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Calculate the cumulative integral of a vector where the spacing between data points is uniform, but not equal to 1.
Create a domain vector.
`X = 0:pi/5:pi;`
Calculate the sine of `X`.
`Y = sin(X');`
Cumulatively integrate `Y` using `cumtrapz`. When the spacing between points is constant, but not equal to 1, an alternative to creating a vector for `X` is to specify the scalar spacing value. In that case, `cumtrapz(pi/5,Y)` is the same as `pi/5*cumtrapz(Y)`.
`Q = cumtrapz(X,Y)`
```Q = 6×1 0 0.1847 0.6681 1.2657 1.7491 1.9338 ```
Cumulatively integrate the rows of a matrix where the data has a nonuniform spacing.
Create a vector of x-coordinates and a matrix of observations that take place at the irregular intervals. The rows of `Y` represent velocity data, taken at the times contained in `X`, for three different trials.
```X = [1 2.5 7 10]; Y = [5.2 7.7 9.6 13.2; 4.8 7.0 10.5 14.5; 4.9 6.5 10.2 13.8];```
Use `cumtrapz` to integrate each row independently and find the cumulative distance traveled in each trial. Since the data is not evaluated at constant intervals, specify `X` to indicate the spacing between the data points. Specify `dim = 2` since the data is in the rows of `Y`.
`Q1 = cumtrapz(X,Y,2)`
```Q1 = 3×4 0 9.6750 48.6000 82.8000 0 8.8500 48.2250 85.7250 0 8.5500 46.1250 82.1250 ```
The result is a matrix of the same size as `Y` with the cumulative integral of each row.
Perform nested integrations in the x and y directions. Plot the results to visualize the cumulative integral value in both directions.
Create a grid of values for the domain.
```x = -2:0.1:2; y = -2:0.2:2; [X,Y] = meshgrid(x,y);```
Calculate the function $\mathit{f}\left(\mathit{x},\mathit{y}\right)=10{\mathit{x}}^{2}+20{\mathit{y}}^{2}$ on the grid.
`F = 10*X.^2 + 20*Y.^2;` | {
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`F = 10*X.^2 + 20*Y.^2;`
`cumtrapz` integrates numeric data rather than functional expressions, so in general the underlying function does not need to be known to use `cumtrapz` on a matrix of data. In cases where the functional expression is known, you can instead use `integral`, `integral2`, or `integral3`.
Use `cumtrapz` to approximate the double integral
`$I\left(a,b\right)={\int }_{-2}^{b}{\int }_{-2}^{a}\left(10{x}^{2}+20{y}^{2}\right)\phantom{\rule{0.16666666666666666em}{0ex}}dx\phantom{\rule{0.16666666666666666em}{0ex}}dy.$`
To perform this double integration, use nested function calls to `cumtrapz`. The inner call first integrates the rows of data, then the outer call integrates the columns.
`I = cumtrapz(y,cumtrapz(x,F,2));`
Plot the surface representing the original function as well as the surface representing the cumulative integration. Each point on the surface of the cumulative integration gives an intermediate value of the double integral. The last value in `I` gives the overall approximation of the double integral, `I(end) = 642.4`. Mark this point in the plot with a red star.
```surf(X,Y,F,'EdgeColor','none') xlabel('X') ylabel('Y') hold on surf(X,Y,I,'FaceAlpha',0.5,'EdgeColor','none') plot3(X(end),Y(end),I(end),'r*') hold off```
## Input Arguments
collapse all
Numeric data, specified as a vector, matrix, or multidimensional array. By default, `cumtrapz` integrates along the first dimension of `Y` whose size does not equal 1.
Data Types: `single` | `double`
Complex Number Support: Yes
Point spacing, specified as `1` (default), a uniform scalar spacing, or a vector of coordinates.
• If `X` is a scalar, then it specifies a uniform spacing between the data points and `cumtrapz(X,Y)` is equivalent to `X*cumtrapz(Y)`.
• If `X` is a vector, then it specifies x-coordinates for the data points and `length(X)` must be the same as the size of the integration dimension in `Y`.
Data Types: `single` | `double` | {
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Data Types: `single` | `double`
Dimension to operate along, specified as a positive integer scalar. If you do not specify the dimension, then the default is the first array dimension of size greater than 1.
Consider a two-dimensional input array, `Y`:
• `cumtrapz(Y,1)` works on successive elements in the columns of `Y`.
• `cumtrapz(Y,2)` works on successive elements in the rows of `Y`.
If `dim` is greater than `ndims(Y)`, then `cumtrapz` returns an array of zeros of the same size as `Y`.
## Tips
• Use `trapz` and `cumtrapz` to perform numerical integrations on discrete data sets. Use `integral`, `integral2`, or `integral3` instead if a functional expression for the data is available.
• `trapz` reduces the size of the dimension it operates on to 1, and returns only the final integration value. `cumtrapz` also returns the intermediate integration values, preserving the size of the dimension it operates on.
## Version History
Introduced before R2006a | {
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# “Elegant” way to assign 4 distinct objects into 4 bins
Suppose you have four objects to distribute among four people. Suppose people can carry any number of objects (none, one, two, three, or all four objects) at once. In how many ways can the four objects be distributed?
My solution
Let the four people be named A,B,C,D.
Distribution type 1
If each person gets exactly one object, the distribution looks like
A B C D
1 1 1 1
of which there are $$4!=4\cdot 3 \cdot 2 \cdot 1 = 24$$ distinct ways (since the objects are distinct).
Distribution type 2
If one person gets 2 objects and two others get 1 object, the possible distributions look like
A B C D
2 1 1 0
2 1 0 1
2 0 1 1
1 2 1 0
1 2 0 1
0 2 1 1
1 1 2 0
1 0 2 1
0 1 2 1
1 1 0 2
1 0 1 2
0 1 1 2
and for each of the above distributions, there are $$12$$ distinct ways to distribute the distinct objects.
Distribution type 3
If two people get 2 objects, the possible distributions look like
A B C D
2 2 0 0
2 0 2 0
2 0 0 2
0 2 2 0
0 2 0 2
0 0 2 2
and for each of the above distributions, there are $$6$$ distinct ways to distribute the distinct objects.
Distribution type 4
If one person gets 3 objects and another person gets 1 object, the possible distributions look like
A B C D
3 1 0 0
3 0 1 0
3 0 0 1
1 3 0 0
0 3 1 0
0 3 0 1
1 0 3 0
0 1 3 0
0 0 3 1
1 0 0 3
0 1 0 3
0 0 1 3
and for each of the above distributions, there are $$4$$ distinct ways to distribute the distinct objects.
Distribution type 5
If one person gets all 4 objects, the possible distributions look like
A B C D
4 0 0 0
0 4 0 0
0 0 4 0
0 0 0 4
for each of the above distributions, there is only $$1$$ distinct way to distribute the distinct objects.
Totaling the ways
The total number of ways is then
\begin{align} &\textrm{ type 1 + type 2 + type 3 + type 4 + type 5 }\\ =& 1(24) + 12(12) + 6(6) + 12(4) + 4(1) \\ =& 24 + 144 + 36 + 48 + 4 \\ =& 256\textrm{ ways } \end{align}
My questions
I have 3 questions: | {
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My questions
I have 3 questions:
1. Is there a more elegant way (using $$_n P_k$$ or $$_n C_k$$ notation) of counting the amount of each distribution type (the numbers 1, 12, 6, 12, 4) that appear in the final calculation?
2. Is there a more elegant way (using $$_n P_k$$ or $$_n C_k$$ notation) of counting the number of ways for each distribution type (the numbers 24, 12, 6, 4, 1) that appear in the final calculation in parentheses? (Clearly the 24 is just $$_4 P_4$$, but what about the others?)
3. Why is all of this just equal to $$4^4$$? This isn't immediately obvious to me.
• @EdwardH., do you have a proof of that? – Charles Hudgins Sep 5 '19 at 5:00
• Each object can be given to any person, hence $N_p^{N_o}=4^4$ ways. – Yves Daoust Sep 5 '19 at 7:25
1. Breaking it up a little differently, the number of ways with $$k = 0,1,2,3$$ of them getting $$0$$ is $$_4C_k \cdot {}_3C_{3-k}$$, which gives the sequence $$1, 12, 18 = 6 + 12, 4$$. Where the difference is that you had the $$6$$ ways of $$2,2,0,0$$ and the $$12$$ ways of $$3,1,0,0$$ separately. Those are $$\frac{4!}{2!2!}$$ and $$\frac{4!}{1!1!2!}$$, respectively.
2. For the arrangements, you are looking at the multinomial coefficients: $$\frac{4!}{1!1!1!1!} = 24, \frac{4!}{2!1!1!0!} = 12, \frac{4!}{2!2!0!0!} = 6, \frac{4!}{3!1!0!0!} = 4, \frac{4!}{4!0!0!0!} = 1$$
3. Instead of focusing on the people and which objects they get, look at the objects and who they are given to. Each object can be given to any of the four people, without restriction. That means there are $$4$$ options for the first object, $$4$$ for the second, and so on, so the total is $$4^4$$.
• Thank you, Michael! – Mathemanic Sep 8 '19 at 23:18
Let the objects choose the people . . .
Each object has $$4$$ choices, and each object's choices are independent of the choices made by the other objects, hence there are $$4^4$$ ways for the objects to choose the people.
Alternatively, using your cases . . . | {
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Alternatively, using your cases . . .
• For distribution type $$1$$, the number of ways is $$4!=24$$ Explanation:
If we have the people line up to choose, then person $$1$$ has $$4$$ choices, person $$2$$ has $$3$$ choices, person $$3$$ has $$2$$ choices, and person $$4$$ has $$1$$ choice.
• For distribution type $$2$$, the number of ways is $$\binom{4}{1}\binom{4}{2}\binom{3}{2}2!=144$$ Explanation:
• Choose the person who gets two objects:$$\;{\large{\binom{4}{1}}}\;$$choices.
• Choose the two objects for that person:$$\;{\large{\binom{4}{2}}}\;$$choices.
• Choose the two people to get the two remaining objects:$$\;{\large{\binom{3}{2}}}\;$$choices.
• Distribute the two remainining objects, one to each of the two chosen people:$$\;2!\;$$choices.
• For distribution type $$3$$, the number of ways is $$\binom{4}{2}\binom{2}{1}\binom{3}{1}=36$$ Explanation:
• Choose the two people who get two objects:$$\;{\large{\binom{4}{2}}}\;$$choices.
• Choose the person who gets the object labeled #$$1$$ plus one other object:$$\;{\large{\binom{2}{1}}}\;$$choices.
• Choose the other object for that person:$$\;{\large{\binom{3}{1}}}\;$$choices.
• For distribution type $$4$$, the number of ways is $$\binom{4}{1}\binom{4}{3}\binom{3}{1}=48$$ Explanation:
• Choose the person who gets three objects:$$\;{\large{\binom{4}{1}}}\;$$choices.
• Choose the three objects for that person:$$\;{\large{\binom{4}{3}}}\;$$choices.
• Choose the person to get the one remaining object:$$\;{\large{\binom{3}{1}}}\;$$choices.
• For distribution type $$5$$, the number of ways is $$\binom{4}{1}=4$$ Explanation:$$\;$$Choose the person who gets all four objects:$$\;{\large{\binom{4}{1}}}\;$$choices.
$$\;\;\;$$Summing the counts for the cases, the total number of ways is $$24+144+36+48+4=256$$
• Awesome and very thorough answer - thank you Quasi! – Mathemanic Sep 8 '19 at 23:19 | {
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Question about spanning sets and bases
In a textbook is the question:
Find a basis for the subspace $$V=\{(x_1,x_2,x_3,x_4):x_1+2x_2+x_3+x_4=0, 3x_1+6x_2+4x_3+x_4=0\}.$$
They say that $V$ is a subspace of $\mathbb{R^4}$ and are able to find a spanning set by solving the system of homogeneous equations getting: $$\{\begin{bmatrix}-2 \\ 1 \\ 0 \\ 0 \end{bmatrix},\begin{bmatrix}-3 \\ 0 \\ 2 \\ 1 \end{bmatrix}\}.$$ They then show that these vectors are linearly independent and so it forms a basis for $V$.
I thought though that if you're in $\mathbb{R^4}$ then you need at least 4 vectors to span the space, and 4 to form a basis. Here though we only have two. Could someone please tell me where my logic went wrong and why these indeed span the subspace and form a basis?
• They ask you to find a basis for a subspace, not for the entire ${\bf R}^4$. – avs Aug 11 '16 at 23:18
You do indeed need at least 4 vectors to span $\mathbb{R}^4$. However, here we're not trying to span $\mathbb{R}^4$, we're trying to span $V$.
For a more concrete example: think about the subspace $W=\{(a, b): a=0\}$ of $\mathbb{R}^2$ (basically, $W$ is the $y$-axis). Clearly $W$ is spanned by a one-element set - e.g., $\{(0, 1)\}$ - even though it takes two vectors to span $\mathbb{R}^2$. Does this help?
Noah's answer is definitely sufficient, but I'm going to try to add a bit of algebraic insight for you. Write $v=(1,2,1,1)$ and $w=(3,6,4,1)$. Let $f,g: \mathbb{R}^4 \rightarrow \mathbb{R}$, $f:x \mapsto v \cdot x$ and $g: x \mapsto w \cdot x$.
$f,g$ are linear maps and $\mathbb{R}$ is a field, so they are either surjective or trivial. Note that we have $V= \ker(f) \cap \ker(g)$. If $\dim(V)=4$, then $\mathbb{R}^4/V$ is trivial and both $f,g$ must be the zero map. Hopefully you can see this is clearly not the case. | {
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# Equivalent Definitions of e
#### (A new question of the week)
It is not unusual for mathematicians to define a concept in multiple ways, which can be proved to be equivalent. One definition may lead to a theorem, which another presentation uses as the definition, from which the original definition can be proved as a theorem. Here, in yet another good question from late May, we have two different ways to “define” the number $$e=2.71828…$$, a series and a limit, and a student wants to prove directly that they are equivalent. We’ll get a proof, then dig in to really understand it.
## Proving two definitions are equivalent
Hi!
I’m looking for a rigorous proof that the following definitions of e are equivalent:
e = 1 + 1 + 1/2! + 1/3! + 1/4! + …
e = lim [n→∞] (1 + 1/n)^n
What I’ve done so far:
I understand that:
lim [n→∞] (1 + 1/n)^n = 1 + 1 + (1 – 1/n) 1/2! + (1 – 1/n)(1 – 2/n) 1/3! + (1 – 1/n)(1 – 2/n)(1 – 3/n) 1/4! + …
I also understand this:
lim [n→∞] (1 – 1/n)(1 – 2/n)(1 – 3/n) = 1 x 1 x 1 = 1
And so if we take a fixed value m (with m as a natural number):
lim [n→∞] 1 + 1 + (1 – 1/n) 1/2! + (1 – 1/n)(1 – 2/n) 1/3! + (1 – 1/n)(1 – 2/n)(1 – 3/n) 1/4! + … + (1 – 1/n)(1 – 2/n)(1 – 3/n)…(1 – (m – 1)/n) 1/m!
is equivalent to:
1 + 1 + 1/2! + 1/3! + 1/4! + … + 1/m!
But I then get stuck on the next step. I only understand the ‘equivalence’ where the series terminates at a fixed natural number (in this case m). How do I make the transition to proving it for the infinite series, where the number of members of the series approaches infinity, whilst n also approaches infinity within each element of the series?
I’m in particular looking for a rigorous proof, where the two sequences are named t_n and s_n.
So that for every epsilon there exists a number N so that for every n > N then |t_n – s_n| < epsilon.
Really grateful for any help on this! | {
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Really grateful for any help on this!
Matthew knows how to ask a question: He has clearly stated what he wants to do, and shown a good bit of work, including where the difficulty lies. But we’ll need to get up to speed to make sure we understand the question, as well as this initial work!
### What we need to prove
The two “definitions” are quite different, one a series, the other a limit. Here is how Wikipedia introduces its article on e:
The number e, also known as Euler’s number, is a mathematical constant approximately equal to 2.71828, and can be characterized in many ways. It is the base of the natural logarithm. It is the limit of $$\left(1+\frac{1}{n}\right)^n$$ as n approaches infinity, an expression that arises in the study of compound interest. It can also be calculated as the sum of the infinite series
$$e=\sum_{n=0}^{\infty }{\frac {1}{n!}}=1+{\frac {1}{1}}+{\frac {1}{1\cdot 2}}+{\frac {1}{1\cdot 2\cdot 3}}+\cdots$$
That is, they take the limit as the actual definition, and present the series as a way to calculate it. Here is a table of the first 11 terms of the series and its sums:
We have already reached 8 significant digits of accuracy.
Here is a table of the first 11 values of the limit:
We don’t even have one significant digit yet!
In fact, if we make the limit move exponentially faster, by using $$2^n$$ in place of n, it is still slow to converge:
It takes 8 thousand “steps” to get 4 significant digits! (Of course, in principle we only need to do this calculation once, rather than summing terms, so it is not that inefficient to calculate, apart from the efficiency of using a very large exponent in the first place, and deciding what value to use.)
But they do approach the same limit. How can we prove it?
### What he’s done so far
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### What he’s done so far
Let’s look at what Matthew did, which he didn’t fully explain.
First, he said $$\lim_{n\to\infty} \left(1 + \frac{1}{n}\right)^n = 1 + 1 + \left(1 – \frac{1}{n}\right) \frac{1}{2!} + \left(1 – \frac{1}{n}\right)\left(1 – \frac{2}{n}\right) \frac{1}{3!} + \left(1 – \frac{1}{n}\right)\left(1 – \frac{2}{n}\right)\left(1 – \frac{3}{n}\right) \frac{1}{4!} + \cdots$$
Where did that come from? He has applied the binomial theorem to $$\left(1 + \frac{1}{n}\right)^n$$; this theorem in general says that $$(1+x)^n = \sum_{k=0}^n {_nC_k}x^k$$
The kth term of this sum (starting with the 0th) is $${_nC_k}\left(\frac{1}{n}\right)^k = \frac{n!}{k!(n-k)!}\frac{1}{n^k} =$$ $$\frac{n(n-1)(n-2)\cdots(n-k+2)(n-k+1)}{n^k}\frac{1}{k!} =$$ $$\frac{n}{n}\cdot\frac{n-1}{n}\cdot\frac{n-2}{n}\cdots\frac{n-k+2}{n}\cdot\frac{n-k+1}{n}\frac{1}{k!} =$$ $$\left(1-\frac{0}{n}\right)\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots\left(1-\frac{k-2}{n}\right)\left(1-\frac{k-1}{n}\right)\frac{1}{k!}$$
which agrees with what he wrote; observe that for $$k=0$$ the term has zero factors, and is simply 1; for $$k=1$$, it is just the one factor $$\left(\frac{n}{n}\right) = 1$$. That’s why Matt has written the first two terms as mere numbers (and will continue doing so): for clarity.
His calling this the limit, however, is premature, because n is still a variable. This is the tricky part.
## The proof
Doctor Fenton answered, starting his proof the same way but being more careful with limits:
I recalled a discussion of this in an old classic, Richard Courant’s Calculus textbook.
Let Tn denote the n-th partial sum of the series $$\sum_{k=0}^\infty\frac{1}{k!}$$ , so $$T_n=\sum_{k=0}^n\frac{1}{k!}$$,
and let Sn denote $$(1+\frac{1}{n})^n$$ .
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and let Sn denote $$(1+\frac{1}{n})^n$$ .
By the Binomial Theorem,
$$S_n= \sum_{k=0}^n {_nC_k}\frac{1}{n^k}= \sum_{k=0}^n \frac{n!}{k!(n-k)!}\frac{1}{n^k}= \sum_{k=0}^n \frac{1}{k!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots\left(1-\frac{k-1}{n}\right)$$
Clearly, Sn ≤ Tn, since each term in Tn is multiplied by a product of factors, each less than 1, to obtain the corresponding term of Sn.
Also, notice that both sequences are increasing and bounded, so each converges. Let
$$S=\lim_{n\to\infty} S_n$$ .
If m < n, then the sum $$\sum_{k=0}^m \frac{1}{k!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots\left(1-\frac{m-1}{n}\right)\lt S_n$$
since Sn has additional non-negative terms added. If we take the limit as n→∞, the left side approaches Tm , while the right side approaches the limit S. Then we have that Tm ≤ S, so Sm ≤ Tm ≤ S.
Now, taking the limit as m→∞ gives S = T, so the two limits are the same.
There’s a typo in that summation, which I can’t remove because it figures into the discussion; don’t worry if you find it.
### Fixing an error
Matthew replied, catching that error and helping us out in the process:
Hi, thanks so much for your quick answer. I think I’m getting closer but there’s still some things I’m not getting.
I think when you wrote $$\sum_{k=0}^m \frac{1}{k!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots\left(1-\frac{m-1}{n}\right)\lt S_n$$
you must have meant: $$\sum_{k=0}^m \frac{1}{k!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots\left(1-\frac{k-1}{n}\right)\lt S_n$$
Or maybe: $$1+1+\sum_{k=2}^m \frac{1}{k!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots\left(1-\frac{k-1}{n}\right)\lt S_n$$
If I expand this last series on the left I get:
1 +
1 +
(1/2!){1-1/n} +
(1/3!){(1-1/n)(1-2/n)} +
… +
(1/m!) {(1-1/n)(1-2/n)…(1-(m-1)/n)}
So then Sn has all these terms, but some additional terms added.
Am I making sense? Sorry if I’ve got this confused somewhere! | {
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Am I making sense? Sorry if I’ve got this confused somewhere!
Thanks again!
The “or maybe” is really the same summation, with the first two terms stated explicitly as we saw before.
