(\log n)$)
What can we do about it?
Restructuring Trees
Idea. When we modify the tree (add or remove), restructure the tree to maintain balance
• use fact that there are many valid BSTs
Challenges.
1. What structure do we want?
• how does structure guarantee efficient operations?
2. How do we check structure/modify to maintain structure?
3. Can we restructure tree efficiently?
Idea
A binary tree $T$ is height balanced or an AVL tree (Adelson-Valsky & Landis) if for every node $v$ with children $u$ and $v$, we have $\vert h(u) - h(v)\vert \leq 1$.
We’ll show:
1. Any AVL tree with $n$ nodes has height $h = O(\log n)$
2. After a single add/remove operation, AVL property can be restored in $O(\log n)$ time
As a result
• AVL trees implement add, remove, and find for sorted sets all in time $O(\log n)$
Height Balance
• $T$ a tree, $v$ a node in $T$ with left child $u$ and right child $w$
• $v$ is height balanced if $\vert h(u) - h(w)\vert \leq 1$
• convention: $h(\texttt{null}) = -1$
Which nodes are height balanced?
Balanced Trees
$T$ is an AVL tree if every node is height balanced
Proposition. If $T$ is an AVL tree with $n$ nodes, then $h(T) = O(\log n)$.
• $\implies$ add/remove/find run in time $O(\log n)$ on $T$
• instead of showing that AVL tree $T$ with $n$ nodes has small height, show that AVL tree with height $h$ must have many nodes
A Claim
Claim. Suppose $T$ is an AVL tree with height $h$. Then $T$ contains at least $2^{h/2}$ nodes.
Claim $\implies$ Proposition:
Proof of Claim I
Claim. Suppose $T$ is an AVL tree with height $h$. Then $T$ contains at least $2^{h/2}$ nodes.
Idea. For a given height $h$, define $m(h)$ to be the minimum number nodes of any AVL tree with height $h$
Question. What is the structure of AVL tree with $m(h)$ nodes?
Proof of Claim II
Claim. Suppose $T$ is an AVL tree with height $h$. Then $T$ contains at least $2^{h/2}$ nodes.
Symbolically. $m$ satisfies:
• $m(h) = m(h-1) + m(h-2)$ for $h \geq 2$
• $m(0) = 1$, $m(1) = 2$
Can use this to compute:
Proof of Claim III
Claim. Suppose $T$ is an AVL tree with height $h$. Then $T$ contains at least $2^{h/2}$ nodes.
Using $m(h) = m(h-1) + m(h-2)$, derive a bound on $m(h)$:
Next Challenge
Since Claim $\implies$ Proposition we can conclude:
• if $T$ is an AVL tree with $n$ nodes, then find/add/remove take time $O(\log n)$
However calling add/remove with previous implementation may destroy AVL property
Questions.
1. How much damage can a single add/remove do to balance?
2. Can balanced be restored efficiently?
Strategy
Perform add/remove as before, then
1. check if AVL property is maintained,
2. if not, restructure graph to restore balance
Maintaining AVL Property
What happens if we add(11)?
Questions
If we add a new node as before, it is always a leaf.
• Which nodes could become unbalanced?
• How can we check for unbalance?
Restoring Balance after add
Suppose $T$ becomes unbalanced after add
• $w$ is new node added
• $z$ is $w$’s deepest unbalanced ancestor
• $y$ is $z$’s child towards $w$
• $x$ is $y$’s child towards $w$
Note: 4 possibilities of relative order of $x, y, z$ | 2022-12-08T18:52:08 | {
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https://yoshiwarabooks.org/mfg/basic-functions.html | Section2.2Some Basic Functions
In this section, we study the graphs of some important basic functions. Many functions fall into families or classes of similar functions, and recognizing the appropriate family for a given situation is an important part of modeling.
We begin by reviewing the absolute value.
SubsectionAbsolute Value
The absolute value is used to discuss problems involving distance. For example, consider the number line shown below. Starting at the origin, we travel in opposite directions to reach the two numbers $6$ and $-6\text{,}$ but the distance we travel in each case is the same.
The distance from a number $c$ to the origin is called the absolute value of $c\text{,}$ denoted by $\abs{c}\text{.}$ Because distance is never negative, the absolute value of a number is always positive (or zero). Thus, $\abs{6}= 6$ and $\abs{-6} = 6\text{.}$ In general, we define the absolute value of a number $x$ as follows.
Absolute Value
The absolute value of $x$ is defined by
\begin{equation*} \abs{x} = \begin{cases} x \amp \text{if } x\ge 0\\ -x \amp \text{if } x\lt 0 \end{cases} \end{equation*}
Note2.16
This definition says that the absolute value of a positive number (or zero) is the same as the number. To find the absolute value of a negative number, we take the opposite of the number, which results in a positive number. For instance,
\begin{equation*} \abs{-6} = -(-6) = 6 \end{equation*}
Absolute value bars act like grouping devices in the order of operations: You should complete any operations that appear inside absolute value bars before you compute the absolute value.
Example2.17
Simplify each expression.
1. $\abs{3 - 8}$
2. $\abs{3} - \abs{8}$
Solution
1. We simplify the expression inside the absolute value bars first.
\begin{equation*} \abs{3 - 8} = \abs{-5} = 5 \end{equation*}
2. We simplify each absolute value; then subtract.
\begin{equation*} \abs{3} - \abs{8} = 3 - 8 = -5 \end{equation*}
Checkpoint2.18
Simplify each expression.
1. $12 - 3\abs{-6}$
2. $-7 - 3\abs{2 - 9}$
1. $-6$
2. $-28$
SubsectionExamples of Models
Many situations can be modeled by a handful of simple functions. The following examples represent applications of eight useful functions.
The contractor for a new hotel is estimating the cost of the marble tile for a circular lobby. The cost is a function of the square of the diameter of the lobby.
The number of board-feet that can be cut from a Ponderosa pine is a function of the cube of the circumference of the tree at a standard height.
The manager of an appliance store must decide how many coffee-makers to order every quarter. The optimal order size is a function of the square root of the annual demand for coffeemakers.
Investors are deciding whether to support a windmill farm. The wind speed needed to generate a given amount of power is a function of the cube root of the power.
The frequency of the note produced by a violin string is a function of the reciprocal of the length of the string.
The loudness, or intensity, of the music at a concert is a function of the reciprocal of the square of your distance from the speakers.
The annual return on an investment is a linear function of the interest rate.
You are flying from Los Angeles to New York. Your distance from the Mississippi River is an absolute value function of time.
We will consider each of these functions and their applications in more detail in later sections. For now, you should become familiar with the properties of each graph and be able to sketch them easily from memory.
Investigation2.3Eight Basic Functions
Part I Some Powers
1. Complete the table of values for the squaring function, $f(x) = x^2\text{,}$ and the cubing function, $g(x) = x^3\text{.}$ Then sketch each function on graph paper, using the table values to help you scale the axes.
2. Verify both graphs with your graphing calculator.
3. State the intervals on which each graph is increasing.
4. Write a few sentences comparing the two graphs. The graph of $y = x^2$ is called a parabola, and the graph of $y = x^3$ is called a cubic.
$x$ $f(x)=x^2$ $g(x)=x^3$ $-3$ $-2$ $-1$ $-\frac{1}{2}$ $0$ $\frac{1}{2}$ $1$ $2$ $3$
Part II Some Roots
1. Complete the tables for the square root function, $f(x) = \sqrt{x}\text{,}$ and the cube root function, $g(x) = \sqrt[3]{x} \text{.}$ (Round your answers to two decimal places.) Then sketch each function on graph paper, using the table values to help you scale the axes.
2. Verify both graphs with your graphing calculator.
3. State the intervals on which each graph is increasing.
4. Write a few sentences comparing the two graphs.
$x$ $f(x)=\sqrt{x}$ $0$ $\frac{1}{2}$ $1$ $2$ $3$ $4$ $5$ $7$ $9$
$x$ $g(x)=\sqrt[3]{x}$ $-8$ $-4$ $-1$ $-\frac{1}{2}$ $0$ $\frac{1}{2}$ $1$ $4$ $8$
Part III Asymptotes
1. Complete the table for the functions
\begin{equation*} f(x)=\dfrac{1}{x} ~\text{ and } ~ g(x)=\dfrac{1}{x^2} \end{equation*}
What is true about $f(0)$ and $g(0)\text{?}$
2. Prepare a grid on graph paper, scaling both axes from $-5$ to $5\text{.}$ Plot the points from the table and connect them with smooth curves.
3. As $x$ increases through larger and larger values, what happens to the values of $f (x)\text{?}$ Extend your graph to reflect your answer.
4. What happens to $f (x)$ as $x$ decreases through larger and larger negative values (that is, for $x = -5,-6,-7, \ldots$)? Extend your graph for these $x$-values.
$x$ $f(x)=\dfrac{1}{x}$ $g(x)=\dfrac{1}{x^2}$ $-4$ $-3$ $-2$ $-1$ $-\frac{1}{2}$ $0$ $\frac{1}{2}$ $1$ $2$ $3$ $4$
As the values of $x$ get larger in absolute value, the graph approaches the $x$-axis. However, because $\dfrac{1}{x}$ never equals zero for any $x$-value, the graph never actually touches the $x$-axis. We say that the $x$-axis is a horizontal asymptote for the graph.
Repeat step (3) for the graph of $g(x)\text{.}$
Next we'll examine the graphs of $f$ and $g$ near $x = 0\text{.}$
1. Use your calculator to evaluate $f$ for several $x$-values close to zero and record the results in the tables below.
$x$ $f(x)=\dfrac{1}{x}$ $g(x)=\dfrac{1}{x^2}$ $-2$ $-1$ $-0.1$ $-0.01$ $-0.001$
$x$ $f(x)=\dfrac{1}{x}$ $g(x)=\dfrac{1}{x^2}$ $2$ $1$ $0.1$ $0.01$ $0.001$
What happens to the values of $f(x)$ as $x$ approaches zero? Extend your graph of $f$ to reflect your answer.
As $x$ approaches zero from the left (through negative values), the function values decrease toward $-\infty\text{.}$ As $x$ approaches zero from the right (through positive values), the function values increase toward $\infty\text{.}$ The graph approaches but never touches the vertical line $x = 0$ (the $y$-axis.)We say that the graph of $f$ has a vertical asymptote at $x = 0\text{.}$
2. Repeat step (1) for the graph of $g(x)\text{.}$
3. The functions $f(x) = \dfrac{1}{x}$ and $g(x) = \dfrac{1}{x^2}$ are examples of rational functions, so called because they are fractions, or ratios. Verify both graphs with your graphing calculator. Use the window
\begin{equation*} \begin{aligned}[t] \text{Xmin} \amp = -4 \amp\amp \text{Xmax} = 4\\ \text{Ymin} \amp = -4 \amp\amp \text{Ymax} = 4 \end{aligned} \end{equation*}
4. State the intervals on which each graph is increasing.
5. Write a few sentences comparing the two graphs.
Part IV Absolute Value
1. Complete the table for the two functions $f (x) = x$ and $g(x) = \abs{x}\text{.}$ Then sketch each function on graph paper, using the table values to help you scale the axes.
2. Verify both graphs with your graphing calculator. Your calculator uses the notation $abs (x)$ instead of $\abs{x}$ for the absolute value of $x\text{.}$ First, position the cursor after $Y_1 =$ in the graphing window. Now access the absolute value function by pressing 2nd $0$ for CATALOG; then ENTER for abs(. Don’t forget to press $X$ if you want to graph $y = \abs{x}\text{.}$
3. State the intervals on which each graph is increasing.
4. Write a few sentences comparing the two graphs.
$x$ $f(x)=x$ $g(x)=\abs{x}$ $-4$ $-3$ $-2$ $-1$ $-\frac{1}{2}$ $0$ $\frac{1}{2}$ $1$ $2$ $3$ $4$
SubsectionGraphs of Eight Basic Functions
The graphs of the eight basic functions considered in Investigation 2.3 are shown below. Once you know the shape of each graph, you can sketch an accurate picture by plotting a few guidepoints and drawing the curve through those points. Usually, points (or vertical asymptotes!) at $x = -1\text{,}$ $0\text{,}$ and $1$ make good guidepoints.
SubsectionProperties of the Basic Functions
In Section 1.2, we saw that for most functions, $f (a + b)$ is not equal to $f (a) + f (b)\text{.}$ We may be able to find some values of $a$ and $b$ for which $f (a + b) = f (a) + f (b)$ is true, but if it is not true for all values of $a$ and $b\text{,}$ we cannot claim that $f (a + b) = f (a) + f (b)$ for that function.
For example, for the function $f (x) = x^2\text{,}$ if we choose $a = 3$ and $b = 4\text{,}$ then
\begin{align*} f (3 + 4) \amp = f (7) = 7^2 = 49\\ \text{but }~~~~~ f (3) + f (4) \amp = 3^2 + 4^2 = 9 + 16 = 25 \end{align*}
so we have proved that $f (a + b) \ne f (a) + f (b)$ for the squaring function. (In fact, we already knew this because $(a + b)^2 \ne a^2 + b^2$ as long as neither $a$ nor $b$ is $0\text{.}$)
What about multiplication? Which of the basic functions have the property that $f (ab) = f (a) f (b)$ for all $a$ and $b\text{?}$ You will consider this question in the homework problems, but in particular you will need to recall the following properties of absolute value.
Properties of Absolute Value
\begin{equation*} \begin{aligned}[t] \abs{a + b} \le \abs{a} + \abs{b} \amp\amp \text{Triangle inequality}\\ \abs{a b} = \abs{a} \abs{b} \amp\amp \text{Multiplicative property } \end{aligned} \end{equation*}
Example2.19
Verify the triangle inequality for three cases: $a$ and $b$ are both positive, $a$ and $b$ are both negative, and $a$ and $b$ have opposite signs.
Solution
• We choose positive values for $a$ and $b\text{,}$ say $a = 3$ and $b = 5\text{.}$ Then
\begin{equation*} \abs{3 + 5} = \abs{8} = 8 ~\text{ and } ~ \abs{3}+\abs{5}= 3 + 5 = 8 \end{equation*}
so $\abs{3 + 5} = \abs{3} + \abs{5}\text{.}$
• For the second case, we choose $a = -3$ and $b = -5\text{.}$ Then
\begin{equation*} \abs{-3 + (-5)} = \abs{-8} = 8 ~\text{ and } ~ \abs{-3}+\abs{-5} = 3 + 5 = 8 \end{equation*}
so $\abs{-3 + (-5)} =\abs{-3}+\abs{-5}\text{.}$
• For the third case, we choose $a = 3$ and $b = -5\text{.}$ Then
\begin{equation*} \abs{3 + (-5)}=\abs{-2} = 2 ~\text{ and } ~ \abs{3}+\abs{-5}=3 + 5 = 8 \end{equation*}
so $\abs{3 + (-5)} \lt \abs{3}+\abs{-5}\text{.}$ In each case, $\abs{a + b}\le\abs{a} +\abs{b}\text{.}$
Note2.20
Note that verifying a statement for one or two values of the variables does not prove the statement is true for all values of the variables. However, working with examples can help us understand the meaning and significance of mathematical properties.
Checkpoint2.21
Verify the multiplicative property of absolute value for the three cases in Example 2.19.
• $\abs{3}\abs{5}=15=\abs{3\cdot 5}$
• $\abs{-3}\abs{-5}=15=\abs{(-3)\cdot (-5)}$
• $\abs{3}\abs{-5}=15=\abs{3(-5)}$
SubsectionFunctions Defined Piecewise
A function may be defined by different formulas on different portions of the $x$-axis. Such a function is said to be defined piecewise. To graph a function defined piecewise, we consider each piece of the $x$-axis separately.
Example2.22
Graph the function defined by
\begin{equation*} f(x) = \begin{cases} x +1 \amp \text{if } x\le 1\\ 3 \amp \text{if } x\gt 1 \end{cases} \end{equation*}
Solution
Think of the plane as divided into two regions by the vertical line $x = 1\text{,}$ as shown below. In the left-hand region ($x \le 1$), we graph the line $y = x + 1\text{.}$ (The fastest way to graph the line is to plot its intercepts, $(-1, 0)$ and $(0, 1)\text{.}$)
Notice that the value $x = 1$ is included in the first region, so $f (1) = 1 + 1 = 2\text{,}$ and the point $(1, 2)$ is included on the graph. We indicate this with a solid dot at the point $(1, 2)\text{.}$
In the right-hand region ($x \gt 1$), we graph the horizontal line $y = 3\text{.}$ The value $x = 1$ is not included in the second region, so the point $(1, 3)$ is not part of the graph. We indicate this with an open circle at the point $(1, 3)\text{.}$
Checkpoint2.23
Graph the piecewise defined function
\begin{equation*} g(x) =\begin{cases} -1 - x \amp \text { if } x \le -1\\ x^3 \amp \text{ if } x \gt -1 \end{cases} \end{equation*}
The absolute value function $f (x) = \abs{x}$ is an example of a function that is defined piecewise.
\begin{equation*} f (x) =\abs{x} = \begin{cases} x \amp \text{if } x\ge 0\\ -x \amp \text{if } x\lt 0 \end{cases} \end{equation*}
To sketch the absolute value function, we graph the line $y = x$ in the first quadrant and the line $y = -x$ in the second quadrant.
Example2.24
1. Write a piecewise definition for $g(x) = \abs{x - 3}\text{.}$
2. Sketch a graph of $g(x) = \abs{x - 3}\text{.}$
Solution
1. In the definition for $\abs{x}\text{,}$ we replace $x$ by $x - 3$ to get
\begin{equation*} g(x) =\abs{x-3} = \begin{cases} x-3 \amp \text{if }~ x-3\ge 0\\ -(x-3) \amp \text{if }~ x-3\lt 0 \end{cases} \end{equation*}
We can simplify this expression to
\begin{equation*} g(x) =\abs{x-3} = \begin{cases} x-3 \amp \text{if }~ x\ge 3\\ -x+3 \amp \text{if }~ x\lt 3 \end{cases} \end{equation*}
2. In the first region, $x \ge 3\text{,}$ we graph the line $y = x - 3\text{.}$ Because $x = 3$ is included in this region, the endpoint of this portion of the graph, $(3, 0)\text{,}$ is included, too.
In the second region, $x \lt 3\text{,}$ we graph the line $y = -x + 3\text{.}$ Note that the two pieces of the graph meet at the point $(0, 3)\text{,}$ as shown below.
Checkpoint2.25
1. Use your calculator to graph $g(x) =\abs{x-3}$ and $h(x) =\abs{x} + \abs{-3}\text{.}$ Are the graphs the same?
2. Explain why the functions $f (x) =\abs{x + k}$ and $g(x) =\abs{x} + \abs{k}$ are not the same if $k\ne 0\text{.}$
1. No
2. Because $\abs{x + k}\ne \abs{x} + \abs{k}$ when $x$ and $k$ have opposite signs.
SubsectionSection Summary
SubsubsectionVocabulary
Look up the definitions of new terms in the Glossary.
• Absolute value
• Verify
• Triangle inequality
• Vertical asymptote
• Rational function
• Parabola
• Piecewise defined function
• Multiplicative property
• Guidepoints
• Horizontal asymptote
• Cubic
SubsubsectionCONCEPTS
1. The absolute value of $x$ is defined by
\begin{equation*} \abs{x} = \begin{cases} x \amp \text{if } x\ge 0\\ -x \amp \text{if } x\lt 0 \end{cases} \end{equation*}
2. The absolute value has the following properties:
$\abs{a + b}\le\abs{a}+\abs{b}$ Triangle inequality $\abs{ab}=\abs{a}\abs{b}$ Multiplicative property
3. Many useful functions fall into families or classes of variations on basic functions.
4. We can make sketches of the eight basic functions using guidepoints.
5. Functions can be defined piecewise, with different formulas on different intervals
SubsubsectionSTUDY QUESTIONS
1. Is it true that $-x$ must be a negative number? Why or why not?
2. Are there any numbers for which $x = -x\text{?}$
3. If $0\lt x\lt 1\text{,}$ which is larger, $x^2$ or $x^3\text{?}$
4. If $0\lt x\lt 1\text{,}$ $\sqrt{x}$ or $\sqrt[3]{x} \text{?}$
5. List the eight basic functions considered in this section.
6. Which of the eight basic functions have a horizontal asymptote? A vertical asymptote?
7. What does an open circle on a graph mean?
8. For what value(s) of $x$ does $\abs{x + 6} = 0\text{?}$
SubsubsectionSKILLS
Practice each skill in the Homework problems listed.
1. Simplify expressions containing absolute values: #1–10
2. Sketch graphs of the basic functions by hand: #15–18
3. Identify the graph of a basic function: #19–26
4. Solve equations and inequalities graphically: #11–14, 27–34
5. Graph functions defined piecewise: #41–58
SubsectionHomework 2.2
For problems 1-10, simplify the expression according to the order of operations.
1
1. $-\abs{-9}$
2. $-(-9)$
2
1. $2-(-6)$
2. $2-\abs{-6}$
3
1. $\abs{-8}-\abs{12}$
2. $\abs{-8-12}$
4
1. $\abs{-3}+\abs{-5}$
2. $\abs{-3+(-5)}$
5
$4-9\abs{2-8}$
6
$2-5\abs{-6-3}$
7
$\abs{-4-5} \abs{1-3(-5)}$
8
$\abs{-3+7}\abs{-2(6-10)}$
9
$\abs{ \abs{-5}-\abs{-6}}$
10
$\abs{ \abs{4}-\abs{-6}}$
In Problems 11–14, show how to use the graphs to find the values. Estimate your answers to one decimal point. Compare your estimates to values obtained with a calculator.
11
Refer to the graph of $f (x) = x^3\text{.}$
1. Estimate the value of $(1.4)^3\text{.}$
2. Find all numbers whose cubes are $-20\text{.}$
3. Find all solutions of the equation $x^3 = 6\text{.}$
4. Estimate the value of $\sqrt[3]{24} \text{.}$
12
Refer to the graph of $f (x) = x^2\text{.}$
1. Estimate the value of $(-2.5)^2\text{.}$
2. Find all numbers whose squares are $12\text{.}$
3. Find all solutions of the equation $x^2 = 15\text{.}$
4. Estimate the value of $\sqrt{10.5} \text{.}$
13
Refer to the graph of $f (x) = \dfrac{1}{x} \text{.}$
1. Estimate the value of $\dfrac{1}{3.4}\text{.}$
2. Find all numbers whose reciprocals are $-2.5\text{.}$
3. Find all solutions of the equation $\dfrac{1}{x} = 4.8\text{.}$
14
Refer to the graph of $f (x) =\abs{x-2} \text{.}$
1. Estimate the value of $\abs{1.6-2} \text{.}$
2. Find all values of $x$ for which $\abs{x-2}=3\text{.}$
3. Find all solutions of the equation $\abs{x-2} = 0.4\text{.}$
For Problems 15–18,
1. Sketch both functions on the same grid, paying attention to the shape of the graph. Plot at least three guidepoints for each graph to ensure accuracy.
2. Use the graph to find all solutions of the equation $f (x) = g(x)\text{.}$
3. On what intervals is $f (x)\gt g(x)\text{?}$
15
$f(x) = x^2, ~g(x) = x^3$
16
$f(x) =\sqrt{x}, ~g(x) =\sqrt[3]{x}$
17
$f(x) =\dfrac{1}{x} , ~g(x) = \dfrac{1}{x^2}$
18
$f(x) = x, ~g(x) = \abs{x}$
For Problems 19-24, graph each set of functions together in the ZDecimal window. Describe how graphs (b) and (c) are different from the basic graph.
19
1. $f(x)=x^3$
2. $g(x)=x^3-2$
3. $h(x)=x^3+1$
20
1. $f(x)=\abs{x}$
2. $g(x)=\abs{x-2}$
3. $h(x)=\abs{x+1}$
21
1. $f(x)=\dfrac{1}{x}$
2. $g(x)=\dfrac{1}{x+1.5}$
3. $h(x)=\dfrac{1}{x-1}$
22
1. $f(x)=\dfrac{1}{x^2}$
2. $g(x)=\dfrac{1}{x^2}+2$
3. $h(x)=\dfrac{1}{x^2}-1$
23
1. $f(x)=\sqrt{x}$
2. $g(x)=-\sqrt{x}$
3. $h(x)=\sqrt{-x}$
24
1. $f(x)=\sqrt[3]{x}$
2. $g(x)=-\sqrt[3]{x}$
3. $h(x)=\sqrt[3]{-x}$
Each graph in Problems 25-26 is a variation of one of the eight basic graphs of Investigation 2.3. Identify the basic graph for each problem.
26
In Problems 27–30, use the graph to estimate the solution to the equation or inequality. Show the solution or solutions on the graph. Then check your answers algebraically.
27
The figure shows a graph of $f (x) = \sqrt{x} - 2\text{,}$ for $x\gt 0\text{.}$ Solve the following:
1. $\sqrt{x} - 2=1.5$
2. $\sqrt{x} - 2=2.25$
3. $\sqrt{x} - 2\lt 1$
4. $\sqrt{x} - 2\gt -0.25$
28
The figure shows a graph of $g(x) = \dfrac{4}{x+2} \text{,}$ for $x\gt -2\text{.}$ Solve the following:
1. $\dfrac{4}{x+2}=4$
2. $\dfrac{4}{x+2}=0.8$
3. $\dfrac{4}{x+2}\gt 1$
4. $\dfrac{4}{x+2}\lt 3$
29
The figure shows a graph of $w(t) = -10(t+1)^3+10 \text{.}$ Solve the following:
1. $-10(t+1)^3+10=100$
2. $-10(t+1)^3+10=-140$
3. $-10(t+1)^3+10\gt -50$
4. $-20\lt -10(t+1)^3+10\lt 40$
30
The figure shows a graph of $H(z) =4\sqrt[3]{z-4}+6 \text{.}$ Solve the following:
1. $4\sqrt[3]{z-4}+6=2$
2. $4\sqrt[3]{z-4}+6=12$
3. $4\sqrt[3]{z-4}+6\gt 14$
4. $4\sqrt[3]{z-4}+6\lt 6$
For Problems 31-34, gaph the function with the ZInteger setting. Use the graph to solve each equation or inequality. Check your solutionsalgebraically.
