text stringlengths 1 2.12k | source dict |
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• The modular argument is not hard. The numbers are all of the form $19,999,999 - 10^k$ so we need $19,999,999 - 10^k \equiv 0 \mod 7$ or $10^k \equiv 5 \mod 7$. As $10$ and $7$ is relativley prime and $7$ is prime the $10^k; 0\le k < 7$ are distinct modulo $7$ and $10^5$ is the only one that works. – fleablood Nov 20 '18 at 17:28
• @fleablood, are you saying, more generally, that if $10$ and $p$ are relatively prime (with $p$ a prime), then $10^k$ for $0\le k\lt p$ are distinct modulo $p$? It's true that $10$ is a primitive root mod $7$, but not because it's relatively prime to $7$. – Barry Cipra Nov 20 '18 at 22:46
• Yeah, I guess I worded it incorrectly. $10\equiv 3$ is a primitive root is what I meant. – fleablood Nov 20 '18 at 23:08
I also verify that the Number 19,899,999 is the smallest number that is divisible by 63 and Also its sum is divisible by 63. Here a program I wrote in Python 3. To find this out by brute forcing :
for i in range(9999990, 19900000, 63):
sum_is = sum(int(d) for d in str(i))
if sum_is%63==0:
print(i)
Prints :
19899999
Also this time I started from 9999990 because this number the last number divisible by 63 that has sum less than 63.
This can also be used to do the same with any number.
Just change the range and 63 as you want.
Hope this Helps!
[EDIT SUMMARY] For more Optimization and better Understanding. Added direct summing instead of sum_digits as request/suggested by @Paul Evans. Also added jump of 63 as Suggested by @Kyle Kanos.
• Instead of the mysterious sum_digits(i) you could write python code: sum(int(d) for d in str(i)) – Paul Evans Nov 20 '18 at 16:01
• You can also save time by starting at 9999990 and incrementing by 63, eliminating the need to test i%63==0. – Kyle Kanos Nov 20 '18 at 19:47
• Yes! That's more optimized. Thanx for advice. – FightWithCode Nov 21 '18 at 13:20 | {
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321 views
For an undirected graph $G=(V, E)$, the line graph $G'=(V', E')$ is obtained by replacing each edge in $E$ by a vertex, and adding an edge between two vertices in $V'$ if the corresponding edges in $G$ are incident on the same vertex. Which of the following is TRUE of line graphs?
1. the line graph for a complete graph is complete
2. the line graph for a connected graph is connected
3. the line graph for a bipartite graph is bipartite
4. the maximum degree of any vertex in the line graph is at most the maximum degree in the original graph
5. each vertex in the line graph has degree one or two
retagged | 321 views
+2
Option B). is true.
+2
yes, Agreed. (y)
0
why not option C?
0
you can check with K(2,2). The line graph obtained is not biparitite.
option B is the right answer
We can solve this question by eliminating options
Option (A) False
Let us take a complete graph of $4$ vertices:
In a line graph, no. of edges is
$\sum_{i=0}^{n}$ ${}^{d_{i}}C_{{2}},$ $d_{i}$is degree of each vertex
=$\frac{3\times 2}{2}+\frac{3\times 2}{2}+\frac{3\times 2}{2}+\frac{3\times 2}{2} = 3\times 4=12$
No. of vertices in line graph = No. of edges in original graph
No. of vertices in line graph $= 6$
So, no. of edges to make complete graph with $6$ vertices $= \frac{6\times 5}{2} = 3\times 5=15$
But for given line graph from complete graph of $4$ vertices we have only $12$ edges.
Option (B) True
1. Smallest line graph for original graph one edge
which is also connected graph
If a graph is connected with more then one edge, it will never be disconnected
Option (C) False
This cannot be 2-colorable and hence is not bipartite.
Option (D) False
Because line graph degree of vertex depends on the attribute
e.g., $\left [ A,B \right ]$ as point in line graph, then degree of this vertex depends on degree of $A$ and degree of $B$ in the original graph.
I'm drawing degree for a point $[AB]$ in line graph.
So, this is wrong. | {
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I'm drawing degree for a point $[AB]$ in line graph.
So, this is wrong.
Option (E) False (wrong as proved in above option (D))
edited
The line graph of a connected graph is connected. If G is connected, it contains a path connecting any two of its edges, which translates into a path in L(G) containing any two of the vertices of L(G). Therefore, option B is correct.
We can also do this question using elimination of options.
edited
0
but your L(G2): e1---e2 it is bipertite right?e1 may be in one partition and e2 on another...it is bipertite i guess.
0
Hi,
I have corrected the example, and updated the file.
0
great example you have updated...proving line of tree is not a tree too :p | {
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fplot3
3-D parametric curve plotter
Syntax
fplot3(funx,funy,funz)
fplot3(funx,funy,funz,tinterval)
fplot3(___,LineSpec)
fplot3(___,Name,Value)
fplot3(ax,___)
fp = fplot3(___)
Description
example
fplot3(funx,funy,funz) plots the parametric curve defined by x = funx(t), y = funy(t), and z = funz(t) over the default interval [-5,5] for t.
example
fplot3(funx,funy,funz,tinterval) plots over the specified interval. Specify the interval as a two-element vector of the form [tmin tmax].
example
fplot3(___,LineSpec) sets the line style, marker symbol, and line color. For example, '-r' specifies a red line. Use this option after any of the previous input argument combinations.
fplot3(___,Name,Value) specifies line properties using one or more name-value pair arguments. For example, 'LineWidth',2 specifies a line width of 2 points.
fplot3(ax,___) plots into the axes specified by ax instead of the current axes. Specify the axes as the first input argument.
example
fp = fplot3(___) returns a ParameterizedFunctionLine object. Use the object to query and modify properties of a specific line. For a list of properties, see ParameterizedFunctionLine Properties.
Examples
collapse all
Plot the 3-D parametric line
$\begin{array}{c}x=\mathrm{sin}\left(t\right)\\ y=\mathrm{cos}\left(t\right)\\ z=t\end{array}$
over the default parameter range [-5 5].
xt = @(t) sin(t); yt = @(t) cos(t); zt = @(t) t; fplot3(xt,yt,zt)
Plot the parametric line
$\begin{array}{c}x={e}^{-t/10}\mathrm{sin}\left(5t\right)\\ y={e}^{-t/10}\mathrm{cos}\left(5t\right)\\ z=t\end{array}$
over the parameter range [-10 10] by specifying the fourth input argument of fplot3.
xt = @(t) exp(-t/10).*sin(5*t); yt = @(t) exp(-t/10).*cos(5*t); zt = @(t) t; fplot3(xt,yt,zt,[-10 10]) | {
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Plot the same 3-D parametric curve three times over different intervals of the parameter. For the first interval, use a line width of 2 points. For the second, specify a dashed red line style with circle markers. For the third, specify a cyan, dash-dotted line style with asterisk markers.
fplot3(@(t)sin(t), @(t)cos(t), @(t)t, [0 2*pi], 'LineWidth', 2) hold on fplot3(@(t)sin(t), @(t)cos(t), @(t)t, [2*pi 4*pi], '--or') fplot3(@(t)sin(t), @(t)cos(t), @(t)t, [4*pi 6*pi], '-.*c') hold off
Plot multiple lines in the same axes using hold on.
fplot3(@(t)t, @(t)t, @(t)t) hold on fplot3(@(t)-t, @(t)t, @(t)-t) hold off
Plot the parametric line
$\begin{array}{c}x={e}^{-|t|/10}\mathrm{sin}\left(5|t|\right)\\ y={e}^{-|t|/10}\mathrm{cos}\left(5|t|\right)\\ z=t.\end{array}$
Assign the parameterized function line object to a variable.
xt = @(t)exp(-abs(t)/10).*sin(5*abs(t)); yt = @(t)exp(-abs(t)/10).*cos(5*abs(t)); zt = @(t)t; fp = fplot3(xt,yt,zt)
fp = ParameterizedFunctionLine with properties: XFunction: @(t)exp(-abs(t)/10).*sin(5*abs(t)) YFunction: @(t)exp(-abs(t)/10).*cos(5*abs(t)) ZFunction: @(t)t Color: [0 0.4470 0.7410] LineStyle: '-' LineWidth: 0.5000 Show all properties
Change the range of parameter values to [-10 10] and change the line color to red.
fp.TRange = [-10 10]; fp.Color = 'r';
For $t$ values in the range $-2\pi$ to $2\pi$, plot the parametric line
$\begin{array}{c}x=t\\ y=t/2\\ z=\mathrm{sin}\left(6t\right).\end{array}$
Add a title, x-axis label, and y-axis label. Additionally, change the view of the axes and display the axes box outline.
xt = @(t)t; yt = @(t)t/2; zt = @(t)sin(6*t); fplot3(xt,yt,zt,[-2*pi 2*pi],'MeshDensity',30,'LineWidth',1); title('x=t, y=t/2, z=sin(6t) for -2\pi<t<2\pi') xlabel('x'); ylabel('y'); view(52.5,30) box on
Access the axes object using gca. Specify the x-axis tick values and associated labels using the XTick and XTickLabel properties of the axes object. Similarly, specify the y-axis tick values and associated labels. | {
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ax = gca; ax.XTick = -2*pi:pi/2:2*pi; ax.XTickLabel = {'-2\pi','-3\pi/2','-\pi','-\pi/2','0','\pi/2','\pi','3\pi/2','2\pi'}; ax.YTick = -pi:pi/2:pi; ax.YTickLabel = {'-\pi','-\pi/2','0','\pi/2','\pi'};
Input Arguments
collapse all
Parametric function for x coordinates, specified as a function handle to a named or anonymous function.
Specify a function of the form x = funx(t). The function must accept a vector input argument and return a vector output argument of the same size. Use array operators instead of matrix operators for the best performance. For example, use .* (times) instead of * (mtimes).
Example: funx = @(t) sin(2*t);
Parametric function for y coordinates, specified as a function handle to a named or anonymous function.
Specify a function of the form y = funy(t). The function must accept a vector input argument and return a vector output argument of the same size. Use array operators instead of matrix operators for the best performance. For example, use .* (times) instead of * (mtimes).
Example: funy = @(t) cos(2*t);
Parametric function for z coordinates, specified as a function handle to a named or anonymous function.
Specify a function of the form z = funz(t). The function must accept a vector input argument and return a vector output argument of the same size. Use array operators instead of matrix operators for the best performance. For example, use .* (times) instead of * (mtimes).
Example: funz = @(t) t;
Interval for parameter t, specified as a two-element vector of the form [tmin tmax].
Axes object. If you do not specify an axes object, then fplot3 uses the current axes (gca).
Line style, marker, and color, specified as a character vector or string containing symbols. The symbols can appear in any order. You do not need to specify all three characteristics (line style, marker, and color). For example, if you omit the line style and specify the marker, then the plot shows only the marker and no line. | {
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Example: '--or' is a red dashed line with circle markers
Line StyleDescription
-Solid line
--Dashed line
:Dotted line
-.Dash-dot line
MarkerDescription
'o'Circle
'+'Plus sign
'*'Asterisk
'.'Point
'x'Cross
'_'Horizontal line
'|'Vertical line
's'Square
'd'Diamond
'^'Upward-pointing triangle
'v'Downward-pointing triangle
'>'Right-pointing triangle
'<'Left-pointing triangle
'p'Pentagram
'h'Hexagram
ColorDescription
y
yellow
m
magenta
c
cyan
r
red
g
green
b
blue
w
white
k
black
Name-Value Pair Arguments
Specify optional comma-separated pairs of Name,Value arguments. Name is the argument name and Value is the corresponding value. Name must appear inside quotes. You can specify several name and value pair arguments in any order as Name1,Value1,...,NameN,ValueN.
Example: 'Marker','o','MarkerFaceColor','red'
The properties listed here are only a subset. For a complete list, see ParameterizedFunctionLine Properties.
Number of evaluation points, specified as a number. The default is 23. Because fplot3 uses adaptive evaluation, the actual number of evaluation points is greater.
Line color, specified as an RGB triplet, a hexadecimal color code, a color name, or a short name.
For a custom color, specify an RGB triplet or a hexadecimal color code.
• An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range [0,1]; for example, [0.4 0.6 0.7].
• A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (#) followed by three or six hexadecimal digits, which can range from 0 to F. The values are not case sensitive. Thus, the color codes '#FF8800', '#ff8800', '#F80', and '#f80' are equivalent.
Alternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes. | {
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Color NameShort NameRGB TripletHexadecimal Color CodeAppearance
'red''r'[1 0 0]'#FF0000'
'green''g'[0 1 0]'#00FF00'
'blue''b'[0 0 1]'#0000FF'
'cyan' 'c'[0 1 1]'#00FFFF'
'magenta''m'[1 0 1]'#FF00FF'
'yellow''y'[1 1 0]'#FFFF00'
'black''k'[0 0 0]'#000000'
'white''w'[1 1 1]'#FFFFFF'
Here are the RGB triplets and hexadecimal color codes for the default colors MATLAB® uses in many types of plots.
[0 0.4470 0.7410]'#0072BD'
[0.8500 0.3250 0.0980]'#D95319'
[0.9290 0.6940 0.1250]'#EDB120'
[0.4940 0.1840 0.5560]'#7E2F8E'
[0.4660 0.6740 0.1880]'#77AC30'
[0.3010 0.7450 0.9330]'#4DBEEE'
[0.6350 0.0780 0.1840]'#A2142F'
Example: 'blue'
Example: [0 0 1]
Example: '#0000FF'
Line style, specified as one of the options listed in this table.
Line StyleDescriptionResulting Line
'-'Solid line
'--'Dashed line
':'Dotted line
'-.'Dash-dotted line
'none'No lineNo line
Line width, specified as a positive value in points, where 1 point = 1/72 of an inch. If the line has markers, then the line width also affects the marker edges.
The line width cannot be thinner than the width of a pixel. If you set the line width to a value that is less than the width of a pixel on your system, the line displays as one pixel wide.
Marker symbol, specified as one of the values listed in this table. By default, the object does not display markers. Specifying a marker symbol adds markers at each data point or vertex.
ValueDescription
'o'Circle
'+'Plus sign
'*'Asterisk
'.'Point
'x'Cross
'_'Horizontal line
'|'Vertical line
'square' or 's'Square
'diamond' or 'd'Diamond
'^'Upward-pointing triangle
'v'Downward-pointing triangle
'>'Right-pointing triangle
'<'Left-pointing triangle
'pentagram' or 'p'Five-pointed star (pentagram)
'hexagram' or 'h'Six-pointed star (hexagram)
'none'No markers
Marker outline color, specified as 'auto', an RGB triplet, a hexadecimal color code, a color name, or a short name. The default value of 'auto' uses the same color as the Color property. | {
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For a custom color, specify an RGB triplet or a hexadecimal color code.
• An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range [0,1]; for example, [0.4 0.6 0.7].
• A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (#) followed by three or six hexadecimal digits, which can range from 0 to F. The values are not case sensitive. Thus, the color codes '#FF8800', '#ff8800', '#F80', and '#f80' are equivalent.
Alternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.
Color NameShort NameRGB TripletHexadecimal Color CodeAppearance
'red''r'[1 0 0]'#FF0000'
'green''g'[0 1 0]'#00FF00'
'blue''b'[0 0 1]'#0000FF'
'cyan' 'c'[0 1 1]'#00FFFF'
'magenta''m'[1 0 1]'#FF00FF'
'yellow''y'[1 1 0]'#FFFF00'
'black''k'[0 0 0]'#000000'
'white''w'[1 1 1]'#FFFFFF'
'none'Not applicableNot applicableNot applicableNo color
Here are the RGB triplets and hexadecimal color codes for the default colors MATLAB uses in many types of plots.
[0 0.4470 0.7410]'#0072BD'
[0.8500 0.3250 0.0980]'#D95319'
[0.9290 0.6940 0.1250]'#EDB120'
[0.4940 0.1840 0.5560]'#7E2F8E'
[0.4660 0.6740 0.1880]'#77AC30'
[0.3010 0.7450 0.9330]'#4DBEEE'
[0.6350 0.0780 0.1840]'#A2142F'
Marker fill color, specified as 'auto', an RGB triplet, a hexadecimal color code, a color name, or a short name. The 'auto' value uses the same color as the MarkerEdgeColor property.
For a custom color, specify an RGB triplet or a hexadecimal color code.
• An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range [0,1]; for example, [0.4 0.6 0.7]. | {
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• A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (#) followed by three or six hexadecimal digits, which can range from 0 to F. The values are not case sensitive. Thus, the color codes '#FF8800', '#ff8800', '#F80', and '#f80' are equivalent.
Alternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.
Color NameShort NameRGB TripletHexadecimal Color CodeAppearance
'red''r'[1 0 0]'#FF0000'
'green''g'[0 1 0]'#00FF00'
'blue''b'[0 0 1]'#0000FF'
'cyan' 'c'[0 1 1]'#00FFFF'
'magenta''m'[1 0 1]'#FF00FF'
'yellow''y'[1 1 0]'#FFFF00'
'black''k'[0 0 0]'#000000'
'white''w'[1 1 1]'#FFFFFF'
'none'Not applicableNot applicableNot applicableNo color
Here are the RGB triplets and hexadecimal color codes for the default colors MATLAB uses in many types of plots.
[0 0.4470 0.7410]'#0072BD'
[0.8500 0.3250 0.0980]'#D95319'
[0.9290 0.6940 0.1250]'#EDB120'
[0.4940 0.1840 0.5560]'#7E2F8E'
[0.4660 0.6740 0.1880]'#77AC30'
[0.3010 0.7450 0.9330]'#4DBEEE'
[0.6350 0.0780 0.1840]'#A2142F'
Example: [0.3 0.2 0.1]
Example: 'green'
Example: '#D2F9A7'
Marker size, specified as a positive value in points, where 1 point = 1/72 of an inch.
Output Arguments
collapse all
One or more ParameterizedFunctionLine objects, returned as a scalar or a vector. You can use these objects to query and modify properties of a specific ParameterizedFunctionLine object. For details, see ParameterizedFunctionLine Properties.
