The last property is a consequence of Property 3 and the fact that matrix multiplication is associative; In the next subsection, we will state and prove the relevant theorems. $$\begin{pmatrix} e & f \\ g & h \end{pmatrix} \cdot \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} ae + cf & be + df \\ ag + ch & bg + dh \end{pmatrix}$$ i.e., (AT) ij = A ji ∀ i,j. The number of rows and columns of a matrix are known as its dimensions, which is given by m x n where m and n represent the number of rows and columns respectively. Multiplicative Identity: For every square matrix A, there exists an identity matrix of the same order such that IA = AI =A. Deﬁnition The transpose of an m x n matrix A is the n x m matrix AT obtained by interchanging rows and columns of A, Deﬁnition A square matrix A is symmetric if AT = A. The proof of Equation \ref{matrixproperties2} follows the same pattern and is … 19 (2) We can have A 2 = 0 even though A ≠ 0. Example 1: Verify the associative property of matrix multiplication … (3) We can write linear systems of equations as matrix equations AX = B, where A is the m × n matrix of coefficients, X is the n × 1 column matrix of unknowns, and B is the m × 1 column matrix of constants. But first, we need a theorem that provides an alternate means of multiplying two matrices. A matrix is an array of numbers arranged in the form of rows and columns. The basic mathematical operations like addition, subtraction, multiplication and division can be done on matrices. MATRIX MULTIPLICATION. The following are other important properties of matrix multiplication. The proof of this lemma is pretty obvious: The ith row of AT is clearly the ith column of A, but viewed as a row, etc. Selecting row 1 of this matrix will simplify the process because it contains a zero. Distributive law: A (B + C) = AB + AC (A + B) C = AC + BC 5. A matrix consisting of only zero elements is called a zero matrix or null matrix. For sums we have. Subsection MMEE Matrix Multiplication, Entry-by-Entry. Let us check linearity. proof of properties of trace of a matrix. The first element of row one is occupied by the number 1 … Properties of transpose Proof of Properties: 1. For the A above, we have A 2 = 0 1 0 0 0 1 0 0 = 0 0 0 0. A diagonal matrix is called the identity matrix if the elements on its main diagonal are all equal to $$1.$$ (All other elements are zero). Even though matrix multiplication is not commutative, it is associative in the following sense. Given the matrix D we select any row or column. Equality of matrices Matrix transpose AT = 15 33 52 −21 A = 135−2 532 1 Example Transpose operation can be viewed as ﬂipping entries about the diagonal. Associative law: (AB) C = A (BC) 4. The determinant of a 4×4 matrix can be calculated by finding the determinants of a group of submatrices. While certain “natural” properties of multiplication do not hold, many more do. Notice that these properties hold only when the size of matrices are such that the products are defined. A square matrix is called diagonal if all its elements outside the main diagonal are equal to zero. If $$A$$ is an $$m\times p$$ matrix, $$B$$ is a $$p \times q$$ matrix, and $$C$$ is a $$q \times n$$ matrix, then $A(BC) = (AB)C.$ This important property makes simplification of many matrix expressions possible. Zero matrix on multiplication If AB = O, then A ≠ O, B ≠ O is possible 3. Multiplicative identity: For a square matrix A AI = IA = A where I is the identity matrix of the same order as A. Let’s look at them in detail We used these matrices Example. More do that these properties hold only when the size of matrices are that! Law: ( AB ) C = A ( B + C ) = +! We have A 2 = 0 1 0 0 = 0 0 0 means of two! ( AB ) C = AC + BC 5 of matrix multiplication notice that properties... 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# Probability concepts - how can balls of same colour be distinguishable?

An urn contains $$6$$ white and $$4$$ black balls. A fair die is rolled and that number of balls are chosen from the urn. Find the probability that the balls selected are white.

I know the basic way to go about solving the problem.

Let $$W$$ be the event of finally drawing all white balls. Let $$P(n)$$ denote the probability of appearance of $$n$$ on the die.

We want: $$P(W) = P(1)P(W\mid 1)+ P(2)P(W\mid 2)+\dots \implies P(W) = \dfrac{1}{6}\left(\sum_{i=1}^6P(W\mid i)\right)$$

Now, I am actually facing trouble in computing $$P(W/i)$$. I saw author's method and in it he has used $$P(W\mid i) = \dfrac{^6C_i}{^{10}C_i}$$

but I fail to understand how that can be true when all white balls are identical and all black balls are identical.

Here, $$^6C_i$$ denotes the combination of $$i$$ different things from 6 different objects, doesn't it? How can that be used here?

• Doesn't it look pretty much as a Hypergeometric distribution were the number of draws is a rv? $K=6$, $N=10$, $n=k$ is a random variable. Does it make any sense to you? – Ramiro Scorolli Nov 14 at 8:13
• I don't know what's a hypergeometric distribution @RamiroScorolli – Abcd Nov 14 at 8:13
• Basically you have $N$ balls(10 balls in total), $K$ represent success (6 white balls), you draw $n$ balls (where n is the number obtained with the dice) and you are expecting $k$ successes. (basically n cause you want all to be white). en.m.wikipedia.org/wiki/Hypergeometric_distribution – Ramiro Scorolli Nov 14 at 8:20
• I'm not completely sure but I think that this could be the $P(W/I)$ you are looking for – Ramiro Scorolli Nov 14 at 8:20
• How about something similar with a smaller amount of balls? If you have two identical black balls and an otherwise similar but white ball in a bag, and you pull out one in random, what's the probability of getting a black ball? Is it 1/2 because there are only two colors (and you couldn't tell the black ones apart)? – ilkkachu Nov 14 at 14:43

• The probability of drawing $$i$$ white balls is the probability of drawing a white ball and then (without replacement) drawing a second white ball, and so on up to $$i$$

• The first ball drawn has a probability of $$\frac{6}{10}$$ of being white

• Given that the first ball drawn is white, the second ball drawn has a probability of $$\frac{5}{9}$$ of being white

• An so on up to the $$i$$th ball having a probability of $$\frac{6-i+1}{10-i+1}$$ of being white

• So the probability all $$i$$ balls drawn are white is $$\frac{6 \times 5 \times \cdots \times(6-i+1)}{10 \times 9 \times \cdots \times(10-i+1)} = \dfrac{\frac{6!}{(6-i)!}}{\frac{10!}{(10-i)!} }= \dfrac{\frac{6!}{(6-i)!i!}}{\frac{10!}{(10-i)!i!}} = \dfrac{^6C_i}{^{10}C_i}$$

• Wow, this is amazing. – Abcd Nov 14 at 8:18
• So does this also imply that considering them as distinct is a sort of "trick" that is always going to work? – Abcd Nov 14 at 8:22
• @Abcd If you are going to use a counting calculation, then you want each event to be of equal probability, and treating the balls as physically distinct even when they look the same does this. Otherwise you risk saying that the probability of all white when $i=6$ might be higher than when $i=4$, since when $i=6$ you can get $4$, $5$ or $6$ white balls while with $i=4$ you can get $0$, $1$, $2$, $3$ or $4$ white balls; you need to be able say what the differing probabilities of the various possible outcomes are – Henry Nov 14 at 8:55

If you draw the $$i$$ balls sequentially $$P(W\mid i)=\frac{6}{10}\cdot\frac{5}{9}\cdots\frac{6-i+1}{10-i+1}$$ If you draw the $$i$$ balls simultaneously $$P(W\mid i)=\frac{\dbinom{6}{i}\cdot\dbinom{4}{0}}{\dbinom{10}{i}}$$ Now, can you convince yourself that these two are equivalent?

Since I don't see it anywhere, let me address this part of your question, rather than specific case:

how that can be true when all blue balls are identical and all black balls are identical.

Here, $$^6C_i$$ denotes the combination of $$i$$ different things from 6 different objects, doesn't it? How can that be used here?

Let's make a simple thought experiment. Let's say that rather than having indistinguishable white and black balls you have 10 balls numbered 1 to 10. Balls numbered 1 to 6 are white and those numbered 7 to 10 are black.

Now you have 6 white balls and 4 black balls that are distinguishable. How does that change your probability?

The numbers on balls change nothing in probability as long as the only thing you're concerned is ball colour. The resulting probability has to be the same as if there were no numbers. We may simply ignore the numbers on the balls and follow the original problem. We still have 6 white balls and 4 black balls so the old approach holds. Whatever other way you count the probability it has to provide the same result.

But now your balls are distinguishable so you can apply methods specific to distinguishable balls, specifically use combination. The results, as shown earlier will be the same.

This is why to indistinguishable balls you can always apply approach "Let's assume the balls are distinguishable..."

Let me answer from a different direction.

The reason you must use the equations which treat the balls as distinguishable is that thousands of repeated experiments have shown that the statistical behavior of all macroscopic objects is that of distinguishable items. There is no "proof" of this, since it's a physical reality. (by comparison, Bosons often do act as indistinguishable, leading to cool stuff in the super-cold regime).

Now, in all statistics, the wording of the question is critical (see arguments about the Monty Hall problem!). If your teacher says "Assume all balls of a given color are truly indistinguishable," then it's a theoretical problem and you basically divide by the number of permutations of distinguishable objects. But unless indistinguishability is specifically given as a premise, objects are distinguishable.


# Math Help - Difficult definite integral

1. ## Difficult definite integral

The question:
$\int_{-1}^{0} \sqrt{t^2 + t^4} dt$

I'm stumped. Not sure how to attempt this.

Any assistance would be great.

2. How about first taking $t^2$ from within the square root, and move it outside as $|t|$:

$\sqrt{t^2+t^4} = \sqrt{t^2(1+t^2)} = |t|\sqrt{1+t^2} = -t\sqrt{1+t^2},$

where $|t|=-t$, because we find ourselves in the range $-1\leq t\leq 0$. This last expression can be more easily integrated. I suggest substitution.

3. Originally Posted by Glitch
The question:
$\int_{-1}^{0} \sqrt{t^2 + t^4} dt$

I'm stumped. Not sure how to attempt this.

Any assistance would be great.

The integrand function is even, so we can as well work with $\int\limits_0^1\sqrt{t^2+t^4}dt=\frac{1}{2}\int\li mits^1_0 2t\sqrt{1+t^2}dt =$ ....take it from here.

Note: the last one is an automatic integral!

Tonio

4. Originally Posted by tonio
The integrand function is even, so we can as well work with $\int\limits_0^1\sqrt{t^2+t^4}dt=\frac{1}{2}\int\li mits^1_0 2t\sqrt{1+t^2}dt =$ ....take it from here.

Note: the last one is an automatic integral!

Tonio
I wonder how You can say: the integrand function is even and immediately after [implicity] write the 'identity'...

$\displaystyle \sqrt{t^{2} + t^{4}} = t\ \sqrt{1+t^{2}}$

It is quite obvious that the second term of the 'identity' (1) is an odd function ...

Kind regards

$\chi$ $\sigma$

5. Originally Posted by HappyJoe
How about first taking $t^2$ from within the square root, and move it outside as $|t|$:

$\sqrt{t^2+t^4} = \sqrt{t^2(1+t^2)} = |t|\sqrt{1+t^2} = -t\sqrt{1+t^2},$

where $|t|=-t$, because we find ourselves in the range $-1\leq t\leq 0$. This last expression can be more easily integrated. I suggest substitution.
I'm a little confused as to why we make |t| = -t. Is it because the area under the curve must be positive? i.e. y >= 0 for all t?

6. Originally Posted by chisigma
I wonder how You can say: the integrand function is even and immediately after [implicity] write the 'identity'...

$\displaystyle \sqrt{t^{2} + t^{4}} = t\ \sqrt{1+t^{2}}$

It is quite obvious that the second term of the 'identity' (1) is an odd function ...

Kind regards

$\chi$ $\sigma$
This really took me by surprise, though I think it doesn't matter here since the even thing was just to change the integral's limits.