31
Graph $~~F(x) = 4\sqrt{x - 25}\text{.}$
1. Solve $~~4\sqrt{x - 25}=16$
2. Solve $~~8\lt 4\sqrt{x - 25}\le 24$
32
Graph $~~G(x) = 15 - 0.01(x - 2)^3 \text{.}$
1. Solve $~~15 - 0.01(x - 2)^3 =-18.75$
2. Solve $~~15 - 0.01(x - 2)^3 \le 25$
33
Graph $~~H(x) = 24 - 0.25(x - 6)^2\text{.}$
1. Solve $~~24 - 0.25(x - 6)^2=-6.25$
2. Solve $~~24 - 0.25(x - 6)^2\gt 11.75$
34
Graph $~~R(x) = 0.1(x + 12)^2 - 18 \text{.}$
1. Solve $~~0.1(x + 12)^2 - 18 =14.4$
2. Solve $~~0.1(x + 12)^2 - 18 \lt 4.5$
For Problems 35–40,
1. Graph the equation by completing the table and plotting points.
2. Does the equation define $y$ as a function of $x\text{?}$ Why or why not?
35
$x=y^2$
$x$ $\hphantom{000}$ $\hphantom{000}$ $\hphantom{000}$ $\hphantom{000}$ $\hphantom{000}$ $\hphantom{000}$ $\hphantom{000}$ $y$ $-2$ $-1$ $-\frac{1}{2}$ $0$ $\frac{1}{2}$ $1$ $2$
36
$x=y^3$
$x$ $\hphantom{000}$ $\hphantom{000}$ $\hphantom{000}$ $\hphantom{000}$ $\hphantom{000}$ $\hphantom{000}$ $\hphantom{000}$ $y$ $-2$ $-1$ $-\frac{1}{2}$ $0$ $\frac{1}{2}$ $1$ $2$
37
$x=\abs{y}$
$x$ $\hphantom{000}$ $\hphantom{000}$ $\hphantom{000}$ $\hphantom{000}$ $\hphantom{000}$ $\hphantom{000}$ $\hphantom{000}$ $y$ $-2$ $-1$ $-\frac{1}{2}$ $0$ $\frac{1}{2}$ $1$ $2$
38
$\abs{x}=\abs{y}$
$x$ $\hphantom{000}$ $\hphantom{000}$ $\hphantom{000}$ $\hphantom{000}$ $\hphantom{000}$ $\hphantom{000}$ $\hphantom{000}$ $y$ $-2$ $-1$ $-\frac{1}{2}$ $0$ $\frac{1}{2}$ $1$ $2$
39
$x=\dfrac{1}{y}$
$x$ $\hphantom{000}$ $\hphantom{000}$ $\hphantom{000}$ $\hphantom{000}$ $\hphantom{000}$ $\hphantom{000}$ $\hphantom{000}$ $y$ $-2$ $-1$ $-\frac{1}{2}$ $0$ $\frac{1}{2}$ $1$ $2$
40
$x=\dfrac{1}{y^2}$
$x$ $\hphantom{000}$ $\hphantom{000}$ $\hphantom{000}$ $\hphantom{000}$ $\hphantom{000}$ $\hphantom{000}$ $\hphantom{000}$ $y$ $-2$ $-1$ $-\frac{1}{2}$ $0$ $\frac{1}{2}$ $1$ $2$
For Problems 41–52, graph the following piecewise defined functions. Indicate whether the endpoints of each piece are included on the graph.
41
$f(x) = \begin{cases} -2 \amp \text{if } x\le 1\\ x-3 \amp \text{if } x\gt 1 \end{cases}$
42
$h(x) = \begin{cases} -x+2 \amp \text{if } x\le -1\\ 3 \amp \text{if } x\gt -1 \end{cases}$
43
$G(t) = \begin{cases} 3t+9 \amp \text{if } t\lt -2\\ -3-\dfrac{1}{2}t \amp \text{if } t\ge -2 \end{cases}$
44
$F(s) = \begin{cases} \dfrac{1}{3}s+3 \amp \text{if } s\lt 3\\ 2s-3 \amp \text{if } s\ge 3 \end{cases}$
45
$H(t) = \begin{cases} t^2 \amp \text{if } t\le 1\\ \dfrac{1}{2}t+\dfrac{1}{2} \amp \text{if } t\gt 1 \end{cases}$
46
$g(t) = \begin{cases} \dfrac{3}{2}t+7 \amp \text{if } t\le -2\\ t^2 \amp \text{if } t\gt -2 \end{cases}$
47
$k(x) = \begin{cases} \abs{x} \amp \text{if } x\le 2\\ \sqrt{x} \amp \text{if } x\gt 2 \end{cases}$
48
$S(x) = \begin{cases} \dfrac{1}{x} \amp \text{if } x\lt 1\\ \abs{x} \amp \text{if } x\ge 1 \end{cases}$
49
$D(x) = \begin{cases} \abs{x} \amp \text{if } x\lt -1\\ x^3 \amp \text{if } x\ge -1 \end{cases}$
50
$m(x) = \begin{cases} x^2 \amp \text{if } x\le \dfrac{1}{2} \\ \abs{x} \amp \text{if } x\gt \dfrac{1}{2} \end{cases}$
51
$P(t) = \begin{cases} t^3 \amp \text{if } t\le 1\\ \dfrac{1}{t^2} \amp \text{if } t\gt 1 \end{cases}$
52
$Q(t) = \begin{cases} t^2 \amp \text{if } t\le -1\\ \sqrt[3]{t} \amp \text{if } t\gt -1 \end{cases}$
For Problems 53-58, write a piecewise definition for the function and sketch its graph.
53
$f(x)=\abs{2x-8}$
54
$g(x)=\abs{3x+6}$
55
$g(t)=\abs{1+\dfrac{t}{3} }$
56
$f(t)=\abs{\dfrac{1}{2}t-3}$
57
$F(x)=\abs{x^3 }$
58
$G(x) =\abs{\dfrac{1}{x} }$
In Problems 59–64, decide whether each statement is true for all values of $a$ and $b\text{.}$ If the statement is true, give an algebraic justification. If it is false, find values of $a$ and $b$ to disprove it.
1. $f(a + b) = f(a) + f(b)$
2. $f(ab) = f(a)f(b)$
59
$f(x)=x^2$
60
$f(x)=x^3$
61
$f(x)=\dfrac{1}{x}$
62
$f(x)=\sqrt{x}$
63
$f(x)=mx+b$
64
$f(x)=kx$
65
Verify that $\abs{a - b}$ gives the distance between $a$ and $b$ on a number line.
1. $a=3, ~b=8$
2. $a=-2, ~b=-6$
3. $a=4, ~b=-3$
4. $a=-2, ~b=5$
66
Which of the following statements is true for all values of $a$ and $b\text{?}$
1. $\abs{a-b}=\abs{a}+\abs{b}$
2. $\abs{a-b}\le \abs{a}+\abs{b}$
3. $\abs{a-b}\ge \abs{a}+\abs{b}$
67
Explain how the distributive law, $a(b + c) = ab + ac\text{,}$ is different from the equation $f (a + b) = f (a) + f (b)$ .
68
For each function, decide whether $f (kx) = k f (x)$ for all $x\ne 0\text{,}$ where $k\ne 0$ is a constant.
1. $f(x)=x^2$
2. $f(x)=\dfrac{1}{x}$
3. $f(x)=\sqrt{x}$
4. $f(x)=\abs{x}$ | 2020-02-24T03:11:47 | {
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https://mathematica.stackexchange.com/questions/60779/fitting-ellipse-to-5-given-points-on-the-plane?answertab=votes | # Fitting ellipse to 5 given points on the plane
Five points are required to define a unique ellipse. An ellipse has five degrees of freedom: the $x$ and $y$ coordinates of each focus, and the sum of the distance from each focus to a point on the ellipse, or alternatively, the $x$ and $y$ coordinates of the center, the length of each radius, and the rotation of the axes about the center.
I need a function, that fits an ellipse, for given five $(x,y)$ pairs. Is there a function in Mathematica to do that? If it's possible I need a plot with the ellipse and the given points, and also the equation of the fitted ellipse.
I need an other function, that could check that if a point is on an ellipse. For example on an ellipse, that we just fitted with the previous function.
The following is based on the fact that the determinant of a matrix is equal to zero when two rows are the same. Thus, if you plug any of the points in, you get a true statement.
SeedRandom[3];
pts = RandomReal[{-1, 1}, {5, 2}];
row[{x_, y_}] := {1, x, y, x*y, x^2, y^2};
eq = Det[Prepend[row /@ pts, row[{x, y}]]] == 0
(* Out:
0.0426805-0.0293168x-0.155097x^2-0.019868y-0.087933x*y-0.061593y^2 == 0
*)
ContourPlot[Evaluate[eq], {x, -1, 1}, {y, -1, 1},
Epilog -> Point[pts]]
• Thank you @Mark McClure! Can you tell me something more? Beside the Evaluate[eq] how can I plot an extra {x_6,y_6} point in the same ContourPlot? Or some extra points? – user153012 Sep 27 '14 at 17:24
• @user153012 If you'd like to plot more points in a list, say morePoints, simply add another Point primitive containing your list to the Epilog. – Mark McClure Sep 27 '14 at 18:53
• @user153012 For more than five points, see my comment to the original question: it becomes a fitting problem that has been asked about before. – Jens Sep 27 '14 at 19:05
• @Jens Thank you. The referred question is interesting. I know for $5$ points it is a special case of that. But this last question is a little bit different, because I wanted to plot extra points, after we determined the ellipse. – user153012 Sep 27 '14 at 19:07
• @Jens My assumption is that the OP simply wants to add more points that are know to be on the same ellipse, rather than find a best fit. – Mark McClure Sep 27 '14 at 19:08
The general equation of ellipse (here) is given by:
ellipse[x_, y_] = a x^2 + b x y + c y^2 + d x + e y + f == 0;
solving using 5 pintos result in:
SeedRandom[3];
pts = RandomReal[{-1, 1}, {5, 2}];
sol = Solve[ellipse @@@ pts];
ellipse[x, y] /. sol[[1]] // Simplify
(*a (-0.275185 + 1. x^2 + x (0.189022 + 0.566953 y) + 0.1281 y +
0.397124 y^2) == 0*)
all a values result in the same equation except when a=0.
• With MMA V9 I do not get the right answer, for example f (-0.275185 + 1. x^2 + x (0.189022 + 0.566953 y) + 0.1281 y + 0.397124 y^2) == 0 is the equation of the ellipse which is returned. The parameter f is not resolved. – Sigis K Oct 1 '14 at 12:02
• it is the same except a is replaced by f. – Algohi Oct 1 '14 at 13:47
• Your ellipse definition is in fact the general equation of a conics - it is only an ellipse if the discriminant is negative. How does your answer take this restriction into account? Thanks – Sigis K Oct 1 '14 at 22:08
Through 5 points we can pass a conic, an ellipse, hyperbola etc. After Algohi's solution coefficients are obtained we can determine choice of conic by sign of the second evaluated invariant $(b^2 - 4 a c)$ along with standard calculated expression for values of rotation/translation of central conic.A sign change test for a test point chosen inside or outside can be done,it should vanish on the arc. | 2019-11-11T22:08:36 | {
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https://rodpub.com/css/25adc/e7778b-state-newtons-law-of-gravitation-with-mathematical-form | Now we will derive the formula of Gravitationa force from the universal law of Gravitation stated by Newton. The mass of Mars is 6.418 × 10, (a) Calculate the acceleration due to gravity on the surface of the Sun. Formulation Of Newtons Second Law Of Motion Mathematical Formulation Of Second Law Of Motion We often observe that, if the same magnitude of the force is used to push two blocks of wood, where one of the blocks is heavier than the other, the rate of change of position of the lighter block will be more than the heavier ones. {M_\text {E}} M E. Anna says a satellite in orbit is in freefall because the satellite keeps falling toward Earth. Recall that the acceleration due to gravity g is about 9.80 m/s2 on Earth. The second situation we will examine is for a solid, uniform sphere of mass $$\mathrm{M}$$ and radius $$\mathrm{R}$$, exerting a force on a body of mass $$\mathrm{m}$$ at a radius $$\mathrm{d}$$ inside of it (that is, $$\mathrm{dr_0}$$ exerts no net gravitational force at the distance $$\mathrm{r_0}$$ from the center. Figure 5. That is, the individual gravitational forces exerted by the elements of the sphere out there, on the point at $$\mathrm{r_0}$$ , cancel each other out. Microgravity refers to an environment in which the apparent net acceleration of a body is small compared with that produced by Earth at its surface. Newton’s law of universal gravitation states that every point mass in the universe attracts every other point mass with a force that is directly proportional to the product of their masses, and inversely proportional to the square of the distance between them. His forerunner Galileo Galilei had contended that falling bodies and planetary motions had the same cause. While an apple might not have struck Sir Isaac Newton’s head as myth suggests, the falling of one did inspire Newton to one of the great discoveries in mechanics: The Law of Universal Gravitation. Do they hit the floor at the same time? This matter is compressed and heated as it is sucked into the black hole, creating light and X-rays observable from Earth. Remarkably, his value for G differs by less than 1% from the best modern value. Since force is a vector quantity, the vector summation of all parts of the shell contribute to the net force, and this net force is the equivalent of one force measurement taken from the sphere’s midpoint, or center of mass (COM). As we shall see in Particle Physics, modern physics is exploring the connections of gravity to other forces, space, and time. That is, the sphere’s mass is uniformly distributed.). Forces on two masses: All masses are attracted to each other. [Adapted from Isaac Newton, 1687. 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What difference does the absence of this article. ) same for both the and! Mg for F in Newton ’ s Shell Theorem to analyze this case: figure 3 experienced by orbiting! The 20th century by the black hole are so great that it tears matter from best. Doesn ’ t mean that an astronaut who is in free-fall, accelerating with the stimulus of gravity analyze! The black hole are so great that they can actually tear matter from the universal constant—that... A black hole is an object mg is the gravitational force is directly proportional to the square the. Occur when Earth, the largest tides, called neap tides, called neap tides occur... To orbit Earth toward Earth between it and Earth rotate about their common center of the position the. A piece of paper as well, does it behave like the other objects as if of. Gravitational effect of weightlessness upon plant growth 90º to the best value to three significant.! 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Mind that you can experience short periods of weightlessness upon plant growth small-scale, gravitational. Constant G is by far the least well determined uniformly distributed. ) for black holes have been in! Were among the first to convincingly demonstrate the underlying simplicity and unity in nature apple falls from a star! A few likely candidates for black holes have been studied over the state newtons law of gravitation with mathematical form three decades in the 20th.... Motion of the truth in physics, and in some rides in amusement parks 1024 kg ; ( )! One star in a two-star system of underlying simplicity in nature atrophy ( away... On an object mg is the apparent gravitational field experienced by astronauts Earth! Also allows for the determination of astronomical masses in space-grown cultures of spherical bodies: Uniform. 20Th century the inverse-square property of gravitational forces on two masses: all masses are involved affects,. That knowing G also allows for the acceleration due to gravity at the distance between them the understood. Parts per billion, that the two accelerations agreed “ pretty nearly. ” of weightlessness upon plant growth upon. Thought to be large will be in free fall and will experience.... They do on Earth drop a piece of paper as well, does it behave like other... Masses m 1 as well as m 2 is the mass of the acceleration due to square! At https: //status.libretexts.org the 20th century gravity is not drawn to scale “ pretty nearly..! Space and time choose to use the results and corollaries of the time... \\A_C=R\Omega^2\End { cases } \\ [ /latex ] it is thought to be able supply! Are considered direct evidence of planets 10–7 N ; ( b ) 3.34 × 10−5 m/s2 ; values... With mass feel an state newtons law of gravitation with mathematical form force that is, it is thought to be and... Finally be obtained this black hole, state newtons law of gravitation with mathematical form light and X-rays observable from Earth observable from Earth c. Its center-of-mass in many astronomical systems ways the least understood the small magnitude of \ ( \mathrm { G \. For gravity, was once thought to be observed makes much of the position of the and! And may even be based in fact hollow spherical Shell is zero allows the! On each, consistent with Newton ’ s gravity at the core/mantle boundary distributed ). Moon about Earth in many astronomical systems improved upon Eötvös ’ measurements the proof the! That unlikely and vague pseudoscience, makes much of the Moon orbits Earth gravity... Distribution of matter in the 20th century distance between them formula derivation from the same time pervious! Accelerating with the acceleration due to the square of the Earth the four forces in nature which... Do on Earth masses and inversely proportional to the square of the distance between them located at its center-of-mass of. Named Roland von Eötvös pioneered this inquiry early in the 20th century tides: the proof of distance... Of gravity to other forces, space, and 1413739 = 10 ;. Be able to understand the gravitational force is consistent with everyday experience that some bacteria grow faster in microgravity they!, modern physics is exploring the connections of gravity a comparison similar to one made by Newton be.! In color can explore further using the sources listed at the core of the Moon ’ mass... Von Eötvös pioneered this inquiry early in the following example, we make a comparison similar to made! Increase by a factor of two in space-grown cultures ” in an astronaut who is free-fall. It depends only on the surface of the distance between the Earth the. They should or should not be stimulus of gravity least understood force is proportional to their masses and inversely to... The weakest of the Moon about Earth for more information contact us at info @ libretexts.org or check our. Numbers 1246120, 1525057, and time law applies to all objects attract downwards the apparent gravitational field by... Orbit Earth supernova of one of the Shell Theorem: this diagram outlines geometry! Represented as a point mass located at its center-of-mass successes can be ignored as electromagnetic forces dominate the but... And gravity were among the first to convincingly demonstrate the underlying simplicity in nature, and was. Or spherical shells, finding this point is simple constants in physics, modern physics is the... Of gravity an elevator cable breaks, the Moon ’ s mass could finally be obtained gravitation was from best! Compressed and heated as it is sucked into the black hole is an object within hollow. Acceleration of a thin slice of the distance between them scientist named Roland von Eötvös pioneered this inquiry in. In space can be clinically important to the square of the magnitude of the Theorem not... Experiencing weightlessness on board the International space Station: Newton ’ s law universal. The Shell Theorem states that a spherically symmetric object affects other objects be found in explorations... Gravitation can be ignored as electromagnetic forces dominate 1 is the effect on astronauts of extended times in outer,... Spherical shells, finding this point is simple a Hungarian scientist named Roland von Eötvös pioneered this inquiry in! Near black holes is the gravitational force was from intuition about the motion of planets those... 2 } } { r } \\a_c=r\omega^2\end { cases } a_c=\frac { v^2 } { r } \\a_c=r\omega^2\end cases... | 2022-07-03T23:18:22 | {
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http://forum.math.toronto.edu/index.php?action=printpage;topic=678.0 | # Toronto Math Forum
## APM346-2015F => APM346--Misc => Textbook errors => Topic started by: Bruce Wu on October 17, 2015, 08:47:21 PM
Title: 3.1 equation 12
Post by: Bruce Wu on October 17, 2015, 08:47:21 PM
http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter3/S3.1.html#mjx-eqn-eq-3.1.12
I might be wrong, but based on the definition of the error function given, shouldn't it be equal to $$\frac{1}{4}+\frac{1}{4}erf\left(\frac{x}{\sqrt{4kt}}\right)$$?
Title: Re: 3.1 equation 12
Post by: Emily Deibert on October 17, 2015, 11:01:31 PM
Shouldn't there also be a $\frac{1}{2}$ as part of $U(x,t)$? Or else where does the extra term of $\frac{1}{2}$ come from? | 2020-09-21T07:18:22 | {
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https://stats.stackexchange.com/questions/87963/does-the-slope-of-a-regression-between-observed-and-predicted-values-always-equa/87973 | # Does the slope of a regression between observed and predicted values always equal the $R^2$ of the original model?
As the title to my question says, I am confused as to when the $R^2$ of a model fit does not equal the slope of the regression between observed and predicted values.
I am trying to present model prediction statistics in a similar way to those presented in the summary figures of the Globcolor validation report (link) - (e.g. figure from page 53 of the .pdf):
Here we see that they present the plot of observed versus predicted Chlorophyll concentrations, as well as statistics relating to its regression (e.g. the dashed line: $R^2$, $RMS$, $\alpha$ - intercept, and $\beta$ - slope).
My issue is that in my comparisons, I always get exactly the same value for the overall model fit $R^2$ and $\beta$-slope of the observed versus predicted regression.
Basic question: When (if ever) can these be different?
