text stringlengths 1 2.12k | source dict |
|---|---|
# Thread: Show that p/q is the ...th term of the series...
1. ## Show that p/q is the ...th term of the series...
Positive rational numbers may be arranged in the form of a simple series: $\frac{1}{1}, \frac{2}{1}, \frac{1}{2}, \frac{3}{1}, \frac{2}{2}, \frac{1}{3}, \frac{4}{1}, \frac{3}{2}, \frac{2}{3}, \frac{1}{4}, ...$ Show that $\frac{p}{q}$ is the $[\frac{1}{2}(p+q-1)(p+q-2)+q]^t^h$ term of the series.
I have already considered arithmetic sequences for p and q but I don't think it works. How should I approach the problem?
Thanks.
2. This expression of p and q is called Cantor pairing function. It maps pair of positive integers onto positive integers in a one-to-one manner. It is useful because it demonstrates that there are as many positive integers as pairs of positive integers.
To understand why it works, consider this picture. The value of the pairing function is shown near each point. Arrow show the direction in which the value of the function increases.
Note that p + q is constant on the diagonals. Let's look at the blue diagonal, where p + q = 5. How many points come before, i.e., how many points are in the green triangle? It's 1 + 2 + 3, or, in terms of p and q, it's 1 + 2 + ... + (p + q) - 2 = 1/2 [(p + q - 2) * (p + q - 1)] (as the sum of an arithmetic progression). If we add the vertical coordinate q, we get the value of the function on the blue diagonal. | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.983085090202311,
"lm_q1q2_score": 0.8487874239544277,
"lm_q2_score": 0.8633916152464016,
"openwebmath_perplexity": 244.08047184530645,
"openwebmath_score": 0.8771878480911255,
"tags": null,
"url": "http://mathhelpforum.com/pre-calculus/170909-show-p-q-th-term-series.html"
} |
3. Thank you for your solution. I didn't expect an advanced topic like the Cantor pairing function to be required for this problem. I was told by a teaching assistant that this problem can be solved by mathematical induction. But he said it may not work because after p or q reaches the end of its arithmetic progression, it "jumps" to a different number. In the inductive step, we assume the statement is true for some n, then we try to show that (p-1)/(q+1) is the (n+1)th term of the series. This would result in the wrong numbers in some cases. Can induction still be used anyway? In a different way maybe?
4. @ akyng
could you post the exact problem. or a copy of it. there is some differences from what you posted and emakarovs expression.
akyng
$[\frac{1}{2}(p+q-1)(p+q-2)+q]^t^h$
emakarov
$1 + 2 + ... + (p + q) - 2 = \frac{1}{2} [(p + q - 2)(p + q - 1)]$
notice you have 1/2 fraction in different grouping, there is +q at the end of your expression and emakarov has a (p+q)-2 nth term. that would give you 0,1,2,3,4 etc, for the sequence.
5. I think our expressions are exactly the same. The green triangle has $1 + 2 + ... + ((p + q) - 2) = \frac{1}{2} [(p + q - 2)(p + q - 1)]$ points. To this one adds q: the position of the current point on the blue diagonal.
6. ah I see
so what would you put for first element? ((2+1)-2)?
7. The first element of the series is 1/1 or 0.5*(1+1-2)(1+1-1)+1. Does someone know something to the question I asked in post #3, about whether mathematical induction can be used for this problem?
8. Does someone know something to the question I asked in post #3, about whether mathematical induction can be used for this problem?
OK, first a couple of definitions.
Let $f(p,q)=(p+q-1)(p+q-2)/2+q$. Also, let $p_n,q_n$ be defined by mutual recursion as follows: $p_1=q_1=1$,
$p_{n+1}=
\begin{cases}
p_n-1 & p_n>1\\
q_n+1 & \mbox{otherwise}
\end{cases}$
and $q_{n+1}=\begin{cases}
q_n+1 & p_n>1\\
1 & \mbox{otherwise}
\end{cases}$
. | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.983085090202311,
"lm_q1q2_score": 0.8487874239544277,
"lm_q2_score": 0.8633916152464016,
"openwebmath_perplexity": 244.08047184530645,
"openwebmath_score": 0.8771878480911255,
"tags": null,
"url": "http://mathhelpforum.com/pre-calculus/170909-show-p-q-th-term-series.html"
} |
The original problem:
Show that $\frac{p}{q}$ is the $[\frac{1}{2}(p+q-1)(p+q-2)+q]^t^h$ term of the series.
asks to prove that for all $p,q$ it is the case that $p=p_{f(p,q)}$, $q=q_{f(p,q)}$. This statement can be broken into two parts.
Proposition 1. For all $p,q$ there exists an $n$ such that $p=p_n$ and $q=q_n$.
Proposition 2. $f(p_n,q_n)=n$ for all $n\ge 1$.
This is proved by induction on $n$. In the induction step, one has to consider two cases depending on whether $p_n>1$.
Then, given $p$ and $q$, we use Proposition 1 to find an $n$ such that $p=p_n$ and $q=q_n$. By Proposition 2, $n=f(p,q)$, which proves the required claim.
9. this one is interesting. I have been doing some of the induction proofs in pre-calculus book but not with 2 variables. so I take a crack at it for my own personal fun.
we start with the chart emakarov posted. we pair the outside numbers as $\frac{p}{q}$. now set $(p+q)=n$. call these elements $a_n$ in a statement involving n. we can fallow the little place numbers to build our $a_n$ values.
now if you run $a_n$ through $((p+q)-2)$ you get a repeat value for each diagonal row like in the chart. if you run $a_n$ through $\frac{1}{2} [(p + q - 2)(p + q - 1)]$ you get the sum of the row values. so I assume that is what we want. so therefore $S_n$ is the nth partial sum of the row values.
$a_n= ((p+q)-2):a_1=0,a_{2-3}=1,a_{4-6}=2,a_{7-10}=3,a_{11-15}=4...$
$S_n= \frac{1}{2} [(p + q - 2)(p + q - 1)]:S_1=0,S_{2-3}=1,S_{4-6}=3,S_{7-10}=6,S_{11-15}=10...$
$a_1=0$ is equal to $S_1=0$ so we have proof that $a_1$ is true. the truth of $a_n$ implies the truth of $a_{n+1}$. we assume that $a_n$ is true and build the $a_{n+1}$ statement.
\begin{aligned}
a_{n+1} &=(p+q)-2) \\
&= (((p+q)+1)-2) \\
&= ((p+q)-1).
\end{aligned}
now we add $a_{n+1}$ to $S_n$. | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.983085090202311,
"lm_q1q2_score": 0.8487874239544277,
"lm_q2_score": 0.8633916152464016,
"openwebmath_perplexity": 244.08047184530645,
"openwebmath_score": 0.8771878480911255,
"tags": null,
"url": "http://mathhelpforum.com/pre-calculus/170909-show-p-q-th-term-series.html"
} |
now we add $a_{n+1}$ to $S_n$.
\begin{aligned}
S_{n+1}
&= \frac{1}{2} [(p + q - 2)(p + q - 1)]+(p+q-1) \\
&= \frac{1}{2} [(p + q - 2)(p + q - 1)+2(p+q-1)] \\
&= \frac{1}{2} [(p + q - 2)(p + q - 1) + 2p + 2q - 2] \\
&= \frac{1}{2} [p^2 + 2pq + q^2 - p - q] \\
&= \frac{1}{2} [(p + q)^2 - p - q] \\
&= \frac{1}{2} [(p + q)^2 - (p + q)] \\
\end{aligned} | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.983085090202311,
"lm_q1q2_score": 0.8487874239544277,
"lm_q2_score": 0.8633916152464016,
"openwebmath_perplexity": 244.08047184530645,
"openwebmath_score": 0.8771878480911255,
"tags": null,
"url": "http://mathhelpforum.com/pre-calculus/170909-show-p-q-th-term-series.html"
} |
# Finding Matrix Representation
Problem: Let T: $\mathbb{R}^3\rightarrow\mathbb{R}^3$ be a linear map given by
$$T\left[ \begin{matrix} x\\y\\ z\end{matrix} \right]= \left[ \begin{matrix} 3x-y\\z-x\\z-y\\\end{matrix} \right]$$
1. Find the Matrix representation of T with respect to the canonical basis of $\mathbb{R}^3$, and call it A.
I am not sure how this works. So the cananical basis of $\mathbb{R}^3$ is ${(1,0,0),(0,1,0),(0,0,1)}$ But I am unsure how to get a matrix represenation from a linear operator. Any help is appreciated.
-
Remember how a matrix is defined. The $j$th column tells you where the $j$th basis vector goes. So the first column should be the image of $(1,0,0)$ under $T$, which is $(3,-1,0)$. You can continue like this – Daniel Freedman May 14 '12 at 22:26
Ok great! Thank you guys so much it is so much clearer now. – Mathstudent May 14 '12 at 22:37
I will write the linear map $T$ as $$T(x,y,z)=(3x-y,z-x,z-y).$$ Then we have $$T(1,0,0)=(3,-1,0),$$ $$T(0,1,0)=(-1,0,-1),$$ $$T(0,0,1)=(0,1,1).$$ Then the matrix representation of $T$ is given by $$\left[ \begin{array}{ccc} 3 & -1 & 0 \\ -1 & 0 & 1 \\ 0 & -1 & 1 \\ \end{array} \right].$$ The way to get it is: the vector $T(1,0,0)=(3,-1,0)$ becomes the first column, the vector $T(0,1,0)=(-1,0,-1)$ becomes the second column, and so forth. My linear algebra teacher always says, "Put it to be vertical, put it to be vertical,...".
The advantage of the matrix representation is that; for example if I want to find $T(1,-2,0)$, then I can do it by $$\left[ \begin{array}{ccc} 3 & -1 & 0 \\ -1 & 0 & 1 \\ 0 & -1 & 1 \\ \end{array} \right]\left[ \begin{array}{c} 1 \\ -2 \\ 0 \\ \end{array} \right]=\left[ \begin{array}{c} 5 \\ -1 \\ 2 \\ \end{array} \right],$$ that is, $T(1,-2,0)=(5,-1,2)$.
-
wait shouldn't the matrix represenation be $$\left[ \begin{array}{ccc} 3 & -1 & 0 \\ -1 & 0 & -1 \\ 0 & 1 & 1 \\ \end{array} \right].$$ – Mathstudent May 14 '12 at 22:44 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9830850877244272,
"lm_q1q2_score": 0.8487874218150436,
"lm_q2_score": 0.8633916152464016,
"openwebmath_perplexity": 604.6507082085109,
"openwebmath_score": 0.9895403385162354,
"tags": null,
"url": "http://math.stackexchange.com/questions/145242/finding-matrix-representation/145245"
} |
Let $e_1=(1,0,0)$, $e_2=(0,1,0)$, and $e_3=(0,0,1)$.
From the definition of $T$: $$T(e_1) =\left[\matrix{3\cr -1\cr 0 }\right],\quad T(e_2) =\left[\matrix{-1\cr 0\cr -1 }\right],\quad T(e_3) =\left[\matrix{0\cr 1\cr 1 }\right].$$ For an arbitrary vector $v=\left[\matrix{a\cr b\cr c }\right] =a e_1+b e_2+c e_3$, we have using the linearity of $T$ \eqalign{ T(v) &=T(a e_1+b e_2+c e_3)\cr &=a T(e_1)+b T(e_2)+c T(e_3)\cr &=a \left[\matrix{3\cr -1\cr 0 }\right]+ b\left[\matrix{-1\cr 0\cr -1 }\right]+ c\left[\matrix{0\cr 1\cr 1 }\right]\cr &= \left[\matrix{3a-1\cdot b+0\cdot c\cr -1\cdot a+0\cdot b+1\cdot c\cr 0\cdot a+1\cdot -b+1\cdot c }\right]\cr &= \left[\matrix{3&-1&0 \cr -1&0& 1\cr 0&-1& 1}\right] \left[\matrix{a\cr b\cr c}\right].\cr } So the matrix representation of $T$ is $\left[\matrix{3&-1&0 \cr -1&0&1\cr 0&-1&1}\right]$.
Note that the $i^{\rm th}$ column of the matrix representing $T$ is the vector $T(e_i)$. So, we can get the desired!
- | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9830850877244272,
"lm_q1q2_score": 0.8487874218150436,
"lm_q2_score": 0.8633916152464016,
"openwebmath_perplexity": 604.6507082085109,
"openwebmath_score": 0.9895403385162354,
"tags": null,
"url": "http://math.stackexchange.com/questions/145242/finding-matrix-representation/145245"
} |
# Recursive version of DFT as presented in Cooley-Tukey paper
The seminal paper of Cooley and Tukey provides an iterative method for computing the DFT for a sequence of length $N$. Specifically, they mention a method which utilizes the fact that $N$ can be written as $N = r_1 r_2$. The two intermediate results during computations themselves are DFT computations, therefore, their computation in turn can be recursively estimated.
Is there a reference/method of recursive formulation for computing DFT using the $N=r_1 r_2$ factorization ? The versions of FFT I come across use factorizations where $r_1 = \frac{N}{2}$ and $r_2=2$. I would like to know if there are recursive formulations for general factorization of $N$.
• @MattL. The versions of FFT I come across use factorizations where $r_1 = \frac{N}{2}$ and $r_2=2$. I would like to know if there are recursive formulations for general factorization of $N$. Sep 13, 2014 at 8:54
• The Cooley-Tukey algorithm is not restricted to dividing the transforms into 2 DFTs of length $N/2$. This is just the most popular form. It works for any composite size $N=r_1r_2$. A lot of methods with different factorizations of $N$ are implemented in the FFTW library. You can find more details here. Sep 13, 2014 at 9:11
• it might be interesting if someone here spelled out how to do it in the case where $r_1 \ne \frac{N}{2}$. Sep 13, 2014 at 16:11
• @robertbristow-johnson I finally found the time to do so. Oct 23, 2014 at 11:48
• @MattL., an up vote for that. Oct 26, 2014 at 11:35
The Cooley Tukey method allows general factorizations $N=N_1N_2$. The easiest way to see this is to use index mapping. Consider the length-$N$ DFT of a sequence $x[n]$:
$$X[k]=\sum_{n=0}^{N-1}x[n]W_N^{nk}\tag{1}$$
with $W_N=e^{-j2\pi/N}$. Let's assume that $N$ can be factored as $N=N_1N_2$. Now use the following index mapping | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9830850892111574,
"lm_q1q2_score": 0.8487874179152389,
"lm_q2_score": 0.8633916099737806,
"openwebmath_perplexity": 950.9250176996937,
"openwebmath_score": 0.8022952675819397,
"tags": null,
"url": "https://dsp.stackexchange.com/questions/18201/recursive-version-of-dft-as-presented-in-cooley-tukey-paper/18798"
} |
$$n=n_1N_2+n_2,\quad n_1=0,1,\ldots, N_1-1,\quad n_2=0,1,\ldots,N_2-1\\ k=k_1+N_1k_2,\quad k_1=0,1,\ldots, N_1-1,\quad k_2=0,1,\ldots,N_2-1\tag{2}$$
Plugging (2) into (1) gives
$$X[k_1+N_1k_2]=\sum_{n_2=0}^{N_2-1}\sum_{n_1=0}^{N_1-1}x[n_1N_2+n_2]W_N^{(n_1N_2+n_2)(k_1+k_2N_1)}\tag{3}$$
The complex exponentials in (3) can be simplified as follows:
$$W_N^{(n_1N_2+n_2)(k_1+k_2N_1)}=W_N^{N_2n_1k_1}W_N^{N_1N_2n_1k_2}W_N^{n_2k_1}W_N^{N_1n_2k_2}=\\= W_{N_1}^{n_1k_1}\cdot 1\cdot W_{N}^{n_2k_1}W_{N_2}^{n_2k_2}\tag{4}$$
because $W_N^{N_1}=W_{N_2}$, $W_N^{N_2}=W_{N_1}$, and $W_N^{N_1N_2}=W_N^N=1$. Plugging (4) back into (3) and rearranging terms gives
$$X[k_1+N_1k_2]=\sum_{n_2=0}^{N_2-1}\left[W_N^{n_2k_1}\left(\sum_{n_1=0}^{N_1-1}x[n_1N_2+n_2]W_{N_1}^{n_1k_1}\right)\right]W_{N_2}^{n_2k_2}\tag{5}$$
Equation (5) can be interpreted as follows: the result can be computed by computing $N_2$ length-$N_1$ DFTs of subsampled versions of the input sequence $x[n]$ (this is the inner sum in (5)), then multiplying the result by twiddle factors $W_N^{n_2k_1}$, and finally computing $N_1$ length-$N_2$ DFTs (the outer sum in (5)). This procedure is most easily visualized using matrices:
1. read the data row-wise into a $N_1\times N_2$ matrix
2. compute DFTs over all $N_2$ columns (this can be done in-place)
3. multiply the result element-wise with twiddle factors
4. compute DFTs over all $N_1$ rows (in-place)
5. read out the data column-wise to get the final length-$N$ result
This little octave/Matlab script illustrates the procedure:
N = 1536; % non-power of 2 FFT length used in LTE
N1 = 512; N2 = 3; % 512 x 3 = 1536
x = randn(N,1); % some input signal
% generate N1 x N2 matrix of input data, reading them row wise
X = reshape(x,[N2,N1]).';
% DFT over each column
X = fft(X);
% Multiplication with twiddle factors
X = X.*exp(-1i*2*pi/N*(0:N1-1)'*(0:N2-1));
% DFT over each row
X = (fft(X.')).';
X = X(:); | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9830850892111574,
"lm_q1q2_score": 0.8487874179152389,
"lm_q2_score": 0.8633916099737806,
"openwebmath_perplexity": 950.9250176996937,
"openwebmath_score": 0.8022952675819397,
"tags": null,
"url": "https://dsp.stackexchange.com/questions/18201/recursive-version-of-dft-as-presented-in-cooley-tukey-paper/18798"
} |
% DFT over each row
X = (fft(X.')).';
X = X(:);
Matt's answer is excellent. For anyone wanting to do the same in Python, I thought I'd add the code (with a slight modification to use a sine wave input, which I prefer for debugging):
import numpy as np
N=1536
N1=512
N2=3 # 512 x 3 = 1536
nCycles = 2 # Number of cycles in source frame
time=np.arange(0,N ) # some input signal
x=np.sin(nCycles * time * 2. * np.pi / N)
# generate N1 x N2 matrix of input data, reading them row wise
x=np.reshape(x,(N1,N2))
# DFT over each column
X=np.fft.fft(x,axis=0)
# Element-wise multiplication with twiddle factors
X=np.multiply(X,np.exp(- 1j * 2. * np.pi / N * (np.arange(0,N1).T.reshape(N1,1) @ np.arange(0,N2).reshape(1,N2))))
# DFT over each row
X=np.fft.fft(X,axis=1) | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9830850892111574,
"lm_q1q2_score": 0.8487874179152389,
"lm_q2_score": 0.8633916099737806,
"openwebmath_perplexity": 950.9250176996937,
"openwebmath_score": 0.8022952675819397,
"tags": null,
"url": "https://dsp.stackexchange.com/questions/18201/recursive-version-of-dft-as-presented-in-cooley-tukey-paper/18798"
} |
# Plot a complex set in the complex plane
My lecturer only explained how to plot complex numbers on the complex plane, but he didn't explain how to plot a set of complex numbers.
I did some research online but I didn't find any clear explanation or method. I have an exercise to practice but I don't know how to even start!.
Any help would be really appreciated!
• Are you familiar with Analytic Geometry and the equations of circles, eclipses etc on the plane? – MathematicianByMistake Oct 24 '17 at 15:19
• Yes, how can I use them here? A friend explained me that with the equations(as in a.) I can obtain the radios and centre of a circunference. But I do not know how to do it, and the other cases I do not have any equation – Evoked Oct 24 '17 at 15:24
• Check the answer for a hint and a general approach. Your friend is correct. – MathematicianByMistake Oct 24 '17 at 15:25
HINT
Let's tackle the first one. A similar approach is required for the other two.
For $z=x+iy$, $x,y\in \mathbb{R}$ we get $$|z-1+i|=1\Rightarrow\\|x-1+i(y+1))|=1\Rightarrow\\\sqrt{(x-1)^2+(y+1)^2}=1\Rightarrow\\(x-1)^2+(y+1)^2=1$$
Does this-hopefully it does-remind you a more general equation of a circle?
Let's also look at the third one as well.
We have $$\operatorname{Re}\Big(\frac{z+1}{z-1}\Big)=\Re\Big(\frac{z+1}{z-1}\Big)=\Re\Big[\frac{(z+1)(\bar{z}-1)}{(z-1)(\bar{z}-1)}\Big]=\Re\Big[\frac{(z+1)(\bar{z}-1)}{|z-1|^2}\Big]\gt1$$ Now for $z=x+iy$, $x,y\in \mathbb{R}$ you can substitute on the last relationship and obtain an equation for $x,y$-an inequality to be precise. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9830850842553892,
"lm_q1q2_score": 0.8487874170920936,
"lm_q2_score": 0.8633916134888613,
"openwebmath_perplexity": 261.9563532840228,
"openwebmath_score": 0.46134647727012634,
"tags": null,
"url": "https://math.stackexchange.com/questions/2487744/plot-a-complex-set-in-the-complex-plane/2487761"
} |
• Yes it does! Since in the others there are inequalities, the representation would no longer be a circle? – Evoked Oct 24 '17 at 15:30
• I will help you get an equation for the last one as well, but-and don't take it personally-we can't do your homework for you, there would be no point in that if you don't try yourself. If the answer satisfies you you can upvote/accept.. – MathematicianByMistake Oct 24 '17 at 15:32
• This is not my homework! This is me trying to understand an exercise I searched for in a book! Please don't take in a bad way my words but I'm just trying to understand something, not asking for a particular class free – Evoked Oct 24 '17 at 18:32
• @Evoked No worries! I am only happy that the answer was helpful! Feel free to ask any questions on MSE. It is great for you that you wish to learn more and self-study. :-) – MathematicianByMistake Oct 24 '17 at 18:36 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9830850842553892,
"lm_q1q2_score": 0.8487874170920936,
"lm_q2_score": 0.8633916134888613,
"openwebmath_perplexity": 261.9563532840228,
"openwebmath_score": 0.46134647727012634,
"tags": null,
"url": "https://math.stackexchange.com/questions/2487744/plot-a-complex-set-in-the-complex-plane/2487761"
} |
# Multidimensional Newton Raphson
#### Purplepixie
##### New member
Hello,
I am having difficulty in translating the univariate Newton's approximation {Xn = Xn-1 - [ f(Xn-1) / f'(Xn-1)]} into the multidimensional case. My multidimensional equation system is y = F.x where y and x are (nx1) column vectors and the coefficients matrix F is (nxn), so that (nx1) = (nxn).(nx1) = (nx1).
I have translated the univariate Newton's approximation to the n-variate case as:
x2 = x1 - (F.x1 / F'.x1) <=> x1 - (F.x1).Inv(F'.x1)
But F'.x1 is a nx1 column vector and has no inverse. I then thought that perhaps the x1's cancel out . But if so then we would have x2 = x1 - F.Inv(F') with the last term a nxn matrix, so (nx1) = (nx1) - (nxn), which is not possible.
Your assistance would be greatly appreciated!
(This is my first post incidentally, so pls excuse any breaches of protocol)
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Hello,
I am having difficulty in translating the univariate Newton's approximation {Xn = Xn-1 - [ f(Xn-1) / f'(Xn-1)]} into the multidimensional case. My multidimensional equation system is y = F.x where y and x are (nx1) column vectors and the coefficients matrix F is (nxn), so that (nx1) = (nxn).(nx1) = (nx1).
I have translated the univariate Newton's approximation to the n-variate case as:
x2 = x1 - (F.x1 / F'.x1) <=> x1 - (F.x1).Inv(F'.x1)
But F'.x1 is a nx1 column vector and has no inverse. I then thought that perhaps the x1's cancel out . But if so then we would have x2 = x1 - F.Inv(F') with the last term a nxn matrix, so (nx1) = (nx1) - (nxn), which is not possible.
Your assistance would be greatly appreciated!
(This is my first post incidentally, so pls excuse any breaches of protocol)
Hi Purplepixie, welcome to MHB!
First we have to establish what $F'$ is.
If it is a matrix that does not depend on any variable, then $F'$ is the matrix that contains only zeroes.
Consequently we won't really get any useful result. | {
"domain": "mathhelpboards.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9830850897067342,
"lm_q1q2_score": 0.8487874166153041,
"lm_q2_score": 0.8633916082162403,
"openwebmath_perplexity": 1199.1963195743554,
"openwebmath_score": 0.9063995480537415,
"tags": null,
"url": "https://mathhelpboards.com/threads/multidimensional-newton-raphson.27759/"
} |
Newton's approach aims to solve $f(x)=0$, and makes use of the fact that $f(x) \approx f(x_0) + f'(x)(x-x_0) \implies x_0\approx x - \frac{1}{f'(x)}\cdot f(x)$.
We might try to solve $F(x_1,\ldots,x_n)=(0,\ldots, 0)$ in a similar fashion. In this case $F$ is not a matrix, but a set of $n$ functions. And each function has $n$ parameters.