Introduced in R2016a | {
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# Number of possible combinations
1. May 14, 2014
### kidkook
I have 16 items and i need to know the number of possible combination for all 16 items. I know the possible number of combination for all 16 items is 1 and the possible number of combinations for 15 of the 16 items is 16 using the formula below:
$$\frac{n!}{r!(n-r)!}.$$
How can i adaprt this formula to calculate for the number of selection from 1 to 16 inclusive without repeating the calculation 16 times.
So n will always be 16 and r will be 1 to 16 inclusive.
2. May 14, 2014
Your wording is a little confusing. Is this the correct interpretation:
You want to know the total number of ways to select at least one item from the 16.
If it is, think about this: you should know the total number of subsets (empty, proper, with the original set) for a set with 16 items. Your question involves finding not the total number but how many there are not counting the empty set.
3. May 14, 2014
### kidkook
Sorry for the confusion. What i have is a dataset with 16 different values. What i'm trying to achieve is to determine how many combination are available without repetition/duplication of a value. So i know the answer is 65536. I achieved this by using the above formula 16 times and substituting the value of r from 1 to 16 etc...and adding the number together.
Number of combinations Value of r
16 1
120 2
560 3
1820 4
4368 5
8008 6
11440 7
12870 8
11440 9
8008 10
4368 11
1820 12
560 13
120 14
16 15
1 16
65535
What i would like to know is...can the formula be adapted to so i don't need to repeat the calculation 16 times. My dataset is likely to increase significantly in the future so it will not be practical to use this method.
4. May 14, 2014
### jbunniii | {
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4. May 14, 2014
### jbunniii
Note that $\frac{n!}{r!(n-r)!}$ is called the binomial coefficient and is often denoted
$${n \choose r} = \frac{n!}{r!(n-r)!}$$
The reason it is called the binomial coefficient is the following property:
$$(x+y)^n = \sum_{r=0}^{n}{n \choose r}x^r y ^{n-r}$$
If we plug in $x=y=1$ then this reduces to
$$2^n = \sum_{r=0}^{n}{n \choose r} = 1 + \sum_{r=1}^{n} {n \choose r}$$
where the second equality follows because ${n \choose 0} = 1$. Therefore,
$$\sum_{r=1}^{n} {n \choose r} = 2^n - 1$$
which is consistent with your calculation since $2^{16} - 1 = 65535$.
5. May 14, 2014
### kidkook
Excellent. This is exactly what i was looking for. A very well documented explaination. Many thanks. | {
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# What is the probability of rolling at least two 6's with 3 Dice and 2 Rolls?
Question: What is the probability of rolling at least two 6's, when rolling 3 dice with two rolls? (with your first roll you keep dice only if they are 6's and roll the remainder for your second roll).
(6*6*6 = 216 = outcomes when rolling 3 dice) (6*6 = 36 = outcomes when rolling 2 dice)
1st roll: (3 dice)
A = 125/216 = 0 sixes
B = 75/216 = 1 six
C = 15/216 = 2 sixes
D = 1/216 = 3 sixes
Outcomes C and D fulfill requirement.
For outcome A: (first roll = no 6's, pick up all dice throw 3 dice again)
2nd roll:
a = 125/216 = 0 sixes
b = 75/216 = 1 six
c = 15/216 = 2 sixes
d = 1/216 = 3 sixes
Outcomes c and d fulfill requirements.
For outcome B: (first roll = one 6, pick up two non 6's and roll again)
2nd roll:
z = 25/36 = 0 sixes
y = 10/36 = 1 six
x = 1/36 = 2 sixes
Outcomes y and x fulfill requirements.
So would the formula below give me my answer?
C + D + Ac + Ad + By + Bx = X
What is the probability of rolling at least two 6's, when rolling 3 dice with two rolls?
If I substituted correctly and did the math correct the answer I got was 22.3%
Is this correct? | {
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If I substituted correctly and did the math correct the answer I got was 22.3%
Is this correct?
• I'm guessing by your odds that you consider ordering important ? – user451844 Jul 29 '17 at 22:20
• I'm not sure if I know what you mean by ordering. – Travis Jul 29 '17 at 22:58
• is (1,2,3) the same as (1,3,2) and (2,3,1) and (2,1,3) and (3,1,2) and (3,2,1) ? if so then it cuts down the number of distinct rolls down to 56. – user451844 Jul 29 '17 at 23:00
• Well let's say we had a blue green and a red die, if the red die shows a 1 and blue shows a 6 and green shows a 6, that would be different than if red showed a 6, green showed a 6, and blue showed a 1. So I guess ordering would matter. The numbers rolled are the same but show on different dice. Both rolls fulfill the two 6's requirement but are different rolls and should be counted as thus. I'm not sure if I'm correct on this, that's why I'm asking but does it makes sense? – Travis Jul 29 '17 at 23:09
• yeah I'm just making sure, because the probability might change depending on conditions like this. of the 56 cases when order didn't matter a full 21 had at least one 6, but only 6 had more than one 6. etc. – user451844 Jul 29 '17 at 23:19
An alternate approach would be to find the probability of the complementary event:
$\textbf{1)}$ The probability of getting no 6's is given by $\big(\frac{5}{6}\big)^3\cdot\big(\frac{5}{6}\big)^3=\big(\frac{5}{6}\big)^6$
$\textbf{2)}$ The probability of getting exactly one 6 is given by
$\hspace{.2 in}\big(\frac{5}{6}\big)^3\cdot3\big(\frac{1}{6}\big)\big(\frac{5}{6}\big)^2+3\big(\frac{1}{6}\big)\big(\frac{5}{6}\big)^2\cdot\big(\frac{5}{6}\big)^2=\big(\frac{1}{2}\big)\big(\frac{5}{6}\big)^5+\big(\frac{1}{2}\big)\big(\frac{5}{6}\big)^4$
Therefore the probability of getting at least two 6's is given by
$\hspace{.2 in} 1-\big(\frac{5}{6}\big)^6-\big(\frac{1}{2}\big)\big(\frac{5}{6}\big)^5-\big(\frac{1}{2}\big)\big(\frac{5}{6}\big)^4=\frac{5203}{23328}\approx.223$ | {
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• From my point of view is your approach the most appropriate one. (+1) – Markus Scheuer Jul 30 '17 at 9:38
Your approach and your calculations are correct. Here is a variation based upon generating functions. We encode the roll of three dice with \begin{align*} (5+t)^3 \end{align*} marking an occurrence of $6$ with $t$ and collecting all other five possibilities with $5$. The probability to get $j$ sixes $0\leq j \leq 3$ in the first roll can be written as \begin{align*} [t^j](5+t)^3\cdot\frac{1}{6^3} \end{align*}
Encoding the second roll with $(5+u)^{3-j}$ we calculate \begin{align*} \sum_{{0\leq j,k\leq 3}\atop{j+k\geq 2}}&[t^j](5+t)^3[u^k](5+u)^{3-j}\cdot\frac{1}{6^{6-j}}\\ &=[t^0](5+t)^3\left([u^2]+[u^3]\right)(5+u)^3\cdot\frac{1}{6^6}\\ &\qquad+[t^1](5+t)^3\left([u^1]+[u^2]\right)(5+u)^2\cdot\frac{1}{6^5}\\ &\qquad+[t^2](5+t)^3\left([u^0]+[u^1]\right)(5+u)^1\cdot\frac{1}{6^4}\\ &\qquad+[t^3](5+t)^3\\ &=\binom{3}{0}5^3\left(\binom{3}{2}5^1+\binom{3}{3}5^0\right)\cdot\frac{1}{6^6}\\ &\qquad+\binom{3}{1}5^2\left(\binom{2}{1}5^1+\binom{2}{2}5^0\right)\cdot\frac{1}{6^5}\\ &\qquad+\binom{3}{2}5^1\left(\binom{1}{0}5^1+\binom{1}{1}5^0\right)\cdot\frac{1}{6^4}\\ &\qquad+\binom{3}{3}5^0\cdot\frac{1}{6^3}\\ &=\frac{125\cdot16}{6^6}+\frac{75\cdot11}{6^5}+\frac{15\cdot6}{6^4}+\frac{1}{6^3}\\ &=\frac{5\,203}{23\,328}\\ &\doteq 0.223 \end{align*} | {
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# How many ways are there to tile a chessboard with 32 knights?
The maximal amount of knights which can be places on a regular $8 \times 8$ chessboard so that no two take each other is $32$ (To see this just notice that a knight on a black square only attacks knights placed on white squares).
I was wondering how many ways there are to tile a regular chessboard with these $32$ knights so that no two take each other.
A generalisation of this question has already been posed here: In how many ways we can place $N$ mutually non-attacking knights on an $M \times M$ chessboard?. However, the answers given include a polynomial approach which cannot be used in the case of an $8 \times 8$ chessboard, because, as an answer states, a computer struggles to compute such a polynomial.
Even if it were quite easy for a computer to compute such a polynomial, in an Olympiad no such computational power is allowed. Hence, I was wondering if there was a pure combinatorics approach for this particular case.
An example of such a tiling is to place all the knights on all the black cells. | {
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An example of such a tiling is to place all the knights on all the black cells.
• You can put the knights on the black squares or on the white squares. This makes two ways. Or am I missing something? – ajotatxe Jul 20 '17 at 11:58
• @ajotatxe Are there no other possibilities? Can you prove this? – Plato Jul 20 '17 at 11:59
• Consider a knight's tour of the chessboard. Obviously, if $32$ of the $64$ squares are occupied by non-attacking knights, then the tour must alternate between occupied and unoccupied squares. This means that there are (at most) two ways to place the $32$ knights. But of course there are two ways, use all the black squares or all the white squares. – bof Jul 20 '17 at 12:11
• In graph-theoretic terms: A cycle graph $C_{2n}$ has exactly two independent sets of size $n.$ Therefore, a Hamiltonian graph of order $2n$ has at most two independent sets of size $n.$ The knight's graph on the $8\times8$ chessboard is a Hamiltonian graph of order $64.$ – bof Jul 20 '17 at 12:20
• The existence of a knight's tour seems crucial here. Actually, in a $2\times 2$ board there is no knight's tour, and there are $6$ ways to put two non-attacking horses on it. – ajotatxe Jul 20 '17 at 12:26 | {
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Consider a knight's tour of the chessboard. Obviously, if $32$ of the $64$ squares are occupied by non-attacking knights, then the tour must alternate between occupied and unoccupied squares, since a horse can only attack a square if the opposite colour. This means that there are (at most) two ways to place the $32$ knights. We can offer examples of these two ways: use all the black squares or all the white squares.
Another way to look at the problem is to consider graph theory. A cycle graph $C_{2n}$ has exactly two independent sets of size $n$. Therefore, a Hamiltonian graph of order $2n$ has at most two independent sets of size $n$. The knight's graph on the $8 \times 8$ chessboard is a Hamiltonian graph of order 64. From here we can conclude in the same manner as in the first solution. | {
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## Rounding Error
#### Problem
A number is rounded to two decimal places and this answer is rounded to one decimal place. If the final number is 0.2 then what is the chance that the original number lies between 0.15 and 0.25?
#### Solution
When rounding to $k$ decimal places (d.p.) we truncate the number at the $k^{th}$ decimal position, but if the next digit is 5 or more then we increase the last digit by 1; this may have a knock-on effect. Consider the following examples of rounding to 2 d.p.:
3.61483 3.61
7.7853301 7.79
0.5982 0.60
Suppose we were told that 0.2 were the result of rounding to one decimal place. The smallest number that would be rounded up to 0.2 is 0.15. However, there is no actual value we can assign to the greatest value that would round down to 0.2. For example, 0.25 would round up to 0.3 and although 0.249 would round down to 0.2, what about 0.2499, or 0.24999, or ... ?
Under these circumstances we define bounds (or limits) in the following way:
• The lower bound is the least value that rounds up to the given number.
• The upper bound is the least value that fails to round down to the given number.
So if 0.2 is the result of rounding to 1 d.p. then the lower and upper bounds are 0.15 and 0.25 respectively.
Working backwards, the lower bound of 0.2 is 0.15 and the lower bound of 0.15 is 0.145. We can easily check this: 0.145 = 0.15 (2 d.p.) and then 0.15 = 0.2 (1 d.p.).
However, we need to tread very carefully when working backwards with the upper bound. Although it is true to say that the upper bound of 0.2 is 0.25, if the original number were rounded to 0.25 (to 2 d.p.), then next process of rounding to 1 d.p. would make it 0.3, not 0.2.
In fact, it is necessary that the result of rounding to 2 d.p. takes it to 0.24; and the upper bound of 0.24 is 0.245.
Hence we are certain that the original number, $x$, lies somewhere in the interval $0.145 \le x \lt 0.245$. | {
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If we are considering if the number lies between 0.15 and 0.25 then it would have to lie in the interval $0.15 \le x \lt 0.245$.
\begin{align}\therefore P(0.15 \le x \lt 0.245 \mid 0.145 \le x \lt 0.245) &= \dfrac{0.245 - 0.15}{0.245 - 0.145}\\\\&= \dfrac{0.095}{0.1}\\\\&= 95\%\end{align}
If instead of rounding to 2 d.p. then to 1 d.p. a number is just rounded to 1 d.p., what is the chance that both methods give the same answer?
Problem ID: 343 (09 Jul 2008) Difficulty: 2 Star
Only Show Problem | {
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# Sum of squares of multiplicities of differences
It's hard to come up with a good title. Let $A$ be a set of $n$ integers, say $A=\{a_1,\ldots,a_n\}$ with $a_1<\cdots<a_n$. We consider all the positive differences $a_j-a_i$ (necessarily $i<j$). There are $n \choose 2$ pairs $(a_i,a_j)$ with $i<j$, but of course, some differences can be repeated, or in other words, have "multiplicity" greater than 1.
For a given $A$, one way to illustrate the differences and their multiplicities is with a "table of differences". Here is one such table for $A=\{0,1,3,5,6\}$, and another for $A=\{1,4,7,10,13\}$.
-| 0 1 3 5 6 - | 1 4 7 10 13
-+---------- --+------------
0| - 1 3 5 6 1 | - 3 6 9 12
1| - 2 4 5 4 | - 3 6 9
3| - 2 3 7 | - 3 6
5| - 1 10| - 3
6| - 13| -
We see that if $A=\{0,1,3,5,6\}$, then the differences $1,2,3,5$ each have multiplicity 2, and the differences $4,6$ each have multiplicity 1. We also see that if $A=\{1,4,7,10,13\}$, then the differences $3,6,9,12$ have multiplicities $4,3,2,1$ respectively.
My question is: What is a tight upper bound on the sum of the squares of the multiplicities? I conjecture that the answer is $1^2+2^2+3^2+\cdots+(n-1)^2$, which occurs if $A$ consists of $n$ integers in arithmetic progression (as in the second example above).
At first, I thought this would be an easy consequence of the following "lemma": The largest multiplicity must be at most $n-1$, the second largest multiplicity must be at most $n-2$, the third largest multiplicity must be at most $n-3$, and so on. However, that "lemma" is false! In the first example above (in which $n=5$), the four largest multiplicities are $2,2,2,2$, which are not bounded above by $4,3,2,1$ respectively.
Note that my conjectured bound grows like $n^3/3$. I'm pretty sure I can find an upper bound that grows like $2n^3/3$, but the gap between those bounds is perplexing me. | {
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-
Your lemma might be true if you modify it to say the largest multiplicity is at most $n-1$, the two largest multiplicities sum to at most $n-1+n-2=2n-3$, the three largest sum to at most $(n-1)+(n-2)+(n-3)=3n-6$, etc. – Gerry Myerson Jun 21 '12 at 1:07
We can show by induction that your bound is the best.
Suppose we have $n$ numbers and the differences have multiplicities $m_1\ge m_2\ge \cdots \ge m_t$. If we now add a new largest number, it creates $n$ distinct differences. If we could freely choose the differences (without constraints based on the other numbers), we could choose up to $n$ of the $m_i$ to increment exactly once, or can add $m_{t+1}=1,m_{t+2}=1,\ldots$ as necessary for differences that do not match.
Since $m_i>m_j\Rightarrow (m_i+1)^2+m_j^2>m_i^2+(m_j+1)^2$ then without constraints on the differences we maximize our sum of squares by incrementing the $n$ largest differences available. Starting with $n=2, m_1=1$, the maximum for $n=3$ is achieved by $m_1\leftarrow 2,m_2=1$, and so on. If we assume that for $n=k$ the maximum is $m_1=(k-1),m_2=(k-2),\ldots,m_{k-1}=1$, then for $k+1$ numbers the maximum is $m_1\leftarrow k,m_2\leftarrow (k-1),\ldots, m_{k-1}\leftarrow 2, m_k=1$. So by induction this pattern gives a bound for all $n$. And, as you pointed out, this bound can be achieved by choosing numbers in an arithmetic sequence.
This line of thought also confirms @Gerry's comment on your Lemma. Starting with one number and adding new largest numbers, the best we can do for the top $M$ multiplicities is to increment some subset of them each time we add a number until we have $M$ numbers, then increment every one of them for each number added after that. This constraint is thus that they add to at most $Mn-M(M+1)/2$, but the $M$th largest can be up to $n-(M+1)/2$.
-
Thinking about it today (everybody's answers and comments certainly helped), I believe there is a relatively short inductive proof that doesn't bother with any lemmas. | {
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My bound is certainly true for $n=2$. When we transition from a set $A=\{a_1<\cdots<a_n\}$ to a set $A'=\{a_1<\cdots<a_n<a_{n+1}\}$, then as Zander mentions, we introduce $n$ differences $a_{n+1}-a_1,\ldots,a_{n+1}-a_n$ that are distinct from each other, so $n$ of the multiplicities increase by 1 (possibly including some multiplicities that were 0).
When we increase a multiplicity from $m$ to $m+1$, we increase the sum of the squares of the multiplicities by $2m+1$. Therefore the total increase of the sum of the squares of the multiplicities has the form $\sum(2m_i+1)$, where there are $n$ terms in the sum, and the $m_i$ are some of the multiplicities with respect to $A$ (and some $m_i$ may be 0).