In fact I should have writtent $\sqrt{t^2+t^4}=|t|\sqrt{1+t^2}$ , but in this case is irrelevant since the

variable t is already positive in the new integration's limits.

Tonio

7. Originally Posted by Glitch
I'm a little confused as to why we make |t| = -t. Is it because the area under the curve must be positive? i.e. y >= 0 for all t?
Let $-u = t$, then $dt = -du$, then:

$\displaystyle \int_{-1}^{0} \sqrt{t^2+t^4}\;{dt} = -\int_{1}^{0}\sqrt{(-u)^2+(-u)^4}\;{du} = -\int_{1}^{0}\sqrt{u^2+u^4}\;{du} = -\int_{1}^{0} \sqrt{t^2+t^4}\;{dt}.$

Now we can write $\sqrt{t^2+t^4} = t\sqrt{1+t^2}$, as our limits are non-negative.


Determine if the alternating series converges absolutely, conditionally or diverges

Trying to determine if this alternating series converges absolutely or conditionally. ATS criteria has been met (terms are positive [ignoring signs] & decreasing, and the lim n->inf = 0, assuming I haven't made a mistake) so I know it's at least convergent but need to prove absolute convergence. However, I believe that if the absolute value of the series is convergent then it is impossible for the original series to be divergent and, thus, the series has to be absolutely convergent. Does that make sense? Thanks for taking a look at this.

The terms, at least after a while, are decreasing in absolute value. However, there is no need to show that. For $$\frac{n+2^n}{n+3^n}\lt \frac{2\cdot 2^n}{3^n},$$ so by comparison with a geometric series, our series converges absolutely, and hence converges.

• This is great but am I correct in saying that if the absolute value of an alternating series is convergent then the original alternating series is also convergent? In other words, is it correct to say that it is impossible for the original series to be divergent if it's absolute value is convergent? – joe schmoe Jul 13 '14 at 22:02
• Yes, you are correct. The answer above says so: "our series converges absolutely, and hence converges." The signs need not actually alternate. For any sequence $(a_n)$, if $\sum|a_n|$ converges, then $\sum a_n$ converges. Of course, the converse need not hold. If $\sum a_n$ converges, then $\sum|a_n|$ need not converge. – André Nicolas Jul 13 '14 at 22:06
• You are welcome. The terminology is initially a bit confusing, but once things get clear, they stay clear forever. – André Nicolas Jul 13 '14 at 22:47

Hint: $$\left|(-1)^{n}\frac{n+2^{n}}{n+3^{n}}\right|=\frac{1+\frac{n}{2^{n}}}{1+\frac{n}{3^{n}}}\left(\frac{2}{3}\right)^{n}$$

• isn't it $\left(\dfrac{3}{2}\right)^{n}$? – Kamster Jul 13 '14 at 21:11
• Perhaps I've made a factoring error. $\frac{n+2^{n}}{n+3^{n}}=\frac{(\frac{n}{2^{n}}+1)2^{n}}{(\frac{n}{3^{n}}+1)3^{n}}=\frac{1+\frac{n}{2^{n}}}{1+\frac{n}{3^{n}}}(\frac{2}{3})^{n}$? – user71352 Jul 13 '14 at 21:13
• Oh no you are completely right, miss read it. I thought you took a $\frac{1}{2^{n}}$ and $\frac{1}{3^{n}}$ for some reason – Kamster Jul 13 '14 at 21:14
• Yea if there is any most common mistake I make in proofs, its algebra mistakes – Kamster Jul 13 '14 at 21:16
• I have the same problem. – user71352 Jul 13 '14 at 21:18


Question

# If $$a_{1}$$, $$a_{2}$$, $$a_{3}$$ are in AP $$a_{2}$$, $$a_{3}$$, $$a_{4}$$ are in GP and $$a_{3}$$, $$a_{4}$$, $$a_{5}$$ are in HP then $$a_{1}$$, $$a_{3}$$, $$a_{5}$$ are in

A
AP
B
GP
C
HP
D
none of these

Solution

## The correct option is A GPGiven $${ a }_{ 1 },{ a }_{ 2 },,{ a }_{ 3 }$$ are in A.PLet's assume that $${ a }_{ 1 }=A-d\\ { a }_{ 2 }=A\\ { a }_{ 3 }=A+d$$similarly $${ a }_{ 2 },{ a }_{ 3 },{ a }_{ 4 }$$ are in G.P$$so\quad { a }_{ 2 }=A\\ { a }_{ 3 }=Ar\\ { a }_{ 4 }=A{ r }^{ 2 }$$$$\Rightarrow A+d=Ar\\ \Rightarrow d=Ar-A$$and finally $${ a }_{ 3 },{ a }_{ 4 },{ a }_{ 5 }$$ are in H.P$$\Rightarrow \dfrac { 1 }{ { a }_{ 3 } } ,\dfrac { 1 }{ { a }_{ 4 } } ,\dfrac { 1 }{ { a }_{ 5 } }$$ are in A.Phence $$\Rightarrow \dfrac { 1 }{ { a }_{ 5 } } =\dfrac { 1 }{ { a }_{ 4 } } +\left( \dfrac { 1 }{ { a }_{ 4 } } -\dfrac { 1 }{ { a }_{ 3 } } \right) \\ \Rightarrow \dfrac { 1 }{ { a }_{ 5 } } =\dfrac { 2 }{ A{ r }^{ 2 } } -\dfrac { 1 }{ Ar } =\dfrac { 2-r }{ A{ r }^{ 2 } } \\ \Rightarrow { a }_{ 5 }=\dfrac { A{ r }^{ 2 } }{ 2-r }$$and $${ a }_{ 1 }=A-\left( Ar-A \right) =2A-Ar=A\left( 2-r \right) \\ { a }_{ 3 }=Ar$$clearly we can see that $${ a }_{ 1 }{ a }_{ 5 }={ { a }_{ 3 } }^{ 2 }$$Thus $${ a }_{ 1 },{ { a }_{ 3 } },{ a }_{ 5 }$$ are in G.PMaths

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# Why is $\sum \limits_{k = 0}^{n} (-1)^{k} k\binom{n}{k} = 0$?

I know that the expansion of $\sum \limits_{k = 0}^{n} (-1)^{k} \binom{n}{k}$ equals to zero. But why is $\sum \limits_{k = 0}^{n} (-1)^{k} k\binom{n}{k}$ also equal to zero for $n \geq 2$?

I've been using the first to derive the second, but it ended with no clue at all. Anyone know about how to derive this formula?

$$\displaystyle \sum_{k = 0}^{n} (-1)^{k} k\binom{n}{k} = 0 .$$

-

Note that $(1-x)^n = \sum_{k=0}^n (-1)^k x^k \binom{n}{k}$. Thus the sum you are interested in is $\left. \frac{\mathrm{d}}{\mathrm{d} x} (1-x)^n \right|_{x=1} = \left. -n (1-x)^{n-1} \right|_{x=1} = -n (1-1)^{n-1}$. Thus it is zero for $n > 1$.

Indeed for $n=1$ is the sum is $-1$, which can be explicitly checked.

-

I would like to give another different proof of the problem the OP proposed. My solution is based on the identity

$$k\binom{n}{k}=n\binom{n-1}{k-1}.$$

Let us first prove this identity: suppose we are given a class of $n$ children and suppose we want to form a team of $k$ people from the class, and moreover we want to elect a captain for our team. We can count the possibilities of doing so in two ways:

First select $k$ people from the class and then elect the captain. Then we have $k$ possibilities for any previously chosen team, so in total $$k\binom{n}{k}$$ ways of proceeding along this path.

But we may also elect first the captain, which can be done in $n$ ways, then form the team, for which we need other $k-1$ children out of $n-1$ remaining. In this other way we count $$n\binom{n-1}{k-1}$$ ways to fulfill our task.

This proves in a combinatorical way the identity which can be however verified by algebraic means.

But then our formula reduces to $$n \sum_{k=0}^{n} (-1)^k\binom{n-1}{k-1}=n\sum_{k=1}^n(-1)^k\binom{n-1}{k-1}=0.$$

-
The first part of your proof can even be shorter. Remember the definition of the binomial by factorials: $\small k \binom{n}{k} = k { n! \over k! (n-k)! } = {n! \over (k-1)! (n-k)! } = n { (n-1)! \over (k-1)! ((n-1)-(k-1))! } = n \binom{n-1}{k-1}$ – Gottfried Helms Oct 25 '11 at 16:15
@Gottfried: I know.. However i'm in deep love with combinatorical double counting proofs so that's why i've chosen this one. However if you see i mentioned the fact that the identity may be proven algebraically. – uforoboa Oct 25 '11 at 16:24

The identity you ask about has a direct algebraic proof using the identity you already know. Let $g(n) = \sum_{k = 0}^{n} (-1)^{k} k\binom{n}{k}$, and let $f(n) = \sum_{k = 0}^{n} (-1)^{k} \binom{n}{k}$. We will show that $g(n+1) = - f(n)$, and thus the fact that $f(n) = [n=0]$ implies $g(n) = -[n=1]$. (Here, [statement] evaluates to $1$ if statement is true and $0$ if statement is false. It's called the Iverson bracket.)

We have $$g(n+1) - g(n) = \sum_k (-1)^{k} k\left(\binom{n+1}{k} - \binom{n}{k}\right) = \sum_k (-1)^{k} k\binom{n}{k-1}$$ $$= \sum_k (-1)^{k+1} (k+1)\binom{n}{k} = -g(n) - f(n).$$ Thus $g(n+1) = -f(n) \Longrightarrow g(n) = - f(n-1) = - [n-1=0] = -[n=1]$.

Generalization. If $g(n) = \sum_{k = 0}^{n} (-1)^{k} \binom{n}{k} b_k$, and $f(n) = \sum_{k = 0}^{n} (-1)^{k} \binom{n}{k} \Delta b_k$ (where $\Delta b_k = b_{k+1} - b_k$), then $g(n) = -f(n-1) + b_0[n=0]$. This relationship can be applied iteratively, starting with the problem above, to show that

$$\sum_{k=0}^n \binom{n}{k} (-1)^k k^{\underline{m}} = (-1)^m m![n=m],$$ and from there to $$\sum_{k=0}^n \binom{n}{k}(-1)^k k^m = \left\{ m \atop n \right\}(-1)^n n!,$$ where $\left\{ m \atop n \right\}$ is a Stirling number of the second kind.

(See, for example, Section 3 of my paper "Combinatorial Sums and Finite Differences," Discrete Mathematics, 307 (24): 3130-3146, 2007.)

-

Here's a purely combinatorial proof that doesn't reduce the sum to the known identity $\sum \limits_{k = 0}^{n} (-1)^{k} \binom{n}{k} = 0$.

The quantity $\binom{n}{k}k$ counts the number of ways to partition people numbered $\{1, 2, \ldots, n\}$ into a chaired committee $A$ of size $k$ and an unchaired committee $B$ of size $n-k$. Given a particular commmittee pair $(A,B)$, let $x$ be the highest-numbered person in either committee who is not the chair of $A$. Move $x$ to the other committee. This mapping is defined for all pairs of committees when $n >1$, is its own inverse (and so is one-to-one), and changes the parity on committee pairs. Thus, for $n > 1$, there are as many committee pairs with even parity as there are with odd parity. In other words, $$\sum_{k = 0}^{n} (-1)^{k} \binom{n}{k} k = 0$$ when $n > 1$.

-


# Determine the position using an iteration method

#### mathmari

##### Well-known member
MHB Site Helper
Hey!!

The function \begin{equation*}f(x)=\frac{x}{e^{x/9}}\cdot \frac{\sin \left (\pi (x-1)\right )}{x-1}\end{equation*} has at exactly one position $\overline{x}>1$ the same value as at the position $x=1$. Determine the position $\overline{x}$ using an iteration method with accuracy of two decimal digits.