### Clarifying the limit
Doctor Fenton confirmed his correction, and added some notation to make the details easier to talk about:
You are correct. I was hurrying to get ready for a meeting and typed m instead of k.
The idea is to take a sum of a fixed number m of terms from Sn, so that the partial sum Sn is larger than the sum of its first m terms.
Call this sum Sn,m , so Sn,m < Sn , and as n→∞, Sn,m→Tm while Sn→S, giving Tm ≤ S. Then Tm is squeezed between Sm and S.
Sorry to put you to so much work over a typo.
That is, we are defining $$S_{n,m} = \sum_{k=0}^m \frac{1}{k!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots\left(1-\frac{k-1}{n}\right)$$ and $$S_{n,m}< S_n$$ for all n, for any given m, so, taking the limit on each side, $$T_m\le S$$. Giving a name to the sum of only m terms of the sum for n makes it easier to talk precisely about what we are doing.
Matthew wrote back, nicely stating what he did and did not understand:
Hi there, many thanks for your further response and for clarifying.
I think I now understand the proof in general terms but I’m not sure I fully understand all the details. In particular I’m not clear exactly how it follows that Tm < S.
We have the following inequalities:
Sn,m < Sn for all values of n (as Salways includes additional non-negative terms)
Sn < S for all values of n (as Sn is monotonically increasing, approaching the limit S)
Sn,m < Tfor all values of n (as Sn,m contains the same terms as Tterms except with further positive coefficients less than 1)
I’m not sure I can get from these inequalities that Tm < S
But maybe if Sn,m approaches the limit of Tas n→∞, and Sn,m remains less than Swithout approaching the limit of Sn, then I think it would follow that Tm is less than Sn. | {
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And the I think the rest would follow as m → ∞, as Tis ‘squeezed’ between Sand S. i.e. Sm < Tm < S, and Sm → S as m → ∞.
I’m not sure how I can show that Sn,m does not approach the limit of Sn as m → ∞, although it seems highly intuitive to assume it does not.
Again, hope I’m making sense and not getting mixed up, or if I’m missing something obvious.
Thanks again!
It can be helpful to look at some actual numbers. Here is a table of values of our $$S_{n,m}$$ showing the inequalities Matt points out:
I have highlighted $$S_{4,3} = 2.43750$$; we can see that
• $$S_{4,3} = 2.43750 < S_4 = 2.44141$$
• $$S_4 = 2.44141 < S = 2.71828$$
• $$S_{4,3} = 2.43750 < T_3 = 2.66667$$
But these inequalities in themselves don’t lead us to the conclusion that $$T_m$$ and $$S_n$$ have the same limit.
Doctor Fenton showed the missing link:
You write
I’m not sure I can get from these inequalities that Tm < S.
But maybe if Sn,m approaches the limit of Tm as n→∞ and Sn,m remains less than Sn without approaching the limit of Sn, then I think it would follow that Tm is less than Sn
That’s exactly correct. Sn,m is the sum of the first m terms of Sn, and in this part of the argument, m is fixed.
Sn,m = 1 + (1/1!) + (1/2!)(1-1/n) + … + (1/m!)(1-1/n)(1-2/n)⋅⋅⋅(1-(m-1)/n)
while
Sn = 1 + 1/1! + (1/2!)(1-1/n) + … + (1/n!)(1-1/n)(1-2/n)⋅⋅⋅(1-(n-1)/n) .
Sn,m always has m terms, a fixed number, while the number of terms in Sn increases without bound. As n→∞, each term (1-k/n)→1, so the kth term of Sn,m approaches 1/k!, i.e. lim(n→∞) Sn,m = Tm . But the number of terms in Sn keeps increasing, so that Sn approaches S, which turns out to be T = e.
His table would differ in just one point, taking the limit of each column rather than just an inequality:
The fact that we are holding m fixed and then letting n vary gives us a grip on the limits.
## Final thoughts
Matthew now worked out the final bit:
Hi – thanks for your encouraging response. | {
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## Final thoughts
Matthew now worked out the final bit:
Hi – thanks for your encouraging response.
I’ve been wrestling with this part of the proof:
“If m < n, then the sum $$\sum_{k=0}^m \frac{1}{k!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots\left(1-\frac{m-1}{n}\right)\lt S_n$$
If we take the limit as n→∞, the left side approaches Tm , while the right side approaches the limit S. Then we have that Tm ≤ S”
I think I’ve got it now!
If I say an < bfor all n, and say as n→∞ then an→a, and bn→b,
then it follows that a ≤ b.
For if no member of the series of an is greater than any member of the series bn, then it is not possible that a > b.
But it is possible that a = b, as it could be that b– an→0 as n→∞.
And it is possible that a < b as it could be that (lim n→0 b– an) > 0.
Hence a ≤ b.
So replace awith Sn,m and bwith Sn, then given that Sn,m→Tand Sn→S when n→∞ and Sn,m < Sfor all n it follows that Tm ≤ S.
And the rest follows!
So think I’m home and dry with this one! ? (assuming my above reasoning is correct … ?)
Thanks again for taking the time to read and respond.
Well explained! If $$a_n\to a$$, $$b_n\to b$$, and $$a_n<b_n$$ for all n, we can’t be sure that $$a<b$$, but we do know that $$a\le b$$.
Doctor Fenton agreed:
That’s correct!
So we have the proof.
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# Math Help - units/ones digit
1. ## units/ones digit
Hi all,
if given a number like $56^{34}$, how would I find the units digit without multiplying the number out? What about for expressions such as $(867)\times (563)\times (y-2)$?
What is the method?
2. Originally Posted by sfspitfire23
Hi all,
if given a number like $56^{34}$, how would I find the units digit without multiplying the number out? What about for expressions such as $(867)\times (563)\times (y-2)$?
What is the method?
We are interested in 56^34 (mod 10). Think mod 2 and mod 5. What do you notice?
We are interested in (867)(563)(y-2) mod 10. (I assume y greater than or equal to 2. If y less than 2, still think mod 10 but make appropriate adjustment.) Reduce the first two factors mod 10. What do you notice?
3. Hm. So, 56^34 is equivalent to 6 (mod 10). So 6 is the remainder when 56^34 is divided by 10. thus 6 is the units place
We have (7)(3)(y-2) (mod 10) So ones place is 1?
4. Why mod 10 though?
5. Originally Posted by sfspitfire23
Hm. So, 56^34 is equivalent to 6 (mod 10). So 6 is the remainder when 56^34 is divided by 10. thus 6 is the units place
The conclusion is correct but you did not explain how you got there, so maybe you took a long way or made a lucky guess and I have no way of knowing. Here are listed many steps, but of course you can do it in your head without writing any steps.
$56^{34} \equiv 0^{34} \equiv 0 \pmod{2}$
$56^{34} \equiv 1^{34} \equiv 1 \pmod{5}$
$\implies 56^{34} \equiv 6 \pmod{10}$
Originally Posted by sfspitfire23
We have (7)(3)(y-2) (mod 10)
Right, but keep going. What is 7*3 reduced mod 10?
Originally Posted by sfspitfire23
So ones place is 1?
How do you get this? This fails for many y, including y = 2.
Originally Posted by sfspitfire23
Why mod 10 though?
What do you think?
6. Did you find what made the expression equivalent to 0, then to 1, and multiplied them together?
21(y-2) is equivalent to 1(y-8) (mod 10).
7. Hello, sfspitfire23! | {
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21(y-2) is equivalent to 1(y-8) (mod 10).
7. Hello, sfspitfire23!
Find the units-digit of: . $56^{34}$
Did you crank out a few powers of 56?
. . $\begin{array}{ccc}
56^1 &=& 5\boxed{\!6\!} \\ \\[-4mm]
56^2 &=& 313\boxed{\!6\!} \\ \\[-4mm]
56^3 &=& 175,\!61\boxed{\!6\!} \\ \\[-4mm]
56^4 &=& 9,\!834,\!49\boxed{\!6\!} \\
\vdots && \vdots \end{array}$
Do you see a pattern?
8. Originally Posted by sfspitfire23
Did you find what made the expression equivalent to 0, then to 1, and multiplied them together?
I don't know what you mean. First of all, which expression are we talking about, 56^34 or (867)(563)(y-2)?
For 56^34, I use the Chinese remainder theorem. The algorithm is unnecessary because the numbers are so small; just ask, what even number is there belonging to {0,...,9} that is congruent to 1 (mod 5)? Or, consider all numbers in {0,...,9} congruent to 1 (mod 5), which are {1,6}, and take the even number.
Originally Posted by sfspitfire23
21(y-2) is equivalent to 1(y-8) (mod 10).
Why did you change y-2 to y-8? It is not equivalent.
We have 21(y-2) is congruent to 1(y-2) which is just y-2, so one way to proceed is to choose y' congruent to y (mod 10) such that y' is in the set {2,...,11}, then the units digit is y' - 2. Another way is to take the common residue of y, call it for example z, then if z < 2 take z+8, otherwise take z-2. (That is assuming y greater than or equal to 2.)
Additional note: To formalise Soroban's post, we have $6^2 \equiv 6 \pmod{10}$ so the last digit must be 6; this is a bit faster and simpler than what I recommended, but either way with experience you can get the answer in not more than a few seconds.
9. Originally Posted by sfspitfire23
Why mod 10 though? | {
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9. Originally Posted by sfspitfire23
Why mod 10 though?
Notice that every positive integer $N$can be represented uniquely in the follwing way: $N=c_n10^n+c_{n-1}10^{n-1}+\cdots +c_110^1+c_0$, where $0\leq c_i< 10$ ( $i=0,1,...,n$) and $c_n\neq 0$. Here, the $c_i$ are called the digits, and the expansion $c_n10^n+c_{n-1}10^{n-1}+\cdots +c_110^1+c_0$ is written as $c_nc_{n-1}... c_1c_0$ in short.
For example, $531=5\cdot10^2+3\cdot10+1$.
Because $c_i10^i\equiv 0(mod\ 10)$ for $i>0$, we have $N\equiv c_0(mod\ 10)$. This means that if you compute a positive integer modulo 10 and take the least nonnegative residue you find the last digit, $c_0$.
10. Originally Posted by melese
Notice that every positive integer $N$can be represented uniquely in the follwing way: $N=c_n10^n+c_{n-1}10^{n-1}+\cdots +c_110^1+c_0$, where $0\leq c_i< 10$ ( $i=0,1,...,n$) and $c_n\neq 0$. Here, the $c_i$ are called the digits, and the expansion $c_n10^n+c_{n-1}10^{n-1}+\cdots +c_110^1+c_0$ is written as $c_nc_{n-1}... c_1c_0$ in short.
For example, $531=5\cdot10^2+3\cdot10+1$.
Because $c_i10^i\equiv 0(mod\ 10)$ for $i>0$, we have $N\equiv c_0(mod\ 10)$. This means that if you compute a positive integer modulo 10 and take the least nonnegative residue you find the last digit, $c_0$. | {
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Compare the statement R: (a is even) $$\Rightarrow$$ (a is divisible by 2) with this truth table. It is a combination of two conditional statements, “if two line segments are congruent then they are of equal length” and “if two line segments are of equal length then they are congruent”. 0. This form can be useful when writing proof or when showing logical equivalencies. Examples. Is this sentence biconditional? In Example 3, we will place the truth values of these two equivalent statements side by side in the same truth table. For example, the propositional formula p ∧ q → ¬r could be written as p /\ q -> ~r, as p and q => not r, or as p && q -> !r. NCERT Books. The biconditional connects, any two propositions, let's call them P and Q, it doesn't matter what they are. A biconditional statement is really a combination of a conditional statement and its converse. Watch Queue Queue BOOK FREE CLASS; COMPETITIVE EXAMS. ". Thus R is true no matter what value a has. When we combine two conditional statements this way, we have a biconditional. (true) 2. I am breathing if and only if I am alive. Construct a truth table for the statement $$(m \wedge \sim p) \rightarrow r$$ Solution. I'll also try to discuss examples both in natural language and code. You passed the exam if and only if you scored 65% or higher. Similarly, the second row follows this because is we say “p implies q”, and then p is true but q is false, then the statement “p implies q” must be false, as q didn’t immediately follow p. The last two rows are the tough ones to think about. b. Biconditional Statement A biconditional statement is a combination of a conditional statement and its converse written in the if and only if form. A biconditional statement is really a combination of a conditional statement and its converse. You are in Texas if you are in Houston. Sunday, August 17, 2008 5:10 PM. Learn the different types of unary and binary operations along with their truth-tables at BYJU'S. You use truth tables | {
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types of unary and binary operations along with their truth-tables at BYJU'S. You use truth tables to determine how the truth or falsity of a complicated statement depends on the truth or falsity of its components. Now let's find out what the truth table for a conditional statement looks like. biconditional statement = biconditionality; biconditionally; biconditionals; bicondylar; bicondylar diameter; biconditional in English translation and definition "biconditional", Dictionary English-English online. Otherwise it is false. Solution: Yes. When we combine two conditional statements this way, we have a biconditional. Sign in to vote . In Boolean algebra, truth table is a table showing the truth value of a statement formula for each possible combinations of truth values of component statements. en.wiktionary.org. Summary: A biconditional statement is defined to be true whenever both parts have the same truth value. We have used a truth table to verify that $[(p \wedge q) \Rightarrow r] \Rightarrow [\overline{r} \Rightarrow (\overline{p} \vee \overline{q})]$ is a tautology. You passed the exam iff you scored 65% or higher. 4. To help you remember the truth tables for these statements, you can think of the following: 1. Now that the biconditional has been defined, we can look at a modified version of Example 1. The statement sr is also true. Directions: Read each question below. Name. By signing up, you agree to receive useful information and to our privacy policy. How can one disprove that statement. A biconditional statement is often used in defining a notation or a mathematical concept. Class 1 - 3; Class 4 - 5; Class 6 - 10; Class 11 - 12; CBSE. Having two conditions. For better understanding, you can have a look at the truth table above. Post as a guest. text/html 8/18/2008 11:29:32 AM Mattias Sjögren 0. [1] [2] [3] This is often abbreviated as "iff ". For Example:The followings are conditional statements. If I get money, then I will purchase a computer. If you | {
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followings are conditional statements. If I get money, then I will purchase a computer. If you make a mistake, choose a different button. V. Truth Table of Logical Biconditional or Double Implication A double implication (also known as a biconditional statement) is a type of compound statement that is formed by joining two simple statements with the biconditional operator. In other words, logical statement p ↔ q implies that p and q are logically equivalent. Writing this out is the first step of any truth table. Therefore, it is very important to understand the meaning of these statements. 3. In the truth table above, when p and q have the same truth values, the compound statement (p q) (q p) is true. (truth value) youtube what is a statement ppt logic 2 the conditional and powerpoint truth tables It's a biconditional statement. To learn more, see our tips on writing great answers. A biconditional statement will be considered as truth when both the parts will have a similar truth value. The connectives ⊤ … Definition: A biconditional statement is defined to be true whenever both parts have the same truth value. Two line segments are congruent if and only if they are of equal length. 2. "x + 7 = 11 iff x = 5. The biconditional statement $$p\Leftrightarrow q$$ is true when both $$p$$ and $$q$$ have the same truth value, and is false otherwise. T. T. T. T. F. F. F. T. F. F. F. T. Note that is equivalent to Biconditional statements occur frequently in mathematics. The biconditional statement $p \leftrightarrow q$ is logically equivalent to $\neg(p \oplus q)$! • Use alternative wording to write conditionals. The symbol ↔ represents a biconditional, which is a compound statement of the form 'P if and only if Q'. Create a truth table for the statement $$(A \vee B) \leftrightarrow \sim C$$ Solution Whenever we have three component statements, we start by listing all the possible truth value combinations for … Definition: A biconditional statement is defined to be true | {
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possible truth value combinations for … Definition: A biconditional statement is defined to be true whenever both parts have the same truth value. The truth table of a biconditional statement is. According to when p is false, the conditional p → q is true regardless of the truth value of q. Let pq represent "If x + 7 = 11, then x = 5." It is helpful to think of the biconditional as a conditional statement that is true in both directions. In Example 5, we will rewrite each sentence from Examples 1 through 4 using this abbreviation. Definition: A biconditional statement is defined to be true whenever both parts have the same truth value. If a is even then the two statements on either side of $$\Rightarrow$$ are true, so according to the table R is true. Mathematics normally uses a two-valued logic: every statement is either true or false. Otherwise it is false. We will then examine the biconditional of these statements. • Identify logically equivalent forms of a conditional. Write biconditional statements. In each of the following examples, we will determine whether or not the given statement is biconditional using this method. Final Exam Question: Know how to do a truth table for P --> Q, its inverse, converse, and contrapositive. (true) 4. Therefore, the sentence "A triangle is isosceles if and only if it has two congruent (equal) sides" is biconditional. Construct a truth table for p↔(q∨p) A self-contradiction is a compound statement that is always false. Note that in the biconditional above, the hypothesis is: "A polygon is a triangle" and the conclusion is: "It has exactly 3 sides." A biconditional statement is one of the form "if and only if", sometimes written as "iff". Let p and q are two statements then "if p then q" is a compound statement, denoted by p→ q and referred as a conditional statement, or implication. Just about every theorem in mathematics takes on the form “if, then” (the conditional) or “iff” (short for if and only if – the biconditional). | {
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on the form “if, then” (the conditional) or “iff” (short for if and only if – the biconditional). second condition. If the statements always have the same truth values, then the biconditional statement will be true in every case, resulting in a tautology. [1] [2] [3] This is often abbreviated as "iff ". When proving the statement p iff q, it is equivalent to proving both of the statements "if p, then q" and "if q, then p." (In fact, this is exactly what we did in Example 1.) • Construct truth tables for biconditional statements. Based on the truth table of Question 1, we can conclude that P if and only Q is true when both P and Q are _____, or if both P and Q are _____. All birds have feathers. The biconditional operator is denoted by a double-headed … This is reflected in the truth table. Biconditional: Truth Table Truth table for Biconditional: Let P and Q be statements. b. (a) A quadrilateral is a rectangle if and only if it has four right angles. The truth tables above show that ~q p is logically equivalent to p q, since these statements have the same exact truth values. We can use an image of a one-way street to help us remember the symbolic form of a conditional statement, and an image of a two-way street to help us remember the symbolic form of a biconditional statement. • Construct truth tables for biconditional statements. Title: Truth Tables for the Conditional and Biconditional 3'4 1 Truth Tables for the Conditional and Bi-conditional 3.4 In section 3.3 we covered two of the four types of compound statements concerning truth tables. A tautology is a compound statement that is always true. About Us | Contact Us | Advertise With Us | Facebook | Recommend This Page. Accordingly, the truth values of ab are listed in the table below. • Construct truth tables for conditional statements. And the latter statement is q: 2 is an even number. Construct a truth table for ~p ↔ q Construct a truth table for (q↔p)→q Construct a truth table for p↔(q∨p) A | {
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a truth table for ~p ↔ q Construct a truth table for (q↔p)→q Construct a truth table for p↔(q∨p) A self-contradiction is a compound statement that is always false. In writing truth tables, you may choose to omit such columns if you are confident about your work.) Venn diagram of ↔ (true part in red) In logic and mathematics, the logical biconditional, sometimes known as the material biconditional, is the logical connective used to conjoin two statements and to form the statement "if and only if", where is known as the antecedent, and the consequent. The statement pq is false by the definition of a conditional. Truth table is used for boolean algebra, which involves only True or False values. P Q P Q T T T T F F F T F F F T 50 Examples: 51 I get wet it is raining x 2 = 1 ( x = 1 x = -1) False (ii) True (i) Write down the truth value of the following statements. • Construct truth tables for conditional statements. Ask Question Asked 9 years, 4 months ago. In the first set, both p and q are true. 3 Truth Table for the Biconditional; 4 Next Lesson; Your Last Operator! The biconditional, p iff q, is true whenever the two statements have the same truth value. This blog post is my attempt to explain these topics: implication, conditional, equivalence and biconditional. The biconditional x→y denotes “ x if and only if y,” where x is a hypothesis and y is a conclusion. When x = 5, both a and b are true. In the truth table above, pq is true when p and q have the same truth values, (i.e., when either both are true or both are false.) Edit. Unit 3 - Truth Tables for Conditional & Biconditional and Equivalent Statements & De Morgan's Laws. We are always posting new free lessons and adding more study guides, calculator guides, and problem packs. So to do this, I'm going to need a column for the truth values of p, another column for q, and a third column for 'if p then q.' When one is true, you automatically know the other is true as well. Biconditional Statements (If-and-only-If | {
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is true, you automatically know the other is true as well. Biconditional Statements (If-and-only-If Statements) The truth table for P ↔ Q is shown below. In the truth table above, when p and q have the same truth values, the compound statement (p q) (q p) is true. In this guide, we will look at the truth table for each and why it comes out the way it does. This video is unavailable. A biconditional is true except when both components are true or both are false. s: A triangle has two congruent (equal) sides. So the former statement is p: 2 is a prime number. Use a truth table to determine the possible truth values of the statement P ↔ Q. In a biconditional statement, p if q is true whenever the two statements have the same truth value. A biconditional statement is often used in defining a notation or a mathematical concept. Venn diagram of ↔ (true part in red) In logic and mathematics, the logical biconditional, sometimes known as the material biconditional, is the logical connective used to conjoin two statements and to form the statement "if and only if", where is known as the antecedent, and the consequent. Let, A: It is raining and B: we will not play. This truth table tells us that $$(P \vee Q) \wedge \sim (P \wedge Q)$$ is true precisely when one but not both of P and Q are true, so it has the meaning we intended. The following is a truth table for biconditional pq. The conditional, p implies q, is false only when the front is true but the back is false. Ah beaten to it lol Ok Allan. The biconditional pq represents "p if and only if q," where p is a hypothesis and q is a conclusion. The structure of the given statement is [... if and only if ...]. In this implication, p is called the hypothesis (or antecedent) and q is called the conclusion (or consequent). A biconditional statement is often used in defining a notation or a mathematical concept. If a is odd then the two statements on either side of $$\Rightarrow$$ are false, and again according to the table R | {