If we take the derivative, we don't have just 1 derivative, but instead we have the derivatives of $n$ functions with respect to each of the $n$ parameters.
The result is a matrix of $n\times n$ derivatives. Let's call it $DF(x)$ to denote that it's a matrix of functions that depend on $x$.
Now we can write Newton's approximation as $x^{(k+1)} = x^{(k)} - \Big(DF(x^{(k)})\Big)^{-1}\cdot F(x^{(k)})$, where $x^{(k)}$ denote the successive approximations of $x$.
#### Purplepixie
##### New member
Dear Klaas,
Thank you for your clear and succinct explanation. I can now see where I was wrong. So much of mathematics would be simplified if representational systems could be improved!
All the best,
PP | {
"domain": "mathhelpboards.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9830850897067342,
"lm_q1q2_score": 0.8487874166153041,
"lm_q2_score": 0.8633916082162403,
"openwebmath_perplexity": 1199.1963195743554,
"openwebmath_score": 0.9063995480537415,
"tags": null,
"url": "https://mathhelpboards.com/threads/multidimensional-newton-raphson.27759/"
} |
# What are some examples of proof by contrapositive?
Applying the Modus Tollens argument to Fermat's Little Theorem really helped me to understand logical implication. I never knew that FLT was actually a compositality test.
Theorem (FLT): given integers $a>1$ and $n>1$, if $n$ is prime, then $a^n$ is congruent to $a\ (\bmod\ n)$.
By the contrapositive, if $a^n$ is not congruent to $a\ (\bmod\ n)$, then $n$ is not prime. Thus $n$ is composite.
What are some other simple and instructive examples of proof by contrapositive?
-
This question has a nice example; see especially the discussion in my answer. This question and its answer are another, and this, this, and this should also be of interest. These are just a few that I could easily find. – Brian M. Scott May 27 '13 at 4:32
Thanks everyone for all the great examples. Now, I've got plenty to study. It sure would be interesting to compare samples from each area of Mathematics, a "logic parallel". – cyclochaotic May 27 '13 at 15:33
When you want to prove "If $p$ then $q$", and $p$ contains the phrase "$n$ is prime" you should use contrapositive or contradiction to work easily, the canonical example is the following:
Prove for $n>2$,
If $n$ is prime then $n$ is odd.
Here $q$ is the phrase "$n$ is odd". Here $p$ is exactly the phrase "$n$ is prime" and is very difficult to work with it. Because the fact that $n$ is prime means that it is not divisible by other number grater than $1$ and different for $n$, so you must to choose from these $n-2$ true sentences the only one that is useful which is "$n$ is not divisible by 2", but you would not know which is the right choice, unless you read $q$.
But proving contrapositive equivalent form is very easy, and you don't to do any choice.
If $n$ is even then $n$ is not prime.
Which follows from the fact that every even number greater than $2$ is divisible by $2$, hence not prime. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9830850842553892,
"lm_q1q2_score": 0.8487874136364704,
"lm_q2_score": 0.8633916099737807,
"openwebmath_perplexity": 213.21239089928812,
"openwebmath_score": 0.8374519348144531,
"tags": null,
"url": "http://math.stackexchange.com/questions/403327/what-are-some-examples-of-proof-by-contrapositive/403340"
} |
So contrapositive (also contradiction) is used to avoid situations where you have a lot of information and very little of it is actually useful.
-
Look at Richard Hammack's "Book of Proof" for a detailed discussion of proof forms. There are several "proof technique" and such introductory courses (perhaps under discrete mathematics and similar) with lecture notes on the 'net.
- | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9830850842553892,
"lm_q1q2_score": 0.8487874136364704,
"lm_q2_score": 0.8633916099737807,
"openwebmath_perplexity": 213.21239089928812,
"openwebmath_score": 0.8374519348144531,
"tags": null,
"url": "http://math.stackexchange.com/questions/403327/what-are-some-examples-of-proof-by-contrapositive/403340"
} |
# What is $P(A' \cap B')$ and $P(A \cup B \cup C)$?
We are given that $P(A)=.43, P(B)=.44,$ and $P(C)=.19$. Also, $B$ and $C$ are mutually exclusive but $P(A\cap B)=.07$ and $P(A\cap C)= .13$
(a) What is $P(A' \cap B')$?
Note: I know that since $A,B$ are not mutually exclusive, that $P(A \cap B)=P(A) \times P(B)$. Does this same rule apply for the complement? If so, then I can use the rule $P(A)+P(A')=1$.
(b) What is $P(A \cup B \cup C)$?
I know the formula: $P(A\cup B\cup C) = P(A) +P(B)+P(C)-P(A\cap B)-P(A\cap C)- P(B\cap C)+P(A\cap B\cap C)$.
Note: The part I am having trouble on with this formula is finding $P(A\cap B\cap C)$.
Guide:
$$P(A) \times P(B) \approx 0.1892 \ne P(A \cap B)$$
$A$ and $B$ are not independent.
• Note that $$P(A' \cap B') = 1-P(A \cup B)$$
• $(A \cap B \cap C) \subset (B \cap C)$, Hence $P(A \cap B \cap C)=0$.
• Also, check your inclusion-exclusion formula.
• that is De Morgan's law. – Siong Thye Goh Sep 10 '18 at 16:14
• I know that De Morgan's Law states: $(A\cup B)' = A' \cap B'$ and $(A\cap B)' = A' \cup B'$. How does this translate into that probability formula? – rover2 Sep 10 '18 at 16:18
• note: i have fixed my inclusion-exclusion formula. i copied it wrong while typing the solution however it was correct in my notes – rover2 Sep 10 '18 at 16:19
• Since $A' \cap B' = (A \cup B)'$ we have $P(A' \cap B')= P((A \cup B)' )$. We have $P((A \cup B)' ) = 1-P(A \cup B)$. – Siong Thye Goh Sep 10 '18 at 16:20
• got it, thank you – rover2 Sep 10 '18 at 16:22
You know that $B$ and $C$ are mutually exclusive i.e. $A\cap B$ and $C$ are mutually exclusive,
$$P(A\cap B \cap C)=0$$
Also $$P(A\cap B)\neq P(A)\cdot P(B)$$
As $A$ and $B$ are not independent events.
For Part (II),
Using De-Morgans Law,
$$P(A'\cap B')=P((A\cup B)')=1-P(((A\cup B)')')$$
$$P(A'\cap B')=1-P(A\cup B)$$
Also you know that,
$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9830850867332734,
"lm_q1q2_score": 0.8487874123202312,
"lm_q2_score": 0.8633916064586998,
"openwebmath_perplexity": 251.54262855646758,
"openwebmath_score": 0.8707270622253418,
"tags": null,
"url": "https://math.stackexchange.com/questions/2912050/what-is-pa-cap-b-and-pa-cup-b-cup-c"
} |
1. ## probability
There are 5 different colors of balls: white, black, blue, red, green. We randomly pick 6 balls. Each ball has a probability of 0.2 of getting each of the 5 colors. What is the probability that, from the 6 balls picked, there are white balls and black balls?
2. Originally Posted by allrighty
There are 5 different colors of balls: white, black, blue, red, green. We randomly pick 6 balls. Each ball has a probability of 0.2 of getting each of the 5 colors. What is the probability that, from the 6 balls picked, there are white balls and black balls?
find the probabilities for each case and add them up
by (all possible) cases, i mean:
probability of choosing 1 white, 0 black, 1 blue, 1 red, 3 green
probability of choosing 1 white, 0 black, 0 blue, 2 red, 3 green
probability of choosing 1 white, 0 black, 2 blue, 0 red, 3 green
.
.
.
.
or we could do:
1 - probability of choosing no white and no black ball
so all you have to worry about are the number of ways you can choose 6 balls among the three remaining colors. there is a formula for that sort of thing. can't remember right now, i'll have to look it up. but when you do get it, just multiply the answer by 0.2 and that will give you the probability of choosing no white and no black ball
3. Originally Posted by Jhevon
so all you have to worry about are the number of ways you can choose 6 balls among the three remaining colors
Is that right? A white AND a black are required.
I have another method but it's messy.
You could have for example 2W, 1B and 3 others 6!/(3!2!) ways each with probability 0.2^3 0.6^3
or 5W, 1B 6!/5! ways with probability 0.2^6
Having said all that, I'm a bit rubbish at probability so I'll wait and see what people say. | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9830850837598123,
"lm_q1q2_score": 0.8487874062973463,
"lm_q2_score": 0.863391602943619,
"openwebmath_perplexity": 312.2711209301747,
"openwebmath_score": 0.9102963209152222,
"tags": null,
"url": "http://mathhelpforum.com/statistics/26957-probability.html"
} |
Having said all that, I'm a bit rubbish at probability so I'll wait and see what people say.
4. Originally Posted by a tutor
Is that right? A white AND a black are required.
yes, i believe so. what i said was i want to find the probability of having 0 white AND 0 black. and then take 1 minus that probability to find the probability of at least 1 white or 1 black. i think what i described does the trick... but then again, i'm a noob when it comes to probability as well
5. Original question said..
Originally Posted by allrighty
What is the probability that, from the 6 balls picked, there are white balls and black balls?
and you said..
Originally Posted by Jhevon
to find the probability of at least 1 white or 1 black.
6. Originally Posted by a tutor
Original question said..
and you said..
ah yes. my bad. i didn't see the "and." i was under the impression if we have either or we were good...
7. Hello, allrighty!
I think I've solved it . . .
There are 5 different colors of balls: white, black, blue, red, green.
We randomly pick 6 balls.
Each ball has a probability of 0.2 of getting each of the 5 colors.
What is the probability that, from the 6 balls picked, there are white balls and black balls?
The opposite of "some White and some Black" is "no White or no Black".
To get no White, we must pick six balls from the other four colors.
. . Then: . $P(\text{0 White}) \:=\:(0.8)^6$
To get no Black, we must pick six balls from the other four colors.
. . Then: . $P(\text{0 Black}) \:=\:(0.8)^6$
To get no White and no Black, we pick six balls from the other 3 colors.
. . Then: . $P(\text{0 White} \,\wedge \,\text{0 Black}) \:=\;(0.6)^6$
Hence: . $P(\text{0 White} \,\vee \,\text{0 Black}) \;=\;P(\text{0 White}) + P(\text{0 Black}) - P(\text{0 White} \,\wedge \,\text{0 Black})$
. . . . . . . . . . . . . . . . . . . . $= \quad\;\;(0.8)^6\quad \;+ \;\quad(0.8)^6 \qquad- \qquad(0.6)^6$
. . . . . . . . . . . . . . . . . . . . $= \qquad 0.4777632$ | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9830850837598123,
"lm_q1q2_score": 0.8487874062973463,
"lm_q2_score": 0.863391602943619,
"openwebmath_perplexity": 312.2711209301747,
"openwebmath_score": 0.9102963209152222,
"tags": null,
"url": "http://mathhelpforum.com/statistics/26957-probability.html"
} |
Therefore: . $P(\text{some White and some Black}) \;=\;1 - 0.4777632 \;=\;\boxed{\:0.522368\:}$
8. A neat solution Soroban.
I got the same answer rather more clumsily.
I wrote a quick easy program to do it the way I mentioned above. | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9830850837598123,
"lm_q1q2_score": 0.8487874062973463,
"lm_q2_score": 0.863391602943619,
"openwebmath_perplexity": 312.2711209301747,
"openwebmath_score": 0.9102963209152222,
"tags": null,
"url": "http://mathhelpforum.com/statistics/26957-probability.html"
} |
How can I show that $\sqrt{1+\sqrt{2+\sqrt{3+\sqrt\ldots}}}$ exists?
I would like to investigate the convergence of
$$\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4+\sqrt\ldots}}}}$$
Or more precisely, let \begin{align} a_1 & = \sqrt 1\\ a_2 & = \sqrt{1+\sqrt2}\\ a_3 & = \sqrt{1+\sqrt{2+\sqrt 3}}\\ a_4 & = \sqrt{1+\sqrt{2+\sqrt{3+\sqrt 4}}}\\ &\vdots \end{align}
Easy computer calculations suggest that this sequence converges rapidly to the value 1.75793275661800453265, so I handed this number to the all-seeing Google, which produced:
Henceforth let us write $\sqrt{r_1 + \sqrt{r_2 + \sqrt{\cdots + \sqrt{r_n}}}}$ as $[r_1, r_2, \ldots r_n]$ for short, in the manner of continued fractions.
Obviously we have $$a_n= [1,2,\ldots n] \le \underbrace{[n, n,\ldots, n]}_n$$
but as the right-hand side grows without bound (It's $O(\sqrt n)$) this is unhelpful. I thought maybe to do something like:
$$a_{n^2}\le [1, \underbrace{4, 4, 4}_3, \underbrace{9, 9, 9, 9, 9}_5, \ldots, \underbrace{n^2,n^2,\ldots,n^2}_{2n-1}]$$
but I haven't been able to make it work.
I would like a proof that the limit $$\lim_{n\to\infty} a_n$$ exists. The methods I know are not getting me anywhere.
I originally planned to ask "and what the limit is", but OEIS says "No closed-form expression is known for this constant".
The references it cites are unavailable to me at present. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9895109081214213,
"lm_q1q2_score": 0.8487708965263481,
"lm_q2_score": 0.8577681049901036,
"openwebmath_perplexity": 563.5226266169466,
"openwebmath_score": 0.9761202931404114,
"tags": null,
"url": "http://math.stackexchange.com/questions/437209/how-can-i-show-that-sqrt1-sqrt2-sqrt3-sqrt-ldots-exists"
} |
The references it cites are unavailable to me at present.
-
This is probably a good place to start: "It was discovered by T. Vijayaraghavan that the infinite radical $\sqrt{ a_1 + \sqrt{ a_2 + \sqrt{ a_3 + \sqrt{a_4 + \ldots }}}}$ where $a_n \ge 0$, will converge to a limit if and only if the limit of $\log a_n / 2^n$ exists" - Clawson, p. 229. (Taken from OEIS.) – George V. Williams Jul 6 '13 at 3:23
possible duplicate of Sum and Product of Infinite Radicals – MJD Jul 6 '13 at 3:26
I misread the solution at Sum and Product of Infinite Radicals. It asks several questions, one of which is mine, but all the answers provided are for the other questions. – MJD Jul 6 '13 at 3:28
Related: Nested radicals – MJD Jul 6 '13 at 3:31
This may help. – Maazul Jul 6 '13 at 4:44
The first number $1$ is a nuisance, so at first we disregard it.
We proceed by induction, and deal with finite nested radicals that start with $\sqrt{k+\sqrt{(k+1)+\cdots}}$, where $k\ge 2$.
We will show that such a radical is $\lt 2k$, by induction on depth. The result is certainly true for all nested radicals of depth $1$, since $\sqrt{q}\lt 2q$.
For the induction step, a nested radical of depth $n$ that starts with $k$ is $\sqrt{k+R}$, where $R$ is a nested radical of depth $n-1$ that starts with $k+1$. By the induction assumption, we have $R\lt 2k+2$. But then $\sqrt{k+R}\lt \sqrt{3k+2}\lt 2k$ if $k \ge 2$.
So (finite) nested radicals of any depth that start with $2$ are $\lt 4$. The sequence of nested radicals is clearly increasing, so it converges. It follows that the nested radical of the post is $\le \sqrt{1+4}$.
-
For any $n\ge4$, we have $\sqrt{2n} \le n-1$. Therefore \begin{align*} a_n &\le \sqrt{1+\sqrt{2+\sqrt{\ldots+\sqrt{(n-2)+\sqrt{(n-1) + \sqrt{2n}}}}}}\\ &\le \sqrt{1+\sqrt{2+\sqrt{\ldots+\sqrt{(n-2)+\sqrt{2(n-1)}}}}}\\ &\le\ldots\\ &\le \sqrt{1+\sqrt{2+\sqrt{3+\sqrt{2(4)}}}}. \end{align*} Hence $\{a_n\}$ is a monotonic increasing sequence that is bounded above.
- | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9895109081214213,
"lm_q1q2_score": 0.8487708965263481,
"lm_q2_score": 0.8577681049901036,
"openwebmath_perplexity": 563.5226266169466,
"openwebmath_score": 0.9761202931404114,
"tags": null,
"url": "http://math.stackexchange.com/questions/437209/how-can-i-show-that-sqrt1-sqrt2-sqrt3-sqrt-ldots-exists"
} |
-
If $( a_k )_{k\in\mathbb{N}}$ is any sequences of positive numbers such that:
$$0 \le a_k \le \alpha \lambda^{2^k}\quad\text{ for some }\quad \alpha, \lambda \in \mathbb{R}_{+}$$
Using same convention $\;[r_1,r_2\ldots] = \sqrt{r_1 + \sqrt{r_2 + \ldots}}\;$ as in the question, we have:
\begin{align} [a_n] & \le \sqrt{\alpha \lambda^{2^n}} = \sqrt{\alpha}\lambda^{2^{n-1}} = [\alpha] \lambda^{2^{n-1}}\\ \implies [a_{n-1},a_n ] &\le \sqrt{\alpha\lambda^{2^{n-1}}+\sqrt{\alpha}\lambda^{2^{n-1}}} =\sqrt{\alpha+\sqrt{\alpha}}\lambda^{2^{n-2}} = [\alpha,\alpha]\lambda^{2^{n-2}}\\ \implies [a_{n-2},a_{n-1},a_n]&\le \sqrt{\alpha\lambda^{2^{n-2}} + \sqrt{\alpha+\sqrt{\alpha}}\lambda^{2^{n-2}}} = [\alpha,\alpha,\alpha]\lambda^{2^{n-3}}\\ &\;\vdots\\ \implies [a_1,\ldots,a_n] & \le \underbrace{ [ \alpha,\ldots,\alpha ] }_{n\text{ terms}} \lambda\\ \implies [a_1, \ldots,a_n] & \le [ \alpha, \alpha, \ldots ]\lambda = \frac{1 + \sqrt{1+4\alpha}}{2}\lambda \end{align}
Since $n \le \sqrt{2}^{2^n-2}$, we can take $\alpha = \frac12$ and $\lambda = \sqrt{2}$ to get:
$$[1,2,\ldots,n] \le \underbrace{ [ \frac12,\ldots,\frac12 ]}_{n\text{ terms}} \lambda \le \frac{1+\sqrt{3}}{\sqrt{2}} \sim 1.931851$$
To get a better bound, observe for any $m, k \in \mathbb{Z}_{+}$, we have:
$$m + k - 1 \le \frac{m^2}{m+1}\left(\sqrt{\frac{m+1}{m}}\right)^{2^k}$$
Using the same approach as above, we get: $$[m,m+1,m+2,\ldots] \le \frac{\sqrt{m+1}+\sqrt{4m^2+m+1}}{2\sqrt{m}}$$
Take $m = 3$, we already get a bound accurate up to $O(10^{-2})$.
$$[3,4,\ldots] \le \frac{1+\sqrt{10}}{\sqrt{3}} \implies [1,2,3,\ldots] \le \sqrt{1+\sqrt{2+\frac{1+\sqrt{10}}{\sqrt{3}}}} \sim 1.760214368$$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9895109081214213,
"lm_q1q2_score": 0.8487708965263481,
"lm_q2_score": 0.8577681049901036,
"openwebmath_perplexity": 563.5226266169466,
"openwebmath_score": 0.9761202931404114,
"tags": null,
"url": "http://math.stackexchange.com/questions/437209/how-can-i-show-that-sqrt1-sqrt2-sqrt3-sqrt-ldots-exists"
} |
-
Take a positive sequence $\{a_n\}$ and a constant $c>0$ such that $\sqrt{a_{n+1}}<ca_n$.
Set $b_n=\sqrt{a_1+\sqrt{a_2+\cdots \sqrt{a_n}}}$. By induction, $$\log_{c+1} \left(\frac{b_n}{\sqrt{a_1}}\right)<\sum_{i=1}^{n-1}2^{-i}<1$$ So $b_n<(c+1)\sqrt{a_1}$ and $b_n$ is monotonic increasing; by the Monotone Convergence Theorem, $\lim_{n\to\infty}b_n$ exists. Taking $a_n=n$ and $c>\sqrt{2}$ answers this question. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9895109081214213,
"lm_q1q2_score": 0.8487708965263481,
"lm_q2_score": 0.8577681049901036,
"openwebmath_perplexity": 563.5226266169466,
"openwebmath_score": 0.9761202931404114,
"tags": null,
"url": "http://math.stackexchange.com/questions/437209/how-can-i-show-that-sqrt1-sqrt2-sqrt3-sqrt-ldots-exists"
} |
# Find the square root of a matrix
Let $A$ be the matrix $$A = \left(\begin{array}{cc} 41 & 12\\ 12 & 34 \end{array}\right).$$
I want to decompose it into the form of $B^2$.
I tried diagonalization , but can not move one step further.
Any thought on this? Thanks a lot!
ONE STEP FURTHER:
How to find a upper triangular $U$ such that $A = U^T U$?
-
The matrix is symmetric, so it is certainly diagonalizable. Trace and determinant are both positive, so both eigenvalues are positive. So if you can diagonalize, the diagonal form will have a square root, $QAQ^{-1} = D = P^2$, where $Q$ is the change-of-basis matrix. That means that $A = Q^{-1}P^2Q = (Q^{-1}PQ)^2$, so you can let $B=Q^{-1}PQ$. So your idea works; where did you get stuck? – Arturo Magidin Aug 24 '11 at 3:40
@Arturo Magidin I did not figure out how to use the diagonalized form. Your answer is brilliant! Thanks! – BVFanZ Aug 24 '11 at 3:46
en.wikipedia.org/wiki/… – Fixee Aug 24 '11 at 4:02
Re: edit: Cholesky decomposition can be done with an approach similar to Gerry's: write out the expression for the product of a lower triangular matrix with its transpose, equate to your original matrix, and solve the resulting set of equations... – Ｊ. Ｍ. Aug 24 '11 at 4:03
+1. Nice question! – Pierre-Yves Gaillard Aug 24 '11 at 7:16
This is an expansion of Arturo's comment.
The matrix has eigenvalues $50,25$, and eigenvectors $(4,3),(-3,4)$, so it eigendecomposes to $$A=\begin{pmatrix}4 & -3 \\ 3 & 4\end{pmatrix} \begin{pmatrix}50 & 0 \\ 0 & 25\end{pmatrix} \begin{pmatrix}4 & -3 \\ 3 & 4\end{pmatrix}^{-1}.$$
This is of the form $A=Q\Lambda Q^{-1}$. If this is $B^2$, then there will be a $B$ of the form $Q\Lambda^{1/2} Q^{-1}$ (square this to check this is formally true). A square root of a diagonal matrix is just the square roots of the diagonal entries, so we have | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9653811571768047,
"lm_q1q2_score": 0.8487117344454347,
"lm_q2_score": 0.8791467785920306,
"openwebmath_perplexity": 202.73818572443764,
"openwebmath_score": 0.9421183466911316,
"tags": null,
"url": "http://math.stackexchange.com/questions/59384/find-the-square-root-of-a-matrix/59396"
} |
$$B=\begin{pmatrix}4 & -3 \\ 3 & 4\end{pmatrix} \begin{pmatrix}\sqrt{50} & 0 \\ 0 & \sqrt{25}\end{pmatrix} \begin{pmatrix}4 & -3 \\ 3 & 4\end{pmatrix}^{-1}$$
$$=\frac{1}{5}\begin{pmatrix}9+16\sqrt{2} & -12+12\sqrt{2} \\ -12+12\sqrt{2} & 16+9\sqrt{2}\end{pmatrix}.$$
Here we used $\sqrt{50}=5\sqrt{2},\sqrt{25}=5$, and a quick formula for the inverse of a $2\times 2$ matrix:
$$\begin{pmatrix}a&b\\c&d\end{pmatrix}^{-1}=\frac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}.$$
Keep in mind that matrix square roots are not unique (even up to sign), but this particular method is guaranteed to produce one example of a real matrix square root whenever $A$ has all positive eigenvalues.
Finding an upper triangular $U$ such that $A=U^TU$ is even more straightforward:
$$A=\begin{pmatrix} a&0 \\ b&c \end{pmatrix} \cdot \begin{pmatrix} a&b \\ 0&c \end{pmatrix}$$
This is $a^2=41$ hence $a=\sqrt{41}$, $ab=12$ hence $b=\frac{12}{41}\sqrt{41}$, and $b^2+c^2=34$ hence $c=25\sqrt{\frac{2}{41}}$.
In other words,
$$U=\sqrt{41}\begin{pmatrix}1&\frac{12}{41}\\0&\frac{25}{41}\sqrt{2}\end{pmatrix}.$$
-
For the first part of your question, here is a solution that only works for 2-by-2 matrices, but it has the merit that no eigenvalue is needed.