So the sum of the squares of the multiplicities increases by $2(\sum m_i) + n$. This can be bounded above by $2S+n$, where $S$ is the sum of all multiplicities with respect to $A$. But $S = {n \choose 2}$, so we have shown that the sum of squares of multiplicities has increased by at most $2{n \choose 2}+n = (n^2-n)+n = n^2$.
- | {
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# Total number of odd numbers in a row of Pascal's triangle is a natural power of two
This seems like a combinatorial problem so there might be something simple that hasn't struck me. Although I do have an idea but I am unable to proceed from it. The statement of the question is:
Prove that in any row of Pascal's triangle, the number of odd elements is $$2^k$$, for some $$k \in \mathbb{N}$$.
I was working on a Pascal's triangle but in a binary format, where two adjacent 1's add up to 0. Something of the sort:
1
1 1
1 0 1
1 1 1 1
1 0 0 0 1
1 1 0 0 1 1
It is a definitive sequence and the thing I liked about it was that you can add up the adjacent 1's in a row to get the number of odd elements, but I haven't been able to generalize this information. I was thinking along the lines of a recursive relation in polynomials whose coefficients would represent the rows of this triangle, then I can just feed 1 into the equation and inductively prove that it is a perfect power of 2.
Can someone help me out on a proof along these lines, if I'm going in the right direction here?
P.S. I know there is the modulo 2 proof but can someone help me generalize that binary pascal's triangle?
• This will be helpful: math.hmc.edu/funfacts/ffiles/30001.4-5.shtml Jan 11 '17 at 13:07
• Okay, cool. But I want to try a proof along the lines of what I was trying. That is a good solution, but I wanna see where this goes. Jan 11 '17 at 13:15
• To get an inductive proof, you may want to use the results from the answers based on Kummer's Theorem. That is, try to show that the number of $1$s in a row of Pascal's Triangle is $2$ to the power of the number of $1$ bits in the binary representation of $n$.
– robjohn
Jan 11 '17 at 17:58
• You may be interested by the fact that Pascal's triangle modulo 2 generates Sierpinski's sieve (mathworld.wolfram.com/SierpinskiSieve.html) Jan 11 '17 at 22:18
## 3 Answers
Similar idea to Andreas Caranti's answer: | {
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## 3 Answers
Similar idea to Andreas Caranti's answer:
Kummer's Theorem says that the number of factors of $$p$$ in $$\binom{n}{k}$$ is the number of carries when adding $$k$$ and $$n-k$$ in base-$$p$$. There are no carries when adding $$k$$ and $$n-k$$ in binary if and only if $$k\veebar n-k=0$$; when that is true, $$n=k\lor n-k$$. This means that $$\binom{n}{k}$$ is odd when the $$1$$ bits of $$k$$ are a subset of the $$1$$ bits of $$n$$. That is, there are $$2^{\sigma_2(n)}$$ odd entries in row $$n$$ of Pascal's Triangle, where $$\sigma_2(n)$$ is the sum of the bits in the binary representation of $$n$$.
Will delete because this is not what OP intends, but a proof via Kummer's Theorem looks rather neat to me.
Write $n, m$ in base $2$ $$n = 2^{k} n_{k} + \dots + 2 n_{1} + n_{0}, \qquad m = 2^{k} m_{k} + \dots + 2 m_{1} + m_{0},$$ with $n_{i}, m_{i} \in \{ 0, 1 \}$.
Then
$\dbinom{n}{m}$ is odd iff there are no carries in the sum in base $2$ $$m + (n - m) = n,$$ that is, iff $m_{i} \le n_{i}$ for each $i$.
Given $n$, there are thus $$\prod_{i=0}^{k} (n_{i} + 1)$$ possibilities for $m$, and each $n_{i} + 1 \in \{ 1, 2 \}$.
This also tells you what is the exponent $k$ in the question - it is the number of $1$'s in the diadic expansion of $n$, or equivalently, the number of distinct powers of $2$ that sum up to $n$.
The simplest (at least simplest to explain) solution (using Lucas's theorem) has not been posted; let me fill that "much needed gap". Lazy as I am, I will mostly copypaste my answer from https://math.stackexchange.com/a/2184460/ , since the argument is essentially the same.
I use the notation $$\mathbb{N}$$ for the set $$\left\{0,1,2,\ldots\right\}$$.
The question asks the following:
Theorem 1. Let $$n \in \mathbb{N}$$. Then, the number of $$i \in \left\{0,1,\ldots,n\right\}$$ such that $$\dbinom{n}{i}$$ is odd is a power of $$2$$.
I shall prove something slightly more general: | {
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I shall prove something slightly more general:
Theorem 2. Let $$n \in \mathbb{N}$$. Let $$q$$ be a real number. Then, there exists a finite subset $$G$$ of $$\mathbb{N}$$ such that \begin{align} \sum\limits\limits_{\substack{i\in\left\{ 0,1,\ldots,n\right\} ;\\\dbinom{n}{i}\text{ is odd}}} q^i = \prod_{g \in G} \left(q^{2^g} + 1\right) . \end{align}
Proof of Theorem 2. Write $$n$$ in the form $$n=a_{k}2^{k}+a_{k-1}2^{k-1}+\cdots+a_{0}2^{0}$$ for some $$k\in\mathbb{N}$$ and $$a_{0},a_{1},\ldots,a_{k}\in\left\{ 0,1\right\}$$. (This is just the base-$$2$$ representation of $$n$$, possibly with leading zeroes.) | {
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Lucas's theorem tells you that if $$i=b_{k}2^{k}+b_{k-1}2^{k-1}+\cdots+b_{0}2^{0}$$ for some $$b_{0} ,b_{1},\ldots,b_{k}\in\left\{ 0,1\right\}$$, then \begin{align} \dbinom{n}{i} &\equiv \dbinom{a_{k}}{b_{k}}\dbinom{a_{k-1}}{b_{k-1}} \cdots\dbinom{a_{0}}{b_{0}}=\prod\limits_{j=0}^{k}\underbrace{\dbinom{a_{j}}{b_{j}} }_{\substack{= \begin{cases} 1, & \text{if }b_{j}\leq a_{j}\\ 0, & \text{if }b_{j}>a_{j} \end{cases} \\\text{(since }a_{j}\text{ and }b_{j}\text{ lie in }\left\{ 0,1\right\} \text{)}}} \\ &=\prod\limits_{j=0}^{k} \begin{cases} 1, & \text{if }b_{j}\leq a_{j}\\ 0, & \text{if }b_{j}>a_{j} \end{cases} = \begin{cases} 1, & \text{if }b_{j}\leq a_{j}\text{ for all }j\text{;}\\ 0, & \text{otherwise} \end{cases} \mod 2 . \end{align} Hence, the $$i\in\mathbb{N}$$ for which $$\dbinom{n}{i}$$ is odd are precisely the numbers of the form $$b_{k}2^{k}+b_{k-1}2^{k-1} +\cdots+b_{0}2^{0}$$ for $$b_{0},b_{1},\ldots,b_{k}\in\left\{ 0,1\right\}$$ satisfying $$\left( b_{j}\leq a_{j}\text{ for all }j\right)$$. Since all these $$i$$ satisfy $$i \in \left\{ 0,1,\ldots,n\right\}$$ (because otherwise, $$\dbinom{n}{i}$$ would be $$0$$ and therefore could not be odd), we can rewrite this as follows: The $$i \in \left\{ 0,1,\ldots,n\right\}$$ for which $$\dbinom{n}{i}$$ is odd are precisely the numbers of the form $$b_{k}2^{k}+b_{k-1}2^{k-1} +\cdots+b_{0}2^{0}$$ for $$b_{0},b_{1},\ldots,b_{k}\in\left\{ 0,1\right\}$$ satisfying $$\left( b_{j}\leq a_{j}\text{ for all }j\right)$$. Since these numbers are distinct (because the base-$$2$$ representation of any $$i\in\mathbb{N}$$ is unique, as long as we fix the number of digits), we thus can substitute $$b_{k}2^{k}+b_{k-1}2^{k-1}+\cdots+b_{0}2^{0}$$ for $$i$$ in the sum $$\sum\limits\limits_{\substack{i\in\left\{ 0,1,\ldots,n\right\} ;\\ \dbinom{n}{i}\text{ is odd}}}q^{i}$$. Thus, this sum rewrites as follows: \begin{align} \sum\limits_{\substack{i\in\left\{ 0,1,\ldots,n\right\} ;\\ \dbinom{n}{i}\text{ is odd}}}q^{i} | {
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\sum\limits_{\substack{i\in\left\{ 0,1,\ldots,n\right\} ;\\ \dbinom{n}{i}\text{ is odd}}}q^{i} &=\underbrace{\sum\limits_{\substack{b_{0},b_{1} ,\ldots,b_{k}\in\left\{ 0,1\right\} ;\\b_{j}\leq a_{j}\text{ for all }j} }}_{=\sum\limits_{b_{0}=0}^{a_{0}}\sum\limits_{b_{1}=0}^{a_{1}}\cdots\sum\limits_{b_{k}=0}^{a_k} }\underbrace{q^{b_{k}2^{k}+b_{k-1}2^{k-1}+\cdots+b_{0}2^{0}}}_{=\left( q^{2^{k}}\right) ^{b_{k}}\left( q^{2^{k-1}}\right) ^{b_{k-1}}\cdots\left( q^{2^{0}}\right) ^{b_{0}}} \\ &=\sum\limits_{b_{0}=0}^{a_{0}}\sum\limits_{b_{1}=0}^{a_{1}}\cdots\sum\limits_{b_{k}=0}^{a_{k} }\left( q^{2^{k}}\right) ^{b_{k}}\left( q^{2^{k-1}}\right) ^{b_{k-1} }\cdots\left( q^{2^{0}}\right) ^{b_{0}} \\ &=\left( \sum\limits_{b_{k}=0}^{a_{k}}\left( q^{2^{k}}\right) ^{b_{k}}\right) \left( \sum\limits_{b_{k-1}=0}^{a_{k-1}}\left( q^{2^{k-1}}\right) ^{b_{k-1} }\right) \cdots\left( \sum\limits_{b_{0}=0}^{a_{0}}\left( q^{2^{0}}\right) ^{b_{0}}\right) \\ &=\left( \sum\limits_{b=0}^{a_{k}}\left( q^{2^{k}}\right) ^{b}\right) \left( \sum\limits_{b=0}^{a_{k-1}}\left( q^{2^{k-1}}\right) ^{b}\right) \cdots\left( \sum\limits_{b=0}^{a_{0}}\left( q^{2^{0}}\right) ^{b}\right) \\ &=\prod\limits_{g=0}^{k}\underbrace{\left( \sum\limits_{b=0}^{a_{g}}\left( q^{2^{g}}\right) ^{b}\right) }_{\substack{= \begin{cases} q^{2^{g}}+1, & \text{if }a_{g}=1;\\ 1 & \text{if }a_{g}=0 \end{cases} \\\text{(since }a_{g}\in\left\{ 0,1\right\} \text{)}}} \\ &=\prod\limits_{g=0}^{k} \begin{cases} q^{2^{g}}+1, & \text{if }a_{g}=1;\\ 1 & \text{if }a_{g}=0 \end{cases} \\ &=\left( \prod\limits_{\substack{g\in\left\{ 0,1,\ldots,k\right\} ;\\a_{g} =1}}\left( q^{2^{g}}+1\right) \right) \underbrace{\left( \prod\limits _{\substack{g\in\left\{ 0,1,\ldots,k\right\} ;\\a_{g}=0}}1\right) }_{=1} \\ &=\prod\limits_{\substack{g\in\left\{ 0,1,\ldots,k\right\} ;\\a_{g}=1}}\left( q^{2^{g}}+1\right) . \end{align} Thus, there exists a finite subset $$G$$ of $$\mathbb{N}$$ such that \begin{align} \sum\limits\limits_{\substack{i\in\left\{ | {
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subset $$G$$ of $$\mathbb{N}$$ such that \begin{align} \sum\limits\limits_{\substack{i\in\left\{ 0,1,\ldots,n\right\} ;\\\dbinom{n}{i}\text{ is odd}}} q^i = \prod_{g \in G} \left(q^{2^g} + 1\right) \end{align} (namely, $$G$$ is the set of all $$g\in\left\{ 0,1,\ldots,k\right\}$$ satisfying $$a_g = 1$$). This proves Theorem 2. $$\blacksquare$$ | {
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Proof of Theorem 1. Theorem 2 (applied to $$q = 1$$) shows that there exists a finite subset $$G$$ of $$\mathbb{N}$$ such that \begin{align} \sum\limits\limits_{\substack{i\in\left\{ 0,1,\ldots,n\right\} ;\\\dbinom{n}{i}\text{ is odd}}} 1^i = \prod_{g \in G} \left(1^{2^g} + 1\right) . \end{align} Consider this $$G$$. Comparing \begin{align} \sum\limits\limits_{\substack{i\in\left\{ 0,1,\ldots,n\right\} ;\\\dbinom{n}{i}\text{ is odd}}} 1^i = \prod_{g \in G} \underbrace{\left(1^{2^g} + 1\right)}_{= 1+1 = 2} = \prod_{g \in G} 2 = 2^{\left|G\right|} \end{align} with \begin{align} \sum\limits\limits_{\substack{i\in\left\{ 0,1,\ldots,n\right\} ;\\\dbinom{n}{i}\text{ is odd}}} \underbrace{1^i}_{=1} & = \sum\limits\limits_{\substack{i\in\left\{ 0,1,\ldots,n\right\} ;\\\dbinom{n}{i}\text{ is odd}}} 1 \\ & = \left(\text{the number of all i\in\left\{ 0,1,\ldots,n\right\} such that \dbinom{n}{i} is odd}\right) \cdot 1 \\ & = \left(\text{the number of all i\in\left\{ 0,1,\ldots,n\right\} such that \dbinom{n}{i} is odd}\right) , \end{align} we obtain \begin{align} \left(\text{the number of all i\in\left\{ 0,1,\ldots,n\right\} such that \dbinom{n}{i} is odd}\right) = 2^{\left|G\right|} . \end{align} Hence, the number of $$i \in \left\{0,1,\ldots,n\right\}$$ such that $$\dbinom{n}{i}$$ is odd is a power of $$2$$. This proves Theorem 1. $$\blacksquare$$ | {
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# Need to prove the sequence $a_n=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}$ converges
I need to prove that the sequence $a_n=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}$ converges. I do not have to find the limit. I have tried to prove it by proving that the sequence is monotone and bounded, but I am having some trouble:
Monotonic:
The sequence seems to be monotone and increasing. This can be proved by induction: Claim that $a_n\leq a_{n+1}$ $$a_1=1\leq 1+\frac{1}{2^2}=a_2$$
Need to show that $a_{n+1}\leq a_{n+2}$ $$a_{n+1}=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}+\frac{1}{(n+1)^2}\leq 1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}+\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}=a_{n+2}$$ Thus the sequence is monotone and increasing.
Boundedness:
Since the sequence is increasing it is bounded below by $a_1=1$. Upper bound is where I am having trouble. All the examples I have dealt with in class have to do with decreasing functions, but I don't know what my thinking process should be to find an upper bound.
Can anyone enlighten me as to how I should approach this, and can anyone confirm my work thus far? Also, although I prove this using monotonicity and boundedness, could I have approached this by showing the sequence was a Cauchy sequence?
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Hint: compare it to another sequence with only powers of two in the denominator. – Jason Polak Mar 18 '13 at 2:58
@Jason: Could you please elaborate on your hint? I don't understand it. – Jonas Meyer Mar 18 '13 at 3:08
@JonasMeyer: The sequence is less than $1 + 1/2^2 + 1/2^2 + 1/4^2 + 1/4^2 + 1/4^2 + 1/4^2 + ... \leq 2$ – Jason Polak Mar 18 '13 at 3:13
@ShuXiaoLi my objection was to the way the comment was presented, i.e., like a solution to the question. A comment like "by the way, the sum is $\pi^2/6$" would have been fine. – Ittay Weiss Mar 18 '13 at 3:17
Your work looks good so far. Here is a hint: $$\frac{1}{n^2} \le \frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n}$$ | {
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To elaborate, apply the hint to get: $$\frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots + \frac{1}{n^2} \le \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots + \left(\frac{1}{n-1} - \frac{1}{n}\right)$$
Notice that we had to omit the term $1$ because the inequality in the hint is only applicable when $n > 1$. No problem; we will add it later.
Also notice that all terms on the right-hand side cancel out except for the first and last one. Thus: $$\frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots + \frac{1}{n^2} \le 1 - \frac{1}{n}$$
Add $1$ to both sides to get: $$a_n \le 2 - \frac{1}{n} \le 2$$
It follows that $a_n$ is bounded from above and hence convergent.
It is worth noting that canceling behavior we saw here is called telescoping. Check out the wikipedia article for more examples.
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Sorry, I'm afraid I'm going to need a bit more of a push. I can't think of how/where to use your hint. Am I supposed to say that $a_n=1+\frac{1}{2^2}+\frac{1}{3^2}+{…}+\frac{1}{n^2}\leq 1+\frac{1}{2^2}+\frac{1}{3^2}+{…}+\frac{1}{n-1}-\frac{1}{n}$ and do something further with that? – user66807 Mar 18 '13 at 4:02
try to apply it more than once. – chx Mar 18 '13 at 4:17
@chx Try to apply it more than once in the "thing" I supposed? – user66807 Mar 18 '13 at 4:22
@user66807, we have $a_n=\sum_{k=2}^{n}\frac{1}{k^2}\le b_n=\sum_{k=2}^{n}\frac{1}{k-1}-\frac{1}{k}$, try to find the first few terms of $b_n$ you will see a pattern (things will start cancelling) then you can find an upper bound of $a_n$. Notice i started the sum from 2 to be able to use Ayman's hint. – i.a.m Mar 18 '13 at 4:32
@user66807, try to sum a few terms of the telescopic series $\,\sum\left(\frac{1}{n-1}-\frac{1}{n}\right)\,$ and see what happens... – DonAntonio Mar 18 '13 at 4:33 | {
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Besides to Ayman's neat answer, you may take $f(x)=\frac{1}{x^2}$ over $[1,+\infty)$ and see that $f'(x)=-2x^{-3}$ and then is decreasing over $[1,+\infty)$. $f(x)$ is also positive and continuous so you can use the integral test for $$\sum_{k=1}^{+\infty}\frac{1}{n^2}$$ to see the series is convergent. Now your $a_n$ is the $n-$th summation of this series.