I have done the following:

We have that \begin{equation*}f(\overline{x})=f(1)\Rightarrow f(\overline{x})-f(1)=0 \Rightarrow g(x):=f(x)-f(1)\end{equation*}

First we have to calculate $f(1)$:
\begin{align*}f(1)&=\lim_{x\rightarrow 1}f(x)=\lim_{x\rightarrow 1}\frac{x}{e^{x/9}}\cdot \frac{\sin \left (\pi (x-1)\right )}{x-1}=\frac{1}{e^{1/9}}\cdot \lim_{x\rightarrow 1}\frac{\sin \left (\pi (x-1)\right )}{x-1}\ \overset{DLH}{ = } \ \frac{1}{e^{1/9}}\cdot \lim_{x\rightarrow 1}\frac{\pi \cos \left (\pi (x-1)\right )}{1}\\ & =\frac{1}{e^{1/9}}\cdot \pi \cos \left (\pi \cdot 0\right )=\frac{1}{e^{1/9}}\cdot \pi =\frac{\pi}{e^{1/9}}\end{align*}

Therefore we get the function \begin{equation*}g(x)=\frac{x}{e^{x/9}}\cdot \frac{\sin \left (\pi (x-1)\right )}{x-1}-\frac{\pi}{e^{1/9}}\end{equation*}

Now we have to apply an iteration method to approximate the root of that function, right? Do we use the Newton's method?

We don't have an interval to which the root will belong, we only know that it is greater than $1$. So do we have to guess such an interval to calculate the first input $x_0$?

#### Opalg

##### MHB Oldtimer
Staff member
Hey!!

The function \begin{equation*}f(x)=\frac{x}{e^{x/9}}\cdot \frac{\sin \left (\pi (x-1)\right )}{x-1}\end{equation*} has at exactly one position $\overline{x}>1$ the same value as at the position $x=1$. Determine the position $\overline{x}$ using an iteration method with accuracy of two decimal digits.

I have done the following:

We have that \begin{equation*}f(\overline{x})=f(1)\Rightarrow f(\overline{x})-f(1)=0 \Rightarrow g(x):=f(x)-f(1)\end{equation*}

First we have to calculate $f(1)$:
\begin{align*}f(1)&=\lim_{x\rightarrow 1}f(x)=\lim_{x\rightarrow 1}\frac{x}{e^{x/9}}\cdot \frac{\sin \left (\pi (x-1)\right )}{x-1}=\frac{1}{e^{1/9}}\cdot \lim_{x\rightarrow 1}\frac{\sin \left (\pi (x-1)\right )}{x-1}\ \overset{DLH}{ = } \ \frac{1}{e^{1/9}}\cdot \lim_{x\rightarrow 1}\frac{\pi \cos \left (\pi (x-1)\right )}{1}\\ & =\frac{1}{e^{1/9}}\cdot \pi \cos \left (\pi \cdot 0\right )=\frac{1}{e^{1/9}}\cdot \pi =\frac{\pi}{e^{1/9}}\end{align*}

Therefore we get the function \begin{equation*}g(x)=\frac{x}{e^{x/9}}\cdot \frac{\sin \left (\pi (x-1)\right )}{x-1}-\frac{\pi}{e^{1/9}}\end{equation*}

Now we have to apply an iteration method to approximate the root of that function, right? Do we use the Newton's method?

We don't have an interval to which the root will belong, we only know that it is greater than $1$. So do we have to guess such an interval to calculate the first input $x_0$?
[DESMOS=-1,4,-1,4]\frac{\left(x\sin\left(\pi\left(x-1\right)\right)\right)}{e^{x/9}(x-1)}[/DESMOS]
The above graph shows that the function repeats its value at $x=1$ when $x$ is somewhere near $1.4$ or $1.5$. So I would take $x_0=1.5$.

Last edited:

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Now we have to apply an iteration method to approximate the root of that function, right? Do we use the Newton's method?

We don't have an interval to which the root will belong, we only know that it is greater than $1$. So do we have to guess such an interval to calculate the first input $x_0$?
Hey mathmari !!

This is an example where Newton-Raphson can have problems if we are not careful.
If we pick a starting value that is too far from the zero, it will likely not converge.
However, a starting value that starts slightly to the right of the zero, such as the 1.5 that Opalg pointed out, should do the job. And it will converge quadratically.
Starting below 1.4 or above 2.1 will likely diverge though.

Alternatively algorithms are bisection and regula falsi.
First we might search for values that are on opposite sides of the x-axis.
That is, we can start with some initial interval, and then either double or half its size until we find both a positive and a negative function value.
The root must then in between those, after which both bisection and regula falsi will find it.

#### mathmari

##### Well-known member
MHB Site Helper
We have \begin{align*}&g(x)=\frac{x\sin \left (\pi(x-1)\right )}{e^{x/9}(x-1)}-\frac{\pi}{e^{1/9}}\\ &g'(x)=\frac{\left [\sin \left (\pi (x-1)\right )+x\pi \cos \left (\pi (x-1)\right )\right ]\left (x-1\right )-x\sin \left (\pi (x-1)\right ) \frac{8+x}{9}}{e^{x/9}(x-1)^2}\end{align*}

Choosing as initial value $x_0=1.5$ we get the following:
\begin{align*}x_1=x_0-\frac{g(x_0)}{g'(x_0)}\approx 1.4259 \\ x_2=x_1-\frac{g(x_1)}{g'(x_1)}\approx 1.4149 \\ x_3=x_2-\frac{g(x_2)}{g'(x_2)}\approx 1.4147\end{align*}
The first two decimal digits are the same as in the previous step and so position that we are looking for is $1.41$.

Is everything correct?

Staff member
Yep.


# Efficient method for computing the properties of a block anti-diagonal matrix

EDIT: the title and content of this question were reformulated for the sake of clarity and to avoid diverting attention from the main issue.

I was trying to answer a question about a system of linear differential equations and it was necessary to find the eigenvalues, eigenvectors and generalized eigenvectors for the following matrix: $$M=\begin{bmatrix}0&E_{\times}\\B_{\times}&0\end{bmatrix},$$ where $E=[E_1,E_2,E_3]^T,B=[B_1,B_2,B_3]$, $E^TB=0$ and $E_\times,B_\times$ are the representation of vectors through antisymmetric matrices: $$E_{\times}= \begin{pmatrix} 0 & -E_{3} & E_{2}\\ E_{3}& 0 & -E_{1}\\ -E_{2} & E_{1} & 0 \\ \end{pmatrix},\qquad B_{\times}= \begin{pmatrix} 0 & -B_{3} & B_{2}\\ B_{3}& 0 & -B_{1}\\ -B_{2} & B_{1} & 0 \\ \end{pmatrix}.$$ With WolframAlpha it is possible to see that all eigenvalues are zero and the kernel has dimension 2. I have tried to find some method to obtain these results without resort to software or brute force (solve a linear system step by step), but I could not.

Then my question is the following: is it possible to determine the null space, eigenvalues and generalized eigenvectors for matrix $M$ efficiently without resort to software or brute force?

It is always possible to compute them the hard way: computing the characteristic polynomial $\det(M-\lambda 1)$, finding its roots, substituting in $(M-\lambda 1)v=0$ to find eigenvectors $v$ and using them to determine the generalized eigenvectors. What I am looking for is a answer that computes the null space, eigenvalues and (generalized) eigenvectors making use of the properties of matrix $M$, not brute force.

• A matrix with all zero eigenvalues is nilpotent, and so, if diagonalizable, is identical zero. – kjetil b halvorsen Feb 7 '18 at 21:10
• – jobe Feb 8 '18 at 1:01

We assume that the system $\{E,B\}$ is linearly independent.

Note that (double cross product) $E_xB_xX=BE^TX-X(E^TB)=BE^TX$, that is, $E_xB_x=BE^T$ and, in the same way, $B_xE_x=EB^T$.

$M[X,Y]^T=[E\times Y,B\times X]^T$ and a basis of $\ker(M)$ is $\{[B,0]^T,[0,E]^T\}$.

Moreover $\det(M-\lambda I_6)=\det(\lambda^2I-BE^T)$; $BE^T$ has rank $1$ and trace $0$; then it is nilpotent and $M$ also.

About the generalized eigenvectors, $M^2=diag(BE^T,EB^T)$ and $M^2[X,Y]^T=0$ can be written $BE^TX=0,EB^TY=0$, that is, $X\in E^{\perp},Y\in B^{\perp}$, that implies that $dim(\ker(M^2))=4$.

It is not difficult to see that $M^3=0$ and we are done.

• What a great answer! Thank you very much for this elegant and very instructive answer. There is only one thing I didn't get it: it is not clear to me why $\det(M-\lambda I_6)=\det(\lambda^2 I-BE^T)$. Can you elaborate a little more? – jobe Feb 9 '18 at 12:38
• Since $B_x$ and $\lambda I_3$ commute, $\det(M-\lambda I)=\det(\lambda ^2I -E_xB_x)=\det(\lambda^2 I-BE^T)$. – loup blanc Feb 9 '18 at 13:05
• I still don't get it. In the first determinant, $\det(M-\lambda I_6)$, we have $6\times 6$ matrices, while in the second one, $\det(\lambda^2 I_3-E_\times B_\times)$, we have $3\times 3$ matrices. Why $\lambda^2$? I think you are using some property I don't know. – jobe Feb 9 '18 at 15:18
• If $CD=DC$, then $\det(\begin{pmatrix}A&B\\C&D\end{pmatrix})=\det(AD-BC)$. – loup blanc Feb 10 '18 at 11:22
• Thank you. I was ignorant of that property. – jobe Feb 11 '18 at 15:58

Very short: You can easily get examples of all ranks less than $n$, when the matrix is $n \times n$. A very simple example, take a matrix with zeros on the diagonal, zeros below, and ones above the diagonal. It is nilpotent, that is, there is some positive integer $k$ such that $A^k=0$.

An extended hint: When all eigenvalues are zero, then, if $x$ is an eigenvector, it will certainly belong to the kernel! So, if there is a basis of eigenvectors then the matrix must be zero.

• Thank you for your answer. Perhaps I was not very clear in my first formulation of this question, so I have reformulated it. The characteristic polynomial is $\lambda^6$ and the kernel has dimension 2, these two facts can be determined through software or brute force. Determination of generalized eigenvectors through brute force for this matrix is not so easy and so I began wondering if there is a more elegant and efficient method to obtain these results. That is what I am looking for as an answer. – jobe Feb 8 '18 at 12:44


# Thread: Show that arccos(...) = ...

1. ## Show that arccos(...) = ...

Hi

Another "show that" problem involving inverse trig. functions.

Show that $\displaystyle arccos(\frac{1}{\sqrt{1+x^{2}}}) = arctan(|x|)$

Since $\displaystyle | \frac{1}{\sqrt{1+x^{2}}} | \leq 1$ , I assume I can take $\displaystyle cos(arccos(\frac{1}{\sqrt{1+x^{2}}})) = \frac{1}{\sqrt{1+x^{2}}}$

But what do I do with my right-side?

Kinda stuck here.

2. Originally Posted by Twig
Hi

Another "show that" problem involving inverse trig. functions.

Show that $\displaystyle arccos(\frac{1}{\sqrt{1+x^{2}}}) = arctan(|x|)$

Since $\displaystyle | \frac{1}{\sqrt{1+x^{2}}} | \leq 1$ , I assume I can take $\displaystyle cos(arccos(\frac{1}{\sqrt{1+x^{2}}})) = \frac{1}{\sqrt{1+x^{2}}}$

But what do I do with my right-side?

