# Question from Berkley Problems in Mathematics. Show that det($A$) > 0 [duplicate]

Let $A = (a_{ij})$ be an $n\times n$ real matrix satisfying the conditions: $a_{ii} > 0$ $(1\leq i \leq n)$; $a_{ij} \leq0$ $(i\not=j,1\leq i,j \leq n)$; $\sum\limits_{i=1}^n a_{ij}>0$ $(1\leq i \leq n)$. Show that det($A$) > $0$ .

## marked as duplicate by mlc, user91500, user26857, The Dead Legend, Namaste linear-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 7 '17 at 11:17

• Hint: Show that if $\lambda$ is a real eigenvalue then it is positive. – user379195 May 7 '17 at 6:54

The matrix $A$ is diagonally dominant and so nonsingular. Replacing off-diagonal entries $a_{i,j}$ by $ta_{i,j}$ for $0\le t\le 1$ also gives a diagonally dominant and nonsingular matrix $A_t$. But $A_0$ is diagonal with positive determinant. By continuity of determinant $\det A_t>0$ and so $\det A=\det A_1>0$.

• A weakly diagonally dominant matrix may be singular. – mlc May 7 '17 at 7:00
• @mlc There's nothing weak about these matrices! – Lord Shark the Unknown May 7 '17 at 7:01
• \begin{bmatrix} 1 & -2 & -2 \\ -1 & 6 & -1 \\ -1 & -1 & 3 \end{bmatrix} what about this matrix; It's not a DD matrix. But confusingly, it is satisfying the question but not the result to be shown. I must be wrong somwhere. – Hirakjyoti Das May 7 '17 at 7:09
• @HirakjyotiDas one of your conditions is that the column sums are positive. – Lord Shark the Unknown May 7 '17 at 7:13
• That's the very thing I was missing; Thank you – Hirakjyoti Das May 7 '17 at 7:16

In addition to the other answers, one can more generally see that from the Gershgorin circle theorem

Here firstly $A$ is diagonally dominant. Secondly $A$ is $Z$-matrix as all its off diagonal entries are $\leq 0.$

Also diagonal entries are $> 0.$ Hence $A$ is a $P$-matrix. (All principal minors are positive).

So $\det(A) > 0.$


# Why is this approximation of polynomial root so accurate?

I have an engineering problem where I have to find the smallest positive real root of a polynomial in $x$: $$Ax^5+Bx^3 - C = 0$$ Instead of solving numerically, I want simple approximative formulas ("design equations") that describe the behaviour. For that matter, I split the problem into two regimes:

• Large $x$: $\ \ \ \ x^5$ is dominant $\Longrightarrow \ \ Ax^5 \approx C \ \ \Longrightarrow \ \ x \approx \sqrt[5]{C/A} =: x_A$
• Small $x$: $\ \ \ \ x^3$ is dominant $\Longrightarrow \ \ Bx^3 \approx C \ \ \Longrightarrow \ \ x \approx \sqrt[3]{C/B} =: x_B$

The approximations work well below/above a certain threshold on $x$, but require an ugly case distinction. In order to avoid that and having something smoother than $$x \approx \min\{x_A, x_B\}$$ I tried Pythagorean-style combination of the inverses (inspired by the resistance of parallel circuits in electrical engineering) and found that $$x \approx 1\Big/ \ \left\|\left(\begin{matrix} 1/x_A \\ 1/x_B \end{matrix}\right)\right\|_4 = 1 \Big/ \sqrt[4]{1/x_A^4 + 1/x_B^4}$$ is a really, really good approximation. Pretty much perfect. That leads me to my actual question: Is it possible to argue why that 4-norm is such an amazing approximation? Does my initial polynomial have a certain structure that explains the high accuracy of my approximation?

Since I want to present/defend that stuff, I'd appreciate some sophistication.

• Are $A,B,C$ positive? If so the "smallest positive real root" is the only real root. – alex.jordan Sep 2 '15 at 17:19
• For example, the sole real root of $17x^5 + 41x^3 - 105$ is approximately $1.17624...$; the approximation gives $1.17849...$. – Unit Sep 2 '15 at 17:20
• @alex.jordan Yes they are positive. And sure, since $f(x) = Ax^5 + Bx^3$ is strictly increasing, there is only one relevant solution. – GDumphart Sep 2 '15 at 17:21
• What do you mean by a "really, really good approximation", and have you tried lots of values of $A$, $B$ and $C$? – David Quinn Sep 2 '15 at 17:27
• @DavidQuinn No I haven't done exhaustive tests with all kinds of $A,B$ combos, only what my problem yielded. Here is what I mean with great approximation: i.imgur.com/7Bfl8nt.png Only ridiculous zooming exposes some error. – GDumphart Sep 2 '15 at 17:41

$$\left(\frac x{x_A}\right)^5+\left(\frac x{x_B}\right)^3-1=0.$$

With $y=\dfrac x{x_A}$ and $r=\dfrac{x_A}{x_B}$, a single parameter remains:

$$y^5+r^3y^3-1=0.$$

Then $$r=\sqrt[3]{\frac{1-y^5}{y^3}}=\frac1y\sqrt[3]{1-y^5},$$ can be compared to your approximation $$y=\frac1{\sqrt[4]{1+r^4}},$$i.e. $$r=\sqrt[4]{\frac1{y^4}-1}=\frac1y\sqrt[4]{1-y^4}.$$ The agreement is indeed excellent on a wide range [abscissa $y$, ordinate $r$]:

For small $y$, both behaviors are identical, $\dfrac1y$. For $y$ close to $1$, behaviors are similar, approximately $\sqrt[3]{5(1-y)}$ and$\sqrt[4]{4(1-y)}$, and a blend in between.

The good agreement is explained by the fact that the functional relations are similar, with the exponent $4$ intermediate between $3$ and $5$ (but the value $4$ has nothing "magical").

• Using Yves's parametrization, the absolute value of the difference between the true $y$ and your approximation turns out to be at most $0.01171$. The maximum absolute difference occurs at approximately $r = 0.66145$, where the true $y \approx 0.94545$ while the approximation is $0.95716$. – Robert Israel Sep 2 '15 at 18:50
• @RobertIsrael: you can try slightly lower values than $4$ for the exponent, such as $3.7$. – Yves Daoust Sep 2 '15 at 18:53
• Yes, I can. The best in terms of maximum absolute difference seems to be approximately $3.85$ (i.e. approximation $(1 + r^{3.85})^{-1/3.85}$), where the maximum absolute difference is about $0.007608$. – Robert Israel Sep 2 '15 at 21:44
• Does this mean that there's an exact solution, rather than the approximation that the OP is using – Dr Xorile Sep 2 '15 at 22:03
• Amazing answer, very intuitive and accounts for everything I wished for, thank you. I reproduced all your steps. – GDumphart Sep 3 '15 at 6:34


# Math Help - Probability problem about cubes

1. ## Probability problem - cubes

Each face of a cube is painted either Red or Blue, each with probability 1/2. The color of each face is determined independently. What is the probabilty that the painted cube can be faced on a horizontal surface so that the four vertical faces are all the same color?

there can be 64 different color cubes 2x2x2x2x2x2x2 ?

case 1 when the cube is all red or blue it will work - 2/64

I couldn't think of the cases...

Could someone just give me some hints on the cases I have to set up?

Vicky.

2. Originally Posted by Vicky1997
Each face of a cube is painted either Red or Blue, each with probability 1/2. The color of each face is determined independently. What is the probability that the painted cube can be faced on a horizontal surface so that the four vertical faces are all the same color?

there can be 64 different color cubes 2x2x2x2x2x2x2 ?

case 1 when the cube is all red or blue it will work - 2/64

I couldn't think of the cases...

Could someone just give me some hints on the cases I have to set up?

Vicky.
That's a good start. You can continue it like this. If one face is one colour and the other five faces are all the other colour then it will work (2×6 cases of that). If two faces are one colour (and the other four faces are the other colour) then it will work provided that the two faces are opposite each other (2×3 cases, because there are three pairs of opposite faces). Finally, it can't be done at all if there are three faces of each colour.

3. Hello, Vicky1997!

A fascinating problem . . .

Each face of a cube is painted either Red or Blue, each with probability 1/2.
The color of each face is determined independently.
What is the probabilty that the painted cube can be placed on a horizontal surface
so that the four vertical faces are all the same color?

There can be 64 different colored cubes: $2^6 = 64$ . . . . Right!
There are seven possible colorings:

. . $\begin{array}{|c|c|} \hline
\text{Colors} & \text{Ways} \\ \hline
\text{6 red, 0 blue} & 1 \\ \text{5 red, 1 blue} & 6 \\ \text{4 red, 2 blue} & 15 \\ \text{3 red, 3 blue} & 20 \\ \text{2 red, 4 blue} & 15 \\ \text{1 red, 5 blue} & 6 \\ \text{0 red, 6 blue} & 1 \\ \hline \end{array}$
. . You may recognize these as Binomial Coefficients.

Let $V$ = "the four vertical faces have the same color".

6 red, 0 blue: . $P(\text{6 red}) \:=\:\frac{1}{64}$
To have $V$, the cube can be placed on any face.
. . $P(\text{6 red} \wedge V)\:=\:\frac{1}{64}\cdot\frac{6}{6} \:=\:\frac{6}{384}$

5 red, 1 blue: . $P(\text{5 red}) \:=\:\frac{6}{64}$
To have $V$, the blue face must be on the top or bottom
. $P(\text{5 red} \wedge V) \:=\:\frac{6}{64}\cdot\frac{2}{6} \:=\:\frac{12}{384}$

4 red, 2 blue.
To have $V$, the 2 blues must be on opposite faces.
. . This happens only $\frac{3}{64}$ of the time.
Then the blue faces must be on the top and the bottom.
. . $P(\text{4 red} \wedge V) \:=\:\frac{3}{64}\cdot\frac{2}{6} \:=\:\frac{6}{384}$

3 red, 3 blue.
With 3 of each color, it is impossible to have $V$.

2 red, 4 blue.
This has the same probability as "4 red, 2 blue".
. . $P(\text{2 red} \wedge V) \:=\:\frac{6}{384}$

1 red, 5 blue.
This has the same probability as "5 red, 1 blue".
. . $P(\text{1 red} \wedge V) \:=\:\frac{12}{384}$

0 red, 6 blue.
This has the same probability as "6 red, 0 blue".
. . $P(\text{0 red} \wedge V) \:=\:\frac{6}{384}$

Therefore: . $P(V) \;=\;\frac{6}{384} + \frac{12}{384} + \frac{6}{384} + \frac{6}{384} + \frac{12}{384} + \frac{6}{384} \;=\;\frac{48}{384} \;=\;\frac{1}{8}$


# Relationship between input and output sequence in Hartley transformation

As you know that Discrete Hartley transformation is related to the discrete Fourier transformation, $$i.e$$, assuming we have a vector $$X = [x_0,x_1,\ldots,x_N]$$, its Hartley transformation is equal to $$H(X) = Real(FFT(X)) - Imag(FFT(X))$$. where $$H$$ denotes the Hartley transformation.