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The problem is we haven't gone over series in class yet, but thanks anyways. – user66807 Mar 18 '13 at 3:41
@user66807: Sorry, I didn't know that. So the Ayman's is more preferable nice and simple approach for you. :-) – Babak S. Mar 18 '13 at 3:43
I hope so, let's just see if I can catch his hint. And thanks. – user66807 Mar 18 '13 at 3:46
+1 This is a good answer, though the OP can't use it yet. – DonAntonio Mar 18 '13 at 4:34
I agree with DonAntonio that this is a good answer! $+1 \;\ddot\smile\;\;$ (I also like the new gravatar!) Good day, to you! – amWhy Mar 18 '13 at 15:08
You can show this geometrically too.
If you take a square, and divide the height into $\frac 12$, $\frac 14$, $\frac 18$, and so forth, doubling the denominator on each deal.
Now take the squares $\frac 12$ and $\frac 13$: these go onto the top shelf.
On the second shelf go $\frac 14$ to $\frac 17$. These fractions are all less than $\frac 14$, so fit onto the second shelf. Likewise, squares from $8$ to $15$ go onto the third shelf, each $\frac 1n$ is smaller than $\frac 18$, and so forth.
Therefore $\sum_{n=2}^{inf}\frac 1n < 1$, and therefore the whole lot is less than two squares.
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Hint: Prove the the following holds for all $n$ by induction.
$$\sum_1^n \frac{1}{k^2} \le 2 - \frac{1}{n}.$$
Is it not uncommon that when proving some inequality by induction, you will first need to strengthen the hypothesis to get the induction to work.
You can use the same technique to bound other values of the zeta function. For example, try showing $\zeta(3)$ is bounded above by $\frac{3}{2}$. | {
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(I am copying my answer from a duplicate question that was closed as a copy of this one since the induction approach is not available here.)
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by the integral test :
$\int^\infty_1 \frac{1}{n^2} dn \le \sum_{i=1}^\infty \frac{1}{n^2}\le 1+\int^\infty_1 \frac{1}{n^2}dn$ .
you can compute the integral so the answer is :
$1 \le \sum_{i=1}^\infty \frac{1}{n^2}\le 2$ .
because : $\int^\infty_1 \frac{1}{n^2} dn =1$
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# probability of rolling a sum of $14$ when rolling a $20$-sided die twice or a $20$-sided die with a $4$-sided die depending on outcome of first roll
A $$20$$-sided die is rolled. If the result is $$10$$ or less, the $$20$$-sided die is rolled again. If the result is $$11$$ or more, a $$4$$-sided die is rolled. In either case, the results are summed after the second die roll. What is the probability that the sum will be $$14$$?
My Approach: there are 10 ways we can get a sum of 14 if the first dice is $$<= 10$$ {$$(10,4),(9,5),(8,6),(7,7),(6,8),(5,9),(4,10),(3,11),(2,12),(1,13)$$} and $$3$$ ways we can get sum of $$14$$ if the first dice has value $$>= 11$$ {$$(11,3),(12,2),(13,1)$$} so the probability is: $$(10/400)$$ + $$(3/80) = 1/16$$
• What goes wrong when you just write it out in the obvious way? – lulu Aug 1 at 18:08
• welcome to MSE. kindly include any attempt. – Siong Thye Goh Aug 1 at 18:12
Hints:
$$Pr(A\cap B) = Pr(A)Pr(B\mid A)$$
$$Pr(B) = Pr(A_1\cap B)+Pr(A_2\cap B)+\dots+Pr(A_n\cap B)$$ where $$A_1,\dots,A_n$$ form a partition of the sample space
Let $$B$$ be the event the sum is $$14$$.
Let $$A_1$$ be the event the first die rolls a number $$1-10$$. Let $$A_2$$ be the event the first die rolls a number $$11-13$$. Let $$A_3$$ be the event the first die rolls a number $$14+$$.
Regardless what was rolled on the first die, so long as it was less than $$14$$, there will be exactly one outcome on the second roll that would let the total sum to $$14$$.
• I've added my approach. Not sure if it's correct. – user3119875 Aug 1 at 18:52
• $\frac{1}{2}\times\frac{1}{20}+\frac{3}{20}\times\frac{1}{4}=\frac{1}{16}$ is correct. There is absolutely no need to write out each of the possibilities however and this causes errors when trying to count things by hand and wastes effort and time. – JMoravitz Aug 1 at 18:56
The probability that you obtain $$14$$ using the $$20$$-sided die on the second role is
$$\frac{10}{20} \cdot \frac{1}{20}$$ | {
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$$\frac{10}{20} \cdot \frac{1}{20}$$
The probability you obtain $$14$$ using the $$4$$-sided die on the second role is
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# Is there any specific convention to how mathematical equations are “read out”?
Often, mathematical information and formulas are shared through text, maybe on books or on websites. If you see the equation:
$$x + 3 = 4y$$
we can easily read it out as "x plus three equals four y". But for some other equations such as:
$$\frac{x+3}{x-7}=y$$
it doesn't seem too obvious on what it should be read out as. I would read it as "x plus three the whole divided by x minus seven equals y", however someone else may read it out differently.
My question is therefore: If in a situation where oral communication of mathematical equations is necessary, does there exist a convention to read out mathematical equations such that one particular equation is read out in only one way, and that it is possible to deduce the equation from the spoken name alone? | {
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• In French, I would read it as "x plus trois sur x moins sept égal y", which would translate as "x plus three over x minus seven equals y". I don't know if "over" really is commonly used in English though. I don't think there is any universal convention though. What I said could also be interpreted as $x+\frac{3}{x} -7 = y$. If I notice that the person I am talking to has gotten it wrong, I would just say it again, insisting on the expression I really meant. – Suzet Jul 12 '18 at 1:55
• "over" is indeed used by some people I know. And here is the point, even in English, the same equation is being read out in two different ways. – Pritt Balagopal Jul 12 '18 at 1:57
• The different groupings / implied parantheses are often expressed via pauses and prosody. – Hagen von Eitzen Jul 12 '18 at 2:02
• "The quotient of [the quantity] $x+3$ and [the quantity] $x-7$ is $y$." To avoid a pile-up of symbols at the end, one could say "$y$ is the quotient of $x+3$ and $x-7$." – Blue Jul 12 '18 at 2:26
• @HagenvonEitzen: On the first day of a stint as a Pre-Calculus substitute teacher, I instructed my students to express groupings using "air parentheses", indicated by raising one's slightly-bent arms over one's head. So, here: "[raise arms] x plus three [lower arms] over [raise arms] ex minus seven [lower arms] is y." The students looked at me like I was crazy, but took to the practice. So earnest were they in this —even prompting each other when someone would forget— that, at the end of my six-week stay, I didn't have the heart to tell them that I'd only been kidding that first day. :D – Blue Jul 12 '18 at 2:33
I would like to elaborate on Hagen von Eitzen's insightful comment:
The different groupings / implied parantheses are often expressed via pauses and prosody.
I think this is exactly right and plays an extremely important role in disambiguating oral mathematics. There is a huge difference between:
"X plus 3, over X minus 7, equals Y"
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"X plus 3, over X minus 7, equals Y"
and
"X, plus 3 over X, minus 7 equals Y"
and
"X plus 3, over X, minus 7 equals Y"
If the cadences and pauses in the sentences are read aloud consistently, I think it is unlikely that anyone would transcribe one of those in place of another.
However, in some cases (particularly for extremely complicated expressions with groupings nested inside other groupings) one may call attention to the grouping using verbal annotations like "the quantity" or "all". For example, the quadratic formula is often read aloud as
• "X equals negative b, plus or minus the square root of the quantity b-squared minus 4ac, all over 2a".
I would read $$\frac{x+3}{x-7}=y$$ as:
Quantity $x$ plus $3$ over quantity $x$ minus $7$ equals $y$. | {
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# Image of compact set under piecewise continuous function
Let $$a,b>0\in\mathbb{R}$$. Let $$U$$ be an domain in $$\mathbb{C}^n$$. Let $$f:[a,b]\longrightarrow U$$ be a piecewise continuous map. Then is $$f[a,b]$$ compact? If not compact, will it be bounded?
Ok. This is in the following context. I am given a piecewise smooth path $$\gamma:[a,b]\longrightarrow U$$. Where $$\gamma(a)=z$$ and $$\gamma(b)=w$$, for given $$z,w\in U$$. We are also given a function $$\alpha:U\times\mathbb{C}^n\longrightarrow \mathbb{R}$$, which is upper semicontinuous. Now it is said that $$t\in[a,b]\longrightarrow \alpha(\gamma(t),\gamma’(t))$$ is bounded and measurable. I wanted to know why the function is bounded. I know that $$\gamma[a,b]$$ is compact. And $$\gamma$$ being upper semicontinuous will attain its maximum on a compact set. But I am not sure about $$\gamma’$$.
• What does "piecewise continuous" mean here? How many pieces? Finitely many? Are the pieces compact? Open? Aug 19, 2020 at 10:51
• What have you tried? Aug 19, 2020 at 10:53
• The image of a compact space under continuous map is always compact. Regardless of what "piecewise" means and what $U$ is (as long as it is Hausdorff). Aug 19, 2020 at 10:57
• Please provide additional context, which ideally explains why the question is relevant to you and the community. Some forms of context include background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc. Aug 19, 2020 at 11:18
• @SahibaArora I have added the context. Aug 19, 2020 at 13:25
Compact, not necessarily: On $$[0,1]$$ let $$f(x) = x, 0\le x<1,$$ $$f(1)=2.$$ Then $$f([0,1]) = [0,1)\cup\{2\}.$$
Bounded, yes: First, a lemma: If $$f$$ is continuous on $$(a,b)$$ and $$f$$ has finite limits at the end points, then $$f(a,b)$$ is bounded. | {
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Proof: Suppose $$\lim_{x\to a^+} f(x)=L,$$ $$\lim_{x\to b^-} f(x)=M.$$ Let $$\epsilon=1.$$ Then there exists $$\delta_a>0, \delta_a<(b-a)/3,$$ such that $$|f(x)-L|<1$$ for $$x\in (a,a+\delta_a).$$ Thus for such $$x,$$
$$|f(x)| = |f(x)-L+L|\le |f(x)-L|+|L| <1+|L|.$$
Similarly, there exists $$\delta_b>0,\delta_b<(b-a)/3,$$ such that $$|f(x)|<1+|M|$$ for $$x\in (b-\delta_b,b).$$ It follows that $$f$$ is bounded on the set $$(a,a+\delta_a)\cup (b-\delta_b,b).$$
Since $$f$$ is continuous on the compact set $$[a+\delta_a,b-\delta_b],$$ $$f([a+\delta_a,b-\delta_b])$$ is compact, hence is bounded. It follows that $$f(a,b)$$ is bounded.
Now suppose $$f$$ is piecewise continuous on $$[a,b].$$ Then there exist points $$a=x_0 such that $$f$$ is continuous on each $$I_k=(x_{k-1},x_k)$$ and has finite limits at the end points of $$I_k.$$ By the lemma, each $$f(I_k)$$ is bounded. The set $$f(\{x_0,\dots x_n\})$$ is also bounded. Therefore
$$f([a,b])=f(I_1)\cup \cdots \cup f(I_n)\cup f(\{x_0,\dots x_n\})$$
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# Is there a quick proof as to why the vector space of $\mathbb{R}$ over $\mathbb{Q}$ is infinite-dimensional?
It would seem that one way of proving this would be to show the existence of non-algebraic numbers. Is there a simpler way to show this?
• A finite dimensional vector space over $\mathbb{Q}$ is countable. – user641 Oct 7 '10 at 2:19
• @Steve: Please add that as an answer so people can upvote. – Aryabhata Oct 7 '10 at 2:50
• Does that mean that a vector space over $\mathbb{Q}$ is finite-dimensional iff the set of the vector space is countable? If so, please prove it. – Elchanan Solomon Oct 7 '10 at 2:55
• @Isaac Your question doesn't require the 'only if' anyway. Steve's observation answers your original question. – yasmar Oct 7 '10 at 3:04
• @Isaac: how is the fact that $\mathbb{R}$ is finite dimensional over itself relevant? – Arturo Magidin Oct 7 '10 at 3:31
As Steve D. noted, a finite dimensional vector space over a countable field is necessarily countable: if $v_1,\ldots,v_n$ is a basis, then every vector in $V$ can be written uniquely as $\alpha_1 v_1+\cdots+\alpha_n v_n$ for some scalars $\alpha_1,\ldots,\alpha_n\in F$, so the cardinality of the set of all vectors is exactly $|F|^n$. If $F$ is countable, then this is countable. Since $\mathbb{R}$ is uncountable and $\mathbb{Q}$ is countable, $\mathbb{R}$ cannot be finite dimensional over $\mathbb{Q}$. (Whether it has a basis or not depends on your set theory). | {
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Your further question in the comments, whether a vector space over $\mathbb{Q}$ is finite dimensional if and only if the set of vectors is countable, has a negative answer. If the vector space is finite dimensional, then it is a countable set; but there are infinite-dimensional vector spaces over $\mathbb{Q}$ that are countable as sets. The simplest example is $\mathbb{Q}[x]$, the vector space of all polynomials with coefficients in $\mathbb{Q}$, which is a countable set, and has dimension $\aleph_0$, with basis $\{1,x,x^2,\ldots,x^n,\ldots\}$.
Added: Of course, if $V$ is a vector space over $\mathbb{Q}$, then it has countable dimension (finite or denumerable infinite) if and only if $V$ is countable as a set. So the counting argument in fact shows that not only is $\mathbb{R}$ infinite dimensional over $\mathbb{Q}$, but that (if you are working in an appropriate set theory) it is uncountably-dimensional over $\mathbb{Q}$. | {
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• Related to the note about uncountable dimension, there are explicit examples of continuum-sized linearly independent sets, as seen in this MathOverflow answer by François G. Dorais: mathoverflow.net/questions/23202/… – Jonas Meyer Oct 7 '10 at 4:18
• Yes: but can one show that there is a basis for $\mathbb{R}$ over $\mathbb{Q}$ without some form of the Axiom of Choice? (There is a difference between exhibiting a large linearly independent subset and exhibiting a basis). – Arturo Magidin Oct 7 '10 at 14:51
• No, one cannot, but without the Axiom of Choice the notion of dimension breaks down (except for finite vs. infinite). Assuming AC (as I virtually always do), the size of a linearly independent set gives a lower bound on the dimension of the vector space, and I think it is wonderful that in this case such "explicit" proof exists that the real numbers have continuum dimension, as opposed to the nice qualitative proof one could give by extending your argument to larger cardinals. – Jonas Meyer Oct 7 '10 at 16:02
• @Jonas: No argument there (with any of your points). – Arturo Magidin Oct 7 '10 at 17:23
• good link @Jonas, thanks – Leon Sep 23 '11 at 0:23
The cardinality argument mentioned by Arturo is probably the simplest. Here is an alternative: an explicit example of an infinite $$\, \mathbb Q$$-independent set of reals. Consider the set consisting of the logs of all primes $$\, p_i.\,$$ If $$\, c_1 \log p_1 +\,\cdots\, + c_n\log p_n =\, 0,\ c_i\in\mathbb Q,\,$$ multiplying by a common denominator we can assume that all $$\ c_i \in \mathbb Z\,$$ so, exponentiating, we obtain $$\, p_1^{\large c_1}\cdots p_n^{\large c_n}\! = 1\,\Rightarrow\ c_i = 0\,$$ for all $$\,i,\,$$ by the uniqueness of prime factorizations. | {
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• @Bill +1 What a nice example. – Adrián Barquero Oct 7 '10 at 4:50
• @Bill. Wow! :-) – Agustí Roig Oct 7 '10 at 6:08
• Is this proof unique to Q or does it generalize to provide explicit examples of reals linearly independent over, e.g., Q(sqrt(2))? Above, Q appears to be "hard-wired" into the proof, as the group of exponents in the prime factorization. – T.. Oct 12 '10 at 7:14
• Can you extend the idea of this proof to get the right dimension? Currently, you only have a countably infinite independent set, but the dimension is size continuum. – JDH Jun 23 '11 at 2:22
• @student this can not happen: the primes are distinct. Put the primes with negative exponent on the other side (exponents become positive). You would find two distinct prime factorizations of the same number. – Student Feb 6 at 19:44
No transcendental numbers are needed for this question. Any set of algebraic numbers of unbounded degree spans a vector space of infinite dimension. Explicit examples of linearly independent sets of algebraic numbers are also relatively easy to write down.
The set $\sqrt{2}, \sqrt{\sqrt{2}}, \dots, = \bigcup_{n>0} 2^{2^{-n}}$ is linearly independent over $\mathbb Q$. (Proof: Any expression of the $n$th iterated square root $a_n$ as a linear combination of earlier terms $a_i, i < n$ of the sequence could also be read as a rational polynomial of degree dividing $2^{n-1}$ with $a_n$ as a root and this contradicts the irreducibility of $X^m - 2$, here with $m=2^n$).
The square roots of the prime numbers are linearly independent over $\mathbb Q$. (Proof: this is immediate given the ability to extend the function "number of powers of $p$ dividing $x$" from the rational numbers to algebraic numbers. $\sqrt{p}$ is "divisible by $p^{1/2}$" while any finite linear combination of square roots of other primes is divisible by an integer power of $p$, i.e., is contained in an extension of $\mathbb Q$ unramified at $p$). | {
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Generally any infinite set of algebraic numbers that you can easily write down and is not dependent for trivial reasons usually is independent. This because the only algebraic numbers for which we have a simple notation are fractional powers, and valuation (order of divisibility) arguments work well in this case. Any set of algebraic numbers where, of the ones ramified at any prime $p$, the amount of ramification is different for different elements of the set, will be linearly independent. (Proof: take the most ramified element in a given linear combination, express it in terms of the others, and compare valuations.)