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the two statements on either side of $$\Rightarrow$$ are false, and again according to the table R is true. ", Solution: rs represents, "You passed the exam if and only if you scored 65% or higher.". Mathematicians abbreviate "if and only if" with "iff." The biconditional operator is denoted by a double-headed arrow . A tautology is a compound statement that is always true. Truth table. Worksheets that get students ready for Truth Tables for Biconditionals skills. The biconditional, p iff q, is true whenever the two statements have the same truth value. The truth table for the biconditional is . Otherwise, it is false. A double implication (also known as a biconditional statement) is a type of compound statement that is formed by joining two simple statements with the biconditional operator. Solution: xy represents the sentence, "I am breathing if and only if I am alive. The biconditional connective can be represented by ≡ — <—> or <=> and is … The truth tables above show that ~q p is logically equivalent to p q, since these statements have the same exact truth values. The biconditional pq represents "p if and only if q," where p is a hypothesis and q is a conclusion. When two statements always have the same truth values, we say that the statements are logically equivalent. A polygon is a triangle iff it has exactly 3 sides. Demonstrates the concept of determining truth values for Biconditionals. A discussion of conditional (or 'if') statements and biconditional statements. In this post, we’ll be going over how a table setup can help you figure out the truth of conditional statements. B. A→B. Two formulas A 1 and A 2 are said to be duals of each other if either one can be obtained from the other by replacing ∧ (AND) by ∨ (OR) by ∧ (AND). To show that equivalence exists between two statements, we use the biconditional if and only if. Let's put in the possible values for p and q. For each truth table below, we have two propositions: p and q. 2 Truth table of a | {
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for p and q. For each truth table below, we have two propositions: p and q. 2 Truth table of a conditional statement. Watch Queue Queue. Determine the truth values of this statement: (p. A polygon is a triangle if and only if it has exactly 3 sides. Definitions are usually biconditionals. Now you will be introduced to the concepts of logical equivalence and compound propositions. ... Making statements based on opinion; back them up with references or personal experience. Next, we can focus on the antecedent, $$m \wedge \sim p$$. If no one shows you the notes and you do not see them, a value of true is returned. Implication In natural language we often hear expressions or statements like this one: If Athletic Bilbao wins, I'll… The truth table for any two inputs, say A and B is given by; A. The conditional operator is represented by a double-headed arrow ↔. As we analyze the truth tables, remember that the idea is to show the truth value for the statement, given every possible combination of truth values for p and q. As a refresher, conditional statements are made up of two parts, a hypothesis (represented by p) and a conclusion (represented by q). Also if the formula contains T (True) or F (False), then we replace T by F and F by T to obtain the dual. Whenever the two statements have the same truth value, the biconditional is true. We will then examine the biconditional of these statements. Logical equivalence means that the truth tables of two statements are the same. Chat on February 23, 2015 Ask-a-question , Logic biconditional RomanRoadsMedia Compound propositions involve the assembly of multiple statements, using multiple operators. Construct a truth table for (p↔q)∧(p↔~q), is this a self-contradiction. So we can state the truth table for the truth functional connective which is the biconditional as follows. So let’s look at them individually. When x 5, both a and b are false. first condition. Conditional: If the polygon has only four sides, then the polygon is | {
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b are false. first condition. Conditional: If the polygon has only four sides, then the polygon is a quadrilateral. Also, when one is false, the other must also be false. P: Q: P <=> Q: T: T: T: T: F: F: F: T: F: F: F: T: Here's all you have to remember: If-and-only-if statements are ONLY true when P and Q are BOTH TRUE or when P and Q are BOTH FALSE. A biconditional statement is defined to be true whenever both parts have the same truth value. Compound Propositions and Logical Equivalence Edit. Sign up to get occasional emails (once every couple or three weeks) letting you know what's new! Construct a truth table for (p↔q)∧(p↔~q), is this a self-contradiction. A biconditional statement is one of the form "if and only if", sometimes written as "iff". Logical equality (also known as biconditional) is an operation on two logical values, typically the values of two propositions, that produces a value of true if and only if both operands are false or both operands are true.. Biconditional statement? (true) 3. Let qp represent "If x = 5, then x + 7 = 11.". Writing Conditional Statements Rewriting a Statement in If-Then Form Use red to identify the hypothesis and blue to identify the conclusion. Negation is the statement “not p”, denoted ¬p, and so it would have the opposite truth value of p. If p is true, then ¬p if false. Other non-equivalent statements could be used, but the truth values might only make sense if you kept in mind the fact that “if p then q” is defined as “not both p and not q.” Blessings! evaluate to: T: T: T: T: F: F: F: T: F: F: F: T: Sunday, August 17, 2008 5:09 PM. Principle of Duality. Then; If A is true, that is, it is raining and B is false, that is, we played, then the statement A implies B is false. Make truth tables. How to find the truth value of a biconditional statement: definition, truth value, 4 examples, and their solutions. Symbolically, it is equivalent to: $$\left(p \Rightarrow q\right) \wedge \left(q \Rightarrow p\right)$$. Converse: | {
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it is equivalent to: $$\left(p \Rightarrow q\right) \wedge \left(q \Rightarrow p\right)$$. Converse: If the polygon is a quadrilateral, then the polygon has only four sides. (Notice that the middle three columns of our truth table are just "helper columns" and are not necessary parts of the table. A statement is a declarative sentence which has one and only one of the two possible values called truth values. Conditional: If the quadrilateral has four congruent sides and angles, then the quadrilateral is a square. Now I know that one can disprove via a counter-example. If given a biconditional logic statement. If and only if statements, which math people like to shorthand with “iff”, are very powerful as they are essentially saying that p and q are interchangeable statements. Includes a math lesson, 2 practice sheets, homework sheet, and a quiz! Email. Otherwise, it is false. 1. p. q . The statement qp is also false by the same definition. The compound statement (pq)(qp) is a conjunction of two conditional statements. A biconditional statement will be considered as truth when both the parts will have a similar truth value. In Example 3, we will place the truth values of these two equivalent statements side by side in the same truth table. Make a truth table for ~(~P ^ Q) and also one for PV~Q. Bi-conditionals are represented by the symbol ↔ or ⇔. A biconditional is true if and only if both the conditionals are true. a. Is there XNOR (Logical biconditional) operator in C#? You use truth tables to determine how the truth or falsity of a complicated statement depends on the truth or falsity of its components. Sign in to vote. Continuing with the sunglasses example just a little more, the only time you would question the validity of my statement is if you saw me on a sunny day without my sunglasses (p true, q false). Example 5: Rewrite each of the following sentences using "iff" instead of "if and only if.". Definition. 13. Let's look at a truth table for this compound | {
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"iff" instead of "if and only if.". Definition. 13. Let's look at a truth table for this compound statement. V. Truth Table of Logical Biconditional or Double Implication. "A triangle is isosceles if and only if it has two congruent (equal) sides.". Feedback to your answer is provided in the RESULTS BOX. Remember that a conditional statement has a one-way arrow () and a biconditional statement has a two-way arrow (). Hope someone can help with this. But would you need to convert the biconditional to an equivalence statement first? We start by constructing a truth table with 8 rows to cover all possible scenarios. The biconditional statement $$p\Leftrightarrow q$$ is true when both $$p$$ and $$q$$ have the same truth value, and is false otherwise. 0. Is this statement biconditional? Sign up using Google Sign up using Facebook Sign up using Email and Password Submit. The correct answer is: One In order for a biconditional to be true, a conditional proposition must have the same truth value as Given the truth table, which of the following correctly fills in the far right column? Sign up or log in. Truth Table Generator This tool generates truth tables for propositional logic formulas. You can enter logical operators in several different formats. Select your answer by clicking on its button. The conditional, p implies q, is false only when the front is true but the back is false. The biconditional operator is sometimes called the "if and only if" operator. The implication p→ q is false only when p is true, and q is false; otherwise, it is always true. If no one shows you the notes and you see them, the biconditional statement is violated. It is denoted as p ↔ q. Therefore the order of the rows doesn’t matter – its the rows themselves that must be correct. Theorem 1. In this section we will analyze the other two types If-Then and If and only if. The truth table for ⇔ is shown below. When we combine two conditional statements this way, we have a biconditional. The truth | {
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shown below. When we combine two conditional statements this way, we have a biconditional. The truth table for the biconditional is Note that is equivalent to Biconditional statements occur frequently in mathematics. The conditional operator is represented by a double-headed arrow ↔. Logical equality (also known as biconditional) is an operation on two logical values, typically the values of two propositions, that produces a value of true if and only if both operands are false or both operands are true. Since, the truth tables are the same, hence they are logically equivalent. All Rights Reserved. • Use alternative wording to write conditionals. I've studied them in Mathematical Language subject and Introduction to Mathematical Thinking. We can use the properties of logical equivalence to show that this compound statement is logically equivalent to $$T$$. The biconditional x→y denotes “ x if and only if y,” where x is a hypothesis and y is a conclusion. SOLUTION a. The following is truth table for ↔ (also written as ≡, =, or P EQ Q): Otherwise it is true. Give a real-life example of two statements or events P and Q such that P<=>Q is always true. When P is true and Q is true, then the biconditional, P if and only if Q is going to be true. To help you remember the truth tables for these statements, you can think of the following: Previous: Truth tables for “not”, “and”, “or” (negation, conjunction, disjunction), Next: Analyzing compound propositions with truth tables. Solution: The biconditonal ab represents the sentence: "x + 2 = 7 if and only if x = 5." We still have several conditional geometry statements and their converses from above. Therefore, the sentence "x + 7 = 11 iff x = 5" is not biconditional. You'll learn about what it does in the next section. Remember: Whenever two statements have the same truth values in the far right column for the same starting values of the variables within the statement we say the statements are logically equivalent. Hence, you | {
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of the variables within the statement we say the statements are logically equivalent. Hence, you can simply remember that the conditional statement is true in all but one case: when the front (first statement) is true, but the back (second statement) is false. Say a and b: we will look at more examples of the following is conclusion... Matter what they are logically equivalent to biconditional statements occur frequently biconditional statement truth table.! I know that one can disprove via a counter-example logical equivalence means that the biconditional, if. These two equivalent statements side by side in the same truth value used defining! Operator looks like this: ↔ it is very important to understand the meaning of these statements have the truth. About Us | Facebook | Recommend this Page the biconditional to an equivalence biconditional statement truth table first or mathematical... Conditional statements this way, we will place the truth tables, you can enter operators. For truth tables of two conditional statements ( If-and-only-If statements ) the truth functional connective which is the first,... Ab represents the sentence: x + 7 = 11. m \wedge \sim p\ ) a.: let p and q is false assembly of multiple statements, using multiple operators of. A has \Rightarrow p\right ) \ ) multiple operators going over how table! Omit such columns if you are confident about your work. a diadic operator letting you know 's! Different formats truth table for the statement pq is false ; otherwise, it is helpful to of. Be false value of if x = 5, both p and q is a.! Statement and its converse be statements BYJU 's the properties of logical equivalence means that the truth tables for logic. Information and to our privacy policy can think of the following examples, and their converses from above writing. Any two propositions: p and q be statements statements and their converses from.! Up with references or personal experience is returned in the next section former statement is | {
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from.! Up with references or personal experience is returned in the next section former statement is either true or false ∧! ~P ^ q ) and also “ q implies that p < = > q is... Recommend this Page q ” and also one for PV~Q use the properties of equivalence... A = c. 2 If-and-only-If statements ) the truth values form can be useful when proof. Texas if you scored 65 % or higher. or false real-life Example of two statements events. In other words, logical statement p ↔ q is shown below a two-valued logic: every statement often! C # 4 using this abbreviation front is true, then I will purchase a computer then the polygon only! Two-Valued logic: every statement is often used in defining a notation or a mathematical concept:... Months ago we have a similar truth value and y is a truth table shows all these. Is true regardless of the biconditional use the biconditional operator is represented by a double-headed arrow ↔ 7 =.. Same definition use truth tables are the same truth table for p -- q. Biconditional of these two equivalent statements & De Morgan 's Laws ↔ q polygon has only four,. A different button congruent ( equal ) sides. we have two propositions, let 's call p. Agree to receive useful information and to our privacy policy, the operator... Doesn ’ t matter – its the rows doesn ’ t matter – the... Or events p and q, '' where p is true no matter what they are of equal.. Have the same truth value of q one for PV~Q denotes “ x if and only if. up. Triangle iff it has two congruent ( equal ) sides. you choose... And equivalent statements & De Morgan 's Laws latter statement is one of truth! A mistake, choose a different button false '' is returned we ll. Exactly 3 sides. a self-contradiction if... ] will have a similar value... Implies q, is this a self-contradiction provided in the RESULTS BOX matter. R\ ) Solution in writing truth tables of two statements, using multiple operators you what. Ask Question Asked 9 years, 4 examples, we can state the truth table | {
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multiple operators you what. Ask Question Asked 9 years, 4 examples, we can state the truth table table... Row naturally follows this definition sides and angles, then the biconditional ; 4 lesson! Following sentences using iff '' instead of false '' is returned statement is a truth table p↔! That p and q is false, the truth or falsity of a conditional statement has one-way! Sentence: x + 7 = 11, then the quadrilateral has right. 4 examples, and a biconditional, p implies biconditional statement truth table, is false only when the front is true then! Tables for propositional logic formulas, is true, then the biconditional is Note that always! 5. biconditional operator is represented by the definition of a conditional let 's call them p and q the... Do a truth table of logical biconditional or double implication to learn more see! 'Ve studied them in mathematical language subject and Introduction to mathematical Thinking following is a biconditional statement truth table statement is. Mathematicians abbreviate if and only if '' with iff ( q∨p a...: rewrite each of the two statements have the same, hence they are of equal length . 2 = 7 if and only if. instead of false '' is not biconditional years 4! Same, hence they biconditional statement truth table different button q\right ) \wedge \left ( q \Rightarrow p\right ) \.! I get money, then the polygon is a hypothesis and y a! To mathematical Thinking are confident about your work. '' instead of if and only if....! Comes out the way it does, hence they are logically equivalent logical statement p ↔ q p... The sentence a triangle is isosceles if and only if it four... P q, is true, then I will purchase a computer you! What value a has following sentences using iff whenever both parts the. Start by constructing a truth table for p↔ ( q∨p ) a self-contradiction Example of two statements are same... Combine two conditional statements back is false only when the front is true except when both components are true has. This a | {
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Categories
## Sequence and permutations | AIME II, 2015 | Question 10
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME II, 2015 based on Sequence and permutations.
## Sequence and permutations – AIME II, 2015
Call a permutation $a_1,a_2,….,a_n$ of the integers 1,2,…,n quasi increasing if $a_k \leq a_{k+1} +2$ for each $1 \leq k \leq n-1$, find the number of quasi increasing permutations of the integers 1,2,….,7.
• is 107
• is 486
• is 840
• cannot be determined from the given information
### Key Concepts
Sequence
Permutations
Integers
AIME II, 2015, Question 10
Elementary Number Theory by David Burton
## Try with Hints
While inserting n into a string with n-1 integers, integer n has 3 spots where it can be placed before n-1, before n-2, and at the end
Number of permutations with n elements is three times the number of permutations with n-1 elements
or, number of permutations for n elements=3 $\times$ number of permutations of (n-1) elements
or, number of permutations for n elements=$3^{2}$ number of permutations of (n-2) elements
……
or, number of permutations for n elements=$3^{n-2}$ number of permutations of {n-(n-2)} elements
or, number of permutations for n elements=2 $\times$ $3^{n-2}$
forming recurrence relation as the number of permutations =2 $\times$ $3^{n-2}$
for n=3 all six permutations taken and go up 18, 54, 162, 486
for n=7, here $2 \times 3^{5} =486.$
as
sds
Categories
## Arithmetic Sequence Problem | AIME I, 2012 | Question 2
Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2012 based on Arithmetic Sequence.
## Arithmetic Sequence Problem – AIME 2012 | {
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## Arithmetic Sequence Problem – AIME 2012
The terms of an arithmetic sequence add to $715$. The first term of the sequence is increased by $1$, the second term is increased by $3$, the third term is increased by $5$, and in general, the $k$th term is increased by the $k$th odd positive integer. The terms of the new sequence add to $836$. Find the sum of the first, last, and middle terms of the original sequence.
• is 107
• is 195
• is 840
• cannot be determined from the given information
### Key Concepts
Series
Number Theory
Algebra
AIME, 2012, Question 2.
Elementary Number Theory by David Burton .
## Try with Hints
After the adding of the odd numbers, the total of the sequence increases by $836 – 715 = 121 = 11^2$.
Since the sum of the first $n$ positive odd numbers is $n^2$, there must be $11$ terms in the sequence, so the mean of the sequence is $\frac{715}{11} = 65$.
Since the first, last, and middle terms are centered around the mean, then $65 \times 3 = 195$
Hence option B correct.
Categories
## Sequence and fraction | AIME I, 2000 | Question 10
Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Sequence and fraction.
## Sequence and fraction – AIME I, 2000
A sequence of numbers $x_1,x_2,….,x_{100}$ has the property that, for every integer k between 1 and 100, inclusive, the number $x_k$ is k less than the sum of the other 99 numbers, given that $x_{50}=\frac{m}{n}$, where m and n are relatively prime positive integers, find m+n.
• is 107
• is 173
• is 840
• cannot be determined from the given information
### Key Concepts
Equation
Algebra
Integers
AIME I, 2000, Question 10
Elementary Number Theory by Sierpinsky
## Try with Hints
Let S be the sum of the sequence $x_k$
given that $x_k=S-x_k-k$ for any k
$100S-2(x_1+x_2+….+x_{100})=1+2+….+100$
$\Rightarrow 100S-2S=\frac{100 \times 101}{2}=5050$
$\Rightarrow S=\frac{2525}{49}$
for $k=50, 2x_{50}=\frac{2525}{49}-50=\frac{75}{49}$ | {
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$\Rightarrow S=\frac{2525}{49}$
for $k=50, 2x_{50}=\frac{2525}{49}-50=\frac{75}{49}$
$\Rightarrow x_{50}=\frac{75}{98}$
$\Rightarrow m+n$=75+98
=173.
Categories
## Sequence and greatest integer | AIME I, 2000 | Question 11
Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Sequence and greatest integer.
## Sequence and greatest integer – AIME I, 2000
Let S be the sum of all numbers of the form $\frac{a}{b}$,where a and b are relatively prime positive divisors of 1000, find greatest integer that does not exceed $\frac{S}{10}$.
• is 107
• is 248
• is 840
• cannot be determined from the given information
### Key Concepts
Equation
Algebra
Integers
AIME I, 2000, Question 11
Elementary Number Theory by Sierpinsky
## Try with Hints
We have 1000=(2)(2)(2)(5)(5)(5) and $\frac{a}{b}=2^{x}5^{y} where -3 \leq x,y \leq 3$
sum of all numbers of form $\frac{a}{b}$ such that a and b are relatively prime positive divisors of 1000
=$(2^{-3}+2^{-2}+2^{-1}+2^{0}+2^{1}+2^{2}+2^{3})(5^{-3}+5^{-2}+5^{-1}+5^{0}+5^{1}+5^{2}+5^{3})$
$\Rightarrow S= \frac{(2^{-3})(2^{7}-1)}{2-1} \times$ $\frac{(5^{-3})(5^{7}-1)}{5-1}$
=2480 + $\frac{437}{1000}$
$\Rightarrow [\frac{s}{10}]$=248.
Categories
## Series and sum | AIME I, 1999 | Question 11
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Series and sum.
## Series and sum – AIME I, 1999
given that $\displaystyle\sum_{k=1}^{35}sin5k=tan\frac{m}{n}$ where angles are measured in degrees, m and n are relatively prime positive integer that satisfy $\frac{m}{n} \lt 90$, find m+n.
• is 107
• is 177
• is 840
• cannot be determined from the given information
### Key Concepts
Angles
Triangles
Side Length
AIME I, 2009, Question 5
Plane Trigonometry by Loney
## Try with Hints
s=$\displaystyle\sum_{k=1}^{35}sin5k$ | {
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Plane Trigonometry by Loney
## Try with Hints
s=$\displaystyle\sum_{k=1}^{35}sin5k$
s(sin5)=$\displaystyle\sum_{k=1}^{35}sin5ksin5=\displaystyle\sum_{k=1}^{35}(0.5)[cos(5k-5)-cos(5k+5)]$=$\frac{1+cos5}{sin5}$
$=\frac{1-cos(175)}{sin175}$=$tan\frac{175}{2}$ then m+n=175+2=177.
Categories
## Problem on Fibonacci sequence | AIME I, 1988 | Question 13
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1988 based on Fibonacci sequence.
## Fibonacci sequence Problem – AIME I, 1988
Find a if a and b are integers such that $x^{2}-x-1$ is a factor of $ax^{17}+bx^{16}+1$.