Kinda stuck here.
let $\displaystyle u = arccos(\frac{1}{\sqrt{1+x^{2}}})$

$\displaystyle cos(u)=(\frac{1}{\sqrt{1+x^{2}}})$

as you can see from the pic

$\displaystyle tan(u) = x$

$\displaystyle arctan(tan(u))=arctan(x)$

$\displaystyle u=arctan(x) = arccos(\frac{1}{\sqrt{1+x^{2}}})$ since

we let
$\displaystyle u=arccos(\frac{1}{\sqrt{1+x^{2}}})$

3. Let $\displaystyle \theta=\arccos\left(\frac1{\sqrt{1+x^2}}\right).$ Then

$\displaystyle \cos\theta\ =\ \frac1{\sqrt{1+x^2}}$

$\displaystyle \implies\ \cos^2\theta\ =\ \frac1{1+x^2}\quad\ldots\,\fbox1$

$\displaystyle \therefore\ \sin^2\theta\ =\ 1-\cos^2\theta\ =\ 1-\frac1{1+x^2}\ =\ \frac{x^2}{1+x^2}\quad\ldots\,\fbox2$

$\displaystyle \therefore\ \tan^2\theta\ =\ \frac{\sin^2\theta}{\cos^2\theta}\ =\ x^2$

$\displaystyle \implies\ \tan\theta\ =\ \pm|x|$

$\displaystyle \implies\ \theta\ =\ \arctan(\pm|x|)\ =\ \pm\arctan|x|$

But since $\displaystyle \arccos y\ge0$ for all $\displaystyle 0<y\le1,$ $\displaystyle \theta\ge0.$ Hence $\displaystyle \arccos\left(\frac1{\sqrt{1+x^2}}\right)=\arctan|x |.$

4. Thanks guys, always nice with a few different approaches.

I think the drawing a triangle solution is something I need to remember.

5. Hello, Twig!

Show that: .$\displaystyle \arccos\left(\frac{1}{\sqrt{1+x^{2}}}\right) \:=\: \arctan(|x|)$
Let: .$\displaystyle y \;=\;\arccos\left(\frac{1}{\sqrt{1+x^2}}\right) \quad\Rightarrow\quad \cos y \:=\:\frac{1}{\sqrt{1+x^2}}$

Then: .$\displaystyle \cos^2\!y \:=\:\frac{1}{1+x^2} \quad\Rightarrow\quad 1-\cos^2\!y \:=\:1 - \frac{1}{1+x^2} \:=\:\frac{x^2}{1+x^2}$

. . Hence: .$\displaystyle \sin^2\!y \:=\:\frac{x^2}{1+x^2} \quad\Rightarrow\quad \sin y \:=\:\frac{\pm x}{\sqrt{1+x^2}}$

Then: .$\displaystyle \tan y \:=\:\frac{\sin y}{\cos y} \;=\;\frac{\dfrac{\pm x}{\sqrt{1+x^2}}}{\dfrac{1}{\sqrt{1+x^2}}}\quad\Ri ghtarrow\quad \tan y \;=\; \pm x$

. . Hence: .$\displaystyle y \:=\:\arctan(\pm x) \;=\;\arctan|x|$

Therefore: .$\displaystyle \arccos\left(\frac{1}{\sqrt{1+x^2}}\right) \;=\;\arctan|x|$


# In an integer program, how I can force a binary variable to equal 1 if some condition holds?

Suppose we have a binary or continuous variable $$x$$, a binary variable $$y$$, and a constant $$b$$, and we want to enforce a relationship like

If $$x \gtreqless b$$, then $$y = 1$$.

How can we write this using one or more linear constraints?

If $$x$$ is binary: Then the "if" condition really means either "$$x = 0$$" or "$$x=1$$".

To enforce "if $$x=0$$ then $$y=1$$": use $$y \ge 1-x.$$ To enforce "if $$x=1$$ then $$y=1$$": use $$y \ge x.$$

If you want to require that $$y=1$$ if and only if the condition holds, then replace the $$\ge$$s above with $$=$$s.

If $$x$$ is continuous: In this case, numerical inaccuracy might produce errors, so be prepared for $$y$$ to be set incorrectly if $$x$$ is close to but on the “wrong” side of $$b$$. To avoid this, you can increase or decrease $$b$$ a little bit to provide some buffer.

To enforce "if $$x < b$$ then $$y=1$$": $$b - x \le My,$$ where $$M$$ is a large constant. The logic is that if $$b - x > 0$$, then $$y$$ must equal 1, and otherwise it may equal 0.

To enforce "if $$x > b$$ then $$y=1$$": $$x - b \le My,$$ with similar logic as above.

To enforce "if $$x = b$$ then $$y=1$$": This one is tricky. I'm not sure my approach is the easiest. (Anyone have a better solution?) We really can't check for $$x=b$$, but we can check for $$b-\delta \le x \le b+\delta$$ for some small $$\delta > 0$$. To do this, we introduce two new binary decision variables.

Let $$z_1$$ be a binary variable that equals 1 if $$x > b - \delta$$, equals 0 if $$x < b - \delta$$, and could equal either if $$x = b - \delta$$. Enforce this definition by adding the following constraints: \begin{alignat}{2} Mz_1 & \ge x - b + \delta\tag1 \\ M(1-z_1) & \ge b - x - \delta\tag2 \end{alignat} The logic is:

• If $$x > b - \delta$$, then (1) forces $$z_1=1$$ and (2) has no effect.
• If $$x < b - \delta$$, then (2) forces $$z_1=0$$ and (1) has no effect.
• If $$x = b - \delta$$, then (1) and (2) have no effect; $$z_1$$ could equal either 0 or 1.

Next, introduce a second binary variable $$z_2$$, which equals 1 if $$x < b + \delta$$, equals 0 if $$x > b + \delta$$, and could equal either if $$x = b + \delta$$. Introduce the following constraints: \begin{alignat}{2} Mz_2 & \ge b - x + \delta\tag3 \\ M(1-z_2) & \ge x - b - \delta\tag4 \end{alignat} The logic is similar:

• If $$x < b + \delta$$, then (3) forces $$z_2=1$$ and (4) has no effect.
• If $$x > b + \delta$$, then (4) forces $$z_2=0$$ and (3) has no effect.
• If $$x = b + \delta$$, then (3) and (4) have no effect; $$z_2$$ could equal either 0 or 1.

From constraints (1)-(4), we can say that if $$z_1=z_2=1$$, then $$b - \delta \le x \le b + \delta$$. Therefore, we can enforce "if $$b - \delta \le x \le b + \delta$$ then $$y=1$$" using: $$y \ge z_1 + z_2 - 1.$$

Note: If your model is relatively large, i.e., it takes a non-negligible amount of time to solve, then you need to be careful with big-$$M$$-type formulations. In particular, you want $$M$$ to be as small as possible while still enforcing the logic of the constraints above.

• In the first part where $x$ is binary, what if I have 2 or more binary variable that leads to the decision of $y$ like $x$ and $z$ when $x=1 \wedge z = 1$ then $y = 1$
– ooo
Jan 28 '20 at 21:22
• Then you'd have to formulate separate constraints in which you enforce the definition of $y$ and then use $y$ as described above. Jan 29 '20 at 14:33
• I didn't get it.
– ooo
Jan 29 '20 at 19:38
• Lets say I have 4 binary variable $a,b,c,d$ if all $a,b,c,d = 1$ the $x = 1$ else $x =0$, then can I write $x \ge 3 - a+b+c+d$
– ooo
Jan 30 '20 at 13:57
• @LarrySnyder610: how small could we set $\delta$ to? Jun 4 '20 at 17:39

Rather than linearising the logical constraint, I would try the logical constraints built in a solver. Gurobi and SCIP both have indicator constraints.

My colleague works with these a lot and he’s finding the indicator constraints in Gurobi perform worse than big-M. He’s in contact with the Gurobi developers so I might be able to get more info if there’s interest.

• I've never tried those; that's a good suggestion. More info would certainly be welcome (maybe in a new Q&A). May 31 '19 at 15:36

To model $$x=b \implies y=1$$, where $$L \le x \le U$$, you can do the following: \begin{align} L y^- + b y + (b+\delta)y^+ \le x &\le (b-\delta) y^- + b y + U y^+\\ y^- + y + y^+ &= 1 \\ y^-, y, y^+ &\in \{0,1\} \end{align} In fact, this formulation also enforces the converse $$y=1 \implies x=b$$.

• If we assume $L=b-\delta$ and $U=b+\delta$, then the first constraint is essentially: $L y^- + b y + Uy^+ \le x \le L y^- + b y + U y^+$ or $x = L y^- + b y + U y^+$.
– EhsanK
Jan 17 '20 at 1:40
• True, but the idea here is that $\delta>0$ is small, so those assumptions on $L$ and $U$ would imply that $x$ is essentially constant. The more useful setting would be when $L\ll b\ll U$. Jan 17 '20 at 1:48


# Expectation of a hitting time

I'm trying to find the expectation of a stopping time. Specifically,

Let $$T_1,...,T_n$$ be i.i.d exponential random variables with mean $$1$$. Let $$S_n = T_1 + ... + T_n$$ denote their partial sum. Define the stopping time,

$$T = \inf\{n \geq 1 : S_n \geq 1\}$$

which is the first time $$S_n$$ exceeds $$1$$. Calculate $$E[T]$$.

Here was my approach. Let $$x = E[T]$$. I want to condition on what happens at the first time $$T_1$$. If $$T_1 \geq 1$$, then the process, $$\{S_n\}$$ has stopped and $$T \equiv 1$$. Otherwise if $$T_1 < 1$$, then after one step, process, $$\{S_n\}$$ will renew again until it reaches $$1$$. So we have the following equation,

$$\begin{eqnarray} x &=& E[T]\\ &=& E[T | T_1 < 1]P(T_1 < 1) + E[T | T_1 \geq 1]P(T_1 \geq 1)\\ &=& (1+x)P(T_1 < 1) + P(T_1 \geq 1)\\ &=& (1+x)(1-e^{-1}) + e^{-1} \end{eqnarray}$$

This results in $$x = e$$.

However, I was told that the answer is $$2$$. They gave a heuristic explanation that $$S_T$$ is distributed as $$1 + S$$ where $$S$$ is an exponential random variable with mean $$1$$. I can see the intuition behind this from the memoryless property, but I can't prove why it is so. I ran three simulations in $$\textsf{R}$$ and got $$x \approx 2.001, 2.0161, 1.9785$$, which seems to confirm that the answer is $$2$$. Can someone explain this result?

Also, why/where did my approach fail?

• When you say $\mathbb{E}[T|T_1 < 1] = 1 + x,$ you're asserting that, in expectation, if the first step doesn't get all the way to $1,$ then the remaining steps have to get all the way to $1$. This is clearly false - if $T_1 = 1/2,$ then $T_2 + \dots T_T$ only have to get up to $1/2$. Since $T_1 > 0$ a.s., we should have $\mathbb{E}[T|T_1 < 1] < 1 + \mathbb{E}[T].$ – stochasticboy321 Apr 13 '19 at 19:00
• I noticed that you wanted a non-heuristic proof - note that $\{T > n\} = \{S_n < 1\}$, since the $S_n$ are non-decreasing. It should be straightforward to figure out $\pi_n := P(S_n<1)$ by establishing a recurrence relation between $\pi_n$ and $\pi_{n-1}$ - you'll likely need the volume of a standard simplex in $n$ dimensions. This will directly give you $P(T = n),$ which you can then use to find the mean. – stochasticboy321 Apr 13 '19 at 21:19
• A more high level argument is from queuing theory - suppose it takes you $\mathrm{Exp}(1)$ time to do a job. How many jobs will you finish in $1$ unit of time? It is a classical result that this number is $\mathrm{Poission}(1)$ distributed. However, intuitively, the mean is simpler to argue - your rate of doing jobs is $1$ per unit, so you, in expectation, should finish one job per unit (a Little's law type argument). You are interested in this number plus one - which job will you be doing when the time unit finishes. – stochasticboy321 Apr 13 '19 at 21:24
• @stochasticboy321 Thank you! I was able to do a brute force calculation along these lines. – Flowsnake Apr 14 '19 at 16:24
• ^That's grand, you're welcome :). You should add an answer below, so that others trying the same problem can have a reference. – stochasticboy321 Apr 15 '19 at 1:59