• I realize you answered this question a loooong time ago, but if you're still around, how do you know that you can write the $n-$th root of $2$ as a linear combination of the $1st$ through $n-1$st roots of $2$? – ALannister Aug 17 '17 at 22:41
• @ALannister I realize you questioned quite a long time ago, but s/he is assuming that there is such $n$ and showing that such assumption leads to contradiction. So in fact, s/he is proving exactly that $n$-th root of $2$ cannot be expressed by a linear combination of previous roots – user160738 May 9 '18 at 17:24
For the sake of completeness, I'm adding a worked-out solution due to F.G. Dorais from his post.
We'll need two propositions from Grillet's Abstract Algebra, page 335 and 640:
Proposition: $[\mathbb{R}:\mathbb{Q}]=\mathrm{dim}_\mathbb{Q}{}\mathbb{R}=|\mathbb{R}|$
Proof: Let $(q_n)_{n\in\mathbb{N_0}}$ be an enumeration of $\mathbb{Q}$. For $r\in\mathbb{R}$, take $$A_r:=\sum_{q_n<r}\frac{1}{n!}\;\;\;\;\text{ and }\;\;\;\;A:=\{A_r;\,r\in\mathbb{R}\};$$ the series is convergent because $\sum_{q_n<r}\frac{1}{n!}\leq\sum_{n=0}^\infty\frac{1}{n!}=\exp(1)<\infty$ (recall that $\exp(x)=\sum_{n=0}^\infty\frac{x^n}{n!}$ for any $x\in\mathbb{R}$). | {
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To prove $|A|=|\mathbb{R}|$, assume $A_r=A_{s}$ and $r\neq s$. Without loss of generality $r<s$, hence $A_s=\sum_{q_n<s}\frac{1}{n!}=\sum_{q_n<r}\frac{1}{n!}+\sum_{r\leq q_n<s}\frac{1}{n!}=A_r+\sum_{r\leq q_n<s}\frac{1}{n!}$, so $\sum_{r\leq q_n<s}\frac{1}{n!}=0$, which is a contradiction, because each interval $(r,s)$ contains a rational number.
To prove $A$ is $\mathbb{Q}$-independent, assume $\alpha_1A_{r_1}+\cdots+\alpha_kA_{r_k}=0\;(1)$ with $\alpha_i\in\mathbb{Q}$. We can assume $r_1>\cdots>r_k$ (otherwise rearrange the summands) and $\alpha_i\in\mathbb{Z}$ (otherwise multiply by the common denominator). Choose $n$ large enough that $r_1>q_n>r_2\;(2)$; we'll increase $n$ two more times. The equality $n!\cdot(1)$ reads $n!(\alpha_1\sum_{q_m<r_1}\frac{1}{m!}+\cdots+\alpha_k\sum_{q_m<r_k}\frac{1}{m!})=0$. Rearranged (via $(2)$ when $m=n$), it reads
$$-\alpha_1\sum_{\substack{m<n\\q_m<r_1}}\frac{n!}{m!}-\cdots-\alpha_k\sum_{\substack{m<n\\q_m<r_k}}\frac{n!}{m!}-\alpha_1 =\alpha_1\sum_{\substack{m>n\\q_m<r_1}}\frac{n!}{m!}+\cdots+\alpha_k\sum_{\substack{m>n\\q_m<r_k}}\frac{n!}{m!}. \tag*{(3)}$$
The left hand side (LHS) of $(3)$ is an integer for any $n$. If $n$ is large enough that $(|\alpha_1|+\cdots+|\alpha_k|)\sum_{m=n+1}^\infty\frac{n!}{m!}<1$ holds (such $n$ can be found since $\sum_{m=n+1}^\infty\frac{n!}{m!}=\frac{1}{n+1}\sum_{m=n+1}^\infty\frac{1}{(n+2)\cdot\ldots\cdot m}\leq\frac{1}{n+1}\sum_{m=n+1}^\infty\frac{1}{(m-n-1)!}\leq\frac{1}{n+1}\exp(1)\rightarrow 0$ when $n\rightarrow\infty$), then the absolute value of RHS of $(3)$ is $<1$, and yet an integer, hence $\text{RHS}(3)=0$. Thus $(3)$ reads $\alpha_1=-\sum_{i=1}^{k}\sum_{m<n,q_m<r_i}\alpha_i\frac{n!}{m!}=0\;(\mathrm{mod}\,n)$. If moreover $n>|\alpha_1|$, this means that $\alpha_1=0$. Repeat this argument to conclude that also $\alpha_2=\cdots=\alpha_k=0$. | {
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Since $A$ is a $\mathbb{Q}$-independent subset, by proposition 5.3 there exists a basis $B$ of $\mathbb{R}$ that contains $A$. Then $A\subseteq B\subseteq\mathbb{R}$ and $|A|=|\mathbb{R}|$ and Cantor-Bernstein theorem imply $|B|=|\mathbb{R}|$, therefore $[\mathbb{R}:\mathbb{Q}]=\mathrm{dim}_\mathbb{Q}{}\mathbb{R}=|\mathbb{R}|$. $\quad\blacksquare$
• Is this uncountable linearly independent set basis ? – user195218 Jun 5 '15 at 2:53
• @user195218 It's not a basis, as it doesn't span the reals, but it is uncountable and linearly independent. – Akiva Weinberger Jul 25 '17 at 19:08
• The only proper answer to the question. Such a shame that it is not the most voted answer. – Akerbeltz May 1 at 21:42
Here is a simple proof that a basis $B$ of $[ \mathbb{R}:\mathbb{Q}]$ has cardinality $|\mathbb{R}|$.
Clearly since $B$ is contained in $R$, $|B| \le| \mathbb{R}|$.
But also $\mathbb{R}=span(B)$ and thus $|\mathbb{R}| =|span(B)| \le |\mathbb{Q}^B|=|B|$. The last equality follows because $B$ is not finite (if it was, then $|\mathbb{R}|=|span(B)| \le |\mathbb{Q}^B|=|\mathbb{N}|$, a contradiction).
Hence $|\mathbb{R}| \le |B| \le| \mathbb{R}|$, so $|\mathbb{R}|=|B|$
• There is an error in what you wrote: $|\mathbb{Q}^B|\geq |2|^{|B|} > |B|$. – Nex Aug 18 '17 at 8:19
• @Next: hmm so is it salvageable what I wrote? I guess I wasn't paying too much attention. – Joshua Benabou Aug 18 '17 at 19:08
Another simple proof:
Take $P=X^n-p$ for a prime $p$.
By Eisenstein's criterion, it is $\mathbb Q[X]$-irreductible. Therefore, the set of algebraic numbers is of infinite dimension over $\mathbb Q$.
Since $\mathbb R$ is bigger, it works for $\mathbb R$ too.
## protected by user26857Nov 25 '15 at 9:30
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\nLicense: Creative Commons<\/a>\n<\/p><\/div>"}, How to Find Least‐Squares Solutions Using Linear Algebra, consider supporting our work with a contribution to wikiHow. MathJax reference. wikiHow is where trusted research and expert knowledge come together. The minimum norm least squares solution is always unique. We can translate the above theorem into a recipe: Recipe 1: Compute a least-squares solution. 6Constrained least squares Constrained least squares refers to the problem of nding a least squares solution that exactly satis es additional constraints. In particular, it leads to the "least squares" method of fitting curves to collections of data. We use cookies to make wikiHow great. Why does regression use least “squares” instead of least “absolute values”? Least Squares. Recall the formula for method of least squares. Does a solution with a minimal norm mean it is a solution that minimizes the residuals? In this case, we're often interested in the minimum norm least squares solution. Parameter A can also be a set of equations that describe the linear least-squares problem. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. I. Problems (PDF) Solutions (PDF) Further Study Eigenvalue Demonstrations* These demonstrations employ Java® applets with voice-over narration by Professor Strang. % In particular, it leads to the "least squares" method of fitting curves to collections of data. It is a set of formulations for solving statistical problems involved in linear regression, including variants for ordinary (unweighted), weighted, and generalized (correlated) residuals. This article has been viewed 4,467 times. How to draw random colorfull domains in a plane? This can be written in terms of an inequality ||y−Xβ^||2≤||y−Xβ||2,{\displaystyle ||\mathbf {y} -X{\hat {\boldsymbol {\beta }}}||^{2}\leq ||\mathbf {y} -X{\boldsymbol {\beta }}||^{2},} where we are minimizing the distance between | {
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||\mathbf {y} -X{\boldsymbol {\beta }}||^{2},} where we are minimizing the distance between y{\displaystyle \mathbf {y} } and Xβ. A = np.array([[1, 2, 1], [1,1,2], [2,1,1], [1,1,1]]) b = np.array([4,3,5,4]) Next, note that minimizing $\| b-Ax \|_{2}^{2}$ is equivalent to minimizing $\| b-Ax \|_{2}$, because squaring the norm is a monotone transform. Remember when setting up the A matrix, that we have to fill one column full of ones. Many calculations become simpler when working with a … On the other hand, if the system is underdetermined, there are infinitely many solutions and thus one can find a solution of minimal norm and this is called the minimum norm solution. Magic. The basic problem is to find the best fit straight line y = ax + b given that, for n 2 f1;:::; Ng, the pairs (xn;yn) are observed. Numerical methods for linear least squares include inverting the matrix of the normal equations and orthogonal decomposition methods. Foundation for someone who tries to learn LA for data science ( and )! Either full/square or non-full/square Q is allowed, i.e, 4, the. Unknowns, also known as overdetermined systems the solution vectors: //www.khanacademy.org/... /v/linear-algebra-least-squares-examples how to find correlations for data... No matter what rank $a$ does n't matter which one you minimize for large matrices appropriate! The x that solves the linear least-squares problem for the matrix of the equation AX=B by solving normal. ” from linear algebra, Fifth Edition, 2016 the bathroom = a T.. The rank of a. s ( min ( M, b is.... Squares ” instead of least squares we ’ ll show later that this indeed gives the minimum solution. Seek the x that minimizes the residuals the example we showed in part 2 dealt with a. The three points the rationale of the matrix is column rank, right -. Be the x such that Ax = b is inconsistent a vector into directional components an! Is obtained as follows a are supported Computing and Technology News on Phys.org,! Could it be a maximum, | {
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they evolve to mathematics Stack Exchange is a least squares estimation, looking it... Equations and orthogonal decomposition methods fact, chose the vector $y$ which forces this term to 0... Downtime early morning Dec 2, compute the length of the normal equation is a least square.. Goal to find a minimum norm least squares and computational aspects of linear equations, there... The least squares include inverting the matrix of the solution for working matrices. View of least-squares regression — the so-called “ linear algebra and started to study this book is to! Have A Good Day In Italian, Nikon D5600 Refurbished Body, Canadian Maple Tree, Golden Tree Feeding Chart, Takahe Vs Pukeko, Digital Ic Design Mcq Questions, Buttermilk Yeast Donuts, " /> | {
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# What does "max" refer to in the definition of a matrix norm?
I am going through my numerical analysis text (Epperson) and came across notation that I don't fully understand. My question is what does max refer to in this definition?
Definition $7.2$ (Matrix Norm): Let $\|\cdot\|$ be a given vector norm defined on $\mathbb{R}^n$. Define the corresponding matrix norm, for matrices $A \in \mathbb{R}^{n\times n}$, by $$\|A\| = \max_{x \neq 0}\frac{\|Ax\|}{\|x\|}$$
• Can you give the exact usage in your question? Feb 12 '17 at 16:35
• Hi David, that's what I'm trying to figure out too... the image in the link is of the definition of a matrix norm. I'm not sure how to read the "max x != 0" term. Feb 12 '17 at 16:37
• It's the maximum value of the expression as the vector $x$ goes through all possible values in $\mathbb{R}^n$ except $x=0$ since the denominator is not defined for this value. Feb 12 '17 at 16:40
• Thanks David for the edit. I'm going to try to wrap my head around this, I'm sure it's simpler than it seems. Feb 12 '17 at 16:56
• @Starcrossed not a problem. The concept will be easy once you wrap your mind around it. For some norms though, it might be very hard to compute. Feb 12 '17 at 17:09
This is simply the extension of a given norm to a matrix. It should be better written: $$\sup \left\{\frac{\|Ax\|}{\|x\|}: x \in \mathbb{R}^n, x\neq 0\right\} = \sup \left\{\|Ax\|:x \in \mathbb{R}^n, \|x\|=1\right\}.$$ That is, go through all normalized vectors, and see where the product $\|Ax\|$ is maximized.
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As an example, let us consider the $L^1$-norm. For some matrix $A \in \mathbb{R}^n$, by definition we have that $$\|A\|_1 = \sup \left\{\|Ax\|_1:x \in \mathbb{R}^n, \|x\|_1=1\right\}.$$ Now, for a given vector $x \in R^n$, we have that $$\displaystyle\|Ax\|_1 = \left\|\begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{pmatrix}\cdot \begin{pmatrix} x_1\\ x_2\\ \vdots\\ x_n \end{pmatrix}\right\|_1 = \left\|\begin{pmatrix} \sum_{j=1}^na_{1j}x_j\\ \sum_{j=1}^na_{2j}x_j\\ \vdots\\ \sum_{j=1}^na_{nj}x_j \end{pmatrix}\right\|_1 = \sum_{i=1}^n\left|\sum_{j=1}^na_{ij}x_j\right|.$$ Thus, we can write the $L^1$-norm of $A$ as $$\displaystyle \|A\|_1 = \sup_{\|x\|=1}\left\{\sum_{i=1}^n\left|\sum_{j=1}^na_{ij}x_j\right|\right\}.$$ We can bound this in the following way, $$\|A\|_1 \leq \sup_{\|x\|=1}\left\{\sum_{i=1}^n\sum_{j=1}^n\left|a_{ij}x_j\right|\right\} = \sup_{\|x\|=1}\left\{\sum_{j=1}^n\left|x_j\right|\sum_{i=1}^n\left|a_{ij}\right|\right\}.$$ Now, consider the fact that $$\displaystyle \sum_{j=1}^n\left|x_j\right|\sum_{i=1}^n\left|a_{ij}\right| \leq \sum_{j=1}^n\left|x_j\right|\max_{1\leq j \leq n}\left\{\sum_{i=1}^n\left|a_{ij}\right|\right\} = \|x\|_1\cdot \max_{1\leq j \leq n}\left\{\sum_{i=1}^n\left|a_{ij}\right|\right\}.$$ Now, since our matrix norm is only concerned with normalized vectors, we have clearly that $$\displaystyle \|A\|_1 \leq \sup_{\|x\|=1}\left\{\|x\|_1\cdot \max_{1\leq j \leq n}\left\{\sum_{i=1}^n\left|a_{ij}\right|\right\}\right\} = \max_{1\leq j \leq n}\left\{\sum_{i=1}^n\left|a_{ij}\right|\right\}.$$ Therefore, we have an upper bound. But this upper bound is clearly attainable, we simply take the $j^*$ corresponding to the column we are maximizing over in the previous equation, and let $x_{j^*} = 1, x_i = 0$ ($i \neq j^*$). Thus, the $L^1$-norm for a matrix is simply the maximum of the absolute column sums. So, as you see, the | {
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the $L^1$-norm for a matrix is simply the maximum of the absolute column sums. So, as you see, the definition may be readily understood, but determining the value of the norm may be less so. And, in many cases, a closed form of the norm doesn't exist, and so you need to numerically solve the optimization (maximization) problem for each given $A$. | {
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The "max" refers to the maximum of the set of numbers $\{\frac{\Vert Ax \Vert}{\Vert x \Vert } , 0\neq x\in \mathbb{R}^n\}$. Note that since this is a set of norms on vectors, then it's a set of non negative numbers.
• Aha okay I think I get it now... basically we are finding a vector x such that we are maximizing the norm, right? Feb 12 '17 at 17:00
• Yes. Just note that we're looking for an $x$ that maximizes the quotient of the norms $\frac{ \Vert Ax \Vert }{ \Vert x \Vert }$
– NSA
Feb 13 '17 at 9:23
It means the maximum of the quantity
$$\frac{||Ax||}{||x||}$$
along all the vectors $x\in\mathbb{R}^n$ that are not zero.
• Thanks Smurf... So to further understand, suppose we are given only one vector x for a linear expression Ax = b, then that vector alone would be the only possible vector to use in finding the max, right? Feb 12 '17 at 16:45
• No, the norm of the matrix is defined as above. If you have an expression $Ax=b$ you still should use the maximum along every $x\neq 0$. Feb 12 '17 at 16:47
• I'm sure it is probably trivially simple to you so I apologize in advance for this dumb question ... wouldn't every max x ≠ 0 in ℝn be infinitely large? Feb 12 '17 at 16:55
• Notice that while it may seem that if you pick and arbitrary large $x$, the numerator increases, so does the denominator. So one "controls" the other and the maximum exists (this has to be proven, I am just giving you the idea of why it works). Feb 12 '17 at 17:16
• @Starcrossed, when the space is infinite dimensional, the quotient can be unbounded. Mar 8 '17 at 10:14 | {
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Expected number of heads on coin flip?
Suppose you were to toss 2 coins each with different probabilities of landing on heads 2 times each, one at a time. So coin 1 tossed twice. Then coin 2 tossed twice. Let N be the total number of heads. To calculate E[N], would you find E[number of heads on coin 1's tosses] and E[number of heads on coin 2's tosses] and sum them together? And to find the var[N] would it be the same concept?
For expectation - yes due to linearity of expectation. As for variance, you can do the same since the each toss (or each event) are independent. Note that it does not matter whether you toss each coin 'one at a time' or the order you do it in due to independence, of course.
Formally, if you let $X_1$ and $X_2$ be the indicator variables for each toss of the first coin ($1$ for heads, $0$ for tails) and the same with $Y_1$ and $Y_2$ for the tosses of the second coin, we have:
$$N=X_1+X_2+Y_1+Y_2$$
Then
$$E(N)=E(X_1+X_2+Y_1+Y_2)$$
$$=E(X_1)+E(X_2)+E(Y_1)+E(Y_2)$$
due to linearity of expectation.