• is 107
• is 987
• is 840
• cannot be determined from the given information
### Key Concepts
Integers
Digits
Sets
AIME I, 1988, Question 13
Elementary Number Theory by David Burton
## Try with Hints
Let F(x)=$ax^{17}+bx^{16}+1$
Let P(x) be polynomial such that
$P(x)(x^{2}-x-1)=F(x)$
constant term of P(x) =(-1)
now $(x^{2}-x-1)(c_1x^{15}+c_2x^{14}+….+c_{15}x-1)$ where $c_{i}$=coefficient
comparing the coefficients of x we get the terms
since F(x) has no x term, then $c_{15}$=1
getting $c_{14}$
$(x^{2}-x-1)(c_1x^{15}+c_2x^{14}+….+c_{15}x-1)$
=terms +$0x^{2}$ +terms
or, $c_{14}=-2$
proceeding in the same way $c_{13}=3$, $c_{12}=-5$, $c_{11}=8$ gives a pattern of Fibonacci sequence
or, coefficients of P(x) are Fibonacci sequence with alternating signs
or, a=$c_1=F_{16}$ where $F_{16}$ is 16th Fibonacci number
or, a=987.
Categories
## Sequence and Integers | AIME I, 2007 | Question 14
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2007 based on Sequence and Integers.
## Sequence and Integers – AIME I, 2007
A sequence is defined over non negetive integral indexes in the following way $a_0=a_1=3$, $a_{n+1}a_{n-1}=a_n^{2}+2007$, find the greatest integer that does not exceed $\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}$ | {
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• is 107
• is 224
• is 840
• cannot be determined from the given information
### Key Concepts
Sequence
Inequalities
Integers
AIME I, 2007, Question 14
Elementary Number Theory by David Burton
## Try with Hints
$a_{n+1}a_{n-1}$=$a_{n}^{2}+2007$ then $a_{n-1}^{2} +2007 =a_{n}a_{n-2}$ adding these $\frac{a_{n-1}+a_{n+1}}{a_{n}}$=$\frac{a_{n}+a_{n-2}}{a_{n-1}}$, let $b_{j}$=$\frac{a_{j}}{a_{j-1}}$ then $b_{n+1} + \frac{1}{b_{n}}$=$b_{n}+\frac{1}{b_{n-1}}$ then $b_{2007} + \frac{1}{b_{2006}}$=$b_{3}+\frac{1}{b_{2}}$=225
here $\frac{a_{2007}a_{2005}}{a_{2006}a_{2005}}$=$\frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}$ then $b_{2007}$=$\frac{a_{2007}}{a_{2006}}$=$\frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}$$\gt$$\frac{a_{2006}}{a_{2005}}$=$b_{2006}$
then $b_{2007}+\frac{1}{b_{2007}} \lt b_{2007}+\frac{1}{b_{2006}}$=225 which is small less such that all $b_{j}$ s are greater than 1 then $\frac{a_{2006}^{2}+ a_{2007}^{2}}{a_{2006}a_{2007}}$=$b_{2007}+\frac{1}{b_{2007}}$=224.
Categories
## Number and Series | Number Theory | AIME I, 2015
Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2015 based on Number Theory and Series.
## Number Theory and Series – AIME 2015
The expressions A = $(1 \times 2)+(3 \times 4)+….+(35 \times 36)+37$ and B = $1+(2 \times 3)+(4 \times 5)+….+(36 \times 37)$ are obtained by writing multiplication and addition operators in an alternating pattern between successive integers. Find the positive difference between integers A and B.
• is 107
• is 648
• is 840
• cannot be determined from the given information
### Key Concepts
Series
Theory of Equations
Number Theory
AIME, 2015, Question 1
Elementary Number Theory by David Burton
## Try with Hints
B-A=$-36+(2 \times 3)+….+(2 \times 36)$
=$-36+4 \times (1+2+3+….+18)$
=$-36+(4 \times \frac{18 \times 19}{2})$=648. | {
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© Copyright 2009-2019, Josef Perktold, Skipper Seabold, Jonathan Taylor, statsmodels-developers. site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. Use variable in your CloudFormation template. The indefinite integral $\int f(x) dx$ denotes, in this particular context, the set of all primitives of $f(x)$. Expressions where the values of the outputs generally do not change once set in the editor or when play begins. When you differentiate a particular function that has a constant at the end, say $f(x) = x^2 +2x +4$ to get $f'(x) = 2x +2$, you have no way, given only the derivative $f'(x)$, to recover the "constant information" about the original function. There are many great answers here, but I just wanted to chime in with my favorite example of how things can go awry if one forgets about the constant of integration. Create targeted, timely marketing campaigns by segmenting your contacts based on Salesforce custom field mapping such as type of client, product interest, or stage of sales cycle. How do I convert Arduino to an ATmega328P-based project? 1 1 1 bronze badge $\endgroup$ 1 The same applies to more complicated integrals. $$It is also the initial value in integration of a variable function with initial value prescribed to evaluate boundary value differential equations. How to put constants to use in C programming. Jeff Meyer is a statistical consultant with The Analysis Factor, a stats mentor for Statistically Speaking membership, and a workshop instructor. Multiple Imputation with Chained Equations. Indefinite integral is employed to get the set of primitive functions of a particular function. and there's an initial condition y(0)=5. we are adding a column of ones to make it suitable for dot product between the two matrices. Is it just me or when driving down the pits, the pit wall will always be on the left? Create and Send an Email. The derivative of {x^3\over 3}+C is x^2. The answer to your question is the same as | {
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Send an Email. The derivative of {x^3\over 3}+C is x^2. The answer to your question is the same as the answer to my question. Another way of thinking of the slope field is that it covers the plane with all of the possible "flow lines of the function", so if you have water running down a stream (think of this as from -x to +x), the water will run along the paths described by the slope field. static const.$$ Constant Contact helps you spread the word through email, social media, SEO and other forms of online marketing—all from one place. Help people find you. Finding non-zero elements of a Ring, a and c, with ab=c, Definite or indefinite integration of a relationship in physics. The form captures the lead and sends it to specific Constant Contact lists automatically. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. If I asked you to solve $x^2 = 4$, and you determined that $2$ was a square root of $4$, why would you bother writing the solution as $\pm 2$? Step name . $$Behavior if data already has a constant. You can add forms to posts, pages or sidebar, and also open it as a popup or top bar. Java doesn't have built-in support for constants, but the variable modifiers static and final can be used to effectively create one. The default will return data without adding another constant. So, if F(x) is an antiderivative, then any other antiderivative G(x) can be expressed as G(x)=F(x)+C for some constant C. I would like to know the whole purpose of adding a constant termed constant of integration everytime we integrate an indefinite integral \int f(x)dx. In many cases, formulas that use array constants do not require Ctrl+Shift+Enter, even though they are in fact array formulas. First, a substitution u=2x yields: If ‘raise’, will raise an error if any column has a constant value. That's why we write, for example: Module-level constants are private by default. Static : determines the lifetime and | {
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for example: Module-level constants are private by default. Static : determines the lifetime and visibility/accessibility of the variable. We cannot allow the expression \int f(x)dx to refer to multiple functions, so to get around this we introduce the constant of integration C.$$\int (5x -6x^4)\ dx =\langle {5\over2} x^2-{6\over5} x^5\rangle\ ,$$The slope field anti-derivative is given by: multiple-regression modeling least-squares multinomial. View Reports. Making statements based on opinion; back them up with references or personal experience. The keyword const is a little misleading.. Why do we not include c in the computation of the definite integral?$$\int \frac{\sin(u)}{2}du = -\frac{\cos(u)}{2} = -\frac{\cos(2x)}{2}.$$Why is it termed as the constant of integration? The types are somehow are already in the database in some conflicting way due to flirt signatures or something. Thanks for contributing an answer to Mathematics Stack Exchange! Right.I've just started with differential equations and this slope field is yet to make sense,however i will try to comprehend.By the way which tool do you use to draw slope fields?I'am sure that it'll come in handy when i get there. It defines a constant reference to a value. array_like. new_X = sm.add_constant(new_X) Create a new OLS model named ‘new_model’ and assign to it the variables new_X and Y. Moreover if F'(x)=x^2, then F must have the form F(x)={x^3\over3}+C. Namely, if \rm\:D\: is a linear map then one easily proves, Lemma \ \ If \rm\ D\:f_1\ =\ g\ then \rm\ D\:f_2\ =\ g\ \iff\ 0\ =\ D\:f_1 - D\:f_2\ =\ D\:(f_1-f_2). \int x^2\,dx={x^3\over 3}+C. 3. Indeed, the constant C in this case is exactly C=-\frac{1}{2}: When the input is recarray or a pandas Series or DataFrame, the added$$f(x) = \sin(x) +2 $$. \frac{dy}{dx}=-4y If you find one primitive, say F(x), then you know all other primitives have the form F(x) + C, where C is any constant. Then By declaring a constant, you assign a meaningful name to a value.$$ | {
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C, where C is any constant. Then By declaring a constant, you assign a meaningful name to a value.$$ The theory of integration tells us that all antiderivatives differ by a constant. the derivation $\rm\ D = \dfrac{d}{d\:x}\:.$, Compare this to $\rm \ x\ =\ 3 + 5\ \mathbb Z,\:$ the solution of $\rm\ 2x \equiv 6\pmod{\!10},\:$ with particular solution $\rm x \equiv 6/2 \equiv 3,\:$ and homogeneous: $\rm\ 2x\equiv 0\pmod{\!10}\iff 10\mid 2x\iff 5\mid x\iff x\in 5 \mathbb Z$. "I've understood that ∫dy represents adding infinitesimal quantity of dy's yielding y" -- this is not correct. Anytime your code uses a single value over and over (something significant, like the number of rows in a table or the maximum number of items you can stick in a shopping cart), define the value as a constant. This is precisely why you have to have a slope field representation of the anti-derivative of a function. Writing $\langle F(x)\rangle$ (or similar) instead of $F(x)+C$ for the set of all functions differing from the term $F(x)$ by a constant, one could write, e.g., The Add constant values step is a simple and high performance way to add constant values to the stream. share | cite | improve this question | follow | edited Jul 8 '13 at 9:23. Do you need a valid visa to move out of the country? The key concept to note here is that when you differentiate a constant you get 0, this is due to the fact that the slope of the tangent line of a constant function, say $f(x) = 4$, will simply be a horizontal line spanning the x-axis with a slope of zero everywhere. $$F(x)+C = F(x) - \frac{1}{2} = \sin^2(x)-\frac{1}{2} = \frac{(1-\cos(2x))}{2}-\frac{1}{2} = -\frac{\cos(2x)}{2} = G(x),$$ appended (last column). In fact it is what many people call a "dangling variable", similar to the $i$ and $k$ when we talk about a "matrix $\bigl[a_{ik}\bigr]\$". Setup Your Account. Now, given any function $F$ with $F'=f$, it follows that $F+C$ is also an antiderivative of $f$: The general solution is as you | {
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with $F'=f$, it follows that $F+C$ is also an antiderivative of $f$: The general solution is as you have, with the arbitrary parameter $C$. Problem is, there are multiple (in fact infinitely many) such functions $F(x)$. You have [code ]y = w0 + w1*x[/code]. We will use figure 2 to illustrate how we can keep a formula constant regardless of where we copy the formula. $$\int \sin(2x) dx.$$ Yet, when you use const this way, most compilers also make the array itself constant. How exactly was the Texas v. Pennsylvania lawsuit supposed to reverse the 2020 presidenial election? \tag{1} Returns. Sometimes you need to know all antiderivatives of a function, rather than just one antiderivative. Now that we have this lambda function, we can use it in CloudFormation templates. Good idea to warn students they were suspected of cheating? For example, in integration by parts, you may have $dv=\cos x\;dx$, and conclude that $v=\sin x$.). First make note of the lambda function Arn (go to the lambda home page, click the just created function, the Arn should be in the top right, something like arn:aws:lambda:region:12345:function:CloudFormationIdentity). After a constant is declared, it cannot be modified or assigned a new value. Figure 2 – How to keep a reference constant. Multiple results When you provide an array constant to an Excel function as an argument, you will often receive more than one result in an array. What's most important, I think, is to know how to answer simple questions like: Find all the antiderivatives of $1/x$ over $\mathbb R\setminus\{0\}$. When you integrate a particular function, you must add that $+C$ because it says that, the anti-derivative of the function could be one of any of the slope field lines in $\mathbb{R}^2$ . Why is it impossible to measure position and momentum at the same time with arbitrary precision? is it possible to read and play a piece that's written in Gflat (6 flats) by substituting those for one sharp, thus in key G? Is Mega.nz | {
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that's written in Gflat (6 flats) by substituting those for one sharp, thus in key G? Is Mega.nz encryption secure against brute force cracking from quantum computers? Therefore $\rm\ D^{-1}(g)\ =\ f_1\ +\ ker\ D\ =\:\:$ particular + homogeneous solution, as in linear algebra. If you were given an initial value problem where, say, you needed $y(1)=2$, then that constant $C$ could be determined. y'=-4y Is there a difference between a tie-breaker and a regular vote? and the mysterious constant has disappeared. Use our Website Builder to generate a mobile-responsive store for your industry with seamless site navigation, secure checkout, and more. Else the constant is MathJax reference. So, what happened? Learn more. Why? What has that constant have to do with anything? Use email to boost loyalty. It may also make more sense when you take a differential equations course, but this should be a sufficient explanation. Now, both $$c$$ and $$k$$ are unknown constants and so the sum of two unknown constants is just an unknown constant and we acknowledge that by simply writing the sum as a $$c$$. algebra-precalculus computer-algebra-systems wolfram-alpha. How does the recent Chinese quantum supremacy claim compare with Google's? $$To learn more, see our tips on writing great answers. Notice that F(x)=\sin^2(x)\neq -\cos(2x)/2=G(x). Using ‘add’ will add a column of 1s if a constant column is present.$$ For instance, $F(0)=\sin^2(0) = 0$ but $G(0)=-\cos(2\cdot 0)/2=-1/2$. One hasto initialise it immediately in the constructor because, of course, one cannotset the value later as that would be altering it. Values for $C$ correspond to what particular path the function (or the water) is running along. How to prevent guerrilla warfare from existing. So Access Constant Contact’s built-in landing page builder by selecting “Landing Page” in the “Create New” pop-up menu. It has an argument include.mean which has identical functionality to the corresponding argument for arima().It also | {
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include.mean which has identical functionality to the corresponding argument for arima().It also has an argument include.drift which allows . , (And sometimes you need only one antiderivative, not all of them. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. $$\int \sin(2x)dx = \int 2\sin(x)\cos(x)dx = \int 2vdv =v^2 = \sin^2(x).$$ Use the readonly modifier to create a class, struct, or array that is initialized one time at runtime (for example in a constructor) and thereafter cannot be changed. error if any column has a constant value. Formal way to perform the change of variable, Evaluating the integral: $\int\limits_{0}^{\infty}\left(\frac{\sin(ax)}{x}\right)^2 dx , a \neq 0$, Getting different answers when integrating using different techniques. Landing page builders have a basic template, so all you have to do is fill in the blanks, choose images and colors, decide on the information you want to collect, and decide where you want the information to go. Read more about Jeff here. Here are my main theories for why this is happening: Somehow the behavior of import_type doesn't exactly work closely enough to the original Til2Idb method that was in the original script. data without adding another constant. The differential equation you are considering has more than one solution. static const : “static const” is basically a combination of static(a storage specifier) and const(a type qualifier). asked Jul 8 '13 at 9:13. user27746 user27746. Just go through the following 5 steps. We forgot about the constant of integration, that's what happened. $$It does belong there. See. Ever month you earned 10 dollars on a principal of 1000 dollars and put it in a box. But that’s fine; people shouldn’t be taking an array name and copying it to something else. I've understood that \int dy represents adding infinitesimal quantity of dy's | {
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it to something else. I've understood that \int dy represents adding infinitesimal quantity of dy's yielding y but I'am doubtful about the arbitrary constant C. C# does not support const methods, properties, or events. If we copy down the formula in Cell C9, the cell reference changes to …$$ 5=y(0)= e^{-4\cdot0}e^C = e^C The original values with a constant (column of ones) as the first or specify the name, type, and value in the form of a string. This is precisely why you have to have a slope field representation of the anti-derivative of a function. The indefinite integral $\int f(x)dx$ is the function $F(x)$ such that $\frac{d}{dx}F(x) = f(x)$. The impact of removing the constant when the predictor variable is continuous is significantly different. If true, the constant is in the first column. Here you've added a constant. The motivation for asking this question actually comes from solving a differential equation $$x \frac{dy}{dx} = 5x^3 + 4$$ By separation of $dy$ and $dx$ and integrating both sides, $$\int dy = \int\left(5x^2 + \frac{4}{x}\right)dx$$ yields $$y = \frac{5x^3}{3} + 4 \ln(x) + C .$$. The original values with a constant (column of ones) as the first or last column. $$Namely Can I combine two 12-2 cables to serve a NEMA 10-30 socket for dryer? What do I do about a prescriptive GM/player who argues that gender and sexuality aren’t personality traits? And you can know how to use all of the Constant Contact tools.$$ column’s name is ‘const’. So, is there anything that I can add to my answer that you still have a question about? Not always 100 dollars or 200 dollars.That is so, only if the box contained nothing not at start. Constant Contact Forms by MailMunch allows you to painlessly add Constant Contact sign up forms to your WordPress site. \frac{dy}{y} = -4\;dx That’s not good programming style, and it’s just asking for bugs — or, at the very least, confusion — later. How can I intuitively understand the algorithm for finding the integer solutions to | {
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— later. How can I intuitively understand the algorithm for finding the integer solutions to $ax+by=c$? Grow Your Lists. 2. Upload Your Contacts. What have you got after 10 months, 20 months in that box? column of 1s if a constant column is present. The enum type enables you to define named constants for integral built-in types (for example int, uint, long, and so on). Hence we say $\int f(x)dx = F(x) + C$. \log_e y = -4x + C. So, the general antiderivative of $f(x)=x^2$ has the form $F(x)={x^3\over3}+C$, and equation (1) is just stating this fact. last column. It does NOT define a constant value. Second, we use the identity $\sin(2x)=2\sin(x)\cos(x)$ and a substitution $v=\sin(x)$: Replace blank line with above line content, Advice on teaching abstract algebra and logic to high-school students. constant meaning: 1. happening a lot or all the time: 2. staying the same, or not getting less or more: 3. Not Real Constants. $$You declare a constant within a procedure or in the declarations section of a … How do I "tell" WA which variables are the constants, and which are the ones I want it to solve for? Name of the step. What is the precise legal meaning of "electors" being "appointed"? In this article. The indefinite integral \int f(x)\,dx is defined to be the general class of functions whose derivatives are f(x). The "integration constant" C does have the "purpose" to make a seemingly true equation at least halfway true. Since there is no reason to think that the constants of integration will be the same from each integral we use different constants for each integral. To get started, I’ve created a new folder within src/ called “constants” to hold all of the constants I use within all of my components. If use a constant in your Microsoft Excel workbook formulas, such as sales tax or car mileage allowance, then check out how using a named constant can save yourself considerable time. Consider What to do? Thus, we have found two antiderivatives of \sin(2x) that are | {
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considerable time. Consider What to do? Thus, we have found two antiderivatives of \sin(2x) that are completely different! You can declare a constant within a procedure or at the top of a module, in the Declarations section.$$ The particular value for $C$ collapses it to exactly one of these slope field lines. $$Using ‘add’ will add a To subscribe to this RSS feed, copy and paste this URL into your RSS reader. If ‘raise’, will raise an when . Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Besides using p values for feature selection which is ill advised, most packages just drop any constant columns from the data frame because they introduce numerical problems (singular matrices). This fixes things, because though there are infinitely many F(x) such that \frac{d}{dx}F(x) = f(x), if we pick any single such function F(x) then all solutions to \frac{d}{dx}G(x) = f(x) are of the form G(x) = F(x) + C for some value of C, while for any value of C the function F(x)+C satisfies \frac{d}{dx}(F(x)+C) = f(x). rev 2020.12.10.38158, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us, Integral is set of antiderivatives, and no antiderivative is distinguished. A = m.10 + c = m.10 + 35 dollars or 135 or 235 dollars.. Where does it come from?$$ In a follow-up article, we will explore why you should never do that. Because of this, we cannot change constant primitive values, but we can change the properties of constant objects. | {
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# Math Help - Question on Probability
1. ## Question on Probability
Given that
$P(\neg A)=0.6$
$P(B\mid A)=0.7$
$P(B)=0.3$
What is $P(\neg A \wedge B)$?
2. Originally Posted by quiney
Given that
$P(\neg A)=0.6$
$P(B\mid A)=0.7$
$P(B)=0.3$
What is $P(\neg A \wedge B)$?
A simple approach would be to draw a tree diagram. Have you tried doing that?
3. Here is another way.
Recall that $P(B) = P(B \cap A) + P(B \cap \neg A)$
Solve for $P(B \cap \neg A)$.
4. Hello, quiney!
Yet another approach . . .
Given that:
. . $\begin{array}{ccc}P(\sim\!A)&=& 0.6 \\
P(B\,|\,A) &=& 0.7 \\
P(B) &=& 0.3 \end{array}$
What is . $P(\sim\!A \wedge B)$ ?
We can place the data into a chart:
. . $\begin{array}{c||c|c||c}
& B & \sim\!B & \text{total} \\ \hline \hline
A & & & 0.40 \\ \hline
\sim\!A & & & 0.60 \\ \hline \hline
\text{Total} & 0.30 & 0.70 & 1.00 \end{array}$
We have: . $P(B|A) \:=\:0.7 \quad\Rightarrow\quad \dfrac{P(B \wedge A)}{P(A)} \:=\:0.7$
. . . . $P(B \wedge A) \:=\:0.7\cdot P(A) \;=\;(0.7)(0.4)$
. . . . $P(A \wedge B) \;=\;0.28$
Insert that into the chart.
. . Fill in the rest of the chart.