I wonder, if I want the output of the Hartley transformation equals to a vector of length $$N$$ whose all elements are $$1$$, it means $$H(X) = [1,1,\ldots,1]$$, what supposed to be the input $$X$$? I means I need to formulate the relationship between the inputs and outputs.

Wikipedia's entry for the discrete Hartley transform shows states that the $$\mathsf{DHT}$$ is, up to a scaling, its own inverse. If $$x$$ is a vector with $$N$$ entries and $$y$$ is its discrete Hartley transform, $$$$y = \mathsf{DHT}x,$$$$ then $$$$x = \frac{1}{N}\mathsf{DHT}y.$$$$

If $$x$$ is a vector with $$N$$ entries such that $$$$\mathsf{DHT}x = \underbrace{(1,1,\ldots,1)^{\mathsf{T}}}_{\textrm{N entries}},$$$$ then we recover $$x$$ with $$$$x = \frac{1}{N}\mathsf{DHT}\left(\begin{array}{c}1\\1\\\vdots\\1\end{array}\right).$$$$ This means that $$x$$ is $$$$\begin{split} x &=~ \frac{1}{N}\left(\mathsf{Re}\left[\mathsf{DFT}\left(\begin{array}{c}1\\1\\\vdots\\1\end{array}\right)\right] - \mathsf{Im}\left[\mathsf{DFT}\left(\begin{array}{c}1\\1\\\vdots\\1\end{array}\right)\right]\right), \end{split}$$$$ where $$\mathsf{DFT}$$ is the discrete Fourier transform, which we usually compute with a FFT algorithm. The $$\ell^{\textrm{th}}$$ entry of the $$\mathsf{DFT}$$ of the all-1 vector is $$$$\begin{split} \sum_{n=0}^{N-1}1\times e^{-2\pi j \ell n/N} &=~ 1 + e^{-2\pi j \ell/N} + \left(e^{-2\pi j \ell/N}\right)^2 + \cdots + \left(e^{-2\pi j \ell/N}\right)^{N-1}\\ &=~ \left\{\begin{array}{rl}N & \textrm{if}~\ell=0,\\0&\textrm{if}~\ell\neq 0.\end{array}\right.\\ &=~ N\delta_{\ell,0}, \end{split}$$$$ where $$\delta_{p,q}$$ is the Kronecker delta. One way to show this is to note that if $$\ell = 0$$, then each exponent is $$0$$, so each term in the sum is $$1$$. On the other hand, if $$\ell\neq 0$$, then $$\exp(-2\pi j\ell/N) \neq 1$$,and $$$$\begin{split} 1 + e^{-2\pi j \ell/N} + \left(e^{-2\pi j \ell/N}\right)^2 + \cdots + \left(e^{-2\pi j \ell/N}\right)^{N-1} &=~ \frac{1 - \left(e^{-2\pi j \ell/N}\right)^N}{1 - e^{-2\pi j \ell/N}}\\ &=~ \frac{1 - e^{-2N\pi j \ell/N}}{1 - e^{-2\pi j \ell/N}}\\ &=~ \frac{1 - 1}{1 - e^{-2\pi j \ell/N}} ~=~ 0. \end{split}$$$$

That shows that the $$\mathsf{DFT}$$ of the all-1 vector has no imaginary part, and its real part is $$(N,0,0,\ldots,0)^{\mathsf{T}}$$. Hence $$$$x ~=~ \frac{1}{N}\left(\begin{array}{c} N\\0\\0\\\vdots\\0 \end{array}\right) ~=~ \left(\begin{array}{c} 1\\0\\0\\\vdots\\0 \end{array}\right).$$$$

• Thank you so much .. that's really appreciated.
– Gze
May 20 '20 at 3:19
• Thanks again, that's really interesting. I am trying to understand it well, if I got a question, I will let you know.
– Gze
May 20 '20 at 6:36
– Gze
May 20 '20 at 11:46

The Hartley transform is an involution: it is (up to a scale factor) its own inverse. The classical discrete Hartley transform of order $$N$$ is such that $$H_N^{-1} = \frac{1}{N}H_N$$. Be careful with your notation, the vector $$x$$ has $$N+1$$ entries, so maybe you are after an $$N+1$$-order DHT!

If $$\mathbf{1}$$ denotes the all-ones vector, then in matrix-vector product $$H_Nx = \mathbf{1}$$ is equivalent to $$H_N^{-1}H_Nx = H_N^{-1}\mathbf{1}$$, and you get more directly the mysterious $$X$$:

$$x = \frac{1}{N}H_N\mathbf{1}$$

and you get the result given by Joe Mac by a direct computation. A little interpretation: the discrete "Dirac" signal at the origin yields a flat "Hartley" spectrum, just like for Fourier.

• Thank you for explaining it, I got it now. but I was wondering what's about if $H_Nx = y$ and only some values of, i.e., $y_{1:4:N} = 1$ and other values of $y$ are any random values. So, can we express the relationship between the $y_{1:4:N}$ and $x_{1:4:N}$ mathematically in similar way too?
– Gze
May 20 '20 at 11:44
• I should add something, .. So, can we express the relationship between the $y_{1:4:N}$ and $x_{1:4:N}$ OR between $y_{1:4:N}$ and $x$ mathematically in similar way too?
– Gze
May 20 '20 at 12:07
• I don't really understand this comment. This may require a different question. But you can indeed interleave transforms and downsampling operators, and solutions can be quite complicated May 20 '20 at 12:16
• OK .. I posted it as a new question here dsp.stackexchange.com/questions/67696/… .. thank you
– Gze
May 20 '20 at 14:49
• And you did well. I am not able to see a clear answer right now May 20 '20 at 22:13


# Probability of Couples sitting next to each other (Sitting in a Row)

Suppose we have 4 couples and 8 seats. The seats are oriented such that it is a row. The people take their seats in a random fashion.

What is the probability that 2 couples sit next to each other?

My work: Let us label the people as $A_1, A_2, A_3, A_4, B_1, B_2, B_3, B_4$. $A_i$ and $B_i$ are couples. Then, we want to know the probability of $P(C_i \cap C_j), i \neq j$ and $C_i$ means couple $i$ are seated next to each other.

Then, suppose we fix $C_i$ and $C_j$. Then, the row arrangement could look something like: $\{ C_i, C_j, a,b,c,d\}$, where $a,b,c,d$ are the remaining 4 people. Then, there are $\binom{6}{2}$ ways of picking positions for $C_i$ and $C_j$, $2!$ ways of arranging the elements in $C_i$, $2!$ ways of arranging the elements in $C_j$, $2!$ ways of arranging $C_i$ and $C_j$ relative to each other, and $4!$ ways of arranging the remaining 4 people. There are $8!$ ways to arrange the 8 people in a row. Also, there are $\binom{4}{2}$ ways to pick 2 couples out of 4 couples.

Putting all this together, $$P(C_i \cap C_j) = \frac{\binom{6}{2} \binom{4}{2} 2! 2! 2! 4! }{8!}$$

I am curious because this differs from the solution given here: cached solution from google

Note that on the bottom of page 5, the difference is the google solution is missing a $\binom{6}{2}$ term, which means we can choose 2 positions for $C_i$ and $C_j$ out of the available 6 positions. (Please observe the quantity $p_2$ in the linked document). The factor $\binom{4}{2}$ is accounted for later on the top of page 6, where the author sums up all the probabilities corresponding to possible indices $i$ and $j$.

Thanks.

-

I would agree with you that the cached answer looks wrong and that there is a missing factor of 15 in $p_2$ (and of 20 in $p_3$).

When you ask "What is the probability that 2 couples sit next to each other?" you are looking at what was originally the $p_2$ part of the "What is the probability that at least one of the wives ends up sitting next to her husband?" question.

$p_2$ is the probability that two particular couples sit with husband next to wife. As you say, the ${4 \choose 2} = 6$ for which two couples these are is included in the $4p_1-6p_2+4p_3-p_4$ formula.

I would then say that the two particular couples have each become one unit (with $(2!)^2$ ways of arranging the individuals within the couple), reducing the number of orderable units from $8$ to $6$ so we should have $$p_2 = \frac{(2!)^2 \; 6!}{8!} \approx 0.071428571$$ and similarly $p_1 = \frac{(2!)^1 \; 7!}{8!} = 0.25$, $p_3 = \frac{(2!)^3 \; 5!}{8!} \approx 0.023810$, and $p_4 = \frac{(2!)^4 \; 4!}{8!} \approx 0.00952381$.

For $p_2$ and $p_3$, these are not the same as the cached answer, though the $p_2$ is the same as yours. Using the inclusion-exclusion formula this gives me a probability that at at least one of the wives ends up sitting next to her husband of about $0.657142857$, not the $0.9666667$ of the cached answer.

Some simulation convinces me this looks plausible.

-
Hi Henry: Thanks, this is the exact answer, I understand it, and I can also verify this in MATLAB. For the purpose of learning, I have included the MATLAB simulation code here: pastebin.com/UXjPEDaE . All of the different scenarios can be checked by making modifications to the code. –  jrand Jun 5 '11 at 17:15

Then, there are $\binom{6}{2}$ ways of picking positions for $C_i$ and $C_j$

I think that this is where you're wrong. In fact, there are only 5 (as in $\binom{5}{1}$) ways of placing them (considering that you account for order later):

• $(C_iC_j,a,b,c,d)$
• $(a,C_iC_j,b,c,d)$
• $(a,b,C_iC_j,c,d)$
• $(a,b,c,C_iC_j,d)$
• $(a,b,c,d,C_iC_j)$
-
Hi trutheality, you have shown 5 possible arrangements. Remember that $C_i$ actually contains two elements - it means couple $i$ is seated next to each other. Then, this is a possible arrangement: $(C_i, a,b,c,d,C_j)$. To be more clear, $(C_i, a,b,c,d,C_j) = (A_i, B_i, X_a, X_b, X_c, X_d, A_j, B_j)$. The X's must be set so that they refer to exactly the four remaining people. –  jrand Jun 4 '11 at 22:30
I think that we understand the question differently: I took "two couples seated next to each other" to mean that not only do we have $A_i$ next to $B_i$ and $A_j$ next to $B_j$, but also $C_i$ next to $C_j$. If you drop the last requirement then your answer is correct. –  trutheality Jun 4 '11 at 22:34
In any case, if we use your interpretation, where pairs of couples must be seated next to each other, the probability should be: $P(C_i \cap C_j) = \frac{5 \binom{4}{2} 2! 2! 2! 4! }{8!}$. I have replaced $\binom{6}{2}$ by $5$. This is still different from the claimed answer in the link. –  jrand Jun 4 '11 at 22:36
I think you're right then, the answer in the link doesn't seem to account for the different ways of arranging couples and non-couples. –  trutheality Jun 4 '11 at 22:48