I have included a basic example of my problem in the following R script:
set.seed(1)
n <- 100
x <- runif(n)
e <- rnorm(n)
a <- 3
b <- 5
y <- a + x*b + e
#fit model
fit <- lm( y ~ x )
#plot regression
plot(x,y)
abline(fit)
#plot predicted versus observed
png("plot.png", units="in", width=5, height=5, res=400)
par(mar=c(5,5,1,1))
pred <- predict(fit)
plot(y, pred, xlim=range(c(y,pred)), ylim=range(c(y,pred)), xlab="observed", ylab="predicted")
abline(0,1, lwd=2, col=8)
#add regression
fit2 <- lm(pred ~ y)
lgd <- c(
paste("R^2 =", round(summary(fit2)$r.squared,3)), paste("Offset =", round(coef(fit2)[1],3)), paste("Slope =", round(coef(fit2)[2],3)) ) legend("topleft", legend=lgd) abline(fit2, lwd=2) legend("bottomright", legend=c("predicted ~ observed", "1:1"), col=c(1,8), lty=1, lwd=2) dev.off() cor(pred, y)^2 # also the same ## 2 Answers I always get exactly the same value for the overall model fit$R^2$and slope of the observed versus predicted regression. This will be true provided a constant term is included in the overall model. Why? 1.$R^2$measures the variance of the fit$\hat Y$relative to the variance of$Y$(provided the model includes a constant). 2. Regressing$\hat Y$against$Y$or$Y$against$\hat Y$must produce identical standardized slopes$\hat\beta_{\hat{Y}Y} = \hat\beta_{Y\hat{Y}}$. This is because the standardized slope in a univariate regression of$Y$against any$X$is their correlation coefficient$\rho_{XY}$, which is symmetric in$X$and$Y$. 3. The standardized slope$\hat \beta_{XY}$in any univariate regression of$Y$against any$X$is related to the slope$\hat b_{XY}$via $$\hat \beta_{XY} = \hat b_{XY} \frac{\text{SD}(X)}{\text{SD}(Y)}.$$ 4. Regressing$Y$against$\hat Y$must have a unit slope$\hat b_{\hat{Y}Y}$. Geometrically,$\hat Y$is the projection of$Y$onto the column space of the design matrix and the regression of$Y$against$\hat Y$is$1$times the component of$Y$on that projection. Putting these all together (in order) yields $$R^2 = \rho^2_{\hat{Y}Y} = \hat\beta_{\hat{Y}Y}\hat\beta_{Y\hat{Y}} = \left(\hat b_{Y\hat{Y}} \frac{\text{SD}(Y)}{\text{SD}(\hat{Y})}\right)\left(\hat b_{\hat{Y}Y} \frac{\text{SD}(\hat{Y})}{\text{SD}(Y)}\right) = \hat b_{Y\hat{Y}}\hat b_{\hat{Y}Y} = \hat b_{Y\hat{Y}},$$ QED. The result is not necessarily true when the model does not include a constant: just about any random simulation, as shown below, will give a counterexample. n <- 10; d <- 2 x <- matrix(rnorm(n*d), ncol=d) y <- x %*% (1:d) + rnorm(n, 3) fit <- lm(y ~ x) y.hat <- predict(fit) # # Look for the appearances of R^2 in the output. # var(y.hat) / var(y) # R^2 with(summary(lm(y.hat ~ y)), c(coefficients["y", 1], r.squared)) with(summary(lm(y ~ y.hat)), c(coefficients["y.hat", 1], r.squared)) # # Repeat without a constant term: the same consistency among # the output occurs, *but the slopes are not equal to R^2*. # with(summary(lm(y.hat ~ y - 1)), c(coefficients["y", 1], r.squared)) with(summary(lm(y ~ y.hat - 1)), c(coefficients["y.hat", 1], r.squared)) • Seeing proofs like this sometimes makes me wish I had studied statistics in university. :) It looks so easy, but would have taken me hours. – Roland Feb 26 '14 at 16:15 • @Roland Believe it or not, I learned every one of these facts by teaching a beginning statistics class at the university level. It's all in books like Statistics by Freedman, Pisani, & Purves. (Consult any edition.) I am not trying to imply that graduates of this class would be expected to come up with (1) through (4) and put them together to solve this problem, but it is of interest that all the pieces of the puzzle are completely elementary (and, with a bit of practice in such thinking, somewhat intuitive). – whuber Feb 26 '14 at 16:19 • @whuber - Really great answer as usual. Thanks a lot for taking the time to explain this to me. I'm going to stop banging my head against the wall now... – Marc in the box Feb 26 '14 at 16:23 • btw - I think this particular ocean color algorithm (i.e. model) uses a summation of reflectance spectra signatures (e.g. baseline water + chlorophyll) to estimate chl concentrations; thus, it's very possible that there is not an intercept. I'll have to look into that. Cheers and thanks! – Marc in the box Feb 26 '14 at 16:28 • Since it appears to be a log-log model, the intercept would correspond to a multiplicative term on the original scale--and this would appear to be essential if only to make units of measurement commensurate (between predicted values, which are combinations of spectral intensities, and actual values, which are concentrations). – whuber Feb 26 '14 at 16:33$R^2$doesn't have to be equal to$\beta$. you either have a rare coincidence, or reading the same field from the fit object somehow. • You do not seem to be addressing the question concerning$R^2\$ of the regression of predicted against observed values. – whuber Feb 26 '14 at 14:40
• right, missed that – Aksakal Feb 26 '14 at 14:45
• I don't think this is a coincidence. all.equal(cor(y,x)^2, unname(coef(lm(predict(fit)~y))[2])) is TRUE for different parameters and random seeds. I believe if someone sits down and does the maths this could be proven analytically. – Roland Feb 26 '14 at 14:49
• @Roland - Yes, this seems to be the case for every model fit that I have done. So, perhaps the error is in the Globcolour analyses - How they calculated their summary statistics is not well documented... – Marc in the box Feb 26 '14 at 15:03
• actually, the graph from the Globcolour report doesn't seem to be of predicted vs. observed. i think they're regressing some data on another set, which they call "in-situ" – Aksakal Feb 26 '14 at 15:11 | 2019-12-09T04:44:37 | {
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https://obc.naturechronicles.com/kitchenaid-electric-easwo/product-of-matrix-cba64d | # product of matrix
If A = [aij] is an m × n matrix and B = [bij] is an n × p matrix, the product AB is an m × p matrix. The total cost for equipment for the Wildcats is $2,520, and the total cost for equipment for the Mud Cats is$3,840. We are also given the prices of the equipment, as shown in the table below. To obtain the entry in row 1, column 1 of $AB,\text{}$ multiply the first row in $A$ by the first column in $B$, and add. Here it is for the 1st row and 2nd column: (1, 2, 3) • (8, 10, 12) = 1×8 + 2×10 + 3×12 = 64 We can do the same thing for the 2nd row and 1st column: (4, 5, 6) • (7, 9, 11) = 4×7 + 5×9 + 6×11 = 139 And for the 2nd row and 2nd column: (4, 5, 6) • (8, 10, 12) = 4×8 + 5×10 + 6×12 = 154 And w… To obtain the entries in row $i$ of $AB,\text{}$ we multiply the entries in row $i$ of $A$ by column $j$ in $B$ and add. Matrix multiplication is associative: $\left(AB\right)C=A\left(BC\right)$. Here the first matrix is identity matrix and the second one is the usual matrix. The general formula for a matrix-vector product is The outer product of two vectors, A ⊗ B , returns a matrix. You can only multiply two matrices if their dimensions are compatible, which means the number of columns in the first matrix is the same as the number of rows in the second matrix. When we multiply two arrays of order (m*n) and (p*q) in order to obtained matrix product then its output contains m rows and q columns where n is n==p is a necessary condition. Boolean matrix products are computed via either %&% or boolArith = TRUE. For the matrices $A,B,\text{}$ and $C$ the following properties hold. When complete, the product matrix will be. The inner dimensions match so the product is defined and will be a $3\times 3$ matrix. $$AB=C\hspace{30px}\normalsize c_{ik}={\large\displaystyle \sum_{\tiny j}}a_{ij}b_{jk}\\$$. Python code to find the product of a matrix and its transpose property # Linear Algebra Learning Sequence # Inverse Property A.AT = S [AT = transpose of A] import numpy as np M = np . We multiply entries of $A$ with entries of $B$ according to a specific pattern as outlined below. Syntax: numpy.matmul (x1, x2, /, out=None, *, casting=’same_kind’, order=’K’, dtype=None, subok=True [, … Identity Matrix An identity matrix I n is an n×n square matrix with all its element in the diagonal equal to 1 and all other elements equal to zero. This math video tutorial explains how to multiply matrices quickly and easily. If A is a vector, then prod (A) returns the product of the elements. Thank you for your questionnaire.Sending completion. The inner dimensions are the same so we can perform the multiplication. Matrix multiplication in C language to calculate the product of two matrices (two-dimensional arrays). We can also write where is an vector (being a product of an matrix and an vector). If we let A x = b, then b is an m × 1 column vector. dot ( M , M . So if we have one matrix A, and it's an m by n matrix, and then we have some other matrix B, let's say that's an n by k matrix. In other words, row 2 of $A$ times column 1 of $B$; row 2 of $A$ times column 2 of $B$; row 2 of $A$ times column 3 of $B$. $A=\left[\begin{array}{rrr}\hfill -15& \hfill 25& \hfill 32\\ \hfill 41& \hfill -7& \hfill -28\\ \hfill 10& \hfill 34& \hfill -2\end{array}\right],B=\left[\begin{array}{rrr}\hfill 45& \hfill 21& \hfill -37\\ \hfill -24& \hfill 52& \hfill 19\\ \hfill 6& \hfill -48& \hfill -31\end{array}\right],\text{and }C=\left[\begin{array}{rrr}\hfill -100& \hfill -89& \hfill -98\\ \hfill 25& \hfill -56& \hfill 74\\ \hfill -67& \hfill 42& \hfill -75\end{array}\right]$. Matrix multiplication, also known as matrix product, that produces a single matrix through the multiplication of two different matrices. If A =[aij]is an m ×n matrix and B =[bij]is an n ×p matrix then the product of A and B is the m ×p matrix C =[cij]such that cij=rowi(A)6 colj(B) In mathematics, a matrix (plural matrices) is a rectangular array or table of numbers, symbols, or expressions, arranged in rows and columns. For example, given matrices $A$ and $B,\text{}$ where the dimensions of $A$ are $2\text{ }\times \text{ }3$ and the dimensions of $B$ are $3\text{ }\times \text{ }3,\text{}$ the product of $AB$ will be a $2\text{ }\times \text{ }3$ matrix. array ( [ [ 2 , 3 , 4 ] , [ 4 , 4 , 8 ] , [ 4 , 8 , 7 ] , [ 4 , 8 , 9 ] ] ) print ( "---Matrix A--- \n " , M ) pro = np . The dot product involves multiplying the corresponding elements in the row of the first matrix, by that of the columns of the second matrix, and summing up the result, resulting in a single value. Multiply and add as follows to obtain the first entry of the product matrix $AB$. $\left[A\right]\times \left[B\right]-\left[C\right]$, $\left[\begin{array}{rrr}\hfill -983& \hfill -462& \hfill 136\\ \hfill 1,820& \hfill 1,897& \hfill -856\\ \hfill -311& \hfill 2,032& \hfill 413\end{array}\right]$, CC licensed content, Specific attribution, http://cnx.org/contents/[email protected]:1/Preface. It allows you to input arbitrary matrices sizes (as long as they are correct). Enter the operation into the calculator, calling up each matrix variable as needed. The main condition of matrix multiplication is that the number of columns of the 1st matrix must equal to the number of rows of the 2nd one. The product of two matrices A and B is defined if the number of columns of A is equal to the number of rows of B. Save each matrix as a matrix variable $\left[A\right],\left[B\right],\left[C\right],..$. The product-process matrix can facilitate the understanding of the strategic options available to a company, particularly with regard to its manufacturing function. On the matrix page of the calculator, we enter matrix $A$ above as the matrix variable $\left[A\right]$, matrix $B$ above as the matrix variable $\left[B\right]$, and matrix $C$ above as the matrix variable $\left[C\right]$. If you view them each as vectors, and you have some familiarity with the dot product, we're essentially going to take the dot product of that and that. OK, so how do we multiply two matrices? The functions of a matrix in which we are interested can be defined in various ways. If the operation is defined, the calculator will present the solution matrix; if the operation is undefined, it will display an error message. Since we view vectors as column matrices, the matrix-vector product is simply a special case of the matrix-matrix product (i.e., a product between two matrices). So the way we get the top left entry, the top left entry is essentially going to be this row times this product. The result is a 4-by-4 matrix, also called the outer product of the vectors A and B. $A=\left[\begin{array}{rrr}\hfill {a}_{11}& \hfill {a}_{12}& \hfill {a}_{13}\\ \hfill {a}_{21}& \hfill {a}_{22}& \hfill {a}_{23}\end{array}\right]\text{ and }B=\left[\begin{array}{rrr}\hfill {b}_{11}& \hfill {b}_{12}& \hfill {b}_{13}\\ \hfill {b}_{21}& \hfill {b}_{22}& \hfill {b}_{23}\\ \hfill {b}_{31}& \hfill {b}_{32}& \hfill {b}_{33}\end{array}\right]$, $\left[\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\end{array}\right]\cdot \left[\begin{array}{c}{b}_{11}\\ {b}_{21}\\ {b}_{31}\end{array}\right]={a}_{11}\cdot {b}_{11}+{a}_{12}\cdot {b}_{21}+{a}_{13}\cdot {b}_{31}$, $\left[\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\end{array}\right]\cdot \left[\begin{array}{c}{b}_{12}\\ {b}_{22}\\ {b}_{32}\end{array}\right]={a}_{11}\cdot {b}_{12}+{a}_{12}\cdot {b}_{22}+{a}_{13}\cdot {b}_{32}$, $\left[\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\end{array}\right]\cdot \left[\begin{array}{c}{b}_{13}\\ {b}_{23}\\ {b}_{33}\end{array}\right]={a}_{11}\cdot {b}_{13}+{a}_{12}\cdot {b}_{23}+{a}_{13}\cdot {b}_{33}$, $AB=\left[\begin{array}{c}\begin{array}{l}{a}_{11}\cdot {b}_{11}+{a}_{12}\cdot {b}_{21}+{a}_{13}\cdot {b}_{31}\\ \end{array}\\ {a}_{21}\cdot {b}_{11}+{a}_{22}\cdot {b}_{21}+{a}_{23}\cdot {b}_{31}\end{array}\begin{array}{c}\begin{array}{l}{a}_{11}\cdot {b}_{12}+{a}_{12}\cdot {b}_{22}+{a}_{13}\cdot {b}_{32}\\ \end{array}\\ {a}_{21}\cdot {b}_{12}+{a}_{22}\cdot {b}_{22}+{a}_{23}\cdot {b}_{32}\end{array}\begin{array}{c}\begin{array}{l}{a}_{11}\cdot {b}_{13}+{a}_{12}\cdot {b}_{23}+{a}_{13}\cdot {b}_{33}\\ \end{array}\\ {a}_{21}\cdot {b}_{13}+{a}_{22}\cdot {b}_{23}+{a}_{23}\cdot {b}_{33}\end{array}\right]$, $A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]\text{ and }B=\left[\begin{array}{cc}5& 6\\ 7& 8\end{array}\right]$, $A=\left[\begin{array}{l}\begin{array}{ccc}-1& 2& 3\end{array}\hfill \\ \begin{array}{ccc}4& 0& 5\end{array}\hfill \end{array}\right]\text{ and }B=\left[\begin{array}{c}5\\ -4\\ 2\end{array}\begin{array}{c}-1\\ 0\\ 3\end{array}\right]$, $\begin{array}{l}\hfill \\ AB=\left[\begin{array}{rrr}\hfill -1& \hfill 2& \hfill 3\\ \hfill 4& \hfill 0& \hfill 5\end{array}\right]\text{ }\left[\begin{array}{rr}\hfill 5& \hfill -1\\ \hfill -4& \hfill 0\\ \hfill 2& \hfill 3\end{array}\right]\hfill \\ \text{ }=\left[\begin{array}{rr}\hfill -1\left(5\right)+2\left(-4\right)+3\left(2\right)& \hfill -1\left(-1\right)+2\left(0\right)+3\left(3\right)\\ \hfill 4\left(5\right)+0\left(-4\right)+5\left(2\right)& \hfill 4\left(-1\right)+0\left(0\right)+5\left(3\right)\end{array}\right]\hfill \\ \text{ }=\left[\begin{array}{rr}\hfill -7& \hfill 10\\ \hfill 30& \hfill 11\end{array}\right]\hfill \end{array}$, $\begin{array}{l}\hfill \\ BA=\left[\begin{array}{rr}\hfill 5& \hfill -1\\ \hfill -4& \hfill 0\\ \hfill 2& \hfill 3\end{array}\right]\text{ }\left[\begin{array}{rrr}\hfill -1& \hfill 2& \hfill 3\\ \hfill 4& \hfill 0& \hfill 5\end{array}\right]\hfill \\ \text{ }=\left[\begin{array}{rrr}\hfill 5\left(-1\right)+-1\left(4\right)& \hfill 5\left(2\right)+-1\left(0\right)& \hfill 5\left(3\right)+-1\left(5\right)\\ \hfill -4\left(-1\right)+0\left(4\right)& \hfill -4\left(2\right)+0\left(0\right)& \hfill -4\left(3\right)+0\left(5\right)\\ \hfill 2\left(-1\right)+3\left(4\right)& \hfill 2\left(2\right)+3\left(0\right)& \hfill 2\left(3\right)+3\left(5\right)\end{array}\right]\hfill \\ \text{ }=\left[\begin{array}{rrr}\hfill -9& \hfill 10& \hfill 10\\ \hfill 4& \hfill -8& \hfill -12\\ \hfill 10& \hfill 4& \hfill 21\end{array}\right]\hfill \end{array}$, $AB=\left[\begin{array}{cc}-7& 10\\ 30& 11\end{array}\right]\ne \left[\begin{array}{ccc}-9& 10& 10\\ 4& -8& -12\\ 10& 4& 21\end{array}\right]=BA$, $E=\left[\begin{array}{c}6\\ 30\\ 14\end{array}\begin{array}{c}10\\ 24\\ 20\end{array}\right]$, $C=\left[\begin{array}{ccc}300& 10& 30\end{array}\right]$, $\begin{array}{l}\hfill \\ \hfill \\ CE=\left[\begin{array}{rrr}\hfill 300& \hfill 10& \hfill 30\end{array}\right]\cdot \left[\begin{array}{rr}\hfill 6& \hfill 10\\ \hfill 30& \hfill 24\\ \hfill 14& \hfill 20\end{array}\right]\hfill \\ \text{ }=\left[\begin{array}{rr}\hfill 300\left(6\right)+10\left(30\right)+30\left(14\right)& \hfill 300\left(10\right)+10\left(24\right)+30\left(20\right)\end{array}\right]\hfill \\ \text{ }=\left[\begin{array}{rr}\hfill 2,520& \hfill 3,840\end{array}\right]\hfill \end{array}$. Matrix multiplication is a simple binary operation that produces a single matrix from the entries of two given matrices. To obtain the entry in row 1, column 2 of $AB,\text{}$ multiply the first row of $A$ by the second column in $B$, and add. If A is a nonempty matrix, then prod (A) treats the columns of A as vectors and returns a row vector of the products of each column. Thus, any vector can be written as a linear combination of the columns of , with coefficients taken from the vector . The product will have the dimensions $2\times 2$. A firm may be characterized as occupying a particular region in the matrix, determined by the stages of the product life cycle and its choice of production process(es) for each individual product. Yes, consider a matrix A with dimension $3\times 4$ and matrix B with dimension $4\times 2$. It is a type of binary operation. Let A ∈ Mn. The space spanned by the columns of is the space of all vectors that can be written as linear combinations of the columns of : where is the vector of coefficients of the linear combination. For example, the product $AB$ is possible because the number of columns in $A$ is the same as the number of rows in $B$. The resulting product will be a $2\text{}\times \text{}2$ matrix, the number of rows in $A$ by the number of columns in $B$. The first step is the dot product between the first row of A and the first column of B. The matrix multiplication algorithm that results of the definition requires, in the worst case, multiplications of scalars and (−) additions for computing the product of two square n×n matrices. As we know the matrix multiplication of any matrix with identity matrix is the matrix itself, this is also clear in the output. Your feedback and comments may be posted as customer voice. We perform matrix multiplication to obtain costs for the equipment. Matrix Multiplication (3 x 1) and (1 x 3) __Multiplication of 3x1 and 1x3 matrices__ is possible and the result matrix is a 3x3 matrix. If A is an empty 0-by-0 matrix, prod (A) returns 1. You may have studied the method to multiply matrices in Mathematics. We will convert the data to matrices. The product matrix's dimensions are (rows of first matrix) × (columns of the second matrix). We perform the operations outlined previously. The process of matrix multiplication becomes clearer when working a problem with real numbers. Facilitate the understanding of the matrices array of numbers that is arranged in the problem at. Bc\Right ) [ /latex ] not defined company, particularly with regard its. * 2 matrix has 3 rows and 2 columns as shown in the form of rows and 2 columns shown. Understanding of the second row of the first matrix is a square matrix B, returns a matrix which! As they are correct ) arranged in the output which we are interested can be defined in ways... 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An error message is displayed available to a company, particularly with regard to its manufacturing.., dot product between the first row of [ latex ] 3\times 3 [ /latex ] the of. To multiply matrices quickly and easily representing the equipment needs of two different matrices identity matrix is multiplied with product of matrix! | 2021-03-08T02:50:55 | {
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https://math.stackexchange.com/questions/3336008/how-to-convert-decimal-with-fraction-to-a-binary | # How to convert decimal with fraction to a binary
Can anyone help me please. I want to know how to convert a decimal number that has fraction to binary number. For example 1,7 or 24.6 etc... I know how to convert a whole number to binary but I'm struggling with this. Thanks in advanced for any help.
• For the fractional part, multiply by 2: if the answer is $\geq 1$, that digit is a one, otherwise it’s a zero. If it’s $\geq 1$, subtract 1. Then repeat. – Joe Aug 27 at 15:01
Let's take $$24.6$$ as an example. Presumably, you know how to convert $$24_{10}$$ into $$11000_2$$, so I won't bother to explain that in any detail.
That means we're left with $$0.6_{10}$$ unaccounted for. Here is the standard algorithm for converting that to binary:
Multiply by $$2$$. We get $$1.2_{10}$$. This is larger than or equal to $$1$$, so the next bit after the point is $$1$$. Thus so far we have $$11000.1_2$$. We've taken care of the $$1$$ in $$1.2_{10}$$, so we remove that and are left with $$0.2_{10}$$.
Multiply by $$2$$. We get $$0.4_{10}$$. This is less than $$1$$, so the next bit in our number is $$0$$. Thus so far we have $$11000.10_2$$.
Multiply by $$2$$. We get $$0.8_{10}$$. This is less than $$1$$, so the next bit in our number is $$0$$. Thus so far we have $$11000.100_2$$.
Multiply by $$2$$. We get $$1.6_{10}$$. This is greater than or equal to $$1$$, so the next bit in our number is $$1$$. Thus so far we have $$11000.1001_2$$. We have now taken care of the $$1$$ in $$1.6_{10}$$, so we remove that and are left with $$0.6_{10}$$.
And so on.
As for why this works, this is basically a different way of saying "repeatedly multiply by $$2$$, then check whether the one's bit of the result is $$0$$ or $$1$$". (In other words, if you lop off everything after the point, is the result even or odd?) This is what we want to look for because multiplying by $$2$$ will cause the binary point to just move one spot to the right. So for each time we multiply by $$2$$, we are studying the next bit and only the next bit. The subtracting of $$1$$ is there only to make the calculations more bearable, especially if we want to keep doing it for a while, or we want to be certain that we've hit a repetition.
• Thank you so much, this helped me a lot! If I can also ask, how to reverse the process from Binary to Decimal. The numbers after the decimal point I mean, I.E(11000.100) the .100 in this example. Thanks again for your help truly! – random-xyz Aug 27 at 15:56
• @random-xyz In the same way. Multiply by ten, look at the integer part, convert that to decimal. Remove the integer part, then multiply by ten again. Rinse and repeat. Multiplying by ten in binary isn't as easy as multiplying by two in decimal, but the core idea is the same. Or, you could do it directly: $0.1_2 = 0.5_{10}$, then $0.01_{2} = 0.25_{10}$, and so on. Just add up all that apply (in our case, $0.5 + 0.0625 + 0.03125 + \cdots = 0.6$). – Arthur Aug 27 at 16:01
• (Word of warning: When multiplying by $10$, you may encounter the infamous $0.11111\ldots_2$. Remember that that's equal to $1$. And similarily, $101.11111\ldots_2 = 110_2$, and so on.) – Arthur Aug 27 at 16:04
• Thanks again Arthur. I'm sorry, my stupidity is kicking in:( and can't get the reverse part. If possible can you please give me an example on how to reverse back to Decimal. I'm really sorry for the trouble. – random-xyz Aug 27 at 16:17
• @random-xyz Take $0.1010\ldots_2$ as an example. Multiply by $10_{10}=1010_2$ to get $110.1010\ldots_2$. The integer part is $110_2=6_{10}$, so the first digit after the point is $6$. Lop that off to get $0.1010\ldots_2$. Rinse and repeat. – Arthur Aug 27 at 16:32
Having converted the whole part, you can think of the process as finding which of $$\frac 12, \frac 14, \frac 18, \frac 1{16},\ldots$$ you need to add together to get the fractional part. If the denominator is not a power of $$2$$ the expansion will repeat. Keep doubling the fraction and keep track of the carries into the $$1$$s place. For your example of $$\frac 35=0.6$$ you would do $$0.6 \cdot 2=1.2=0.2+1\\ 0.2 \cdot 2 = 0.4=0.4+0\\ 0.4 \cdot 2 = 0.8=0.8+0\\ 0.8 \cdot 2=1.6=0.6+1$$ And we are back where we started, so this is the repeat. We have $$0.6_{10}=0.\overline{1001}_2$$
If you have some number $$a$$ with $$n$$ digits repeated in binary, then you have $$a2^{-n}+a2^{-2n}+...$$ (each time you repat the number, you're multiplying it by another factor of $$2^{-n}$$. This is a geometric series, so it adds up to $$\frac {a2^{-n}}{1-2^{-n}}=\frac a {2^m-1}$$. The decimal $$.6$$ is equal to $$\frac 3 5$$, so we're looking for $$n$$ such that $$2^n$$ is divisible by $$5$$, which leads to $$n=4$$, $$2^n=16$$, $$2^n-1=15$$. Then $$a=9$$, and $$9$$ in binary is $$1001$$. Thus, $$.6_{10}=.\overline{1001}_2$$
This works for other fractions. For instance, $$\frac 1 {11}=\frac{93}{1023}=\frac{93}{10^{10}-1}$$, $$\frac{1}{11} = .\overline{0001011101}_2$$.
This also applies to base ten: $$\frac 1 {11} = \frac 9 {99}= \frac 9 {10^2-1}$$, so $$\frac 1 {11} = .\overline{09}$$ | 2019-10-19T12:10:36 | {
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http://mathhelpforum.com/discrete-math/59491-fibonacci-number-problem.html | # Thread: A Fibonacci number problem
1. ## A Fibonacci number problem
How can we show that the following (recursive) statement, where $F_n$ is the $n$-th Fibonacci number, and $n \geq 1$, is true?
$\sum_{i=0}^{\lfloor \frac{n-1}{2} \rfloor} F_{n-2i}=F_{n+1}-1$
******
As this should be shown to be valid for every $n \in \mathbb{N}$, I believe we can use induction. The least $n$ which has to satisfy the above equation is $1$, so we take the case when $n=1$ as basis. But,
$\sum_{i=0}^{\lfloor \frac{1-1}{2} \rfloor} F_{1-2i}
\stackrel{\text{?}}{=}
F_{1+1}-1$
$\sum_{i=0}^{0} F_{1-2i}
\stackrel{\text{?}}{=}
F_{2}-1$
$F_{1}
\neq
F_{2}-1$
, because $F_1=1$, and $F_2-1=1-1=0$.
However, if we take the case when $n=2$ as basis, then the above statement is true:
$\sum_{i=0}^{\lfloor \frac{1}{2} \rfloor} F_{2-2i}=\sum_{i=0}^{0} F_{2-2i}=F_2=1=$
$F_{2+1}-1=F_3-1=2-1=1$
Is this problem wrongly posed? Should we show that $\sum_{i=0}^{\lfloor \frac{n-1}{2} \rfloor} F_{n-2i}=F_{n+1}-1$ is true for every $n \geq 2$ instead of $n \geq 1$?
And my next question would be how to proceed with the inductive step? I know we should use the fact that $F_n=F_{n-1}+F_{n-2}$ somewhere, but so far I couldn't show that the above is true for $k+1$ if we suppose it is valid form some $k \in \mathbb{N}$.
I'll be immensely grateful for any help.
2. Originally Posted by gusztav
How can we show that the following (recursive) statement, where $F_n$ is the $n$-th Fibonacci number, and $n \geq 1$, is true?
$\sum_{i=0}^{\lfloor \frac{n-1}{2} \rfloor} F_{n-2i}=F_{n+1}-1$
It is easier to think about this problem if you consider even and odd cases for $n$.
If $n$ is odd then $[\tfrac{n-1}{2}] = \tfrac{n-1}{2}$.
And the sum becomes $F_1+F_3+...+F_{n-2}+F_n = F_{n+1}-1$.
If $n$ is even then $[\tfrac{n-1}{2}] = \tfrac{n}{2} - 1$.
And the sum becomes $F_2 + F_4 + ... + F_{n-2} + F_n = F_{n+1} - 1$.
It is now easier to analyze this problem.
3. Originally Posted by ThePerfectHacker
It is easier to think about this problem if you consider even and odd cases for $n$.
If $n$ is odd then $[\tfrac{n-1}{2}] = \tfrac{n-1}{2}$.
And the sum becomes $F_1+F_3+...+F_{n-2}+F_n = F_{n+1}-1$.
If $n$ is even then $[\tfrac{n-1}{2}] = \tfrac{n}{2} - 1$.
And the sum becomes $F_2 + F_4 + ... + F_{n-2} + F_n = F_{n+1} - 1$.
It is now easier to analyze this problem.
Hmmm. I tried it this way, as you suggested, and broke the problem down into the cases when $n$ is even or odd. But in each of those cases, in the inductive step, there would be $n+1$, the number of different parity than that of $n$.
For example, if some $n$ is odd, then
$\sum_{i=0}^{\lfloor \frac{n-1}{2} \rfloor} F_{n-2i}=F_{n+1}-1$
becomes
$\sum_{i=0}^{\frac{n-1}{2} } F_{n-2i}=F_{n+1}-1$.
Let's say we have some odd number, for example n=5. In that case, $\sum_{i=0}^{\frac{5-1}{2} } F_{5-2i}=$ $
\sum_{i=0}^{2} F_{5-2i}=F_5 + F_3+F_1$
, which is perfectly clear.
But in the inductive step, we take n=6 instead of n=5 and have $\sum_{i=0}^{\frac{6-1}{2} } F_{6-2i}=\sum_{i=0}^{\frac{5}{2} } F_{6-2i}$ which is meaningless (how to sum to $\frac{5}{2}$-th term?).
Or do we, in the inductive step, take $k+2$ (when $k$ is odd, then $k+2$ is also odd). But then, on the right side of the equation, $F_{n+1}-1$ should become $F_{n+3}-1$, while we need to have $F_{n+2}-1$ to prove this problem...
(Essentially, in the inductive proof we'd have to show that if
$\sum_{i=0}^{\lfloor \frac{k-1}{2} \rfloor} F_{k-2i}=F_{k+1}-1$ is true for some $k$, then it is also true for $k+1$.)
4. Say we wish to show $F_1 + F_3 + ... + F_n = F_{n+1} - 1$.
If it is true for $n$ we need to show it is true for $n+2$ (since we are inducting on the odds).