Also, as $X_1,X_2,Y_1,Y_2$ are independent events, we have
$$Var(N)=Var(X_1+X_2+Y_1+Y_2)$$
$$=Var(X_1)+Var(X_2)+Var(Y_1)+Var(Y_2)$$
The general rules being used here are:
$$E(aX+bY)=aE(X)+bE(Y)\tag{linearity of expectation}$$
and if $X$ and $Y$ are independent,
$$Var(aX+bY)=a^2Var(X)+b^2Var(Y)$$ | {
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and if $X$ and $Y$ are independent,
$$Var(aX+bY)=a^2Var(X)+b^2Var(Y)$$
• I just have a general question. I had a practice problem like this but with given probabilities for each coin. Let's assume N=X+Y. X is number of heads on coin 1, and Y is the number of heads on coin 2. I tried calculating the variance by doing E[(X+Y)^2] - E^2[(X+Y)] but received a negative number. But when I calculated the variances of each coin individually and then summing it, I received a positive number. Why does this happen? Basically I'm asking, why doesn't E[(X+Y)^2] - E^2[(X+Y)] =var(X)+var(Y)? Aug 10, 2017 at 21:18
• Could you give the probabilities in particular or edit your question to include your calculation? Aug 10, 2017 at 21:23
• Never mind, I see what I did wrong. When squaring E[(X+Y)^2], I did not calculate it correctly. Aug 10, 2017 at 21:35
• Letting the probabilities be $p_1$ and $p_2$, we have $E[(X+Y)^2] - (E[X+Y])^2=(0\times(1-p_1)^2(1-p_2)^2+1\times(1-p_1)^2p_2(1-p_2)+4\times(1-p_1)^2p_2^2+1\times p_1(1-p_1)(1-p^2)^2+4\times p_1(1-p_1)p_2(1-p_2)+9\times p_1(1-p_1)p_2^2+4\times p_1^2(1-p_2)^2+9\times p_1^2p_2(1-p_2)+16\times p_1^2p_2^2)-4(p_1+p_2)^2$ Aug 10, 2017 at 21:35 | {
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Long answer: It sounds like the two coins in your example do not affect each other in any way. In other words, it sounds like the number of heads on one is independent of how many heads happen for the other. For any two random variables, whether independent or not, we do indeed have that $E[X+Y]=E[X]+E[Y]$. Furthermore, when the variables are independent, we also have that $E[XY]=E[X]E[Y]$. This allows us to reason as follows. \begin{align} Var[X+Y] &= E[((X+Y)-E(X+Y))^2]\\ &=E[(X-E[X]+Y-E[Y])^2]\\ &=E[(X-E[X])^2] +2E[(X-E[X])(Y-E[Y])] + E[(Y-E[Y])^2]\\ &=Var[X] + Var[Y] + 2 E[XY -XE[Y] -YE[X] + E[X]E[Y]]\\ &= Var[X] + Var[Y] + 2(E[XY] -E[X]E[Y] - E[X]E[Y] + E[X]E[Y])\\ &= Var[X] + Var[Y] + 2(E[X]E[Y] -E[X]E[Y] - E[X]E[Y] + E[X]E[Y])\\ &= Var[X] + Var[Y] +2 \cdot 0\\ &= Var[X] + Var[Y] \end{align} So the variance of the sum of independent random variables is the sum of their variance. | {
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# Implicit differentiation misunderstanding
I'm trying to see why my textbook's solution is correct and mine isn't.
"Find an expression in terms of $x$ and $y$ for $\displaystyle \frac{dy}{dx}$, given that $x^2+6x-8y+5y^2=13$
First, the textbook's solution, which I understand and agree with fully:
Now my similar solution, for which I don't see my error:
Differential: $$2x+6-8\frac{dy}{dx}+10y\frac{dy}{dx}=0$$ $$\frac{dy}{dx}(10y-8)=-2x-6$$ $$\frac{dy}{dx}=\frac{-2x-6}{10y-8}=\frac{-x-3}{5y-4}$$
So I end up with the negative of the correct solution, because I moved the $(2x+6)$ to the RHS and the textbook author moved the other part instead. I would have thought it would produce an equivalent answer?
Thanks!
-
HINT $\rm\displaystyle\ \ \frac{-A}{-B}\ =\ \frac{A}B\qquad$ – Bill Dubuque Jun 4 '11 at 18:14
As you now know, your answer is correct. But there is kind of a convention in writing mathematical expressions, that fewer minus signs is generally better. – André Nicolas Jun 4 '11 at 18:22
@user6312 duly noted, thanks! :) – Danny King Jun 4 '11 at 18:24
Fractions can be written multiple ways.
$$\frac{-x-3}{(5y-4)} = \frac{x+3}{4-5y}$$
In general $$\frac{a-b}{c-d}=\frac{b-a}{d-c}$$
This is just multiplying both the top and the bottom by −1. In other words, your answer and the books differ by multiplication of $$\frac{-1}{-1} = 1$$
-
Oh! When you put it like that, it becomes very clear. Thank you! – Danny King Jun 4 '11 at 18:05
Your answers agree. Note that: $$\frac{-x-3}{5y-4} = \frac{-(x+3)}{-(-5y+4)} = \frac{x+3}{4-5y}.$$
-
I don't know, what's wrong with your solution. Notice, that
$${-x-3\over5y-4}={3+x\over4-5y}$$
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# Number of simple edge-disjoint paths needed to cover a planar graph
Let $G=(V,E)$ be a graph with $|E|=m$ of a graph class $\mathcal{G}$. A path-cover $\mathcal{P}=\{P_1,\ldots,P_k\}$ is a partition of $E$ into edge-disjoint simple paths. The size of the cover is $\sigma(\mathcal{P})=k$. I am interested in upper and lower bounds for the quantity
$$\max_{G\in \mathcal{G}} \quad \min_{\mathcal{P} \text{ is path-cover of G}} \sigma(\mathcal{P})/m.$$
In other words, how large is the inverse of the average path length in the smallest path-cover in the worst case.
The graphs classes $\mathcal{G}$ I am interested in are (1) planar 3-connected graphs, and (2) triangulations.
There is a simple observation for a lower bound: In every odd-degree vertex one path has to start/end. So when all vertices have odd degree there any path-cover needs at least $m/6+1$ paths (a triangulation has $3|V|-6$ edges). You get the same result by noticing, that $(n-1)/2$ paths have to pass through a vertex of degree $n-1$.
Is a path-cover with $\frac{m}{6}$ paths always possible for planar graphs?
I am aware of a few results covering planar graphs with paths of length $3$.
Here is an example of a path-cover of the graph of the icosahedron. | {
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Here is an example of a path-cover of the graph of the icosahedron.
-
This is a great question; don't know why it hasn't received more attention or upvotes. – joriki Dec 12 '12 at 10:10
I don't understand the lower bound. Since a triangulation has $m=3n-6$ edges, a triangulation with all odd-degree vertices requires at least $n/2=(m+6)/6=m/6+1$ paths. For instance, $K_4$ needs $2$ (not $1$) paths to cover it. – mjqxxxx Dec 13 '12 at 18:28
I am not sure if I would be able to do it, but somebody should try generating a big list of $\sigma(\mathcal{P})$'s for a bunch of the graphs in question. Maybe some data will help us see a pattern. – Alexander Gruber Dec 19 '12 at 2:40
@A.Schulz Maybe the lower bound $m/6$ is weak for some classes of graphs. For instance, for the wheel graphs $W_n$, consisting of the cycle of length $n-1$ and the additional center vertex joined with all vertices of the cycle. Also there are Moon-Moser graphs $T_k$ (see Section 2 of the article “Long Cycles in 3-Connected Graphs” by Guantao Chen and Xingxing Yu), having no cycles of length greater than $(7/2) n^{\log_3 2}$. Maybe these graphs also have no small path-covers. – Alex Ravsky May 27 '13 at 3:47
A related question is that of Linear Arboricity, but the results for it are based upon the maximum degree of vertices. See for instance mimuw.edu.pl/~kowalik/papers/LinearArboricity.pdf – jp26 Jan 27 at 21:26
I will show that $\frac{m}{6}$ is not enough for 3-connected planar graphs. The following is inspired by fedja's comment (and also some nice discussion with Krystal Guo).
Consider the prism graph, which we will denote $P_n$. We say it has $2n$ vertices and $3n$ edges. Then in this case $\frac{m}{6} = \frac{n}{2}$. We will show that $\sigma(P_n) =n$. It is clear that $n$ paths is enough.
Consider the following graphs which we denote $T_n$:
and so on.. Note that $T_n$ has $3n - 1$ edges and $2n+2$ vertices. | {
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and so on.. Note that $T_n$ has $3n - 1$ edges and $2n+2$ vertices.
It is not too hard to show that $\sigma(T_n) = n+1$ (and this is the best possible). One can show by induction, considering the left most edge between the upper and lower paths.
Let $p_1 , \ldots , p_k$ be a path cover of $P_n$. Now $P_n$ is two copies of $C_n$ with a matching, $M$, in between them. If every $e \in M$ is actually some $p_i$, then we have $k \geq n+2$.
So we may assume there is a $e = (u,v)$ such that the path, $p_i$, that contains $e$, contains at least one more edge, wolog say $(v,w)$.
$(i)$ The vertex $u$ is also contained in another edge of $p_i$ (other than $e$). Then $(V(P_n) , E(P_n) \setminus E(p_i))$ is $T_n$ minus at most two paths (actually, exactly two: the two paths start with precisely the two edges incident to $e$, respectively). In this case $k \geq n+1 - 2 + 1 = n$ (the last $+1$: don't forget to count $p_i$).
$(ii)$ The vertex $u$ is not contained in any other edge of $p_i$. Then let $p_j$ be another path that contains $u$. Then $(V(P_n) , E(P_n) \setminus (E(p_i)\cup E(p_j))$ is a $T_n$ minus at most 3 paths ($p_i$ is one path in $T_n$ and $p_j$ is at most two paths in $T_n$). It follows that $k \geq n + 1 - 3 +2= n$.
Thus we obtain $\sigma(P_n) = n$, as desired.
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+0
im stuck in this koobits !
0
412
5
Find the sum of the following series by pairing two numbers to make ten(s). 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
Guest Nov 20, 2015
#2
+18927
+15
Find the sum of the following series by pairing two numbers to make ten(s).
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
$$\begin{array}{rcrcrcrcrcrcrcrcrcrcr} \hline \hline 1 &+& 3 &+& 5 &+& 7 &+& 9 &+& 11 &+& 13 &+& 15 &+& 17 &+& 19 & \\ \hline \hline 1 & & & & & & & & & & & & & & & & &+& 19 & =& 20 \\ \hline & & 3 & & & & & & & & & & & & &+& 17 & & & =& 20 \\ \hline & & & & 5 & & & & & & & & &+& 15 & & & & & =& 20 \\ \hline & & & & & & 7 & & & & &+& 13 & & & & & & & =& 20 \\ \hline & & & & & & & & 9 &+& 11 & & & & & & & & & =& 20 \\ \hline \hline & & & & & & & & & & & & & & & & & & & =& 100\\ \hline \hline \end{array}$$
( 1 + 19 ) + ( 3 + 17 ) + ( 5 + 15 ) + ( 7 + 13 ) + ( 9 + 11 ) = 20 + 20 + 20 + 20 + 20 = 100
heureka Nov 20, 2015
edited by heureka Nov 20, 2015
edited by heureka Nov 20, 2015
Sort:
#1
+5
(1+9) + (3+7) + (5+15) + (7+13) + (11+19)
10 + 10 + 20 + 20 + 30 = 90
Guest Nov 20, 2015
#2
+18927
+15
Find the sum of the following series by pairing two numbers to make ten(s).
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
$$\begin{array}{rcrcrcrcrcrcrcrcrcrcr} \hline \hline 1 &+& 3 &+& 5 &+& 7 &+& 9 &+& 11 &+& 13 &+& 15 &+& 17 &+& 19 & \\ \hline \hline 1 & & & & & & & & & & & & & & & & &+& 19 & =& 20 \\ \hline & & 3 & & & & & & & & & & & & &+& 17 & & & =& 20 \\ \hline & & & & 5 & & & & & & & & &+& 15 & & & & & =& 20 \\ \hline & & & & & & 7 & & & & &+& 13 & & & & & & & =& 20 \\ \hline & & & & & & & & 9 &+& 11 & & & & & & & & & =& 20 \\ \hline \hline & & & & & & & & & & & & & & & & & & & =& 100\\ \hline \hline \end{array}$$
( 1 + 19 ) + ( 3 + 17 ) + ( 5 + 15 ) + ( 7 + 13 ) + ( 9 + 11 ) = 20 + 20 + 20 + 20 + 20 = 100
heureka Nov 20, 2015
edited by heureka Nov 20, 2015
edited by heureka Nov 20, 2015
#3
+91773
0
What a fabulous diagram Heureka, | {
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What a fabulous diagram Heureka,
Melody Nov 20, 2015
#4
0
Two great answers, but one of them must be wrong.
The first says 90, the second 100.
Guest Nov 20, 2015
#5
0
Yep One of them is wrong....and it's my answer the answer is 100...I made a typo. The 4th term in my sequence should read (17+13) NOT (7 +13)
Guest Nov 21, 2015
5 Online Users
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### How fast is a comet moving when it crosses Earth's orbit?
• Is it about the same as Earth's orbital speed?
It depends. An object's speed is related to the length of it's orbit, and this varies greatly. See Kepler's Laws.
• Walter Correct answer
8 years ago
1. Comets don't cross Earth's orbit really. Orbits are one-dimensional objects and their chance of crossing in 3D space is 0. Henceforth, I consider a comet at distance 1AU from the Sun.
2. What's the maximum speed of a returning comet at 1AU from the Sun? This can be easily worked out from the orbial energy
$$E = \frac{1}{2}v^2 - \frac{GM_\odot}{r},\qquad\qquad(*)$$
which is conserved along the orbit ($v$ and $r$ the Heliocentric speed and distance). For a returning comet, $E<0$ and the speed cannot exceed the escape speed (which occurs for $E=0$)
$$v_{\rm escape}^2 = 2\frac{GM_\odot}{r}.$$
The speed of the Earth can be worked out from the Virial theorem, according to which the orbital averages of the kinetic and potential energies, $T=\frac{1}{2}v^2$ and $W=-GM_\odot/r$, satisfy
$2\langle T\rangle + \langle W\rangle=0.$
For a (near-)circular orbit (such as Earth's), $r$ is constant and we have
$v^2_{\rm Earth} = GM_\odot/r.$
Thus, at $r$=1AU
$$v_{\rm escape} = \sqrt{2} v_{\rm Earth}$$
as already pointed out by Peter Horvath. Non-returning comets have local speed exceeding the escape speed.
3. Can a comet near Earth have a speed similar to Earth's orbital speed?. Let's assume a comet with the same speed as Earth at $r$=1AU and work out the consequences. Such a comet must have the same orbital energy as the Earth and, since
$$E = -\frac{GM_\odot}{2a}\qquad\qquad(**)$$
with $a$ the orbital semimajor axis, must also have $a=1AU$ and the same orbital period as Earth, i.e. one year. Moreover, the comet's apohelion satisfies
$$r_{\rm apo}\le 2a = 2{\rm AU}.$$
Such comets don't exist AFAIK. Most returning comets have much longer periods than 1 year. | {
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4. What's the typical speed of a returning comet when at distance 1AU from the Sun? To work out this question, let's parameterise the comet's orbit by its period $P=2\pi\sqrt{a^3/GM_\odot}$. From this relation we immediately get
$$\frac{a_{\rm comet}}{\rm AU} = \left(\frac{P_{\rm comet}}{\rm yr}\right)^{2/3}.$$
From equations ($*$) and ($**$), we can then find
$$v_{\rm comet}(r=1{\rm AU}) = \sqrt{2-\left(\frac{P_{\rm comet}}{\rm yr}\right)^{-2/3}}\,v_{\rm Earth}.$$
In the limit of $P\to\infty$, this recovers our previous result $v_{\rm comet}\to v_{\rm escape}$. For typical period of $\sim$70yr, the speed of the comet is close to this number.
5. Finally, I shall coment that all this only relates to the magnitude of the orbital velocity (speed), but not to its direction. Comets are typically on highly eccentric orbits and, when at $r$=1AU, move in quite a different direction than Earth, even if their speed is only slightly larger. So the relative speed $|\boldsymbol{v}_{\rm comet}-\boldsymbol{v}_{\rm Earth}|$ of a comet with respect to Earth can be anyting between about 10 and 70 km/s.
This is a great answer! You're better than Wolfram! This is exactly the information and tools that I was looking for. Thanks very much. Doug.
who is Wolfram?
@Walter the answer to "**Who** is Wolfram" is either Stephen Wolfram or perhaps Wolfram Research. But of course user38715 was referring to wolframalpha.com
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# Solution of $x^2 + s(x)\cdot x - n = 0$, with $s(x)$ is the sum of digits of $x$.
This problem comes from an programming competition website, but I'd interested in analyze it from mathematics prespective. Given this problem below, we must create a program that could give us the correct output from the given input. The program must run under the time limit that provided by the system.
## The Problem
Let define function $s$ that take input from positive integers, such that $s(x)$ is the sum of the digit of $x$.
For example, $$s(1024) = 1 + 0 + 2 + 4 = 7$$
Given some integer constant $n$ as the input of the program. Now, find positive integer solution of $x$ in equation
$$x^2 + s(x) \cdot x - n = 0.$$
If more than one solution exist, choose the lowest one. If the solution doesn't exist, give output -1.
Time Limit of the program: 2 second.
Range of the input $n$: $[1, 10^{18}]$
## The Solution
In finding the solution, we could iterate from some range of possible value of x, and try each of these value whether it satisfy the equation or not.
To improve the run time of the program, we must choose the shortest range for x.
## My Question
1. What is the shortest range of possible value of x that we could iterate to find the answer of the problem above?
2. Is there a better way to find the solution without iterating from the set possible value of x?
## My Attempt:
Consider that $$x^2 + s(x) \cdot x - n = 0$$ $$\Leftrightarrow x ( x + s(x) ) = n$$
and since both $x$ and $(x + s(x))$ are positive integers and the product of both number is $n$, we can conclude that $x$ and $(x + s(x))$ is a pair of positive factor of $n$.