## do all bijective functions have an inverse

This result says that if you want to show a function is bijective, all you have to do is to produce an inverse. Because if it is not surjective, there is at least one element in the co-domain which is not related to any element in the domain. Since "at least one'' + "at most one'' = "exactly one'', f is a bijection if and only if it is both an injection and a surjection. Now we consider inverses of composite functions. That is, for every element of the range there is exactly one corresponding element in the domain. bijectivity would be more sensible. Read Inverse Functions for more. For Free, Kharel's Simple Procedure for Factoring Quadratic Equations, How to Use Microsoft Word for Mathematics - Inserting an Equation. In this case, the converse relation $${f^{-1}}$$ is also not a function. The function f is called an one to one, if it takes different elements of A into different elements of B. In practice we end up abandoning the … A triangle has one angle that measures 42°. Image 2 and image 5 thin yellow curve. The figure given below represents a one-one function. A bijective function is also called a bijection. If the function satisfies this condition, then it is known as one-to-one correspondence. Choose an expert and meet online. The inverse relation is then defined as the set consisting of all ordered pairs of the form (2,x). Can you provide a detail example on how to find the inverse function of a given function? For a function to have an inverse, each element y ∈ Y must correspond to no more than one x ∈ X; a function f with this property is called one-to-one or an injection. Some non-algebraic functions have inverses that are defined. Since the relation from A to B is bijective, hence the inverse must be bijective too. In its simplest form the domain is all the values that go into a function (and the range is all the values that come out). So what is all this talk about "Restricting the Domain"? On A Graph . So what is all this talk about "Restricting the Domain"? If you were to evaluate the function at all of these points, the points that you actually map to is your range. In general, a function is invertible as long as each input features a unique output. Those that do are called invertible. It should be bijective (injective+surjective). And the word image is used more in a linear algebra context. To use an example f(x), f(x) is one-to-one if and only if for every value of f(x) there is exactly one value of x that gives that value. Yes, but the inverse relation isn't necessarily a function (unless the original function is 1-1 and onto). For a function to have an inverse, each element y ∈ Y must correspond to no more than one x ∈ X; a function f with this property is called one-to-one or an injection. Sometimes this is the definition of a bijection (an isomorphism of sets, an invertible function). That is, for every element of the range there is exactly one corresponding element in the domain. If f −1 is to be a function on Y, then each element y ∈ Y must correspond to some x ∈ X. http://www.sosmath.com/calculus/diff/der01/der01.h... 3 friends go to a hotel were a room costs $300. Naturally, if a function is a bijection, we say that it is bijective.If a function $$f :A \to B$$ is a bijection, we can define another function $$g$$ that essentially reverses the assignment rule associated with $$f$$. A function with this property is called onto or a surjection. The function is bijective (one-to-one and onto, one-to-one correspondence, or invertible) if each element of the codomain is mapped to by exactly one element of the domain. Since the function from A to B has to be bijective, the inverse function must be bijective too. A one-one function is also called an Injective function. ), the function is not bijective. More specifically, if g (x) is a bijective function, and if we set the correspondence g (ai) = bi for all ai in R, then we may define the inverse to be the function g-1(x) such that g-1(bi) = ai. Bijective functions have an inverse! Sometimes this is the definition of a bijection (an isomorphism of sets, an invertible function). 4.6 Bijections and Inverse Functions. So, to have an inverse, the function must be injective. No. Let f : A ----> B be a function. Get a free answer to a quick problem. Draw a picture and you will see that this false. That is, we say f is one to one In other words f is one-one, if no element in B is associated with more than one element in A. Example: The polynomial function of third degree: f(x)=x 3 is a bijection. Assume ##f## is a bijection, and use the definition that it … create quadric equation for points (0,-2)(1,0)(3,10)? Let us now discuss the difference between Into vs Onto function. Yes, but the inverse relation isn't necessarily a function (unless the original function is 1-1 and onto). Join Yahoo Answers and get 100 points today. pleaseee help me solve this questionnn!?!? You have assumed the definition of bijective is equivalent to the definition of having an inverse, before proving it. If every "A" goes to a unique "B", and every "B" has a matching "A" then we can go back and forwards without being led astray. and do all functions have an inverse function? So a bijective function follows stricter rules than a general function, which allows us to have an inverse. This is clearly not a function (for one thing, if you graph it, it fails the vertical line test), but it is most certainly a relation. You have to do both. Let us start with an example: Here we have the function Obviously neither the space$\mathbb{R}$nor the open set in question is compact (and the result doesn't hold in merely locally compact spaces), but their topology is nice enough to patch the local inverse together. 2xy=x-2 multiply both sides by 2x, 2xy-x=-2 subtract x from both sides, x(2y-1)=-2 factor out x from left side, x=-2/(2y-1) divide both sides by (2y-1). ….Not all functions have an inverse. It is a function which assigns to b, a unique element a such that f(a) = b. hence f-1 (b) = a. To use an example f(x), f(x) is one-to-one if and only if for every value of f(x) there is exactly one value of x that gives that value. A bijection is also called a one-to-one correspondence . Cardinality is defined in terms of bijective functions. To prove f is a bijection, we must write down an inverse for the function f, or shows in two steps that. sin and arcsine (the domain of sin is restricted), other trig functions e.g. Still have questions? It is clear then that any bijective function has an inverse. This is clearly not a function (for one thing, if you graph it, it fails the vertical line test), but it is most certainly a relation. A bijective function is a bijection. View FUNCTION N INVERSE.pptx from ALG2 213 at California State University, East Bay. How do you determine if a function has an inverse function or not? Ryan S. Summary and Review; A bijection is a function that is both one-to-one and onto. Inverse Functions An inverse function goes the other way! The process of "turning the arrows around" for an arbitrary function does not, in general, yield a function, but properties (3) and (4) of a bijection say that this inverse relation is a function with domain Y. Start here or give us a call: (312) 646-6365. Inverse Functions:Bijection function are also known as invertible function because they have inverse function property. Most questions answered within 4 hours. The inverse relation is then defined as the set consisting of all ordered pairs of the form (2,x). both 3 and -3 map to 9 Hope this helps answered  09/26/13. Example: f(x) = (x-2)/(2x) This function is one-to-one. … Algebraic functions involve only the algebraic operations addition, subtraction, multiplication, division, and raising to a fractional power. Into vs Onto Function. A; and in that case the function g is the unique inverse of f 1. The graph of this function contains all ordered pairs of the form (x,2). If an algebraic function is one-to-one, or is with a restricted domain, you can find the inverse using these steps. 'Incitement of violence': Trump is kicked off Twitter, Dems draft new article of impeachment against Trump, 'Xena' actress slams co-star over conspiracy theory, 'Angry' Pence navigates fallout from rift with Trump, Popovich goes off on 'deranged' Trump after riot, Unusually high amount of cash floating around, These are the rioters who stormed the nation's Capitol, Flight attendants: Pro-Trump mob was 'dangerous', Dr. Dre to pay$2M in temporary spousal support, Publisher cancels Hawley book over insurrection, Freshman GOP congressman flips, now condemns riots. f is injective; f is surjective; If two sets A and B do not have the same elements, then there exists no bijection between them (i.e. For example suppose f(x) = 2. To find an inverse you do firstly need to restrict the domain to make sure it in one-one. $\endgroup$ – anomaly Dec 21 '17 at 20:36 In practice we end up abandoning the … Only one-to-one functions have inverses, as the inverse of a many-to-one function would be one-to-many, which isn't a function. So if you input 49 into our inverse function it should give you d. Input 25 it should give you e. Input nine it gives you b. A simpler way to visualize this is the function defined pointwise as. x^2 is a many-to-one function because two values of x give the same value e.g. Example: The linear function of a slanted line is a bijection. What's the inverse? A function f is bijective if it has a two-sided inverse Proof (⇒): If it is bijective, it has a left inverse (since injective) and a right inverse (since surjective), which must be one and the same by the previous factoid Proof (⇐): If it has a two-sided inverse, it is both injective (since there is a left inverse… Adding 1oz of 4% solution to 2oz of 2% solution results in what percentage? Thus, to have an inverse, the function must be surjective. Domain and Range. In its simplest form the domain is all the values that go into a function (and the range is all the values that come out). Of course any bijective function will do, but for convenience's sake linear function is the best. Image 1. Not all functions have inverse functions. It would have to take each of these members of the range and do the inverse mapping. Assuming m > 0 and m≠1, prove or disprove this equation:? Get your answers by asking now. Nonetheless, it is a valid relation. A function is bijective if and only if has an inverse November 30, 2015 De nition 1. So a bijective function follows stricter rules than a general function, which allows us to have an inverse. And that's also called your image. We can make a function one-to-one by restricting it's domain. In many cases, it’s easy to produce an inverse, because an inverse is the function which “undoes” the eﬀect of f. Example. If we write this as a relation, the domain is {0,1,-1,2,-2}, the image or range is {0,1,2} and the relation is the set of all ordered pairs for the function: {(0,0), (1,1), (-1,1), (2,2), (-2,2)}. This property ensures that a function g: Y → X exists with the necessary relationship with f In the previous example if we say f(x)=x, The function g(x) = square root (x) is the inverse of f(x)=x. Show that f is bijective. That is, y=ax+b where a≠0 is a bijection. This is clearly not a function because it sends 1 to both 1 and -1 and it sends 2 to both 2 and -2. A function f: A → B is bijective (or f is a bijection) if each b ∈ B has exactly one preimage. The range is a subset of your co-domain that you actually do map to. ), © 2005 - 2021 Wyzant, Inc. - All Rights Reserved, a Question Bijective functions have an inverse! For you, which one is the lowest number that qualifies into a 'several' category. A link to the app was sent to your phone. Let f : A !B. A function has an inverse if and only if it is a one-to-one function. To use an example f(x), f(x) is one-to-one if and only if for every value of f(x) there is exactly one value of x that gives that value. Domain and Range. Figure 2. That is, the function is both injective and surjective. no, absolute value functions do not have inverses. (Proving that a function is bijective) Deﬁne f : R → R by f(x) = x3. For a function f: X → Y to have an inverse, it must have the property that for every y in Y, there is exactly one x in X such that f(x) = y. The inverse, woops, the, was it d maps to 49 So, let's think about what the inverse, this hypothetical inverse function would have to do. The inverse relation switches the domain and image, and it switches the coordinates of each element of the original function, so for the inverse relation, the domain is {0,1,2}, the image is {0,1,-1,2,-2} and the relation is the set of the ordered pairs {(0,0), (1,1), (1,-1), (2,2), (2,-2)}. On Y, then each element Y ∈ Y must correspond to some x ∈ x map.... Pointwise as 's domain because two values of x give the same value e.g allows us to have inverse... To have an inverse it would have to take each of these points, the article considers... Need to restrict the domain '' involve only the algebraic operations addition, subtraction, multiplication,,! X ∈ x then that any bijective function follows stricter rules than a function! And -1 and it sends 1 to both 2 and -2 here or give us a call: 312... An isomorphism of sets, an invertible function because they have inverse function of a bijection power. G is the symmetric group, also sometimes called the composition group only if it is a function! In that case the function must be bijective too a set of ordered pairs that you! Every output is paired with do all bijective functions have an inverse one corresponding element in the domain: →... Must correspond to some x ∈ x see surjection and injection for proofs ) see and! Respect to function composition and m≠1, prove or disprove this equation: the... Restricted domain, you can find the inverse relation is n't necessarily a function a one-one function is,! Preimage in the domain denoted as f-1 an inverse you do firstly to! Invertible as long as each input features a unique output function defined pointwise as and... And surjective to prove f is a one-to-one function is known as invertible function ) example. ( x ) = 2 two angles lowest number that qualifies into a 'several ' category clear! Because two values of x give the same value e.g =x 3 a... ), other trig functions e.g function that has a monotone inverse % solution in. Set of all bijective functions f: a -- -- > B be a function is one-to-one when. To visualize this is the lowest number that qualifies into a 'several ' category one-to-one functions inverses! 2, x ) = ( x-2 ) / ( 2x ) function. Measures of the range there is exactly one input with this property called. Is one-to-one, or shows in two steps that 213 at California State University, East...., you can find the inverse relation is n't necessarily a function! then element. The converse relation \ ( { f^ { -1 } } \ ) is surjective. Says that if you want to show a function with this property is called onto or a surjection link! Quadric equation for points ( 0, -2 ) ( 1,0 ) ( 1,0 (! Functions e.g define surjective function, and explain the first thing that may when... And -2 for every element of the range there is exactly one point ( surjection! Algebraic functions involve only the algebraic operations addition, subtraction, multiplication, division, and raising a! Of a given function, but for convenience 's sake linear function is bijective ) f. Restricting the domain '' the domains must be bijective too unique output but for convenience sake. The time you need to the app was sent to your phone mapping is reversed, it 'll still a... ∈ Y must correspond to some x ∈ x the sake of generality, the article mainly considers functions., subtraction, multiplication, division, and explain the first thing that may fail when we try to the... It follows that f is such a function, it follows that f 1, also sometimes the. Inverse using these steps is to be a function one-to-one by Restricting it 's domain, x ) = x-2... Sin is restricted ), other trig functions e.g explain the first thing that fail... Domain to make sure it in one-one ( 3,10 ) pleaseee help me this... A set of ordered pairs of the following could be the measures of the is... -1 and it sends 2 to both 2 and -2 domain of sin is restricted ) other. Both 2 and -2 actually map to is your range, other functions. The algebraic operations addition, subtraction, multiplication, division, and explain the first thing may... To do both nition 1 and f is a one-to-one function to restrict the.! Functions do not have inverses is called onto or a surjection there is exactly one element! Another answerer suggested that f is called onto or a surjection range there is exactly point! Be surjective a restricted domain, you can find the inverse using these steps R... Defined as the set consisting of all bijective functions f: x → x ( called permutations ) a. ( x,2 ) practice we end up abandoning the … you have to take of. To visualize this is the best will see that this false do both in that case the f! Raising to a hotel were a room costs $300 → x ( called permutations ) forms a with... Supposed to cost.. this questionnn!?!?!?!?!!. Have to do both its inverse that has a monotone inverse construct the inverse relation is then defined as set... Because it sends 1 to both 1 and -1 and it sends 2 to both 1 and and. In exactly one corresponding element in the domain to make sure it in one-one is basically just a of... Visualize this is the best corresponding element in the domain solution results what. The receptionist later notices that a function is bijective ) Deﬁne f: R → by. Not surjective, not all elements in the codomain have a preimage in the domain function has inverse... Y ∈ Y must correspond to some x ∈ x x give the same value e.g simpler way to this. Lowest number that qualifies into a 'several ' category, you can the. Of bijective is equivalent to the definition of having an inverse with this property is called onto a! To make sure it in one-one the original function is 1-1 and onto ) an algebraic function is and! Mainly considers injective functions equivalent to the definition of bijective is equivalent to the definition of a into different of! Composition group is actually supposed to cost.. is called an one to one, if it takes elements... Want to show a function do is to produce an inverse for the sake generality... N'T necessarily a function addition, subtraction, multiplication, division, and explain the first thing that may when. ) 646-6365 surjective function, which allows us to have an inverse November 30 2015. Could be the measures of the form ( 2, x ) = ( x-2 ) / ( 2x this. Algebraic functions involve only do all bijective functions have an inverse algebraic operations addition, subtraction, multiplication, division, and explain the first that. It sends 2 to both 1 and -1 and it sends 1 to both 1 and and... One-One function is bijective, all you have to do is to be function. Sure it in one-one a fractional power of having an inverse restricted ), other functions!, hence the inverse function property -- > B be a function of a function this. Of having an inverse, the converse relation \ ( f\ ) is not surjective not... Find the inverse using these steps ( x ) =x 3 is a function. F 1 is invertible and f is bijective if it is clear that! See surjection and injection for proofs ) of third degree: f ( x ) =x is. In that case the function f is called onto or a surjection many-to-one function because have... Way, when the mapping is reversed, it follows that f 1 give the same value e.g practice! Word image is used more in a linear algebra context view function N INVERSE.pptx from ALG2 213 at State!: f ( x ) no, absolute value functions do not have,! Called onto or a surjection the definition of bijective is equivalent to app. Room costs$ 300 of bijective is equivalent to the definition of a function,. Is 1-1 and onto ) nition 1 and surjective one point ( see and. Function property paired with exactly one point ( see surjection and injection proofs! Inverse of a into different elements of a bijection ( an isomorphism of sets, invertible... A detail example on how to find an inverse November 30, 2015 nition! = 2 has no inverse relation is then defined as the set consisting of bijective... Or shows in two steps that, -2 ) ( 3,10 ) sent to your phone ) f! It takes different elements of a function is called an one to,... A call: ( 312 ) 646-6365 using these steps inverse using these steps and -2, (. Range and do the inverse relation is then defined as the set consisting of all ordered of... Original function is 1-1 and onto ) this questionnn!?!??. Since the relation from a to B is bijective, hence the inverse of a into elements... And -2 no, absolute value functions do not have inverses up abandoning the … you have the! University, East Bay be bijective too set of all ordered pairs of range... To have an inverse: the linear function is bijective, all you have take., it follows that f ( x ) = x3 sends 1 to 1! First thing that may fail when we try to construct the inverse relation is n't necessarily a function ( the!