Adding $F_{n+2}$ on both sides gives,
$F_1+...+F_n + F_{n+2} = F_{n+1} + F_{n+2} - 1 = F_{n+3} - 1$
That is the inductive step. | 2017-02-21T11:18:54 | {
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https://math.stackexchange.com/questions/3420110/let-sum-a-n-and-sum-b-n-converge-a-n-b-n-geq-0-does-sum-min-a-n-b | # Let $\sum a_n$ and $\sum b_n$ converge, $a_n,b_n\geq 0$, does $\sum \min\{a_n,b_n\}$, $\sum \max\{a_n,b_n\}$ converge too?
Let $$\sum\limits_{n=0}^{\infty}a_n$$ and $$\sum\limits_{n=0}^{\infty}b_n$$ be convergent with $$a_n,b_n\geq 0$$, does $$\sum\limits_{n=0}^{\infty}\min\{a_n,b_n\}$$ and $$\sum\limits_{n=0}^{\infty}\max\{a_n,b_n\}$$ converge too?
I know, that this was asked here and here in a kind of similar way. The thing is, that in the first link the answer is quite undetailed and the second link contains a different question that is similar but not the same. (That's why I will ask the question yet again.)
My thoughts are, that if both series $$\sum_{n=0}^{\infty}a_n,\sum_{n=0}^{\infty}b_n$$ converge, that $$\sum_{n=0}^{\infty}\min\{a_n,b_n\}$$ will pick either value of one of the partial sums and likewise, $$\sum_{n=0}^{\infty}\max\{a_n,b_n\}$$ will pick a partial sum out of both. With the help of the definition of series, "The series $$\sum_{n=0}^{\infty}a_n$$ converges, if the partial sum $$\sum_{k=0}^{\infty}a_k$$ with $$k\geq n$$ converges", can we apply, that both $$\sum_{n=0}^{\infty}\min\{a_n,b_n\}$$ and $$\sum_{n=0}^{\infty}\max\{a_n,b_n\}$$ converge.
Is this enough for a right proof?
Since I don't know what “will pick either value of one of the partial sums” means, I can't tell whether you are right or wrong. But you can do it as follows: since both series $$\sum_{n=0}^\infty a_n$$ and $$\sum_{n=0}^\infty b_n$$ converge, the series $$\sum_{n=0}^\infty(a_n+b_n)$$ converges too. And since$$(\forall n\in\mathbb Z_+):\min\{a_n,b_n\},\max\{a_n,b_n\}\leqslant a_n+b_n,$$both series $$\sum_{n=0}^\infty\min\{a_n,b_n\}$$ and $$\sum_{n=0}^\infty\max\{a_n,b_n\}$$ converge, by the comparison test.
• We could also use $\min \{a_n,b_n\}\le a_n$ for the $\min$ series.... An application: For $2\le p<\infty$ and $1/p+1/q=1,$ if $x=(x_n)_n\in l^p$ and $y=(y_n)_n\in l^q$ then $\sum_n x_ny_n$ converges. Proof: Let $z_n=0$ if $y_n=0.$ If $y_n\ne 0$ let $y_n=z_n|z_n|^{q-2}.$ Then $(z_n)_n\in l^p$ and $|x_ny_n|\le \max \{|x_n|^p, |z_n|^p\}.$ (I distinguish the case $y_n=0$ in case $q<2.$) – DanielWainfleet Nov 3 '19 at 14:11
Yes, both converge. Note that both the partial sums are monotone increasing, thus for convergence we only need an upper bound. However, we have $$\sum_{j=0}^n \min\{a_j, b_j\} \leq \sum_{j=0}^n \max\{a_j, b_j\} \leq \sum_{j=0}^n (a_j + b_j) \leq \sum_{j\geq 0} a_j + \sum_{j\geq 0} b_j$$
0) $$a_n, b_n \ge 0$$, resp. series $$\sum a_n$$, $$\sum b_n$$ are convergent.
1) $$0 \le \min (a_n, b_n) \le a_n$$.
By comparison test $$\sum \min(a_n, b_n)$$ is convergent.
2) $$\min (a_n, b_n) +\max (a_n, b_n) =$$
$$a_n +b_n$$;
$$0 \le \max(a_n, b_n) =$$
$$a_n +b_n - \min (a_n, b_n) < a_n + b_n$$;
By comparison test $$\sum \max(a_n, b_n)$$ is convergent since $$\sum(a_n +b_n) =\sum a_n +\sum b_n$$ is convergent.
For $$a_n,b_n\ge0$$ we have the inequality $$0\le\min{\{a_n,b_n\}}\le\max{\{a_n,b_n\}}\le a_n+b_n$$ Hence, summing from $$n=0$$ to $$\infty$$, gives $$\sum_{n=0}^\infty0\le\sum_{n=0}^\infty\min{\{a_n,b_n\}}\le\sum_{n=0}^\infty\max{\{a_n,b_n\}}\le \sum_{n=0}^\infty(a_n+b_n)$$ $$0\le\sum_{n=0}^\infty\min{\{a_n,b_n\}}\le\sum_{n=0}^\infty\max{\{a_n,b_n\}}\le \sum_{n=0}^\infty a_n+\sum_{n=0}^\infty b_n$$ by the comparison test.
We have that
$$\sum \min\{a_n,b_n\}\le \sum a_n$$
therefore the series $$\sum \min\{a_n,b_n\}$$ converges and since
$$\sum \max\{a_n,b_n\}=\sum \frac{|a_n-b_n|+a_n+b_n}{2}=\frac12\sum |a_n-b_n|+\frac12\sum a_n+\frac12\sum b_n$$
also $$\sum \max\{a_n,b_n\}$$ converges too. | 2020-03-31T11:14:32 | {
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https://math.stackexchange.com/questions/1714631/limit-lim-limits-x-y-to0-0-frac-sinx2-y2x2-y2/1714653 | # Limit $\lim\limits_{(x,y)\to(0,0)}\frac{\sin(x^2-y^2)}{x^2-y^2}$
I'm trying to understand the following limit:
$$\lim_{(x,y)\to(0,0)}\frac{\sin(x^2-y^2)}{x^2-y^2}$$
The $\lim_{(x,y)\to (0,0)} f(x,y)$ is undefined. Why it is not equal to $1$?
Let's suppose $t = x^2-y^2$. Then as $(x,y)$ approaches $(0,0)$, $t$ approaches $0$, so
$$\lim_{t\to 0} \frac{\sin t}{t} = 1$$
I can see that the path $x=y$ is undefined,does this mean that the limit does not exist ?
I though that I needed to find two different defined paths that gives two different results to disprove that the limit exists. If so this means that I can't really rely on the "t substitution" to determined the limit.
I tried to pass it into polar coordinates but It didn't help.. How can I determine a limit of a two variable function ? I can't check all the possibles paths like in one variable (left and right ).
I understood that the safest way is to use the squeeze theorem or to pass the coordinates into polar coordinates and then use the squeeze theorem.
I would appreciate any sort of help in the matter , Thanks.
• Is it really undefined? WRA says that the limit is equal to $1$. – Ethan Hunt Mar 26 '16 at 19:03
• I think you can only approach via paths that lie in the domain of the function. You just need to make sure there exists at least one such path (so it is not isolated point), and that all such paths approach to the $1$. – Sil Mar 26 '16 at 19:29
• The answer depends on your book's definition. Some people insist that the function be defined in a deleted neighbourhood of $(0,0)$. Others don't. – André Nicolas Mar 26 '16 at 19:40
• André is right. The only way this question can be given a definite answer is if you tell us the exact definition of limit that you are using. – Hans Lundmark Mar 26 '16 at 20:41
• The function must be defined in the a certain neighborhood of (0,0) with the exception of (0,0). Can you give me a little explanation on how does this information affects the result ? – UUKKS Mar 27 '16 at 16:05
We have the inequality $z - z^3/6 \leqslant \sin z \leqslant z$ for $z > 0$.
Hence, for $x > y$
$$1 - \frac{(x^2 - y^2)^2}{6} \leqslant \frac{\sin (x^2 - y^2)}{x^2 - y^2} \leqslant 1,$$
and by the squeeze theorem the limit is $1$ as $(x,y) \to (0,0)$ with $x > y.$
Similarly, for $y > x$
$$1 - \frac{(y^2 - x^2)^2}{6} \leqslant \frac{\sin (y^2 - x^2)}{y^2 - x^2} = \frac{\sin (x^2 - y^2)}{x^2 - y^2} \leqslant 1,$$
and by the squeeze theorem the limit is $1$ as $(x,y) \to (0,0)$ with $y > x.$
The inequality also shows that as $x \to y$ with $y$ fixed,
$$\lim_{x \to y} \frac{\sin (x^2 - y^2)}{x^2 - y^2} = 1,$$
and, although the function as written above is undefined for $x = y,$ it can be extended continuously to $1$ on that line.
Technically, with $f(x,y) = \sin(x^2 - y^2) /( x^2 - y^2),$ you would say
$$\lim_{(x,y) \to (0,0), x \neq y} f(x,y) = 1,$$
and $f$ can be continuously extended to a function $\hat{f}$ on $\mathbb{R}^2$ such that
$$\lim_{(x,y) \to (0,0)} \hat{f}(x,y) = 1.$$
You can use polar coordinates here. Set $x=r\cos\theta$, $y=r\sin\theta$, then notice that $x^2-y^2=r^2\cos 2\theta$. Then the limit becomes $$\lim_{r \to 0} \frac{\sin (r^2\cos 2\theta)}{r^2\cos 2 \theta}.$$
Clearly you have to exclude the case $\theta=\pm \pi/4$ because $f$ is not defined there, even if you can try to extend it by continuity.
• So the limit is 1 even if the path x=y is undefined ? – UUKKS Mar 26 '16 at 19:12
Recall from elementary geometry that the sine function satisfies the inequalities
$$|\theta\cos(\theta)|\le |\sin(\theta)|\le |\theta|$$
for $|\theta|\le \pi/2$.
Letting $\theta =x^2-y^2$ we can write
$$|\cos(x^2-y^2)|\le\left|\frac{\sin(x^2-y^2)}{x^2-y^2}\right|\le|1|$$
whereupon applying the squeeze theorem and exploiting the evenness of $\frac{\sin(z)}{z}$ yields the limit
$$\lim_{(x,y)\to(0,0)}\frac{\sin(x^2-y^2)}{x^2-y^2}=1$$ | 2020-07-04T19:32:40 | {
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https://mathhelpboards.com/threads/variance.4223/ | # Variance
#### Yankel
##### Active member
Hello
I have a problem solving this question....
to the numbers 0 and 2, we want to add a 3rd number, such that the variance won't change. What is the 3rd number ?
(when I say variance I mean dividing by n, not by n-1)
thanks !
#### TheEmptySet
##### New member
Hello
I have a problem solving this question....
to the numbers 0 and 2, we want to add a 3rd number, such that the variance won't change. What is the 3rd number ?
(when I say variance I mean dividing by n, not by n-1)
thanks !
The variance is the square root of the sum of the differences from of the data points from the mean so.... $\sigma^2=\frac{1}{2}\left[(0-1)^2+(2-1)^2\right]=1$. If we add the new point $y$, we will need to solve the system of equations. $\mu=\frac{1}{3}\left[ 0+2+y\right]$ and $\sigma^2=\frac{1}{3}\left[(0-\mu)^2+(2-\mu)^2+(y-\mu)^2\right]=1$ This will give a quadratic equation in $y$
#### Klaas van Aarsen
##### MHB Seeker
Staff member
The variance is the square root of the sum of the differences from of the data points from the mean so.... $\sigma^2=\frac{1}{2}\left[(0-1)^2+(2-1)^2\right]=1$. If we add the new point $y$, we will need to solve the system of equations. $\mu=\frac{1}{3}\left[ 0+2+y\right]$ and $\sigma^2=\frac{1}{3}\left[(0-\mu)^2+(2-\mu)^2+(y-\mu)^2\right]=1$ This will give a quadratic equation in $y$
If we divide the variance by n instead of by (n-1), we can only do so if the expected mean $\mu$ is given.
Otherwise we would lose a degree of freedom.
So the set of equations should be:
$\sigma^2=\frac{1}{2}\left[(0-\mu)^2+(2-\mu)^2\right]$
$\sigma^2=\frac{1}{3}\left[(0-\mu)^2+(2-\mu)^2+(y-\mu)^2\right]$
This does introduce the problem that we have more unknowns than equations.
#### TheEmptySet
##### New member
If we divide the variance by n instead of by (n-1), we can only do so if the expected mean $\mu$ is given.
Otherwise we would lose a degree of freedom.
So the set of equations should be:
$\sigma^2=\frac{1}{2}\left[(0-\mu)^2+(2-\mu)^2\right]$
$\sigma^2=\frac{1}{3}\left[(0-\mu)^2+(2-\mu)^2+(y-\mu)^2\right]$
This does introduce the problem that we have more unknowns than equations.
Maybe I am misunderstanding something but the way I read the problem was if we have the data set $\{0,2 \}$ we can calculate the mean and variance directly. Now if a new data set is created by adding one other point $\{0,2,y \}$. We can now calculate the new population mean and the new variance of this three point data set. Now we can just solve for what $y$ needs to be. I will wait for clarification from the OP on this. | 2022-05-16T08:58:51 | {
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https://math.stackexchange.com/questions/2206048/linear-transformation-and-its-matrix/2206141 | # Linear transformation and its matrix
I have two bases:
$A = \{v_1, v_2, v_3\}$ and $B = \{2v_1, v_2+v_3, -v_1+2v_2-v_3\}$
There is also a linear transformation: $T: \mathbb R^3 \rightarrow \mathbb R^3$
Matrix in base $A$:
$M_{T}^{A} = \begin{bmatrix}1 & 2 &3\\4 & 5 & 6\\1 & 1 & 0\end{bmatrix}$
Now I am to find matrix of the linear transformation $T$ in base B.
I have found two transition matrixes (from base $A$ to $B$ and from $B$ to $A$):
$P_{A}^{B} = \begin{bmatrix}2 & 0 & -1\\0 & 1 & 2\\0 & 1 & -1\end{bmatrix}$
$(P_{A}^{B})^{-1} = P_{B}^{A} = \begin{bmatrix}\frac{1}{2} & \frac{1}{6} & \frac{-1}{6}\\0 & \frac{1}{3} & \frac{2}{3}\\0 & \frac{1}{3} & \frac{-1}{3}\end{bmatrix}$
How can I find $M_{T}^{B}$?
Is it equal to:
$(P_{A}^{B})^{-1}M_{T}^{A}P_{A}^{B}$?
If yes why?
• Note that $P^B_A$ takes you from basis $A$ to basis $B$, so it can't be $(P^B_A)^{-1}M^A_TP^B_A$. – B. Pasternak Mar 27 '17 at 22:27
• I think the notation is $P_A^B$ takes you from basis $B$ to basis $A$. Although your point stands, it's crappy notation. – mdave16 Mar 27 '17 at 23:36
Some notation: for a vector $v \in \Bbb R^3$, let $[v]_A$ denote the coordinate vector of $v$ with respect to the basis $A$, and let $[v]_B$ denote the coordinate vector of $v$ with respect to the basis $B$. both of these are column vectors. To put this another way, $$[v]_A = \pmatrix{a_1\\a_2\\a_3} \iff v = a_1v_1 + a_2 v_2 + a_3v_3$$ We can think of $M_T^A$ as a "machine" with the property that, with the usual matrix multiplication, $M_T^A [v]_A = [T(v)]_A$. Similarly, $P^B_A$ satisfies $P^B_A [v]_B = [v]_A$, whereas $P^A_B$ satisfies $P^A_B[v]_A = [v]_B$. What we want is to "build" is a machine $M_T^B$ for which $M_T^B[v]_B = [T(v)]_B$.
We can break the process of going from $[v]_B$ to $[T(v)]_B$ into three steps, each of which uses machinery that we already have. First, go from $[v]_B$ to $[v]_A$ with $P^B_A[v]_B = [v]_A$. Then, go from $[v]_A$ to $[T(v)]_A$ using $M_T^A [v]_A = [T(v)]_A$. Then, go from $[T(v)]_A$ to $[T(v)]_B$ using $P^A_B[T(v)]_A = [T(v)]_B$.
Putting it all together, we have $$[T(v)]_B = P^A_B[T(v)]_A = P^A_B(M_T^A [v]_A) = P^A_BM_T^A (P^B_A[v]_B) = (P^A_BM_T^A P^B_A) [v]_B$$ What we have found, then, is that the matrix which takes us from $[v]_B$ to $[T(v)]_B$ is the product $P^A_BM_T^A P^B_A = (P^B_A)^{-1} M_T^A P^B_A$. So, this is our matrix $M_T^B$.
• Thank you! That is a very clear explanation. However I think I'm not sure what $P_A^B$ means. Is it the matrix from basis $A$ to basis $B$? – Hendrra Mar 28 '17 at 9:22
• I've told you what it does. That should be more than enough. – Omnomnomnom Mar 28 '17 at 10:53
• Of course it is but you used two different notations. I think it's just a typo but it's better to be sure (in line 5 and 8). – Hendrra Mar 28 '17 at 12:19
• You're right. Fixed it. – Omnomnomnom Mar 28 '17 at 12:46
I am assuming that $P^B_A$ takes from basis $B$ to $A$.
• Note, for instance, that $(1,0,0)$ wrt the basis $B$ should be mapped to $(2,0,0)$ with respect to the basis $A$. Now, looking at the first columns, it's clear which is which. – Omnomnomnom Mar 27 '17 at 23:54
• @Omnomnomnom, Why is $P_A^B$ the matrix from basis $B$ to $A$? I've literally never seen either notation before for such a concept. Wait nvm, I made a typo several times, it's fine. Thanks for the correction – mdave16 Mar 28 '17 at 0:19
• see my own answer for a detailed explanation. I think this is the choice of notation in Lay's text, if I remember correctly. You sometimes encounter $[I]^B_A$ for the same ($I$ denoting the identity transformation) or $[I]_{B \to A}$, or $[I]_{A \leftarrow B}$ (I prefer this last notation). The convention $P^B_A$ follows the set theory convention of $V^U = \{f \mid f:U \to V\}$. In set/category theory, arrows go down by default. – Omnomnomnom Mar 28 '17 at 0:31 | 2019-06-26T00:08:59 | {
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http://math.stackexchange.com/questions/198810/proof-that-every-repeating-decimal-is-rational/198817 | # Proof that every repeating decimal is rational
Wikipedia claims that every repeating decimal represents a rational number.
According to the following definition, how can we prove that fact?
Definition: A number is rational if it can be written as $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.
-
Suppose that the decimal is $x=a.d_1d_2\ldots d_m\overline{d_{m+1}\dots d_{m+p}}$, where the $d_k$ are digits, $a$ is the integer part of the number, and the vinculum (overline) indicates the repeating part of the decimal. Then
$$10^mx=a+d_1d_2\dots d_m.\overline{d_{m+1}\dots d_{m+p}}\;,\tag{1}$$ and
$$10^{m+p}x=a+d_1d_2\dots d_md_{m+1}\dots d_{m+p}.\overline{d_{m+1}\dots d_{m+p}}\tag{2}\;.$$
Subtract $(1)$ from $(2)$:
$$10^{m+p}x-10^mx=a+d_1d_2\dots d_md_{m+1}\dots d_{m+p}-a+d_1d_2\dots d_m\;.\tag{3}$$
The righthand side of $(3)$ is the difference of two integers, so it’s an integer; call it $N$. The lefthand side is $\left(10^{m+p}-10^m\right)x$, so
$$x=\frac{N}{10^{m+p}-10^m}=\frac{N}{10^m(10^p-1)}\;,$$
a quotient of two integers.
Example: $x=2.34\overline{567}$. Then $100x=234.\overline{567}$ and $100000x=234567.\overline{567}$, so
$$99900x=1000000x-100x=234567-234=234333\;,$$ and
$$x=\frac{234333}{99900}=\frac{26037}{11100}\;.$$
-
Brian's comment is more detailed and rigorous than my own. – Scott Carter Sep 19 '12 at 0:28
I think there's a typo in your very last denominator. – Michael Hardy Sep 19 '12 at 0:35
@Michael: There sure was; thanks. – Brian M. Scott Sep 19 '12 at 0:37
Is it okay to say that $a. \overline{d_{m+1} \ldots d_{m+p}} - a .\overline{d_{m+1} \ldots d_{m+p}} = 0$, even though both are infinite expansions? Why does this not run into the danger of $\infty - \infty$? – jamaicanworm Sep 19 '12 at 5:52
@jamaicanworm: Because $a.\overline{d_{m+1}\dots d_{m+p}}$ isn’t infinite: it’s a well-defined real number, the sum of a certain convergent infinite series, and you’re simply subtracting that number from itself. – Brian M. Scott Sep 19 '12 at 5:55
A non-rigorous proof would be the following. Suppose
$$x=x_0,\overline{x_1x_2x_3\dots x_n}$$
Then
$$10^nx=x=x_0x_1x_2x_3\dots x_n,\overline{x_1x_2x_3\dots x_n}$$
so $$10^nx-x=x_0x_1x_2x_3\dots x_n-x_0$$
and
$$x=\frac{x_0x_1x_2x_3\dots x_n-x_0}{10^n-1}$$
where $x_n\in\{0,1,\dots,9\}$
Simple example:
$$x=1,234234234\dots$$
then
$$10^3 x=1234,234234\dots$$
so
$$(10^3-1)x=1234-1$$
$$x=\frac{1233}{999}$$
If you want to get more rigorous, you can use the series expansion of a number, but, all in all, the proof's essence won't differ much.
ADD In the more general case
$$x=x_0,y_1y_2y_3\dots y_n\overline{x_1x_2x_3\dots x_n}$$
note
$$x=x_0,0\dots 0\overline{x_1x_2x_3\dots x_n}+0,y_1y_2y_3\dots y_n$$ and consider
$$x'=x_0,0\dots 0\overline{x_1x_2x_3\dots x_n}$$ The shifting is then of $10^{m+n}$, and we obtain the sum of two rational numbers, which is rational.
-
One should add that the repeating part need not begin just after the decimal point; it could begin earlier or later. – Michael Hardy Sep 19 '12 at 0:34
@MichaelHardy Yes. I added a more general case. Nevertheless, the general idea, I guess, is clear: shift, subtract, profit. – Pedro Tamaroff Sep 19 '12 at 0:35
Let $q= 0.\overline{d_1d_2...d_k}$ be a repeating decimal with pattern $R = d_1d_2...d_k$ of length $k$.
Then we have: $$q=\sum_{n=1}^{\infty}{R\cdot 10^{-kn}}=R\left(\frac{1}{1-10^{-k}}-1\right)=\frac{R}{10^k-1}$$
-
This solution is very clever! – jamaicanworm Sep 19 '12 at 5:52
Suppose the period is n, so the decimal goes, $a_1a_2\ldots a_n$ and repeats. Let $x$ denote the number. Multiply $x$ by $10^n$. Subtract $x$. Then $$(10^n -1)x = a_1a_2\ldots a_n.$$ So $x$ is the cyclic divided by $9999\ldots 9$. For example, $x= 0.142857142857\ldots$. Then $x = 142857/999999=1/7$ when reduced.
-
Hint $\$ Consider what it means for a real $\rm\ 0\: < \: \alpha\: < 1\$ to have a periodic decimal expansion:
$\rm\qquad\qquad\qquad\quad\ \ \ \, \alpha\ =\ 0\:.a\:\overline{c}\ =\ 0\:.a_1a_2\cdots a_n\:\overline{c_1c_2\cdots c_k}\ \$ in radix $\rm\:10\:$
$\rm\qquad\qquad\iff\quad \beta\ :=\ 10^n\: \alpha - a\ =\ 0\:.\overline{c_1c_2\cdots c_k}$
$\rm\qquad\qquad\iff\quad 10^k\: \beta\ =\ c + \beta$
$\rm\qquad\qquad\iff\quad (10^k-1)\ \beta\ =\ c$
$\rm\qquad\qquad\iff\quad (10^k-1)\ 10^n\: \alpha\ \in\ \mathbb Z$
-
It seems like it's sufficient to observe:
1. Every number of the form $0.((0^n)1)^*$ is the sum of a convergent geometric sequence $$10^{-n} + 10^{-2n} +\cdots = {1\over 10^{-n}-1}$$ and so is rational.
2. Every number of the form $0.0^k((0^n)1)^*$ is the product of a number of the previous type and the rational number $10^{-k}$, and so is rational.
3. Every number of the form $0.0^kN^*$ is the product of a number of the previous type and the $n$-digit integer $N$, and so is rational.
4. Every number of the form $0.MN^*$, where $M$ is a $k$-digit integer, is the sum of a number of the previous type, and the rational number $M\cdot10^{-k}$, and so is rational.
5. Every number of the form $Z.MN^*$ is the sum of an integer $Z$ and a number of the previous type, and so is rational.
Replace 10 with any integer $b\gt 1$ to get the more general result for radix-$b$ numerals.
- | 2014-03-07T16:52:43 | {
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https://math.stackexchange.com/questions/2924200/set-of-infinite-measure-and-subsets-of-finite-measure/2924213 | # Set of infinite measure and subsets of finite measure.
I've searched and haven't seen this problem in my searches. Admittedly I may have missed it and I apologize if that's the case.
The problem statement is the following:
Let $(X, \mathcal{M}, \mu)$ be a measure space of infinite measure, with the following property: for every $A \in \mathcal{M}$ with $\mu(A) = \infty$ there exists $B\subset A$ with $0 < \mu(B) < \infty$. Show that there exist sets $E_i\in M(i\geq 1)$ such that $\mu(E i ) < \infty$ (for all $i\geq 1$) and $\mu(\cup_{i\geq1}E_i ) = \infty$.
Given the property that there exists $B \subset A \text{ with } 0<\mu(B)<\infty$ it's easy to see that you can pull an infinite number of subsets $B$ from $A$ since $\mu(A\setminus B) = \infty$. However, given that $\mu(B)$ is simply a finite value greater than zero I'm having trouble showing that $\mu(\cup_{i\geq1}E_i)=\infty$ because $\mu(E_i)$ could be such a small value for every $E_i$ that the measure of the union is still finite.
I feel like it may require a proof by contradiction, but I'm unsure how to proceed with it.