Following @stochasticboy321's approach, we want to find $$P(T > n) = P(S_n < 1)$$. Since $$S_n = T_1 + ... + T_n \sim$$ Gamma($$n,1$$), we have,
$$P(S_n < 1) = \frac{1}{\Gamma(n)}\int_0^1x^{n-1}e^{-x}dx = 1 - \frac{\Gamma(n,1)}{\Gamma(n)}$$
where $$\Gamma(n,1)$$ is the incomplete Gamma function. To get this expression, I used the nice identity found here. Finally,
$$E[T] = 1 + \sum\limits_{n=1}^\infty P(T > n) = 1 + \sum\limits_{n=1}^\infty P(S_n<1) = 1+ \sum\limits_{n=1}^\infty\left(1 - \frac{\Gamma(n,1)}{\Gamma(n)}\right)$$
I have no idea how to evaluate the sum in closed form, but a computation in Wolfram Alpha seems to suggest it converges to $$1$$. Thus, $$E[T] = 2$$ (at least by conjecture from this numerical computation).
• If we set the integral above to $I_n,$ then by integration by parts, and using $\Gamma(n) = (n-1) \Gamma(n-1) = (n-1)!,$ we get for $n \ge 1,$ $$P(T > n) = \frac{I_n}{\Gamma(n)} = \frac{(n-1) I_{n-1}}{(n-1) \Gamma(n-1)} - \frac{e^{-1}}{{(n-1)!}} = P(T > n-1) - \frac{e^{-1}}{(n-1)!}$$ But then $$P(T = n) = P(T> n-1) - P(T > n) = \frac{e^{-1}}{(n-1)!}.$$ We immediately have that $T \overset{\mathrm{law}}= 1 + Z,$ where $Z \sim \mathrm{Pois}(1),$ and so has mean $2$. Implicitly this also solves the series above (by summation by parts). – stochasticboy321 Apr 15 '19 at 19:35


+0

+1
104
16
+365

Guys please solve this question and let me know the correct option

please its a request here the link of question if you cant see it (https://ibb.co/ZWftpcR)

Mar 22, 2021
edited by kes1968  Mar 22, 2021

#1
+9279
+2

(Assuming that  log  means natural log)

y   =   5log x

Take the log of both sides

log y   =   log( 5log x )

Now we can apply the rule that says:  log(xy)  =  y log x

log y   =   log x log 5

Divide both sides by  log 5

log y / log 5   =   log x

Do the reverse of natural log to both sides (make both sides the exponent of e )

e^( log y / log 5  )   =   e^log x

Now the right side simplifies to just  x

e^( log y / log 5  )   =   x

So we have...

$$x\ =\ e^{\frac{\log y}{\log 5}}\\~\\ x\ =\ (e^{\log y})^{\frac{1}{\log 5}}\\~\\ x\ =\ y^{\frac{1}{\log 5}}$$

Mar 22, 2021
#2
+365
+2

brother what you have found is the same function just in terms y , but inverse function is symmetrical about y=x axis ,

also f(f^-1(x))=x which is not satisfied ,

by the way if i do the same operation and interchange x and y i get 2 and 4 option , that satifies both conditions

Mar 22, 2021
#3
+9279
+3

The inverse function can be graphed by:    $$y=x^{\frac{1}{\log5}}$$

Here's a graph:

https://www.desmos.com/calculator/aijpzpouk1

In other words...

If   $$f(x)=5^{\log x}$$    then the inverse is   $$f^{-1}(x)=x^{\frac{1}{\log 5}}$$

And...

$$f(f^{-1}(x))\ =\ f(x^{\frac{1}{\log 5}})\ =\ 5^{\log(x^{\frac{1}{\log 5}})}\ =\ 5^{\frac{\log x}{\log 5}}\ =\ 5^{\log_5 x}\ =\ x$$

The options leave it in the form that is solved for  x, and so I left it like that to match the options

Mar 22, 2021
#4
+365
+2

brother so what you feel (as a math expert) should be the right answer to the given question ,

Mar 22, 2021
#5
+9279
+1

I'm not sure what you meant by "by the way if i do the same operation and interchange x and y i get 2 and 4 option , that satifies both conditions"....but if this didn't answer your question then please feel free to ask for more clarification!

And I think the answer is option 1:  $$x=y^\frac{1}{\log 5}$$

Mar 22, 2021
#6
+365
+3

brother , if you see your second answer and change it in terms of x , wont you get option 2 and option 4 ,

Mar 22, 2021
#7
+9279
+1

Hmm...actually....I see what you mean.... (maybe I made the question harder than it has to be!)

I take back my original answer!! Now I think the answer is option 4

Mar 22, 2021
#8
+365
+1

brother do you really believe its option 4 or just to keep my heart , please clarify if you still believe the answer is 1 , if yes then please prove the same to me as well

Mar 22, 2021
#9
+1740
+1

For computing the inverse function, the plan I know is
1. interchange x & y

2. solve for y

but actually, after 1. you already have a term for the inverse function. It's just not written in the "usual" way, wich is y=f(x).

Answer for is exactly what you get after interchanging x&y, so the correct answer is answer 4.

Mar 22, 2021
#10
+9279
+2

I do agree Probolobo, but then the confusing thing is that

$$5^{\log y}\ =\ y^{\log 5}$$

Which means option 2 and option 4 are the same function and so are equivalent....

Mar 22, 2021
#12
+1740
+1

That's correct, then there are actually 2 correct answers, 2 & 4. Didn't see that equivalence on first glance ;)

Probolobo  Mar 22, 2021
#11
+365
+1

so whats the final answer so that i can challenge the answer key?

Mar 22, 2021
#13
+9279
+1

If I had to guess right now, I would guess option 4. My next guess would be option 1.

But I am honestly not sure!! I think this is a bad question.

Can you let us know when you find out the "correct" answer according to the answer key?

hectictar  Mar 22, 2021
#14
+112907
+3

I think you are making hard work of it

the inverse of

$$y=5^{logx}$$

is simply

$$x=5^{logy}$$

You just have to switch the x and y over.

there would be restrictions on x and on y but the question isn't worrying about that.

Here is the graphs

https://www.desmos.com/calculator/4ftfa2bny7

See they are reflections of each other about y=x

Mar 22, 2021
#15
+365
+2

another question , another superb explanation from mod

Melody you are simply awesome ! thanks for the help mate

kes1968  Mar 22, 2021
edited by kes1968  Mar 22, 2021
#16
+112907
+1

You are welcome Kes

Melody  Mar 22, 2021


# Five people get on the elevator that stops at five floors. In how many ways they can get off?

Five people get on the elevator that stops at five floors. In how many ways they can get off? For example, one person gets off at the 1st floor, two will get off at the third, and the remaining two at the fifth floor. In how many ways they can get off at different floors? Now, consider that people in elevator have names, say A,B,C,D and E., assuming that, for example, the case A on the 1st floor is different from the case B on the 1st floor. Answer the previous questions with this assumption.

I was told that there are 4 questions, but I'm not really sure about it.

1. People don't have names and get off
2. People don't have names and get off at different floors
3. People have names and get off
4. People have names and get off at different floors.

Do the names matter?

I think I can answer one question out of 4 - the number of all outcomes should be $5^5=3125$. I am confused about the rest though

• I think I can answer one question out of 4 - the number of all outcomes should be 5^5=3125. I am confused about the rest though. – Gabriel Oct 6 '17 at 16:33
• Wait, there are 4 questions here? I only see two: one where you do differentiate between the 5 people and one where you don't. And yes, for the second one you indeed get $5^5$. – Bram28 Oct 6 '17 at 16:36
• Do you know about Stars and Bars? – Bram28 Oct 6 '17 at 16:37
• Thank you!! I was told that there are 4 questions, but I'm not really sure about it. 1. People don't have names and get off, 2. People don't have names and get off at different floors, 3. People have names and get off, 4. People have names and get off at different floors. Do the names matter? And thank you, I will definetely read about Stars and Bars!! – Gabriel Oct 6 '17 at 16:52

1. This has been answered in other answers: ${9\choose4}$.

2. Since there are $5$ people and $5$ floors there is just $1$ way for them to get off at different floors: $1$ per floor.

3. You already found $5^5=3125$.

4. At each floor exactly one of $A$, $\ldots$, $E$ gets off. This can be done in $5\cdot4\cdot3\cdot2=5!=120$ ways.

For the first question, use the basic 'stars and bars' method.

We can represent your example scenario ("For example, one person gets off at the 1st floor, two will get off at the third, and the remaining two at the fifth floor.") as:

$$*||**||**$$

The people are the 'stars' and the separators between the different floors are the 'bars'.

Another example: one person getting off on each floor would be:

$$*|*|*|*|*$$

And everyone getting off on the 3rd floor would be:

$$||*****||$$

See how this works?

Now, with $4$ 'bars' to divide the people into the $5$ groups/floors, and those bars taking up $4$ of the $9$ possible positions in this 'symbol string' of bars and stars, you get

$$9 \choose 4$$

possibilities.

• Thank you so much for the explanation!! I think I understand how it works now. – Gabriel Oct 6 '17 at 16:54
• @Gabriel You're welcome! Your instructor did not teach 'stars and bars' yet? ... Because this was a typical stars and bars question! :) – Bram28 Oct 6 '17 at 16:56
• Sometimes they just give the formula derived from stars and bars which is $\dbinom{n+r-1}{r-1}$ in this case $n=5$ and $r=5$ of course, the integer solutions for the equation $x_1+x_2+x_3+x_4+x_5=5$ are considered there are no left and right boundries (luckily). Of course giving the formula without the logic is problematic but I haven't seen one problem that required the logic rather than the formula. – Deniz Tuna Yalçın Oct 6 '17 at 20:21

The question could also be solved using a well known formula derived from stars and bars which is; $$\dbinom{n+r-1}{r-1}$$ where we distribute $r$ equally accepted objects to $n$ spaces with or without restrictions like $\geq1$ or $\leq9$, this formula is also used for solving linear equations (in integers) that's why this works for combinatoric problems, so without making it longer;$$x_1+x_2+x_3+x_4+x_5=5$$ so we apply the formula, we can think that this is equivalent to distributing five equally accepted apples to five children for instance, the result is $$\dbinom{5+5-1}{5-1}=126$$ this statement could be proven using alternate ways, we should open a post about it too.


# Have you ever used Scratch pad like this?

Find the root of following equation.

$4y^3-2700(1-y)^4=0$

Above equation can be solved using newton raphson iterative method with initial approximation to be unity.

1) First Let $f(y)=4y^3-2700(1-y)^4$ We have to find root of above function.

2)Find derivative of $f(y)$. Here,$f^{'}(y)=12y^2+10800(1-y)^3$

3)Find $y_n$ using newton raphson method $y_0=1$ $y_{n+1}=y_n-\frac{f(y_n)}{f^{'}(y_n)}$ $y_{n+1}=y-\frac{4y^3-2700(1-y)^4}{12y^2+10800(1-y)^3}$

You can use calculator or scratchpad to calculate $y_1,y_2,y_3,....$.If this equation has real solution ,these values $y_1,y_2,y_3,....$ would converge to root of $f(y)$.

How to do such iterations smartly?

1) $\color{#69047E}{\text{On first line of scratch pad write y=1}}$

2) $\color{#D61F06}{\text{On second line write}}$

$y=y-\frac{4y^3-2700(1-y)^4}{12y^2+10800(1-y)^3}$

$\color{#20A900}{\text{You will see new value of y as result.}}$

3) $\color{#EC7300}{\text{From line 3, continue pasting}}$

$y=y-\frac{4y^3-2700(1-y)^4}{12y^2+10800(1-y)^3}$

until you get a constant value.If your initial approximation is not too away from actual solution,you would get constant value in no more than 10 steps.

By this method you can solve most of the transcendental equation.

$\color{#3D99F6}{\text{By using scratchpad, try finding sum of initial 10 fibonacci number.}}$

Note by Aamir Faisal Ansari
5 years, 4 months ago

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Yes !i have used it this way.Sometimes in Recursion problems and like that.