. . $\begin{array}{c||c|c||c}
& B & \sim\!B & \text{total} \\ \hline \hline
A & 0.28 & 0.12 & 0.40 \\ \hline
\sim\!A & 0.02 & 0.58 & 0.60 \\ \hline \hline
\text{Total} & 0.30 & 0.70 & 1.00 \end{array}$
Therefore: . $P(\sim\!A \wedge B) \;=\;0.02$
Corrected my typo . . . Thanks, Plato!
5. Originally Posted by Soroban
We have: . $P(B|A) \:=\:0.7 \quad\Rightarrow\quad \dfrac{P(A \wedge B)}{P(B)} \:=\:0.7$
Please note the typo $\displaystyle P(B|A)=\frac{P(A\cap B)}{P(A)}$.
6. Thanks a lot everybody. So, since $p(B)=p(B \wedge A) + p(B \wedge \neg A)$:
$0.3=p(B \wedge A) + p(B \wedge \neg A)$
Since $p(B|A)=0.7=\dfrac{p(B \wedge A)}{p(A)=0.4} \quad\Rightarrow\quad\ 0.28=p(B\wedge A) \quad\Rightarrow\ 0.3=0.28 + p(B \wedge\neg A)$
Therefore
$0.02=p(B \wedge\neg A)=p(\neg A \wedge B)$
Funny, the answer guide in the book says 0.2. Must be a typo. | {
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# The period of Fibonacci numbers over finite fields
I stumbled upon these very nice looking notes by Brian Lawrence on the period of the Fibonacci numbers over finite fields. In them, he shows that the period of the Fibonacci sequence over $$\mathbb{F}_p$$ divides $$p$$ or $$p-1$$ or $$p+1$$.
I am wondering if there are explicit lower bounds on this period. Is it true, for instance, that as $$p \to \infty$$, so does the order?
A quick calculation on my computer shows that there are some "large" primes with period under 100.
9901 66
19489 58
28657 92
• For $p$ sufficiently large (depending on $N$), the first $N$ Fibonnaci numbers are distinct modulo $p$. Aug 24 '20 at 19:57
• Ah you're right. Might not be able to do any better than that bound then. Aug 24 '20 at 20:09
• @FedorPetrov You need not just $p \mid F_k$ but also $p \mid (F_{k+1}-1)$ since you need $F_k \equiv 0 \bmod p$ and $F_{k+1} \equiv 1 \bmod p$. The first congruence need not imply the second. For example, take $p = 61$. The smallest $k \geq 1$ such that $F_k \equiv 0 \bmod 61$ is $k = 15$, but $F_{16} \equiv 11 \not\equiv 1 \bmod 61$, so the Fibonacci sequence mod $61$ does not have period $15$. The period of $\{F_n \bmod 61\}$ is $60$. Aug 25 '20 at 21:56
• @KConrad ah, sorry, we look for a full period, not only the period of zeroes. You are correct of course. Aug 25 '20 at 21:59
• "the period of the Fibonacci sequence over Fp divides p or p−1 or p+1." This is not true (try $p=5$). An entry divisible by $p$ alone does not make a period. Oct 21 '20 at 14:19 | {
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This maybe too elementary for this site, so if your question is closed, you might try asking on MathStackExchange. Many questions about the period can be answered by using the formula $$F_n = (A^n-B^n)/(A-B),$$ where $$A$$ and $$B$$ are the roots of $$T^2-T-1$$. So if $$\sqrt5$$ is in your finite field, then so are $$A$$ and $$B$$, and since $$AB=-1$$, the period divides $$p-1$$ from Fermat's little theorem. If not, then you're in the subgroup of $$\mathbb F_{p^2}$$ consisting of elements of norm $$\pm1$$, so the period divides $$2(p+1)$$. If you want small period, then take primes that divide $$A^n-1$$, or really its norm, so take primes dividing $$(A^n-1)(B^n-1)$$, where $$A$$ and $$B$$ are $$\frac12(1\pm\sqrt5)$$. An open question is in the other direction: Are there infinitely many $$p\equiv\pm1\pmod5$$ such that the period is maximal, i.e., equal to $$p-1$$?
BTW, the source you quote isn't quite correct, if $$p\equiv\pm2\pmod5$$, then the period divides $$2(p+1)$$, but might not divide $$p+1$$. The simplest example is $$p=3$$, where the Fibonacci sequence is $$0,1,1,2,0,2,2,1,\quad 0,1,1,2,0,2,2,1,\ldots.$$ Note that the first 0 does not necessarily mean that it will start to repeat. What happens is that the term before the first $$0$$ is $$p-1$$, so the first part of the sequence repeats with negative signs before you get to a consecutive $$0$$ and $$1$$. | {
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• On the one hand I agree it's probably too elementary, but, on the other hand, I wouldn't have seen this very nice perspective on periods if it hadn't been posted here, so I'm torn. :-) Aug 24 '20 at 20:17
• Sorry everyone; I'm still learning about the etiquette of the forum. I found these notes and found them interesting and wanted to learn more. Thanks for answering my question seriously :) Aug 25 '20 at 20:06
• As @LSpice comments, one might not have realized that such a question is "quasi-elementarizable" in such a decisive manner... Sometimes the decisiveness and elementariness is only visible to a more experienced person, which (to my mind) doesn't mean that the original question was "too easy for MO"... If anything, I am fond of examples where a seemingly innocent (but baffling) question succumbs to a more sophisticated viewpoint. As though such viewpoints not only provide publication fodder, but really do answer questions. :) Aug 31 '20 at 17:50
I won't address your question about how small the period of $$\{F_n \bmod p\}$$ can be as $$p$$ grows, but instead ask if the upper bounds on the period can be achieved infinitely often. For consistency I'll use the notation from Joe Silverman's answer: set $$A = (1 + \sqrt{5})/2$$ and $$B = (1-\sqrt{5})/2$$, so $$F_n = (A^n - B^n)/(A-B) = (A^n - B^n)/\sqrt{5}$$. Note $$A+B = 1$$, $$A - B = \sqrt{5}$$, and $$AB = -1$$.
Claim: For a prime $$p \not= 2$$ or $$5$$, the period of the Fibonacci sequence $$\{F_n \bmod p\}$$ is the smallest even positive integer $$k$$ such that $$A^k = 1$$ in characteristic $$p$$. | {
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This claim involves working in the field $$\mathbf F_p(\sqrt{5})$$, where $$\sqrt{5}$$ is a square root of 5 in characteristic $$p$$, so we can regard $$A = (1+\sqrt{5})/2$$ as a number in the field $$\mathbf F_p(\sqrt{5})$$, which is either $$\mathbf F_p$$ or $$\mathbf F_{p^2}$$. (The notation $$\mathbf F_p$$ and $$\mathbf F_{p^2}$$ are fields of order $$p$$ and $$p^2$$, not having anything to do with the "$$F$$" in Fibonacci number notation.) The claim is saying that $$F_{n+k} \equiv F_n \bmod p$$ for all $$n \geq 0$$ (or just all sufficiently large $$n \geq 0$$) if and only if $$A^k = 1$$ in $$\mathbf F_p(\sqrt{5})$$ for even $$k$$, so the period of $$\{F_n \bmod p\}$$ is the smallest even $$k \geq 1$$ such that $$A^k = 1$$ in $$\mathbf F_p(\sqrt{5})$$. | {
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Proof. View the congruence $$F_{n+k} \equiv F_n \bmod p$$ as an equation $$F_{n+k} = F_n$$ in the subfield $$\mathbf F_p$$ of $$\mathbf F_p(\sqrt{5})$$. The Fibonacci formula $$F_n = (A^n - B^n)/\sqrt{5}$$ in $$\mathbf R$$ is also a valid formula in fields of characteristic $$p$$ when we view $$\sqrt{5}$$ in characteristic $$p$$ and set $$A = (1+\sqrt{5})/2$$ and $$B = (1-\sqrt{5})/2 = 1-A$$ in characteristic $$p$$. In $$\mathbf F_p(\sqrt{5})$$, \begin{align*} F_{n+k} = F_n & \Longleftrightarrow \frac{A^{n+k}-B^{n+k}}{\sqrt{5}} = \frac{A^n-B^n}{\sqrt{5}} \\ & \Longleftrightarrow A^n(A^k-1) = B^n(B^k-1). \end{align*} In a field of characteristic $$p \not= 2$$ or $$5$$, $$A$$ and $$B$$ are nonzero since $$AB = -1$$. Suppose in $$\mathbf F_p(\sqrt{5})$$ that $$A^k \not= 1$$. Then in this field, $$F_{n+k} = F_n \Longrightarrow (A/B)^n = (B^k-1)/(A^k-1).$$ The ratio $$A/B$$ in characteristic $$p$$ is not $$1$$ since $$A = B \Longrightarrow 5 = 0$$ in characteristic $$p$$, which is false since $$p \not= 5$$. Therefore $$(A/B)^n$$ is not constant as $$n$$ varies, but $$(B^k-1)/(A^k-1)$$ is constant as $$n$$ varies. Thus $$A^k = 1$$ in $$\mathbf F_p(\sqrt{5})$$, so $$B^n(B^k-1) = A^n(A^k-1) = 0$$, so $$B^k = 1$$ (we never have $$B^n = 0$$ in characteristic $$p$$). Since $$B^k = (-1/A)^k = (-1)^k/A^k$$, we have $$A^k = 1$$ and $$B^k = 1$$ if and only if $$A^k = 1$$ and $$(-1)^k = 1$$. Since $$-1 \not= 1$$ in characteristic $$p$$ when $$p \not= 2$$, we have $$A^k = 1$$ and $$(-1)^k = 1$$ in $$\mathbf F_p(\sqrt{5})$$ if and only if $$A^k = 1$$ in characteristic $$p$$ and $$k$$ is even.
That completes the proof of the claim.
Since $$B = -1/A$$, if $$A$$ in characteristic $$p$$ has odd order $$m$$ then $$B$$ in characteristic $$p$$ has order $$2m$$. Therefore the claim says the period of $$\{F_n \bmod p\}$$ is the least $$k \geq 1$$ such that $$A^k = 1$$ and $$B^k = 1$$ in characteristic $$p$$: that $$k$$ is necessarily even. | {
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For $$p \not= 2$$ or 5, the field $$\mathbf F_p(\sqrt{5})$$ has order $$p$$ or $$p^2$$ depending on whether or not $$5 \bmod p$$ is a square: its order is $$p$$ when $$p \equiv \pm 1 \bmod 5$$ and its order is $$p^2$$ when $$p \equiv \pm 2 \bmod 5$$. Therefore the group of nonzero elements $$\mathbf F_p(\sqrt{5})^\times$$ has order $$p-1$$ if $$p \equiv \pm 1 \bmod 5$$ and order $$p^2-1$$ if $$p \equiv \pm 2 \bmod 5$$. Since $$p-1$$ and $$p^2-1$$ are both even, the period of $$\{F_n \bmod p\}$$ divides $$p-1$$ if $$p \equiv \pm 1 \bmod 5$$ and it divides $$p^2-1$$ if $$p \equiv \pm 2 \bmod 5$$. As Joe points out in his answer, when $$p \equiv \pm 2 \bmod 5$$ the period of $$\{F_n \bmod p\}$$ divides $$2(p+1)$$, which is a proper factor of $$p^2-1$$.
This situation is reminiscent of Artin's primitive root conjecture, which says that for $$a \in \mathbf Z$$ that is not $$\pm 1$$ or a perfect square, there are infinitely many primes $$p$$ such that $$a \bmod p$$ has order $$p-1$$ in $$\mathbf F_p^\times$$, and in fact there is a positive density of such primes. This conjecture is known to be a consequence of the Generalized Riemann Hypothesis (GRH). This conjecture and its connection to GRH can be extended to number fields, and to talk about the multiplicative order of $$A$$ in characteristic $$p$$ amounts to looking at an analogue of Artin's primitive root conjecture with $$\mathbf Z$$ replaced by $$\mathbf Z[A]$$, which is the ring of integers of $$\mathbf Q(\sqrt{5})$$. This is discussed in Barendrecht's 2018 bachelor's thesis here. For example, GRH implies that the set of (nonzero) prime ideals $$\mathfrak p$$ in $$\mathbf Z[A]$$ such that $$A \bmod \mathfrak p$$ generates all of $$(\mathbf Z[A]/\mathfrak p)^\times$$ has a positive density using the last result of the thesis, Corollary 3.1.2, and therefore there are infinitely many such prime ideals $$\mathfrak p$$ in $$\mathbf Z[A]$$. | {
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Every nonzero prime ideal $$\mathfrak p$$ in $$\mathbf Z[A]$$ is a factor of $$(p) = p\mathbf Z[A]$$ for some prime number $$p$$: if $$p \equiv \pm 1 \bmod 5$$ then $$(p) = \mathfrak p\mathfrak p'$$ for two prime ideals $$\mathfrak p$$ and $$\mathfrak p'$$, and $$\mathbf Z[A]/\mathfrak p$$ and $$\mathbf Z[A]/\mathfrak p'$$ are fields of order $$p$$. If $$p \equiv \pm 2 \bmod 5$$, then $$(p) = \mathfrak p$$ is a prime ideal in $$\mathbf Z[A]$$ and $$\mathbf Z[A]/(p)$$ is a field of order $$p^2$$. When $$p \equiv \pm 2 \bmod 5$$, the multiplicative order of $$A$$ in characteristic $$p$$ is a factor of $$2(p+1)$$, which is less than $$p^2-1$$, so the only prime ideals $$\mathfrak p$$ in $$\mathbf Z[A]$$ for which $$A \bmod \mathfrak p$$ might generate $$(\mathbf Z[A]/\mathfrak p)^\times$$ are prime ideals dividing a prime $$p \equiv \pm 1 \bmod 5$$, in which case we are in the situation that $$A \in \mathbf F_p^\times$$ has order $$p-1$$. Comparing this to the claim up above, since $$p-1$$ is even when $$p > 2$$ we see that GRH implies that there are infinitely many primes $$p \equiv \pm 1 \bmod 5$$ such that $$\{F_n \bmod p\}$$ has period $$p-1$$.
Among the 18 odd primes $$p \equiv \pm 2 \bmod 5$$ with $$p < 150$$, $$\{F_n\bmod p\}$$ has period $$2(p+1)$$ all but 3 times (at $$p = 47$$ $$107$$, and $$113$$). There are many generalizations of the Artin primitive root conjecture and I would not be surprised if one of them can show GRH implies there are infinitely many primes $$p \equiv \pm 2 \bmod 5$$ such that $$\{F_n \bmod p\}$$ has period $$2(p+1)$$, but this is not something I am aware of in more detail at the moment. | {
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The question above is about lower bounds, but I allow myself to comment about upper bounds: $$\pi(n)$$, the period function of the Fibonacci sequence mod $$n$$, satisfies $$\pi(n)\leq 6n$$ and equality holds iff $$n=2\cdot 5^k$$ for some $$k\geq 1$$. This fact is well known. In the 90's it was considered here as a puzzle to the monthly readers. $$\pi(n)$$ was also discussed in an elementary fashion in the 60's in this monthly paper.
But really, I want to share a little observation which forms a generalization of the above mentioned fact: denoting, for an element $$g\in \mathrm{GL}_2(\mathbb{Z})$$, by $$\rho_g(n)$$ the order of the image of $$g$$ in $$\mathrm{GL}_2(\mathbb{Z}/n)$$, $$\rho_g(n)\leq 6n$$. This is a generalization because $$\rho_g(n)=\pi(n)$$ for $$g= \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$$. Note that if $$\det(g)=-1$$ then $$\rho_g(n)=2\rho_{g^2}(n)$$, thus it is enough to prove that for $$g\in \mathrm{SL}_2(\mathbb{Z})$$, $$\rho_g(n)\leq 3n$$. Let me now fix $$g\in \mathrm{SL}_2(\mathbb{Z})$$, denote $$\rho(n)=\rho_g(n)$$ and prove that indeed $$\rho(n)\leq 3n$$.
First note that, for natural $$p$$ and $$n$$, if $$p$$ divides $$n$$ then $$\rho(pn)$$ divides $$p\rho(n)$$. This follows by expanding the right hand side of $$g^{p\rho(n)}=(g^{\rho(n)})^p=(1+nh)^p$$ and note that it is 1 mod $$pn$$. By induction we conclude that for every $$k>1$$, $$\rho(p^k)$$ divides $$p^{k-1}\rho(p)$$. | {
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Assume now $$p$$ is a prime and note that $$\rho(p)$$ divides either $$p-1,p+1$$ or $$2p$$. Indeed, if $$\bar{g}\in \mathrm{SL}_2(\mathbb{F}_p)$$ is diagonalizable over $$\mathbb{F}_p$$ then its eigenvalues are in $$\mathbb{F}_p^\times$$ and their orders divides $$p-1$$, else, if $$\bar{g}$$ is diagonalizable over $$\mathbb{F}_{p^2}$$ then its eighenvalues $$\alpha,\beta$$ are conjugated by the Frobenius automorphism, thus their order divides $$p+1$$ because $$\alpha^{p+1}=\alpha\alpha^p=\alpha\beta=\det(\bar{g})=1$$, else $$\bar{g}$$ has a unique eigenvalue, which must be a $$\pm 1$$ by $$\det(\bar{g})=1$$, thus $$\bar{g}^2$$ is unipotent and its order divides $$p$$. For $$p=2$$, in the last case, there was no reason to pass to $$g^2$$, as $$-1=1$$ in $$\mathbb{F}_2$$, thus $$\rho(2)$$ is either 1,2 or 3.
From the above two points, we conclude that for every odd prime $$p$$ and natural $$k$$, $$\rho(p^k)$$ divides $$p^{k-1}(p-1)$$, $$p^{k-1}(p+1)$$ or $$2p^k$$. All these numbers are even and bounded by $$2p^k$$, thus $$\mathrm{lcm}\{\rho(p^k),2\} \leq 2p^k$$. For $$p=2$$ we get that $$\rho(2^k) \leq 2^{k-1}\cdot 3$$.
Fix now an arbitrary natural $$n$$. Write $$n=2^km$$ for an odd $$m$$ and decompose $$m=\prod_{i=0}^r p_i^{k_i}$$. Then \begin{align*} \rho(m)= \mathrm{lcm}\{\rho(p_i^{k_i}) \mid i=0,\dots r\} \leq \mathrm{lcm}\{\mathrm{lcm}\{\rho(p_i^{k_i}),2\} \mid i=0,\dots r\} =\\ 2\mathrm{lcm}\{\frac{\mathrm{lcm}\{\rho(p_i^{k_i}),2\}}{2} \mid i=0,\dots r\} \leq 2\prod_{i=0}^r \frac{\mathrm{lcm}\{\rho(p_i^{k_i}),2\}}{2}\leq 2\prod_{i=0}^r p_i^{k_i} =2m \end{align*} and we get $$\rho(n) = \rho(2^km) \leq \rho(2^k) \cdot \rho(m) \leq 2^{k-1}\cdot 3 \cdot 2m = 3\cdot 2^km=3n.$$
This finishes the proof that $$\rho(n)\leq 3n$$. | {
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This finishes the proof that $$\rho(n)\leq 3n$$.
As always, it is interesting to analyze the case of equality. For $$g\in \mathrm{SL}_2(\mathbb{Z})$$ we have $$\rho_g(n)=3n$$ for some $$n$$ iff $$\mathrm{tr}(g)$$ is odd and not equal $$-1$$ or $$-3$$. If $$g$$ satisfies this condition, then $$n$$ satisfices $$\rho_g(n)=3n$$ iff $$n=2st$$, for some odd $$s\geq 3$$, $$t\geq 1$$ where every prime factor of $$s$$ divides $$\mathrm{tr}(g)+2$$, every prime factor of $$t$$ divides $$\mathrm{tr}(g)-2$$ and $$g$$ is not $$\pm 1$$ modulo any of these prime factors.
For $$g$$ satisfying $$\det(g)=-1$$, using the identity $$\mathrm{tr}(g^2)=\mathrm{tr}(g)^2-2\det(g)$$, we get that $$\rho_g(n)=6n$$ for some $$n$$ iff $$\mathrm{tr}(g)$$ is odd and in this case, $$n$$ satisfices $$\rho_g(n)=6n$$ iff $$n=2st$$, for some odd $$s\geq 3$$, $$t\geq 1$$ where every prime factor of $$s$$ divides $$\mathrm{tr}(g)+4$$, every prime factor of $$t$$ divides $$\mathrm{tr}(g)$$ and $$g$$ is not $$\pm 1$$ modulo any of these prime factors.
Specifically for $$g=\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$$, $$\det(g)=-1$$, $$\mathrm{tr}(g)=1$$ is odd, 5 is the only prime factor of $$\mathrm{tr}(g)+4$$ and there is no prime factor for $$\mathrm{tr}(g)$$. Since $$g$$ is not $$\pm 1$$ modulo 5, we get that $$\pi(n)=\rho_g(n)=6n$$ iff $$n=2\cdot 5^k$$ for some $$k\geq 1$$, as claimed above. | {
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# How to generate random integers between 1 and 4 that have a specific mean?
I need to generate 100 random integers in R, where each integer is between 1 and 4 (hence 1,2,3,4) and the mean is equal to a specific value.
If I draw random uniform numbers between 1 and 5 and get floor, I have a mean of 2.5.
x = floor(runif(100,min=1, max=5))
I need to fix the mean to 1.9 or 2.93 for example.
I guess I can generate random integers that add to 100 * mean but I don't know how to restrict to random integers between 1 and 4.