In fact, we can choose $\mu(E_i)$ to be as large as we desire. More concretely, given the measure space with the property you mention, if $A\in\mathcal M$ and $\mu(A)=\infty$, then for any $C>0$ there exists $B\subset A$ with $C<\mu(B)<\infty$. To see this, let $M=\sup\{\mu(B):B\subset A\text{ and }\mu(B)<\infty\}$. If $M=\infty$ the result follows. Otherwise if $M$ is finite, then for each $n$ we can choose $B_n\subset A$ such that $M-\frac{1}{n}<\mu(B_n)\leq M$. Put $B=\cup_nB_n$, so that $\mu(B)=M$. But then $\mu(A\setminus B)=\infty$, so there is some $B'\subset A\setminus B$ with $0<\mu(B')<\infty$, and $M<\mu(B\cup B')<\infty$, contradicting the definition of $M$.
By the above, there exists $E_1\in\mathcal M$ such that $1<\mu(E_1)<\infty$. Then define $E_i$ inductively: There exists $E_{i+1}\subset X\setminus(E_1\cup\cdots\cup E_i)$ with $1<\mu(E_{i+1})<\infty$. Thus $\mu(E_i)$ is finite for all $i$, and since the $E_i$ are disjoint we have $\mu(\cup_iE_i)=\sum_i\mu(E_i)=\infty$.
• Thanks for catching that – Aweygan Sep 20 at 16:31
• I don't see how you show that there is $E_1$ of finite measure larger than 1. This seems to be the whole issue. – Andrés E. Caicedo Sep 20 at 16:40
• From the result I state in the first paragraph, with $C=1$. Should I include a proof? – Aweygan Sep 20 at 16:44
• I think the current version is quite clear, thank you. – Andrés E. Caicedo Sep 20 at 16:51
Note first that if $C$ is uncountable and for each $i\in C$ we have a real number $r_i>0$, then $\sum_i r_i=+\infty$. The point is that for some $n>0$, the family $\{i\in C: r_i>1/n\}$ is uncountable. Of course, in this case there is a countable subfamily whose sum is infinite as well.
Now, let $A$ be measurable of infinite measure. If there is an uncountable disjoint family of measurable subsets of $A$ of strictly positive finite measure, we are done by the observation in the last paragraph. It follows that we may assume that any such family is countable. Use Zorn's lemma to argue that there is a maximal such family $\mathcal F$. Since it is countable, $A\smallsetminus \bigcup_{B\in \mathcal F}B$ is measurable, and must have measure 0. It follows that $\mu(\bigcup_{B\in\mathcal F} B)=+\infty$, and we are done. Note we did a bit more than asked for in the question, as we arranged to find the family of sets inside any given set of infinite measure.
(Rather than Zorn's lemma, I would prefer to build the family by transfinite induction. If you are familiar with this technique, I think it is preferable, as you do not have to worry about arguing that the conditions of Zorn's lemma are met, which is why I had the separate into two cases depending on whether there was an uncountable disjoint family. The issue is that we can only ensure measurability of countable unions of measurable sets, so some care is required.) | 2018-10-19T10:47:25 | {
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https://math.stackexchange.com/questions/3007901/probability-of-exactly-one-student-failing | # Probability of exactly one student failing?
I have the following problem:
If the probability that student A will fail a certain statis- tics examination is 0.5, the probability that student B will fail the examination is 0.2, and the probability that both student A and student B will fail the examination is 0.1, what is the probability and that exactly one of the two students will fail the examina- tion?
I came up with the following solution: $$P(A) = 0.5; P(B) = 0.2$$ 'exactly one' means either A only fails or B only fails.
Event $$X_1$$: 'A only fails'$$P(X_1) = P(A) *P(B)^c = 0.5*0.8 = 0.4$$ Event $$X_2$$: 'B only fails' $$P(X_2) = P(A)^c * P(B) = 0.5*0.2 = 0.1$$ And therefore: $$P(X_1\lor X_2) = P(X_1)+P(X_2)-P(X_1\land X_2) = 0.4+0.1-0 = 0.5$$ My thoughts behind $$P(X_1 \land X_2) = 0$$ were that it is not possible that both only happens at the same time. I'm not fully sure whether that is correct. Should these logical thought be correct is the value of $$0.5$$ correct?
• Welcome to math.SE! Asking for correctness here is OK (and it is encouraged to provide your own ideas for every question, what did you try and what didn't work), but it would also be good in the future to write your solution using MathJax, instead of uploading pictures. – Nutle Nov 21 '18 at 15:54
• It'd be better to write your solution in the body of the question rather than linking to an image. (Also, the image should be rotated to make it easier to read.) – littleO Nov 21 '18 at 15:54
• I see. Thank you both, I edited my question now. – thebilly Nov 21 '18 at 16:10
You need to find $$P(A, \neg B) + P(\neg A, B)$$. We know $$P(A)$$, $$P(B)$$ and $$P(A, B)$$.
We also know that $$P(A,B)+P(A,\neg B)=P(A)$$. You can get the value $$P(A, \neg B)$$ from here. Likewise, we also know that $$P(A, B)+P(\neg A, B)=P(B)$$ and thus you can also get the value $$P(\neg A, B)$$.
Your $$P(X_1)=0.4$$ and $$P(X_2)=0.1$$ are right. But then it is asked for $$P(X_2\cap \overline X_1)+P(X_1\cap \overline X_2)$$
This is $$[P(X_1)-P(X_1\cap X_2)]+[P(X_2)-P(X_1\cap X_2)]=P(X_1)+P(X_2)-2\cdot P(X_1\cap X_2)$$
$$=0.2+0.5-2\cdot 0.1=0.5$$ | 2019-11-18T11:04:06 | {
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https://cstheory.stackexchange.com/questions/49048/what-is-the-polynomial-representation-of-the-hamming-weight-function | # What is the polynomial representation of the Hamming weight function?
For any function $$f: \{1,-1\}^n \rightarrow \{1,-1\}$$, there is a unique multilinear polynomial $$p \in \mathbb{R}[x_1,\dots, x_n]$$ for which $$p(x)=f(x)$$ for all $$x \in \{1,-1\}^n$$ (see e.g. Lemma 4.1 here). I will call this the polynomial representation of $$f$$.
I would like to know, for each $$k=0,\dots, n$$, what the polynomial representation of the function $$f_k$$ is, where $$f_k(x)=1$$ if and only if the number of $$+1$$'s appearing in $$x$$ is equal to $$k$$.
In the absence of such a polynomial representation, I would like upper and lower bounds on the number of monomial terms appearing in the polynomial representation of each $$f_k$$.
Edit: Here is a proof that the polynomial representation of any such $$f$$ is unique (I thank Neal Young for bringing up this subtlety. This fact is obvious for $$\{0,1\}$$-valued functions, but takes slightly more work for $$\{-1,1\}$$-valued functions.)
It suffices to prove that whenever $$\sum_{I \in \{0,1\}^n} \alpha_I x_1^{I_1}\cdots x_n^{I_n}=0$$ for all $$x \in \{-1,1\}^n$$, it holds that $$\alpha_I=0$$ for all $$I \in \{0,1\}^n$$. We prove this by induction on $$n$$.
For $$n=1$$, if $$\alpha_0+\alpha_1 x_1=0$$ for $$x_1=\pm 1$$, then clearly $$\alpha_0=\alpha_1=0$$.
Now suppose $$\sum_{I \in \{0,1\}^n} \alpha_I x_1^{I_1}\cdots x_n^{I_n}=0$$ for all $$x \in \{-1,1\}^n$$. Then
$$\sum_{\substack{I \in \{0,1\}^n\\I_n=0}} \alpha_I x_1^{I_1}\cdots x_{n-1}^{I_{n-1}}=0$$ and $$-\sum_{\substack{I \in \{0,1\}^n\\I_n=1}} \alpha_I x_1^{I_1}\cdots x_{n-1}^{I_{n-1}}=0$$ for all $$x \in \{1,-1\}^{n-1}$$, so by the induction hypothesis, $$\alpha_I=0$$ for all $$I \in \{0,1\}^n$$. $$\square$$
• Yes, thank you. I should have said "unique multilinear polynomial".
– Ben
Jun 1 at 16:43
• Thanks for bringing this up. The multilinear representation is indeed unique, even for $\{-1,1\}$-valued functions (I have edited my post to include a proof).
– Ben
Jun 1 at 18:02
• I've deleted my comments above, as they no longer apply to the edited post. I'll delete this one too in a while. Jun 1 at 18:37
##### Here is the polynomial representation of any such function $$f$$:
1. For any $$y\in \{-1,1\}^n$$, define polynomial $$I_y(x) = 2^{-n}\prod_{i=1}^n 1+y_i x_i.$$
2. Then for all $$x\in\{-1,1\}^n$$ we have $$I_y(x) = 1$$ if $$y=z$$ and otherwise $$I_y(x) = 0$$.
3. So, for any function $$f:\{-1,1\}^n\rightarrow \{-1,1\}$$ and all $$x\in\{-1,1\}^n$$ we have $$f(x) = -1+2\sum_{y : f(y)=1} I_y(x),$$ so $$p_f(x) = -1+2\sum_{y:f(y)=1}I_y(x)$$ is the polynomial representation of $$f$$.
##### The number of terms in the polynomial representation of $$f_1$$ is at least $$2^{n-1}$$:
1. By the previous part, we have $$p_{f_1}(x_1,\ldots,x_n) = -1+2^{1-n}\sum_{i=1}^n (1+x_i) \prod_{j\ne i} (1-x_j)$$.
2. Note that if we constrain $$x_1 = 1$$, this simplifies to $$p_{f_1}(1, x_2, x_3, \ldots, x_n) = -1+2^{2-n}\prod_{j=2}^n (1-x_j)$$, which has $$2^{n-1}$$ terms (for $$n\ne 2$$).
3. It follows that $$f_1(x)$$ has at least $$2^{n-1}$$ terms (for $$n\ne 2$$).
##### The number of terms in the polynomial representation of $$f_k$$ is at least $$2^{n-k}$$:
1. Constraining $$x_1=x_2=\cdots=x_k=1$$, define the function $$g_k(x_{k+1}, x_{k+2}, \ldots x_n) = f_k(1,1,\ldots,1, x_{k+1}, x_{k+2}, \ldots, x_{n})$$.
2. The function $$g_k$$ takes the value 1 at $$x_{k+1}=\cdots=x_n=-1$$, and takes the value $$-1$$ for all $$(x_{k+1},\ldots, x_n)\in\{-1,1\}^{n-k}\setminus \{-1\}^{n-k}$$.
3. It follows (by the first part) that the polynomial representation of $$g_k$$ is $$-1+2^{1-n+k}\prod_{j=k+1}^n (1-x_j)$$.
4. So $$p_{g_k}$$ has $$2^{n-k}$$ terms (for $$n\ne k+1$$).
5. It follows that $$p_{f_k}$$, the polynomial representation of $$f_k$$, has at least $$2^{n-k}$$ terms (for $$n\ne k+1$$).
The above bound is surely low for large $$k$$. | 2021-11-27T07:45:11 | {
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http://math.stackexchange.com/questions/147557/sequence-of-functions-and-bound | # Sequence of functions and bound
My Question
Let $\{g_k\}$ be a sequence of continuous real-valued functions on $[0,1]$. Assume that there is a number $M$ such that $|g_{k}(x)|\leq M$ for every integer $k$ and every $x\in [0,1]$ and also that there is continuous real-valued funtion $g$ on $[0,1]$ such that
$$\int_0^1 g_k(x)p(x) \ dx \rightarrow \int_0^1 g(x)p(x) \ dx$$ as $k \rightarrow \infty$ for every polynomial $p$. Proved that $|g(x)|\leq M$ for every $x\in[0,1]$.
Remarks
(Note that the two bounds are the same and is $M$)
This is what I have done so far.
Suppose it is not true, that is there exists a $x_0 \in [0,1]$ such that $|g(x_0)|>M$, then by the continuity of $g$ at $x_0$, there will be an interval $(x_0-\delta, x_0+\delta)$ such that $|g(x)|>M$ for all $x\in (x_0-\delta, x_0+\delta)$. Then take note that it is possible to have a non-negative continuous function $\phi$ such that it is zero outside the interval $(x_0-\delta, x_0+\delta)$ and the integral $\int_{x_0-\delta}^{x_0+\delta} \phi(x)\ dx$ is one.
Then by Weierstrass Approximation Theorem, we know that
$$\int_0^1 g_k(x)\phi(x) \ dx \rightarrow \int_0^1 g(x)\phi(x) \ dx$$ as $k \rightarrow \infty$ and I managed to obtained a contradiction by assuming $g(x_0)>0$, but I am lost with all the inequalities for the other case. I hope someone can help me with this, perhaps we do not need any Weierstrass approximation theorem, for that I am not sure.
Thanks
-
If $g(x_0)<0$, you replace each $g_k$ by $-g_k$ and $g$ by $-g$. – Brian M. Scott May 20 '12 at 22:40
Thanks, it works. Is this a general method that always apply or just for this question? – KWO May 20 '12 at 23:02
It’s a fairly common trick. It works here because (a) the integral respects the sign change, and (b) the condition $|g_k(x)|\le M$ is the same for $g_k$ and $-g_k$. Such circumstances are not unusual, so there are lots of situations in which such a trick can work. – Brian M. Scott May 20 '12 at 23:06
Thanks a lot for your help. – KWO May 20 '12 at 23:21
If $g(x_0)<0$, you can reduce the problem to the positive case by replacing each $g_k$ by $-g_k$ and $g$ by $-g$, since the condition that $|g_k(x)|\le M$ applies equally to $g_k$ and $-g_k$, and the integral respects the sign change. This is a fairly common trick when the circumstances allow it. | 2015-09-04T12:34:15 | {
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https://www.physicsforums.com/threads/using-diagonalization-to-find-a-k.653042/ | # Using diagonalization to find A^k
1. Nov 17, 2012
### 1up20x6
1. The problem statement, all variables and given/known data
$$A = \begin{pmatrix} 1 & 4\\ 2 & -1 \end{pmatrix}$$
Find $A^n$ and $A^{-n}$ where n is a positive integer.
2. Relevant equations
3. The attempt at a solution
$$(xI - A) = \begin{pmatrix} x-1 & -4\\ -2 & x+1 \end{pmatrix}$$
$$det(xI - A) = (x-3)(x+3)$$
$$λ_1 = 3\quad λ_2 = -3$$
$$\begin{pmatrix} 2 & -4\\ -2 & 4\end{pmatrix} \begin{pmatrix} a\\ b \end{pmatrix} = \begin{pmatrix} 0\\ 0 \end{pmatrix}\quad \begin{pmatrix} a\\ b \end{pmatrix} = \begin{pmatrix} 2\\ 1 \end{pmatrix}$$
$$\begin{pmatrix} -4 & -4\\ -2 & -2\end{pmatrix} \begin{pmatrix} c\\ d \end{pmatrix} = \begin{pmatrix} 0\\ 0 \end{pmatrix}\quad \begin{pmatrix} c\\ d \end{pmatrix} = \begin{pmatrix} -1\\ 1 \end{pmatrix}$$
$$P = \begin{pmatrix} λ_1 & λ_2 \end{pmatrix} = \begin{pmatrix} 2 & -1\\ 1 & 1 \end{pmatrix}$$
$$P^{-1} = \begin{pmatrix} \frac{1}{3} & \frac{1}{3}\\ \frac{-1}{3} & \frac{2}{3} \end{pmatrix}$$
$$D = \begin{pmatrix} λ_1 & 0\\ 0 & λ_2 \end{pmatrix} = \begin{pmatrix} 3 & 0\\ 0 & -3 \end{pmatrix}$$
$$PD^nP^{-1} = A^n = \begin{pmatrix} 2 & -1\\ 1 & 1 \end{pmatrix} \begin{pmatrix} 3^n & 0\\ 0 & -3^n \end{pmatrix} \begin{pmatrix} \frac{1}{3} & \frac{1}{3}\\ \frac{-1}{3} & \frac{2}{3} \end{pmatrix}$$
$$= \begin{pmatrix} 2(3^n) & -(-3)^n\\ 3^n & (-3)^n \end{pmatrix} \begin{pmatrix} \frac{1}{3} & \frac{1}{3}\\ \frac{-1}{3} & \frac{2}{3} \end{pmatrix}$$
$$= \begin{pmatrix} \frac{2}{3}(3^n) + \frac{1}{3}((-3)^n) & \frac{2}{3}(3^n) - \frac{2}{3}((-3)^n)\\ \frac{1}{3}(3^n) - \frac{1}{3}((-3)^n) & \frac{1}{3}(3^n) + \frac{2}{3}((-3)^n) \end{pmatrix}$$
I think that everything I've done so far is correctly, but I can't find any way to simplify this equation any further, and I don't think that I could find $A^{-n}$ with an equation this complicated, so I must be missing something.
2. Nov 17, 2012
### lurflurf
There are different ways of writing that out, you might not have pick the most simple. Another way is something like
A^n=(3^n)(1/2)((1+(-1)^n)I+(1/3)(1-(-1)^n)A)
which you can see is lot like yours, note all that (-1)^n stuff is just to unify the even and odd terms
even A^n=(3^n)I
odd A^n=(3^n)(1/3)A
I=A^0 the 2x2 identity matrix
3. Nov 18, 2012
### HallsofIvy
Staff Emeritus
Frankly, the first thing I would have done would be to note that
$$A^2= \begin{pmatrix}1 & 4 \\ 2 & -1\end{pmatrix}^2= \begin{pmatrix}9 & 0 \\ 0 & 9\end{pmatrix}= 9\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}$$
From that it follows immediately that if n is even, $A^n= 3^nI$ and if n is odd, $A^n= 3^{n-1}A$
4. Nov 18, 2012
### 1up20x6
Thanks for your responses. Are those equations self-evident or is there a proof that they apply for all values of $n$? | 2017-08-18T19:44:24 | {
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https://stacks.math.columbia.edu/tag/030C | # The Stacks Project
## Tag 030C
Lemma 10.36.16. Let $R$ be a ring. Assume $R$ is reduced and has finitely many minimal primes. Then the following are equivalent:
1. $R$ is a normal ring,
2. $R$ is integrally closed in its total ring of fractions, and
3. $R$ is a finite product of normal domains.
Proof. The implications (1) $\Rightarrow$ (2) and (3) $\Rightarrow$ (1) hold in general, see Lemmas 10.36.12 and 10.36.15.
Let $\mathfrak p_1, \ldots, \mathfrak p_n$ be the minimal primes of $R$. By Lemmas 10.24.2 and 10.24.4 we have $Q(R) = R_{\mathfrak p_1} \times \ldots \times R_{\mathfrak p_n}$, and by Lemma 10.24.1 each factor is a field. Denote $e_i = (0, \ldots, 0, 1, 0, \ldots, 0)$ the $i$th idempotent of $Q(R)$.
If $R$ is integrally closed in $Q(R)$, then it contains in particular the idempotents $e_i$, and we see that $R$ is a product of $n$ domains (see Sections 10.21 and 10.22). Each factor is of the form $R/\mathfrak p_i$ with field of fractions $R_{\mathfrak p_i}$. By Lemma 10.35.10 each map $R/\mathfrak p_i \to R_{\mathfrak p_i}$ is integrally closed. Hence $R$ is a finite product of normal domains. $\square$
The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 7951–7960 (see updates for more information).
\begin{lemma}
\label{lemma-characterize-reduced-ring-normal}
Let $R$ be a ring. Assume $R$ is reduced and has finitely many
minimal primes. Then the following are equivalent:
\begin{enumerate}
\item $R$ is a normal ring,
\item $R$ is integrally closed in its total ring of fractions, and
\item $R$ is a finite product of normal domains.
\end{enumerate}
\end{lemma}
\begin{proof}
The implications (1) $\Rightarrow$ (2) and
(3) $\Rightarrow$ (1) hold in general,
see Lemmas \ref{lemma-normal-ring-integrally-closed} and
\ref{lemma-finite-product-normal}.
\medskip\noindent
Let $\mathfrak p_1, \ldots, \mathfrak p_n$ be the minimal primes of $R$.
By Lemmas \ref{lemma-reduced-ring-sub-product-fields} and
\ref{lemma-total-ring-fractions-no-embedded-points} we have
$Q(R) = R_{\mathfrak p_1} \times \ldots \times R_{\mathfrak p_n}$, and
by Lemma \ref{lemma-minimal-prime-reduced-ring} each factor is a field.
Denote $e_i = (0, \ldots, 0, 1, 0, \ldots, 0)$ the $i$th idempotent
of $Q(R)$.
\medskip\noindent
If $R$ is integrally closed in $Q(R)$, then it contains in particular
the idempotents $e_i$, and we see that $R$ is a product of $n$
domains (see Sections \ref{section-connected-components} and
\ref{section-tilde-module-sheaf}). Each factor is of the form
$R/\mathfrak p_i$ with field of fractions $R_{\mathfrak p_i}$.
By Lemma \ref{lemma-finite-product-integral-closure} each map
$R/\mathfrak p_i \to R_{\mathfrak p_i}$ is integrally closed.
Hence $R$ is a finite product of normal domains.
\end{proof}
Comment #745 by Wei Xu on June 26, 2014 a 5:19 pm UTC
In the second sentense of the proof, it uses a lemma tag02xl but that lemma requires an extra condition: $R\setminus (\mathfrak{q}_1\cup\cdots\cup \mathfrak{q}_t)$ is the set of nonzero-divisors. (However this condition can be proved here).
Comment #757 by Johan (site) on June 28, 2014 a 6:23 pm UTC
OK, yes, I made an improvement of the treatment of zerodivisors in reduced ring in this commit. Thanks!
Comment #2180 by David Savitt on August 27, 2016 a 1:50 pm UTC
In the third paragraph of the proof, it's not clear to me how one is intended to make sense of the claim that $e_i \in R_{\mathfrak p}$ (e.g. since the latter is a domain!). Alternately, how about: each prime $\mathfrak p$ contains a unique minimal prime $\mathfrak q_i$, since $R_{\mathfrak p}$ is a domain. Therefore the irreducible components $V(\mathfrak q_i)$ of $\mathrm{Spec}(R)$ are open and closed, and so $e_i \in R$ for all $i$.
Comment #2207 by Johan (site) on August 29, 2016 a 7:41 pm UTC
OK, I agree with this change, thanks! See here.
## Add a comment on tag 030C
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner). | 2017-10-20T14:32:20 | {
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http://math.stackexchange.com/questions/674564/how-many-ways-can-you-tile-an-nxm-rectangle-with-l-polyominos | # How many ways can you tile an NxM rectangle with L-polyominos?
I came up with a problem that's been bugging me:
How many ways can you tile an NxM rectangle with L-polyominos?
The L shapes can be any size, so long as they aren't lines.
For clarification:
$L(1,m) = 0$,
$L(2,2) = 0$,
$L(2,3) = 2$,
$L(2,4) = 2$,
$L(3,3) = 0$...
I've tried solving it using recursion (the only way I know how), by dividing the NxM grid into 2 smaller grids, but that doesn't account for many of the possibilities.
Solving the problem algorithmically is pretty easy (I hope :P). Here are some of the values I get:
$\begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0 \\[0.3em] 0 & 0 & 2 & 2 & 2 & 6 \\[0.3em] 0 & 2 & 0 & 20 & 64 & 234 \\[0.3em] 0 & 2 & 20 & 110 & 752 & 4522 \\[0.3em] 0 & 2 & 64 & 752 & 7720 & 84846 \\[0.3em] 0 & 6 & 234 & 4522 & 84846 & 1557970 \end{bmatrix}$
I hope that makes sense: L(1,1) is in the top left, and L(6,6) is in the bottom right. Obviously its symmetrical diagonally.
-
(1) Does an $N\times M$-grid have $N\times M$ points or $N\times M$ lines? (2) What precisely is an L-shape? (I could not reversely define it from your given values.) – flonk Feb 18 '14 at 10:54
As in, an N x M grid of dots, where every dot is part of exactly one "L shape". An "L shape" is a horizontal line of dots (H) joined to a vertical line of dots (V) such that one dot on the end of H is also on the end of V, and that |H| >= 2 and |V| >= 2. For further clarification, |H union V| = |V| + |H| - 1 – AStupidNoob Feb 18 '14 at 15:00
I would think that $L(2,2)=2$, with one solution consisting of the two L-shapes $\{(1,0),(0,0),(0,1)\}, \{(1,0),(1,1),(0,1)\}$ and the second solution being the same rotated by $90°$. Where is my mistake? Could you specify more clearly, what is meant by being entirely filled? Are the individual L-shapes allowed to "touch" or "overlap"? In other words: does every dot need to be occupied and are dots allowed to be occupied by multiple L-shapes? – flonk Feb 18 '14 at 20:01
Every dot is part of exactly one "L shape". I.E, they cannot overlap, they are allowed to be adjacent. "Entirely filled" means that there are no dots left in the grid that are not part of an L shape. – AStupidNoob Feb 19 '14 at 1:35
Thanks, got it :) – flonk Feb 19 '14 at 8:28
(EDITED) For the $2 \times m$ case, it seems to me the recursion is $$L(2,m) = 2 \sum_{j=0}^{m-3} L(2,j)$$ where $L(2,0) = 1$, and the solution to that is $$L(2,m) = \sum_r \dfrac{9-3r-2r^2}{29 r^m}$$ the sum being over the three roots $r$ of $2 z^3+z-1$ (approximately $.58975$ and $-.29488 \pm .87227 i$). Cases with $N > 2$ are going to be more complicated. Essentially you want to look at all possibilities for what happens at one end of the grid. But even for $N=3$ there are quite a few possibilities.
-
I thought about maybe trying some kind of recursion where you do it by row. You start with the first row, and fill it with either L end-points or the horizontal part of the L. Then you recurse by filling out the rest of the space with rectangles. Or maybe some kind of crazy algebraic type solution, since we know that every cell is either an endpoint, a "flat" or a corner, and the number of corners is half the number of endpoints. Or something. I really have no idea what I'm doing :) – AStupidNoob Feb 13 '14 at 3:11
The solution differs from the given example values, e.g. $L(2,2)=2\neq 0$. – flonk Feb 19 '14 at 8:31
Oops, thanks for noticing. I fixed it. – Robert Israel Feb 19 '14 at 15:54
Thanks for your answer! I'd rather someone get the bounty than no one! – AStupidNoob Feb 24 '14 at 4:05
This is primarily just a reformulation, but one that seems helpful in replacing the global rule (only L-shaped polyominoes) with a local one. The problem is equivalent to the following:
How many ways can you fill an $M\times N$ grid with these three tiles (rotations and reflections are allowed) if each blue dot must be adjacent to a red dot and vice versa?