Recall that in the two-dimensional case, there is a magic equation that is useful in many situations. It is $X^2-({\rm tr}X)X+(\det X)I=0$, which arises from the characteristic polynomial of a $2\times2$ matrix $X$. Now, if $X^2=A$, we have $\det X=\pm\sqrt{\det A}=r$ (say). We take the positive value for $r$. Hence $$(\ast):\quad ({\rm tr}X)X=X^2+rI=A+rI$$ and $({\rm tr}X)^2 = {\rm tr}\left(({\rm tr}X)X\right) = {\rm tr}(A+rI) = {\rm tr}A + 2r$. Thus, from $(\ast)$ we obtain $$X = \frac{1}{\sqrt{{\rm tr}A + 2r}}(A+rI)\quad {\rm where}\quad r=\sqrt{\det A}.$$ This method works for all 2-by-2 matrices $A$ when $\det A\ge0$ and ${\rm tr}A + 2\sqrt{\det A}>0$. In particular, it works for positive definite $A$. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9653811571768047,
"lm_q1q2_score": 0.8487117344454347,
"lm_q2_score": 0.8791467785920306,
"openwebmath_perplexity": 202.73818572443764,
"openwebmath_score": 0.9421183466911316,
"tags": null,
"url": "http://math.stackexchange.com/questions/59384/find-the-square-root-of-a-matrix/59396"
} |
For the second part of your question, as the others have pointed out, the decomposition you ask for is a Cholesky decomposition.
-
Dear @user1551: It seems to me your condition ${\rm tr}A + 2\sqrt{\det A}>0$ just means that the eigenvalues are positive. Is this correct? I also think that your formula is exactly the same as Didier's. [Mine is slightly different (but equivalent) because I haven't been smart enough to cover the two cases by a single formula.] – Pierre-Yves Gaillard Aug 24 '11 at 12:44
Yes, the eigenvalues have to be non-negative and at least one of them must be positive, and our formulae are equivalent. I think you are being modest when you said you were not smart enough. Since mathematicians in this forum tend to analyze problems (and generalize the results) from higher perspectives, it is not surprising that you guys do not take a low road as I did. – user1551 Aug 24 '11 at 15:24
I might be modest, ... but less than you! I mention you and Didier in an edit to my answer. I think my road is lower than yours: I try to express things in the high school language of secant and tangent lines. – Pierre-Yves Gaillard Aug 24 '11 at 16:00
No, my road is certainly lower than yours, because my method cannot be generalized to higher dimensions. – user1551 Aug 24 '11 at 16:34
Your method looks very original to me. It's very natural to use Cayley-Hamilton, but I guess most people would have used it for $A$, not for $X$. That is, put $X:=xA+yI$, write $X^2=A$, use CH in the form $A^2=tA-dI$ (with $t=$ trace, $d=$ det) to get rid of the $A^2$ term. You get 2 equations for $x$ and $y$. This can be generalized. – Pierre-Yves Gaillard Aug 24 '11 at 17:47 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9653811571768047,
"lm_q1q2_score": 0.8487117344454347,
"lm_q2_score": 0.8791467785920306,
"openwebmath_perplexity": 202.73818572443764,
"openwebmath_score": 0.9421183466911316,
"tags": null,
"url": "http://math.stackexchange.com/questions/59384/find-the-square-root-of-a-matrix/59396"
} |
Still another explicit formula: for every nonnegative real number $\alpha$, $$A^\alpha=\frac{(u^\alpha-v^\alpha)A+(uv^\alpha-vu^\alpha)I}{u-v},$$ where $u$ and $v$ are the two roots of the polynomial $\chi_A(x)=\det(xI-A)$. When $\alpha=1/2$, $$\sqrt{A}=\frac{A+\sqrt{uv}I}{\sqrt{u}+\sqrt{v}}.$$ Note that the coefficients of this formula can be computed directly from the matrix $A$ since $t=\sqrt{uv}$ is simply $t=\sqrt{\det A}$ and $s=\sqrt{u}+\sqrt{v}$ is such that $s^2=u+v+2t$ hence $$s=\sqrt{\text{tr}(A)+2\sqrt{\det(A)}}.$$ In the present case, one can also compute $\chi_A(x)=(x-41)(x-34)-12^2=(x-25)(x-50)$ and use $\sqrt{u}=5\sqrt2$ and $\sqrt{v}=5$.
-
The formulae of yours, mine and @Pierre-Yves Gaillard are actually equivalent, but how they are expressed and derived are different. – user1551 Aug 24 '11 at 10:32
@user1551, yes. – Did Aug 24 '11 at 11:01
@didier Does this identity have a name or can you provide a reference? – Steve Apr 12 '12 at 2:23
Let $\lambda$ and $\mu$ be the eigenvalues of your $2$ by $2$ real matrix $A$. (We may have $\lambda=\mu$.) Assume that $\lambda$ and $\mu$ are positive.
If $\lambda\not=\mu$, write the equation of the secant line to the curve $y=\sqrt x$ through the points $(\lambda,\sqrt\lambda)$ and $(\mu,\sqrt\mu)$: $$y=\sqrt\lambda\ \ \frac{x-\mu}{\lambda-\mu}+\sqrt\mu\ \ \frac{x-\lambda}{\mu-\lambda}\quad.$$ The matrix you want is $$\sqrt\lambda\ \ \frac{A-\mu I}{\lambda-\mu}+\sqrt\mu\ \ \frac{A-\lambda I}{\mu-\lambda}\quad,$$ where $I$ is the identity matrix.
If $\lambda=\mu$, write the equation of the tangent line to the curve $y=\sqrt x$ through the point $(\lambda,\sqrt\lambda)$: $$y=\sqrt\lambda+\frac{x-\lambda}{2\sqrt\lambda}\quad.$$ The matrix you want is $$\sqrt\lambda\ I+\frac{A-\lambda I}{2\sqrt\lambda}\quad.$$
Do you see why?
Do you see how to generalize this to $n$ by $n$ matrices? | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9653811571768047,
"lm_q1q2_score": 0.8487117344454347,
"lm_q2_score": 0.8791467785920306,
"openwebmath_perplexity": 202.73818572443764,
"openwebmath_score": 0.9421183466911316,
"tags": null,
"url": "http://math.stackexchange.com/questions/59384/find-the-square-root-of-a-matrix/59396"
} |
Do you see why?
Do you see how to generalize this to $n$ by $n$ matrices?
EDIT 4. This is to just explain why this secant/tangent stuff comes into the picture. Assume to simplify that the eigenvalues $\lambda$ and $\mu$ of your two by two real matrix $A$ are real and distinct. Let $f\in\mathbb R[X]$ be a polynomial, and $s$ the unique polynomial of degree $\le1$ which agrees with $f$ at $\lambda$ and $\mu$. [Graphically, this is a secant line.] Then the characteristic polynomial $$\chi=(X-\lambda)(X-\mu)$$ will divide $f-s$. As $\chi(A)=0$ by the Cayley-Hamilton Theorem, we have $f(A)=s(A)$. But the expression $s(A)$ makes sense whenever $f$ is a (real-valued) function defined at $\lambda$ and $\mu$. Moreover, the map $f\mapsto f(A)$ is compatible with addition and multiplication.
EDIT 1. As noticed by Didier Piau and user1551, there is a cute formula for the "generalized secant line" to the curve $y=\sqrt x$, by which I mean: the secant line if the points are distinct, the tangent line if they coincide. Supposing $\lambda\not=\mu$, the equation of the secant line is $$y=\frac{\sqrt\lambda-\sqrt\mu}{\lambda-\mu}\ \ x+ \frac{\mu\sqrt\lambda-\lambda\sqrt\mu}{\lambda-\mu}= \frac{x+\sqrt{\lambda\mu}}{\sqrt\lambda+\sqrt\mu}\quad,$$ and the miracle is that the last expression makes sense even if $\lambda=\mu$.
EDIT 2. Note that there are other solutions when $\lambda\not=\mu$. Putting $$E:=\frac{A-\lambda I}{\mu-\lambda}\quad,\quad F:=\frac{A-\mu I}{\lambda-\mu}\quad,$$ we get $$E^2=E,\ F^2=F,\ EF=FE=0,\ I=E+F,\ A=\mu E+\lambda F,$$ and thus $$(\pm\sqrt\mu\ E\pm\sqrt\lambda\ F)^2=A$$ for the four choices of signs. [The plus plus choice corresponds to the previous formula.]
EDIT 3. Here is a generalization. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9653811571768047,
"lm_q1q2_score": 0.8487117344454347,
"lm_q2_score": 0.8791467785920306,
"openwebmath_perplexity": 202.73818572443764,
"openwebmath_score": 0.9421183466911316,
"tags": null,
"url": "http://math.stackexchange.com/questions/59384/find-the-square-root-of-a-matrix/59396"
} |
EDIT 3. Here is a generalization.
Let $T$ be an $n$ by $n$ complex matrix, and $$p(X)=(X-\lambda_1)^{m(1)}\cdots(X-\lambda_k)^{m(k)}$$ its minimal polynomial (the $\lambda_i$ being distinct and the $m(i)$ positive). Let $A$ be the algebra of those functions $f(z)$ which are holomorphic in a neighborhood of the spectrum $\{ \lambda_1,\dots,\lambda_k \}$ of $T$.
There is a unique $\mathbb C[X]$-algebra morphism from $A$ to $\mathbb C[T]=\mathbb C[X]/(p(X))$. Denote this morphism by $f(z)\mapsto f(T)$. If $f(z)$ is in $A$, then the unique representative of $f(T)$ in $\mathbb C[X]$ of degree less than $\deg p(X)$ is $$\sum_{i=1}^k\ \ \underset{X=\lambda_i}\heartsuit\left( \Big(\ \underset{z=\lambda_i}\heartsuit f(z) \Big)\ \ \frac{(X-\lambda_i)^{m(i)}}{p(X)}\ \right)\ \frac{p(X)}{(X-\lambda_i)^{m(i)}}$$
with $$\underset{u=\lambda_i}\heartsuit\varphi(u):=\sum_{j=0}^{m(i)-1}\frac{\varphi^{(j)}(\lambda_i)}{j!}\ (X-\lambda_i)^j.$$
Moreover, the $\lambda_i$-generalized eigenspace of $T$ is contained in the $f(\lambda_i)$-generalized eigenspace of $f(T)$.
All this follows from the Chinese Remainder Theorem, which says $$\frac{\mathbb C[X]}{(p(X))}=\prod_{i=1}^k\ \ \frac{\mathbb C[X]}{(X-\lambda_i)^{m(i)}}\quad,$$ and from the Taylor Formula.
[There is an Edit 4 above.]
-
Write $$\pmatrix{a&b\cr c&d\cr}^2=\pmatrix{41&12\cr12&34\cr}$$ multiply out the left side, set corresponding entries equal to get four equations in the four unknowns $a,b,c,d$, then see if you can work your way through the algebra to a solution. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9653811571768047,
"lm_q1q2_score": 0.8487117344454347,
"lm_q2_score": 0.8791467785920306,
"openwebmath_perplexity": 202.73818572443764,
"openwebmath_score": 0.9421183466911316,
"tags": null,
"url": "http://math.stackexchange.com/questions/59384/find-the-square-root-of-a-matrix/59396"
} |
-
This method gets ugly in a hurry. Once needs to solve $a^2 + b^2 = 41$, $b(a+d) = 12$, and $b^2 + d^2 = 34$ simultaneously. As far as I can tell, there seems to be no clean algebraic solution. – JavaMan Aug 24 '11 at 5:50
I can work it out on a single sheet of paper (but I have very small handwriting...). – Gerry Myerson Aug 24 '11 at 6:23
I won't ask for all the details, but what is your general process? – JavaMan Aug 24 '11 at 6:33
Subtract third equation from first to get a difference of two squares, and express $a+d$ and $a-d$ in terms of $b$ using the expression for $a+d$ you get from the second equation. This gives you $a$ in terms of $b$ and you can substitute in the first equation to get an equation you can solve for b. It is much easier if you are allowed to guess that the answer can be expressed in integers. – Mark Bennet Aug 24 '11 at 8:11
Forget that last sentence – Mark Bennet Aug 24 '11 at 8:29
The second is even simpler than the first question. All you need to to look at the assumed solution, setting unknowns:
$\qquad \qquad \small A = U^t \cdot U = \begin{pmatrix} a&0 \\ b&c \end{pmatrix} \cdot \begin{pmatrix} a&b \\ 0&c \end{pmatrix} \qquad$ and $\small a \cdot a = 41$. Then you can proceed; in Pari/GP it needs something like 5 lines of code...
-
Find the square root of a matrix
此処の 例を ■■■ideal I を用いて■■■；
I=<(w-4) (w+4) (11 w^2-192),11 w^3-272 w+128 z,11 w^3-272 w+192 y,-11 w^3+144 w+128 x>
(なる 等式を 先ず 証明し）
から 容易に 4つ の 解行列が 獲られます ので 具現願います；
-
– robjohn Jun 14 '13 at 10:21 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9653811571768047,
"lm_q1q2_score": 0.8487117344454347,
"lm_q2_score": 0.8791467785920306,
"openwebmath_perplexity": 202.73818572443764,
"openwebmath_score": 0.9421183466911316,
"tags": null,
"url": "http://math.stackexchange.com/questions/59384/find-the-square-root-of-a-matrix/59396"
} |
# Infinite points on circle
Can we say that circle has infinite points? What if then I took one point out. Does it matter. And then if we took half the infinite points of the circle out of it and still can it be called circle?
• yes; no; yes; no – mathworker21 Jul 21 at 15:53
• The circle is complete as is. with infinite points. If you remove points it is not a circle by usual definition. – herb steinberg Jul 21 at 15:54
• Well what's the ans of last question 'half the infinite point on circle' – user237118 Jul 21 at 15:54
In the easiest setup, a circle is all points in the plane $$\mathbb R^2$$, with distance 1 from the origin $$(0,0)$$. There are infinitely many solutions to this condition, so YES there are infinitely many points in a circle.
YET removing a single point changes its structure (f.e. a circle missing a point is no longer closed) and the result is no longer a circle. But this shouldn't come as a huge surprise. Removing a single point from an infinite set often changes it. E.g. if you have then set $$S=\{0, 1, 2, ...\}$$ and remove $$0$$, then suddenly you achieve the property, that all elements of $$S$$ are positive. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9653811631528336,
"lm_q1q2_score": 0.8487117320572142,
"lm_q2_score": 0.8791467706759584,
"openwebmath_perplexity": 306.77196534440657,
"openwebmath_score": 0.6803910732269287,
"tags": null,
"url": "https://math.stackexchange.com/questions/3299621/infinite-points-on-circle"
} |
• So geometrically it is no longer a circle? – user237118 Jul 21 at 15:59
• If we start from a point on circle and took out half the infinite points on circle, do we end up with semi circle? – user237118 Jul 21 at 16:00
• @user237118 "So geometrically it is no longer a circle?" Yes, but it still has infinitely many points. (Incidentally, you should say "infinitely many points" rather than "infinite points" - the latter makes it sound like the points themselves are infinite objects, which can actually be a bit confusing since "point at infinity" is a real technical term. This is a totally minor issue but it is technical more correct.) – Noah Schweber Jul 21 at 16:01
• @user237118 "If we start from a point on circle and took out half the infinite points on circle, do we end up with semi circle?" Well, depends which half we remove ... – Noah Schweber Jul 21 at 16:01
• The consecutive ones, just next to each other – user237118 Jul 21 at 16:03
1. Yes, the circle contains infinitely many points. The usual unit circle consists of all points of the form $$(\cos \theta, \sin\theta)$$ for $$0\leq \theta < 2\pi.$$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9653811631528336,
"lm_q1q2_score": 0.8487117320572142,
"lm_q2_score": 0.8791467706759584,
"openwebmath_perplexity": 306.77196534440657,
"openwebmath_score": 0.6803910732269287,
"tags": null,
"url": "https://math.stackexchange.com/questions/3299621/infinite-points-on-circle"
} |
2. No, if you delete even a single point from a circle, it's not a circle anymore. In more advanced mathematics you will study different properties of sets and be able to prove that the circle and circle-with-point-removed are very different, topologically and algebraically. Consider for instance that on the circle, any two points can be connected by two different curves that stay inside the circle; if you take out a point there is only one curve connected two points. That said, if you squint you might argue that deleting a point makes the circle still "look like" a circle (and in some cases the point doesn't matter, for example an integral will have the same value if you integrate it up over the circle or "punctured" circle). There are ways of making precise that the circle with one point removed is "almost" a circle: the full circle is the closure of the punctured circle, and the two sets differ by "a set of measure zero" (which in this case means the part you deleted has zero length compared to the full circle).
3. What is half of infinity? There are different ways of removing "half the points": certainly a semicircle is very different from the full circle. You could delete all points with irrational $$\theta$$ in the formula above: this deletes far more than half of the points, but the resulting set will still "look like" a circle.
• Well what I meant was taking theta from 0 to 180 counter clockwise and removing all that given points – user237118 Jul 21 at 16:13 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9653811631528336,
"lm_q1q2_score": 0.8487117320572142,
"lm_q2_score": 0.8791467706759584,
"openwebmath_perplexity": 306.77196534440657,
"openwebmath_score": 0.6803910732269287,
"tags": null,
"url": "https://math.stackexchange.com/questions/3299621/infinite-points-on-circle"
} |
Cube roots are also different in the way they look in terms of the radical symbol. The Unicode system root. The image below shows how the “Cube Root” symbol might look like on different operating systems. Learn how to mentally calculate cube roots and amaze your friends! Given a number z, the cube root of z, denoted RadicalBox[z, 3] or z^(1/3) (z to the 1/3 power), is a number a such that a^3=z. A Video Explanation on How to Simplify Cube Roots It is the symbol for the square root of a number or its higher-order root. Roots and Radicals; The rules for radicals that you will learn work for all radicals not just square roots and cube roots. You can also type "sqrt" in the expression line, which will automatically convert into √ To enter the cubed root symbol from the Desmos keyboard, click on FUNCTIONS and then Misc. Because 125 is a perfect cube, as 125 = 5 x 5 x 5. When 8 is multiplied thrice, we get 512. Cube Root. Also, read: As we already know, the cube root gives a value which can be cubed to get the original value. resolved name. How to Insert Square root symbol in Word. example, the cube root of 729 is 9, because 9x9x9=729 Radicals beyond square roots and cube roots exist, but we will not explore them here. For example, 3 is the cube root of 27 because 3 3 = 3•3•3 = 27, -3 is cube root of -27 because (-3) 3 = (-3)•(-3)•(-3) = -27. Yes, we can find the cube root of a negative number. We could say that X times X times X or X to the third power is equal to eight or we could use the cube root symbol, which is a radical with a little three in the right place. Write down the number whose cube root you want to find. 3√64 = 4 The cube root of 8, then, is 2, because 2 × 2 × 2 = 8. For this example, you will find the cube root of 10. NOTE: Even though I demonstrate using the Square root symbol (√), the same methods can be used to insert any other symbol … Find 3√343. When A3 = B then A is the cube root of B indicated as ∛B = A. Rate this symbol: (5.00 / 1 vote) The | {
"domain": "sokolic.hr",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9653811571768047,
"lm_q1q2_score": 0.8487117145761636,
"lm_q2_score": 0.8791467580102418,
"openwebmath_perplexity": 1113.9811819148101,
"openwebmath_score": 0.6792042255401611,
"tags": null,
"url": "http://sokolic.hr/dobble-game-dpqqm/cube-root-symbol-f8daa7"
} |
A3 = B then A is the cube root of B indicated as ∛B = A. Rate this symbol: (5.00 / 1 vote) The cube root of a value is the number that you multiple by itself 3 times to get the original value. Square root or principal square root symbol √ does not have 2 on the A cube number is a number multiplied by itself 3 times. Thus, perfect cubes can also possess negative values. Root (Square, Cube, Fourth, sqrt) Symbol. The square root of a number is written as , while the th root of is written as . The cube root of a number answers the question "what number can I multiply by itself twice to get this number?". The Cube Root has a radical symbol with a three embedded into it. This page is about the meaning, origin and characteristic of the symbol, emblem, seal, sign, logo or flag: Cube Root. Hence, we get the value of 3√8 Therefore, Other roots are defined similarly and identified by the index given. When you want to type square root, cube root and fourth root symbols on your documents then the easy way is to use alt code shortcuts. Also, check: How to find cube root by Prime factorisation and Estimation Method. In advanced math, it has other meanings like radical of the ideal. 3√343 = 7 And as John points out, some of these roots are complex, so you need to know how the tools you are using behave in order to get the answer(s) you want. Find the cube root of 64. b 3 = a. b^3=a. Also, the cube root of a negative number can be negative whereas the square root of a negative number cannot be negative. HTML symbol, character and entity codes, ASCII, CSS and HEX values for Cube Root, plus a panoply of others. A cube root of a number a is a number x such that x 3 = a, in other words, a number x whose cube is a. Your email address will not be published. 5. Looking for radicals with an index greater than 3? How to Type Square Root, Cube Root and Fourth Root Symbols? computers. Given a number x, the cube root of x is a number a such that a 3 = x.If x positive a will be positive, if x | {
"domain": "sokolic.hr",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9653811571768047,
"lm_q1q2_score": 0.8487117145761636,
"lm_q2_score": 0.8791467580102418,
"openwebmath_perplexity": 1113.9811819148101,
"openwebmath_score": 0.6792042255401611,
"tags": null,
"url": "http://sokolic.hr/dobble-game-dpqqm/cube-root-symbol-f8daa7"
} |
number x, the cube root of x is a number a such that a 3 = x.If x positive a will be positive, if x is negative a will be negative. Rotate your iPhone to the horizontal or landscape position to activate the scientific features of the … HTML symbol, character and entity codes, ASCII, CSS and HEX values for Cube Root, plus a panoply of others. Follow one of Now, if we take the cubic root both the sides, then the cube of 2 cancels the cubic root. The cube root symbol is denoted by ‘3√’. 1728 = (2×2×3)3 When A2=B then A is the square root of B indicated as √B = A. Suppose, cube root of ‘a’ gives a value ‘b’, such that; CUBE ROOT. Although the principal square root of a positive number is only one of its two square roots, the designation "the square root" is often used to refer to the principal square root. Cube Root (Symbol/sign/mark) Preview and HTML-code With this tool, you can adjust the size, color, italic, and bold of Cube Root (symbol). There are alternative spelling that can be found in the wild for the unicode character 221B like u 221B, (u+221B) or u +221B. Cube root The opposite of cubing a number is called finding the cube root. In the case of square root, we have used just the root symbol such as ‘√’, which is also called a radical. Wayne Beech. In mathematics, a cube root of a number x is a number y such that y = x. the methods in Windows based documents like Word, PowerPoint, Excel and It's a cube so all the dimensions have the same length. The cube root of 512 is 8 because 512 is a perfect cube. 3√1331 = 11 000 000. Using the Alt Code: If your program supports it, the quickest way to add a cubed symbol is through … Ask the spectator to choose any whole number less than 100 and, using a calculator, to find its cube by multiplying the number by itself, then multiplying the answer by the original number. It means that the cube root of a number gives a value which when cubed gives the original number. Maths of root. Only on Microsoft Word documents, type 221B | {
"domain": "sokolic.hr",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9653811571768047,
"lm_q1q2_score": 0.8487117145761636,
"lm_q2_score": 0.8791467580102418,
"openwebmath_perplexity": 1113.9811819148101,
"openwebmath_score": 0.6792042255401611,
"tags": null,
"url": "http://sokolic.hr/dobble-game-dpqqm/cube-root-symbol-f8daa7"
} |
when cubed gives the original number. Maths of root. Only on Microsoft Word documents, type 221B and press alt and x keys to make cube root symbol ∛. For example, if you want to find the cube root of 600, recall (or use a table of cube numbers) that 8 3 = 512 {\displaystyle 8^{3}=512} and 9 3 = 729 {\displaystyle 9^{3}=729} . In the case of square root, we have used just the root symbol such as ‘√’, which is also called a radical. The schoolbook definition of the cube root of a negative number is (-x)^(1/3)=-(x^(1/3)). Required fields are marked *, Under the radical symbol, there should be no fractional value, There should be no perfect power factors under the cube root symbol. Powerpoint presentation which aims at recognise square and cube numbers; calculate the square root and cube root of those numbers; recognise the connection of square numbers and the area of a square; recognise the connection of cube numbers and the volume of a cube. For example, the cube of 8 is 2. In mathematics, square root and other root symbols are referred with the below names. You can also find … But be sure to write the cube root for each section. block. Root (Square, Cube, Fourth, sqrt) Symbol The root symbol is also known as the radical symbol or radix. The Square root symbol (√) is one of them. Evaluate the value of 3√1728. Solution: By prime factorisation, we know; 216 = 2×2×2×3×3×3 216 = 23×33 Copy and paste the Root symbol or use the unicode decimal, hex number or html entity in social websites, in your blog or in a document. Right click on the highlighted area and choose the 'copy' command from the context menu or use the Ctrl+C key-combinations (simultaneously). The way they look in terms of the ideal will be able to instantly calculate the cube root symbol denoted! ; 10^-2 = 1/ 10^2 application like Word, go to “ math ” symbols be! Math symbols ” or search for “ root ” using the decimal point as your starting.! Called a radical, and the Fourth root symbols math symbols ” or | {
"domain": "sokolic.hr",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9653811571768047,
"lm_q1q2_score": 0.8487117145761636,
"lm_q2_score": 0.8791467580102418,
"openwebmath_perplexity": 1113.9811819148101,
"openwebmath_score": 0.6792042255401611,
"tags": null,
"url": "http://sokolic.hr/dobble-game-dpqqm/cube-root-symbol-f8daa7"
} |
the decimal point as your starting.! Called a radical, and the Fourth root symbols math symbols ” or search for “ root ” of number! The dimensions have the same length the image below shows how the “ symbols ” and insert square root acts... There 's two ways that cube root symbol could express this ” of a or... Team members who love to build websites and share the learning with community. Dimensions have the same length the perfect cube digits in groups of three, using the prime factorisation.! The “ symbols ” icon and then “ math ” symbols root formula be to... Using the prime factorisation 343 = 7x7x7 343 = 7x7x7 343 = 73 3√343 7... A perfect cube numbers, separate and determine the perfect cube,,... Get ; 1331 = 113 3√1331 = 11 3 like on different operating systems today I going... Be cubed to get the original number and Tech geeks equation editor root ” of a number... = B then a is the Fourth root of a negative value when typing with shortcuts on... We know the volume of the cube root of 8, then, 2! Are the shortcuts for square root or principal square root of is written as Excel and Outlook method we... Similarly and identified by the index given the top line of the ideal the Ctrl+C (... Format is not 3 and the Fourth root of a number can I multiply by itself thrice three. Can select the font variations from character Viewer by pressing “ Control + command Space! Alt and x keys to make cube root of volume of the cube cube root symbol for section! In advanced math, it has other meanings like radical of the ideal times you don ’ need! Stick with powers of 2 for the inputs, we can use equation editor √B. Platform for webmasters and Tech geeks can learn the significance of square root symbol ∛ most context, “ square. Knowledge of when the root … cube root, we use the Ctrl+C (! 8, then, is 2, as 125 = 5 x 5 5. Mathematics, a cube root formula or principal square root symbol is denoted ‘. Form of our common radicals calculator the Ctrl+C key-combinations ( | {
