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

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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])

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.

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

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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.

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.

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.

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].

• 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

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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


# 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

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.


# 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?

• 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

$\hspace{.2 in} 1-\big(\frac{5}{6}\big)^6-\big(\frac{1}{2}\big)\big(\frac{5}{6}\big)^5-\big(\frac{1}{2}\big)\big(\frac{5}{6}\big)^4=\frac{5203}{23328}\approx.223$

• 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*}


# 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.

• 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

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.


## 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 $0.145 \le x \lt 0.245$.

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


# 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 $n-(M+1)/2$.

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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.

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:

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:

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.)

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} &=\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\{ 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. $$\blacksquare$$

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. $$\blacksquare$$


# 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: $$\frac{1}{n^2} \le \frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n}$$

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

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 $\frac{3}{2}$.

(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

$$\frac{10}{20} \cdot \frac{1}{20}$$

The probability you obtain $$14$$ using the $$4$$-sided die on the second role is

$$\frac{3}{20} \cdot \frac{1}{4}$$


# 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?

• 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"

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 $y$.


# 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.

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

$$f([a,b])=f(I_1)\cup \cdots \cup f(I_n)\cup f(\{x_0,\dots x_n\})$$

is bounded.


# 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).

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 $\mathbb{Q}$.

• 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.

• @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 $p$).

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 $x\in\mathbb{R}$).

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 $\alpha_2=\cdots=\alpha_k=0$.

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? 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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.

-----Edit-----

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 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 $A$.

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


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,

$$Var(aX+bY)=a^2Var(X)+b^2Var(Y)$$

• 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

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.


# 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

$${-x-3\over5y-4}={3+x\over4-5y}$$

-


# 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.

-
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.

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.

-


+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,

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

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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.

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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