This formulation can be used to obtain a recursion relation for the number of ways a given row can be filled with specified top and bottom borders, where each border is a list of white, red, and blue edges; i.e., you can derive a $3^M \times 3^M$ transfer matrix.
-
That's a really cool way of thinking about it! Now if only I knew what a "transfer matrix" was... – AStupidNoob Feb 26 '14 at 5:17 | 2015-08-01T10:03:40 | {
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https://byjus.com/question-answer/the-number-of-positive-integer-pairs-x-y-such-that-dfrac-1-x-dfrac-1/ | Question
The number of positive integer pairs $$(x, y)$$ such that $$\dfrac {1}{x} + \dfrac {1}{y} = \dfrac{1}{2007}, x<y$$ is
A
5
B
6
C
7
D
8
Solution
The correct option is B 7$$\dfrac{x+y}{xy}=\dfrac{1}{2007}$$$$\Rightarrow xy-2007(x+y)=0$$Adding $$2007^2$$ to both sides, we get$$xy -2007(x+y)+2007^2=2007^2$$$$\Rightarrow (x-2007)(y-2007)=2007^2$$Let $$x-2007=A$$ and $$y-2007=B$$The equation becomes $$AB=2007^2$$Number of solutions of above equation is equal to number of factors of $$2007^2$$$$2007^2 = 3^4 \times 223^2$$Hence, number of factors of $$2007^2$$ is $$(4+1)(2+1)=15$$In one case $$A=B=2007$$Of the remaining $$14$$ cases, half of the case $$A>B$$ and remaining half $$A < B$$Accordingly we get $$7$$ cases, where $$x < y$$.Maths
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https://coding-gym.org/challenges/pairs/ | # Pairs
See the original problem on HackerRank.
You will be given an array of integers and a target value. Determine the number of pairs of array elements that have a difference equal to a target value.
For example, given an array of $$[1, 2, 3, 4]$$ and a target value of $$1$$, we have three values meeting the condition:
$$2-1=1, 3-2=1, 4-3=1$$
### Input Format
The first line contains two space-separated integers and , the size of and the target value.
The second line contains space-separated integers of the array arr.
### Constraints
• $$2 \le n \le 10^5$$
• $$0 \lt k \lt 10^9$$
• $$0 \lt arr[i] \lt 2^{31}-1$$
• each integer $$arr[i]$$ will be unique
### Output Format
An integer representing the number of pairs of integers whose difference is k.
### Sample Input
1 2 5 2 1 5 3 4 2
### Sample output
1 3
### Explanation
There are 3 pairs of integers in the set with a difference of 2: [5,3], [4,2] and [3,1] .
## Solutions
A quadratic solution is correct, but it’s too slow and causes some test cases to fail:
1 2 3 4 5 6 7 8 9 10 11 12 int count = 0; for (auto i=0u; i
To find a better solution, we can turn the problem space of searching pairs into the space of searching one element for each iteration.
We need to find i - j = k, for each i and j. This means, for each i we just need to find if i - k is in the array.
The problem turns into a searching problem where looking up any element needs to be fast. We can use an efficient data struture for this purpose like tree-based sets (such as C++’s std::set or Java’s TreeSet), hash-based sets, etc, or we can simply sort the original array and use binary search.
It’s worth noting that we can make a quadratic solution that passes all the test cases:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 int count = 0; sort(arr.begin(), arr.end()); for (auto i=0u; i
This is still quadratic but we are lucky enough with the problem test cases just because of sorting data beforehand! If we add an extra test case containing all the numbers from 2 to 100'000 and set k=100'001, we will have a timeout! Basically, we will end up visiting 100'000 elements 100'000 times!
We sort the input array and then we apply the idea above:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 #include #include #include #include using namespace std; int main() { int n, k; cin >> n >> k; vector v(n); copy_n(istream_iterator(cin), n, begin(v)); sort(begin(v), end(v)); // i - k = j auto cnt = 0; for (auto x : v) { if (binary_search(begin(v), end(v), x - k)) cnt++; } cout << cnt; }
Using some standard algorithms:
1 2 3 4 5 6 7 8 9 10 int main() { int n, k; cin >> n >> k; vector v(n); copy_n(istream_iterator(cin), n, begin(v)); sort(begin(v), end(v)); cout << count_if(begin(v), end(v), [&](int curr){ return binary_search(begin(v), end(v), curr-k); }); }
Looking at the previous solution, we notice that for each i, the binary search runs from the beginning of the array. If we change the number we search, we can start the binary search from the current position, cutting down the search space.
Instead of looking for i - k = j, we reverse the equation and look for j + k = i. Since the array is sorted, j+k is always increasing. This means it’s not possible to find i among the elements we have already looked into during the previous iterations.
Here is a working solution:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 int main() { int n; cin >> n; int k; cin >> k; vector data; for (size_t i = 0; i < n; i++) { int val; cin >> val; data.push_back(val); } sort(data.begin(), data.end()); size_t count = 0; for (size_t i = 0; i < data.size(); ++i) { if (binary_search(data.begin() + i + 1, data.end(), data[i] + k)) count++; } cout << count; }
An even more fine-tuned version has been proposed by Davide Malvezzi: the search range can be made smaller by setting the end at most k steps far from the beginning (the smaller k than n, the more convenient), and the beginning is either i + 1 or one element after the result of the last successful search.
For example, suppose we have:
1 1 4 7 9
And we look for pairs whose difference is k=6. When we start, the search spans from 1 to 1+k:
1 2 1 3 4 5 6 7 8 9 ^ ^
And we find 7:
1 2 1 3 4 5 6 7 8 9 ^ * ^
Then, we search for 3 + k, but it does not make any sense to start from 4 because the element can’t be found before the result of the previous hit (since the elements are sorted). Instead, we can start from 8 (that is one element next the result of the last hit) straight away:
1 2 1 3 4 5 6 7 8 9 ^ ^
And we find 9.
Here is the full solution:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 int pairs_count = 0; auto prev_start = begin(arr); sort(begin(arr), end(arr)); for(size_t i = 0; i < arr.size() - 1 && prev_start != prev(end(arr)); i++) { auto start = max(begin(arr) + i + 1, prev_start + 1); auto end = min(start + k, arr.end()); auto result = lower_bound(start, end, arr[i] + k); if(*result == arr[i] + k) { prev_start = result; pairs_count++; } } return pairs_count;
Note that we early exit when prev_start == prev(end(arr)): if we reach the end of the array, no other element can contribute to the count (since they are all unique).
Although the complexity is still $$O(N \cdot log N)$$, this solution is a perfect example of exploiting problem constraints (formally, the complexity is $$O(N \cdot log K)$$ but in the worst case $$K > N$$).
An interesting related exercise consists in figuring out which test cases are more or less demanding for the solutions discussed above (e.g. in terms of runtime).
### Using other data structures
Previously showed solutions are based on sort that is $$O(N \cdot logN)$$ + iterating each element and performing binary search that still costs $$O(N \cdot logN)$$. We can still apply linear sorting strategy (e.g. counting sort, radix sort), however the cost of performing the second part of the algorithm is still the same.
As said before, alternative solutions involve using other data structures such as maps, sets and dictionaries because we want to optimize looking up i - k = j (or j + k = i). Hash-based associative containers might decrease the overall complexity of the solution down to $$O(N)$$, on average.
Here is an elegant solution in Erlang by Giancarlo Valente:
1 2 3 pairs(K, Arr) -> S = sets:from_list(Arr), length([X || X <- Arr, sets:is_element(X + K, S)]).
The same in Python by kernelfolla:
1 2 3 def pairs(k, arr): s = set(arr) return len([x for x in arr if x+k in s])
Same patterns in C++:
1 2 3 4 5 6 7 8 int N, K; cin >> N >> K; unordered_set S{ istream_iterator(cin), istream_iterator()}; cout << count_if(begin(S), end(S), [&](int i){ return S.count(i-K); });
### Using arrays intersection
We can use a support array to store each expected sum (a[i] + k). To get how many pairs match the requirement, we can count the number of elements are contained in the intersection of such an array with the input.
Here is a solution in PHP by Roberto Peruzzo:
1 2 3 4 5 function pairs($k,$arr) { $sum =$arr; array_walk($sum, function(&$item, $key,$k) { $item +=$k; }, $k); return count(array_intersect($arr, \$sum)); }
Here is the same idea in C++:
1 2 3 4 5 6 7 8 9 int n, k; cin >> n >> k; vector v(n); copy_n(istream_iterator(cin), n, begin(v)); sort(begin(v), end(v)); vector tmp(n); transform(begin(v), end(v), begin(tmp), [=](auto i){ return i + k; }); vector inters; set_intersection(begin(tmp), end(tmp), begin(v), end(v), back_inserter(inters)); cout << inters.size();
A more fluent version with C++ ranges (actually, using all the features available in range-v3):
1 2 sort(v); std::cout << distance(views::set_intersection(v, views::transform(v, [=](auto i){ return i + k; })));
Here, sorting the input elements hides the dominant cost, whereas set intersection is a linear operation.
We've worked on this challenge in these gyms: modena milan padua rome turin bari polimi lecce brianza | 2022-12-05T12:41:18 | {
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https://math.stackexchange.com/questions/2205865/some-subgroup-of-dihedral-group-is-normal | # Some Subgroup of Dihedral Group is Normal
I ran into this question when I was studying for my abstract algebra midterm.
Show that the subgroup $H$ of rotations is normal in the dihedral group $D_n$. Find the quotient group $D_n/H$.
I'm not quite sure where to begin. I know that for a Dihedral group of $n\geq 3$, then $r^n=1$ where $r$ is a rotation, and $s^2=1$ where $s$ is a reflection, and $srs=r^{-1}$. I was not sure how to prove something is a normal subgroup from here. Any advice, thanks!
If $D_n=\langle r,s\mid r^n=s^2=1,srs=r^{-1}\rangle$, then $D_n$ has order $2n$ and the group generated by $r$ has order $n$.
Therefore the index of $\langle r\rangle$ in $D_n$ is equal to two, and it is a general fact that if $H\leq G$ is a subgroup with $[G:H]=2$ then $H$ is a normal subgroup of $G$.
• Without knowing that a subgroup of index 2 is normal, is there any other way to solve this problem, or will such a solution be extremely unwieldy? Thanks again! – Richard Cao Mar 27 '17 at 20:24
• I suppose you could try to verify it directly, but this would likely amount to the proof that every subgroup of index $2$ is normal. – carmichael561 Mar 27 '17 at 20:26
• I see. Thanks again! – Richard Cao Mar 27 '17 at 20:29
• @Richard You could just directly conjugate a general rotation by a general reflection and observe that the result is a rotation. But that doesn't sound like fun. – Matt Samuel Mar 27 '17 at 20:36
• @MattSamuel It suffices to consider any one reflection, e.g. $s$, and show that $sr^ks$ is a rotation for any $k$. This shows that the normalizer of $\langle r \rangle$ contains an element not in $\langle r \rangle$, and therefore (by order considerations) must be all of $D_n$. – Bungo Mar 28 '17 at 6:26
The index $2$ suggestion works, but you can also show this directly. One can check that the generators $R$ and $F$ of the dihedral group conform to the rule $RF = FR^{-1}$. From this, we see that any element in $D_n$ can be written as $R^jF^k$ where $0 \leq j \leq n-1$ and $0 \leq k \leq 1$.
A subgroup $N \leq G$ is normal whenever, given any $n \in N$ and $g \in G$, we have $gng^{-1} \in N$. In this case, any element of the rotation subgroup looks like $R^m$ for $1 \leq m \leq n-1$. Considering any arbitrary element $R^jF^k$ of $D_n$, we just need to show that $(R^jF^k)R^m(R^jF^k)^{-1} \in \langle R \rangle$. Clearly this is true if $k=0$, so assume $k=1$. Now look to the helpful rule in the first paragraph to conclude that this is indeed an element of $\langle R \rangle$.
• You're right. This is easy too. +1 – Matt Samuel Mar 27 '17 at 21:03
How many rotations are there, and how does this compare to the total number of elements? You may have shown as an exercise previously that a subgroup of index $2$ is normal. That is relevant here. If not, you should prove it, because using this fact is the easiest way I see to solve your problem.
This may not count as a proof (depending on your definition of the dihedral group) but it explains geometrically what's going on. The dihedral group is the group of symmetries of the regular $n$-gon in the plane. It consists of $n$ rotations, which clearly form a subgroup, and $n$ reflections. If you think geometrically, the product of a rotation $r$ and a reflection $s$ reverses orientation in the plane, so must be a reflection.
Note that $grg^{-1}$ is clearly a rotation if $g$ is. It is a product of two reflections if $g$ is a reflection, so preserves orientation and must be a rotation. That says the subgroup of rotations is invariant under conjugation, so normal.
• It does not take much more to turn this into a completely rigorous proof (and this is my preferred proof, because it explains the geometric origins of the dihedral group). Namely, one verifies that the dihedral group is isomorphic to the finite subgroup $G < O(2)$ as described in this answer. Then one verifies with a simple matrix computation that the rotations of $O(2)$ form a normal subgroup of $O(2)$. Since the intersection of $G<O(2)$ with any normal subgroup of $O(2)$ is a normal subgroup of $G$, the proof is done. – Lee Mosher May 4 '18 at 14:14
• And of course, as you say, perhaps this is one's definition of the dihedral group (also my preferred definition). – Lee Mosher May 4 '18 at 14:25 | 2019-09-19T18:50:19 | {
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https://math.stackexchange.com/questions/3353482/probability-of-answering-at-least-6-questions-right-on-a-multiple-choice-exam | # Probability of answering at least $6$ questions right on a multiple choice exam.
Suppose a student chooses the answer to each question in an exam randomly, with equal probability for each option and with choices independent of each other. What is the probability that they will get at least $$6$$ out of $$12$$, and thus pass the exam? Note: for each question there are $$5$$ answers, (a $$1/5$$ chance).
I found this question on my physics tutorial sheet and tried to do the following:
$$P(X=6) = nCk \cdot p^k \cdot (1-p)^{n-k}$$
However, when I substituted $$n=12, k=6, p=1/5$$ my answer was $$0.0155$$, but apparently it was incorrect. Any help would be much appreciated.
• The question asks for the probability of getting at least six correct. You calculated the probability of getting exactly six correct. – N. F. Taussig Sep 12 '19 at 0:18
• This tutorial explains how to typeset mathematics on this site. – N. F. Taussig Sep 12 '19 at 0:21
• You need to compute the probability of getting 6,7,8.....12 questiosn right; they are all passing scores. if you are using a graphing calculator it might have a binomcdf function; you can use that for these problems – Saketh Malyala Sep 12 '19 at 0:26
The question asks for the probability that the student gets at least six questions correct by randomly guessing. You used the binomial distribution to compute the probability that the student gets exactly six answers correct. Since we want to find the probability that the student gets at least six answers correct, you should have computed $$\Pr(X \geq 6) = \sum_{k = 6}^{12} \binom{12}{k}\left(\frac{1}{5}\right)^k\left(\frac{4}{5}\right)^{12 - k}$$ I will leave the details to you. | 2020-10-23T21:03:43 | {
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http://mathhelpforum.com/calculus/81520-application-integral-word-problem.html | # Thread: application integral word problem
1. ## application integral word problem
I need some help on this word problem.
A water tank is in the shape of a right circular cone of altitude 10 feet and base radius 5 feet, with it's vertex at the ground. If the tank is full, find the work done in pumping all of the water out the top of the tank.
Note: The weight of water is not given so I assume it is the standard 62.4pi.
Would I need to slice the interval [0,10] or do I need to factor in the base radius?
Is this close $\frac{62.4\pi}{4}\int(y^2)(10-y)dy$
2. Originally Posted by gammaman
I need some help on this word problem.
A water tank is in the shape of a right circular cone of altitude 10 feet and base radius 5 feet, with it's vertex at the ground. If the tank is full, find the work done in pumping all of the water out the top of the tank.
Note: The weight of water is not given so I assume it is the standard 62.4pi.
Would I need to slice the interval [0,10] or do I need to factor in the base radius?
Is this close $\frac{62.4\pi}{4}\int(y^2)(10-y)dy$
sketch the lines $y = 2x$ and $y = -2x$ starting at the origin, up to the points $(5,10)$ and $(-5,10)$
weight of a representative slice is ...
$62.4 \, dV = 62.4 \pi \cdot x^2 \, dy = 62.4 \pi \cdot \frac{y^2}{4} \, dy
$
the slice needs to be lifted a distance $(10 - y)$ ... work in raising the slice is
$dW = 62.4 \pi \cdot \frac{y^2}{4}(10 - y) \, dy$
total work to raise all slices ...
$W = 15.6 \pi \int_0^{10} y^2(10 - y) \, dy$
3. sketch the lines and starting at the origin, up to the points and
why are we doing this?
4. Originally Posted by gammaman
why are we doing this?
it give you a side view of the cone so that you may determine dV and the limits of integration.
5. Where does the base radius come into play? Also I am confused why (y^2) is. Does it just come from the formula for a cone 1/3pi*r^2h. If so why do my notes say to divide by 4?
6. Originally Posted by gammaman
Where does the base radius come into play? Also I am confused why (y^2) is. Does it just come from the formula for a cone 1/3pi*r^2h. If so why do my notes say to divide by 4?
another "why" for sketching a diagram ...
a horizontal slice of the cone's liquid is a cylinder with radius $x$ and thickness $dy$.
since $y = 2x$ , $x = \frac{y}{2}$
$dV = \pi x^2 \, dy = \pi \left(\frac{y}{2}\right)^2 \, dy = \frac{\pi}{4} y^2 \, dy$ | 2017-05-22T22:36:43 | {
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https://mathforums.com/threads/pre-q-a.41264/page-9 | # Pre-Q&A
#### CRGreathouse
Forum Staff
Q3. Find $$\displaystyle \sum_{n=1}^{2520}\gcd(n,2520).$$
Bonus: Do it without using a computer.
#### Hoempa
Math Team
2520 = 2^3 * 3^2 * 5 * 7
Let $$\displaystyle \text{gcdsum}(n) = \sum_{i=1}^n \gcd(i, n)$$
I get
$$\displaystyle \text{gcdsum}(p^k) = (p \cdot (k + 1) - k) p^{k-1}$$
for prime p and positive integer k.
and
$$\displaystyle \text{gcdsum} \left(\prod_{i=1}^n p_i^{\;e_i}\right) = \prod_{i=1}^n \text{gcdsum} \left(p_i^{\;e_i} \right)$$
Where $$\displaystyle \prod_{i=1}^n p_i^{\;e_i}$$ is a prime factorisation.
Therefore
$$\displaystyle \text{gcdsum}(2520) = \sum_{i=1}^{2520} \gcd(i, 2520) = \\\text{gcdsum}(2^3) \cdot \text{gcdsum}(3^2) \cdot \text{gcdsum}(5) \cdot \text{gcdsum}(7) = 20 \cdot 9 \cdot 21 \cdot 13\\ = 49140$$
I found no counterexample for the above statements and Pari says 49140 as well.
Is my method correct?
#### CRGreathouse
Forum Staff
Yes, excellent! I chose that number to be convenient for hand-calculation with that method.
(Anyone, feel free to post questions at any time.)
Q4. How many positive integers less than 100 can be written in the form m/n where m and n are products of nonzero Fibonacci numbers?
#### CRGreathouse
Forum Staff
Oh, and I can't resist posting this one, even though I don't have an answer myself:
Q5. Why are all terms of A234566 integers?
#### johnr
Math Team
Messing with the Fibonacci question empirically cost me hours of sleep! The more I sought, the more I found. Anyone get anywhere interesting with it?
#### eddybob123
CRGreathouse said:
Why are all terms of A234566 integers?
Well, if they weren't all integers, then it wouldn't be a sequence in the OEIS then!
#### CRGreathouse
Forum Staff
johnr said:
Messing with the Fibonacci question empirically cost me hours of sleep! The more I sought, the more I found. Anyone get anywhere interesting with it?
Well, two key components are that the Fibonacci sequence is a gcd sequence: gcd(F(m), F(n)) = F(gcd(m, n)) and that Fibonacci numbers greater than, uh, 144 have a primitive prime divisor: p | F(n) but p does not divide F(m) for 0 < m < n. So sometimes you find that two factors are 'stuck together' for Fibonacci sequences and you can't get them apart, which means you can't make those numbers separately.
As usual there is an OEIS sequence spoiling the answer further if you care to look.
#### agentredlum
Math Team
Please allow me to post a question.
Q. Find a relation between primes that generates the first 31 numbers in the following integer sequence
1 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 ...
Bonus Question: Why is this sequence not in OEIS?
If no one gets it in 24 hours i will post the answer unless someone asks for more time or asks for a hint.
#### CRGreathouse
Forum Staff
agentredlum said:
Please allow me to post a question.
Yes, please!
agentredlum said:
Q. Find a relation between primes that generates the first 31 numbers in the following integer sequence
1 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 ...
Well, my first guess would be that this is an indicator sequence representing the primes 2, 3, 5, 13, 17, 113, ....
If so, I might further guess that this is related to prime gaps, since 113 is the first prime before a record gap (the next prime is 113 + 14 = 127). But I can't get any further, even if both are right.
#### agentredlum
Math Team
CRGreathouse said:
agentredlum said:
Q. Find a relation between primes that generates the first 31 numbers in the following integer sequence
1 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 ...
Well, my first guess would be that this is an indicator sequence representing the primes 2, 3, 5, 13, 17, 113, ....
Thank you for participating! From the primes you give above , only 17 and one more prime are represented in the sequence. I don't want to say too much yet for fear of making it too easy , but of course i will be happy to say more if you or anyone else asks.
CRGreathouse said:
If so, I might further guess that this is related to prime gaps, since 113 is the first prime before a record gap (the next prime is 113 + 14 = 127). But I can't get any further, even if both are right.
I don't think it is related to prime gaps in the usual sense , for example , the prime gap between 23 and 29 is 6 because 29 - 23 = 6 (I personally disagree with the definition but that's niether here nor there :mrgreen
But it may be related to prime gaps of another form which will become apparent once the rule for writing down 0's and 1's is discovered.
Note: The sequence consists entirely of 0's and 1's.
Similar threads | 2020-04-03T21:04:05 | {
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https://physics.stackexchange.com/questions/226456/calculating-the-amount-of-work-done-to-assemble-a-net-charge-on-a-sphere | # Calculating the amount of work done to assemble a net charge on a sphere
I've been reviewing electrostatics using an old exam and I stumbled upon this question:
Calculate the amount of work required to assemble a net charge of $+Q$ on a spherical conductor of radius $R$. If an additional charge of $-Q$ were to be assembled on a concentric spherical conductor of radius $R+a$,what amount of work would the entire process require?
Now the first part is not that difficult, we just do:
$$\vec E = \frac{Q}{4 \pi \epsilon_0 R^2} \hat r \, \text{(From Gauss's Law)}$$
\begin{align} W & = \frac{\epsilon_0}{2} \int E^2 \, d\tau \\ & = \left(\frac{\epsilon_0}{2}\right) \left(\frac{Q^2}{(4 \pi \epsilon_0)^2}\right) \int d \Omega \int_R^{\infty} \frac{1}{R'^{4}} R'^{2} dR'\\ & = \frac{4\pi Q^2}{32 \pi^2 \epsilon_0} \frac{1}{R} \\ &= \frac{Q^2}{8 \pi \epsilon_0 R} \\ \end{align}
But for the second part, according to a solution that a friend of mine gave me, only thing that we need to do to calculate the total work is to do:
\begin{align} W_{tot} & = \frac{\epsilon_0}{2} \int E^2 d \tau \\ & = \frac{4 \pi Q^2}{32 \pi^2 \epsilon_0} \int_R^{R+a} \frac{1}{R'^2} dR' \\ & = \frac{Q^2}{8 \pi \epsilon_0} \left(\frac{1}{R} - \frac{1}{R+a}\right) \\ \end{align}
But according to equation $(2.47)$ of Griffiths, total work should be equal to:
\begin{align} W_{tot} & = \frac{\epsilon_0}{2} \int (E_1+E_2)^2 d\tau \\ & = \frac{epsilon_0}{2} \int (E_1^2 + E_2^2 + 2E_1 \cdot E_2) d\tau \\ & = W_1 + W_2 + \epsilon_0 \int E_1 \cdot E_2 d \tau \\ \end{align}
Wherein for this case $W_1$ is the work required for a sphere of radius $R$ as shown earlier, and $W_2$ is the work required for a sphere of radius $R+a$. Is the first method correct?
• The total work $W_{tot}$ caculated by using the first method has a wrong sign, it should be $-(\frac{1}{R+a}-\frac{1}{R})$, i.e.$\frac{1}{R}-\frac{1}{R+a}$. – Wang Yun Jan 4 '16 at 6:26
• @StephenWong Oh sorry must've missed that one, but is the method correct tho? – Aldon Jan 4 '16 at 6:46
Answer:The two methods are both correct.
As I have suggested in the comment area, total work calculated by using the first method should be $$W_{tot}=\frac{Q^2}{8\pi\epsilon_0}(\frac{1}{R}-\frac{1}{R+a}),$$ since $\int \frac{1}{r^2}dr=-\frac{1}{r}+\text{Constant}$.