- 5 years, 4 months ago

By using scratchpad, try finding sum of initial 10 fibonacci number.

- 5 years, 4 months ago


# pre calculus

posted by joe

cube roots of 64i?

1. MathMate

Convert the given number into the form
a+bi = α(cos(β)+i sin(β))
where
α = sqrt(a^2+b^2)
β = sin-1(b/α)
If the number is plot in the complex (Z) plane, it will be more evident.
For example, 64i will have
α=64 (the distance from origin)
β=90° (sin-164/64=1)
The equivalent angles are:
90°
450°
810°
(we only need 3)

Divide each of these angles by , to get β1, β2, & β3. And take α'=α^(1/3)=64^(1/3)=4
The three cube roots are then
z1=α'(cos(β1)+isin(β1)
=4(cos(30°)+isin(30°))
z2=4(cos(150°)+isin(150°))
z3=4(cos(270°)+isin(270°))

Check by expanding z1^3, z2^3 and z3^3 to get back 64i.

2. MathMate

β=90° (sin-164/64=sin-11=π/2)

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# What is the minimum number of rotations about axes in a plane that can describe an arbitrary rotation in 3D?

I'm trying to decompose an arbitrary rotation of a 3D sphere into a series of rotations about any axes that lie in the equatorial plane.

I know I can factor the arbitrary rotation into three rotations in the equatorial plane using Euler angle formulas. Can I factor into two rotations if I allow any choice of axis in the equatorial plane? If it exists, is this factorization unique?

## migrated from mathoverflow.netAug 2 '16 at 14:05

This question came from our site for professional mathematicians.

• There is a unique rotation mapping one unit vector into a distinct vector. Take one of the vectors of your basis, and apply this rotation. Now hopefully you can see that once one axis is coincidental, you only need another rot. in the orthogonal plane. This means at two rotations are necessary and sufficient in general. – Real Aug 2 '16 at 3:48
• what do you understand by equatorial plane (is it the xy-plane)? – Manfred Weis Aug 2 '16 at 7:22
• @ManfredWeis Yes by equatorial plane I mean the xy-plane. – Talon Chandler Aug 2 '16 at 12:28
• Talon, I thought the answer is yes and that I have a simple proof for it. But the question might not be appropriate for this site. If you ask over at Mathematics StackExchange and then let me know here (with a link) when you have posted, I'd be happy to give more details. – user43208 Aug 2 '16 at 13:49
• On second thought, let me migrate it for you... – user43208 Aug 2 '16 at 14:05

My answer is that you can.

If you know about quaternions, then you know that a rotation about a unit vector $v \in \mathbb{R}^3$ through an angle $\theta$ can be accomplished by regarding elements $w = (a, b, c) \in \mathbb{R}^3$ as purely quaternionic elements $ai + bj + ck$, and then mapping $w \mapsto uwu^{-1}$ where $u = \cos(\frac{\theta}{2})\cdot 1 + \sin(\frac{\theta}{2})v$. Such $u$ are precisely quaternions of unit norm.

Let us denote the rotation above by $R_u$ (so $R_u(w) := uwu^{-1}$). Notice that a composition of two rotations $R_u \circ R_v$ is just $R_{u v}$; this follows by associativity of quaternionic multiplication.

So we can translate your problem into the following: for any rotation $R_w \in SO(3)$ given by a nonzero quaternion $w$, show that there exist quaternions $u, v$ in the linear span of $1, i, j$ such that $u v = w$. For these $u, v$ describe rotations about vectors in the equatorial plane spanned by $i, j$.

Now this is not difficult. Let us write $u = a + bi + cj$ and $v = a' + b'i + c'j$. We compute

$$(a + bi + cj)(a' + b'i + c'j) = aa' - bb' - cc' + (ab' + a'b)i + (ac' + a'c)j + (bc' - b'c)k$$

and so given a nonzero quaternion $w = p + qi + rj + sk$, our task is to cook up parameters $a, b, c, a', b', c'$ such that

$$p = aa' - bb' - cc'$$

$$q = a b' + a'b$$

$$r = ac' + a'c$$

$$s = bc' - b'c$$

In fact, let's make our life easier and simply set $b = 0, b' = 1$. Then $a = q$, $c = -s$, and we can solve for $a', c'$ in the linear system

$$p = qa' + sc'$$ $$r = -sa' + qc'$$

provided the determinant $q^2 + s^2$ is nonzero, i.e., provided one of $q, s$ is nonzero. If both $q, s = 0$, then of course $w = p + rj$ corresponds to a rotation about the vector $j$ which is already in the equatorial plane, so there was nothing to do in this case.

Once you have solutions $u = a + bi + cj, v = a' + b'i + c'j$ to the equation $uv = w$ in your hands, the rotations $R_u, R_v$ are rotations about the vectors $bi + cj$ and $b'i + c'j$, respectively. If you want the rotation angles, then normalize $u$ and $v$ (i.e. divide by their norms $(a^2 + b^2 + c^2)^{1/2}$ and $(a')^2 + (b')^2 + (c')^2)^{1/2}$) and use the fact that $R_u = R_{\frac{u}{\|u\|}}$. You can then read off the desired angles by writing e.g. $\frac{u}{\|u\|} = \cos(\frac{\theta}{2}) + \sin(\frac{\theta}{2})\frac{bi + cj}{(b^2 + c^2)^{1/2}}$, so for instance $\cos(\frac{\theta}{2}) = \frac{a}{(a^2 + b^2 + c^2)^{1/2}}$.

By the way, this solution also shows that the decomposition is in general non-unique. For example, we could have also chosen $c = 0, c' = 1$ and arrive at quaternionic solutions.

• Do we need to constrain $u$, $v$, and $w$ to be unit quaternions so they represent rotations? If this is the case, then we have 7 equations with 6 unknowns which might have a unique solution. – Talon Chandler Aug 2 '16 at 18:10
• No, you don't need to, but you can assume they are unit quaternions if you like because $R_u = R_{\frac{u}{\|u\|}}$ for any nonzero quaternion $u$. If $w$ is constrained to the unit sphere in the quaternions (3-dimensional $S^3$) and $u, v$ are each constrained to the unit sphere in the span of $\{1, i, j\}$ (so $(u, v)$ parametrizes a space that looks like $S^2 \times S^2$, which is 4-dimensional), then we are asking about surjectivity of a multiplication map $S^2 \times S^2 \to S^3$, so 3 equations in 4 variables by my count. – user43208 Aug 2 '16 at 18:24
• Okay that makes sense. What if we consider only infinitesimal rotations? Can we uniquely decompose an arbitrary rotation into infinitesimal rotations in the plane spanned by $i,j$? In this case we can make the approximations: $$\cos\left(\frac{\theta}{2}\right) \approx 1$$ $$\sin\left(\frac{\theta}{2}\right) \approx \frac{\theta}{2}$$ So $p=a=a'=1$ and we have four equations with three unknowns which gives a unique solution---i.e. an infinitesimal rotation about an arbitrary axis can be decomposed uniquely into two infinitesimal rotations about axes in the span of $i,j$. – Talon Chandler Aug 2 '16 at 19:28
• Well, the answer to the third sentence in your last comment would be 'no'. But why is getting uniqueness important to you? I think in fact the method I gave you gives virtual uniqueness of a solution if you constrain the first rotation $R_u$ to rotate about $j$. That is: unit quaternions $u$ of the form $a + bj$ form a circle $S^1$, and the unit $v$ for rotations about a second axis in the $ij$-plane form a sphere $S^2$, and the multiplication map $S^1 \times S^2 \to S^3: (u, v) \mapsto w = uv$ is onto, and unless $w$ itself is a rotation about $j$, only one pair $(u, v)$ satisfies $w = uv$. – user43208 Aug 2 '16 at 20:35
• I'm interested in uniqueness because I'm working on a reconstruction problem. I have measurements that correspond to a known rotation about an arbitrary axis, and I want to reconstruct the set of rotations in the $ij$-plane that gave rise to the "total" rotation. I'm hoping to find the conditions under which this reconstruction is possible. Can you expand on why we can't decompose an arbitrary infinitesimal rotation into two unique infinitesimal rotations in the $ij$-plane? – Talon Chandler Aug 2 '16 at 21:03

The answer is no, because when letting $v$ be an arbitrary direction about which to rotate by an angle $\rho$, that rotation can only then be realized with a rotation around an axis in the xy-plane if $v^Tz=0$; that shows that at least three rotations are necessary

1. rotate about $v\times z$ to bring $v$ into the $xy$-plane
2. rotate about the image of $v$ in the $xy$-plane
3. rotate the image of $v$ back around $v\times z$ to bring the image back to the original direction of $v$
• I agree that this set of rotations in the xy-plane can describe an arbitrary rotation, but how can we be sure there isn't a shortcut with just two rotations? – Talon Chandler Aug 2 '16 at 12:32


How can I find the time when two people agree to meet if both of their watches have an offset?

Chemist116

The problem is as follows:

Jenny and Vincent agreed to meet at the library at $6\,p.m$. Both synchronized their watches at midnight ($\textrm{0 hours}$). We know that Vincent's watch is not functioning correctly and gets ahead of the real time $\textrm{50 seconds}$ each hour and Jenny's watch gets delayed from the real time $\textrm{50 seconds}$ each hour. Vincent arrived to the library $\textrm{15 minutes}$ before the agreed time accoring to his watch and Jenny $\textrm{15 minutes}$ late by looking at her watch. Using this information, find how long did Vincent waited Jenny?

The alternatives according to my book are as follows:

$\begin{array}{ll} 1.&\textrm{15 min}\\ 2.&\textrm{0 min}\\ 3.&\textrm{25 min}\\ 4.&\textrm{60 min}\\ \end{array}$

This part I'm stuck at exactly how should I make up an equation which can relate both offsets. Can someone help me here? I'm assuming that by $\textrm{6 p.m}$ the time which would be seen by Vincent will be:

$18\times 50= 900\,s$

which would be $60$ minutes

But Vincent seen in his watch was:

$\textrm{5:45 pm}$

hence until 5 pm would be:

$17\times 50=850\,s$

$56\frac{2}{3}$ minutes

and Vincent's watch would have seen:

$\textrm{5:56:40 pm}$

At this point I could try guessing reducing the number of minutes until adjusting the time which will be seen by Vincent and this would be same for Jenny, but it doesn't seem something which can be effective. Can someone help me here? How exactly can I find what is being requested?

skeeter

Math Team
50 seconds an hour = 5 minutes in 6 hours

Vincent’s watch runs fast, which means the real time he arrives is earlier than 17:45.

Jenny’s watch runs slow, which means the real time she arrives is later than 18:15.

The only answer choice that makes sense is the 60 minutes Vincent waited.

I worked out the problem using proportions, where T represents the actual time

for Jenny ...
T/18.25 = 72/71

for Vincent ...
T/17.75 = 72/73

topsquark

Chemist116

50 seconds an hour = 5 minutes in 6 hours

Vincent’s watch runs fast, which means the real time he arrives is earlier than 17:45.

Jenny’s watch runs slow, which means the real time she arrives is later than 18:15.

The only answer choice that makes sense is the 60 minutes Vincent waited.

I worked out the problem using proportions, where T represents the actual time

for Jenny ...
T/18.25 = 72/71

for Vincent ...
T/17.75 = 72/73
This is the major problem which I had with this question. It mentioned that until his time was 17:45, not 18:00. How would be the difference between those fractions? Can you help me with that part?


# Combinations: How many handshakes?

I need help with the following question which I cannot seem to solve:

17 students are sitting in a circle. Each person shakes hands with everyone but his/her neighbours. How many handshakes have been exchanged?

My approach: no. of ways $= 1 + 2 + 3 + ... + 14 = 7(15) = 105$.

Apparently my answer is wrong (correct ans is 119). But I can't seem to understand why. Could someone please explain?