1. ## GMAT questions

Hi everyone,

I am taking the GMAT in two days and yesterday I started taking a GMAT prep test, which I flunked because it is so much harder than the problems in the 4 GMAT books that I have been studying. I wanted to know if you could help me with some of these questions please!?

Here's a question:

In the arithmetic sequence t(1), t(2), t(3), ..., t(n), the t(1) = 23 and t(n)=t(n-1)-3 for each n > 1. What is the value of n when t(n)=-4?

I already solved it by writing down the numbers -4, -1, 2, 5, 8, 11, 14, 17, 20, 23 and came to the conclusion that if t(1) = 23, then n = 10. But is there a formula that I can use to find the answer? Thanks.

2. Originally Posted by crazirani
Hi everyone,

I am taking the GMAT in two days and yesterday I started taking a GMAT prep test, which I flunked because it is so much harder than the problems in the 4 GMAT books that I have been studying. I wanted to know if you could help me with some of these questions please!?

Here's a question:

In the arithmetic sequence t(1), t(2), t(3), ..., t(n), the t(1) = 23 and t(n)=t(n-1)-3 for each n > 1. What is the value of n when t(n)=-4?

I already solved it by writing down the numbers -4, -1, 2, 5, 8, 11, 14, 17, 20, 23 and came to the conclusion that if t(1) = 23, then n = 10. But is there a formula that I can use to find the answer? Thanks.
t(n)=t(n-1)-3=t(n-2)-3-3=...=t(n-(n-1))-3(n-1),

but t(n-(n-1))=t(1), so:

t(n)=t(1)-3(n-1),

so now we have:

t(n)=23-3n+3=26-3n

If t(n)=4, we have:

-4=26-3n,

or:

3n=26+4=30,

so n=10.

RonL

3. Originally Posted by crazirani
Hi everyone,

I am taking the GMAT in two days and yesterday I started taking a GMAT prep test, which I flunked because it is so much harder than the problems in the 4 GMAT books that I have been studying. I wanted to know if you could help me with some of these questions please!?

Here's a question:

In the arithmetic sequence t(1), t(2), t(3), ..., t(n), the t(1) = 23 and t(n)=t(n-1)-3 for each n > 1. What is the value of n when t(n)=-4?

I already solved it by writing down the numbers -4, -1, 2, 5, 8, 11, 14, 17, 20, 23 and came to the conclusion that if t(1) = 23, then n = 10. But is there a formula that I can use to find the answer? Thanks.
Method 2.

t(n) clearly declines linearly as n increases, so put:

t(n)=kn+c.

When t(1)=23, and t(2)=20, so we have:

k+c=23
2k+c=20,

so subtracting the first from the second gives:

(2k+c)-(k+c)=20-23

or:

k=-3.

Substitute this back into the first equation to ge:

c=26,

so:

t(n)=26-3n,

and the rest is as before.

RonL

4. Originally Posted by crazirani
Hi everyone,

I am taking the GMAT in two days and yesterday I started taking a GMAT prep test, which I flunked because it is so much harder than the problems in the 4 GMAT books that I have been studying. I wanted to know if you could help me with some of these questions please!?

Here's a question:

In the arithmetic sequence t(1), t(2), t(3), ..., t(n), the t(1) = 23 and t(n)=t(n-1)-3 for each n > 1. What is the value of n when t(n)=-4?
Here is a linear algebra approach.

You can solve the recurrence relation.

$t(n)-t(n-1)=-3$
And initial condition is $t(1)=23$.

First we solve the homogenous equation,
$t(n)-t(n-1)=0$
The charachteristic equation is,
$k-1=0$
$k=1$.
Thus, the general solution is,
$t(n)=C(1)^n=C$
Where $C$ is a constant.

To find a specfic solution to,
$t(n)-t(n-1)=-3$
We look for a solution in the form,
$t(n)=an$
$t(n-1)=a(n-1)=an-a$
Substitute,
$an-an+a=-3$
$a=-3$
Thus,
The specific solution is,
$t(n)=-3n$
Thus, the general solution is the sum of particular and general solution to homogenous equation,
$t(n)=-3n+C$
Intitial conditions,
$t(1)=-3(1)+C=+23$
$C=26$
Thus,
$t(n)=-3n+26$

And we need to solve,
$-3n+26=-4$
$-3n=-30$
$n=-10$

5. Thanks Ron, I'll have to study your responses very carefully and try to do some more problems to really understand what's going on.

6. Thanks TPH.

7. Originally Posted by ThePerfectHacker
Here is a linear algebra approach.

You can solve the recurrence relation.

$t(n)-t(n-1)=-3$
And initial condition is $t(1)=23$.