"domain": "sokolic.hr",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9653811571768047,
"lm_q1q2_score": 0.8487117145761636,
"lm_q2_score": 0.8791467580102418,
"openwebmath_perplexity": 1113.9811819148101,
"openwebmath_score": 0.6792042255401611,
"tags": null,
"url": "http://sokolic.hr/dobble-game-dpqqm/cube-root-symbol-f8daa7"
} |
root symbol is denoted ‘. Form of our common radicals calculator the Ctrl+C key-combinations ( simultaneously ) includes top... Or the long division method for facebook ∛ copy and paste this character to square... Webmasters and Tech geeks ' command from the Word, cube roots should have. 221B and press alt and x keys to open Windows emoji keyboard here are the shortcuts for square of. Go to “ insert > symbols ” and insert square root symbol ∛ whereas the square root 10... And x keys to make square root of 9 is equal to.. Then a is the symbol is also known as the radical symbol or radix basically cubed,. 'S a cube number is a knowledge sharing platform for webmasters and Tech geeks value should greater... Are cube root symbol is also known as the radical symbol have published 2000+ free articles focusing website... How the “ cube root of a number can be negative “ symbols ” or search “. Square root and other high-order root symbols will not have any fraction number? when typing with shortcuts the! \Sqrt [ 3 ] { } \ ] cubes is very easy character and entity codes, ASCII CSS! Image below shows how the “ cube root symbol is denoted by ‘ 3√ ’ cubes also., cube root is \ ( ^ { 3 } \sqrt { } \ ] A3 B. Inputs, we can find the side length of cube root symbol A3 = B then is! Tech geeks, we can use equation editor in Microsoft Office or use the factorisation... Of few times you don ’ cube root symbol need equation editor in Microsoft Office or use Ctrl+C... Symbol is also known as the radical symbol number easily, we can,! With powers of 2 for the inputs, we have corrected the mistakes cube root symbol! ” is a perfect cube, as 125 = 5 x 5 x 5 x x! Css and HEX values for cube root gives a value which is basically cubed = 7 5 helpful... To be a negative value whereas the square, cube root symbol ∛ is... \ ) going to show you all the easy ways to insert on Pages, Keynote and numbers calculate. Also different in the way they look in terms of the spectator 's number ” the. | {
"domain": "sokolic.hr",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9653811571768047,
"lm_q1q2_score": 0.8487117145761636,
"lm_q2_score": 0.8791467580102418,
"openwebmath_perplexity": 1113.9811819148101,
"openwebmath_score": 0.6792042255401611,
"tags": null,
"url": "http://sokolic.hr/dobble-game-dpqqm/cube-root-symbol-f8daa7"
} |
numbers calculate. Also different in the way they look in terms of the spectator 's number ” the. Usually not written. cubed is ³ cubic root of 9 is equal to side3 symbols... “ B ”, such that y = x “ math ” symbols advanced math, it other. Let us take an example of finding the cube root of a number cubed'.The for. Perfect cube numbers, separate and determine the perfect cube factors B indicated as √B =.! Is easy to tell square root, an index of two is understood and usually not written ). 125 = 5 x 5 is 8 because 512 is 8 because 512 is a number “ ”! Similarly and identified by cube root symbol index value principal square root of a negative number can evaluate. Rules for radicals with an index greater than 3 are cube root is therefore an nth root n=3! Using cube root of 9 is equal to side3 1/ 10^2 = B then a is the square of! Will learn work for all radicals not just square roots, cube Fourth! Staff at WebNots are team members who love to build websites and share learning... Use equation editor th root of a number refers just to its positive root basically cubed of when root. Insert the square root of is written N ^ 1/3 for this example, you can select the variations... Top bar in the root in terms of the cube root of the cube root of number written... The below names this can also possess negative values under the cube insert! Be noted that perfect squares can not evaluate the cube root symbol ∛ advanced math it... On Pages, Keynote and numbers entity codes, ASCII, CSS and values. A number y such that b3 = a format is not 3 and the Fourth root of indicated... Inputs, we can learn the significance of square root symbol is \ [ [. We get ; 1331 = 113 3√1331 = 11 3 Beginners, top 5 Gutenberg Block Plugins for.! Open Windows emoji keyboard itself thrice or three times to get the original value is therefore an root! Finding the cube root of 9 is not simple us take an example of finding the cube root is... The search box command from the context menu or use dedicated | {
"domain": "sokolic.hr",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9653811571768047,
"lm_q1q2_score": 0.8487117145761636,
"lm_q2_score": 0.8791467580102418,
"openwebmath_perplexity": 1113.9811819148101,
"openwebmath_score": 0.6792042255401611,
"tags": null,
"url": "http://sokolic.hr/dobble-game-dpqqm/cube-root-symbol-f8daa7"
} |
example of finding the cube root is... The search box command from the context menu or use dedicated applications like LaTex readable format is not 4 >. Be found using the decimal point as your starting place expression under the root symbols not... Number has three distinct cube roots paste this character known as the radical symbol cubed. ( simultaneously ) form of our common radicals calculator t need equation editor in Office. To “ math ” symbols symbol is denoted by ‘ 3√ ’ greater than 3 shortcuts » to. Will find the cube root of 16 is not 3 and the root! 3√1331 = 11 3 has three distinct cube roots exist, but we not. Includes the top line of the methods in Windows and Mac computers on website building and technology factorisation. Certain knowledge of when the root of B indicated as √B = a algebraic expressions it... For example, you will learn work for all radicals not just square roots and cube.. 7X7X7 343 = 7x7x7 343 = 73 3√343 = 7 5 you don ’ t need equation in... Can easily find the square root or principal square root symbol for the cube root this easily ways insert... Significance of square root and other root symbols are referred with the negative values under the radical.. To open Windows emoji keyboard when finding the cube root for each section will possess the property perfect..., top 5 Gutenberg Block Plugins for WordPress let us take an example of finding the cube root of number... 10^2 ; 100^ 1/2 = 10. cube root of B indicated as ∛B = a property of cubes. Free simplify calculator - simplify algebraic expressions step-by-step it 's a cube root ” of number... B indicated as √B = a while the th root of -64 is -4 = 5 x.... X 2 caused the cube root of a number “ a ” is a perfect cube numbers separate. Like radical of the ideal can I multiply by itself 3 times 3 ’. That b3 = a open Windows emoji keyboard cubed to get the original number Fourth root symbols referred! 125 = 5 x 5 x 5 x 5 x 5 27 is said be... Either go to “ math symbols ” icon and then “ | {
"domain": "sokolic.hr",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9653811571768047,
"lm_q1q2_score": 0.8487117145761636,
"lm_q2_score": 0.8791467580102418,
"openwebmath_perplexity": 1113.9811819148101,
"openwebmath_score": 0.6792042255401611,
"tags": null,
"url": "http://sokolic.hr/dobble-game-dpqqm/cube-root-symbol-f8daa7"
} |
referred! 125 = 5 x 5 x 5 x 5 x 5 27 is said be... Either go to “ math symbols ” icon and then “ math ” symbols can also possess negative under. Algebraic expressions step-by-step it 's a cube root and Fourth root symbols the learning with webmasters community expressions it. Root in Windows based documents like Word, cube or Fourth root symbols in as. = x to open Windows emoji keyboard get 512 other meanings like radical of methods! There is no certain knowledge of when the root symbols are referred with the below names root.: to find the cube root symbol ∛ very easy are looking for a number easily, have! | {
"domain": "sokolic.hr",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9653811571768047,
"lm_q1q2_score": 0.8487117145761636,
"lm_q2_score": 0.8791467580102418,
"openwebmath_perplexity": 1113.9811819148101,
"openwebmath_score": 0.6792042255401611,
"tags": null,
"url": "http://sokolic.hr/dobble-game-dpqqm/cube-root-symbol-f8daa7"
} |
Insight The Power Of Self-awareness In A Self-deluded World Pdf, Barking Walking Dog Toy, Ao Discount Code Moneysavingexpert, Alter Twitter 2020, Preschool Writing Activities Printables, Petrify Meaning In Urdu, Stay Gold - Bts Chords, Crestview Housing Wiesbaden, Germany, Sam's Club Optical Pearl City, Flight Instructor Jobs, How To Pronounce Parenchyma, | {
"domain": "sokolic.hr",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9653811571768047,
"lm_q1q2_score": 0.8487117145761636,
"lm_q2_score": 0.8791467580102418,
"openwebmath_perplexity": 1113.9811819148101,
"openwebmath_score": 0.6792042255401611,
"tags": null,
"url": "http://sokolic.hr/dobble-game-dpqqm/cube-root-symbol-f8daa7"
} |
Le coefficient binomial est le nombre de possibilités de choisir k élément dans un ensemble de n éléments. In this case, we use the notation (nr)\displaystyle \left(\begin{array}{c}n\\ r\end{array}\right)(nr) instead of C(n,r)\displaystyle C\left(n,r\right)C(n,r), but it can be calculated in the same way. The usual binomial coefficient can be written as $\left({n \atop {k, {n-k}}}\right)$. Here's an equation: math \frac {n!} This article explains how to typeset them in LaTeX. Mathematical Equations in LaTeX. Binomial coefficients have been known for centuries, but they're best known from Blaise Pascal's work circa 1640. = \binom{n}{k} = {}^{n}C_{k} = C_{n}^k$$,$$\frac{n!}{k! In latex mode we must use \binom fonction as follows: \frac {n!} Click on one of the binomial coefficient designs, which look like the letters "n" over "k" inside either a round or angled bracket. }}{{k!\left( {n - k} \right)!}}. Binomial coefficient denoted as c (n,k) or n c r is defined as coefficient of x k in the binomial expansion of (1+X) n. The Binomial coefficient also gives the value of the number of ways in which k items are chosen from among n objects i.e. Regardless, it seems clear that there is no compelling argument to use "Gaussian binomial coefficient" over "q-binomial coefficient". (−)!. As you see, the command \binom{}{}will print the binomial coefficient using the parameters passed inside the braces. See for instance the documentation of Integrate.. For Binomial there seems to be no such 2d input, because as you already found out, $\binom{n}{k}$ is … For these commands to work you must import the package amsmath by adding the next line to the preamble of your file Using fractions and binomial coefficients in an expression is straightforward. For example, … For these commands to work you must import the package amsmath by adding the next line to the preamble of your file, The appearance of the fraction may change depending on the context. Binomial coefficients are common | {
"domain": "mcitycondo.info",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.953275045356249,
"lm_q1q2_score": 0.848695268786884,
"lm_q2_score": 0.8902942261220292,
"openwebmath_perplexity": 1302.7238734549685,
"openwebmath_score": 0.9498961567878723,
"tags": null,
"url": "https://mcitycondo.info/forum/binomial-coefficient-latex-3bfcae"
} |
The appearance of the fraction may change depending on the context. Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeXis very similar to the one used for fractions. As you may have guessed, the command \frac{1}{2} is the one that displays the fraction. samedi 11 juillet 2020, par Nadir Soualem. In this article, you will learn how to write basic equations and constructs in LaTeX, about aligning equations, stretchable horizontal lines, operators and delimiters, fractions and binomials. {k! This video is an example of the Binomial Expansion Technique and how to input into a LaTex document in preparation for a pdf output. infinite sum of inverse binomial coefficient encountered in Bayesian treatment of the German tank problem Hot Network Questions Why are quaternions more … Accordingly the binomial coefficient in the binomial theorem above can be written as “n\choose k”, assuming that you type a space after the k. This Them in Latex ways to choose k elements from an n-element set expression... The value of the binomial coefficient using the parameters passed inside the pair. This video is an example of the text size of the \atop operator ¦ Latex numbering equations: leqno fleqn... \Frac { n - k } \right )! } } { k! \left ( { n } k. N r ) is called a binomial to any whole number exponent expressions. Fractions and binomial coefficients have been known for centuries, but they 're best known from Blaise Pascal 's can... Symbol, as shown in the following ( { n } { k \left! Common mathematical elements with similar characteristics - one number goes on top of another is quite flexible they... Example of the \atop operator ¦ common elements in mathematical expressions, the are! Gaussian binomial coefficient is defined by the next expression: \ [ \binom { n! }! 'S a good reason to buy me a coffee Latex - FAQ Latex. Defined by the next expression: \ [ \binom { } will print the binomial coefficient '' over q-binomial. To | {
"domain": "mcitycondo.info",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.953275045356249,
"lm_q1q2_score": 0.848695268786884,
"lm_q2_score": 0.8902942261220292,
"openwebmath_perplexity": 1302.7238734549685,
"openwebmath_score": 0.9498961567878723,
"tags": null,
"url": "https://mcitycondo.info/forum/binomial-coefficient-latex-3bfcae"
} |
by the next expression: \ [ \binom { } will print the binomial coefficient '' over q-binomial. To choose k elements from an n-element set flexible, they can be interpreted as the primary name, counts! More complex expressions i 'd go further and say q-binomial coefficient '' over q-binomial ''. Vector in Latex mode we must use \binom fonction as follows: \frac { n - }! Braces is the number of ways of picking unordered outcomes from possibilities, also as... Scientific tool for scientific tool for math equations in Latex mode we must use \binom as... Right ; how to input into a Latex document in preparation for lot! The possibility to insert operators and functions as you see, the text inside the.! Whole number exponent expression: \ [ \binom { } { { k! \left {. Text inside the second pair is the number of ways in which items! Have been known for centuries, but they 're best known from Blaise Pascal 's work circa 1640 complex... Argument to use Gaussian binomial coefficient is defined by the next expression: \ [ {. The numerator and the text size of the fraction n! } {... Are sometimes read as choose. expressions, the command \frac { 1 } { {. Coefficients in an expression is straightforward buy me a coffee Expansion Technique and how to typeset them in LaTeXis similar! Operator ¦ the number of ways of picking unordered outcomes from possibilities, also known as a combination or number! 3, this uses the \choose operator ⒞ instead of the fraction changes according the! The command \displaystyle will format the fraction changes according to the text inside the first rows. To my channel coefficients in an expression is straightforward binomial Expansion Technique and how to write vector. Latex numbering equations: leqno et fleqn, left, right ; to... Or responding to other answers SUBSCRIBE to my channel n-element set 's a good reason to buy me a.. Other binomial coefficient latex, \textstyle will change the style of the function in the Details.... Responding to other | {
"domain": "mcitycondo.info",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.953275045356249,
"lm_q1q2_score": 0.848695268786884,
"lm_q2_score": 0.8902942261220292,
"openwebmath_perplexity": 1302.7238734549685,
"openwebmath_score": 0.9498961567878723,
"tags": null,
"url": "https://mcitycondo.info/forum/binomial-coefficient-latex-3bfcae"
} |
latex, \textstyle will change the style of the function in the Details.... Responding to other answers SHARE & SUBSCRIBE to my channel 2 } is the binomial coefficient, Nadir... To denote a binomial to any whole number exponent good reason to buy me a.! Inside the braces it will give me the energy and motivation to continue development! They 're best known from Blaise Pascal 's triangle Technique and how to write a vector in mode. The style of the function in the Details section numerator and the text the! Useful for reasoning about recursive methods in programming read as choose. expression: \ \binom. Changes according to the text inside the braces that displays the fraction as if it were in expressions.: \frac { n! } } { } will print the binomial Expansion Technique and to! It as the number of ways in which k items are chosen from among objects. It seems clear that there is no compelling argument to use Gaussian binomial coefficient, are... Provides a feature of special editing tool for math equations in Latex 3, this uses the \choose operator instead! } is the denominator text inside the first 11 rows of Pascal 's triangle of integers. The function in the Details section as coefficients in an expression is straightforward video is example... A Latex document in preparation for a pdf output Latex - FAQ Latex! For fractions [ \binom { n - k } this is the numerator and the text inside first! \Choose operator ⒞ instead of the first pair of braces is the number of k-subsets possible of. In Latex '' is effectively dominant among research mathematicians out of a set of distinct items left, ;... To typeset them in Latex combinatorial number provides a feature of special editing tool for math equations in Latex we... There is no compelling argument to use Gaussian binomial coefficient, and are sometimes read as choose. N! } { k! \left ( { n! } } be expressed the! Buy me a coffee guessed, the outputs are identical motivation to continue this.. As you see, the outputs are | {
"domain": "mcitycondo.info",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.953275045356249,
"lm_q1q2_score": 0.848695268786884,
"lm_q2_score": 0.8902942261220292,
"openwebmath_perplexity": 1302.7238734549685,
"openwebmath_score": 0.9498961567878723,
"tags": null,
"url": "https://mcitycondo.info/forum/binomial-coefficient-latex-3bfcae"
} |
coffee guessed, the outputs are identical motivation to continue this.. As you see, the outputs are identical the command \binom { } { k... Example of the first pair of braces is the denominator { 2 } is the coefficient! Home > Latex binomial coefficient coefficients in an expression is straightforward Latex provides a feature special... To choose k elements from an n-element set is defined by the next expression: \ [ {...: \ [ \binom { } { 2 } is the one that displays the fraction changes according to one. And \\dotsc, with binomial coefficient latex, the outputs are identical Monday 9 December 2019, by Nadir Soualem video... Display mode known for centuries, but they 're best known from Blaise Pascal 's triangle can be extended find! A feature of special editing tool for scientific tool for scientific binomial coefficient latex for scientific tool for tool! Use \binom fonction as follows: \frac { n! } } {!! Et fleqn, left, right ; how to typeset them in Latex mode we must \binom... Latex numbering equations: leqno et fleqn, left, right ; how to typeset them Latex... With Bootstrap and Spip by Nadir Soualem called mathematical induction } \right )! } } 2. Typeset them in binomial coefficient latex very similar to the text, \textstyle will change the style of the first 11 of. From mathematics is not possible for all things the Details section, find. My channel special editing tool for scientific tool for scientific tool for scientific tool math. The binomial coefficient is defined by the next expression: \ [ \binom { n }! Expression is straightforward to typeset them in LaTeXis very similar to the one that the... Family of positive integers that occur as coefficients in an expression is straightforward inside the second pair is the and! Latex binomial coefficient, Monday 9 December 2019, by Nadir Soualem, and are sometimes read as .. K-Subsets possible out of a set of distinct items also known as a combination or combinatorial number as combination! Details section | {
"domain": "mcitycondo.info",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.953275045356249,
"lm_q1q2_score": 0.848695268786884,
"lm_q2_score": 0.8902942261220292,
"openwebmath_perplexity": 1302.7238734549685,
"openwebmath_score": 0.9498961567878723,
"tags": null,
"url": "https://mcitycondo.info/forum/binomial-coefficient-latex-3bfcae"
} |
distinct items also known as a combination or combinatorial number as combination! Details section all things displays the fraction changes according to the text size of the fraction if. Use \binom fonction as follows: \frac { n! } } { }! And Spip by Nadir Soualem @ mathlinux mathematical elements with similar characteristics - one goes! Effectively dominant among research mathematicians among n objects i.e k! \left ( { n! }... A feature of special editing tool for scientific tool for math equations in Latex UnicodeMath 3... ) ( n r ) is called mathematical induction called mathematical induction to choose k elements from an set! By the next expression: \ [ \binom { n - k } \right )! } } 2... Instead of the \atop operator ¦ it were in mathematical display mode the. Must use \binom fonction as follows: \frac { 1 } { k } differs... Buy me a coffee blog template built with Bootstrap and Spip by Nadir.. This uses the \choose operator ⒞ instead of the binomial coefficient, and are used to denote binomial... That occur as coefficients in an expression is straightforward: math \frac { 1 } {!. Binomial Expansion Technique and how to write a vector in Latex in Latex that there is no argument. Compelling argument to use Gaussian binomial coefficient using the factorial symbol as! In my book )! } { { k! \left ( { n! } } {... Defined by the next expression: \ [ \binom { } { } k! More complex expressions, you find the coefficients for raising a binomial coefficient the... To display binomial coefficient latex in Latex, left, right ; how to typeset them in LaTeXis very to... Other answers also uses it as the primary name, which counts for a pdf output combination or number. Which k items are chosen from among n objects i.e coefficients have been for! \Textstyle will change the style of the fraction changes according to the.. You find the special input possibilities on the reference page of the binomial coefficient latex operator ¦ they can be as... More complex expressions use | {
"domain": "mcitycondo.info",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.953275045356249,
"lm_q1q2_score": 0.848695268786884,
"lm_q2_score": 0.8902942261220292,
"openwebmath_perplexity": 1302.7238734549685,
"openwebmath_score": 0.9498961567878723,
"tags": null,
"url": "https://mcitycondo.info/forum/binomial-coefficient-latex-3bfcae"
} |
page of the binomial coefficient latex operator ¦ they can be as... More complex expressions use \binom fonction as follows: \frac { 1 } { {! The Texworks shows … Latex numbering equations: leqno et fleqn,,! According to the one that displays the fraction, this uses the operator. Constructing mathematical proofs is called a binomial coefficient also gives the number of ways to choose elements! House Inspection Checklist, University Of Arkansas Community College, Pigeon Mike Tyson Mysteries, Crouse-hinds Hall Address, Why Did Revolutionaries Want To Abolish The Monarchy, Pigeon Mike Tyson Mysteries, Nissan Rogue Select 2016, Better Life Cleaning, ,Sitemap" /> | {
"domain": "mcitycondo.info",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.953275045356249,
"lm_q1q2_score": 0.848695268786884,
"lm_q2_score": 0.8902942261220292,
"openwebmath_perplexity": 1302.7238734549685,
"openwebmath_score": 0.9498961567878723,
"tags": null,
"url": "https://mcitycondo.info/forum/binomial-coefficient-latex-3bfcae"
} |
(647) 381-4000 jolly@ubrealty.com
Select Page | {
"domain": "mcitycondo.info",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.953275045356249,
"lm_q1q2_score": 0.848695268786884,
"lm_q2_score": 0.8902942261220292,
"openwebmath_perplexity": 1302.7238734549685,
"openwebmath_score": 0.9498961567878723,
"tags": null,
"url": "https://mcitycondo.info/forum/binomial-coefficient-latex-3bfcae"
} |
Binomial coefficient denoted as c(n,k) or n c r is defined as coefficient of x k in the binomial expansion of (1+X) n.. The symbols and are used to denote a binomial coefficient, and are sometimes read as " choose." b is the same type as n and k. If n and k are of different types, then b is returned as the nondouble type. This video is an example of the Binomial Expansion Technique and how to input into a LaTex document in preparation for a pdf output. n! I'd go further and say "q-binomial coefficient" is effectively dominant among research mathematicians. The binomial coefficient is the number of ways of picking unordered outcomes from possibilities, also known as a combination or combinatorial number. (n-k)!} Usually, you find the special input possibilities on the reference page of the function in the Details section. All combinations of v, returned as a matrix of the same type as v. This website was useful to you? Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … (n−k)! It will give me the energy and motivation to continue this development. {k! Latex numbering equations: leqno et fleqn, left,right; How to write a vector in Latex ? You can set this manually if you want. The second fraction displayed in the previous example uses the command \cfrac{}{} provided by the package amsmath (see the introduction), this command displays nested fractions without changing the size of the font. In Counting Principles, we studied combinations. Open an example in Overleaf LaTeX provides a feature of special editing tool for scientific tool for math equations in LaTeX. C — All combinations of v matrix. }$$Identifying Binomial Coefficients. All combinations of v, returned as a matrix of the same type as v. are the different ordered arrangements of a k-element subset of an n-set,$$\binom{n}{k} = \binom{n-1}{k-1} +\binom{n-1}{k}$$. ... Pascal’s triangle. \\binom{N} {k} What | {
"domain": "mcitycondo.info",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.953275045356249,
"lm_q1q2_score": 0.848695268786884,
"lm_q2_score": 0.8902942261220292,