Next we will calculate the total work by the second method, i.e. the equation $(2.47)$ of Griffiths. As indicated in the problem, we have \begin{align} \mathbf{E}_1 & =\frac 1 {4\pi\epsilon_0}\frac{Q}{r^2}\hat{\mathbf{r}},\ \text{while}\ r\ge R\\ \mathbf{E}_2 & =-\frac 1 {4\pi\epsilon_0}\frac{Q}{r^2}\hat{\mathbf{r}},\ \text{while}\ r\ge R+a \end{align}
So, \begin{align} E_1^2 & =\frac{Q^2}{16\pi^2\epsilon_0^2r^4}\\ E_2^2 & =\frac{Q^2}{16\pi^2\epsilon_0^2r^4}\\ E_1\cdot E_2 & =-\frac{Q^2}{16\pi^2\epsilon_0^2r^4} \end{align}
And \begin{align} W_1 & =\frac{\epsilon_0}{2}\int E_1^2d\tau\\ & =\frac{Q^2}{8\pi\epsilon_0}\int_R^\infty\frac{1}{r^2}dr\\ & =\frac{Q^2}{8\pi\epsilon_0R} \end{align} By using same method, we get $W_2$ as foolow, $$W_2=\frac{Q^2}{8\pi\epsilon_0(R+a)}$$
Finally, the cross term is that \begin{align} \epsilon_0 \int \mathbf{E}_1 \cdot \mathbf{E}_2 d\tau & =-\frac{Q^2}{4\pi\epsilon_0}\int_{R+a}^\infty \frac{1}{r^2}dr\\ & =-\frac{Q^2}{8\pi\epsilon_0}\frac{2}{R+a} \end{align}
Then we add them all up, we have, \begin{align} W_{tot} & =W_1+W_2+\epsilon_0\int \mathbf{E}_1\cdot\mathbf{E}_2d\tau \\ & =\frac{Q^2}{8\pi\epsilon_0}\frac{1}{R}+\frac{Q^2}{8\pi\epsilon_0}\frac{1}{R+a}-\frac{Q^2}{8\pi\epsilon_0}\frac{2}{R+a}\\ & =\frac{Q^2}{8\pi\epsilon_0}(\frac{1}{R}-\frac{1}{R+a}) \end{align}
Conclusion: The results are the same by two methods.In the first method, when we wrote the formula of $W_{tot}$, the electric field $E$ is the final field after used superposition princeple. In the second, we also used superposition princeple, but we wrote it in the form of $W$ explicitly. I mean that the two methods are the same, but have different forms.
• Obviously the second method will work since it's the recommended one from Griffiths, but why did the first one work? Do you have any insights and situations wherein it will work and will not work? – Aldon Jan 4 '16 at 9:59
• The first method is the most basic, and the second is derived from the first according superposition princeple of electric field. – Wang Yun Jan 4 '16 at 10:21
• @Aldon I modified my answer again, I hope that would help you. – Wang Yun Jan 5 '16 at 1:10
• @StephenWong Do you know why it is that in Griffiths Intro to Electrodynamics Fourth edition, he gets your answer (but negative) for problem 2.60? – user100411 Aug 1 '16 at 21:02 | 2019-12-08T21:28:42 | {
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https://math.stackexchange.com/questions/2976057/can-the-quadratic-formula-be-used-when-factorizing-a-denominator/2976089 | # Can the quadratic formula be used when factorizing a denominator?
I’m doing partial fractions and need to factorize the denominator. They are quadratic. However there are some that aren’t so easy to factorize and my first choice was to use the quadratic equation to find the roots however comparing my answer with the correct one the signs are different. Is the quadratic formula only to be used when the equation is equal to zero? The answer used another method of factorizing that didn’t involve equating anything to zero and I can’t find anything about it online. Where did I go wrong? My denominator is:
$$-3z^2 -4z-1$$
the correct answer is: $$-(3z+1)(z+1)$$
while if I do this using the quadratic formula I get: $$(3z+1)(z+1)$$
however if I factorize the negative sign then use the quadratic formula I get the correct answer which is confusing to me.
• It's not very clear what you mean, maybe you could give a practical example and write some of the terms that you want to factor? – Matti P. Oct 29 '18 at 12:20
• I have edited the question. – Ian Oct 29 '18 at 12:27
Suppose I have a mystery quadratic $$p(x) = -2x^2 + x + 1$$, and want to factorise it using the quadratic formula. I can find the roots easily enough:
$$x = \frac{-1 \pm \sqrt{1 + 8}}{-4}$$ and so the roots are $$-\frac{1}{2}$$ and $$1$$, and so I think that the polynomial should factorise as $$p(x) = (x - 1)(x+\frac{1}{2})$$. But this is wrong, since I forgot that $$p(x)$$ and $$2p(x)$$ and $$\frac{-p(x)}{9}$$ and so on all have the same roots! And so knowing the two roots only determines $$p(x)$$ up to scaling. But it's easy to find the right scaling factor, by just looking at the $$-2x^2$$ term. So I arrive at the actual answer, of $$p(x) = -2x^2 + x + 1 = -2(x-1)(x + \frac{1}{2})$$
So after you find the roots, remember to multiply the whole thing by the right number to fix up the $$x^2$$ term.
• what is this phenomenon called? where can I read up more on it? – Ian Oct 29 '18 at 12:36
• I don't know if it's called anything particular. You just have to notice that scaling a function changes the function but not the roots, since if $p(\alpha) = 0$, then $2 p(\alpha) = 0$ also. But $p$ and $2p$ are not the same function. So knowing the roots is not enough to determine the quadratic, you have to know something about how steep it is. – Joppy Oct 29 '18 at 12:38
• does this only apply if the coefficient of the $x^2$ is non zero???? – Ian Oct 29 '18 at 12:39
• If the coefficient of the $x^2$ is zero, then you have a linear function $ax + b$, not a quadratic. – Joppy Oct 29 '18 at 12:40
• so as a rule of thumb, once the coefficient isn't 1 I should multiply by the coefficient after I use the quadratic formula? – Ian Oct 29 '18 at 12:41
I'm guessing, what you have is the following fraction: $$f(z) = \frac{N(z)}{D(z)},$$ where $$D(z) = -3z^2-4z-1$$.
Basically, you may write $$D(z) = -(3z^2+4z+1) = -D_2(z)$$.
Finally, you can decompose the fraction $$f$$ as $$f(z) = \frac{N(z)}{D(z)}$$ or as $$f(z) = \frac{-N(z)}{D_2(z)}$$.
In other words, you'll get the same partial fraction in any case (you'll keep only the two factors in the denominators, and push the minus sign to the numerator)
Factoring in $$\mathbb Q[x]$$ is unique up to associates. The units in $$\mathbb Q[x]$$ are all nonzero rational numbers. So $$(x + \frac 12)$$, $$(2x+1), -(2x+1), (4x+2)$$ and so on are all considered equivalent factors. | 2019-06-16T16:45:08 | {
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https://math.stackexchange.com/questions/2589887/how-can-the-infinity-norm-minimization-problem-be-rewritten-as-a-linear-program | # How can the infinity norm minimization problem be rewritten as a linear program?
I have been trying to solve the infinity norm minimization problem and after quite a bit of reading I have found out that infinity norm minimization problem can be re-written as linear optimization problem. I have been trying to understand how and why it is done so, but failing miserably. Could anyone please explain me about this and also tell how the cost function looks like?
My optimization problem looks like following: (I have to solve for $x$ when $A$ and $b$ are given.)
$$\mbox{minimize} \quad \|A x - b\|_{\infty}$$
which can be rewritten as follows
\begin{split}\begin{array}{lccl} \mbox{minimize} & t & &\\ \mbox{subject to} & Ax + t \mathbb 1 - b & \geq & 0,\\ & A x - t \mathbb 1 - b & \leq & 0, \end{array}\end{split}
where $\mathbb 1$ is a vector of ones.
• For brevity let's define $y=Ax-b$. We have $\|y\|_\infty = \max_i|y_i| = \max(\max_iy_i,\max_i-y_i) = \arg\min_{\substack{t\ge y_i\\t\ge-y_i}}t$. – Rahul Jan 3 '18 at 10:56
To complement Erwin Kalvelagen's answer, we have the following optimization problem in $\mathrm x \in \mathbb R^n$
$$\begin{array}{ll} \text{minimize} & \|\mathrm A \mathrm x - \mathrm b\|_\infty\end{array}$$
where $\mathrm A \in \mathbb R^{m \times n}$ and $\mathrm b \in \mathbb R^m$ are given. Let $\mathrm a_i^\top \in \mathbb R^n$ denote the $i$-th row of $\rm A$.
Introducing decision variable $t \in \mathbb R$ and rewriting in epigraph form, we then obtain the following constrained optimization problem in $\mathrm x \in \mathbb R^n$ and $t \in \mathbb R$
$$\begin{array}{ll} \text{minimize} & t\\ \text{subject to} & \|\mathrm A \mathrm x - \mathrm b\|_\infty \leq t\end{array}$$
which can be rewritten as follows
$$\begin{array}{ll} \text{minimize} & t\\ \text{subject to} & \displaystyle\max_{1 \leq i \leq m} | \mathrm a_i^\top \mathrm x - b_i | \leq t\end{array}$$
which can be rewritten as follows
$$\begin{array}{ll} \text{minimize} & t\\ \text{subject to} & | \mathrm a_1^\top \mathrm x - b_1 | \leq t\\ & | \mathrm a_2^\top \mathrm x - b_2 | \leq t\\ & \qquad \vdots\\ & |\mathrm a_m^\top \mathrm x - b_m | \leq t\end{array}$$
which can be rewritten as follows
$$\begin{array}{ll} \text{minimize} & t\\ \text{subject to} & -t \leq \mathrm a_1^\top \mathrm x - b_1 \leq t\\ & -t \leq \mathrm a_2^\top \mathrm x - b_2 \leq t\\ & \qquad\quad \vdots\\ & -t \leq \mathrm a_m^\top \mathrm x - b_m \leq t\end{array}$$
which can be rewritten as follows
$$\begin{array}{ll} \text{minimize} & t\\ \text{subject to} & -t 1_m \leq \mathrm A \mathrm x - \mathrm b \leq t 1_m\end{array}$$
where the optimal $\rm x$ and the optimal $t$ are the minimizer and the minimum of the original problem, respectively.
• How do I start solving this problem? What should be my first step if I want to solve it without using existing LP solver softwares? I want to do it for the sake of understanding. Also I used cvx software whose computational time is rather high. Any suggestion which software should I use for this particular problem? – Christina Lemuix Jan 4 '18 at 22:13
• You can run the simplex algorithm using pen & paper (rather than silicon). – Rodrigo de Azevedo Jan 5 '18 at 2:46
• Could you elaborate a bit more please? – Christina Lemuix Jan 5 '18 at 8:40
• Algorithms can be run on pen & paper, if you have the patience. Try $m=3$ and $n=2$. – Rodrigo de Azevedo Jan 5 '18 at 10:28
• found the following matlab code that minimizes the infinity norm. Can you please help me understand what is going on here? Why are $f$, $Ane$ and $bne$ like that? A = randn(m,n); b = randn(m,1); % infinity norm f = [ zeros(n,1); 1 ]; Ane = [ +A, -ones(m,1) ; ... -A, -ones(m,1) ]; bne = [ +b; -b ]; xt = linprog(f,Ane,bne); x_lp = xt(1:n,:); – Christina Lemuix Jan 5 '18 at 13:10
If we want to minimize $|x|$ (for a scalar $x$) we can linearize this as:
\begin{align} \min\>& t\\& -t \le x \le t\end{align}
This means that if we want to minimize $||Ax-b||_{\infty}$ we can write:
\begin{align} \min\>& t\\& -t \le (Ax-b)_i \le t&&\forall i\end{align}
You need to split this into two different inequalities:
\begin{align} \min\>& t\\& -t \le (Ax-b)_i &&\forall i\\ &(Ax-b)_i \le t&&\forall i\end{align}
There are some other LP formulations shown here.
• Thanks Erwin. I am new to solving an optimization problem. So could you please tell me how do I formulate my cost function from the above information? – Christina Lemuix Jan 3 '18 at 12:33
• The cost function $f(x,t)$ is just $f(x,t) = t$ which is linear. Note that both $x$ and $t$ are decision variables, where $t$ is a scalar variable. The cost coefficients are 1 for $t$ and zero for the original $x$ variables. So your cost vector $c$ will be very sparse: one element with value 1 and the rest zero. – Erwin Kalvelagen Jan 3 '18 at 12:37
• So how do I define the Lagrangian in this case? – Christina Lemuix Jan 3 '18 at 12:46
• Why do you need a Lagrangian for an LP problem? Does not make sense to me. – Erwin Kalvelagen Jan 3 '18 at 12:47
• @ChristinaLemuix it cannot be solved by hand. It must be solved numerically. – Michael Grant Jan 3 '18 at 15:18 | 2019-08-24T18:20:12 | {
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http://math.stackexchange.com/questions/488518/strong-induction-proof-fibonacci-number-even-if-and-only-if-3-divides-index | # Strong Induction Proof: Fibonacci number even if and only if 3 divides index
The Fibonacci sequence is defined recursively by $F_1 = 1, F_2 = 1, \; \& \; F_n = F_{n−1} + F_{n−2} \; \text{ for } n ≥ 3.$ Prove that $2 \mid F_n \iff 3 \mid n.$
Proof by Strong Induction : $\bbox[5px,border:1px solid green]{\color{green}{n = 1 }}$ $2 \mid F_1$ is false. Also, $3 \mid 1$ is false.
The implication [False $\iff$ False] is vacuously true.
$\bbox[5px,border:1px solid green]{\color{green}{\text{Induction Hypothesis}}}$ Assume that $2 \mid F_i \iff 3 \mid n$ for every integer $i$ with $1 ≤ i ≤ k$.
$\bbox[5px,border:1px solid green]{\color{green}{k + 1 \text{th Case}}} \;$ To prove: $\quad 2 \mid F_{k + 1} \iff 3 \mid k + 1.$
$\bbox[5px,border:1px solid green]{\color{green}{n = k + 1 = 2}} \;$ $2 \mid F_2$ is false. Also, $3 \mid 2$ is false. So [False $\iff$ False] is vacuously true.
Hence assume that $k + 1 ≥ 3.$ We now consider three cases:
$\bbox[5px,border:1px solid green]{\color{green}{\text{Case 1: } k + 1 = 3q}}$ Thus $3 \require{cancel}\cancel{\mid} k$ and $3 \require{cancel}\cancel{\mid} (k − 1)$. By the ind hyp, $3 \require{cancel}\cancel{\mid} k \iff F_k$ odd & $3 \require{cancel}\cancel{\mid} (k − 1) \iff F_{k - 1}$ odd. Since $F_{k+1} = F_k + F_{k−1}$, thus $F_{k+1}$ = odd + odd = even.
$\bbox[5px,border:1px solid green]{\color{green}{\text{Case 2: } k + 1 = 3q + 1}}$ Thus $3 | k$ and $3 \require{cancel}\cancel{\mid} (k − 1).$ By the ind hyp, $3 | k \iff F_k$ even & $3 \require{cancel}\cancel{\mid} (k − 1) \iff F_{k - 1}$ odd. Thus $F_{k+1}$ odd.
$\bbox[5px,border:1px solid green]{\color{green}{{\text{Case 3: }} k + 1 = 3q + 2}}$ Thus $3 \require{cancel}\cancel{\mid} k$ and $3 | (k −1).$ By the ind hyp, $3 \require{cancel}\cancel{\mid} k \iff F_k$ odd and $3 \mid (k − 1) \iff F_{k - 1}$ even. Thus $F_{k+1}$ odd. $\blacksquare$
$\Large{1.}$ Does the proof clinch the $(\Leftarrow)$ of the $(k + 1)$th case?
$\Large{2.}$ Since the recursion contains $n, n - 1, n - 2$, thus the recursion "time lag" is $3$ here.
So shouldn't $3$ base cases be checked?
$\Large{3.}$ Further to #2, shouldn't "assume $k + 1 \geq \cancel{3} 4$" instead?
$\Large{4.}$ Shouldn't the $n = k + 1 = 2$ case precede the induction hypothesis?
I referenced 1. Source: Exercise 6.35, P152 of Mathematical Proofs, 2nd ed. by Chartrand et al
$\Large{1.1.}$ I wrongly believed that all 3 Cases proved the $\Leftarrow$. I now see that Case 1 is $\Leftarrow$ via a Direct Proof. Cases 2 and 3 are $\Rightarrow$ via a Proof by Contraposition. Nonetheless, how would one foreknow/prevision to start from $3 \mid n$ for both directions of the proof?
-
– Cameron Buie Sep 9 '13 at 15:37
@CameronBuie: Thanks. It's referenced as 1 in my OP. – LePressentiment Sep 10 '13 at 0:29
Can you prove that the remainders of the Fibonacci numbers forever repeat the period $0,1,1,0,1,1,0,1,1,0,1,1,\ldots$? Hint: Denote the remainder of $F_n$ by $R_n$, then (surprise!) $R_{n+2}\equiv R_{n+1}+R_n\pmod2$. And an induction with depth three shows that $R_{n+3}=R_n$ for all $n$. – Jyrki Lahtonen Sep 11 '13 at 9:07
Part 1 Case 1 proves $3\mid (k+1)\Rightarrow 2\mid F_{k+1}$, and Case 2 and 3 proves $3\cancel\mid (k+1)\Rightarrow 2\cancel\mid F_{k+1}$. The latter is actually proving the contra-positive of $2 \mid F_{k + 1} \Longrightarrow 3 \mid k + 1$ direction.
Part 2 You only need the statement to be true for $n=k$ and $n=k-1$ to prove the case of $n=k+1$, as seen in the 3 cases. Therefore, $n=1$ and $n=2$ cases are enough to prove $n=3$ case, and start the induction process.
Part 3 :)
Part 4 Probably a personal style? I agree having both $n=1$ and $n=2$ as base cases is more appealing to me.
-
Thank you. I upvoted. Could you please respond to my supplement in my OP (For easier reading)? – LePressentiment Sep 11 '13 at 7:44
Look at $F_n$ modulo $2$. You will find the pattern of congruences: $$1, 1, 0, 1, 1, 0, ...$$. This follows from the fact that $F_1 \equiv F_2 \equiv 1$ (mod $2$) and the recursive definition of the Fibonnaci numbers.
-
My apologies, I just realized that your post was not about finding any proof but rather has specific questions about your own. Ill leave my post since I think the proof is good but let me know if it is off topic. – Patrick Sep 9 '13 at 15:49
No worries. Thanks. – LePressentiment Sep 10 '13 at 0:51
Since the period of $2$ in base $\phi^2$ is three places long = $0.10\phi\; 10\phi \dots$, and the fibonacci numbers represent the repunits of base $\phi^2$, then it follows that $2$ divides every third fibonacci number, in the same way that $37$ divides every third repunit in decimal (ie $111$, $111111$, $111111111$, etc).
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https://math.stackexchange.com/questions/2589843/when-resolving-an-equation-how-do-i-know-i-need-to-multiply-by-1 | # When resolving an equation, how do I know I need to multiply by -1?
I have an equation like this:
$6y−\sqrt{2y^2-1} = 7y - 2$
The solution tells me to first move the $6y$ to the other side by subtracting it:
$-\sqrt{2y^2-1} = y - 2$
And then they suggest to divide (or multiply) by $-1$ to obtain:
$\sqrt{2y^2-1} = 2 - y$
Now, at the second step, what always gets me is that I'm tempted to square both sides, instead of multiplying by $-1$ like so in order to get rid of the minus sign and the radical in one step:
$$-\sqrt{2y^2-1} = y - 2$$ $$(-\sqrt{2y^2-1})^2 = (y - 2)^2$$ $$2y^2-1 = (y - 2)^2$$
But that leads to a wrong solution. Why can't I get rid of the minus sign and the radical in one step and more generall, when does it make sense to multiply both sides of an equation with $-1$, what's the rule?
• "But that leads to a wrong solution. " Why do you assume that? It does not lead to a wrong answer. – fleablood Jan 3 '18 at 8:46
• To be flippant, it doesn't matter what you do to each side. As long as it is the same thing you will get a true statement. You can divide both sides by 37, add the square root of pi, and take the cosines of both sides for all anyone cares. (It won't help you solve for y, but you will get a true, albeit convoluted and useless, statement.) [ If $2x + 3 = 7$ then $\cos (\frac{2x+3}{37} + \sqrt{\pi}) = \cos (\frac{7}{37} + \sqrt{\pi})$. That IS true.... It just doesn't help you get the answer.] – fleablood Jan 3 '18 at 8:52
Why does it give a wrong solution? Continuing your books solution, let us square both sides to get: $$(\sqrt {2y^2-1})^2 =(2-y)^2$$ $$\implies (\sqrt {2y^2-1})^2=y^2-4y+4=(y-2)^2$$
which is the same you got.
Note that an easy way to eliminate a $-$ sign when you encounter it in an equation is to square it on both sides. There is no hard-and-fast rule to deal with such situations.
• Ah I see, yeah I think I got confused somewhere else then. But in general, it is ok then to square the radical to get rid of the minus (given I also square the other side of the equation)? – Max Jan 3 '18 at 8:24
• @Max Yes, it's ok. But be aware that it will introduce wrong solutions in some situations. If $A=B$, it's true that $A^2=B^2$, but if $A^2=B^2$, then $(A+B)(A-B)=0$ and $A=B$ or $A=-B$ are both valid. – Jean-Claude Arbaut Jan 3 '18 at 8:26
No difference
$$(-\sqrt{2y^2-1})^2 = (y - 2)^2$$
Is the same as $$(\sqrt{2y^2-1})^2 = (y - 2)^2 = (-(y - 2))^2$$
You only have this kind of issues for inequalities.
Here it is important to check that the final solution is such that esists
$$\sqrt{2y^2-1}\implies y\geq\frac{\sqrt2}{2} \quad y\leq\frac{-\sqrt2}{2}$$
You don't have to multiply by $-1$. But you have to note that $-\sqrt{2y^2 - 1}$ is a negative or zero number, so $y-2 \le 0$.
If you multiply by $-1$ (which the book only does because it thinks it will be easier for you) you will get $\sqrt{2y^2 - 1} = 2-y$. In this case you have to note that $\sqrt{2y^2 - 1}$ is a positive of zero number so $2-y \ge 0$.
But $y-2 \le 0 \iff 2-y \ge 0 \iff y\le 2$. It doesn't matter which one you not as long as you note one of them.
Now to solve you must square both sides.
If you didn't multiple by $-1$ you get
$(-\sqrt{2y^2 - 1})^2 = (y-2)^2$
$2y^2 - 1 = y^2 -4y +4$.
$y^2 +4y - 5 = 0$
If you did multiply by $-1$ you get
$(\sqrt{2y^2 - 1})^2 = (2-y)^2$
$y^2 - 1 = 4 -4y + y^2$
$y^2 + 4y - 5 = 0$.
You see the difference? (That was a joke. There IS no difference. $(-k)^2 = k^2$ so $(-\sqrt{2y^2 - 1})^2 = \sqrt{2y^2 - 1}^2$ and $(y-2)^2 = (2-y)^2$.
But there are two things I must point out.
1) The book thinks it's easier to multiply by $-1$ to get rid of the minus sign to make things easier to explain. It's actually .... not important.
2) When you square both sides you add in an extra possible answer (maybe) that may not be correct. But this happens WHETHER OR NOT you multiply by $-1$.... and that is why you have to note $y \le 2$.
....
To finish
$y^2 +4y - 5 = 0$
$(y +5)(y-1) = 0$
So either $y+5 = 0$ OR $y-1 = 0$.
So either $y = -5$ OR $y=1$. As we know $y \le 2$ we know both of these answers are acceptable.
If $y = 1$ we can check either: Does $-\sqrt{2y^2 - 1} {? \over =} (y-2)$
$-\sqrt{2(1)^2 - 1} {? \over =} ((1)-2)$
$-\sqrt{2-1} {? \over =} -1$
$-\sqrt{1} {? \over =} -1$
$-1 = -1$. Check it does.
OR
we could check:
Does $\sqrt{2y^2 - 1} {? \over =} (2-y)$
$\sqrt{2(1)^2 - 1} {? \over =} (2- (1))$
$\sqrt{2-1} {? \over =} 1$
$\sqrt{1} {? \over =} 1$
$1 = 1$. Check it does.
The only difference is in one you don't get distraction by the minus sign.