• The reason why this is wrong is as follows: the first person has 14 people to shake hands with. The second person as well has 14 people to shake hands with, since he wouldn't shake hands with his neighbour anyway! So, the sum is 14+(14+13+...+1), which does add up to 119. – Sanchises Feb 16 '15 at 19:49

Each of 17 people shakes hands with 14 people (all except themselves and their 2 neighbors), so there are

$$\frac{17\times 14}{2} = 119$$

handshakes (dividing by 2 to account for symmetry, as you would otherwise count "$A$ shaking hands with $B$" and "$B$ shaking hands with $A$" as distinct events).

• Does it mean that we don't care about how they sit, say in a circle or in a row? – Nighty Feb 16 '15 at 15:04
• @LeeKM: The difference is that in a circle, everyone has two neighbors, while in a row, the first and last person have only one each. – user139000 Feb 16 '15 at 15:05
• Thanks! Now I know that I misunderstood the quuestion. I thought that the handshakes were to be made with distinct persons. – Donald Feb 16 '15 at 21:44

Here's another way to think about it.

If everyone shook hands with everyone you'd have $\frac {17*16}2 = 136$ handshakes (divide by 2 because the above is double counting A shaking hands with B and so forth).

However, there are $\frac {17*2}2 = 17$ handshakes that aren't happening because neighbors aren't shaking hands (2 neighbors for each of the 17 people, again divided by 2 to remove duplicates).

So $136-17 = 119$

Pew's response is a little more direct, but finding the number by calculating the total possibilities minus the "not allowed" interactions is sometimes a little more intuitive for some people.

You can think of this in graph theory as the following:

$G$ has $17$ vertices with degree $14$, since there are no loops(people self handshaking), and no vertices are adjacent to the vertices beside them. So total degree is $17\times 14$.

We know that degree is equal to $2\times\text{number of edges}$ and hence there are $\frac{17\times 14}{2}=119$ edges in total, where edges represent handshakes.

As is typical for problems with big numbers, you should always resort to a smaller number if you can't solve the full problem.

Here, 17 people at the table is a bit hard to imagine immediately.

How many handshakes now?

There are zero.

A B

C D

A can't shake with B or C, B can't with A or D, D can't with B or C, C can't with A or D--only A & D and B & C can shake.

So 2 shakes:

A B
X
C D

5 people is when it gets interesting, and when you should be able to see the pattern:

      B

A       C

D   E

Everyone has two people they can't shake with, leaving two shakes per person--but we overcount by just multiplying 5 & 3, so we divide by two.

This is easiest to see by trying to draw the graph mentioned by @Commitingtoachalleng -- draw a line between any people who can shake hands:

        B
/ \
A-----------C   (also C-D & A-E--a 5-point star)
/     \
D       E

So we hypothesize the answer is $\frac{n(n-3)}{2}$. Note that this formula holds for $n=3$ and $n=4$ as well, as we'd hope!

One final way to see this is to look again at the graphs--

you might notice that they're always complete graphs (every vertex connected to every other vertex) with the outer edges removed.

Since there are $\frac{n(n-1)}{2}$ edges in a complete graph on $n$ vertices (which you can confirm yourself by a similar process), and $n$ outer edges,

there must be $\frac{n(n-1)}{2} - n = \frac{n^2-n-2n}{2} = \frac{n(n-3)}{2}$ handshakes.

However you cut it, there are $\frac {17 \cdot 14} 2 = 119$ total handshakes.

• We should also point out that the second option (working backwards from complete graphs) is the graph analogue of @Duncan's solution – MichaelChirico Feb 17 '15 at 6:42

I will explain for 6, same logic applies for all the numbers.

first, if 6 persons shakes will all in the party, then number of shakes will be:

A B C D E F

A can shake with only rest of 5 persons(B,C,D,E,F) i.e, right side ppl. = 5

B can shake with only his right sides(C,D,E,F) ppl(not with A bcos,A already shook with B just now. so no repeat) = 4

C can shake with only his right sides(D,E,F) ppl(not with A&B bcos,A&B already shook with C just now. so no repeat) = 3

D can shake with only his right sides(E,F) ppl(not with left side ppl bcos, they already shook with D just now. so no repeat) = 2

E can shake with only his right sides(F) ppl(not with left side ppl bcos, they already shook with E just now. so no repeat) = 1

as all of the persons shook hand with F, He dont shake with any one now.

So total possible shakes for 6 persons are = 5+4+3+2+1 => 15

now in questions it is given that, No one shakes hand with neighbors.

Imagine ppls sitting around table

        A   B
F       C
E   D

Please treat above image as 6 Ppl around circle : So now,

Invalid Shakes:

A cannot shake with B,F . I am representing it as below : A -no- B & F => 2 invalid shakes.
B -no- C only. Bcos B -no- A is covered as part of first case.
=> 1 invalid shake
C -no- D only. => 1 invalid shake
D -no- E only. => 1 invalid shake
E -no- F only. => 1 invalid shake
F dont shake with A is already covered as part of first invalid shakes.

so total invalid shakes are 2+1+1+1+1 = 6

so for 6 ppl: out of total 15 , 6 are invalid. So 15-6 = 9

on the same line :

for 17 ppl: out of total (16+15+...+1)=136 , 17 are invalid. So 136-17 = 119

shortcut : For N ppl, Total shakes are (N-1)(first term + last term)/2 Arithmetic progression sum. out of which Invalids are N.
So valid shakes => (Total - N)

• What is the point of posting such a complex answer to a question with a simple and accepted answer posted almost four years ago? – José Carlos Santos Oct 31 '18 at 17:23


# Is this function injective / surjective?

A question regarding set theory.

Let $g\colon P(\mathbb R)\to P(\mathbb R)$, $g(X)=(X \cap \mathbb N^c)\cup(\mathbb N \cap X^c)$

that is, the symmetric difference between $X$ and the natural numbers.

We are asked to show if this function is injective, surjective, or both.

I tried using different values of $X$ to show that it is injective, and indeed it would seem that it is, I can't find $X$ and $Y$ such that $g(X)=g(Y)$ and $g(X) \neq g(Y)$ but how do I know for sure? How do I formalize a proof for it?

Regarding surjective: I think that it is.

We can take $X=\mathbb R - \mathbb N$ and we get that $g(X)=\mathbb R$

What do I do about the injective part?

-
are you able to express in formula what injective or surjective means? –  Robert.Sie Nov 28 '13 at 14:14

Hint:

$$g\circ g = \mathrm{id}_{\mathcal{P}(\mathbb{R})}, \quad \text{ that is, } \quad \forall X \in \mathcal{P}(\mathbb{R}).\ g(g(X)) = X.$$

I hope this helps $\ddot\smile$

-

We can view subsets of $X$ as functions $X \to \{0,1\}$: 0 if the point is not in the set, It is commonly very useful to rewrite questions about subsets as questions about functions.

What form does $g$ take if we do this rewrite?

Terminology wise, if $S \subseteq X$, then $\chi_S$ is the function described above:

$$\chi_S(x) = \begin{cases} 0 & x \notin S \\ 1 & x \in S \end{cases}$$

It is called the characteristic function of $S$.

-
0 if $a \in X\cap \mathbb N$, 1 otherwise –  Oria Gruber Nov 28 '13 at 14:19
That's not right. To be more clear, $g(\chi_S)$ is a function. To describe this function, we could do so by writing down the equation that says what its value at a point $x$ is: i.e. what is $g(\chi_S)(x)$? –  Hurkyl Nov 28 '13 at 14:22

HINT: Recall that symmetric difference is associative and for every set $B$, $B\triangle B=\varnothing$.

-
And why was this downvoted? –  Asaf Karagila Nov 28 '13 at 22:04


# Find sum of series $\sum _{ n=1 }^{ \infty }{ { (-1) }^{ n+1 }\left( \frac { n+2 }{ { n }^{ 5 } } \right) }$ correct to 3 decimal places.

This is a question I came across while studying on Khan Academy:

Find the sum of the series $\sum _{ n=1 }^{ \infty }{ { (-1) }^{ n+1 }\left( \frac { n+2 }{ { n }^{ 5 } } \right) }$ correct to three decimal places.

I solved this by calculating ${ S }_{ 1 },{ S }_{ 2 },{ S }_{ 3 },...\\$ until I had a close enough value (bad method, I know). Anyway, here is the solution steps given in the answer:

Note that this is a convergent alternating series. The Alternating Series Estimation Theorem states that the error bound in the sum of the first $n$ terms is the absolute value of the first omitted term; i.e., $\left| { a }_{ n+1 } \right|$. To obtain three-digit accuracy, we need to find $n$ such that $\left| { a }_{ n+1 } \right| <\ 0.0005$, or equivalently $\frac { n+3 }{ { (n+1) }^{ 5 } } <0.0005$.

When $n=6$, we have $\left| { a }_{ n+1 } \right| =\frac { n+3 }{ { (n+1) }^{ 5 } } =\frac { 9 }{ 16807 } \approx .000535$; that is not accurate enough.

When $n=7$, we have $\left| { a }_{ n+1 } \right| =\frac { n+3 }{ { (n+1) }^{ 5 } } =\frac { 10 }{ 32768 } \approx .000305$; that will give a final result within $0.0005$.

Thus we need to find the sum of the first seven terms of the series.

$$\sum _{ n=1 }^{ 7 }{ { (-1) }^{ n+1 }\left( \frac { n+2 }{ { n }^{ 5 } } \right) \approx 2.8914 }$$

Hence, $2.891$ is accurate to three decimal places.

What I don't understand about this answer is if the estimate is $2.8914$ and the error bound is $0.000305$, I think the actual sum can be close to $2.8914+0.000305\approx 2.891705\approx 2.892$. Then the result above is not correct.

Does this answer take into account the fact that the $(7+1)th$ term is negative? Or is there something about rounding and error bounds that I don't understand?

Thank you.

• Indeed, 2.8914 means between 2.89135 and 2.89145 and 0.000305 is less than 0.0004 hence the sum is between 2.89135-0.0004 and 2.89145, which implies that 2.891 has three correct decimal places. (The same estimates for n=6 show it was not necessary to go to n=7.) – Did Oct 21 '15 at 7:27
• You are right, it is the fact that $\sum_1^7$ is an overestimate that enables us to conclude that $2.891$ is correct to 3 decimal places. – André Nicolas Oct 21 '15 at 7:28
• To the op: If you want it back I'll undelete my answer, but it seems unnecessary given you're already pretty much there yourself. – Adam Hughes Oct 21 '15 at 7:32
• @Did Thank you. But do you mean if somehow the estimate is an underestimate, the sum can be between 2.89135 and 2.89145+0.0004. Then 2.892 would be correct, right? – user52139 Oct 21 '15 at 7:49
• @AdamHughes Thank you very much. I actually need it back if you don't mind, please. – user52139 Oct 21 '15 at 7:52

I would read the question the same way you did-find a three decimal value that you know the true sum of the series rounds to. We know from the alternating series theorem that the truncation error is smaller than the first neglected term in absolute value and of the same sign. There is no truncation error that we can guarantee will be small enough, because we could be poised right on the rounding boundary, with the computed answer something like $2.8915$. Here, because after seven terms we have $2.8914$ and we know the eighth term is negative, the correct answer is in the range $(2.8911,2.8914)$ and this entire range rounds to $2.891$

• +1, warning about problem with rounding boundary is well placed here. This answer might be further improved by pointing out the problem of numerically finding any decimal of $\sum_{n=0}^\infty \frac 1{2^n}$. – Ennar Oct 21 '15 at 10:38

This is an alternating series, so the error is estimated by the last term. Since you want the error to be less than $10^{-3}$ in size, you need only choose $n$ so that ${n+2\over n^5}<10^{-3}$ since the terms are decreasing.

Solving we see $n+2<10^{-3}n^5$. We claim this is so for $n\ge 7$ as

$$7+2=9<16.807=7^5\cdot 10^{-3}.$$

For $n>7$ we see that by induction we may conclude this with the inductive step being

$$(n+1)+2<10^{-3}n^5+1< 10^{-3}(n+1)^5,$$

Then adding up the first $6$ terms gives you $2.891$ to three decimal places.