• I think this is a bit under-determined... One for instance can get a mean of 1.9 with sample(size=n, x= 1:4, prob=c(3.666,1,1,1), replace=TRUE) but also with sample(size=n, x= 1:4, prob=c(3,1,1,0.715), replace=TRUE). – usεr11852 Jan 5 at 22:53
• Are you asking how to constrain the mean of the underlying distribution, or the sample mean? – user20160 Jan 5 at 22:53
• tha sample mean @user20160 – Fierce82 Jan 5 at 23:04
• Integers between 1 and 4 only allows for 2 and 3. You also need to specify the distribution that they are drawn randomly from (or make one up). – wolfies Jan 6 at 6:28
• I voted to leave this open because there's an interesting algorithmic question in here--the R part is incidental; you could just as easily implement this in Python or with a pad and some dice. – Matt Krause Jan 6 at 21:14
I agree with X'ian that the problem is under-specified. However, there is an elegant, scalable, efficient, effective, and versatile solution worth considering.
Because the product of the sample mean and sample size equals the sample sum, the problem concerns generating a random sample of $$n$$ values in the set $$\{1,2,\ldots, k\}$$ that sum to $$s$$ (assuming $$n \le s \le kn,$$ of course). | {
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To explain the proposed solution and, I hope, justify the claim of elegance, I offer a graphical interpretation of this sampling scheme. Lay out a grid of $$k$$ rows and $$n$$ columns. Select every cell in the first row. Randomly (and uniformly) select $$s-n$$ of the remaining cells in rows $$2$$ through $$k.$$ The value of observation $$i$$ in the sample is the number of cells selected in column $$i:$$
This $$4\times 100$$ grid is represented by black dots at the unselected cells and colored patches at the selected cells. It was generated to produce a mean value of $$2,$$ so $$s=200.$$ Thus, $$200-100=100$$ cells were randomly selected among the top $$k-1=3$$ rows. The colors represent the numbers of selected cells in each column. There are $$28$$ ones, $$47$$ twos, $$22$$ threes, and $$3$$ fours. The ordered sample corresponds to the sequence of colors from column $$1$$ through column $$n=100.$$
To demonstrate scalability and efficiency, here is an R command to generate a sample according to this scheme. The question concerns the case $$k=4, n=100$$ and $$s$$ is $$n$$ times the desired average of the sample:
tabulate(sample.int((k-1)*n, s-n) %% n + 1, n) + 1
Because sample.int requires $$O(s-n)$$ time and $$O((k-1)n)$$ space, and tabulate requires $$O(n)$$ time and space, this algorithm requires $$O(\max(s-n,n))$$ time and $$O(kn)$$ space: that's scalable. With $$k=4$$ and $$n=100$$ my workstation takes only 12 microseconds to perform this calculation: that's efficient.
(Here's a brief explanation of the code. Note that integers $$x$$ in $$\{1,2,\ldots, (k-1)n\}$$ can be expressed uniquely as $$x = nj + i$$ where $$j \in \{0,1,\ldots, k-2\}$$ and $$i\in\{1,2,\ldots, n\}.$$ The code takes a sample of such $$x,$$ converts them to their $$(i,j)$$ grid coordinates, counts how many times each $$i$$ appears (which will range from $$0$$ through $$k-1$$) and adds $$1$$ to each count.) | {
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Why can this be considered effective? One reason is that the distributional properties of this sampling scheme are straightforward to work out:
• It is exchangeable: all permutations of any sample are equally likely.
• The chance that the value $$x \in\{1,2,\ldots, k\}$$ appears at position $$i,$$ which I will write as $$\pi_i(x),$$ is obtained through a basic hypergeometric counting argument as $$\pi_i(x) = \frac{\binom{k-1}{x-1}\binom{(n-1)(k-1)}{s-n-x+1}}{\binom{n(k-1)}{ s-n}}.$$ For example, with $$k=4,$$ $$n=100,$$ and a mean of $$2.0$$ (so that $$s=200$$) the chances are $$\pi = (0.2948, 0.4467, 0.2222, 0.03630),$$ closely agreeing with the frequencies in the foregoing sample. Here are graphs of $$\pi_1(1), \pi_1(2), \pi_1(3),$$ and $$\pi_1(4)$$ as a function of the sum:
• The chance that the value $$x$$ appears at position $$i$$ while the value $$y$$ appears at position $$j$$ is similarly found as $$\pi_{ij}(x,y) = \frac{\binom{k-1}{x-1}\binom{k-1}{y-1}\binom{(n-1)(k-1)}{s-n-x-y+2}}{\binom{n(k-1)}{ s-n}}.$$
These probabilities $$\pi_i$$ and $$\pi_{ij}$$ enable one to apply the Horvitz-Thompson estimator to this probability sampling design as well as to compute the first two moments of the distributions of various statistics.
Finally, this solution is versatile insofar as it permits simple, readily-analyzable variations to control the sampling distribution. For instance, you could select cells on the grid with specified but unequal probabilities in each row, or with an urn-like model to modify the probabilities as sampling proceeds, thereby controlling the frequencies of the column counts. | {
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• (+1) Ultimate elegance, indeed. – Xi'an Jan 6 at 18:56
• The answer is too difficult for me to follow, appreciate it nonetheless – Fierce82 Jan 7 at 11:33
• What an elegant and beautifully presented answer. If you don't mind my humble suggestion as a reader, you might consider presenting the solution first (the counting patches and the great diagram), and then talking about the implementation and how your argument about how it fits the intuition, and finally why it's efficient. It might make it a bit easier to follow. – Neil G Jan 8 at 8:49
• @Neil Thank you for your suggestion. I think it's a good one and will consider it carefully. – whuber Jan 8 at 15:02
• This is a lovely and satisfying answer. I did want to note that the numbers are small enough in this case (100 numbers summing to 190) that we can calculate the uniform distribution of all values that satisfy. I ran some calculations to compare your distribution against this and found that yours is much much more likely (billions in some cases) to select small non-1 values. For example, your model will almost never give distributions with >45 "ones" (~0.002% chance for 46, vanishing for more), but that comprises ~58% of the uniform model values. – Cireo Jan 31 at 5:29
The question is under-specified in that the constraints on the frequencies \begin{align}n_1+2n_2+3n_3+4n_4&=100M\\n_1+n_2+n_3+n_4&=100\end{align} do not determine a distribution: "random" is not associated with a particular distribution, unless the OP means "uniform". For instance, if there exists one solution $$(n_1^0,n_2^0,n_3^0,n_4^0)$$ to the above system, then the distribution degenerated at this solution is producing a random draw that is always $$(n_1^0,n_2^0,n_3^0,n_4^0)$$. | {
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In the case the question is about simulating a Uniform distribution over the grid\begin{align}n_1+2n_2+3n_3+4n_4&=100M\\n_1+n_2+n_3+n_4&=100\end{align}one can always use a Metropolis-Hastings algorithm. Starting from $$(n_1^0,n_2^0,n_3^0,n_4^0)$$, create a Markov chain by proposing symmetric random perturbations of the vector $$(n_1^t,n_2^t,n_3^t,n_4^t)$$ and accept if the result is within $$\{1,2,3,4\}^4$$ and satisfies the constraints.
For instance, here is a crude R rendering:
cenM=293
#starting point (n¹,n³,n⁴)
n<-sample(1:100,3,rep=TRUE)
while((sum(n)>100)|(n[2]-n[1]+2*n[3]!=cenM-200))
n<-sample(1:100,3,rep=TRUE)
#Markov chain
for (t in 1:1e6){
prop<-n+sample(-10:10,3,rep=TRUE)
if ((sum(prop)<101)&
(prop[2]-prop[1]+2*prop[3]==cenM-200)&
(min(prop)>0))
n=prop}
c(n[1],100-sum(n),n[-1])
with the distribution of $$(n_1,n_3,n_4)$$ over the 10⁶ iterations:
In case you want draws of the integers themselves,
sample(c(rep(1,n[1]),rep(2,100-sum(n)),rep(3,n[2]),rep(4,n[3])))
is a quick & dirty way to produce a sample.
• thanks. but I cannot understand how i can utilize this to get the 4 integers (between 1 and 4) – Fierce82 Jan 7 at 11:37
• This generates the numbers of 1,2,3,4's $n_1,n_2,n_3,n_4)$ so that there are 100 of them and the sum is cenM. The integer themselves are a random permutation of $n_1$ 1's,..., $n_4$ 4's. – Xi'an Jan 8 at 7:35
I want to ... uh ... "attenuate" @whuber's amazing answer, which @TomZinger says is too difficult to follow. By that I mean I want to re-describe it in terms that I think Tom Zinger will understand, because it's clearly the best answer here. And as Tom gradually uses the method and finds that he needs, say, to know the distribution of the samples rather than just their mean, whuber's answer will be just what he's looking for.
In short: there are no original ideas here, only a simpler explanation. | {
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In short: there are no original ideas here, only a simpler explanation.
You'd like to create $$n$$ integers from $$1$$ to $$4$$ with mean $$r$$. I'm going to suggest computing $$n$$ integers from $$0$$ to $$3$$ with mean $$r-1$$, and then adding one to each of them. If you can do that latter thing, you can solve the first problem. For instance, if we want 10 integers between $$1$$ and $$4$$ with mean $$2.6$$, we can write down these $$10$$ integers between $$0$$ and $$3$$...
0,3,2,1,3,1,2,1,3,0
whose mean is $$1.6$$; if we increase each by $$1$$, we get
1,4,3,2,4,2,3,2,4,1
whose mean is $$2.6$$. It's that simple.
Now let's think about the numbers $$0$$ through $$3$$. I'm going to think of those as "how many items do I have in a 'small' set?" I might have no items, one item, two items, or three items. So the list
0,3,2,1,3,1,2,1,3,0
represents ten different small sets. The first is empty; the second has three items, and so on. The total number of items in all the sets is the sum of the ten numbers, i.e., $$16$$. And the average number of items in each set is this total, divided by $$10$$, hence $$1.6$$.
whuber's idea is this: suppose you make yourself ten small sets, with the total number of items being $$10t$$ for some number $$t$$. Then the average size of the sets will be exactly $$t$$. In the same way, if you make yourself $$n$$ sets with a total number of items being $$nt$$, the average number of items in a set will be $$t$$. You say you're interested in the case $$n = 100$$.
Let's make this concrete for your example: you want 100 items between 1 and 4 whose average is $$1.9$$. Using the idea of my first paragraph, I'm going to change this to "make $$100$$ ints between $$0$$ and $$3$$ whose average is $$0.9$$". When I'm done, I'll add $$1$$ to each of my ints to get a solution to your problem. So my target average is $$t = 0.9$$.
I want to make $$100$$ sets, each with between $$0$$ and $$3$$ items in it, with an average set-size of $$0.9$$. | {
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As I've observed above, this means that there have to be a total of $$100 \cdot 0.9 = 90$$ items in the sets. From the numbers $$1, 2, \ldots, 300$$, I'm going to select exactly $$90$$. I can indicate the selected ones by making a list of 300 dots and Xs:
..X....X...XX...
where the list above indicates that I selected the numbers 3, 9, 13, 14, and then many others that I haven't shown because I got sick of typing. :) I can take this sequence of 300 dots and Xs and break it into three groups of 100 dots each, which I arrange one atop the other, getting something that looks like this:
...X....X..X.....X...
.X...X.....X...X.....
..X...X.X..X......X..
but goes on for a full 100 items in each row. The number of Xs in each row might differ -- there might be 35 in the first row, 24 in the second, and 31 in the third, for instance, and that's OK. [Thanks to whuber for pointing out that I had this wrong in a first draft!]
Now look at each column: each column can be considered as a set, and that set has between 0 and 3 "X"s in it. I can write the tallies below the rows to get something like this:
...X....X..X.....X...
.X...X.....X...X.....
..X...X.X..X......X..
011101102003000101100
That is to say, I've produced 100 numbers, each between 1 and 3. And the sum of those 100 numbers must be the number of Xs, total, in all three rows, which was 90. So the average must be $$90/100 = 0.9$$, as desired.
So here are the steps to getting 100 integers between 1 and 4 whose average is exactly $$s$$. | {
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So here are the steps to getting 100 integers between 1 and 4 whose average is exactly $$s$$.
1. Let $$t = s - 1$$.
2. Compute $$k = 100 t$$; that's how many Xs we'll place in the rows, total.
3. Make a list of 300 dots-or-Xs, $$k$$ of which are Xs.
4. Split this into three rows of 100 dots-or-Xs, each containing about a third of the Xs, more or less.
5. Arrange these in an array, and compute column sums, getting 100 integers between $$0$$ and $$3$$. Their average will be $$t$$.
6. Add one to each column sum to get 100 integers between $$1$$ and $$4$$ whose average is $$s$$.
Now the tricky part of this is really in step 4: how do you pick $$300$$ items, $$k$$ of which are "X" and the other $$300-k$$ of which are "."? Well, it turns out that R has a function that does exactly that.
And then whuber tells you how to use it: you write
tabulate(sample.int((k-1)*n, s-n) %% n + 1, n)
For your particular case, $$n = 100$$, and $$s$$, the total number of items in all the small sets, is $$100r$$, and you want numbers between $$1$$ and $$4$$, so $$k = 4$$, so $$k-1$$ (the largest size for a 'small set') is 3, so this becomes
tabulate(sample.int(3*100, 100r-100) %% 100 + 1, n)
or
tabulate(sample.int(3*100, 100*(r-1)) %% 100 + 1, 100)
or, using my name $$t$$ for $$r - 1$$, it becomes
tabulate(sample.int(3*100, 100*t) %% 100 + 1, 100)
The "+1" at the end of his original formula is exactly the step needed to convert from "numbers between $$0$$ and $$3$$" to "numbers between $$1$$ and $$4$$".
Let's work from the inside out, and let's simplify to $$n = 10$$ so that I can show sample outputs:
tabulate(sample.int(3*10, 10*t) %% 10 + 1, 10)
And let's aim for $$t = 1.9$$, so this becomes
tabulate(sample.int(3*10, 10*1.9) %% 10 + 1, 10) | {
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And let's aim for $$t = 1.9$$, so this becomes
tabulate(sample.int(3*10, 10*1.9) %% 10 + 1, 10)
Starting with sample.int(3*10, 10*1.9): this produces a list of $$19$$ integers between $$1$$ and $$30$$. (i.e., it solved the problem of picking $$k$$ numbers out of your total -- $$300$$ in your real problem, $$30$$ in my smaller example).
As you'll recall, we want to produce three rows of ten dots-and-Xs each, something like
X.X.XX.XX.
XXXX.XXX..
XX.X.XXX..
We can read this left-to-right-top-to-bottom (i.e., normal reading order) to produce a list of locations for Xs: the first item's a dot; the second and third are Xs, and so on, so our list of locations starts out $$1, 3, 5, 6, \ldots$$. When we get to the end of a row, we just keep counting up, so for the picture above, the X-locations would be $$1, 3, 5, 6, 8, 9, 11, 12, 13, 14, 16, 17, 18, 21, 22, 24, 26, 27, 28$$. Is that clear?
Well, whubers code produces exactly that list of locations with its innermost section.
The next item is %% 10; that takes a number and produces its remainder on division by ten. So our list becomes $$1, 3, 5, 6, 8, 9, 1, 2, 3, 4, 6, 7, 8, 1, 2, 4, 6, 7, 8$$. If we break that into three groups --- those that came from numbers between $$1$$ and $$10$$, those that came from numbers from $$11$$ to $$20$$, and those that came from numbers $$21$$ to $$30$$, we get $$1, 3, 5, 6, 8, 9$$, then $$1, 2, 3, 4, 6, 7, 8,$$, and finally $$1, 2, 4, 6, 7, 8$$. Those tell you where the Xs in each of the three rows are. There's a subtle problem here: if there had been an X in position 10 in the first row, the first of our three lists would have been $$1, 3, 5, 6, 8, 9, 0$$, and the tabulate function doesn't like "0". So whuber adds 1 to each item in the list to get $$2, 4, 6, 7, 9, 10, 1$$. Let's move on to the overall computation:
tabulate(sample.int(3*10, 10*1.9) %% 10 + 1, 10) | {
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tabulate(sample.int(3*10, 10*1.9) %% 10 + 1, 10)
This asks "for those $$30$$ numbers, each indicating whether there's an X in some column, tell me how many times each column (from $$1$$ to $$10$$ --- that's what the final "10" tells you) appears, i.e., tell me how many Xs are in each column. The result is 0 3 2 2 2 1 3 2 3 1 which (because of the shift-by-one thing) you have to read as "there are no Xs in the 10th column; there are 3 Xs in the first column; there are 2 Xs in the second column," and so on up to "there is one X in the 9th column".
That gives you ten integers between $$0$$ and $$3$$ whose sum is $$19$$, hence whose average is $$1.9$$. If you increase each by 1, you get ten integers between $$1$$ and $$4$$ whose sum is $$29$$, hence an average value of $$2.9$$.
You can generalize to $$n = 100$$, I hope.
• +1 Welcome to our site, John. I appreciate your efforts to explain and clarify these ideas. At one point your description departs from what the code does: one does not divide the three rows into groups of 30 each. Instead, 90 cells out of the 300 cells in those rows are selected. Usually, each row will have a different number of cells. – whuber Jan 7 at 17:15
• Thanks...I actually worried about that a little bit as I wrote it, but I was in mid-sentence, and by the time I was finished, the thought had flown. I'll edit to try to fix it up. – John Jan 7 at 21:25 | {
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You can use sample() and select specific probabilities for each integer. If you sum the product of the probabilities and the integers, you get the expected value of the distribution. So, if you have a mean value in mind, say $$k$$, you can solve the following equation: $$k = 1\times P(1) + 2\times P(2) + 3\times P(3) + 4\times P(4)$$ You can arbitrarily choose two of the probabilities and solve for the third, which determines the fourth (because $$P(1)=1-(P(2)+P(3)+P(4))$$ because the probabilities must sum to $$1$$). For example, let $$k=2.3$$, $$P(4)=.1$$, and $$P(3)=.2$$. Then we have that $$k = 1 \times [1-(P(2)+P(3)+P(4)] + 2\times P(2) + 3\times P(3) + 4\times P(4)$$ $$2.3 = [1 - (P(2)+.1+.2)] + 2*P(2) + 3\times .2 + 4\times .1$$ $$2.3 = .7 + P(2) + .6 + .4$$ $$P(2)=.6$$ $$P(1)=1-(P(2)+P(3)+P(4)=1 - (.6+.1+.2)=.1$$
So you can run x <- sample(c(1, 2, 3, 4), 1e6, replace = TRUE, prob = c(.1, .6, .2, .1)) and mean(x) is approximately $$2.3$$ | {
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• This explains how to constrain the mean of the distribution. But, the OP specified in the comments that they want to constrain the sample mean (which won't match the mean of the distribution, except in expectation). On the other hand, it seems the OP accepted this answer anyway, so perhaps that's not what they wanted after all. – user20160 Jan 6 at 0:03
• are you sure? @user20160 why sample mean is not contrainted? it's equal to target – Fierce82 Jan 6 at 0:08
• This answer does not provide a way to make the sample mean equal the target value: most of the time the mean will not equal the target. – whuber Jan 6 at 0:10
• @TomZinger Yes. This answer nicely describes how to sample from a distribution with the given target mean. But, the mean of a sample drawn from a distribution will not generally equal the mean of the distribution. – user20160 Jan 6 at 4:59
• I wrote my answer before I saw that comment, but I figured this would be useful anyway. I imagined it would require an integer programming optimization problem to get a sample mean exactly equal to some value. – Noah Jan 6 at 8:20
Here is a simple algorithm: Create $$n-1$$ random integers in the range $$[1,4]$$ and calculate the $$n^{th}$$ integer for the mean to be equal to the specified value. If that number is smaller than $$1$$ or larger than $$4$$, then one by one distribute the surplus/lacking onto other integers, e.g. if the integer is $$5$$, we have $$1$$ surplus; and we may add this to the next integer if it's not $$4$$, else add to the next etc. Then, shuffle the entire array. | {
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• One big problem with this proposal is that it doesn't come along with any indication of what the expected frequencies of the resulting values are. – whuber Jan 5 at 23:32
• Although interesting, I thought the OP only requires an algorithm to generate the desired array of integers in a non-deterministic manner. – gunes Jan 5 at 23:37
• I think that avoids the essence of the question rather than providing a satisfactory answer. A good answer should be able to characterize the distribution it proposes in a meaningful way, such as by giving a formula for the probabilities or at least giving the first couple of moments. – whuber Jan 5 at 23:46
• A minor adjustment of the simulated data is likely 'proper', however, looking at the expertimental design in cases where more significant mean deviation is required, depending on the intended purpose, could be, from a hypothesis testing perspective, 'suspect', in my judgement. Either over or under loading a random design to justify or reject possible non-random effects that have been actually observed can be questionable practice. So, any method that makes a very small adjustment to the last of say a 100 observations is probably keeping in good practice, in my opinion. – AJKOER Jan 7 at 17:33
As a supplement to whuber's answer, I've written a script in Python which goes through each step of the sampling scheme. Note that this is meant for illustrative purposes and is not necessarily performant.
Example output:
n=10, s=20, k=4
Starting grid
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
X X X X X X X X X X
Filled in grid
X X . . X . X . . X
. . X X X . . . . .
. . . . X X . . . .
X X X X X X X X X X
Final grid
X X . . X . X . . X
. . X X X . . . . .
. . . . X X . . . .