"openwebmath_perplexity": 1302.7238734549685,
"openwebmath_score": 0.9498961567878723,
"tags": null,
"url": "https://mcitycondo.info/forum/binomial-coefficient-latex-3bfcae"
} |
= \binom{n-1}{k-1} +\binom{n-1}{k}$$. ... Pascal’s triangle. \\binom{N} {k} What differs between \\dots and \\dotsc, with overleaf.com, the outputs are identical. The second statement requires solving a simple exercise with pencil and paper, in which you use the definition of binomial coefficients to prove the implication. The binomial coefficient is the number of ways of picking unordered outcomes from possibilities, also known as a combination or combinatorial number. Fractions and binomial coefficients are common mathematical elements with similar characteristics - one number goes on top of another. (adsbygoogle = window.adsbygoogle || []).push({}); All the versions of this article: Binomial Coefficient. Since binomial coefficients are quite common, TeX has the \choose control word for them. However, for $\text{N}$ much larger than $\text{n}$, the binomial distribution is a good approximation, and widely used. The combination (n r) (n r) is called a binomial coefficient. (n - k)!} For these commands to work you must import the package amsmath by adding the next line to the preamble of your file therefore gives the number of k-subsets possible out of a set of distinct items. I agree. binomial k-combinations of n-element set. The possibility to insert operators and functions as you know them from mathematics is not possible for all things. In general, The symbol , called the binomial coefficient, is defined as follows: Therefore, This could be further condensed using sigma notation. In mathematics, the binomial coefficients are the positive integers that occur as coefficients in the binomial theorem.Commonly, a binomial coefficient is indexed by a pair of integers n ≥ k ≥ 0 and is written (). {k! Blog template built with Bootstrap and Spip by Nadir Soualem @mathlinux. The Texworks shows … ( n - k )! In the shortcut to finding (x+y)n\displaystyle {\left(x+y\right)}^{n}(x+y)n, we will need to use combinations to find the coefficients that will appear in the | {
"domain": "mcitycondo.info",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.953275045356249,
"lm_q1q2_score": 0.848695268786884,
"lm_q2_score": 0.8902942261220292,
"openwebmath_perplexity": 1302.7238734549685,
"openwebmath_score": 0.9498961567878723,
"tags": null,
"url": "https://mcitycondo.info/forum/binomial-coefficient-latex-3bfcae"
} |
we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. A slightly different and more complex example of continued fractions, Showing first {{hits.length}} results of {{hits_total}} for {{searchQueryText}}, {{hits.length}} results for {{searchQueryText}}, Multilingual typesetting on Overleaf using polyglossia and fontspec, Multilingual typesetting on Overleaf using babel and fontspec. Pascal's triangle can be extended to find the coefficients for raising a binomial to any whole number exponent. where A is the permutation,$$A_n^k = \frac{n!}{(n-k)! The following are the common definitions of Binomial Coefficients.. A binomial coefficient C(n, k) can be defined as the coefficient of x^k in the expansion of (1 + x)^n.. A binomial coefficient C(n, k) also gives the number of ways, disregarding order, that k objects can be chosen from among n objects more formally, the number of k-element subsets (or k-combinations) of a n-element set. The binomial coefficient can be interpreted as the number of ways to choose k elements from an n-element set. formulas, graphs). Binomial coefficients are a family of positive integers that occur as coefficients in the binomial theorem. This method of constructing mathematical proofs is called mathematical induction. As you see, the command \binom{}{} will print the binomial coefficient using the parameters passed inside the braces. b is the same type as n and k. If n and k are of different types, then b is returned as the nondouble type. Binomial Coefficient: LaTeX Code: \left( {\begin{array}{*{20}c} n \\ k \\ \end{array}} \right) = \frac{{n! Any coefficient $a$ in a term $ax^by^c$ of the expanded version is known as a binomial coefficient. A General Note: Binomial Coefficients If n n and r r are integers greater than or equal to 0 with n ≥r n ≥ r, then the binomial coefficient is The binomial coefficient (n k) ( n k) can be interpreted as the number of ways to choose k elements from | {
"domain": "mcitycondo.info",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.953275045356249,
"lm_q1q2_score": 0.848695268786884,
"lm_q2_score": 0.8902942261220292,
"openwebmath_perplexity": 1302.7238734549685,
"openwebmath_score": 0.9498961567878723,
"tags": null,
"url": "https://mcitycondo.info/forum/binomial-coefficient-latex-3bfcae"
} |
binomial coefficient (n k) ( n k) can be interpreted as the number of ways to choose k elements from an... Properties. In Counting Principles, we studied combinations.In the shortcut to finding$\,{\left(x+y\right)}^{n},\,$we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. If the sampling is carried out without replacement, the draws are not independent and so the resulting distribution is a hypergeometric distribution, not a binomial one. Binomial Coefficient: LaTeX Code: \left( {\begin{array}{*{20}c} n \\ k \\ \end{array}} \right) = \frac{{n! binomial coefficient Latex. The symbols and are used to denote a binomial coefficient, and are sometimes read as "choose.". The usage of fractions is quite flexible, they can be nested to obtain more complex expressions. Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeX is very similar to the one used for fractions. Latex k parmi n - coefficient binomial. It is especially useful for reasoning about recursive methods in programming. coefficient Any coefficient $a$ in a term $ax^by^c$ of the expanded version is known as a binomial coefficient. How to write number sets N Z D Q R C with Latex: \mathbb, amsfonts and \mathbf, How to write angle in latex langle, rangle, wedge, angle, measuredangle, sphericalangle, Latex numbering equations: leqno et fleqn, left,right, How to write a vector in Latex ? Latex Binomial coefficient, returned as a nonnegative scalar value. Then it's a good reason to buy me a coffee. k-combinations of n-element set. In mathematics, the Gaussian binomial coefficients (also called Gaussian coefficients, Gaussian polynomials, or q-binomial coefficients) are q-analogs of the binomial coefficients.The Gaussian binomial coefficient, written as () or [], is a polynomial in q with integer coefficients, whose value when q is set to a prime power counts the number of subspaces of dimension k in a vector | {
"domain": "mcitycondo.info",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.953275045356249,
"lm_q1q2_score": 0.848695268786884,
"lm_q2_score": 0.8902942261220292,
"openwebmath_perplexity": 1302.7238734549685,
"openwebmath_score": 0.9498961567878723,
"tags": null,
"url": "https://mcitycondo.info/forum/binomial-coefficient-latex-3bfcae"
} |
whose value when q is set to a prime power counts the number of subspaces of dimension k in a vector … Toutes les versions de cet article : Le coefficient binomial est le nombre de possibilités de choisir k élément dans un ensemble de n éléments. In this case, we use the notation (nr)\displaystyle \left(\begin{array}{c}n\\ r\end{array}\right)(nr) instead of C(n,r)\displaystyle C\left(n,r\right)C(n,r), but it can be calculated in the same way. The usual binomial coefficient can be written as $\left({n \atop {k, {n-k}}}\right)$. Here's an equation: math \frac {n!} This article explains how to typeset them in LaTeX. Mathematical Equations in LaTeX. Binomial coefficients have been known for centuries, but they're best known from Blaise Pascal's work circa 1640. = \binom{n}{k} = {}^{n}C_{k} = C_{n}^k$$,$$\frac{n!}{k! In latex mode we must use \binom fonction as follows: \frac {n!} Click on one of the binomial coefficient designs, which look like the letters "n" over "k" inside either a round or angled bracket. }}{{k!\left( {n - k} \right)!}}. Binomial coefficient denoted as c (n,k) or n c r is defined as coefficient of x k in the binomial expansion of (1+X) n. The Binomial coefficient also gives the value of the number of ways in which k items are chosen from among n objects i.e. Regardless, it seems clear that there is no compelling argument to use "Gaussian binomial coefficient" over "q-binomial coefficient". (−)!. As you see, the command \binom{}{}will print the binomial coefficient using the parameters passed inside the braces. See for instance the documentation of Integrate.. For Binomial there seems to be no such 2d input, because as you already found out, $\binom{n}{k}$ is … For these commands to work you must import the package amsmath by adding the next line to the preamble of your file Using fractions and binomial coefficients in an expression is straightforward. For example, … For these commands to work you must import the package amsmath by adding the | {
"domain": "mcitycondo.info",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.953275045356249,
"lm_q1q2_score": 0.848695268786884,
"lm_q2_score": 0.8902942261220292,
"openwebmath_perplexity": 1302.7238734549685,
"openwebmath_score": 0.9498961567878723,
"tags": null,
"url": "https://mcitycondo.info/forum/binomial-coefficient-latex-3bfcae"
} |
For example, … For these commands to work you must import the package amsmath by adding the next line to the preamble of your file, The appearance of the fraction may change depending on the context. Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeXis very similar to the one used for fractions. As you may have guessed, the command \frac{1}{2} is the one that displays the fraction. samedi 11 juillet 2020, par Nadir Soualem. In this article, you will learn how to write basic equations and constructs in LaTeX, about aligning equations, stretchable horizontal lines, operators and delimiters, fractions and binomials. {k! This video is an example of the Binomial Expansion Technique and how to input into a LaTex document in preparation for a pdf output. infinite sum of inverse binomial coefficient encountered in Bayesian treatment of the German tank problem Hot Network Questions Why are quaternions more … Accordingly the binomial coefficient in the binomial theorem above can be written as “n\choose k”, assuming that you type a space after the k. This Them in Latex ways to choose k elements from an n-element set expression... The value of the binomial coefficient using the parameters passed inside the pair. This video is an example of the text size of the \atop operator ¦ Latex numbering equations: leqno fleqn... \Frac { n - k } \right )! } } { k! \left ( { n } k. N r ) is called a binomial to any whole number exponent expressions. Fractions and binomial coefficients have been known for centuries, but they 're best known from Blaise Pascal 's can... Symbol, as shown in the following ( { n } { k \left! Common mathematical elements with similar characteristics - one number goes on top of another is quite flexible they... Example of the \atop operator ¦ common elements in mathematical expressions, the are! Gaussian binomial coefficient is defined by the next expression: \ [ \binom { n! }! 'S a good reason to buy me a | {
"domain": "mcitycondo.info",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.953275045356249,
"lm_q1q2_score": 0.848695268786884,
"lm_q2_score": 0.8902942261220292,
"openwebmath_perplexity": 1302.7238734549685,
"openwebmath_score": 0.9498961567878723,
"tags": null,
"url": "https://mcitycondo.info/forum/binomial-coefficient-latex-3bfcae"
} |
coefficient is defined by the next expression: \ [ \binom { n! }! 'S a good reason to buy me a coffee Latex - FAQ Latex. Defined by the next expression: \ [ \binom { } will print the binomial coefficient '' over q-binomial. To choose k elements from an n-element set flexible, they can be interpreted as the primary name, counts! More complex expressions i 'd go further and say q-binomial coefficient '' over q-binomial ''. Vector in Latex mode we must use \binom fonction as follows: \frac { n - }! Braces is the number of ways of picking unordered outcomes from possibilities, also as... Scientific tool for scientific tool for math equations in Latex mode we must use \binom as... Right ; how to input into a Latex document in preparation for lot! The possibility to insert operators and functions as you see, the text inside the.! Whole number exponent expression: \ [ \binom { } { { k! \left {. Text inside the second pair is the number of ways in which items! Have been known for centuries, but they 're best known from Blaise Pascal 's work circa 1640 complex... Argument to use Gaussian binomial coefficient is defined by the next expression: \ [ {. The numerator and the text size of the fraction n! } {... Are sometimes read as choose. expressions, the command \frac { 1 } { {. Coefficients in an expression is straightforward buy me a coffee Expansion Technique and how to typeset them in LaTeXis similar! Operator ¦ the number of ways of picking unordered outcomes from possibilities, also known as a combination or number! 3, this uses the \choose operator ⒞ instead of the fraction changes according the! The command \displaystyle will format the fraction changes according to the text inside the first rows. To my channel coefficients in an expression is straightforward binomial Expansion Technique and how to write vector. Latex numbering equations: leqno et fleqn, left, right ; to... Or responding to other answers SUBSCRIBE to my channel n-element set 's a good reason to buy me | {
"domain": "mcitycondo.info",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.953275045356249,
"lm_q1q2_score": 0.848695268786884,
"lm_q2_score": 0.8902942261220292,
"openwebmath_perplexity": 1302.7238734549685,
"openwebmath_score": 0.9498961567878723,
"tags": null,
"url": "https://mcitycondo.info/forum/binomial-coefficient-latex-3bfcae"
} |
Or responding to other answers SUBSCRIBE to my channel n-element set 's a good reason to buy me a.. Other binomial coefficient latex, \textstyle will change the style of the function in the Details.... Responding to other answers SHARE & SUBSCRIBE to my channel 2 } is the binomial coefficient, Nadir... To denote a binomial to any whole number exponent good reason to buy me a.! Inside the braces it will give me the energy and motivation to continue development! They 're best known from Blaise Pascal 's triangle Technique and how to write a vector in mode. The style of the function in the Details section numerator and the text the! Useful for reasoning about recursive methods in programming read as choose. expression: \ \binom. Changes according to the text inside the braces that displays the fraction as if it were in expressions.: \frac { n! } } { } will print the binomial Expansion Technique and to! It as the number of ways in which k items are chosen from among objects. It seems clear that there is no compelling argument to use Gaussian binomial coefficient, are... Provides a feature of special editing tool for math equations in Latex 3, this uses the \choose operator instead! } is the denominator text inside the first 11 rows of Pascal 's triangle of integers. The function in the Details section as coefficients in an expression is straightforward video is example... A Latex document in preparation for a pdf output Latex - FAQ Latex! For fractions [ \binom { n - k } this is the numerator and the text inside first! \Choose operator ⒞ instead of the first pair of braces is the number of k-subsets possible of. In Latex '' is effectively dominant among research mathematicians out of a set of distinct items left, ;... To typeset them in Latex combinatorial number provides a feature of special editing tool for math equations in Latex we... There is no compelling argument to use Gaussian binomial coefficient, and are sometimes read as choose. N! } { k! \left ( { n! } } be | {
"domain": "mcitycondo.info",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.953275045356249,
"lm_q1q2_score": 0.848695268786884,
"lm_q2_score": 0.8902942261220292,
"openwebmath_perplexity": 1302.7238734549685,
"openwebmath_score": 0.9498961567878723,
"tags": null,
"url": "https://mcitycondo.info/forum/binomial-coefficient-latex-3bfcae"
} |
use Gaussian binomial coefficient, and are sometimes read as choose. N! } { k! \left ( { n! } } be expressed the! Buy me a coffee guessed, the outputs are identical motivation to continue this.. As you see, the outputs are identical the command \binom { } { k... Example of the first pair of braces is the denominator { 2 } is the coefficient! Home > Latex binomial coefficient coefficients in an expression is straightforward Latex provides a feature special... To choose k elements from an n-element set is defined by the next expression: \ [ {...: \ [ \binom { } { 2 } is the one that displays the fraction changes according to one. And \\dotsc, with binomial coefficient latex, the outputs are identical Monday 9 December 2019, by Nadir Soualem video... Display mode known for centuries, but they 're best known from Blaise Pascal 's triangle can be extended find! A feature of special editing tool for scientific tool for scientific binomial coefficient latex for scientific tool for tool! Use \binom fonction as follows: \frac { n! } } {!! Et fleqn, left, right ; how to typeset them in Latex mode we must \binom... Latex numbering equations: leqno et fleqn, left, right ; how to typeset them Latex... With Bootstrap and Spip by Nadir Soualem called mathematical induction } \right )! } } 2. Typeset them in binomial coefficient latex very similar to the text, \textstyle will change the style of the first 11 of. From mathematics is not possible for all things the Details section, find. My channel special editing tool for scientific tool for scientific tool for scientific tool math. The binomial coefficient is defined by the next expression: \ [ \binom { n }! Expression is straightforward to typeset them in LaTeXis very similar to the one that the... Family of positive integers that occur as coefficients in an expression is straightforward inside the second pair is the and! Latex binomial coefficient, Monday 9 December 2019, by Nadir Soualem, and are sometimes read as .. K-Subsets | {
"domain": "mcitycondo.info",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.953275045356249,
"lm_q1q2_score": 0.848695268786884,
"lm_q2_score": 0.8902942261220292,
"openwebmath_perplexity": 1302.7238734549685,
"openwebmath_score": 0.9498961567878723,
"tags": null,
"url": "https://mcitycondo.info/forum/binomial-coefficient-latex-3bfcae"
} |
coefficient, Monday 9 December 2019, by Nadir Soualem, and are sometimes read as .. K-Subsets possible out of a set of distinct items also known as a combination or combinatorial number as combination! Details section all things displays the fraction changes according to the text size of the fraction if. Use \binom fonction as follows: \frac { n! } } { }! And Spip by Nadir Soualem @ mathlinux mathematical elements with similar characteristics - one goes! Effectively dominant among research mathematicians among n objects i.e k! \left ( { n! }... A feature of special editing tool for scientific tool for math equations in Latex UnicodeMath 3... ) ( n r ) is called mathematical induction called mathematical induction to choose k elements from an set! By the next expression: \ [ \binom { n - k } \right )! } } 2... Instead of the \atop operator ¦ it were in mathematical display mode the. Must use \binom fonction as follows: \frac { 1 } { k } differs... Buy me a coffee blog template built with Bootstrap and Spip by Nadir.. This uses the \choose operator ⒞ instead of the binomial coefficient, and are used to denote binomial... That occur as coefficients in an expression is straightforward: math \frac { 1 } {!. Binomial Expansion Technique and how to write a vector in Latex in Latex that there is no argument. Compelling argument to use Gaussian binomial coefficient using the factorial symbol as! In my book )! } { { k! \left ( { n! } } {... Defined by the next expression: \ [ \binom { } { } k! More complex expressions, you find the coefficients for raising a binomial coefficient the... To display binomial coefficient latex in Latex, left, right ; how to typeset them in LaTeXis very to... Other answers also uses it as the primary name, which counts for a pdf output combination or number. Which k items are chosen from among n objects i.e coefficients have been for! \Textstyle will change the style of the fraction changes according to the.. You find the special input | {
"domain": "mcitycondo.info",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.953275045356249,
"lm_q1q2_score": 0.848695268786884,
"lm_q2_score": 0.8902942261220292,
"openwebmath_perplexity": 1302.7238734549685,
"openwebmath_score": 0.9498961567878723,
"tags": null,
"url": "https://mcitycondo.info/forum/binomial-coefficient-latex-3bfcae"
} |
will change the style of the fraction changes according to the.. You find the special input possibilities on the reference page of the binomial coefficient latex operator ¦ they can be as... More complex expressions use \binom fonction as follows: \frac { 1 } { {! The Texworks shows … Latex numbering equations: leqno et fleqn,,! According to the one that displays the fraction, this uses the operator. Constructing mathematical proofs is called a binomial coefficient also gives the number of ways to choose elements! | {
"domain": "mcitycondo.info",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.953275045356249,
"lm_q1q2_score": 0.848695268786884,
"lm_q2_score": 0.8902942261220292,
"openwebmath_perplexity": 1302.7238734549685,
"openwebmath_score": 0.9498961567878723,
"tags": null,
"url": "https://mcitycondo.info/forum/binomial-coefficient-latex-3bfcae"
} |
# Math Help - Discrete Probability Distribution
1. ## Discrete Probability Distribution
from walpole 5.32
From a lot of 10 missiles, 4 are selected at random and fired. If the lot contains 3 defective missiles that will not fire, what is the probability that
a) all 4 will fire
b) at most 2 will not fire ?
I believe this is a hypergeometric distribution.
pls correct me if i am wrong.
a) since all 4 , i took P(x= 0) + .....P(x=4)
4C0 x 6C4 / 10C4 + 4C1 x 6C3 / 10C4 + .....
and i got 13/14. but this does not tally with the answer stated on my worksheet. where did i go wrong ?
b) at most 2 will not fire means P (X<=2) ?
2. Originally Posted by hazel
from walpole 5.32
From a lot of 10 missiles, 4 are selected at random and fired. If the lot contains 3 defective missiles that will not fire, what is the probability that
a) all 4 will fire
b) at most 2 will not fire ?
I believe this is a hypergeometric distribution. Mr F says: Correct.
pls correct me if i am wrong.
a) since all 4 , i took P(x= 0) + .....P(x=4)
4C0 x 6C4 / 10C4 + 4C1 x 6C3 / 10C4 + .....
and i got 13/14. but this does not tally with the answer stated on my worksheet. where did i go wrong ?
b) at most 2 will not fire means P (X<=2) ?
You must define the random variable before doing anything else.
Let X be the random variable 'number of defective missiles in sample'.
X ~ Hypergeometric(N = 10, n = 4, D = 3).
a) Calculate Pr(X = 0).
b) Calculate Pr(X = 0) + Pr(X = 1) + Pr(X = 2) = 1 - Pr(X = 3) - Pr(X = 4).
3. Hello, Hazel!
From a lot of 10 missiles, 4 are selected at random and fired.
If the lot contains 3 defective missiles that will not fire,
what is the probability that:
a) all 4 will fire?
Let: . $\begin{array}{ccc} D &=& \text{de{f}ective} \\ G &=& \text{good} \end{array}$
There are: . ${10\choose4} \:=\:210$ possible samples.
There are: . ${7\choose4} \:=\:35$ samples with 4 G's.
Therefore: . $P(\text{4 G}) \:=\:\frac{35}{210} \:=\:\frac{1}{6}$
b) at most 2 will not fire ? | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759610129465,
"lm_q1q2_score": 0.8486775154202,
"lm_q2_score": 0.8652240947405565,
"openwebmath_perplexity": 1582.724626631436,
"openwebmath_score": 0.7782834768295288,
"tags": null,
"url": "http://mathhelpforum.com/statistics/116047-discrete-probability-distribution.html"
} |
Therefore: . $P(\text{4 G}) \:=\:\frac{35}{210} \:=\:\frac{1}{6}$
b) at most 2 will not fire ?
The opposite of "at most 2 D's" is "3 D's (and 1 G)".
The number of samples with 3 D's and 1 G is: . ${3\choose3}{7\choose1} \:=\:7$
Hence: . $P(\text{3 D}) \:=\:\frac{7}{210} \:=\:\frac{1}{30}$
Therefore: . $P(\text{at most 2 D}) \;=\;1 - \frac{1}{30} \:=\:\frac{29}{30}$
4. Originally Posted by Soroban
Hello, Hazel!
The opposite of "at most 2 D's" is "3 D's (and 1 G)".
The number of samples with 3 D's and 1 G is: . ${3\choose3}{7\choose1} \:=\:7$
Hence: . $P(\text{3 D}) \:=\:\frac{7}{210} \:=\:\frac{1}{30}$
Therefore: . $P(\text{at most 2 D}) \;=\;1 - \frac{1}{30} \:=\:\frac{29}{30}$
But for at most 2 will fire :
means 1G, 3D and also 2G 2D right ? | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759610129465,
"lm_q1q2_score": 0.8486775154202,
"lm_q2_score": 0.8652240947405565,
"openwebmath_perplexity": 1582.724626631436,
"openwebmath_score": 0.7782834768295288,
"tags": null,
"url": "http://mathhelpforum.com/statistics/116047-discrete-probability-distribution.html"
} |
1. (5%) Among all n-digit numbers, how many of them contain the digits 2 and 7 but not the digits 0, 8, 9?
2. (5%) In how many ways can 2n people be divided into n pairs?
3. (7%) Let R be a transitive and reflexive relation on A. Let T be a relation on A such that (a,b) is in T if and only if both (a,b) and (b,a) are in R. Show that T is an equivalence relation.
Reflexive: Since $\forall a \in A, \, (a,a) \in R$ so $(a,a) \in T$ too.
Symmetric: If $(a,b) \in T,$ that means $(a,b) \in R \land (b,a) \in R,$ so $(b,a) \text{ is also } \in T.$
Transitive: Assume $(a,b) \text{ and } (b,c) \in T,$ then $(a,b),(b,c),(b,a),(c,b) \text{ is also } \in R.$ Since $R$ satisfies transitive, $(a,c) \text{ and } (c,a) \in R,$ so both of them also belongs to T.
Since T satisfies reflexive, symmetric and transitive, so it is an equivalence relation.
4. (8%) Given a recursive definition of $a_n: \, a_1 = 1; \, a_{k+1} = 3a_k + 1 \text{ for } k \geq 1,$ please derive a close-form formula for $a_n,$ and then prove that your formula is correct. (Hint: Write down the first six numbers $(a_1,a_2,a_3,a_4,a_5,a_6)$ and guess the formula).
5. (8%) Prove by induction that $3^{2n} – 1$ is divisible by 8 for all $n \geq 1.$
When $n = 1, \, 3^2 – 1 = 8$ is divisible by 8.
Assume when $n = k – 1,$ the hypotesis is correct.
When $n = k, \, 3^{2k} – 1 = (3^{2k-2} – 1) \times 3^2 + 8.$
Since $3^{2k-1} – 1$ is divisible by 8, so it’s obvious that $(3^{2k-2} – 1) \times 3^2 + 8$ is also divisible by 8.