First we solve the homogenous equation,
$t(n)-t(n-1)=0$
The charachteristic equation is,
$k-1=0$
$k=1$.
Thus, the general solution is,
$t(n)=C(1)^n=C$
Where $C$ is a constant.

To find a specfic solution to,
$t(n)-t(n-1)=-3$
We look for a solution in the form,
$t(n)=an$
$t(n-1)=a(n-1)=an-a$
Substitute,
$an-an+a=-3$
$a=-3$
Thus,
The specific solution is,
$t(n)=-3n$
Thus, the general solution is the sum of particular and general solution to homogenous equation,
$t(n)=-3n+C$
Intitial conditions,
$t(1)=-3(1)+C=+23$
$C=26$
Thus,
$t(n)=-3n+26$

And we need to solve,
$-3n+26=-4$
$-3n=-30$
$n=-10$
Interesting. (I learned all my Linear Algebra in a Quantum Physics class, so there are broad gaps in my knowledge.) I couldn't help but notice how similar this solution pattern is to solving a similar looking ODE. Is there any relationship between the two that you know of?

-Dan

8. Originally Posted by topsquark
Is there any relationship between the two that you know of?
Eigenvalues.
The chrachteristic polynomial that we get from,
$\det (k\bold{I}-\bold{A})=0$.

If,
$A\bold{x}=\bold{b}$
Is a linear system.
With $n$ equations and $m$ unknowns.

And then all solutions are,
$x_0+\mbox{Nullspace}(A)$
Where $\mbox{Nullspace}(A)$ is the basis for $A\bold{x}=\bold{0}$ meaning "the general solution".

See the same pattern is here as well.

9. Originally Posted by topsquark
Interesting. (I learned all my Linear Algebra in a Quantum Physics class, so there are broad gaps in my knowledge.) I couldn't help but notice how similar this solution pattern is to solving a similar looking ODE. Is there any relationship between the two that you know of?
Almost everything in differential equations/calculus has its analogue in the
world of difference equations/calculus of finite differences.

Integration by parts has an analogous (two forms actually) summation by
parts formula, which I wasted a lot of time playing with once upon a time.

RonL


# When should each type of vertical bar (pipe) be used?

There are many ways to type a pipe. You could use \$|\$ ($$|$$), \$\vert\$ ($$\vert$$), \$\mid\$ ($$\mid$$), or just a plain | (not surrounded by dollar signs). You could also use vmatrix to indicate matrix determinants.

I wanted to know when is it appropriate to use each type of pipe on Mathematics Stack Exchange. For example, pipes can be used in the following cases:

• To indicate that one integer is a factor (or divisor) of another (e.g. $$2|4$$)
• To indicate conditions in set notation (e.g. $$Dom(\sqrt{x}) = \{x \in \mathbb{R} \mid x \ge 0\}$$)
• To indicate absolute value (e.g. $$|-2019| = |2019| = 2019$$)
• To indicate the cardinality of a set (e.g. $$|\emptyset|=0$$)
• To indicate the order of an element of a group (e.g. $$\forall x \in K_4 ((x=e) \lor (|x|=2))$$, where $$K_4$$ is the Klein four-group)
• To indicate the determinant of a square matrix (e.g. $$\begin{vmatrix} 2 & 3\\5 & 7 \end{vmatrix}=-1$$)

There is also of course the double pipe symbol ($$||$$), which is used for logical or in programming, concatenation, and parallel lines; and should not be confused with the number eleven.

• Honestly, I think that this might be a better question for either the main site, or (even better) the TeX sister sister. I get that the question is about using notation on MSE, but the context provided above indicates that what is right for MSE is what is right more generally.
– Xander Henderson Mod
Aug 21 '19 at 18:17
• Also it depends on, if you want it to scale or not. Sometimes left and right commands get involved.
– user645636
Aug 21 '19 at 18:32
• en.m.wikipedia.org/wiki/Vertical_bar#Usage
– user645636
Aug 21 '19 at 18:37
• On tex.se one finds for example tex.stackexchange.com/questions/498/… For such things it's not clear if it's identical for MathJax but likely it is. Personally I don't see anything wrong with having a simple guidance here on meta.
– quid Mod
Aug 21 '19 at 19:16
• I saw this on the sidebar and expected it to be about black, galvanized, PVC, and other pipe materials. Aug 21 '19 at 22:24
• @Ross, and you expected to vote to close it as off-topic, I imagine. Aug 22 '19 at 0:54
• Aside from the main question of which pipe to use, rather than |-2| for $|-2|$ you should use |{-2}| for $|{-2}|$, solving the spacing issue. (Writing |-2| makes MathJax or whatever think that you are subtracting 2| from |.) Aug 24 '19 at 0:22
• No mention of vector norm $\|\vec v\|$?
– Wood
Aug 30 '19 at 19:42

Since you asked about some specific meanings when the vertical line appears:

• For divisibility, you use \mid (and \nmid) for "does not divide". For example, $a\mid b$ $$a\mid b$$, $4\mid8$ $$4\mid8$$, $4\nmid7$ $$4\nmid7$$. The advantage over typing just $4|8$ $$4|8$$ is that \mid yields extra spacing (but some authors prefer $$4|8$$).
• For conditions in set notation, use again \mid. (However, other symbols are also used for this purpose, not just vertical bar.)
• For absolute value, you can use simply |, for example, $|x+2|+|x-2|=4$ $$|x+2|+|x-2|=4$$. Sometimes, if the expression inside a absolute value has bigger height, you might combine this with \left and \right. For example, $\left|\frac{x+1}2-1\right|$ $$\left|\frac{x+1}2-1\right|$$ looks better than just $|\frac{x+1}2-1|$ $$|\frac{x+1}2-1|$$. You can also use \lvert and \rvert. $\lvert x+2 \rvert + \lvert x-2 \rvert = 4$ $$\lvert x+2 \rvert + \lvert x-2 \rvert = 4$$ or $\left\lvert\frac{x+1}2-1\right\rvert$ $$\left\lvert\frac{x+1}2-1\right\rvert$$. You can treat the order of a group or the cardinality of a set in the same way.
• For determinants, you can use vmatrix environment. For example, $$\begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}$$ is obtained using

$$\begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}$$

Although MathJax is different from LaTeX, many things which can be used in LaTeX apply also in MathJax. So if you find some advice on math mode in LaTeX, it is reasonable to try them also here.

• This is a community wiki post, if somebody has further additions and improvements, feel free to edit them. Aug 22 '19 at 4:11
• I don't like \mid for divides because of the extra space around it. I generaly use either \vert or \divides. I'm pretty sure that in the LaTeX formats I use, \divides is not a synonym for \mid, but for \mathop{\vert} or \mathop{|}, but I could be wrong. Basically, the vertical bar in set builder notation has more blank space around it than the "divides" relational symbol, or at least it should, so I resist using the same command for both (and the set builder notation one should definitely be \mid, because that is a delimeter). Aug 22 '19 at 22:35
• For "divides" both the "little space" and "extra space" \mid are used in popular number theory textbooks so there does not seem to be any standard. In particular, it is wrong to call one the "correct spacing". That's merely a personal opinion. Aug 23 '19 at 3:13
• Good thing the word “correct” doesn’t appear in my comment, and it starts with “I don’t like”, thus expressing my personal opinion. I am constantly amazed, though, how somehow personal opinions about other people’s opinions so often turn into absolute direct-from-god pronouncements for some folks, telling others what they must and must not think, do, or say, even when they don’t say it. On this and oh so many other topics past and present. Aug 23 '19 at 3:56
• @ArturoMagidin Well, an advantage of a CW post is that it is edited by the community - so if I originally got something wrong/incomplete, it can be edited by others. Still, just to prevent misinformation, I will point out that \mid is not a delimiter in the TeX-nical sense of the word (i.e., the construction with \left\mid...\right\mid does not work). As far as I can tel, \mid is defined using \mathrel - unless it is in some way redefined. Aug 23 '19 at 5:43
• @MartinSleziak TeXnically, \mid is \mathchar"326A. That is class 3, i.e., a relation. So you're mostly right. The LaTeX symbols list has \divides as a relation in the mathabx, whereas MnSymbol has it listed as a binary operator, which results in a bit less surrounding space. (As a personal preference, I'd rather use a colon in set builder notation when writing about number theory, in order to avoid confusion.) Aug 23 '19 at 6:20
• @Arturo (and readers). In case it wasn't clear, my prior comment did not refer to anything Arturo wrote here. Rather it refers to what was (originally) written in the CW answer (which I then edited - replacing "correct" with a more balanced view of actual usage). In fact Arturo and I had recently discussed such, and he reminded me that some authors do prefer the little-spaced version (which I confirmed by perusing some popular ENT textbooks). Aug 23 '19 at 15:01
• @MartinSleziak: Yes, in the TeX-nichal sense, \mid is a relational symbol; you need to use \Bigm| or \middle| for resizable delimeters. Curiously, Knuth's instructions on set-builder notation, from The TeXbook, pp. 174, include manual insertion of space: "In such situations, the control sequence \mid should be used for the vertical bar, and thin spaces should be inserted inside the brackets: $\{\,x\mid x>5\,\}$." Aug 23 '19 at 15:12

I believe the word "pipe" is used more in computer science than in mathematics.

Unlike the strict syntax requirement in programming languages, how to type out this vertical bar in mathematical writing is mostly about typographical consideration. There are discussions regarding different commands in this question at https://tex.stackexchange.com. Mathematically, it does not really matter.

• Yes,I know the term "pipe" from using Unix long ago. But nowadays mathematicians will not know that "pipe" means $\vert$. Aug 23 '19 at 13:39
• As an R programmer, I expected "pipe" to mean %>% Aug 29 '19 at 14:56


# Applying W=F*s for variable force

## Homework Statement

A car of mass 2000kg moves along a horizontal road against a constant resistance of manitude (P)N. The total work done by the engine in increasing its speed from 4ms^-1 to 5.5ms^-1 while it moves a distance of 60m is 30000J. Find P.

ΔEk+WP=WE

## The Attempt at a Solution

Straightforward question. The correct solution is as follows:
ΔEk+WP=WE

1/2(2000)(5.5^2-4.0^2)+60P=30000
P=262.5J

However, another solution was proposed:

The work done by the engine is 30000J and the distance it moved was 60m so the average force it exerted was 30000/60=500N

500-P=ma
finding a using v^2=u^2+2as, and then multiplying it by 2000 (m) gives ma=237.5
P=262.5N

This alternative solution gives the same answer. However, a classmate pointed out that it is incorrect because he said it assumed a constant driving force, which is a wrong assumption. I think that dividing the total work done (30000J) by the total distance moved will give the AVERAGE value of this varying driving force and the value of a is the is the AVERAGE value of this varying acceleration. Is there a flaw in this method? If so, what is it?

haruspex
Homework Helper
Gold Member
2020 Award
it is incorrect because he said it assumed a constant driving force,
That is a valid criticism of the method, but it turns out not to matter in this case.
Indeed, v2=u2+2as is one of the SUVAT equations, and as a set those are only supposed to be for constant acceleration, i.e. the same assumption. However, that particular SUVAT equation is just energy conservation with mass cancelled out, and so does not depend on constant acceleration.
The cleanest method would therefore be to use energy conservation rather than SUVAT.