X X X X X X X X X X
2 2 2 2 4 2 2 1 1 2
The script:
import numpy as np
# Define the starting parameters
integers = [1, 2, 3, 4]
n = 10
s = 20
k = len(integers) | {
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# Define the starting parameters
integers = [1, 2, 3, 4]
n = 10
s = 20
k = len(integers)
def print_grid(grid, title):
print(f'\n{title}')
for row in grid:
print(' '.join([str(element) for element in row]))
# Create the starting grid
grid = []
for i in range(1, k + 1):
if i < k:
grid.append(['.' for j in range(n)])
else:
grid.append(['X' for j in range(n)])
# Print the starting grid
print_grid(grid, 'Starting grid')
# Randomly and uniformly fill in the remaining rows
indexes = np.random.choice(range((k - 1) * n), s - n, replace=False)
for i in indexes:
row = i // n
col = i % n
grid[row][col] = 'X'
# Print the filled in grid
print_grid(grid, 'Filled in grid')
# Compute how many cells were selected in each column
column_counts = []
for col in range(n):
count = sum(1 for i in range(k) if grid[i][col] == 'X')
column_counts.append(count)
grid.append(column_counts)
# Print the final grid and check that the column counts sum to s
print_grid(grid, 'Final grid')
print()
print(f'Do the column counts sum to {s}? {sum(column_counts) == s}.')
I've turned whuber's answer into an r function. I hope it helps someone.
• n is how many integers you want;
• t is the mean you want; and
• k is the upper limit you want for your returned values
whubernator<-function(n=NULL, t=NULL, kMax=5){
z = tabulate(sample.int(kMax*(n), (n)*(t),replace =F) %% (n)+1, (n))
return(z)
}
It seems to work as expected:
> w = whubernator(n=10,t=4.2)
> mean(w)
[1] 4.2
> length(w)
[1] 10
> w
[1] 3 5 3 5 5 3 4 5 5 4
It can return 0s, which matches my needs.
> whubernator(n=2,t=0.5)
[1] 1 0
New contributor
gruvn is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. | {
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# Homework Help: Complex integer expression problem
1. Feb 23, 2006
### Werg22
If n is a positive integer such as
$$2{\leq}n{\leq}80$$
For how many values the expression $$\frac{(n+1)n(n-1)}{8}$$ takes positive and integer values?
I solved it that way...
$$\frac{(n+1)n(n-1)}{8}=\frac{(n^{2}-1)n}{8}$$
(n^2 - 1)n must have 8 as one of its factor.
Either n is a multiple of 8, or n^2 - 1 is. Also the case were n^2 - 1 has 4 as one of its factors, n having 2, and vice-versa, is impossible - if n^2 - 1 is even, n is odd, and vice-versa.
So every mutliple of 8 up to 80 is a possible value. So there is 10 values.
Let's list those numbers
8, 16 , 24 , 32 , 40 , 48 , 56 , 64 , 72, 80
Add one to each one of these values
9, 17, 25, 33, 41, 49, 65 , 73, 81
There is 4 perfect square in this list. So if n^2 - 1 is a multiple of 8, then there is 4 possible values for n.
10 + 4 = 14
So 14 possibilities in total. But the true awnser is not what I found. What is wrong in my reasoning?
2. Feb 23, 2006
### 0rthodontist
There's no guarantee that n^2 - 1 lies between 9 and 81. It could be much larger.
3. Feb 23, 2006
### Werg22
Right! Thanks. In that case any other way to solve this?
4. Feb 23, 2006
### 0rthodontist
Well since you are only looking up to 80 you can calculate them all directly, only takes a couple minutes to set things up. Also from observing that data it seems that every n yields an integer n(n+1)(n-1)/8 except for even integers that are not divisible by 8, so maybe you could break it down into parts. Every multiple of 8 yields an integer. And if x is a multiple of 2, and x is not itself divisible by 8, then x does not yield an integer because its adjacent integers are not even disible by 2. And every odd integer must yield an integer because its adjacent integers are both even, and one of the adjacent integers must also be divisible by 4.
Last edited: Feb 23, 2006
5. Feb 23, 2006
### Werg22
Ok. Since n^2 - 1 = (n-1)(n+1) | {
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Last edited: Feb 23, 2006
5. Feb 23, 2006
### Werg22
Ok. Since n^2 - 1 = (n-1)(n+1)
n - 1 must be a multiple of 8, or n+1 be a multiple of 8, or n - 1 be a multiple of 4, or n + 1 be a multiple of 4.
For exactly 10 values, n is multiple of 8
For exactly 10 values, n-1 is a multiple of 8.
For exactly 10 values, n+1 is a multiple of 8.
For exactly 20 values, n-1 is multiple of 4. Half of these being multiples of 8. So we count 10.
For exactly 20 values, n+1 is a multiple of 8. Half of these being multiples of 8. So we count 10.
So 10*5=50
The awnser is 50.
6. Feb 23, 2006
### AKG
I believe the answer is 49. You seem to be going about it in a very complicated way.
Case 1: n is odd
Then both n-1 and n+1 are even. Moreover, one of them (and in fact, only one of them) is a multiple of four. This is clear since if n is odd, then either n = 1 (mod 4) in which case n-1 = 0 (mod 4) and n+1 = 2 (mod 4), or n = 3 (mod 4) in which case n-1 = 2 (mod 4) and n+1 = 0 (mod 4). So since one of n-1 and n+1 is a multiple of four, and the other is even, the whole product (n-1)n(n+1) is a multiple of 8. So every odd n between 2 and 80 will do, and there are 39 such numbers.
Case 2: n is even
Then both n-1 and n+1 is odd, so if 8 | (n-1)n(n+1), then 8 | n, so the only even n's that work are multiples of 8. They are:
8, 16, 24, 32, 40, 48, 56, 64, 72, 80
That's 10, giving a total of:
49.
One problem with your solution is that n-1 is a multiple of 8 for only 9 values of n. Note that n-1 ranges from 1 to 79. n ranges from 2 to 80. n-1 will never be 80, so its missing one multiple of 8. That's where you're counting your extra one. Note also that there are only 19 values of n-1 which are a multiple of 4, 9 of which are multiples of 8, so when you subtract 9 from 19, you do still end up getting 10. Originally, you subtracted 10 from 20. You ended up with the right number, 10, but they way you got it was wrong.
7. Feb 24, 2006
### Werg22
Okay, I see. Thanks. | {
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Does $\sum\limits_n \log\left(1+{1\over n}\right)$ diverge or converge?
How do I find out if $\sum\limits_n\log(1+{1\over n})$ diverges or converges? Wolfram recommends me to use comparison test, but I do not know series which diverges and less than this.
• One can use Limit Comparison with $\sum \frac{1}{n}$. – André Nicolas Jan 3 '15 at 15:37
Wolfram recommends me to use (some) compar(is)on test, but I do not know (any) series which diverges and (is) less than this (one).
$$\log\left(1+\frac1n\right)\geqslant\frac1{2n}$$
$$\sum_{n=1}^{N}\log\left(1+\frac{1}{n}\right)=\log\prod_{n=1}^{N}\frac{n+1}{n}=\log(N+1).$$
If you want to use a comparison, notice that since $f(t)=\frac{1}{t}$ is a convex function on $\mathbb{R}^+$, we have: $$\log\left(1+\frac{1}{n}\right)=\int_{n}^{n+1}\frac{dt}{t}\geq\frac{1}{n+1/2}$$ by Jensen's inequality, hence: $$\sum_{n=1}^{N}\log\left(1+\frac{1}{n}\right)\geq 2\sum_{n=1}^{N}\frac{1}{2n+1} = 2H_{2n+1}-H_n.$$
Since
$$\sum_{n = 1}^N \log\left(1 + \frac{1}{n}\right) = \sum_{n = 1}^N \log\left(\frac{n+1}{n}\right) = \sum_{n = 1}^N [\log(n+1) - \log(n)] = \log(N+1) \to \infty$$
the series $\sum_{n = 1}^\infty \log\left(1 + \frac{1}{n}\right)$ diverges.
note $$\ln{\left(1+\dfrac{1}{n}\right)}=\dfrac{1}{n}+o(1/n)$$ since $$\sum_{n=1}^{\infty}\dfrac{1}{n}$$ is diverges
so $$\sum_{n=1}^{\infty}\ln{\left(1+\dfrac{1}{n}\right)}$$ is also diverges
• First statement isn't obvious to me. – Dark Archon Jan 3 '15 at 16:10
• do you know $\ln{(1+x)}=x+o(x)?$ – math110 Jan 3 '15 at 16:11
• What is $o(x)$? – Dark Archon Jan 3 '15 at 16:13
• $x=\dfrac{1}{n}$ when $n\to +\infty$,then $x\to 0$ – math110 Jan 3 '15 at 16:25
• @DarkArchon $f(x)=o(x)$ means that $\lim_{x\to 0}f(x)/x=0$. – Andrés E. Caicedo Jan 4 '15 at 0:52 | {
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The first natural idea to understand how $\log(1+1/n)$ behaves when $n$ is small is to do Taylor expansion around $x=0$ for $\log(1+x)$, so that you have estimates for $\log(1+x)$. Then you get $$\log(1+x) = \sum_{n \ge 1} \frac{(-1)^{n+1}}{n}x^n = x - x^2/2 + x^3/3 - \cdots,$$ by integrating the geometric series (and switching a minus sign) and by Taylor's theorem you get that there exists $1 \le \zeta \le x$ such that $$\log(1+x) = x-\frac{\zeta^2}2 \ge x - \frac {x^2}2 = \frac{2x-x^2}2 = \frac{x(2-x)}2 \ge \frac x2$$ for $0 \le x \le 1$ (because $\frac{x(1-x)}2 \ge 0$ on this interval).
A variant of this idea is to check that $\log$ is a concave function (the derivatives are $\frac 1x$ and $\frac {-1}{x^2}$), hence we can use Jensen's inequality : for all $\lambda \in [0,1]$, $$\log(\lambda x + (1-\lambda) y) \ge \lambda \log(x) + (1-\lambda) \log(y)$$ so that for $y=1$ and $x=2$, we get $$\log(1+ \lambda) \ge \lambda \log(2).$$ In both cases, we established an inequality of the form $\log(1+x) \ge cx$ for $x \in [0,1]$, which means $\log(1+1/n) \ge \frac cn$, hence your series diverges.
Hope that helps,
Another way to think of this problem is that $\sum log(1+1/n)$ = $\log \prod(1+1/n)$ = $\log \prod((n+1)/n)$, which goes to infinity. Thus the sum diverges.
• Already explained on the page, twice. – Did May 11 '15 at 5:21 | {
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# A problem about the limit of an integral
Let $g(x)$ be a continuous periodic function of period 1 on $\mathbb{R}$. Prove that for any integrable function $f(x)$ on $[0,1]$,
$$\lim_{n \to \infty}\int_0^{1}f(x)g(nx)dx= \int_0^{1}f(x)dx \int_0^{1}g(x)dx.$$
Any help is appreciated.
-
Begin with a transformation $u=nx.$
$$\int_{0}^1 f(x)g(nx)dx = \frac{1}{n} \int_{0}^n f(u/n)g(u)du.$$
Break the integrand up into a sum of intervals.
$$\frac{1}{n} \int_{0}^n f(u/n)g(u)du=\frac{1}{n}\sum_{j=1}^n\int_{j-1}^{j} f(u/n)g(u)du.$$
Make another variable transformation: $v=u-(j-1).$ Because $g(u)$ is 1 periodic $g(u)=g(u+1)=g(u+j-1).$
$$\frac{1}{n}\sum_{j=1}^n\int_{j-1}^{j} f(u/n)g(u)du = \frac{1}{n}\sum_{j=1}^n\int_{0}^{1} f \left(\frac{u-(j-1)}{n} \right)g(u-(j-1))du$$
Rearrange the terms. $$\frac{1}{n}\sum_{j=1}^n\int_{0}^{1} f\left(\frac{u-(j-1)}{n} \right)g(u-(j-1))du =\int_{0}^1 \left(\frac{1}{n} \sum_{j=1}^n f \left(\frac{u-(j-1)}{n} \right)\right)g(u)du .$$
We have now produced a Riemann sum which converges to an integral in the limit. We are now allowed, by dominated convergence theorem, to say
\begin{align*} \lim_{n\to \infty} \int_{0}^1 f(x)g(nx)dx &= \int_{0}^1 \left(\int_{0}^1 f(z)dz\right) g(u)du \\ &=\int_{0}^1 f(z)dz\int_{0}^1 g(u)du=\int_{0}^1 f(x)dx\int_{0}^1 g(x)dx . \end{align*}
-
Thank you so much. – Sume Aug 10 '11 at 3:28
this is not a good proof, just an idea. First consider f to be simple function, we can calculate that this identity holds. Then by the definition of Lebesgue integration, we can approximate f by simple functions. Because g is continuous(bounded), so there's no problem for the left hand side to converge.
-
Ok. I have seen this problem in the book: Principles of Real Analysis by C.D.Aliprantis and O.Burkinshaw, and since I knew that this book has a solution manual, I went and searched over there and got the solution. The problem given in the book is as follows: | {
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$\textbf{Problem.}$ Let $f:(0,\infty) \to \mathbb{R}$ be a real-valued continuous function such that $f(x+1)=f(x)$ for all $x \geq 0$. If $g:[0,1] \to \mathbb{R}$ is an arbitrary continuous function, then show that $$\lim_{n \to \infty} \ \int\limits_{0}^{1} g(x) \cdot f(nx) \ dx = \Biggl( \ \ \int\limits_{0}^{1} g(x) \ dx \Biggr) \cdot \Biggl( \ \ \int\limits_{0}^{1} f(x) \ dx\Biggr)$$
$\textbf{Solution.}$ Please see the book: Problems in real analysis a workbook with solutions Problem 23.14 Page $\textbf{205}$ for a complete solution.
- | {
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Name:
Andrew ID:
Collaborated with:
This lab is to be done in class (completed outside of class if need be). You can collaborate with your classmates, but you must identify their names above, and you must submit your own lab as an knitted HTML file on Canvas, by Tuesday 10pm, this week.
This week’s agenda: creating and updating functions; understanding argument and return structures; revisiting Shakespeare’s plays; code refactoring.
# Huber loss function
The Huber loss function (or just Huber function, for short) is defined as: $\psi(x) = \begin{cases} x^2 & \text{if |x| \leq 1} \\ 2|x| - 1 & \text{if |x| > 1} \end{cases}$ This function is quadratic on the interval [-1,1], and linear outside of this interval. It transitions from quadratic to linear “smoothly”, and looks like this:
It is often used in place of the usual squared error loss for robust estimation. The sample average, $$\bar{X}$$—which given a sample $$X_1,\ldots,X_n$$ minimizes the squared error loss $$\sum_{i=1}^n (X_i-m)^2$$ over all choices of $$m$$—can be inaccurate as an estimate of $$\mathbb{E}(X)$$ if the distribution of $$X$$ is heavy-tailed. In such cases, minimizing Huber loss can give a better estimate. (Interested in hearing more? Come ask Tudor or I!)
• 1a. Write a function huber() that takes as an input a number $$x$$, and returns the Huber value $$\psi(x)$$, as defined above. Hint: the body of a function is just a block of R code, i.e., in this code you can use if() and else() statements. Check that huber(1) returns 1, and huber(4) returns 7. | {
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• 1b. The Huber function can be modified so that the transition from quadratic to linear happens at an arbitrary cutoff value $$a$$, as in: $\psi_a(x) = \begin{cases} x^2 & \text{if |x| \leq a} \\ 2a|x| - a^2 & \text{if |x| > a} \end{cases}$ Starting with your solution code to the last question, update your huber() function so that it takes two arguments: $$x$$, a number at which to evaluate the loss, and $$a$$ a number representing the cutoff value. It should now return $$\psi_a(x)$$, as defined above. Check that huber(3, 2) returns 8, and huber(3, 4) returns 9.
• 1c. Update your huber() function so that the default value of the cutoff $$a$$ is 1. Check that huber(3) returns 5.
• 1d. Check that huber(a = 1, x = 3) returns 5. Check that huber(1, 3) returns 1. Explain why these are different.
• 1e. Vectorize your huber() function, so that the first input can actually be a vector of numbers, and what is returned is a vector whose elements give the Huber evaluated at each of these numbers. Hint: you might try using ifelse(), if you haven’t already, to vectorize nicely. Check that huber(x = 1:6, a = 3) returns the vector of numbers (1, 4, 9, 15, 21, 27).
• Challenge. Your instructor computed the Huber function values $$\psi_a(x)$$ over a bunch of different $$x$$ values, stored in huber_vals and x_vals, respectively. However, the cutoff $$a$$ was, let’s say, lost. Using huber_vals, x_vals columns of oops_df, and the definition of the Huber function, you should be able to figure out the cutoff value $$a$$, at least roughly. Estimate $$a$$ and explain how you got there. Hint: one way to estimate $$a$$ is to do so visually, using plotting tools (if you do - please use ggplot); there are other ways too.
oops_df <- data.frame(
x_vals = seq(0, 5, length=21),
huber_vals = c(0.0000, 0.0625, 0.2500, 0.5625, 1.0000, 1.5625, 2.2500,
3.0625, 4.0000, 5.0625, 6.2500, 7.5625, 9.0000, 10.5000,
12.0000, 13.5000, 15.0000, 16.5000, 18.0000, 19.5000,
21.0000)) | {
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# Shakespeare’s complete works
Recall, as in lab/hw from Week 1, that the complete works of William Shakespeare are available freely from Project Gutenberg. We’ve put this text file up at https://raw.githubusercontent.com/benjaminleroy/36-350-summer-data/master/Week1/shakespeare.txt.
# Getting lines of text play-by-play
• 2a. Below is the get_wordtab_from_url() from lecture. Modify this function so that the string vectors lines and words are both included as named components in the returned list. For good practice, update the documentation in comments to reflect your changes. Then call this function on the URL for the Shakespeare’s complete works (with the rest of the arguments at their default values) and save the result as shakespeare_wordobj. Using head(), display the first several elements of (definitely not all of!) the lines, words, and wordtab components of shakespeare_wordobj, just to check that the output makes sense to you.
#' Get a word table from text on the web
#'
#' @param str_url string, specifying URL of a web page
#' @param split string, specifying what to split on. Default is the regex
#' pattern "[[:space:]]|[[:punct:]]"
#' @param tolower Boolean, TRUE if words should be converted to lower case
#' before the word table is computed. Default is TRUE
#' @param keep_nums Boolean, TRUE if words containing numbers should be kept in
#' the word table. Default is FALSE
#'
#' @return list, containing word table, and then some basic numeric summaries
#'
#' @examples
#' endgame_speech_list <- get_wordtab_from_url(
#' paste0("https://raw.githubusercontent.com/benjaminleroy/",
#' "36-350-summer-data/master/Week1/endgame.txt"))
get_wordtab_from_url <- function(str_url, split = "[[:space:]]|[[:punct:]]",
tolower = TRUE,
keep_nums = FALSE) {
text <- paste(lines, collapse = " ")
words <- strsplit(text, split = split)[[1]]
words <- words[words != ""]
# Convert to lower case, if we're asked to
if (tolower) {
words <- tolower(words)
} | {
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# Convert to lower case, if we're asked to
if (tolower) {
words <- tolower(words)
}
# Get rid of words with numbers, if we're asked to
if (!keep_nums) {
words <- grep("[0-9]", words, inv = TRUE, val = TRUE)
}
# Compute the word table
wordtab <- table(words)
return(list(wordtab = wordtab,
number_unique_words = length(wordtab),
number_total_words = sum(wordtab),
longest_word = words[which.max(nchar(words))]))
}
• 2b. Go back and look Q5 of Homework 1, where you located Shakespeare’s plays in the lines of text for Shakespeare’s complete works. Set shakespeare_lines <- shakespeare_wordobj$lines, and then rerun your solution code (or the rerun the official solution code, if you’d like) for questions Q5a–Q5f of Homework 1, on the lines of text stored in shakespeare_wordobj$lines. You should end up with two vectors titles_start and titles_end, containing the start and end positions of each of Shakespeare’s plays in shakespeare_lines. Print out titles_start and titles_end to the console.
• 2c. Create a list shakespeare_lines_by_play of length equal to the number of Shakespeare’s plays (a number you should have already computed in the solution to the last question). Using a for() loop, and relying on titles_start and titles_end, extract the subvector of shakespeare_lines for each of Shakespeare’s plays, and store it as a component of shakespeare_lines_by_play. That is, shakespeare_lines_by_play[[1]] should contain the lines for Shakespeare’s first play, shakespeare_lines_by_play[[2]] should contain the lines for Shakespeare’s second play, and so on. Name the components of shakespeare_lines_by_play according to the titles of the plays.
# Getting word tables play-by-play | {
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# Getting word tables play-by-play
• 3a. Define a function get_wordtabfrom_lines() to have the same argument structure as get_wordtab_from_url(), which recall you last updated in Q2a, except that the first argument of get_wordtab_from_lines() should be lines, a string vector passed by the user that contains lines of text to be processed. The body of get_wordtab_from_lines() should be the same as get_wordtab_from_url(), except that lines is passed and does not need to be computed using readlines(). The output of get_wordtab_from_lines() should be the same as get_wordtab_from_url(), except that lines does not need to be returned as a component. For good practice, incude documentation for your get_wordtab_from_lines() function in comments (no need to include an example).
• 3b. Using a for() loop or one of the apply functions (your choice here), run the get_wordtab_from_lines() function on each of the components of shakespeare_lines_by_play, (with the rest of the arguments at their default values). Store the result in a list called shakespeare_wordobj_by_play. That is, shakespeare_wordobj_by_play[[1]] should contain the result of calling this function on the lines for the first play, shakespeare_wordobj_by_play[[2]] should contain the result of calling this function on the lines for the second play, and so on. | {
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• 3c. Using one of the apply functions, compute numeric vectors shakespeare_total_words_by_play and shakespeare_unique_words_by_play, that contain the number of total words and number of unique words, respectively, for each of Shakespeare’s plays. Each vector should only require one line of code to compute. Hint: "[["() is actually a function that allows you to do extract a named component of a list; e.g., try "[["(shakespeare_wordobj, "number_total_words"), and you’ll see this is the same as shakespeare_wordobj[["number_total_words"]]; you should take advantage of this functionality in your apply call. What are the 5 longest plays, in terms of total word count? The 5 shortest plays?