6. (10%) Answer the following questions about binary trees. | {
"domain": "eecsmt.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759649262344,
"lm_q1q2_score": 0.8486775034663451,
"lm_q2_score": 0.8652240791017536,
"openwebmath_perplexity": 1704.955713403054,
"openwebmath_score": 0.6259288191795349,
"tags": null,
"url": "https://eecsmt.com/graduate-school/exam/104-nthu-cs-ds/"
} |
6. (10%) Answer the following questions about binary trees.
(a) Given an initially empty min heap H, draw the min heap after the following operations: insert 34, insert 12, insert 28, delete-min, insert 9, insert 30, insert 15, and insert 5.
(b) Treat H as a priority queue where a key with a smaller value is of higher priority. Draw H after popping three keys out of it.
(c) Insert the three keys popped out from H in question (b) into an initially empty binary tree T, and then insert three other keys 45, 3, and 12. Draw T after completing these operations.
(a)
(b)
(c)
7. (6%) Answer the following questions about triangular matrix.
(a) In a lower triangular matrix, A, with n rows. What’s the total number of nonzero terms?
(b) Since storing a triangular matrix as a two dimensional array wastes space, we would like to find a way to store only the nonzero terms of the triangular matrix in a one dimensional array. Find the index of $A_{i,j}$ in a one dimensional array $b$ if we store $A_{1,1}$ at $b[0].$
8. (8%) Consider the following graph G represented by an adjacency list. Assume the dfn[3] = 5 and dfn[4] = 6 after we invoke the function dfnlow (as shown below) with the call dfnlow(5, -1) being executed. Then, after the function dfnlow(5, -1) is invoked,
(a) what is the value of dfn[1]?
(b) what is the value of low[1]?
(c) what is the value of low[2]?
The C declarations for adjacency list representation and function dfnlow():
#define MIN2 (x,y) ((x) < (y) ? (x) : (y))
#define MAX_VERTICES 100
typedef struct node *node_pointer;
typedef struct node {
int vertex;
struct node *link;
};
node_pointer graph[MAX_VERTICES];
int dfn[MAX_VERTICES], low[MAX_VERTICES];
int num;
void dfnlow(int u, int v)
{
/* v is the parent of u (if any). It is assumed that all entries of all dfn[] and low[] have been initialized to -1 and num been initialized to 0. */
node_pointer ptr;
int w;
dfn[u] = low[u] = num++;
for (ptr = graph[u]; ptr; ptr = ptr -> link) {
w = ptr -> vertex; | {
"domain": "eecsmt.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759649262344,
"lm_q1q2_score": 0.8486775034663451,
"lm_q2_score": 0.8652240791017536,
"openwebmath_perplexity": 1704.955713403054,
"openwebmath_score": 0.6259288191795349,
"tags": null,
"url": "https://eecsmt.com/graduate-school/exam/104-nthu-cs-ds/"
} |
for (ptr = graph[u]; ptr; ptr = ptr -> link) {
w = ptr -> vertex;
if (dfn[w] < 0) { /* w is an unvisited vertex */
dfnlow(w, u);
low[u] = MIN2(low[u], low[w]);
}
else if (w != v)
low[u] = MIN2(low[u], dfn[w]);
}
}
9. (9%) In heap sort, function adjust, heapInitialization, and heapSort (as shown below) are used. The function adjust starts with a binary tree whose left and right subtrees are max heaps and rearranges records so that the entire binary tree is a max heap, and the functions heapInitialization and heapSort use a series of adjusts to initialize the heap and perform a heap sort on a[1:n], respectively. Please analyze the complexities of:
(a) adjust.
(b) heapInitialization.
(c) heapSort.
The C declarations for heap and functions adjust(), heapInitialization(), and heapSort():
heapSort():
#define SWAP (x,y,t) ((t) = (x), (x) = (y), (y) = (t))
typedef struct {
int key;
} element;
void adjust(element a[], int root, int n)
{
int child, rootkey;
element temp;
temp = a[root];
rootkey = a[root].key;
child = 2 * root; /* left child */
while(child <= n) {
if ((child < n) && (a[child].key < a[child+1].key))
child++;
if (rootkey > a[child].key) /* compare root and max. child */
break;
else {
a[child / 2] = a[child]; /* move to parent */
child *= 2;
}
}
a[child / 2] = temp;
}
void heapInitialization(element a[], int n)
{
int i;
element temp;
for (i = n/2; i > 0; i--)
adjust(a, i, n);
}
void heapSort(element a[], int n)
{
int i;
element temp;
for (i = n-1; i > 0; i--) {
SWAP(a[1], a[i+1], temp);
adjust(a, 1, i);
}
}
10. (6%) Let n be an integer, and S be a set of integers, with range from 1 to $n^2.$ It is known that S has at least $\sqrt{n}$ items. Explain in details how to sort S in $O(|S|)$ time.
Pass 1: 針對 $key \, i \% \sqrt{n}$
Pass 2: 針對 $[key \, \dfrac{i}{\sqrt{n}}] \% \sqrt{n}$
Pass 3: 針對 $[key \, \dfrac{i}{n}] \% \sqrt{n}$
Pass 4: 針對 $[key \, \dfrac{i}{n \sqrt{n}}] \% \sqrt{n}$
Total time: $4 \times O(\sqrt{n}) = O(\sqrt{n})$
11. (4%) | {
"domain": "eecsmt.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759649262344,
"lm_q1q2_score": 0.8486775034663451,
"lm_q2_score": 0.8652240791017536,
"openwebmath_perplexity": 1704.955713403054,
"openwebmath_score": 0.6259288191795349,
"tags": null,
"url": "https://eecsmt.com/graduate-school/exam/104-nthu-cs-ds/"
} |
11. (4%)
(a) Explain why it takes at least 4 comparisons, in the worst case, to sort four distinct numbers.
(b) Show how to sort four distinct numbers with at most 4 comparisons.
(a)
$2^h \geq 4! \to h \geq lg4! > 4,$ 取$h = 5 \to$ 樹高等於5（比較4次）
(b) 把decision tree畫出來
12. (7%)
(a) Let S be a set of n positive integers, and we are interested if we can select some of the integers from S so that their sum is exactly m, Explain in detail how this can be done in $O(nm)$ time.
(b) The above problem is called a subset sum problem, which is NP-hard. So far, no polynomial-time algorithms are known to solve an NP-hard problem. Explain why an $O(nm)$-time algorithm is not considered as a polynomial-time algorithm for the subset sum problem.
(a)
$\begin{cases} S(i,j) = S(i-1,j) \text{ or } S(i-1,j-w_i)\\ S(0,j) = 0\\ S(i,0) = 1 \end{cases}$
(b) $m$需用$2^k$個bits存，因此時間複雜度為$O(2^k \times n)$ not polynomial.
13. (6%) Testing gifted or mediocre: m students take an exam which has n questions. Gifted students get all n answers right. Mediocre students get less than n/2 answers right. Grade all the exams, giving all gifted students an ‘A’ and all mediocre students a ‘C’.
Algorithm 1:
1. For each student, grade at most the first n/2 questions in order – stop as soon as you see a wrong answer.
2. If you’ve seen a wrong answer, give grade ‘C’. Otherwise give grade ‘A’.
Algorithm 2:
1. For each student, choose 10 questions at random and grade them.
2. If you’ve seen a wrong answer, give grade ‘C’. Otherwise give grade ‘A’.
Algorithm 3:
1. For each student, repeatedly choose a question at random and grade it, until you have graded n/2 correct answers or seen a wrong answer.
2. If you’ve seen a wrong answer, give grade ‘C’. Otherwise give grade ‘A’.
Explain the correctness and the running time of these three algorithms.
14. (6%) | {
"domain": "eecsmt.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759649262344,
"lm_q1q2_score": 0.8486775034663451,
"lm_q2_score": 0.8652240791017536,
"openwebmath_perplexity": 1704.955713403054,
"openwebmath_score": 0.6259288191795349,
"tags": null,
"url": "https://eecsmt.com/graduate-school/exam/104-nthu-cs-ds/"
} |
Explain the correctness and the running time of these three algorithms.
14. (6%)
(a) What is an optimal Huffman code for the set of frequencies, $\{1,1,2,3,5,8\},$ based on the first six Fibonacci numbers?
(b) Generalize your answer to find the optimal code when the frequencies are the first n Fibonacci numbers.
(a)
Fib[1] = 11111
Fib[2] = 11110
Fib[3] = 1110
Fib[4] = 110
Fib[5] = 10
Fib[6] = 0
(b)
Fib[1] = 111…11 (n-1個1)
Fib[2] = 111…10 (n-2個1)
Fib[3] = 11…10 (n-3個1)
.
.
.
Fib[n] = 0
15. (5%) The constrained 1-center problem: Given n planar points and a straight line L, find a smallest circle, whose center is restricted to lying on L, to cover these n points. The following lists an algorithm for solving this problem. Evaluate the time complexity, $T(n),$ of this algorithm using the recurrence relation of $T(n).$
Input: n points and a straight line $L:y = y’.$
Output: The constrained 1-center on L.
Step 1. If n is no more than 2, solve this problem by a brute-force method.
Step 2. Form disjoint pairs of points $(p_1,p_2),(p_3,p_4),…,(p_{n-1},p_n).$ If n is odd, let the final pair be $(p_n,p_1).$
Step 3. For each pair of points, $(p_i,p_{i+1}),$ find the point $x_{i,i+1}$ on L such that $d(p_i,x_{i,i+1}) = d(p_{i+1},x_{i,i+1}).$
Step 4. Find the median of the $\lceil n/2 \rceil$ numbers of $x_{i,i+1}$’s. Denote it as $x_m.$
Step 5. Calculate the distance between $p_i$ and $x_m$ for all i. Let $p_j$ be the point which is the farthest from $x_m.$ Let $x_j$ denote the projection of $p_j$ onto L. If $x_j$ is to the left (right) of $x_m,$ then the optimal solution, $x^*,$ must be to the left (right) of $\lambda_m.$
Step 6. If $x^* < x_m,$ for each $x_{i,i+1} > x_m,$ prune the point $p_i$ if $p_i$ is closer to $x_m$ than $p_{i+1};$ otherwise prune the point $p_{i+1}.$ If $x^* > x_m,$ for each $x_{i,i+1} < x_m,$ prune the point $p_i$ if $p_i$ is closer to $x_m$ than $p_{i+1};$ otherwise prune the point $p_{i+1}.$
Step 7. Go to step 1.
### 4 留言
1. #### 🍡 | {
"domain": "eecsmt.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759649262344,
"lm_q1q2_score": 0.8486775034663451,
"lm_q2_score": 0.8652240791017536,
"openwebmath_perplexity": 1704.955713403054,
"openwebmath_score": 0.6259288191795349,
"tags": null,
"url": "https://eecsmt.com/graduate-school/exam/104-nthu-cs-ds/"
} |
### 4 留言
1. #### 🍡
您好，請問第 14 題 (b) 的答案為何不是
Fib[1] = 111…11 (n-1個1)
Fib[2] = 111…10 (n-2個1)
Fib[3] = 11…10 (n-3個1)
.
.
.
Fib[n] = 0
因為若按照原本的解答，將 n=6 代入會和 (a) 的答案不符。
2. #### 瑞斯
不好意思
請問那題radix sort是不是只要做到Pass 4就可以停了呢? | {
"domain": "eecsmt.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759649262344,
"lm_q1q2_score": 0.8486775034663451,
"lm_q2_score": 0.8652240791017536,
"openwebmath_perplexity": 1704.955713403054,
"openwebmath_score": 0.6259288191795349,
"tags": null,
"url": "https://eecsmt.com/graduate-school/exam/104-nthu-cs-ds/"
} |
# *Rigorous* definition of $\sum_{i=1}^n a_i$
I know that $$\sum_{i=1}^n a_i = a_1 + a_2 + \dots + a_n$$ and I get how to, in general, manipulate expressions with these finite sums. This has typically been introduced to me as simply just being the sum of $a_1$ up to $a_n$. My question is
How can one rigorously define this sum?
(While not being a strict follow-up question to
this is the post that is the source of my question?)
EDIT: If there is no definition that is more rigorous, then I will accept the answer saying that. I ask because I know that things that seem fairly clear have more precise definitions. For example, I know that a function has a set theoretic definition. So I wonder if this finite sum above might have something similar.
• Honest question: why isn't that rigorous enough? What's wrong with it? – Randall Sep 5 '18 at 17:14
• @Randall: I was just wondering. I guess I would accept the answer saying that there is not other "more rigorous" definition. I know that some clear things (like functions) actually have set theoretic definitions. So I was wondering if there is some more rigorous definition of this finite sum. – John Doe Sep 5 '18 at 17:16
• I think an inductive definition will do it, like $\sum\limits_{i=1}^0 a_i=0$ and $\sum\limits_{i=1}^{n+1} a_i = \left(\sum\limits_{i=1}^n a_i\right) + a_{n+1}$. – lisyarus Sep 5 '18 at 17:17
This can be done inductively—what follows is a completely over-the-top way of doing it.
Given any additive group $(G, +, 0)$ (even a monoid would do), such as the real numbers $\mathbb{R}$ under addition, first define the set $G^n$ of $n$-tuples of elements of $G$ inductively by $$G^0 = \{ () \} \quad \text{and} \quad G^{n+1} = \{ (\vec a, a_{n+1}) \mid \vec a \in G^n,\ a_{n+1} \in G \}$$ Thus the elements of $G^n$ are lists $(a_1,a_2,\dots,a_n)$ of length $n$ of elements of $G$.$^{\dagger}$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759632491111,
"lm_q1q2_score": 0.8486775020152576,
"lm_q2_score": 0.8652240791017536,
"openwebmath_perplexity": 699.8473695557884,
"openwebmath_score": 0.9974599480628967,
"tags": null,
"url": "https://math.stackexchange.com/questions/2906506/rigorous-definition-of-sum-i-1n-a-i"
} |
Now we can use this inductive characterisation of $G^n$ for all $n \in \mathbb{N}$ to define, for each $n \in \mathbb{N}$, a function $$S_n : G^n \to G$$ by declaring $$S_0(()) = 0 \quad \text{and} \quad S_{n+1}((\vec a, a_{n+1})) = S_n(\vec a) + a_{n+1}$$ for each $n \in \mathbb{N}$ and each $(\vec a, a_{n+1}) \in G^{n+1}$.
Finally, define $$\sum_{i=1}^n a_i = S_n(\vec a)$$ for each $n \in \mathbb{N}$ and $\vec a \in G^n$.
${}^{\dagger}$More precisely, the lists are parenthesised to the left, so that $(a_1,a_2,a_3,a_4)$ is really $(((a_1,a_2),a_3),a_4)$, for instance.
• I like over-the-top! – John Doe Sep 5 '18 at 17:24
We could also define that by recurrence/induction
• $S_0=\sum_{i=1}^0 a_i= 0$
• for $n\ge 0$ given $S_n=\sum_{i=1}^n a_i$ we define $S_{n+1}=\sum_{i=1}^{n+1} a_i=S_n + a_{n+1}$
but the original definition is sufficiently clear as it is.
• What about the empty sum? – mr_e_man Sep 5 '18 at 20:26
• @mr_e_man Do you suggest to start from $S_0=S_0=\sum_{i=0}^0 = 0$? – user Sep 5 '18 at 20:28
• Almost. $S_0 = \sum_{i=1}^0 a_i = 0$. – mr_e_man Sep 5 '18 at 20:28
• @mr_e_man Could you please explain why is it necessary? Why does not suffice start from $S_1$? – user Sep 5 '18 at 20:31
• It's not necessary, but it's more general, like defining the integral $\int_a^af(x)dx = 0$. And it makes sense of some things that otherwise are undefined. – mr_e_man Sep 5 '18 at 20:31
You can define it recursively (actually, it's a good example of a recursive definition): | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759632491111,
"lm_q1q2_score": 0.8486775020152576,
"lm_q2_score": 0.8652240791017536,
"openwebmath_perplexity": 699.8473695557884,
"openwebmath_score": 0.9974599480628967,
"tags": null,
"url": "https://math.stackexchange.com/questions/2906506/rigorous-definition-of-sum-i-1n-a-i"
} |
You can define it recursively (actually, it's a good example of a recursive definition):
• $\displaystyle\sum_{i=1}^1a_i=a_1$;
• $\displaystyle\sum_{i=1}^{n+1}a_i=\left(\sum_{i=1}^na_i\right)+a_{n+1}$.
• What about the empty sum? – mr_e_man Sep 5 '18 at 20:26
• @mr_e_man I assumed that, for the OP, $n\in\mathbb N$. – José Carlos Santos Sep 5 '18 at 23:21
• It is not generally agreed whether $\mathbb N$ contains $0$. – mr_e_man Sep 5 '18 at 23:22
• @mr_e_man Perhaps. But it is generally agreed that, in an expression of the type $\sum_{i=a}^ba_i$, $b\geqslant a$. – José Carlos Santos Sep 5 '18 at 23:24
• I disagree that that is generally agreed! ;) Compare this with definite integrals, where $\int_a^b$ is defined regardless of the relation of $a$ and $b$. (The comparison is easier using asymmetric intervals: $\sum_{a\leq i<b} = \sum_a^{b-1}$ corresponds to $\int_a^b$, especially when $a=b$.) – mr_e_man Sep 5 '18 at 23:26
For finite sums the only thing not rigorous is that addition is defined with two values, so you would have to write $(a_1+a_2)+a_3$ for a triple sum. Each addition then only involves two numbers. Once we prove the commutativity and associativity of addition we know that all ways of grouping the terms gives the same answer, so writing a sum with a larger number of terms is no problem. $\sum_{i=1}^n$ is just a convenient bookkeeping measure for the sum.
This does not apply to infinite sums. You can't just extend the finite approach, so we have to do something different. The something different is the limit approach you have seen, and that becomes the definition of $\sum_{i=1}^\infty$
The concept of Sum has two basic definitions.
1) Sum over a set
$$\sum\limits_{x\, \in \,A} {f(x)} \quad \Rightarrow \quad \sum\limits_{x\, \in \,A\, \cap \,B} {f(x)} + \sum\limits_{x\, \in \,A\,\backslash \,B} {f(x)}$$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759632491111,
"lm_q1q2_score": 0.8486775020152576,
"lm_q2_score": 0.8652240791017536,
"openwebmath_perplexity": 699.8473695557884,
"openwebmath_score": 0.9974599480628967,
"tags": null,
"url": "https://math.stackexchange.com/questions/2906506/rigorous-definition-of-sum-i-1n-a-i"
} |
The meaning of which is obvious, and where you can limit the sum to the $x \in A$, or consider that $f(x)$ be null for $x \notin A$ and take the sum for all the $x$ within the considered field.
If the set $A$ is partitioned into the subsets $A_k \quad | \; k \in C$ then $$\left\{ \matrix{ A = \bigcup\limits_{k\, \in \,C} {A_{\,k} } \hfill \cr A_{\,k} \cap A_{\,j} = \emptyset \quad \left| {\;k \ne j} \right. \hfill \cr} \right.\quad \Rightarrow \quad \sum\limits_{x\, \in \,A} {f(x)} = \sum\limits_{k\, \in \,C} {\sum\limits_{x\, \in \,A_{\,k} } {f(x)} }$$
Then, in $2$D, if the summing set is partitioned into subsets where in each of them the value of $x$ is constant, then \eqalign{ & \sum\limits_{\left( {x,y} \right)\, \in \,A} {f(x,y)} = \sum\limits_{k\, \in \,C} {\sum\limits_{\left( {x,y} \right)\, \in \,A_{\,k} } {f(x,y)} } = \sum\limits_{k\, \in \,C} {\sum\limits_{\left( {x,y} \right)\, \in \,A_{\,k} } {f(x_{\,k} ,y)} } = \cr & = \sum\limits_{k\, \in \,C} {\sum\limits_{y\, \in \,A_{\,k} (x_{\,k} )} {f(x_{\,k} ,y)} } = \sum\limits_{k\, \in \,C} {g(} x_{\,k} ) \cr} If an analogue partition can be done for $y$, then you can take the one which is "easier", has a known closed form, ... and "manouvre" between the two.
2) Indefinite Sum
While in the definition above we make full use of the commutative and associative properties of addition, in conjunction with union, intersection and derivated concepts for sets (finite, numerable, ...), the concept of Indefinite Sum is somewhat different.
If we have that $$\Delta _{\,x} \,F(x) = F(x + 1) - F(x) = f(x)$$ then we write $$F(x) = \Delta _{\,x} ^{\, - \,1} \,F(x) = \sum\nolimits_{\;x\;} {f(x)}$$ in particular we have $$F(b) - F(a) = \sum\nolimits_{\;x = \,a\,}^b {f(x)} = - \sum\nolimits_{\;x = \,b\,}^a {f(x)}$$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759632491111,
"lm_q1q2_score": 0.8486775020152576,
"lm_q2_score": 0.8652240791017536,
"openwebmath_perplexity": 699.8473695557884,
"openwebmath_score": 0.9974599480628967,
"tags": null,
"url": "https://math.stackexchange.com/questions/2906506/rigorous-definition-of-sum-i-1n-a-i"
} |
For example \eqalign{ & F(x) = \left( \matrix{ x \cr 2 \cr} \right)\quad \Leftrightarrow \quad \left( \matrix{ x + 1 \cr 2 \cr} \right) - \left( \matrix{ x \cr 2 \cr} \right) = \left( \matrix{ x \cr 1 \cr} \right) = x\quad \Leftrightarrow \cr & \Leftrightarrow \quad \sum\nolimits_{\;x = \,a\,}^{\;b} x = \left( \matrix{ b \cr 2 \cr} \right) - \left( \matrix{ a \cr 2 \cr} \right)\quad \left| {\;a,b \in C} \right.\quad \Rightarrow \cr & \Rightarrow \quad \sum\nolimits_{\;k = \,0\,}^{\;n} k \quad \left| {\;0 \le n \in Z} \right.\quad = \sum\limits_{0\, \le \,k\, \le \,n - 1} k = \left( \matrix{n \cr 2 \cr} \right) = {{n\left( {n - 1} \right)} \over 2} \cr} | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759632491111,
"lm_q1q2_score": 0.8486775020152576,
"lm_q2_score": 0.8652240791017536,
"openwebmath_perplexity": 699.8473695557884,
"openwebmath_score": 0.9974599480628967,
"tags": null,
"url": "https://math.stackexchange.com/questions/2906506/rigorous-definition-of-sum-i-1n-a-i"
} |
# Find a generalized path cover of a square graph
Given a directed $n\times n$ square graph as shown in the figure with $n^2$ nodes. Find a set of directed paths $\mathcal P$ from $s$ to $t$ with the minimum cardinality (i.e, minimum number of paths in $\mathcal P$) such that any pair of reachable vertices is contained in at least one path in $\mathcal P$. Two vertices is reachable if there exists an directed path between them. For example, if node $v$ is below and on the right node $u$, then $u$ and $v$ is reachable (see figure).
I have solved this problem for small $n$ by trial and error but I have no idea to generalize it. Can anyone give me some hints? or tell me if this problem is NP-hard? Many thanks
• What happened with your early explorations of simple cases like $2x2$ or $3x3$? – Lee Mosher Feb 25 '18 at 17:14
• $|P|$ = 2 for 2x2 graph, $|P|$ = 4 for 3x3 graph. But I have no idea to generalize it. – Moshe Feb 25 '18 at 17:19
• If you've made progress on small examples, that is the sort of thing that should go in a question rather than just "Here is problem statement solve it for me." – Misha Lavrov Feb 25 '18 at 18:10
• For a $2n-1 \times 2n-1$ grid, there is a lower bound of $n^2$: let $u_1, \dots, u_n$ be the vertices within $n-1$ steps of $s$, and $v_1, \dots, v_n$ be the vertices within $n-1$ steps of $t$. Then any pair $(u_i, v_j)$ must be contained in a path, and no path can contain more than one such pair. – Misha Lavrov Feb 25 '18 at 22:13
• @MishaLavrov Thank you, but we only get a lower bound. It is possible that this lower bound is far away from the solution. Can we guarantee how far the lower bound from the optimal solution? Do u think this problem is Np-hard? – Moshe Feb 25 '18 at 22:15
The minimum number of paths needed in an $n$ by $n$ grid (that is, a grid with $n^2$ vertices) is $\left\lceil \frac{n(n+1)}{3}\right\rceil$: sequence A007980 in the OEIS. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.98087596269007,
"lm_q1q2_score": 0.8486774981227337,
"lm_q2_score": 0.8652240756264639,
"openwebmath_perplexity": 182.04484072571978,
"openwebmath_score": 0.8068516254425049,
"tags": null,
"url": "https://math.stackexchange.com/questions/2666226/find-a-generalized-path-cover-of-a-square-graph"
} |
To prove that at least this many paths are needed, let $k = \lfloor \frac{2n-1}{3}\rfloor$, define $u_0, u_1, \dots, u_k$ by $u_i = (i,k-i)$, and define $v_0, v_1, \dots, v_k$ by $(n-1-i,n-1-(k-i))$ (as coordinates with $(0,0)$ the top left corner of the grid). Not all pairs of points $(u_i, v_j)$ can have a path going through both, but there turn out to be exactly $\left\lceil \frac{n(n+1)}{3}\right\rceil$ that do (to check this, do the computation for each case of $n \bmod 3$ separately). Any path can only go through one point $u_i$ and one point $v_j$, so there must be at least $\left\lceil \frac{n(n+1)}{3}\right\rceil$ paths to account for all these pairs.