Since $x$ is a factor of $n$, we now have the first possible range of $x$, that is $[1, n]$.
But, iterating $x$ over this range gives bad performance to the program. For small value of $n$, it won't matter, but for $n \geq 10^{17}$, the program will reach it's time limit.
If I could recall correctly, we have this theorem | {
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If I could recall correctly, we have this theorem
$$x_1 \cdot x_2 = x$$
and
$$x_1 \leq x_2$$
implies that
$$x_1 \leq \sqrt{x} \leq x_2$$
And since $x (x + s(x)) = n$ and $x \leq x + s(x)$
we have $x \leq \sqrt{n} \leq x + s(x)$
Now we have the second range: $[1, \sqrt{n}]$
But, still, this range also give bad performance for larger $n$.
My last attempt is by considering the biggest possible value of $s(x)$, that is the digit of $x$ is all $9$ and $x$ less than or equals to the maximum value of $n$, $10^{18}$.
$$s(x) \leq s(999,999,99 \cdots 9,999) = 9 \cdot 18 = 162$$
Hence, we have
$$x + s(x) \leq x + 162$$
so that, $$\sqrt{n} \leq x + s(x) \leq x + 162 \leq \sqrt{n} + 162$$ hence, $$\sqrt{n} - 162 \leq x \leq \sqrt{n}$$
Now, we have better range that will suffice for all value of $x$. $$l_1 = \max \{ 1, \sqrt{n} - 162 \}$$ $$l_2 = \sqrt{n}$$ the range is $[l_1, l_2]$
But, I though there must be a better solution to this problem. Hence I try $$l_1 = \max \{ 1, \sqrt{n} - 162/2 \}$$ And that range is also correct. But I can't justify my finding.
So, hence my third question:
1. Why $l_1 = \max \{ 1, \sqrt{n} - 162/2 \}$ also gives us correct solution?
-
While not a full answer to your questions, here is a trick that will allow you to reduce the collection of numbers you have to try (and so an answer to question 2).
Because $10\equiv 1 \pmod 9$, we have $10^k \equiv 1 \pmod 9$ for every $k\in \mathbb N$. If we write a number in terms of its digits, $x=\sum a_k 10^k$, then since $\sum a_k 10^k \equiv \sum a_k \pmod 9$, we have $s(x)\equiv x \pmod 9$. This fact can be used to give a way to check arithmetic, among other things.
Therefore, if $x^2 + s(x)x = n$, then $2x^2 \equiv n \pmod 9$. Since $(2)(5)=10 \equiv 1 \pmod 9$, we can multiply to get $x^2 \equiv 5n \pmod 9$.
The squares modulo 9 are $0, 1, 4, 7$, so unless the remainder of dividing $5n$ by $9$ is one of these numbers, there are NO solutions. | {
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If the remainder is one of those numbers, you need to know the square roots modulo 9 to know what values of $x$ to try.
For this, we have the following.
$0^2\equiv 3^2 \equiv 6^2 \equiv 9 \pmod 9$, $1^2\equiv 8^2 \equiv 1 \pmod 9$, $2^2\equiv 7^2 \equiv 4 \pmod 9$ and $4^2 \equiv 5^2 \equiv 7 \pmod 9$.
The upshot of this is that if $n$ is a multiple of $9$, then you only have to check the multiples of $3$ (reducing the number of things you have to check by a factor of $3$). Otherwise, depending on the modulus of $5n\bmod 9$, you either have to check NO numbers (5 of the remaining cases), or you have to check $2/9$ of the numbers (3 of the remaining cases). Thus, whatever range you are working in, you have at least a speed up by a factor of $3$, if not more.
Another comment, which gives a small speedup over what you already have but answers question 3. As you observed, $x<\sqrt n$, and the number of digits of $x$ is at most $\lceil \log_{10} x \rceil$, and since $s(x)$ is at most $9$ times the number of digits of $x$, we have $s(x)<9 \lceil \log_{10} \sqrt n \rceil=9\lceil \frac{1}{2}\log_{10} n \rceil$.
Therefore, given $n$, you can limit your search for $x$ to $[\sqrt n - 9\lceil \frac{1}{2}\log_{10} n \rceil, \sqrt n]$. The factor of $1/2$ here answers question 3. Note, we could probably shave off $1$ or $2$ numbers from the range by being more careful with rounding, but I don't see any obvious reason we could shave off more. If we combine this with the beginning of my answer, you have to check at most $3\lceil \frac{1}{2}\log_{10} n \rceil$ numbers to find your solution, if it exists, and no numbers a good portion of the time when no solution exists.
-
I think $(2)(5)=10 \equiv 1 \pmod 5$ in pharagraph 3 should be $(2)(5)=10 \equiv 1 \pmod 9$. – Kamal Hajjaj Isa Oct 28 '12 at 14:58
@KamalHajjajIsa: Yes, that is correct. Thank you. – Aaron Oct 29 '12 at 3:07
I like the Idea of using $\pmod 9$. – Kamal Hajjaj Isa Oct 29 '12 at 4:56 | {
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Since $x<\sqrt n$ and $n\le10^{18}$, we have that $x<10^9$. Then $x$ has at most $9$ digits and $s(x)\le9\times9=81$. This implies that $x^2+81\,x-n\ge0$. It follows easily that we must have $$x\ge\frac{-81+\sqrt{81^2+4\,n}}{2}=\sqrt{n+(81/2)^2}-81/2.$$
-
This only answers the third question
If you use the general quadratic equation formula for the equation $ax^2+bx+c=0$ $$\displaystyle x= \frac {-b \pm \sqrt{b^2-4ac}}{2}$$ It follows from $x^2 + s(x)\cdot x - n = 0$ that $$\displaystyle x= \frac {-s(x) \pm \sqrt{s^2(x)+4n}}{2}$$ Since all the variables are positive and obviously $\sqrt{s^2(x)+4n}>s(x)$ , one can remove the negative sign and have $$\displaystyle x= \frac {-s(x) + \sqrt{s^2(x)+4n}}{2}$$ If one extends the range of values of $n$ to $0$, we have that the lowest possible value of $s(x)$ is $0$ and that $s(x) \ge 0$ which implies that $$\displaystyle x \ge \frac {-s(x) + \sqrt{4n}}{2} = \frac {-s(x)}{2} + \frac {2\sqrt{n}}{2} \ge \frac {-162}{2} + \sqrt n, \space \space (-s(x) \ge -162)$$ and there you have your inequality $$x \ge \sqrt n - \frac {162}{2}$$
Extending the range to $0$ doesn't change the validity of the inequality and helps to simplify it. You've been working with the maximum value of $s(x)$, using the minimum value helps in this case.
-
Beside the excellent modulo 9 tricks Aaron gave you, if $n$ passes his tests the easiest thing is to try the range of $1 \le s(x) \le 162$ with the proper $s(x) \pmod 9$. Just plug each one into the quadratic formula and solve for $x$. If it doesn't come out integral, go on to the next. If it does, check the sum of digits against the assumed $s(x)$. If it matches, success and report it. If not, go on to the next. At most $50$ numbers to try, which should fit in 2 seconds easily. | {
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# Probability Problems And Solutions | {
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S. The bet can either be on red or on blue and the amount of each bet is arbitrary. Thus, the probability that a randomly chosen piece is highlighted is 1/2. Work on the probability problems on pages 1-4 and 1-5 of the lab manual 2. The probability of event A is the number of ways event A can occur divided by the total number of possible outcomes. A pair of fair, standard dice are rolled. Solution to Problem 1. ECE302 Spring 2006 HW5 Solutions February 21, 2006 3 Problem 3. The correct reasoning is to calculate the conditional probability. SolutionStep 1 of 5We have to verify the given function is a probability mass function or notAnd we have to find the following probabilitiesHere each probability value is greater. As with any quantum system, the wave functions give the probability amplitude for finding the electron in a particular region of space, and these amplitudes are used to compute actual probabilities associated with measurements of the electron's position. Probability. And the probability of getting a head is equal to the probability of getting a tail. 316 that an audit of a retail business. nclearly there are 12 possible outcomes (6 x 2) nthere are 3 possible outcomes for an odd die and a head. Probability of Mutually Exclusive Events: If two events, A and B, are mutually exclusive, then the probability that either A OR B occurs is… P(A or B) = P(A) + P(B) Mutually Exclusive Events: If two events, A and B, are mutually exclusive, then that means that if A occurs, than B cannot, and vice versa. Then, n(E) = 2. Make sure you understand your probability questions is all we can say. Please be aware, however, that the handbook might contain, and almost certainly contains, typos as well as incorrect or inaccurate solutions. Q x = P(female lives to age x) = number of female survivors at age x 100,000. Calculating probability with percentages is a common topic learned in the K-12 years and is useful throughout your life. Solution of exercise 1. The remaining | {
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WBC Problem-page 10. Ignore problem 2 and 3 since they are about continuous distributions. Browse through all study tools. The solution is provided for each exercise. Use features like bookmarks, note taking and highlighting while reading Probability Problems and Solutions. More Problems on probability and statistics are presented. This Probability Pdf we are Providing is free to download. Probability on Numbers Worksheet These Probability Worksheets will produce problems with simple numbers between 1 and 50. One of the toughest and probably the most prestigious undergraduate competition in the world. All class notes posted in Part II above excluding. Devore Chapter 1. Divisibility intuition. Practice exam 1 and solution Practice exam 2 and solution Midterm 1 solutions Midterm 2 practice exams: Practice exam 1 and solution Practice exam 2 and solution Topics for Midterm 2: All of Chapter 3, Sections 1-3 of Chapter 4, Sections 1-3 of Chapter 5, Sections 1,2,3,4,6,7,9 of Chapter 6. Arrange - this is how many 'WAYS' something can be arranged. The remaining sections each investigate a different game, providing an overview of the rules, some discussion of basic strategy, and a series of problems and solutions. Experiment 1: A spinner has 4 equal sectors. nthe sample space for the last experiment would be all the ordered pairs in the form (d,c), where d represents the roll of a die and c represents the flip of a coin. Using the same data from the compound event video, I make a diagram to illustrate compound event probability and how to avoid double counting data when finding probability. Example 6 (Parts Inspection) Consider the parts problem again, but now assume that a one-year warranty is given for the parts that are shipped to customers. , the probability of getting the first two spheres as blue while the third sphere is red is same as the probability of getting the second and third spheres as blue while the first sphere is red As Jennifer pointed out, Baye's theorem | {
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and third spheres as blue while the first sphere is red As Jennifer pointed out, Baye's theorem can be used. The area of the shaded region to the left of z in the diagram above demonstrates these problems. This is a problem. Y1 - 1967/1/1. In a certain game, players toss a coin and roll a dice. Sugar daddies together with glucose little ones are believed as the perfect flings that can bring individuals along. Probability problem on Dice shortcut tricks are very important thing to know for your exams. PROBABILITY. I need the full detailed solution of the following 6 probability questions. Number of consonants ¡n English alphabets = 21. BIOSTATISTICS DESCRIBING DATA, THE NORMAL DISTRIBUTION 1. The long run probability of player A winning all the coins is. How does this impact the probability of some other A. Make sure you understand your probability questions is all we can say. The manual states that the lifetime T of the product, defined as the amount of time (in years) the product works properly until it breaks down, satisfies P(T ≥ t) = e − t 5, for all t. Let P(D) be the probability that the letter is in DOGS. This is a problem. This content was COPIED from BrainMass. to access the Math Probability menu. Solved by Expert Tutors. Solved examples with detailed answer description, explanation are given and it would be easy to understand. Solution of exercise 1. The wikipedia page claims that likelihood and probability are distinct concepts. The initial table listed in problem description is a table of probabilities p(1), p(2), , p(20). Miracle Mountain. And the probability of getting a head is equal to the probability of getting a tail. In a philosophical sense it can even be said that the risk does not have a real existence in the present, but it only exists in the future. Two and three-digit subtraction. Probability of choosing 2nd chocobar = 3/7. Often used in market research studies, it is useful in business to predict sales, scores and other numerical data based | {
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research studies, it is useful in business to predict sales, scores and other numerical data based off research that is either conducted or gathered at a secondary. (If it comes up six, you get$6. We then look at some word problems. Devore Chapter 1. It follows that the probability of an ace or a spade = 4+13 52 = 17 52. nclearly there are 12 possible outcomes (6 x 2) nthere are 3 possible outcomes for an odd die and a head. This post provides practice problems to reinforce the concept of correlation coefficient discussed in this post in a companion blog. ‘The Democrats will probably win the next election. For example, a weather forecast that predicts a 75 percent chance of snow is using probability to communicate that people should be prepared to deal with snowy conditions. Our Probability solvers include the most commonly used probability distributions, such as the normal, exponential, Poisson, binomial, and uniform distributions. MIDTERM 1 solutions and Histogram. More Problems on probability and statistics are presented. ) whereas ex-post solutions try to gather auxiliary information about non-respondents which is then used to calculate a probability of response for different population sub. Brilliant Premium. (a) Show that if X and Y are conditionally independent given Z and X and Z are conditionally independent given Y then X is independent of (Y,Z). 01 of probability is being shifted from $2400 to$0. It can be a finite set, a countable set (a set whose elements can be put in a sequence, in fact a finite set is also. Most of the problems require very little mathematical background to solve. Z for 90th percentile=1. Chapter 4 - Continuous Random Variables and Probability Distributions. Probability - math word problems Probability is the measure of the likeliness that an event will occur. Q2) Generate a Punnett Square for a heterozygous individual crossed with a heterozygous individual. Standard probability themes like coins, spinners, number cubes, and marbles | {
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heterozygous individual. Standard probability themes like coins, spinners, number cubes, and marbles are featured along with some other situations. Probability Questions are provided with detailed answers to every question. We also discuss some applications of probability theory to computing, including systems for making likely inferences from data and a class of useful algorithms that work “with high probability” but are not guaranteed to work all the time. what is the probability of generating a random number which has at least one digit = 1 for example 115. Then, as they advance, students create models for probability questions and ultimately use probability to make decisions. Probability Calculators and Solvers Our site contains an ample array of Probability Calculators that can greatly help you with all of your academic needs. The probability that a student works part time is P(PT)=0. A Problem With Pearls. Get Textbook Solutions and 24/7 study help for Statistics And Probability Step-by-step solutions to problems over 34,000 ISBNs Find textbook solutions Close. The area of the shaded region to the left of z in the diagram above demonstrates these problems. 5 1 − p = 0. • Probability and Statistics for Engineering and the Sciences by Jay L. Objective: I know how to solve probability word problems. Press [2] to evaluate a permutation or press [3] to evaluate a combination. Problem 1. Now practice our shortcut tricks on probability problem on coin and read examples. 1 • The random variable X has probability density function fX (x) = ˆ cx 0 ≤ x ≤ 2, 0 otherwise. Listed in the following table are practice exam questions and solutions, and the exam questions and solutions. But, as humans, we know the world is abnormal. A persuasive problem statement consists of three parts: 1) the ideal, 2) the reality, and 3) the consequences for the reader of the feasibility report. Learn and practice basic word and conditional probability aptitude questions with shortcuts, useful | {
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Learn and practice basic word and conditional probability aptitude questions with shortcuts, useful tips to solve easily in exams. This book is ideal for an upper-level undergraduate or graduate level introduction to probability for math, science, engineering and. The binomial probability calculator will calculate a probability based on the binomial probability formula. This is the number of times the event will occur. Using the same data from the compound event video, I make a diagram to illustrate compound event probability and how to avoid double counting data when finding probability. The problem concerns a game of chance with two players who have. u p is probability of a success and q = 1 - p is probability of a failure u Consider a game where the player bats 4 times: H probability of 0/4 = (0. Browse through all study tools. Probability Questions and answers PDF with solutions. Sarah has two children and we know that she has a daughter. Solution: Let’s think of this like a decision tree. Get Textbook Solutions and 24/7 study help for Statistics And Probability Step-by-step solutions to problems over 34,000 ISBNs Find textbook solutions Close. In other words, the probability of winning for a player is the ratio of the number of coins the player starts with to the total numbers of coins. Standard probability themes like coins, spinners, number cubes, and marbles are featured along with some other situations. The bet can either be on red or on blue and the amount of each bet is arbitrary. The probability of any event E is number of outcomes in E total number of outcomes in the sample space This probability is denoted by P E n E n S This probability is called classical probability and it uses the sample space S. Then, if x is the number of blue balls in urn 2,. Example Question #2 : How To Find The Probability Of An Outcome If x is chosen at random from the set (4, 6, 7, 9, 11) and y is chosen at random from the set (12, 13, 15, 17) then what is the probability | {
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6, 7, 9, 11) and y is chosen at random from the set (12, 13, 15, 17) then what is the probability that xy is odd?. Geometric Probability. P (A n B) = A and B = A x B P (A u B) = A or B = A + B. Find the probability that they are of same colour. when the selection of one does not affect the probability of a…. Conditional Probability In die and coin problems, unless stated otherwise, it is assumed coins and dice are fair and repeated trials are independent. • Probability and Statistics for Engineering and the Sciences by Jay L. At a picnic, Julio reaches into an ice-filled cooler containin…. The solutions are aimed to make your understanding clear. In non-technical parlance, "likelihood" is usually a synonym for "probability," but in statistical usage there is a clear distinction in perspective: the number that is the probability of some observed outcomes given a set of parameter values is regarded as the likelihood of the set of parameter values given the. Let E 1,E 2,E 3 be events. (Solution): Probability Problems. Sample Space. You need at most one of the three textbooks listed below, but you will need the statistical tables. The function f ( x ) is called a probability density function for the continuous random variable X where the total area under the curve bounded by the x -axis is equal to `1. Solutions Manual A First Course in Probability 9th Edition Sheldon Ross. Solution: Gambler’s Ruin Problem. It includes the list of lecture topics, lecture video, lecture slides, readings, recitation problems, recitation help videos, tutorials with solutions, and a problem set with solutions. "} BarChart3D[{win, lose}] The probability of winning is 30. We write P(AjB) = the conditional probability of A given B. If X and Y are independent uniform (0;1) random variables, show that [Filename: Assignment7_solutions I. Two Games in a Row. To navigate from one page of exercises to another, you will use the right navigation bar. If the first four marbles drawn are red, what is | {