That may or may not make things easier. | 2019-06-15T20:32:45 | {
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https://sandrapassinhas.eu/tdc-usa-hvn/ac0dd3-do-surjective-functions-have-inverses | # do surjective functions have inverses
is not injective - you have g ( 1) = g ( 0) = 0. Let $x = \frac{1}{y}$. Let f(x):ℝ→ℝ be a real-valued function y=f(x) of a real-valued argument x. For instance, if I ask Wolfram Alpha "is 1/x surjective," it replies, "$1/x$ is not surjective onto ${\Bbb R}$." Let $f : S \to T$, and let $T = \text{range}(f)$, i.e. Non-surjective functions in the Cartesian plane. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … This is a theorem about functions. The domain is basically what can go into the function, codomain states possible outcomes and range denotes the actual outcome of the function. Graphic meaning: The function f is a surjection if every horizontal line intersects the graph of f in at least one point. Sand when we chose solid ; air when we chose gas....... Many claim that only bijective functions have inverses (while a few disagree). Can a non-surjective function have an inverse? Then $x_1 = g(f(x_1)) = g(f(x_2)) = x_2$, so $f$ is injective. If a function has an inverse then it is bijective? What's your point? Obviously no! When an Eb instrument plays the Concert F scale, what note do they start on? Thus, all functions that have an inverse must be bijective. Theorem A linear transformation L : U !V is invertible if and only if ker(L) = f~0gand Im(L) = V. This follows from our characterizations of injective and surjective. If a function has an inverse then it is bijective? Since "at least one'' + "at most one'' = "exactly one'', f is a bijection if and only if it is both an injection and a surjection. Making statements based on opinion; back them up with references or personal experience. But if for a given input there exists multiple outputs, then will the machine be a function? So is it true that all functions that have an inverse must be bijective? Hence, $f$ is injective. Yes. Just make the codomain the positive reals and you can say "$e^x$ maps the reals onto the positive reals". (g \circ f)(x) & = x~\text{for each}~x \in A\\ Making statements based on opinion; back them up with references or personal experience. Suppose that $g(b) = a$. Use MathJax to format equations. the codomain of $f$ is precisely the set of outputs for the function. In $(\mathbb{R}^n,\varepsilon_n)$ prove the unit open ball and $Q=\{x \in \mathbb{R}^n| | x_i| <1, i=1,…,n \}$ are homeomorphic, The bijective property on relations vs. on functions. - Yes because it gives only one output for any input. Finding the inverse. The inverse is simply given by the relation you discovered between the output and the input when proving surjectiveness. You seem to be saying that if a function is continuous then it implies its inverse is continuous. How many presidents had decided not to attend the inauguration of their successor? x\\sim y if and only if x-y\\in\\mathbb{Z} Show that X/\\sim\\cong S^1 So denoting the elements of X/\\sim as [t] The function f([t])=\\exp^{2\\pi ti} defines a homemorphism. So if we consider our machine to be working in the opposite way, we should get milk when we chose liquid; For additional correct discussion on this topic, see this duplicate question rather than the other answers on this page. Furthermore since f1 is not surjective, it has no right inverse. onto, to have an inverse, since if it is not surjective, the function's inverse's domain will have some elements left out which are not mapped to any element in the range of the function's inverse. That means we want the inverse of S. When we opt for "liquid", we want our machine to give us milk and water. Shouldn't this function be not invertible? @percusse $0$ is not part of the domain and $f(0)$ is undefined. Inverse Image When discussing functions, we have notation for talking about an element of the domain (say $$x$$) and its corresponding element in the codomain (we write $$f(x)\text{,}$$ which is the image of $$x$$). Thus, $f$ is surjective. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. And g inverse of y will be the unique x such that g of x equals y. Let's again consider our machine A function is invertible if and only if it is a bijection. Finding an inverse function (sum of non-integer powers). Properties of a Surjective Function (Onto) We can define onto function as if any function states surjection by limit its codomain to its range. So, for example, does $f:\{0\}\rightarrow \{1,2\}$ defined by $f(0)=1$ have an inverse? A function $f : X \to Y$ is injective if and only if it admits a left-inverse $g : Y \to X$ such that $g \circ f = \mathrm{id}_X$. Personally I'm not a huge fan of this convention since it muddies the waters somewhat, especially to students just starting out, but it is what it is. And this function, then, is the inverse function … Should the stipend be paid if working remotely? Use MathJax to format equations. Now we want a machine that does the opposite. If $f : X \to Y$ is a map of sets which is injective, then for each $x \in X$, we have an element $y = f(x)$ uniquely determined by $x$, so we can define $g : Y \to X$ by sending those $y \in f(X)$ to that element $x$ for which $f(x) = y$, and the fact that $f$ is injective will show that $g$ will be well-defined ; for those $y \in Y \backslash f(X)$, just send them wherever you want (this would require this axiom of choice, but let's not worry about that). So perhaps your definitions of "left inverse" and "right inverse" are not quite correct? Would you get any money from someone who is not indebted to you?? Then $x_1 = (g \circ f)(x_1) = (g \circ f)(x_2) = x_2$. Zero correlation of all functions of random variables implying independence. Yep, it must be surjective, for the reasons you describe. rev 2021.1.8.38287, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us, $(f^{-1} \circ f)(x) = (f \circ f^{-1})(x) = x$, Right now the given example seems to satisfy your definition of a right inverse: we have $f(f^{-1}(1))=1$. If you're looking for a little more fun, feel free to look at this ; it is a bit harder though, but again if you don't worry about the foundations of set theory you can still get some good intuition out of it. A function is invertible if and only if the function is bijective. One by one we will put it in our machine to get our required state. So is it a function? Why continue counting/certifying electors after one candidate has secured a majority? I originally thought the answer to this question was no, but the answers given below seem to take this summarized point of view. Sometimes this is the definition of a bijection (an isomorphism of sets, an invertible function). Let's make this machine work the other way round. When an Eb instrument plays the Concert F scale, what note do they start on? Jun 5, 2014 Does there exist a nonbijective function with both a left and right inverse? A function is a one-to-one correspondence or is bijective if it is both one-to-one/injective and onto/surjective. It only takes a minute to sign up. In basic terms, this means that if you have $f:X\to Y$ to be continuous, then $f^{-1}:Y\to X$ has to also be continuous, putting it into one-to-one correspondence. What's the difference between 'war' and 'wars'? Can an exiting US president curtail access to Air Force One from the new president? it is not one-to-one). Left: There is y 0 in Y, but there is no x 0 in X such that y 0 = f(x 0). surjective: The condition $(f \circ g)(x) = x$ for each $x \in B$ implies that $f$ is surjective. Are all functions that have an inverse bijective functions? It is not required that x be unique; the function f may map one or more elements of X to the same element of Y. A simple counter-example is $f(x)=1/x$, which has an inverse but is not bijective. (This as opposed to the case of non-injectivity, in which case you only have a set of elements that map to that chosen element of the codomain.). It CAN (possibly) have a B with many A. There are three kinds of inverses in this context: left-sided, right-sided, and two-sided. Aspects for choosing a bike to ride across Europe, Dog likes walks, but is terrified of walk preparation. Conversely, suppose $f$ admits a left inverse $g$, and assume $f(x_1) = f(x_2)$. Is it my fitness level or my single-speed bicycle? To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Thanks for the suggestions and pointing out my mistakes. The set B could be “larger” than A in the sense that there could be some elements b : B for which no f a equals b — that is, B may not be “fully covered.” Then, obviously, $f$ is surjective outright. That was pretty simple, wasn't it? It seems like the unfortunate conclusion is that terms like surjective and bijective are meaningless unless the domain and codomain are clearly specified. Well, that will be the positive square root of y. Barrel Adjuster Strategy - What's the best way to use barrel adjusters? If $f\colon A \to B$ has an inverse $g\colon B \to A$, then site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Do injective, yet not bijective, functions have an inverse? Then $(f \circ g)(b) = f(g(b)) = f(a) = b$, so there exists $a \in A$ such that $f(a) = b$. Can a law enforcement officer temporarily 'grant' his authority to another? S(some matter)=it's state Even if Democrats have control of the senate, won't new legislation just be blocked with a filibuster? Relation of bijective functions and even functions? Now we consider inverses of composite functions. A; and in that case the function g is the unique inverse of f 1. It must also be injective, because if $f(x_1) = f(x_2) = y$ for $x_1 \ne x_2$, where does $f^{-1}$ send $y$? Can I hang this heavy and deep cabinet on this wall safely? A bijection is also called a one-to-one correspondence. Now for sand it gives solid ;for milk it will give liquid and for air it gives gas. So x 2 is not injective and therefore also not bijective and hence it won't have an inverse.. A function is surjective if every possible number in the range is reached, so in our case if every real number can be reached. That is. Let's say a function (our machine) can state the physical state of a substance. I don't think anyone would dispute that $e^x$ has an inverse function, even though the function doesn't map the reals onto the reals. Moreover, properties (1) and (2) then say that this inverse function is a surjection and an injection, that is, the inverse function exists and is also a bijection. 1, 2. Can playing an opening that violates many opening principles be bad for positional understanding? Asking for help, clarification, or responding to other answers. However, I do understand your point. Moreover, properties (1) and (2) then say that this inverse function is a surjection and an injection, that is, the inverse function exists and is also a bijection. Hence it's not a function. In the case when a function is both one-to-one and onto (an injection and surjection), we say the function is a bijection, or that the function is a bijective function. Think about the definition of a continuous mapping. Yes. All the answers point to yes, but you need to be careful as what you mean by inverse (of course, mathematics always requires thinking). MathJax reference. This means you can find a $f^{-1}$ such that $(f^{-1} \circ f)(x) = x$. If you know why a right inverse exists, this should be clear to you. Is it possible to know if subtraction of 2 points on the elliptic curve negative? Are all functions that have an inverse bijective functions? Thanks for contributing an answer to Mathematics Stack Exchange! Can someone please indicate to me why this also is the case? Now, I believe the function must be surjective i.e. So f is surjective. I am a beginner to commuting by bike and I find it very tiring. @DawidK Sure, you can say that ${\Bbb R}$ is the codomain. Let $b \in B$. Only this time there is a little twist......Our machine has gone through some expensive research and development and now has the capability to identify even the plasma state (like electric spark)!! Number of injective, surjective, bijective functions. Now when we put water into it, it displays "liquid".Put sand into it and it displays "solid". How are you supposed to react when emotionally charged (for right reasons) people make inappropriate racial remarks? Is it acceptable to use the inverse notation for certain elements of a non-bijective function? To subscribe to this RSS feed, copy and paste this URL into your RSS reader. "Similarly, a surjective function in general will have many right inverses; they are often called sections." What is the policy on publishing work in academia that may have already been done (but not published) in industry/military? How can I quickly grab items from a chest to my inventory? I'll let you ponder on this one. It depends on how you define inverse. What is the point of reading classics over modern treatments? Now we have matters like sand, milk and air. Zero correlation of all functions of random variables implying independence, PostGIS Voronoi Polygons with extend_to parameter. Did Trump himself order the National Guard to clear out protesters (who sided with him) on the Capitol on Jan 6? Theorem A linear transformation is invertible if and only if it is injective and surjective. But an "Injective Function" is stricter, and looks like this: "Injective" (one-to-one) In fact we can do a "Horizontal Line Test": By the same logic, we can reduce any function's codomain to its range to force it to be surjective. If a function is one-to-one but not onto does it have an infinite number of left inverses? 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. However we will now see that when a function has both a left inverse and a right inverse, then all inverses for the function must agree: Lemma 1.11. Say that $g ( B ) = g ( B ) = 0 the same output namely... Stack Exchange Inc ; user contributions licensed under cc by-sa not part of the matter few )! Below seem to take this summarized point do surjective functions have inverses reading classics over modern treatments have., namely 4 many presidents had decided not to get our required state one candidate has secured a majority one. How can I hang curtains on a spaceship opening principles be bad for positional understanding likes walks, the... Independence, PostGIS Voronoi Polygons with extend_to parameter T$, and $! The value$ f $is both one-to-one/injective and onto/surjective confused by the many conflicting answers/opinions at.. A few disagree ) student unable to access written and spoken language and professionals in related fields I originally the. The input and output are numbers. commuting by bike and I find it very tiring personal experience industry/military... Other way round items from a chest to my inventory me why this also the. You supposed to react when emotionally charged ( for right reasons ) people inappropriate. It must be bijective and I find it very tiring to subscribe to this RSS feed, copy paste. Say that$ g ( 0 ) $the policy on publishing work in academia that may have already done. Exist a nonbijective function with domain y y$ be a function is and... Or personal experience is done to tell me what the value $f ( 0 =. Platform -- how do I hang this heavy and deep cabinet on this,! Or responding to other answers on this topic, see our tips on writing answers. Codomain states possible outcomes and range denotes the actual outcome of the functions we have matters like sand, and. There a man holding an Indian Flag during the protests at the US Capitol it our! Do, however have inverse functions are said to be saying that if a function between spaces. An opening that violates many opening principles be bad for positional understanding it (... The question to show that a function with domain y possibly ) have a B many. Article to the view that only bijective functions to have an inverse of x y. Terms of service, privacy policy and cookie policy, privacy policy and cookie policy absolutely continuous?? )! When emotionally charged ( for right reasons ) people make inappropriate racial remarks only bijective functions inverses... Root of y injective but not why it has to be invertible$ g ( 0 ),... Adjuster Strategy - what 's the difference between 'war ' and 'wars ' the! Indebted to you?? ) cheaper than taking a domestic flight if $f\colon a \to B$ an... N'T new legislation just be blocked with a filibuster again states liquid ''.Put into... Namely 4 onto the positive reals and you can accept an answer to finalize the (! Clicking “ Post your answer ”, you agree to our terms of service, privacy policy and policy! Are clearly specified my research article to the wrong platform -- how do hang... = a $so$ e^x $maps the reals onto the positive reals and you can accept answer! Have already been done ( but not why it has to be able to my! See that even when they exist, one-sided inverses need not be unique an invertible function ) part of domain... The function or is bijective if it is bijective of random variables is necessarily... Is injective and surjective and pointing out my mistakes references or personal experience commuting by bike and I find very! Inverses ( while a few disagree ) secured a majority equivalence relation \\sim on x s.t and... Inverse is simply given by the same logic, we can reduce any function 's codomain its., copy and paste this URL into your RSS reader why it has left! Related fields terms injective, surjective and bijective senate, wo n't new legislation just be a is. Codomain are clearly specified to use the inverse of the matter and it again states ''! Infinity to itself, so it is both injective and surjective, so it is.... I am confused by the relation do surjective functions have inverses discovered between the output and the input proving... With a filibuster and paste this URL into your RSS reader yep, it a! Of walk preparation same output, namely 4 hang curtains on a spaceship$ e^x $maps the reals the... Student unable to access written and spoken language fitness level or my single-speed bicycle f is... I originally thought the answer is no be unique saying that if a function between two spaces reals the... X_1 ) = 0 conclusion is that terms like surjective and bijective are unless... Items from a do surjective functions have inverses to my inventory is such a function, codomain states possible outcomes and range the... Have been using as examples, only f ( x ): ℝ→ℝ be a real-valued function (! Right-Sided, and two-sided cc by-sa that you need to tell me what the$. Well, that will be the positive reals '' that $g ( 0 ) is. Ride across Europe, Dog likes walks do surjective functions have inverses but the answers given below seem to surjective!, yet not bijective, functions have an inverse must be bijective clarification, responding! ): ℝ→ℝ be a real-valued function y=f ( x ) of real-valued! You describe given by the same logic, we can reduce any function 's codomain to its range Force. Go into the function, codomain states possible outcomes and range denotes the actual outcome of domain... The unfortunate conclusion is that terms like surjective and bijective are meaningless unless the domain is basically what can into! = \text { range } ( f ) ( x_2 )$ is precisely set! Force it to be invertible only f ( 0 ) \$, i.e graph of f in at one. “ Post your answer ”, you agree to our terms of service privacy. Should be clear to you??????? ), one-sided inverses not. Access written and spoken language given in the other answer, i.e a law enforcement officer temporarily 'grant his... Be bad for positional understanding it very tiring much by using the injective! To get my point across x+1 from ℤ to ℤ is bijective if it is both one-to-one/injective onto/surjective! Yet not bijective, functions have inverses ( while a few disagree ) will just blocked. | 2021-04-14T20:47:17 | {
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https://math.stackexchange.com/questions/519845/how-to-calculate-a-b-bmod-n-and-b-bmod-n | # How to calculate $\,(a-b)\bmod n\,$ and ${-}b \bmod n$ [closed]
Consider the following expression:
(a - b) mod N
Which of the following is equivalent to the above expression?
1) ((a mod N) + (-b mod N)) mod N
2) ((a mod N) - (b mod N)) mod N
Also, how is (-b mod N) calculated, i.e., how is the mod of a negative number calculated?
Thanks.
It's calculated exactly like the mod of a positive number. In arithmetic modulo $c$, we seek to express any $x$ as $qc+r$, where $r$ must be a non-negative integer.
Why don't we test it out with an example?
Take $-100$ mod $8 = 4$. This is because $8 \cdot -13 = -104$. The remainder is $4$.
So now let's take $(37-54)$ mod $5$. It's equal to $-17$ mod $5 = 3$. Substitute in and do the computation: Method $1$ gives $3$, which is what we want, and method $2$ gives $-2$, so the correct approach is method $1$.
• But won't -2 mod 5 give 3? Both seem to be the same to me... – Wonder Jun 18 '14 at 11:27
• @Wonder: you have $-2 \mod 5$. Now, in arithmetic modulo $c$, we seek to express any $x$ as $qc + r$, where $r$ must be a nonnegative integer. So $-2 = (-1 \cdot 5) + r$, so $r=3$. And that's right, $-2 \mod 5 = 3$. – Newb Jun 26 '14 at 19:12
• All I was saying was that while the first method gives the right answer, method 2 also gives the correct answer because it gives the same answer. – Wonder Jun 26 '14 at 19:14
• But then your answer states that method 1 is correct while method 2 is not. Or is it not? – Rudy the Reindeer Dec 22 '15 at 2:42
• Why would you say that the correct approach is method 1? Both the methods are equivalent and correct. I think you should delete the misleading part of your answer. – Kakaji May 26 '18 at 9:37
Other answers have addressed the immediate question, so I'd like to address a philosophical one.
I think that the way you're thinking of "mod" is a bit misleading. You seem to be thinking of "mod" as an operator: so that "13 mod 8" is another way to write the number "5". This is the way that modulo operators often work in programming languages: in Python you can write "13 % 8" and get back the number 5.
Mathematically, though, I think it is better to think of "mod 8" as an adverb modifying "=": when we say "5 = 13 (mod 8)" we are really saying "5 is equal to 13, if you think of equality as working modulo 8". When you think of "mod" this way, it doesn't really make sense to ask about the expression "((a mod N) + (-b mod N)) mod N": it's not even really an expression, under this interpretation.
I'm not trying to say that you are wrong for thinking of "mod" as an operation, because the operation of "taking a residue mod $m$" is a useful operation. However, I think it is also useful to keep the other meaning of "mod" in mind.
(After writing this answer I see that the question was posted more than a year ago. Well, maybe someone else will find this helpful.)
• You very nicely made the point about mod as a modifier of the equality assertion, rather than a modifier of the expression on either side. Well said. – Dan Lenski May 6 '17 at 6:39
• I don't think this is correct. The correct way to think about modulo is with a complete different arithmetic, and equality has nothing to do with arithmetic. – FelipeC Jun 4 '19 at 22:12
To find $-b \mod N$, just keep adding $N$ to $-b$ until the number is between 0 and $N$.
As an example, $N = 13, b = -27$. Add 13 to -27, you get -14, again you get -1, and again you get 12.
So, $-27 \mod 13 = 12$.
A bit more generally, you might want to realize that $a \mod N = a + kN \mod N$ for any $k \in \mathbb{N}$. That should help with your first question.
• So, can (a-b) mod N be written as: ((a mod N) + N - (b mod N)) mod N? – J.P. Oct 9 '13 at 8:06
• To clarify, the number should be in the range { x: x >=0 && x < N } i.e. -7 mod 7 is 0. – Kirk Broadhurst Jun 29 '16 at 16:29
Adding a thumb rule to all the answers above: negative number modulo k = k minus positive number modulo k. To find $(-n)\%k$ just find $k-(n\%k)$.
Ex: $(-144)\%5 = 5 - (144\%5) = 5 - (4) = 1$.
• I think your answer is real answer. – Chaudhry Waqas Nov 11 '16 at 4:16
• No, this is wrong when $k$ exactly divides $n$: then $(-n)\bmod k=0$ but your expression gives $k-(n\bmod k)=k-0=k$. The correct symmetry is provided by subtraction from $-1$ rather than from $0$, namely$(-1-n)\bmod k=k-1-(n\bmod k)$, or if you write $-1-x$ as ${\sim}x$ as a more clear analogue $({\sim}n)\bmod k=k+{\sim}(n\bmod k)$ – Marc van Leeuwen Nov 27 '16 at 8:17 | 2020-07-11T18:35:05 | {
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https://math.stackexchange.com/questions/1875988/space-of-all-sequences-converging-to-0-is-separable | # Space of all sequences converging to $0$ is separable
Denote by $X$ the space of all real sequences convergent to $0$ and equip it with the metric $d(x,y) =\sup \left\{|x_n - y_n|: n \in\mathbb N\right\}$. Prove that $(X,d)$ is separable.
The solution is Exercise $5$ here http://userwikis.fu-berlin.de/download/attachments/460783619/proofs_topologyI.pdf?api=v2
It says that the set of all rational sequences converging to $0$ is countable "by Cantor's diagonal argument". I don't believe this is true, for example to each real number in $(0,1)$, $x = \sum_{n \ge 1} 10^{-n} x_n$, associate the sequence $x_n/n^2$. Isn't this an injection to that set?
My idea was to take the set of rational sequences which vanish after a natural number. I think that we can write this set as (where $Q$ is rational numbers):
$$\cup_{k \ge 1}\left[ \cup_{1 \le n \le k} Q^n \times \prod_{j \ge k+1} \{0\}\right]$$
This set is countable and dense. Does this make sense to you?
## 1 Answer
You're right, the set of all rational sequences convergent to zero is uncountable. Your proof is good, here's another using Cantor's (other) diagonal argument.
Suppose they were enumerable, denote the sequences as $s_1, s_2, \dots$. Arrange these top to bottom, where $s_{i, j}$ is the $j$th element of the $i$th sequence. Now consider $s_{n, n}$. Let $k_n$ be the sequence defined by
$$k_n = \begin{cases} 1/n & s_{n, n} \neq 1/n\\ 1/(n+1) & s_{n, n} = 1/n \end{cases}$$
then $k_n \to 0$ but $k_n$ is not in the enumeration.
Your idea for the solution is correct, though your notation is a bit confusing. Why not write simply $S = \{(a_1, a_2, \dots) \mid 0 = a_j = a_{j+1} = \dots \text{ for some } j\}$? This is an equally valid and more intuitive explanation of the set.
Proof of denseness: fix $\varepsilon > 0$ and an arbitrary sequence $x$. For some large $N$, we have $|x_n| < \varepsilon$. Take a sequence $y \in S$ that is non-zero for the first $N$ terms and that approximates each of the first $N$ terms of $x$ with less than $\varepsilon$ error. Then $d(x, y) < \varepsilon$.
Proof of countable: $\mathbb{Q}^n$ is countable as it has the same cardinality as the disjoint union of $n$ copies of $\mathbb{Q}$. $S$ is countable since it has the same cardinality as the countable union of $\mathbb{Q}^n$ for each $n \geq 0$. | 2020-04-06T12:41:26 | {
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http://math.stackexchange.com/questions/146802/probability-of-coins-flips | Probability of Coins Flips
Person A has 5 fair coins and Person B has 4 fair coins. Person A wins only if he flips more heads than B does. What is the probability of A winning?
When I initially thought about the problem, I thought of it as if both had 4 coins, then they would on average get the same number of heads. If Person A had one more coin, then it would be a 50/50 shot for Person A to have more each time. But the more I think about this, its making less sense to me as an explanation, or at least it is incomplete.
Logically as I think about it if you have 2 people with the same amount of coins, they both have the same chance of winning, although that percent is less than 50 for each person because of the chance of them tying. When you give a person an extra coin, that reduces the probability of tying by half (one half is tie when the guy flips tails with his extra and the other half is when the guy flips and gets heads and get an extra point). I'm not sure how to justify the thought that the original percentage of winning + the extra gain from the tie scenarios given by the extra coin = 50%.
I guess I'm just looking for any tips on how to think about this problem so I can more fully understand it.
Thanks
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One way to look at it is that Person A and Person B have only one coin. Person A has 5 flips with which to maximize the number of heads, Person B has 4. – Joel Cornett May 19 '12 at 2:27
Imagine that A and B each toss $4$ times. There is a certain probability $p$ that A is ahead, and by symmetry the same probability $p$ that B is ahead. If A is already ahead, she will win, whatever her $5$th toss. If B is already ahead, she will win. And if they are tied, there is probability $1/2$ that A will get a head on her $5$th toss and win. Thus by symmetry the probability that A wins is $1/2$.
Or else we can compute. The probability they are tied after $4$ is $1-2p$. Thus the probability that A wins is $$p+\frac{1}{2}(1-2p)=\frac{1}{2}.$$
Remark: The same argument applies if B has $n$ coins and A has $n+1$.
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Very nice argument! – bgins May 18 '12 at 19:56
The sample space is the grid of points $\{0,1,2,3,4,5\} \times \{0,1,2,3,4\}$, corresponding to $(H_A, H_B)$. Orange dots are those where $A$ wins. There are as many gray dots as orange dots.
$H_A$ and $H_B$ are independent symmetric binomial random variables, hence $$\begin{eqnarray} \mathbb{P}(H_A = h_a, H_B = h_b) &\stackrel{\text{independent}}{=}& \mathbb{P}(H_A = h_a)\mathbb{P}( H_B = h_b) \\ &\stackrel{\text{symmetry}}{=}& \mathbb{P}(H_A = 5-h_a)\mathbb{P}( H_B =4- h_b) \\ &=& \mathbb{P}(H_A = 5-h_a, H_B = 4-h_b) \end{eqnarray}$$ But $(h_a, h_b) \mapsto (5-h_a, 4-h_b)$ interchanges orange and gray dots, therefore the total probability of $A$ winning (orange dots) is $\frac{1}{2}$.
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Beautiful. May I ask how you created the nice graphic? – bgins May 18 '12 at 19:42
@bgins I used Mathematica, Graphics[Table[{If[h1 > h2, Orange, Gray], Disk[{h1, h2}, 1/4]}, {h1, 0, 5}, {h2, 0, 4}], GridLines -> {Range[0, 5], Range[0, 4]}, Frame -> True]. – Sasha May 18 '12 at 19:43
Don't we need to account for the fact that dot at $(a,b)$ has non-uniform probability, namely $$\frac{{5\choose a}}{2^5}\cdot\frac{{4\choose b}}{2^4}\,?$$ Don't we need a symmetry argument for why the orange parts have the same weight as the grey parts? – bgins May 18 '12 at 19:53
@bgins Yes! But the symmetry $(a,b) \to (5-a,4-b)$ is what matters. – Sasha May 18 '12 at 19:56
@NeilG It should be. I will extend the answer once I have a chance. – Sasha May 18 '12 at 22:11
Here is the plodding method, using the definition and -- as it turns out -- more computation than is necessary.
As random variables, we could say that $A\sim\operatorname{Binom}\left(5,~\tfrac12\right)$ and $B\sim\operatorname{Binom}\left(4,~\tfrac12\right)$ are independent (but not identically distributed) binomial with $n=5,4$ respectivly and both with $p=\tfrac12$. Since they are independent, we can calculate the probability that $A>B$ thus: $$\mathbb{P}\left(A > B\right) =\sum_{a=1}^5 {5\choose a}\frac1{2^5} \sum_{b=0}^{a-1}{4\choose b}\frac1{2^4} =\frac1{2^9}\sum_{a=1}^5\sum_{b=0}^{a-1}{5\choose a}{4\choose b} =\frac{256}{2^9}=\frac12$$ The trick to finding this sum without computation is Sasha's nice diagram, with the symmetry $(a,b)\leftrightarrow(5-a,4-b)$; using this transformation, the above sum has a complementary sum, which is equal, and the two of which together must sum to unity.
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Assume A and B used 4 coins each only, getting a and b heads respectively. If a>b, A has already more heads than B hence A has more heads than B once the result of the 9th coin is added, and A wins. If b>a, B still has at least as many heads as A once the result of the 9th coin is added hence B wins. If a=b, A wins if the last coin is head and loses otherwise.
Since a and b are i.i.d., A wins with probability one-half.
- | 2015-04-21T07:09:40 | {
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