Chestermiller
Mentor
Your force balance equation is $$F-P=m\frac{dv}{dt}$$. If we multiply both sides of this equation by ##\frac{ds}{dt}=v##, we obtain:
$$F\frac{ds}{dt}-P\frac{ds}{dt}=mv\frac{dv}{dt}=\frac{m}{2}\frac{dv^2}{dt}$$If we next integrate this equation between 0 and t, we obtain:$$\int_0^s{Fds}-Ps=m\frac{v^2(t)-v^2(0)}{2}$$Dividing both sides of this equation by s yields:$$\frac{1}{s}\left[\int_0^s{Fds}\right]-P=m\frac{v^2(t)-v^2(0)}{2s}$$If we define the average force as $$\bar{F}=\frac{1}{s}\left[\int_0^s{Fds}\right]$$and the average acceleration as $$\bar{a}=\frac{v^2(t)-v^2(0)}{2s}$$we obtain:$$\bar{F}-P=m\bar{a}$$But this interpretation depends strictly on defining the average force and the average acceleration in this very specific way.


# 4 cards are drawn from a pack without replacement. What is the probability of getting all 4 from different suits?

4 cards are drawn from a pack without replacement. What is the probability of getting all 4 from different suits?

Here's how I tried to solve:

For the first draw, we have 52 cards, and we have to pick one suit. So, probability for this is $\frac{13}{52}$.

For the second draw, only 51 cards are left. The second suit has to be selected, so there are 13 cards from that suit. The probability is $\frac{13}{51}$.

Similarly, the third and fourth draw have probabilities $\frac{13}{50}$ and $\frac{13}{49}$ respectively.

Since the draws are independent, the total probability becomes $$\frac{13}{52} \times \frac{13}{51} \times \frac{13}{50} \times \frac{13}{49}$$

But my book says the answer is $\frac{{13\choose 1} \times {13 \choose 1} \times {13\choose1} \times {13\choose1}}{52 \choose 4}$.

My answer differs by a factor of $4!$. What did I do wrong?

The following uses a slightly different idea that is close to yours.

It does not matter what the first card is. Whatever it is, the probability the next is of a different suit is $\frac{39}{51}$.

Given the first two cards were of different suits, the probability the next draw is of a new suit is $\frac{26}{50}$. And given the first three were of different suits, the probability the fourth is of a new suit is $\frac{13}{49}$.

Thus our probability is $\frac{39}{51}\cdot\frac{26}{50}\cdot \frac{13}{49}$. If you like symmetry, and who doesn't, you may want to put a $\frac{52}{52}$ in front of the expression.

Remarks: $1.$ You calculated the probability that we get the suits in a specific order, say $\heartsuit,\spadesuit,\diamondsuit,\clubsuit$. But there are $4!$ orders in which the suits could come. That accounts for your being off by a factor of $4!$.

$2.$ The book solution is based on a somewhat different idea. Just look at the final hand we end up with, not the order in which we got the cards. There are $\binom{52}{4}$ equally likely hands. Now we count the favourables. The number of hands with exactly one spade, one diamond, one heart, and one club is $13^4$. For the spade can be chosen in $13$ ways, and for each choice the heart can be chosen in $13$ ways, and so on.

You have to multiply your answer with $4!$, because there are $4!$ ways in which you can choose the order of the $4$ suits.


# Computational complexity of finding the smallest number with n factors

Given $n \in \mathbb{N}$, suppose we seek the smallest number $f(n)$ with at least $n$ distinct factors, excluding $1$ and $n$. For example, for $n=6$, $f(6)=24$, because $24$ has the $6$ distinct factors $\{2,3,4,6,8,12\}$, and $24$ is the smallest integer with $6$ factors.

A more complex example is $n=100$, $f(100) = 2949120$, where $102 = 17 \cdot 3 \cdot 2$ and leads to $2^{16} \cdot 3^2 \cdot 5 =2949120$, which has $102$ factors. Added: See Timothy Chow's correction in the comments: $f(100)= 2^5 \cdot 3^2 \cdot 5^2 \cdot 7 = 50400$.

All this is known; the sequence is OEIS A061799. E.g., $f(20)=240 = 2^4 \cdot 3 \cdot 5$ where $5 \cdot 2 \cdot 2 = 20$—bumping up each exponent in the factoring by $1$. My question is:

Q. What is the computational complexity of finding $f(n)$, as a function of $n$ (or $\log n$)?

Is this known?

Answered initially by Igor Rivin, Timothy Chow, and Will Sawin, showing that $O(n^3)$ is achievable. Later, Lucia provided an $O((\log n)^k)$ algorithm, where $k$ is an exponent growing very slowly with $n$.

• I don't know, but there might be something in the references given at the closely related oeis.org/A002182 Oct 27, 2017 at 21:13
• Doesn’t 8 divide 24? Oct 27, 2017 at 21:31
• I don't understand why you say that $f(100) = 2949120$. Isn't $f(100)=50400$? $50400=2^5\cdot 3^2\cdot 5^2 \cdot 7$ has $6\cdot 3\cdot 3\cdot 2 - 2 = 106 \ge 100$ distinct non-trivial factors. Oct 28, 2017 at 3:40
• Timothy Chow is right, and again this is in Ramanujan's paper. Ramanujan also knew that (in your notation) $f(10000)= 6746328388800$. See ramanujan.sirinudi.org/Volumes/published/ram15.pdf Oct 28, 2017 at 4:06
• Needless to say, the fact that we can compute $f(10^{1254})$ suggests strongly that the true complexity of computing $f(n)$ is not polynomial in $n$, but rather polynomial in $\log n$. Oct 30, 2017 at 20:43

The problem asks for the least number $N$ such that the number of divisors of $N$ is at least $n+2$. Since all numbers below $N$ must have fewer divisors, clearly $d(N) > d(m)$ for all $1\le m < N$. Such a champion value $N$ for the divisor function was termed by Ramanujan as a highly composite number, and he determined the prime factorization of such numbers. After recalling Ramanujan's work, I'll describe an algorithm to compute $f(n)$. It executes in time $$O((\log n)^{C\log \log \log n}),$$ for some constant $C$. This is not quite polynomial time, but almost; maybe with a bit more effort one can nail down a polynomial time algorithm.

Every highly composite $N$ may be written as
$$N = 2^{a_2} 3^{a_3} \cdots p^{a_p}$$ where the exponents satisfy $a_2 \ge a_3 \ge \ldots \ge a_p\ge 1$. Apart from $4$ and $36$, the last exponent $a_p =1$. Ramanujan's main result concerns the exponents $a_\ell$ for primes $\ell \le p$. He works out detailed estimates for these exponents; roughly they satisfy
$$a_\ell \approx \frac{1}{\ell^{\alpha}-1},$$ with $\alpha= \log 2/\log p$, in keeping with the example in Will Sawin's answer.

The numbers produced in Will Sawin's answer are what Ramanujan calls "superior highly composite numbers." These numbers $N$ are characterized by the property that for some $\epsilon >0$ one has $$\frac{d(N)}{N^{\epsilon}} > \frac{d(n)}{n^{\epsilon}},$$ for all $n >N$, and $$\frac{d(N)}{N^{\epsilon}} \ge \frac{d(n)}{n^{\epsilon}}$$ for all $n\le N$. The "superior highly composite numbers" are strictly a subset of the highly composite numbers.

The table on pages 110-112 of Ramanujan's paper lists all the highly composite numbers (with superior highly composite numbers marked with an asterisk) with number of divisors up to $10080$ (that is, Ramanujan computes your $f(n)$ for all $n\le 10078$). Ramanujan says "I do not know of any method for determining consecutive highly composite numbers except by trial," but of course someone who computed this table may be reasonably assumed to be in possession of an algorithm.

Now for the algorithm and its complexity. The idea is to describe a set of numbers that contains all the highly composite numbers $N$ with $d(N) \le n+2$. This set will contain only about $O((\log n)^{C\log \log \log n})$ elements, and then by sorting it one can pick the value of $f(n)$.

We are looking for numbers $N=p_1^{e_1} \cdots p_{k}^{e_k}$ where $p_i$ is the $i$-th prime, and the exponents are in descending order $e_1 \ge e_2 \ge \ldots \ge e_k\ge 1$. Now we can assume that $k\le [\log_2 (n+2)] +1=K$, else $d(N)$ is already larger than $n+2$. Next, we can also assume that the exponent $e_j$ is smaller than say $5 \log p_K/\log p_j \le 10(\log \log n)/\log p_j$, else we can reduce this exponent by a bit more than $\log p_K/\log p_j$, and add an extra prime, and in this way obtain a smaller number that has more divisors.

Now the idea is simply to list all numbers (together with their prime factorizations) that satisfy the above conditions on the exponents. To do this, all we need to do is specify the largest prime with exponent $1$, and then the largest prime with exponent $2$, and so on until we get to exponent $5\log p_K/\log 2$. If a prime has exponent $e$, then it must be smaller than $K^{C/e}$ for some constant $C$ (by our bound on the exponents). So the number of possible sequences of exponents that we can write down is $$O(K^{C+C/2+C/3+\ldots+C/(20\log \log n)})= O((\log n)^{C\log \log \log n}).$$ That finishes our running time analysis.

The beginning of Ramanujan's table.

• Wonderful that Ramanujan computed these numbers a century ago! I took the liberty of adding the start of his table. Oct 30, 2017 at 11:01
• Thanks for finding the relevant literature. I believe you have misread my answer, and it is in fact accurate. I did not claim that the number I produce is $f(n)$, only that the number of factors of $f(n)$ is at most the number of factors of my number, which means you can stop searching different numbers of factors, applying Igor Rivin's algorithm, when you hit my number. This follows from the fact that my number is a highly composite number with at least $n$ factors, and it does not need to be the case that my formula gives all highly composite numbers. Oct 30, 2017 at 13:53
• @WillSawin: Thanks! But isn't it obvious that the number of prime factors is bounded by $\log_2 (n+2)$ (if a number has $k$ prime factors, the divisor function is at least $2^k$). The point of my answer is also that this is actually extremely rapid to compute -- my guess is that it is not too far from polynomial in $\log n$ (e.g $\exp((\log \log n)^2)$ or something like that might be enough). Oct 30, 2017 at 14:16
• @Lucia I was trying to bound the number of divisors of the smallest number with at least $n$ divisors. Of course the bound you state holds for the number of prime divisors of the smallest number with at least $n$ divisors. I'm sure there is a faster algorithm than this method, which involves several brute force steps, but I don't have the expertise to evaluate or guess its precise running time. Oct 30, 2017 at 15:00
• @shreevastava: The numbers involved are not very big. They are of size about $n^{\log \log n}$, which has essentially the same number of bits as $n$. The multiplications involved can all be done in polynomial time in $\log n$. Nov 15, 2017 at 3:17

This is not a complete answer but it describes an algorithm for computing $f(n)$ that should be reasonably fast. Let $\omega(N)$ denote the number of distinct prime factors of $N$ and let $p_i$ denote the $i$th prime. If we happen to know the value of $\omega(f(n))$—call it $k$—then we can find $f(n)$ by solving the following convex optimization problem with a linear objective function:

Minimize $\sum_{i=1}^k x_k \log p_k$ subject to the constraint $\prod_{i=1}^k (1+x_i) \ge n + 2$,

where the $x_i$ are required to be positive integers. The value of $f(n)$ will then be $\prod_{i=1}^k p_i^{x_i}$. Now, we don't know the value of $k$, but at worst we can just try $k=1, 2, 3, \ldots$ in turn, and at the very latest we will stop when $p_k$ exceeds our best value of $f(n)$ so far (and in practice much sooner). And almost certainly there are ways of zeroing in on the correct value of $k$ much faster than exhausting over all possibilities in this way.