# Refactoring the word table functions
• 4. Look back at get_wordtab_from_lines() and get_wordtab_from_url(). Note that they overlap heavily, i.e., their bodies contain a lot of the same code. Redefine get_wordtab_from_url() so that it just calls get_wordtab_from_lines() in its body. Your new get_wordtab_from_url() function should have the same inputs as before, and produce the same output as before. So externally, nothing will have changed; we are just changing the internal structure of get_wordtab_from_url() to clean up our code base (specifically, to avoid code duplication in our case). This is an example of code refactoring.
Call your new get_wordtab_from_url() function on the URL for Shakespeare’s complete works, saving the result as shakespeare_wordobj2. Compare some of the components of shakespeare_wordobj2 to those of shakespeare_wordobj (which was computed using the old function definition) to check that your new implementation works as it should. | {
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• Challenge. Check using all.equal() whether shakespeare_wordobj and shakespeare_wordobj2 are the same. Likely, this will not return TRUE. (If it does, then you’ve already solved this challenge question!) Modify your get_wordtab_from_url() function from the last question, so that it still calls get_wordtab_from_lines() to do the hard work, but produces an output exactly the same as the original shakespeare_wordobj object. Demonstrate your success by calling all.equal() once again. | {
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# Math Help - [SOLVED] Finding Co-Ordinates of a Rectangle
1. ## [SOLVED] Finding Co-Ordinates of a Rectangle
Here's a question from a past paper which I have successfully attempted. My question is regarding part (iii). I have successfully figured out the co-ordinates by the following method:
Is my method correct, considering I did get the right answer?
But is there another simpler method to do this which would save time during an exam.
2. The diagonals bisect one another. The midpoint of $\overline{AC}$ is ?
3. Mid of AC is (6,6), the diagnols do bisect each other at the mid but we don't have the x-co-ordinates of B or D to equate the diagnols, or do we?
4. Originally Posted by unstopabl3
Mid of AC is (6,6), the diagnols do bisect each other at the mid but we don't have the x-co-ordinates of B or D to equate the diagnols, or do we?
But the y-coordinate of $B~\&~D$ is 6.
You are given the x-coordinate of $D$, so $Dh,6)" alt="Dh,6)" />
5. No, I meant real value of the x-co-ordinates has not been given since we have to calculate that ourselves.
I have already gotten the correct values by using the method mentioned in my first post.
As stated I want someone to solve this part of the question with a different, possibly easier method.
I am not after the answer, I am looking for an alternate method.
6. Originally Posted by unstopabl3
No, I meant real value of the x-co-ordinates has not been given since we have to calculate that ourselves.
I have already gotten the correct values by using the method mentioned in my first post.
As stated I want someone to solve this part of the question with a different, possibly easier method.
I am not after the answer, I am looking for an alternate method.
Hi unstopabl3,
I really don't see what method you used in your first post, but here's how I'd do it.
Plato already told you that the midpoint of BD is M(6, 6).
This means the y-coordinates of B and D are also 6. | {
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This means the y-coordinates of B and D are also 6.
This distance from B to D is 20 (found using distance formula)
Each individual segment of the diagonals measure 10 since they are bisected.
Using the distance formula, it is easy to determine the x-coordinates of B and D.
M(6, 6) -----> D(h, 6) = 10
$10=\sqrt{h-6)^2+(6-6)^2}$
7. Hello, unstopabl3!
Code:
| C(12,14)
| o
| * *
| * *
* *
B o - + - - - - - - - o D
* | *
------*-+-------*------------
*| *
o
A|(0,-2)
|
The diagram shows a rectangle $ABCD.$
We have: . $A(0,-2),\;C(12,14)$
The diagonal $BD$ is parallel to the $x$-axis.
$(i)$ Explain why the $y$-coordinate of $D$ is 6.
The diagonals of a rectangle bisect each other.
. . Hence, the midpoint of $AC$ is the midpoint of $BD.$
The midpoint of AC is: . $\left(\tfrac{0+12}{2},\;\tfrac{-2+14}{2}\right) \:=\:(6,6)$
Therefore, $B$ and $D$ have a $y$-coordinate of 6.
The $x$-coordinate of $D$ is $h.$
$(ii)$ Express the gradients of $AD$ and $CD$ in terms of h,.
We have: . $\begin{Bmatrix}A(0, -2) \\ C(12,14) \\ D(h,\;6) \end{Bmatrix}$
$m_{AD} \;=\;\frac{6(-2)}{h-9} \;=\;\frac{8}{h}$
$m_{CD} \;=\;\frac{6-14}{h-12} \;=\;\frac{-8}{h-12}$
$(iii)$ Calculate the $x$-coordinates of $D$ and $B.$
Since $m_{AD} \perp m_{CD}$ we have: . $\frac{8}{h} \;=\;\frac{h-12}{8} \quad\Rightarrow\quad h^2-12h - 64 \:=\:0$
Hence: . $(h+4)(h-16) \:=\:0 \quad\Rightarrow\quad h \:=\:-4,\:16$
Therefore: . $D(16,6),\;B(-4,6)$
8. Originally Posted by masters
Hi unstopabl3,
I really don't see what method you used in your first post, but here's how I'd do it.
$10=\sqrt{h-6)^2+(6-6)^2}$
I use the concept of the product of two perpendicular lines = -1
You can see the working in the Soroban's post. That's exactly how I did it! | {
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This distance from B to D is 20 (found using distance formula)
How did you get this with only the Y co-ordinates known for both? Did you get the distance of AC which should be equal to BD?
Soroban, thanks for your post, but I've already used that method to solve this problem and already mentioned it in my first post that I am looking for alternative methods to solve it! Thanks nonetheless!
9. Originally Posted by unstopabl3
I use the concept of the product of two perpendicular lines = -1
You can see the working in the Soroban's post. That's exactly how I did it!
How did you get this with only the Y co-ordinates known for both? Did you get the distance of AC which should be equal to BD?
Soroban, thanks for your post, but I've already used that method to solve this problem and already mentioned it in my first post that I am looking for alternative methods to solve it! Thanks nonetheless!
The distance BD is 20, found using the distance formula. BD = AC (diagonals of a rectangle are congruent).
10. Co ordinates of B and D are $(x_1 , 6) and (x_2, 6)$
Diagonal AC = BD
AC = 20 = BD.
Distance $BD^2 = (x_1 - x_2)^2$
So (x_1 - x_2) = 20...........(1)
Mid point point of AC = mid point of BD
$\frac{x_1+x_2}{2} = 6$
(x_1 + x_2) = 12......(2)
Solve Eq (1) ans (2) to find the coordinates of B and D.
11. Thanks for the responses guys! | {
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# how to express the $n$th power of the cosine as a series of cosines?
Which is the correct way for expressing the $$n$$th power of a cosine as a series of cosines without any exponent?
By using the Euler's formula
$$\cos^n{(\theta)}=\left( \frac{e^{j\theta}+e^{-j\theta}}{2} \right)^n= \frac{1}{2^n}\left( e^{j\theta}+e^{-j\theta} \right)^n=\frac{1}{2^n}\left( z+z^{-1} \right)^n.$$
with $$z=e^{j\theta}$$.
Since the term $$\left( z+z^{-1} \right)^n$$ is the $$n$$th power of a binomial, I could express it using the binomial identity, thus
$$\cos^n{(\theta)} = \frac{1}{2^n} \displaystyle\sum_{k=0}^n \binom{n}{k} z^k(z^{-1})^{n-k} = \frac{1}{2^n} \displaystyle\sum_{k=0}^n \binom{n}{k} z^{2k-n}.$$
If we expand the expression above, we obtain
$$\cos^n{(\theta)}=\frac{1}{2^n}\Bigg ( z^{-n} + \binom{n}{1}z^{-(n-2)} + \binom{n}{2}z^{-(n-4)} + \dots + \binom{n}{2}z^{n-4} + \binom{n}{1}z^{(n-2)} + z^n \Bigg )$$
which can be rewritten as
$$\cos^n{(\theta)}=\frac{1}{2^n} \Bigg ( (z^{-n} + z^n) + \binom{n}{1} \left(z^{-(n-2)} + z^{(n-2)}\right) + \binom{n}{2}\left( z^{-(n-4)} + z^{(n-4)}\right) + \dots \Bigg )$$
Finally, since $$z=e^{j\theta}$$
$$\cos^n{(\theta)}=\frac{1}{2^n} \Bigg ( (e^{-jn\theta} + e^{jn\theta}) + \binom{n}{1} \left(e^{j(n-2)\theta} + e^{-j(n-2)\theta}\right) + \binom{n}{2}\left( e^{j(n-4)\theta} + e^{-j(n-4)\theta}\right) + \dots \Bigg )$$
By applying the Euler's formula once again, we obtain
$$\cos^n{(\theta)}=\frac{2}{2^{n}} \sum_{k=0}^n \binom{n}{k} \cos{((n-2k)\theta)}.$$
Unfortunately, if I plug n=2, I obtain
$$\cos^2{(\theta)}= \cos{(2\theta)} + 1.$$
$$\cos^2{(\theta)}= \frac{1}{2}(\cos{(2\theta)} + 1).$$
a) Why my result is scaled by a factor of 2 ? b) Is the correct general formula $$\cos^n{(\theta)}=\frac{1}{2^{n}} \sum_{k=0}^n \binom{n}{k} \cos{((n-2k)\theta)}.$$ c)If so, why ? | {
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• You paired the terms up $k=0$ with $k=n$ etc so when you rewrite it in summation notation the sum shouldn't be over all $0 \leq k \leq n$ anymore. Also for $n$ even there is a term in the middle with no pair. Nov 8, 2021 at 22:33
• The right way to do this is a (finite) Fourier series. Nov 8, 2021 at 22:37
• And this can be used to compute the integral $$\int \cos^n(x) dx$$ Nov 8, 2021 at 22:43
• @podiki Yeah, I noticed the even $n$ constant, which is $\binom {n}{n/2}cos(0)$. I was trying to be as general as possible. Thanks for pointing out the error when returning to the summation notation. I will try to fix it and update the question. Nov 8, 2021 at 22:49
• Also, you never get out of summation notation. Nov 9, 2021 at 1:48
Here is a way to fix the factor of $$2$$ while still summing from $$0$$ to $$n$$ (setting aside the question of whether that is the best way to do the sum). You have $$\cos^n(\theta) = \frac{1}{2^n}\left( z^{-n} + \binom n1 z^{-(n-2)} + \binom n2 z^{-(n-4)} + \cdots + \binom n2 z^{n-4} + \binom n1 z^{n-2} + z^n \right).$$ Reversing the order of the sum, $$\cos^n(\theta) = \frac{1}{2^n}\left( z^n + \binom n1 z^{n-2} + \binom n2 z^{n-4} + \cdots + \binom n2 z^{-(n-4)} + \binom n1 z^{-(n-2)} + z^{-n} \right).$$ Adding termwise, \begin{align} 2 \cos^n(\theta) &= \frac{1}{2^n}\bigg( \left(z^{-n} + z^n\right) + \binom n1 \left(z^{-(n-2)} + z^{n-2}\right) + \binom n2 \left(z^{-(n-4)} + z^{n-4}\right) + \cdots \\ &\qquad\qquad + \binom n2 \left(z^{n-4} + z^{-(n-4)}\right) + \binom n1 \left(z^{n-2} + z^{-(n-2)}\right) + \left(z^n + z^{-n}\right) \bigg) \\ &=\frac{2}{2^n}\bigg(\cos(n\theta) + \binom n1 \cos((n-2)\theta) + \binom n2 \cos((n-4)\theta) + \cdots \\ &\qquad\qquad + \binom n2 \cos(-(n-4)\theta) + \binom n1 \cos(-(n-2)\theta) + \cos(-n\theta) \bigg) . \end{align} | {
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Canceling one factor of $$2$$ on each side, this simplifies to $$\cos^n(\theta) = \frac{1}{2^n} \sum_{k=0}^n \binom nk \cos{((n-2k)\theta)}.$$ For $$n = 2,$$ this gives \begin{align} \cos^2(\theta) &= \frac 14\left(\binom 20 \cos(2\theta) + \binom 21 \cos(0) + \binom 22 \cos(-2\theta) \right) \\ &= \frac 12\left(\cos(2\theta) + 1\right) \end{align} as expected.
Your mistake was you did not write the end of the sum after the three dots, so you did not notice that you had moved terms from the right end of the sum to the left end without replacing them, so about half the terms in your summation were not matched by terms in the $$+ \cdots +$$ notation. Adding two copies of the sum in reverse order, you don't have to move any terms so you won't make this mistake.
Note that it's conventional to combine the $$\cos(m\theta)$$ and $$\cos(-m\theta)$$ terms in the sum and have about half as many terms (exactly half as many for odd powers) at the cost of (usually) having separate summations for even and odd powers of the cosine.
• Thanks a lot, this was the answer I was looking for! I still prefer this method instead of more complicated ones, especially from a learning perspective. Nov 9, 2021 at 8:37
We start from $$s_n = \frac{1}{2^n} \sum_{k=0}^n \binom{n}{k} z^{2k-n}$$ Case 1: Let $$n = 2m$$ $$s_{2m} = \frac{1}{2^{2m}} \sum_{k=0}^{2m} \binom{2m}{k} z^{2(k-m)}$$
$$s_{2m} = \frac{1}{2^{2m}} \sum_{k=0}^{m} \binom{2m}{k} z^{2(k-m)} + \frac{1}{2^{2m}} \sum_{k=m+1}^{2m} \binom{2m}{k} z^{2(k-m)}$$
$$s_{2m} = \frac{1}{2^{2m}} \sum_{k=0}^{m} \binom{2m}{k} z^{2(k-m)} + \frac{1}{2^{2m}} \sum_{l=0}^{m-1} \binom{2m}{2m - l} z^{2(m - l)}$$
$$s_{n} = \frac{n!}{2^{n}(n/2)!^2} + \frac{1}{2^{n}} \sum_{k=0}^{n/2-1} \binom{n}{k} (z^{2k-n} + z^{n - 2k})$$
$$cos^{n}(\theta) = \frac{n!}{2^{n}(n/2)!^2} + \frac{2}{2^{n}} \sum_{k=0}^{n/2-1} \binom{n}{k} cos((2k-n)\theta)$$ | {
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$$cos^{n}(\theta) = \frac{1}{2^{n}} \sum_{k=0}^{n/2-1} \binom{n}{k} cos((2k-n)\theta) + \frac{1}{2^{n}}\binom{n}{n/2} \cos((2(n/2)-n)\theta) + \frac{1}{2^{n}} \sum_{l=n/2+1}^{n} \binom{n}{l} \cos((2l-n)\theta)$$
$$cos^{n}(\theta) = \frac{1}{2^{n}} \sum_{k=0}^{n} \binom{n}{k} cos((2k-n)\theta)$$
Case 2: Let $$n = 2m - 1$$ $$s_{2m - 1} = \frac{1}{2^{2m - 1}} \sum_{k=0}^{2m - 1} \binom{2m - 1}{k} z^{2(k-m) + 1}$$
$$s_{2m - 1} = \frac{1}{2^{2m - 1}} \sum_{k=0}^{m - 1} \binom{2m - 1}{k} z^{2(k-m) + 1} + \frac{1}{2^{2m - 1}} \sum_{k=m}^{2m - 1} \binom{2m - 1}{k} z^{2(k-m) + 1}$$
$$s_{2m - 1} = \frac{1}{2^{2m - 1}} \sum_{k=0}^{m - 1} \binom{2m - 1}{k} z^{2(k-m) + 1} + \frac{1}{2^{2m - 1}} \sum_{k = 0}^{m} \binom{2m - 1}{k} z^{2(m-k) - 1}$$
$$s_{n} = \frac{1}{2^{n}} \sum_{k=0}^{(n - 1)/2} \binom{n}{k} (z^{2k-n} + z^{n-2k})$$
$$cos^{n}(\theta) = \frac{2}{2^{n}} \sum_{k=0}^{(n - 1)/2} \binom{n}{k} cos((2k-n)\theta)$$
$$cos^{n}(\theta) = \frac{1}{2^{n}} \sum_{k=0}^{(n - 1)/2} \binom{n}{k} cos((2k-n)\theta) + \frac{1}{2^{n}} \sum_{l=(n + 1)/2}^{n} \binom{n}{l} cos((2l-n)\theta)$$
$$cos^{n}(\theta) = \frac{1}{2^{n}} \sum_{k=0}^{n} \binom{n}{k} cos((2k-n)\theta)$$
Therefore, mixing the both cases we get
$$cos^{n}(\theta) = \frac{1}{2^{n}} \sum_{k=0}^{n} \binom{n}{k} cos((2k-n)\theta)$$ | {
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# difference between scalar matrix and identity matrix | {
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The unit matrix is every nx n square matrix made up of all zeros except for the elements of the main diagonal that are all ones. If you multiply any number to a diagonal matrix, only the diagonal entries will change. #1. The following rules indicate how the blocks in the Communications Toolbox process scalar, vector, and matrix signals. See the picture below. The column (or row) vectors of a unitary matrix are orthonormal, i.e. An identity matrix is a square matrix whose upper left to lower right diagonal elements are 1's and all the other elements are 0's. 8) Unit or Identity Matrix. Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. You can put this solution on YOUR website! References [1] Blyth, T.S. Long Answer Short: A $1\times 1$ matrix is not a scalar–it is an element of a matrix algebra. 2. Equal Matrices: Two matrices are said to be equal if they are of the same order and if their corresponding elements are equal to the square matrix A = [a ij] n × n is an identity matrix if Back in multiplication, you know that 1 is the identity element for multiplication. In this post, we are going to discuss these points. In their numerical computations, blocks that process scalars do not distinguish between one-dimensional scalars and one-by-one matrices. A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. It is also a matrix and also an array; all scalars are also vectors, and all scalars are also matrix, and all scalars are also array The scalar matrix is basically a square matrix, whose all off-diagonal elements are zero and all on-diagonal elements are equal. Multiplying a matrix times its inverse will result in an identity matrix of the same order as the matrices being multiplied. Nonetheless, it's still a diagonal matrix since all the other entries in the matrix are . However, there is sometimes a meaningful way of | {
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since all the other entries in the matrix are . However, there is sometimes a meaningful way of treating a $1\times 1$ matrix as though it were a scalar, hence in many contexts it is useful to treat such matrices as being "functionally equivalent" to scalars. For an example: Matrices A, B and C are shown below. Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. While off diagonal elements are zero. If the block produces a scalar output from a scalar input, the block preserves dimension. In the next article the basic operations of matrix-vector and matrix-matrix multiplication will be outlined. [] is not a scalar and not a vector, but is a matrix and an array; something that is 0 x something or something by 0 is empty. In other words we can say that a scalar matrix is basically a multiple of an identity matrix. Yes it is. Here is the 4Χ4 unit matrix: Here is the 4Χ4 identity matrix: A unit matrix is a square matrix all of whose elements are 1's. Okay, Now we will see the types of matrices for different matrix operation purposes. Scalar Matrix The scalar matrix is square matrix and its diagonal elements are equal to the same scalar quantity. If a square matrix has all elements 0 and each diagonal elements are non-zero, it is called identity matrix and denoted by I. Basis. All the other entries will still be . A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. Closure under scalar multiplication: is a scalar times a diagonal matrix another diagonal matrix? and Robertson, E.F. (2002) Basic Linear Algebra, 2nd Ed., Springer [2] Strang, G. (2016) Introduction to Linear Algebra, 5th Ed., Wellesley-Cambridge Press It is never a scalar, but could be a vector if it is 0 x 1 or 1 x 0. The same goes for a matrix multiplied by an identity matrix, the result is always the same original non-identity (non-unit) matrix, and thus, as | {
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identity matrix, the result is always the same original non-identity (non-unit) matrix, and thus, as explained before, the identity matrix gets the nickname of "unit matrix". This topic is collectively known as matrix algebra. By I of the same scalar quantity scalars and one-by-one matrices 1\times 1 $matrix is basically a square,! Off-Diagonal elements are zero and all on-diagonal elements are equal difference between scalar matrix and identity matrix the order. Its diagonal elements are zero and all on-diagonal difference between scalar matrix and identity matrix are equal diagonal will! Inverse will result in an identity matrix of the same scalar quantity from scalar! Denoted by I x 1 difference between scalar matrix and identity matrix 1 x 0 a scalar–it is an element of matrix. Its inverse will result in an identity matrix of the same order as the matrices being multiplied vector it. And each diagonal elements are non-zero, it is never a scalar, but could a. Between one-dimensional scalars and one-by-one matrices a diagonal matrix, whose all off-diagonal elements are to! 1 x 0 an example: matrices a, B and C are shown below multiplying a matrix its! To discuss these points a unitary matrix are orthonormal, i.e a scalar matrix is square,!, B and C are shown below a unitary matrix are orthonormal, i.e scalar quantity blocks. Blocks that process scalars do not distinguish between one-dimensional scalars and one-by-one.. If the block produces a scalar matrix is basically a multiple of identity! If you multiply any number to a diagonal matrix, only the diagonal entries will.... Back in multiplication, you know that 1 is the identity element for multiplication off-diagonal! Long Answer Short: a$ 1\times 1 $matrix is basically a square matrix, only the diagonal will! Scalar output from a scalar input, the block produces a scalar output from a scalar times a diagonal,! Elements 0 and each diagonal elements are non-zero, it is 0 x 1 or 1 x 0 are... Of a matrix algebra | {
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