# Does the Bezout GCD equation hold in a UFD?

I'm wondering when Bezout's Theorem is valid.

I know when the ring $R$ is a Euclidean Domain or a Principal Ideal Domain there is always $\gcd(a,b)$ and I can find $x,y \in R$ such that $\gcd(a,b) = ax + by.$

I know when the ring $R$ is a Unique Factorization Domain, there is always $\gcd(a,b)$. My question is, in this case can I find $x,y \in R$ such that $\gcd(a,b) = ax + by.$ ?

Now if $R$ is just a commutative ring, if there is $d \in R$ such that $<a>+<b> = <d>$ I know that $d = \gcd(a,b)$. In this case, can I find $x,y \in R$ such that $\gcd(a,b) = ax + by$ ?

• In $\mathbb Z[x]$, a UFD, $2$ and $x$ have GCD $1$, but there is no solution to $2f(x)+xg(x)=1$. More generally, a UFD in which Bezout is true is always a principal ideal domain. I think. – Thomas Andrews Jul 13 '17 at 15:30

It is not necessarily true in a UFD. For instance, if $k$ is a field, then the polynomial ring $k[s,t]$ is a UFD. But $\gcd(s,t)=1$, and there do not exist $x$ and $y$ such that $sx+ty=1$.

For your second question, the answer is yes by definition. Indeed, by definition, $\langle a\rangle+\langle b\rangle$ is the set of elements of the form $ax+by$. So if $d\in \langle a\rangle+\langle b\rangle$, then there exist $x,y\in R$ such that $d=ax+by$.

The converse holds as well: if $d=\gcd(a,b)$ can be written in the form $ax+by$, then that means $d\in \langle a\rangle+\langle b\rangle$, and it follows that $\langle a\rangle+\langle b\rangle=\langle d\rangle$.

Thus a commutative ring has the property that the GCD of any two elements $a$ and $b$ can be written in the form $ax+by$ iff the sum of any two principal ideals is principal. By induction on the number of generators, this is equivalent to any finitely generated ideal being principal. Such a ring is known as a Bezout ring. A Noetherian Bezout ring is the same thing as a principal ideal ring, but there exist non-Noetherian examples as well. For instance, the ring of holomorphic functions on a connected open subset of $\mathbb{C}$ is a Bezout domain but has ideals which are not finitely generated.

The rings where it is valid Bézout's lemma are known as Bézout rings. In the case of integral domains rings we have the domains known as Bézout domains.

In general, a UFD doesn't have to be a Bézout domain. The classical example is $\Bbb{Z}[x]$, as Thomas Andrews pointed out in his comment. Theoretically there is a reason for which the above is true. We have the following:

Theorem 1: If $D$ is both a UFD and a Bézout domain, then $D$ is a PID.

Proof: See exercise 11 of section 8.3 of the book "Abstract Algebra" by Dummit and Foote.

By this theorem, if every UFD were also a Bézout domain, we would conclude that every UFD is a PID, and this is false (look at $\Bbb{Z}[x]$ again), as you surely know.

More generally, for Bézout domains we have the following result:

Theorem 2: Let $D$ be a Bézout domain. TFAE:

i) $D$ is a PID.

ii) $D$ is Noetherian.

iii) $D$ is a UFD.

iv) $D$ satisfies the ACCP (ascending chain condition on principal ideals).

v) $D$ is an atomic domain.

Proof: This is theorem 46 given in Pete L. Clark's notes on factorization in integral domains

Your second question has been also answered, but let me include a general version of what you wanted to prove.

Theorem 3: Let $a_1,a_2,\ldots a_n$ be nonzero elements of a commutative ring $R$. Then $a_1,a_2,\ldots a_n$ have a greatest common divisor $d$, expressible in the form $$d=r_1a_1+r_2a_2+\cdots +r_na_n$$ if only if the ideal $(a_1,a_2,\ldots a_n)$ is principal.

Note that $(a_1,a_2,\ldots a_n)$ is the same as $(a_1)+(a_2)+\cdots +(a_n)$.

Proof: This is theorem 6-3 in Burton's book "First Course in Rings and Ideals".

There are simple examples of non-Bezout UFDs, e.g. any UFD of dimension $$> 1$$, e.g. $$\Bbb Z[x,y],\,$$ by the below equivalent conditions for a UFD to be Bezout (or directly: $$\,\gcd(x,y)=1\,$$ but $$\,xf+y\,g = 1\Rightarrow 0 = 1\,$$ via eval at $$\,x=0=y)$$.

Theorem $$\rm\ \ \ TFAE\$$ for a $$\rm UFD\ D$$

$$(1)\ \$$ prime ideals are maximal if nonzero,  i.e. $$\rm\ dim\,\ D \le 1$$
$$(2)\ \$$ prime ideals are principal
$$(3)\ \$$ maximal ideals are principal
$$(4)\ \ \rm\ gcd(a,b) = 1\, \Rightarrow\, (a,b) = 1,$$ i.e.  coprime $$\Rightarrow$$ comaximal
$$(5)\ \$$ $$\rm D$$ is Bezout, i.e. all ideals $$\,\rm (a,b)\,$$ are principal.
$$(6)\ \$$ $$\rm D$$ is a $$\rm PID$$

Proof $$\$$ (sketch of $$\,1 \Rightarrow 2 \Rightarrow 3 \Rightarrow 4 \Rightarrow 5 \Rightarrow 6 \Rightarrow 1)\$$ where $$\rm\,p_i,\,P\,$$ denote primes $$\neq 0$$

$$(1\Rightarrow 2)$$ $$\rm\ \ p_1^{e_1}\cdots p_n^{e_n}\in P\,\Rightarrow\,$$ some $$\rm\,p_j\in P\,$$ so $$\rm\,P\supseteq (p_j)\, \Rightarrow\, P = (p_j)\:$$ by dim $$\le1$$
$$(2\Rightarrow 3)$$ $$\$$ max ideals are prime, so principal by $$(2)$$
$$(3\Rightarrow 4)$$ $$\ \rm \gcd(a,b)=1\,\Rightarrow\,(a,b) \subsetneq (p)$$ for all max $$\rm\,(p),\,$$ so $$\rm\ (a,b) = 1$$
$$(4\Rightarrow 5)$$ $$\ \ \rm c = \gcd(a,b)\, \Rightarrow\, (a,b) = c\ (a/c,b/c) = (c)$$
$$(5\Rightarrow 6)$$ $$\$$ Ideals $$\neq 0\,$$ in Bezout UFDs are generated by an elt with least #prime factors
$$(6\Rightarrow 1)$$ $$\ \ \rm (d) \supsetneq (p)$$ properly $$\rm\Rightarrow\,d\mid p\,$$ properly $$\rm\,\Rightarrow\,d\,$$ unit $$\,\rm\Rightarrow\,(d)=(1),\,$$ so $$\rm\,(p)\,$$ is max

As for the gcd as the ideal sum in a PID, that follows using "contains = divides", viz.

$$c\mid a,b\iff (c)\supseteq (a),(b)\iff (c)\supseteq (a)\!+\!(b)\!=\!(d)\iff c\mid d$$

And, yes, we have $$\,d\in (a)+(b) = aR + bR\,$$ $$\Rightarrow\, d = ar + br'$$ for some $$\,r,r'\in R$$


# Find the capacitance of the capacitor

Given: Consider a capacitor connected to a battery of voltage V . Let the capacitor have an area A, and a distance L between the plates. Assume that the capacitor has a layer of linear dielectric (of dielectric constant κ, so that ε = κε0) of thickness L/2 on the lower plate.

I found the capacitance to be $C=\frac{2\kappa\epsilon_0 A}{L}$. Was I correct about the separation distance? In more common examples of a capacitor, they claim that the plates are flat with nothing in between. In this case, the linear dielectric is as tall as half the total distance between the plates. Am I correct in using the remaining free space, namely $L/2$, as the separation distance in the equation? For reference, the equation is:

$$C=\frac{\kappa \epsilon_0 A}{d}$$

Instant check that shows your answer is wrong: set $\kappa = 1$ as if there is no dielectric, and you don't recover the parallel plate formula for a gap of $L$.

How to solve this problem:

Let the voltage across the capacitor be $V$ and the voltages across the top and bottom be $V_t$ and $V_b$ respectively, then $$V=V_t+V_b,$$ which implies that $$1/C=1/C_t+1/C_b$$ since $V=Q/C$.

Thus essentially this system consists of two capacitors in series. One capacitor is the top gap, and one is the bottom part. Use the formula for plate capacitors twice, with the separation being $L/2$ in both cases, then get the total capacitance by using the formula for two capacitors in series. Some algebra is involved.

You can check the answer you get, again, by setting $\kappa=1$, and seeing if you recover the usual plate formula.

• Why do you say there are 2 capacitors in series? The problem states there is only 1. – whatwhatwhat Oct 20 '16 at 23:35
• @whatwhatwhat - think of the bottom part and the top part as two capacitors as a way to solve the problem. – Suzu Hirose Oct 20 '16 at 23:36
• So then for the top section, there is a buildup of positive charge on the top plate and negative charge on the surface of the dielectric? – whatwhatwhat Oct 20 '16 at 23:42
• @whatwhatwhat - sorry my explanation was not very clear. There is an electric field between the two plates of the capacitor generated by the charges on the plates. The capacitance of this system of two plates is got by considering that the sum of the voltage over the two plates must be equal to the total voltage across the whole capacitor. I will edit the answer to make it clearer. – Suzu Hirose Oct 20 '16 at 23:48
• Suzu Hirose is completely right! – freecharly Oct 21 '16 at 0:12

The $\mathbf{D}$ field between the plates is (as usual, neglecting fringing effects) uniform and normal to the plane of the plates (assume in the z direction for simplicity) and is given by

$$\mathbf{D} = \frac{Q}{A}\hat{\mathbf{z}}$$

where the plates have charge $Q$ and $-Q$ respectively. The electric field (between the plates) within the dielectric is

$$\mathbf{E_\kappa} = \frac{\mathbf{D}}{\kappa \epsilon_0}$$

and the electric field (between the plates) in air is

$$\mathbf{E} = \frac{\mathbf{D}}{\epsilon_0}$$

Thus, the potential difference between the plates is

$$V = \left(\frac{L}{2}\mathbf{E_\kappa} + \frac{L}{2}\mathbf{E}\right)\cdot\hat{\mathbf{z}} = \left(\frac{L}{2}\frac{\mathbf{D}}{\kappa \epsilon_0} + \frac{L}{2}\frac{\mathbf{D}}{\epsilon_0}\right)\cdot\hat{\mathbf{z}} = \left(\frac{L}{2}\frac{Q}{A\kappa\epsilon_0} + \frac{L}{2}\frac{Q}{A\epsilon_0} \right)$$

The capacitance is then

$$C = \frac{Q}{V} = \frac{1}{\frac{L/2}{A\kappa\epsilon_0} + \frac{L/2}{A\epsilon_0}} = \frac{1}{\frac{1}{C_1}+ \frac{1}{C_2}}$$

which is the formula for series connected capacitors.

• Better explanation than my answer of why it is like two capacitors in series. – Suzu Hirose Oct 21 '16 at 1:22
• So I'm still treating this as 1 capacitor correct? My friend was helping with this and she said that one way to solve the problem is to treat the system as 2 capacitors, but I couldn't visualize where the 1st capacitor begins/ends and likewise for the 2nd one. – whatwhatwhat Oct 22 '16 at 18:59
• @whatwhatwhat, correct, it is just one capacitor however, it is instructive and intuitively 'nice' that we would (ideally) get the same result if we found the equivalent capacitance of two series connected capacitors with the same plate area $A$ and spacing $L/2$ but with different dielectrics; one with $\epsilon_0$ and the other with $\kappa \epsilon_0$ – Alfred Centauri Oct 22 '16 at 22:46