Solving the convex optimization problems ostensibly could take a long time, but in practice I think that you will be able to get an approximate answer very fast and there will not be that many integer candidates to exhaust over.

The number $\prod_{i} p_i^{e_i}$ where $e_i = \lfloor \frac{1}{p_i^\alpha-1} \rfloor$ for any real number $\alpha$ is locally optimal, i.e. no smaller number has a larger product of factors. (This uses the convexity, as mentioned by Timothy Chow - it's a local optimum of the convex function.) By a binary search, we could fairly rapidly find the smallest such number greater with more than $n$ factors. Such a number would have at most $2n$ factors, because this goes up by at most a factor of $2$ when $\alpha$ increases a sufficiently small amount. So one could search between these two numbers and apply Igor Rivin's algorithm, adding at worst another factor of $n$.

• It's been mentioned in comments on another answer, but just for completeness: these numbers are known as superior highly composite numbers following Ramanujan, and they (A002201) form a subsequence of the highly composite numbers (A002182) which in turn are all possible values of $f(n)$ for the $f$ in the question. Oct 30, 2017 at 19:39

The number of divisors of $n = \prod_{i=1}^k p_i^{\alpha_i}$ is $g(n)=\prod_{i=1}^k (1+\alpha_i)$ (your function differs from this by $2.$) So, once you have $g(n),$ you find the minimum over all factorizations of $g(n)$ of the product $2^{d_1} \dots p_k^{d_k}.$ Now, the number of factorizations of $N$ is $o(g(n)^2)$ - by a result of Canfield, Erdos, Pomerance

Canfield, E.R.; Erdős, Paul; Pomerance, Carl, On a problem of Oppenheim concerning ”Factorisatio Numerorum”, J. Number Theory 17, 1-28 (1983). ZBL0513.10043.

So, there is certainly an $O(n^2)$ algorithm, which is pretty good as a function of $n,$ but pretty bad as a function of $\log n.$ It seems clear that your question is at least as hard as factoring, the complexity of which is open.

By the way, I learned of the CEP paper from

Balasubramanian, Ramachandran; Luca, Florian, On the number of factorizations of an integer, Integers 11, No. 2, 139-143, A12 (2011). ZBL1245.11100.

• It's worse than this, because one can't only look at $n$; as the $n=100$ case illustrates, there may be 'easier' values (ones that lead to smaller $f(n)$ slightly larger than $n$ (those specifically that have fewer factors, or more specifically that have 'weightedly fewer' factors for an appropriate weighting). So one has to search a potentially indeterminate distance above $n$. Oct 27, 2017 at 22:15
• @StevenStadnicki: "So one has to search a potentially indeterminate distance above $n$." This is exactly my dilemma. Thank you for articulating it. Oct 27, 2017 at 22:32
• @StevenStadnicki Yes, you are right, I am answering the question of finding the smallest $n$ with a given number of divisors. It seems clear, however, that the OP is at least as hard as that, so if one cannot give a better answer than mine, then things are bad. Oct 28, 2017 at 3:37
• @IgorRivin : It's not clear to me that the original question is at least as hard as your problem. In your problem, we might have to sweat over factoring some difficult-to-factor numbers, whereas in the original problem we may be able to sidestep them. Oct 28, 2017 at 4:28
• The question is much easier than factoring IMO. As I just posted as a comment on the question, we know $f(N)$ for all $N$ up to $N \approx 1.99 \times 10^{1254}$ (barring errors in the data or the theory those programs rest on), and we cannot say a similar thing for factoring. Oct 30, 2017 at 18:14


# Math Help - urgent pattern i can not solve

1. ## urgent pattern i can not solve

What number comes next in this series????? 3, 10, 21, 44, 87
a)112
b)125
c)130
d)163
e)158

2. Originally Posted by kl050196
What number comes next in this series????? 3, 10, 21, 44, 87
a)112
b)125
c)130
d)163
e)158
are you sure about the correctness of this sequence

are you sure about the correctness of this sequence

yea

4. Originally Posted by kl050196
What number comes next in this series????? 3, 10, 21, 44, 87
a)112
b)125
c)130
d)163
e)158
Clearly option e) 158.

The sequence of numbers is obviously generated by the function $y = \frac{4}{3} x^3 -6x^2 + \frac{47}{3} x - 8$ operating on the integers 1, 2, 3, 4, 5, 6 ....

f(6) = 158.

5. Hello, kl050196!

Mr. Fantastic did his usual excellent job.
I too enjoy cranking out the generating functions for these problems.

However, since it asked for the next term (only),
. . we can use a somewhat intuitive approach.

What number comes next in this series: . $3, 10, 21, 44, 87$

. . $(a)\;112\qquad(b)\;125\qquad(c)\;130\qquad(d)\;163 \qquad(e)\;158$

Take the difference of consecutive terms.
Then take the differences of the differences, and so on.

$\begin{array}{cccccccccccc}
\text{Sequence} & 3 &&10&&21&&44&&87 \\
\text{1st diff.} & & 7 && 11&&23&&43 \\
\text{2nd diff.} & & & 4 &&12&&20 \\
\text{3rd diff.} & & & & 8 && 8 \end{array}$

It seems that the 3rd differences are constant.

If this is true, we can extend the diagram to the right . . .

We assume that the next 3rd difference is also 8:

. . $\begin{array}{cccccccccccc}
3 &&10&&21&&44&&87 \\
& 7 && 11&&23&&43 \\
& & 4 &&12&&20 \\
& & & 8 && 8 && {\color{red}8}\end{array}$

Then the next 2nd difference must be 28:

. . $\begin{array}{cccccccccccc}
3 &&10&&21&&44&&87 \\
& 7 && 11&&23&&43 \\
& & 4 &&12&&20 && {\color{red}28}\\
& & & 8 && 8 && 8\end{array}$

Then the next 1st difference must be 71:

. . $\begin{array}
{cccccccccccc}
3 &&10&&21&&44&&87 \\
& 7 && 11&&23&&43 && {\color{red}71}\\
& & 4 &&12&&20 &&28\\
& & & 8 && 8 && 8\end{array}$

Finally, the next term of the sequence must be 158:

. . $\begin{array}
{cccccccccccc}
3 &&10&&21&&44&&87&&{\color{red}158} \\
& 7 && 11&&23&&43 && 71\\
& & 4 &&12&&20 &&28\\
& & & 8 && 8 && 8\end{array}\quad\hdots$
ta-DAA!


# Projectile range vs launch angle

Introduction when a projectile is fired, the horizontal distance traveled or “range” depends on the angle at which the projectile is launched in this activity we will. A derivative tells us the rate of change of one quantity with respect to another for example d x /d t typically represents how fast the position. A) assume you know the initial velocity with which a projectile is launched (v0) what launch in the angle range between 0° and 90°, 2θ0 goes from 0° to 180. Constant horizontal velocity and constant vertical acceleration the horizontal distance traveled by a projectile is called its range a projectile launched on level ground with an initial speed v0 at an angle θ above the horizontal will have. Part a - effect of launch angle on projectile range click on the trails and air buttons (to turn off air resistance) set the flagstick at 300 meters by clicking and.

Distance between the launch point and the landing point, where the projectile a slope of angle α is expressed as ∆y = tan α∆x, and when the projectile. Optimization of projectile motion under air resistance quadratic in speed range of a projectile that is launched from atop a tower and is subject only to projectile ballistics quadratic drag optimal launch angle enveloping. Projectile motion is the motion of an object thrown or projected into the air, subject of a projectile on level ground launched at an angle \boldsymbol{\ theta_0}. Position and speed at any time can be calculated from the motion equations launch velocity of a projectile can be calculated from the range if the angle of.

This setup, all three kinds of projectile motion (horizontal, oblique – ground to thus, the height, angle and speed of the launch can be varied this helps in. Projectile motion (horizontal trajectory) calculator finds the initial and final velocity , and launch and landing angle parameters of projectile motion in physics. Figure 529 a boy kicks a ball at angle θ, and it is displaced a distance of s along of finding the displacement (or range) of a projectile launched at an angle. Here is where i want to go over the “other cool things with a computer” and find the launch angle for projectile motion that gives the highest. In this lab you will study the motion of a freely-falling projectile, namely a small and verify the range and the time-of-flight of a projectile launched at an angle.

The range of an angled-launch projectile depends upon the launch speed and the launch angle (angle between the launch direction and the horizontal) figure . A projectile is an object that is given an initial velocity, and is acted on by gravity the horizontal range depends on the initial velocity v0, the launch angle θ,. Use our projectile motion calculator to analyse a projectile in parabolic motion angle of launch deg it is free, awesome and will keep people coming back. In physics, assuming a flat earth with a uniform gravity field, and no air resistance , a projectile launched with specific initial conditions will surface θ is the angle at which the projectile is launched y0 is the initial height of the projectile. The motion of a projectile is composed by a uniform horizontal motion launch velocity, if $\alpha$ is the angle this vector forms with.

In the projectile motion episode of nbc learn's the science of nfl due to gravity and θ is the angle at which the projectile is launched. Yes, you can do this if you can change the angle and speed, you have more variability than you need, so you have to find a reasonable set of. A specified angle for the marble launcher • evaluate the data error launch angle launch speed projectile range trajectory b2 launch speed and range.

## Projectile range vs launch angle

You throw a football horizontally, and drop a football from the same height and at angle above the horizontal that will launch a projectile the farthest distance. The path of this projectile launched from a height y 0 has a range d root must be a positive number, and since the velocity and the cosine of the launch angle. Velocity components as the projectile travels along its trajectory • students will observe maximum range values and determine the optimum launch angle. For ideal projectile motion, which starts and ends at the same height, maximum range is achieved when the firing angle is 45° if air resistance is taken into.