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to another, you will use the right navigation bar. If the first four marbles drawn are red, what is the probability the next marble drawn will not be red? 7. The probability that a student belongs to a club is P(C)=0. Conditional Probability In die and coin problems, unless stated otherwise, it is assumed coins and dice are fair and repeated trials are independent. This is, however, wrong, because given that heads came up, it is more likely that the two-headed coin was chosen. by electronjohn. If we toss a fair die, what. Objective: I know how to solve probability word problems. Addition and subtraction word problems. AU - Haji-Sheikh, A. Here is a set of 14 GMAT probability questions, all in the Problem Solving style on the test, collected from a series of blog articles. The actual probability of finding the particle is given by the product of the wavefunction with its complex conjugate (like the square of the amplitude for a complex function). Theory of Statistics c 2000–2020 James E. IB Math – Standard Level – Probability Practice - MarkScheme Alei - Desert Academy C:\Users\Bob\Documents\Dropbox\Desert\SL\6StatProb\TestsQuizzesPractice\SLProbPractice1. Probability Problems. P (A n B) = A and B = A x B P (A u B) = A or B = A + B. Probability MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. Feel Free to Learn 316,387 views. Probability puzzles require you to weigh all the possibilities and pick the most likely outcome. PROBABILITY AND STATISTICS FOR ENGINEERS LESSON INSTRUCTIONS The lecture notes are divided into chapters. Review of probability theory • Definitions (informal) -Probabilities are numbers assigned to events that indicate "how likely" it is that the event will occur when a random experiment is performed -A probability law for a random experiment is a rule that assigns probabilities to the events in the experiment. Chapter 4 - Continuous Random Variables and Probability Distributions. This section provides materials for a lecture | {
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Random Variables and Probability Distributions. This section provides materials for a lecture on discrete random variables, probability mass functions, and expectations. Multiplication Rule of Probability The addition rule helped us solve problems when we performed one task and wanted to know the probability of two things happening during that task. The bet can either be on red or on blue and the amount of each bet is arbitrary. Q2) Generate a Punnett Square for a heterozygous individual crossed with a heterozygous individual. What Are Example Statistics and Probability Problems and Their Solutions? Credit: Echo/Cultura/Getty Images Two examples of probability and statistics problems include finding the probability of outcomes from a single dice roll and the mean of outcomes from a series of dice rolls. This is, however, wrong, because given that heads came up, it is more likely that the two-headed coin was chosen. Practice calculating conditional probability, that is, the probability that one event occurs given that another event has also occurred. Conditional Probability In die and coin problems, unless stated otherwise, it is assumed coins and dice are fair and repeated trials are independent. 01 of losing everything in going to A therefore looms large in peoples minds. The simulation was repeated by performing 50 repetitions of the experiment. starting the probability problems and solutions to door every daylight is agreeable for many people. Find the probability that a student takes calculus, given that he or she is on the dean’s list. No need to worry if you miss class or just need some extra help. All class notes posted in Part II above excluding. (If it comes up six, you get \$6. A probability experiment has four possible outcomes: e 1, e 2, e 3. Practice Problems - Try specific problems and see the solution. Think Stats: Probability and Statistics for Programmers is a textbook for a new kind of introductory prob-stat class. Complete Solutions Manual to | {
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is a textbook for a new kind of introductory prob-stat class. Complete Solutions Manual to Probability by Jim Pitman. It takes a computational approach, which has several advantages: • Students write programs as a way of developing and testing their un-derstanding. Chapter 1 covers this theory at a fairly rapid pace. Find the probability that the number on the drawn ticket is a multiple of 3 or 7. We break down problems on video in a step-by-step easy to follow format. Finding the Median in a pdf : S2 Edexcel. The probability of two or more heads is, therefore: Probability = 4 8 = 1 2 We solved this problem by first enumerating the set of possibl e outcomes, known as the Sample Space, and then by deciding which of these outcomes satisfied the cr iterion of containing 'two or more heads'. Certainly X is binomial with n = 40 and p = 0. More Problems on probability and statistics are presented. P(X) gives the probability of successes in n binomial trials. Example of Using a Contingency Table to Determine Probability. Problem Solving: Find a Pattern What Is It? Finding a Pattern is a strategy in which students look for patterns in the data in order to solve the problem. Generate integer from 1-8 with equal probability. Problem 4 (Solution on p. Topics covered range from measure and integration theory to functional analysis and basic concepts of probability; the interplay between measure theory and topology; conditional probability and expectation; the central limit theorem; and strong laws of large numbers in terms of martingale theory. Calculate the value of a in a normal distribution with a mean of 4 and a standard deviation of 2 for which:. Solution : Sample space : {(1, 1)(1, 2)(1, 3)(1. Review of probability theory • Definitions (informal) -Probabilities are numbers assigned to events that indicate "how likely" it is that the event will occur when a random experiment is performed -A probability law for a random experiment is a rule that assigns probabilities to the | {
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is performed -A probability law for a random experiment is a rule that assigns probabilities to the events in the experiment. The actual probability of finding the particle is given by the product of the wavefunction with its complex conjugate (like the square of the amplitude for a complex function). Pedigrees and probability problems 1-28 and 1-30. It includes the list of lecture topics, lecture video, lecture slides, readings, recitation problems, recitation help videos, tutorials with solutions, and a problem set with solutions. Chapter 1 covers this theory at a fairly rapid pace. Solution; Determine the value of $$c$$ for which the function below will be a probability density function. In a binomial distribution the probabilities of interest are those of receiving a certain number of successes, r, in n independent trials each having only two possible outcomes and the same probability, p, of success. a Show that $$f\left( x \right)$$ is a probability density function. the number of ways several people can stand in a line if 2 particular people must — or must not — stand next to each other). (321 problems) IMO Shortlisted Problems. Probability of choosing 1 icecream out of a total of 6 = 4/6 = 2/3. Playing cards probability problems based on a well-shuffled deck of 52 cards. Chegg Solution Manuals are written by vetted Chegg Statistics And Probability experts, and rated by students - so you know you're getting high quality answers. Probability puzzles require you to weigh all the possibilities and pick the most likely outcome. The post in the companion blog shows how to evaluate the covariance and the correlation coefficient of two continuous random variables and. Importantly, the solution for the problem of points became known as an expectation value (the average expected value). Suppose that a good part fails within the first year with probability 0. This is essentially a 1D scattering problem. At a picnic, Julio reaches into an ice-filled cooler containin…. 1 | {
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a 1D scattering problem. At a picnic, Julio reaches into an ice-filled cooler containin…. 1 Problem 4E. Drawing/Picking/Choosing Single/One ball from a bag/urn/box - Probability - Problems Solutions. Brilliant Premium. Probability questions pop up all the time. Fully solved Aptitude questions with detailed answer description and explanation. Practice Problems - Try specific problems and see the solution. IB Math - Standard Level - Probability Practice - MarkScheme Alei - Desert Academy C:\Users\Bob\Documents\Dropbox\Desert\SL\6StatProb\TestsQuizzesPractice\SLProbPractice1. Probability of problem getting solved = 1 - (5/7) x (3/7) x (5/9) = (122/147) Example 9: Find the probability of getting two heads when five coins are tossed. Problem 1. ECE302 Spring 2006 HW5 Solutions February 21, 2006 3 Problem 3. The higher the probability of an event, the more certain that the event will occur. Enter r, the number of items selected from the set, and press [ENTER] to display the result. Playing cards probability problems based on a well-shuffled deck of 52 cards. also known as the birthday problem, states that in a random group of 23 people, there is about a 50 percent chance that two people have the same birthday. Test your understanding with practice problems and step-by-step solutions. Learn and practice basic word and conditional probability aptitude questions with shortcuts, useful tips to solve easily in exams. The mathematics field of probability has its own rules, definitions, and laws, which you. Conditional Probability: It is known that a student who does his online homework on aregular basishas a chance of83 percentto get a good grade (A or B); but the chance drops to58 percentif he doesn't do the homework regularly. An agent sells life insurance policies to five equally aged, healthy people. Students were encouraged to prepare a 4x6 inch notecard to use for reference during each exam. Behind one is a expensive car, but behind the other two are goats. when the | {
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during each exam. Behind one is a expensive car, but behind the other two are goats. when the selection of one does not affect the probability of a…. Importantly, the solution for the problem of points became known as an expectation value (the average expected value). " Understanding how to calculate these percentages with real numbers of people and things. Basic concept on drawing a card: In a pack or deck of 52 playing cards, they are divided into 4 suits of 13 cards each i. and define the function. Now practice our shortcut tricks on probability problem on coin and read examples. Drawing/Picking/Choosing Single/One ball from a bag/urn/box - Probability - Problems Solutions. Homework 2 from last time OLD HW2 solutions Practice midterm and Midterm 2015 solutions. The text-books listed below will be useful for other courses on probability and statistics. Cards of Spades and clubs are black cards. (See diagram. 1) Basics of Electromagnetics Part I – Download MCQs from here. Solution the the Second WBC Problem-page 10. Multiplicative Principle for Counting n The total number of outcomes is the product of the possible outcomes at each step in the sequence n if a is selected from A, and b selected from B… n n (a,b) = n(A) x n(B) q (this assumes that each outcome has no influence on the next outcome) n How many possible three letter words are there? q you can choose 26 letters for each of the three positions, so. GMAT Advanced Probability Problems By Mike MᶜGarry on January 3, 2014 , UPDATED ON January 15, 2020, in GMAT Math In the following probability problems, problems #1-3 function as a set, problems #4-5 are another set, and problems #6-7 are yet another set. by electronjohn. Homework 2 from last time OLD HW2 solutions Practice midterm and Midterm 2015 solutions. Solution; Determine the value of $$c$$ for which the function below will be a probability density function. Test your understanding with practice problems and step-by-step solutions. Probability theory is | {
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Test your understanding with practice problems and step-by-step solutions. Probability theory is the most directly relevant mathematical background, and it is assumed that the reader has a working knowledge of measure-theory-based probability theory. Midterm 2016 solutions Combinatorics practice problem set and solutions. Bayes' theorem describes the probability of occurrence of an event related to any condition. For instance, let’s say the probability that you will get to class on time is 2/3. In non-technical parlance, "likelihood" is usually a synonym for "probability," but in statistical usage there is a clear distinction in perspective: the number that is the probability of some observed outcomes given a set of parameter values is regarded as the likelihood of the set of parameter values given the. Finding the Median in a pdf : S2 Edexcel. Thanks for the A2A. At a picnic, Julio reaches into an ice-filled cooler containin…. Three circle Venn Diagrams are a step up in complexity from two circle diagrams. Possible Solutions to Probability Problems Consider the experiment of selecting a card from an ordinary deck of 52 playing cards. The PDF is the density of probability rather than the probability mass. 2 Counting Assignments One of the simplest but most important counting problems deals with a list of. Probability MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. The higher the probability of an event, the more certain that the event will occur. by mathemagician. Throw 2 dices simultaneously. Then, we'll teach you all of the basic rules and formulas you need to know as well as give you tips for studying probability and approaching GRE probability questions on test day. Problem 1 : Two dice are rolled, find the probability that the sum is. For i = 1,2, let R i = event that a red ball is drawn from urn i and let B i = event that a blue ball is drawn from urn i. 𝗧𝗼𝗽𝗶𝗰: conditional probability problems with solutions 𝗦𝘂𝗯𝗷𝗲𝗰𝘁: Engineering | {
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is drawn from urn i. 𝗧𝗼𝗽𝗶𝗰: conditional probability problems with solutions 𝗦𝘂𝗯𝗷𝗲𝗰𝘁: Engineering Mathematics 𝗧𝗼 𝗕𝗨𝗬 𝗻𝗼𝘁𝗲𝘀 𝗼𝗳. So how will you make the determination of how to solve a conditional probability problem? As a general rule, problems where information is given in a table are best solved by using the formula. The point is that the order of events doesn't affect with respect to conditional probability. After solving all ten math questions write down total time taken by you to solve those questions. Notice that the phenotypic ratio for dominant/recessive is identical to Q1 (3/4 Dominant, 1/4 recessive) except you now have to realize your mutation is Dominant and wild type is recessive. Midterm 2016 solutions Combinatorics practice problem set and solutions. Calculate the sample mean. P(F AND P) = d. Sarah has two children and we know that she has a daughter. Playing cards probability problems based on a well-shuffled deck of 52 cards. many new problems and solutions, a number of improvements in the pre-sentation, and corrections. PROBABILITY AND STATISTICS FOR ENGINEERS LESSON INSTRUCTIONS The lecture notes are divided into chapters. Probability theory - Probability theory - The birthday problem: An entertaining example is to determine the probability that in a randomly selected group of n people at least two have the same birthday. Probability is the chance that an event will occur. Solution: (F) Probability always lie within 0 to 1. Probability_Problems Based on Coins#LESSON-2 - Duration: 17:39. If we toss a fair die, what. $$∴ P(E)={n(E)\over n(S)} = {2\over52}={1\over26}$$ Learn and practice these probability problems with solutions and you can ask me anything in the comment section if you have any doubts regarding probability problems and it's solutions. To get a feeling for PDF, consider a continuous random variable. To link to this page, copy the following code to your site:. The text-books listed below will be useful for other courses on probability | {
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code to your site:. The text-books listed below will be useful for other courses on probability and statistics. Topics covered range from measure and integration theory to functional analysis and basic concepts of probability; the interplay between measure theory and topology; conditional probability and expectation; the central limit theorem; and strong laws of large numbers in terms of martingale theory. [Syllabus] [Textbooks] [Grading] [Reading assignment] [Homework problems and solutions] [Tests] [Knowledge Milestones]. to access the Math Probability menu. Finding the Median in a pdf : S2 Edexcel. Practice Problems - Try specific problems and see the solution. The question is incomplete, that is, it doesn’t provide information regarding with the level of probability that you are familiar with. PROBLEM 13 : Consider a rectangle of perimeter 12 inches. Conroy and a 5=6 chance of not rolling a six, after which the number of rolls we expect to throw is the same as when we started. This Collection of problems in probability theory is primarily intended for university students in physics and mathematics departments. Finding the mean in a Normal Distribution : Statistics S1 Edexcel June 2013 Q6 (a) ExamSolutions - youtube Video. It can be a finite set, a countable set (a set whose elements can be put in a sequence, in fact a finite set is also. Most probability problems on the GRE involve independent events. A problem is given to three students whose chances of solving it are 1/2,. "} BarChart3D[{win, lose}] The probability of winning is 30. The process to solve the problem is rarely straightforward and takes practice to perfect. This is a problem. No need to worry if you miss class or just need some extra help. According to recent data, the probability of a person living in these conditions for 30 years or. Colavito’s homepage for solutions. The initial table listed in problem description is a table of probabilities p(1), p(2), , p(20). Total Number of ways = 2 5 = | {
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in problem description is a table of probabilities p(1), p(2), , p(20). Total Number of ways = 2 5 = 32. The point is that the order of events doesn't affect with respect to conditional probability. MIDTERM 2 solutions and Histogram. The incubation periods of a random sample of 7 HIV infected individuals is given below (in years): 12. Advanced Techniques. A Problem With Pearls. Normal distribution probability : Statistics S1 Edexcel June 2013 Q6 (b) : ExamSolutions - youtube Video. The probability mass function of is but and Therefore, the probability mass function can be written as which is the probability mass function of a Bernoulli random variable. Then, n(E) = 2. Rolling the Dice. The empirical probability of three consecutive female births is. 1, Median 22. ) For a different problem, allow every one of n people to place an even bet on the color of his hat. In the Options dialog box, select the Show Iteration Results check box to see the values of each trial solution, and then click OK. Most of the problems require very little mathematical background to solve. Problem Solving: Find a Pattern What Is It? Finding a Pattern is a strategy in which students look for patterns in the data in order to solve the problem. The probability (chance) is a value from the interval 0;1> or in percentage (0% to 100%) expressing the occurrence of some event. P (A n B) = A and B = A x B P (A u B) = A or B = A + B. DISCRETE PROBABILITY DISTRIBUTIONS 27. Miracle Mountain. Geometric Probability. Life or Death? The Emperor's Proposition. We want to generate a random number between 100 and 499. The probability of a success, denoted by p, remains constant from trial to trial and repeated trials are independent. Probability Questions and answers PDF with solutions. Probability MCQ Questions and answers with easy and logical explanations. Please do not email me completed problem sets. Multiplying by multiples of 10. John has a special die that has one side with a six, two sides with twos | {
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by multiples of 10. John has a special die that has one side with a six, two sides with twos and three sides with ones. At any particular time period, both outcomes cannot be achieved together so […]. Solution of exercise 1. This is a beta distribution. We want to generate a random number between 100 and 499. The probability equals 31,8 %. The question is incomplete, that is, it doesn’t provide information regarding with the level of probability that you are familiar with. There is one desired outcome and six possible outcomes. random variables, and some notation. Normal distribution probability : Statistics S1 Edexcel June 2013 Q6 (b) : ExamSolutions - youtube Video. 74% of the X values are less than three standard deviations from the mean. (Opens a modal) Random number list to run experiment. Practice Problems - Try specific problems and see the solution. Here is a set of 14 GMAT probability questions, all in the Problem Solving style on the test, collected from a series of blog articles. This book is ideal for an upper-level undergraduate or graduate level introduction to probability for math, science, engineering and. Solution : Sample space : {(1, 1)(1, 2)(1, 3)(1. Probability Exam Questions with Solutions by Henk Tijms1 December 15, 2013 This note gives a large number of exam problems for a first course in prob-ability. Each trial results in an outcome that may be classified as a success or a failure (hence the name, binomial);. Playing cards probability problems based on a well-shuffled deck of 52 cards. com: Many students are also using our Free Statistics Lab Manual. I require the probability that after these n stretches he is at a distance between rand r+ rfrom his starting point 0. A Music Survey was carried out to find out what types of music a group of people liked. Assume there is a beam of particles with definite momentum coming in from the left and assume there is no flux of particles coming from the right. One of the famous problems that motivated | {
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