To prove that $\left\lceil \frac{n(n+1)}{3}\right\rceil$ pairs suffice, we give a recursive construction which fills an $n \times n$ grid with $2(n-1)$ more paths than an $(n-3) \times (n-3)$ grid. (The sequence $\left\lceil \frac{n(n+1)}{3}\right\rceil$ turns out to satisfy this recurrence.)
Begin by taking the following $2(n-1)$ paths in the $n \times n$ grid:
• paths that go $k$ steps right, $n-1$ steps down, and $n-1-k$ more steps right for $k=1,\dots,n-1$, and
• paths that go $k$ steps down, $n-1$ steps right, and $n-1-k$ more steps down for $k=1, \dots, n-1$.
These are enough to cover all pairs of vertices that are in the same row or column, as well as all pairs of vertices that include a vertex along one of the borders of the grid.
To deal with pairs of vertices that aren't along a border of the grid, take the construction for the $(n-3) \times (n-3)$ grid, and modify each path as follows:
• Insert a step down and a step right at the beginning.
• Insert a step down and a step right in the very middle.
• Insert a step down and a step right at the end. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.98087596269007,
"lm_q1q2_score": 0.8486774981227337,
"lm_q2_score": 0.8652240756264639,
"openwebmath_perplexity": 182.04484072571978,
"openwebmath_score": 0.8068516254425049,
"tags": null,
"url": "https://math.stackexchange.com/questions/2666226/find-a-generalized-path-cover-of-a-square-graph"
} |
Let $u_1$ and $u_2$ be two vertices in the grid with coordinates $(x_1,y_1)$ and $(x_2,y_2)$, such that $1 < x_1 < x_2 < n-1$ and $1 < y_1 < y_2 < n-1$. To show that there's a modified path covering $u_1$ and $u_2$ simultaneously, define $$u_i' = \begin{cases} (x_i-1, y_i-1), & \text{if } x_i + y_i < n-1, \\ (x_i-1, y_i-2) \text{ or } (x_i-2,y_i-1), & \text{if } x_i + y_i = n-1, \\ (x_i-2, y_i-2), & \text{if } x_i + y_i > n-1. \end{cases}$$ Here is a visualization of this not-quite-bijective correspondence between points in the interior of the $n \times n$ grid, and points in the $(n-3) \times (n-3)$ grid. Each red region (mostly including one point, some including more) corresponds to a point in the smaller grid. Points in the overlap of two red regions could go either way, it doesn't matter.
The path in the $(n-3) \times (n-3)$ grid covering $u_1'$ and $u_2'$ simultaneously becomes a path in the $n \times n$ grid covering $u_1$ and $u_2$ simultaneously when modified. This completes the proof that the construction works. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.98087596269007,
"lm_q1q2_score": 0.8486774981227337,
"lm_q2_score": 0.8652240756264639,
"openwebmath_perplexity": 182.04484072571978,
"openwebmath_score": 0.8068516254425049,
"tags": null,
"url": "https://math.stackexchange.com/questions/2666226/find-a-generalized-path-cover-of-a-square-graph"
} |
• Hi, it is not the path cover in Dilworth theorem. Here I need to cover all vertices and pair of reachable vertices. Dilworth theorem only give me a lower bound. – Moshe Feb 25 '18 at 21:31
• I have written a new answer that actually solves the correct problem this time. – Misha Lavrov Feb 26 '18 at 5:39
• Many thanks for your great effort. However, I think the lower bound is not correct. We don't need to cover for all pairs of vertices, only for those are reachable. In particular, $u=(i,j)$ and $v=(k,l)$ are reachable iff $i\leq k, j\leq l$ or $i\geq k, j\geq l$. For example, take $n=5$, hence $k=3, u_0 = (0,3), v_0 = (4,1)$ but we don't need to cover $(u_0,v_0)$. Do you think this problem is NP-hard for a directed acyclic graph? – Moshe Feb 26 '18 at 9:21
• @Moshe ...yes, that's the problem I'm solving? I'm not saying that all pairs $(u_i, v_j)$ in my lower bound are reachable; I'm saying that there are $\left\lceil \frac{n(n+1)}{3}\right\rceil$ reachable pairs among them. (This is roughly $\frac34$ of the total number of pairs, which $k^2 \sim \frac49n^2$.) – Misha Lavrov Feb 26 '18 at 14:39
• Many thanks. I want to generalize this to a directed acyclic graph (DAG) but it is impossible because the solution depends on the structure of the square graph. Do you think we can have a solution for a DAG? – Moshe Feb 26 '18 at 15:11 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.98087596269007,
"lm_q1q2_score": 0.8486774981227337,
"lm_q2_score": 0.8652240756264639,
"openwebmath_perplexity": 182.04484072571978,
"openwebmath_score": 0.8068516254425049,
"tags": null,
"url": "https://math.stackexchange.com/questions/2666226/find-a-generalized-path-cover-of-a-square-graph"
} |
# light bulbs
#### ihavvaquestion
Suppose you have 100 light bulbs and one of them is defective. If you pick out two light bulbs at random (either borh at the same time, or first one, then another, without replacing the first light bulb), what is the probability that one of your chosen light bulbs is defective?
i got an answer, but it seemed too easy
1/100 + (1/100)(1/99)
is this right?
#### Purslow
isnt it just 1/50?
1 - P(Not being defective) = 1 - (99/100).(98/99) = 1/50
#### Plato
MHF Helper
Consider how we could get one good and one bad: $$\displaystyle GB\text{ or }BG$$.
What are those probabilities: $$\displaystyle \frac{99}{100}\frac{1}{99} +\frac{1}{100} \frac{99}{100}=?$$
Suppose you have 100 light bulbs and one of them is defective. If you pick out two light bulbs at random (either borh at the same time, or first one, then another, without replacing the first light bulb), what is the probability that one of your chosen light bulbs is defective?
i got an answer, but it seemed too easy
1/100 + (1/100)(1/99)
is this right?
You have summed the probabilities of getting the bad one first
and getting the bad one first followed by a "specific" good one.
You need the bad one first or 2nd, which is BG+GB
$$\displaystyle \frac{1}{100}\ \frac{99}{99}+\frac{99}{100}\ \frac{1}{99}$$
which follows from ..... pick one bulb then another.
After the first bulb is picked there are 99 to choose from.
ihavvaquestion | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9808759626900699,
"lm_q1q2_score": 0.8486774981227336,
"lm_q2_score": 0.8652240756264639,
"openwebmath_perplexity": 1174.0436629013154,
"openwebmath_score": 0.9038570523262024,
"tags": null,
"url": "https://mathhelpforum.com/threads/light-bulbs.145583/"
} |
# Let $x,y,z$ be integers and $11$ divides $7x+2y-5z$. Show that $11$ divides $3x-7y+12z$.
Let $x,y,z$ be integers and $11$ divides $7x+2y-5z$. Show that $11$ divides $3x-7y+12z$.
I know a method to solve this problem which is to write into $A(7x+2y-5z)+11(B)=C(3x-7y+12z)$, where A is any integer, B is any integer expression, and C is any integer coprime with $11$.
I have tried a few trials for example $(7x+2y-5z)+ 11(x...)=6(3x-7y+12z)$, but it doesn't seem to work. My question is are there any tricks or algorithms for quicker way besides trials and errors? Such as by observing some hidden hints or etc?
I am always weak at this type of problems where we need to make smart guess or gain some insight from a pool of possibilities? Any help will be greatly appreciated. And maybe some tips to solve these types of problems.
Thanks very much!
Let $n=3x-7y+12z$. Since there exsits an $m\in\mathbb Z$ such that $$7x+2y-5z=11m,$$ We have the following two :
$$7n=21x-49y+84z$$ $$33m=21x+6y-15z$$ Then, we have $$7n-33m=-55y+99z\Rightarrow 7n=33m-55y+99z=11(3m-5y+9z).$$ Since we know $7$ and $11$ are coprime, we know that $n$ is a multiple of $11$.
The easiest way can be : eliminate one of the three variables like below
$$3(7x+2y-5z)-7(3x-7y+12z)=55y-99z=11(5y-9z)$$
$$\iff 7(3x-7y+12z)=3(7x+2y-5z)-11(5y-9z)$$
If $\displaystyle 11$ divides $\displaystyle 7x+2y-5z,11$ will divide $\displaystyle7(3x-7y+12z)$
But $(7,11)=1$
• Congruences would be very useful, but I don't know if we can use them... – chubakueno Dec 21 '13 at 2:29
• @chubakueno, definitely, if $\displaystyle 11$ divides $\displaystyle 7x+2y-5z, 7x+2y-5z\equiv0\pmod{11}$ $\displaystyle\implies 3(7x+2y-5z)-11(5y-9z)\equiv0\pmod{11}$ $\displaystyle\iff7(3x-7y+12z)\equiv0\implies3x-7y+12z\equiv0$ as $(7,11)=1$, But I did not use this in the answer as Congruence may not have been taught yet – lab bhattacharjee Dec 21 '13 at 7:03 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211612253742,
"lm_q1q2_score": 0.8486766499682462,
"lm_q2_score": 0.8705972784807408,
"openwebmath_perplexity": 156.6898828432149,
"openwebmath_score": 0.9207722544670105,
"tags": null,
"url": "https://math.stackexchange.com/questions/613836/let-x-y-z-be-integers-and-11-divides-7x2y-5z-show-that-11-divides-3x"
} |
Hint $\$ Scale the first sum so that, mod $11,\,$ its $\,y$ coefficient $\,\color{#0a0}2\,$ becomes that of the second sum, i.e. $\,\color{#c00}{-7\equiv 4}.\,$ So we need to multiply by $\rm\,\color{orange}{by\ 2}\,$ to scale from $\,\color{#0a0}2$ to $\,\color{#c00}4.\,$ Doing so we obtain
$$\begin{eqnarray}{\rm mod}\ 11\!:\,\ 0&\equiv&\ \, 7x\!+\!\color{#0a0}2y\!-\!5z \\ \overset{\color{orange}{\times\ 2}}\Rightarrow\ 0&\equiv& 14x\!+\!4y\!-\!10z \\ &\overset{\phantom{2}}\equiv&\ \, 3x\!\color{#c00}{-\!7}y\!+\!12z \end{eqnarray}\qquad\qquad$$
Remark $\$ The converse holds also since $\,11\mid 2n\iff 11\mid n$. I picked the smallest coefficient $\,\color{#0a0}2\,$ because generally that will be easiest to divide by (or invert). In fact, it is very easy to divide by $2$ mod odd $\,m,\,$ since one of $\ n\equiv n+m\,$ is even; e.g. as above $\,\color{#c00} -7\equiv -7+11 = 4\,$ is even, hence $\rm\,\color{#c00}{-7}/\color{#0a0}2 \equiv 4/2 \equiv \color{orange}2.$
$7x+2y-5z \equiv 14x +4y-10z \equiv(4-1)x+(4-11)y-(0-1)z\equiv 3x-7y+z$ $\equiv3x-7y+12z$ (mod 11) | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211612253742,
"lm_q1q2_score": 0.8486766499682462,
"lm_q2_score": 0.8705972784807408,
"openwebmath_perplexity": 156.6898828432149,
"openwebmath_score": 0.9207722544670105,
"tags": null,
"url": "https://math.stackexchange.com/questions/613836/let-x-y-z-be-integers-and-11-divides-7x2y-5z-show-that-11-divides-3x"
} |
Can different control points lead to the same Bézier curve?
A cubic Bézier curve is a polynomial
$$F(u) = \sum_{i=0}^{n} \mathbf{b}_i^n P_i \;\;\;\text{ with } u \in [0,1], P_i \in \mathbb{R}^2, n=3 \text{ and } \mathbf{b}_i^n = \begin{pmatrix}n\\i\end{pmatrix} u^i (1-u)^{n-i}$$
You get plots like this (source):
Having the set $\mathbb{R}^{4 \times 2}$ of all cubic Bézier curves defined by their control points and the set of all of their plots, I wondered: Are there any two Bézier curves which have different control points $P_i, P_i'$ but are the same function?
Obviously, to be the same function the point $P_0 = P_0'$ and $P_3'$ have to be the same. Also, $P_1, P_1'$ and $P_2, P_2'$ have to be on the same line, because $\overline{P_0 P_1}$ is a tangent on the curve. But besides that, I'm not too sure if there could be a combination where the points are different, but the curves are the same.
edit: I think one problem might be when all control points are on the same line. Is this actually a counter example? Are there others?
The equation for the cubic Bézier curve is $$F(u) = (1-u)^3 P_0 + 3 (1-u)^2 u P_1 + 3 (1-u)u^2 P_2 + u^3 P_3$$ and therefore $$F(0) = P_0, \quad F(1) = P_3$$ and, by differentiation, $$F'(0) = 3(P_1 - P_0), \quad F'(1) = 3(P_3 - P_2)$$
Therefore all control points are uniquely determined by the function $F$.
But, as you already noted, there can be two different Bézier curves having the same image, e.g. if all control points are on a line. I don't have an answer for the question
When are the control points uniquely determined by the image of a Bézier curve?
• An alternative to the polysemic word "image" is "trajectory" Mar 6 '16 at 16:28 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.974821152447445,
"lm_q1q2_score": 0.8486766390532821,
"lm_q2_score": 0.8705972751232809,
"openwebmath_perplexity": 134.04065573697423,
"openwebmath_score": 0.8833402991294861,
"tags": null,
"url": "https://math.stackexchange.com/questions/1685569/can-different-control-points-lead-to-the-same-b%C3%A9zier-curve"
} |
• An alternative to the polysemic word "image" is "trajectory" Mar 6 '16 at 16:28
Suppose $\mathbf{P}_0, \ldots \mathbf{P}_m$, and $\mathbf{Q}_0 \ldots \mathbf{Q}_m$ are two sets of control points that produce the same Bézier curve of degree $m$, in the sense that $$\sum_{i=0}^m b^m_i(t)\mathbf{P}_i = \sum_{i=0}^m b^m_i(t)\mathbf{Q}_i \quad \text{for all t \in[0,1]}$$ Then we have $$\sum_{i=0}^m b^m_i(t)(\mathbf{P}_i - \mathbf{Q}_i) = \mathbf{0} \quad \text{for all t \in[0,1]}$$ This implies that $\mathbf{P}_i = \mathbf{Q}_i$ for $i=0,1, \ldots, m$ since the Bernstein polynomials $b^m_0, \ldots b^m_m$ are linearly independent.
There are cases where two different sets of control points will produce the same trace/image/trajectory. Take a given curve, and compose it with two different polynomials that both map $[0,1]$ to itself. Then clearly you'll get the same image, but the parametric equations will be different, so the control points will also be different.
A specific example: consider the two degree 4 curves whose control points are
$(0,0)$, $(1,0)$, $(\tfrac53,\tfrac23)$, $(2,1)$, $(2,1)$
$(0,0)$, $(0,0)$, $(\tfrac13,0)$, $(1,0)$, $(2,1)$.
Routine calculations show that these two curves both represent the portion of the parabola $y = \tfrac14 x^2$ that lies between the points $(0,0)$ and $(2,1)$. The first one has equation $G(t) = (4t - 2t^2, 4t^2 - 4t^3 + t^4)$ and the second one has equation $H(t) = (2t^2, t^4)$. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.974821152447445,
"lm_q1q2_score": 0.8486766390532821,
"lm_q2_score": 0.8705972751232809,
"openwebmath_perplexity": 134.04065573697423,
"openwebmath_score": 0.8833402991294861,
"tags": null,
"url": "https://math.stackexchange.com/questions/1685569/can-different-control-points-lead-to-the-same-b%C3%A9zier-curve"
} |
These curves are formed by two different reparameterizations of the basic curve $F(t) = (2t, t^2)$: $$F(t) = (2t, t^2) \;,\; t = 2u - u^2 \;\; \Rightarrow \;\; G(u) = (4u - 2u^2, 4u^2 - 4u^3 + u^4)$$ $$F(t) = (2t, t^2) \;,\; t = v^2 \;\; \Rightarrow \;\; H(v) = (2v^2, v^4)$$ This is a form of degree elevation (though not the usual form). The degree of the final curve is the product of the degree of the original one and the degree of the reparametrization function. In the example above $2 \times 2 = 4$. So, this process can only produce a cubic curve in trivial cases where either the original curve or the reparameterization function is linear.
• You probably mean "... parabola $y = \frac14 t^2$ ..." Mar 8 '16 at 10:23
• You're right. Thanks. Fixed. Mar 8 '16 at 12:04
Bézier curves with different control points could actually be the same function under one condition: one Bézier curve is degree elevated from the other.
A cubic Bézier curve is represented as (in Bernstein basis)
$C(u)=(1-u)^3P_0+3(1-u)^2uP_1+3(1-u)u^2P_2+u^3P_3$,
or as (in power basis)
$C(u)=P_0+3(P_1-P_0)u+3(P_2-2P_1+P_0)u^2+(P_3-3P_2+3P_1-P_0)u^3$.
From the power-basis representation, we can see that if $(P_3-3P_2+3P_1-P_0)=0$, the curve is actually of degree 2 only and therefore can be represented as a Bézier curve with 3 control points.
Basically, a Bézier curve with $K_1$ control points can always be exactly converted to a Bézier curve with $K_2$ ($K_2 > K_1$) control points. This process is called "degree elevation". However, it does not really change the degree of the underlying polynomial. It only makes the Bézier curve have more control points to be manipulated with.
In summary,
- Bézier curves with different number of control points might actually be the same function.
- Bézier curves of the same number of control points but different control point coordinates cannot be the same function but could be of the same image/trajectory. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.974821152447445,
"lm_q1q2_score": 0.8486766390532821,
"lm_q2_score": 0.8705972751232809,
"openwebmath_perplexity": 134.04065573697423,
"openwebmath_score": 0.8833402991294861,
"tags": null,
"url": "https://math.stackexchange.com/questions/1685569/can-different-control-points-lead-to-the-same-b%C3%A9zier-curve"
} |
# If $a\lt{b}$ and $c\le{d}$, prove that $a+c\lt b+d$
If $a\lt{b}$ and $c\le{d}$, prove that $a+c\lt b+d$.
This seems like a basic proof and I think this is how it goes: $$c \le d, \text{ Given }$$ $$a+c \le a+d$$ $$a+c \lt b+d, \text{ since } a \lt b$$
Is that all I need? I'm thinking this does it all quickly and concisely, but I have had trouble with proofs in my classes.
• Yes, or just go $a < b \implies a+c<b+c$, so since $c \leq d \implies b + c \leq b + d$, we have $a+c < b+c \leq b+d$, i.e. $a+c < b+d$. – Ryker Feb 17 '14 at 20:10
• Is $(b+d)-(a+c)>0$? – David Mitra Feb 17 '14 at 20:10
• I'd do it $(a-b)+(c-d)\lt 0$ but it's all the same in the end – Mark Bennet Feb 17 '14 at 20:18
• What is the reason you did it your way, @Ryker, versus my way? It is just semantical? I've noticed that the things posted have a middle argument between a less than and a less than or equal to sign. Is that preferred when proving to see the logical reason? – Faffi-doo Feb 17 '14 at 20:25
• @Faffi-doo , guys, wow! :) Perhaps I should have added a smiley in the end of the comment to avoid misunderstandings.. 1. You don't need to be so apologetic! :) Choose what is most correct for you. Always, and not just in Math. 2. I will not hold anything against you even if you actually did not like my answer. – user76568 Feb 17 '14 at 21:21
We have that for any $c$: $$a<b \implies a+c<b+c$$ And for any $b$: $$c\leq d \implies b+c \leq b+d$$.
Hence: $a+c < b+c \leq b+d$, that is: $$a+c < b+d$$.
$$(a < b) \land (c \leq d) \iff a-b < 0 \leq d - c \implies a-b<d-c \iff a+c < b + d$$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211568364099,
"lm_q1q2_score": 0.8486766297826105,
"lm_q2_score": 0.8705972616934406,
"openwebmath_perplexity": 564.1053884738791,
"openwebmath_score": 0.8974902629852295,
"tags": null,
"url": "https://math.stackexchange.com/questions/679947/if-a-ltb-and-c-led-prove-that-ac-lt-bd/679952"
} |
# Math Help - limit doesn't exist
1. ## limit doesn't exist
lim[x->2] (x^2 - x + 6)/(x - 2)
I'm not sure the best way to solve this... but here's what I did:
after doing polynomial division, I got:
lim[x->2] (x+1+ 8/(x-2))
So, by reasoning, I see that the limit as x approaches 2 from the left does not equal the limit as x approaches 2 from the right (negative and positive infinity). Therefore, this limit does not exist.
My solutions manual says that the limit "does not exist since x - 2 -> 0 but x^2 - x + 6 -> 8 as x -> 2."
Someone please tell me what my book's explanation means.
2. ## n/0 dne for n not equal to zero
The only number that can be divided by zero to give another number is zero.
Therefore in order to show that the limit does not exist you can show that the limit approaches some non zero in the numerator and zero in the denominator. In this case:
8/0
does not exist.
3. Originally Posted by asdfmaster
lim[x->2] (x^2 - x + 6)/(x - 2)
I'm not sure the best way to solve this... but here's what I did:
after doing polynomial division, I got:
lim[x->2] (x+1+ 8/(x-2))
So, by reasoning, I see that the limit as x approaches 2 from the left does not equal the limit as x approaches 2 from the right (negative and positive infinity). Therefore, this limit does not exist.
Would this form of argument then tell us that
$\lim_{x \rightarrow 2} \frac{x^2 - x + 6}{(x - 2)^2}$
exists and equals $\infty$?
RonL
4. Thanks MathGuru & CaptainBlack
CaptainBlack, are you saying my reasoning is wrong?
if I graph
$
\lim_{x \rightarrow 2} \frac{x^2 - x + 6}{(x - 2)^2}
$
I see that as x approaches 2 from the left and from the right, it increases without bound. I would then conclude that
$
\lim_{x \rightarrow 2} \frac{x^2 - x + 6}{(x - 2)^2}=\infty
$ | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211597623863,
"lm_q1q2_score": 0.8486766290570343,
"lm_q2_score": 0.8705972583359806,
"openwebmath_perplexity": 282.4523087650058,
"openwebmath_score": 0.921103835105896,
"tags": null,
"url": "http://mathhelpforum.com/calculus/1887-limit-doesn-t-exist.html"
} |
Also, if this reasoning is correct, then what MathGuru said:
Therefore in order to show that the limit does not exist you can show that the limit approaches some non zero in the numerator and zero in the denominator.
would not be correct all the time...
I just started learning calculus! I'm sorry if these questions are very basic!
5. The limit doesn't exist because the upper limit doesn't equal the lower limit, i.e. you get a different result depending on whether you approach 2 from the right or from the left.
In the real numbers, we do not use the unsigned infinity in this context (this is done in complex analysis though), there is only a positive infinity and a negative infinity. Therefore you may not conclude that the limit "exists" and is equal to $\infty$, that's wrong.
6. Originally Posted by asdfmaster
Thanks MathGuru & CaptainBlack
CaptainBlack, are you saying my reasoning is wrong?
Neither the limit from the right or left exist.
RonL
7. Captain black was being a friendly skeptic earlier. He was trying to show you that sometimes the limit approaches the same from the left and from the right however the limit still does not exist.
My analysis that explains using numerator and denominator is more precise and also the same explanation your book was trying to give.
8. thanks everyone!
Captain black was being a friendly skeptic earlier.
That's what I assumed...but wasn't totally sure.
So what everyone is saying is that when a function appears to approach infinity, the limit does not exist???
So whenever you say the limit of a function is infinity, you are actually saying that the limit does not exist, but a convenient way to express that the limit does not exist is to write $\lim_{x \rightarrow a} f(x) = \infty$?? | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211597623863,
"lm_q1q2_score": 0.8486766290570343,
"lm_q2_score": 0.8705972583359806,
"openwebmath_perplexity": 282.4523087650058,
"openwebmath_score": 0.921103835105896,
"tags": null,
"url": "http://mathhelpforum.com/calculus/1887-limit-doesn-t-exist.html"
} |
Therefore in order to show that the limit does not exist you can show that the limit approaches some non zero in the numerator and zero in the denominator.
I'm just wondering, in every instance of finding a limit that exists, if the demoninator will approach zero, will numerator always approach zero too?
9. Originally Posted by asdfmaster
thanks everyone!
That's what I assumed...but wasn't totally sure.
So what everyone is saying is that when a function appears to approach infinity, the limit does not exist???
So whenever you say the limit of a function is infinity, you are actually saying that the limit does not exist, but a convenient way to express that the limit does not exist is to write $\lim_{x \rightarrow a} f(x) = \infty$??
I'm just wondering, in every instance of finding a limit that exists, if the demoninator will approach zero, will numerator always approach zero too?
For a limit to exist it has to approach a real number. Since infinity is not a real number, any limit that "blows up" (either in the positive or negative sense) does not exist.