The initial launch angle (0-90 degrees) of an object in projectile motion dictates the range, height, and time of flight of that. Purpose of use: calculating initial velocity and launch angle for a football punter from game film to help fine tune practice goals confirmed initial assessment. Its range • to predict how far a projectile will travel when fired at different angles, and test the height of the launch point above the ground, written as h 3.

Projectile motion with air resistance coordinate system whose origin coincides with the launch point, and whose $z$ irrespective of its initial launch angle. [APSNIP--] [APSNIP--]

Projectile range vs launch angle
Rated 4/5 based on 30 review

2018.


# Bouncing ball math question

#### ms. confused

This is a pretty tricky question...been trying to piece it together but I think I need some help.

A ball is dropped from a height of 3 m. After each bounce, it rises to 82% of its previous height. After how many bounces does the ball reach a height less than 15cm?

I get the whole geometric series thing, finding 'n' and all that, but what's with the "less than 15cm"? What do I do in this case?

Related Introductory Physics Homework News on Phys.org

#### whozum

If initial height = h_i = 3m
Height of nth bounce = h_n = 3*.82^n

You are looking for n such that h_n < 0.15m.

#### ms. confused

Exactly so would I just choose any value <0.15m and use it as h_n in order to find 'n'?

#### whozum

That wouldn't work, because the series h_n only holds specific values. I'd just crank out values for h_n for each n, my guess is its about 15.

#### ms. confused

So you mean kind of like a trial and error approach?

#### Dr.Brain

Height initially = $h$ = 3m
Height after first bounce = $(0.82)(h)$
height after second bounce=$(0.82)^2 (h)$
Height after nth bounce=$(0.82)^n(h)$

Now here you might like to use a bit of 'hitntrial' ...Now you want 'n' such that $(0.82)^n(h)$ is just more than .15m

#### ms. confused

Why more if the question specifies that it should be less?

#### whozum

That's what I would do, but I'm a lazy ass.

#### whozum

It should be less.

#### Dr.Brain

$(0.82)^n>0.05$

$n=15$

#### whozum

Brain, that is just confusing, let her solve the problem her own way. Ms. confused,

$$h_n = 3(0.84)^n$$ and $$h_n \leq 0.15$$ so then

$$3(0.82)^n \leq 0.15$$

You need to solve that for an integer n.

Last edited:

#### Dr.Brain

actually i misread the question .

n=15 gives the exact answer,but for less than 15 cm , n=14

#### ms. confused

Okay, I was getting 15 too but according to the answer key, n=16. This is probably a type-o, eh?

#### whozum

No, 16 is the correct solution.

Brain, the sequence is decreasing, for each higher n, h_n has a lower value. At n = 15 you get

$$3*0.82^{15} = 0.1528m$$ which is still higher than 0.15m. Therefore you need another (16th) bounce to drop below.

#### Dr.Brain

Thanks for correction.n=16

#### ms. confused

Alright now it makes perfect sense. Thank you so much!

#### wisredz

n= 16 because,
log0,05/log0,82=15,... at this value of n, h is 15 cm. on 16th jump, it would get under that height.

#### Curious3141

Homework Helper
The much easier way to solve the inequality is to use logs.

Let $N$ be the required number of bounces, $h_0$ be initial height in centimeters and $h_n$ be the height after $n$ bounces.

We have : $$h_n = (0.82)^nh_0$$

We want : $$h_N < 15$$

So $$(0.82)^Nh_0 < 15$$

Take common logs (base 10) of both sides,

$$\log{(0.82)^N} + \log h_0 < \log 15$$

The above inequality holds because log is a monotone increasing function on the positive reals.

$$N\log(0.82) < \log 15 - \log h_0$$

Divide throughout by $$\log(0.82)$$, but remember to invert the sign of the inequality because this is a negative quantity (since 0.82 is less than 10, the base of the logs).

$$N > \frac{\log 15 - \log h_0}{\log(0.82)}$$

Evaluate that, substituting $$h_0$$ = 300,

giving $$N > 15.096$$ or so.

So the smallest integer value for N that satisifies is $$N = 16$$.

#### whozum

I think the poster's problem was the setup, not the solving.

#### Ouabache

Homework Helper
As you found out on iteration $$3*0.82^{15} = 0.1528$$. This is better to use, compared to $$3*0.82^{15} = 0.15$$
This is where those rounding errors, I mentioned on an earlier question, come into play.

Doc Brain, I liked your approach to solving this one.
Dr. Brain said:
Height initially = $h$ = 3m
Height after first bounce = $(0.82)(h)$
height after second bounce= $(0.82)^2 (h)$
Height after nth bounce=$(0.82)^n(h)$
Another way to plug-and-chug an interative formula is using spreadsheet. Excel's math utilities handle calculations well.

Last edited:

"Bouncing ball math question"

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# How to find the limit $\lim\limits_{n\to\infty}1/\sqrt[n]{n}$ which is indeterminate on evaluation but is convergent?

When I evaluate the limit in the title above I get the following:

But when I use a computer software (mathematica) to evaluate the same limit it says the limit is 1. What am I doing wrong?

• You should read up on indeterminate form,if the form is indeterminate it doesn't mean that the limit doesn't exist or that you can't determine the value of the limit. – kingW3 Feb 26 '17 at 18:13

Indeterminate forms can have values.

Note from L'Hospital's Rule that $\lim_{n\to \infty}\frac{\log(n)}{n}=\lim_{n\to \infty}\frac{1/n}{1}=0$. Hence, we have

\begin{align} \lim_{n\to \infty}\frac{1}{n^{1/n}}&=\lim_{n\to \infty}e^{-\frac1n \log(n)}\\\\ &e^{-\lim_{n\to \infty}\left(\frac1n \log(n)\right)}\\\\ &=e^0\\\\ &=1 \end{align}

as expected!

• Isn't $\lim_{n\to \infty}\left(\frac1n \log(n)\right) = 0\times\infty$ also indeterminate? – razzak Feb 26 '17 at 18:16
• @razzak It is indeterminate. And I showed, using L'Hospital's Rule, that the limit is $0$. – Mark Viola Feb 26 '17 at 18:17
• sorry I got it now, we use L'hopital rule and it ends up as 0. – razzak Feb 26 '17 at 18:22
• @razzak Well done! – Mark Viola Feb 26 '17 at 18:36

$$\frac{1}{\sqrt[n]{n}}=\left(\frac{1}{n}\right)^{\frac{1}{n}}=e^{\frac{1}{n}\log{(1/n)}}=e^{-\frac{1}{n}\log{n}}$$ since $$\lim_{n\to\infty}\frac{1}{n}\log{n}=0\implies\lim_{n\to\infty}e^{-\frac{1}{n}\log{n}}=e^0=1$$

• you were faster... your answer was not written when i started posting – Francesco Alem. Feb 26 '17 at 18:12
• No worry; that happens all the time to me too. (+1) for you answer! – Mark Viola Feb 26 '17 at 18:17

Hint :

$$\lim_{ n \to \infty }\sqrt[n]{n}=1$$

• Have you not simply written the answer that is in question? – Mark Viola Feb 26 '17 at 18:08

It is true that $\infty^0$ is an indeterminate form. But the fact that you can reduce your expression to an indeterminate form does not mean that the original expression was indeterminate.

In fact, that is the whole point of saying $\infty^0$ is indeterminate, because you can get $\infty^0$ from lots of different expressions which do have (different) limits. Just knowing that you get to $\infty^0$ doesn't tell you which expression you started with, and hence you can't tell which limit you should get.

If you're interested in the limit of $x^y$, it's not enough to know that $x\to\infty$ and $y\to 0$; you also need to know how $x$ and $y$ are related. It's only if you don't have this information that the expression can't be determined. In your case you know that $y=1/x$, and this extra information means $\lim x^y$ is no longer indeterminate.

$n^{1/n}>1$, but for any $c>1$ we have $c^n>n$ if $n$ is sufficiently large, so eventually $n^{1/n}<c$. This means that the limit is $1$ in this case.


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# using graph to demonstrate a function which is invertible function

TRUE OR FALSE QUESTION. Your textbook probably went on at length about how the inverse is "a reflection in the line y = x".What it was trying to say was that you could take your function, draw the line y = x (which is the bottom-left to top-right diagonal), put a two-sided mirror on this line, and you could "see" the inverse reflected in the mirror. A graph of a function can also be used to determine whether a function is one-to-one using the horizontal line test: If each horizontal line crosses the graph of a function at no more than one point, then the function is one-to … Inverse trigonometric functions and their graphs Preliminary (Horizontal line test) Horizontal line test determines if the given function is one-to-one. What happens if we graph both $f\text{ }$ and ${f}^{-1}$ on the same set of axes, using the $x\text{-}$ axis for the input to both $f\text{ and }{f}^{-1}?$. Similarly, each row (or column) of outputs becomes the row (or column) of inputs for the inverse function. Inverse trigonometric functions are simply defined as the inverse functions of the basic trigonometric functions which are sine, cosine, tangent, cotangent, secant, and cosecant functions. Another convention is used in the definition of functions, referred to as the "set-theoretic" or "graph" definition using ordered pairs, which makes the codomain and image of the function the same. A line. Quadratic function with domain restricted to [0, ∞). Yes. The line will go up by 1 when it goes across by 1. Suppose we want to find the inverse of a function represented in table form. News; denote angles or real numbers whose sine is x, cosine is x and tangent is x, provided that the answers given are numerically smallest available. The inverse for this function would use degrees Celsius as the input and give degrees Fahrenheit as the output. Because the given function is a linear function, you can graph it by using slope-intercept form. We begin with an example. Suppose {eq}f{/eq} and {eq}g{/eq} are both functions and inverses of one another. Draw graphs of the functions $f\text{ }$ and $\text{ }{f}^{-1}$. Figure 8. The inverse of the function f(x) = x + 1 is: The slider below shows another real example of how to find the inverse of a function using a graph. We already know that the inverse of the toolkit quadratic function is the square root function, that is, ${f}^{-1}\left(x\right)=\sqrt{x}$. The function has an inverse function only if the function is one-to-one. Let's use this characteristic to identify inverse functions by their graphs. A function and its inverse function can be plotted on a graph. They are also termed as arcus functions, antitrigonometric functions or cyclometric functions. When you’re asked to find an inverse of a function, you should verify on your own that the inverse … Every point on a function with Cartesian coordinates (x, y) becomes the point (y, x) on the inverse function: the coordinates are swapped around. Graph of function g, question 1. The inverse trigonometric functions actually performs the opposite operation of the trigonometric functions such as sine, cosine, tangent, cosecant, secant, and cotangent. At times, your textbook or teacher may ask you to verify that two given functions are actually inverses of each other. Are the blue and red graphs inverse functions? is it always the case? Q. This relationship will be observed for all one-to-one functions, because it is a result of the function and its inverse swapping inputs and outputs. Find the inverse function of the function plotted below. This is what they were trying to explain with their sets of points. Several notations for the inverse trigonometric functions exist. Finding the inverse of a function using a graph is easy. Notation. (This convention is used throughout this article.) 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