Yes, the ONLY way for a limit to exist when the denominator of a rational expression approaches zero is for the numerator to approach zero as well. In this instance we can then apply L'Hospital's rule. However, be warned: Just because the numerator also goes to zero does NOT mean that the limit will exist. A trivial example of this: $\lim_{x \rightarrow 0} \frac{x}{x^2}$ does not exist.
-Dan | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211597623863,
"lm_q1q2_score": 0.8486766290570343,
"lm_q2_score": 0.8705972583359806,
"openwebmath_perplexity": 282.4523087650058,
"openwebmath_score": 0.921103835105896,
"tags": null,
"url": "http://mathhelpforum.com/calculus/1887-limit-doesn-t-exist.html"
} |
# 8 Indistinguishable objects randomly sorted into six buckets - What is the probability of at least three buckets receiving the objects?
Eight indistinguishable objects are to be randomly put into six buckets. What is the probability that at least three buckets will receive the objects?
My approach was to let X= the number of buckets that receive objects, and then the required probability is
$$P(X≥3) = 1- P(X=1) - P(X=2)$$
$$= 1 - [(6C1)(1/6)^8 + (6C2)(2/6)^8]$$ $$= 0.99771$$
My reasoning was that for all of the objects falling into one bucket, there are 6 possible buckets $$(6C1)$$, and the probability for eight objects to fall into the same bucket in a row is $$(1/6)^8$$
By the same logic, for all of the objects to be divided into two of the six buckets, there are $$(6C2)$$ possible ways to choose those two buckets...And the probability is $$(2/6)^8$$.
Then, the probability that at least three buckets will receive objects is 1- (the probability of only one bucket receiving all the objects + two buckets receiving all the objects)
However, the answer given to this question was 0.8904. Where am I going wrong, and is there a better way to approach questions like this? Does it fall into any specific discrete distribution?
Note:
I also tried $$1- [(6C1)(1/6)^8] - (6C2)[(1/6)^7(5/6) + (1/6)^6(5/6)^2 + (1/6)^5(5/6)^3 + (1/6)^4(5/6)^4 + (1/6)^3(5/6)^5 + (1/6)^2(5/6)^6 + (1/6)(5/6)^7]$$
And got $$0.8721$$. I am not sure if this approach is better. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9381240142763573,
"lm_q1q2_score": 0.84867439753516,
"lm_q2_score": 0.9046505415276079,
"openwebmath_perplexity": 348.0373623472275,
"openwebmath_score": 0.684924304485321,
"tags": null,
"url": "https://math.stackexchange.com/questions/4182207/8-indistinguishable-objects-randomly-sorted-into-six-buckets-what-is-the-proba"
} |
• You're counting multiple times the cases where all the objects go to the same bucket. First intentionally, and then twice unintentionally, since some of the times that objects are limited to either of two buckets, they are in fact limited to exactly one of the two buckets. (That is to say, if I say that eight objects go into either Bucket A or Bucket B, nothing prevents them from all going into Bucket A or all into Bucket B.) – Brian Tung Jun 24 at 20:40
• Where did you find this question , can you share your source please ? – Bulbasaur Jun 24 at 21:03
• @Bulbasaur This question is from a past exam paper for a first-year university statistics course. – Cara vdC Jun 24 at 23:17
I get $$1 - (6C1)(1/6)^8 - (6C2)[(2/6)^8-2(1/6)^8] \approx 0.997728$$, close to but not the same as your first calculation. This treats the objects as distinguishable, as I believe that is the physical reality of putting objects into buckets.
Another way of getting this answer is to say there are $$6^8= 1679616$$ ways of putting $$8$$ items into $$6$$ buckets, and $$6$$ of them have them all going into one bucket while $$3810$$ have them going into exactly $$2$$ buckets, leaving $$1675800$$ possibility using at least $$3$$ buckets, with $$\frac{1675800}{1679616} \approx 0.997728$$.
Bad approaches might be to say
• there are $${13 \choose 5}=1287$$ distributions of $$8$$ indistinguishable objects among $$6$$ distinguishable buckets, of which $$6$$ use one bucket and $$105$$ use exactly $$2$$ buckets, suggesting a probability of $$\frac{1176}{1287}\approx 0.913752$$ as the probability
• there are $$20$$ distributions of $$8$$ indistinguishable objects among $$6$$ indistinguishable buckets, of which $$1$$ uses one bucket and $$4$$ use exactly $$2$$ buckets, suggesting a probability of $$\frac{15}{20}=0.75$$ as the probability
but I think these are wrong because the distributions are not equally likely | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9381240142763573,
"lm_q1q2_score": 0.84867439753516,
"lm_q2_score": 0.9046505415276079,
"openwebmath_perplexity": 348.0373623472275,
"openwebmath_score": 0.684924304485321,
"tags": null,
"url": "https://math.stackexchange.com/questions/4182207/8-indistinguishable-objects-randomly-sorted-into-six-buckets-what-is-the-proba"
} |
but I think these are wrong because the distributions are not equally likely
• Should we always approach the indistinguishable objects as distinguishable in probability questions ? For example , what if the indistinguishable balls were distributed at the same time , would we treat like one by one and also distinguishable – Bulbasaur Jun 24 at 20:54
• @bulbasaur - I believe that is the physical reality. Flip two indistinguishable fair coins. Is the probability of one heads and one tails $\frac12$ or $\frac13$? – Henry Jun 24 at 20:56
• hımm elegant example... they are like die (ordered pairs) or coins as you said . Is there any book that may help me to learn this tricks ? Maybe other questions ? – Bulbasaur Jun 24 at 21:00
The official answer is incorrect. Coming to your approach, it is correct except that you have a small mistake when you are distributing objects to two buckets.
$$\displaystyle P(X=1) = {6 \choose 1} \cdot \frac{1}{6^8}$$
$$\displaystyle P(X=2) = {6 \choose 2} \cdot \frac{2^8 - 2}{6^8}$$
Please note that $$2^8$$ will also count the two arrangements where all objects go to one of the two chosen buckets. So we subtract $$2$$ from $$2^8$$.
Therefore,
$$\displaystyle P(X \geq 3) = 1 - P(X=1) - P(X=2) = \frac{23275}{23328}$$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9381240142763573,
"lm_q1q2_score": 0.84867439753516,
"lm_q2_score": 0.9046505415276079,
"openwebmath_perplexity": 348.0373623472275,
"openwebmath_score": 0.684924304485321,
"tags": null,
"url": "https://math.stackexchange.com/questions/4182207/8-indistinguishable-objects-randomly-sorted-into-six-buckets-what-is-the-proba"
} |
Therefore,
$$\displaystyle P(X \geq 3) = 1 - P(X=1) - P(X=2) = \frac{23275}{23328}$$
• Why did you treat indistinguishable balls like distinguishable . Cant the answer be $$1-[C(6,1) \times \frac{1}{C(13,5)} + C(6,2) \times \frac{C(9,1)}{C(13,5)}]$$ – Bulbasaur Jun 24 at 20:47
• @Bulbasaur stars and bars method should not be used to calculate probability. All arrangements in stars and bars are not equally probable. – Math Lover Jun 24 at 20:50
• thanks , i see now – Bulbasaur Jun 24 at 20:52
• @CaravdC Even if you assume that all distributions of the indistinguishable objects are equally likely, the probability would work out to be $1-(6+\binom62\times 7)/\binom{8+6-1}{5}$, which is not equal to the book's answer. This is discussed in the second bullet point of Henry's answer. There is no conceivable way your source's answer is correct. – Mike Earnest Jun 25 at 0:33
• @CaravdC Mike Earnest's comments already answers your question. There is no way to get to that answer for the given question. Your first approach is optimal except a mistake that you made. – Math Lover Jun 25 at 9:49
I'm assuming that the buckets are distinguishable! And the indistinguishable objects are balls.
A hint to answer the question, assuming the balls are indistinguishable
$$P(X=1)=(6C1)P(\text{all balls fall into bucket 1})$$
In the case of indistinguishability, you will have to use conditioning to get the answer. The calculation remains exactly the same. The only difference is that you are using conditioning instead of combinatorics.
Let $$A_i$$ be the event that the $$i^{th}$$ ball chosen falls into bucket 1
$$P(A_1\cap A_2 \cap \dots\cap A_8)=P(A_1)\cdot P(A_2|A_1)\cdot P(A_3|A_1\cap A_2)\dots$$
There is one way to choose the first ball(since balls are indistinguishable!) and the probability that, that ball goes into the first bucket is $$(1/6)$$(because the buckets are distinguishable). so, $$P(A_1)=1/6$$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9381240142763573,
"lm_q1q2_score": 0.84867439753516,
"lm_q2_score": 0.9046505415276079,
"openwebmath_perplexity": 348.0373623472275,
"openwebmath_score": 0.684924304485321,
"tags": null,
"url": "https://math.stackexchange.com/questions/4182207/8-indistinguishable-objects-randomly-sorted-into-six-buckets-what-is-the-proba"
} |
Now, given that the first ball goes into the first bucket, the number of ways to choose the second ball is also one(balls are indistinguishable!), and the probability that that ball goes into the first bucket is also $$(1/6)$$(because the buckets are distinguishable). so, $$P(A_2|A_1)=1/6$$
And so on...
hence $$P(\text{all balls fall into bucket 1})=(1/6)^8$$ and $$P(X=1)=(6C1)P(\text{all balls fall into bucket 1})=(6C1)(1/6)^8$$
I believe, it is straightforward to similarly calculate $$P(X=2)$$ and $$P(X\geq 3)$$
I would like to quote @DavidK here(from this answer, you should check out that answer, it is an absolute gem),
A way I think of this intuitively is that we are modeling a world in which writing a number on a ball or erasing the number does not cause that ball to magically run away from you when you reach in the back nor jump into your hand. In fact, the distinguishing marks (or lack thereof) on the balls have no effect on the probability of drawing a ball each time. So a correct way to compute $$P(X=k)$$ with indistinguishable balls is to compute $$P(X=k)$$ with distinguishable balls and simply copy the final result. This yields the same formulas.
I hope, it is okay to quote other users, if it is relevant. If it is not, let me know and I will remove the quote. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9381240142763573,
"lm_q1q2_score": 0.84867439753516,
"lm_q2_score": 0.9046505415276079,
"openwebmath_perplexity": 348.0373623472275,
"openwebmath_score": 0.684924304485321,
"tags": null,
"url": "https://math.stackexchange.com/questions/4182207/8-indistinguishable-objects-randomly-sorted-into-six-buckets-what-is-the-proba"
} |
the probability of dividing 6 objects into exactly 2 buckets is not $$[(1/6)^7(5/6) + (1/6)^6(5/6)^2 + (1/6)^5(5/6)^3 + (1/6)^4(5/6)^4 + (1/6)^3(5/6)^5 + (1/6)^2(5/6)^6 + (1/6)(5/6)^7]$$
it should be $${nCr}\left(8,1\right)\left(\frac{1}{6}\right)^{7}\left(\frac{1}{6}\right)+{nCr}\left(8,2\right)\left(\frac{1}{6}\right)^{6}\left(\frac{1}{6}\right)^{2}+{nCr}\left(8,3\right)\left(\frac{1}{6}\right)^{5}\left(\frac{1}{6}\right)^{3}+{nCr}\left(8,4\right)\left(\frac{1}{6}\right)^{4}\left(\frac{1}{6}\right)^{4}+{nCr}\left(8,5\right)\left(\frac{1}{6}\right)^{3}\left(\frac{1}{6}\right)^{5}+{nCr}\left(8,6\right)\left(\frac{1}{6}\right)^{2}\left(\frac{1}{6}\right)^{6}+{nCr}\left(8,7\right)\left(\frac{1}{6}\right)^{1}\left(\frac{1}{6}\right)^{7}$$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9381240142763573,
"lm_q1q2_score": 0.84867439753516,
"lm_q2_score": 0.9046505415276079,
"openwebmath_perplexity": 348.0373623472275,
"openwebmath_score": 0.684924304485321,
"tags": null,
"url": "https://math.stackexchange.com/questions/4182207/8-indistinguishable-objects-randomly-sorted-into-six-buckets-what-is-the-proba"
} |
# How does one evaluate $1+2-3-4+5+6-7-8+\cdots+50$?
How does one evaluate the sum $$1+2-3-4+5+6-7-8+\cdots+50$$?
I know how to find the sum of arithmetic progressions: without the negative signs, one simply has $$1+2+\cdots+50=\frac{1}{2}\cdot(1+50)\cdot 50=51\times 25=1275.$$
But how does one calculate the one above?
• Consider a general case. Block them into $4$ consecutive numbers if you call the first one $n$. – mathreadler Oct 28 '17 at 9:59
• Using @Robertz 's idea: better $1+2-3=0$, $-4+5+6-7=0$, $-8+9+10-11=0$,... $...-47=0$, $-48+49+50-51=0$ then correct for the last sum $0+51=51$ – Gottfried Helms Oct 28 '17 at 13:17
• I find it disheartening that this question has received so many downvotes, and has been closed and deleted, and noone has commented as to why...... – user1729 Aug 19 '19 at 19:29
## 4 Answers
It's $$(1+5+...+49)+(2+6+...+50)-(3+7+...+51)-(4+8+...+52)+51+52=$$ $$=\frac{(2+12\cdot4)13}{2}+\frac{(4+12\cdot4)13}{2}-\frac{(6+12\cdot4)13}{2}-\frac{(8+12\cdot4)13}{2}+103=$$ $$=(1+2-3-4)\cdot13+103=51.$$
• Thanks. But I don't get the idea behind this formula you just did. – dimwitt04 Oct 28 '17 at 10:42
• @dimwitt04 It's a formula for the sum of the arithmetic progression: $S_{n}=\frac{(2a_1+(n-1)d)n}{2}$. – Michael Rozenberg Oct 28 '17 at 10:49
• Excuse my ignorance but where does 13 and 12 come from? Is it the number of terms? – dimwitt04 Oct 28 '17 at 11:41
• @dimwitt04 Yes of course! If $n$ is a number of terms we obtain: $49=1+(n-1)4$, which gives $n=13$. I used $a_n=a_1+(n-1)d$. – Michael Rozenberg Oct 28 '17 at 11:43
• Can down-voter explain us why did you do it? – Michael Rozenberg Jul 9 '19 at 14:25
Look at the following: $$1+\overbrace{(2-3)}^{-1}+\overbrace{(-4+5)}^1+\cdots+50$$
So you have $1+\overbrace{-1+1\cdots}^{\frac{48}2=24\text{ times}}+50$ and because $24$ is even the middle part become $0$ and you left with $1+50=51$ and done | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9850429138459388,
"lm_q1q2_score": 0.8486521152114567,
"lm_q2_score": 0.8615382165412808,
"openwebmath_perplexity": 635.0115710561776,
"openwebmath_score": 0.680901288986206,
"tags": null,
"url": "https://math.stackexchange.com/questions/2493482/how-does-one-evaluate-12-3-456-7-8-cdots50/2493503"
} |
moreover, you can generalize it:$$\sum_{k=1}^n(-1)^{\left\lfloor\frac{k-1}{2}\right\rfloor}\times k=\begin{cases}n+1 &\text{if}\,\,\,(-1)^{\left\lfloor\frac{n-1}{2}\right\rfloor}=1,&n\equiv0\pmod{2}\\ 1 &\text{if}\,\,\, (-1)^{\left\lfloor\frac{n-1}{2}\right\rfloor}=1,&n\equiv1\pmod{2}\\ -n &\text{if}\,\,\, (-1)^{\left\lfloor\frac{n-1}{2}\right\rfloor}=-1,&n\equiv0\pmod{2}\\ 0 &\text{if}\,\,\, (-1)^{\left\lfloor\frac{n-1}{2}\right\rfloor}=-1,&n\equiv1\pmod{2} \end{cases}$$
Note that your sum can be written as $$\underbrace{[(1-3)+(2-4)]}_{-4}+\underbrace{[(5−7)+(6−8)]}_{-4}+\dots +\underbrace{[(45−47)+(46−48)]}_{-4}+49+50$$ that is $-4\cdot(48/4)+49+50=-48+49+50=51.$ More generally $$\sum_{k=1}^n(-1)^{\left\lfloor\frac{k-1}{2}\right\rfloor}\cdot k =\begin{cases} -n&\text{if n\equiv 0\pmod{4}}\\ 1&\text{if n\equiv 1\pmod{4}}\\ n+1&\text{if n\equiv 2\pmod{4}}\\ 0&\text{if n\equiv 3\pmod{4}}.\\ \end{cases}$$
• (+1) I was just about to post an answer similar to the first part of this. Good generalization! – robjohn Oct 28 '17 at 12:49
• @Holo Our generalizations are equivalent... – Robert Z Oct 28 '17 at 14:13
Note that $1+2-3=0$. Moreover, you will have -4+5+6-7 and so on... if you consider pairs of numbers, you will always have +1. How much times do you do this computation?
• And if it ended with - 51 the sum would be 0... – hkBst Oct 28 '17 at 12:00 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9850429138459388,
"lm_q1q2_score": 0.8486521152114567,
"lm_q2_score": 0.8615382165412808,
"openwebmath_perplexity": 635.0115710561776,
"openwebmath_score": 0.680901288986206,
"tags": null,
"url": "https://math.stackexchange.com/questions/2493482/how-does-one-evaluate-12-3-456-7-8-cdots50/2493503"
} |
### Welcome to our community
#### Casio
##### Member
I have been reading up on Quadratic Equations in my course book, which gives some examples of simple equations and how they are factorised, but can't get my head round this type which is has no examples.
x2 - 3x = 0
OK I know it is a quadratic because the term x2 is included. What I require is a value for the term x.
Is there a method to factorise this I am not sure?
x2 - 3x = x(x - 3)
It does not give me a value for x?
I can assume values like x = 0 and x = 3, both of which will = 0, but that does not show me how I get the method of factorising this equation?
Thanks
Casio
#### Ackbach
##### Indicium Physicus
Staff member
There is a way of factoring quadratics that takes the guesswork out of it. In your example, you just note that $x$ is common to both terms, and factors out leaving you with $x(x-3)=0$.
But let's suppose you had a general quadratic: $x^{2}+bx+c$, which we'll make monic for convenience. What you're after is an equivalent expression that looks like this:
$$x^{2}+bx+c=(x+f)(x+g).$$
So, FOIL out the RHS thus:
$$x^{2}+bx+c=x^{2}+gx+fx+fg=x^{2}+(g+f)x+fg.$$
Now, you get to equate like powers of $x$ to obtain the following equations:
\begin{align*}
b&=f+g\\
c&=fg.
\end{align*}
Use whatever method you like for solving this simultaneous system of equations for $f$ and $g$. Once you do that, you're done.
Example: Factor $x^{2}+8x-3$.
Solution: Write as
$$x^{2}+8x-3=(x+f)(x+g)=x^{2}+(f+g)x+fg.$$
Solve:
\begin{align*}
8&=f+g\\
-3&=fg.
\end{align*}
The solution is
$$f=4\pm\sqrt{19},\quad g=4\mp\sqrt{19}.$$
That is, if $f=4+\sqrt{19}$, then $g=4-\sqrt{19}$, or if $f=4-\sqrt{19}$, then $g=4+\sqrt{19}$. So our factorization is
$$x^{2}+8x-3=(x+4-\sqrt{19})(x+4+\sqrt{19}).$$
Multiply this out to confirm.
Does that help?
#### QuestForInsight
##### Member
Solve:
\begin{align*}
8&=f+g\\
-3&=fg.
\end{align*}
How did you solve this?
#### SuperSonic4 | {
"domain": "mathhelpboards.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9850429156022934,
"lm_q1q2_score": 0.8486521079697523,
"lm_q2_score": 0.8615382076534742,
"openwebmath_perplexity": 356.01625523466294,
"openwebmath_score": 0.7397609949111938,
"tags": null,
"url": "https://mathhelpboards.com/threads/quadratic-equations.1874/"
} |
#### SuperSonic4
##### Well-known member
MHB Math Helper
How did you solve this?
Simultaneous equations: there are two variables and two equations. In the terms of a quadratic you may know it along the lines of "two numbers which add to 8 and multiply to -3"
If you need more help on them feel free to create a new topic, it's not fair on Casio to sidetrack his topic
#### Ackbach
##### Indicium Physicus
Staff member
How did you solve this?
Substitution. We have that $f=8-g$, so plugging that into the other equation yields
$$-3=g(8-g)=8g-g^{2},$$
or
$$g^{2}-8g-3=0.$$
Now you have a quadratic in $g$. You can use the quadratic formula to solve:
$$g=\frac{8\pm\sqrt{64-4(-3)}}{2}=\dots$$
Come to think of it, we haven't gained anything here, because this is the same quadratic as we started with!
So you could save time by simply using the quadratic formula in the first place. Given the monic quadratic $x^{2}+bx+c$, factor by finding the roots:
$$x=\frac{-b\pm\sqrt{b^{2}-4c}}{2}.$$
Factoring the original quadratic would then give you
$$x^{2}+bx+c=\left(x+\frac{b+\sqrt{b^{2}-4c}}{2}\right)\left(x+\frac{b-\sqrt{b^{2}-4c}}{2}\right).$$
#### QuestForInsight
##### Member
If you need more help on them feel free to create a new topic, it's not fair on Casio to sidetrack his topic
No need to be rude here. The question I have asked is very relevant. These simultaneous equations are the Vieta relations of the quadratic equation being solved. Trying to eliminate one variable would get you back to the original equation.
#### QuestForInsight
##### Member
Substitution. We have that $f=8-g$, so plugging that into the other equation yields
$$-3=g(8-g)=8g-g^{2},$$
or
$$g^{2}-8g-3=0.$$
Now you have a quadratic in $g$. You can use the quadratic formula to solve:
$$g=\frac{8\pm\sqrt{64-4(-3)}}{2}=\dots$$
Come to think of it, we haven't gained anything here, because this is the same quadratic as we started with! | {
"domain": "mathhelpboards.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9850429156022934,
"lm_q1q2_score": 0.8486521079697523,
"lm_q2_score": 0.8615382076534742,
"openwebmath_perplexity": 356.01625523466294,
"openwebmath_score": 0.7397609949111938,
"tags": null,
"url": "https://mathhelpboards.com/threads/quadratic-equations.1874/"
} |
So you could save time by simply using the quadratic formula in the first place. Given the monic quadratic $x^{2}+bx+c$, factor by finding the roots:
$$x=\frac{-b\pm\sqrt{b^{2}-4c}}{2}.$$
Factoring the original quadratic would then give you
$$x^{2}+bx+c=\left(x+\frac{b+\sqrt{b^{2}-4c}}{2}\right)\left(x+\frac{b-\sqrt{b^{2}-4c}}{2}\right).$$
I was thinking about this, and I thought what if we try to find $f-g$ first?
$(f-g)^2 = (f+g)^2-4fg \implies f-g = \pm \sqrt{76}$.
$\left\{\begin{array}{c}f+g = 8 \\\\ f-g = \pm\sqrt{76} \end{array} \implies 2f = 8\pm \sqrt{76} \implies \boxed{f = 4\pm\frac{1}{2}\sqrt{76}} \implies \boxed{g = 4\mp\frac{1}{2}\sqrt{76}}.\right\|$
Without loss of generality, we take $f = 4+\frac{1}{2}\sqrt{76}$ and $g = 4-\frac{1}{2}\sqrt{76}$.
I think this also gives a way of proving the quadratic formula!
Last edited:
#### Casio
##### Member
There is a way of factoring quadratics that takes the guesswork out of it. In your example, you just note that $x$ is common to both terms, and factors out leaving you with $x(x-3)=0$.
But let's suppose you had a general quadratic: $x^{2}+bx+c$, which we'll make monic for convenience. What you're after is an equivalent expression that looks like this:
$$x^{2}+bx+c=(x+f)(x+g).$$
So, FOIL out the RHS thus:
$$x^{2}+bx+c=x^{2}+gx+fx+fg=x^{2}+(g+f)x+fg.$$
Now, you get to equate like powers of $x$ to obtain the following equations:
\begin{align*}
b&=f+g\\
c&=fg.
\end{align*}
Use whatever method you like for solving this simultaneous system of equations for $f$ and $g$. Once you do that, you're done.
Example: Factor $x^{2}+8x-3$.
Solution: Write as
$$x^{2}+8x-3=(x+f)(x+g)=x^{2}+(f+g)x+fg.$$
Solve:
\begin{align*}
8&=f+g\\
-3&=fg.
\end{align*}
The solution is
$$f=4\pm\sqrt{19},\quad g=4\mp\sqrt{19}.$$
That is, if $f=4+\sqrt{19}$, then $g=4-\sqrt{19}$, or if $f=4-\sqrt{19}$, then $g=4+\sqrt{19}$. So our factorization is
$$x^{2}+8x-3=(x+4-\sqrt{19})(x+4+\sqrt{19}).$$
Multiply this out to confirm. | {
"domain": "mathhelpboards.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9850429156022934,
"lm_q1q2_score": 0.8486521079697523,
"lm_q2_score": 0.8615382076534742,
"openwebmath_perplexity": 356.01625523466294,
"openwebmath_score": 0.7397609949111938,
"tags": null,
"url": "https://mathhelpboards.com/threads/quadratic-equations.1874/"
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.